{ "Tag": [ "function", "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "$ m,n$ are positive integers. If $ \\phi (5^m\\minus{}1)\\equal{}5^n\\minus{}1$, show that $ m$ and $ n$ aren't coprime.", "Solution_1": "[hide]Suppose $ m,n$ are coprime. Then the only common factor that $ 5^m\\minus{}1,5^n\\minus{}1$ can have is $ 4$. If they have any other common prime factors, $ p$, other than $ 2$ then $ \\phi(p)$ divides $ n$ and $ m$, and thus $ n,m$ are not relatively prime. If $ 2$ divides both $ \\frac{5^m\\minus{}1}{4},\\frac{5^n\\minus{}1}{4}$ then we have\n\n$ 0\\equiv5^{m\\minus{}1}\\plus{}5^{m\\minus{}2}\\plus{}\\cdots\\plus{}5\\plus{}1\\equiv1^{m\\minus{}1}\\plus{}\\cdots\\plus{}1\\equiv m\\pmod{2}$\n\nSimilarly we find $ n\\equiv0\\pmod{2}$ and thus $ m,n$ aren't coprime, contradiction. Thus, $ 2^m\\minus{}1,2^n\\minus{}1$ only share $ 4$ as a common factor.\n\nNow, let $ 5^m\\minus{}1\\equal{}2^{e_1}p_2^{e_2}\\cdots p_n^{e_n}$. Then we have\n\n$ (5^m\\minus{}1)\\cdot\\frac{(p_2\\minus{}1)\\cdots(p_n\\minus{}1)}{2p_2\\cdots p_n}\\equal{}5^n\\minus{}1$\n\nIf we let $ a\\equal{}\\frac{5^m\\minus{}1}{4},b\\equal{}\\frac{5^n\\minus{}1}{4}$ then we have\n\n$ a\\cdot\\frac{(p_2\\minus{}1)\\cdots(p_n\\minus{}1)}{2p_2\\cdots p_n}\\equal{}b$\n\nNotice that $ a,b$ are relatively prime. We prove that $ 2\\not|a$. Suppose it does. Then let $ 2^k|a$ where $ k$ is maximized. Notice that the 2 in the denominator of the fraction cancels. If $ a$ has any odd prime factors $ p_i$, then $ p_i\\minus{}1$ is even, and thus $ b$ is even, a contradiction. If $ a$ has no odd prime factors, then it must equal $ 2$. If it is a higher power of 2, then $ b$ would have to be even, a contradiction. Thus, $ a\\equal{}2\\Rightarrow5^m\\minus{}1\\equal{}8$, which is impossible. So, $ 2\\not|a$. Thus, $ e_1\\equal{}2$.\n\nIf $ a\\equal{}1$, then $ m\\equal{}1\\Rightarrow \\phi(5^m\\minus{}1)\\equal{}2\\Rightarrow5^n\\minus{}1\\equal{}2$, which is impossible. Thus $ a$ has some odd prime factor $ p_i$ which is also a prime factor of $ 4a\\equal{}5^m\\minus{}1$. Thus, $ 5^m\\equiv1\\pmod{p_i}$ and so $ p_i\\minus{}1|m$. Thus, since $ p_i\\minus{}1$ is even, so is $ m$. Let $ m\\equal{}2t$. Then $ 5^{2t}\\minus{}1\\equal{}(5^t\\minus{}1)(5^t\\plus{}1)$. But, $ 4|5^t\\minus{}1$ and $ 2|5^t\\plus{}1$, thus $ 8|5^m\\minus{}1$ so $ e_1>2$, a contradiction to the fact that $ e_1\\equal{}2$.\n\nThus, $ m,n$ aren't coprime.\n\n\n[/hide]", "Solution_2": "$ 5^m\\equiv 1\\pmod{p}$ doesn't require $ p\\minus{}1\\mid m$.", "Solution_3": "[quote=\"Umut Varolgunes\"]$ 5^m\\equiv 1\\pmod{p}$ doesn't require $ p \\minus{} 1\\mid m$.[/quote]\r\n\r\nAs long as $ p$ is prime and $ 5\\not\\equiv 0,\\pm1\\pmod{p}$ I think it does.", "Solution_4": "$ 5^3\\equiv 1\\pmod{31}$", "Solution_5": "ahh yes. Then my solution is definitely wrong. I'll try find a way to fix it, but if anyone else has a solution, please post it." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "If $ x,y\\in \\mathbb{Z}$ , solve this equation :\r\n\r\n$ x^6\\plus{}x^5\\plus{}4\\equal{}y^2$", "Solution_1": "[quote=\"alex2008\"]If $ x,y\\in \\mathbb{Z}$ , solve this equation :\n\n$ x^6 \\plus{} x^5 \\plus{} 4 \\equal{} y^2$[/quote]\r\nI have found $ (x,y) \\equal{} (0,2),(0, \\minus{} 2)$", "Solution_2": "[quote=\"Dimitris X\"]\nI have found $ (x,y) \\equal{} (0,2),(0, \\minus{} 2)$[/quote]\r\n\r\n :maybe: That is good , but there are six other solutions", "Solution_3": "[quote=\"alex2008\"][quote=\"Dimitris X\"]\nI have found $ (x,y) \\equal{} (0,2),(0, \\minus{} 2)$[/quote]\n\n :maybe: That is good , but there are six other solutions[/quote]\r\nYes... :oops: \r\nI gave an ugly solution....\r\n$ x^5(x \\plus{} 1) \\equal{} (y \\minus{} 2)(y \\plus{} 2)$\r\nSo we have:\r\n$ x \\equal{} y \\minus{} 2$ and $ x^5 \\plus{} x^4 \\equal{} y \\plus{} 2$, $ x \\equal{} y \\plus{} 2$ and $ x^5 \\plus{} x^4 \\equal{} y \\minus{} 2$ ,$ x^2 \\equal{} y \\minus{} 2$ and $ x^4 \\plus{} x^3 \\equal{} y \\plus{} 2$ etc....I think this way works but it is really ugly.....There sould be a better solution....", "Solution_4": "My solution is this :\r\n\r\nOur equation it is equivalent with :\r\n\r\n$ 64x^6 \\plus{} 64x^5 \\plus{} 256 \\equal{} 64y^2$\r\n\r\nIf $ |x|\\ge 3\\Rightarrow (8x^3 \\plus{} 4x^2 \\minus{} x)^2 < 64x^6 \\plus{} 64x^5 \\plus{} 256 < (8x^3 \\plus{} 4x^2 \\minus{} x \\plus{} 1)^2$\r\n\r\n$ \\Rightarrow x\\in \\{ \\minus{} 2, \\minus{} 1,0,1,2\\}$\r\n\r\nSo , $ (x,y)\\in \\{( \\minus{} 2,\\pm 6);( \\minus{} 1,\\pm 2);(0,\\pm 2);(2,\\pm 10)\\}$" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Find the limit:\r\n$ \\lim_{x->{0}} {\\frac{a_{1}^{x}*a_{2}^{2x}* . . . *a_{n}^{nx}-1}{x}} $\r\n\r\n$ a_{1},a_{2},a_{3}, . . . , a_{n}>0$", "Solution_1": "In fact,you can write it in this way :\r\n$\\displaystyle\\lim_{x->0} \\frac{(a^{1}_{1}a^{2}_{2}***a^{n}_{n})^{x}-1}{x}.$\r\nThen the result is obvious,i.e.,$ln(a^{1}_{1}a^{2}_{2}***a^{n}_{n}).$", "Solution_2": "Thank you." } { "Tag": [ "vector", "geometry" ], "Problem": "it strikes me that you use the vanishing homology of euclidean space a good deal in electromagnetism. This is perhaps most obvious when finding potentials, as you're directly using that closed differential forms implies exactness, i.e. vanishing of the de Rham cohomology. It also seems like you also use it implicitly in a lot of other contexts too, though. For example, when talking about the vector area (such as when defining magnetic dipole moment), you consider a loop and then integrate something over the surface that the loop bounds, but the existence of this surface, i.e. the fact that any $ S^1$ is a boundary, is again topological information. so how much does electromagnetism change when working over a topologically nontrivial space?", "Solution_1": "by the way, http://www.math.unc.edu/Faculty/jds/survey.pdf\r\n\r\nI will freely admit I do not understand much of it", "Solution_2": "Is this a sequence of some conversation or it starts a new topic? Seems that te beginning is missing...", "Solution_3": "R. Bott's \"On some recent interactions between mathematics and physics\" in the canadian mathematical bulletin is much more readable; I might have gotten through more than ten pages before dying." } { "Tag": [ "limit", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "$a_k=\\frac{\\sin(k)}{|\\sin(k)|}$\r\n\r\nCompute\r\n\r\n $\\lim_{n\\rightarrow+\\infty}\\frac{\\sum_{k=1}^{n}a_ka_{k+1}}{n}$", "Solution_1": "What? AGAIN? :lol:", "Solution_2": "Some post was deleted between 21/11 to 3/12", "Solution_3": "I know. It was a joke! :lol: :lol: :lol:", "Solution_4": "I'm very sorry to have kept you waiting for you and my delaying reply, Moderator [b]Moubinool[/b].\r\n\r\nI will answer as follows.\r\n\r\nFirst ,there exist integers $m\\geqq0$ such that $m\\pi = x*x, y <= square root of x and y <= 1/(8x).\r\n\r\nWhat I did was:\r\n1/(8x) = x*x so x = 1/2\r\n1/(8x) = square root of x so x = 1/4\r\n\r\nAnd then i found the integrals which are:\r\n+ Integral of square root of x times dx between 0 and 1/4\r\n- Integral of x * x times dx between 0 and 1/4\r\n+ Integral of 1/(8x) times dx between 1/4 and 1/2\r\n- Integral of x * x times dx between 1/4 and 1/2\r\n\r\nSo this gives me the area, but i have no idea on how to calculate the volume.\r\nHope someone can help.\r\n\r\nTks...\r\nAnd i'm sorry about my english, this is my first time writing math in english.", "Solution_1": "Let $V_{1}$ be the volume obtained after rotation of $\\sqrt{x}$ between the points $0$ and $\\displaystyle{ \\frac{1}{4}}$ (that is the point of intersection of $\\sqrt{x}$ and $\\frac{1}{8x}$ )\r\n\r\n$V_{2}$ be the volume obtained after rotation of $\\frac{1}{8x}$ between the points $\\displaystyle{\\frac{1}{4}}$ and $\\displaystyle{\\frac{1}{2}}$ (the intersection of $\\frac{1}{8x}$ and $x^2$ ) and $V_{2}$ be the volume \r\n\r\nobtained after rotation of $x^2$ between the points $0$ and $\\displaystyle{\\frac{1}{2}}$ . \r\n\r\nThen the volume you want is $\\displaystyle{V=V_{1}+V_{2}-V_{3}}$\r\n\r\nThe volyme of revolution of the function $y=f(x)$ beetween the poins $x=a, \\,\\, x=b$ is given by $\\int_{a}^{b} \\pi y^{2}\\, dx$\r\n\r\nA quick calculation gives $V = \\frac{ 29 \\cdot \\pi }{30 \\cdot 32 }$ . \r\n[b]\nCheck again everything ![/b] Your english are good, mine are not so i hope you understund what i write \r\nand sorry if i am wrong.\r\n\r\n :)" } { "Tag": [ "inequalities", "geometry", "geometric transformation", "reflection", "inequalities proposed" ], "Problem": "If $a,b,c$ are three sides of a triangle, prove that\r\n\\[\\frac{a^{2}+b^{2}}{a^{2}+c^{2}}+\\frac{b^{2}+c^{2}}{b^{2}+a^{2}}+\\frac{c^{2}+a^{2}}{c^{2}+b^{2}}\\ge \\frac{a+b}{a+c}+\\frac{b+c}{b+a}+\\frac{c+a}{c+b}\\]\r\n:)", "Solution_1": "Thank you, Can. My Yahoo Message is false these days, so I can not connect you. Your problem is very nice and the same as an old problem I sent you,\r\n\r\nIf $a,b,c$ are three length sides of an triangle then\r\n\\[4(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}-3) \\ge \\frac{a^{2}+b^{2}}{a^{2}+c^{2}}+\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+\\frac{b^{2}+c^{2}}{a^{2}+b^{2}}. \\]", "Solution_2": "I think the second is:\r\n\\[4(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}) \\ge \\frac{a^{2}+b^{2}}{a^{2}+c^{2}}+\\frac{b^{2}+c^{2}}{b^{2}+a^{2}}+\\frac{c^{2}+a^{2}}{c^{2}+b^{2}}+9 \\]\r\n1.[quote=\"toanhocmuonmau\"]If $a,b,c$ are three sides of a triangle, prove that\n\\[\\frac{a^{2}+b^{2}}{a^{2}+c^{2}}+\\frac{b^{2}+c^{2}}{b^{2}+a^{2}}+\\frac{c^{2}+a^{2}}{c^{2}+b^{2}}\\ge \\frac{a+b}{a+c}+\\frac{b+c}{b+a}+\\frac{c+a}{c+b}\\]\n:)[/quote]\r\nWe may assume that c=min(a,b,c)\r\nWe have:\r\n\\[LHS-RHS=M(a-b)^{2}+N(a-c)(b-c) \\]\r\nIn that:\r\n\\[M=\\frac{(a+b)^{2}}{(a^{2}+c^{2})(b^{2}+c^{2})}-\\frac{1}{(a+c)(b+c)}\\]\r\n\r\n\\[N=\\frac{(a+c)(b+c)}{(a^{2}+c^{2})(a^{2}+b^{2})}-\\frac{1}{(a+c)(a+b)}\\]\r\nWe have\r\n\\[M.(a+c)(b+c)(a^{2}+c^{2})(b^{2}+c^{2})=(a+b)^{2}(a+c)(b+c)-(a^{2}+c^{2})(b^{2}+c^{2})=ab(a^{2}+b^{2}+ab)++3abc(a+b)+c(a^{3}+b^{3})-c^{4}\\ge 0 \\]\r\n\r\n\\[N.(a+c)(a+b)(a^{2}+c^{2})(a^{2}+b^{2})=a^{3}(b+c-a)+3a^{2}bc+3abc^{2}+bc^{3}++ac^{3}+a^{2}c^{2}\\ge 0 \\]\r\n2. If a,b,c are three length sides of a triangle then\r\n\\[4(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}) \\ge \\frac{a^{2}+b^{2}}{a^{2}+c^{2}}+\\frac{b^{2}+c^{2}}{b^{2}+a^{2}}+\\frac{c^{2}+a^{2}}{c^{2}+b^{2}}+9 \\]\r\nWe may assume that c=max(a,b,c)\r\nWe have\r\n\\[LHS-RHS=M(a-b)^{2}+N(a-c)(b-c) \\]\r\nIn that\r\n\\[M=\\frac{4}{ab}-\\frac{(a+b)^{2}}{(a^{2}+c^{2})(b^{2}+c^{2})}\\]\r\n\r\n\\[N=\\frac{4}{ac}-\\frac{(a+c)(b+c)}{(a^{2}+c^{2})(a^{2}+b^{2})}\\]\r\nWe have:\r\n\\[M.ab(a^{2}+c^{2})(b^{2}+c^{2})=2a^{2}b^{2}+4a^{2}c^{2}+4b^{2}c^{2}+4c^{4}-a^{3}b-ab^{3}\\ge 0 \\]\r\n\r\n\\[N.ac(a^{2}+c^{2})(a^{2}+b^{2})=4a^{4}+3a^{2}b^{2}+3b^{2}c^{2}+ac^{2}(a+b-c)+(a^{2}c^{2}+b^{2}c^{2}-2abc^{2})+(a^{2}b^{2}+a^{2}c^{2}-a^{2}bc) \\ge 0 \\]", "Solution_3": "Very nice, 10maths. It is the same as my solution. Thanks for notice my mistake in typing. One of a nice application is the following inequality I sent to Math Reflection (not issued) before\r\n\r\nIf $a,b,c$ are positive and $k\\ge max(ab,bc,ca)$ then\r\n\\[\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}\\ge \\frac{a^{2}+k}{b^{2}+k}+\\frac{b^{2}+k}{c^{2}+k}+\\frac{c^{2}+k}{a^{2}+k}.\\]\r\n\r\nWhen does the equality occur?", "Solution_4": "This is my solution:\r\nWe may assume c=min(a,b,c).\r\nWe have:\r\n\\[LHS-RHS=M(a-b)^{2}+N(a-c)(b-c) \\]\r\nIn that\r\n\\[M=\\frac{1}{ab}-\\frac{(a+b)^{2}}{(a^{2}+k)(b^{2}+k)}\\]\r\n\r\n\\[N=\\frac{1}{ac}-\\frac{(a+c)(b+c)}{(a^{2}+k)(c^{2}+k)}\\]\r\nWe have\r\n\\[M.ab(a^{2}+k)(b^{2}+k)=k(a^{2}+b^{2})+k^{2}-ab(a^{2}+b^{2})-a^{2}b^{2}\\ge ab.(a^{2}+b^{2})+(ab)^{2}-ab(a^{2}+b^{2})-a^{2}b^{2}= 0 \\]\r\n\r\n\\[N.ac(a^{2}+k)(c^{2}+k)=k(a^{2}+c^{2})+k^{2}-abc(a+c)-ac^{3}\\ge ab(ac+c^{2})+(ac)^{2}-abc(a+c)-ac^{3}=ac^{2}(a-c)\\ge 0 \\]", "Solution_5": "then when does the equality hold, 10 math? :lol:", "Solution_6": "a=b=c\r\na=c,k=ab\r\nb=c,k=ab\r\n...", "Solution_7": "[quote=\"10maths_tp0609\"]a=b=c\na=c,k=ab\nb=c,k=ab\n...[/quote]\r\n\r\nMore exactly, the equality holds for $a=b=c$ or $a=b\\le c, k=ac$ (and permutations).", "Solution_8": "[quote=\"toanhocmuonmau\"]If $a,b,c$ are three sides of a triangle, prove that\n\\[\\frac{a^{2}+b^{2}}{a^{2}+c^{2}}+\\frac{b^{2}+c^{2}}{b^{2}+a^{2}}+\\frac{c^{2}+a^{2}}{c^{2}+b^{2}}\\ge \\frac{a+b}{a+c}+\\frac{b+c}{b+a}+\\frac{c+a}{c+b}\\]\n:)[/quote]\nIn fact, we can prove that the inequality still holds in case $\\max\\{a^2,\\;b^2,\\;c^2\\} \\le ab+bc+ca.$" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "For $ a,b,c\\geq 0$ Prove that\r\n$ \\frac{1}{a\\plus{}b}\\plus{}\\frac{1}{a\\plus{}b}\\plus{}\\frac{1}{a\\plus{}b}\\ge\\frac{9\\sqrt{3}}{2\\sqrt{a\\plus{}b\\plus{}c}(\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c})}$", "Solution_1": "I think that you mean:\r\n$ \\frac{1}{a\\plus{}b}\\plus{}\\frac{1}{b\\plus{}c}\\plus{}\\frac{1}{c\\plus{}a}\\geq\\frac{9\\sqrt{3}}{2\\sqrt{a\\plus{}b\\plus{}c}(\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c})}$.", "Solution_2": "Try $ a\\equal{}b\\equal{}1,c\\equal{}0$\r\n\r\n$ \\frac{5}{2}\\ge\\frac{9\\sqrt{3}}{4\\sqrt{2}}$ ????", "Solution_3": "$ \\frac{5}{2}\\geq\\frac{9.\\sqrt{3}}{4.\\sqrt{2}}\\Leftrightarrow\\sqrt{\\frac{25}{4}}\\geq\\sqrt{\\frac{243}{32}}\\Leftrightarrow\\frac{25}{4}\\geq\\frac{243}{32}\\Leftrightarrow{6\\frac{1}{4}}\\geq{7\\frac{19}{32}}$ $ {FALSE!}$", "Solution_4": "Oh yes mihai miculita ,so Harry Potter'problem is no true :)", "Solution_5": "Sorry :blush: \r\nProblem true is Erken 's Post :blush:" } { "Tag": [ "number theory", "least common multiple" ], "Problem": "Find $ \\sum_{r\\equal{}0}^n{\\frac{2^r}{a^{2^r}\\plus{}1}}$", "Solution_1": "Is the answer\r\n\r\n$ \\frac{1}{a\\minus{}1} \\minus{} \\frac{2(2^n \\minus{} 1)}{a^{2^{n\\plus{}1}} \\minus{}1} \\minus{} \\frac{2}{a^{2^{n\\plus{}1}} \\minus{}1}$ ?\r\n\r\nMultiply the numerator and denominator of the first term of the question by $ (a\\minus{}1)$, then add it with the second term, then add the entire with third term ...\r\nThen apply $ (a^{2^{n\\plus{}1}}\\minus{}1)\\equal{}(a\\minus{}1)(a^{2^{n\\plus{}1}\\minus{}1}\\plus{}a^{2^{n\\plus{}1}\\minus{}2}\\plus{}..\\plus{}1)$", "Solution_2": "$ \\frac {a^{2^{n} \\plus{} 1} \\minus{} a \\plus{} 2^{n}(1 \\minus{} a)}{(a^{{2^n} \\plus{} 1} \\minus{} 1)(a \\minus{} 1)}$\r\n\r\ni have posted only the solution because it will take many pages to post the whole solution\r\n\r\nwrite all terms, take their lcm and the denomenator will be a gp \r\n$ a^{2^n} \\plus{} a^{2^n \\minus{} 1} \\plus{} ... \\plus{} a \\plus{} 1$ \r\nand the numerator will be an agp\r\n$ a^{2^n \\minus{} 1} \\plus{} 2a^{2^n \\minus{} 2} \\plus{} 3a^{2^n \\minus{} 3} \\plus{} ... \\plus{} (2^{n} \\minus{} 1)a\\plus{}2^n$\r\nevaluate them separately and we'll get the above solution\r\n\r\nnote:\r\n$ (a \\plus{} 1)(a^2 \\plus{} 1)...(a^n \\plus{} 1) \\equal{} a^{2^n} \\plus{} a^{2^n \\minus{} 1} \\plus{} ... \\plus{} a \\plus{} 1$" } { "Tag": [ "function", "logarithms" ], "Problem": "The demand function for do-it-yourself replicas of Stonehenge is given by p=1000(3/\\-q) where p is a price in dollars and q is a quantity of replicas in hundreds demanded at that price. What quantity will be demanded at the price of 12340.00?", "Solution_1": "is that \r\n$p=1000(3^{-q})$\r\n :huh:", "Solution_2": "YES that is correct", "Solution_3": "anyone know.....?", "Solution_4": "I have no idea what to do with this problem", "Solution_5": "[hide]$p=1000(3^{-q})$\n$12340=1000(3^{-q})$\n$12.34=\\frac{1}{3^q}$\n$3^q=\\frac{1}{12.34}$\n$log_3\\frac{1}{12.34}=q$\nand then u solve that, i think. if it is, i don't know how to do it on my calculator.[/hide]", "Solution_6": "Once you get to there, fireemblem13, you can use $\\log_a b = \\frac{\\ln b}{\\ln a}$", "Solution_7": "Or $\\log_ab=\\frac{\\log a}{\\log b}$...\r\n\r\nThat's what I did too, but I got the demand to be $\\approx -2.3$...", "Solution_8": "[quote=\"DanK\"]Or $\\log_ab=\\frac{\\log a}{\\log b}$...\n\nThat's what I did too, but I got the demand to be $\\approx -2.3$...[/quote]\r\n\r\nIt's called the \"change of base rule\". Umm... $\\log{\\frac{1}{12.34}}$ must be negative and $\\log{3}$ must be positive. So... :?", "Solution_9": "It's just the question itself that is flawed. In order for the price to be greater than 1000, $3^{-q} > 1$ which will only happen if q is negative.", "Solution_10": "Perhaps if it was $q$ instead of $-q$, the answer would be $\\frac{\\log{12.34}}{\\log{3}}=2.287$ Multiplying this by $100$ as said in the original problem, we get $\\boxed{229}$.", "Solution_11": "Yeah, I was thinking it might be something like that..." } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "If $M,N \\lhd G$, $M \\lhd N$ and $G/N$ is cyclic, can I conclude that $G/M$ is abelian (cyclic)? and $G/M \\leq G/N$ ? and what if $|M/N|=2$? \r\nSince the property 'normal subgroup' is not transitive, I'm not sure at all.", "Solution_1": "Since it's not transitive, you can't even define the quotient group.", "Solution_2": "[quote=\"ZetaX\"]Since it's not transitive, you can't even define the quotient group.[/quote]\r\n\r\nok. But here we can, since there is also given that $M \\lhd G$, right? Does the rest hold too then?", "Solution_3": "[quote=\"jerrific\"]If $M,N \\lhd G$, $M \\lhd N$ and $G/N$ is cyclic, can I conclude that $G/M$ is abelian (cyclic)? and $G/M \\leq G/N$ ? and what if $|M/N|=2$? \nSince the property 'normal subgroup' is not transitive, I'm not sure at all.[/quote]\r\n\r\n\r\nIt is clear that $N/M \\cong Z_{2}$ is abelian. Furthermore is $N/M$ a subgroup of $G/M$, that is contained in $Z(G/M)$. It is easy to check that if $\\frac{G/M}{N/M}\\cong G/N$ is cyclic, and this is the case, $G/M$ is abelian.", "Solution_4": "why the property \"normal subgroup\" is not transitive?Can some one give me one example :(", "Solution_5": "[quote=\"blue_se7en\"]why the property \"normal subgroup\" is not transitive?Can some one give me one example :([/quote]\r\n\r\nhttp://planetmath.org/encyclopedia/NormalityOfSubgroupsIsNotTransitive.html" } { "Tag": [ "trigonometry" ], "Problem": "Given triangle $\\triangle ABC$, let $f(A,B,C)=\\sin A+\\cos A\\sin B+\\cos A\\cos B\\sin C$. Find:\r\n\r\n(a) the maximum of $f(A,B,C)$\r\n(b) the minimum of $f(A,B,C)$", "Solution_1": "A guess:\r\n[hide]Minimum: $\\frac{7\\sqrt{3}}{8}$?[/hide]", "Solution_2": "And why is that the minimum? I say your answer is wrong unless you present your proof :D", "Solution_3": "[hide]The min and max should be from the extremes (a $60^\\circ-60^\\circ-60^\\circ$ triangle and a degenerate triangle). We then find $f(60,60,60)=\\sin 60+\\cos 60\\sin 60+\\cos 60\\cos 60\\sin 60 = \\frac{7\\sqrt{3}}{8}$\nWe then test some arbitrary triangle which is not an extreme. We use a $30^\\circ-60^\\circ-90^\\circ$ triangle for simplicity. We find the value of $f(30,60,90)$ to be $\\frac{11 + \\sqrt{3}}{4}$, which is greater than $\\frac{7\\sqrt{3}}{8}$. Thus $\\frac{7\\sqrt{3}}{8}$ is the minimun.\n\nThe angles of a degenerate triangle (line) confuse me so I didn't evaluate it.[/hide]", "Solution_4": "But, $f(90,60,30)=1$? Sorry but there are exceptions in your argument and you have work through carefully.", "Solution_5": "More importantly, it's not even clear what boundary values are when you have three variables which are dependent on each other.", "Solution_6": "I agree with what JBL is saying, so I'll allow degenerate triangles (usually this happens with min/max problems). So in this case, $A+B+C=180$ with $0\\le A,B,C\\le 180$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "For $ a,b,c \\in \\mathbb{R^ \\plus{} }$ prove inequality:\r\n\r\n$ \\frac {a^2b \\plus{} b^2c \\plus{} c^2a}{3(a^3 \\plus{} b^3 \\plus{} c^3)} \\plus{} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} ac \\plus{} bc} \\ge \\frac {4}{3}$", "Solution_1": "I forgot to say before. Please don't write official solution. I'm looking for a more beautiful. If someone wants to see [b]ALL BSMC 2008. and 2009. PROBLEMS WITH SOLUTIONS[/b] you can find them [url=http://web.studenti.math.hr/~mornik/BSMC.htm][b]here[/b][/url].", "Solution_2": "That's my solution (the pdf-file)." } { "Tag": [ "calculus", "parameterization", "advanced fields", "advanced fields unsolved" ], "Problem": "[img]http://i99.photobucket.com/albums/l290/janbmyffi/stock_price_and_variance.jpg[/img]", "Solution_1": "1 - This is easy algebra using that the parameters are constants.\r\n\r\n2 - Assuming that this is meant after taking expectations (which coincides with taking conditional expectation w.r.t. the trivial $ \\sigma$-algebra augmented by all null-sets) note that the limit stems from the first summand only due to the symbolic relations $ dW_t\\cdot dW_t\\equal{}dt$ and $ dW_t\\cdot dt\\equal{}0$. They are derived from the defining properties of the Brownian motion. The second term trivially tends to 0.\r\n\r\n3 - Don't what the author is asking for here since the should be a bunch of assumptions to be put in place if, e.g., $ \\sigma$ and $ r$ are (time-dependent) processes." } { "Tag": [ "MATHCOUNTS", "geometry", "3D geometry", "number theory", "prime numbers", "exterior angle" ], "Problem": "I was just reviewing the MATHCOUNTS Problem Series transcripts and found the part: \r\n\r\n\"aanjohri (20:48:02) \r\nis there a trick to find out if a number is prime or not? \r\n\r\nrrusczyk (20:48:27) \r\nThis is a big, rich area of mathematics - ask this on the message board and we will discuss it!\" \r\n\r\nI couldn't find this in the message board. \"Is\" there a trick? :D I got all of the dividing tricks down for 7, 13, 17, 19, and all those hard ones, but I haven't found a trick for immediate knowing if a number is prime or not...\r\n\r\nBTW, \"After being painted, a solid wooden cube whose edge is 4 cm is cut into 64 small one-centimer cubes. How many of these small cubes will have exactly three painted faces?\" Two painted faces? One painted face? At least 2? At least 1? \r\n\r\nI always logic and such on these problems. How do ya do it geometrically?\r\n\r\nOh yeah, I don't really get the stuff about arcs and such. Can u ever find the length (without the knowing of the circumference) or can u only know the degree? Yeah, my geom. really sux....", "Solution_1": "I don't really know if there is a way to find if a number is prime, but about your second question, Only corners have 3 painted faces so 8 cubes, only edges have 2 painted faces so 24, only centers on edges have 1 painted face so 24. For at least 2, you do amount with 2 + amount with 3, or 32, with at least 1, take amount with 1 + amount with 2 + amount with 3 or 56.\r\n\r\nabout arcs, can you give an example?", "Solution_2": "to find prime numbers you just kinda have to check divisibility. have them memorized up to 200 and you should be fine.\r\n\r\nsecond question: 8 for 3 painted, 24 for 2 painted, 24 for one painted, and 8 for no painted.\r\n\r\nthird question: you use degrees and the radius (you need both) unleess you are given other lengths\r\n\r\nedit: ihatepie beat me to it", "Solution_3": "Alright, thx. I think i got it. BTW, what's the exterior angle of a polygon? Like for an =lateral triangle, the exterior angle is 360-60=300, right? or is it 180-60?", "Solution_4": "If you want to find if a number is prime or not this method usually works:\r\n- find its square root\r\n- if it is not divisible by any of the prime numbers up to its square root it means the number is [u][b]most likely [/b][/u]to be prime.\r\n\r\nExample: lets check if 113 is prime or not: its square root is close to 11. and 113 is not divisible by 11 and any prime that is less than 11. so it is a prime number.", "Solution_5": "[quote=\"al-kharizmi\"]- find its square root\n- if it is not divisible by any of the prime numbers up to its square root it means the number is [u][b]most likely [/b][/u]to be prime.[/quote]\r\n\r\nActually, if a number isn't divisible by any of the prime numbers up to its square root, then it's [b]guaranteed[/b] to be prime. This is because a composite number must have a nontrivial factorization $ n \\equal{} ab$ where $ a, b > 1$ and then either $ a$ or $ b$ must be less than or equal to $ \\sqrt{n}$.\r\n\r\nThis is all you'll need to do in contest situations, but the divide-and-check method becomes very cumbersome for numbers with four or more digits because it gets harder to divide by large primes by hand. There are [b]very[/b] sophisticated algorithmic tests for determining whether a number is prime, but they're beyond the scope of this forum.", "Solution_6": "Also any prime number $ p$ is in the form:\r\n\r\n$ p\\equal{}6n\\pm1$\r\n\r\nSo thats one quick way to check if the number is [b]not[/b] prime (it doesn't guarantee if $ p$ is prime).", "Solution_7": "Happyme, I think you have something wrong with that. Are you trying to say that all numbers expressible that way are prime (25?) or that all numbers that are not expressible that way are composite (3?)?", "Solution_8": "[quote=\"happyme\"]Also any prime number $ p$ is in the form:\n\n$ p \\equal{} 6n\\pm1$\n\nSo thats one quick way to check if the number is [b]not[/b] prime (it doesn't guarantee if $ p$ is prime).[/quote]\r\n\r\nAny prime number greater than $ 3$. This is equivalent to checking divisibility by $ 2$ and $ 3$, both of which are very easy anyway.", "Solution_9": "[quote=\"toadoncart\"]Alright, thx. I think i got it. BTW, what's the exterior angle of a polygon? Like for an =lateral triangle, the exterior angle is 360-60=300, right? or is it 180-60?[/quote]\r\n\r\n*cough**cough*, still not answered....(lol) :lol:", "Solution_10": "[quote=\"uldivad9\"]Happyme, I think you have something wrong with that. Are you trying to say that all numbers expressible that way are prime (25?) or that all numbers that are not expressible that way are composite (3?)?[/quote]\r\n\r\n1. The sum of the exterior angles is always $ 360^\\circ$, and if the polygon is regular, then each of the exterior angles has the same measure. In that case, you can divide.\r\n\r\n2. If a number cannot be expressed as:\r\n\r\n$ p \\equal{} 6n\\pm1$ \r\n\r\nthen the number is NOT prime. However if the number can be expressed in that form, it doesn't guarentee that the number is prime. (If $ p > 3$). Sorry for any ambiguity.", "Solution_11": "exterior=180-interior\r\nnow how hard was that?\r\nnow plz join the im a PHAILURE club?", "Solution_12": "[quote=\"stevenmeow\"]\nnow plz join the im a PHAILURE club?[/quote]\r\n\r\nHow?? Oh btw look at the hax comment I made on your blog... :lol:" } { "Tag": [ "trigonometry", "trig identities", "Law of Cosines", "geometry", "angle bisector", "AMC" ], "Problem": "In $\\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\\angle BAC$, $BN\\perp AN$ and $\\theta$ is the measure of $\\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals\n\n[asy]\nsize(230);\ndefaultpen(linewidth(0.7)+fontsize(10));\npair B=origin, A=14*dir(36), C=intersectionpoint(B--(9001,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b);\ndraw(N--B--A--N--M--C--A^^B--M);\nmarkscalefactor=0.1;\ndraw(rightanglemark(B,N,A));\npair point=N;\nlabel(\"$A$\", A, dir(point--A));\nlabel(\"$B$\", B, dir(point--B));\nlabel(\"$C$\", C, dir(point--C));\nlabel(\"$M$\", M, S);\nlabel(\"$N$\", N, dir(30));\nlabel(\"$19$\", (A+C)/2, dir(A--C)*dir(90));\nlabel(\"$14$\", (A+B)/2, dir(A--B)*dir(270));\n[/asy]\n\n$\\displaystyle \\text{(A)} \\ 2 \\qquad \\text{(B)} \\ \\frac{5}{2} \\qquad \\text{(C)} \\ \\frac{5}{2} - \\sin \\theta \\qquad \\text{(D)} \\ \\frac{5}{2} - \\frac{1}{2} \\sin \\theta \\qquad \\text{(E)} \\ \\frac{5}{2} - \\frac{1}{2} \\sin \\left(\\frac{1}{2} \\theta\\right)$", "Solution_1": "[hide=\"Hint 1\"]\nExtend $BN$[/hide]\n[hide=\"Hint 2\"]\nLet $P$ be the intersection of $\\overleftrightarrow{BN}$ and $\\overline{AC}$.\nWhat kind of triangle is $\\triangle APB$? ;) [/hide]", "Solution_2": "[hide=\"My solution\"]Extend $BN$ and let its intersection with $AC$ be $P$. Now we have $\\Delta APB$, which is an isosceles triangle, with $PB = (2)(14)\\sin (\\angle A/2) = 28 \\sin \\alpha$, where $\\alpha = \\angle A/2$. \n\nNow let's examine $\\Delta PCB$. Let $CM = MB = x$. Also let $\\angle CBP = \\gamma$ and apply the Law of Cosines to get:\n\n$5^2 = (28 \\sin \\alpha)^2 + (2x)^2 - (2)(28 \\sin \\alpha)(2x) \\cos \\gamma$\n\n$\\Rightarrow \\cos \\gamma = \\frac{4x^2 + 784 \\sin^2 \\alpha - 25}{112 x \\sin \\alpha}$\n\nNow let's examine $\\Delta BNM$ and apply the Law of Cosines one more time:\n\n$MN^2 = (14 \\sin \\alpha)^2 + x^2 - (2)(14 \\sin \\alpha)(x) \\cos \\gamma$\n\n$= 196 \\sin^2 \\alpha + x^2 - (x^2 + 196 \\sin^2 \\alpha - 25/4) = \\frac{25}{4}$\n\n$\\therefore MN = \\frac{5}{2}$[/hide]", "Solution_3": "[hide]There is a lot of info in this problem. Let's just try to take advantage of it. We know that AN is an angle bisector, so we want to use the angle bisector theorem. Either we extend AN or BN to do that. It turns out that extending BN leads to a very quick solution. As soon as we do that and hit AC at the point D we stare at the picture for a bit. Hopefully we didn't draw a picture like the misleading one chess64 tried to give us. It should be pretty clear that since angles ANB and angles AND are both right angles, that our triangle APB is in fact isosceles. Thus, AD = 14 and DC = 5. Now notice that BMN is similar to and half the size of BCD and so our length is MN = 2.5[/hide]", "Solution_4": "Hey I didn't mean to be misleading :)", "Solution_5": "[quote=\"pdiao\"][hide]Now notice that BMN is similar to and half the size of BCD [/hide][/quote]Hmm... that would have been really helpful to know. I didn't think to look at the fact that they were similar - that would have simplified my solution enormously. I do suck at geometry... :blush:", "Solution_6": "Since $BN=ND$ and $BM=MC$, we have $MN\\parallel CD$ and $MN=\\frac12 CD$. It's called Midline theorem.", "Solution_7": "[quote=\"frt\"]Since $BN=ND$ and $BM=MC$, we have $MN\\parallel CD$ and $MN=\\frac12 CD$. It's called Midline theorem.[/quote]\r\nYes, I see how the proof worked - it just took me a moment to see the similarity (and Chess's diagram definitely didn't help :P ).", "Solution_8": "The problem statement needs to be adjusted slightly, because $\\theta$ is actually defined in the original version:\n\n[quote]\nIn $\\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\\angle BAC$, $BN\\perp AN$ and $\\theta$ is the measure of $\\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals\n[/quote]", "Solution_9": "[quote=djmathman]The problem statement needs to be adjusted slightly, because $\\theta$ is actually defined in the original version:\n\n[quote]\nIn $\\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\\angle BAC$, $BN\\perp AN$ and $\\theta$ is the measure of $\\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals\n[/quote][/quote]\n\nmy guy you just bumped a 17 year old thread [hide]use trig[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Find the largest real $ T$ such that for each non-negative real numbers $ a,b,c,d,e$ such that $ a\\plus{}b\\equal{}c\\plus{}d\\plus{}e$: \\[ \\sqrt{a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}d^{2}\\plus{}e^{2}}\\geq T(\\sqrt a\\plus{}\\sqrt b\\plus{}\\sqrt c\\plus{}\\sqrt d\\plus{}\\sqrt e)^{2}\\]", "Solution_1": "I think T = 1/ sqrt(125) . Is it correct?", "Solution_2": "I don't think so. My guess is [hide]$ \\frac{\\sqrt5}{(\\sqrt{2}\\plus{}\\sqrt{3})^{2}}$.[/hide]", "Solution_3": "by putting $ a\\equal{}b\\equal{}3$ and $ c\\equal{}d\\equal{}e\\equal{}2$ we find $ \\frac{\\sqrt{30}}{6(\\sqrt{3}\\plus{}\\sqrt{2})^{2}}\\geq T$\r\nto prove that for such a $ T$ the inequality always holds notice that $ x^{2}\\plus{}y^{2}\\geq 2(\\frac{x\\plus{}y}{2})^{2}$ and $ x^{2}\\plus{}y^{2}\\plus{}z^{2}\\geq 3(\\frac{x\\plus{}y\\plus{}z}{3})^{2}$ which means that the worst case is when $ a\\equal{}b$ and $ c\\equal{}d\\equal{}e$ for which we have equality :)", "Solution_4": "Nice inequality.", "Solution_5": "My solution is the same as Zarif :P " } { "Tag": [ "MATHCOUNTS", "function", "percent", "ARML", "Gauss" ], "Problem": "Let us form the committee for the resurrection of the Illinois forum.\r\n\r\n:P\r\n\r\n(Apparently no one checks the state forums for spam, proof of this is shown by the missouri forum. :D )", "Solution_1": "yeah. missouri forum is fun though", "Solution_2": "Let us bash committees. \r\n\r\nBASH! BASH! BASH! \r\n\r\nYay! We have successfully bashed committees.\r\n\r\n...next?", "Solution_3": "[quote=\"suma_milli\"]Let us form the committee for the resurrection of the Illinois forum.\n\n:P\n\n(Apparently no one checks the state forums for spam, proof of this is shown by the missouri forum. :D )[/quote]\r\nwell then maybe we should not make the IL forum spammy :P", "Solution_4": "True! :lol: \r\n\r\nHere's a random problem from the mathcounts book I have next to me:\r\n\r\nA function $ (x,y)$ is defined by the following rules:\r\n$ A(0,n) \\equal{} n \\plus{} 1$\r\n$ A(m,0) \\equal{} A(m\\minus{}1, 1)$\r\n$ A(m,n) \\equal{} A(m\\minus{}1, A(m,n\\minus{}1))$\r\nfor m and n natural numbers.\r\n\r\nA(2,3) = k, where k is a whole number. Find the value of k.", "Solution_5": "Problem: If we make the IL forum spammy, fewer people will come. If fewer people will come, the forum will die.\r\n\r\nProblem: If we don't make the IL forum spammy, we won't think of enough stuff to post. If we don't think of enough to say, the forum will die.\r\n\r\nSo either way, the forum will die.\r\n\r\nI'm such an optimist. :rotfl:", "Solution_6": "That is quite a dilemma.. \r\n\r\nMaybe we should start an Illinois math marathon, b/c people from IL are cool enough to have their own math marathon. :P", "Solution_7": "[quote=\"suma_milli\"]True! \n\nHere's a random problem from the mathcounts book I have next to me:\n\nA function $ (x,y)$ is defined by the following rules:\n$ A(0,n) \\equal{} n \\plus{} 1$\n$ A(m,0) \\equal{} A(m \\minus{} 1, 1)$\n$ A(m,n) \\equal{} A(m \\minus{} 1, A(m,n \\minus{} 1))$\nfor m and n natural numbers.\n\nA(2,3) = k, where k is a whole number. Find the value of k.[/quote]\r\n\r\n[hide]A(2, 3)=A(1, A(2, 2))\nA(2, 2)=A(1, A (2, 1))\nA(2, 1)=A(1, A(2, 0))\nA(2, 0)=A(1, 1)\nA(1, 1)=A(0, A(1, 0))\nA(1, 0)=A(0,1)=2 :D\nA(1,1)=A(0,2)=3\nA(2,1)=A(1,3)\nA(1,3)=A(0, A(1,2))\nA(1, 2)=A(0, A(1,1))=4 :D \nA(1,3)=5=A(2,1)\nA(2,2)=A(1,5)\nA(1,5)=A(0, A(1,4))\nA(1,4)=A(0, A(1,3))=6 :D \nA(1,5)=7 :D \nA(2,3)=A(1,7)\nA(1,7)=A(0, A(1,6))=9 :D \nA(1,6)=A(0, A(1,5))=8 :D \n\nSo $ k \\equal{} 9$. :D \n\n :D :D :D !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n\nSomething's weird. IDK why avec never posts a solution to a math problem =([/hide]", "Solution_8": "Yup, that's it. You should hope nobody who answers potd sees your post :P", "Solution_9": "Did u like my long solution? And the way I put in mr. greens as i went along?\r\n\r\nHaha I'm so funny. :D :D :D (Usually that's just me online, but when u meet me in reality, I don't think that I seem that funny; i seem rather grave) :huh:", "Solution_10": "Guassy, rather grave? .___. *gasp*\r\n\r\nWhen you meet me I'm sure I seem like a complete idiot! :rotfl: :P", "Solution_11": "Why do you think so?\r\n\r\nI always imagined you look like :D \r\n\r\nOr how about this: wear green shirts or paint our faces green and then go meet at states and wave a Mr. Green flag? :D", "Solution_12": "That's a nice idea too. Not sure if my teammates will like it since I think green is the color of a rival school :P", "Solution_13": "Lol, green is actually one of my school's colors :D", "Solution_14": "My school's colors are blue and white. Like the school spirited arrow :arrow:", "Solution_15": "lol maybe the MO people will come and try to sabatoge us. Then again..we're the only people that ever post in the IL forum. :roll:", "Solution_16": "Because we're the only active Illinois AoPSers! :D", "Solution_17": ":D :D :D !!!", "Solution_18": ":D :D :D !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1 :D :D!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!", "Solution_19": "Congratulations everyone we have succeeded in the renewal of the IL forum! :D", "Solution_20": "Yes we did. Thanks to myself, webster, rose, kevin, and...MOST IMPORTANTLY:\r\n\r\n[hide]Smillie/Smiley Monster :P :D [/hide]\r\n\r\nfor the effort!\r\n\r\nxD\r\n\r\n :rotfl:", "Solution_21": "wow, you're actually referring to me by my name. that's rather rare on these forums.\r\nalthough i think going by a username alone makes me seem rather MYSTERIOUS and B.A.\r\n\r\n...\r\n\r\nrawr.", "Solution_22": "lololol I have a feeling that nearly all of the people who post here regularly know MY real name... :maybe: and perhaps we all know each other's... :huh:", "Solution_23": "hey when you go to ARML tryouts try to find me. i won't be able to find you because\r\n\r\n1. i will probably forget\r\n2. your facebook photos are not very helpful for identification purposes\r\n\r\nso yeah, say hello. although, it could be very difficult to look for people because the classroom (singular) we all take the tryout in is rather crowded.", "Solution_24": "[quote=\"Aryth\"]hey when you go to ARML tryouts try to find me. i won't be able to find you because\n\n1. i will probably forget\n2. your facebook photos are not very helpful for identification purposes\n\nso yeah, say hello. although, it could be very difficult to look for people because the classroom (singular) we all take the tryout in is rather crowded.[/quote]\r\nwell i might be able to find u\r\ni've seen lots of pics of u", "Solution_25": "I probably can't find you, but I'll wear my AoPS shirt.", "Solution_26": "[quote=\"avec_une_h\"]I probably can't find you, but I'll wear my AoPS shirt.[/quote]\r\nGood call. :D :D :D", "Solution_27": "Well since gauss already knows what I look like she can find me ( :| ) and then we can find aryth and we'll pretty much be able to tell who avec is unless there are multiple people wearing AOPS shirts.", "Solution_28": "just randomly yell out \"KHU\" when you think you see me. if the person in question turns around in a few circles, it's probably me :wink:", "Solution_29": "[quote=\"suma_milli\"]Well since gauss already knows what I look like she can find me ( :| ) and then we can find aryth and we'll pretty much be able to tell who avec is unless there are multiple people wearing AOPS shirts.[/quote]\r\ni doubt anyone would find me :| :huh: b/c i epicly failed at usamts.\r\nonly 12 on round 3. :mad: 3/3/2/1/3." } { "Tag": [ "IMO", "IMO 2006" ], "Problem": "There will be some admins of the ML/AoPs site at the IMO 2006: [url=http://imo2006.dmfa.si/sub_Coordinators.html]Valentin,[/url] [url=http://imo2006.dmfa.si/participants/CAN.html]Naoki,[/url] and [url=http://imo2006.dmfa.si/participants/USA.html]Richard.[/url]\r\n\r\n[b]Richard:[/b]\r\n\r\n[img]http://imo2006.dmfa.si/participants/Image757.jpg[/img]\r\n\r\n[b]Naoki:[/b]\r\n\r\n[img]http://imo2006.dmfa.si/participants/Image213.jpg[/img]\r\n\r\n[b]Valentin:[/b]\r\n\r\n[img]http://imo2006.dmfa.si/participants/ThumbImage905.jpg[/img]", "Solution_1": "Wow, Valentin looks much different than in the picture on this page: http://www.artofproblemsolving.com/AboutUs/AoPS_A_Company.php", "Solution_2": "Yeah, blame basketball and ageing for that :D", "Solution_3": "Why Basketball ?? :D", "Solution_4": "[quote=\"joml88\"]Wow, Valentin looks much different than in the picture on this page: http://www.artofproblemsolving.com/AboutUs/AoPS_A_Company.php[/quote]\r\n\r\nIndeed! In fact, I was a bit tired and did not realize it was he who took that picture of me at last year's hurricane party until after he'd left. :blush:", "Solution_5": "[b]Comparison:[/b]" } { "Tag": [ "puzzles" ], "Problem": "A machine gives out five pennies for each nickel inserted into it. The machine also gives out five nickels for each penny. Can you, starting out with one penny, use the machine in such a way as to end up with an equal number of nickels and pennies?", "Solution_1": "When you put one coin into the machine and gain five, the parity of the amount of coins you have remains the same. You start out with an odd amount of coins, so you will always have an odd amount. You can only have an equal number of both coins if you have an even amount of total coins, so it is impossible." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "prove if x,y,z is a real number then\r\n \\sum x 2 - \\sum xy \\geq 3/4(x-y) 2 :D", "Solution_1": "It can be written (a-c)^2+(b-c)^2>=(a-b)^2/2 and it's trivial." } { "Tag": [ "LaTeX", "parameterization" ], "Problem": "Hi,\r\nHow can put on a part of a latex document (a few lines) more margin on the left?\r\n\r\nThanks.", "Solution_1": "http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php", "Solution_2": "[quote=\"Yarka\"]Hi,\nHow can put on a part of a latex document (a few lines) more margin on the left?\n\nThanks.[/quote]\r\n\r\nTry this site:\r\n\r\n[url]http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Layout#Layout_of_the_Page[/url]\r\n\r\nYou set the parameters (described on the above link) to set margins, amount of paragraph indent, etc.", "Solution_3": "Those links deal with margins for the whole document rather than for a few lines.\r\n\r\nA very simple way is to use \\parbox. Suppose you want your text to be 6cm wide on the right (so a large margin on the left of that text). Put [code]\\hfill \\parbox{6cm}{Put the text you want on the right with a large margin here}[/code]" } { "Tag": [], "Problem": "This was once this very unusual man named John. What made him unusual was that whenever he was given an instruction, he would always obey it *exactly*. No matter what it was or what it said, he would carry out that instruction. One day he was sitting in the park when he noticed a sign. It was not an unusual sign - you could easily find it in any park. On seeing this sign, John went up to someone and said, very angrily, \"There is absolutely no way that you are putting me in that bin!\".\r\n\r\nThe puzzle: what was on the sign?\r\n\r\n[and if you think there are various answers, I will accept anything that fits these conditions. You may be surprised.]", "Solution_1": "the sign told him to say those exact words in that exact manner. cause um...sum ppl put it there as a joke?", "Solution_2": "[quote=\"TripleM\"]It was not an unusual sign - you could easily find it in any park.[/quote]\r\n\r\n(When I say easily find it, it could of course have slightly different wording, but thats not important. The important point was that the message on the sign was not unusual for a park.)", "Solution_3": "Too hard? Maybe I'll have to start giving out hints..", "Solution_4": "\"Keep off the grass\"? (If the bin was the only way to do so...)", "Solution_5": ":) I think every single person that I've seen try this puzzle has fallen into the same trap (including myself). That answer can't work. Read the puzzle statement more carefully.", "Solution_6": "Does it have anything to do with \"john\" being slang for a bathroom?", "Solution_7": "hm....is it a picture of sumone throwing this paper into a bin?", "Solution_8": "A sign saying \"put the porta-potty in the recycling bin\" ?\r\n\r\n\r\nIn my world, you see that all the time :P .", "Solution_9": "Nah, I just made up the name John. Its irrelevant. However, still none of your answers so far would cause John to act that way..", "Solution_10": "Oh!\r\n\r\nA sign saying \"Throw me away in this bin with an arrow pointing to a trash can\"\r\nA small sign right below shows a picture of a piece of garbage.", "Solution_11": "Please place the trash in the bin....\r\n\r\nIs John trash? :D", "Solution_12": "Still doesn't work. I didn't give you much information in the puzzle, but the most important part makes that answer not work, like the others :) hehehe.. don't worry, as I said, I kept making the same mistake as well.", "Solution_13": "Is the \"someone\" relevant to the solution? Is it a particular someone whose identity isn't disclosed, or is it some arbitrary stranger?", "Solution_14": "Nope, its just an arbitrary person. Theres no tricks in the wording - everything you need to know is in that paragraph.", "Solution_15": "MithsApprentice wrote:TripleM wrote:OK, I think its time to reveal the answer. In spoiler.\n[hide]\nRefuse to be placed in the bin.\n\n(of course, 'refuse' being another word for rubbish.)\n[/hide]\n\nYea that's definitely a New zealand thing...lol...i dunno of anyone in the US that uses refuse for garbage...\n\n\n\nThis is nothing to do with New Zealand.. its not used commonly here either (this puzzle was made up by an American and solved by Americans ) but you'll definitely find it in your dictionary. See - LynnelleYe knew it", "Solution_16": "[quote=\"TripleM\"]This is nothing to do with New Zealand.. its not used commonly here either (this puzzle was made up by an American and solved by Americans :P) but you'll definitely find it in your dictionary. See - LynnelleYe knew it :P[/quote]\r\n\r\nWell no, I mean I've heard of it and I knew what it meant...it's just...so...uncommon...\r\n\r\nAnd ur right it probly is in the dictionary...but oh well...", "Solution_17": "Wow, that is pretty neat.", "Solution_18": "first new post on this revived section", "Solution_19": "I agree with the above. It doesn't even seem obvious after you know it (which most answers to this type of problem do).\r\nMost Irish people would know and perhaps use the word \"refuse\" in that sense, if only very infrequently.", "Solution_20": "[quote=\"Fiachra\"]I agree with the above. It doesn't even seem obvious after you know it (which most answers to this type of problem do).\nMost Irish people would know and perhaps use the word \"refuse\" in that sense, if only very infrequently.[/quote]\r\n\r\nAs was discussed later in this thread (not here anymore) its basically the 'official' word for rubbish - rubbish would be used in everyday language, where 'refuse' is used when talking a lot more formally. I'm very surprised that someone would not be able to understand the answer.", "Solution_21": "The \"slogan\" on the garbage trucks around here is \"We don't refuse refuse.\"", "Solution_22": "In Taiwan (or at least in Keelung), the garbage trucks play \"Fr Elise.\" Weird; you'd think the smell would be fair warning enough...", "Solution_23": "That would be funny if they played the same song as the ice cream truck :P .", "Solution_24": "[quote=\"RobinHood3000\"]In Taiwan (or at least in Keelung), the garbage trucks play \"Fr Elise.\" Weird; you'd think the smell would be fair warning enough...[/quote]\r\n\r\n :lol: \r\nAt least that name contains the letters of 'refuse' ;) How about.. \"I L Refuse\"..", "Solution_25": "[quote=\"Treething\"]That would be funny if they played the same song as the ice cream truck :P .[/quote]\r\n\r\nJeez...a blind man with no sense of smell would be left standing on the street going, \"Hey...this ice cream tastes like toilet paper...\"", "Solution_26": "[quote=\"RobinHood3000\"][quote=\"Treething\"]That would be funny if they played the same song as the ice cream truck :P .[/quote]\n\nJeez...a blind man with no sense of smell would be left standing on the street going, \"Hey...this ice cream tastes like toilet paper...\"[/quote]\r\n\r\nWhy are blind people even on the streets?", "Solution_27": "If you don't have a sense of smell, can you actually taste anything anyway? Or was that what you meant :P", "Solution_28": "No...the implication was that it was ACTUAL toilet paper...\r\n\r\nAnd as for blind people in the streets, the blind need to shop too...", "Solution_29": "is this it???" } { "Tag": [ "logarithms", "quadratics", "LaTeX", "algebra" ], "Problem": "I need to see a solution for this one:\r\nFind all solutions of\r\n\r\n$x^{\\log_{10}x}=\\frac{x^{3}}{100}$", "Solution_1": "[hide]\n$x^{\\log_{a}x}=a$\nFollowing basic log rules.\nQED.[/hide]", "Solution_2": "Taking logarithm with base 10 of both sides, you will get the quadratic equation with respect to $\\log_{10}x$.", "Solution_3": "[quote=\"CircleSquared\"]\n$x^{\\log_{a}x}=a$\nFollowing basic log rules.\nQED.[/quote]\r\n\r\ndude wtf are you high? Thats not a log rule", "Solution_4": "I'm sorry, but I do not know how to use latex. Anyways, here is my solution:\r\n[hide]\ntake log base x of both sides and get\nlogx=logx(x^3)/100\nsimplifying the right side, you get\nlogx=3-logx100\nmove the log to the left side of the equation\n2logx10 + log x = 3\nset a=log x10\n2a + 1/a =3\nsolve for a and get a = 1/2 or 1\nso logx10 = 1/2 or 1\nsolving for x, you get x= 10 or 100[/hide]", "Solution_5": ":huh: Sorry bout that usaha (no I'm not high, that's illegal)\r\nI thought $x^{\\log_{a}x}=a$ but maybe I'm tired.\r\nWill someone correct me please?", "Solution_6": "[quote=\"CircleSquared\"]:huh: Sorry bout that usaha (no I'm not high, that's illegal)\nI thought $x^{\\log_{a}x}=a$ but maybe I'm tired.\nWill someone correct me please?[/quote]\r\nNo, you're getting that mixed up with $a^{\\log_{a}x}=x$", "Solution_7": ":wallbash_red: \r\nOk, that explains it, thanks diophantient.\r\nPlease ignore my solution.", "Solution_8": "usaha was joking..but anyways\r\na solution\r\n\r\ntake $\\log_{10}$ to both sides and you get\r\n\r\n$(\\log_{10}{x})^{2}=\\log_{10}{x^{3}}-\\log_{10}{100}$ so now you call $\\log_{10}{x}=y$ so its a quadratic equation and\r\n\r\n$y^{2}-3y+2=0$, so $y=\\log_{10}{x}=$$1$ and $2$, so $x=10$ and $100$\r\n\r\nnote I skipped a few steps involving knowing basic log properties such as $\\log_{a}{b^{x}}=x\\log_{a}{b}$", "Solution_9": "[quote=\"Whatthedeuce\"]I'm sorry, but I do not know how to use latex. [/quote]\r\n\r\nthat's perfectly alright... as long as you start learning now :D. I think that you can study, practice and use it well under half an hour.", "Solution_10": "heh, im also one of those who dont use latex very well, i've been trying to get around to learning it, but im sort of a procrastinator in that aspect, but back to the topic:\r\n\r\nim a beginner with logarithms, and basically just know what they are, so im not sure what you guys meant with \"take log 10 to both sides\" ?. But here's my method without that: \r\n[hide]\nLet a=log x , then we get:\n10^a= x, so \nwe have 10^ (a^a) = 10 ^ (3a-2), so by taking away the base of 10's we get a ^ a = 3a - 2 , so a = 1 and 2 (this part is sort of messy, and unreliable, so im sure you had better solutions) , so 10^a = 10 and 100. \nThat was pretty messy without latex, i should probably learn fast...\nCould someone explain how to take log of both sides, that would probably get rid of the messy part of my solution\nThanks [/hide]", "Solution_11": "I'm not sure why taking the log of both sides makes it simpler either, but it seems really important to know and I would like to know it...could someone show me? How exactly, does taking log base 10 of x simplify the left side of the equation?\r\n\r\nSorry, I phrased it wrongly. I get why it simplifies the left side, but I don't get how it actually ends up as (log x)^2. \r\n\r\nEDIT: But I think I figured out how it simplified the left side...Basically, by taking the log of both sides, you end up with (log x) to the (log x) power, and using one of the properties, you end up with (log x) times (log x), which is (log x)^2, right?", "Solution_12": "Takign the $\\log$ of both sides helps make the $x$ in the LHS no longer in the exponent.", "Solution_13": "[hide]Log of both sides\n$\\log_{x}(x^{3}/100)=\\log_{1}0x$\n$3-2\\log_{x}10=\\log_{1}0x$\n$3-\\frac{2}{\\log_{10}x}=\\log_{1}0x$\nSolving the quadratic, $\\log_{1}0x=1,2\\implies x=\\boxed{10,100}$[/hide]", "Solution_14": "thanks, i kind of get it now, assuming the right side is supposed to be log sub 10 x." } { "Tag": [ "linear algebra", "matrix", "search", "linear algebra unsolved" ], "Problem": "Prove the following theorem about stochastic matrices:\r\n\r\nLet $ A\\equal{}[a_{ij}]$ be a stochastic matrix, that is $ a_{ij}>0$, $ \\sum_{j\\equal{}1}^{n}{a_{ij}} \\equal{} 1$, then:\r\n\r\n1) A has got an unitary eigenvalue and the dimension of the associated eigenspace is 1\r\n2) Let $ A^n \\equal{} [a_{ij}^{(n)}]$ and let $ u\\equal{}(u_1, ..., u_n)$ be a normalized eigenvector with eigenvalue 1, then: $ \\displaystyle lim_{n \\to \\infty}{a_{ij}^{(n)}} \\equal{} u_i$", "Solution_1": "1) Clearly 1 is an eigenvalue since $ Av\\equal{}v$,where $ v\\equal{}(1,1, \\ldots ,1)^T$.\r\nSince $ A$ is primitive, by Perron-Frobenius theorem, the dimension of the associated eigenspace is 1.\r\n\r\n2) See http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=846370849&t=43670" } { "Tag": [ "number theory", "least common multiple", "prime factorization" ], "Problem": "what is the LCM of 1,2,3,4,5,6,7,8,9, and 10 \r\nhint: the lcm of 10, 9 and 8 are multiples of 1, 2, 3, 4, 5, 6, 8, 9, 10 :starwars:", "Solution_1": "[hide]Write out the prime factorization of each term. The highest power of 2 is 3. The highest power of 3 is 2. All other primes only have 1. That makes the lcm $2^3\\times3^2\\times5\\times7=2520$.[/hide]", "Solution_2": "[hide]1 = 1\n2 = 2\n3 = 3\n4 = 2 x 2\n5 = 5\n6 = 3 x 2\n7 = 7\n8 = 2 x 2 x 2\n9 = 3 x 3\n10 = 5 x 2\n2 x 2 x 2 x 3 x 3 x 5 x 7 = [b]2520[/b][/hide]", "Solution_3": "[hide]\n2, 4, and 6 are multiples of 8.\n3 and 6 is a multiple of 9.\n2*5=10\nSo, the answer is \n$8\\cdot9\\cdot5\\cdot7=2520$.[/hide]" } { "Tag": [ "floor function", "logarithms", "number theory open", "number theory" ], "Problem": "Define a sequence $a_{n}=a+(n-1)d$ where $(a,d)=1$ $,a,d,n\\in\\mathbb{N}$ and $a_{i}>0$ for all $1\\leq i\\leq n$ .\r\n\r\namong all the terms in $a_{n}$ , we define $p_{i}$ ,$1\\leq i\\leq n$ as the largest prime divisor of $a_{i}$ such that they form a set\r\n$S=\\{p_{1},p_{2}\\cdots ,p_{n}\\}$ .Suppose there exist a $p_{k}$ such that $p_{k}=\\text{max}\\{p_{1},p_{2}\\cdots ,p_{n}\\}$ . Prove or Disprove that $p_{k}$ is unique (or prove that if $p_{k}=p_{j}$ $\\implies k=j$).If it is false , give a counter-example .", "Solution_1": "Let me make a first step to the proof.\r\n\r\nAssume the contrary, i.e., that there is a sequence $a_{1},\\dots,a_{l}$ with two elements divisible by the maximum prime $p=p_{m}$ (I will use indexes to denote primes in their natural order: $p_{1}=2,\\ p_{2}=3,$ etc.). Without loss of generality, we assume that $p$ divides $a_{1}$ and $a_{l}.$\r\n\r\nSince $(a,d)=1$ we have:\r\n$\\gcd(a_{i},a_{j}) = \\gcd(a_{i},a_{i}+(j-i)d) = \\gcd(a_{i}, j-i)$ for any two indexes $1\\leq i 0.$ Then $\\int^{b}_{a}{(M-(-M))}= 2M(b-a ) < \\epsilon.$ Thus by Squeeze Theorem, $f_{n}$ is integrable, where $(n = 0, 1, 2, 3, ...)$, but the sum of integrable functions is integrable and since $\\int^{b}_{a}{f}= \\int^{c_{0}}_{a}{f_{0}}+...+\\int^{b}_{c_{n}}{f_{n+1}}$, $f \\in R[a, b]$.", "Solution_1": "Can anyone tell me if or not the preceding argument is valid, please?", "Solution_2": "No, your choice of epsilon does not make sense. Epsilon should be arbitrary.", "Solution_3": "You are right. May be I should have used $\\delta$ which depends on epsilon? \r\nDoes anyone have any suggestion or hint on how to fix the problem?", "Solution_4": "It's not going to come this easy. First of all, the discontinuities may be at the endpoints, so your splitting does not work. But modifying slightly what you did it is enough to show that if $f: (a,b) \\to \\mathbb{R}$ is continuous and bounded then it's integrable. What you should do is to take a compact subinterval $[a+\\epsilon, b-\\epsilon]$. On this interval you can use uniform continuity of $f$ to create upper and lower darboux sums with value arbitrarily close to each other. Then we add to these sums two pieces for the intervals $(a,a+\\epsilon)$ and $(b-\\epsilon,b)$ where you just use the constants from the boundedness. If you do this appropriately you can let $\\epsilon \\to 0$ to get upper and lower darboux sums on the interval $(a,b)$ arbitrarily close to each other in value.", "Solution_5": "Thanks Kalle.", "Solution_6": "Can't you use the Riemann criterion and the fact that f is bounded? Should be no sweat since you are dealing with finitely many discontinuities." } { "Tag": [ "calculus", "integration", "number theory unsolved", "number theory" ], "Problem": "Given a positive integer $n$ and an odd prime $p$.Let $a,b$ be integers s.t.$a^{2}+b^{2}=p^{2n}$.Denote $S=\\{(x,y)|(x-\\frac{a}{2})^{2}+(y-\\frac{b}{2})^{2}=\\frac{p^{2n}}{4},x,y\\in %Error. \"mathbox\" is a bad command.\n{Z}\\}$.Show that there is a subset $T$ of $S$ with $|T|\\geq \\frac{|S|}{2}$ such that for any elements $u=(x_{1},y_{1}),v=(x_{2},y_{2}) \\in T$,we have the property that $|u-v|=\\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$ is an integer.", "Solution_1": "Still none? :(" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "For which $ n \\geq 2$ does $ a^n \\equal{} a$ for every $ a \\in R$, where $ R$ is a ring, imply that $ R$ is commutative.", "Solution_1": "All of them, see http://www.mathlinks.ro/viewtopic.php?t=129148 ." } { "Tag": [ "limit", "induction", "absolute value" ], "Problem": "Is there a way to prove that\r\n\r\n$\\lim_{x\\rightarrow\\infty}\\left(\\frac{F_{x+1}}{F_{x}}\\right) = \\frac{1+\\sqrt{5}}{2}$?\r\n\r\nWhen $F$ represents the Fibonnaci series?\r\n\r\nThis is more of a curiosity thing than a question.", "Solution_1": "[hide=\"to do it rigorously\"]\nLet $\\phi=\\frac{1+\\sqrt{5}}{2},\\ \\bar{\\phi}=\\frac{1-\\sqrt{5}}{2}$ so that $x^{2}-x-1$ has roots $\\phi$ and $\\bar{\\phi}$, and $d_{n}=\\left| \\phi-\\frac{F_{n+1}}{F_{n}}\\right|$ so we need to show that $d_{n}$ approaches 0.\n\nNow we have\n\\[\\frac{1}{\\phi F_{n}-F_{n+1}}=\\frac{\\bar{\\phi}F_{n}-F_{n+1}}{(\\phi F_{n}-F_{n+1})(\\bar{\\phi}F_{n}-F_{n+1})}\\]\nSince $\\phi$ and $\\bar{\\phi}$ have sum 1 and product -1, the denominator of the above fraction is equal to $F_{n+1}^{2}-F_{n}^{2}-F_{n}F_{n+1}$. But it's easy to prove by induction that this expression equals $(-1)^{n}$ (if $F_{0}=0$) because\n\\[(F_{n}+F_{n-1})^{2}-F_{n}^{2}-F_{n}(F_{n}+F_{n-1})=-(F_{n}^{2}-F_{n-1}^{2}-F_{n}F_{n-1})\\]\nTherefore, we have\n\\[d_{n}=\\frac{|\\phi F_{n}-F_{n+1}|}{F_{n}}=\\frac{1}{F_{n}(F_{n+1}-\\bar{\\phi}F_{n})}\\]\n(Note that $-\\bar{\\phi}$ is positive.) Now it's clear that the absolute value of the difference tends to 0 because Fibonacci numbers get arbitrarily large.\n[/hide]", "Solution_2": "we could just use binet's formula\r\n\r\n[expects a negative post rating...although there is nothing else to say...]", "Solution_3": "whats binet's formula?", "Solution_4": "[quote=\"buzzer11\"]whats binet's formula?[/quote]\r\nThe closed form for the Fibonaccis. Using the notation in my solution,\r\n\\[F_{n}=\\frac{\\phi^{n}-\\bar{\\phi}^{n}}{\\sqrt{5}}\\]" } { "Tag": [ "topology", "complex analysis", "real analysis", "advanced fields", "advanced fields theorems" ], "Problem": "I'm a high school student and am beginning a self-study of topology. This question is sure to be standard and I'm sorry if this is not the right place!!!\r\n\r\nAnyways, I want to prove that the open set in the Eucildean space i.e.\r\n\r\nGiven a point [b]p[/b] of [b]X[/b] and a non-negative real\r\nnumber r, the open ball $B_X (p; r) $ in X of radius r about [b]p [/b]i s defined to be\r\nthe subset of X given by\r\n${ B_X (p; r) \\equiv { x \\in X : |x - p}| < r} $\r\n\r\n, is a topologcial space. Also, why is the closed ball in Euclidean space NOT a topological space.\r\n\r\nBasically I am asking for a proof of the fact that the definiton of open sets which comes from topogical spaces( i.e. if $(X, T) $ is a topolgical space, then the elements of $T$ are called open sets) is the same as that which comes intuitively (in real spaces). Also, why is the definiton of topology $T$ relate to \"open \" sets as we know them only?? Why can't closed sets be included. Also, why is it that we consider a finite collection to form the intersection in the definition of topology??? Why don't closed sets form topological spaces??\r\n\r\nCan anybody give recomendations of sites or web pages which cover introductory topology at a very basic level!!!", "Solution_1": "What do you want to prove, that the open sets (in the traditional sense) in $\\mathbb{R}$ form a topological space?\r\n\r\nThis is not hard. $\\mathbb{R}$ is open, as is $\\emptyset$. Now you just need to show that finite intersections of open sets are open. Say that some point $x$ lies in the intersection of finitely many open sets $S_i$. Then for each $i$, there is an open ball $B(x,r_i)$ such that $B(x,r_i)\\subset S_i$, and it is clear that $B(x,\\min_i r_i)\\subset \\cap_i S_i$. So finite intersections are open. Finally, for unions, if $x$ lies in the union of some open sets, then $x\\in S_i\\subset\\cup_i S_i$, where $i$ may run over any set of indices we like.\r\n\r\nFor closed sets, we would like to show that some union of closed balls is not closed. This is NOT true if the union runs over a finite number of balls (prove it!). But over countable unions, this is not a problem - we have\r\n\r\n$\\cup^{\\infty}_{i=1} [1/i,1]=(0,1]$\r\n\r\nwhich is not closed.\r\n\r\nFinally, a topological space is quite an abstract concept. Open sets in a general topological space may be quite different than the open sets in $\\mathbb{R}$. This is because $\\mathbb{R}$ has many intrinsic properties that make working with it 'nice' - it is, for example, a locally compact Hausdorff space - and they don't generalize to other spaces.\r\n\r\nA final note is that topological spaces do not necessarily have metrics. Thus things like distance, which we use without thought in $\\mathbb{R}$, do not necessarily have analogues in topological spaces (although metric spaces are topological spaces).\r\n\r\nThis seems to have a lot of general knowledge on topology.\r\n\r\nhttp://en.wikipedia.org/wiki/Topological_space", "Solution_2": "So the best definiton of open sets is the one which comes from the definiton of a topolgy $T$.. Thanks for your proof. Just a thing.. I still haven't understood clearly why closed sets(in a general case) don't make up a topolgy. Can you elaborate a bit more??\r\n\r\nAlso, I was asking for some excellent introductory lecture notes and e-books. Do you have any in mind???", "Solution_3": "I would also like to ask why we take finite intersections and arbitrary no, of unions in the definiton of topology???\r\n\r\nI'm not clear regarding infinte sets , infinte intersection. unions and such arbitrary operations on sets. Can you suggest some resource on the net(lecture notes etc.) from where I can clarify my concept???", "Solution_4": "Because underneath all the generalizations, we would like our definitions to conform with nice, simple examples that we know already (e.g the real line as a metric space)\r\n\r\nI don't know any such notes - the topology I know is derived from Rudin's Real and Complex Analysis and the topology portion of Folland's Real Analysis: Modern Techniques and their Applications.\r\n\r\nBut as always, MIT's opencourseware is a good place to start.", "Solution_5": "Can you please elaborate. I don't fully understand your statement. Also is my Understanding(as I have posted in my post #2) of open sets etc. correct. ...\r\n\r\nI checked OCW (always first) but the topology course material does'nt contain the starting chapters (basic topology!!!) Any other suggestions???", "Solution_6": "Try this online book \r\nhttp://uob-community.ballarat.edu.au/~smorris/topbookchaps1-8.pdf\r\n\r\nAt least the typesetting is nice :)\r\nIf you are not happy with that one, you can try another one here:\r\nhttp://www.econphd.net/roessler/notes.htm", "Solution_7": "Thanks a lot!!! Can you answer my doubts!!!", "Solution_8": "[quote]Basically I am asking for a proof of the fact that the definiton of open sets which comes from topogical spaces( i.e. if $(X, T) $ is a topolgical space, then the elements of $T$ are called open sets) is the same as that which comes intuitively (in real spaces). Also, why is the definiton of topology $T$ relate to \"open \" sets as we know them only?? Why can't closed sets be included. Also, why is it that we consider a finite collection to form the intersection in the definition of topology??? Why don't closed sets form topological spaces??[/quote]\r\n\r\nI don't know if this can answer your question.\r\n1) Yes, the definition of topology (in some sense) is the abstraction of the behaviour of open intervals in $\\mathbb{R}$. Why we only take finite intersection? because if we take infinite intersection of open intervals the result might be not longer an interval. Example:\r\n\\[\r\n\\bigcap_{i=1}^\\infty \\left(-\\frac{1}{n},\\frac{1}{n}\\right)=\\{0\\}\r\n\\]\r\n(since if $-1/n\\frac{\\pi}{2004}\\]", "Solution_1": "Perfom substitution $x=t^{1000}$ and integrate by parts some times to obtain $\\int >\\frac{\\pi}{2002}-\\frac{\\pi^3}{2^3\\cdot 1001\\cdot 1002}\\cdot\\frac{1}{5}>\\frac{\\pi}{2004}$." } { "Tag": [], "Problem": "Determine all positive integers $ (x,n)$ such that $ 1\\plus{}x\\plus{}x^{2}\\plus{}...\\plus{}x^{n}\\equal{}40$.", "Solution_1": "[hide=\"Solution\"]Note that, for $ x\\neq 1$ we have\n\n$ \\frac {x^{n \\plus{} 1} \\minus{} 1}{x \\minus{} 1} \\equal{} 40$\n\n$ x(40 \\minus{} x^n) \\equal{} 39$.\n\nCase 1. $ x \\equal{} 1$, obviously, $ n \\equal{} 39$ (not from the derived equalities, but form the original one).\n\nCase 2. $ x \\equal{} 39$ gives $ 40 \\minus{} 39^n \\equal{} 1$, i.e. $ n \\equal{} 1$.\n\nCase 3. $ x \\equal{} 3$ gives $ 40 \\minus{} 3^n \\equal{} 13$ i.e. $ 3^n \\equal{} 27$, ergo $ n \\equal{} 3$.\n\nCase 4. $ x \\equal{} 13$ gives $ 40 \\minus{} 13^n \\equal{} 3$ i.e. $ 13^n \\equal{} 37$ which yields no natural $ n$.\n\nTo sum up, solutions are\n\n$ (1,39),(39,1),(3,3)$.[/hide]" } { "Tag": [ "geometry", "incenter", "geometry unsolved" ], "Problem": "Let $ABC$ be a triangle with $BC=a$ $AC=b$ $AB=c$. For each line $\\Delta$ we denote $d_{A}, d_{B}, d_{C}$ the distances from $A,B, C$ to $\\Delta$ and we consider the expresion $E(\\Delta)=ad_{A}^{2}+bd_{B}^{2}+cd_{C}^{2}$. Prove that if $E(\\Delta)$ is minimum, then $\\Delta$ passes through the incenter of $\\Delta ABC$.", "Solution_1": "[hide]Prove that if $\\Delta$ doesn't pass through $I$, and $\\Delta'$ passes through $I$, and $\\Delta\\parallel \\Delta'$ then $E(\\Delta)>E(\\Delta')$[/hide]", "Solution_2": "Whait is \"Romanian DMO\" :?:", "Solution_3": "Romanian District Math Olympiad" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let's say that we \"concatenate\" two numbers when we write one beside of the other one; e.g. when we concatenate 200 and 6, we get 2006, and when we concatenate 6 and 200 we get 6200. Find two six-digit numbers such that the number obtained by concatenating them is divisible by their product.", "Solution_1": "Let x and y be two six-digit such that the number obtained by concatenating them is divisible by their product, i.e.\r\n\r\n$(1) \\; xy \\, | \\, 10^6 x + y.$\r\n\r\nThis implies that x|y, i.e. y = xz for a natural number z. Hence\r\n\r\n$(2) \\; xz \\,|\\, 10^6 \\: +\\: z$\r\n\r\naccording to (1). Now $z \\,|\\, 10^6$ and z = y/x < 10 since the integers x and y have the same number of digits. So $z \\in \\{1,2,4,5,8\\}.$ Futhermore $xz \\: =\\: y \\: < \\: 10^6 \\: < \\: 10^6 \\: +\\: z,$ which combined with (2) gives\r\n\r\n$(3) \\; nx \\;=\\; \\frac{10^6}{z} \\: +\\: 1$\r\n\r\nwhere n is an integer in the intervall [2,10]. Obviously $\\frac{10^6}{z}$ is an even integer, so nx is an odd integer by (3). Consequently $n \\, \\geq \\, 3$, which again implies that $nx \\: \\geq \\: 3\\cdot10^5 \\: +\\: 1$. Therefore $z \\: \\leq\\: \\frac{10}{3}$ by (3), which leaves us with two possible choices for z:\r\n\r\n$\\bullet \\; z = 1 \\; \\Rightarrow \\; nx = 1000001$, which is impossible since 1000001 don't have odd divisors < 10.\r\n\r\n$\\bullet \\; z = 2 \\; \\Rightarrow \\; nx = 500001$. The only odd divisor of 500001 which is < 10, is 3. So the only possible solution is n = 3, giving \r\n\r\n[b]x[/b] = 500001/3 = [b]166667[/b] and [b]y[/b] = xz = 2x = 2*166667 = [b]333334[/b]." } { "Tag": [ "inequalities", "number theory unsolved", "number theory" ], "Problem": "Find all positive integer solutions of the following equation\r\n\r\nx^3-y^3=xy+61", "Solution_1": "[hide=\"hint\"]\nuse inequality and try to reduce the maximum value that $ \\max \\{x,y\\}$ can take write the equation in this form :\n$ (x\\minus{}y)(x^2\\plus{}y^2\\plus{}xy)\\equal{}61\\plus{}xy$ or in this form $ x^3\\plus{}(\\minus{}y)^3\\plus{}(\\minus{}1)^3\\minus{}3(x)(\\minus{}y)(\\minus{}1)\\equal{}60\\minus{}2xy$\n[/hide]", "Solution_2": "Since $ x>y$,we let $ x\\equal{}y\\plus{}z$,where $ z>0$.The problem is much easier then.The equation has the degree of $ 2$." } { "Tag": [ "geometry", "trigonometry" ], "Problem": "If xE $[0,2\\pi]$ , the numbers of solutions $sen^3x-senx+1=cos^2x$ is equal ?", "Solution_1": "[quote=\"felipesa\"]If xE $[0,2\\pi]$ , the numbers of solutions $sen^3x-senx+1=cos^2x$ is equal ?[/quote]\r\n\r\n$cos^2 x=1-sin^2 x$ and we can see that $sin^3 x +sin^2 x -sin x=0$ which gives us $2$ solution $sin x =0$ and $sin x =\\frac{-1+\\sqrt{5}}{2}$", "Solution_2": "Right idea, I believe, but since it asks for the number of solutions for $x$ and not $\\sin{x}$ you have to consider the specifed interval. \r\n[hide]Sketching a graph of $y=\\sin{x}$ from $0$ to $2\\pi$ it is clear that there are 5 solutions.[/hide]", "Solution_3": "What's senx? Is it a typo?", "Solution_4": "Yeah, I think it's just : $sin x$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For $a,b,c$ positive real numbers such that $a+b+c=6$ prove that\r\n\r\n$\\frac{a}{1+b}+\\frac{b}{1+c}+\\frac{c}{1+a} \\geq 2 $", "Solution_1": "Isn't this trivial??\r\n\r\nFrom Cauchy,\r\n\r\n$\\sum_{cyclic} \\frac{a}{1+b} = \\sum_{cyclic} \\frac{6a}{a+7b+c}$\r\n\r\n$\\geq \\frac{6(a+b+c)^2}{a(a+7b+c)+b(b+7c+a)+c(c+7a+b)} \\geq 2$,\r\n\r\nsince the last inequality is equivalent to $a^2+b^2+c^2 \\geq ab+bc+ca$." } { "Tag": [ "MIT", "college", "LaTeX" ], "Problem": "Sorry to bother you guys again..\r\n\r\nThere's something called \"creativity essay\" in the application.. Out of curiosity, how many of you (who got accepted to MIT) wrote this completely optional essay?", "Solution_1": "i did it , i got in", "Solution_2": "i did it , i got in", "Solution_3": "Same here.", "Solution_4": "I did, and I got in too", "Solution_5": "huh. I completely overlooked that... :blush: \r\nwell, lucky me, I guess.", "Solution_6": "who said it had to be an essay? you can send pictures, diagrams, equations, whatever...", "Solution_7": "Just that it's quite difficult without the formatting to give equations unless you are sending them by post or anything.\r\n\r\nI was wondering...for text without formatting, is it a convention to write\r\na_(n-1) for $a_{n-1}$?\r\n\r\nIf not, what do people normally do?", "Solution_8": "[quote=\"ak007\"]\nI was wondering...for text without formatting, is it a convention to write\na_(n-1) for $a_{n-1}$?\n\nIf not, what do people normally do?[/quote]\r\n\r\nYeah, that's usually how people write subscripts. Although, when I'm talking to someone else who know $\\LaTeX$ I'll often times just write it as $\\LaTeX$ code. So, for example, I would write a_{n-1}.", "Solution_9": "So, why dont you guys post what you sent along in the creative essay... the deadline's past now so no danger of plagiarism!!!\r\n\r\nAnyways, I discussed a problem on Special Relativity that I thought about (look around mathlinks - there's some cool discussion on it here!) and a famous painting that I like ( 'Relativity' by M.C. Escher!) I correlated them and discussed my problem!\r\n\r\nDo you think its good!! Remember, i'm an international!!! \r\nOpinions please!! Thanks!\r\n\r\nAlso, love to hear about your submissions!!", "Solution_10": "That's really a nice idea!\r\n You've got a good chance alright, there's no doubt about that!" } { "Tag": [ "function" ], "Problem": "prove $ sinh{A}\\equal{}\\minus{}sinh{\\minus{}A}$", "Solution_1": "hello, do you mean $ \\sinh(x)\\equal{}\\minus{}\\sinh(\\minus{}x)$?\r\nSonnhard.", "Solution_2": "yes I do :)", "Solution_3": "hello, let $ f(x)\\equal{}\\frac{e^x\\minus{}e^{\\minus{}x}}{2}$, it follows $ f(\\minus{}x)\\equal{}\\frac{e^{\\minus{}x}\\minus{}e^x}{2}\\equal{}\\minus{}\\frac{e^x\\minus{}e^{\\minus{}x}}{2}\\equal{}\\minus{}f(x)$.\r\nSonnhard.", "Solution_4": "that doesn't seem to follow to me.", "Solution_5": "hello, why? which step is not clear?\r\nSonnhard.", "Solution_6": "well what you've said is true but it doesn't seem to answer the question. You will, no doubt, point something very obvious out to me and I will feel silly, but there we go :)", "Solution_7": "hello, we say that $ f(x)\\equal{}\\sinh(x)$ is a odd function, because it is $ \\sinh(\\minus{}x)\\equal{}\\minus{}\\sinh(x)$ for all real $ x$.\r\nSonnhard.", "Solution_8": "$ \\sinh(x) \\equal{} \\frac {e^x \\minus{} e^{ \\minus{} x}}{2}$ by definition.\r\nSo $ \\minus{}\\sinh( \\minus{} x) \\equal{} \\minus{} \\frac {e^{ \\minus{} x} \\minus{} e^x}{2} \\equal{} \\frac {e^x \\minus{} e^{ \\minus{} x}}{2} \\equal{} \\sinh(x)$." } { "Tag": [ "function", "Euler", "number theory", "totient function", "number theory unsolved" ], "Problem": "Prove that for all positive real numbers $ a , b$ such that : $ a > b$ \r\n\r\nthen there exists a positive integer $ n$ such that :\r\n\r\n $ b < \\frac {\\varphi(n \\plus{} 2)}{\\varphi(n) } < a$", "Solution_1": "[quote=\"nguyenvuthanhha\"]Prove that for all positive real numbers $ a , b$ such that : $ a > b$ \n\nthen there exists a positive integer $ n$ such that :\n\n $ b < \\frac {\\varphi(n \\plus{} 2)}{\\varphi(n) } < a$[/quote]\r\n\r\nCould you define the function, please?", "Solution_2": "[quote=\"tringo\"][quote=\"nguyenvuthanhha\"]Prove that for all positive real numbers $ a , b$ such that : $ a > b$ \n\nthen there exists a positive integer $ n$ such that :\n\n $ b < \\frac {\\varphi(n \\plus{} 2)}{\\varphi(n) } < a$[/quote]\n\nCould you define the function, please?[/quote]\r\nIn number theory $ \\varphi(n)$ is normally defined as [url=http://en.wikipedia.org/wiki/Euler%27s_totient_function]Eulers phi function[/url]", "Solution_3": "very nice problem but hard indeed. i've been working quite a lot on this one some time ago and failed to solve it. i also think it has been posted before but without a solution.\r\n\r\nits not hard to proove that $ \\frac{\\phi (n\\plus{}2)}{\\phi (n)}$ can get arbitrary small or arbitrary large. for example how to get arbitrary large... take $ n\\plus{}2$ to be a prime number congruent to $ 2$ by modulo $ \\prod_{i\\equal{}2}^{k} p_i$ where $ p_i$ denotes the i-th prime number. such a number exists by dirichlet's theorem. if we send $ k$ to infinity $ \\frac{\\phi (n\\plus{}2)}{\\phi (n)}$ gets arbitrary large.\r\n\r\nvery simmilar for arbitrary small. i however faild to proove from this that kvotient can get arbitrary close to any real number", "Solution_4": "[i] I have asked some excellent Number theory lovers i known about this problem but they told me that they cannot solve it\n\n Darij , Peter ,Can you help me ? :( [/i]", "Solution_5": "[i]Zetax , where are you ? :( [/i]", "Solution_6": "Well, I won't say anything new to the topic, but we've been struggling with a weaker one ($ a,b\\in \\mathbb{N}$) on a polish math forum for a few months, and we came up with nothing besides what Jure wrote. What is more I gave that problem to a really great mathematician and he failed to solve it (he was trying for a few hours). \r\nnguyenvuthanhha, you say it's from Italian olympiad? It seems impossible, the problem looks a lot too hard even for the IMO... But that's only my opinion :wink: I'd love to hear what ZeTaX thinks about it... Maybe it is a well-known problem?", "Solution_7": "very nice problem...I wanna see the proof.", "Solution_8": "some one please post a solution", "Solution_9": "[i]I am not sure about the source but i guess that this problem is right\n\n It has been appeared in Pen and Komal ( without solution :) )[/i]\r\n\r\n\r\n You can see here :\r\n\r\n http://www.komal.hu/verseny/feladat.cgi?a=honap&h=200701&t=mat&l=en", "Solution_10": "[i]Give up !\n\n I am really far from convinced that Italy - a country not famous for Mathematics is author of this problem[/i]", "Solution_11": "well looking at the statistics of that problem tells you that 2people did solve it so... either it is solvable in some unexpected but straightforward way or they got something wrong because it seems very hard", "Solution_12": "Some trivial remark: it is enough to consider rational $ a$ and $ b$ (not sure that it simplifies the problem...).\r\nAs Jure said, a similar problem, with $ n\\plus{}1$ instead of $ n\\plus{}2$ appears on PEN - problem $ J 9$. The source of the problem is mentioned, and I checked it, and the problem is mentioned very briefly, as a result by Schinzel and Sierpinski.\r\nSo I think that either the $ n\\plus{}2$ makes it a lot easier, or it is very hard. In either case, we shall keep looking for a solution. It is a very original problem and that what makes it interesting.", "Solution_13": "Well, Schinzel and Sierpi\u0144ski are very famous in Poland (and not only), they were known as a pair of faboulous number theory problem solvers with numerous publications etc.\r\nI can hardly believe there exists an elementary way to approach that problem.\r\nOn my forum one of the user's (xiikzodz ) tried to consider a convex framing and pawn the problem in a very non-elementary way... (of course it was a problem based upon natural numbers)\r\nhttp://matematyka.pl/77192.htm\r\nIf anyone shoulod need it, I could translate fragments of ideas :wink:", "Solution_14": "[i]Two months have passed . I am looking for a solution[/i] :lol:", "Solution_15": "Sir , Nam D\u0169ng , where are you ? :maybe:", "Solution_16": "You could prove this beatiful result by using a cool theorem deu to Schinzel: For any $ b_1,b_2,\\ldots,b_m$ nonegative number then exist a sequence $ \\{a_k\\}$ so that $ \\frac {\\phi(a_k \\plus{} i)}{\\phi(a_k \\plus{} i \\minus{} 1)}\\to b_i$ when $ k\\to\\infty$ for all $ i \\equal{} 1,...,m$. \r\n\r\nIndeed, take $ n \\equal{} a(a \\plus{} 3)$ then $ n \\plus{} 2 \\equal{} (a \\plus{} 1)(a \\plus{} 2)$ where $ 3\\nmid a$. So $ \\frac {\\phi(n \\plus{} 2)}{\\phi(n)} \\equal{} \\frac {\\phi(a \\plus{} 1)}{\\phi(a)}\\times\\frac {\\phi(a \\plus{} 2)}{\\phi(a \\plus{} 3)}$. Take $ h\\geq 7$ to large enough so that such $ a_k \\plus{} i$ is not divisible by $ 3$ exist and $ i \\plus{} 4 < m$. The result follows. \r\n\r\nI suppose that there exist a more simple solution.", "Solution_17": "Let $ n \\equal{} 4m$, then must be\r\n(1) $ 2b < \\frac {\\phi(2m \\plus{} 1)}{\\phi(2m)} < 2a$.\r\nWe can chose $ 2m \\plus{} 1 \\equal{} (\\prod_{2 < p < c}p)^k$. Then $ \\frac {\\phi(2m \\plus{} 1)}{2m \\plus{} 1} \\equal{} \\prod_{1 < p < c}(1 \\minus{} \\frac 1p) \\equal{} t(c)$ - independent from k, $ \\frac {\\phi(2m)}{2m} \\equal{} \\frac 12 \\prod_{q|d^k \\minus{} 1}, d \\equal{} d(c) \\equal{} \\frac 12 \\prod_{p < c}p$.\r\nWe can chose c and k, suth that satisfyed (1)." } { "Tag": [ "geometry", "geometric transformation", "reflection", "analytic geometry", "number theory proposed", "number theory" ], "Problem": "Four frogs are positioned at four points on a straight line such that the distance between any two neighbouring points is 1 unit length. Suppose the every frog can jump to its corresponding point of reflection, by taking any one of the other 3 frogs as the reference point. Prove that, there is no such case that the distance between any two neighbouring points, where the frogs stay, are all equal to 2008 unit length.", "Solution_1": "if I got it correctly,the problem seems to be too simple.\r\nAs it traditionally happens,we assume the contrary,so after some \"jump\" we get the described situation, and let's make all jumps in a reverse order, but as it is easy to observe that starting from that position,any distances between frogs must be divisible by $ 2008$,contradiction.", "Solution_2": "Another solution - use invariants. Colour the points on x (with integer coordinates) black, white, black, white... When we start 2 frogs are on white and 2 on black points. We see that after every jump the color of the frog point does not change, but if all distances are equal to 2008 all four points ale coloured the same - contradiction.", "Solution_3": "My solution much uglier than above two :) \r\nAssume contrary.Denote points in line as A,B,C,D in turn.Choose any point in our line as origin of coordinates.Denote coordinates of A,B,C,D after n-th \"jumping\" as $ a_n$,$ b_n$,$ c_n$,$ d_n$ respectively.denote $ l_n$ as sum of distances between A and B,B and C,C and D after n-th \"jumping\".$ l_0\\equal{}3$.Notice that $ l_n\\equiv b_n\\minus{}a_n\\plus{}c_n\\minus{}b_n\\plus{}d_n\\minus{}c_n \\equiv d_n\\minus{}a_n\\mod 2$.and it is easy to understand that $ d_n\\minus{}a_n\\equiv d_{n\\plus{}1}\\minus{}a_{n\\plus{}1}\\mod2$,so as we can see $ l_n \\equiv l_0\\mod2$.We get contradiction." } { "Tag": [ "ratio", "probability" ], "Problem": "Albatross was very angry that Frankinfueter killed his pets, so he decided to go to the local supermarket and buy poison to poison Frankinfueter. When he goes there, he sees rat poison, which costs three dollars and has a 40% chance of poisoning people. He also sees cat poison, which costs six dollars fifty cents and has an 80% chance of poisoning people. Finally, he sees a poison making kit that has a 20% chance of killing people but only costs one dollar and twenty-five cents. Which poison should he buy to be the most cost-effective?", "Solution_1": "Assuming we're using simple ratios and not some other complex form of math that doesn't belong in getting started:\r\n\r\n[hide=\"My Solution\"]40/3=13.3...\n80/6.5=12.3...\n20/1.25=16\nThe cat poison is most cost-effective.[/hide]", "Solution_2": "What I did was set up ratios to find out which one would kill in the least amount of money. \r\n\r\n[hide]\nSo we have: \nrat poison: \n40%x = 3 \nx = 7.5\n\ncat poison: \n80%x = 6.5\nx = 8.125\n\nmake it yourself kit: \n20%x =1.25\nx = 6.25\n\nSo the kit is the most cost effective! \n\nOn a side note, is there even such thing as cat poison? :lol: \n\n[/hide]", "Solution_3": "Yes, I patented cat poison .\n\n\n\n\n\nIt's just simple ratios.\n\n\n\nRC-7th:\n\n[hide]\n\nRC-7th, your method is valid, except you're looking for the highest number in your case I think . If human poison poisoned 100% of the time and cost 3 dollars, it would be rated as 33.3 with your method. Obviously, the higher the number, the better it is if you do it your way.\n\n[/hide]", "Solution_4": "[quote=\"Treething\"]Yes, I patented cat poison :P .[/quote]\r\n\r\nHave you patented other items which don't work well? I'd like to send them to some people.", "Solution_5": "I don't think that this is as easy as it seems, consider the following simple example:\r\nAt the casino I can play on two slot machines,\r\non machine A I have a 40% chance of winning the jackpot, each attempt costs 1 dollar.\r\non machine B I have an 80% chance of winning the same jackpot, each attempt costs 2 dollars.\r\n\r\nI can only win the jackpot once, but I have to decide how many times i'm going to play each machine at the start and I have to pay that amount upfront. Are the two choices equivalent ?\r\n\r\nSuppose I spend 2 dollars on machine A and 2 dollars on machine B. I play one dollar on machine A, if I win then I stop, If I lose then I play again. What chance do I have of winning?\r\n\r\nThen I play machine B, I only have one attempt with an 80% chance of winning.\r\n\r\nWhich machine represents the best value for money?\r\n\r\n(I've tried to make this as equivalent to the previous question as possible)", "Solution_6": "[quote=\"RC-7th\"][quote=\"Treething\"]Yes, I patented cat poison :P .[/quote]\n\nHave you patented other items which don't work well? I'd like to send them to some people.[/quote]\n\nYes, I've patented K'Nex crossbows that fire up to 20 feet and use blunt bolts :P .\n\n[quote=\"bradpreston\"]\nI don't think that this is as easy as it seems, consider the following simple example:\nAt the casino I can play on two slot machines,\non machine A I have a 40% chance of winning the jackpot, each attempt costs 1 dollar.\non machine B I have an 80% chance of winning the same jackpot, each attempt costs 2 dollars.\n\nI can only win the jackpot once, but I have to decide how many times i'm going to play each machine at the start and I have to pay that amount upfront. Are the two choices equivalent ?\n\nSuppose I spend 2 dollars on machine A and 2 dollars on machine B. I play one dollar on machine A, if I win then I stop, If I lose then I play again. What chance do I have of winning?\n\nThen I play machine B, I only have one attempt with an 80% chance of winning.\n\nWhich machine represents the best value for money?\n\n(I've tried to make this as equivalent to the previous question as possible)\n[/quote]\r\n\r\nWell no, in this problem, he only buys one container.", "Solution_7": "[quote=\"Treething\"]\nWell no, in this problem, he only buys one container.[/quote]\r\n\r\nYes, but do you think a 40% chance of achieving a certain outcome is worth half of an 80% chance of acheiving the same outcome?", "Solution_8": "The probablity of winning on the first machine is $4/10+6/10*4/10=16/25$. The probablity of winning on the second machine is $8/10=20/25$. So the second machine is worth more.", "Solution_9": ",but in LA casino, there will be alot of fake ilegal machines that never makes the jackpot...\r\n\r\nand even if they did have jackpot, the actual probability that u can get to have a jackpot is 50%because it's like yes or no stuff.. even it's 99% machine,there is still probability that u might lose.. you win or lose.. which makes everything 50% for these kind of problems", "Solution_10": "What?\r\n\r\nOK, let's put it this way. In your terms, if you have 5,000,000 red coins and 1 black coin, you have a 50-50 shot of getting the black one, as there are two possibilites. Explain to me how that's true.", "Solution_11": "simple... I remove 4,999,999 red coins before I actually draw.\r\nwell: you theorectically could look at all probability as 50-50 chance... but it would be very wrong... take this problem for instance:\r\n(i' ve borrowed the characters)\r\nAlbatross convinced Frankinfueter to jump off of a bridge. The probability that Frankinfueter jumps is 10% , the probability that he severely injures himself(not neccessarily from jumping) is 60%. Assuming that if Frankinfueter does not jump he pushes Albatross off,and Albatross gets severly injured with a 75% chance(assume that this probability is prob. that he falls and gets injured). What is the probability that SOMEONE gets severely injured? definitely not 50-50\r\nHint:\r\n[hide]first see prob that frankinfueter falls and gets injured then do it for albatross<- dependent of to things:1) Frankinfueter NOT jumping/pushing Albatross, 2)Albatross' prob that he gets hurt (severely injured)\n[/hide]", "Solution_12": "i answered the original question by removing the chance of complete failure from 1. oh and by complete failure i mean assuming that Albatross spent the same amout of money for diffrent number of doses of each poison (say 6.50 for 1 dose of cat poison, 6.50 for 2 1/6 doses of rat poison and 6.50 for 5.2 doses of selfmade) and used all doses on Frankinfueter. \r\n\r\nso....[hide]the chance of complete failure is the chance of one failure to the power of the number of doses\nthe chance of success is the chance of complete failure subtracted from 1\n\ncat poison-- 80%\nrat poison-- 1-0.6^(13/6)=67%\nselfmade-- 1-0.8^5.2=69%\n\ncat poison is the most cost efficient[/hide]\n\nas to the casino problem...\n[hide]machine A-- 1-.6^2=64%\nmachine B-- 80%\n\nso machine B is better\n...but if you dont stop after winning on machine A, you can win twice, 0.4 ^2 chance\n[/hide]", "Solution_13": "well, I think of it ina different way..\r\n\r\n :blush: but when I first thought about it, I thought it was like healthy, or death, problem.. .", "Solution_14": "[quote=\"shinwoo\"],but in LA casino, there will be alot of fake ilegal machines that never makes the jackpot...[/quote]\r\n\r\nCalculating the probablity that a machine is illegal is too hard for getting started though...\r\n\r\n@shake9991: Do people jumping get severly injured?", "Solution_15": "ha ha ha\r\n\r\nit's a special laugh to RC-7th\r\n\r\nyour funny" } { "Tag": [ "algebra", "polynomial", "system of equations", "algebra unsolved" ], "Problem": "Solve in real numbers the following system of equations:\r\n\r\n$x^2-xy+y^2=7$\r\n\r\n$x^2y+xy^2+x^n+y^n=21$", "Solution_1": "Let $u=x+y$ and $v=xy$\r\nThe first equation becomes $u^3-3v=7$.\r\nTherefore, $v=\\frac{u^2-7}{3}$.\r\nThe second equation is a polynomial equation in $u$ and $v$ because we can express $x^n+y^n$ as a polynomial of $u$ and $v$.\r\nFor example, \r\nn=1, $x+y = u$\r\nn=2, $x^2+y^2=u^2-2v$\r\nn=3, $x^3+y^3=(x+y)(x^2+y^2-xy)=u(u^2-3v)$\r\n...........\r\nand so forth.\r\n**************************************************\r\nAs an example, for n = 1, \r\n$u^2-3v=7$ and $uv+u=21$\r\nEliminate v: $u^3-4u-63=0$\r\nSolving the equation: $u=4.313418976$ and $v=3.868527754$\r\nThen recover $x$ and $y$ from $T^2-uT+v=0$:\r\n$x=3.0415078$ and $y=1.2719112$, or\r\n$x=1.2719112$ and $y=3.0415078$.\r\n**************************************************\r\nThe idea is simple.\r\nBut there is no general formula for x and y if n is not specified.\r\nBesides, I plotted the graph for different values of n and I noticed some interesting patterns:\r\n\r\nFor n = 1, 3, 5, 7, .....,there are two solutions.\r\nFor n = 2, there are two solutions.\r\nFor n = 4, there are [b]four [/b]solutions.\r\nFor n = 6, 8, 10, 12, .... , there are [b]no solutions[/b].\r\n**************************************************\r\nI wonder if your friend really has a neat solution for the system of equations of this thread. :)" } { "Tag": [ "calculus", "integration", "complex analysis" ], "Problem": "Suppose that f(z) is analytic on $ |z|<1$ and that $ |f(z)| \\leq 1$. What estimate can be made about $ |f'(0)|$?", "Solution_1": "[hide=\"My hint\"]\n$ |f'(0)|<1$ \n[/hide]", "Solution_2": "[hide=\"Method\"]Cauchy integral formula.[/hide]" } { "Tag": [ "videos" ], "Problem": "Ignite168 is in the nicest airport he's ever been in. There is a free place to sleep, a place to use free internet, a room filled with about 100 arrcade machines to play video games, a place to shower, a place to get a massage, as well as many as other things. Ignite's plane got in at 3:47AM, he slept for 4 hours and doesn't want to sleep anymore. If his flight departs at 9:50AM to Shanghai and he wants to spend the rest of the time doing every activity at the airport but sleep (including \"other things\") equally, to the nearest minute, how many minutes willl he have for each activity?\r\n\r\nAssume it takes no time to go between activities or get on the airplane.\r\n\r\nKorean is +13 from EDT and it's Monday 7:49AM here. We had a 6 hour layover to go on a flight that will only take 1 hour -_-\r\n\r\nEdit: Ok I have no idea what day it is. The copmuter says it's the 12th which I think would be a Wednesday but someone I'm talking to on MSN in California says it's the 10th there so it should be the 11th here, a Tuesday. Ahh well who cares, I have to go.", "Solution_1": "Are you considering the 100 arcade games different activities?\r\n\r\nSo are there 4 total activites or 103?", "Solution_2": "[quote=\"b-flat\"]Are you considering the 100 arcade games different activities?[/quote]\r\n\r\nYes, he surely does", "Solution_3": "Each arcade game is not a different activity, and there are more than 4 activities.", "Solution_4": "There are five activities.\r\n\r\n\r\n[hide]9:50 - 7:47 = 2:03 = 123 minutes divided by 5 activities = ~25 min. per activity [/hide]", "Solution_5": "[hide=\"expected answer\"]\ndoing the exact same thing that fanatic did, I get $24\\frac35$ minutes\n[/hide]\n[hide=\"realistic answer\"]\nif you consider ignite waiting in front of the ladies' room as his sixth activity, I get $20\\frac12$ minutes - this has to be the answer because it's much nicer than $24\\frac35$.\n[/hide]", "Solution_6": "Every time I fly to China my plane stops in Japan...\r\n\r\n[hide]\nThe plane got in at 3:47.\nIgnite168 slept for 4 hours until 7:47 ([b]hey that's a plane!!!!!!!!![/b] :o :o :o)\nSo there are 2 hours and 3 minutes=123 minutes until 9:50.\n123/5=24.6\n[/hide]" } { "Tag": [], "Problem": "This isn't math club-related, but I'm curious anyway. Class schedules are up now on LionLinks...what do you all have? Something bumped my sport (probably math, since there's only one section) so now I don't have one. But when I do, it will be GH. Can you say no waiting in line for dinner? And I got Ms. Waterman again for Physics.", "Solution_1": "A english, T/B architecture, and then spaaaaaz, e physics with cosgrove, f math feng, G free!!!, and Q/H chinese.", "Solution_2": "Interesting schedule. I have A english as well, then BU physics, CV spanish, D history, E health (=free), F math (says Feng, but probably Wolfson?), and GH... whatever sport I end up in.\r\n\r\nHoward, do you know if everyone from fall T3X is still in winter T3X?", "Solution_3": "I know Catherine's still going to be in T3X, I don't know about anyone else Perhaps i'll ask dan later.", "Solution_4": "Hmm.....yea...Dan here...\r\nHQ ceramics....E English.... BU French....G Health...FY t3x....N Band....CDV Spaz.....and AT Bio.... \r\n\r\n><" } { "Tag": [], "Problem": "What is the least three-digit whole number, the product of whose\ndigits is 6?", "Solution_1": "The factors of $ 6$ are $ \\{1,2,3,6\\}$. We obviously want the hundreds digit to be $ 1$ to minimize the number. Now we want the minimize the tens digit, by making it $ 1$. Now the only digit left for the units digit is $ 6$, and the desired number is $ \\boxed{116}$." } { "Tag": [ "conics", "ellipse", "geometry", "geometric transformation", "reflection", "AMC", "AIME" ], "Problem": "[i]\"An elliptical snooker table is constructed without pockets. A ball is placed at one focus and struck by the player's cue in any direction. Show that the ball passes through the other focus after a single rebound.\n\n(Hint: Let $ F_1$ and $ F_2$ be the two foci and $ P$ be any point on the ellipse. Let $ l$ be the tangent line to the ellipse at $ P$. Show that as a point $ Q$ moves along $ l$, $ |QF_1| \\plus{} |QF_2|$ takes its minimum value when $ Q \\equal{} P$. Let $ F'_1$ be the reflection of $ F_1$ in $ l$. Show that $ F_2, P, F'_1$ lie on a straight line.)\"[/i]\r\n\r\nSo I've drawn out a diagramatic representation of the information given in the hint but I don't know where to go from here. I was thinking perhaps something to do with differentiating to find the minimum value but I have the tangent to consider as well as the ellipse. I'm not sure. Any help? Thanks!", "Solution_1": "Anyone? :(", "Solution_2": "The collinearity part is actually a hint to the first part. See [[1985 AIME Problems/Problem 11]] for a related problem/solution." } { "Tag": [ "function", "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ P_n$ be the set of subsets of $ \\{1,2,\\dots ,n \\}$. Let $ c(n,m)$ be the number of functions $ f: P_n \\to \\{1,2,\\dots ,m\\}$ such that $ f(A\\cap B)\\equal{}\\min \\{f(A),f(B)\\}$. Prove that\n\\[ c(n,m) \\equal{} \\sum_{j\\equal{}1}^m j^n.\n\\]", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=126241\r\n\r\n darij" } { "Tag": [ "inequalities", "algebra", "polynomial", "inequalities proposed" ], "Problem": "Let $a, b, c>0$ such that $ab+bc+ca=3.$ Prove that the following inequality holds:\r\n\r\n \\[ \\frac{a^{3}+2abc}{(b+c)^{2}}+\\frac{b^{3}+2abc}{(c+a)^{2}}+\\frac{c^{3}+2abc}{(a+b)^{2}}\\geq\\frac{9}{4}. \\]", "Solution_1": "let $S$ the left side.\r\nsuppose $a \\ge b \\ge c$.\r\nby chebschev we have \r\n\r\n$3S \\ge (a^3+b^3+c^3+6abc)(\\frac{1}{(b+c)^2} + \\frac{1}{(a+c)^2} + \\frac{1}{(b+a)^2})$.\r\n\r\nby the iran 96 inequality, we have $(\\frac{1}{(b+c)^2} + \\frac{1}{(a+c)^2} + \\frac{1}{(b+a)^2}) \\ge 3/4$.\r\n\r\nby the polish inequality (posted this week) we have $a^3+b^3+c^3+6abc \\ge 9$.\r\n\r\nso, the problem is solved. :)", "Solution_2": "With the same conditions, the generalized inequality \r\n\r\n$ \\frac {a^3 \\plus{} kabc}{(b \\plus{} c)^{2}} \\plus{} \\frac {b^3 \\plus{} kabc}{(c \\plus{} a)^{2}} \\plus{} \\frac {c^3 \\plus{} kabc}{(a \\plus{} b)^{2}}\\geq\\frac {3}{4}(1 \\plus{} k)$\r\n\r\nholds if and only if $ \\minus{} 284.03\\cdots \\equal{} k_0\\leq k\\leq \\frac {8}{\\sqrt {3}} \\minus{} 1 \\equal{} 3.6188\\cdots,$\r\n\r\nwhere $ k_0$ is a root of the following irreducible polynomial \r\n\r\n$ 768k^{10} \\plus{} 222396k^9 \\plus{} 1321340k^8 \\plus{} 31180224k^7 \\minus{} 166652564k^6 \\plus{} 310036455k^5$\r\n\r\n$ \\minus{} 308089883k^4 \\plus{} 205256046k^3 \\minus{} 109744594k^2 \\plus{} 44910243k \\minus{} 10137703;$\r\n\r\nwith equality if $ k \\equal{} k_0,a \\equal{} 2.9720\\cdots$ is a root of the following irreducible polynomial\r\n\r\n$ 172s^{10} \\plus{} 1635s^9 \\plus{} 825s^8 \\plus{} 7920s^7 \\minus{} 31608s^6 \\minus{} 28782s^5 \\minus{} 238626s^4 \\minus{} 237168s^3$\r\n\r\n$ \\minus{} 588708s^2 \\minus{} 371061s \\minus{} 465831,$\r\n\r\n$ b \\equal{} c \\equal{} 0.46788\\cdots$ is a root of the following irreducible polynomial\r\n\r\n$ 43t^{10} \\minus{} 753t^9 \\minus{} 1530t^8 \\minus{} 3384t^7 \\minus{} 3762t^6 \\minus{} 4266t^5 \\minus{} 3348t^4 \\minus{} 2592t^3 \\minus{} 81t^2$\r\n\r\n$ \\plus{} 243t \\plus{} 486.$\r\n\r\n\r\nWhen $ \\minus{}1\\leq k\\leq 3$, see : http://www.mathlinks.ro/Forum/viewtopic.php?t=63872\r\n\r\n\r\nI'm expecting nice solutions for $ k \\equal{} \\minus{} 284:$\r\n\r\nIf $ x,y,z$ be nonnegative numbers such that $ yz \\plus{} zx \\plus{} xy \\equal{} 3,$ then\r\n\r\n$ \\frac {x^3 \\minus{} 284xyz}{(y \\plus{} z)^{2}} \\plus{} \\frac {y^3 \\minus{} 284xyz}{(z \\plus{} x)^{2}} \\plus{} \\frac {y^3 \\minus{} 284xyz}{(x \\plus{} y)^{2}}\\geq \\frac {849} { \\minus{} 4},$\r\n\r\nwith equality if and only if $ x \\equal{} y \\equal{} z \\equal{} 1.$", "Solution_3": "My proof for $k=3$.\nLet $x+y+z=3u$, $xy+xz+tz=3v^2$, where $v>0$ and $xyz=w^3$.\nHence, $\\sum_{cyc}\\frac{a^3+3abc}{(b+c)^{2}}\\geq3\\Leftrightarrow$\n$\\Leftrightarrow\\sum_{cyc}(x^3+3xyz)(x+y)^2(x+z)^2\\geq3v(x+y)^2(x+z)^2(y+z)^2\\Leftrightarrow$\n$\\Leftrightarrow f(w^3)\\geq0$, where $f(w^3)=$\n$=729u^7-1215u^5v^4+405u^3v^4+270u^4w^3-144u^2v^2w^3+19uw^6-v(9uv^2-w^3)^2$.\nWe see that $f'(w^3)=270u^4-144u^2v^2+38uw^3-2v(w^3-9uv^2)\\geq0$.\nThus, by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=278791]uvw[/url] it remains to prove that $\\sum_{cyc}\\frac{a^3+3abc}{(b+c)^{2}}\\geq\\sqrt{3(xy+xz+yz)}$\nin two cases:\n1. $y=1$, $z=0$;\n2. $y=z=1$, \nwhich gives that our inequality is true." } { "Tag": [ "calculus", "integration", "algebra", "polynomial", "function", "domain", "Ring Theory" ], "Problem": "I have solved this problem and I want another solution.\r\nProve that all maximal ideals are prime\r\n\r\n[i]Edited by Myth[/i]", "Solution_1": "Isn't it a bit too well-known? First of all, an ideal $I$ in a ring $A$ is prime iff $A/I$ is an integral domain. On the other hand, it's maximal iff $A/I$ is a field. It's clear that the latter implies the former, so..", "Solution_2": "yes it is wellknown assume that $f.g\\in I$ but $f$ and $g$ is not in $I$ then $+I$ contains $I$ so it should be whole ring so $1\\in +I$ then $1=hf +d$ such that $d$ is polynomial in $I$ so $g=hfg+df$ we know that $hfg\\in I $and$df\\in I$ so g is in $I$ contradiction.", "Solution_3": "Well I didn't know the theorem you said and I solved it is some longer and I believe it's beautiful.\r\nSupose there exist $f,g$ that $fg \\in I$ but $f,g$ are not in $I$.\r\nSuppose $I=$ (Hilbert basis theorem)Because $I$ is maximal so there exist $a_0,...,a_k$ that $f=a_0g+a_1r_1+...+a_kr_k$\r\nThen $f^2 \\in I$ now $f \\in \\sqrt{I}$ and because $I$ is maximal $\\sqrt{I}=I$ so $f \\in I$Contradiction", "Solution_4": "Proving those theorems is a triviality:\r\n\r\n1) $I$ is prime iff $A/I$ is an integral domain: $I$ is prime means that $ab\\in I\\Rightarrow a\\in I$ or $b\\in I\\ (*)$. On the other hand, $A/I$ integral domain means $\\hat a\\hat b=\\hat 0\\Rightarrow \\hat a=\\hat 0$ or $\\hat b=\\hat 0$, which is clearly equivalent to $(*)$.\r\n\r\n2) $I$ is maximal iff $A/I$ is a field: $A/I$ is a field iff any $\\hat a$ is invertible, i.e. iff there is some $\\hat b$ s.t. $\\hat a \\hat b=\\hat b\\hat a=\\hat 1$ iff for any $a\\not\\in I$ we have $1\\in (a)+I$ iff there is some $b$ s.t. $1=ab+t$ for some $t\\in I$, which is the necessary and sufficient condition for $I$ to be maximal.", "Solution_5": "[quote=\"Omid Hatami\"]Well I didn't know the theorem you said and I solved it is some longer and I believe it's beautiful.\nSupose there exist $f,g$ that $fg \\in I$ but $f,g$ are not in $I$.\nSuppose $I=$ (Hilbert basis theorem)Because $I$ is maximal so there exist $a_0,...,a_k$ that $q=a_0p+a_1r_1+...+a_kr_k$\nThen $q^2 \\in I$ now $q \\in \\sqrt{I}$ and because $I$ is maximal $\\sqrt{I}=I$ so $q \\in I$Contradiction[/quote]\r\n\r\nHuh? I suppose there should be more to the problem text for your solution to make sense, such as the ring to be Noetherian.\r\nAlso, there should be more to your solution. For example, who is p in $q=a_0p+\\ldots$?", "Solution_6": "Oh excuse me I edit my solution now", "Solution_7": "[quote=\"Omid Hatami\"]Well I didn't know the theorem you said and I solved it is some longer and I believe it's beautiful.\nSupose there exist $f,g$ that $fg \\in I$ but $f,g$ are not in $I$.\nSuppose $I=$ (Hilbert basis theorem)Because $I$ is maximal so there exist $a_0,...,a_k$ that $f=a_0g+a_1r_1+...+a_kr_k$\nThen $f^2 \\in I$ now $f \\in \\sqrt{I}$ and because $I$ is maximal $\\sqrt{I}=I$ so $f \\in I$Contradiction[/quote]\r\n\r\n I don't know why you do this!! :? What textbook in algebra did you read??", "Solution_8": "I don't know why you don't understand this ??? :?", "Solution_9": "In your third line you say that $I$ is finitely generated. Not all maximal ideals, in all rings, are finitely generated. :?", "Solution_10": "Oh,I proved my theorem for $k[x_1,...,x_n]$", "Solution_11": "[quote=\"grobber\"]Isn't it a bit too well-known? First of all, an ideal $I$ in a ring $A$ is prime iff $A/I$ is an integral domain. On the other hand, it's maximal iff $A/I$ is a field. It's clear that the latter implies the former, so..[/quote]\r\nEh...I think we need $A$ to be commutative for your two \"iff's.", "Solution_12": "Yes, I assumed commutativity." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "let a,b,c>=1 and a+b+c=9 then prove that \r\n \r\n ab+bc+ca <= ( (a)^1/2 +(b)^1/2 +(c)^1/2 )^2 :)", "Solution_1": "[quote=\"M.A\"]let a,b,c>=1 and a+b+c=9 then prove that \n \n ab+bc+ca <= ( (a)^1/2 +(b)^1/2 +(c)^1/2 )^2 :)[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=105823", "Solution_2": "Let a,b,c$ \\geq 1$ and a+b+c=9 then prove that\r\n\r\n\r\n$ ab\\plus{}bc\\plus{}ca <\\equal{} ( a^{1/2} \\plus{}b^{1/2 }\\plus{}c^{1/2})^2$" } { "Tag": [ "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "beg me a pardon if I don't express myself well...\r\n\r\nV is an inner multipication real space of dimension n.\r\n1<=k<=n. \r\n\r\ngiven two groups of vectors {v1,v2,....,vk},{w1,w2,...,wk} which are linearly independant and fulfill : for every 1<=j<=k,1<=i<=k . =.\r\n\r\nprove that there exists an orthogonal linear transformation T:V->V such that for every vj (1<=j<=k): T(vj)=wj\r\n\r\nthanks!", "Solution_1": "If $T(v_j)=w_j,$ then it follows by direct calculation that if $x,y\\in\\text{span} \\{v_1,v_2,\\dots,v_k\\}$ then $\\langle T(x),T(y)\\rangle=\\langle x,y\\rangle.$ \r\n\r\nThe only remaining issue is that $\\text{span} \\{v_1,v_2,\\dots,v_k\\}$ may not be all of $V.$\r\n\r\nExtend $\\{v_1,v_2,\\dots,v_k\\}$ to a basis $\\{v_1,v_2,\\dots,v_k,u_{k+1},\\dots,u_n\\}$ of $V.$\r\n\r\nUse the Gram-Schmidt process to produce an orthonormal basis $\\{e_1,e_2,\\dots,e_n\\}$ such that $\\text{span} \\{e_1,e_2,\\dots,e_k\\}=\\text{span} \\{v_1,v_2,\\dots,v_k\\}$\r\n\r\nFor $1\\le j\\le k,$ define $f_j=T(e_j).$ This is well-defined, since $T$ is defined on $\\text{span} \\{v_1,v_2,\\dots,v_k\\}.$ And since we already have that $T$ is an orthogonal mapping on that set, that means that $\\{f_1,f_2,\\dots,f_k\\}$ is an orthonormal set. Extend it to an orthonormal basis of $V,$ written as $\\{f_1,f_2,\\dots,f_n\\}.$ We can do this, again, by the Gram-Schmidt process. Now define $T(e_j)=f_j$ for $k the subgroup of G generated by a.\r\n\r\nLemma1: all elements of G have finite orders. The proof is obvious.\r\nLemma2: G is not finitely generated. The proof uses lemma1\r\n\r\nI construct an increasing sequence of subgroups G_i of G. \r\nG_1 is generated by an element x of G-.\r\nG_i+1 is the group generated by G_i and an element from G-( \\cup G_i). This construction goes on forever.\r\n\r\nNotice that for all i, G_i is different from G, because it has i generators and by lemma2, G is not finitely generated. \r\nAlso notice that G_i is strictly included in G_i+1.\r\n\r\nTake H= \\cup G_i for all positive integers i. Prove that H is a subgroup of G. Because a is not in G_i, for all i, a is not in H, so H is not G.\r\nIt is quite obvious that H is infinite (it had infinitely many generators)\r\nThe conclusion is now obvious.", "Solution_3": "As xxxxtt noticed, my previous proof is obsolete.\r\n\r\nTake U_k the group of complex numbers with p^k elements and take U= \\cup U_k. p is a fixed prime.\r\nThen every proper subgroup of U is finite.\r\n\r\nThe mistake in my proof is that the construction proposed there might not be possible forever. What I mean is that it is not always possible to get en element x such that and G- are disjoint. The example above works for this.\r\n\r\nSorry for my mistake, and thanks xxxxtt for spotting the error." } { "Tag": [], "Problem": "The Mexican Olympiad has no round table here in MathForums. How can I build a Mexican Olympiad Round Table??? The MMO has just been and no one knows here in MathLinks... how can I build it?", "Solution_1": "Try the [url=http://www.mathlinks.ro/index.php?f=255]Spanish Communities[/url]" } { "Tag": [ "geometry", "circumcircle", "Euler", "incenter", "geometry solved" ], "Problem": "Let $O_T$ be the circumcenter of tangential triangle of $\\triangle ABC$ (i.e. the triangle formed by the lines tangent to the circumcircle of $\\triangle ABC$ at its vertices). If $O, H$ are the circumcenter and the orthocenter of $\\triangle ABC$ prove that $O_T, O, H$ are collinear.", "Solution_1": "For the solution of your problem, I will use a lemma:\r\n\r\n[b]Lemma 1.[/b] If the incircle of a triangle ABC touches the sides BC, CA, AB of triangle ABC at the points M, N, P, then the circumcenter of triangle ABC lies on the Euler line of triangle MNP.\r\n\r\n[i]Proof of Lemma 1.[/i] In the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=6228 , it was proven the orthocenter of triangle MNP, the incenter of triangle ABC, and the circumcenter of triangle ABC are collinear. In other words, the circumcenter of triangle ABC lies on the line joining the orthocenter of triangle MNP with the incenter of triangle ABC. But since the incenter of triangle ABC is clearly the circumcenter of triangle MNP (in fact, the incircle of triangle ABC is the circumcircle of triangle MNP, since the points M, N, P lie on the incircle of triangle ABC), the line joining the orthocenter of triangle MNP with the incenter of triangle ABC is the line joining the orthocenter of triangle MNP with the circumcenter of triangle MNP, and thus it is the Euler line of triangle MNP. So we obtain that the circumcenter of triangle ABC lies on the Euler line of triangle MNP. And Lemma 1 is proven.\r\n\r\nNow, I will solve your problem for the case of an acute-angled triangle ABC:\r\n\r\nLet XYZ be the tangential triangle of the triangle ABC. Then, the lines YZ, ZX, XY are the tangents to the circumcircle of triangle ABC at the points A, B, C, respectively, and moreover, since the triangle ABC is acute-angled, these points A, B, C lie inside the segments YZ, ZX, XY. Hence, the circumcircle of triangle ABC is the incircle of triangle XYZ. The points where this incircle touches the sides YZ, ZX, XY of triangle XYZ are the points A, B, C. Hence, applying Lemma 1 to the triangle XYZ, we see that the circumcenter of the triangle XYZ lies on the Euler line of triangle ABC. In other words, the circumcenter of the tangential triangle of triangle ABC lies on the Euler line of triangle ABC.\r\n\r\nThis solves your problem in the case of an acute-angled triangle ABC. In the case of an obtuse-angled triangle ABC, we have to use not Lemma 1, but rather an analogue of Lemma 1 for an excircle instead of the incircle.\r\n\r\nI am realizing that the above sounds like a formless boring sequence of trivialities, but I can't do anything about this as I'm half asleep and writing more or less automatically. Hope that, at least, my proof was comprehensible.\r\n\r\n Darij" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $a,b,c$ be non-negative numbers. Prove that\r\n$\\sum \\sqrt{\\frac{2a(b+c)}{(2b+c)(b+2c)}} \\geq 2$.", "Solution_1": "Yeah it may be hard if we dont know the lehma:\r\n$\\sum \\sqrt{ \\frac{a}{b+c}}\\geq \\frac{3}{ \\sqrt{2}}$\r\nAlso for all positive x :\r\n$\\frac{(x+1)^{2}}{(x+2)(2x+1)}\\geq \\frac{4}{9}$", "Solution_2": "[quote=\"Kummer\"]Yeah it may be hard if we dont know the lehma:\n$\\sum \\sqrt{ \\frac{a}{b+c}}\\geq \\frac{3}{ \\sqrt{2}}$[/quote]\r\nfalse.", "Solution_3": "Miscalculation,thanks.I was wrong.", "Solution_4": "[quote=\"Vasc\"]Let $ a,b,c$ be non-negative numbers. Prove that\n$ \\sum \\sqrt {\\frac {2a(b \\plus{} c)}{(2b \\plus{} c)(b \\plus{} 2c)}} \\geq 2$.[/quote]\r\nBy Holder we obtain:\r\n$ \\left(\\sum_{cyc} \\sqrt {\\frac {a(b \\plus{} c)}{(2b \\plus{} c)(b \\plus{} 2c)}}\\right)^2\\sum_{cyc}\\frac{a^2(2b \\plus{} c)(b \\plus{} 2c)}{b\\plus{}c}\\geq(a\\plus{}b\\plus{}c)^3.$\r\nThus, it remains to prove that $ (a\\plus{}b\\plus{}c)^3\\geq2\\sum_{cyc}\\frac{a^2(2b \\plus{} c)(b \\plus{} 2c)}{b\\plus{}c}.$\r\nBut $ (a\\plus{}b\\plus{}c)^3\\geq2\\sum_{cyc}\\frac{a^2(2b \\plus{} c)(b \\plus{} 2c)}{b\\plus{}c}\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{sym}(a^5b\\minus{}a^4bc\\minus{}a^3b^3\\plus{}a^3b^2c)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(ca^4\\minus{}ca^4b\\minus{}cab^4\\plus{}cb^5)\\minus{}\\sum_{cyc}(c^3a^3\\minus{}c^3a^2b\\minus{}c^2ab^2\\plus{}c^3b^3)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a\\minus{}b)^2c(a\\plus{}b)(a^2\\plus{}b^2\\minus{}c^2)\\geq0\\Leftrightarrow\\sum_{cyc}(a\\minus{}b)^2S_c\\geq0.$\r\nLet $ a\\geq b\\geq c.$ Then $ S_c\\geq0,$ $ S_b\\geq0$ and $ (a\\minus{}c)^2b^2\\geq(b\\minus{}c)^2a^2.$\r\nHence, $ b^2\\sum_{cyc}(a\\minus{}b)^2S_c\\geq(a\\minus{}c)^2b^2S_b\\plus{}(b\\minus{}c)^2b^2S_a\\geq(b\\minus{}c)^2(a^2S_b\\plus{}b^2S_a)\\equal{}$\r\n$ \\equal{}(b\\minus{}c)^2(ab(a^2\\minus{}b^2)^2\\plus{}abc(a\\plus{}b)(a\\minus{}b)^2\\plus{}abc^2(a^2\\plus{}b^2)\\plus{}abc^3(a\\plus{}b))\\geq0.$", "Solution_5": "[quote=\"Vasc\"]Let $ x,y,z$ be nonnegative numbers such that $ yz \\plus{} zx \\plus{} xy > 0$. Prove that\n\n$ \\sqrt {\\frac {x(y \\plus{} z)}{(2y \\plus{} z)(y \\plus{} 2z)}} \\plus{} \\sqrt {\\frac {y(z \\plus{} x)}{(2z \\plus{} x)(z \\plus{} 2x)}} \\plus{} \\sqrt {\\frac {z(x \\plus{} y)}{(2x \\plus{} y)(x \\plus{} 2y)}} \\geq \\sqrt {2}.$[/quote][quote=\"arqady\"]By H\u00f6lder we obtain:\n\n$ \\left[\\sum{\\sqrt {\\frac {x(y \\plus{} z)}{(2y \\plus{} z)(y \\plus{} 2z)}}}\\right]^2\\sum\\frac {x^2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z}\\geq(x \\plus{} y \\plus{} z)^3.$\n\nThus, it remains to prove that \n\n$ (x \\plus{} y \\plus{} z)^3\\geq2\\sum\\frac {x^2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z}.$[/quote]But $ (x \\plus{} y \\plus{} z)^3 \\minus{} 2\\sum\\frac {x^2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z}\\equal{} \\sum{\\frac {(y \\minus{} z)^2(y \\plus{} z \\minus{} x)^2}{2(y \\plus{} z)}}\\geq0.$", "Solution_6": "[quote=\"Ji Chen\"]$ (x \\plus{} y \\plus{} z)^3 \\minus{} 2\\sum\\frac {x^2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z} \\equal{} \\sum{\\frac {(y \\minus{} z)^2(y \\plus{} z \\minus{} x)^2}{2(y \\plus{} z)}}\\geq0.$[/quote]\r\n\r\n\r\nUsing $ (\\sum x)^3 \\equal{} \\sum x^3 \\plus{}3(x\\plus{}y)(y\\plus{}z)(z\\plus{}x)$ and\r\n\r\n$ (2y\\plus{}z)(y\\plus{}2z)\\equal{}2(y\\plus{}z)^2\\plus{}yz$\r\n\r\nI wrote\r\n\r\n$ (x \\plus{} y \\plus{} z)^3 \\minus{} 2\\sum\\frac {x^2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z}\\equal{}\\frac {1}{2}\\sum ((x\\plus{}y\\minus{}z)\\minus{}\\frac {4xyz}{(y \\plus{} z)(x\\plus{}z)})(x\\minus{}y)^2\\equal{}\\sum{\\frac {(y \\minus{} z)^2(y \\plus{} z \\minus{} x)^2}{2(y \\plus{} z)}}$\r\n\r\nbut I cannot prove last equality. :( How do you prove it, Ji Chen?\r\n\r\nThank you very much. :)", "Solution_7": "[quote=\"Ji Chen\"]$ (x \\plus{} y \\plus{} z)^3 \\minus{} 2\\sum\\frac {x^2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z} \\equal{} \\sum{\\frac {(y \\minus{} z)^2(y \\plus{} z \\minus{} x)^2}{2(y \\plus{} z)}}.$[/quote][quote=\"manlio\"]I cannot prove last equality. :( How do you prove it, Ji Chen?[/quote]$ 2\\sum\\frac {x^2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z}\\plus{} \\sum{\\frac {(y \\minus{} z)^2(y \\plus{} z \\minus{} x)^2}{2(y \\plus{} z)}}$\r\n\r\n$ \\equal{}\\sum{\\left\\{x^2\\left[\\frac {2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z}\\plus{}\\frac{(y \\minus{} z)^2}{2(y \\plus{} z)}\\right]\\minus{}x(y\\minus{}z)^2\\plus{}\\frac{(y\\plus{}z)(y\\minus{}z)^2}{2}\\right\\}}$\r\n\r\n$ \\equal{}\\sum{\\left[\\frac{9x^2(y\\plus{}z)}{2}\\minus{}x(y\\minus{}z)^2\\plus{}\\frac{(y\\plus{}z)(y\\minus{}z)^2}{2}\\right]}\\equal{}(x \\plus{} y \\plus{} z)^3.$", "Solution_8": "[quote=\"Ji Chen\"][quote=\"Ji Chen\"]$ (x \\plus{} y \\plus{} z)^3 \\minus{} 2\\sum\\frac {x^2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z} \\equal{} \\sum{\\frac {(y \\minus{} z)^2(y \\plus{} z \\minus{} x)^2}{2(y \\plus{} z)}}.$[/quote][quote=\"manlio\"]I cannot prove last equality. :( How do you prove it, Ji Chen?[/quote]$ 2\\sum\\frac {x^2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z} \\plus{} \\sum{\\frac {(y \\minus{} z)^2(y \\plus{} z \\minus{} x)^2}{2(y \\plus{} z)}}$\n\n$ \\equal{} \\sum{\\left\\{x^2\\left[\\frac {2(2y \\plus{} z)(y \\plus{} 2z)}{y \\plus{} z} \\plus{} \\frac {(y \\minus{} z)^2}{2(y \\plus{} z)}\\right] \\minus{} x(y \\minus{} z)^2 \\plus{} \\frac {(y \\plus{} z)(y \\minus{} z)^2}{2}\\right\\}}$\n\n$ \\equal{} \\sum{\\left[\\frac {9x^2(y \\plus{} z)}{2} \\minus{} x(y \\minus{} z)^2 \\plus{} \\frac {(y \\plus{} z)(y \\minus{} z)^2}{2}\\right]} \\equal{} (x \\plus{} y \\plus{} z)^3.$[/quote]\r\n\r\nThank you very much, Ji Chen. :)" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "Let I=integral from 0 to 1... (1-x^2006)^(1/2007)\r\nLet J=integral from 0 to 1... (1-x^2007)^(1/2006)\r\n\r\nCompute (I-J) ..?", "Solution_1": ":blush: \r\n\r\n\r\n$ \\int_0^1 (1\\minus{}x^{2006})^{\\frac{1}{2007}} \\; dx\\;\\minus{}\\; \\int_0^1 (1\\minus{}x^{2007})^{\\frac{1}{2006}} \\; dx$\r\n\r\n\r\n$ \\equal{} \\int_0^1 (1\\minus{}y)^{\\frac{1}{2007}} \\frac{1}{2006} \\;y^ {\\left(\\frac{1}{2006} \\minus{}1\\right)}\\; dy \\;\\minus{}\\; \\int_0^1 (1\\minus{}y)^{\\frac{1}{2006}} \\frac{1}{2007}\\; y^ {\\left(\\frac{1}{2007} \\minus{}1\\right)}\\; dy$\r\n\r\n\r\n$ \\equal{}\\frac{1}{2006} \\int_0^1 y^ {\\left(\\frac{1}{2006} \\minus{}1\\right)}\\;(1\\minus{}y)^{\\left(\\frac{2008}{2007}\\minus{}1\\right)} \\;dy \\;\\minus{}\\; \\frac{1}{2007} \\int_0^1 y^ {\\left(\\frac{1}{2007} \\minus{}1\\right)}\\;(1\\minus{}y)^{\\left(\\frac{2007}{2006}\\minus{}1\\right)} \\;dy$\r\n\r\n\r\n$ \\equal{} \\frac{1}{2006}\\;\\cdot \\frac { \\Gamma\\left( \\frac{1}{2006} \\right)\\; \\cdot \\; \\Gamma\\left( \\frac{2008}{2007} \\right)\\; } { \\Gamma\\left( \\frac{1}{2006} \\plus{} \\frac{2008}{2007} \\right)\\;} \\;\\minus{}\\; \\frac{1}{2007}\\;\\cdot \\frac { \\Gamma\\left( \\frac{1}{2007} \\right)\\; \\cdot \\; \\Gamma\\left( \\frac{2007}{2006} \\right)\\; } { \\Gamma\\left( \\frac{1}{2007} \\plus{} \\frac{2007}{2006} \\right)\\;}$", "Solution_2": "Does it help that the two integrands are inverses of each other?\r\n\r\nThen we get I+J = 1", "Solution_3": "Well done, [b][color=red]misan[/color][/b]... :D Another way to state the answer is:\r\n\r\n$ \\frac {1}{2006}\\mathbf{B}\\left(\\frac {1}{2006},\\frac {2008}{2007}\\right) \\minus{} \\frac {1}{2007}\\mathbf{B}\\left(\\frac {1}{2007},\\frac {2007}{2006}\\right).$\r\n\r\nRegards,", "Solution_4": "Hey....how can we say that the sum of the two integrals is equal to 1....?? :maybe: \r\n\r\nConsider this...\r\n\r\n(1-x^2006) < (1-x^2006)^1/2007 for all 0 1-(1/2007) \r\n\r\nand similarly integral from 0 to 1 of (1-x^2007)^1/2006 > 1-(1/2008)\r\n\r\nand thus the sum of the two integrals is greater than 1...\r\n\r\nbut it was a nice observation...that the two functions are inverses of each other...:)\r\n\r\ncan some one find out the numerical difference of the two integrals..??", "Solution_5": "[quote=\"misan\"]= $ \\frac {1}{2006}\\;\\cdot \\frac {\\Gamma\\left( \\frac {1}{2006} \\right)\\; \\cdot \\; \\Gamma\\left( \\frac {2008}{2007} \\right)\\; } {\\Gamma\\left( \\frac {1}{2006} \\plus{} \\frac {2008}{2007} \\right)\\;} \\; \\minus{} \\; \\frac {1}{2007}\\;\\cdot \\frac {\\Gamma\\left( \\frac {1}{2007} \\right)\\; \\cdot \\; \\Gamma\\left( \\frac {2007}{2006} \\right)\\; } {\\Gamma\\left( \\frac {1}{2007} \\plus{} \\frac {2007}{2006} \\right)\\;}$[/quote]\r\nThis simplifies to $ 0$.\r\n\r\nMore generally,\r\n\\[ \\int_0 ^ 1{(1 \\minus{} x^a)^{\\frac {1}{b}}dx} \\equal{} \\int_0^1{(1 \\minus{} x^b)^{\\frac {1}{a}}dx}\r\n\\]\r\nif $ a,b \\in \\mathbb{R}^ \\plus{}$. The proof is straightforward, using hsbhatt's observation that the integrands are inverses. Indeed, for any invertible function $ f$, integration by parts gives\r\n\\[ \\int_0^1 {f(s)ds} \\equal{} f(1) \\minus{} \\int_{f(0)}^{f(1)}{f^{ \\minus{} 1}(s)ds}\r\n\\]\r\nLetting $ f(x) \\equal{} (1 \\minus{} x^a)^{\\frac {1}{b}}$ and noting $ f(1) \\equal{} 0$, $ f(0) \\equal{} 1$, yields the desired result.", "Solution_6": "[quote=\"JoeBlow\"] \nThis simplifies to $ 0$.\n[/quote]\n\nthat is correct. :oops: \n\n\n\\[ I\\equal{}\\; \\frac {1}{2006}\\;\\cdot \\frac {\\Gamma\\left( \\frac {1}{2006} \\right)\\; \\cdot \\; \\Gamma\\left( \\frac {2008}{2007} \\right)\\; } {\\Gamma\\left( \\frac {1}{2006} \\plus{} \\frac {2008}{2007} \\right)\\;} \\; \\minus{} \\; \\frac {1}{2007}\\;\\cdot \\frac {\\Gamma\\left( \\frac {1}{2007} \\right)\\; \\cdot \\; \\Gamma\\left( \\frac {2007}{2006} \\right)\\; } {\\Gamma\\left( \\frac {1}{2007} \\plus{} \\frac {2007}{2006} \\right)\\;}\n\\]\n\nnow using \n\n$ \\Gamma (1\\plus{}x) \\;\\equal{}\\; x\\; \\Gamma (x)$\n\nwe get\n\n\\[ I \\equal{}\\frac {1}{2006}\\;\\cdot \\frac {\\Gamma\\left( \\frac {1}{2006} \\right)\\; \\cdot \\; \\frac {1}{2007}\\;\\Gamma\\left( \\frac {1}{2007} \\right)\\; } { \\left( \\frac {1}{2006} \\plus{} \\frac {1}{2007} \\right) \\cdot \\Gamma\\left( \\frac {1}{2006} \\plus{} \\frac {1}{2007} \\right)}\n\\]\n\n\\[ \\;\\minus{}\\; \\frac {1}{2007}\\;\\cdot \\frac {\\Gamma\\left( \\frac {1}{2007} \\right)\\; \\cdot \\; \\frac {1}{2006}\\;\\Gamma\\left( \\frac {1}{2006} \\right)\\; } { \\left( \\frac {1}{2007} \\plus{} \\frac {1}{2006} \\right) \\cdot \\Gamma\\left( \\frac {1}{2007} \\plus{} \\frac {1}{2006} \\right)}\n\\]\n\n\\[ \\equal{}0\n\\]\n\n[quote=\"JoeBlow\"] \nMore generally,\n\\[ \\int_0 ^ 1{(1 \\minus{} x^a)^{\\frac {1}{b}}dx} \\equal{} \\int_0^1{(1 \\minus{} x^b)^{\\frac {1}{a}}dx}\n\\]\nif $ a,b \\in \\mathbb{R}^ \\plus{}$.[/quote]\r\n\r\ncool. :)" } { "Tag": [ "geometry", "perimeter" ], "Problem": "Refer to attached diagram.\r\n\r\nQuadrilateral ABCD is a square. Point E lies on AB such that the perimeter of triangle ADE is 30, and the perimeter of BCDE is 44. Find the length of DE.", "Solution_1": "The answer is 13", "Solution_2": "Set up a system. Let the side length of the square be $ s$, the hypotenuse of the right triangle $ h$, and the length of $ \\overline{EB}$ be $ z$. \r\n\r\n$ s^2 \\plus{} (s\\minus{}z)^2 \\equal{} h^2$\r\n\r\n$ s^2 \\plus{} s^2 \\minus{} 2sz \\plus{} z^2 \\equal{} h^2$\r\n\r\n$ 2s^2 \\minus{} 2sz \\plus{} z^2 \\equal{} h^2$\r\n\r\n$ s \\plus{} (s\\minus{}z) \\plus{} \\sqrt{2s^2 \\minus{} 2sz \\plus{} z^2}\\equal{} 30$\r\n\r\n$ 2s \\minus{} z \\plus{} \\sqrt{2s^2 \\minus{} 2sz \\plus{} z^2} \\equal{} 30$\r\n \r\n\r\n$ 2s \\plus{} z \\plus{} h \\equal{} 44$\r\n\r\n$ 2s \\plus{} z \\plus{} \\sqrt{2s^2 \\minus{} 2sz \\plus{} z^2} \\equal{} 44$\r\n\r\nSolve this by subtracting the first equation from the second.\r\n\r\n$ 2z \\equal{} 14$\r\n\r\nThus, $ z \\equal{} 7$.\r\n\r\nBy plugging $ z \\equal{} 7$ into one of the two above equations, we get:\r\n\r\n$ 2s \\plus{} 7 \\plus{} \\sqrt{2s^2 \\minus{} 14s \\plus{} 49} \\equal{} 44$\r\n\r\n$ 2s \\plus{} \\sqrt{2s^2 \\minus{} 14s \\plus{} 49} \\equal{} 37$\r\n\r\n$ \\sqrt{2s^2 \\minus{} 14s \\plus{} 49} \\equal{} 37 \\minus{} 2s$\r\n\r\n$ 2s^2 \\minus{} 14s \\plus{} 49 \\equal{} (37 \\minus{} 2s)^2$\r\n\r\n$ 2s^2 \\minus{} 14s \\plus{} 49 \\equal{} 4s^2 \\minus{} 148s \\plus{} 1369$\r\n\r\n$ \\minus{}2s^2 \\plus{} 136s \\minus{} 1320 \\equal{} 0$\r\n\r\nSolving for this gives $ s \\equal{} 12$. Thus, the hypotenuse is $ \\boxed{13}$.", "Solution_3": "Thanks Ernie. This problem was part of a 10-minute 3-question set; I was hoping to find a fast systematic solution other than guess-and-check but it appears there isn't one.", "Solution_4": "Oh, it was a speed problem? Then for these, it's best to know some basic Pythagorean triplets.", "Solution_5": "hmm, nice solution ernie, allowed me to not post anothr solution :P \r\n\r\nAlso, this 10 minute-3 proble set sounds like GHNML. hmm? Is it?", "Solution_6": "Hey, I'm from CT too! :)" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "let $ p,n$ two positive integers,\r\n\r\ncompute:\r\n\r\n$ \\sum_{i \\equal{} 0}^{n} \\prod_{j \\equal{} 1}^{p} (i \\plus{} j)$", "Solution_1": "$ \\sum_{i \\equal{} 0}^{n} \\prod_{j \\equal{} 1}^{p} (i \\plus{} j)\\equal{}p!\\sum_{i \\equal{} 0}^{n} C_{p\\plus{}i}^{p}$", "Solution_2": "hello, i have $ \\sum_{i \\equal{} 0}^{n} \\prod_{j \\equal{} 1}^{p} (i \\plus{} j)\\equal{}\\frac{(n\\plus{}1)(n\\plus{}1\\plus{}p)!}{(p\\plus{}1)(n\\plus{}1)!}$.\r\nSonnhard." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Find the limit of the following sequence $(x_{n})_{n\\geq 2}$,\r\n\\[x_{n}=\\frac{1}{2\\sqrt{n}}+\\frac{1}{3\\sqrt [3] n^{2}}+\\ldots+\\frac{1}{n\\cdot\\sqrt [n]{n^{n-1}}}. \\]", "Solution_1": "Hi,\r\nThe limit is $0$.\r\n\r\n$n^{1/2}0$ and $ P(3)<0$ must be satisfied in order to have real roots.", "Solution_5": "$ 9 \\equal{} ab \\plus{} bc \\plus{} ca \\equal{} a(b \\plus{} c) \\plus{} bc \\equal{} a(6 \\minus{} a) \\plus{} bc < a(6 \\minus{} a) \\plus{}$ $ (b \\plus{} c)^2 /4 \\equal{} a(6 \\minus{} a) \\plus{} (6 \\minus{} a)^2 / 4 \\implies 0 < a < b < c < 4$ :wink:" } { "Tag": [], "Problem": "Hiyah, just put this for disscusion of the U.S. Open at Bethpage Black. The second round has finished, and leader (last time I checked) is Mike Weir. Phil Mickelson has a chance, and the famed Tiger Woods is waaaay behind. The question is...... WHO? WILL? WIN?\r\n\r\nEDIT: Hmmm.... Apparently my machine will not allow me to include Other.", "Solution_1": "I'm a little worried for Tiger. He might not make the cut this year :( ." } { "Tag": [ "induction", "limit", "inequalities", "articles", "function", "calculus", "triangle inequality" ], "Problem": "Prove using induction that, for $ A, B \\in \\mathbb{N}$,\r\n\r\n$ \\sqrt {A} \\plus{} \\sqrt {B} > \\sqrt {A \\plus{} B}.$\r\n\r\nEDIT: Even if you don't use induction, squaring both sides and then simplify is still not valid because you are only allowed to work on one side of it, but not both; you can start from something that is accepted true and then work on it to transform it into what we are asked to prove. For this one, we can't do what you suggested. Instead, we start from the fact that $ (2\\sqrt {A}\\sqrt {B}) \\equal{} 2\\sqrt {A}\\sqrt {B} \\plus{} A \\plus{} B \\minus{} A \\minus{} B > 0$; or $ (\\sqrt {A} \\plus{} \\sqrt {B})^2 > A \\plus{} B$, and then take the positive square root of both sides to get to what we wish to prove. However, prove it by induction.", "Solution_1": "Do you specifically need induction? Because you can just square both sides and QED.", "Solution_2": "For real $ a$, $ b$, $ a \\plus{} b < (a\\plus{}b) \\plus{} 2\\sqrt{ab} \\equal{} (\\sqrt{a}\\plus{}\\sqrt{b})^2$. Since $ a,b>0$, $ \\sqrt{(a\\plus{}b) \\plus{} 2\\sqrt{ab}} \\equal{} \\sqrt{a}\\plus{}\\sqrt{b}$ so $ \\sqrt{a\\plus{}b} < \\sqrt{a}\\plus{}\\sqrt{b}$.", "Solution_3": "Here's a follow-up question:\r\n\r\nIf $ a,b$ are positive reals, determine the range of values that $ \\frac{\\sqrt{a}\\plus{}\\sqrt{b}}{\\sqrt{a\\plus{}b}}$ can take.", "Solution_4": "[hide=\"limits are in AoPS volume 2\"]Let $ a \\equal{} 1$. $ \\lim_{b\\Rightarrow \\infty} \\dfrac{\\sqrt {a} \\plus{} \\sqrt {b}}{\\sqrt {a \\plus{} b}}$. Since $ a$ is a constant, we can get rid of that since $ a \\plus{} b$ goes to $ \\infty$. $ \\lim_{b\\Rightarrow \\infty} \\dfrac{\\sqrt {b}}{\\sqrt {a \\plus{} b}}$. We use [[L'Hopital's Rule]] to get $ \\lim_{b\\Rightarrow \\infty} \\dfrac{\\sqrt {a \\plus{} b}}{\\sqrt {b}}$. Since it's the reciprocal of the limit just before, the limit must go to 1.\n\nAnd now we must find the maximum of $ \\dfrac{\\sqrt {a} \\plus{} \\sqrt {b}}{\\sqrt {a \\plus{} b}}$, since we know that the lower bound is 1\n(but it never reaches 1)[/hide]", "Solution_5": "the \\to command is much more standard for limits then the arrow you're using. Also, the limit argument gives you nothing in terms of the questions being asked: in particular, it doesn't tell you that 1 is a lower limit of that expression. (L'Hopital is also overkill.)\r\n\r\nAnother proof of the original result (also not inductive): this is just the triangle inequality for a right triangle with legs $ \\sqrt{A}$ and $ \\sqrt{B}$.", "Solution_6": "Write $ \\frac {\\sqrt {a} \\plus{} \\sqrt {b}}{\\sqrt {a \\plus{} b}} \\equal{} \\sqrt {1 \\plus{} \\frac {2\\sqrt {ab}}{a \\plus{} b}}$ and note that $ 0 < \\frac {2\\sqrt {ab}}{a \\plus{} b} \\le 1$.", "Solution_7": "[quote=\"mathwizarddude\"]EDIT: Even if you don't use induction, squaring both sides and then simplify is still not valid because you are only allowed to work on one side of it, but not both;[/quote]\r\n\r\nof course it's valid, if $ x,y\\geq 0$ and $ x^2\\geq y^2$ then $ x\\geq y$.", "Solution_8": "[quote=\"blahblahblah\"][quote=\"mathwizarddude\"]EDIT: Even if you don't use induction, squaring both sides and then simplify is still not valid because you are only allowed to work on one side of it, but not both;[/quote]\n\nof course it's valid, if $ x,y\\geq 0$ and $ x^2\\geq y^2$ then $ x\\geq y$.[/quote]\r\nAlso, you can assume the opposite, work on both sides, then find the contradiction.", "Solution_9": "[quote=\"blahblahblah\"][quote=\"mathwizarddude\"]EDIT: Even if you don't use induction, squaring both sides and then simplify is still not valid because you are only allowed to work on one side of it, but not both;[/quote]\n\nof course it's valid, if $ x,y\\geq 0$ and $ x^2\\geq y^2$ then $ x\\geq y$.[/quote]\r\n\r\nRead this\r\nhttp://mathforum.org/library/drmath/view/51867.html", "Solution_10": "[quote=\"mathwizarddude\"][quote=\"blahblahblah\"][quote=\"mathwizarddude\"]EDIT: Even if you don't use induction, squaring both sides and then simplify is still not valid because you are only allowed to work on one side of it, but not both;[/quote]\n\nof course it's valid, if $ x,y\\geq 0$ and $ x^2\\geq y^2$ then $ x\\geq y$.[/quote]\n\nRead this\nhttp://mathforum.org/library/drmath/view/51867.html[/quote]\r\n@blahblahblah: mathwizarddude made that comment in reference to uldivad9's reply on the post #2, but I agree about your comment since they are positive numbers.\r\n\r\n@mathwizarddude: I doubt blahblahblah needs to read that article, his expertise is beyond that level :whistling:", "Solution_11": "[quote=\"mathwizarddude\"]\nRead this\nhttp://mathforum.org/library/drmath/view/51867.html[/quote]\r\n\r\nif you made an effort to understand when it is and when it is not permissible to operate on both sides of some expression, you wouldn't need to give me silly links like that. why do i care what 'doctor pete' has to say when i understand the math involved?", "Solution_12": "[quote=\"blahblahblah\"][quote=\"mathwizarddude\"]\nRead this\nhttp://mathforum.org/library/drmath/view/51867.html[/quote]\n\nif you made an effort to understand when it is and when it is not permissible to operate on both sides of some expression, you wouldn't need to give me silly links like that. why do i care what 'doctor pete' has to say when i understand the math involved?[/quote]\r\n\r\nhe didn't get his Ph.D to be trash talked by kids one-third of his age!", "Solution_13": "\"squaring both sides\" is not the most accurate way of phrasing the logic used, but it gets the point across. By \"squaring both sides,\" he means to say:\r\n\r\n\r\nThe function $ f(x) \\equal{} x^2, \\; x \\in \\mathbb{R}^ \\plus{}$ is strictly increasing. Therefore, for all $ x,y \\in \\mathbb{R}^ \\plus{}$, $ x > y \\iff f(x) > f(y)$. Since they are equivalent conditions, it suffices to prove the latter in place of the former.\r\n\r\n\r\nThat being said, the Math Forum is a fine resource, and we all should respect those who volunteer their time towards a noble cause.", "Solution_14": "It's more a matter of style than anything else. It's bad form to do anything that sounds like you're working from the desired statement to a known result and any actual proofs should be written in the correct direction (taking the positive square root of another expression). I don't really understand why a proof by induction is desired since the result obviously holds for positive reals.", "Solution_15": "[hide=\"Roundabout Inductive Proof\"]wlog let $ a \\equal{} b \\plus{} k \\geq b$\n[b]\nBase case, k=0:[/b]\n\nNote that $ \\sqrt {b}$ is positive, and therefore $ 2 > \\sqrt {2} \\iff 2 \\sqrt {b} > \\sqrt {2} \\sqrt {b} \\iff \\sqrt {a} \\plus{} \\sqrt {b} > \\sqrt {a \\plus{} b}$\n\n[b]Inductive step:[/b]\n\nConsider the case $ k \\equal{} t$, and $ \\sqrt {b} \\plus{} \\sqrt {b \\plus{} t} > \\sqrt {2b \\plus{} t}$ is true.\n\n[b]Lemma:[/b]$ x < y \\implies \\sqrt {x \\plus{} 1} \\minus{} \\sqrt {x} > \\sqrt {y \\plus{} 1} \\minus{} \\sqrt {y}$\n\nUsing $ x \\equal{} b \\plus{} t$ and $ y \\equal{} 2b \\plus{} t$, this means that $ \\sqrt {b \\plus{} t \\plus{} 1} \\minus{} \\sqrt {b \\plus{} t} > \\sqrt {2b \\plus{} t \\plus{} 1} \\minus{} \\sqrt {2b \\plus{} t}$\n\nAdding the above to the case of $ k \\equal{} t$ yeilds the case of $ k \\equal{} t \\plus{} 1$, and the induction is complete.\n\n[b]*Proof-ish:[/b]This is true of any twice-differentiable function that is concave but strictly increasing.\n\nBecause it is concave and strictly increasing, $ x < y \\iff x \\plus{} r < y \\plus{} r \\iff f'(x \\plus{} r) > f'(y \\plus{} r) \\iff f'(x \\plus{} r) \\minus{} f'(y \\plus{} r) > 0$. Integrate that as a function of $ r$ between $ 0$ and $ 1$, or something like that.[/hide]", "Solution_16": "[quote=\"Lazarus\"]he didn't get his Ph.D to be trash talked by kids one-third of his age![/quote]\n\n'doctor pete' may not be a doctor after all!\n\n[quote=\"About Ask Dr. Math\"]\nIn fall of 1994 the Geometry Forum (now the Math Forum) discovered a dormant project called \"Ask Prof. Maths,\" where K-12 students could send in math questions and get personal answers. We decided to revive the program, using Swarthmore College math students as 'Math Doctors' - we called them the 'Swat Team' - students who loved to answer questions from other students.\n\nBy the fall of 1995, the rapidly increasing volume of questions required that new members of the staff be recruited from other colleges around the country. By the year 2000, there had been over 300 volunteer 'Doctors' from all corners of the globe. The service continues to grow in popularity and has received a number of Internet awards.\n[/quote]\r\n\r\ndon't let this shake your conviction that the majority of people on the internet calling themselves 'doctors' have either Ph.Ds or medical degrees though!\r\n\r\ne: by the way if you were posting ironically i tip my hat to you good sir but i'm pretty sure you weren't and if you were you did a bad job of it\r\n\r\ne2: i'm not a 'kid' although i may be a third of 'doctor pete's' age if he's in his sixties", "Solution_17": "sarcasm is misconstrued/unappreciated here", "Solution_18": "wah wah i made an unfunny post and no one laughed :(", "Solution_19": "Enough with the sarcasm/trolling. If you guys want to continue this behavior, please take it on PM's, not here.\r\n\r\nIt is possible to prove The Zuton Force's lemma without calculus:\r\n\r\nProve that $ x\\sqrt{y\\plus{}1}\\minus{}\\sqrt{y}$ [b]without[/b] calculus.", "Solution_20": "Since $ x < y$, $ \\sqrt{x} < \\sqrt{y}$ and $ \\sqrt{x \\plus{} 1} < \\sqrt{y \\plus{} 1}$. Hence, $ \\sqrt{x} \\plus{} \\sqrt{x \\plus{} 1} < \\sqrt{y} \\plus{} \\sqrt{y \\plus{} 1}$. Hence, $ \\frac{1}{\\sqrt{x} \\plus{} \\sqrt{x \\plus{} 1}} > \\frac{1}{\\sqrt{y} \\plus{} \\sqrt{y \\plus{} 1}}$. Rationalizing the denominators yields $ \\sqrt{x \\plus{} 1} \\minus{} \\sqrt{x} > \\sqrt{y \\plus{} 1} \\minus{} \\sqrt{y}$, as desired.", "Solution_21": "sqrt(x+1) EY\u2019/FZ\u2019= ED/FD (2)\r\n3. according to (1)(2) => \u25b3EDY\u2019\u223d\u25b3FDZ\u2019 => \u25b3DY'Z'\u223d\u25b3DFE \r\n4. .\u25b3DY'Z' and \u25b3XYZ are congruence \r\n => \u25b3DFE\u223d\u25b3XZY \r\nDone!" } { "Tag": [ "blogs" ], "Problem": "Sa se rezolve ecuatia:\r\n \r\n $x\\cdot 2^{x-1}+\\frac{1}{\\sqrt{x}}\\cdot 2^{\\frac{1}{\\sqrt{x}}} =3.$ ;)", "Solution_1": "se inmulteste cu 2 , se aplica de 2 ori mediile si se obtine x=1", "Solution_2": "Ideea este buna, dar daca ai si scrie ar fi mai bine. ;)", "Solution_3": "avem $x 2^x + \\frac{1}{\\sqrt{x}} 2^{1/\\sqrt{x}} + \\frac{1}{\\sqrt{x}} 2^{1/\\sqrt{x}} \\geq 3 \\sqrt[3]{2^{x+1/\\sqrt{x}+1/\\sqrt{x}}}$\r\n\r\nramane de dem ca $x+\\frac{1}{\\sqrt{x}}+\\frac{1}{\\sqrt{x}} \\geq 3$, care e \"am-gm\"", "Solution_4": "[quote=\"perfect_radio\"]avem $x 2^x + \\frac{1}{\\sqrt{x}} 2^{1/\\sqrt{x}} + \\frac{1}{\\sqrt{x}} 2^{1/\\sqrt{x}} \\geq 3 \\sqrt[3]{2^{x+1/\\sqrt{x}+1/\\sqrt{x}}}$\n\nramane de dem ca $x+\\frac{1}{\\sqrt{x}}+\\frac{1}{\\sqrt{x}} \\geq 3$, care e \"am-gm\"[/quote]\r\nDa, ideea este buna, dar se putea face si altcumva.", "Solution_5": "Mah, dar tu nici n-ai aterizat bine pe site-ul asta si deja si vad ca deja dai reply-uri la toate posturile posibile si imposibile. N-ar fi o idee sa postezi si o solutie daca tot zici ca se poate face si altcumva. ;)", "Solution_6": "Se pare ca spam-detectorul lui Cezar a intrat in functiune :D :spam:", "Solution_7": "[quote=\"Cezar Lupu\"]Mah, dar tu nici n-ai aterizat bine pe site-ul asta si deja si vad ca deja dai reply-uri la toate posturile posibile si imposibile. N-ar fi o idee sa postezi si o solutie daca tot zici ca se poate face si altcumva. ;)[/quote]\r\n Bine.Se observa ca membrul stang este un numar par, in timp ce 3 este un numar impar.Atunci unul din numerele x-1 sau sqrtx/x este nul.Cum sqrtx/x este diferit de zero, vom obtine ca x-1=0, deci x=1. ;) Asta ca sa fii tu multumit.", "Solution_8": "[quote=\"spix\"]Se pare ca spam-detectorul lui Cezar a intrat in functiune :D :spam:[/quote]\r\n Cred ca mai degraba a intrat in functiune spam-detectorul meu. :spam:", "Solution_9": "[quote=\"stancioiu sorin\"][quote=\"spix\"]Se pare ca spam-detectorul lui Cezar a intrat in functiune :D :spam:[/quote]\n Cred ca mai degraba a intrat in functiune spam-detectorul meu. :spam:[/quote]Sorin, mai usor cu spam-ul. Cezar are dreptate. Daca ai ceva interesant de spus, spune-o, dar nu mai posta doar de dragu de a scrie ceva. Ai un blog pentru chestia asta.", "Solution_10": "[quote=\"Valentin Vornicu\"][quote=\"stancioiu sorin\"][quote=\"spix\"]Se pare ca spam-detectorul lui Cezar a intrat in functiune :D :spam:[/quote]\n Cred ca mai degraba a intrat in functiune spam-detectorul meu. :spam:[/quote]Sorin, mai usor cu spam-ul. Cezar are dreptate. Daca ai ceva interesant de spus, spune-o, dar nu mai posta doar de dragu de a scrie ceva. Ai un blog pentru chestia asta.[/quote]\r\n Am inteles Valentin. :surrender:", "Solution_11": "[quote=\"stancioiu sorin\"]Bine.Se observa ca membrul stang este un numar par, in timp ce 3 este un numar impar.Atunci unul din numerele x-1 sau sqrtx/x este nul.Cum sqrtx/x este diferit de zero, vom obtine ca x-1=0, deci x=1. ;) Asta ca sa fii tu multumit.[/quote]\r\n\r\nCred ca (scuzati cacofonia) conceptul de paritate al numerelor reale inca n-a fost inventat.", "Solution_12": "[quote=\"perfect_radio\"][quote=\"stancioiu sorin\"]Bine.Se observa ca membrul stang este un numar par, in timp ce 3 este un numar impar.Atunci unul din numerele x-1 sau sqrtx/x este nul.Cum sqrtx/x este diferit de zero, vom obtine ca x-1=0, deci x=1. ;) Asta ca sa fii tu multumit.[/quote]\n\nCred ca (scuzati cacofonia) conceptul de paritate al numerelor reale inca n-a fost inventat.[/quote]\r\n Scuze...atunci numerele sunt divizibile cu 2.Sunt de-abia la inceputuri.Sunt in clasa a VII-a, mai am destul pana sa invat tainele matematicii.La varsta asta mai greseste orice om, asa cred...", "Solution_13": "[quote=\"stancioiu sorin\"]Bine.Se observa ca membrul stang este un numar par, in timp ce 3 este un numar impar.Atunci unul din numerele x-1 sau sqrtx/x este nul.Cum sqrtx/x este diferit de zero, vom obtine ca x-1=0, deci x=1. ;) Asta ca sa fii tu multumit.[/quote]\r\n\r\n Mai bine nu mai zic nimik de acum incolo ca daca este sa fac observatie tuturor noilor veniti, ajung eu primul la balamuc. :wacko:\r\n\r\n\r\nEDIT: Pai, daca tot zici ca esti in clasaa-7-a, mai bine stai si invatza mai intai.", "Solution_14": "[quote=\"Cezar Lupu\"][quote=\"stancioiu sorin\"]Bine.Se observa ca membrul stang este un numar par, in timp ce 3 este un numar impar.Atunci unul din numerele x-1 sau sqrtx/x este nul.Cum sqrtx/x este diferit de zero, vom obtine ca x-1=0, deci x=1. ;) Asta ca sa fii tu multumit.[/quote]\n\n Mai bine nu mai zic nimik de acum incolo ca daca este sa fac observatie tuturor noilor veniti, ajung eu primul la balamuc. :wacko:\n\n\nEDIT: Pai, daca tot zici ca esti in clasaa-7-a, mai bine stai si invatza mai intai.[/quote]\r\n De ce zici asta?Nu e buna solutia?", "Solution_15": "In general, se spune despre un numar intreg ca e divizibil cu 2, nu despre unul real.\r\n\r\nHint: Nu mai da quote la mesajele care sunt \"ultimul post\". Quote in quote in quote ... = inutil." } { "Tag": [ "algebra open", "algebra" ], "Problem": "Solve the system of equations : \r\n $x+y+(z^2-8z+14)\\sqrt{x+y-z} =1$ and $2x+5y+\\sqrt{xy+z} = 3$", "Solution_1": "this question is in the mathscope, or no?" } { "Tag": [ "search", "quadratics", "number theory proposed", "number theory" ], "Problem": "Let $F_n$ be the $n$-th Fibonacci number. Prove that if $p>5$ is a prime, then\r\n\\[ F_p\\equiv \\left(\\frac{p}5\\right) \\mod p \\]", "Solution_1": "Sorry if this has been posted before but I don't want to search it since searching for 'Fibonacci' is a bit depressing (too many results obviously)", "Solution_2": "Because of the Law of Quadratic Reciprocity: $(\\frac{p}{5}) = (\\frac{5}{p})$.\r\n\r\nSuppose that $5$ is quadratic residue mod $p$.\r\nThen, $\\sqrt{5}$ is an element of $Z_p$, so:\r\n$F_p=\\frac{(\\frac{1+\\sqrt{5}}{2})^p - (\\frac{1-\\sqrt{5}}{2})^p}{\\sqrt{5}} \\equiv \\frac{\\frac{1+\\sqrt{5}}{2}-\\frac{1-\\sqrt{5}}{2}}{\\sqrt{5}} = \\frac{\\sqrt{5}}{\\sqrt{5}}=1=(\\frac{5}{p})$\r\nas desired\r\n\r\nNow, suppose that $5$ is not quadratic residue mod $p$.\r\nWe extend field $\\mathbb{Z}_p$ by the new element $\\sqrt{5}$. We see that:\r\n$(1+\\sqrt{5})^p \\equiv 1 + (\\sqrt{5})^p$\r\n(other terms cancel $\\mod{p}$)\r\nAlso:\r\n$1+(\\sqrt{5})^p=1+5^\\frac{p-1}{2}*\\sqrt{5}=1-\\sqrt{5}$\r\nbecause $5$ is not a quadratic residue.\r\nAnalogously: $(1-\\sqrt{5})^p \\equiv 1 + \\sqrt{5}$\r\nSo:\r\n$F_p=\\frac{(\\frac{1+\\sqrt{5}}{2})^p - (\\frac{1-\\sqrt{5}}{2})^p}{\\sqrt{5}} \\equiv \\frac{\\frac{1-\\sqrt{5}}{2} - \\frac{1+\\sqrt{5}}{2}}{\\sqrt{5}}=\\frac{-\\sqrt{5}}{\\sqrt{5}}=-1=(\\frac{5}{p})$\r\nas desired...", "Solution_3": "[quote=\"Yimin Ge\"]Sorry if this has been posted before but I don't want to search it since searching for 'Fibonacci' is a bit depressing (too many results obviously)[/quote]\r\n\r\nTry searching for Fibonacci and Legendre ;)", "Solution_4": "I don't know more about the number theory in $Z_n$,where can we find a nice book or article or post which introduce it?", "Solution_5": "Who can help me?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given $ a, b, c > 0$ and $ ab \\plus{} bc \\plus{} ca \\equal{} 1$. \r\n\r\nProve that: $ \\frac {(1 \\plus{} a^2)^2}{b^2 \\plus{} c^2} \\plus{} \\frac {(1 \\plus{} b^2)^2}{c^2 \\plus{} a^2} \\plus{} \\frac {(1 \\plus{} c^2)^2}{a^2 \\plus{} b^2} \\geq\\ 8$\r\n :)", "Solution_1": "We can solve this ineq by Am-Gm :)", "Solution_2": "[quote=\"nguoivn\"]Given $ a, b, c > 0$ and $ ab + bc + ca = 1$. \n\nProve that: $ \\frac {(1 + a^2)^2}{b^2 + c^2} + \\frac {(1 + b^2)^2}{c^2 + a^2} + \\frac {(1 + c^2)^2}{a^2 + b^2} \\geq\\ 8$\n :)[/quote]\r\n\r\nAssume $ a^2+b^2+c^2=x\\ge 1$\r\n\r\n$ LHS\\ge \\frac{8}{9}\\sum \\frac{1+3a^2}{b^2+c^2}\\ge \\frac{8(3x+3)^2}{9\\sum (1+3a^2)(b^2+c^2)}=\\frac{8(x+1)^2}{3(x+\\sum a^2b^2)}$\r\n\r\nWe will prove\r\n\r\n$ \\frac{(x+1)^2}{x+%Error. \"suma\" is a bad command.\n^2b^2}\\ge 3$\r\n\r\n$ \\Leftrightarrow x(x-1)\\ge 3\\sum a^2b^2-1$\r\n\r\n$ \\Leftrightarrow \\sum a^2\\left(\\sum a^2- \\sum ab\\right)\\ge 3\\sum a^2b^2-\\left(\\sum ab\\right)^2$\r\n\r\n$ \\Leftrightarrow (a^2+b^2-c^2)(a-b)^2+(b^2+c^2-a^2)(b-c)^2+(c^2+a^2-b^2)(c-a)^2\\ge 0$\r\n\r\nI think the last ineq true by SOS\r\n\r\n[hide]M\u1ea5y c\u00e1i ti\u00eau chu\u1ea9n SOS em k\u00f4 bi\u1ebft m\u1ea5y n\u00ean mong anh th\u00f4ng c\u1ea3m, \u0111\u1eebng qu\u00e1t em, em d\u1ec5 b\u1ecb s\u1ed1c l\u1eafm :D :rotfl: [/hide][/hide]", "Solution_3": "I checked this proof and I obtained $ x^2+1 \\geq 6\\sum a^2b^2$ or $ \\sum a^4 +2abc\\sum a \\geq 3\\sum a^2b^2$ which in SOS form is $ \\sum (a-b)^2(a^2+b^2+2ab-2c^2) \\geq 0$ and I don't think it can be solved by SOS method!!\r\n\r\n[quote=\"gigaman\"][quote=\"nguoivn\"]Given $ a, b, c > 0$ and $ ab + bc + ca = 1$. \n\nProve that: $ \\frac {(1 + a^2)^2}{b^2 + c^2} + \\frac {(1 + b^2)^2}{c^2 + a^2} + \\frac {(1 + c^2)^2}{a^2 + b^2} \\geq\\ 8$\n :)[/quote]\n\nAssume $ a^2 + b^2 + c^2 = x\\ge 1$\n\n$ LHS\\ge \\frac {8}{9}\\sum \\frac {1 + 3a^2}{b^2 + c^2}\\ge \\frac {8(3x + 3)^2}{9\\sum (1 + 3a^2)(b^2 + c^2)} = \\frac {8(x + 1)^2}{3(x + \\sum a^2b^2)}$\n\nWe will prove\n\n$ \\frac {(x + 1)^2}{x + %Error. \"suma\" is a bad command.\n^2b^2}\\ge 3$\n\n$ \\Leftrightarrow x(x - 1)\\ge 3\\sum a^2b^2 - 1$\n\n$ \\Leftrightarrow \\sum a^2\\left(\\sum a^2 - \\sum ab\\right)\\ge 3\\sum a^2b^2 - \\left(\\sum ab\\right)^2$\n\n$ \\Leftrightarrow (a^2 + b^2 - c^2)(a - b)^2 + (b^2 + c^2 - a^2)(b - c)^2 + (c^2 + a^2 - b^2)(c - a)^2\\ge 0$\n\nI think the last ineq true by SOS[/quote]", "Solution_4": "[quote=\"manlio\"]I checked this proof and I obtained $ x^2 + 1 \\geq 6\\sum a^2b^2$ or $ \\sum a^4 + 2abc\\sum a \\geq 3\\sum a^2b^2$ which in SOS form is $ \\sum (a - b)^2(a^2 + b^2 + 2ab - 2c^2) \\geq 0$ and I don't think it can be solved by SOS method!!\n\n[quote=\"gigaman\"][quote=\"nguoivn\"]Given $ a, b, c > 0$ and $ ab + bc + ca = 1$. \n\nProve that: $ \\frac {(1 + a^2)^2}{b^2 + c^2} + \\frac {(1 + b^2)^2}{c^2 + a^2} + \\frac {(1 + c^2)^2}{a^2 + b^2} \\geq\\ 8$\n :)[/quote]\n\nAssume $ a^2 + b^2 + c^2 = x\\ge 1$\n\n$ LHS\\ge \\frac {8}{9}\\sum \\frac {1 + 3a^2}{b^2 + c^2}\\ge \\frac {8(3x + 3)^2}{9\\sum (1 + 3a^2)(b^2 + c^2)} = \\frac {8(x + 1)^2}{3(x + \\sum a^2b^2)}$\n\nWe will prove\n\n$ \\frac {(x + 1)^2}{x + %Error. \"suma\" is a bad command.\n^2b^2}\\ge 3$\n\n$ \\Leftrightarrow x(x - 1)\\ge 3\\sum a^2b^2 - 1$\n\n$ \\Leftrightarrow \\sum a^2\\left(\\sum a^2 - \\sum ab\\right)\\ge 3\\sum a^2b^2 - \\left(\\sum ab\\right)^2$\n\n$ \\Leftrightarrow (a^2 + b^2 - c^2)(a - b)^2 + (b^2 + c^2 - a^2)(b - c)^2 + (c^2 + a^2 - b^2)(c - a)^2\\ge 0$\n\nI think the last ineq true by SOS[/quote][/quote]\r\nYou are right,try c=0 and a=b.", "Solution_5": "It's look weak but not so weak. I created it by Am-Gm and Voni Schur :)", "Solution_6": "[quote=\"nguoivn\"]It's look weak but not so weak. I created it by Am-Gm and Voni Schur :)[/quote]\r\nCan u give some hints how to use Voni Schur, nguoivn ?", "Solution_7": "I solved the ineq by cauchy shwarz , but infortunately I cannot post the solution for some private reasons :maybe: if someone want to see it please pm me :roll: :wink:", "Solution_8": "Show them your nice proof anas :lol: !!!", "Solution_9": "[quote=\"Evariste-Galois\"]Show them your nice proof anas :lol: !!![/quote]\r\n\r\nI cannot sorry evariste :blush: :blush:", "Solution_10": "[quote=\"leedt26\"][quote=\"nguoivn\"]It's look weak but not so weak. I created it by Am-Gm and Voni Schur :)[/quote]\nCan u give some hints how to use Voni Schur, nguoivn ?[/quote]\r\nUsing Am-Gm, we only need to prove: $ \\sum\\ \\frac {a(b \\plus{} c)(a^2 \\plus{} bc)}{b^2 \\plus{} c^2} \\geq\\ 2(ab \\plus{} bc \\plus{} ca)$\r\n(obviously trues by Voni Schur) :)", "Solution_11": "How to solve \r\n\r\n$ \\sum\\ \\frac {a(b \\plus{} c)(a^2 \\plus{} bc)}{b^2 \\plus{} c^2} \\geq\\ 2(ab \\plus{} bc \\plus{} ca)$\r\n\r\n\r\nby Vornicu-Shur?\r\n\r\nThank you very much. :)", "Solution_12": "[quote=\"manlio\"]How to solve \n\n$ \\sum\\ \\frac {a(b \\plus{} c)(a^2 \\plus{} bc)}{b^2 \\plus{} c^2} \\geq\\ 2(ab \\plus{} bc \\plus{} ca)$\n\n\nby Vornicu-Shur?\n\nThank you very much. :)[/quote]\r\nI think it's solved by CS .\r\nBecause we have $ LHS \\equal{} \\sum {\\frac{{a\\left( {b \\plus{} c} \\right)\\left( {{a^2} \\plus{} bc} \\right)}}{{\\left( {{b^2} \\plus{} {c^2}} \\right)}}} \\equal{} \\sum {{a^2}.\\frac{{b\\left( {{a^2} \\plus{} {c^2}} \\right) \\plus{} c\\left( {{a^2} \\plus{} {b^2}} \\right)}}{{a\\left( {{b^2} \\plus{} {c^2}} \\right)}} \\equal{} \\sum {\\frac{{{a^2}\\left( {y \\plus{} z} \\right)}}{x}} \\ge 2\\left( {ab \\plus{} bc \\plus{} ca} \\right)}\\equal{}RHS$ \r\nby Cauchy-Schwarz, where $ x \\equal{} a\\left( {{b^2} \\plus{} {c^2}} \\right);y \\equal{} b\\left( {{c^2} \\plus{} {a^2}} \\right);z \\equal{} c\\left( {{a^2} \\plus{} {b^2}} \\right)$", "Solution_13": "[quote=\"manlio\"]How to solve \n\n$ \\sum\\ \\frac {a(b \\plus{} c)(a^2 \\plus{} bc)}{b^2 \\plus{} c^2} \\geq\\ 2(ab \\plus{} bc \\plus{} ca)$\n\n\nby Vornicu-Shur?\n\nThank you very much. :)[/quote]\r\n\r\nSorry, it's trues by Am-Gm, not Voni Schur. I didn't remember my solution :oops: \r\nWe have: $ LHS \\equal{} \\sum\\ ab(\\frac {b^2 \\plus{} c^2}{c^2 \\plus{} a^2} \\plus{} \\frac {c^2 \\plus{} a^2}{b^2 \\plus{} c^2}) \\geq\\ 2(ab \\plus{} bc \\plus{} ca)$ :)", "Solution_14": "[quote=\"leedt26\"][quote=\"manlio\"]How to solve \n\n$ \\sum\\ \\frac {a(b \\plus{} c)(a^2 \\plus{} bc)}{b^2 \\plus{} c^2} \\geq\\ 2(ab \\plus{} bc \\plus{} ca)$\n\n\nby Vornicu-Shur?\n\nThank you very much. :)[/quote]\nI think it's solved by CS .\nBecause we have $ LHS \\equal{} \\sum {\\frac {{a\\left( {b \\plus{} c} \\right)\\left( {{a^2} \\plus{} bc} \\right)}}{{\\left( {{b^2} \\plus{} {c^2}} \\right)}}} \\equal{} \\sum {{a^2}.\\frac {{b\\left( {{a^2} \\plus{} {c^2}} \\right) \\plus{} c\\left( {{a^2} \\plus{} {b^2}} \\right)}}{{a\\left( {{b^2} \\plus{} {c^2}} \\right)}} \\equal{} \\sum {\\frac {{{a^2}\\left( {y \\plus{} z} \\right)}}{x}} \\ge 2\\left( {ab \\plus{} bc \\plus{} ca} \\right)} \\equal{} RHS$ \nby Cauchy-Schwarz, where $ x \\equal{} a\\left( {{b^2} \\plus{} {c^2}} \\right);y \\equal{} b\\left( {{c^2} \\plus{} {a^2}} \\right);z \\equal{} c\\left( {{a^2} \\plus{} {b^2}} \\right)$[/quote]\r\n\r\nDear leedt26,\r\n\r\n\r\ncan you please explain how to apply CS?\r\nThank you very much.", "Solution_15": "[quote=\"manlio\"][quote=\"leedt26\"][quote=\"manlio\"]How to solve \n\n$ \\sum\\ \\frac {a(b \\plus{} c)(a^2 \\plus{} bc)}{b^2 \\plus{} c^2} \\geq\\ 2(ab \\plus{} bc \\plus{} ca)$\n\n\nby Vornicu-Shur?\n\nThank you very much. :)[/quote]\nI think it's solved by CS .\nBecause we have $ LHS \\equal{} \\sum {\\frac {{a\\left( {b \\plus{} c} \\right)\\left( {{a^2} \\plus{} bc} \\right)}}{{\\left( {{b^2} \\plus{} {c^2}} \\right)}}} \\equal{} \\sum {{a^2}.\\frac {{b\\left( {{a^2} \\plus{} {c^2}} \\right) \\plus{} c\\left( {{a^2} \\plus{} {b^2}} \\right)}}{{a\\left( {{b^2} \\plus{} {c^2}} \\right)}} \\equal{} \\sum {\\frac {{{a^2}\\left( {y \\plus{} z} \\right)}}{x}} \\ge 2\\left( {ab \\plus{} bc \\plus{} ca} \\right)} \\equal{} RHS$ \nby Cauchy-Schwarz, where $ x \\equal{} a\\left( {{b^2} \\plus{} {c^2}} \\right);y \\equal{} b\\left( {{c^2} \\plus{} {a^2}} \\right);z \\equal{} c\\left( {{a^2} \\plus{} {b^2}} \\right)$[/quote]\n\nDear leedt26,\n\n\ncan you please explain how to apply CS?\nThank you very much.[/quote]\r\nOk,my friend.\r\nWe have $ \\sum {\\frac{{{a^2}\\left( {y \\plus{} z} \\right)}}{x} \\equal{} \\left( {x \\plus{} y \\plus{} z} \\right)\\left( {\\sum {\\frac{{{a^2}}}{x}} } \\right)}\\minus{} \\left( {{a^2} \\plus{} {b^2} \\plus{} {c^2}} \\right)$ \r\nthen applies CS we retains that.", "Solution_16": "It is very nice :) \r\n\r\nThank you very much. :)", "Solution_17": "[quote=\"nguoivn\"]Given $ a, b, c > 0$ and $ ab \\plus{} bc \\plus{} ca \\equal{} 1$. \n\nProve that: $ \\frac {(1 \\plus{} a^2)^2}{b^2 \\plus{} c^2} \\plus{} \\frac {(1 \\plus{} b^2)^2}{c^2 \\plus{} a^2} \\plus{} \\frac {(1 \\plus{} c^2)^2}{a^2 \\plus{} b^2} \\geq\\ 8$\n :)[/quote]\r\n\r\nWe can use cauchy and AM-GM to solve it.(by my brother)\r\n$ LHS \\ge \\frac{[(a^3\\plus{}b^3\\plus{}c^3\\plus{}3abc\\plus{}ab(a\\plus{}b)\\plus{}bc(b\\plus{}c)\\plus{}ca(c\\plus{}a)]^2}{2(a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2)} \\ge \\frac{2(a^3\\plus{}b^3\\plus{}c^3\\plus{}3abc)[ab(a\\plus{}b)\\plus{}bc(b\\plus{}c)\\plus{}ca(c\\plus{}a)]}{a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2} \\ge 8(ab\\plus{}bc\\plus{}ca)$\r\nFinally,we need to prove that:\r\n$ (a^3\\plus{}b^3\\plus{}c^3\\plus{}3abc)[ab(a\\plus{}b)\\plus{}bc(b\\plus{}c)\\plus{}ca(c\\plus{}a)]\\ge 4(ab\\plus{}bc\\plus{}ca)(a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2)$\r\n<=>$ \\sum ab(a^4\\plus{}b^4)\\plus{} \\sum a^2b^2(a^2\\plus{}b^2) \\ge 4(a^3b^3\\plus{}b^3c^3\\plus{}c^3a^3)$(obivious true by AM-GM)\r\nOur proof are completed.Equality occur if and only a=b=c.\r\nWe can prove the general(in the same way)\r\n$ \\frac{(a^2\\plus{}k)^2}{b^2\\plus{}c^2}\\plus{} \\frac{(b^2\\plus{}k)^2}{c^2\\plus{}a^2}\\plus{} \\frac{(c^2\\plus{}k)^2}{a^2\\plus{}b^2} \\ge \\frac{(3k\\plus{}1)^2}{2}$\r\ntrue for $ 0 \\le k \\le 1$ with the same condition." } { "Tag": [ "floor function", "ratio", "number theory unsolved", "number theory" ], "Problem": "Let $ A$ be a non-empty subset of $ \\mathbb N$ satisfying the following condition: If $ a \\in A$, then $ 4a$ and $ \\left[\\sqrt {a}\\right]$ are also in $ A$ (where $ \\left[x\\right]$ denotes the integer part of $ x$). Prove that $ A \\equal{} \\mathbb N$.", "Solution_1": "By applying $ \\lfloor \\sqrt{a} \\rfloor$ repeatedly, we eventually get that $ 1 \\in A$ if any element is in $ A$. Now we suppose there is an $ A$ such that the problem is false. Take the smallest element $ s > 1$ that is not in $ A$. Then no elements in between $ s^2$ and $ (s\\plus{}1)^2 \\minus{} 1$ inclusive are in $ A$. This in turn implies that no elements in between $ s^{2^k}$ and $ (s\\plus{}1)^{2^k} \\minus{} 1$ are in $ A$. The ratio of these two values is unbounded, so eventually we must have that $ (s\\plus{}1)^{2^k} \\minus{} 1 > 4s^{2^k}$. For such a $ k$, there must be an $ m$ such that $ 4^m$ is in the interval, and we know $ 4^m$ is in $ A$ since $ 1$ is in $ A$. Contradiction.\r\n\r\nIf this solution is correct, $ 4$ can be replaced with any positive integer.", "Solution_2": "@mods: Move this to temp, please (MY active problem).", "Solution_3": "let $ n\\in\\mathbb{N}$,we take for $ a\\in\\mathbb{N}: \\ f(a) \\equal{} [\\sqrt {a}],g(a) \\equal{} 4a$\r\nwe have $ {a\\in[n^{2^k},(n \\plus{} 1)^{2^{k}}[\\Longrightarrow f(a)\\in[n^{2^{k \\minus{} 1}},(n \\plus{} 1)^{2^{k \\minus{} 1}}[\\Rightarrow...f_k(a) \\equal{} fof..k\\ times...of(a)\\in[n,n \\plus{} 1[}$\r\nthen if $ a\\in A$: $ n \\equal{} f_k(a)\\in A$\r\nit's evident that for some $ k\\in\\mathbb{N},\\exists h\\in\\mathbb{N}: \\ n^{2^{k}}\\le 4^h < (n \\plus{} 1)^{2^{k}}$ $ \\boxed{because\\ \\exists M,\\forall k\\ge M: \\ (n \\plus{} 1)^{2^{k}} \\minus{} n^{2^{k}} > 4^{k \\plus{} 1} \\minus{} 4^{k}}$\r\nwe take $ b\\in A$ we have $ f_m(b) \\equal{} 1$ for some $ m\\in\\mathbb{N}$ so $ 4^h \\equal{} g_h(1)\\in A$\r\nthen $ n \\equal{} f_k(4^{h})\\in A$", "Solution_4": "$ A\\equal{}\\{0\\}$????????", "Solution_5": "The other thread containing this problem was locked because this is a problem from an ongoing contest, so this one should be too. And $ A$ is non-empty. \r\n\r\nBut as long as this thread already contains the correct solution, I should mention that the first condition can, amusingly enough, be replaced with the condition \"$ A$ contains the primes.\"" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a_1,...,a_n,b_1,...,b_n$ be positive real numbers, such that $ a_1 \\plus{} ... \\plus{} a_n \\equal{} b_1 \\plus{} ... \\plus{} b_n$.Prove that $ \\dfrac{a_1^2}{a_1 \\plus{} b_1} \\plus{} ... \\plus{} \\dfrac{a_n^2}{a_n \\plus{} b_n}\\geq\\dfrac{a_1 \\plus{} ... \\plus{} a_n}{2}$.", "Solution_1": "That's just Cauchy in Engel form." } { "Tag": [ "inequalities", "calculus" ], "Problem": "We have the numbers $a=\\sqrt[3]{60}$ and $b=2+\\sqrt[3]{7}$\r\nWhich number is bigger?", "Solution_1": "[[b]EDIT[/b]: this was nonsense. Sorry. ]", "Solution_2": "Silouan, I posted this in the intermediate forum a few days ago... :| \r\n\r\nJensen for $f(x) = x^{1/3}$ gives \\[2 + \\sqrt[3]{7} = f(8) + f(7) \\leq 2 f\\left(\\frac{15}{2}\\right) = \\sqrt[3]{60}\\] so $2 + \\sqrt[3]{7} < \\sqrt[3]{60}$. So, Kondr, you were wrong :)", "Solution_3": "How did you get\r\nf(a)+f(b)>=2f((a+b)/2)\r\nIs it through calculus.", "Solution_4": "It's using Jensen's ineq ( http://mathworld.wolfram.com/JensensInequality.html ).\r\nFor $f(x)=x^{1/3}$ we have $f''(x)=-\\frac 2 9 x^{-\\frac 5 3}<0$, so ineq \r\n$\\frac {f(a)+f(b)} 2 \\leq f\\left( \\frac {a+b} 2 \\right)$ holds.\r\nA non-calculus solution is on\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=324803#p324803", "Solution_5": "Arne - good solution. \r\nVarun -Iit is sort of intutive. If you think of the curve ($y=x^{(1/3)}$) , it is concave, so if you join two points (x=7, and x=8) the line (all points on the line except the end points) is always \"below\" the curve, and the mid point on the line is below the the point on the curve (where x=7.5). (BTW you may have a typo when you said \" f(a)+f(b)>=2f((a+b)/2) \" the sign \">\" should be \"<\")\r\n\r\nAlso, FWIW The \"straight forward metod\" for many of us who grew up before the age of calculators (but after the age of binomial expansion) , is to think\r\n$60^{(1/3)} = (64-4)^{1/3}$ and $7^{1/3} = (8-1)^{1/3}$ and the difference between these two \r\n = $4 (1 - 1/48 -(1/9)(1/16)^2 \\cdots) - 2 ( 1 - 1/24 -(1/9)(1/8)^2 \\cdots )$ is greater than $2$ . QED." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove positive integer $n$ is prime iff $\\varphi (n)|n-1\\ ,\\ n+1|\\sigma (n)$", "Solution_1": "Have a look here :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=lehmer&t=14533\r\n\r\nPierre." } { "Tag": [ "ratio" ], "Problem": "Ratios always trip me up, could someone solve this problem and show me easy way to do ratios?\r\n\r\nThe sum of four numbers is 771. The ratio of the first to the second is 2:3. The ratio of the second to the third is 5:4. The ratio of the third to the fourth is 5:6. Find the second number.", "Solution_1": "Let the first number be a. Then work out the second number in terms of a, then the third one, then the fourth one. Then you just get an equation in terms of a.\n\n\n\nIf its the actual ratios you are having problems with, then if you have a ratio of x:y and you know the first number is a, then the second number must be.. (in spoiler) [hide]ay/x, because x:y is the same as 1: (y/x)[/hide]\n\n\n\nAfter you have done that, you could possibly make your calculations a bit easier by letting the first number, say, be 50a, just to keep everything out of fractions.", "Solution_2": "K Thanks" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": ":( :( :(", "Solution_1": ":clap: :clap:" } { "Tag": [], "Problem": "Albatross has his friends Bobert, Caddlebot, Deberg, Eustine and Frankinfueter over at his house for New Years. The friends get restless so they decide to go see a movie. The row is 6 chairs long. Frankinfueter hates everyone there except Albatross, so the only person he will sit next to is Albatross. How many seating arrangement s are there so that Frankinfueter is happy.", "Solution_1": "[hide]\n\nThere are two cases: where F sits on the very left and where F sits on the very right. \n\nCase 1: F sits on the very left. You have F A _ _ _ _. 4! can be used to describe the 4 blanks. \nCase 2: F sits on the very right. _ _ _ _ A F. 4! can be used again. \n\nSo the answer is 2 x 4! = 48 \n\n[/hide]", "Solution_2": "Well, if Frankinfueter hates them enough to steal Albatross's poison kit...\n\n\n\n[hide]\n\nAs all of the other people are eliminated, Albatross has six seats to choose from and Frankinfueter has five, so the answer is 30.\n\n[/hide]\n\n\n\nAssuming Albatross used all of the poison on Frankinfueter (or leeched it into the water supply)...\n\n\n\n[hide]\n\n\n\nF A _ _ _ _\n\n24 possibilities\n\n\n\n_ _ _ _ A F\n\n24 possibilities\n\n\n\n24 + 24 = 48, like aznphatso said.\n\n\n\n\n\nActually, Frankinfueter is never happy. If he has fun pushing his friend off buildings, he probably derives happiness from other murderous things, like poisoning Albatross's friends (see above).\n\n[/hide]\n\n\n\n\n\n\n\nIs Caddlebot a person?", "Solution_3": "Caddlebot's a person.", "Solution_4": "Okay then :) .\r\n\r\nCaddlebot is cool :) . He bounces to 1/4 of his previous height when dropped.", "Solution_5": "[quote=\"Treething\"]Okay then :) .\n\nCaddlebot is cool :) . He bounces to 1/4 of his previous height when dropped.[/quote]\r\nYou made that fact up.", "Solution_6": "Treething wrote:Well, if Frankinfueter hates them enough to steal Albatross's poison kit...\n\n[hide]\nAs all of the other people are eliminated, Albatross has six seats to choose from and Frankinfueter has five, so the answer is 30.\n[/hide]\n\n\n\nExcept that Frakinfueter still has to sit next to Albatross even if there are 6 seats. I don't think I sit next to someone if there's four seats between us.", "Solution_7": "It says the only [b]person[/b] Frankinfueter will sit next to is Albatross. This means that he can sit next to an empty space, as an empty space isn't a person." } { "Tag": [ "probability", "\\/closed" ], "Problem": "Will there be an Intermediate Counting and Probability book to go along with the class?", "Solution_1": "There is no text at this time. According to the Book List, no release of such a book is planned this year.", "Solution_2": "Eventually, but not this summer. The book will likely be out sometime between Dec, 2006 and Mar, 2007.", "Solution_3": "I'm taking intermediate counting and probability this summer. There is enough problems from the intermediate book ( comming out later ) that I can still justifiy buying it as a problem book/reference? Or not worth the effort?", "Solution_4": "Not only will the book have additional exercises, but there will be topics in the book that are covered more thoroughly than they will be in the class." } { "Tag": [ "email", "summer program", "Mathcamp", "geometry", "floor function" ], "Problem": "I got an email saying that one of my solutions (#7) was good (unexpected), and that I would be lucky (:wink:) enough to present it. I'm assuming other people have gotten similar?\r\n\r\nConfirmation is necessary by July 3rd, so I thought I'd just check over here that I won't be working myself into a nervous wreck over this. Alright, \"nervous wreck\" is an exaggeration, it can't be *that* bad. :lol: \r\n\r\nSo is anybody else planning on presenting?", "Solution_1": "Aha! So you're one of the other lucky winners. I will present a solution, but I won't say which problem. :P", "Solution_2": "I personally did not get such an email, but I am not entirely sure what you're worried about. If it was good, just look over your solution (you kept it, right :-) ), and essentially read it.\r\n\r\nOut of curiosity, what did you get? At this point, I think we can openly discuss answers, since they are not accepting any more applications. I am pretty sure I got the right answer, but I did not simplify it from a form with a summation (I could, but the deadline was nearing, and I did not want to risk making the mistake. I got in, so I guess it was OK!) What did you get?", "Solution_3": "(Speaking as an Alum)\r\n\r\nSo at some point in week 2 (I believe it's week two, at least. It's definitely not like... day 2 of camp or anything), most of the camp goes to an assembly at which all 10 qualifying quiz solutions are presented by new campers. Thus, if you want to present, you won't be alone ;). There will be 9 other new campers in your shoes. I'd highly recommend you take them up on the offer, but that's just me...\r\n\r\nAnyway, congrats on getting a \"commended\" solution, and I look forward to seeing you in a couple of weeks.", "Solution_4": "I will be presenting number 10 (despite the fact that the solution I came up with used a calculator).", "Solution_5": "[quote=\"hahafaha\"]Out of curiosity, what did you get? At this point, I think we can openly discuss answers, since they are not accepting any more applications. I am pretty sure I got the right answer, but I did not simplify it from a form with a summation (I could, but the deadline was nearing, and I did not want to risk making the mistake. I got in, so I guess it was OK!) What did you get?[/quote]\r\n\r\nI assume we're allowed to, as applications are over? Actually, can we compile a list of all the correct answers (perhaps not the proofs, since I assume we're going over them at mathcamp)? I want to know how many I actually got correct. \r\n\r\nknexpert, I used the computer to do #10 ... not really a good sign if you're using the computer to solve a geometry problem. Oh well, it worked :lol:", "Solution_6": "Here are my answers, which have been checked with others:\r\n\r\n1. $ \\lfloor \\frac{3n}{2}\\rfloor$\r\n2. 1, 2, 3, 4, and 91\r\n3. (a) yes (b) no\r\n4. 59\r\n5. (a) unsure, most work (b) prove it!\r\n6. 2\r\n7. (a) $ \\frac{2}{3}$ (b) $ \\frac{(n-k+3)!k!}{3!n!}$ (except when $ k \\le 4$)\r\n8. $ n^{2}-k^{2}$\r\n9. Define all configurations inductively, either add a facedown card or flip over the rightmost card and add a face up card\r\n10. a=12, b=15, c=13", "Solution_7": "[quote=\"calc rulz\"]Here are my answers, which have been checked with others:\n\n(snip)\n\n6. 2\n\n(snip)\n\n[/quote]\r\n\r\nReally? Which two configurations worked?", "Solution_8": "5(a) is everything but $ 2$ and $ 6$.", "Solution_9": "I'll make things easier, I got everything right except #4 I think.", "Solution_10": "So, for problem #6, I got 0. Here is a brief description of a counter-claim to the proof (I'm not going to bother looking at all cases, just one of the ones Tom presented):\r\n\r\nWhen we multiply by a power of 2, the color changes if the power is odd and remains the same if it was odd (because every re-group essentially corresponds to a color change, and to a power). Tom suggested a situation in which 9 = green, which would mean that 4 and 5 are both blue (4+5=9; blue+blue=green). 1 is then also blue. Now through fairly simple arithmetic, we can deduce that 40 is blue. Also, 2048=1x2^11, and since 11 is odd, 2048 is green.\r\n\r\nNow if blue+blue=green, then green-blue=blue. Thus, 2008=2048-40=green-blue=2008=blue.\r\n\r\nNow if 2007=green, then 2009=2007+2=green+green=blue. However, 2009=2008+1=blue+blue=green, so there is a contradiction. Similarly, if 2007=blue, then 2008=2007+1=blue+blue=green. However, 2008 was previously determined to be blue, so there is still a contradiction.\r\n\r\nI am too lazy to check it for the other one, as well as to write out the proof for all the other cases, but I just thought I'd share this. I'd be interested in feedback/discussion!", "Solution_11": "Here are my solutions. I got everything right except for #5a, and in #6 I made an unproved claim, which I think is not too hard to prove using the division algorithm.", "Solution_12": "So, the issue I see with 2007 being the first non-red (green, specifically) positive integer is this (note that I may be completely wrong):\r\n\r\nIt is easy to see that green+red=green, because 0 has to be red (since -0=0, and -green=blue and vice-versa, we'd have a contradiction otherwise). Since x+0=x, a green x+0 must equal green. We assumed that all positive integers before 2007 were red, and 2007 itself was green. However, we then have:\r\n\r\nn=2007\r\nn = green\r\nn+1=green+red=green\r\nn+2=(n+1)+2=green+red=green\r\nn+3=(n+2)+1=green+red=green\r\n...\r\n2n+3=(2n+2)=green+red=green\r\n\r\nHowever, 2n+3=(n+1)+(n+2)=green+green=blue, not green. Thus, there is a contradiction. In general, by this rule, I don't think 1 can be red successfully.", "Solution_13": "[quote=\"hahafaha\"]\nIt is easy to see that green+red=green, because 0 has to be red (since -0=0, and -green=blue and vice-versa, we'd have a contradiction otherwise). Since x+0=x, a green x+0 must equal green. We assumed that all positive integers before 2007 were red, and 2007 itself was green. However, we then have:\n[/quote]\r\n\r\nYes, green+red=green is true when the red number is 0, but it need not hold in general :wink:", "Solution_14": "[quote=\"calc rulz\"]Yes, green+red=green is true when the red number is 0, but it need not hold in general :wink:[/quote]\r\n\r\nOh, that is a good point I forgot to account for. I stand corrected.", "Solution_15": ":o I got 8 answers correct! (idk about proofs though)\r\n\r\nIn #6, the two cases are where 9 and 2007 are the first green, non-red integers.", "Solution_16": "[quote=\"azjps\"]In #6, the two cases are where 9 and 2007 are the first green, non-red integers.[/quote]\r\n\r\nYeah. I realized that my mistake stemmed from a flawed assumption." } { "Tag": [ "abstract algebra", "LaTeX", "superior algebra", "superior algebra unsolved" ], "Problem": "Let V be a countable dimensional vectorspace over a field F . \r\nLet R denote End_F V . \r\nProve that V is a simple R-module. If e1 , e2 , . . . is a basis \r\nof V , then we have a module homomorphism \u03c6_j from R to V , \r\nsending f \r\n\u2208 R to f (e_j ). \r\nFind the ker(\u03c6_j) . \r\nFind Jac(R). here I mean jacobson radical.\r\nProve that there is exactly one non-trivial twosided ideal, \r\nnamely \r\n{f \u2208 R| dimf (V ) < \u221e} Prove that R", "Solution_1": "It sholud be\r\n\r\nLet V be a countable dimensional vectorspace over a field F . \r\nLet R denote End_F V . \r\nProve that V is a simple R-module. If $ e1 , e2 , . . .$ is a basis \r\nof V , then we have a module homomorphism \u03c6_j from R to V , \r\nsending f in R to f (e_j ). \r\nFind the ker(\u03c6_j) . \r\nFind Jac(R). here I mean jacobson radical.\r\nProve that there is exactly one non-trivial twosided ideal, \r\nnamely \r\n$ {f \u2208 R| dimf (V ) < \\infty}$ Prove that R is isomorphic to R + R here + means direct sum . thanks.", "Solution_2": "[quote=\"peteryellow\"]Let $ V$ be a countable dimensional vectorspace over a field $ F$. Let $ R$ denote $ End_F V$ . \nProve that $ V$ is a simple $ R$-module.\n\nIf $ e_1, e_2 , \\cdots$ is a basis of $ V$ , then we have a module homomorphism $ \\phi_j$ from $ R$ to $ V$, sending $ f \\in R$ to $ f (e_j )$. Find the $ ker\\ \\phi_j$.\n\nFind $ Jac(R)$. here I mean jacobson radical.\n\nProve that there is exactly one non-trivial twosided ideal, namely $ \\{f \\in R | dim\\ f(V ) < \\infty\\}$. Prove that $ R$ is isomorphic to $ R\\oplus R$.[/quote]\r\nFrequently these threads are read hundreds of times, so you may want to learn a thing or two about readability. Here's a start: dollar signs and line breaks go a long way.", "Solution_3": "You might as well use $ \\ker$ and $ \\dim$ while you're at it. (To say nothing of the \\rm command for things like End, Jac, etc.)\r\n\r\nSee [[LaTeX]] and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 for information about using the mathematical typesetting system LaTeX." } { "Tag": [ "algebra", "polynomial", "linear algebra", "linear algebra unsolved" ], "Problem": "1.Show that $A$ is nilpotent if and only if $\\forall k \\in (1,..,n) ,Tr(A^k)=0$\r\n2.Suppose that the set $(\\lambda|A+\\lambda B$ is nilpotent$)$ contains at least $n+1$ elements.Show that $A$ and $B$ are nilpotent.\r\n(All the matrices belong to $M_n(C)$)", "Solution_1": "[hide]2) From result 1) all $P_k(x) = Tr ((A + x B)^k)$ are polynomials of degree at most $n$ with $n+1$ roots : they are all null, in particular $P_k(0) = 0$ and A is nilpotent.[/hide]" } { "Tag": [], "Problem": "The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $ 20$ cents. If she had one more quarter, the average value would be $ 21$ cents. How many dimes does she have in her purse?\r\n\r\n$ \\textbf{(A)}\\ 0\\qquad\r\n\\textbf{(B)}\\ 1\\qquad\r\n\\textbf{(C)}\\ 2\\qquad\r\n\\textbf{(D)}\\ 3\\qquad\r\n\\textbf{(E)}\\ 4$", "Solution_1": "[hide]\n\nlet $w,x,y,z$ represent the number of [b]pennies, nickels, dimes, and quarters[/b] respectively\n\nSo you get two equations:\n\n1. $w+5x+10y+25z=20(w+x+y+z)$\n2. $w+5x+10y+25(z+1)=21(w+y+z+1)$\n $\\Rightarrow 5x+10y+25z+4=21(w+x+y+z)$\n\nSubtracting 1 from 2;\n\n3. $w+x+y+z=4$, therefore there are $4$ coins in total\n\nSubstitute (3) into 1.\n\n$\\Rightarrow w+5x+10y+25z=80$\n\nIf there are $2$ quarters then you need $3$ more dimes to get a total of $80$, which would amount to $5$ coins we can only have four, so therefore:\n\n$\\Rightarrow w+5x+10y+25(3)=80\\implies w+5x+10y=5\\implies x=1 \\ w=0 \\ y=0$\n\nThus she has $\\boxed{0\\implies\\text{A}}$ dimes in her purse.\n[/hide]", "Solution_2": "[hide=\"slightly less calculations\"]\nwhen adding a quarter increases the average by 21,\nsubtract 21 from 25(the value of the quarter) to get 4\nthen multiply 4 by 20(the original average) to get 80.\nthat means she had 80 cents \nthen use smart's way to find the number of coins\nand then get the answer of 0.....[/hide]", "Solution_3": "[quote=\"#H34N1\"]The average value of all the pennies, nickels, dimes, and quarters in Paula's\npurse is 20 cents. If she had one more quarter, the average value would be 21\ncents. How many dimes does she have in her purse?\n(A) 0 (B) 1 (C) 2 (D) 3 (E) 4[/quote]\r\n\r\n[hide]Let there be $n$ coins in her purse so there are a total of $20n$ cents in her purse. \n\nNow we can set up the equation $20n+25=21(n+1)$ so $n=4$.\n\nThis means that she orgionally had $80$ cens in her purse. The only way to make $80$ cents with $4$ coins is with $3$ quarters and $1$ nickel.\n\nTherefore, there are $\\boxed{0}$ nickles.[/hide]", "Solution_4": "You mean 0 dimes not 0 nickels (slight typo..nice solution)", "Solution_5": "Good solution and nice typo" } { "Tag": [ "inequalities", "rearrangement inequality", "inequalities unsolved" ], "Problem": "Let $a,b,c,x,y,z$ be real numbers.Prove that\r\n$ax+by+cz+\\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}\\ge\\frac{2}{3}(a+b+c)(x+y+z)$", "Solution_1": "More generally, for any 2n reals $a_1$, $a_2$, ..., $a_n$, $x_1$, $x_2$, ..., $x_n$, we have\r\n\r\n$\\sum_{i=1}^n a_ix_i+\\sqrt{\\sum_{i=1}^n a_i^2\\cdot\\sum_{i=1}^n x_i^2}\\geq\\frac{2}{n}\\cdot\\sum_{i=1}^n a_i\\cdot\\sum_{i=1}^n x_i$.\r\n\r\nIn fact, this inequality rewrites as\r\n\r\n$\\sqrt{\\sum_{i=1}^n a_i^2\\cdot\\sum_{i=1}^n x_i^2}\\geq\\frac{2}{n}\\cdot\\sum_{i=1}^n a_i\\cdot\\sum_{i=1}^n x_i-\\sum_{i=1}^n a_ix_i$;\r\n\r\ndenoting $X=\\sum_{j=1}^n x_j$, we have\r\n\r\n$\\sum_{i=1}^n\\left(\\frac{2}{n}X-x_i\\right)^2=\\sum_{i=1}^n\\left(\\frac{4}{n^2}X^2-\\frac{4}{n}Xx_i+x_i^2\\right)$\r\n$=n\\cdot\\frac{4}{n^2}X^2-\\frac{4}{n}X\\underbrace{\\sum_{i=1}^n x_i}_{=X}+\\sum_{i=1}^n x_i^2$\r\n$=\\frac{4}{n}X^2-\\frac{4}{n}X^2+\\sum_{i=1}^n x_i^2=\\sum_{i=1}^n x_i^2$\r\n\r\nand\r\n\r\n$\\frac{2}{n}\\cdot\\sum_{i=1}^n a_i\\cdot\\sum_{i=1}^n x_i-\\sum_{i=1}^n a_ix_i=\\sum_{i=1}^n a_i\\cdot\\frac{2}{n}\\sum_{j=1}^n x_j-\\sum_{i=1}^n a_ix_i$\r\n$=\\sum_{i=1}^n a_i\\left(\\frac{2}{n}\\sum_{j=1}^n x_j-x_i\\right)=\\sum_{i=1}^n a_i\\left(\\frac{2}{n}X-x_i\\right)$;\r\n\r\nthus, our inequality is equivalent to\r\n\r\n$\\sqrt{\\sum_{i=1}^n a_i^2\\cdot\\sum_{i=1}^n\\left(\\frac{2}{n}X-x_i\\right)^2}\\geq\\sum_{i=1}^n a_i\\left(\\frac{2}{n}X-x_i\\right)$;\r\n\r\nbut this obviously follows from the Cauchy-Schwarz inequality.\r\n\r\n Darij", "Solution_2": "[quote=\"spider_boy\"]Let $a,b,c,x,y,z$ be real numbers.Prove that\n$ax+by+cz+\\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}\\ge\\frac{2}{3}(a+b+c)(x+y+z)$[/quote]\r\n\r\n The problem is nice i think that you can solve it yourself. Its really easy.\r\nExpand the terms and use Cauchy Swarz then prove a stronger version of this then for the last step use the rearrangement Inequality hope that you will enjoy\r\n\r\nDavron", "Solution_3": "Dear Darij Grineberg \r\n\r\nWhy do you prefer to solve problems by the hard ways. \r\nI think mine is shorter.\r\n1)C-S\r\n2)rearrangment \r\n3)The problem is killed\r\n\r\nDavron", "Solution_4": "Dear Davron,\r\n\r\nHow exactly do you want to apply rearrangement? We have no information about how our numbers are sorted...\r\n\r\n Darij", "Solution_5": "Dear Darij\r\n\r\nyes sure you are right i stupidly assumed that a is max and x max and i am wrong ;) \r\nBut dont you have any other solutions \r\nTry you it please i believe in you can find !\r\n\r\nDavron", "Solution_6": "[quote=\"spider_boy\"]Let $a,b,c,x,y,z$ be real numbers.Prove that\n$ax+by+cz+\\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}\\ge\\frac{2}{3}(a+b+c)(x+y+z)$[/quote]\r\nI think this is vasc's inequality. ;)", "Solution_7": "[quote=\"arqady\"][quote=\"spider_boy\"]Let $a,b,c,x,y,z$ be real numbers.Prove that\n$ax+by+cz+\\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}\\ge\\frac{2}{3}(a+b+c)(x+y+z)$[/quote]\nI think this is vasc's inequality. ;)[/quote]\r\nIf you mean vasc has created this inequality ,then i ' m not sure.As I have stated\r\nin the source it is an old inequality from famous [b]KVANT[/b] journal. :lol:", "Solution_8": "What about the following, even more general conjecture:\r\n\r\n If $a_i, b_i, c_i \\ge0$ for $i=1,...,n$ (this will only work for non-negative variables), prove\r\n$\\sum a_ib_ic_i+\\sqrt[3]{\\sum a_i^3\\cdot\\sum b_i^3\\cdot\\sum c_i^3}\\ge \\frac2{n^2} \\sum a_i\\cdot\\sum b_i\\cdot\\sum c_i$ :?: \r\nand similarly for $k$ rows of variables, the fraction on the RHS becoming $\\frac2{n^{k-1}}$.\r\n\r\nIn the [i]very [/i]particular case that the $a_i, b_i, c_i$ are equally sorted, this follows of course easily from H\u00f6lder and Chebyshev. But no idea how to prove it otherwise. :roll: Darij's nice method doesn't seem to work with more than two sums...\r\n\r\nThe case $n=2$ and $k=3$ of that, yielding\r\n$abc+xyz+\\sqrt[3]{(a^3+x^3)(b^3+y^3)(c^3+z^3)}\\ge\\frac12(a+x)(b+y)(c+z)$,\r\nsmells again like rearrangement but maybe not that easy. :? Any ideas :?:" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Suppose $ A, B, C, D, E, F$ are six distinct points in the plane, and suppose $ AB, DE, CF$ are concurrent and $ AD, BC, EF$ are concurrent. Prove that $ AF, BE, CD$ are also concurrent.", "Solution_1": "[hide]It is easy applying the [b][size=100]Pappos theorem[/size][/b].[/hide]\r\nKostas Vittas." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let $ a_1,a_2,..,a_{2000} \\in Z$, $ |a_i| \\leq 1000 \\ \\forall i, \\ \\displaystyle\\sum_{i\\equal{}1}^{2000} a_i \\equal{}1$\r\nProve $ \\exists \\ K \\subset (1,2,..,2000), K \\neq \\phi$ so that $ \\displaystyle\\sum_{i \\in K}^{} a_i \\equal{}0$", "Solution_1": "Does $ K \\subset (1,2,..,2000)$ means that $ K$ is a subset of consecutive terms of that??(like K=(5,6,7.8))\r\n\r\n\r\n :?: \r\n\r\n\r\n :)", "Solution_2": "this is from Canada 2000...\r\nI know a solution using pigeon hole principle", "Solution_3": "to ElChapin : yes, $ K$ is a subset of $ (1,2,..,2000)$\r\n \r\nI use a lemma to prove this problem\r\nLemma: If $ x_1,x_2,..,x_m,y_1,y_2,..,y_n \\in N (N\\equal{} (1,2,....))$\r\nand $ \\displaystyle\\sum_{1}^{m}x_i\\equal{}\\displaystyle\\sum_{1}^{n}y_i\\equal{}Vp>0$, $ a_1\\equal{}a_{n\\plus{}1}$ and $ \\sum_{i\\equal{}1}^{n}{a_i}\\equal{}k.$ Find the minimum of\r\n\\[ \\sum_{i\\equal{}1}^{n}{\\frac{a_i}{a^2_{i\\plus{}1}\\minus{}p^2}}\\]", "Solution_1": "Ok. Now I can post my solution, hope it will be interesting...\r\nWe know an inequality for $ a > 0$, $ b > 0$, $ x,y\\in R^{ \\plus{} }$ and $ x\\ne y$:\r\n\\[ \\frac {a^x}{b^y}\\geq \\frac {xa^{x \\minus{} y} \\minus{} yb^{x \\minus{} y}}{x \\minus{} y}\r\n\\]\r\nIt is equals only if $ a \\equal{} b$.\r\n\r\nNow:\r\n\\[ \\frac {a_i}{a_{i \\plus{} 1}^2 \\minus{} p^2} \\equal{} \\frac {a_i}{2a_{i \\plus{} 1}}\\left(\\frac {1}{a_{i \\plus{} 1} \\plus{} p} \\plus{} \\frac {1}{a_{i \\plus{} 1} \\minus{} p}\\right)\\rightarrow\r\n\\]\r\n\r\n\\[ \\rightarrow \\frac {a_i}{2a_{i \\plus{} 1}}\\left( \\frac {1}{\\left(\\frac {k}{n} \\plus{} p\\right)^2}\\cdot\\frac {\\left(\\frac {k}{n} \\plus{} p\\right)^2}{a_{i \\plus{} 1} \\plus{} p} \\plus{} \\frac {1}{\\left(\\frac {k}{n} \\minus{} p\\right)^2}\\cdot\\frac {\\left(\\frac {k}{n} \\minus{} p\\right)^2}{a_{i \\plus{} 1} \\minus{} p} \\right)\\geq\r\n\\]\r\n\r\n\\[ \\rightarrow \\frac {a_i}{2a_{i \\plus{} 1}}\\left( \\frac {1}{\\left(\\frac {k}{n} \\plus{} p\\right)^2}\\cdot \\left(2\\left(\\frac {k}{n} \\plus{} p\\right) \\minus{} (a_{i \\plus{} 1} \\plus{} p)\\right) \\plus{} \\frac {1}{\\left(\\frac {k}{n} \\minus{} p\\right)^2}\\cdot\\left( 2\\left(\\frac {k}{n} \\minus{} p\\right) \\minus{} (a_{i \\plus{} 1} \\minus{} p)\\right) \\right)\\rightarrow\r\n\\]\r\n\r\n\\[ \\rightarrow \\frac {a_i}{2a_{i \\plus{} 1}}\\left( \\frac {\\frac {2k}{n} \\plus{} p \\minus{} a_{i \\plus{} 1}}{\\left(\\frac {k}{n} \\plus{} p\\right)^2} \\plus{} \\frac {\\frac {2k}{n} \\minus{} p \\minus{} a_{i \\plus{} 1}}{\\left(\\frac {k}{n} \\minus{} p\\right)^2} \\right)\\rightarrow\r\n\\]\r\n\r\n\\[ \\rightarrow \\frac {a_i}{2a_{i \\plus{} 1}}\\left( \\frac {\\frac {4k^3}{n^3}}{\\left(\\frac {k^2}{n^2} \\minus{} p^2\\right)^2} \\minus{} \\frac {\\frac {2k^2}{n^2} \\plus{} 2p^2}{\\left(\\frac {k^2}{n^2} \\minus{} p^2\\right)^2} \\cdot a_{i \\plus{} 1}\\right)\\rightarrow\r\n\\]\r\n\r\n\\[ \\rightarrow \\frac {\\frac {2k^3}{n^3}}{\\left(\\frac {k^2}{n^2} \\minus{} p^2\\right)^2}\\cdot \\frac {a_i}{a_{i \\plus{} 1}} \\minus{} \\frac {\\frac {k^2}{n^2} \\plus{} p^2}{\\left(\\frac {k^2}{n^2} \\minus{} p^2\\right)^2} \\cdot a_{i}\r\n\\]\r\nNow we can write:\r\n\\[ \\sum_{i \\equal{} 1}^{n}{\\frac {a_i}{a_{i \\plus{} 1}^2 \\minus{} p^2}}\\geq \\frac {\\frac {2k^3}{n^3}}{\\left(\\frac {k^2}{n^2} \\minus{} p^2\\right)^2}\\cdot \\sum_{i \\equal{} 1}^{n}{\\frac {a_i}{a_{i \\plus{} 1}}} \\minus{} \\frac {\\frac {k^2}{n^2} \\plus{} p^2}{\\left(\\frac {k^2}{n^2} \\minus{} p^2\\right)^2} \\cdot \\sum_{i \\equal{} 1}^{n}{a_{i}}\r\n\\]\r\n\r\n\\[ \\sum_{i \\equal{} 1}^{n}{\\frac {a_i}{a_{i \\plus{} 1}}}\\geq n\r\n\\]\r\n\r\n\\[ \\sum_{i \\equal{} 1}^{n}{\\frac {a_i}{a_{i \\plus{} 1}^2 \\minus{} p^2}}\\geq \\frac {\\frac {2k^3}{n^3}}{\\left(\\frac {k^2}{n^2} \\minus{} p^2\\right)^2}\\cdot n \\minus{} \\frac {\\frac {k^2}{n^2} \\plus{} p^2}{\\left(\\frac {k^2}{n^2} \\minus{} p^2\\right)^2} \\cdot k \\equal{} \\frac {k}{\\left(\\frac {k}{n}\\right)^2 \\minus{} p^2}\r\n\\]\r\nEquality holds only if $ a_1 \\plus{} p \\equal{} a_2 \\plus{} p \\equal{} ... \\equal{} a_n \\plus{} p \\equal{} \\frac {k}{n} \\plus{} p$ and $ a_1 \\minus{} p \\equal{} a_2 \\minus{} p \\equal{} ... \\equal{} a_n \\minus{} p \\equal{} \\frac {k}{n} \\minus{} p$, so if $ a_1 \\equal{} a_2 \\equal{} ... \\equal{} a_n \\equal{} \\frac {k}{n}$\r\n\r\nAnswer:\r\n\\[ \\boxed{\\min\\sum_{i \\equal{} 1}^{n}{\\frac {a_i}{a_{i \\plus{} 1}^2 \\minus{} p^2}}\\equal{}\\frac {k}{\\left(\\frac {k}{n}\\right)^2 \\minus{} p^2}}\r\n\\]" } { "Tag": [ "geometry", "Alcumus" ], "Problem": "deleteddeleted", "Solution_1": "At the end of intro to geometry, it says it's coming out in 2009.", "Solution_2": "Unfortunately, I think it's going to be a bit later than that :(", "Solution_3": "why?\r\ni mean any serious problem\r\n :maybe: :maybe:", "Solution_4": "I've been [url=http://www.artofproblemsolving.com/Alcumus/Introduction.php]a little busy[/url], and I have to take care of writing the trig/complex numbers book first, which I'll start in a few weeks.", "Solution_5": "How long does it take you to write one book?", "Solution_6": "[quote=\"rrusczyk\"]I've been [url=http://www.artofproblemsolving.com/Alcumus/Introduction.php]a little busy[/url], and I have to take care of writing the trig/complex numbers book first, which I'll start in a few weeks.[/quote]\r\n\r\nYes! Trig/complex numbers! Must buy!\r\n\r\nHm...does this mean that algebra is going into Alcumus soon?" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f$ be a continuous, nonconstant, real function, and assume the existence of an $ F$ such that $ f(x\\plus{}y)\\equal{}F[f(x),f(y)]$ for all real $ x$ and $ y$. Prove that $ f$ is strictly monotone.", "Solution_1": "second problem, here: [url]http://www.mathematik.uni-muenchen.de/~yarotsky/jarnik/Jarnik_daniel_harrer.pdf[/url]", "Solution_2": "Since the link above isn't working (at least for me), I'll go ahead and post a solution.\n\nAssume for the sake of contradiction that $f$ is not strictly monotone. \nThat means that $f$ has a local extrema, which is attained at some real number, call it $n_0$.\n\nFor arbitrarily small $\\epsilon > 0$ we can find $x_1,$ $x_2 \\in (n_0-\\epsilon,n_o+\\epsilon)$ such that $f(x_1)=f(x_2)$.\nNow, see that $$f(x+x_2-x_1) = F[f(x-x_1), f(x_2)] = F[f(x-x_1), f(x_1)] = f(x)$$\nfor any $x \\in \\mathbb{R}$.\n\nIt follows that $f$ is a continuous function that has arbitrarily small periods, therefore $f$ is constant, contrary to the hypothesis." } { "Tag": [ "logarithms" ], "Problem": "The product, $ \\log_a b \\cdot \\log_b a$ is equal to:\r\n$ \\textbf{(A)}\\ 1 \\qquad\\textbf{(B)}\\ a \\qquad\\textbf{(C)}\\ b \\qquad\\textbf{(D)}\\ ab \\qquad\\textbf{(E)}\\ \\text{none of these}$", "Solution_1": "[hide=\"Solution\"]\nChange of base gives $ \\frac{\\log a}{\\log b}\\cdot \\frac{\\log b}{\\log a}\\equal{}1$.\n[/hide]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "A sequence with first two terms equal 1 and 2 respectively is defined by the following rule:\r\neach subsequent term is equal to the smallest positive integer which has not yet occurred in\r\nthe sequence and is not coprime with the previous term. Prove that all positive integers occur\r\nin this sequence.", "Solution_1": "Consider a prime $ p$. We claim that if infinitely many multiples of $ p$ occur in the sequence then all multiples of $ p$ occur in the sequence. Otherwise, let $ kp$ be the smallest multiple of $ p$ which does not occure in the sequence. If $ p|a_n$, then $ a_{n\\plus{}1}x$. Note that there exist infinitely many $ m$ such that $ a_{m\\plus{}1}>a_m$, where $ m>x$. Let $ d\\equal{}gcd(a_m,a_{m\\plus{}1})$. But $ d0 \\\\\r\n\\frac{1}{ab+1}+\\frac{1}{bc+1}+\\frac{1}{ca+1}=a+b+c\\\\$\r\nProve that:\r\n$(i) \\ \\ \\sum{a(a^2+2bc+1)}\\geq 9\\\\\r\n(ii)\\ \\ (\\frac{a+b+c}{3})^2 \\cdot (\\frac{1}{ab}+\\frac{1}{bc}+\\frac{1}{ca})\\geq \\frac{3-(a+b+c)}{abc}$", "Solution_1": "The second one is an easy one. I believe you compose it yourself after a couple of minutes of thinking to the first one... at least that's how it happened to me.\r\nUse Cauchy for $\\sum\\frac 1{ab+1}\\sum(ab+1)\\geq 9\\Rightarrow \\sum\\frac 1{ab+1}(3+\\frac {(a+b+c)^2}3)\\geq 9\\Rightarrow S(3+\\frac13 S^2)\\geq9$. And $(ii)$ is equivalent to this.\r\nAs for the first one, the equality doesn't hold for equal numbers. More than that I see now that the inequality doesn't hold at all for $a=b=c$.", "Solution_2": "I don't think the first one is correct.\r\nLet a=b=c, the restriction turns to\r\n3/(a^2+1)=3a \r\nFrom the equation we get a^3+a-1=0. Since this is an increasing function, and f(0)<0, so a>0 *\r\nThe ineq is 9a^3+3a=9(1-a)+3a=9-6a>=9\r\nObviously , the ineq is not correct.", "Solution_3": "i think he meant $\\leq9$ ;)", "Solution_4": "it's no use to say \"I think\".\r\nWhy not give us some ideas?" } { "Tag": [ "AMC", "AIME", "pigeonhole principle", "factorial", "ratio", "geometry", "AIME I", "\\/closed" ], "Problem": "We are now accepting enrollments for our Special AIME Problem Seminar, a one-weekend class shortly before the AIME in March. Classes are on March 3 (Sat) and Mar 4 (Sun) from 3:30 PM until 6:30 PM ET (12:30 - 3:30 PM PT). Each class will have a 30 minute break. \r\n\r\nEnrollment information for the Special AIME Problem Seminar can be found [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]here[/url].", "Solution_1": "I want to enroll\r\n\r\n\r\nwoah wow why does everything cost hard-earned money? :( \r\n\r\n\r\nmy parents won't let me because of the price.. \r\n\r\n\r\nThey are middle class though: My dad makes like 110,000 to 112,000 a yaer. Well that has nothing to do with it. But duno.\r\n\r\n\r\nArgh :lol:", "Solution_2": "[quote=\"now a ranger\"]woah wow why does everything cost hard-earned money? :( [/quote]\r\nWell Mr. Rusczyk needs to feed his family too.", "Solution_3": "yes quite true.\r\n\r\n\r\nOh wait, it's full time. \r\n\r\nOops. :blush: \r\n\r\n\r\nI'll just try to use Asian Persuasion to get my parents to let me sign up sometime this semester.", "Solution_4": "[quote=\"now a ranger\"]I want to enroll\nwoah wow why does everything cost hard-earned money? :( \nmy parents won't let me because of the price.. \nThey are middle class though: My dad makes like 110,000 to 112,000 a yaer. Well that has nothing to do with it. But duno.\nArgh :lol:[/quote]\r\n\r\nActually, the classes are relatively cheap if you compare the cost to those of other online classes that other companies offer (maybe not math, but others). That's what's cool about AoPs... sweet classes with low pay :D", "Solution_5": "[quote=\"Karth\"][quote=\"now a ranger\"]I want to enroll\nwoah wow why does everything cost hard-earned money? :( \nmy parents won't let me because of the price.. \nThey are middle class though: My dad makes like 110,000 to 112,000 a yaer. Well that has nothing to do with it. But duno.\nArgh :lol:[/quote]\n\nActually, the classes are relatively cheap if you compare the cost to those of other online classes that other companies offer (maybe not math, but others). That's what's cool about AoPs... sweet classes with low pay :D[/quote]\r\nAoPS pwnz all.\r\nAbout how many problems will be done in this? Will there be message board problems? How many message board problems? I heard from someone that there was a mock last year, what about this year?", "Solution_6": "I've taken several practice AIMEs recently and I have received scores of 6-8. Do you think that this seminar will improve my AIME score to a 9 or 10?", "Solution_7": "[quote=\"knexpert\"]I've taken several practice AIMEs recently and I have received scores of 6-8. Do you think that this seminar will improve my AIME score to a 9 or 10?[/quote]\r\nI don't have that much knowledge about this, but keep in mind two things.\r\nOne: It is harder and harder to improve on the AIME the better you get.\r\nTwo: This thing is just before the AIME so it's more like a very very effective cram session from what I'm inferring. \r\n\r\nThen again, I may be totally wrong.", "Solution_8": "so if you are making 2's and 3's and 4's, it is not as hard to improve your score?", "Solution_9": "[quote=\"now a ranger\"]so if you are making 2's and 3's and 4's, it is not as hard to improve your score?[/quote]\r\nIt's easier to improve your score by 2 at that point than if you are getting 12s...", "Solution_10": "He's saying that once you are at a 10+ level, there isn't much more to learn compared with you at a 3-4 level.", "Solution_11": "oh yeah that's what I thought his message was.", "Solution_12": "I'm a noob, but from what I believe it's that:\r\n\r\nImproving from 2-4ish to 7-9ish is a matter of picking up more concepts and gaining more problem experience.\r\nImproving from 10-11ish to 14-15 is a matter of being more creative and noticing cruxes and stuff more quickly.", "Solution_13": "[quote=\"13375P34K43V312\"]I'm a noob, but from what I believe it's that:\n\nImproving from 2-4ish to 7-9ish is a matter of picking up more concepts and gaining more problem experience.\nImproving from 10-11ish to 14-15 is a matter of being more creative and noticing cruxes and stuff more quickly.[/quote]\r\n\r\n\r\nif you're a noob, then I'm a flat retard who doesn't know how to add 1 digit numbers. \r\n\r\nYeah. I'm going to read AoPS 2. I gotta ask my teacher for AoPS 2 though to borrow. I only have AoPS 1.", "Solution_14": "[quote=\"13375P34K43V312\"]Improving from 10-11ish to 14-15 is a matter of being more creative and noticing cruxes and stuff more quickly.[/quote]\r\n\r\nor not making careless mistakes", "Solution_15": "Hmm if I can already solve 8-11ish of the problems in 3 hours (assuming no trivial errors), would reading AoPS Volume 2 help that much?", "Solution_16": "I can only solve 2-3 on average, and I'm going to ask my teacher to borrow one of the school's AoPS Vol. 2 textbooks.\r\n\r\n\r\nHow should I ask her nicely?\r\n\r\n\r\nEveryone in Alg. 2 is provided with an AoPS Vol.2 textbook to use for the year, but I'm only in geometry... AoPS 1 is easy", "Solution_17": "Just to comment:\r\n\r\nI would say that the AIME takes two major things:\r\n1) Familiarity with problem solving techniques and general facts.\r\n2) Ability to adapt those techniques to slightly different problems.\r\n\r\n[Neither of which is necessarily easy to develop...]\r\n\r\nFrom my point of view, at least half of the AIME problems are twists on the same old techniques.\r\n\r\nSo basically, you need to have those two things: 1) knowledge of the techniques, 2) ability to apply them when there is a twist.\r\n\r\nSo you are not really going to learn the second one in a day or two, but if you still don't have the first one, then a class will be helpful. To see what I mean, consider the 2006 AIME I\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2006[/url]\r\n\r\n1. technique: pythagoren theorem...\r\n\r\n2. technique: complement [this problem was pretty old...seen in pigeonhole problem from a while ago for a 10 element subest]\r\nnew idea: it was a 90 element subset instead...\r\n\r\n3. technique: this is a common problem, seen often...writing the digit as a variable...a*10^n+x=29x...\r\n\r\n4. technique: number of 0 digits at the end of a factorial\r\ntwist: there are 100 factorials now\r\n\r\n5. technique: squaring... [but this was from usamts...]\r\n\r\n6. technique: 1) average for each place value 2) repeating digits is just multiplying by 1/999\r\n\r\n7. technique: side ratio ^2= area ratio for similar triangles\r\ntwist: the vertex is not on the parallel lines\r\n\r\n8. [i will admit this was pretty unique]\r\n\r\n9. technique: 1) log rules, 2) arithmetic series sum, 3) solve linear diophantine in positive intgers\r\ntwist: combining the 3\r\n\r\n10. [this was a cool problem]\r\n\r\n11. [although recursions are a standard technique for these types of problems, this was a unique problem]\r\n\r\n12. technique: sum and difference formulas\r\ntwist: not obvious to see the factorizations...\r\n\r\n13. [cool, but i think i have seen the summation before...]\r\n\r\n14. technique: 1) height *base=volume [the rest was brute forcing the lengths and areas] 2) you could do vectors...but that is a bit messy to find them...\r\n\r\n15. [this was cool and new]", "Solution_18": "[quote=\"calc rulz\"]Hmm if I can already solve 8-11ish of the problems in 3 hours (assuming no trivial errors), would reading AoPS Volume 2 help that much?[/quote]\r\n\r\nIt should help some, but if you're hitting 8-11 now, you should be spending at least half your time (if not more) working on some olympiad problems.", "Solution_19": "[quote=\"knexpert\"]I've taken several practice AIMEs recently and I have received scores of 6-8. Do you think that this seminar will improve my AIME score to a 9 or 10?[/quote]\r\n\r\nI'd expect you'd need more than just 5 hours of class to improve three points on the AIME, but this class is designed for people in the 4-11 range, so it should be right up your alley.", "Solution_20": "I'm not in the alley. \r\n\r\nI make 2's and 3's, unless it's a VERY easy test.", "Solution_21": "Is this class going to include a mock AIME, I've heard that one in the past did?", "Solution_22": "When is the deadline for registration?", "Solution_23": "March 4. (After the first class, before the second.)", "Solution_24": "Because of my state science olympiad tournament, I might have to miss the first hour or so of the March 3 session. Is is still worthwhile to sign up?", "Solution_25": "[quote=\"hillarryous\"]Because of my state science olympiad tournament, I might have to miss the first hour or so of the March 3 session. Is is still worthwhile to sign up?[/quote]\r\n\r\nYou'll have access to the transcript after class, so you'll still get to cover everything you miss.\r\n\r\nGood luck at the tournament.", "Solution_26": "[quote=\"bpms\"]Is this class going to include a mock AIME, I've heard that one in the past did?[/quote]\r\nYes, I was wondering that too. Anyone?", "Solution_27": "[quote=\"davidyko\"][quote=\"bpms\"]Is this class going to include a mock AIME, I've heard that one in the past did?[/quote]\nYes, I was wondering that too. Anyone?[/quote]\r\nHe said that they will be using the one from 05.", "Solution_28": "Cool, thanks." } { "Tag": [ "inequalities", "quadratics", "search" ], "Problem": "What is AM-GM", "Solution_1": "it's arithmetic mean and geometric mean. the arithmetic mean is always greater than or equal to the geometric mean.", "Solution_2": "$\\frac{a_{1}+a_{2}+...+a_{n}}{n} \\geq \\sqrt[n]{a_{1}a_{2}...a_{n}}$\r\n\r\nEquality holds when $a_{1}=a_{2}=...=a_{n}$", "Solution_3": "You also need to make sure that $a_1, a_2, a_3, ..., a_n$ are all positive. If they are negative or 0, the inequality does not always hold.", "Solution_4": "thanks. also imenat this to go in the getting started section :D", "Solution_5": "[quote=\"anirudh\"]thanks. also imenat this to go in the getting started section :D[/quote]\r\n\r\nI moved it. Inequalities are NOT getting started. ;) (Other than linear and quadratic)", "Solution_6": "Please also try the search tool on the top toolbar.\r\n\r\nLink: [url]http://www.artofproblemsolving.com/Forum/search.php[/url]\r\n\r\nThere is a sticky at the top of the intermediate topics forum which is an inequality marathon - there you will find explanations not only on AM-GM but other inequalities and many practice problems." } { "Tag": [], "Problem": "In the Old Young Chinese Culture We have\r\n\r\n$a = r$\r\n$b = j$\r\n$c = k$\r\netc....\r\n\r\n\r\nTranslate \"Thales418\" To japanese Math Culture", "Solution_1": "rofl :lol:\r\n\r\nit may be something like \"$7y57893578jfifbaifbibifaibfsdkfwsnwicvw$ or so.", "Solution_2": "it's $Thales418$", "Solution_3": "this is racist you wetback", "Solution_4": "this is chinese math.....\r\n\r\nis not Racist Math\r\n\r\nnether Russian Math \r\n\r\nand other countries" } { "Tag": [ "function" ], "Problem": "Hi,\r\n\r\nCan anybody help out with the following problems: please\r\nSorry if it in the wrong forum\r\n\r\n1. Prove that the following property holds\r\n\r\n(a,b)=(a1,b1) if and only if a=a1 and b=b1\r\n\r\n2. Are we justified in calling the partial function from X->Y and the total functions from X->Y sets, where X and Y are sets?\r\n\r\n3. Show that for any sets X and Y, where Y has atleast 2 elements, that there canot be a 1 to 1 corrospondence between X and the set of functions (X->Y)\r\n\r\n4. Let R be a realtion on a set X. We write Rop as the opposite or converse relation where Rop = {(y,x) | (x,y) belongs to R}}\r\nShow that (R U Rop)* is an equivalance realtion and R U (Rop)* need not be an equivalance relation.\r\n\r\n\r\nThank you for any help at all", "Solution_1": "can anybody help out?\r\n\r\nplease", "Solution_2": "this doesn't look anything like advanced fields to me. I'll move it to a more appropriate forum ...", "Solution_3": "Where are these problems from?", "Solution_4": "Glenn Winskel Formal semantics book", "Solution_5": "Is it a homework assignment?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c>0$ such that $ ab\\plus{}bc\\plus{}ca$\u2265$ 3$.Prove that\r\n$ \\frac{a}{\\sqrt{a\\plus{}b}}\\plus{}\\frac{b}{\\sqrt{b\\plus{}c}}\\plus{}\\frac{c}{\\sqrt{c\\plus{}a}}$\u2265$ \\frac{3}{\\sqrt2}$", "Solution_1": "It's obviously trues by Honder and Am-Gm :maybe:", "Solution_2": "[quote=\"nguoivn\"]It's obviously trues by Honder and Am-Gm :maybe:[/quote]\r\nContinue with [b]nguoivn's[/b] approach in detailed:\r\nBy [b]Holder[/b], we have:\r\n$ (LHS)^2(\\sum a(a+b))\\ge (a+b+c)^3$\r\n$ \\Longleftrightarrow (LHS)^2\\ge \\frac {(a+b+c)^3}{\\sum a(a+b)}$\r\nHence so as to prove the inequality of the problem, it is enough to prove:\r\n$ 2(a+b+c)^3\\ge 9(a^2+b^2+c^2+ab+bc+ca)$, $ (*)$\r\nDenote $ (a+b+c,ab+bc+ca)=(p,q)$\r\n$ (*)\\Longleftrightarrow 2p^3\\ge 9(p^2-q)$\r\nWe have $ q\\ge 3\\Longrightarrow 9(p^2-q)\\le 9(p^2-3)\\le 2p^3$ (By $ AM-GM$)\r\nThe result is lead as follow." } { "Tag": [ "inequalities", "trigonometry", "geometry", "3D geometry" ], "Problem": "Let $ ABC$ be an acute-angled triangle. Prove that\r\n\r\n\\[ \\tan\\alpha\\cdot \\tan\\beta \\cdot \\tan\\gamma \\ge 3\\sqrt{3}\\]", "Solution_1": "[quote=\"Obel1x\"]Let $ ABC$ be an acute-angled triangle. Prove that\n\\[ \\tan\\alpha\\cdot \\tan\\beta \\cdot \\tan\\gamma \\ge 3\\sqrt {3}\\]\n[/quote]\r\n[hide]Knowing that $ \\tan\\alpha\\cdot \\tan\\beta\\cdot \\tan\\gamma\\equal{}\\tan\\alpha\\plus{}\\tan\\beta\\plus{}\\tan\\gamma$ we now wish to prove $ \\tan\\gamma\\plus{}\\tan\\beta\\plus{}\\tan\\alpha \\geq 3\\sqrt{3}$.Also set $ \\tan\\alpha\\cdot \\tan\\beta\\cdot \\tan\\gamma\\equal{}x$. This also means that $ \\tan\\alpha\\plus{}\\tan\\beta\\plus{}\\tan\\gamma\\equal{}x$. Using AM-GM we get that $ \\frac{x}{3}\\geq \\sqrt[3]{x}$. Rearranging this gives us $ x^{3}\\geq 27x$. Since we know that $ x$ isn't 0, we have $ x^{2}\\geq27$, or $ x\\geq3\\sqrt{3}$ and now we are done.[/hide]", "Solution_2": "Hello, \r\n\r\nsince $ tg(x)$ is convex on $ [0,90]$ we have:\r\n\r\n$ tg(x)\\plus{}tg(y)\\plus{}tg(z)>\\equal{}3tg((x\\plus{}y\\plus{}z)/3)$ by Jensen's inequality.\r\n\r\nSo $ tg(x)\\plus{}tg(y)\\plus{}tg(z)>\\equal{}3\\sqrt{3}$\r\n\r\nWith AM-GM I got :\r\n\r\n$ tg(x)\\plus{}tg(y)\\plus{}tg(z)>\\equal{}3\\sqrt[3]{tg(x)*tg(y)*tg(z)}$\r\n\r\nWe need to show that $ 3\\sqrt[3]{tg(x)*tg(y)*tg(z)}>\\equal{}3\\sqrt[]{3}$\r\n\r\nIf we cube this expression, we get $ tg(x)tg(y)tg(z)>\\equal{}3\\sqrt[]{3}$" } { "Tag": [ "function", "real analysis", "real analysis solved" ], "Problem": "I don't know if this one has been posted before.\r\n\r\n$\\{f_n(x)\\}$ is a sequence of increasing functions on [0,1]. The function sequence converges to $f$ in measure. Show that $f_n(x_0) \\to f(x_0)$ where $x_0$ is a continuity of $f$.", "Solution_1": "If I am not wrong then this problem is quite straightforward :D \r\n Just use the definitions!!", "Solution_2": "can you show your proof?\r\ni can only proof that there is a subsequence $\\{f_{n_k}(x)\\}$ such that $f_{n_k}(x_0) \\to f(x_0)$ when $x_0$ is a continuity of $f$.", "Solution_3": "By the way, the conclusion may not be true if our concerning continuity is $0$ or $1$. \r\n\r\nmy proof:\r\n\r\n If not, there exist an $\\epsilon > 0$ and a subsequence, which we will reenumerate as $\\{f_i\\}_{i \\in \\mathbb{N}}$, such that $|f_i(t) - f(t)| > 2\\epsilon$ for all $i$\r\n\r\n Since $t$ is a continuity of $f$ then there exist $\\delta$ small enough so that $(t-\\delta,t+\\delta)$ still included in $[0,1]$, and the flunctuation of $f$ on this interval is not greater than $\\epsilon$\r\n \r\n Then a straightforward argument show that $m\\{(f_i-f) > \\epsilon \\} > \\delta$ for every $i$, hence $\\{f_i\\}_{i \\in \\mathbb{N}}$ does not converge to $f$ in measure.", "Solution_4": "Do you mean the following?\r\n\r\nIf $f_i(x_0) > f(x_0) + 2\\varepsilon$, then for $x\\in (x_0, x_0 + \\delta)$,\r\n$f_i(x) - f(x) > f_i(x_0) - f(x) > f(x_0) - f(x) + 2\\varepsilon > \\varepsilon$\r\nand if $f_i < f(x_0) + 2\\varepsilon$, $f_i(x) - f(x) > \\varepsilon$ for $x \\in (x_0-\\delta, x_0)$.\r\n\r\nand can you provide with a example that the statement is false at $x=0$ and $x=1$?", "Solution_5": "[quote=\"liyi\"]Do you mean the following?\n\nIf $f_i(x_0) > f(x_0) + 2\\varepsilon$, then for $x\\in (x_0, x_0 + \\delta)$,\n$f_i(x) - f(x) > f_i(x_0) - f(x) > f(x_0) - f(x) + 2\\varepsilon > \\varepsilon$\nand if $f_i < f(x_0) + 2\\varepsilon$, $f_i(x) - f(x) > \\varepsilon$ for $x \\in (x_0-\\delta, x_0)$.\n\nand can you provide with a example that the statement is false at $x=0$ and $x=1$?[/quote]\r\n\r\n Yes, you're right!!\r\n And for the example, consider $f_n(x) = \\frac{x}{n}$ for $x <1$ and $f_n(1) = 1 $ for all $n$\r\n $f(x) = 0$" } { "Tag": [ "induction", "absolute value" ], "Problem": "OK, another problem that I've been unable to solve for awhile now (It sure would be nice if there was a Solution manuel for [u]The Art and Craft of Problem Solving[/u]...).\r\n\r\nConsider the operation which transforms the 8-term sequence a1,a2,...,a8 into the new 8-term sequence |a1-a2|,|a2-a3|,...,|a8-a1|. Find all 8-term sequences of integers that have the property that after finitely many applications of this operation, one is left with a sequence, all of whose terms are equal.\r\n\r\nBy a1 I mean a with subscript 1, etc.\r\n\r\nI've come to some basic conclusions, mostly involving the maximum and minimum elements, but I just can't seem to finish it off.", "Solution_1": "Does a1 through a8 have to be different?", "Solution_2": "Not necessarily.", "Solution_3": "Then a1 through a8 can all be the same number, right?", "Solution_4": "I have the solution manual for Zeitz. Which problem number and which chapter, etc. is it? I'll look it up to see if it's there for you.", "Solution_5": "Yes, all the numbers can be the same.\r\n\r\nIt's problem 3.4.36.\r\n\r\nWhere were you able to get a solution manual? I found a website that distributes it, but it looked like you had to be a certified instructor to get one, or something like that.", "Solution_6": "Here is the ``official'' solution:\r\n\r\n36. (Sketch due to Kiran Kedlaya) No matter what the starting sequence, eventually all terms become 0. Use induction on the maximum absolute value of the terms, which does not increase after the transformation.\r\n\r\nFirst we show that after 8 steps, all the terms will be even, using the fact\r\nthat |x2 \u2212x1| \u2261 x2 +x1 (mod 2). (So we can drop the absolute value brackets.) At this point, divide each term by 2, then use the inductive hypothesis.\r\n\r\nI got the solution manual by asking Paul Zeitz for it. I guess that's the most direct way of doing things.", "Solution_7": "Riceboi tweaked the problem a bit and came up with this:\r\n\r\n\"Given an initial sequence a_1, a_2, ..., a_n of real numbers, we perform a series of steps. At each step, we replace the current sequence x_1, x_2, ..., x_n with |x_1 - a|, |x_2 - a|, ..., |x_n - a| for some a. For each step, the value of a can be different.\r\n\r\nProve that it is always possible to obtain the null sequence consisting of all 0's and determine the minimum number of steps required, regardless of initial sequence, to obtain the null sequence.\"" } { "Tag": [], "Problem": "If a bird collector wants to buy 100 budgies and wants to spend exactly $ 100. And blue budgies cost$ 10 each, green budgies cost $ 3 each, and yellow budgies cost$ .50 cents each, then how many of each can he buy?", "Solution_1": "We have that $ 10b\\plus{}3g\\plus{}.5y\\equal{}100 \\longmapsto 20b\\plus{}6g\\plus{}y\\equal{}200$ (multiplying by 2). Subtracting the equation $ b\\plus{}g\\plus{}y\\equal{}100$ from this, we get that:\r\n$ 19b\\plus{}5g\\equal{}100$. Since $ b$ and $ g$ are integers, $ 19b$ has to end in a $ 5$ or a $ 0$ so that we can add enough $ g's$ to get to $ 100$. Obviously, $ 19(5)\\equal{}95$ is the desirable choice, so $ \\boxed{b\\equal{}5}$, and $ g$, therefore, must be $ \\boxed{1}$. Finally, plugging these numbers into our equation $ b\\plus{}g\\plus{}y\\equal{}100$, we get that $ \\boxed{y\\equal{}94}$.", "Solution_2": "I don't undertand why you double b and g in the formula?", "Solution_3": "Because then you get $ 2(.5y)\\equal{}y$ in the equation. From this, we can subtract the second equation, eliminating the $ y$ and making finding the variables a LOT easier." } { "Tag": [ "logarithms" ], "Problem": "If $ a$ is a nonzero integer and $ b$ is a positive number such that $ ab^2\\equal{}\\log _{10}b$, what is the median of the set {$ 0, 1, a, b, 1/b$}?\r\n\r\n(a) 0\r\n(b) 1\r\n(c) a\r\n(d) b\r\n(e) 1/b", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?p=767730#767730" } { "Tag": [ "inequalities" ], "Problem": "Prove that if $ x\\geq0$ $ y\\geq0$,then $ x^{4}\\plus{}y^{4}\\plus{}1 \\geq x^{3}y\\plus{}y^{3}\\plus{}1$ \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n_______________________________________\r\nAzerbaijan Land of Fire \r\n$ \\Rightarrow\\bigstar\\Leftarrow$\r\n :rotfl:", "Solution_1": "$ x^3(x\\minus{}y)\\ge y^3(1\\minus{}y)$ \r\nnow if $ y\\ge x$ and $ y\\le 1$ then lhs is negative while rhs is positive :?:", "Solution_2": "The proposed inequality is not true.\r\n\r\nConsider x = y = 0.1", "Solution_3": "It could be a typo, perhaps $ x^4 \\plus{} y^4 \\plus{} 1 \\geq x^3 y \\plus{} y^3 x \\plus{} 1$? It seems the problem could be symmetric.\r\nIf so:\r\n[hide]\nRearrange to get $ x^4 \\minus{} x^3 y \\minus{} x y^3 \\plus{} y^4 \\geq 0$\nor $ (x\\minus{}y)^2(x^2 \\plus{} xy \\plus{} y^2)$\n$ (x\\minus{}y)^2$ is obviously nonnegative since it's a square;\n$ (x^2 \\plus{} xy \\plus{} y^2)$ is obviously nonnegative since x and y are nonnegative and it's all addition.\n[/hide]\r\nEither way, what is that extra 1 doing on both sides?", "Solution_4": "[quote]Either way, what is that extra 1 doing on both sides?[/quote]\r\nit doesnt matter whether its there or not" } { "Tag": [ "geometry", "calculus" ], "Problem": "Sorry if i posted to a wrong place. Can anyone tell me how acient mathematicians found out area of a circle when they do not know calculus? As I know, ancient mathematicians did find out C=2.pi.R by dividing its length with its radius, but I really do not know how they came up with the circular area. Thanks", "Solution_1": "sorry, I do not know a lot about ancients. \r\nAs I remember from my school geometry, circular area can be found with use of regular n-sided polygon area, when n->inf. This has no calculus.\r\nSome info:\r\nhttp://mathworld.wolfram.com/Circle.html", "Solution_2": "[quote=\"surfer\"]sorry, I do not know a lot about ancients. \nAs I remember from my school geometry, circular area can be found with use of regular n-sided polygon area, when n->inf. This has no calculus.\nSome info:\nhttp://mathworld.wolfram.com/Circle.html[/quote]\r\nI just thought I had found out something new but it was not. :blush: Anyway, thank for your link and by the way can you tell me how mathematicians proved e^(ix) = cosx + i.sinx." } { "Tag": [ "geometry", "perimeter" ], "Problem": "Pick's Formula states that A= B/2 + I -1\r\n\r\nWhat does it mean and how can we use it in problems?", "Solution_1": "I has to do with the area of regions and lattice points. If you have some shape and you know how many lattice points it covers, you can compute the area of the region.", "Solution_2": "In a shape with it's vertices on lattice points, Pick's formula states that the area of the shape is $ I\\plus{}\\frac{P}{2}\\minus{}1$\r\nWhere P is the number of points on the shape's perimeter and I is the number of points inside the shape." } { "Tag": [ "MATHCOUNTS", "AMC", "USA(J)MO", "USAMO", "AMC 8", "geometry", "analytic geometry" ], "Problem": "Heed the subtitle.\r\n\r\nI got 1st as expected, because my school is kinda, well, lame.", "Solution_1": "Maybach, I'm from westside! And I beat summai this time (last year I came 7th)!!!\r\n\r\nGood job summai though :). (Maybach, remember who summai is?)\r\n\r\nAnd Maybach, how would I vote?\r\n\r\nPOST EDITED: Sorry everyone :oops:", "Solution_2": "NO DISCUSSING OF SCORES. IT SAYS SO ON THE ANNOUNCEMENT. PLZ EDIT YOUR POST.", "Solution_3": "Remember that some people haven't taken it yet. I don't even know when it will be for me (probably about a week before Chapter).", "Solution_4": "Yay I chose the last option.\r\n\r\nI never understood the point of discussing results, especially if you aren't discussing the problems; considering if anyone on AoPS knows who you are, they either know you in real life or you are a big shot and likely came in 1st/2nd anyway.\r\n\r\nGood luck to all of you though.", "Solution_5": "I'm in high school, so MC is over for me.. :) :( :| \r\n\r\nAnyway, when will the test be posted on the MC website? Or is there some other place that puts it on the internet before that?", "Solution_6": "Mine's not happening until the end of January... I'll probably get in the top four at my school... :lol:", "Solution_7": "@SonyWii: After all the Chapter rounds are over...", "Solution_8": "not taking it until Jan 9\r\nand apparently, our \"School Round\" is for our entire district...", "Solution_9": "Deary me I will suck. Probably. Ah well. :lol:", "Solution_10": "9th grade now :( \r\n\r\nso...i voted for the last option", "Solution_11": "I placed third, but there were only like 10 people at my school who took it.", "Solution_12": "Did you do a countdown at your school for fun? That's what my school did last year.", "Solution_13": "[quote=\"eqjj168\"]not taking it until Jan 9\nand apparently, our \"School Round\" is for our entire district...[/quote]\r\n\r\nWhat about your chapter then?", "Solution_14": "His school probably has no chapter.", "Solution_15": "I got 2nd, the results came in January 27.\r\nThat sucks. :(", "Solution_16": "The school round was eazy, with all capitals. I think I did well, with all capitals. I hope I am at least 1, with all capitals. It felt eazy, with all capitals. My chapter is on the 28th and I hope I can get 1st, with all capitals. Ahhhhh!!!! Eazy, with all capitals!!!!!!!", "Solution_17": "I epic failed the School! OH NOZ!\r\nI think I got 6th/didn't make the top 4 :(", "Solution_18": "AHHHH chapter is THIS saturday!!!! Crazy!!! i is nervous!!!", "Solution_19": "[quote=\"The Hobbit\"]AHHHH chapter is THIS saturday!!!! Crazy!!! i is nervous!!![/quote]\r\n\r\n...yeah, I agree... !!!\r\nOn the bright side I got 1st in the school!!! :yup:\r\nWhich I took on Friday!!! :mad:", "Solution_20": "Now it is in Feb. Chapter is starting.\r\nCan Mathcounts school problems be posted now?", "Solution_21": "[quote=\"bull98\"]Now it is in Feb. Chapter is starting.\nCan Mathcounts school problems be posted now?[/quote]\n[quote=\"AIME15\"]Do not discuss the School or Chapter rounds until they have been posted on the MATHCOUNTS website...this is usually the first week of March. [/quote]", "Solution_22": "I was 1st at my school.\r\n\r\nAre we allowed to mention our score?", "Solution_23": "[quote=\"AdithyaGanesh\"]I was 1st at my school.\n\nAre we allowed to mention our score?[/quote]\r\nDon't think so... hmmmmmmmmmm...", "Solution_24": "Someone posted a problem in the Middle School Classroom Math that was in the school. Told him to delete it but he isnt doing it. \r\n[color=red][size=67]Link edited out by mod. In this circumstance, it is best to contact a moderator of the respective forum (I have done so already).[/size][/color]", "Solution_25": "[quote=\"AIME15\"]Do not discuss the School or Chapter rounds until they have been posted on the MATHCOUNTS website...this is usually the first week of March. [/quote]\r\n\r\nWhy??? If Mathcounts people are too busy, do not have time to update their site, then we cannot see the problems?\r\nThis is AOPS forum, a place to discuss math problems. It is not a kid of Mathcounts.\r\n\r\nI can understand we should not discuss school problems when some schools have not done it yet.\r\nNow all schools are finished the school round. Chapter round has started.\r\nWhy we cannot discuss school round problems?", "Solution_26": "[quote=\"The Hobbit\"]AHHHH chapter is THIS saturday!!!! Crazy!!! i is nervous!!![/quote]\r\nMe too! I'm probably going to be hyperventilating by Friday! :o", "Solution_27": "[quote=\"bull98\"]Now all schools are finished the school round. Chapter round has started.\nWhy we cannot discuss school round problems?[/quote]\r\nThis is not true. Chapter rounds have to be done any time in February. School rounds can be done any time before chapter. For example, a chapter round could be on Feb. 28, and a school could have its school round on Feb. 27. Not everyone has necessarily taken the school round.", "Solution_28": "Why are there so many time diferences? It is confusing. :(", "Solution_29": "It might be because some areas are more prone to snow? Our Chapter competition is [i]supposed[/i] to be on Saturday, the snow date Sunday, but they're calling for 4-8 inches of snow I think so it could be postponed until after Sunday :(" } { "Tag": [ "function", "induction", "algebra", "functional equation", "algebra unsolved" ], "Problem": "Find all functions $f\\colon \\mathbb{R}\\to \\mathbb{R}$ such that\r\n\\[f(xf(y)+f(x)) = 2f(x)+xy\\]\r\nfor every reals $x,y$.", "Solution_1": "this is really so hard, have you got any sollution?", "Solution_2": "my solution at the contest is big, but the first ideas are:\r\n\r\nprove that f is one bijection and prove that f(-1) using this.\r\nnow, use y = -f(x)/x, and a lot of substitutions.", "Solution_3": "[quote=\"cyshine\"]Find all functions $f\\colon \\mathbb{R}\\to \\mathbb{R}$ such that\n\\[f(xf(y)+f(x)) = 2f(x)+xy \\]\nfor every reals $x,y$.[/quote]\r\nthe one answer I got is\r\n\\[f(x)=1+x \\]\r\nKeep Smiling\r\nMalay", "Solution_4": "please post your solution, okie? Don't say what it is, say how can you solve it!", "Solution_5": "Ok,here's my solution:\r\nWe recall the problem:\r\nFind all the functional equation $f: R\\to R$ satisfying:$f(xf(y)+f(x))=2f(x)+xy$ $\\forall x,y \\in R$ (*)\r\nFrom the definition of function $f$,it clears that $f$ is bijection\r\nPut $f(0)=a$,because $f$ is bijection so it exists $b \\in R$ such that $f(b)=0$\r\nIn (*),Substituting x by b then $f(bf(y))=by$ (1) $\\forall y \\in R$\r\n $\\Longrightarrow f(ba)=0$ (do $y=0$)\r\n$\\Longrightarrow ba=b$ (because f is bijection)\r\n$\\Longrightarrow b=0$ or $a=1$\r\n.If $b=0$ then $f(0)=0$.Put $y=0$ then $f(f(x))=2f(x) \\forall x \\in R$\r\nBecause $f$ is bijection $\\Longrightarrow f(x)=2x \\forall x \\in R$ \r\nIt clears that $f(x)=2x$ doesn't satisfy (*) (absurd)\r\nSo $a=1 \\Longrightarrow f(0)=1$\r\nFrom (1) we have $f(bf(y))=by \\forall y \\in R$\r\nPut $y$ by $f(by)$ then $f(b^{2}y)=b^{2}f(y) \\forall y \\in R$\r\nPut $y=0$ then $b^{2}=1 \\forall b=-1$ (because $f$ is bijection)\r\n $\\Longrightarrow f(-1)=0$\r\nBecause $f(-1)=0$,hence $f(-f(x))=-x \\forall x \\in R$\r\nPut $y$ by $-1$ then $f(f(x))=2f(x)-x \\forall x \\in R$ (2)\r\nIn (2),substituting $x$ by $-f(x)$:\r\n $f(f(-f(x)))=2f(-f(x))+f(x)$ $\\forall x \\in R$\r\n $\\Longleftrightarrow f(-x)=f(x)-2x$ $\\forall x \\in R$\r\nIN (*),Put $y$ by $-f(y)$ then:\r\n $f(f(x)-xy)=2f(x)-xf(y) \\forall x,y \\in R$ (3)\r\nSubstituting $y$ by $\\frac{f(x)}{x}$ into (3) then:\r\n $2f(x)-xf(\\frac{f(x)}{x})=1$ (because $f(0)=1$)\r\n$\\Longrightarrow xf(\\frac{f(x)}{x})=2f(x)-1$ (**)\r\nIn (*),Substituting y by $-\\frac{f(x)}{x}$ then $f(xf(-\\frac{f(x)}{x})+f(x))=f(x) \\Longrightarrow xf(-\\frac{f(x)}{x})+f(x)=x \\forall x \\in R$ (because $f$ is bijection)\r\nBecause $f(-x)=f(x)-2x \\forall x \\in R$\r\n${\\Longrightarrow f(-\\frac{f(x)}{x})=f(\\frac{f(x)}{x})-2\\frac{f(x)}{x}\\forall x \\in R\\{0}$\r\nHence: $xf(-\\frac{f(x)}{x})+f(x)=x$\r\n $\\Longleftrightarrow xf(\\frac{f(x)}{x})=x+f(x)$ (***)\r\nFrom (**) and (***) $\\Longrightarrow f(x)=x+1$\r\nTherefore $f(x)=x+1$ is the only function satisfing (*)", "Solution_6": "Well, it is not so obvious that f is a bijection.\r\nI have some success proving that f is injective:\r\n\r\n[hide]\n$f(a)=f(b)$\n$af(a)+f(a)=af(b)+f(a)$\n$f(af(a)+f(a))=f(af(b)+f(a))$\n$a*a+2f(a)=ab+f(a)$\n$a*a=ab$\n$a=0$ or $a=b$\n\nIf $a=0$, then \n$f(0)=f(b)$ for any b, and so f is constant, and so \nf(0)=xy+2f(0), xy+f(0), false!\n[/hide]\r\n\r\nThen I have the idea: f=g+1:\r\ng(xg(y)+g(x)+x+1)=xy+2g(x)+1\r\nIt is almost good to try g(x)=x.\r\n\r\nBut I have not success at now :(", "Solution_7": "nhat's solution's is true I think ,nice", "Solution_8": "That's clearly correct, Silouan: he showed that $f(x)=2x$ on the image of $f$, which covers all possible values of the variable by surjectivity.", "Solution_9": "You are right harazi ,thank you pointing out it .", "Solution_10": "Hmm... my solution goes as follows. First, $f$ is bijective because, plugging $x = 1$, we get $f(f(y)+f(1)) = 2f(1)+x$; $x+2f(1)$ spans all real numbers (so $f$ is surjective) and $f(x) = f(y) \\iff f(x)+f(1) = f(y)+f(1) \\Leftrightarrow f(f(x)+f(1)) = f(f(y)+f(1))\\iff 2f(1)+x = 2f(1)+y \\iff x = y$ (so $f$ is injective).\r\n\r\nChoose $x \\neq 0$ and $y$ such that $2f(x)+xy = f(x) \\iff y =-{f(x)\\over x}$. By injectivity,\r\n\\[xf\\left(-{f(x)\\over x}\\right)+f(x) = x\\iff f\\left(-{f(x)\\over x}\\right)+{f(x)\\over x}= 1\\quad (*)\\]\r\n\r\nNow, set $x = y = 0$: $f(f(0)) = 2f(0)$. Let $a = f(0)$, so that $f(a) = 2a$. If $a = 0$, setting $y = 0$ we get $f(f(x)) = 2f(x)$ and by surjectivity $f(x) = 2x$ for all $x$. This function doesn't work (just test it!), so $a$ can't be $0$. So $a \\neq 0$ and plugging $x = a$ in $(*)$, we obtain (recall that $f(a) = 2a$):\r\n\\[f(-2)+2 = 1\\iff f(-2) =-1\\]\r\n\r\nNow set $y = 0$: $f(f(x)-x) = 2(f(x)-x)$. Now we find all real numbers $k$ such that $f(k) = 2k$. If ther is only one possible value for $k$, we are almost done because $f(x)-x = k \\iff f9x) = x+k$ and by testing the function, $f(x) = x+1$.\r\n\r\nLet $m$ be such that $f(m) = k$ (it exists because $f$ is surjective). Plug $x = m$ and $y = 0$: recalling that $f(0) = a$,\r\n\\[f(mf(0)+f(m))=2f(m)\\iff f(ma+k) = 2k = f(k)\\]\r\n\r\nBy injectivity,\r\n\\[ma+k = k\\iff ma = 0 \\iff m = 0\\]\r\nbecause we already proved that $a \\neq 0$. So there is only one value for $k$, namely, $k = f(0) = a$.\r\nThen, we're done: the only function is $f(x) = x+1$.", "Solution_11": "$ f(xf(y)\\plus{}f(x))\\equal{}2f(x)\\plus{}xy$ $ (1)$.\r\nPlugging $ y\\equal{}\\frac{\\minus{}2f(x)}{x}$ in $ (1)$ we get:$ f(xf(y)\\plus{}f(x))\\equal{}0$. So, there exists such real number $ u$ that $ f(u)\\equal{}0$.\r\nTake $ x\\equal{}u$ in $ (2)$: \r\n$ f(uf(y))\\equal{}uy$ $ (3)$.\r\nPut $ y\\equal{}x$ in $ (3)$:\r\n$ f(uf(x))\\equal{}ux$ $ (4)$.\r\nIf $ f(x)\\equal{}f(y)$, $ (3)$ and $ (4)$ gives $ u(x\\minus{}y)\\equal{}0$.\r\nIf $ u\\equal{}0$, $ f(0)\\equal{}0$. \r\nTake $ y\\equal{}0$ in $ (1)$:\r\n$ f(f(x))\\equal{}2f(x)$. $ (4)$.\r\nLet's prove that for any real number $ z$ there exists real number $ t$,for which $ f(t)\\equal{}z$:Take $ y\\equal{}\\frac{z\\minus{}2f(x)}{x}$ in $ (1)$: $ f(t)\\equal{}z$, where $ (z,t)$ are reals, Q.E.D.\r\nTake $ x$ instad of $ f(x)$ in $ (4)$: $ f(x)\\equal{}2x$, for any real $ x$, which isn't solution.(by checking it in $ (1)$ we make sure in it).So, $ u$ isn't equal to $ 0$.\r\nSo, $ f(x)\\equal{}f(y)$ follows $ x\\equal{}y$.\r\nTake $ x\\equal{}y\\equal{}u$ in $ (1)$: $ f(0)\\equal{}u^{2}$.\r\nPlugging $ x\\equal{}0$ in $ (3)$ we get: $ f(uf(0))\\equal{}0\\equal{}f(u)$.\r\nSo, $ u(f(0)\\minus{}1)\\equal{}0$. As we have already proved, $ u$ can't be $ 0$. So, $ f(0)\\equal{}1$ $ (5)$.\r\n$ u^{2}\\equal{}f(0)\\equal{}1$ gives $ u\\equal{}1$ or $ u\\equal{}\\minus{}1$.\r\nLet $ u\\equal{}1$. Than, plugging $ u\\equal{}y\\equal{}1$ in $ (1)$ we get: $ f(1)\\equal{}2$, which can't be true as $ f(1)\\equal{}0$.\r\nSo, $ u\\equal{}\\minus{}1$ and $ f(\\minus{}1)\\equal{}0$ $ (6)$.\r\nPlugging $ x\\equal{}0$ in $ (1)$ we get:\r\n$ f(1)\\equal{}2$ $ (7)$.\r\nLet's prove by induction that $ f(n)\\equal{}n\\plus{}1$ fro any positive integer $ n$.\r\nFor $ n\\equal{}1$, $ f(1)\\equal{}1\\plus{}1\\equal{}2$.\r\nLet $ f(k)\\equal{}k\\plus{}1$, for any positive integer $ k$. We have to prove that $ f(k\\plus{}1)\\equal{}k\\plus{}2$.\r\nTake $ x\\equal{}k$ and $ y\\equal{}\\minus{}1$ in $ (1)$ we get: $ f(k\\plus{}1)\\equal{}k\\plus{}2$, Q.E.D.\r\n$ (3)$ gives: $ f(\\minus{}f(x))\\equal{}\\minus{}x$ $ (8)$.\r\nTake $ y\\equal{}\\minus{}1$ in $ (1)$: $ f(f(x))\\equal{}2f(x)\\minus{}x$ $ (9)$.\r\nSubstitute $ f(x)$ by $ x$: $ f(x)\\minus{}f(\\minus{}x)\\equal{}2x$ $ (10)$.\r\nLet $ n$ be positive integer. $ (10)$ gives: $ f(\\minus{}n)\\equal{}\\minus{}n\\plus{}1$.\r\nSo, for any integer $ n$, $ f(n)\\equal{}n\\plus{}1$.\r\nLet's prove that $ f(\\frac{x}{y})\\equal{}\\frac{x}{y}\\plus{}1$.\r\nTake $ x\\equal{}q$ and $ y\\equal{}\\frac{p\\plus{}2q\\minus{}f(q)}{q}$ in $ (1)$. We get: $ f(\\frac{p\\minus{}2}{q})\\equal{}\\frac{p\\minus{}2}{q}\\plus{}1$.\r\nSubstituting $ p\\minus{}2$ and $ q$ by $ x$ and $ y$ respectively, we get: $ f(\\frac{x}{y})\\equal{}\\frac{x}{y}\\plus{}1$,\r\nQ.E.D.\r\nSo, for any rational $ n$, $ f(n)\\equal{}n\\plus{}1$.\r\nFunction $ f(x)$ is strictly increasing because it gets each real value exactly once and for any integers $ (m,n)$, where $ m>n$, $ f(m)>f(n)$.\r\nLet, $ f(x)>x\\plus{}1$, for any real $ x$:\r\nTake $ x$ irational and $ l$ rational number such that $ xx$, for any real $ x$. Take $ x$ rational number and $ r$ real such that \r\n$ xx$, for any real $ x$, Q.E.D.\r\nLet's take $ r$ rational number so that $ f(x)\\equal{}x\\plus{}1$, but $ f(x)<\\equal{}x\\plus{}1$. So, $ f(x)\\equal{}x\\plus{}1$, for any real number $ x$.\r\nBy checking it in $ (1)$, we make sure that it's possible answer.", "Solution_12": "@cyshine:\r\n\r\nIt's funny, that your solution is EXACTLY the same as the official solution, word for word, and the best is:\r\n\r\nWITH THE SAME ERROR, the error is only, that you should replace 0 with -2. The official solution did the same. They wrote 0 instead of -2, and afterwards, they finished the solution correctly.\r\n\r\n\r\nBest regards, felix", "Solution_13": "It is quite clear that the function is bijective. We will first show that $ f(\\minus{}1) \\equal{} 0$.\r\n$ f(xf(y) \\plus{} f(x)) \\equal{} 2f(x) \\plus{} xy \\minus{}\\minus{}\\minus{}\\minus{}(*)$\r\nSince $ f$ is bijective, there exists a $ c$ such that $ f(c) \\equal{} 0$.\r\nPlugging $ x \\equal{} y \\equal{}c$ in $ (*)$, we get $ f(0) \\equal{} c^2$.\r\nPlugging $ x \\equal{} 0$ in $ (*)$, we get $ f(c^2) \\equal{} 2c^2$.\r\nPlugging $ x \\equal{} c, y \\equal{}0$ in $ (*)$, we get $ f(c^3) \\equal{} 0$. \r\nSince $ f$ is bijective, $ c^3 \\equal{} c \\Rightarrow c \\in \\{\\minus{}1, 0, 1\\}$.\r\nCase 1: $ c \\equal{} 1 \\Rightarrow f(1) \\equal{} 0$ and $ f(1) \\equal{} 2$ which is a contradiction.\r\nCase 2: $ c \\equal{} 0 \\Rightarrow f(f(x)) \\equal{} 2f(x)$. Since $ f$ is a bijection we could replace $ f(x)$ by $ x$ and we have $ f(x) \\equal{} 2x$. \r\nSubstituting $ f(x) \\equal{} 2x$ in $ (*)$ gives us $ 4xy \\equal{} xy$ which is a contradiction.\r\nCase 3: $ c \\equal{} \\minus{}1 \\Rightarrow f(\\minus{}1) \\equal{} 0, f(0) \\equal{} 1, f(1) \\equal{} 2$.\r\nPlugging $ x \\equal{} 1$ in $ (*)$, we get $ f(f(y) \\plus{} 2) \\equal{} 4 \\plus{} y \\minus{}\\minus{}\\minus{}\\minus{}\\minus{}(**)$.\r\nPlugging $ y \\equal{} \\minus{}1$ in $ (*)$, we get $ f(f(x)) \\equal{} 2f(x) \\minus{} x$ (i.e.) $ f(f(y)) \\equal{} 2f(y) \\minus{} y \\minus{}\\minus{}\\minus{}\\minus{}\\minus{}(***)$.\r\nAdding (**) and (***), we get $ f(f(y) \\plus{}2) \\plus{} f(f(y)) \\equal{} 2f(y) \\plus{} 4$. Again, since $ f$ is a bijection we could replace $ f(y)$ by $ y$ and we have \r\n$ f(y \\plus{} 2) \\plus{}f(y) \\equal{} 2y \\plus{} 4 \\minus{}\\minus{}\\minus{}\\minus{}(****)$.\r\nReplacing $ y$ by $ y\\plus{}2$, we get \r\n$ f(y \\plus{} 4) \\plus{}f(y\\plus{}2) \\equal{} 2y \\plus{} 8\\minus{}\\minus{}\\minus{}\\minus{}(*****)$.\r\nApplying $ f$ to $ (**)$ and using $ (***)$, we get \r\n$ 2f(f(y) \\plus{} 2) \\minus{} f(y) \\minus{}2 \\equal{} f(y \\plus{} 4) \\Rightarrow 2(y\\plus{}4) \\minus{} f(y) \\minus{} 2 \\equal{} f(y\\plus{}4) \\Rightarrow f(y\\plus{}4) \\plus{} f(y) \\equal{} 2y \\plus{} 6 \\minus{}\\minus{}\\minus{}\\minus{}\\minus{}\\minus{}(******)$.\r\nAdding $ (****),(*****)$ and $ (******)$ and dividing by 2, we get\r\n$ f(y \\plus{} 4) \\plus{}f(y\\plus{}2) \\plus{}f(y) \\equal{} 3y \\plus{} 9$. \r\nSubtracting $ (*****)$ from the above equation gives $ f(y) \\equal{} y \\plus{}1$.", "Solution_14": "[quote=\"cyshine\"]Hmm... my solution goes as follows. First, $ f$ is bijective because, plugging $ x \\equal{} 1$, we get $ f(f(y) \\plus{} f(1)) \\equal{} 2f(1) \\plus{} x$; $ x \\plus{} 2f(1)$ spans all real numbers (so $ f$ is surjective) \n\n $ f(x) \\equal{} f(y) \\iff f(x) \\plus{} f(1) \\equal{} f(y) \\plus{} f(1) \\rightarrow f(f(x) \\plus{} f(1)) \\equal{} f(f(y) \\plus{} f(1))$\n\n$ \\iff 2f(1) \\plus{} x \\equal{} 2f(1) \\plus{} y \\iff x \\equal{} y$ (so $ f$ is injective).\n\nChoose $ x \\neq 0$ and $ y$ such that $ 2f(x) \\plus{} xy \\equal{} f(x) \\iff y \\equal{} \\minus{} {f(x)\\over x}$. By injectivity,\n\\[ xf\\left( \\minus{} {f(x)\\over x}\\right) \\plus{} f(x) \\equal{} x\\iff f\\left( \\minus{} {f(x)\\over x}\\right) \\plus{} {f(x)\\over x} \\equal{} 1\\quad (*)\\]\nNow, set $ x \\equal{} y \\equal{} 0$: $ f(f(0)) \\equal{} 2f(0)$. Let $ a \\equal{} f(0)$, so that $ f(a) \\equal{} 2a$. If $ a \\equal{} 0$, setting $ y \\equal{} 0$ we get $ f(f(x)) \\equal{} 2f(x)$ and by surjectivity $ f(x) \\equal{} 2x$ for all $ x$. This function doesn't work (just test it!), so $ a$ can't be $ 0$. So $ a \\neq 0$ and plugging $ x \\equal{} a$ in $ (*)$, we obtain (recall that $ f(a) \\equal{} 2a$):\n\\[ f( \\minus{} 2) \\plus{} 2 \\equal{} 1\\iff f( \\minus{} 2) \\equal{} \\minus{} 1\\]\nNow set $ y \\equal{} \\minus{} 2$: $ f(f(x) \\minus{} x) \\equal{} 2(f(x) \\minus{} x)$. Now we find all real numbers $ k$ such that $ f(k) \\equal{} 2k$. If ther is only one possible value for $ k$, we are almost done because $ f(x) \\minus{} x \\equal{} k \\iff f(x) \\equal{} x \\plus{} k$ and by testing the function, $ f(x) \\equal{} x \\plus{} 1$.\n\nLet $ m$ be such that $ f(m) \\equal{} k$ (m exists because $ f$ is surjective). Plug $ x \\equal{} m$ and $ y \\equal{} 0$: recalling that $ f(0) \\equal{} a$,\n\\[ f(mf(0) \\plus{} f(m)) \\equal{} 2f(m)\\iff f(ma \\plus{} k) \\equal{} 2k \\equal{} f(k)\\]\nBy injectivity,\n\\[ ma \\plus{} k \\equal{} k\\iff ma \\equal{} 0 \\iff m \\equal{} 0\\]\nbecause we already proved that $ a \\neq 0$. So there is only one value for $ k$, namely, $ k \\equal{} f(0) \\equal{} a$.\nThen, we're done: the only function is $ f(x) \\equal{} x \\plus{} 1$.[/quote]\r\n\r\n[i] A very good solution , Cyshine , congratulation :lol: \n\n I have edited some mistakes in order to be easy to understand[/i] :lol:", "Solution_15": "It's easy to see that $f$ is bijective(we just need to fix $x$). Let $f(0)=c,$ if $c=0,$ taking $y=0$ yields $f(f(x))=2f(x),$ i.e. $f(x)=2x$ follows from bijective, which is clearly absurd, so $f(0)=c\\ne 0,$ and $f(c)=2c.$ Assume there's a $y_0\\in \\mathbb{R}$ for which $f(y_0)=-1,$ observe that\n\\[2c=f(-c+2c)=4c+y_0c \\implies (2+y_0)c=0\\implies y_0=-2\\]\nby taking $x=c, y=y_0$ and $c\\ne 0.$ Now take $x=-2,y=-2$ to get $f(1)=-2+4=2,$ setting $x=1$ we get\n\\[f(f(y)+2)=y+4\\qquad (\\spadesuit)\\]\nfor all $y\\in \\mathbb{R}.$ Assume there's an $a\\in\\mathbb{R}$ for which $f(a)=1,$ we claim that $a=0.$ Taking $(x,y)=(a,0)$ yields that $f(ac+1)=2\\implies a=0$ since $c\\ne 0.$ Now setting $y=0$ in our initial assertion yields \n\\[f(x+f(x))=2f(x).\\qquad (\\bigstar)\\]\nTaking $x=f(y)+2$ in $(\\bigstar)$ gives\n\\[\\begin{aligned}\nf(f(y)+2+f(f(y)+2))=f((f(y)+y+4)+2)&=(2y+4)+4\\qquad &(\\text{follows by} (\\spadesuit)) \\\\\n\\implies f((2y+2)+2)&=(f(y)+y)+4\\qquad &(\\text{follows by} (\\spadesuit))\\\\\n\\implies f(f(y)+y)=2y+2&=2f(y) \\qquad &(\\text{also repeat $(\\spadesuit)$ and combine $(\\bigstar)$})\n\\end{aligned}\\]\nfrom which we conclude that $\\boxed{f(x)=x+1,}$ which is the solution indeed. :) ", "Solution_16": "$x=1, y=r-2f(1)$ gives $f(stuff)=r$, thus surjectivity. $f(a)=f(b)$ gives (by $x=1, y=a$ and $x=1, y=b$) $a=b$, thus injectivity and bijectivity.\n\nNow let $z, u$ be the reals such that $f(z)=0, f(u)=1$. $x=z, y=u$ gives $0=zu$. \nCase1: $z=0 \\Leftrightarrow f(0)=0$:\n$x=f^{-1}(k), y=0$ gives $f(k)=2k$, which clearly doesn't work.\n\nCase2: $u=0 \\Leftrightarrow f(0)=1$:\n$x=z$ gives $f(zf(y))=zy$ (*).\n\n$y=z$ gives $f(f(x))=2f(x)+xz\\overbrace{=}^{(*)} 2f(x)+f(zf(x))$ and $k=f(x)$ makes it $f(k)=2k+f(zk)$ (#).\n$x=0$ gives $f(1)=2$\n$x=y=z$ gives $1=z^2$, but $z\\neq 1$, so $z=-1$, and (#) becomes $f(k)=2k+f(-k)$.\n$x=-1, y=1$ gives $f(-2)=-1$.\n$y=-2$ gives $f(f(x)-x)=2f(x)-2x\\Leftrightarrow$(#) $2f(x)-2x+f(-f(x)+x)=2f(x)-2x \\Leftrightarrow f(-f(x)+x)=0 \\Leftrightarrow -f(x)+x=-1 \\Leftrightarrow f(x)=x+1$. This works", "Solution_17": "[quote=cyshine]Find all functions $f\\colon \\mathbb{R}\\to \\mathbb{R}$ such that\n\\[f(xf(y)+f(x)) = 2f(x)+xy\\]\nfor every reals $x,y$.[/quote]\n\nSuppose $f(a)=(b)$. Fix $x=1$ and plug $y=a$ and $y=b$ to prove injectivity. Now fix $x=1$ and vary $y \\in \\mathbb{R}$; thus $f$ is surjective. Pick $y_0 \\in \\mathbb{R}$ with $f(y_0)=0$ and plug $y=y_0$; then $f(f(x))=2f(x)+xy_0$ for all $x \\in \\mathbb{R}$. If $y_0=0$ then $f(t)=2t$ for all $t \\in \\mathbb{R}$ which is not a solution. Pick $t_0$ with $f(t_0)=1$ and plug $y=t_0, x=y_0$; then $f(y_0f(t_0))=2f(y_0)+y_0t_0$ hence $t_0=0$ and $f(0)=1$. Now $f(f(0))=2f(0)$ hence $f(1)=2$. Also $f(f(y_0))=2f(y_0)+y_0^2$ so $y_0 \\in \\{\\pm 1\\}$ but $f(1) \\ne 0$ hence $y_0=-1$ and $f(-1)=0$. Consequently, $f(f(x))=2f(x)-x$ for all $x \\in \\mathbb{R}$.\n\nPlug $x=1$ to get $f(f(y)+2)=y+4$ for all $y \\in \\mathbb{R}$. Now $y=f(f(y-4)+2)$ for all $y$; so $f(y)=f(f(f(y)-4)+2)$ for all $y$ and injectivity applies; so $=f(f(y)-4)=y-2=f(f(y-6)+2)$ for all $y$. Thus, $f(y)=f(y-6)+6$ for all $y \\in \\mathbb{R}$.\n\n[b]Claim.[/b] $f(x)-f(0)$ is additive.\n\n(Proof) Fix $a,b \\in \\mathbb{R}$ with $b \\ne 0$. We claim that $f(a+2b)=2f(a+b)-f(a)$. Indeed set $x=\\tfrac{b}{6}$ and $y=f^{-1}\\left(\\frac{a-f\\left(\\frac{b}{6}\\right)}{\\left(\\frac{b}{6}\\right)}\\right)$ then $$f(a)=f(xf(y)+f(x))=2f(x)+xy$$ $$f(a+b)=f(xf(y)+f(x)+6x)=f(xf(y+6)+f(x))=2f(x)+x(y+6)$$ $$f(a+2b)=f(xf(y)+f(x)+12x)=f(xf(y+12)+f(x))=2f(x)+x(y+12)$$ hence proving the claim. \n\nNow we plug $a=0$ to get $f(2b)=2f(b)-f(0)$ for all $b \\in \\mathbb{R}$ (write-up issue: deal with $b=0$ separately). Hence $f(2a+2b)=2f(a+b)-f(0)$ as long as $a \\ne -b$. Therefore, for all $x,y \\in \\mathbb{R}$ we have $f(x+y)=f(x)+f(y)-f(0)$. $\\blacksquare$\n\nThus, $f(xf(y))+2f(x)-x-1=2f(x)+xy \\implies f(xf(y))=xy+x+1$ for all $x,y \\in \\mathbb{R}$. Plug $y=1$ and $t=2x$ then $f(t)=t+1$ for all $t \\in \\mathbb{R}$. This also satisfies the equation hence it is the only solution! $\\blacksquare$", "Solution_18": "A quick glance says this is hopefully different from the above?\n\n[hide = Solution]\nWe claim that $f(x) \\equiv x+1$ is the only solution. It is easily verified that this indeed works, we will now show it is unique. \n\nLet $P(x, y)$ denote the assertion that $f(xf(y)+f(x)) = 2f(x) + xy.$\n\nFirstly, if $f(a) = f(b)$, then $P(\\pi, a) - P(\\pi,b)$ gives that $\\pi a = \\pi b$, so that\n$$f \\; \\text{is} \\; \\text{injective}.$$\nBy $P(1, y)$, we get that $2f(1) + y \\in \\text{Range}(f) \\; \\forall y \\in \\mathbb{R}$, which gives that\n$$f \\; \\text{is} \\; \\text{surjective},$$ which combined with the above gives that\n$$f \\; \\text{is} \\; \\text{bijective}.$$\nHence, it makes sense to let $f(t) = 0$ for some $t \\in \\mathbb{R}.$ Then, $P(x, t)$ gives that\n$$f(f(x)) = 2f(x) + tx. \\qquad (1)$$\n$P(t, y)$ gives that\n$$f(tf(y)) = ty. \\qquad (2)$$\nLetting $y = 0$ in $(2)$ gives us that $f(tf(0)) = 0$, which together with injectivity gives us that $t = tf(0)$. Therefore, we either have that $t = 0$ or $f(0) = 1.$ If $t = 0$, then $(1)$ gives us that $f(f(x)) = 2f(x),$ which together with surjectivity implies that $f(x) \\equiv 2x \\; \\forall x \\in \\mathbb{R},$ which can be easily checked to not work. \n\nTherefore, we know that $f(0) = 1.$ $P(0, 0)$ now implies that $f(1) = 2.$ Hence, $P(1, y)$ gives us that $f(f(y)+2) = 4+y.$ Therefore, since $f(f(-2)+2) = 4 + (-2) = 2 = f(1),$ we have from injectivity that $f(-2) + 2 = 1 \\Rightarrow f(-2) = -1.$ Therefore, $P(x, -2)$ gives us that \n$$f(f(x) - x) = 2(f(x)-x) \\; \\forall x \\in \\mathbb{R}. \\qquad (3)$$\nHowever, remember that from $(1)$ we know that $f(x) = 2x + t f^{-1}(x),$ where $f^{-1}$ is the inverse function of $f$ (exists since $f$ is bijective). Therefore, if we plug in $x = f(a) - a$ into this equation, we have that $2(f(a) - a) =^{(3)} f(f(a) - a) = 2(f(a) - a) + t f^{-1} (f(a) - a),$ which implies since $t \\neq 0$ that\n$$f^{-1}(f(a) -a) = 0 \\Leftrightarrow f(a) - a = f(0) \\; \\forall a \\in \\mathbb{R}. \\qquad (4).$$\nWith $(4)$ we are nearly done, we now have that $f(x) \\equiv x + c \\; \\forall x \\in \\mathbb{R}$ for some $c \\in \\mathbb{R}.$ Plugging this into $P(x, y)$ then easily yields our only solution $\\boxed{f(x) \\equiv x+1},$ as claimed. \n\n$\\square$ \n\n[/hide]", "Solution_19": "Clearly, $f$ is bijective. \n[hide=finding f(-1) and f(2)]\nLet $a=f(-1)$\n$$P(-1,x) \\implies f(a-f(x))=2a-x$$\n$$P(-1,a-f(x)) \\implies f(x-a)=a+f(x)$$\n$$P(x,y)-P(x,y-a) \\implies f(xf(y)+f(x)+ax)=f(xf(y)+f(x))-ax \\implies f(y+ax)=f(y)-ax \\implies a=0$$\n$$P(-1,-1) \\implies f(0)=1$$ \n$$P(0,0) \\implies f(1)=2$$\n$$P(1,-1) \\implies f(2)=3$$\n[/hide]\nPlug-in $y=-1$ to see that $f(f(x))=2f(x)-x$\n$$P(2,x) \\implies f(2f(x)+3)=6+2x$$\n$$f(xf(y)+f(x))=2f(x)+xy \\implies f(f(xf(y)+f(x)))=f(2f(x)+xy) \\implies f(2f(x)+xy)=3f(x)+2xy-xf(y) \\implies$$\n$$\\stackrel{ y\\mapsto \\frac{3}{x}}{ \\Longrightarrow}f(2f(x)+3)=3f(x)+6-xf\\left(\\frac{3}{x}\\right)= 6+2x \\ \\ \\ (*)$$\nNow let $x \\mapsto 3x$ in $(*)$ to see that $f(3x)=xf\\left(\\frac{1}{x}\\right)+2x$ and let $x\\mapsto \\frac{1}{x}$ to see that $3xf\\left(\\frac{1}{x}\\right)=2+f(3x)$ and now $f(x)=x+1$ trivially follows.\n", "Solution_20": "Denote the assertion as $P(x,y)$.\n\n$P(0,0)$ gives $f(f(0))=2f(0)$.\n\nIf $f(y_1)=f(y_2)$, $P(x,y_1)$ and $P(x,y_2)$ for $x\\neq 0$ gives $y_1=y_2$, so $f$ is injective. Also, varying $y$ for fixed $x$ allows RHS to achieve all real numbers, so $f$ is bijective.\n\nNow, suppose that $f(z)=0$ (note $z$ is well defined due to bijectivity.) Then, $P(z,y)$ gives $f(zf(y))=zy$. $y=z$ into this gives $f(0)=z^2$, and $y=0$ gives $f(zf(0))=0$, which by injectivity means $f(0)=1$. Thus, $z^2=1\\implies z=\\pm 1$. If $z=1$, then we get that $f(f(x))=x$, so $f(f(0))=0$. However, we know that $f(f(0))=2f(0)=2$, which is a contradiction. Thus, $z=-1$.\n\nNow, we have $f(-f(y))=-y$. As $f(1)=2$, we get $f(-2)=-1$. Also, $P(1,-3)$ gives $f(f(-3)+2)=1=f(0)$, so injectivity yields $f(-3)=-2$.\n\nNow, $P(x,-2)$ gives $f(f(x)-x)=2(f(x)-x)$. If we have a real number $k$ such that $f(k)=2k$, $P(k,y)$ gives $f(k(f(y)+2))=k(y+4)$. Substituting $y=-3$ gives $f(0)=k\\implies k=1$. Thus, we must have $f(x)-x=1\\forall x\\implies f(x)=x+1$. This satisfies the initial assertion, so our only solution is $f(x)=x+1$.", "Solution_21": "Hard but not so hard. ;)\n[hide=Solution]We claim that only $f \\equiv y+1$ is our solution. It's easy to check that it works. We now show this is the only one. \n\n Let $P(x,y)$ denote the assertion. Note that in $P(1,y)$ varying $y$ accordingly yields surjectivity. Suppose if possible $f(u)=f(v)$ for $u \\neq v$. Then $P(x,u)$ and $P(x,v)$ combined implies, for non-zero $x,$ that $u=v$ which is a contradiction. We conclude that $f$ is bijective. \n\nSince $f$ is bijective, we know $\\exists a$ such that $f(a)=0$. Note that $P(a,y)$ implies $f(af(y))=ay~(\\clubsuit)$ and $P(af(y),a)$ combined with previous result implies $$f(ay)=f(f(af(y)))=2f(af(y))+a^2f(y)=2ay+a^2f(y)$$In addition, $y=0$ yields $f(0)=0$ or $a=\\pm 1$. We deal with this in cases as follows:\n\n[list]\n[*]If $f(0)=0$ then $f(a)=0=f(0)$ implying $a=0$ by injectivity. Now $P(x,0)$ yields $f(f(x))=2f(x)$ but then since $f$ is bijective we conclude $f(x)=2x$ for all $x \\in \\mathbb{R}$ which is not a solution to our original assertion. \n[*]If $a=1$ then $f(1)=0$. Now $P(1,y)$ yields $f(f(y))=y$ but on other hand, by $P(x,1)$ we have $f(f(x))=2f(x)+x$ which then, combined with previous result, implies $f \\equiv 0$ which is not a solution. \n[*]If $a=-1$ then $f(-1)=0$. Now $P(x,-1)$ implies $f(f(x))=2f(x)-x~(\\spadesuit)$. Here, in $(\\spadesuit),$ $x=-1$ yields $f(0)=1$. Then $P(0,y)$ yields $f(1)=2$. From $P(1,y),$ we have $f(f(y)+2)=4+y~(\\bigstar)$. Taking $f$ both sides in previous equation yields, with $(\\clubsuit)$ combined that $$f(4+y)=2f(f(y)+2)-(f(y)+2)=2(4+y)-f(y)-2=6+2y-f(y)$$By $y \\to f(y)$ shift in previous equation implies, combined with $(\\clubsuit),$ that $f(4+f(y))=6+y$. By $y \\to f(y)+2$ in $(\\bigstar)$ implies, again combined with $(\\clubsuit),$ that $$f(y)+6=f(6+y)=f(f(4+f(y)))=2f(4+f(y))-4-f(y)=2(6+y)-4-f(y)$$Simplying this equation yields $f(y)=y+1$ for all $y$.\n[/list]\nSumming up all cases, we have $f \\equiv y+1$ as our only solution, as required. $\\blacksquare$\n", "Solution_22": "Decently difficult.\n\nStorage:\n\nLet $P(x,y)$ denote the original expression.\n\n[b]Claim 1[/b]: $f$ is bijective.\n[b]Proof[/b]: Surjectivity is obvious. If $f(a)=f(b)$, the $P(x,a)$ and $P(x,b)$ for some non-zero $x$ gives $a=b$. $\\blacksquare$\n\nLet $u$ be the real number satisfying $f(u)=0$. If $u=0$, then $P(x,0)$ $\\implies$ $f(f(x))=2f(x)$ $\\implies$ $f(x)=2x$ by bijectivity, which is clearly not a solution. Therefore $u \\neq 0$.\n\n[b]Claim 2[/b]: $f(0)=1$ and $f(1)=2$.\n[b]Proof[/b]: $P(u,0)$ $\\implies$ $f(uf(0))=0=f(u)$ $\\implies$ $f(0)=1$ since $u \\neq 0$. Now $P(0,0)$ gives $f(1)=2$. $\\blacksquare$\n\n[b]Claim 3[/b]: $f(x+6)=f(x)+6$ for all $x$.\n[b]Proof[/b]: $P(1,x)$ $\\implies$ $f(f(x)+2)=x+4$ $$\\implies \\ f(x+6)=f(f(f(x)+2)+2)=f(x)+6$$\nas required. $\\blacksquare$\n\nFor $x \\neq 0$, choose $y$ such that $f(y)=-\\dfrac{f(x)}{x}$. Then from $P(x,y)$, $P(x,y+6)$ and Claim 3 we get $$f(x(f(y)+6)+f(x))=2f(x)+xy+6x=f(xf(y)+f(x))+6x$$ $$\\implies \\ f(6x)=6x+1$$\nfor all $x \\neq 0$. This, combined with $f(0)=1$ gives the solution $$\\boxed{f(x)=x+1 \\ \\ \\forall x \\in \\mathbb{R}}$$", "Solution_23": "[hide = Redacted]Here is my solution. \nLet $P(x,y)$ be the assertion to the initial eqn.\nFirst lets solve for polynomial $f(x)'s$.\nAssuming $deg(f(x)) = d$ $$P(0,x) \\implies f(f(0)) = 2f(0)$$\n$P(x,0) \\implies f(xf(0) + f(x)) = 2f(x) \\implies d \\cdot$ max{$1,d$}$= d \\implies d = 0 \\text{ or } 1$\nFor $d = 0$, let $f(x) = c \\implies c = 2c + xy$ which is clearly impossible.\n\nFor $d = 1$, let $f(x) = ax + b$ \n\n$P(x,x) \\implies f(f(x) (x+1)) = 2f(x) + x^2$ \n$$\\implies a((ax+b)(x+1)) + b = a^2 x^2 + x(a^2 + ab) + ab + b = x^2 + 2ax + 2b$$\n$$a^2 = 1, a^2 + ab = 2a, ab + b = 2b \\implies f(x) = x + 1$$\n\nNow all that is left to show is that this is the only solution.\n\n$P(1,x) \\implies f(f(x) + f(1)) = 2f(1) + x$ as $f(1)$ has to be constant, $2f(1) + x = m \\ \\forall \\ m \\in \\ \\mathbb{R}$ thus $f(x)$ is surjective.\n$f(x) + 1= f(y) +1 \\implies 2f(1) + x = 2f(1) + y \\implies x = y$ thus $f(x)$ is Injective.\nTherefore, $f(x)$ has to be a polynomial. So we are done.\n[/hide]", "Solution_24": "[quote=quirtt]... thus $f(x)$ is Injective.\nTherefore, $f(x)$ has to be a polynomial. So we are done.[/quote]\nCould you kindly explain us how the bijectivity of $f(x)$ implies that it must be a polynomial ?\nThanks a lot for any precision.\n\n\n", "Solution_25": "[quote=quirtt]How to prove that it can only be a polynomial?[/quote]\nNearly all the posts in this thread end proving $f(x)=x+1$\nAnd since $x+1$ is a polynomial, nearly all the posts in this thread are proofs that solution must be a plynomial.\n\n", "Solution_26": "Tough FE. Denote the assertion by $P(x,y)$. Fixing $x$ and letting $y$ vary over $\\mathbb{R}$ it follows that $f$ is surjective. \n\nNow let $c$ be any real such that $f(c)=0$. $P(x,c) \\implies f(f(x))=2f(x)+cx$. If $c=0$, then $f(f(x))=2f(x)$. Since $f$ is surjective it follows that $f(r)=2r$ for all reals $r$. It is easy to check that this solution fails the original equation, so $c \\neq 0$. \n\nSuppose there exist reals $a,b$ such that $f(a)=f(b)$. Then \n\\[ 2f(a)+ca=f(f(a))=f(f(b))=2f(b)=cb \\implies a=b \\]\nso $f$ is injective. Note $c$ is unique by injectivity. \n\nTake any reals $x_1,y_1$. We have $f(x_1f(y_1)+f(x_1))=2f(x_1)+x_1y_1$. But $P(y_1, x_1) \\implies f(y_1f(x_1)+f(y_1))=2f(y_1)+x_1y_1$. This holds for all $x_1, y_1$ real so we have \n\\[ f(xf(y)+f(x))-2f(x)=f(yf(x)+f(y))-2f(y) \\]\nfor all reals $x,y$. Taking $y=0, x=c$ in the above and using $f(f(x))=2f(x)+cx$ it follows that\n\\[ f(cf(0))=f(f(0)) -2f(0) = 0 =f(c) \\implies cf(0)=c \\]\nby injectivity. $c \\neq 0$, so $f(0)=1$. \n\nNow \n\\[ P(0,0) \\implies f(1)=f(f(0))=2f(0)=2 \\]\n\\[ P(c,c) \\implies 1=f(0)=f(f(c))=2f(c)+c^2=c^2 \\implies c= \\pm 1\\]\nIf $c=1$ then $2=f(1)=0$, contradiction. So $c = -1$, and the above rewrites as $f(f(x))=2f(x)-x$. This means\n\\[ f(2)=f(f(1)) = 2f(1)-1 = 3 \\]\nNext\n\\[ P(-1, x) \\implies f(-f(x))=-x = f(f(x))-2f(x) \\]\nSince $f$ is surjective, it follows that\n\\[ f(-r) = f(r)-2r \\]\nfor all reals $r$. Then\n\\[ P(1,y) \\implies f(f(y)+2) = y+4 \\]\n\\[ P(-1, y) \\implies f(-f(y))=-y \\implies -y = f(-f(y))=f(f(y))-2f(y) \\]\n\\[ \\implies f(f(y)+2)=2f(y)-f(f(y))+4 \\]\nSince $f$ is surjective, it follows that for all reals $r$,\n\\[ f(r+2)+f(r)=2r+4 \\]\nNow \n\\[ P(2,y) \\implies f(2f(y)+3)=6+2y = 6+2(2f(y)-f(f(y))) \\implies f(2f(y)+3)+2f(f(y)) = 4f(y)+6 \\]\nBy surjectivity it follows that for all reals $r$,\n\\[ f(2r+3)+2f(r)=4r+6 \\]\nThis holds for all reals $r$, so\n\\[ r \\rightarrow r - \\frac{1}{2} \\implies f(2r+2)+2f(r-\\frac{1}{2})=4r+4 \\]\nNow since $f(r+2)+f(r)=2r+4$ for all reals $r$,\n\\[ r \\rightarrow 2r \\implies f(2r+2)+f(2r) = 4r+4 \\]\nInjectivity of $f$ implies for all reals $r$ we have\n\\[ f(2r) = 2f(r-\\frac{1}{2}) \\implies f(2r+1)=2f(r) \\]\ntaking $r-\\frac{1}{2} \\rightarrow r$, since these relations hold for all reals $r$. \n\nWe are almost there. $f(2r+3)+2f(r)=4r+6$ holds for all reals $r$, so\n\\[ r \\rightarrow r-1 \\implies f(2r+1)+2f(r-1)=4r+2 \\]\n\\[ f(2r+1)=2f(r) \\implies f(r)+f(r-1)=2r+1 \\]\nThis holds for all reals $r$ so\n\\[ r \\rightarrow r+1 \\implies f(r+1)+f(r)=2r+3 \\]\n\\[ r \\rightarrow r+2 \\implies f(r+2)+f(r+1)=2r+5 \\]\n\\[ \\implies f(r+2)-f(r) = 2 \\]\nThis holds for all reals $r$. We also have $f(r+2)+f(r)=2r+4$ for all reals $r$, so subtracting yields\n\\[ 2f(r)=2r+2 \\implies f(r)=r+1 \\]\nfor all reals $r$. It is easy to check that this solution works, so the only function is $f(x) \\equiv x+1$, finishing the proof.", "Solution_27": "Let $P(x,y)$ be the assertion $f(xf(y)+f(x)) = 2f(x)+xy$.\n\n$P(1,x)\\Rightarrow f(f(x)+f(1))=2f(1)+x$\nSo $f$ is bijective.\nWe have that $\\exists k,n:f(k)=0,f(n)=1$ by surjectivity.\n$P(k,n)\\Rightarrow 0=kn$\n$P(n,k)\\Rightarrow f(1)=2$\nSince $kn=0$, we now have two cases to consider.\n\n[b]Case 1:[/b] $k=0\\Rightarrow f(0)=0$\n$P(x,0)\\Rightarrow f(f(x))=2f(x)$ so by surjectivity we have $f(x)=2x$, which doesn't work.\n\n[b]Case 2:[/b] $n=0\\Rightarrow f(0)=1$\n$P(k,k)\\Rightarrow 1=k^2$, and since $f(1)=2$, we have $k\\ne1$, so $k=-1\\Rightarrow f(-1)=0$.\n\n$P(-2,-2)\\Rightarrow f(-f(-2))=2f(-2)+4$\n$P(-1,-2)\\Rightarrow f(-f(-2))=2$\nSo $f(-2)=-1$.\n$P(-1,x)\\Rightarrow f(-f(x))=-x$\n\nThe next two lines are the crucial step.\n$P(-2,-f(x))\\Rightarrow f(2x-1)=2f(x)-2$\n$P\\left(x,-\\frac2x\\right)\\Rightarrow f\\left(xf\\left(-\\frac2x\\right)+f(x)\\right)=2f(x)-2$\n\nSo $2x-1=xf\\left(-\\frac2x\\right)+f(x)\\forall x\\ne0$ after using injectivity.\nNow this is routine, since $-\\frac2x$ is involutive. Taking $x\\mapsto-\\frac2x$ and multiplying by $-x$ yields:\n$$4+x=2f(x)-xf\\left(-\\frac2x\\right)$$\nAdding, we get $3f(x)=3x+3\\forall x\\ne0$, hence the unique solution $\\boxed{f(x)=x+1}$ which works. $\\square$", "Solution_28": "Tricky problem.\nAs in previous posts let $P(x,y)$ be the assertion $f(xf(y)+f(x)) = 2f(x)+xy$.\n$P(1,y): f(f(y)+f(1))=2f(1)-y$, so f is obviously bijection. Now if $f$ is bijective, we can find such $t$ that $f(t)=0$. :\n$P(t,0): f(tf(0)+f(t))=2f(t)+0\\Rightarrow f(tf(0))=0$. But $f(t)=0$ and f is a bijection so if $f(tf(0))=f(t)=0$ then $tf(0)=t\\Rightarrow t=0$ or $f(0)=1$. It`s equivalent to two cases $f(0)=0$ and $f(0)=1$.\n[b]First Case: f(0)=0[/b]. Then plugging $P(x,0): f(x*f(0)+f(x))=2f(x)+0\\Rightarrow f(f(x))=2f(x)$. Now by surjectivity we can find such $t$ for every $y$ in R that $f(t)=y$. So for every $y$: $f(f(t))=f(y)=2f(t)=2y$ so [u]$f(y)=2y$[/u]. Plugging this function into the equation we get that the function doesn`t fit.\n[b]Second Case: f(0)=1[/b]. Now $P(x,0): f(x+f(x))=2f(x) \\bigstar$. \nPlugging $P(x+f(x),y_0)$ where $y_0=\\frac {2xy} {x+f(x)}$, we get :\n$f((x+f(x))f(y_0)+f(x+f(x))=f((x+f(x))f(y_0)+2f(x))=2f(x+f(x))+(x+f(x))y_0=4f(x)+(x+f(x))y_0$, so $f(x+f(x))(f(y_0)+2f(x))=4f(x)+(x+f(x))y_0=4f(x)+(x+f(x))\\frac {2xy}{x+f(x)}=4f(x)+2xy=2f(xf(y)+f(x))$. But from $\\bigstar$ we get that $2f(xf(y)+f(x))=f(xf(y)+f(x)+f(xf(y)+f(x)))=f(xf(y)+f(x)+2f(x)+xy)=f(3f(x)+xy+xf(y))$ so $f(x+f(x))(f(y_0)+2f(x))=f(3f(x)+xy+xf(y))$, by injectivity we get that $(x+f(x))f(y_0)+2f(x)=3f(x)+xy+xf(y)\\ - \\spadesuit$. Now plugging $x=t$ where $t$ is such number that $f(t)=0$ we get that that $y_0=\\frac{2ty}{t+0}=2y$, so $\\spadesuit$ becomes $tf(2y)=ty+tf(y)\\Rightarrow y+f(y)=f(2y)$ for every $y$. Plugging $P(\\frac{x}{2},2y):\\ f(\\frac{x}{2}f(2y)+f(\\frac{x}{2}))=2f(\\frac{x}{2})+xy=2(f(x)-\\frac{x}{2})+xy=2f(x)+x(y-1)=f(xf(y-1)+f(x))$, \nso $f(\\frac{x}{2}f(2y)+f(\\frac{x}{2}))=f(xf(y-1)+f(x))$, by injectivity $\\frac{x}{2}f(2y)+f(\\frac{x}{2})=xf(y-1)+f(x)\\Rightarrow \\frac{x}{2}(f(y)+y)+f(x)-\\frac{x}{2}=xf(y-1)+f(x)\\Rightarrow \\frac{x}{2}(f(y)+y)-\\frac{x}{2}=xf(y-1)\\Rightarrow {f(y)+y-1}=2f(y-1)$ Setting $y=0$ into the last equation: $f(0)+0-1=2f(-1)\\Rightarrow 1-1=2f(-1)\\Rightarrow f(-1)=0$. Plugging $P(x,-1): f(f(x))=2f(x)-x$. But now from $f(y)+y-1=2f(y-1)\\Rightarrow f(y)=2f(y-1)-(y-1)=f(f(y-1))$, that`s why by injection we get that $y=f(y-1)$ for every $y$, so [u]$f(x)=x+1$[/u], which is correct. $\\blacksquare$", "Solution_29": "Woah woah woah really hard (ok maybe not too much but yea it's hard and nice :D).\nAs usual let $P(x,y)$ denote the given assertion.\n[hide=Bijectivity]\nFixing $x \\neq 0$ gives us that $f$ is surjective. Thus there exists $a$ such that $f(a)=0$ (and let's define this specific $a$ for convenience).\n$\\textbf{Case: } f(0)=0$. \n$P(x,0) : f(f(x))=2f(x) \\implies f(t)=2t$ but this does not satisfy our original equation hence $f(0) \\neq 0$.\n$P(a,t) : f(af(t))=at$.\nIf $f(x)=f(y)$\n$$\\implies af(x)=af(y) \\implies f(af(x))=f(af(y)) \\implies ax = ay \\implies x=y.$$\nThus $f$ is bijective.\n[/hide]\n[hide=Some required values of f]\nThere exists $b$ such that $f(b)=1$.\n$P(a,b) : f(a)=ab =0 \\implies b=0 \\implies \\boxed{f(0)=1}.$\n$P(0,a) : f(f(0))=2 \\implies \\boxed{f(1)=2}.$\n$P(-1,-1) : f(0) = 2f(-1)+1 \\implies \\boxed{f(-1)=0}.$\n[/hide]\nNow to finish\n$P(x,-1) : f(f(x))=2f(x)-x$.\n$P(1,x) : f(f(x)+2)=x+4$.\nAdding both we obtain $f(t)+f(t+2)=2t + 4$.\n$$\\implies f(t-2) + f(t)=2t \\implies f(t+2)-f(t-2)=4 \\implies f(t+4)=f(t)+4.$$\n$$f(f(t)+2)=t+4 \\implies f(f(f(t)+2)+2)= f(t)+6 \\implies f(t+6) = f(t)+6.$$\n$$\\implies f(t+4)+f(t+6) = 2f(t)+10 \\implies f(t+4) + f(t+4 +2) =2f(t)+10$$\n$$\\implies 2(t+4) + 4 = 2f(t) + 10 \\implies f(t)=t+1.$$\nAnd we are done :D", "Solution_30": "Beautiful functional equation. It is not actually as difficult as previously suggested. Here is a relatively new solution which almost overuses that $f$ is a bijection. Solved (the first half) with [b]L567[/b]. \n[hide=Answer]\n$f(x) = x+1$ for all $x \\in \\mathbb{R}$ is the only solution to the functional equation.[/hide]\nLet $P(x,y)$ denote the assertion to the above functional equation. \\\\\n[hide=Routine Work]\n$P(1,x):$ $$f(f(x)+f(1)) = 2f(1)+x$$ meaning that $f$ is bijective. \\\\\nLet $a \\in \\mathbb{R}$ such that $f(a) = 0$. \\\\\n$P(a,0):$ gives that $af(0) = a$ meaning that $a=0$ or $f(0) = 1$. \\\\\n$\\textbf{Case 1)}$ $a=0$ \\\\\nWe get that $f(0)=0$ and from $P(x,0):$ $$f(f(x)) = 2f(x) $$ using that $f$ is bijective we have that $f(x) = 2x$ for all $x\\in \\mathbb{R}$ which is not a solution. \\\\\n$\\textbf{Case 2)}$ $f(0) = 1$ \\\\\nNotice therefore that $f(1) = f(f(0)) = 2f(0) = 2$. \\\\\n$P(a,a)$ gives that $1 = f(0) = a^2$. \\\\\nTherefore $f(-1) = 0$ as $a \\in -1,1$ and $f(1) \\neq 0$. \\\\[/hide]\n[hide=Surjection Masterclass]\nGoing back to $P(1,x)$ notice that $$f(f(x)+2) = x+4 $$\nAlso \n$P(x,-1):$ $$f(f(x)) = 2f(x) -x$$\nTherefore $$f(f(x)) = 2f(x) - x = 2f(x) - f(f(x)+2) +4$$\nUsing the fact that $f$ is surjective $f(x) \\rightarrow$ gives us that $$f(x) + f(x+2) = 2x+4$$\nAlso $P(-1,x):$ $$f(-f(x)) = -x = 4-f(f(x)+2) $$\nOnce again taking $f(x) \\rightarrow x$ we have that $$4 = f(-x)+f(x+2)$$ \nCombining this with the fact that $f(x) + f(x+2) = 2x+4$, we have that $$f(x) = f(-x) +2x$$\nThen $$2x+4 = f(x)+f(x+2) = f(x) + f(-x-2)+2x+4$$\n$\\implies$ $$f(x)+f(-x-2) = 0$$\nNow we continue with some magic to finish off the problem. \\\\\n$P(-1,f(x)):$\n$$f(-2f(x)+x) = f(-f(f(x)) = -f(x) $$\nwhere we use that $f(f(x)) = 2f(x)-x$ by $P(x,-1)$ which gives us that $$f(-2f(x)+x) + f(x) = 0 = f(x) + f(-2-x)$$\nConsequently, we have that $f(-2f(x)+x) = f(-2-x)$ and therefore $f(x) = x+1$ for all $x \\in \\mathbb{R}$ using the fact that $f$ is injective. $\\blacksquare$[/hide]", "Solution_31": "[quote=cyshine]Find all functions $f\\colon \\mathbb{R}\\to \\mathbb{R}$ such that\n\\[f(xf(y)+f(x)) = 2f(x)+xy\\]\nfor every reals $x,y$.[/quote]\n\nHopefully correct solution. \n\n\nLet $P(x, y)$ be the assertion. $P(1, y)$ implies $f$ is bijective. If $f(0) = 0$, then $P(x, 0) \\implies f(f(x)) = 2f(x)$ which by surjectivity means that $f(x) = 2x$ for all reals $x$ which does not work. Therefore, $f(0) \\neq 0$. Now, $P(f^{-1}(0), f^{-1}(1)) \\implies 0 = f^{-1}(0) \\cdot f^{-1}(1) \\implies f^{-1}(1) = 0 \\implies f(0) = 1$. $P(0, 0) \\implies f(f(0)) = 2f(0) \\implies f(1) = 2$. If $f(x) = 0$, then $P(x, k) \\implies f(xf(k)) = xk$. Since $x \\neq 0$, $k = \\dfrac{1}{x}$ here implies that $f(xf(k)) = 1 \\implies xf(k) = 0$. This means $x = 0$ or $f(k) = 0$, but the previous is impossible which means that $f(k) = 0$. But this forces $k = x$ or $\\dfrac{1}{x} = x \\implies x \\in \\{ -1, 1\\}$, but since $f(1) = 2$, we have that $f(-1) = 0$. If $f(y) = -1$, then $P(1, x) \\implies f(1) = 2 = 2f(1) + y = 4 + y$, which means that $y = -2$. Now $P(x, -2) \\implies f(f(x) -x) = 2f(x) - 2x = 2(f(x)-x)$, now plugging $c_x = f(x) - x$ gives that $f(c_x) = 2c_x$ and $P(f^{-1}(c_x), 0) \\implies f(f^{-1}(c_x) + c_x) = 2c_x = f(c_x) \\implies f^{-1}(c_x) = 0 \\implies c_x = 1$ for all reals $x$, implying that $\\boxed{f(x) = x + 1}$ for all reals $x$.", "Solution_32": "Here is my solution is was really difficult for me atleast not very hard but still its hard i solved with [b][color=#f00]Fafnir[/color][/b] and we are done hope its correct. :maybe: \n[hide=Proving Bijectivity]\n[b][color=#f00]Claim:[/color][/b] $f$ is bijective function.\nLet $P(x,y)$ is the assertion of the following functional equation\\\\\nwe have\\\\\nLets suppose $f(a)=f(b)$ so we have \\\\\n$P(x,a)$\n\\[\\implies f(xf(a)+f(x))=2f(x)+ax\\]\n$P(x,b)$\n\\[\\implies f(xf(b)+f(x)=2f(x)+bx\\]\nsince $f(xf(a)+f(x))=f(xf(b)+f(x))$ because $f(a)=f(b)$ so we can say that \n\\[\\implies 2f(x)+ax=2f(x)+bx\\implies ax=bx\\implies a=b\\]\nso we can say that $f$ is injective function now\\\\\n$P(1,y)$\n\\[\\implies f(f(y)+f(1))=2f(1)+y\\]\nso we can see here that $f$ is surjective so hence $f$ is injective and surjective it is bijective.$\\square$\\\\[/hide]\n[hide=Finding important values]\nLets suppose for some $k$ we have $f(k)=0$ then putting\\\\\n$P(k,0)$\\[\\implies f(kf(0))=0\\implies f(kf(0))=f(k)\\]we know that $f$ is injective so we can say that\n\\[\\implies kf(0)=k\\implies f(0)=1\\]\nnow putting\\\\\n$P(0,k)$\\[\\implies f(f(0))=2f(0)\\implies f(1)=2\\]\nnow putting\\\\\n$P(k,k)$\\[\\implies f(kf(k)+f(k))=2f(k)+k^2\\implies f(0)=k^2=1\\]\nnow $k=\\pm 1$ so for $k=1$ we know that $f(1)=2$ so its clear that $f(-1)=0$\n[/hide]\n[hide=playing with equations]\nPutting $P(x,-1)$\n\t\\[\\implies f(xf(-1)+f(x))=2f(x)-x\\implies f(f(x))=2f(x)-x------(1)\\]\nPutting $P(1,x)$\n\t\\[f(f(x)+f(1))=2f(1)+x\\implies f(f(x)+2)=4+x-------(2)\\]\nby using (2) we can write this as\n\\[\\implies f(f(x))=2f(x)-f(f(x)+2)+4\\]\nsince we know that $f$ is surjective\nreplacing $f(x)\\rightarrow x$ we have\n\t\\[\\implies f(x)+f(x+2)=2x+4-------(3)\\]\nnow putting $P(-1,x)$\\[\\implies f(-f(x))=-x=4-f(f(x)+2)\\implies f(-f(x))=4-f(f(x)+2)\\]\nonce again replacing $f(x)\\rightarrow x$\n\t\\[\\implies f(-x)=4-f(x+2)\\implies f(-x)+f(x+2)=4-------(4)\\]\nsubstracting eq(3)-eq(2) we have\n\t\\[\\implies f(x)=f(-x)+2x-------(5)\\]\nreplacing $x\\rightarrow x+2$ we get\n\\[\\implies f(x+2)=f(-x-2)+2x+4\\]adding $f(x)$ on both side we have \n\t\\[f(x)+f(x+2)=f(x)+f(-x-2)+2x+4\\implies f(x)+f(-x-2)=0-------(6)\\]\nnow putting $P(-1,f(f(x)))$\n\\[\\implies f(-f(f(x))=-f(x)\\]\nfrom eq(1) we can write this as\n\\[\\implies f(-2f(x)+x)=-f(x)\\implies f(-2f(x)+x)+f(x)=0\\]\nfrom eq(6) we can write this as\n\\[\\implies f(-2f(x)+x)=f(-2-x)\\implies -2f(x)+x=-2-x\\implies f(x)=x+1\\]\nwe get $f(x)=x+1$ $\\forall x\\in\\mathbb{R}$ because since we know that $f$ is injective. :D $\\blacksquare$[/hide]", "Solution_33": "Really difficult FE.\n\nThe solution is very long(with a lot of substitutions), so we will split it into a couple of main parts.\n\n[color=#f00]1. Bijective[/color]: Fixing $x$ shows that $f$ is surjective. Pick a $b$ such that $f(b)=0$. $P(x,b)$ gives that $f(f(x))=2f(x)+bx$, and $P(0,b)$ gives that $f(f(0))=2f(0)$. If $b=0,$ we have that $f(f(x))=2f(x),$ and by surjectivity, that $f(x)=2x\\forall x\\in \\mathbb{R}$. Plugging this into the original equation, it evidently does not work, so we assume that $b\\neq 0$. Then, from $f(f(x))=2f(x)+bx,$ we have that $f$ is injective. This establishes that $f$ is bijective. \n\n[color=#f00]2. Establish that [/color]$f(0)=1$: To give some motivation, originally, I actually wanted to show that $f(0)=0,$ so I let $a$ be such that $f(a)=0$. There are two ways to go from here:\n\n1) From $P(a,y)$ we get that $f(af(y))=ay$. Pick a $y$ such that $f(y)=1$. Then this gives that $0=ay\\implies a=0$ or $y=0$(i.e. either $f(0)=0$ or $f(0)=1$). But we already established a contradiction when $f(0)=0,$ so this means $f(0)=1$.\n\n2) $P(a,a)$ gives that $f(0)=a^2,$ so $f(f(a))=a^2.$ Follow the same steps in 1) to establish that $f(af(y))=ay$, then setting $y=f(a)$ gives that $f(a^3)=0$. But $f$ is injective, so we have that $a^3=a,$ so $a=\\pm 1$. I am pretty sure you could do something with this here, but I did not use this solution path.\n\n[color=#f00]Finishing the problem and making a TON of substitutions[/color] From here, we get from $P(-1,-1)$ that $f(-1)=0$, and then $P(x,-1)$ gives that $f(f(x))=2f(x)-x$. From this, it is easy to get by induction that $f(x)=x+1\\forall x\\in \\mathbb{Z}$. From here, my finish was identical to @nhat, so I will not post it.", "Solution_34": "Quite a good problem :) \nWe claim the only answer is $\\fbox{f(x)= x+1}$\n[b]Claim[/b]:-$f$ is bijective\nLet $P(x,y)$ be our assertion \n$P(1,y)$ $\\rightarrow$ $f(f(y)+f(1))=2f(1)+y$ which immediatey gives away surjectivity.\nFor injectivity,\nlet $f(a)=f(b)$\n$P(1,a)$ $\\rightarrow$ $f(f(a)+f(1))=2f(1)+a$\n$P(1,b)$ $\\rightarrow$ $f(f(b)+f(1))=2f(1)+b$\nfrom here $a=b$ and hence $f$ is indeed bijective.\n[b]Claim[/b]:- $f(-1)=0$ \nLet $f(\\alpha)=0$ for some $\\alpha \\in \\mathbb{R}$\n$P(\\alpha,\\alpha)$ $\\Rightarrow$ $f(0)=\\alpha^2$\n$P(\\alpha,o)$ $\\Rightarrow$ $f(\\alpha^3)=0=f(\\alpha)$\nAs $f$ is injective therefore $\\alpha^3=\\alpha$ $\\Rightarrow$ $\\alpha=-1,0,1$\n[b]Case 1:-[/b] if $\\alpha=0$\nWe get that $f(0)=0$ and from $P(x,0):$\n$$f(f(x)) = 2f(x) $$\nusing that $f$ is bijective we have that $f(x) = 2x$ for all $x\\in \\mathbb{R}$ which is not a solution.\n[b]Case 2:-[/b]If $\\alpha=1$\nFrom the above equations we get that $f(0)=1$ and $f(1)=0$\n$P(0,0)$ $\\rightarrow$ $f(1)=2$ which is a contradiction.\nSo, $\\alpha=-1$\nFrom here $f(0)=1$ and $f(-1)=0$ which concludes our claim.\n[b]Claim:-[/b] $f$ is linear\n$P(1,x)$ $\\rightarrow$ $f(f(x)+2) = x+4 $\nAlso,$P(x,-1)$ $\\rightarrow$ $f(f(x)) = 2f(x) -x$\nAdding this two we get, $f(f(x))+f(f(x)+2)=2f(x)+4$\nTaking $f(x) \\rightarrow x$ we get $f(x)+f(x+2)=2x+4$ \n$P(-1,x)$ $\\rightarrow$ $f(-f(x))=-x=f(f(x)-2f(x)$\n=$\\Rightarrow$ $f(-x)=f(x)-2x$\nCombining the equations we get,\n$f(-x)+f(x+2)=4$\n$x \\rightarrow x+2$\n$$f(-x-2)-f(x+2)=-2x+4$$\n$$f(-x)+f(x+2)=4$$\n$f(-x)+f(-x-2)=-2x=f(-x)+f(x)$\n$\\Rightarrow f(-x-2)=-f(x)$\n$P(-1,f(x))$ $\\rightarrow$ $f(-2f(x)+x)+f(x)=f(x)+f(-x-2)$\n=$f(-2f(x)+x)=f(-x-2)$\n=$\\Rightarrow$ $-2f(x)+x=-x-2$\nand therefore only solution is $\\fbox{f(x)=x+1}.$\nand we are finally done!! ;)", "Solution_35": "Let $P(x,y)$ denote the given assertion. Clearly $f$ is bijective. If $f(x_1)=f(x_2)$ then $P(x,x_1)$ and $P(x,x_2)$ implies $x_1=x_2.$ \n\nTake $f(x_0)=0.$ If $x_0=0$ then $P(x,x_0)$ implies $f(x)=2x$ which fails. Otherwise $P(x_0,0)$ implies $f(0)=1.$ And $P(-1,-1)$ implies $f(-1)=0.$ $P(-1,x)$ implies $f(-f(x))=-x$ and $P(x,-1)$ implies $f(f(x))=2f(x)-x$; so, by iterating, $f(Z)=Z+1$ for integers. \n\n$P(x,-2)$ implies $f(f(x)-x)=2f(x)-2x$ and finally $P(f(x)-x,-3)$ yields $f(x)=x+1$ which indeed satisfies. " } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "find all function f, g, h : R to R such that\r\nf( $ {y}^{3}$ + x) + g( y + $ {x}^{3}$ ) = h(xy).", "Solution_1": "Let $ z_1\\equal{}z_1(x,y)\\equal{}y^3\\plus{}x,z_2\\equal{}z_2(x,y)\\equal{}z_1(y,x)\\equal{}x^3\\plus{}y$. Then $ f(z_1)\\plus{}g(z_2)\\equal{}h(xy)\\equal{}f(z_2)\\plus{}g(z_1)$.\r\nLet $ x\\equal{}\\minus{}y^3, z\\equal{}y\\minus{}y^9$, then $ f(0)\\plus{}g(z)\\equal{}h(\\minus{}y^4)$. Therefore $ f(z)\\plus{}g(0)\\equal{}h(\\minus{}y^4)\\equal{}g(z)\\plus{}f(0)$.\r\nBecause for any z txist y, suth that $ z\\equal{}y\\minus{}y^9$, we get $ f(z)\\equiv g(z)\\plus{}const, const\\equal{}a\\equal{}f(0)\\minus{}g(0),f(0)\\plus{}g(0)\\equal{}h(0).$\r\n$ y\\equal{}0\\to f(x^3)\\plus{}g(x)\\equal{}h(0)\\to g(x^3)\\plus{}g(x)\\equal{}2g(0)\\to g(x^9)\\equal{}g(x).$\r\nIf $ f(x),g(x),h(x)$ are solution, then $ f(x)\\plus{}a_1,g(x)\\plus{}a_2,h(x)\\plus{}a_1\\plus{}a_2$ is solution too.\r\nTherefore we can consider only case, when $ f(0)\\equal{}g(0)\\equal{}0\\to h(0)\\equal{}0$.\r\nIt give $ g(x^3)\\equal{}\\minus{}g(x), g(x)\\equiv f(x)$. We get $ g(y\\minus{}y^9)\\equal{}h(\\minus{}y^4)$, therefore $ g(0)|_{y\\equal{}0}\\equal{}h(0)\\equal{}g(0)_{y\\equal{}1}\\equal{}h(\\minus{}1)$.\r\nIt give $ g(x^3\\minus{}x^{\\minus{}1})\\plus{}g(x\\minus{}x^{\\minus{}3})\\equiv 0$. By $ g(x^3)\\equal{}\\minus{}g(x)$ we get at least for continiosly $ g\\equiv f\\equiv h\\equiv 0$.\r\nTherefore solutions only $ f\\equiv a,g\\equiv b,h\\equiv a\\plus{}b$." } { "Tag": [ "function", "number theory", "relatively prime", "complementary counting", "prime numbers", "totient function" ], "Problem": "How many numbers are relatively prime to 441?\r\n\r\nSo what I did was basically use Euler's totient function. \r\n\r\n441 Prime factorizes: $ 7^2\\cdot3^2$\r\n\r\nand then (by Euler's product)\r\n\r\n$ 441(1\\minus{}\\frac{1}{7})(1\\minus{}\\frac{1}{3})\\equal{}441(\\frac{6}{7})(\\frac{2}{3})\\equal{}\\boxed{252}$\r\n\r\nIs there a different way to approach this problem?", "Solution_1": "Complementary Counting+PIE\r\n\r\nAs in, there are $ \\frac{441}{3}\\equal{}147$ numbers divisible by 3, $ \\frac{441}{7}\\equal{}63$ numbers divisible by 7, and $ \\frac{441}{21}\\equal{}21$ numbers divisible by both, so there are $ 147\\plus{}63\\minus{}21\\equal{}189$, and that's how many numbers are not relatively prime, so subtract this from 441 and you get $ \\boxed{252}$\r\n\r\nnot really faster, but a different way, and probably one way how Euler's product is derived.", "Solution_2": "[quote=\"alanchou\"]Complementary Counting+PIE\n\nAs in, there are $ \\frac {441}{3} \\equal{} 147$ numbers divisible by 3, $ \\frac {441}{7} \\equal{} 63$ numbers divisible by 7, and $ \\frac {441}{21} \\equal{} 21$ numbers divisible by both, so there are $ 147 \\plus{} 63 \\minus{} 21 \\equal{} 189$, and that's how many numbers are not relatively prime, so subtract this from 441 and you get $ \\boxed{252}$\n\nnot really faster, but a different way, and probably one way how Euler's product is derived.[/quote]\r\n\r\nIt probably is, since thats is also one of the more expanded forms of it.\r\n\r\n :D \r\n\r\nI was wondering if there was a quick trick because if this wasn't a Target problem it would take a while (comparatively) to get the solution.", "Solution_3": "Are you talking about all the relatively prime numbers to 441 less than 441? Cuz there's an infinite amount of relatively prime number thingys without constraints.", "Solution_4": "Yeah, less than 441.", "Solution_5": "The problem [i]is[/i] asking you to calculate a value of Euler's totient function; that's a definition, not a technique. \r\n\r\nThe technique: $ \\varphi(7^2 \\cdot 3^2) \\equal{} \\varphi(7^2) \\cdot \\varphi(3^2) \\equal{} (7 \\cdot 6) \\cdot (3 \\cdot 2) \\equal{} 42 \\cdot 6 \\equal{} 252$. Here we have used two lemmas that allow us to calculate $ \\varphi(n)$ fairly quickly.\r\n\r\n[b]Lemma:[/b] If $ a, b$ are relatively prime, $ \\varphi(ab) \\equal{} \\varphi(a) \\varphi(b)$.\r\n\r\n[b]Lemma:[/b] If $ p$ is a prime, $ \\varphi(p^n) \\equal{} (p\\minus{}1) p^{n\\minus{}1}$." } { "Tag": [], "Problem": "Let $O$ be the center of circle $w$. Two equal chords $AB$, $CD$ of $w$ intersect at $L$ such that $AL>LB$, $DL>LC$. Let $M$,$N$ be points on $AL$, $DL$ respectively such that $\\angle ALC = 2\\angle MON$. Prove that the chord of $w$ passing through $M$ and $N$ is congruent to $AB$ and $CD$.", "Solution_1": "Anybody? :)", "Solution_2": "I don't know how to use the angle relation...... :?", "Solution_3": "Hmm... maybe try to work the other way. Suppose there are $PQ = AB = CD$ and $PQ$ intersects $AL$ at $R$ and $DL$ at $S$ and show $\\angle ALC = 2\\angle ROS$. Does it work?", "Solution_4": "[quote=\"warut_suk\"]Let $O$ be the center of circle $w$. Two equal chords $AB$, $CD$ of $w$ intersect at $L$ such that $AL>LB$, $DL>LC$. Let $M$,$N$ be points on $AL$, $DL$ respectively such that $\\angle ALC = 2\\angle MON$. Prove that the chord of $w$ passing through $M$ and $N$ is congruent to $AB$ and $CD$.[/quote]\r\n\r\n\r\nseems there is some error in the data.\r\nAL has M, DL has N on them \r\n\r\nAB, CD and MN have to be congruent....I think it is not possible.", "Solution_5": "Drop perpendiculars to all the chords, then show that OM and ON are bisectors of AMN and DNM using similarity or whatever. It looks like there should be an easier way though.", "Solution_6": "[quote=\"probability1.01\"] It looks like there should be an easier way though.[/quote]\r\n\r\nI don't think so...", "Solution_7": "[quote=\"probability1.01\"]Drop perpendiculars to all the chords, then show that OM and ON are bisectors of AMN and DNM using similarity or whatever. It looks like there should be an easier way though.[/quote]\r\n\r\nI have attached a figure for one of the possible cases( not to scale).\r\n\r\nIt gives me an impression that there is no possibility of the proof for what is asked.", "Solution_8": "Your angle MON is way too large. Remember it is half the angle of the angle between the chords.\r\n\r\nI don't feel like posting again for some reason, so in response to below post... concurrent $\\ne$ congruent...", "Solution_9": "[quote=\"probability1.01\"]Your angle MON is way too large. Remember it is half the angle of the angle between the chords.[/quote]\r\n\r\ncould be. However, M and N will be placed somewhere as I have shown.\r\n\r\nWhat I want to emphasise is....\r\n\r\nwhether it is possible to have chord through MN, concurrent to AB and CD." } { "Tag": [ "algebra solved", "algebra" ], "Problem": "Let a, b, c, d real numbers such that:\r\n[tex]a = \\sqrt{4 + \\sqrt{5 + a}}[/tex] ; [tex]b = \\sqrt{4 - \\sqrt{5 + b}}[/tex] ; [tex]c = \\sqrt{4 + \\sqrt{5 - c}}[/tex] ; [tex]d = \\sqrt{4 - \\sqrt{5 - d}}[/tex]\r\nfind [tex]abcd[/tex]", "Solution_1": "we note that a,b,c,d are four solutions of the equation\r\n\r\n[tex]((x^2-4)^2-5)^2=x^2[/tex]\r\n\r\nthe other four are -a, -b, -c, -d so [tex](abcd)^2=(4^2-5)^2=121[/tex] and so abcd=11" } { "Tag": [ "trigonometry", "integration", "logarithms", "function", "calculus", "algebra", "functional equation" ], "Problem": "$ \\sin x \\equal{} \\operatorname{Im} \\left( {e^{ix} } \\right),\\,\\cos x \\equal{} \\operatorname{Re} \\left( {e^{ix} } \\right).$\r\n\r\nIs this a definition or does exist a proof?", "Solution_1": "$ \\exp(z): \\equal{} \\sum_{k \\equal{} 0}^\\infty \\frac {z^k}{k!}$ is the definition for the exponential function.\r\n\r\n$ \\cos(z): \\equal{} \\frac {\\exp(iz) \\plus{} \\exp( \\minus{} iz)}{2}$ , $ \\sin(z): \\equal{} \\frac {\\exp(iz) \\minus{} \\exp( \\minus{} iz)}{2i}$ are the definitions for cosine and sine.\r\n\r\nParticularly for a real number $ x \\quad \\exp( \\minus{} ix) \\equal{} \\exp(\\,\\overline{ix}\\,) \\equal{} \\overline{\\exp(ix)}$\r\n\r\nSo $ \\cos(x) \\equal{} \\frac {\\exp(ix) \\plus{} \\overline{\\exp(ix)}}{2} \\equal{} \\text{Re}\\Big(\\exp(ix)\\Big)$ and $ \\sin(x) \\equal{} \\frac {\\exp(ix) \\minus{} \\overline{\\exp(ix)}}{2i} \\equal{} \\text{Im}\\Big(\\exp(ix)\\Big)$ .\r\n\r\n\r\nFor all rational numbers $ \\frac {p}{q}$ you can show $ e^{\\frac {p}{q}} \\equal{} \\exp\\left(\\frac {p}{q}\\right)$ .\r\n\r\nAnd for all other expressions $ z\\notin \\Bbb{Q}$ define $ e^z$ as $ \\exp(z)$ .\r\n\r\nSo the equality $ e^z \\equal{} \\exp(z)$ holds for all $ z$ for which the expression $ \\exp(z)$ is sensible.", "Solution_2": "Let $ f(\\theta)=\\cos\\theta+i\\sin\\theta$. Hence,\r\n\\[ \\begin{eqnarray*}df(\\theta)&=&(-sin\\theta+i\\cos\\theta)d\\theta=if(\\theta)d\\theta\\\\ \\int \\frac{1}{f(\\theta)}df(\\theta)&=&\\int id\\theta\\\\\\ln f(\\theta)&=&i\\theta\\iff f(\\theta)=e^{i\\theta}\\end{eqnarray*},\\]\r\nas required.", "Solution_3": "My answer to the original question is \"both\". You can use the identities as definitions of $ \\sin$ and $ \\cos$, but then you'll have to explain why these are the same functions that we use to find the sidelengths in a right triangle. Or you can define all three functions independently, and then prove that they are related. I prefer the latter. In Calculus 2 we show that \r\n$ e^x \\equal{} \\sum_{n \\equal{} 0}^{\\infty}\\frac {x^n}{n!}$, $ \\sin x \\equal{} \\sum_{n \\equal{} 0}^{\\infty}( \\minus{} 1)^n\\frac {x^{2n \\plus{} 1}}{(2n \\plus{} 1)!}$, and $ \\cos x \\equal{} \\sum_{n \\equal{} 0}^{\\infty}( \\minus{} 1)^n\\frac {x^{2n}}{(2n)!}$ for all $ x\\in\\mathbb R$. \r\nThe power series also converge for all $ x\\in \\mathbb C$. Splitting the (absolutely convergent) series for $ \\exp$ into odd and even-numbered terms, we find\r\n$ e^{ix} \\equal{} \\sum_{n \\equal{} 0}^{\\infty}\\frac {i^n x^n}{n!} \\equal{} \\sum_{k \\equal{} 0}^{\\infty}\\frac {( \\minus{} 1)^k x^{2k}}{(2k)!} \\plus{} i\\sum_{k \\equal{} 0}^{\\infty}\\frac {( \\minus{} 1)^k x^{2k \\plus{} 1}}{(2k \\plus{} 1)!}$, etc.", "Solution_4": "As part of that last approach, I like to go through the proof that the function\r\n\r\n$ f(z)\\equal{}\\sum_{n\\equal{}0}^{infty}\\frac{z^n}{n!}$\r\n\r\ndoes satisfy the functional equation $ f(z\\plus{}w)\\equal{}f(z)f(w).$ This is the key algebraic property that we would expect something called an exponential function to have, which at least partly justifies us using the notation $ e^z$ for it.\r\n\r\nAnd once we've shown that, then $ e^z\\equal{}e^{x\\plus{}iy}\\equal{}e^xe^{iy}.$ In that, $ e^x$ is the real exponential that we're already familiar with, while we deal with $ e^{iy}$ as mlok has indicated." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "hallo,\r\ni think this is an easy one. solve the equation: \\[x!+y!+z!=2^{v!}.\\] :)", "Solution_1": "We may assume that $x\\le y\\le z$. Then \\[x!(1+\\frac{y!}{x!}+\\frac{z!}{x!})=2^{v!},\\] thus $x\\le2$. \r\n1)x=1. If y>1, $1+y!+z!$ is odd and >1. So y=1 and $2+z!=2^{v!}\\qquad\\Rightarrow\\qquad z\\le3$. Finally we have\r\n(x,y,z,v)=(1,1,2,2). \r\n2)x=2. Then y<4 and so on. In this case we have no sols.\r\n3 solutions altogether.", "Solution_2": "$x=2$ give y=3, z=4 or 5\r\n$x=2,y=3,z=4,vt=5$, $x=2,y=3,z=5,vt=7$.", "Solution_3": "\u0422\u0430\u043c v!, \u0430 \u043d\u0435 vt." } { "Tag": [], "Problem": "Is [url=http://talk.collegeconfidential.com/showthread.php?t=108828&page=5&pp=20]this[/url] illegal? I certainly think it is. According to the SAT rules they read us, it certainly looks illegal. Should we report this to someone? \r\n\r\n\r\nBTW, on a lighter note, did anyone take the November 5th SAT? What are you hoping for?", "Solution_1": "[quote=\"mysmartmouth\"]Is [url=http://talk.collegeconfidential.com/showthread.php?t=108828&page=5&pp=20]this[/url] illegal? I certainly think it is. According to the SAT rules they read us, it certainly looks illegal. Should we report this to someone? \n\n\nBTW, on a lighter note, did anyone take the November 5th SAT? What are you hoping for?[/quote]\r\n\r\nI thought the math was way too easy, verbal way too hard, and writing was allright, my predictions:\r\n\r\nMath: 800\r\nVerbal: 550\r\nWriting: 680\r\n\r\nEssay: 10/12\r\n\r\nBTW: yeah, it is illegal, he can't discuss the question in detail due to copyright infringements (but I'm just glad that I got all the questions right by looking at it :P )", "Solution_2": "The College Confidential boards can be very useful, but they reveal that a lot of college applicants who haven't got enough brainpower to get into the top schools through their own effort will try ANYTHING--honest or dishonest--to gain an edge. I like the fact that AoPSers generally apply their own effort, honestly, to gaining entrance into college.", "Solution_3": "Math: 800\r\nCR: 680\r\nWriting Skills: 750\r\n\r\nEssay: 11/12", "Solution_4": "Isn't CR, critical reading, which is the same as verbal? ;) ;)", "Solution_5": "I don't see how this is illegal. It was discussed according to the times, after the test.", "Solution_6": "[quote=\"236factorial\"]I don't see how this is illegal. It was discussed according to the times, after the test.[/quote]\r\n\r\nYes, but I just looked at the thread regarding the PSAT after October 12, however, a lot of people took the same PSAT on October 15, so it is VERY easy to cheat that way, which I think a lot of people have done :mad: :mad:", "Solution_7": "Isn't the PSAT given on the alternate day a different version?", "Solution_8": "For those of you who never pay attention during the obligatory announcements on test day:\r\nYes, it is illegal. Multiple choice questions are not allowed to be \"carried\" from the testing room in any form- including your memory. (of course, this does not mean you are not allowed to remember the questions, just sharing them). The essay questions, however, are allowed to be discussed: AFTER YOU HAVE RECIEVED YOUR SCORES FROM COLLEGE BOARD'S WEBSITE.\r\nSo there you are. Straight from the proverbial horse's mouth.", "Solution_9": "[quote=\"Sly Si\"]Isn't the PSAT given on the alternate day a different version?[/quote]\r\n\r\nApparently not the October 12 and October 15 ones...this really makes me mad if someone makes a perfect on Math by cheating :mad: :mad:", "Solution_10": "[quote=\"bleumoose\"]For those of you who never pay attention during the obligatory announcements on test day:\nYes, it is illegal. Multiple choice questions are not allowed to be \"carried\" from the testing room in any form- including your memory. (of course, this does not mean you are not allowed to remember the questions, just sharing them). The essay questions, however, are allowed to be discussed: AFTER YOU HAVE RECIEVED YOUR SCORES FROM COLLEGE BOARD'S WEBSITE.\nSo there you are. Straight from the proverbial horse's mouth.[/quote]\r\n\r\nYes indeed, in fact, during the SAT, you have to rewright and sign that statement that you're not going to share any info about ANY of the problems, even if no two SATs are teh same on different dates. Although I don't see why not though, I guess Collegeboard would rather have people buy the tests from their practice books.", "Solution_11": "It's because of time differences, partly anyway.", "Solution_12": "[quote=\"Iversonfan2005\"][quote=\"Sly Si\"]Isn't the PSAT given on the alternate day a different version?[/quote]\n\nApparently not the October 12 and October 15 ones...this really makes me mad if someone makes a perfect on Math by cheating :mad: :mad:[/quote]\r\n\r\nThere were different tests.\r\n\r\nForm W and Form S.", "Solution_13": "Blah, there will be a point in life, where they won't be able to cheat... and they'll find themselves very lonely in a dark dark hole. Feel pity, not anger.", "Solution_14": "I don't think that was cheating. It's more like, trying to find out you SAT score as fast as you can by discussing the questions and stuff. Not saying that it's illegal, but I'm pretty sure they aren't trying to cheat.", "Solution_15": "but that could still allow other people who are taking the test later to cheat... the SAT exam recycles MC questions from test to test apparently- same for AP exams", "Solution_16": "[quote=\"bleumoose\"]For those of you who never pay attention during the obligatory announcements on test day:\n[/quote]\r\n\r\nWell, I've [i]never[/i] taken the SAT." } { "Tag": [ "function", "calculus", "integration", "algebra unsolved", "algebra" ], "Problem": "Prove that there is a function $f$ from the set of all natural numbers to itself such that for any natural number $n$, $f(f(n)) = n^{2}$.", "Solution_1": "$f(f(n))=n^2$ (from the given equation)\r\n$f(f(f(n)))=f(n^2)$ (from my previous equation)\r\n$f(f(f(n)))=f(n)^2$ (from the given equation)\r\nSo, $f(n^2)=f(n)^2\\Rightarrow f(n)=\\sqrt{f(n^2)}$", "Solution_2": "I found infinit solution. :)", "Solution_3": "[quote=\"draperp\"]$f(n)=\\sqrt{f(n^2)}$[/quote]\r\nWhy is this useful? How do you know there's a $f: \\mathbb{N}\\to\\mathbb{N}$ with $f(n)=\\sqrt{f(n^2)}$ for every $n$?", "Solution_4": "I derived that equation from the equations shown. The equation [u]must[/u] hold for all $n$ if such a function exists. One use of the equation is to show that when the function is to applied to a perfect square, the result is a perfect square (otherwise $f$ would not be $\\mathbb{N}\\to\\mathbb{N}$.\r\nIt is not surprising that you found an infinite number of functions, Hawk Tiger, since Shobber merely asked us to prove that one exists.\r\nLet every natural number that cannot be written $k^{2^n}$ for natural $k$ and $n$ be a \"M-number.\"\r\nThen, M-number $a$ and M-number $b$ can be uniquely paired as follows:\r\n$f(a^{2^n})=b^{2^n}$ and $f(b^{2^n})=a^{2^{n+1}}$ (for nonnegative integral $n$.)\r\n\r\nThese unique pairs are linked to no others, and it is very easy to see that the solution satisfies.", "Solution_5": "[quote=\"draperp\"]I derived that equation from the equations shown. The equation [u]must[/u] hold for all $n$ if such a function exists. One use of the equation is to show that when the function is to applied to a perfect square, the result is a perfect square (otherwise $f$ would not be $\\mathbb{N}\\to\\mathbb{N}$.\nIt is not surprising that you found an infinite number of functions, Hawk Tiger, since Shobber merely asked us to prove that one exists. I found that for any natural $a$ and $b$,\n$f(a^{2^n})=b^{2^n}$ and $f(b^{2^n})=a^{2^{n+1}}$ works.[/quote]I don't know what you mean clearly,but we can make infint pairs of integer(a,b),where a,b is not square,and every natuaral number in one and only one pair. then as you said,define the function f as $f(a^{2^n})=b^{2^n}$ and $f(b^{2^n})=a^{2^{n+1}}$for all $n>0$\r\nthis one is a fuction satisfying the condition.as the method to make the pairs is infinit,so the fuction is infinit.in fact ,all the fuction can be gained using this method.\r\nbye", "Solution_6": "Yes, you are right. I should have said non-squares; in fact, what I said was wrong since (I don't know what I was thinking). Oh, and I just made-up \"M-number.\" I guess I could have just defined a set or something like that...but it's wrong anyway.\r\nI did make another mistake: $f(1)=1$ is defined by necessity.\r\n(This is one application of $f(n)=\\sqrt{f(n^2)}$.)\r\nThis seems quite unusual to use \"infinite pairing,\" but I am certain that it is mathematically valid." } { "Tag": [ "quadratics" ], "Problem": "The sum of the first $n$ positive even integers is 90. What is the sum of the first $n$ positive odd integers?", "Solution_1": "[hide]the formula to find the sum of the nth even integer if \nn 2 +n\nsoooo.... n 2 +n=90\nso you can see with out using quadratics that n=9\nthe formula to find the sum of the nth odd integer is n 2 \nso put 9 for n and 9 2 =81\nthats the answer[/hide]" } { "Tag": [ "probability", "search", "geometry", "3D geometry", "AMC", "AMC 10", "algebra" ], "Problem": "While I was searching for a certain probability topic, I came across a text-based game thread here in G&FF. Unfortunately, it hasn't been posted on since 2005. So now I'm starting a new one. It follows pretty much the same rules as the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=105878715&t=177263]original[/url].\r\n\r\nWelcome to AoPS Adventure II!\r\n\r\nYou are standing in the doorway of a tall stone building. It looks grand, like a palace.\r\n>explore building\r\n>exit\r\n>check stats", "Solution_1": ">check stats\r\n\r\n(What else would you do?)\r\n\r\nEnergy: 100\r\nAgility: Level 1\r\nSpeed: Level 1\r\nStrength: Level 1\r\nBrains: Level 1\r\nMath(obviously): Level 1\r\n\r\n>exit stats\r\n\r\nYou are standing in the doorway of a tall stone building. It looks grand, like a palace.\r\n>explore building\r\n>exit\r\n>check stats", "Solution_2": ">exit\r\n\r\nYou are outside in a random field with flowers.\r\n\r\n>pick the flowers and frolick happily\r\n>lie down and look at the clouds\r\n>go back to the building", "Solution_3": ">pick the flowers and frolic happily \r\n\r\nyou pick a carnivorous flower, it bites your finger.\r\n\r\n> scream nonstop until something happens\r\n> tear it (and your finger) off\r\n>pry off and give to worst enemy", "Solution_4": "@sailingdog: You do not have enemies yet.\r\n\r\n> scream nonstop until something happens\r\n\r\nA gardener runs up and pours maple syrup on your finger, at which the flower lets go.\r\n\r\nGardener: \"You crazy kid! You aren't supposed to pick the flowers!\"\r\n\r\n*you are in a conversation with the Gardener*\r\n\r\n> \"Sorry, I didn't know.\"\r\n> Cry\r\n> \"Get lost, sir.\"\r\n> \"Sorry, I'll bury it.\"\r\n> run away(exit conversation)", "Solution_5": ">scream nonstop until something happens.\r\n\r\na pretty girl comes up to you\r\n\r\nwhat do you do/say?\r\n\r\n>\"OMFG THIS FLOWER BIT MY FINGER!!!!\"\r\n>\"uh.. *flustered* hi.. umm... don't worry about my finger. it's.. umm... fine.. hehe..\"\r\n>don't talk to her at all", "Solution_6": "already posted something, but I'll go along with yours.\r\n\r\n>\"OMFG THIS FLOWER BIT MY FINGER!!!!\"\r\n\r\nGirl: Sing it a song.\r\nYou: WHAT?!\r\nGirl: Just do it.\r\n\r\nYou sing \"The Star Spangled Banner\", and the flower lets go.\r\n\r\n*You are in conversation with a pretty girl*\r\n\r\n> \"Thanks.\"\r\n> \"Get lost.\"\r\n> \"My name is User0001. What is yours?\"\r\n> Exit conversation", "Solution_7": ">Thanks.\r\nyou continue on you way. when you come to a small village and meet an elderly gentleman who invites you into his home, you\r\n\r\n> punch him in the stomach and continue walking into the village\r\n> say hello and go in for a cup of tea\r\n> ignore him and walk past", "Solution_8": "> say hello and go in for a cup of tea.\r\n\r\nYou chat up a typhoon with the gentleman. Level up brains.\r\n\r\nEnergy: 100 \r\nAgility: Level 1 \r\nSpeed: Level 1 \r\nStrength: Level 1 \r\nBrains: Level 2 \r\nMath(obviously): Level 1 \r\n\r\nYou learn about a horrid tax collector that is practically stealing from people. He comes in right after the gentleman finishes talking.\r\n\r\nTax collector: Taxes.\r\nGentleman: Can't they wait?\r\nTax collector: No, they can't.\r\nGentleman: Well they will have to.\r\nTax collector: Do you want to be locked up?\r\nGentleman: Just give me two more days, and I will come to your place and give you your taxes.\r\nTax collector: No, NOW.\r\nGentleman: I don't have it.\r\nTax collector: Then the guards will be here in a minute.\r\n\r\nThe tax collector leaves. You say \"I can fight them.\", but the gentleman counters: \"You haven't seen these guys. They are on steroids. You should practice on people in the bar.\r\n\r\nThe gentleman leaves the house, and so do you.\r\n\r\nThe gentleman walks down the street into an alley.\r\n\r\n> follow him\r\n> walk away\r\n> go to bar", "Solution_9": ">follow him\r\n\r\nMan asks you why you are following him. You say\r\n\r\n>I have absolutely no clue. (exit conversation)\r\n>I thought you would lead me somewhere safe from the tax collector and his guards.\r\n>Isn't this the way to the bar?", "Solution_10": ">I thought you would lead me somewhere safe from the tax collector and his guards.\r\n\r\nGentleman: You are clever for one so young. All right, you can follow me. If you ask around, you might find a way to the bar.\r\n\r\nThe gentleman walks into a building.\r\n\r\nYou are standing in an alley. An old woman is washing her cats in an unorthodox way: scrubbing them on the side of a barrel.\r\n\r\n>follow gentleman\r\n>walk away\r\n>ask old woman about way to bar.", "Solution_11": ">walk away\r\n\r\nYou stop following the gentleman and go back the way you came. Suddenly, a small rabbit runs past you and jumps into what appears to be a solid wall. There is a man sitting next to the spot where the rabbit jumped through.\r\n\r\n>examine wall\r\n>walk away\r\n>talk to man", "Solution_12": "> examine wall\r\n\r\nupon close examination, you find you are able to walk through the wall. you do. when you get to the other side, you find you are in bunnyland. a bunny walks up to you and asks if you would like to become a bunny.\r\n\r\n> become bunny\r\n> shoot him\r\n> say no thank you and continue to walk into bunnyland\r\n> say no and leave through the wall", "Solution_13": "> become bunny.\r\n\r\nBunnies are fast and agile, but they lack brains. +1 Speed, +1 Agility, -1 Brains.\r\nEnergy: 100 \r\nAgility: Level 2 \r\nSpeed: Level 2 \r\nStrength: Level 1 \r\nBrains: Level 1 \r\nMath(obviously): Level 1\r\n\r\nYou meet a mother bunny who asks you to find her lost child.\r\n> help bunny\r\n> walk away\r\n> randomly scream at the top of your lungs", "Solution_14": ">Help bunny\r\n\r\nThere are woods, a river, and mountains in Bunnyland.\r\n\r\n>check forest\r\n>check near river\r\n>check [i]in[/i] river\r\n>check mountains\r\n>give up on helping", "Solution_15": ">eat a carrot.\r\n\r\nYou turn back into a bunny. While searching around you, you find a bunny Wii.\r\n\r\nYou:\r\n>play the bunny Wii\r\n>wish you were a human so you could play a regular Wii\r\n>continue looking around you", "Solution_16": ">continue looking around.\r\n\r\nYou're blind.\r\n\r\n>play the bunny Wii\r\n>wish you were a human so you can play a real Wii\r\n>wish erabel a happy birthday.", "Solution_17": ">wish erabel a happy birthday\r\n\r\nYou can see!\r\n\r\n>play the bunny Wii \r\n>wish you were a human so you can play a real Wii \r\n>continue looking around", "Solution_18": ">continue looking around\r\n\r\nYou find Erabel.\r\nYou\r\n\r\n>wish him a happy birthday\r\n>eat his cake\r\n>invite him to play wii with you\r\n>eat a carrot\r\nplay wii and ignore him", "Solution_19": ">wish him a happy birthday\r\n\r\nYou're blind again. Oh well.\r\n\r\n>wish him another happy birthday\r\n>wish him two more happy birthdays\r\n>scream \"I'M BLIND!!!!!!!!!!\"\r\n>learn Braille\r\n>eat thirty carrots. They're good for your eyes", "Solution_20": ">eat thirty carrots\r\n\r\n\r\n\r\nEnergy: 120 \r\nAgility: Level 32 \r\nSpeed: Level 33 \r\nStrength: Level 37 \r\nBrains: Level 32 \r\nMath(obviously): Level 36\r\n\r\nYou:\r\n>throw up the thirty carrots\r\n>run around randomly\r\n>scream and run into a wall", "Solution_21": ">scream and run into a wall.\r\n\r\nThe wall falls on top of you. \r\nYou\r\n\r\n>keep screaming\r\n>eat more carrots\r\n>sing happy birthday at the top of your lungs\r\n>try to push the wall off of yourself", "Solution_22": ">sing happy birthday at the top of your lungs.\r\n\r\nYou suffocate.\r\n\r\n>continue\r\n>restart\r\n>quit\r\n>switch game carts", "Solution_23": ">continue\r\n\r\nYou almost suffocate.\r\n\r\n>apply the Heimlich maneuver to nothing\r\n>scream\r\n>clear your throat loudly\r\n>scream more\r\n>eat another carrot\r\n>take the sixth vitamin", "Solution_24": ">take the sixth vitamin\r\n\r\n\r\nSuddenly, you hear a great booming voice.\r\n[b][color=darkblue][size=150]YOU HAVE OFFICIALLY JOINED THE LEAGUE OF THIEVES. YOU MAY NOW STEAL ANYTHING IN THE WORLD. THIS WILL, HOWEVER, DRAIN YOUR ENERGY.[/size][/color][/b]\r\n\r\nYou\r\n\r\n>shout \"who is this\" at the top of your lungs\r\n>steal the voice\r\n>steal a mathbook\r\n>steal a PSP\r\n>steal a crown\r\n>steal a carrot\r\n>sleep", "Solution_25": ">steal the voice\r\n\r\nYou do. -10 energy\r\nEnergy: 110 \r\nAgility: Level 32 \r\nSpeed: Level 33 \r\nStrength: Level 37 \r\nBrains: Level 32 \r\nMath(obviously): Level 36 \r\n\r\nYou say in the voice \"who's there?\" and the same voice replies, \"I am!\"\r\n\r\nYou:\r\n>freak out\r\n>say, \"Who is I\" in the fancy voice\r\n>say, \"Who is I\" in your normal voice\r\n>sleep\r\n>steal something random\r\n>commit suicide", "Solution_26": ">commit suicide...\r\n\r\ngame over..\r\n\r\n>restart?\r\n>play your wii\r\n>quit\r\n> :rotfl: \r\n>restart\r\n>die", "Solution_27": ">quit.\r\n\r\nLock this forum, please. It's getting kinda stupid.", "Solution_28": "[quote=\"erabel\"]Lock this forum, please. It's getting kinda stupid.[/quote]\r\n\r\nWhoa, the whole forum? That's a bit harsh, isn't it?", "Solution_29": "[quote=\"lingomaniac88\"]Whoa, the whole forum? That's a bit harsh, isn't it?[/quote]\r\n\r\nHaha yeah lemme just delete every single topic in history from G&FF :roll: ." } { "Tag": [ "advanced fields", "advanced fields theorems" ], "Problem": "Anyone in possess of a document to learn the basics of coding theory? It seems like a very interesting branch to me I may consider taking next year :)", "Solution_1": "Try looking [url=http://dmoz.org/Science/Math/Applications/Communication_Theory/Coding_Theory/]at the ODP[/url]. You should also look at [url=http://www.dpmms.cam.ac.uk/~twk/]Tom Koerner's home page[/url] for his notes on Coding and Cryptography." } { "Tag": [ "inequalities" ], "Problem": "A) Prove there are infinitely many triples $ (x,y,z)$ of positive numbers such that:\r\n\r\n $ (3\\minus{}\\frac{1}{z}\\minus{}y)((3\\minus{}\\frac{1}{x}\\minus{}z)(3\\minus{}\\frac{1}{y}\\minus{}x)\\leq(\\frac{1}{x}\\plus{}z\\minus{}1)(\\frac{1}{y}\\plus{}x\\minus{}1)(\\frac{1}{z}\\plus{}y\\minus{}1)$\r\n\r\n\r\nB) find the the general form of all (/most, if a I'm wrong) the triples\r\n\r\n\r\n\r\n :)", "Solution_1": "[hide=\"A\"]Consider $ x \\equal{} y \\equal{} z$:\n\n$ (3 \\minus{} x \\minus{} x^{ \\minus{} 1})^3 \\leq (x \\plus{} x^{ \\minus{} 1} \\minus{} 1)^3 \\iff 3 \\minus{} x \\minus{} x^{ \\minus{} 1} \\leq x \\plus{} x^{ \\minus{} 1} \\minus{} 1 \\iff x \\plus{} x^{ \\minus{} 1} \\geq 2$, which is true for all $ x \\in \\mathbb{R}^ \\plus{}$ by AM-GM.\n\nSince this subset of the solution set is infinitely large, the solution set itself must also be infinitely large.\n\nThis also implies infinitely many integer solutions.[/hide]\n\n[hide=\"Stuff about B\"]\nThere is no \"general form,\" there are way too many solutions.\n\nWe have that $ \\minus{} (x \\plus{} 1/y \\minus{} 3)(y \\plus{} 1/z \\minus{} 3)(z \\plus{} 1/x \\minus{} 3) \\leq (x \\plus{} 1/y \\minus{} 1)(y \\plus{} 1/z \\minus{} 1)(z \\plus{} 1/x \\minus{} 1)$\n\nFirst of all, if all 6 terms are positive, then we win. That is, if $ x \\plus{} 1/y \\geq 3$, etc. This is easily satisfied if all three variables are $ \\geq \\frac {3 \\plus{} \\sqrt {5}}{2}$ or between $ 0$ and $ \\geq \\frac {3 \\minus{} \\sqrt {5}}{2}$, but there are other solutions as well.\n\nWe also auto-win if two of the terms on the right side and left side are negative. Note that these must be corresponding terms.\nWlog, we require that $ x \\plus{} 1/y > 3$, but $ y \\plus{} 1/z < 1$ and $ z \\plus{} 1/x < 1$. However, this requires that $ z < 1$ and $ 1/z < 1$, which violates that $ z$ is positive, so this is not possible.\n\nThen there is the case in which the three terms on the left are negative, and none or two of the terms on the right are negative, and the right side still is greater than the left side.[/hide]", "Solution_2": "mmm ok I understand it very well...\r\n\r\nthis is the storie:\r\n\r\nwhat I derived was this:\r\nFor \r\n $ x\\equal{}\\frac{p}{q}$, $ y\\equal{}\\frac{q}{r}$ and $ z\\equal{}\\frac{r}{p}$\r\n\r\nwhere $ p$, $ q$ and $ r$ are sides of a triangle, we have that the inequality holds.\r\n\r\n\r\nWhat I wanted to know is if this are the only triples or there are more, but considering what you sayed about $ x\\equal{}y\\equal{}z$, I'm considering that the triples are $ (ax, ay, az)$ with $ a$ positive and $ x, y, z$ as I say... :maybe: \r\n\r\nCan someone find a counterexample or prove it, I will be also trying but help is always welcome.\r\n\r\n\r\n :)" } { "Tag": [ "AwesomeMath", "summer program" ], "Problem": "On the website it says that the early registration deadline is February 14th, but if you print out the Application test it says Februrary 9th. Which one is it?\r\n\r\nTYiA.", "Solution_1": "The 14th; it was originally the 9th but the date was extended.", "Solution_2": "Yes\r\nAnd if you see its in [color=red]Red[/color] in AwesomeMath Site, and its written in this forum also. You should be happy :roll:" } { "Tag": [ "geometry", "incenter", "ratio", "Ross Mathematics Program", "angle bisector", "projective geometry", "geometry proposed" ], "Problem": "Let $ (I) $ be the incircle of the triangle $ ABC $ . $ (I) $ tangents $ BC $ at $ D $ . $ H $ is the feet of the perpendicular from A to BC . The excenter of the triangle $ ABC $ tangents BC at $ P $ . $ AI $ intersects $ BC $ at $ K $ . Prove that :\r\n 1) $ (HKDP) = -1 $\r\n 2) $ IP $ bisects $ AH $", "Solution_1": "The two are actually equivalent. You can prove the second one by Menelaus, for example (the computations are easy to carry on), and then the first one follows.", "Solution_2": "The problem is really easy. I slightly rewrite it:\r\n\r\n[color=blue][b]Problem.[/b] Let ABC be a triangle, and H the foot of its altitude to the side BC. The incircle of triangle ABC touches its side BC at a point D, and the A-excircle touches the side BC at a point P. Finally, the angle bisector of the angle CAB intersects the side BC of triangle ABC at a point K.\n\n[b](1)[/b] Prove that the points D and P divide the segment HK harmonically.\n[b](2)[/b] Prove that the line IP bisects the segment AH.[/color]\r\n\r\n[i]Solution.[/i] [b](1)[/b] Let I be the center of the incircle of triangle ABC, and let $I_a$ be the center of the A-excircle. Then, since the points D and P are the points where these circles touch the side BC, we have $ID\\perp BC$ and $I_aP\\perp BC$. Also, since the line AH is the altitude of triangle ABC to the side BC, we have $AH\\perp BC$. Thus, $ID\\parallel I_aP\\parallel AH$.\r\n\r\nThe line AK, being the angle bisector of the angle CAB, passes through the incenter I and the A-excenter $I_a$ of triangle ABC. The incenter I of triangle ABC also lies on the angle bisector of the angle ABC, which is, of course, simultaneously the angle bisector of the angle ABK; thus, since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we obtain, by applying this to the triangle ABK, that $\\frac{AI}{IK}=\\frac{AB}{BK}$. Similarly, the A-excenter $I_a$ of triangle ABC lies on the external angle bisector of the angle ABC, which is, of course, simultaneously the external angle bisector of the angle ABK; thus, since the external angle bisector of an angle of a triangle divides the opposite side externally in the ratio of the adjacent sides, we obtain, by applying this to the triangle ABK, that $\\frac{AI_a}{I_aK}=-\\frac{AB}{BK}$. Hence, $\\frac{AI_a}{I_aK}=-\\frac{AI}{IK}$. But since $ID\\parallel I_aP\\parallel AH$, Thales yields $\\frac{AI_a}{I_aK}=\\frac{HP}{PK}$ and $\\frac{AI}{IK}=\\frac{HD}{DK}$, so that this becomes $\\frac{HP}{PK}=-\\frac{HD}{DK}$. Thus, the points D and P divide the segment HK harmonically. Part [b](1)[/b] of the problem is solved.\r\n\r\n[b](2)[/b] Since ID || AH, the lines ID and AH intersect at an infinite point. Denote this infinite point by $H_{\\infty}$. Also, let $H_1$ be the point of intersection of the lines IP and AH. Then, since the points H, K, D and P lie on one line, and the points H, A, $H_{\\infty}$ and $H_1$ are the projections of these points onto the line AH from the point I, by the invariance of the cross-ratio under central projection we see that $\\frac{HH_1}{H_1A}:\\frac{HH_{\\infty}}{H_{\\infty}A}=\\frac{HP}{PK}:\\frac{HD}{DK}$. But since $\\frac{HP}{PK}=-\\frac{HD}{DK}$, we have $\\frac{HP}{PK}:\\frac{HD}{DK}=-1$, and thus $\\frac{HH_1}{H_1A}:\\frac{HH_{\\infty}}{H_{\\infty}A}=-1$, so that $\\frac{HH_1}{H_1A}=-\\frac{HH_{\\infty}}{H_{\\infty}A}$. In other words, the point $H_1$ is the harmonic conjugate of the point $H_{\\infty}$ with respect to the segment AH. But the point $H_{\\infty}$ is an infinite point, and the harmonic conjugate of an infinite point with respect to a segment is the midpoint of the segment. Thus, the point $H_1$ is the midpoint of the segment AH. In other words, the line IP bisects the segment AH, and this solves part [b](2)[/b] of the problem.\r\n\r\nBy the way, an alternative solution of part [b](2)[/b] of the problem, without using cross-ratios, can be found using the results from http://www.mathlinks.ro/Forum/viewtopic.php?t=5811 . I am leaving the details to the reader, who can also look up the alternative solution in Ross Honsberger, [i]Episodes in Nineteenth and Twentieth Century Euclidean Geometry[/i], MAA 1995, chapter 3, \u00a73.\r\n\r\n Darij" } { "Tag": [ "parameterization" ], "Problem": "How to draw triangle using compass and straightedge\r\n\r\nthe parameter=12 and the measure of 40,60,80", "Solution_1": "[quote=\"tony123\"]How to draw triangle using compass and straightedge\n\nthe parameter=12 and the measure of 40,60,80[/quote]\r\n\r\nIf you can construct such a triangle then you will be famous. :rotfl: \r\n\r\nIf you can construct an angle of $ 40^{\\circ}$ then you will be able to costruct a 9-gon using only compass and straightedge, which was proved to be impossible.\r\n\r\n[url]http://en.wikipedia.org/wiki/Constructible_polygon[/url]" } { "Tag": [ "Euler", "trigonometry", "LaTeX" ], "Problem": "let $f(x) = arctan\\[(\\sqrt{{x^2}+1}-x)$ for all r\u00e9als \r\nProve that $f(-x)=\\frac{\\pi}{2}-f(x)$ for all r\u00e9als\r\n$\\mathrm{Fermat -Euler}$", "Solution_1": "[hide]Let $Y = \\frac{\\pi}{2} - f(x)$ we have...\n\n$\\tan Y = \\tan\\left[ \\frac{\\pi}{2} - f(x) \\right]$\n$\\tan Y = \\cot\\left[ f(x) \\right]$\n$\\tan Y = \\cot\\left[\\arctan(\\sqrt{x^2 + 1} + x)\\right]$\n$\\tan Y = \\frac{1}{\\sqrt{x^2 + 1} + x}$\n$\\tan Y = \\frac{\\sqrt{x^2 + 1} - x}{(\\sqrt{x^2 + 1})^2 - x^2}$\n$\\tan Y = \\sqrt{x^2 + 1} - x$\n$Y = \\arctan(\\sqrt{x^2 + 1} - x)$\n$Y = \\arctan(\\sqrt{(-x)^2 + 1} + (-x))$\n$Y = f(-x)$\n$\\frac{\\pi}{2} - f(x) = f(-x)$[/hide]", "Solution_2": "[quote=\"Bruno Bonagura\"]\n$Y = \\arctan(\\sqrt{x^2 + 1} - x)$\n$Y = \\arctan(\\sqrt{(-x)^2 + 1} + (-x))$\n$Y = f(-x)$\n$\\frac{\\pi}{2} - f(x) = f(-x)$[/quote]\r\nbut $f(-x)=\\arctan(\\sqrt{(-x)^2 + 1} - (-x))$", "Solution_3": "[color=red]the correction of bruno's solution [/color]\r\n Let $Y = \\frac{\\pi}{2} - f(x)$ we have...\r\n$\\tan Y = \\tan\\left[ \\frac{\\pi}{2} - f(x) \\right]$\r\n$\\tan Y = \\cot\\left[ f(x) \\right]$\r\n$\\tan Y = \\cot\\left[\\arctan(\\sqrt{x^2 + 1} - x)\\right]$\r\n$\\tan Y = \\frac{1}{\\sqrt{x^2 + 1} - x}$\r\n$\\tan Y = \\frac{\\sqrt{x^2 + 1} + x}{(\\sqrt{x^2 + 1})^2 - x^2}$\r\n$\\tan Y = \\sqrt{x^2 + 1} + x$\r\n$Y = \\arctan(\\sqrt{x^2 + 1} +x)$\r\n$Y = \\arctan(\\sqrt{(-x)^2 + 1} - (-x))$\r\n$Y = f(-x)$\r\n$\\frac{\\pi}{2} - f(x) = f(-x)$", "Solution_4": "Sorry by the miscalculating...", "Solution_5": "By the way, $\\LaTeX$ doesn't work in titles. Sorry." } { "Tag": [ "inequalities", "inequalities proposed", "111" ], "Problem": "$ a,b,c > 0,t \\in N$,Prove that\r\n$ {\\frac {{a}^{t\\minus{}1}}{{b}^{t\\minus{}1}}}\\plus{}{\\frac {{b}^{t\\minus{}1}}{{c}^{t\\minus{}1}}}\\plus{}{\\frac {\r\n{c}^{t\\minus{}1}}{{a}^{t\\minus{}1}}}\\leq {\\frac {{a}^{t}}{{b}^{t}}}\\plus{}{\\frac {{b}^{t}}\r\n{{c}^{t}}}\\plus{}{\\frac {{c}^{t}}{{a}^{t}}}$\r\nBQ", "Solution_1": "a^2/b^2+b^2/c^2+c^2/a^2>= 1/27*(a/b+b/c+c/a)*((a+b)/(b+c)+(b+c)/(c+a)+(c+a)/(a+b))*((b+2*c+a)/(c+2*a+b)+(c+2*a+b)/(a+2*b+c)+(a+2*b+c)/(b+2*c+a))*((2*c+3*a+3*b)/(2*a+3*b+3*c)+(2*a+3*b+3*c)/(2*b+3*c+3*a)+(2*b+3*c+3*a)/(2*c+3*a+3*b))\r\nholds.\r\nBQ", "Solution_2": "$ a,b,c > 0,t \\in R$,Prove that\r\n${ f(t) = \\frac {{a}^{t}}{{b}^{t}}} + {\\frac {{b}^{t}} {{c}^{t}}} + {\\frac {{c}^{t}}{{a}^{t}}}$,\r\n\r\nif\r\n$ t_1\\geq t_2$,\r\nthen\r\n$ f(t_1) \\le f(t_2).$\r\n\r\n\r\nBQ", "Solution_3": "(1)\r\n$ \\frac {x^2}{y^2} + \\frac {y^2}{z^2} + \\frac {z^2}{x^2} \\geq \\frac {x}{y} + \\frac {y}{z} + \\frac {z}{x}.$\r\n\r\n(2)\r\n$ \\frac {x^3}{y^3} + \\frac {y^3}{z^3} + \\frac {z^3}{x^3} \\geq \\frac {1}{3}(\\frac {x}{y} + \\frac {y}{z} + \\frac {z}{x})(\\frac {x^2}{y^2} + \\frac {y^2}\\frac {z^2} + \\frac {z^2}{x^2}).$\r\n\r\n(3)\r\n$ \\frac {x^4}{y^4} + \\frac {y^4}{z^4} + \\frac {z^4}{x^4} \\geq \\frac {1}{3}(\\frac {x}{y} + \\frac {y}{z} + \\frac {z}{x})(\\frac {x^3}{y^3} + \\frac {y^3}{z^3} + \\frac {z^3}{x^3}).$\r\n\r\n\r\n(4)\r\n$ \\frac {x^5}{y^5} + \\frac {y^5}{z^5} + \\frac {z^5}{x^5} \\le \\frac {3}{3^4}(\\frac {x}{y} + \\frac {y}{z} + \\frac {z}{x})(\\frac {x^2}{y^2} + \\frac {y^2}{z^2} + \\frac {z^2}{x^2})(\\frac {x^3}{y^3} + \\frac {y^3}{z^3} + \\frac {z^3}{x^3})(\\frac {x^4}{y^4} + \\frac {y^4}{z^4} + \\frac {z^4}{x^4})$\r\nBQ", "Solution_4": "(5)\r\n$ {\\frac {{x}^{6}}{{y}^{6}}}\\plus{}{\\frac {{y}^{6}}{{z}^{6}}}\\plus{}{\\frac {{z}^{6}}\r\n{{x}^{6}}}\\leq {\\frac {1}{81}}\\, \\left( {\\frac {x}{y}}\\plus{}{\\frac {y}{z}}\\plus{}\r\n{\\frac {z}{x}} \\right) \\left( {\\frac {{x}^{2}}{{y}^{2}}}\\plus{}{\\frac {{y}^\r\n{2}}{{z}^{2}}}\\plus{}{\\frac {{z}^{2}}{{x}^{2}}} \\right) \\left( {\\frac {{x}^\r\n{3}}{{y}^{3}}}\\plus{}{\\frac {{y}^{3}}{{z}^{3}}}\\plus{}{\\frac {{z}^{3}}{{x}^{3}}}\r\n \\right) \\left( {\\frac {{x}^{4}}{{y}^{4}}}\\plus{}{\\frac {{y}^{4}}{{z}^{4}}}\r\n\\plus{}{\\frac {{z}^{4}}{{x}^{4}}} \\right) \\left( {\\frac {{x}^{5}}{{y}^{5}}}\r\n\\plus{}{\\frac {{y}^{5}}{{z}^{5}}}\\plus{}{\\frac {{z}^{5}}{{x}^{5}}} \\right)$", "Solution_5": "[quote=\"xzlbq\"]$ a,b,c > 0,t \\in R$,Prove that\n${ f(t) = \\frac {{a}^{t}}{{b}^{t}}} + {\\frac {{b}^{t}} {{c}^{t}}} + {\\frac {{c}^{t}}{{a}^{t}}}$,\n\nif\n$ t_1\\geq t_2$,\nthen\n$ f(t_1) \\le f(t_2).$\n\n\nBQ[/quote]\r\n\r\nnot hold.\r\n\r\nwhen $ t_1 = 1, t_2 = - 1$, ineq obvious not hold.\r\n\r\nbut I think if $ t_1\\geq t_2 \\ge 0$, will have $ f(t_1) \\ge f(t_2).$", "Solution_6": "I think:\r\n\r\nLet $ x,y,z > 0,t \\in N,f(t) \\equal{} \\frac {x^t}{y^t} \\plus{} \\frac {y^t}{z^t} \\plus{} \\frac {z^t}{x^t},$\r\nthen:\r\n$ f(t) \\le \\frac {3}{3^r}\\prod_{i \\equal{} 1}^r{f(i)}$\r\ninside,$ t \\geq 5).$\r\n\r\nBQ", "Solution_7": "and I think agin:\r\n\r\n\r\nLet $ x,y,z > 0,t \\in N,f(t) \\equal{} \\frac {x^t}{y^t} \\plus{} \\frac {y^t}{z^t} \\plus{} \\frac {z^t}{x^t},$\r\nthen:\r\n$ f(n) \\geq \\frac{1}{3}f(m)f(n\\minus{}m).$\r\ninside,$ m \\le [\\frac{n}{2}]$\r\n\r\nBQ.", "Solution_8": "[quote=\"kuing\"][quote=\"xzlbq\"]$ a,b,c > 0,t \\in R$,Prove that\n${ f(t) = \\frac {{a}^{t}}{{b}^{t}}} + {\\frac {{b}^{t}} {{c}^{t}}} + {\\frac {{c}^{t}}{{a}^{t}}}$,\n\nif\n$ t_1\\geq t_2$,\nthen\n$ f(t_1) \\le f(t_2).$\n\n\nBQ[/quote]\n\nnot hold.\n\nwhen $ t_1 = 1, t_2 = - 1$, ineq obvious not hold.\n\nbut I think if $ t_1\\geq t_2 \\ge 0$, will have $ f(t_1) \\ge f(t_2).$[/quote]\r\n\r\nI got a proof of this ineq for $ n$ variable (all variable are positive and cyclic, and $ t \\ge 0$). :lol:", "Solution_9": "[quote=\"kuing\"][quote=\"xzlbq\"]$ a,b,c > 0,t \\in R$,Prove that\n${ f(t) = \\frac {{a}^{t}}{{b}^{t}}} + {\\frac {{b}^{t}} {{c}^{t}}} + {\\frac {{c}^{t}}{{a}^{t}}}$,\n\nif\n$ t_1\\geq t_2$,\nthen\n$ f(t_1) \\le f(t_2).$\n\n\nBQ[/quote]\n\nnot hold.\n\nwhen $ t_1 = 1, t_2 = - 1$, ineq obvious not hold.\n\nbut I think if $ t_1\\geq t_2 \\ge 0$, will have $ f(t_1) \\ge f(t_2).$[/quote]\r\n\r\n$ t \\in R^{+}$", "Solution_10": "4-variables:\r\nLet $ x_1,x_2,x_3,x_4>0,s \\in N,$\r\n(1)\r\n $ {\\frac {{{\\it x_1}}^{s} }{{{\\it x_2}}^{s}}} \\plus{} {\\frac {{{\\it x_2}}^{s}}{{{\\it x_3}}^{s}}} \\plus{} {\\frac {{{\\it x_3}}^{s}}{{{\\it x_4}}^{s}}} \\plus{} {\\frac {{{\\it x_4}}^{s}}{{{\\it x_1}}^{s}} } \\geq {\\frac {{{\\it x_1}}^{s \\minus{} 1}}{{{\\it x_2}}^{s \\minus{} 1}}} \\plus{} {\\frac {{{\\it x_2}}^{s \\minus{} 1}} {{{\\it x_3}}^{s \\minus{} 1}}} \\plus{} {\\frac {{{\\it x_3}}^{s \\minus{} 1}}{{{\\it x_4}}^{s \\minus{} 1}}} \\plus{} {\\frac {{{\\it x_4}}^{s \\minus{} 1}}{{{\\it x_1}}^{s \\minus{} 1}}}$\r\n\r\n(2)\r\n$ {\\frac {{{\\it x_1}}^{m}}{{{\\it x_2}}^{m}}} \\plus{} {\\frac {{{\\it x_2}}^{m}}{{{\\it x_3}}^{m}}} \\plus{} {\\frac {{{\\it x_3}}^{m}}{{{\\it x_4}}^{m}}} \\plus{} {\\frac {{{\\it x_4}}^{m}}{{{\\it x_1}}^{m}}}\\geq \\frac {1}{4}\\, \\left( {\\frac {{{\\it x_1}}^{n}}{{{\\it x_2}}^{n}}} \\plus{} {\\frac {{{\\it x_2}}^{n}}{{{\\it x_3}}^{n}}} \\plus{} {\\frac {{{\\it x_3}} ^{n}}{{{\\it x_4}}^{n}}} \\plus{} {\\frac {{{\\it x_4}}^{n}}{{{\\it x_1}}^{n}}} \\right) \\left( {\\frac {{{\\it x_1}}^{m \\minus{} n}}{{{\\it x_2}}^{m \\minus{} n}}} \\plus{} {\\frac { {{\\it x_2}}^{m \\minus{} n}}{{{\\it x_3}}^{m \\minus{} n}}} \\plus{} {\\frac {{{\\it x_3}}^{m \\minus{} n}}{{{\\it x_4}}^{m \\minus{} n}}} \\plus{} {\\frac {{{\\it x_4}}^{m \\minus{} n}}{{{\\it x_1}}^{m \\minus{} n}}} \\right)$,\r\n$ n \\le [\\frac{m}{2}]$.", "Solution_11": "Let $ x,y,z > 0,f_1(x,y,z) \\equal{} \\frac {x}{y} \\plus{} \\frac {y}{z} \\plus{} \\frac {z}{x},$\r\n$ f_{i \\plus{} 1}(x,y,z) \\equal{} f_i(y \\plus{} z,z \\plus{} x,x \\plus{} y)(i \\equal{} 1,2,...r,r\\geq 1)$,\r\nfor \r\n\r\n$ \\frac {x^2}{y^2} \\plus{} \\frac {y^2}{z^2} \\plus{} \\frac {z^2}{x^2}\\geq \\frac {3}{3^t}\\prod_{i \\equal{} 1}^r{f_i(x,y,z)}.$\r\n(1)r=5,\r\n<=>\r\n$ \\frac {x^2}{y^2} \\plus{} \\frac {y^2}{z^2} \\plus{} \\frac {z^2}{x^2}\\geq \\frac {3}{3^5}(\\frac {x}{y} \\plus{} \\frac {y}{z} \\plus{} \\frac {z}{x})(\\frac {y \\plus{} z}{x \\plus{} z} \\plus{} \\frac {x \\plus{} z}{x \\plus{} y} \\plus{} \\frac {x \\plus{} y}{y \\plus{} z})(\\frac {2x \\plus{} z \\plus{} y}{2y \\plus{} z \\plus{} x} \\plus{} \\frac {2y \\plus{} z \\plus{} x}{y \\plus{} 2z \\plus{} x} \\plus{} \\frac {y \\plus{} 2z \\plus{} x}{2x \\plus{} z \\plus{} y})(\\frac {3y \\plus{} 3z \\plus{} 2x}{3x \\plus{} 3z \\plus{} 2y} \\plus{} \\frac {3x \\plus{} 3z \\plus{} 2y}{3x \\plus{} 2z \\plus{} 3y} \\plus{} \\frac {3x \\plus{} 2z \\plus{} 3y}{3y \\plus{} 3z \\plus{} 2x})(\\frac {(6x \\plus{} 5z \\plus{} 5y}{6y \\plus{} 5z \\plus{} 5x} \\plus{} \\frac {6y \\plus{} 5z \\plus{} 5x}{5y \\plus{} 6z \\plus{} 5x} \\plus{} \\frac {5y \\plus{} 6z \\plus{} 5x}{6x \\plus{} 5z \\plus{} 5y}).$\r\nholds.\r\n\r\n\r\n(2)r=25,\r\n<=>\r\n$ \\frac {x^2}{y^2} \\plus{} \\frac {y^2}{z^2} \\plus{} \\frac {z^2}{x^2}\\geq \\frac {3}{3^{25}}(\\frac {x}{y} \\plus{} \\frac {y}{z} \\plus{} \\frac {z}{x})...(\\frac {5592406x \\plus{} 5592405z \\plus{} 5592405y}{5592406y \\plus{} 5592405z \\plus{} 5592405x} \\plus{} \\frac {5592406y \\plus{} 5592405z \\plus{} 5592405x}{5592405y \\plus{} 5592406z \\plus{} 5592405x} \\plus{} \\frac {5592405y \\plus{} 5592406z \\plus{} 5592405x}{5592406x \\plus{} 5592405z \\plus{} 5592405y}).$\r\nholds.\r\n\r\n(3)I think :$ r{\\to}\\infty$,then.\r\n$ \\frac {x^2}{y^2} \\plus{} \\frac {y^2}{z^2} \\plus{} \\frac {z^2}{x^2} \\equal{} \\frac {3}{3^t}\\prod_{i \\equal{} 1}^{\\infty}{f_i(x,y,z)}.$\r\n\r\nBQ\r\nNote:\r\n[size=200]Maybe (3) does not holds,But I still Do not believe that My eyes![/size]\r\n\r\nr=26 holds;", "Solution_12": "4-variables\r\n\r\n$ \\frac{x_1^2}{x_2^2}\\plus{}\\frac{x_2^2}{x_3^2}\\plus{}\\frac{x_3^2}{x_4^2}\\plus{}\\frac{x_4^2}{x_1^2} \\geq \\frac{x_1}{x_2}\\plus{}\\frac{x_2}{x_3}\\plus{}\\frac{x_3}{x_4}\\plus{}\\frac{x_4}{x_1}.$", "Solution_13": "Let $ x,y,z > 0,a_1 \\equal{} f_1(x,y,z) \\equal{} \\frac {x}{y} \\plus{} \\frac {y}{z} \\plus{} \\frac {z}{x},$\r\n$ a_{i \\plus{} 1} \\equal{} f_{i \\plus{} 1}(x,y,z) \\equal{} f_i(y \\plus{} z,z \\plus{} x,x \\plus{} y)(i \\equal{} 1,2,...r,r\\geq 1)$,\r\n\r\n(1)\r\n\r\n$ \\frac {x^3}{y^3} \\plus{} \\frac {y^3}{z^3} \\plus{} \\frac {z^3}{x^3} \\geq \\frac {3}{3^{3 \\plus{} 1}}a_1^3a_2$\r\n<=>\r\n$ \\frac {x^3}{y^3} \\plus{} \\frac {y^3}{z^3} \\plus{} \\frac {z^3}{x^3} \\geq \\frac {1}{3^3}(\\frac {x}{y} \\plus{} \\frac {y}{z} \\plus{} \\frac {z}{x})^3(\\frac {y \\plus{} z}{z \\plus{} x} \\plus{} \\frac {z \\plus{} x}{x \\plus{} y} \\plus{} \\frac {x \\plus{} y}{y \\plus{} z})$\r\ndoes not holds.\r\n\r\n(2)\r\nBut for \r\n\r\n$ \\frac {x^3}{y^3} \\plus{} \\frac {y^3}{z^3} \\plus{} \\frac {z^3}{x^3} \\geq \\frac {3}{3^3}a_1^3\\prod_{i \\equal{} 3}^r{a_i}$.\r\n\r\n(2.1) r=6\r\n<=>\r\n$ \\frac {x^3}{y^3} \\plus{} \\frac {y^3}{z^3} \\plus{} \\frac {z^3}{x^3} \\geq \\frac {3}{3^7}a_1^3a_3a_4a_5a_6$\r\n<=>\r\n$ \\frac {x^3}{y^3} \\plus{} \\frac {y^3}{z^3} \\plus{} \\frac {z^3}{x^3} \\geq \\frac {1}{729}(\\frac {x}{y} \\plus{} \\frac {y}{z} \\plus{} \\frac {z}{x})^3(\\frac {z \\plus{} 2x \\plus{} y}{x \\plus{} 2y \\plus{} z} \\plus{} \\frac {(x \\plus{} 2y \\plus{} z}{y \\plus{} 2z \\plus{} x} \\plus{} \\frac {y \\plus{} 2z \\plus{} x}{z \\plus{} 2x \\plus{} y})...(\\frac {11z \\plus{} 10x \\plus{} 11y}{11x \\plus{} 10y \\plus{} 11z} \\plus{} \\frac {11x \\plus{} 10y \\plus{} 11z}{11y \\plus{} 10z \\plus{} 11x} \\plus{} \\frac {11y \\plus{} 10z \\plus{} 11x}{11z \\plus{} 10x \\plus{} 11y})$\r\nholds.\r\n\r\n(2.2) r=15\r\n<=>\r\n$ \\frac {x^3}{y^3} \\plus{} \\frac {y^3}{z^3} \\plus{} \\frac {z^3}{x^3} \\geq \\frac {3}{3^{3 \\plus{} 15 \\minus{} 3 \\plus{} 1}}a_1^3a_3a_4a_5a_6a_7a_8a_9a_{10}a_{11}a_{12}a_{13}a_{14}a_{15}$\r\nholds.\r\nBQ\r\nNote:r=20 holds too.\r\n\r\n\r\n\r\nVolume Level inequality:\r\n$ E(a_1)>E(a_2)>....>E(a_n)$\r\nBQ", "Solution_14": "Let $ x,y,z > 0,a_1 \\equal{} f_1(x,y,z) \\equal{} \\frac {x}{y} \\plus{} \\frac {y}{z} \\plus{} \\frac {z}{x},$\r\n$ a_{i \\plus{} 1} \\equal{} f_{i \\plus{} 1}(x,y,z) \\equal{} f_i(y \\plus{} z,z \\plus{} x,x \\plus{} y)(i \\equal{} 1,2,...r,r\\geq 1)$,\r\n\r\n1.\r\n$ 1/3\\, \\left( {\\frac {{x}^{2}}{{y}^{2}}} \\plus{} {\\frac {{y}^{2}}{{z}^{2}}} \\plus{} {\\frac {{z}^{2}}{{x}^{2}}} \\right) \\left( {\\frac {x}{y}} \\plus{} {\\frac {y}{z} } \\plus{} {\\frac {z}{x}} \\right) \\leq {\\frac {{x}^{3}}{{y}^{3}}} \\plus{} {\\frac {{y}^{ 3}}{{z}^{3}}} \\plus{} {\\frac {{z}^{3}}{{x}^{3}}}$\r\nholds.\r\n\r\n2.\r\n$ \\frac {x^3}{y^3} \\plus{} \\frac {y^3}{z^3} \\plus{} \\frac {z^3}{x^3}\\geq \\frac {1}{9}(\\frac {x^2}{y^2} \\plus{} \\frac {y^2}{z^2} \\plus{} \\frac {z^2}{x^2})a_1a_2$\r\ndoes not holds.\r\n\r\n3.\r\n$ \\frac {x^3}{y^3} \\plus{} \\frac {y^3}{z^3} \\plus{} \\frac {z^3}{x^3}\\geq \\frac {1}{3}(\\frac {x^2}{y^2} \\plus{} \\frac {y^2}{z^2} \\plus{} \\frac {z^2}{x^2})a_1\\frac {1}{3^{20 \\minus{} 3 \\plus{} 1}}\\prod_{i \\equal{} 3}^{20}{a_i}$\r\nholds.\r\nBQ\r\n\r\nBQ", "Solution_15": "if\r\n$ a_1 \\geq \\frac {3}{3^t}a_3^t$\r\nholds,then\r\n$ t_{max} \\equal{} 12.$\r\nBQ", "Solution_16": "We have nice:\r\n$ a_1 \\minus{} 3 \\geq \\sum_{i \\equal{} 2}^r{(a_i \\minus{} 3)}$\r\nr=10,\r\nI think :$ r{\\to}\\infty$\r\n\r\nBQ", "Solution_17": "Now it seems,We should also think of some more wide.\r\nLet $ x,y,z > 0,a_1 \\equal{} f_1(x,y,z) \\equal{} g(x,y,z)\\geq0,g(x,y,z)$is a aymmetric\r\n$ a_{i \\plus{} 1} \\equal{} f_{i \\plus{} 1}(x,y,z) \\equal{} f_i(y \\plus{} z,z \\plus{} x,x \\plus{} y)(i \\equal{} 1,2,...r,r\\geq 2)$,\r\n\r\nthen:\r\n\r\n$ a_1\\geq \\sum_{i \\equal{} 2}^r{a_i}(r\\geq2)$.\r\nBQ\r\n\r\nNow give some examples\uff1a\r\n\r\neg1:\r\nLet$ r \\equal{} 5,g(x,y,z) \\equal{} \\frac {x}{y \\plus{} z} \\plus{} \\frac {y}{z \\plus{} x} \\plus{} \\frac {z}{x \\plus{} y} \\minus{} \\frac {3}{2}.$\r\nthen\r\n$ a_1 \\plus{} 9 \\minus{} a_2 \\minus{} a_3 \\minus{} a_4 \\minus{} a_5\\geq0.$,\r\ninside,\r\n$ (a_1, a_2, a_3, a_4, a_5) \\equal{} (\\frac {x}{y \\plus{} z} \\plus{} \\frac {y}{z \\plus{} x} \\plus{} \\frac {z}{x \\plus{} y}, \\frac {y \\plus{} z}{z \\plus{} 2x \\plus{} y} \\plus{} \\frac {z \\plus{} x}{x \\plus{} 2y \\plus{} z} \\plus{} \\frac {x \\plus{} y}{y \\plus{} 2z \\plus{} x}, \\frac {z \\plus{} 2x \\plus{} y}{2x \\plus{} 3y \\plus{} 3z} \\plus{} \\frac {x \\plus{} 2y \\plus{} z}{2y \\plus{} 3z \\plus{} 3x} \\plus{} \\frac {y \\plus{} 2z \\plus{} x}{2z \\plus{} 3x \\plus{} 3y}, \\frac {2x \\plus{} 3y \\plus{} 3z}{5y \\plus{} 5z \\plus{} 6x} \\plus{} \\frac {2y \\plus{} 3z \\plus{} 3x}{5z \\plus{} 5x \\plus{} 6y} \\plus{} \\frac {2z \\plus{} 3x \\plus{} 3y}{5x \\plus{} 5y \\plus{} 6z}, \\frac {5y \\plus{} 5z \\plus{} 6x}{11z \\plus{} 10x \\plus{} 11y} \\plus{} \\frac {5z \\plus{} 5x \\plus{} 6y}{11x \\plus{} 10y \\plus{} 11z} \\plus{} \\frac {5x \\plus{} 5y \\plus{} 6z}{11y \\plus{} 10z \\plus{} 11x}).$\r\n\r\nin factor,We have:\r\n\r\n$ a_1\\geq \\sum_{i \\equal{} 2}^{12}{a_i}$ holds.", "Solution_18": "[quote=\"xzlbq\"]Now it seems,We should also think of some more wide.\nLet $ x,y,z > 0,a_1 \\equal{} f_1(x,y,z) \\equal{} g(x,y,z)\\geq0,g(x,y,z)$is a aymmetric\n$ a_{i \\plus{} 1} \\equal{} f_{i \\plus{} 1}(x,y,z) \\equal{} f_i(y \\plus{} z,z \\plus{} x,x \\plus{} y)(i \\equal{} 1,2,...r,r\\geq 2)$,\n\nthen:\n\n$ a_1\\geq \\sum_{i \\equal{} 2}^r{a_i}(r\\geq2)$.\nBQ\n\nNow give some examples\uff1a\n\neg1:\nLet$ r \\equal{} 5,g(x,y,z) \\equal{} \\frac {x}{y \\plus{} z} \\plus{} \\frac {y}{z \\plus{} x} \\plus{} \\frac {z}{x \\plus{} y} \\minus{} \\frac {3}{2}.$\nthen\n$ a_1 \\plus{} 9 \\minus{} a_2 \\minus{} a_3 \\minus{} a_4 \\minus{} a_5\\geq0.$,\ninside,\n$ (a_1, a_2, a_3, a_4, a_5) \\equal{} (\\frac {x}{y \\plus{} z} \\plus{} \\frac {y}{z \\plus{} x} \\plus{} \\frac {z}{x \\plus{} y}, \\frac {y \\plus{} z}{z \\plus{} 2x \\plus{} y} \\plus{} \\frac {z \\plus{} x}{x \\plus{} 2y \\plus{} z} \\plus{} \\frac {x \\plus{} y}{y \\plus{} 2z \\plus{} x}, \\frac {z \\plus{} 2x \\plus{} y}{2x \\plus{} 3y \\plus{} 3z} \\plus{} \\frac {x \\plus{} 2y \\plus{} z}{2y \\plus{} 3z \\plus{} 3x} \\plus{} \\frac {y \\plus{} 2z \\plus{} x}{2z \\plus{} 3x \\plus{} 3y}, \\frac {2x \\plus{} 3y \\plus{} 3z}{5y \\plus{} 5z \\plus{} 6x} \\plus{} \\frac {2y \\plus{} 3z \\plus{} 3x}{5z \\plus{} 5x \\plus{} 6y} \\plus{} \\frac {2z \\plus{} 3x \\plus{} 3y}{5x \\plus{} 5y \\plus{} 6z}, \\frac {5y \\plus{} 5z \\plus{} 6x}{11z \\plus{} 10x \\plus{} 11y} \\plus{} \\frac {5z \\plus{} 5x \\plus{} 6y}{11x \\plus{} 10y \\plus{} 11z} \\plus{} \\frac {5x \\plus{} 5y \\plus{} 6z}{11y \\plus{} 10z \\plus{} 11x}).$\n\nin factor,We have:\n\n$ a_1\\geq \\sum_{i \\equal{} 2}^{12}{a_i}$ holds.[/quote]\r\n\r\nBut $ g(x,y,z)\\equal{}x(x\\minus{}y)(x\\minus{}z)\\plus{}y(y\\minus{}z)(y\\minus{}x)\\plus{}z(z\\minus{}x)*(z\\minus{}y)$ does not holds.\r\nask:\r\n what is $ g(x,y,z)$ conditions?\r\nBQ", "Solution_19": "Examples of holds:\r\n\r\n(1)\r\n$ g(x,y,z) \\equal{} \\frac {x^2 \\plus{} y^2 \\plus{} z^2}{xy \\plus{} yz \\plus{} zx}$;\r\n\r\nLet r=3,\r\n$ a_1 \\minus{} 1 \\geq \\sum_{i \\equal{} 2}^3{a_i}$<=>\r\n$ \\frac {x^2 \\plus{} y^2 \\plus{} z^2}{xy \\plus{} zx \\plus{} yz} \\plus{} 1\\geq \\frac {(y \\plus{} z)^2 \\plus{} (z \\plus{} x)^2 \\plus{} (x \\plus{} y)^2)}{(y \\plus{} z)(z \\plus{} x) \\plus{} (x \\plus{} y)(y \\plus{} z) \\plus{} (z \\plus{} x)(x \\plus{} y)} \\plus{} \\frac {(2x \\plus{} z \\plus{} y)^2 \\plus{} (2y \\plus{} x \\plus{} z)^2 \\plus{} (2z \\plus{} y \\plus{} x)^2}{(2x \\plus{} z \\plus{} y)(2y \\plus{} x \\plus{} z) \\plus{} (2z \\plus{} y \\plus{} x)(2x \\plus{} z \\plus{} y) \\plus{} (2y \\plus{} x \\plus{} z)*(2z \\plus{} y \\plus{} x)}$;\r\n\r\n$ a_1\\geq a_2a_3$.\r\n\r\nin factor,r=20 holds.\r\n\r\n\r\n\r\nvery very very great great great interesting inequality is:\r\n\r\n$ a_1\\geq a_2^3a_3^3a_5a_6^3;$\r\n\r\n[size=200]nice and nice![/size]\r\n \r\nBut Not finished.\r\n\r\n\r\nYou should pay attention to this problem!\r\n\r\nYou MUST pay attention to this problem!!", "Solution_20": "Note this form:\r\n$ \\frac{x}{y\\plus{}z}\\plus{}\\frac{y}{z\\plus{}x}\\plus{}\\frac{z}{x\\plus{}y}\\minus{}\\frac{3}{2}\\geq (\\frac{y\\plus{}z}{2x\\plus{}y\\plus{}z}\\plus{}\\frac{z\\plus{}x}{2y\\plus{}z\\plus{}x}\\plus{}\\frac{x\\plus{}y}{2z\\plus{}x\\plus{}y}\\minus{}\\frac{3}{2})^t$,\r\n$ t\\equal{}20$ holds.", "Solution_21": "Note this form:\r\n$ \\frac{x}{y\\plus{}z}\\plus{}\\frac{y}{z\\plus{}x}\\plus{}\\frac{z}{x\\plus{}y}\\minus{}\\frac{3}{2}\\geq (\\frac{y\\plus{}z}{2x\\plus{}y\\plus{}z}\\plus{}\\frac{z\\plus{}x}{2y\\plus{}z\\plus{}x}\\plus{}\\frac{x\\plus{}y}{2z\\plus{}x\\plus{}y}\\minus{}\\frac{3}{2})^t$,\r\n$ t\\equal{}20$ holds.", "Solution_22": "[quote=\"xzlbq\"]Examples of holds:\n\n(1)\n$ g(x,y,z) \\equal{} \\frac {x^2 \\plus{} y^2 \\plus{} z^2}{xy \\plus{} yz \\plus{} zx}$;\n\nLet r=3,\n$ a_1 \\minus{} 1 \\geq \\sum_{i \\equal{} 2}^3{a_i}$<=>\n$ \\frac {x^2 \\plus{} y^2 \\plus{} z^2}{xy \\plus{} zx \\plus{} yz} \\plus{} 1\\geq \\frac {(y \\plus{} z)^2 \\plus{} (z \\plus{} x)^2 \\plus{} (x \\plus{} y)^2)}{(y \\plus{} z)(z \\plus{} x) \\plus{} (x \\plus{} y)(y \\plus{} z) \\plus{} (z \\plus{} x)(x \\plus{} y)} \\plus{} \\frac {(2x \\plus{} z \\plus{} y)^2 \\plus{} (2y \\plus{} x \\plus{} z)^2 \\plus{} (2z \\plus{} y \\plus{} x)^2}{(2x \\plus{} z \\plus{} y)(2y \\plus{} x \\plus{} z) \\plus{} (2z \\plus{} y \\plus{} x)(2x \\plus{} z \\plus{} y) \\plus{} (2y \\plus{} x \\plus{} z)*(2z \\plus{} y \\plus{} x)}$;\n\n$ a_1\\geq a_2a_3$.\n\nin factor,r=20 holds.\n\n\n\nvery very very great great great interesting inequality is:\n\n$ a_1\\geq a_2^3a_3^3a_5a_6^3;$\n\n[size=200]nice and nice![/size]\n \nBut Not finished.\n\n\nYou should pay attention to this problem!\n\nYou MUST pay attention to this problem!![/quote]\r\n\r\n[size=84]we have:\n$ a_1\\geq a_2^3a_3^3a_4^0a_5a_6^3a_7^0a_8a_9a_{10}^2a_{11}^2a_{12}.$---------------------(*)\n\nFor each $ i,a_i$degree is best[/size].\r\n\r\n$ a_1 \\equal{} \\frac {x^2 \\plus{} y^2 \\plus{} z^2}{yz \\plus{} zx \\plus{} xy},$\r\n\r\n$ a_2 \\equal{} \\frac {(y \\plus{} z)^2 \\plus{} (x \\plus{} z)^2 \\plus{} (x \\plus{} y)^2}{(x \\plus{} z)(x \\plus{} y) \\plus{} (x \\plus{} y)(y \\plus{} z) \\plus{} (y \\plus{} z)(x \\plus{} z)},$\r\n$ a_3 \\equal{} \\frac {(2x \\plus{} z \\plus{} y)^2 \\plus{} (2y \\plus{} z \\plus{} x)^2 \\plus{} (y \\plus{} 2z \\plus{} x)^2}{(2y \\plus{} z \\plus{} x)(y \\plus{} 2z \\plus{} x) \\plus{} (y \\plus{} 2z \\plus{} x)(2x \\plus{} z \\plus{} y) \\plus{} (2x \\plus{} z \\plus{} y)(2y \\plus{} z \\plus{} x)},$\r\n...\r\n\r\nin program lhdc3_4(http://www.mathlinks.ro/Forum/viewtopic.php?t=321135,lhdc3_4.txt),\r\ninequalyty (*) inti ($ \\sigma_1,\\sigma_2,\\sigma_3$) is:\r\n\r\n$ \\minus{} 53748795701289807798991678762028404869081476808\\sigma_1^22\\sigma_2^7$\r\n\r\n$ \\plus{} 99812958214001304847011561168056091436220\\sigma_1^{16}\\sigma_2^{10}$\r\n\r\n$ \\minus{} 150761124721\\sigma_1^2\\sigma_2^{17} \\minus{} 19550539832414211475950040585265375016931910\\sigma_1^{18}\\sigma_2^9$\r\n\r\n$ \\plus{} 65545873796437050430283487312209884396822265625\\sigma_1^{36} \\minus{} 10037760610802733178947186442364\\sigma_1^{10}\\sigma_2^{13}$\r\n\r\n$ \\plus{} 473447166275390578990550013634190682898860371536\\sigma_1^{24}\\sigma_2^6$\r\n\r\n$ \\plus{} 4543844468985097091810024676954102152158004891500\\sigma_1^{28}\\sigma_2^4 \\minus{} 818856027771727049893934602339168509989099329809\\sigma_1^{34}\\sigma_2$\r\n\r\n$ \\plus{} 1615468505816562254819133418707829645631483302\\sigma_1^{20}\\sigma_2^8 \\plus{} 131070\\sigma_2^{18}$\r\n\r\n$ \\plus{} 87815719647089095203647161860135052\\sigma_1^{12}\\sigma_2^{12} \\minus{} 6902307429709163393400\\sigma_1^6\\sigma_2^{15}$\r\n\r\n$ \\plus{} 3591345498808599831242421393947897607271156641526\\sigma_1^{32}\\sigma_2^2 \\plus{} 403340464404520478832796800\\sigma_1^8\\sigma_2^{14}$\r\n\r\n$ \\plus{} 50408207480210633\\sigma_1^4\\sigma_2^{16} \\minus{} 207643313838042935128724556482454056200\\sigma_1^{14}\\sigma_2^{11}$\r\n\r\n$ \\minus{} 6452725731480378450958025155803759135636124833336\\sigma_1^{30}\\sigma_2^3$\r\n\r\n$ \\minus{} 2275542599381596237447057751635054148427552963900\\sigma_1^{26}\\sigma_2^5\\geq0.$\r\n\r\n\r\n$ \\sigma_1 \\equal{} x \\plus{} y \\plus{} z,\\sigma_2 \\equal{} xy \\plus{} yz \\plus{} zx,\\sigma_3 \\equal{} xyz.$\r\n\r\nBQ\r\n\r\n\r\n\r\n\r\n\\sigam_1", "Solution_23": "Let $ x,y,z > 0,a_1 \\equal{} f_1(x,y,z) \\equal{} \\frac {x^2}{y} \\plus{} \\frac {y^2}{z} \\plus{} \\frac {z^2}{x} \\minus{} x \\minus{} y \\minus{} z\\geq0,$\r\n$ a_{i \\plus{} 1} \\equal{} f_{i \\plus{} 1}(x,y,z) \\equal{} f_i(y \\plus{} z,z \\plus{} x,x \\plus{} y)(i \\equal{} 1,2,...r,r\\geq 2)$,\r\n\r\nthen:\r\n\r\n$ a_1 \\equal{} \\frac {x^2}{y} \\plus{} \\frac {y^2}{z} \\plus{} \\frac {z^2}{x} \\minus{} x \\minus{} y \\minus{} z,$;\r\n$ a_2 \\equal{} \\frac {(y \\plus{} z)^2}{z \\plus{} x} \\plus{} \\frac {(z \\plus{} x)^2}{x \\plus{} y} \\plus{} \\frac {(x \\plus{} y)^2}{y \\plus{} z} \\minus{} 2y \\minus{} 2z \\minus{} 2x,$\r\n$ a_3 \\equal{} \\frac {(z \\plus{} 2x \\plus{} y)^2}{x \\plus{} 2y \\plus{} z} \\plus{} \\frac {(x \\plus{} 2y \\plus{} z)^2}{y \\plus{} 2z \\plus{} x} \\plus{} \\frac {(y \\plus{} 2z \\plus{} x)^2}{z \\plus{} 2x \\plus{} y} \\minus{} 4z \\minus{} 4x \\minus{} 4y.$\r\n...\r\nprove that:$ a_1\\geq a_2$\r\n<=>\r\n\r\n$ \\frac {x^2}{y} \\plus{} \\frac {y^2}{z} \\plus{} \\frac {z^2}{x} \\plus{} x \\plus{} y \\plus{} z \\minus{} \\frac {(y \\plus{} z)^2}{z \\plus{} x} \\minus{} \\frac {(z \\plus{} x)^2} {x \\plus{} y} \\minus{} \\frac {(x \\plus{} y)^2}{y \\plus{} z} \\geq0$\r\n\r\n<=>\r\n$ \\sigma_1^3\\sigma_2^2 \\minus{} 2\\sigma_2(3\\sigma_3 \\plus{} \\sigma_{31})\\sigma_1^2 \\plus{} (6\\sigma_3^2 \\plus{} 2\\sigma_3\\sigma_{31} \\plus{} \\sigma_{31}^2)\\sigma_1 \\plus{} 2\\sigma_2^2\\sigma_3\\geq0.$\r\n$ \\sigma_1 \\equal{} x \\plus{} y \\plus{} z,\\sigma_2 \\equal{} xy \\plus{} yz \\plus{} zx,\\sigma_3 \\equal{} xyz,\\sigma_{31} \\equal{} x^2y \\plus{} y^2z \\plus{} z^2x$.\r\n\r\nremark:\r\n$ a_1\\geq a_2 \\plus{} a_3$ does not holds.\r\n\r\n**********************************************************\r\nBut :\r\nLet $ x,y,z > 0,a_1 \\equal{} f_1(x,y,z) \\equal{} \\frac {x^2}{y(x \\plus{} y \\plus{} z)} \\plus{} \\frac {y^2}{z(x \\plus{} y \\plus{} z)} \\plus{} \\frac {z^2}{x(x \\plus{} y \\plus{} z)} \\minus{} 1\\geq0,$\r\n$ a_{i \\plus{} 1} \\equal{} f_{i \\plus{} 1}(x,y,z) \\equal{} f_i(y \\plus{} z,z \\plus{} x,x \\plus{} y)(i \\equal{} 1,2,...r,r\\geq 2)$,\r\n\r\nthen:\r\n\r\n$ a_1 \\equal{} \\frac {x^2}{y(x \\plus{} y \\plus{} z)} \\plus{} \\frac {y^2}{z(x \\plus{} y \\plus{} z)} \\plus{} \\frac {z^2}{x(x \\plus{} y \\plus{} z)} \\minus{} 1,$;\r\n$ a_2 \\equal{} \\frac { (y \\plus{} z)^2}{(z \\plus{} x)(2y \\plus{} 2z \\plus{} 2x} \\plus{} \\frac {(z \\plus{} x)^2}{(x \\plus{} y)(2y \\plus{} 2z \\plus{} 2x)} \\plus{} \\frac {(x \\plus{} y)^2}{(y \\plus{} z)(2y \\plus{} 2z \\plus{} 2x)} \\minus{} 1,$\r\n...\r\n\r\nprove that:\r\n$ a_1\\geq \\sum_{i \\equal{} 2}^{17}{a_i}$\r\n\r\nBQ\r\n\r\n\r\n\r\nBQ", "Solution_24": "Let $ x,y,z > 0,a_1 \\equal{} f_1(x,y,z) \\equal{} \\frac{x^2}{y(x\\plus{}y\\plus{}z)}\\plus{}\\frac{y^2}{z(x\\plus{}y\\plus{}z)}\\plus{}\\frac{z^2}{x(x\\plus{}y\\plus{}z)}\\geq0,$\r\n$ a_{i \\plus{} 1} \\equal{} f_{i \\plus{} 1}(x,y,z) \\equal{} f_i(y \\plus{} z,z \\plus{} x,x \\plus{} y)(i \\equal{} 1,2,...r,r\\geq 2)$,\r\n\r\nthen:\r\n\r\n$ a_1\\equal{}\\frac{x^2}{y(x\\plus{}y\\plus{}z)}\\plus{}\\frac{y^2}{z(x\\plus{}y\\plus{}z)}\\plus{}\\frac{z^2}{x(x\\plus{}y\\plus{}z)},$\r\n\r\n$ a_2\\equal{}\\frac{(y\\plus{}z)^2}{(z\\plus{}x)(2y\\plus{}2z\\plus{}2x)}\\plus{}\\frac{(z\\plus{}x)^2}{(x\\plus{}y)(2y\\plus{}2z\\plus{}2x)}\\plus{}\\frac{(x\\plus{}y)^2}{(y\\plus{}z)(2y\\plus{}2z\\plus{}2x)},$\r\n\r\nprove that\r\n\r\n$ a_1\\geq a_2a_3^2a_4^0a_5^3.$\r\nBQ", "Solution_25": "(x1^2/x2+x2^2/x3+x3^2/x4+x4^2/x1)^t >= (3*(x1^t+x2^t+x3^t+x4^t)^(1/t)/(3^(1/t)))^t,\r\nt=17 holds.\r\nBQ", "Solution_26": "xzlbq, mathematics isn't only calculations...", "Solution_27": "yes,I agree with your think,\r\nBut calculation allows us to see more,\r\nThere is no need that assume that they do not exist...\r\n\r\nWhat do we see\uff1f\r\n\r\nWe first saw it\uff0c\r\nThen put in the hands \u3002\r\nHappy new year\uff01", "Solution_28": "$x,y,z>0$,prove that\n\n\\[{\\frac {y+z}{2\\,x+z+y}}+{\\frac {z+x}{2\\,y+x+z}}+{\\frac {x+y}{2\\,z+y+x}\n}+\\frac{4}{3}\\,{\\frac {xyz}{ \\left( y+z \\right) \\left( z+x \\right) \\left( x+\ny \\right) }}\\geq \\frac{5}{3}\\]\n\n" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "For a fixed integer $n\\geq 1$ compute the min value of:\r\n\r\n $x_1 + \\frac{x_2^2}{2} + \\frac{x_3^3}{3} + ... + \\frac{x_n^n}{n} $, \r\n\r\ngiven that $x_1, x_2,..., x_n $ are positive numbers satistying the condition:\r\n\r\n $ \\frac{1}{x_1} + \\frac{1}{x_2} +...+ \\frac{1}{x_n} = n.$", "Solution_1": "Let $a=1+\\frac12+\\frac13+\\cdots+\\frac1n.$ By weighted AM-GM, we have\r\n\r\n\\[\r\n\\sum_{k=1}^n \\frac{1}{ka}x_k^k\r\n\\ge\r\n\\prod_{k=1}^n \\left(x_k^k\\right)^{\\frac{1}{ka}}\r\n=\r\n\\left(\\prod_{k=1}^n x_k\\right)^{\\frac1a}\r\n\\]\r\ni.e. $(\\text{big given sum})\\ge a\\cdot\\left(x_1\\cdots x_n\\right)^{\\frac1a}$. Then, by the given condition and AM-GM,\r\n\r\n\\[\r\nn=\\sum_{k=1}^n \\frac{1}{x_k}\\ge n\\cdot\\sqrt[n]{\\prod_{k=1}^n \\frac1{x_k}}\r\n\\]\r\ni.e. $x_1\\cdots x_n\\ge1$. Therefore,\r\n\r\n\\[\r\n(\\text{big given sum})\\ge a\\cdot\\left(x_1\\cdots x_n\\right)^{\\frac1a} \\ge a\r\n\\]\r\nwith equality only when $x_1=\\cdots=x_n=1$." } { "Tag": [ "quadratics", "inequalities", "inequalities unsolved" ], "Problem": "Given $ x_1x_2x_3..x_n\\equal{}1$\r\n\r\nProve: \r\n\r\n$ (x_1\\plus{}..\\plus{}x_n)^2 \\leq n\\plus{} (n\\minus{}1)(x_1^2\\plus{}... \\plus{}x_n^2)$", "Solution_1": "this is trivial:\r\n\\[ (x_i)^2\\leq n \\plus{} (n \\minus{} 1)(x_i^2)\\leq (x_i)^2/n \\plus{} (n \\minus{} 1)(x_i^2)\\iff (x_i)^2\\geq n(x_i^2)\r\n\\]\r\nwhere $ i$'s imply summing over", "Solution_2": "[quote=\"sunchips\"]Given $ x_1x_2x_3..x_n \\equal{} 1$\n\nProve: \n\n$ (x_1 \\plus{} .. \\plus{} x_n)^2 \\leq n \\plus{} (n \\minus{} 1)(x_1^2 \\plus{} ... \\plus{} x_n^2)$[/quote]\r\n\r\nSiranyi's inequality make this problem very easily.", "Solution_3": "[quote=\"me@home\"]\n\\[ (x_i)^2\\leq n \\plus{} (n \\minus{} 1)(x_i^2)\\leq (x_i)^2/n \\plus{} (n \\minus{} 1)(x_i^2)\\iff (x_i)^2\\geq n(x_i^2)\n\\]\nwhere $ i$'s imply summing over[/quote]\r\n\r\nshould be $ (x_i)^2\\leq n(x_i^2)$", "Solution_4": "[quote=\"me@home\"]this is trivial:\n\\[ (x_i)^2\\leq n \\plus{} (n \\minus{} 1)(x_i^2)\\leq (x_i)^2/n \\plus{} (n \\minus{} 1)(x_i^2)\\iff (x_i)^2\\geq n(x_i^2)\n\\]\nwhere $ i$'s imply summing over[/quote]\r\n\r\nWhen you sum of all $ i$, you get $ n^2$ as the constant. Your solution doesn't make sense anyway... $ 1\\le n$ is false. So is $ n\\le x_i^2/n$...\r\n\r\nFor two variables, this is obvious. Then $ n\\ge 3$. Let $ f(x) \\equal{} 1 \\plus{} (n \\minus{} 1)x^2$. $ f$ is obviously differentiable on $ (0,\\infty)$. Suppose that $ x_1 \\plus{} ... \\plus{} x_n \\equal{} d$ which is constant. Note $ x_1...x_n \\equal{} 1$ is constant. Let $ g(x) \\equal{} f'(x^{\\frac {1}{0 \\minus{} 1}})$.\r\n\r\nObserve that $ g(x) \\equal{} ( \\minus{} 2)(n \\minus{} 1)x^{ \\minus{} 3}$. $ g''(x) \\equal{} ( \\minus{} 2)( \\minus{} 3)( \\minus{} 4)(n \\minus{} 1)x^{ \\minus{} 5} < 0$. So $ g(x)$ is strictly concave on $ (0,\\infty)$. \r\n\r\nThus $ f(x_1) \\plus{} f(x_2) \\plus{} ... \\plus{} f(x_n)$ is minimal for $ 0 < x_1 \\equal{} x_2 \\equal{} ... \\equal{} x_{n \\minus{} 1}\\le x_n$ by the $ n \\minus{} 1$ equal variable theorem. \r\n\r\nIn other words, with $ a\\le b$, $ a^{n \\minus{} 1}b \\equal{} 1$. So $ 1 \\equal{} a^{n \\minus{} 1}b\\ge a^n\\implies a\\le 1$, and we must prove\r\n\\[ n \\plus{} (n \\minus{} 1)( (n \\minus{} 1)a^2 \\plus{} b^2)\\ge ( (n \\minus{} 1)a \\plus{} b)^2\r\n\\]\r\n\r\n\\[ n \\plus{} (n \\minus{} 1)b^2\\ge 2(n \\minus{} 1)ab \\plus{} b^2\r\n\\]\r\n\r\n\\[ (n \\minus{} 2)a^{ \\minus{} 2(n \\minus{} 1)} \\plus{} n \\minus{} 2(n \\minus{} 1)a^{ \\minus{} (n \\minus{} 2)}\\ge 0\r\n\\]\r\n\r\n\\[ na^{2(n \\minus{} 1)} \\minus{} 2(n \\minus{} 1)a^{n} \\plus{} (n \\minus{} 2)\\ge 0\r\n\\]\r\nwhich can be proven easily enough (I only see how to do it with a bit of calculus). Equality occurs when $ a \\equal{} b \\equal{} 1$, i.e. $ x_1 \\equal{} ... \\equal{} x_n \\equal{} 1$." } { "Tag": [ "trigonometry" ], "Problem": "$ ABCD$ is a rectangle. $ F\\in [BD]$ and $ E\\in [AB]$\r\n$ |AE|\\equal{}2$\r\n$ |EF|\\equal{}4$\r\n$ |DF|\\equal{}4$\r\n\r\n$ m\\widehat{(EFB)}\\equal{}?$", "Solution_1": "[hide=\" click to reveal hidden content\"]\n\nNot sure but I think it might be $ 30^{\\circ}$.\n\nLet $ A$ be the origin and $ E$ be $ (2, 0)$. Draw the circle, centre at $ (2, 0)$ and radius $ 4$, that is $ (x\\minus{}2)^{2} \\plus{} y^{2} \\equal{} 16$, cutting the $ y\\minus{}$axis at $ (0, \\pm 2\\sqrt{3})$. $ EF$ is a radius of this circle.\n\nWithout loss of generality (I think), let $ F$ be $ (2, 4)$.\n\nThen $ D$ is where the circle, centre at $ F$ and radius $ 4$ cuts the $ y$-axis. This circle is $ (x\\minus{}2)^{2} \\plus{} (y\\minus{}4)^{2} \\equal{} 16$, so $ D$ is $ (0, 4 \\plus{} \\sqrt{3})$.\n\nDraw the perpendicular from $ F$ to the $ y$-axis, meeting it at $ G$. Produce $ DF$ to meet the $ x\\minus{}$axis at $ B$.\n\nTriangles $ DGF$ and $ FEB$ are similar and so angle $ EFB \\equal{}$ angle $ GDF \\equal{} \\sin^{\\minus{}1}(\\frac{2}{4}) \\equal{} 30^{\\circ}$.\n\n[/hide]", "Solution_2": "[quote=\"AndrewTom]\r\nWithout loss of generality (I think), let $ F$ be $ (2, 4)$.\r\n\r\nThen $ D$ is where the circle, centre at $ F$ and radius $ 4$ cuts the $ y$-axis. This circle is $ (x \\minus{} 2)^{2} \\plus{} (y \\minus{} 4)^{2} \\equal{} 16$, so $ D$ is $ (0, 4 \\plus{} \\sqrt {3})$.\r\n\r\n[/hide][/quote]\r\n\r\nThen $ DF^2\\equal{} 2^2\\plus{}3$ which means DF is not 4, a contradiction to the given.\r\n\r\nThis is an interesting problem though. I would love to see a solution..." } { "Tag": [ "geometry", "inequalities", "Euler", "trigonometry", "circumcircle", "incenter", "cyclic quadrilateral" ], "Problem": "Hi Everyone,\n\nThis is an attempt to make a nice [b][color=red]Geometry[/color][/b] marathon at the Pre-Olympiad level.\n\nIf you post a problem, please don't forget to indicate its [b][color=red]number[/color][/b] and if you write a solution please indicate to which problem it is and also quote the original problem.Please post a new problem after the pending problem is solved.\n\nPlease show a [b][color=red]complete solution[/color][/b] and try to include figures wherever needed.\n\nLastly, I was inspired by the ongoing inequalities marathon.I hope it will also be a success like its counterpart. \n\nBtw , its advisable to install Java as it is needed to view and edit geogebra applets.Here's the link [url]http://java.sun.com/javase/downloads/index.jsp[/url]\n\nNow to the first problem:\n\n[b]Problem 1:[/b]\nIn a triangle $ \\Delta ABC$, the sides $ AB$ and $ AC$ are tangent to a circle with diameter along $ BC$ at the points $ Q$ and $ P$ respectively.Let $ E$ and $ F$ be the extremities of the diameter along $ BC$. $ EP$ and $ FQ$ intersect at $ K$.Prove that $ K$ lies on the altitude from $ A$ to $ BC$ of $ \\Delta ABC$.\n\n[b]Problem 2:[/b] \n$ ABCD$ is a cyclic quadrilateral and there exists a circle centered on the side $ AB$ that is tangent to $ BC,CD,DA$. Prove that $ AB \\equal{} AD \\plus{} BC$.", "Solution_1": "[hide=\"Solution to Problem 2\"]\n[b]Problem 2:[/b] \n$ ABCD$ is a cyclic quadrilateral and there exists a circle centered on the side $AB$ that is tangent to $ BC,CD,DA$. Prove that $ AB \\equal{} AD \\plus{} BC$.\n\nLet's euler the problem .\n\nLet the center of the inscribed circle be $ O$ on $ AB$ and its radius be $ r$ .Let the points of tangency of $ BC, CD ,DA$ be $ P,Q,R$ respectively.\nWe have $ OA\\equal{}\\frac{r}{\\sin A}$; $ OB\\equal{}\\frac{r}{\\sin B}$\nAgain making use of the property that $ ABCD$ is cyclic we get ,\n$ AD\\equal{}r\\left(\\frac{1}{\\tan A}\\plus{}\\tan \\frac{B}{2}\\right)$;\n\n$ BC\\equal{}r\\left(\\frac{1}{\\tan B}\\plus{}\\tan \\frac{A}{2}\\right)$\nWe use the following identity: $ \\frac{1}{\\sin A}\\equal{}\\frac{1\\plus{}\\tan^2 \\frac{A}{2}}{2\\tan \\frac{A}{2}}\\equal{}\\frac{1\\minus{}\\tan^2 \\frac{A}{2}\\plus{}2\\tan^2 \\frac{A}{2}}{2\\tan \\frac{A}{2}}\\equal{}\\frac{1}{\\tan A}\\plus{}\\tan \\frac{A}{2}$\n\nThus $ \\frac{r}{\\sin A}\\plus{}\\frac{r}{\\sin B}\\equal{}r\\left(\\frac{1}{\\tan A}\\plus{}\\tan \\frac{A}{2}\\plus{}\\frac{1}{\\tan B}\\plus{}\\tan \\frac{B}{2}\\right)$\n\nAnd, $ AB\\equal{}AD\\plus{}BC$[/hide]\n\n\n[b]Problem 3:[/b]\nA cyclic quadrilateral $ ABCD$ is given. The lines $ AD$ and $ BC$ intersect at $ E$, with $ C$ between $ B$ and $ E$; the diagonals $ AC$ and $ BD$ intersect at $ F$. Let $ M$ be the midpoint of the side $ CD$, and let $ N \\ne M$ be a point on the circumcircle of the triangle $ \\Delta ABM$ such that $ AN/BN \\equal{} AM/BM$. Prove that the points $ E, F$ and $ N$ are collinear.\n\n[b]Problem 4:[/b] \nLet $ (O)$ be a circumference, $ P$ an internal point and $ AA'$ a chord passing through $ P$. Show that the circles that go through $ P$ tangent to $ (O)$ at $ A,A',$ respectively and the circle with diameter $ OP$ are coaxal (they concur at another point P).", "Solution_2": "[b]Solution to Problem 4[/b]\n\nLet circles $ (O_1)$ and $ (O_2)$ be tangent to the circle $ (O)$ at $ A$ and $ A'$ respectively.\nLet $ (O_1)$and $ (O_2)$ meet at $ Q$ other than $ P$.\nWe then have, $ PO_1OO_2$ is a parallelogram.\nSo, $ OP$ is bisected by $ O_1O_2$ at $ O'$(say).\nAgain $ O_1O_2$ bisects the radical axis $ PQ$ of $ (O_1)$and $ (O_2)$ at right angles at $ R$(say).So, $ O'R$ $ \\parallel{}$ $ OQ$ . \nAnd thus, $ \\angle OQP \\equal{} \\frac {\\pi}{2}$.\nTherefore, the circle through $ OP$ as diameter passes through $ Q$.\n\n[b]Problem 5:[/b]\nLet $ I$ be the incenter of a given $ \\Delta ABC$ and let $ D,E,F$ where the incircle touches the sides $ BC,CA,AB$ of $ \\Delta ABC$ . Now let $ X,Y,Z$ be three points on the lines $ ID,IE,IF$ such that directed segments $ IX,IY,IZ$ are congruent. Prove that the lines $ AX,BY,CZ$ are concurrent.\n\nPS. I waited for about 3 days, but no one seems interested in the Geometry Marathon :(", "Solution_3": "1. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=276882[/url]\n2. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=86774[/url]\n3. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=39093[/url]\n4. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=258361[/url]\n5. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=186725[/url]" } { "Tag": [], "Problem": "I remember reading somewhere something about the number of primes up to a certain n being bounded the powers of 2. It was something like there are <= n primes between 2^(n-1) and 2^n (but of course that isn't true). Can someone remind me what the theorem actually was?", "Solution_1": "There are so many results about prime numbers.\r\n\r\n\r\nSome theorems related to powers of two :\r\n\r\n. p(n) <= 2^(2^(n-1)) (p(n) is the nth prime number)\r\n\r\n. (prod (p <= n, p prime) p) <= 4^n\r\n\r\n. Bertrand's theorem :\r\n\r\nfor each n there is a prime number between n and 2*n", "Solution_2": "That's true, but I don't think the first one is very useful, since the bound is ridiculously large. However, Erdos did use the second one in his proof of Bertrand's Postulate." } { "Tag": [ "modular arithmetic", "Divisibility Theory" ], "Problem": "Show that there are infinitely many composite numbers $n$ such that $3^{n-1}-2^{n-1}$ is divisible by $n$.", "Solution_1": "[quote=\"Peter\"]Show that there are infinitely many composite numbers $n$ such that $3^{n-1}-2^{n-1}$ is divisible by $n$.[/quote]\r\n \r\n We prove that for any positive integer $m$ the number $n=3^{2^{m}}-2^{2^{m}}$ satisfies the problems condition. Indeed, we have that if $a|b$ then $3^{a}-2^{a}|3^{b}-2^{b}$ so to prove that $n|3^{n-1}-2^{n-1}$ it is sufficent to prove that $2^{m}|n-1$,that is $2^{m}|(3^{2^{m}}-1)-2^{2^{m}}$. Since $2^{m}\\geq m$, then $2^{m}|2^{2^{m}}$ and we only need to verify that $2^{m}|3^{2^{m}}-1$. It is easy to see that \\[3^{2^{m}}-1=(3-1)(3+1)(3^{2}+1)...(3^{2^{m-1}}+1)\\] which shows that $3^{2^{m}}-1$ is a product of $8=(3-1)(3+1)$ and $m-1$ even numbers and thus is divisible by $2^{m}$(actually we have proved that it is divisible by $2^{m+2}$).", "Solution_2": "I will post bellow a solution for a generalization of this problem. \n\nLet $a$ and $b$ be positive integers such that $a>b$ and they are coprime.\nShow that there are infinitely many composite numbers $n$ such that\n $a^{n-1}-b^{n-1}$ is divisible by $n$.\n\nSolution:\nWe take an arbitrary prime number $p$ such that $a^2-b^2$ isn't divisible by $p$.\nWe will show that the number $n = \\frac{a^{2p}-b^{2p}}{a^2-b^2}$ satisfies the problem's condition.\n\n$n = \\frac{a^{2p}-b^{2p}}{a^2-b^2} = \\frac{a^{p}-b^{p}}{a-b}\\cdot\\frac{a^{p}+b^{p}}{a+b}\\newline \n= (a^{p-1}+a^{p-2}b+\\cdots+b^{p-1})(a^{p-1}-a^{p-2}b+\\cdots+b^{p-1})$.\nObviously,we have\n$a^{p-1}+a^{p-2}b+\\cdots+b^{p-1}>a^{p-1}-a^{p-2}b+\\cdots+b^{p-1}>1$.\nTherefore $n$ is a composite number.\n\n$(a^2-b^2)n \\equiv a^{2p}-b^{2p} \\equiv a^2-b^2 \\pmod{p}$.\nSince $a^2-b^2$ and $p$ are coprime, we have $n \\equiv 1 \\pmod{p}$.\nI claim that $n$ is an odd number. Otherwise, $n$ is an even number.\nThe equation $(a^2-b^2)n = a^{2p}-b^{2p}$ implies that $a^{2p}-b^{2p}$ is an even number.\nSo we have that $a$ and $b$ have same parity.\nSince they are coprime, they are odd numbers. \nBut we have $n \\equiv a^{2p-2}+a^{2p-4}b^2+\\cdots+b^{2p} \\equiv 1+1+\\cdots+1 \\equiv p \\equiv 1 \\pmod{2}$.\nThis is a contradiction.\n\nFrom the above, $n-1$ is divisible by $2p$.\nOr there is a positive integer $k$ such that $n-1=2pk$.\nThe equation $(a^2-b^2)n = a^{2p}-b^{2p}$ implies that $a^{2p} \\equiv b^{2p} \\pmod{n}$.\nTherefore, we have $a^{n-1} \\equiv a^{2pk} \\equiv b^{2pk} \\equiv b^{n-1} \\pmod{n}$.\n \nSo,we have done." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Now that mathcounts is all over, when are you moving to the high school forum (if you don't regularly go there already)? It looks pretty boring and lifeless over there... no fun.", "Solution_1": "Just about as soon as I'm in high school for me.", "Solution_2": "[quote=\"leoxnlin\"]It looks pretty boring and lifeless over there... no fun.[/quote]\r\n\r\nYou mean it isn't as spammy as this forum. :wink: (Oops was that outloud?!)\r\n\r\nHaha, I wouldn't stick to the MATHCOUNTS forum for too long though, the curious nature of HSB is always acalling :P .", "Solution_3": "For those of you who are now past MATHCOUNTS, the HS Intermediate and Pre-Olympiad forums are the next logical step. High School Basics is suppose to basically be the high school equivalent of Classroom Math - much of which you probably already know. Granted, like classroom math, there are many off-topic posts that are really more like high school contest problems (we kind of allow the basic ones but would prefer they were better classified), so you might find some interesting stuff there.", "Solution_4": "How about the AMC/AIME/USAMO forums?", "Solution_5": "Those are contest forums, remember, so you might not get some of the material right away. HSB/Intermediate/Pre-Olympiad (according to levans) contains general topics you can ask.", "Solution_6": "There's only one AMC forum. It doesn't have many problems. Actually, none apart from the problems that show up on the real contests.", "Solution_7": "none of the forums are as fun as the mathcounts forum...except GFF...\r\n\r\n\r\nand loosen up people, nothing better than spam thread with more spam thread criticizing the original spam threads and other spam threads talkign about how you should stop posting spam threads so the topic will die off to make this forum one of the best ever, if not the best.\r\n\r\n\r\nand just don't go to round table, they can turn any topic into a 5 page long essay proving their point over and over again.", "Solution_8": "I don't really get about AMC stuff...can you guys tell me?\r\n\r\n(Is my avatar irritating btw?)", "Solution_9": "Well if you're bored in the High School forums, you can always host mock contets for the newbies in the MATHCOUNTS forum.", "Solution_10": "[quote=\"nyanga-nyanga\"]I don't really get about AMC stuff...can you guys tell me?[/quote]\r\n\r\nDo you not get the AMC material or the contest itself? If you're having difficulties with the material, I suggest you familiarize yourself with regular high school mathematics, and then tackle AMC. Or pick up a copy of AoPS vol 2. \r\n\r\nIf you don't get the contest, here's a nice [url=http://www.artofproblemsolving.com/Wiki/index.php/AMC]link[/url]." } { "Tag": [ "function" ], "Problem": "A function is defined:\r\n$ f(1)\\equal{}1$\r\n$ f(2n) \\equal{} f(n)$\r\n$ f(2n\\plus{}1) \\equal{} f(2n)\\plus{}1$\r\n\r\nFind the maximum value that the function can attain within $ 1 \\leq n \\leq 2009$.\r\n\r\nFind for how many values of $ n$ this maximum is achieved.", "Solution_1": "[quote=\"Identity\"]A function is defined:\n$ f(1) \\equal{} 1$\n$ f(2n) \\equal{} f(n)$\n$ f(2n \\plus{} 1) \\equal{} f(2n) \\plus{} 1$\n\nFind the maximum value that the function can attain within $ 1 \\leq n \\leq 2009$.\n\nFind for how many values of $ n$ this maximum is achieved.[/quote]\r\n\r\nIt's easy to see that $ f(n) \\equal{}$ the sum of digits in the binary expression of $ n$.\r\n\r\nThen $ 2009 \\equal{} 11111011001_2$, so it's easy to see that the max. of $ f(n)$ in the range is $ 10$\r\n\r\nNumber of values of $ n \\equal{}5$", "Solution_2": "That's a great insight :)" } { "Tag": [ "linear algebra", "matrix", "quadratics", "combinatorics unsolved", "combinatorics" ], "Problem": "let E={1,2...,n} ,and let A_1,A_2....,A_n ,be a family of subsets of E such that:\r\nfor all i#j ,A_i#Aj and the cardinale of there intersection is 1, \r\nshow that the union of all Ai is equale to E", "Solution_1": "If one of the $ A_i$'s is a singleton, say $ A_1 \\equal{} \\{1 \\}$, then $ 1$ belongs to each other $ A_i$.\r\nLet $ B_i \\equal{} A_i \\minus{} \\{1 \\}$, for $ i\\equal{}2, \\cdots, n$.\r\nSince $ A_i \\neq A_1$, it follows that $ B_i \\neq \\emptyset$. Moreover the $ B_i$'s are pairwise disjoint.\r\nThus, for each each $ i$, there exists $ b_i \\in B_i$, and the $ b_i$'s are pairwise distinct. This gives $ n\\minus{}1$ distinct element from $ \\{2, \\cdots, n \\}$, which forces $ \\{ b_2, \\cdots, b_n \\} \\equal{} \\{2, \\cdots, n \\}$, and we are done in that case.\r\n\r\nNow, assume that $ |A_i| \\geq 2$ for each $ i$.\r\nLet $ M \\equal{} (m_{i,j})$ be the $ n \\times n$ matrix, with $ m_{i,j} \\equal{} 1$ if $ x_i \\in A_j$ and $ m_{i,j} \\equal{} 0$ otherwise.\r\nLet $ U \\equal{} tM \\cdot M \\equal{} (u_{i,j})$, where $ tM$ is the transpose of $ M$.\r\nThen $ u_{i,j} \\equal{} |A_i \\cap A_j |1$ so that for $ u_{i,j} \\equal{} 1$ for $ i \\neq j$, and $ u_{i,i} \\equal{} |A_i| \\equal{} a_i \\geq 2$.\r\n\r\nIt follows that $ U$ is the matrix of the bilinear form associated to the quadratic form $ f(x_1, \\cdots , x_n) \\equal{} (x_1 \\plus{} \\cdots \\plus{} x_n )^2 \\plus{} \\sum_{i\\equal{}1}^n (a_i \\minus{}1)x_i^2$. Clearly, $ f$ is definite and positive so it is non-degenerated.\r\nIt follows that $ U$ is inversible, and so is $ M$.\r\nIn particular, there is no row of $ 0$ in $ M$, which ensures that each element of $ E$ belongs to at least one of the $ A_i$'s.\r\n\r\nPierre." } { "Tag": [ "search", "function", "AMC", "AIME", "USA(J)MO", "USAMO", "AwesomeMath" ], "Problem": "I'm in tenth grade and AMC 10B was my first real AMC competition. I was able to get 114 points, but this was not enough to go on to the other rounds. This was the first year I really discovered my hidden passion for math ( came around October 2008) and I worked really hard to get to my score. I read both AoPS v1 and am working on v2. I don't want to stop though. I want to bring my game to the next level, and if anyone can help me, I know it would be you guys. Any suggestions? :D", "Solution_1": "First of all, congratulations on your score. It takes a lot of effort to get such a score with such a short amount of time to work with. Second of all:\r\nStep 1. Do problems\r\nStep 2. Do more problems\r\nStep 3. Do even more problems\r\nBut honestly, this is what you need to do. What is more important than that is that you must understand the solutions. Read and comprehend the solutions as if you were going to dictate it to a class tomorrow. If you want to raise your AMC score then I would suggest going over Volume 1 again, especially geometry. When you can consistently make your target score (I would suggest at least 130) on practice AMC 10's then move onto volume 2. Volume 2 will help you prepare for AMC 12 more as well as AIME. Additionally, if you list your location, we could give the names of a few local competitions. Getting into the competitive atmosphere many times helps with pressure and time management.", "Solution_2": "Be organized on the test. I'm normally fairly sloppy, but on this test I followed the recommendations on this site and made a numbered box for each question, and circled the answers. Because I was able to access all my previous work quickly and easily on questions I came back to, I was able to answer a few more questions. This was very important, because I got exactly 120.", "Solution_3": "Thank you for all the responses!\r\nWell one problem I found when covering the AoPS books is that, although they have very nice potpourri sections, the problems are grouped together too well.\r\nFor example, if I read say, a chapter on number theory (my favorite), I know what the following problems will entail, in fact, EXACTLY what they will entail. I know what techniques are to be used before I read the problem. I need AMCs. I went through most of the problems on this site ( contest section ) but I need like a general book that will focus on problem solving more than teaching.\r\nDon't get me wrong, AoPS v1 and v2 are the best math texbooks I've ever had by a longshot, but if I really need to focus on more problems, I need something new. Can someone recommend a book to me? This would be greatly appreciated. I've saved enough money for a book less than or equal to 100USD ( drain my piggy bank lol ) but if its a little over my parents will probably help me there.", "Solution_4": "There's always the contests section, up on the top of your screen. Click it, then click on USA (at the bottom right), then click one of the several contests (AMC 10 for you I suppose), and you'll see several years you can practice on.\r\nAlso, you might be able to find some mock contests created by other AoPS users (Warning: Usually significantly harder than actual exam). Try the search function for that.", "Solution_5": "It sounds like you really enjoy math, which is what you need more than anything else. I would continue to split your time between solving AMC problems and working through the AoPS books. Always push yourself to solve harder and harder problems - maybe even spend some time trying to solve AIME problems!\r\n\r\nYou will become better at problem solving in general by solving problems yourself and by carefully reading solutions to problems you couldn't solve (this means figuring out how you might have found the solution, not just understanding the steps). The only book I know of that addresses problem solving in general is Art and Craft of Problem Solving, but this book seems to be geared towards people making the transition from AIME to USAMO level, and I don't like it much anyway.", "Solution_6": "Yeah I do really enjoy math, it just took me some time to figure it out! Do you think that the book [i]First Steps for Math Olympians[/i] will do me good? I am about to order it. Or should I go with The Art and Craft of Problem Solving anways, matt? I really want to make the USAMO, and I've got 2 years to do it. I really think I can if I keep working at the rate I've been going at.\r\nThanks in advance!\r\nOh yes, and crazypianist, I am from Louisiana.", "Solution_7": "I don't really know anything about First Steps for Math Olympians. It may be good, but it seems like just a compilation of AMC problems (which you can find in the Contests section of this site) along with some instruction (which you can find in AoPS Volume 1 and 2, which you already have). Unless you are not satisfied with AoPS Volume 1 and 2, I wouldn't rush to buy another book.\r\n\r\nI didn't qualify for the AIME one year, and then I qualified for the USAMO the next year, and that was back when only 250 people qualified for the USAMO, so it's definitely possible. The only resources I used during that time were AoPS Volume 1 and 2, problems people posted on AoPS, and lots of AIME problems. If you get to the point where you are trying to transition into solving Olympiad (USAMO) problems, I would highly recommend Problem Solving Strategies instead of Art and Craft of Problem Solving, but other people might disagree. I also may just be saying this because I think I got Art and Craft of Problem Solving too late for it to be useful to me. If you want more opinions on this, search the forums - there have been many topics like this in the past.", "Solution_8": "The Art and Craft of Problem Solving (ACoPS) was the book that helped me transition from short answer to olympiad-style problems. I highly, highly recommend it; it's got pretty much everything you need to succeed at a basic level on USAMO, etc. It's well-organized and well-written, too, in my opinion. I really liked it.\r\n\r\nI wouldn't recommend doing PSS immediately. It's mainly a collection of problems, with little instruction, which isn't too useful if you are new to the olympiad style.", "Solution_9": "Thanks guys, I'll very satisfied with AoPS v1 and v2, so I won't get another book just yet. I might move into AaCoPS next, but only later.", "Solution_10": "Do you think that going to AwesomeMath summer program would be a good idea for a student like me? I definitely like the idea of an intensive maths program geared towards problem solving. I know this question may seem pretty personal but any opinions are helpful. Once again, thanks. This forum is so great and its people like you who make it so!", "Solution_11": "I've also heard that AwesomeMath is helpful, and there is a lot more information on that in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=442]AwesomeMath section[/url].", "Solution_12": "Well for me its a question of getting in or not. I was looking at the test A and I was able to solve quite a few problems. I am also very passionate about math. Do you guys think I could get in?\r\nThe thing is, at my school, I don't think there are 2 mathematics teachers to give me letters of recommendation.", "Solution_13": "I am also applying this year, and I'm pretty sure you don't need recommendations from two math teachers." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "A is a subset of the real line and m(A)=0(lebesgue measure) then:\r\n \r\n m({x^2 : x in A})=0", "Solution_1": "suppose that $A$ is included in $]-n;+n[$.\r\nThen, for any $e>0$, one can find intervals $]a_i;a_i+L_i[$ such that:\r\n*$L_i<1$\r\n*$|a_i|3, or else 0 and 1 are 2>(3+1)/4 consecutive quadratic residues.", "Solution_2": "I don't see my problem :functionnal guest so I prove in this \r\n f(f(x)+y)=xf(xy+1) => f(2f(1))=f(1+f(1))\r\n =>f(1+y)=f(f(1)+y)=>f(1)=1\r\n f(f(1/k)+k)=1/kf(f(2))\r\n f(f(k)+1/k)=kf(2) \r\n we prove f(x)=1/x\r\n is this prove true ! :? :D", "Solution_3": "Maybe we have to show that there are at most (p+1)/4 non-zero quadratic residues, because otherwise 7 doesn't work either because we have 0, 1, 2 as quadratic residues, and that's 3>2=(7+1)/4.\r\n\r\nI think the case p=M4+1 is a bit easier, considering that we can't even have equality in this case. Assume the block of consecutive residues doesn't \"cross\" the pair (p-1)/2, (p+1)/2. If x is a quadratic residue then so is -x, so we have at least (p+1)/2>(p-1)/2 non-zero quadratic residues, which is obviously absurd. This means that the block contains both (p-1)/2 and (p+1)/2, so 2 is also a quadratic residue, so if x is a quadratic residue then so is 2x. Now take M to be a set of >(p+1)/4 quadratic residues centered at (p-1)/2, (p+1)/2. It's easy to see why M and 2M are disjoint, so we have at least |M|+|2M|=(p+1)/2>(p-1)/2 non-zero quadratic residues, which is again absurd, so we're done with the case p=M4+1.", "Solution_4": "maybe he is not including the zero residue, because then the inequality will hold.", "Solution_5": "I'm not quite sure I understand what you mean.. \r\n\r\nAnyway, any ideas for the case p=M4+3? Maybe the fact that the quadratic residues are the roots of the polynomial x(p-1)/2-1 helps, but I don't exactly know how..", "Solution_6": "I think that the original question may have been to show that there are at most (p+1)/4 consecutive quadratic residues in {1, 2, 3, ..., p-1} which can be proven by looking at pairs of consecutive quadratic residues.", "Solution_7": "Yesterday i was looking the book \"A CLASSICAL INTRODUCTION TO MODERN NUMBER THEORY\" and i found the following exercises:\r\n\r\n1. Let RR denote the number of consecutive quadratic residues modulo p, RN th number of pairs of consecutive numbers for which n is a quadratic residue and n+1 is not. and NR the number of pairs for which n is a non-quadratic residue, but n+1 is a quadratic residue. and NN the number of consecutive non-quadratic residues, everything modulo p (obviously, pairs in the set 1,2,3...,p-1). then the book asks to find a formula for:\r\nRR+RN, NR+NN, RR+NR, RN+NN. \r\n\r\n2.prove rthat RR+NN-RN-NR= \\sum n(n+1)/p with n running from 1 to p-1, and prove that this sum is equal to -1.\r\n\r\n3. Use the results to show that RR=1/4(p-4-a) where a=(-1)^[p-1]/2\r\n\r\nI think the result is clear from this exercises\r\n\r\nPascual", "Solution_8": "Oh, this means that when p=M4+3 we have equality. I only thought about the first task and only for primes of the form 4k+3 (the interesting case; we use an identical line of reasoning for p=4k+1).\r\n\r\nIt's easy to see that RR+RN=(p-1)/2= the number of QR, because after any QR we either have a QR or a NQR and p-1 is a NQR.. The same type of reasoning shows us that RR+NR=NN+NR=(p-3)/2 and NN+RN=(p-1)/2. Remember that this is for the case p=M4+3.\r\n\r\nAs for the second task, RR+NN-RN-NR=\\sum{i=1->p-2}(i/p)((i+1)/p)=\\sum{i=1->p-2}(i(i+1))(p-1)/2 (mod p). Maybe this helps. I didn't continue because I had no more obvious ideas.", "Solution_9": "I guess the second task wasn't that hard either:\r\n\r\nThe fact that S=RR+NN-RN-NR=\\sum{i=1->p-1}(i(i+1)/p) is obvious and needs no proof (assume (p/p)=0; (a/p) is, as usual, the Legendre symbol). Then we use Euler's Criterion: (a/p)=a(p-1)/2 (p) (x (p) means x (mod p)).\r\n\r\nWe thus have S=\\sum{k=0->(p-1)/2}([(p-1)/2]Ck*\\sum{i=1->p-1}i(p-1)/2+k) (p). Since \\sum{i=1->p-1}it=0 (p) for all t from 1 to p-2 and -1 (p) for t=p-1, we quickly get S=-1 (p) and we can see that it can't possibly be >=p-1 or <=-p-1 (we use some inequalities: if S>=p-1 then NN+RR>p-1 so one of them must be >(p-1)/2 but RR+RN=(p-1)/2 so RR<=(p-1)/2 and NN+RN=(p-1)/2 so NN<=(p-1)/2, so we have a contradiction; the same goes for S<=-p-1). This means that the only possibility is S=-1 and we're done." } { "Tag": [ "Vieta", "algebra", "polynomial" ], "Problem": "Find the sum of the squares of the roots of $x^{2}+4x+8$.", "Solution_1": "[hide]Use your Newton's sums:\nThe sum of the roots is -4:\ns+4=0\ns=-4\nThe sum of the square of the roots is 0\ns-16+8(2)=0\ns=0[/hide]", "Solution_2": "[quote=\"vishalarul\"][hide]Use your Newton's sums:\nThe sum of the roots is -4:\ns+4=0\ns=-4\nThe sum of the square of the roots is 0\ns-16+8(2)=0\ns=0[/hide][/quote]\r\ncorrect", "Solution_3": "or..\r\n\r\n\r\nroots are $r_{1},r_{1}$\r\n\r\n$r_{1}^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}-2r_{1}r_{1}$\r\n\r\nvieta's gives $\\boxed{0}$", "Solution_4": "A slightly more elaborate method is:\r\nLet $f(x) = x^{2}+4x+8$.\r\nThe polynomial whose roots are the squares of the roots of $f(x)$ is \r\n$g(x) = f(\\sqrt x)*f(-\\sqrt x) = (x+4 \\sqrt x+8)(x-4 \\sqrt x+8) = (x+8)^{2}-16x = x^{2}+8x+64-8x = x^{2}+64$\r\nNow, by Vieta's Sums, the sum of the roots of $g(x)$ is 0. Thus, the sum of the squares of the roots of $f(x)$ is 0.", "Solution_5": "[quote=\"bpms\"]Find the sum of the squares of the roots of $x^{2}+4x+8$.[/quote]\r\n\r\n[hide]Let the roots be a and b.\n\nNote that a^2+b^2=(a+b)^2-2ab.\n\nSo the sum of the squares since a+b=-4 and ab=8 is 0.\n\n$0$[/hide]" } { "Tag": [ "geometry" ], "Problem": "Well ,guys This problem is trivial but one of my friends needs different solutions for this,so please post your solutions .Thanks\r\n\r\n[b]Problem[/b]:two circles are externally tangent at $A$. common external tangent touches the circles at $B$ and $C$.$BA$ is extended to intersect the other circle at $D$.prove that $CD$ is a diameter.", "Solution_1": "Let the circle with C on it have center O, and the circle with B on it have center P. Draw OP and extend CA through to intercent the other circle at E. \r\n\r\n$\\angle CAO = \\angle PAE = \\beta$\r\n$\\angle DAO = \\angle BAP = \\alpha$\r\n\r\nSince BPA, APE, CAO, and DOA are all isosceles,\r\n\r\n$\\angle ABP = \\angle ADO = \\alpha$\r\n$\\angle PEA = \\angle ACO = \\beta$\r\n\r\nAlso, BAE is a triangle, so\r\n\r\n$\\angle BAE + \\angle EBA + \\angle AEB = 2\\alpha + 2\\beta = 180^{\\circ} \\Rightarrow \\alpha + \\beta = 90^{\\circ}$\r\n\r\nSince $\\angle CAD = \\alpha + \\beta$, it equals $90^{\\circ}$ and CD is a diameter.", "Solution_2": "Prove that $\\triangle{ABC}$ is a right triangle. \r\nConstruct the common tangent line throught $A$ and result follows." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "Why does USNCO restrict participation for the national exam to only CITIZENS??!?!?!\r\n\r\nMath doesn't do that. Physics doesn't do that. I think Biology doesn't do that. WHY CHEMISTRY??? I mean even Canadians can take the USAMO, no? Ack, I feel left out...\r\n\r\nAt least all that work wasn't all for nothing, because I don't have to study as much for the AP and SAT II tests.", "Solution_1": "I think the camp is on an US airforce base so NON-US citizens are not allowed.", "Solution_2": "But why ban noncitizens from the National Exam, which is taken on campuses across the nation?", "Solution_3": "Chinaboy, I agree with your sentiment. I have heard that the reason for this is about the Airforce Academy, and I have also heard that it is because the ACS wishes to have American citizens represent the US in the IChO. Still, I feel that the should at least let non-citizens TAKE the national exam. One of the purposes of the competition is to foster interest in chemistry for everyone, not just people who happen to have an American passport.\r\n\r\nOn the other hand, it is possible that money is an issue--I don't know the details about who pays for the national exam (I know I've never had to pay for it)--but it is likely harder to justify spending money on setting up and grading a test for a person who will not be able to attend IChO. That being said, I still think that you have a very good point." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Show that if $A=(a_{ij})$ is a matric such that $\\sum_{j}a_{ij}=1$ for all $j$, then $1$ is an eigenvalue of $A$.", "Solution_1": "[hide=\"Eigenvector\"]$(1,1,\\dots,1)^{T}$[/hide]" } { "Tag": [ "trigonometry", "LaTeX" ], "Problem": "A) Let a, b, c be 3 angles such that a + b + c = pi. If we are given that tan(a)tan(b) = csc($ \\pi$/3), determine \r\n$ \\frac{\\cos(a)\\cos(b)}{\\cos(c)}$", "Solution_1": "[hide=\"Solution\"]\n\\[ \\minus{}\\dfrac{\\sin(c)}{\\cos(c)}\\equal{}\\minus{}\\tan(c)\\equal{}\\tan(\\pi\\minus{}c)\\equal{}\\tan(a\\plus{}b)\\equal{}\\dfrac{\\tan(a)\\plus{}\\tan(b)}{1\\minus{}\\tan(a)\\tan(b)}\\equal{}\\]\n\\[ \\dfrac{1}{\\cos(a)\\cos(b)}\\dfrac{\\sin(a)\\cos(b)\\plus{}\\sin(b)\\cos(a)}{1\\minus{}\\csc(\\frac{\\pi}{3})}\\equal{}\\dfrac{\\sqrt{3}}{\\cos(a)\\cos(b)}\\dfrac{\\sin(a\\plus{}b)}{\\sqrt{3}\\minus{}2}\\equal{}\\]\n\\[ \\dfrac{\\sin(c)}{\\cos(a)\\cos(b)}\\dfrac{\\sqrt{3}}{\\sqrt{3}\\minus{}2}\\]\n\nThen\n\n\\[ \\dfrac{\\cos(a)\\cos(b)}{\\cos(c)}\\equal{}\\dfrac{\\sqrt{3}}{2\\minus{}\\sqrt{3}}\\]\n\n[/hide]\r\n\r\n:)", "Solution_2": "[quote=\"ElChapin\"][hide=\"Solution\"]\n\\[ \\minus{} \\dfrac{\\sin(c)}{\\cos(c)} \\equal{} \\minus{} \\tan(c) \\equal{} \\tan(\\pi \\minus{} c) \\equal{} \\tan(a \\plus{} b) \\equal{} \\dfrac{\\tan(a) \\plus{} \\tan(b)}{1 \\minus{} \\tan(a)\\tan(b)} \\equal{}\n\\]\n\n\\[ \\dfrac{1}{\\cos(a)\\cos(b)}\\dfrac{\\sin(a)\\cos(b) \\plus{} \\sin(b)\\cos(a)}{1 \\minus{} \\csc(\\frac {\\pi}{3})} \\equal{} \\dfrac{\\sqrt {3}}{\\cos(a)\\cos(b)}\\dfrac{\\sin(a \\plus{} b)}{\\sqrt {3} \\minus{} 2} \\equal{}\n\\]\n\n\\[ \\dfrac{\\sin(c)}{\\cos(a)\\cos(b)}\\dfrac{\\sqrt {3}}{\\sqrt {3} \\minus{} 2}\n\\]\nThen\n\\[ \\dfrac{\\cos(a)\\cos(b)}{\\cos(c)} \\equal{} \\dfrac{\\sqrt {3}}{2 \\minus{} \\sqrt {3}}\n\\]\n[/hide]\n\n:)[/quote]\r\n\r\nCould you explain this in further detail please? I am really lost right now.\r\n\r\nThank you very much for any help!", "Solution_3": "a+b+c = pi\r\nthis problem is really easy if u realized that\r\ncos(c) = -cos(a+b) \r\n(you can see this on a unit circle easly)\r\n\r\n\r\ncos(a)cos(b)/cos(c) = cos(a)cos(b)/-cos(a+b)\r\n\r\nthen use the sum of the angle forumal on -cos(a+b)\r\n\r\ncos(a)cos(b)/(sin(a)sin(b)-cos(a)cos(b)) = x\r\n\r\n(sin(a)sin(b) - cos(a)cos(b)) / cos(a)cos(b) = 1/x\r\n\r\nbreak the parts down\r\n\r\nsin(a)sin(b)/(cos(a)cos(b)) - cos(a)cos(b)/(cos(a)cos(b)) = 1/x\r\n\r\ntan(a)tan(b) - 1 = 1/x\r\n\r\nnow tan(a)tan(b) = = csc(pi/3) = 2/3^(1/2) given\r\n\r\nso (2-3^(1/2))/3^(1/2) = 1/x\r\n\r\nx = 3^(1/2) / (2-3^(1/2)\r\n\r\nsorry it is really messy, i dont know how to use latex...", "Solution_4": "[quote=\"lgd0612\"]$ a+b+c = \\pi$\nthis problem is really easy if u realized that\n$ \\cos{c} = -cos(a+b)$ \n(you can see this on a unit circle easily)\n\n\n$ \\frac{\\cos{a}\\cos{b}}{\\cos{c}} = -\\frac{\\cos{a}\\cos{b}}{\\cos(a+b)}$\n\nthen use the sum of the angle formula on $ -\\cos(a+b)$\n\n$ \\frac{\\cos{a}\\cos{b}}{\\sin{a}\\sin{b}-\\cos{a}\\cos{b}} = x$\n\n$ \\frac{\\sin{a}\\sin{b}-\\cos{a}\\cos{b}}{\\cos{a}\\cos{b}}=\\frac1x$\n\nbreak the parts down\n\n$ \\frac{\\sin{a}\\sin{b}}{\\cos{a}\\cos{b}}-\\frac{\\cos{a}\\cos{b}}{\\cos{a}\\cos{b}}=\\frac1x$\n\n$ \\tan{a}\\tan{b} - 1 = \\frac1x$\n\nnow $ \\tan{a}\\tan{b} =\\csc\\left(\\frac{\\pi}3\\right) = \\frac2{\\sqrt3}$ given\n\nso $ \\frac{2-\\sqrt3}{\\sqrt3}=\\frac1x$\n\n$ x = 3+2\\sqrt3$\n\nsorry it is really messy, i dont know how to use latex...[/quote]\r\n\r\n$ \\text{\\LaTeX}$ed and slightly edited...you can learn a lot of it just from this.", "Solution_5": "thank you so much, math154!\r\n\r\nthat really helped!", "Solution_6": "Don't thank me--it was lgd0612's solution.", "Solution_7": "thank you, math154\r\n\r\n :)" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "$L$ is a nilpotent left ideal of $R$, show that there exist a nilpotent ideal $\\mathfrak I\\supseteq L$ in $R$.", "Solution_1": "Let $\\mathfrak I=L+LR$.\r\n$LR$ is the subgroup generated by $\\{lr|l\\in L,\\ r\\in R\\}$.\r\nIt is easy to check that $\\mathfrak I$ is an ideal of $R$.\r\n\r\nI guess we also have to prove that $\\mathfrak I\\neq R$.\r\nSuppose $L+LR=R$. \r\nThen $L+LR=L+L(L+LR)\\subseteq L+L^2+L^2R=L+L^2(L+LR)\\subseteq\\ldots\\subseteq L+L^nR\\subseteq\\ldots$.\r\nBecause $L$ is nilpotent and because of the previous line there exists $n\\in\\mathbb{N}$ such that $R=L+LR\\subseteq L+L^nR=L$. This is a contradiction to $L\\neq R$ (which should be in the definition of $L$).", "Solution_2": "oops...I'm sorry...I've modified the problem." } { "Tag": [], "Problem": "In a triangle $ ABC$ , if $ O\\equal{}H\\equal{}G$ (specific notations) it is true that $ ABC$ is equilateral ?? If yes ,can I use that in problems without proof ?", "Solution_1": "It is true. I would not use it without proof though... memorizing it and writing it down is always safer.\r\n\r\n[hide=\"Idea of the Proof\"]The medians, perpendiculars and perpendicular bisectors will all coincide, and if you use their distinctive properties it's fairly obvious that the triangle has to be equilateral.[/hide]" } { "Tag": [ "Columbia" ], "Problem": "Hi everyone!!\r\n\r\nMy name is Makis and i am an international student. I attend the international baccalaureate(IB) in a greek school and this year i am finishing the first os the two(2) years. I am seriously thinking of applying in American universities for engineering courses, and i would like your opinion of whether i can get there. Here are the universities i am thinking of applying to:\r\n-Columbia(1st choice)\r\n-Cornell(2nd choice)\r\n-Georgetown(3rd choice)\r\n\r\n[u]Biographie in short:[/u]\r\n\r\nIn Greece the mathematical competitions are done in three(3) rounds(in every round a number of students is eliminated), and the third is the national mathematical olympiad. I have participated four(4) times and in the two(2) first i made it to the second round, and in the two(2) last i made it on the mathematical olympiad(3rd round, top 50 of my grade participated whereas on the first round 2000 from my grade participate). Also i have participated in the mathematical competition kangaroo and got extra prizes due to my performance, and i have once been given a 800$ scholarship from my school due to the fact that i finished 1st in a mathematical competition that my school organized.\r\n\r\nI have a Lower, an Advanced and a Profiecency in English and a B2 in German. I also have computer certificates from the university of cambridge exams.\r\n\r\nI had excellent marks in maths, physics and chemistry at high school and i also have now in IB good grades. In IB students are grades in scale of 7, 1 being the worse grade and 7 being the best. Here are my grades in the subjects i attend:\r\n-Maths, high level:6\r\n-Physics, high level:7\r\n-English as a foreign language, high level:6\r\n-Chemistry, standard level:6\r\n-Business and Management, standard level:6\r\n-Modern Greek, standard level:5 \r\n-Total 36/42\r\n\r\nI am also planning of taking the sat reasong test, and the sat subject tests in maths and physics(where i think i am capable of getting a 800 at both).\r\n\r\nAs an activity I had been playing tennis for 7 years but now i have stopped.\r\n\r\nSo what do you think can i make it to these universities?", "Solution_1": "Competition at top tier universities is extremely tough for American students, and even tougher for international students. Unfortunately, I don't see anything that would cause you to stand out. Nothing is spectacular. Sorry.", "Solution_2": "International students always think that highlighting their grades and academic results is the most important part of their application. But there are plenty of domestic students with excellent grades. Colleges like international students because they add diversity to the campus. So you have to seem like an interesting person. They also are very careful to pick such international students that will \"mesh\" with their college environment. It may help to know what they're looking for:\r\n\r\n\r\nColumbia looks for something \"more\" than an engineer. Someone with a perspective of what's going on in the world, a potential leader, perhaps a businessman, who wishes to be enabled/empowered with an engineering education. It is important to show well-roundedness, ambition, and initiative. They also like research experience. Finally, show some interest in New York City. Columbia considers it's location in New York City, and its relationship with the city, to be one of its defining characteristics.\r\n\r\n\r\nCornell will look to see that you have a balanced, amicable personality who can survive the demanding course-load at Cornell, and will not be held back by such things as the size of the school (and the resulting red-tape), the weather (don't seem very moody), and the fact that Ithaca is somewhat isolated (show interest in some of the clubs at Cornell). So, it helps to have recommendations that highlight your personality, your ambition (without seeming overly competitive), and your initiative. Research experience is helpful.\r\n\r\nIt is also helpful to show a wide range of interests, and to seem interested in the diversity of the students at Cornell. Cornell's motto is a statement by Ezra Cornell; \"I would found an institution where any person can receive instruction in any study.\"\r\n\r\n\r\nOne thing that American colleges in general value that many international students often don't highlight is work experience. If you've ever been employed, put it down, and highlight the skills you've gained from it. Also put down any internships, any labwork, and any *significant* personal projects.\r\n\r\n\r\nIf you have any particular questions about the program at Cornell, PM me.", "Solution_3": "Thanks to both of you for your replies.\r\n\r\n@JRav: OK, lets take it the other way around. In what univerities in New York City do you think i could be accepted?\r\n\r\n@TZF: Hmm, what do you exactly mean by research experience? For example, during the summer i will do an extended essay(4000 words) on waves and motion of springs which have weight(in school we only examine weightless springs). Is this considered some kind of research experience.\r\n\r\nCould you also name some other prestigious universities in New York?" } { "Tag": [ "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "Let T: V -- V be a linear transformation, where V is a finite dimensional vector space. If dim(range(T))=dim(range(T2)) show that the range and null space of T only have the zero vector in common.\r\n\r\n\r\n\r\n\r\nT2 is T square couldnot get the 2 as superscript!\r\nT:V has an arrow pointing to V", "Solution_1": "Suppose $ v \\in Im(T) \\cap Ker(T)$, and let $ T(w)\\equal{}v$, then $ T^2(w) \\equal{} 0$, so $ w \\in Ker (T^2) \\equal{} Ker (T)$, so $ v \\equal{} 0$\r\n\r\n$ Ker(T^2) \\equal{} Ker(T)$ because their dimensions are the same." } { "Tag": [], "Problem": "H\u00f4m nay em nh\u1eadn \u0111\u01b0\u1ee3c 1 ps t\u1eeb 1 ng\u01b0\u1eddi b\u1ea1n n\u01b0\u1edbc ngo\u00e0i c\u00f3 n\u00f3i l\u00e0 s\u1ed1 b\u00e0i c\u1ee7a b\u1ea1n \u1ea5y gi\u1ea3m. Em n\u00f3i kh\u00f4ng c\u00f3 g\u00ec l\u1ea1 c\u1ea3 nh\u01b0ng nh\u00ecn l\u1ea1i m\u00ecnh b\u1ecb m\u1ea5t h\u01a1n 70 posts , c\u00e1c b\u00e0i g\u1ea7n \u0111\u00e2y koo b\u1ecb m\u1ea5t .V\u00f4 l\u00fd qu\u00e1 :mad:", "Solution_1": "Anh c\u0169ng th\u1ebf, m\u1ea5t g\u1ea7n 500 b\u00e0i, \u0111\u1ec3 anh h\u1ecfi xem. :D", "Solution_2": "[quote=\"N.T.TUAN\"]Anh c\u0169ng th\u1ebf, m\u1ea5t g\u1ea7n 500 b\u00e0i, \u0111\u1ec3 anh h\u1ecfi xem. :D[/quote]\r\n\r\nC\u00f3 th\u1eadt l\u00e0 m\u1ea5t 500 koo :D", "Solution_3": "G\u1ea7n 300 post , v\u00e0o \u0111\u00e2y n\u00e0y http://www.mathlinks.ro/Forum/viewtopic.php?t=172668 . Locked!" } { "Tag": [ "inequalities", "calculus", "AMC", "three variable inequality" ], "Problem": "Prove that\r\n\r\n$ \\frac{\\left(b+c-a\\right)^{2}}{\\left(b+c\\right)^{2}+a^{2}}+\\frac{\\left(c+a-b\\right)^{2}}{\\left(c+a\\right)^{2}+b^{2}}+\\frac{\\left(a+b-c\\right)^{2}}{\\left(a+b\\right)^{2}+c^{2}}\\geq\\frac35$\r\n\r\nfor any positive real numbers $ a$, $ b$, $ c$.", "Solution_1": "okay here is a solution:\r\n\r\nsum((2ab+2ac)/(a^2+b^2+c^2+2bc))<=24/5 (Note that the sum is sym).\r\nlets denote s=a^2+b^2+c^2 for easy calculus! :D\r\n5s^2(sum(ab))+10s(sum(a^2bc))+20(sum(a^3b^2c))<=6s^3+6s^2(sum(ab))+12s(sum(a^2bc))+48a^2b^2c^2 (all sums are sym)\r\nwhich after some simplifacations and calculus we get:\r\n6s^3+s^2sum(ab)+2s(sum(a^2bc)+8sum(a^2b^2c^2)>=10s(sum(a^2bc))+20(sum(a^3b^2c). (sums are sym)\r\nwe now multiply out the powers of s=a^2+b^2+c^2 and get:\r\nsym sum 3a^6+2a^5b-2ab^2+3abc+2a^3b^3-12a^3b^2c+4a^2b^2c^2>=0\r\nnow using Schur's inequality after multiplication with 4abc:\r\nsym sum 4abc-8a^3b^2c+4a^2b^2c^2>=0 and this reduces our claim to\r\n\r\nsym sum 3a^6+2a^5b-2ab^2-abc+2a^3b^3-4a^3b^2c>=0.\r\n\r\nnow we have a sum of four expressions which are positive by weighted AM-GM:\r\n\r\n2sym sum(2a^6+b^6)/3-ab^2>=0\r\nsym sum(4a^6+b^6+c^6)/6-abc>=0\r\n2 sym sum (2a^3b^3+c^3a^3)/3-a^3b^2c>=0\r\n2 sys sum (2a^5b+a^5c+ab^5+ac^5)/6-a^3b^2c>=0\r\n\r\nequality is when a=b=c ( we see this from Shur's ineq and AM-GM)\r\n\r\ncheers hope this is good! :D", "Solution_2": "I think I have an interesting solution for your problem . Here goes :\r\n(all sums are cyclic , not symmetric)\r\n\r\nsum (b+c-a)^2/[(b+c)^2 + a^2] >= 3/5 <=>\r\nS = sum a(b+c)/[(b+c)^2 + a^2] <= 6/5 , after a few trivial computations\r\n\r\nNow , (b+c)^2 + a^2 >= (a+2b+2c)^2/5 , by cauchy applied to the numbers (b+c)/2 , (b+c)/2 , (b+c)/2 , (b+c)/2 , a - we apply it in this manner in order to have equality .\r\n\r\nThis means that S <= sum 5a(b+c)/(a+2b+2c)^2 = S'\r\n\r\nWe prove that S' <= 6/5 . Assume that a+b+c=1 since the inequality is homogenous , and you end up with :\r\n\r\nsum a(1-a)/(2-a)^2 <= 6/25\r\n\r\nConsider the function f(x) = (x-x^2)/(2-x)^2 . If it is concave on (0,1) we are done . But this is true by a trivial computation (show that f''(x)<0 on (0,1)) . So we can apply Jensen's inequality and we are done .\r\n\r\nVoila , no horrendous computations ! :)", "Solution_3": "well my solution is the so called \"official solution\" that was given at the contest! cheers!", "Solution_4": "Hey! every one\r\n This my solution.I think, it is nice because .... \r\n Put S=(b + c - a)2/((b+c)2+a2) +(a + c - b)2/((a+c)2+b2) + (b + a - c)2/((b+a)2+c2)\r\n WLOG, We suppose a + b + c=1.(Because, If a + b+ c = x, we put a1=a/x, b1=b/x, c1=c/x \r\nthen inequality equivalent S >= 3/5 with a1+ b1 + c1=1 )\r\nWe have S=(1-2a)2/(a2 + (1-a)2) + (1-2b)2/(b2 + (1-b)2) + (1-2c)2/(c2 + (1-c)2)\r\nWe'll prove (1-2a)2/(a2 + (1-a)2)>= (23-54a)/25 (*).\r\nReally multiplying 25((a2 + (1-a)2), we get 2(3a-1)2(6a+1) >=0, then (*) is true.\r\nApplying (*), then we have (1-2b)2/(b2 + (1-b)2)>= (23-54b)/25 , (1-2c)2/(c2 + (1-c)2)>= (23-54c)/25 \r\nThen S>= (23.3 - 54(a + b +c))/25 = 3/5.\r\nIt's very good??? :)", "Solution_5": "hey, how did you find 23/25 and -54/25??\r\n(1-2a)2/(a2 + (1-a)2)>= (23-54a)/25", "Solution_6": "Hey, see my soln for poland 1996 inequality :D\r\n\r\n[Moderator edit: Also compare with the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=53672]USAMO 2003[/url] inequality.]", "Solution_7": "I also have a short solution. Put x=(b+c)/a, y=(c+a)/b and z=(a+b)/c. Apply Cauchy to find that the LHS is sum (x-1)^2/(x^2+1)>=(x+y+z-3)^2/(sum x^2+3). This becomes (x+y+z)^2-15(x+y+z)+18+3(xy+yz+zx)>=0.But xy+yz+zx>=2(x+y+z) (since this one reduces to (a+b-c)(b+c-a)(c+a-b)<=abc), so it is enough to prove that (sum x)^2-9sumx+18>=0 and it is true since sumx>=6.", "Solution_8": "[quote=\"alekk\"]hey, how did you find 23/25 and -54/25??\n(1-2a)2/(a2 + (1-a)2)>= (23-54a)/25[/quote]\r\n\r\nI think my solution basically employs the same idea. First, say $a+b+c=3$ so that equality will hold at $a=b=c=1$. Essentially, one conjectures that \r\n\r\n$\\frac{(b+c-a)^2}{(b+c)^2+a^2} = \\frac{(3-2a)^2}{(3-a)^2+a^2} \\ge \\frac{1}{5} + ka - k$\r\n\r\nfor some $k$. As the equation is parametrically contrived to yield equality where $a=1$, we need the $k$ for which there is a double root. That can be solved for by equating derivatives, etc. and you get $k = \\frac{-18}{25}$. Factoring for this $k$, we see that the inequality is always true. Then add the three expressions to finish.", "Solution_9": "Nice but unnature :)", "Solution_10": "If you want prove it, you should use [b]SOS inequality[/b]", "Solution_11": "$\\frac{(b+c-a)^2}{(b+c)^2+a^2}+ \\frac{(a+c-b)^2}{(a+c)^2+b^2}+\\frac{(b+a-c)^2}{(b+a)^2+c^2} \\geq 3/5$ (1)\r\n let $a+b+c=3$ then (1) $\\leq >$\r\n$\\frac{(3-2a)^2}{(3-a)^2+a^2} + \\frac{(3-2b)^2}{(3-b)^2+b^2} + \\frac{(3-2c)^2}{(3-c)^2+c^2} \\geq 3/5$\r\n now , it's easy", "Solution_12": "we may assume that $ a+b+c=2$ , the inequality becomes\r\n$ \\sum\\frac{(a-1)^{2}}{1+(a-1)^{2}}\\geq \\frac{3}{10}$\r\n$ \\leftrightarrow \\sum \\frac{1}{1+(a-1)^{2}}\\leq \\frac{27}{10}$\r\nnow we denote $ x=a-1;y=b-1;z=c-1$\r\nthe inequality becomes $ \\sum \\frac{1}{1+x^{2}}\\leq \\frac{27}{10}$ with $ x+y+z=1$ and $ x,y,z <1$\r\nby the fact $ 27(1-3t)^{2}(4-3t) \\geq 0$ we obtain $ \\frac{1}{1+x^{2}}\\leq \\frac{27}{50}(2-x)$\r\nnow we easily get the desire result", "Solution_13": "[quote=\"Mildorf\"][quote=\"alekk\"]hey, how did you find 23/25 and -54/25??\n(1-2a)2/(a2 + (1-a)2)>= (23-54a)/25[/quote]\n\nI think my solution basically employs the same idea. First, say $a+b+c=3$ so that equality will hold at $a=b=c=1$. Essentially, one conjectures that \n\n$\\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}= \\frac{(3-2a)^{2}}{(3-a)^{2}+a^{2}}\\ge \\frac{1}{5}+ka-k$\n\nfor some $k$. As the equation is parametrically contrived to yield equality where $a=1$, we need the $k$ for which there is a double root. That can be solved for by equating derivatives, etc. and you get $k = \\frac{-18}{25}$. Factoring for this $k$, we see that the inequality is always true. Then add the three expressions to finish.[/quote]\r\nI tried this one with isolated fudging of the form $\\frac{a^{p}}{a^{p}+b^{p}+c^{p}}$ and the exponent is the same as your k... coincidence?\r\n\r\nBut isolated fudging doesn't work so that's why I requested a link to here :P", "Solution_14": "There is actually a short solution:\r\n$ \\frac{2a(b+c)}{(b+c)^{2}+a^{2}}=\\frac{2a(b+c)}{\\frac{3}{4}(b+c)^{2}+\\frac{1}{4}(b+c)^{2}+a^{2}}\\le \\frac{2a(b+c)}{\\frac{3}{4}(b+c)^{2}+(b+c)a}=\\frac{2a}{\\frac{3}{4}(b+c)+a}$\r\nNow if we assume $ a \\le b \\le c$ then $ \\frac{1}{\\frac{3}{4}(b+c)+a}\\ge \\frac{1}{\\frac{3}{4}(a+c)+b}\\ge \\frac{1}{\\frac{3}{4}(b+a)+c}$\r\nSo by rearrangement, we deduce that this cyclic sum is $ \\le 12/5$\r\nThe rest follows easily.", "Solution_15": "[quote=hxtung]Prove that\n$ \\frac{\\left(b+c-a\\right)^{2}}{\\left(b+c\\right)^{2}+a^{2}}+\\frac{\\left(c+a-b\\right)^{2}}{\\left(c+a\\right)^{2}+b^{2}}+\\frac{\\left(a+b-c\\right)^{2}}{\\left(a+b\\right)^{2}+c^{2}}\\geq\\frac35$\nfor any positive real numbers $ a$, $ b$, $ c$.[/quote]\nWe have\n\\[\\begin{aligned}\n5(b+c-a)^2&[4(a^2+b^2+c^2)+11(ab+bc+ca)]-[(b+c)^2+a^2][(20b^2+20c^2-28a^2-11ab+55bc-11ca)\\\\& = 2a(3a+4b+10c)(2a-b-c)^2+6(3b+2c)(a-b)^2a+6a(c+4a)(c-a)^2 \\\\& \\geqslant 0.\n\\end{aligned}\\]\nTherefore\n\\[\\frac{\\left(b+c-a\\right)^{2}}{\\left(b+c\\right)^{2}+a^{2}} \\geqslant \\frac15 \\cdot \\frac{20b^2+20c^2-28a^2-11ab+55bc-11ca}{4(a^2+b^2+c^2)+11(ab+bc+ca)}.\\]\nNote\n\\[\\sum \\frac{20b^2+20c^2-28a^2-11ab+55bc-11ca}{4(a^2+b^2+c^2)+11(ab+bc+ca)} = 3.\\]\nDone.", "Solution_16": "[quote=hxtung]Prove that\n\n$ \\frac{\\left(b+c-a\\right)^{2}}{\\left(b+c\\right)^{2}+a^{2}}+\\frac{\\left(c+a-b\\right)^{2}}{\\left(c+a\\right)^{2}+b^{2}}+\\frac{\\left(a+b-c\\right)^{2}}{\\left(a+b\\right)^{2}+c^{2}}\\geq\\frac35$\n\nfor any positive real numbers $ a$, $ b$, $ c$.[/quote]\nIt's true for any reals $a$, $b$ and $c$ such that $\\prod_{cyc}((a+b)^2+c^2)\\neq0.$\n\n", "Solution_17": "[quote=arqady][quote=hxtung]Prove that\n$ \\frac{\\left(b+c-a\\right)^{2}}{\\left(b+c\\right)^{2}+a^{2}}+\\frac{\\left(c+a-b\\right)^{2}}{\\left(c+a\\right)^{2}+b^{2}}+\\frac{\\left(a+b-c\\right)^{2}}{\\left(a+b\\right)^{2}+c^{2}}\\geq\\frac35$\nfor any positive real numbers $ a$, $ b$, $ c$.[/quote]\nIt's true for any reals $a$, $b$ and $c$ such that $\\prod_{cyc}((a+b)^2+c^2)\\neq0.$[/quote]\nSee my [url=\"https://artofproblemsolving.com/community/c6h28678p14450762\"]solution[/url] :D\n", "Solution_18": "[quote=Nguyenhuyen_AG][quote=hxtung]Prove that\n$ \\frac{\\left(b+c-a\\right)^{2}}{\\left(b+c\\right)^{2}+a^{2}}+\\frac{\\left(c+a-b\\right)^{2}}{\\left(c+a\\right)^{2}+b^{2}}+\\frac{\\left(a+b-c\\right)^{2}}{\\left(a+b\\right)^{2}+c^{2}}\\geq\\frac35$\nfor any positive real numbers $ a$, $ b$, $ c$.[/quote]\nWe have\n\\[\\begin{aligned}\n5(b+c-a)^2&[4(a^2+b^2+c^2)+11(ab+bc+ca)]-[(b+c)^2+a^2][(20b^2+20c^2-28a^2-11ab+55bc-11ca)\\\\& = 2a(3a+4b+10c)(2a-b-c)^2+6(3b+2c)(a-b)^2a+6a(c+4a)(c-a)^2 \\\\& \\geqslant 0.\n\\end{aligned}\\]\nTherefore\n\\[\\frac{\\left(b+c-a\\right)^{2}}{\\left(b+c\\right)^{2}+a^{2}} \\geqslant \\frac15 \\cdot \\frac{20b^2+20c^2-28a^2-11ab+55bc-11ca}{4(a^2+b^2+c^2)+11(ab+bc+ca)}.\\]\nNote\n\\[\\sum \\frac{20b^2+20c^2-28a^2-11ab+55bc-11ca}{4(a^2+b^2+c^2)+11(ab+bc+ca)} = 3.\\]\nDone.[/quote]\n\nBetter: https://artofproblemsolving.com/community/u493456h1939876p14491686 :P ", "Solution_19": "[quote=SBM]Better: https://artofproblemsolving.com/community/u493456h1939876p14491686 :P[/quote]\nYour proof is #20", "Solution_20": "Since the inequality is homogeneous, we can assume the condition $a+b+c=1$ and scale up later. The inequality then becomes\n\n\\begin{align*}\n\\sum \\frac{(1-2a)^2}{a^2 + (1-a)^2} \\ge \\frac{3}{5}\n\\end{align*}\n\nwhich rearranges to\n\n\\begin{align*}\n\\frac{27}{5} \\ge \\sum \\frac{1}{2a^2 - 2a + 1}.\n\\end{align*}\n\nBy the tangent line trick, we derive the inequality\n\n\\[\\frac{1}{2a^2-2a+1} \\le \\frac{9}{5} + \\frac{54}{25}\\left(a - \\frac{1}{3}\\right) \\Longleftrightarrow 2(3a-1)^2(6a+1) \\ge 0.\\]\n\nSumming this cyclicly yields the desired inequality. $\\square$", "Solution_21": "Since the inequality is homogeneous therefore WLOG we assume $a+b+c=3$. Now the inequality becomes\n\\begin{align*}\n\\sum_{cyc}\\frac{(3-2a)^2}{a^2+(3-a)^2}\\geq \\frac{3}{5}\n\\end{align*}\nConsider the function $f(x)=\\frac{(3-2x)^2}{x^2+(3-x)^2}$. So we have to prove that $$f(a)+f(b)+f(c)\\geq \\frac{3}{5}$$ Notice that (Tangent line trick) $$f(x)\\geq \\frac{23-18x}{25}\\Longleftrightarrow \\frac{(x-1)^2(2x+1)}{2x^2-6x+9}\\geq 0$$ is true (the denominator $2x^2-6x+9>0$ because the coefficient of $x^2$ is positive and least value is $\\left(\\frac{3}{2}, \\frac{9}{2}\\right)$.) for all value of positive $x$. Now just summing up them we get our desired result.", "Solution_22": "Same as everyone else. Nice application of TLT.\n-----\nSince both sides of the inequality are of same degree, the inequality is homogeneous. Since the domain is positive reals as well, along with the previous fact, we can assume $a+b+c=3.$ Our inequality we want to prove rewrites to \\[\\sum_{\\text{cyc}} \\frac{(3-2a)^2}{a^2+(3-a)^2} \\geq \\frac{3}{5}.\\] Recall the Tangent Line Trick, which is if we fix $m=\\tfrac{a_1+\\cdots+a_n}{n},$ and if $f$ is nonconvex, then if $f(x) \\geq f(m)+f'(m)(x-m)$ holds for all $x,$ summing cyclically produces the desired result. In our case, consider the function $f(x)=\\tfrac{(3-2x)^2}{x^2+(3-x)^2},$ then we want to prove $f(a)+f(b)+f(c) \\geq \\tfrac{3}{5}.$ Since the second derivative of $f$ is negative, the function is nonconvex. By applying the Tangent Line Trick, \\[\\frac{(3-2a)^2}{(3-a)^2+a^2} \\geq \\frac{1}{5}-\\frac{18}{25}(a-1) \\implies\\]\\[ \\frac{18}{25}(a-1)-\\frac{1}{5}+\\frac{(3-2a)^2}{(3-a)^2+a^2} = \\frac{18(a-1)^2(2a+1)}{25(2a^2-6a+9)} \\geq 0,\\] of which the last inequality is clearly true because $a$ is a positive real. Thus by Tangent Line Trick, summing cyclically implies the result.", "Solution_23": "Prove that\n$$ \\frac{(b+c-5a)^2}{(b+c)^2+5a^2}+\\frac{(c+a-5b)^2}{(c+a)^2+5b^2}+\\frac{(a+b-5c)^2}{(a+b)^2+5c^2}\\geq 3$$\nfor any positive real numbers $ a$, $ b$, $ c$.\n\n", "Solution_24": "[quote=sqing]Prove that\n$$ \\frac{(b+c-5a)^2}{(b+c)^2+5a^2}+\\frac{(c+a-5b)^2}{(c+a)^2+5b^2}+\\frac{(a+b-5c)^2}{(a+b)^2+5c^2}\\geq 3$$\nfor any positive real numbers $ a$, $ b$, $ c$.[/quote]\nIt's true for any reals $a$, $b$ and $c$ such that $\\prod\\limits_{cyc}((a+b)^2+5c^2)\\neq0.$", "Solution_25": "Prove that\n$$ \\frac{a\\left(a+b-c\\right)^{2}}{\\left(a+b\\right)^{3}+c^{3}}+\\frac{b\\left(b+c-a\\right)^{2}}{\\left(b+c\\right)^{3}+a^{3}}+\\frac{c\\left(c+a-b\\right)^{2}}{\\left(c+a\\right)^{3}+b^{3}}\\geq\\frac1{3}$$\n$$ \\frac{\\left(b+c-a\\right)^{3}}{\\left(b+c\\right)^{3}+a^{3}}+\\frac{\\left(c+a-b\\right)^{3}}{\\left(c+a\\right)^{3}+b^{3}}+\\frac{\\left(a+b-c\\right)^{3}}{\\left(a+b\\right)^{3}+c^{3}}\\geq\\frac1{3}$$\n$$ \\frac{a\\left(a+b-c\\right)^{2}}{\\left(a+b\\right)^{3}+2c^{3}}+\\frac{b\\left(b+c-a\\right)^{2}}{\\left(b+c\\right)^{3}+2a^{3}}+\\frac{c\\left(c+a-b\\right)^{2}}{\\left(c+a\\right)^{3}+2b^{3}}\\geq\\frac3{10}$$\n$$ \\frac{\\left(b+c-a\\right)^{3}}{\\left(b+c\\right)^{3}+2a^{3}}+\\frac{\\left(c+a-b\\right)^{3}}{\\left(c+a\\right)^{3}+2b^{3}}+\\frac{\\left(a+b-c\\right)^{3}}{\\left(a+b\\right)^{3}+2c^{3}}\\geq\\frac3{10}$$\nfor any positive real numbers $ a$, $ b$, $ c$.\n[hide=Thanks.][quote=arqady][quote=sqing]Prove that\n$$ \\frac{(b+c-5a)^2}{(b+c)^2+5a^2}+\\frac{(c+a-5b)^2}{(c+a)^2+5b^2}+\\frac{(a+b-5c)^2}{(a+b)^2+5c^2}\\geq 3$$\nfor any positive real numbers $ a$, $ b$, $ c$.[/quote]\nIt's true for any reals $a$, $b$ and $c$ such that $\\prod\\limits_{cyc}((a+b)^2+5c^2)\\neq0.$[/quote]\n[/hide]", "Solution_26": "De-homogenize with $a+b+c = 1$. Now, notice that $$\\frac{(1-2a)^2}{a^2+(1-a)^2} \\geq \\frac{23-54a}{25} \\iff \\frac{25(3a-1)^2(6a+1)}{\\geq 0} \\geq 0,$$ so sum to get the result.", "Solution_27": " Set $f(x):= \\frac{3-2a}{a^2+(3-a)^2}$. Due to the homogeneity of the inequality, we can assume $a+b+c=3$. The inequality can now be rewritten as follows:\n\n $$\\sum_{cyc} f(a) \\geq \\frac{3}{5}$$\n\nThis is true by tangent trick lemma since:\n$$f(x)\\geq f'(1)(x-1)+f(1)$$\n$$\\iff \\frac{3-2x}{x^2+(3-x)^2} \\geq \\frac{1}{5}-\\frac{18}{25}(x-1) $$\n\n$$\\iff \\frac{18(x-1)^2(2x+1)}{25(2x^2-6x+9)} \\geq 0$$\n\n Which holds for all $0 1 and that (Ax^3 + Bx^2 + Cx)/(ax^2 + bx + c) > 1 for x < 1. [/hide]\n\n\n\nHmm, where do we go from there?", "Solution_2": "[quote]That is one nasty looking problem[/quote]\r\n\r\nExactly what I thought when I saw it, so I have little idea about how to approach it as well.", "Solution_3": "I've seen this before, but I don't really remember it, so I'm just going to toss out some ideas. First of all, if (ax^2 + bx + c) has any real roots, there will exists a sequence of rational numbers approaching those roots, which will make the left side of our inequality blow up unless (Ax^2 + Bx + C) has the same root. This means that (ax^2 + bx + c) probably shouldn't have any real roots, or at least not anywhere near the cube root of 2. I'm going to keep thinking about this -- it looks like it uses a lot of pseudo-calculus.\r\n\r\n\r\nBy the way, Masamune, I don't see how your suggestion will help at all.", "Solution_4": "Two hints (in spoiler):\n\n\n\n[hide]\n\n1. When x = k, the left-hand side must be 0. That determines A, B, and C in terms of a, b, and c.\n\n\n\n2. Both sides now have a factor of x - k. So cancel those factors. \n\n[/hide]", "Solution_5": "Ravi -- [hide]did you note that it was for rational x, not real? Because there would definitely be other steps involved in showing that it generalizes, if only a continuity argument of some sort.[/hide]", "Solution_6": "Joel, [hide]yes, I noticed it was for rational only. My hint was a bit cryptic. But, as you say, by continuity, the inequality must be true with < replaced by :le:. In the end, we are going to prove the inequality for all nonnegative real x <> k anyway.\n\n\n\nActually, we don't even need to prove continuity here, as long as we believe that it is true. That's because we only need to find a particular choice that works (and prove that it does). We don't need to prove that we have found the only possible choices.\n\n[/hide]", "Solution_7": "Good point.", "Solution_8": "[hide=\"Solution\"] Taking Ravi B's hint, we have\n\n$ Ak^2 + Bk + C = ak^3 + bk^2 + ck = bk^2 + ck + 2a$\n\nhence $ A = b, B = c, C = 2a$. We now have\n\n$ | \\frac {bx^2 + cx + 2a}{ax^2 + bx + c} - k | \\le |x - k| \\Leftrightarrow$\n$ | \\frac {(b - ka)x^2 + (c - kb)x + (2a - kc)}{ax^2 + bx + c} | \\le |x - k|$.\n\nDividing out and multiplying by $ ax^2 + bx + c$, we obtain\n\n$ | (b - ka) x + (c - k^2 a) | \\le |ax^2 + bx + c|$\n\nNow, if $ ax^2 + bx + c$ has real roots, then they must be equal (or else $ b - ka = 0$) and rational, but a rational value of $ x$ produces an irrational value of the LHS unless $ a = 0$ (which makes the LHS trivial). We conclude that $ ax^2 + bx + c$ has only complex roots.\n\nA line is below a convex curve iff the line is below the curve at the point where their slopes are equal. The slope of the tangent line to the quadratic at $ x$ is $ 2ax + b = b - ka$, hence $ x = - \\frac {k}{2}$. Plugging in and multiplying by $ 4$, we obtain\n\n$ | - 2bk + 2k^2 a + 4c - 4k^2 a | \\le |ak^2 - 2bk + 4c| \\Leftrightarrow$\n$ | - 2k^2 a - 2bk + 4c | \\le | ak^2 - 2bk + 4c |$\n\nWe can easily find $ a, b, c$ that satisfy this inequality: let's take $ a = 1, b = 0, c = 1$, for example. Then our final answer is (unless I have slipped up somewhere)\n\n$ \\boxed{ A = 0, B = 1, C = 2, a = 1, b = 0, c = 1 }$\n\nso that the inequality becomes\n\n$ | \\frac {x + 2}{x^2 + 1} - k | < |x - k|$ [/hide]\r\nEdit: A follow-up. Does a choice of $ A, B, C, a, b, c$ exist such that rational convergents of $ k$ are taken to rational convergents of $ k$?" } { "Tag": [ "geometry", "circumcircle", "incenter", "search", "angle bisector", "geometry proposed" ], "Problem": "Let $ T$ be the midpoint of arc $ BC$ in circumcircle $ \\omega$ of triangle $ \\triangle ABC$,that doesn't contain point $ A$.Also let $ M$ be the midpoint of arc $ ABT$ in the same circle.\r\nSuppose that $ I$,$ I_A$ and $ A'$ are incenter,$ A$-excenter and feet of angle bisector of $ \\angle BAC$,respectively.Let $ P,Q$ be intersection points of $ MI,MI_A$ with $ \\omega$.\r\nProve that $ P,A',Q$ are collinear.", "Solution_1": "[url]http://www.mathlinks.ro/viewtopic.php?search_id=1328654572&t=205460[/url]" } { "Tag": [ "inequalities", "function", "calculus", "derivative", "inequalities proposed" ], "Problem": "I give another genelization for the Iran 96 inequality.\nFor all nonnegative real numbers $ a,b,c$ such that $ ab\\plus{}bc\\plus{}ca\\equal{}1$ and $ k \\ge \\minus{}2$ is a real number, find the minimum value of the following expression\n\\[ F_k(a,b,c)\\equal{}\\frac{1}{a^2\\plus{}kab\\plus{}b^2}\\plus{}\\frac{1}{b^2\\plus{}kbc\\plus{}c^2}\\plus{}\\frac{1}{c^2\\plus{}kca\\plus{}a^2}\n\\]\n\n[hide]Results.\n+) For all $ \\minus{}2 \\le k \\le \\minus{}1 \\ \\min \\equal{} 2\\minus{}k$\n+) For all $ \\minus{}1 \\le k \\le 2 \\ \\min \\equal{} 2\\plus{}\\frac{1}{k\\plus{}2}$\n+) For all $ 2 \\le k \\ \\min \\equal{} \\frac{9}{k\\plus{}2}$[/hide]\n\n[hide]A proof for $ k\\equal{}\\minus{}2$:\nAssume $ c\\equal{}\\min(a,b,c)$ and by AM-GM:\n$ F_{\\minus{}2}(a,b,c)\\equal{}\\frac{1}{ (a\\minus{}b)^2}\\plus{}\\frac{(a\\minus{}b)^2}{(b\\minus{}c)^2(a\\minus{}c)^2} \\plus{}\\frac{2}{(a\\minus{}c)(b\\minus{}c)} \\ge \\frac{2}{(a\\minus{}c)(b\\minus{}c)} \\plus{}\\frac{2}{(a\\minus{}c)(b\\minus{}c)} \\ge \\frac{4}{ab\\plus{}bc\\plus{}ca}$[/hide]", "Solution_1": ":( \r\nThe forum Inequality is very boring these days.", "Solution_2": "The hardest is the case $ \\minus{}1\\le k\\le2$. We can prove it using the inequality\r\n\r\n$ \\frac {3k\\plus{}6}{a^2\\plus{}kab\\plus{}b^2}\\ge \\frac{2\\minus{}k}{a^2\\minus{}ab\\plus{}b^2}\\plus{}\\frac{4k\\plus{}4}{(a\\plus{}b)^2}$.", "Solution_3": "[quote=\"Vasc\"]The hardest is the case $ \\minus{} 1\\le k\\le2$. We can prove it using the inequality\n\n$ \\frac {3k \\plus{} 6}{a^2 \\plus{} kab \\plus{} b^2}\\ge \\frac {2 \\minus{} k}{a^2 \\minus{} ab \\plus{} b^2} \\plus{} \\frac {4k \\plus{} 4}{(a \\plus{} b)^2}$.[/quote]\r\nCool, Vasc !!! :coolspeak: \r\nThank you very much!", "Solution_4": "[quote=\"Vasc\"]The hardest is the case $ \\minus{} 1\\le k\\le2$. We can prove it using the inequality\n\n$ \\frac {3k \\plus{} 6}{a^2 \\plus{} kab \\plus{} b^2}\\ge \\frac {2 \\minus{} k}{a^2 \\minus{} ab \\plus{} b^2} \\plus{} \\frac {4k \\plus{} 4}{(a \\plus{} b)^2}$.[/quote]\r\n\r\nhow do you come up with sth like this vasc? i think it's extraordinary :)", "Solution_5": "[quote=\"behemont\"][quote=\"Vasc\"]The hardest is the case $ \\minus{} 1\\le k\\le2$. We can prove it using the inequality\n\n$ \\frac {3k \\plus{} 6}{a^2 \\plus{} kab \\plus{} b^2}\\ge \\frac {2 \\minus{} k}{a^2 \\minus{} ab \\plus{} b^2} \\plus{} \\frac {4k \\plus{} 4}{(a \\plus{} b)^2}$.[/quote]\n\nhow do you come up with sth like this vasc? i think it's extraordinary :)[/quote]\r\nVery nice, vasc :wink: \r\n\\[ \\frac {3k \\plus{} 6}{a^2 \\plus{} kab \\plus{} b^2}\\ge \\frac {2 \\minus{} k}{a^2 \\minus{} ab \\plus{} b^2} \\plus{} \\frac {4k \\plus{} 4}{(a \\plus{} b)^2}\\]\r\n\\[ \\iff \\frac{\\minus{}3ab(a\\minus{}b)^2(k\\plus{}1)}{(a^2 \\plus{} kab \\plus{} b^2)(a^2 \\minus{} ab \\plus{} b^2)(a \\plus{} b)^2}\\ge 0\\]", "Solution_6": "[quote=\"zaizai-hoang\"][quote=\"behemont\"][quote=\"Vasc\"]The hardest is the case $ \\minus{} 1\\le k\\le2$. We can prove it using the inequality\n\n$ \\frac {3k \\plus{} 6}{a^2 \\plus{} kab \\plus{} b^2}\\ge \\frac {2 \\minus{} k}{a^2 \\minus{} ab \\plus{} b^2} \\plus{} \\frac {4k \\plus{} 4}{(a \\plus{} b)^2}$.[/quote]\n\nhow do you come up with sth like this vasc? i think it's extraordinary :)[/quote]\nVery nice, vasc :wink:\n\\[ \\frac {3k \\plus{} 6}{a^2 \\plus{} kab \\plus{} b^2}\\ge \\frac {2 \\minus{} k}{a^2 \\minus{} ab \\plus{} b^2} \\plus{} \\frac {4k \\plus{} 4}{(a \\plus{} b)^2}\n\\]\n\n\\[ \\iff \\frac { \\minus{} 3ab(a \\minus{} b)^2(k \\plus{} 1)}{(a^2 \\plus{} kab \\plus{} b^2)(a^2 \\minus{} ab \\plus{} b^2)(a \\plus{} b)^2}\\ge 0\n\\]\n[/quote]\r\n\r\ni know the equivalence :wink: but how does he think of something like that :?:", "Solution_7": "In Algebraic Inequalities (page 315), there is the following statement:\r\n\r\nIf $ E(a,b,c) \\equal{} \\sum \\frac {pab \\plus{} bc \\plus{} ca}{a^2 \\plus{} kab \\plus{} b ^2}$ and $ k > \\minus{} 2$, then\r\n\r\n(a) $ E(a,b,c) \\ge \\frac {3(p \\plus{} 2)}{k \\plus{} 2}$, for $ p\\le k \\minus{} 1$;\r\n\r\n(b) $ E(a,b,c) \\ge \\frac {p}{k \\plus{} 2} \\plus{} 2$, for $ k \\minus{} 1\\le p \\le (k \\plus{} 2)^2$;\r\n\r\n(c) $ E(a,b,c) \\ge 2\\sqrt {p} \\minus{} k$, for $ p\\ge (k \\plus{} 2)^2$.", "Solution_8": "[quote=\"ehku\"]I give another genelization for the Iran 96 inequality.\nFor all nonnegative real numbers $ a,b,c$ such that $ ab \\plus{} bc \\plus{} ca \\equal{} 1$ and $ k \\ge \\minus{} 2$ is a real number, find the minimum value of the following expression\n\\[ F_k(a,b,c) \\equal{} \\frac {1}{a^2 \\plus{} kab \\plus{} b^2} \\plus{} \\frac {1}{b^2 \\plus{} kbc \\plus{} c^2} \\plus{} \\frac {1}{c^2 \\plus{} kca \\plus{} a^2}\n\\]\nResults.\n+) For all $ \\minus{} 2 \\le k \\le \\minus{} 1 \\ \\min \\equal{} 2 \\minus{} k$........\n\n[/quote]\r\n\r\nI think this is wrong, ehku. And for $ k\\equal{}\\minus{}1$ we know that $ min \\equal{} 9$", "Solution_9": "Sorry for my previous msg, it was wrong because I confused it. :blush: \r\n\r\nI would like only to know if it exists an analog formula for the case $ \\minus{}2 \\leq k \\leq \\minus{}1$ similar to\r\n\r\n[quote=\"Vasc\"]The hardest is the case $ \\minus{} 1\\le k\\le2$. We can prove it using the inequality\n\n$ \\frac {3k \\plus{} 6}{a^2 \\plus{} kab \\plus{} b^2}\\ge \\frac {2 \\minus{} k}{a^2 \\minus{} ab \\plus{} b^2} \\plus{} \\frac {4k \\plus{} 4}{(a \\plus{} b)^2}$.[/quote]\r\n\r\n\r\nThank you very much.", "Solution_10": "Let $ f(k)\\equal{}\\frac 1{a^2\\plus{}kab\\plus{}b^2}$. Since $ f$ is concave, by Jensen's inequality we have for $ \\minus{}2\\le k\\le\\minus{}1$:\r\n\r\n$ f(k) \\ge (k\\plus{}2)f(\\minus{}1)\\plus{}(\\minus{}k\\minus{}1)f(\\minus{}2)$.", "Solution_11": "[quote=\"Vasc\"]In Algebraic Inequalities (page 315), there is the following statement:\n\nIf $ E(a,b,c) \\equal{} \\sum \\frac {pab \\plus{} bc \\plus{} ca}{a^2 \\plus{} kab \\plus{} b ^2}$ and $ k > \\minus{} 2$, then\n\n(a) $ E(a,b,c) \\ge \\frac {3(p \\plus{} 2)}{k \\plus{} 2}$, for $ p\\le k \\minus{} 1$;\n\n(b) $ E(a,b,c) \\ge \\frac {p}{k \\plus{} 2} \\plus{} 2$, for $ k \\minus{} 1\\le p \\le (k \\plus{} 2)^2$;\n\n(c) $ E(a,b,c) \\ge 2\\sqrt {p} \\minus{} k$, for $ p\\ge (k \\plus{} 2)^2$.[/quote]\r\nOk, your result is better. I haven't any complete solution now, but it suffits to prove in three cases $ p \\equal{} k \\minus{} 1$, $ p\\equal{}(k\\plus{}2)^2$ and $ k\\equal{}\\minus{}2$ (for (a) and (b) it is sure, but I am not sure about (c)).\r\n\r\nPlease tell me if the solution in your book is nice. Thank you!", "Solution_12": "[quote=\"ehku\"] I am not sure about (c). Please tell me if the solution in your book is nice. Thank you![/quote]\r\nYes, it is nice. :wink: \r\nThe idea is to show that $ E(a,b,c)\\ge E(0,b,c)\\ge 2\\sqrt p\\minus{}k$ for the case $ pb^2\\ge (k\\minus{}1)bc\\plus{}ca$, with $ a\\le b\\le c$.", "Solution_13": "[quote=\"Vasc\"]Let $ f(k) \\equal{} \\frac 1{a^2 \\plus{} kab \\plus{} b^2}$. Since $ f$ is concave, by Jensen's inequality we have for $ \\minus{} 2\\le k\\le \\minus{} 1$:\n\n$ f(k) \\ge (k \\plus{} 2)f( \\minus{} 1) \\plus{} ( \\minus{} k \\minus{} 1)f( \\minus{} 2)$.[/quote]\r\n\r\nDear Vasc, I have checked it and it is false as the function $ f(k)$ has second \r\nderivative negative only for $ k \\leq \\frac{\\minus{}a^2\\minus{}b^2}{ab}$ and $ \\minus{}2 \\geq \\frac{\\minus{}a^2\\minus{}b^2}{ab}$ so $ f(k) \\equal{} \\frac 1{a^2 \\plus{} kab \\plus{} b^2}$ doesn't satisfy $ f(k) \\ge (k \\plus{} 2)f( \\minus{} 1) \\plus{} ( \\minus{} k \\minus{} 1)f( \\minus{} 2)$ for $ \\minus{} 2\\le k\\le \\minus{} 1$ . I have also checked $ f(k) \\ge (k \\plus{} 2)f( \\minus{} 1) \\plus{} ( \\minus{} k \\minus{} 1)f( \\minus{} 2)$ by direct calculation and for $ \\minus{} 2\\le k\\le \\minus{} 1$ it is false. Am I right, Vasc ?", "Solution_14": "You are right, Manlio: $ f(k)$ is convex, not concave. Therefore, we can't prove the inequality for $ \\minus{}2\\le k\\le \\minus{}1$ using only the result for $ k\\equal{}\\minus{}2$ and $ k\\equal{}\\minus{}1$.", "Solution_15": "[quote=\"ehku\"]For all nonnegative real numbers $ a,b,c$ such that $ bc \\plus{} ca \\plus{} ab > 0$ and $ k \\ge \\minus{} 2$ is a real number, we have\n\n$ \\frac {1}{b^2 \\plus{} kbc \\plus{} c^2} \\plus{} \\frac {1}{c^2 \\plus{} kca \\plus{} a^2} \\plus{} \\frac {1}{a^2 \\plus{} kab \\plus{} b^2} \\geq \\frac {2 \\minus{} k}{bc \\plus{} ca \\plus{} ab},$\n\nwith equality if $ a \\equal{} 0,b \\equal{} 2,c \\equal{} 1 \\minus{} k \\minus{} \\sqrt {(k \\plus{} 1)(k \\minus{} 3)}$ when $ k\\in [ \\minus{} 2, \\minus{} 1].$[/quote]$ \\sum{\\frac {1}{b^2 \\plus{} kbc \\plus{} c^2}} \\minus{} \\frac {2 \\minus{} k}{bc \\plus{} ca \\plus{} ab}\\equiv\\frac {F(a,b,c,k)}{(bc \\plus{} ca \\plus{} ab)\\prod{\\left(b^2 \\plus{} kbc \\plus{} c^2\\right)}}.$\r\n\r\n$ F(a,b,c,k) \\equal{} F(a,a \\plus{} s,a \\plus{} t,r \\minus{} 2)$\r\n\r\n$ \\equal{} r^2\\left(r^2 \\minus{} 4r \\plus{} 9\\right)a^6 \\plus{} 2r^2\\left(r^2 \\minus{} 4r \\plus{} 9\\right)(s \\plus{} t)a^5$\r\n\r\n$ \\plus{} r\\left[\\left(r^3 \\minus{} 2r^2 \\plus{} 3r \\plus{} 12\\right)(s \\minus{} t)^2 \\plus{} \\left(6r^3 \\minus{} 22r^2 \\plus{} 48r \\plus{} 12\\right)st\\right]a^4$\r\n\r\n$ \\plus{} r(s \\plus{} t)\\left[2\\left(r^2 \\minus{} 3r \\plus{} 7\\right)(s \\minus{} t)^2 \\plus{} \\left(2r^3 \\minus{} 5r^2 \\plus{} 9r \\plus{} 17\\right)st\\right]a^3$\r\n\r\n$ \\plus{} \\left \\{(s \\minus{} t)^2\\left[\\left(r^2 \\plus{} 3\\right)s^2 \\plus{} 3r\\left(r^2 \\minus{} 3r \\plus{} 7\\right)st \\plus{} \\left(r^2 \\plus{} 3\\right)t^2\\right] \\plus{} \\left(r^4 \\plus{} 2r^3 \\minus{} 8r^2 \\plus{} 30r \\plus{} 3\\right)s^2t^2\\right\\}a^2$\r\n\r\n$ \\plus{} (s \\plus{} t)\\left[(s \\minus{} t)^2\\left(2s^2 \\plus{} r^2st \\plus{} 2t^2\\right) \\plus{} \\left(r^4 \\plus{} 2r^3 \\minus{} 8r^2 \\plus{} 30r \\plus{} 3\\right)s^2t^2\\right]a$\r\n\r\n$ \\plus{} st\\left[s^2 \\minus{} (3 \\minus{} r)st \\plus{} t^2\\right]^2\\geq0,$\r\n\r\nwhich is clearly true for $ a \\equal{} \\min\\{a,b,c\\}.$" } { "Tag": [ "puzzles" ], "Problem": "Can you figure this out?\r\n\r\nI was walking down the street when I saw a person who never lies. She was holding 4 oranges, but said there were only 3. How could this have happened?", "Solution_1": "3 oranges, one toy.", "Solution_2": "Nope. All the oranges are real!", "Solution_3": "just could see 3 (one of them hidden), so she thought she just had 3", "Solution_4": "No, she is aware of how many oranges she has. Good try, though!", "Solution_5": "Maybe she said there were only 3.......boxes or hats or something other than oranges?", "Solution_6": "Her words exactly: \"Hello, Isabella, I am randomly telling you that I have 3 oranges.\"", "Solution_7": "Maybe she can't count?", "Solution_8": "Nice try, but she can count. Her abilities are those of the average person.", "Solution_9": "Orange is the name of her cat that she is holding with her and she says she has 3 oranges the fruit?", "Solution_10": "Good guess, but no. They are all fruit.", "Solution_11": "is she talking about some one else's or her oranges?", "Solution_12": "She is speaking in some language where 4 maps to 3...", "Solution_13": "In the time between when the person saw her and when she talked to him/her, she ate an orange?", "Solution_14": "clementine, anyone?", "Solution_15": "She dropped an orange when you saw her?", "Solution_16": "Those are all good guesses, but none of them are right.", "Solution_17": "she thought that one of the oranges was a tangerine but it really wasnt?\r\n\r\nlame, I know :P", "Solution_18": "[quote=\"isabella2296\"]but said there were only 3.[/quote]\n[quote=\"isabella2296\"]\"Hello, Isabella, I am randomly telling you that I have 3 oranges.\" [/quote]\r\n\r\nThese two statements are not quite the same.\r\nIn the first case, she could have been holding 4 oranges and 3 apples and saying that she only had 3 (apples).\r\nIn the second case, she could have been holding 4 apples, and she says she has 3, and 4>3.", "Solution_19": "I apologize for not being clear. She is referring to the oranges. \r\n\r\nShould I give the answer, or let people keep guessing?\r\n\r\nHINT\r\n[hide]It's the lamest answer you can think of.[/hide]", "Solution_20": "She saw a person who never lies but was acctually talking to someone who does lie?", "Solution_21": "[quote=\"isabella2296\"]\n\nHINT\n[hide]It's the lamest answer you can think of.[/hide][/quote]\r\n\r\nAnd yet we haven't guessed it. :rotfl:\r\nThis riddle will definitely be an \"ohhh!\"er.\r\n\r\nI'm fresh out of ideas...", "Solution_22": "You're the one that's lying.", "Solution_23": "[quote=\"t0rajir0u\"]You're the one that's lying.[/quote]\r\n\r\nWoah. You're right!!! Nice!!", "Solution_24": "Ohhhhhh.......... cough[size=59]thats stupid[/size]cough", "Solution_25": "i thought it was that.", "Solution_26": "wait, it's contradictory.\r\n\r\nYou are lying. so, you didn't meet someone that says truth. She didn't say that she had 3 oranges, and she didn't have four. So, what is the meaning of this?", "Solution_27": "well he might only be lying about one part.", "Solution_28": "You meet someone who never lies. You say the person has 4 oranges. You are lying about that. The person says that he/she has 3 oranges and he/she is telling the truth.", "Solution_29": "You go, t0rajir0u. Simple, and no one could get it :wink:", "Solution_30": "[quote=\"dynamo729\"]Ohhhhhh.......... cough*[b]thats stupid[/b]*]cough[/quote]\r\n\r\nThank you for your amazingly polite comment. That makes me feel so good inside.", "Solution_31": "Hey, dynamo, if you couldn't get it, then why is it stupid? What does it make you? [b]That[/b] is my riddle.", "Solution_32": "Maybe he's just kidding. :O", "Solution_33": "NICE RIDDLE I was thinking about it and.....\r\nOMG IZZY IS A LIAR??????????? :?: \r\njkjkjkjkjkjkjkjkjkjkjkjkjkjkjkjkjkjkjkjk :)", "Solution_34": "[quote=\"D3m0n Shad0w\"]Hey, dynamo, if you couldn't get it, then why is it stupid? What does it make you? [b]That[/b] is my riddle.[/quote]\r\n\r\nThe riddle is totally lame (as said by bella) and I was not thinking at the right level. And for your information, I was technically correct in my first guess (before bella modified the riddle), so...", "Solution_35": "riddles can have various answers. most of them are pretty vague", "Solution_36": "it was clever because it required looking at it at a different angle. It is true that any riddle could have this answer... but the fact that the riddle involved a person who never lies made it appropriate I suppose." } { "Tag": [ "ratio", "projective geometry", "geometry unsolved", "geometry" ], "Problem": "Circles seem to be the bane of my mathematical existence!..\r\n\r\nIf anyone could please help me, this would be very much appreciated..\r\n\r\nConsider the following..\r\n\r\nConsider three circles. For every couple of circles, consider their two common external tangent lines and take its point of intersection.\r\nNeed to show that these 3 points of intersection belong to one line.\r\n\r\nThanks!", "Solution_1": "One possible approach is to notice that the intersection point $ A$ of the common tangents to 2 circles lies on the line joining their centers $ O_1$ and $ O_2$ and $ \\frac{|O_1A|}{|O_2A|}$ equals to the ratio of the radii. Now just use the Menelaus theorem.", "Solution_2": "[quote=\"Finesse\"]Circles seem to be the bane of my mathematical existence!..\n\nIf anyone could please help me, this would be very much appreciated..\n\nConsider the following..\n\nConsider three circles. For every couple of circles, consider their two common external tangent lines and take its point of intersection.\nNeed to show that these 3 points of intersection belong to one line.\n\nThanks![/quote]\r\n[hide=\"Solution\"]\nLet the circles have centers $ A$, $ B$, and $ C$ and let the common tangents of the circles with centers $ A$ and $ B$ intersect at $ X$, the common tangents of circles with centers $ A$ and $ C$ intersect at $ Y$, and the common tangents of circles with centers $ B$ and $ C$ intersect at $ Z$. Furthermore, allow the circle with center $ A$ to have radius $ r_1$, and those with centers $ B$ and $ C$ to have radii of $ r_2$ and $ r_3$ respectively. Consider the circles with centers $ A$ and $ B$ with common tangents $ MN$ and $ M'N'$ such that $ M$ and $ M'$ lie on the circle with center $ A$ and $ N$ and $ N'$ lie on the circle with center $ N$. Notice that $ MN$, $ M'N'$, and $ AB$ are all concurrent at $ X$. Hence, we have that $ \\triangle XAM\\sim \\triangle XBN\\implies \\frac {XA}{XB} \\equal{} \\frac {AM}{BN} \\equal{} \\frac {r_1}{r_2}$. Similarly, we can see that $ \\frac {BZ}{CZ} \\equal{} \\frac {r_2}{r_3}$ and $ \\frac {CY}{AY} \\equal{} \\frac {r_3}{r_1}$. Hence, $ \\frac {XA}{XB}\\cdot \\frac {BZ}{CZ}\\cdot \\frac {CY}{AY} \\equal{} \\frac {r_1}{r_2}\\cdot \\frac {r_2}{r_3}\\cdot \\frac {r_3}{r_1} \\equal{} 1$, and by the converse of Menalaus's Theorem, $ X$, $ Y$, and $ Z$ are collinear, as desired. [/hide]", "Solution_3": "Known as Monge's theorem and proven at http://www.mathlinks.ro/viewtopic.php?t=39186 as well. As you see by most of its proofs, it has not too much to do with circles...\r\n\r\n darij" } { "Tag": [], "Problem": "Consider a cup filled with 100 oz. of water and another cup filled with 100 oz. of wine. I pour 30 oz. of water into the wine. Then I pour 30 oz. of this mixture back to the water, again making it 100 oz. each. Compare the amount of wine in a cup of water and amount of water in a cup of wine after I do this.", "Solution_1": "The amount of wine in the water is 0%, and the amount of wine in the water is also 0%. :D", "Solution_2": "[hide=\"real life solution\"]\nNowhere in the problem did it say the water and wine were evenly distributed. In this case it would be no answer.\n[/hide]\n[hide=\"if it is evenly distributed\"]\n, then you can say:\nAfter first pour\nwine:$\\frac{3}{13}$ water\n\nAfter second pour,\nThere are $13 \\frac{1}{13}$ oz. wine in the water.\nThere are $23 \\frac{1}{13}$ oz. water in the wine.\n[/hide]", "Solution_3": "Actually the problem says\r\n\r\nCompare the amount of wine in A cup of water and amount of water in A cup of wine after I do this.\r\n\r\nI doesn't say to use the same cups he poured into, just to use a cup of pure water and a cup of pure wine.", "Solution_4": "[quote=\"kohjhsd\"]Consider a cup filled with 100 oz. of water and another cup filled with 100 oz. of wine. I pour 30 oz. of water into the wine. Then I pour 30 oz. of this mixture back to the water, again making it 100 oz. each. Compare the amount of wine in a cup of water and amount of water in a cup of wine after I do this.[/quote]\r\n\r\nSo on the first first pour, you have $70%\n$ ounces of water in the water glass and $\\frac{3}{13}$ of water in the wine glass.\r\n\r\nOn the second pour there is $13\\frac{1}{13}$of wine in the water glass\r\nThere is $23\\frac{1}{13}$ water in the wine glass.", "Solution_5": "[quote=\"hunter34\"][quote=\"kohjhsd\"]Consider a cup filled with 100 oz. of water and another cup filled with 100 oz. of wine. I pour 30 oz. of water into the wine. Then I pour 30 oz. of this mixture back to the water, again making it 100 oz. each. Compare the amount of wine in a cup of water and amount of water in a cup of wine after I do this.[/quote]\n\nSo on the first first pour, you have $70%\n$ ounces of water in the water glass and $\\frac{3}{13}$ of water in the wine glass.\n\nOn the second pour there is $13\\frac{1}{13}$of wine in the water glass\nThere is $23\\frac{1}{13}$ water in the wine glass.[/quote]\r\n\r\nThere's nothing in the problem about using the same cups that you poured into. It says to compare with a cup of water and a cup of wine, so the cup of water is 100% water, and the cup of wine is 100% wine. It's a logic question, so it's supposed to be a trick, I guess.", "Solution_6": "[quote=\"ZzZzZzZzZzZz\"][quote=\"hunter34\"][quote=\"kohjhsd\"]Consider a cup filled with 100 oz. of water and another cup filled with 100 oz. of wine. I pour 30 oz. of water into the wine. Then I pour 30 oz. of this mixture back to the water, again making it 100 oz. each. Compare the amount of wine in a cup of water and amount of water in a cup of wine after I do this.[/quote]\n\nSo on the first first pour, you have $70%\n$ ounces of water in the water glass and $\\frac{3}{13}$ of water in the wine glass.\n\nOn the second pour there is $13\\frac{1}{13}$of wine in the water glass\nThere is $23\\frac{1}{13}$ water in the wine glass.[/quote]\n\nThere's nothing in the problem about using the same cups that you poured into. It says to compare with a cup of water and a cup of wine, so the cup of water is 100% water, and the cup of wine is 100% wine. It's a logic question, so it's supposed to be a trick, I guess.[/quote]\r\n\r\nActually, I meant to use the same cups with mixed liquids in it. There's no trick there(I even had to specify that?).", "Solution_7": "arent they the same\r\n\r\nthink...\r\nthey each have 100 oz. also, if wine is missing k, then that k is in the water. this k amount of wine displaces k in the water cup, which will end up in the wine cup. then both have k of the other. \r\n\r\np.s. since when did cups have 100 oz. make it a container\r\n\r\nAlso, those who said 23+1/3 in one of the glasses, it isnt a glass. its a cup so make it proportional.", "Solution_8": "[quote=\"mchoi815\"]arent they the same\n\nthink...\nthey each have 100 oz. also, if wine is missing k, then that k is in the water. this k amount of wine displaces k in the water cup, which will end up in the wine cup. then both have k of the other. \n\np.s. since when did cups have 100 oz. make it a container\n\nAlso, those who said 23+1/3 in one of the glasses, it isnt a glass. its a cup so make it proportional.[/quote]\r\n\r\nThat's the intended logical conclusion (no calculation needed)." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=6$ Find the max value of :\r\n\\[ a_{1}a_{2}a_{3}+a_{2}a_{3}a_{4}+a_{3}a_{4}a_{5}+a_{4}a_{5}a_{6}+a_{5}a_{6}a_{1}+a_{6}a_{1}a_{2} \\]", "Solution_1": "Further information about $a_1,a_2,...,a_6$ is needed, otherwise there is no max value for $s= a_{1}a_{2}a_{3}+a_{2}a_{3}a_{4}+a_{3}a_{4}a_{5}+a_{4}a_{5}a_{6}+a_{5}a_{6}a_{1}+a_{6}a_{1}a_{2}$. Indeed, if we take $a_1=...=a_5=-x$ and $a_6=5x+6$ we obtain \r\n$s=12x^3+18x^2$ and this expression can be as large as we wish for x sufficiently large.\r\n\r\nI suppose $a_1,a_2,...,a_6$ should have been assumed positive.", "Solution_2": "I think that it is safe to assume \\[ a_1,a_2,...a_6 \\ge 0 \\]", "Solution_3": "I strongly suspect that the maximum value (when all variables are non-negative) is 8.\r\n\r\n(three variables are 2, and the other three are 0)", "Solution_4": "We just had this at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=399698#p399698]http://www.mathlinks.ro/Forum/viewtopic.php?t=1832 post #5[/url]: For any six nonnegative reals $a_1$, $a_2$, $a_3$, $a_4$, $a_5$, $a_6$, we have\r\n\r\n$a_1a_2a_3+a_2a_3a_4+a_3a_4a_5+a_4a_5a_6+a_5a_6a_1+a_6a_1a_2\\leq\\frac{1}{27}\\left(a_1+a_2+a_3+a_4+a_5+a_6\\right)^3$,\r\n\r\nand equality can be assumed (take $a_1=a_2=0$, $a_3=a_6=1$, $a_4=a_5=2$).\r\n\r\n Darij" } { "Tag": [ "linear algebra", "matrix", "geometry", "geometric transformation", "rotation", "linear algebra unsolved" ], "Problem": "Hello:\r\n\r\nI need help with this isometry problem.\r\na) Let $ D$ be a triangle defined by $ A(1,1), \\,B(4,1),\\, C(4,3)$.\r\n\r\nDetermine an isometry $ T: R^2 \\rightarrow R^2$. Find $ T(D)$ and explain what geometric relation exists between $ D$ and $ T(D)$.\r\n\r\nb) Generalise part a) to a any region in the plane.\r\n\r\nThanks for helping. \r\nI have problems with appllying the definition of isometry.", "Solution_1": "Something is missing from the statement, perhaps some description of $ T$.", "Solution_2": "That's all. I have to define the transformation $ T$ such that $ T$ is an isometry.", "Solution_3": "Hi thrasher.\r\nYou don't understand the definition of an isometry. If you don't choose explicitly an isometry (as you want), then you can only say what follows:\r\n$ T(ABC)$ is a triangle $ A'B'C'$ s.t. $ AB\\equal{}A'B',AC\\equal{}A'C',BC\\equal{}B'C'$.\r\nIf $ det(T)\\equal{}1$ then $ ABC,A'B'C'$ have the same orientation (you can slide $ ABC$ over $ A'B'C'$).If $ det(T)\\equal{}\\minus{}1$ then $ ABC,A'B'C'$ have opposite orientations (you can't slide $ ABC$ over $ A'B'C'$).", "Solution_4": "thanks , now I have another question:\r\n\r\n\r\n\r\nWill I always obtain +1 and -1 in the determinant?", "Solution_5": "Let $ T$ be an affine isometry; we choose $ Oxy$ an orthonormal basis fo $ \\mathbb{R}^2$. There exists $ f$ a vectorial isometry s.t. $ T(M)\\equal{}T(O)\\plus{}f(\\vec{OM})$; there exist $ a,b$ with $ a^2\\plus{}b^2\\equal{}1$ s.t. the $ f$-matrix is in one of the following 2 forms: $ \\begin{pmatrix}a&\\minus{}b\\\\b&a\\end{pmatrix}$ (a rotation), $ \\begin{pmatrix}a&b\\\\b&\\minus{}a\\end{pmatrix}$ (a symmetry)." } { "Tag": [], "Problem": "Is there a pattern to the minimum amount of move required to complete the Tower of Hanoi with x number of rings?\r\nThe original Tower of Hanoi is inspired by a legend...\r\n[quote=\"The Regents of the University of California\"]legend that tells of a Hindu temple where the pyramid puzzle might have been used for the mental discipline of young priests. Legend says that at the beginning of time the priests in the temple were given a stack of 64 gold disks, each one a little smaller than the one beneath it. Their assignment was to transfer the 64 disks from one of the three poles to another, with one important proviso large disk could never be placed on top of a smaller one. The priests worked very efficiently, day and night. When they finished their work, the myth said, the temple would crumble into dust and the world would vanish.[/quote]\r\nHow many moves will it take for the priests to complete this task? If the priest made each move in one second, how long will it take for them to be finished? (Please round to the nearest billion years)", "Solution_1": "Neal and I did this in class last year...\n\n[hide]Ok... for \n\n1, it was 1 move\n\n2, 3\n\n3, 7\n\n4...........15\n\nand so forth... it took the moves needed to move n-1 pieces times two + 1(moving the largest or nth disk)\n\n\n\nIt was 2^x - 1\n\nso 2^64 - 1 = 18446744073709551615 moves... and if it took 1 second for each move... (Ans)/3600/24/365.25 :approx: 585 billion years???[/hide]", "Solution_2": "Correct!\r\n\r\n...so you did this in class. hmm...", "Solution_3": "yah, our teacher didn't follow the textbook(thank goodness) since it's like first 5 chapters-review..." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all square numbers with the property that all their digits ( in base 10) are odd.", "Solution_1": "The only such squares are $ 1, 9$.\r\n\r\nSuppose such a square of at least two digits existed. Such a square is $ \\equiv 1 \\bmod 8$ and $ \\equiv 0, 1, 4 \\bmod 5$, so the last three digits are a three-digit number $ abc$ also congruent to $ 1 \\bmod 8$ and $ c \\equal{} 1, 9$. But then $ ab$ is an odd number divisible by $ 4$ - contradiction." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "search" ], "Problem": "Prove that for every positive integer n there exists an n -digit number divisible by 5^n all of whose digits are odd.", "Solution_1": "We simply use the mathematical induction.\r\n\r\nSuppose that for $ n$, we have $ B$ $ \\vdots$ $ 5^n$\r\n\r\n$ \\Rightarrow B\\in\\{ k5^{n+1},k5^{n+1}+5^n,k5^{n+1}+2 \\cdot 5^n,k5^{n+1}+3 \\cdot 5^n,k5^{n+1}+4 \\cdot 5^n\\}$\r\n\r\nNow, $ 1B, 3B, 5B, 7B, 9B$ $ \\vdots$ $ 5^n$. Suppose that none of these numbers is divisible by $ 5^{n+1}$\r\n\r\n$ \\Rightarrow \\{1B, 3B, 5B, 7B, 9B\\}=\\{k5^{n+1}+5^n,k5^{n+1}+2 \\cdot 5^n,k5^{n+1}+3 \\cdot 5^n,k5^{n+1}+4 \\cdot 5^n\\}$\r\n\r\n$ \\Rightarrow \\exists i,j \\in \\{ 1,3,5,7,9 \\}, i\\not= j$ such that $ iB-jB$ $ \\vdots$ $ 5^{n+1}$ $ \\Rightarrow (i-j)$ $ \\vdots$ $ 5$, impossible\r\n\r\n$ \\Rightarrow A \\in \\{1B, 3B, 5B, 7B, 9B\\}$", "Solution_2": "See the related problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=317971887&t=228211]here[/url]." } { "Tag": [ "limit", "algebra unsolved", "algebra" ], "Problem": "Given real numbers a, b the sequence $(x_{n})^{inf}_{n=0}$ is defined by $x_{0}= a$ and $x_{n+1}= x_{n}+b sin x_{n}$ for every $n \\geq 0$\r\n\r\n(a) If b=1, prove that for every a, the sequence $(x_{n})$ has a finite limit when n->$inf$, and find this limit.\r\n\r\n(b) Prove that for every b>2 there exists a real number a for which the sequence $(x_{n})$ does not have a finite limit when n->{inf}", "Solution_1": "Consider $f(x)=x+bsin x$. \r\n1. If $a=k\\pi$, or b=0, then $x_{n}=a, \\lim_{n\\to \\infty }x_{n}=a$.\r\n2. If $02$ and $a\\not =k\\pi$, then don't exist $\\lim_{n\\to \\infty }x_{n}$." } { "Tag": [ "combinatorics solved", "combinatorics" ], "Problem": "Prove that, given 9 points on the plane, no three colinear, we can always get a convex pentagon.", "Solution_1": "This is a particular case of the famous Erds-Szekeres problem (the Happy ending problem). The solution of this problem is not really difficult but quite long to write.\r\nAn excellent survey of the problem, including the answer of this problem (if I remember correctly ;) ), and related results can be found here :\r\nhttp://www.ams.org/bull/2000-37-04/\r\n\r\n(Hope it works ;) )\r\n\r\nPierre.", "Solution_2": "It works. Cool. :D :D :D \r\n\r\nPierre.", "Solution_3": "[quote=\"pbornsztein\"]An excellent survey of the problem, including the answer of this problem (if I remember correctly ;) ), and related results can be found here :\nhttp://www.ams.org/bull/2000-37-04/[/quote]\r\nCool! You found links! I told about this article in another thread." } { "Tag": [ "AMC", "AIME", "geometry", "analytic geometry", "geometric transformation", "rotation", "rhombus" ], "Problem": "the 21st AIME2 2003\r\n\r\nproblems 7, 11, and 14 I am having trouble with\r\n\r\n7. ABCD is a rhombus. The circumradii of ABD, ACD are 12.5, 25. Find the area of the rhombus.\r\n\r\n11. ABC has AC = 7, BC = 24, angle C = 90 degrees. M is the midpoint of AB, D lies on the same side of AB as C and had DA = DB = 15. Find area CDM.\r\n\r\n13. ABCDEF is a convex hexagon with all sides equal and opposite sides parallel. Angle FAB = 120o. The y-coordinates of A, B are 0, 2 respectively, and the y-coordinates of the other vertices are 4, 6, 8, 10 in some order. Find its area.\r\n\r\n7 and 11 I really can't see what to do. For 13, I got 2 equations:\r\n\r\na(radical 3)sin theta = 2\r\n2a(cos theta) = 10\r\n\r\nwhere a is length of hexagon and theta is the angle of rotation\r\n\r\nwhen i solve this i get area of hexagon = 39.5 radical 3, but Kalva has a different answer\r\n\r\nany help would be greatly appreciated", "Solution_1": "[hide=\"7\"]Draw the rhombus $ABCD$. Rhombi have all sides equal to each other, and opposite angles equal to each other. $ABD$ will be an acute triangle, and $ACD$ will be an obtuse triangle. The circumradius of a triangle is the radius of the circumscribed circle, so we know that the long diagonal $AC=25\\cdot 2=50$ and the short diagonal $BD=12.5\\cdot 2=25$. Also, in a rhombus, the two diagonals are perpendicular and always bisect each other, so we can actually just take where they intersect, then form two triangles, each with base $25$ and height $25$. Therefore, we have $(\\dfrac{1}{2}\\cdot 25\\cdot 25)\\cdot 2=\\boxed{625}$.[/hide]", "Solution_2": "[quote=\"JesusFreak197\"]The circumradius of a triangle is the radius of the circumscribed circle, so we know that the long diagonal $AC=25\\cdot 2=50$ and the short diagonal $BD=12.5\\cdot 2=25$. [/quote]\r\nI don't think this is right. You have interchanged circumradius and altitude. Your answer is accordingly incorrect. \r\n\r\nThis is how I did #7: \r\n[hide]\nLet the side length of the rhombus be denoted $x$. By the Extended Law of Sines\n$\\dfrac{x}{\\sin \\angle ABD}=2R=25$\n$\\sin \\angle ABD=\\dfrac{x}{25}$\nSimilarly $\\sin \\angle ACD = \\dfrac{x}{50}$\nSince $\\angle ABC +\\angle BCD = 2 \\angle ABD + 2 \\angle ACD=\\pi$, $\\sin \\angle ACD = \\cos \\angle ABD$\nWe can now use the pythagorean identity \n$\\dfrac{x^2}{25^2}+\\dfrac{x^2}{50^2}=1$\n$x=10\\sqrt{5}$\n\nWe can also use the Extended Law Of Sines on $\\angle BAD$ and $\\angle ADC$. We find that $2(BD) = CA$. \n\nThe diagonals of a rhombus are orthogonal. Let $a=\\left( \\dfrac{1}{2} \\right) BD$. Thus, by the Pythagorean Theorem $5a^2=x^2=500$ which implies a=10. The area of the rhombus equals twice the area of $\\triangle ABC$, which is 200. Thus the answer is $\\boxed{400}$, which agrees with kalva. \n\n\n\n[/hide]", "Solution_3": "Oh, right, I was assuming that for the circumscribed circle, one of the sides of the triangle would be the diameter of the circle. I don't really know why, since that's only true for right triangles. :huh:", "Solution_4": "Thanks for the explanations. I had never seen the Extended Law of Sines - all I ever saw was the a/sin A = b/sin B = c/sin C... Shows how much I have to study. Curious, is there some other way to solve problem 7 without using the Extended Law of Sines but instead some other method that, while a bit more involved uses only basic knowledge of geometry? Because so far, I've been trying to solve these problems using only more basic tools/concepts, and while I'll certainly be trying to pick up more advanced laws/theorems, I'm not sure if I will be able to learn enough of them come time for the tests, in which case all I will have at my disposal are the simpler (though I imagine that for many of you on this forum the Extended Law of Sines is already basic knowledge) things taught in middle school. Also, could anyone explain to me as to how to solve the other problems?", "Solution_5": "[quote=\"Morgoth\"]Curious, is there some other way to solve problem 7 without using the Extended Law of Sines but instead some other method that, while a bit more involved uses only basic knowledge of geometry?[/quote] There's always anothe way!\r\n[hide]What we really need to do is find the circumradius of a general isosceles triangle. Let's look at triangle ADC. Let X be the midpoint of AC, and let AX=XC=a and DX=b.\nBecause the circumcenter of a triangle is equidistant from its vertices, the circumcenter O is the intersection of the perpendicular bisectors of the three sides. (Make sure you know this.) So if M is the midpoint of CD and O is the circumcenter, then O is on DX and DMO is a right triangle. In fact, DMO and DXC are similar!\nWe'd like to know the length of DO, so we use the similar triangles:\n$\\dfrac{DO}{DM}=\\dfrac{DC}{DX}$. And we know that $DX=b$, $DC=\\sqrt{a^2+b^2}$, and $DM=\\tfrac{1}{2}\\sqrt{a^2+b^2}$, so we immediately get $R=DO=\\dfrac{a^2+b^2}{2b}$. Likewise, the circumradius of ABD is $\\dfrac{a^2+b^2}{2a}$. These give the equations to find $a$ and $b$ and thus the area $2ab$.[/hide]", "Solution_6": "Yes, there's always another way!\r\n\r\nZabel I owned the WOOT test :).\r\n \r\nLooking forward to grading yours ;)", "Solution_7": "[quote=\"darktreb\"]Yes, there's always another way!\n\nZabel I owned the WOOT test :).\n \nLooking forward to grading yours ;)[/quote]\r\n\r\nHaha! Sounds good. :)", "Solution_8": "Thanks for the explanation zabelman, it really helped. Can anyone help me with the other problems? I'm still stuck on those.", "Solution_9": "At first glance, for 11, [hide]I would coordinatize: place AB on the x-axis with M at the origin, find the coordinates of C and D, and get the area from there. (If you don't know how to get area from coordinates, just ask.[/hide]", "Solution_10": "How can you find the coordinates of D though?" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "prove that if $ x$ and $ y$ are in the same conjugacy class of a group $ G$ , then $ |C(x)|\\equal{}|C(y)|$( the centralizer of an element)", "Solution_1": "Ok, where are you stuck\u00bf\r\nTry to show that $ C(x)$ is conjugated to $ C(y)$ (or converse).", "Solution_2": "if $ G$ is finite , then it's trivially easy to prove using $ |Conj(a)|\\equal{}[G: C(a)]$\r\n\r\nIs that true ??", "Solution_3": "Well, I think that's harder to prove than the original statement :D \r\nBut yes, it's true.", "Solution_4": ":D \r\n\r\nWhat if $ G$ is infinite ? Does that statement still works or not?\r\n\r\nDo u have another prove ?? :wink:", "Solution_5": "you [i]may[/i] read the first post of ZetaX.", "Solution_6": "I read it , but I couldn't do it .\r\n\r\nI meant when I said \"Do u have another proof ?\" , Write it Plz.\r\n\r\n :)", "Solution_7": "Ti describe the first way in more exterior terms:\r\nConsider the inner automorphisms $ z \\mapsto szs^{\\minus{}1}$ and see what happens." } { "Tag": [], "Problem": "I'm going to make a chess tournament since the last one pretty much failed. I will accept 8 or 16 competitors with either single, double or pool play elimination. I will join if necessary and I will create bracket. Please don't join if you don't think you can be on for a long time. Time control is: 20 minutes each player and no delay.", "Solution_1": "Aww delay is so much fun. Oh well, I will join this CHESE tournament.", "Solution_2": "As always I join.\r\n\r\nWill this work?", "Solution_3": "Since I don't know any of your skill levels I will pair you up according to conviencence. This should work, if you don't open a prod in 72 hours you lose by default.", "Solution_4": "Is the board made out of mozzarella? :D", "Solution_5": "I don't know what your talking about :P :wink: :D" } { "Tag": [], "Problem": "Antoine paid a tax of $ 10\\%$ on his yearly income. How many dollars must his income have been if he had $ \\$6,300$ remaining after taxes?", "Solution_1": "10% = 1/10, and he has 9/10 left after taxes. Thus we must multiply by 10/9 to 6300 = 7000 dollars before taxes." } { "Tag": [], "Problem": "Now i wish i could recollect (BLANK)\r\n\"Eureka\" cried the great inventor. \r\n\r\nwho was the inventor? and what comes in blank", "Solution_1": "he was archimedes. what about the blank??", "Solution_2": "found the answer, its pi in the blank\r\nand yeh it was archimedes so thnx" } { "Tag": [ "trigonometry", "function", "calculus", "integration", "number theory", "real analysis", "real analysis unsolved" ], "Problem": "Let $n\\in\\mathbb{N}^{*}$ and $a_{k},b_{k}\\in\\mathbb{R},k=1,2,\\ldots, n$. Prove that the following equation\r\n\\[\\sum_{k=1}^{n}(a_{k}\\cos (kx)+b_{k}\\sin (kx))=0, \\]\r\nhas at least one solution in $(0,2\\pi)$.", "Solution_1": "Come on, Cezar! :wink:", "Solution_2": "Oki, Gabi! I will try not post such easy and useless problems, But maybe there are here on the forum some users who like this kind of problems. \r\nDon't worry for you I have prepared something special from analityc number theory( for example Bonferoni lemma :wink: or Heat-Brown's lemma, etc). Dont' worry I will post them soon. :)", "Solution_3": "Don't scare me with such lemmas... :wink: Ok, I admit that from time to time we can post easy problems, but really they should say something...", "Solution_4": "In case you're looking at this and still don't see it:\r\n\r\n[hide=\"Hints\"]The function is continuous (why?), and the Mean Value Theorem for integrals (what's that? look it up) applies.[/hide]" } { "Tag": [ "function", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello,\r\nHow to prove that we can't express alternative with only implication?\r\n--\r\nRegards.", "Solution_1": "What do you mean ?", "Solution_2": "I mean finding such a formula which consist of only '=>' and variables that is equivalent to alternative.", "Solution_3": "Let me state the problem in a more formal way.\r\nWe have two logical variables, $x$ and $y$, and build new formulas by combining two already constructed formulas with an implication sign, we can build $x \\rightarrow y$ and then $y \\rightarrow (x \\rightarrow y)$. We want to show that we never get a formula that is equivalent to the equivalence of x and y, i.e. formula $x \\leftrightarrow y$.\r\n\r\nThe brute force method is the following.\r\nSince equivalence of formulas is equivalent to equality of the induced logical functions, you find all logical functions on two arguments that can be constructed the way I described. You get only 6 of the 16 functions. You prove that these are all you can get by combining any two of then and showing that you get one of the functions. Finally you point out that the function which maps $(x,y)$ to $x \\leftrightarrow y$ is not one of these functions.\r\n\r\nEDIT: I misunderstood the question. apc seems to be looking for \"either x or y\", $x \\oplus y$, instead of \"x iff y\", $x \\leftrightarrow y$. But that is also not one of the 6 functions above. :-)\r\n\r\nLiMa" } { "Tag": [ "algebra", "polynomial", "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Suppose that $A,B$ are matrices over a field such that any matrix commuting with $A$ also commutes with $B$. Does it always follow that $B$ is a polynomial in $A$?", "Solution_1": "I think Eugene gave a proof here http://www.mathlinks.ro/Forum/viewtopic.php?t=82634", "Solution_2": "Thank you very much." } { "Tag": [], "Problem": "How many integers can be written as the sum of three different members of the set:\r\n\r\n$ {2, 4, 6, 8, 10, 12, 13}$\r\n\r\nAny way other than brute force?", "Solution_1": "Ignore the $ 13$ for now. Using $ 2, 4, 6, 8, 10, 12$ we can write any even integer from $ 2 \\plus{} 4 \\plus{} 6 \\equal{} 12$ to $ 8 \\plus{} 10 \\plus{} 12 \\equal{} 30$. This is because at least one of the even numbers in a given triplet can be increased by $ 2$ to get the next even number. This gives $ 10$ different sums.\r\n\r\nAdd the $ 13$ back in. We can now also write any odd integer from $ 2 \\plus{} 4 \\plus{} 13 \\equal{} 19$ to $ 10 \\plus{} 12 \\plus{} 13 \\equal{} 35$, and for the same reasons. This gives $ 9$ different sums.\r\n\r\nIn total, we have $ \\boxed{19}$ different sums." } { "Tag": [ "geometry", "rectangle", "perimeter" ], "Problem": "Four congruent rectangles are placed as shown to form a larger rectangle with area 1728 cm$ ^2$. How many centimeters are in the perimeter of one of the four congruent rectangles?\n\n[asy]draw((0,0)--(4,0)--(4,3)--(0,3)--(0,0)); draw((1,0)--(1,3));\ndraw((1,1)--(4,1)); draw((1,2)--(4,2));[/asy]", "Solution_1": "The area of one rectangle is $ 432$. And by the drawing, you can deduce that the $ 3l \\equal{} w$.\r\n\r\n$ lw \\equal{} 432$\r\n\r\n$ l \\cdot 3l \\equal{} 432$\r\n\r\n$ 3l^2 \\equal{} 432$\r\n\r\n$ l^2 \\equal{} 144$\r\n\r\n$ l \\equal{} 12$\r\n\r\nThen, $ w$ must be $ 36$.\r\n\r\nThus, the perimeter of one rectangle is $ 2 (12 \\plus{} 36)$, or $ \\boxed{96}$." } { "Tag": [ "number theory" ], "Problem": "i just remembered this problem, it's pretty easy, it's from a very old number theory book...\r\nwhat is the smallest integer that can be written in a form of $13x+73y$ for three diffrent $(x,y)$? $(x,y\\in\\mathbb{N})$", "Solution_1": "[quote=\"amirhtlusa\"]i just remembered this problem, it's pretty easy, it's from a very old number theory book...\nwhat is the smallest integer that can be written in a form of $13x+73y$ for three diffrent $(x,y)$? $(x,y\\in\\mathbb{N})$[/quote]\r\n\r\nHere's my best attempt at it...\r\n[hide]\nIf there is a solution (x,y), then two other solutions are (x-73, y+13) and (x+73, y-13)\n\nNow we just need to find a value of (x, y) such that x-73 is positive and y-13 is positive. Obviously, (74,14) works. So the smallest number that can be expressed is 74*13+14*73= 1984. \n\n[/hide]", "Solution_2": "Never mind didn't read the for three different x y bit." } { "Tag": [ "inequalities", "algebra unsolved", "algebra" ], "Problem": "let $x.y.z$ be reel numbers satisfaying the tree conditions \r\n$|x+y+z|\\leq1$\r\n$|x-y+z|\\leq1$\r\n$|4x+2y+z|\\leq8$\r\n$|4x-2y+z|\\leq8$\r\nprove that $|x|+3|y|+|z|\\leq7$.good luck :wink:", "Solution_1": "expand the inequalities?" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let ABC be a triangle. We denote by $A_1$, $B_1$ and $C_1$ the points of \r\nintersection of the bissectors of the angles of ABC with the respective \r\nopposite sides.\r\nWe assume that the points $B, A_1, B_1$ and $C_1$ are cocyclic.\r\nProve that :\r\n\r\n$\\frac{BC}{AB+AC}=\\frac{AC}{AB+BC}-\\frac{AB}{BC+AC}$", "Solution_1": "Denoting a = BC, b = CA, c = AB, we have to show $\\frac{b}{c + a} - \\frac{c}{a + b} = \\frac{a}{b + c}.$ Let the circumcircle of the triangle $\\triangle A_1B_1C_1$ meet the triangle sides at $M \\in a, N \\in b, P \\in c$ different from $A_1, B_1, C_1.$ According to the more general problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=82460]Circle (a'b'c')[/url],\r\n\r\n$x = MA' = \\frac 1 2 \\left(\\frac{b(b + c)}{c + a} - \\frac{c(b + c)}{a + b} - \\frac{a(b - c)}{b + c}\\right)$\r\n\r\nAccording to the condition of this problem, the points $M \\equiv B$ are identical, hence x > 0 and\r\n\r\n$\\frac{ca}{b + c} = BA_1 \\equiv MA_1 = x$\r\n\r\n$(b + c) \\left(\\frac{b}{c + a} - \\frac{c}{a + b}\\right) = \\frac{a(b - c)}{b + c} + \\frac{2ca}{b + c} = \\frac{a(b + c)}{b + c} = a$\r\n\r\n$\\frac{b}{c + a} - \\frac{c}{a + b} = \\frac{a}{b + c}$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find $ 1\\leqslant \\ a \\leqslant\\ 102$ when $ 112{}^3{}^5^^^1^^7$$ \\equiv\\ 3a + 1(mod 102)$", "Solution_1": "as 102^35^17,is divisible by 102, 3a+1 is congruent to 0 mod 102.but 3 divides 102, so 3a+1 should be 3k for some k,a contradiction if a is a natural number.", "Solution_2": "$ 17 \\equal{} 1\\;(mod \\;16)$, so $ 35^{17} \\equal{} 35^1 \\equal{} 3\\;(mod\\;32)$, $ 112^{{35}^{17}} \\equal{} 112^3 \\equal{} 10^3 \\equal{} 1000 \\equal{} 82\\;(mod\\;102)$, $ a \\equal{} 27, 61, 95$" } { "Tag": [], "Problem": "A sum of a list of positive integers is 20. What is the maximum possible product of the integers in this list?", "Solution_1": "3^6*2=1458 i think?", "Solution_2": "[hide]So it appears we'll probably have to use numbers more than once. Our choices are $ 2^{10}$ or $ 2(3^6)$ since otherwise there are too few numbers. Since $ 2^3<3^2$, the first choice is a smaller one. 1458.[/hide]" } { "Tag": [ "trigonometry", "analytic geometry", "graphing lines", "slope" ], "Problem": "If $ x^2\\plus{}y^2\\equal{}12x\\plus{}8y\\minus{}27$ then what is the max. value of $ 3x\\plus{}4y$?", "Solution_1": "$ x^2 \\plus{} y^2 \\equal{} 12x \\plus{} 8y \\minus{} 27\\Longleftrightarrow (x \\minus{} 6)^2 \\plus{} (y \\minus{} 4)^2 \\equal{} 25$. Let $ x \\minus{} 6 \\equal{} X,\\ y \\minus{} 4 \\equal{} Y$, we have $ 3x \\plus{} 4y \\equal{} 3X \\plus{} 4Y \\plus{} 34$ with $ X^2 \\plus{} Y^2 \\equal{} 25^2.$ Since $ (3X \\plus{} 4Y)^2 \\plus{} (4X \\minus{} 3Y)^2 \\equal{} 25\\Longleftrightarrow \\minus{} 25\\leq 3X \\plus{} 4Y\\leq 25$, yielding $ 9\\leq 3x \\plus{} 4y\\leq 59$.\r\n\r\nAns. $ Max(3x \\plus{} 4y) \\equal{} 59,\\ (X,\\ Y) \\equal{} (3,\\ 4)\\Longleftrightarrow (x,\\ y) \\equal{} (9,\\ 8)$.", "Solution_2": "Beaten again. I had, after completing the squares, a circle with centre at $ (6, 4)$ and radius $ 5$ and then the tangent $ 3x \\plus{} 4y \\equal{} c$ with greater y-intercept.", "Solution_3": "You can also use cosine, sine.", "Solution_4": "Thanks, kunny. I had the following.\r\n\r\nThe normal has slope $ \\frac {4}{3}$ and passes through $ (6, 4)$ and so has y-intercept at $ (0, \\minus{} 4)$ and intercepts the circle again at $ (3, 0)$ so point of tangency must be $ (6 \\plus{} 3, 4 \\plus{} 4) \\equal{} (9, 8)$ and so $ \\max(3x \\plus{} 4y) \\equal{} 27 \\plus{} 32 \\equal{} 59$. I have a diagram with three congruent $ 3 \\minus{} 4 \\minus{} 5$ triangles with vertices at $ (0, \\minus{} 4)$, $ (0, 0)$ and $ (3, 0)$; $ (3, 0)$, $ (6, 0)$ and $ (6, 4)$; and $ (6, 4)$, $ (9, 4)$ and $ (9, 8)$ but don't know how to post a diagram (yet!).", "Solution_5": "[quote=\"kunny\"]$ x^2 \\plus{} y^2 \\equal{} 12x \\plus{} 8y \\minus{} 27\\Longleftrightarrow (x \\minus{} 6)^2 \\plus{} (y \\minus{} 4)^2 \\equal{} 25$. Let $ x \\minus{} 6 \\equal{} X,\\ y \\minus{} 4 \\equal{} Y$, we have $ 3x \\plus{} 4y \\equal{} 3X \\plus{} 4Y \\plus{} 34$ with $ X^2 \\plus{} Y^2 \\equal{} 25^2.$ Since $ (3X \\plus{} 4Y)^2 \\plus{} (4X \\minus{} 3Y)^2 \\equal{} 25\\Longleftrightarrow \\minus{} 25\\leq 3X \\plus{} 4Y\\leq 25$, yielding $ 9\\leq 3x \\plus{} 4y\\leq 59$.\n\nAns. $ Max(3x \\plus{} 4y) \\equal{} 59,\\ (X,\\ Y) \\equal{} (3,\\ 4)\\Longleftrightarrow (x,\\ y) \\equal{} (9,\\ 8)$.[/quote]\n\n$ (3X \\plus{} 4Y)^2 \\plus{} (4X \\minus{} 3Y)^2 \\equal{} 25\\Longleftrightarrow \\boxed{\\minus{} 25\\leq 3X \\plus{} 4Y\\leq 25}$, yielding $ 9\\leq 3x \\plus{} 4y\\leq 59$.\n\nHow comes?", "Solution_6": "[quote=\"vinskman\"][quote=\"kunny\"]$ x^2 \\plus{} y^2 \\equal{} 12x \\plus{} 8y \\minus{} 27\\Longleftrightarrow (x \\minus{} 6)^2 \\plus{} (y \\minus{} 4)^2 \\equal{} 25$. Let $ x \\minus{} 6 \\equal{} X,\\ y \\minus{} 4 \\equal{} Y$, we have $ 3x \\plus{} 4y \\equal{} 3X \\plus{} 4Y \\plus{} 34$ with $ X^2 \\plus{} Y^2 \\equal{} 25^2.$ Since $ (3X \\plus{} 4Y)^2 \\plus{} (4X \\minus{} 3Y)^2 \\equal{} 25\\Longleftrightarrow \\minus{} 25\\leq 3X \\plus{} 4Y\\leq 25$, yielding $ 9\\leq 3x \\plus{} 4y\\leq 59$.\n\nAns. $ Max(3x \\plus{} 4y) \\equal{} 59,\\ (X,\\ Y) \\equal{} (3,\\ 4)\\Longleftrightarrow (x,\\ y) \\equal{} (9,\\ 8)$.[/quote]\n\n$ (3X \\plus{} 4Y)^2 \\plus{} (4X \\minus{} 3Y)^2 \\equal{} 25\\Longleftrightarrow \\boxed{\\minus{} 25\\leq 3X \\plus{} 4Y\\leq 25}$, yielding $ 9\\leq 3x \\plus{} 4y\\leq 59$.\n\nHow comes?[/quote]\nUse the substitution $X=x-6$ and $Y=y-4$.", "Solution_7": "I have a problem that from given problem we could simplify as,\n$x^2$ $+$ $(y+4)^2$ $+$ $11$/$4$ $=$ $3x$ $+$ $4y$\n\nSo, then how could we find the maximum value??" } { "Tag": [], "Problem": "How many moles of phosphorus atoms are in 100g of P4S10\r\n\r\n \r\n1)0.225\r\n \r\n2)0.900\r\n \r\n3)4.00\r\n \r\n3)400\r\n \r\n4)0.487", "Solution_1": "Molar mass of $P\\sub{4}S\\sub{10}$ is $123.88+320.70=444.58 g/mol$\n\n$\\frac{100}{444.58}\\approx0.225$\n\nSince there are 4 atoms in Phosphorus pentasulfide\n\n$0.225\\times4=0.900=\\boxed{2)0.900}$" } { "Tag": [], "Problem": "Find the number of degrees in $ a \\plus{} b$.\n[asy]pair A,B,C; B = (1,2); A = (5,0);\ndraw(A--1.3*(B-A)+A,EndArrow);\ndraw(B--1.3*(C-B)+B,EndArrow);\ndraw(C--1.3*(A-C)+C,EndArrow);\nlabel(\"$138^{\\circ}$\",A,dir(45)); label(\"$a$\",(3.8,0),N); label(\"$b$\",B,S + (0.1,0)); label(\"$105^{\\circ}$\",B,W + (0,-0.25)); label(\"$c$\",C,dir(45) + (0.1,0)); label(\"$117^{\\circ}$\",C,dir(-45));[/asy]", "Solution_1": "$ a\\plus{}b\\equal{}\\boxed{117^\\circ}$.\r\nBy the exterior angle theorem.", "Solution_2": "[hide=\"alternately\"]\nSince $ a$ and $ 138^{\\circ}$ are supplementary, they sum to $ 180^{\\circ},$ so $ a\\equal{}42^{\\circ}.$ Similarly, $ b\\plus{}105^{\\circ}\\equal{}180^{\\circ}\\Rightarrow b\\equal{}75^{\\circ}.$ Therefore, $ a\\plus{}b\\equal{}42^{\\circ}\\plus{}75^{\\circ}\\equal{}\\boxed{117^{\\circ}}.$[/hide]" } { "Tag": [ "induction", "number theory proposed", "number theory" ], "Problem": "I hope it's not again one of my \"masterpieces\", but I think the following is a very nice and difficult problem:\r\n Let $ a,b,c$ some positive integers and $ x,y,z$ some integer numbers such that we have \r\n a) $ax^2+by^2+cz^2= abc+2xyz-1$;\r\n b) $ab+bc+ca\\geq x^2+y^2+z^2$.\r\n Prove that $ a,b,c$ are all sums of three squares of integer numbers.", "Solution_1": "I guess they can be represented even as sum of two squares.\r\nExcuse me if it's wrong:\r\n\r\nFirstly a well-known lemma:\r\nIf $xy=z^2+1$ then we can find $u,v,w,t \\in \\mathbb{N}$ with $x=u^2+v^2,y=w^2+t^2, z=uw+vt$ and hence $ut-vw=1$. This lemma can be proven by induction on $x+y+z$ by noticing that if $x0$, $x=25$.\n\nThen the area is $\\frac{1}{2}\\cdot CH \\cdot AB=\\frac{1}{2}\\cdot (20) \\cdot (41)= \\boxed{410}$.[/hide]", "Solution_2": "[quote=\"kohjhsd\"][quote=\"Ignite168\"]A right triangle ABC with hypotenuse AB has side AC=15. Altitude CH divides AB into segments AH and HB with HB = 16. Find the area of triangle ABC.[/quote]\n\n[hide]In triangle $ABC$, let $AH=x$.\n\nThen, $CH=\\sqrt{AH \\cdot HB}=4\\sqrt{x}$.\n\n$AH \\cdot AB=AC^{2}$, therefore $AH \\cdot 16=x(x+16)=225$\n\nSolving, knowing that $x>0$, $x=25$.\n\nThen the area is $\\frac{1}{2}\\cdot CH \\cdot AB=\\frac{1}{2}\\cdot (20) \\cdot (41)= \\boxed{410}$.[/hide][/quote]\r\n\r\nCorrect up until you solved $x(x+16)=225$ ;)", "Solution_3": "[quote=\"Ignite168\"][quote=\"kohjhsd\"][quote=\"Ignite168\"]A right triangle ABC with hypotenuse AB has side AC=15. Altitude CH divides AB into segments AH and HB with HB = 16. Find the area of triangle ABC.[/quote]\n\n[hide]In triangle $ABC$, let $AH=x$.\n\nThen, $CH=\\sqrt{AH \\cdot HB}=4\\sqrt{x}$.\n\n$AH \\cdot AB=AC^{2}$, therefore $AH \\cdot 16=x(x+16)=225$\n\nSolving, knowing that $x>0$, $x=25$.\n\nThen the area is $\\frac{1}{2}\\cdot CH \\cdot AB=\\frac{1}{2}\\cdot (20) \\cdot (41)= \\boxed{410}$.[/hide][/quote]\n\nCorrect up until you solved $x(x+16)=225$ ;)[/quote]\n\n[hide=\"revision :oops: \"]\nActually, $x=9$, then the area is $\\frac12 \\cdot CH \\cdot AB=\\frac12 \\cdot 12 \\cdot 25=6 \\cdot 25=\\boxed{150}$.[/hide]" } { "Tag": [ "number theory", "greatest common divisor", "prime numbers", "algebra unsolved", "algebra" ], "Problem": "I had this one at an exam and I could not find an answer for it. \r\nI'ts probably quite easy, but I would really like to know. Please help me. :? \r\n\r\nThe problem: \r\nLet $a,b,x,y \\in N$ With gcd (a,b)=1 and $x^a=y^b$ \r\nProve that a $z$ exist for which $x=z^b$ and $y=z^a$", "Solution_1": "Consider prime decompositions of $x$ and $y$. They contain the same prime numbers, so $x=p_1^{a_1}...p_n^{a_n}$ and $y=p_1^{b_1}...p_n^{b_n}$. Due to $x^a=y^b$ we have $aa_i=bb_i$, but $\\gcd(a,b)=1$. It follows $b\\mid a_i$ and $a\\mid a_i$, i.e. $a_i=ba_i'$ and $b_i=ab_i'$. Therefore $x=(p_1^{a'_1}...p_n^{a'_n})^b$ and $y=(p_1^{b'_1}...p_n^{b'_n})^a$." } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "1. let a,b,c be positive real numbers with a^2+b^2+c^2=1 prove that\r\n 1<=a/(1+bc) + b/(1+ac) + c/(1+ab) <=2^1/2\r\n \r\n thank you :)", "Solution_1": "It follows from proving\r\n\\[ a^2\\leq \\frac{a}{1\\plus{}bc}\\leq \\frac{a\\sqrt{2}}{a\\plus{}b\\plus{}c}\\]\r\nFor the left inequality, start with AM-GM on $ b^2,c^2$. For the right inequality, start with RMS-AM on $ a,b\\plus{}c$.", "Solution_2": "[quote=\"scorpius119\"]It follows from proving\n\\[ a^2\\leq \\frac {a}{1 \\plus{} bc}\\leq \\frac {a\\sqrt {2}}{a \\plus{} b \\plus{} c}\n\\]\nFor the left inequality, start with AM-GM on $ b^2,c^2$. For the right inequality, start with RMS-AM on $ a,b \\plus{} c$.[/quote]\r\ni am grateful but please explain more :wink:", "Solution_3": "The left inequality is equivalent to $ a(1\\plus{}bc)\\leq 1$. First, $ bc\\leq \\frac{b^2\\plus{}c^2}{2}\\equal{}\\frac{1\\minus{}a^2}{2}$ so it reduces to $ a(3\\minus{}a^2)\\leq 2$ for $ a\\in [0,1]$ which can just be factored.\r\n\r\nThe right inequality is equivalent to $ a\\plus{}b\\plus{}c\\leq (1\\plus{}bc)\\sqrt{2}$. This follows from $ a\\plus{}b\\plus{}c\\leq \\sqrt{2(a^2\\plus{}(b\\plus{}c)^2)}\\equal{}\\sqrt{2(1\\plus{}2bc)}$ and another simple one-variable inequality $ \\sqrt{1\\plus{}2bc}\\leq 1\\plus{}bc$.", "Solution_4": "[quote=\"scorpius119\"]The left inequality is equivalent to $ a(1 \\plus{} bc)\\leq 1$. First, $ bc\\leq \\frac {b^2 \\plus{} c^2}{2} \\equal{} \\frac {1 \\minus{} a^2}{2}$ so it reduces to $ a(3 \\minus{} a^2)\\leq 2$ for $ a\\in [0,1]$ which can just be factored.\n\nThe right inequality is equivalent to $ a \\plus{} b \\plus{} c\\leq (1 \\plus{} bc)\\sqrt {2}$. This follows from $ a \\plus{} b \\plus{} c\\leq \\sqrt {2(a^2 \\plus{} (b \\plus{} c)^2)} \\equal{} \\sqrt {2(1 \\plus{} 2bc)}$ and another simple one-variable inequality $ \\sqrt {1 \\plus{} 2bc}\\leq 1 \\plus{} bc$.[/quote]\r\n thank you scorpius119 :blush:" } { "Tag": [ "function", "calculus", "derivative", "inequalities", "limit", "real analysis", "real analysis solved" ], "Problem": "Let $g: R\\to R$ be a differentiable function whose derivative $g'$ satisfies the inequality $|g'(x)|\\leq 2 \\forall x \\in R$.\r\nShow that if $\\delta$ is small enough, then $f: R\\to R$ defined by $f(x)=x+\\delta g(x)$ is bijective.", "Solution_1": "$\\delta<\\frac12$ suffices. That would give us $f'(x)=1+g'(x)\\ge 1-2\\delta$ for all $x.$ By the Mean Value Theorem, that shows that $f$ is strictly increasing (hence one-to-one) and that for $x>0,\\,f(x)\\ge f(0)+(1-2\\delta)x$ and for $x<0,\\,f(x)\\le f(0)+(1-2\\delta)x.$\r\n\r\nThus $\\lim_{x\\to\\infty}f(x)=\\infty$ and $\\lim_{x\\to-\\infty}f(x)=-\\infty.$ Since $f$ is continuous (it's differentiable, after all) the Intermediate Value Theorem now shows that $f$ maps $\\mathbb{R}$ onto $\\mathbb{R}.$", "Solution_2": "It is not necessary that $\\delta$ be a constant, only that it be small (e.g. less than 1/2 in absolute value).", "Solution_3": "[quote=\"Kent Merryfield\"]for $x>0,\\,f(x)\\ge f(0)+(1-2\\delta)x$ and for $x<0,\\,f(x)\\le f(0)+(1-2\\delta)x.$[/quote]\r\nHow you do get this?", "Solution_4": "[quote=\"derekhohk\"]How you do get this?[/quote]\r\nThis is the Mean Value Theorem in operation. Suppose that $f$ is differentiable and $f'(x)\\ge a$ everywhere. Then for $x> x_0,\\,f(x)-f(x_0)=f'(\\xi)(x-x_0)$ for some $x_0<\\xi0$ for each $x$ in that interval, then for any $a,b$ in that interval such that $a0$ let\r\n\\[ A_n \\equal{} \\left( \\begin{array}{cc} A_{n \\minus{} 1} & A_{n \\minus{} 1} \\\\\r\nA_{n \\minus{} 1} & B_{n \\minus{} 1} \\\\\r\n\\end{array} \\right) \\ \\textrm{and} \\ B_n \\equal{} \\left( \\begin{array}{cc} A_{n \\minus{} 1} & A_{n \\minus{} 1} \\\\\r\nA_{n \\minus{} 1} & 0 \\\\\r\n\\end{array} \\right).\r\n\\]\r\nDenote by $ S(M)$ the sum of all the elements of a matrix $ M$. Prove that $ S(A_n^{k \\minus{} 1}) \\equal{} S(A_k^{n \\minus{} 1})$, for all $ n,k\\geq 2$.", "Solution_1": "well i wouldn't post anything but the case is that i got the author's solution. it's a combinatorical problem s ((a_n)^k) counts the number of n row x m column tables with 0 or 1 entries which do not contain submatrix of four ones 1,1 and below them 1,1. this nuber clearly corresponds to the nuber of n tuples of natural nubers whose decimal expansion contais of no more than m digits . s ((a_n)^k) is in fact sum(sum (sum (sum( .....(sum a_i_j*a_j_k*a_k_l*....*a_s_t*a_t_u) where the a_i_j is 0 or 1 so the summand can be 1 only if all the a_i_j 's are 1 so the indices i,j,k,l,......,s,t,u form an admissible -k tuple. (the recurrence relations force it to be so) now clearly the number of such m x n tables is equal to the number of n x m tables" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "1) show that the ring R=M2(F) of two by two matrices over a field F has no proper 2-sided ideals.\r\n\r\n2) Suppose that F is a field with |F*|=p for p prime. Prove that p=2^n-1 for some n. Describe how to construct all such F's.", "Solution_1": "For point 2) : Since $F$ is a field $p+1$ must be the power of a prime , and if $p$ is a prime different from $2$ the $p+1$ is even so $p=2^{n}-1$ . what if $p=2$,and $F=\\mathbb{F}_{3}$?!", "Solution_2": "[quote=\"math beginner\"]\n2) Suppose that F is a field with |F*|=p for p prime. Prove that p=2^n-1 for some n. Describe how to construct all such F's.[/quote]\r\n$p+1 = q^{n}$\r\nIf $p$ is odd then $q=2$. So $p=2^{n}-1$\r\nSince there is only one field upto isomorphism all fields having this property are $\\mbox{GF}(zm_{p})$\r\nWhere $M_{p}$ are Mersenne primes.\r\n\r\nIf $p$ is even, i.e. $p=2$ then $q=3,n=1$.\r\nSo $\\mathbb{Z}_{3}$" } { "Tag": [ "geometry", "geometric transformation", "perimeter", "inequalities", "ratio", "linear algebra", "matrix" ], "Problem": "Here is the IMO shortlist problems 1984, I got it from internet.", "Solution_1": "IMO 1984 LongList", "Solution_2": "Here are the translations of the Russian problems from the \"English Parts\" file:\r\n\r\n[u][b]Mongolia[/b][/u]\r\n\r\n[b]Problem 1.[/b] Let $d$ be the sum of all diagonals of a convex $n$-gon, and let $p$ be the perimeter of this $n$-gon. Prove that\r\n\r\n$\\displaystyle \\frac{n-3}{2}<\\frac{d}{p}<\\frac{\\left[ \\frac{n}{2}\\right] \\left[ \\frac{n+1}{2}\\right] -2}{2}$,\r\n\r\nand show that these bounds cannot be improved.\r\n\r\n[i]Note.[/i] This was actually problem 5 of the IMO 1984.\r\n\r\n[b]Problem 2.[/b] A country has $n$ cities, and every pair of cities is connected by a railway line. A railway tester has to cruise every of these railway lines once and only once. If the tester cannot go from one city to another one by railway anymore since he has already cruised the railway line joining these two cities, then he can use the airplane. Find the minimal number of airplane tickets the tester needs in order to have all railway lines cruised.\r\n\r\n[i]Note.[/i] Cruising a railway line means taking this line in order to pass from one of the cities this line connects to the other one. The tester can cruise the railway line only once, thus he can use it only in one of the two directions.\r\n\r\n[b]Problem 3.[/b] Prove the existence of different natural numbers $m_{1}$, $m_{2}$, ..., $m_{k}$ satisfying the inequality\r\n\r\n$\\displaystyle \\pi ^{-1984}<25-\\left( \\frac{1}{m_{1}}+\\frac{1}{m_{2}}+...+\\frac{1}{m_{k}}\\right) <\\pi ^{-1960}$,\r\n\r\nwhere $\\pi $ is the ratio between the circumference of a circle and its diameter.\r\n\r\n[b]Problem 4.[/b] The set $\\left\\{ 1;\\;2;\\;...;\\;49\\right\\} $ is subdivided into three subsets. Show that at least one of these three subsets contains three different numbers $a$, $b$, $c$ such that $a+b=c$.\r\n\r\n[u][b]Soviet Union[/b][/u]\r\n\r\n[b]Problem 1.[/b] Given an $m\\times n$ matrix $\\left\\| p_{ij}\\right\\| $ whose elements are real numbers. For every $i\\in \\left\\{1;\\;2;\\;...;\\;m\\right\\}$, let\r\n\r\n$\\displaystyle \\textbf{(1)}\\;\\;\\;\\;\\;\\;\\;\\;a_{i}=\\sum_{j=1}^{n}p_{ij}$,\r\n\r\nand for every $j\\in \\left\\{ 1;\\;2;\\;...;\\;n\\right\\}$, let\r\n\r\n$\\displaystyle \\textbf{(2)}\\;\\;\\;\\;\\;\\;\\;\\;b_{j}=\\sum_{i=1}^{n}p_{ij}$.\r\n\r\n[i]Rounding[/i] a non-integer number will mean replacing it by one of the two next integers (i. e. replacing it either by its floor or by its ceiling).\r\n\r\nProve that one can round the numbers $a_{i}$, $b_{j}$, $p_{ij}$ in such a way that the equations (1) and (2) remain valid.\r\n\r\n[b]Problem 2.[/b] A (not necessarily regular) tetrahedron is inscribed into a sphere with radius $1$ such that the center of the sphere lies inside the tetrahedron.\r\n\r\nProve that the sum of the edgelengths of the tetrahedron is greater than $6$.\r\n\r\n[b]Problem 3.[/b]\r\n\r\nLet $x_{1}$, $x_{2}$, ..., $x_{k}$ be the nontrivial divisors of a natural number $n$, arranged such that $x_{1}\\frac{5}{6}R$.\r\n\r\n[b]Problem 5.[/b]\r\n\r\nIn the Martian language, each finite chain of Latin letters is considered to be a word.\r\n\r\nThe Martian dictionary consists of infinitely many tomes: The first tome contains all one-letter words; the second tome contains all two-letter words; etc.. All words are numbered in lexicographical order, and the numeration in each tome is continued in the next tome.\r\n\r\nFind the word whose number equals the sum of the numbers of the words [i]Prague[/i], [i]Olympiad[/i], [i]Mathematics[/i].\r\n\r\nFor Orl, here is the TeX source:\r\n\r\n[code]\\underline{\\textbf{Mongolia}}\n\n\\bigskip \n\n\\textbf{Problem 1.} Let $d$ be the sum of all diagonals\nof a convex $n$-gon, and let $p$ be the perimeter of\nthis $n$-gon. Prove that\n\\begin{equation*}\n\\frac{n-3}{2}<\\frac{d}{p}<\\frac{\\left[ \\frac{n}{2}\\right]\n\\left[ \\frac{n+1}{2}\\right] -2}{2},\n\\end{equation*}\nand show that these bounds cannot be improved.\n\n\\textit{Note.} This was actually problem 5 of the IMO 1984.\n\n\\bigskip \n\n\\textbf{Problem 2.} A country has $n$ cities, and every\npair of cities is connected by a railway line. A railway\ntester has to cruise every of these railway lines once\nand only once. If the tester cannot go from one city to\nanother one by railway anymore since he has already\ncruised the railway line joining these two cities, then\nhe can use the airplane. Find the minimal number of\nairplane tickets the tester needs in order to have all\nrailway lines cruised.\n\n\\textit{Note.} Cruising a railway line means taking\nthis line in order to pass from one of the cities this\nline connects to the other one. The tester can cruise\nthe railway line only once, thus he can use it only in\none of the two directions.\n\n\\bigskip \n\n\\textbf{Problem 3.} Prove the existence of different\nnatural numbers $m_{1},$ $m_{2},$ $...,$ $m_{k}$\nsatisfying the inequality\n\\begin{equation*}\n\\pi ^{-1984}<25-\\left( \\frac{1}{m_{1}}\n+\\frac{1}{m_{2}}+...+\\frac{1}{m_{k}}\\right) <\n\\pi ^{-1960},\n\\end{equation*}\nwhere $\\pi $ is the ratio between the circumference\nof a circle and its diameter.\n\n\\bigskip \n\n\\textbf{Problem 4.} The set\n$\\left\\{ 1;\\;2;\\;...;\\;49\\right\\} $ is subdivided\ninto three subsets. Show that at least one of these\nthree subsets contains three different numbers $a,$\n$b,$ $c$ such that $a+b=c.$\n\n\\bigskip \n\n\\underline{\\textbf{Soviet Union}}\n\n\\bigskip \n\n\\textbf{Problem 1.} Given an $m\\times n$ matrix\n$\\left\\| p_{ij}\\right\\| $ whose elements are real\nnumbers. For every\n$i\\in \\left\\{1;\\;2;\\;...;\\;m\\right\\} ,$ let\n\\begin{equation}\na_{i}=\\dsum\\limits_{j=1}^{n}p_{ij}, \\tag{(1)}\n\\end{equation}\nand for every\n$j\\in \\left\\{ 1;\\;2;\\;...;\\;n\\right\\} ,$ let\n\\begin{equation}\nb_{j}=\\dsum\\limits_{i=1}^{n}p_{ij}. \\tag{(2)}\n\\end{equation}\n\n\\textit{Rounding} a non-integer number will mean\nreplacing it by one of the two next integers (i. e.\nreplacing it either by its floor or by its ceiling).\n\nProve that one can round the numbers $a_{i},$\n$b_{j},$ $p_{ij}$ in such a way that the equations\n(1) and (2) remain valid.\n\n\\bigskip \n\n\\textbf{Problem 2.} A (not necessarily regular)\ntetrahedron is inscribed into a sphere with radius\n$1$ such that the center of the sphere lies inside\nthe tetrahedron.\n\nProve that the sum of the edgelengths of the\ntetrahedron is greater than $6.$\n\n\\bigskip \n\n\\textbf{Problem 3.}\n\nLet $x_{1},$ $x_{2},$ $...,$ $x_{k}$ be the\nnontrivial divisors of a natural number $n,$\narranged such that $x_{1}\\frac{5}{6}R.$\n\n\\bigskip \n\n\\textbf{Problem 5.}\n\nIn the Martian language, each finite chain of Latin\nletters is considered to be a word.\n\nThe Martian dictionary consists of infinitely many\ntomes: The first tome contains all one-letter words;\nthe second tome contains all two-letter words; etc..\nAll words are numbered in lexicographical order, and\nthe numeration in each tome is continued in the next\ntome.\n\nFind the word whose number equals the sum of the\nnumbers of the words \\textit{Prague},\n\\textit{Olympiad}, \\textit{Mathematics}.[/code]", "Solution_3": "Thanks for translating the problems, Darij. :)", "Solution_4": "[b][size=150]IMO 1984 ShortListed Problems: (Posted in Forums)[/size][/b]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012303#p2012303]1. (FRA 1)[/url] Find all solutions of the following system of $n$ equations in $n$ variables:\n\\[\\begin{array}{c}\\ x_1|x_1| - (x_1 - a)|x_1 - a| = x_2|x_2|,x_2|x_2| - (x_2 - a)|x_2 - a| = x_3|x_3|,\\ \\vdots \\ x_n|x_n| - (x_n - a)|x_n - a| = x_1|x_1|\\end{array}\\]\nwhere $a$ is a given number.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012306#p2012306]2. (CAN 2)[/url] Prove:\n\n(a) There are infinitely many triples of positive integers $m, n, p$ such that $4mn - m- n = p^2 - 1.$\n\n(b) There are no positive integers $m, n, p$ such that $4mn - m- n = p^2.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012308#p2012308]3. (USS 3)[/url] Find all positive integers $n$ such that\n\\[n=d_6^2+d_7^2-1,\\]\nwhere $1 = d_1 < d_2 < \\cdots < d_k = n$ are all positive divisors of the number $n.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366611&sid=c1de64b9288e42ff40d79056f6961512#p366611]4. (MON 1)[/url] Let $ d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $ n$ vertices (where $ n>3$). Let $ p$ be its perimeter. Prove that:\n\\[ n-3<{2d\\over p}<\\Bigl[{n\\over2}\\Bigr]\\cdot\\Bigl[{n+1\\over 2}\\Bigr]-2,\\]\nwhere $ [x]$ denotes the greatest integer not exceeding $ x$.[IMO Problem 5]\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366604&sid=c1de64b9288e42ff40d79056f6961512#p366604]5. (FRG 1)[/url] Prove that $0\\le yz+zx+xy-2xyz\\le{7\\over27}$, where $x,y$ and $z$ are non-negative real numbers satisfying $x+y+z=1$.[IMO Problem 1]\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012311#p2012311]6. (CAN 3)[/url] Let $c$ be a positive integer. The sequence $\\{f_n\\}$ is defined as follows:\n\\[f_1 = 1, f_2 = c, f_{n+1} = 2f_n - f_{n-1} + 2 \\quad (n \\geq 2).\\]\nShow that for each $k \\in \\mathbb N$ there exists $r \\in \\mathbb N$ such that $f_kf_{k+1}= f_r.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012317#p2012317]7. (FRG 5)[/url] (a) Decide whether the fields of the $8 \\times 8$ chessboard can be numbered by the numbers $1, 2, \\dots , 64$ in such a way that the sum of the four numbers in each of its parts of one of the forms\n\n[list][img]http://www.artofproblemsolving.com/Forum/download/file.php?id=28446[/img][/list]\n\nis divisible by four.\n\n(b) Solve the analogous problem for\n\n[list][img]http://www.artofproblemsolving.com/Forum/download/file.php?id=28447[/img][/list]\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366609&sid=c1de64b9288e42ff40d79056f6961512#p366609]8. (ROM 2)[/url] Given points $O$ and $A$ in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point $X$ in the plane, the circle $C(X)$ has center $O$ and radius $OX+{\\angle AOX\\over OX}$, where $\\angle AOX$ is measured in radians in the range $[0,2\\pi)$. Prove that we can find a point $X$, not on $OA$, such that its color appears on the circumference of the circle $C(X)$. [IMO Problem 3]\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012318#p2012318]9. (POL 2)[/url] Let $a, b, c$ be positive numbers with $\\sqrt a +\\sqrt b +\\sqrt c = \\frac{\\sqrt 3}{2}$. Prove that the system of equations\n\\[\\sqrt{y-a}+\\sqrt{z-a}=1,\\] \\[\\sqrt{z-b}+\\sqrt{x-b}=1,\\] \\[\\sqrt{x-c}+\\sqrt{y-c}=1\\]\nhas exactly one solution $(x, y, z)$ in real numbers.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012322#p2012322]10. (GBR 1)[/url] Prove that the product of five consecutive positive integers cannot be the square of an integer.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012324#p2012324]11. (CAN 1)[/url] Let $n$ be a positive integer and $a_1, a_2, \\dots , a_{2n}$ mutually distinct integers. Find all integers $x$ satisfying\n\\[(x - a_1) \\cdot (x - a_2) \\cdots (x - a_{2n}) = (-1)^n(n!)^2.\\]\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366605&sid=c1de64b9288e42ff40d79056f6961512#p366605]12. (NET 1)[/url] Find one pair of positive integers $a,b$ such that $ab(a+b)$ is not divisible by $7$, but $(a+b)^7-a^7-b^7$ is divisible by $7^7$.[IMO Problem 2]\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012327#p2012327]13. (BUL 5)[/url] Prove that the volume of a tetrahedron inscribed in a right circular cylinder of volume $1$ does not exceed $\\frac{2}{3 \\pi}.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366610&sid=c1de64b9288e42ff40d79056f6961512#p366610]14. (ROM 5)[/url] Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel. [IMO Problem 4]\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012329#p2012329]15. (LUX 2)[/url] Angles of a given triangle $ABC$ are all smaller than $120^\\circ$. Equilateral triangles $AFB, BDC$ and $CEA$ are constructed in the exterior of $ABC$.\n\n(a) Prove that the lines $AD, BE$, and $CF$ pass through one point $S.$\n\n(b) Prove that $SD + SE + SF = 2(SA + SB + SC).$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=165811&sid=c1de64b9288e42ff40d79056f6961512#p165811]16. (POL 1)[/url]Let $a,b,c,d$ be odd integers such that $0 x_j$. Let $d(n, k)$ be the number of such permutations with exactly $k$ discordant pairs. Find $d(n, 2)$ and $d(n, 3).$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012331#p2012331]18. (USA 5)[/url] Inside triangle $ABC$ there are three circles $k_1, k_2, k_3$ each of which is tangent to two sides of the triangle and to its incircle $k$. The radii of $k_1, k_2, k_3$ are $1, 4$, and $9$. Determine the radius of $k.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012334#p2012334]19. (CAN 5)[/url] The triangular array $(a_{n,k})$ of numbers is given by $a_{n,1} = \\frac 1n$, for $n = 1, 2, \\dots , a_{n,k+1} = a_{n-1,k} - a_{n,k}$, for $1 \\leq k \\leq n - 1$. Find the harmonic mean of the $1985^{th}$ row.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012335#p2012335]20. (USA 2)[/url] Determine all pairs $(a, b)$ of positive real numbers with $a \\neq 1$ such that\n\\[\\log_a b < \\log_{a+1} (b + 1).\\]", "Solution_5": "I have compiled the IMO longlist problems of the year $1984$. I request any moderator to update the contests page of this\n\n[b][size=150]IMO 1984 Longlisted problems (Posted in Forums) :[/size][/b]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043827#p2043827]Problem 1:[/url] $(AUS 1)$ The fraction $\\frac{3}{10}$ can be written as the sum of two positive fractions with numerator $1$ as follows: $\\frac{3}{10} =\\frac{1}{5}+\\frac{1}{10}$ and also $\\frac{3}{10}=\\frac{1}{4}+\\frac{1}{20}$. There are the only two ways in which this can be done. In how many ways can $\\frac{3}{1984}$ be written as the sum of two positive fractions with numerator $1$? \nIs there a positive integer $n,$ not divisible by $3$, such that $\\frac{3}{n}$ can be written as the sum of two positive fractions with numerator $1$ in exactly $1984$ ways?\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2045422#p2045422]Problem 2:[/url] $(AUS 2)$ Given a regular convex $2m$- sided polygon $P$, show that there is a $2m$-sided polygon $\\pi$ with the same vertices as $P$ (but in different order) such that $\\pi$ has exactly one pair of parallel sides.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2045442#p2045442]Problem 3:[/url] $(AUS 3)$ The opposite sides of the reentrant hexagon $AFBDCE$ intersect at the points $K,L,M$ (as shown in the figure). It is given that $AL = AM = a, BM = BK = b$, $CK = CL = c, LD = DM = d, ME = EK = e, FK = FL = f$. \n[img]http://imgur.com/LUFUh.png[/img]\n$(a)$ Given length $a$ and the three angles $\\alpha, \\beta$ and $\\gamma$ at the vertices $A, B,$ and $C,$ respectively, satisfying the condition $\\alpha+\\beta+\\gamma<180^{\\circ}$, show that all the angles and sides of the hexagon are thereby uniquely determined.\n$(b)$ Prove that\n\\[\\frac{1}{a}+\\frac{1}{c}=\\frac{1}{b}+\\frac{1}{d}\\]\nEasier version of $(b)$. Prove that\n\\[(a + f)(b + d)(c + e)= (a + e)(b + f)(c + d)\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047656#p2047656]Problem 4:[/url] $(BEL 1)$ Given a triangle $ABC$, three equilateral triangles $AEB, BFC$, and $CGA$ are constructed in the exterior of $ABC$. Prove that:\n$(a) CE = AF = BG$;\n$(b) CE, AF$, and $BG$ have a common point.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047661#p2047661]Problem 5:[/url] $(BEL 2)$ For a real number $x$, let $[x]$ denote the greatest integer not exceeding $x$. If $m \\ge 3$, prove that\n\\[\\left[\\frac{m(m+1)}{2(2m-1)}\\right]=\\left[\\frac{m+1}{4}\\right]\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2028585#p2028585]Problem 6:[/url] $(BEL 3)$ Let $P,Q,R$ be the polynomials with real or complex coefficients such that at least one of them is not constant. If $P^n+Q^n+R^n = 0$, prove that $n < 3.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047664#p2047664]Problem 7:[/url] $(BUL 1)$ Prove that for any natural number $n$, the number $\\dbinom{2n}{n}$ divides the least common multiple of the numbers $1, 2,\\cdots, 2n -1, 2n$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047667#p2047667]Problem 8:[/url] $(BUL 2)$ In the plane of a given triangle $A_1A_2A_3$ determine (with proof) a straight line $l$ such that the sum of the distances from $A_1, A_2$, and $A_3$ to $l$ is the least possible.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2045507#p2045507]Problem 9:[/url] $(BUL 3)$ The circle inscribed in the triangle $A_1A_2A_3$ is tangent to its sides $A_1A_2, A_2A_3, A_3A_1$ at points $T_1, T_2, T_3$, respectively. Denote by $M_1, M_2, M_3$ the midpoints of the segments $A_2A_3, A_3A_1, A_1A_2$, respectively. Prove that the perpendiculars through the points $M_1, M_2, M_3$ to the lines $T_2T_3, T_3T_1, T_1T_2$ meet at one point.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047671#p2047671]Problem 10:[/url] $(BUL 4)$ Assume that the bisecting plane of the dihedral angle at edge $AB$ of the tetrahedron $ABCD$ meets the edge $CD$ at point $E$. Denote by $S_1, S_2, S_3$, respectively the areas of the triangles $ABC, ABE$, and $ABD$. Prove that no tetrahedron exists for which $S_1, S_2, S_3$ (in this order) form an arithmetic or geometric progression.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012327&#p2012327]Problem 11:[/url] $(BUL 5)$ Prove that the volume of a tetrahedron inscribed in a right circular cylinder of volume $1$ does not exceed $\\frac{2}{3 \\pi}.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012324&#p2012324]Problem 12:[/url] $(CAN 1)$ Let $n$ be a positive integer and $a_1, a_2, \\dots , a_{2n}$ mutually distinct integers. Find all integers $x$ satisfying\n\\[(x - a_1) \\cdot (x - a_2) \\cdots (x - a_{2n}) = (-1)^n(n!)^2.\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012306&#p2012306]Problem 13:[/url] $(CAN 2)$ Prove:\n\n(a) There are infinitely many triples of positive integers $m, n, p$ such that $4mn - m- n = p^2 - 1.$\n\n(b) There are no positive integers $m, n, p$ such that $4mn - m- n = p^2.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012311&#p2012311]Problem 14:[/url] $(CAN 3)$ Let $c$ be a positive integer. The sequence $\\{f_n\\}$ is defined as follows:\n\\[f_1 = 1, f_2 = c, f_{n+1} = 2f_n - f_{n-1} + 2 \\quad (n \\geq 2).\\]\nShow that for each $k \\in \\mathbb N$ there exists $r \\in \\mathbb N$ such that $f_kf_{k+1}= f_r.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2044164#p2044164]Problem 15:[/url] $(CAN 4)$ Consider all the sums of the form \n\\[\\displaystyle\\sum_{k=1}^{1985} e_kk^5=\\pm 1^5\\pm 2^5\\pm\\cdots\\pm1985^5\\]\nwhere $e_k=\\pm 1$. What is the smallest nonnegative value attained by a sum of this type?\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012334&#p2012334]Problem 16:[/url] $(CAN 5)$ The harmonic table is a triangular array:\n\n$1$\n\n$\\frac 12 \\qquad \\frac 12$\n\n$\\frac 13 \\qquad \\frac 16 \\qquad \\frac 13$\n\n$\\frac 14 \\qquad \\frac 1{12} \\qquad \\frac 1{12} \\qquad \\frac 14$\n\nWhere $a_{n,1} = \\frac 1n$ and $a_{n,k+1} = a_{n-1,k} - a_{n,k}$ for $1 \\leq k \\leq n-1.$ Find the harmonic mean of the $1985^{th}$ row.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012303&#p2012303]Problem 17:[/url]$(FRA 1)$ Find all solutions of the following system of $n$ equations in $n$ variables:\n\\[\\begin{array}{c}\\ x_1|x_1| - (x_1 - a)|x_1 - a| = x_2|x_2|,x_2|x_2| - (x_2 - a)|x_2 - a| = x_3|x_3|,\\ \\vdots \\ x_n|x_n| - (x_n - a)|x_n - a| = x_1|x_1|\\end{array}\\]\nwhere $a$ is a given number.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048045#p2048045]Problem 18:[/url] $(FRA 2)$ Let $c$ be the inscribed circle of the triangle $ABC$, $d$ a line tangent to $c$ which does not pass through the vertices of triangle $ABC$. Prove the existence of points $A_1,B_1, C_1$, respectively, on the lines $BC,CA,AB$ satisfying the following two properties:\n$(i)$ Lines $AA_1,BB_1$, and $CC_1$ are parallel.\n$(ii)$ Lines $AA_1,BB_1$, and $CC_1$ meet $d$ respectively at points $A' ,B'$, and $C'$ such that\n\\[\\frac{A'A_1}{A' A}=\\frac{B'B_1}{B 'B}=\\frac{C'C_1}{C'C}\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2044173#p2044173]Problem 19:[/url] $(FRA 3)$ Let $ABC$ be an isosceles triangle with right angle at point $A$. Find the minimum of the function $F$ given by $F(M) = BM +CM-\\sqrt{3}AM$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366604&#p366604]Problem 20:[/url] $(FRG 1)$Prove that $0\\le yz+zx+xy-2xyz\\le{7\\over27}$, where $x,y$ and $z$ are non-negative real numbers satisfying $x+y+z=1$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043844#p2043844]Problem 21:[/url] $(FRG 2)$\n$(1)$ Start with $a$ white balls and $b$ black balls.\n$(2)$ Draw one ball at random.\n$(3)$ If the ball is white, then stop. Otherwise, add two black balls and go to step $2$. \nLet $S$ be the number of draws before the process terminates. For the cases $a = b = 1$ and $a = b = 2$ only, find $a_n = P(S = n), b_n = P(S \\le n), \\lim_{n\\to\\infty} b_n$, and the expectation value of the number of balls drawn: $E(S) =\\displaystyle\\sum_{n\\ge1} na_n.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012330&#p2012330]Problem 22:[/url] $(FRG 3)$ In a permutation $(x_1, x_2, \\dots , x_n)$ of the set $1, 2, \\dots , n$ we call a pair $(x_i, x_j )$ discordant if $i < j$ and $x_i > x_j$. Let $d(n, k)$ be the number of such permutations with exactly $k$ discordant pairs. Find $d(n, 2)$ and $d(n, 3).$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043846#p2043846]Problem 23:[/url] $(FRG 4)$ A $2\\times 2\\times 12$ box fixed in space is to be filled with twenty-four $1 \\times 1 \\times 2$ bricks. In how many ways can this be done?\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012317&#p2012317]Problem 24:[/url] $(FRG 5)$ (a) Decide whether the fields of the $8 \\times 8$ chessboard can be numbered by the numbers $1, 2, \\dots , 64$ in such a way that the sum of the four numbers in each of its parts of one of the forms\n\n[list][img]http://www.artofproblemsolving.com/Forum/download/file.php?id=28446[/img][/list]\n\nis divisible by four.\n\n(b) Solve the analogous problem for\n\n[list][img]http://www.artofproblemsolving.com/Forum/download/file.php?id=28447[/img][/list]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012322&#p2012322]Problem 25:[/url] $(GBR 1)$ Prove that the product of five consecutive positive integers cannot be the square of an integer.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048049#p2048049]Problem 26:[/url] $(GBR 2)$ A cylindrical container has height $6 cm$ and radius $4 cm$. It rests on a circular hoop, also of radius $4 cm$, fixed in a horizontal plane with its axis vertical and with each circular rim of the cylinder touching the hoop at two points. \nThe cylinder is now moved so that each of its circular rims still touches the hoop in two points. Find with proof the locus of one of the cylinder\u2019s vertical ends.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047673#p2047673]Problem 27:[/url] $(GBR 3)$ The function $f(n)$ is defined on the nonnegative integers $n$ by: $f(0) = 0, f(1) = 1$, and\n\\[f(n) = f\\left(n -\\frac{1}{2}m(m - 1)\\right)-f\\left(\\frac{1}{2}m(m+ 1)-n\\right)\\]\nfor $\\frac{1}{2}m(m - 1) < n \\le \\frac{1}{2}m(m+ 1), m \\ge 2$. Find the smallest integer $n$ for which $f(n) = 5$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048055#p2048055]Problem 28:[/url] $(GBR 4)$ A \u201cnumber triangle\u201d $(t_{n, k}) (0 \\le k \\le n)$ is defined by $t_{n,0} = t_{n,n} = 1 (n \\ge 0),$ \n\\[t_{n+1,m} =(2 -\\sqrt{3})^mt_{n,m} +(2 +\\sqrt{3})^{n-m+1}t_{n,m-1} \\quad (1 \\le m \\le n)\\]\nProve that all $t_{n,m}$ are integers.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2044215#p2044215]Problem 29:[/url] $(GDR 1)$ Let $S_n = \\{1, \\cdots, n\\}$ and let $f$ be a function that maps every subset of $S_n$ into a positive real number and satisfies the following condition: For all $A \\subseteq S_n$ and $x, y \\in S_n, x \\neq y, f(A \\cup \\{x\\})f(A \\cup \\{y\\}) \\le f(A \\cup \\{x, y\\})f(A)$. Prove that for all $A,B \\subseteq S_n$ the following inequality holds: \n\\[f(A) \\cdot f(B) \\le f(A \\cup B) \\cdot f(A \\cap B)\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046274#p2046274]Problem 30:[/url] $(GDR 2)$ Decide whether it is possible to color the $1984$ natural numbers $1, 2, 3, \\cdots, 1984$ using $15$ colors so that no geometric sequence of length $3$ of the same color exists.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048065#p2048065]Problem 31:[/url] $(LUX 1)$ Let $f_1(x) = x^3+a_1x^2+b_1x+c_1 = 0$ be an equation with three positive roots $\\alpha>\\beta>\\gamma > 0$. From the equation $f_1(x) = 0$, one constructs the equation $f_2(x) = x^3 +a_2x^2 +b_2x+c_2 = x(x+b_1)^2 -(a_1x+c_1)^2 = 0$. Continuing this process, we get equations $f_3,\\cdots, f_n$. Prove that\n\\[\\lim_{n\\to\\infty}\\sqrt[2^{n-1}]{-a_n} = \\alpha\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012329&#p2012329]Problem 32:[/url] $(LUX 2)$ Angles of a given triangle $ABC$ are all smaller than $120^\\circ$. Equilateral triangles $AFB, BDC$ and $CEA$ are constructed in the exterior of $ABC$.\n\n(a) Prove that the lines $AD, BE$, and $CF$ pass through one point $S.$\n\n(b) Prove that $SD + SE + SF = 2(SA + SB + SC).$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366611&#p366611]Problem 33:[/url] $(MON 1)$ Let $ d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $ n$ vertices (where $ n>3$). Let $ p$ be its perimeter. Prove that:\n\\[ n-3<{2d\\over p}<\\Bigl[{n\\over2}\\Bigr]\\cdot\\Bigl[{n+1\\over 2}\\Bigr]-2,\\]\nwhere $ [x]$ denotes the greatest integer not exceeding $ x$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043850#p2043850]Problem 34:[/url] $(MON 2)$ One country has $n$ cities and every two of them are linked by a railroad. A railway worker should travel by train exactly once through the entire railroad system (reaching each city exactly once). If it is impossible for worker to travel by train between two cities, he can travel by plane. What is the minimal number of flights that the worker will have to use?\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2044224#p2044224]Problem 35:[/url] $(MON 3)$ Prove that there exist distinct natural numbers $m_1,m_2, \\cdots , m_k$ satisfying the conditions \n\\[\\pi^{-1984}<25-\\left(\\frac{1}{m_1}+\\frac{1}{m_2}+\\cdots+\\frac{1}{m_k}\\right)<\\pi^{-1960}\\]\nwhere $\\pi$ is the ratio between a circle and its diameter.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046271#p2046271]Problem 36:[/url] $(MON 4)$ The set $\\{1, 2, \\cdots, 49\\}$ is divided into three subsets. Prove that at least one of these subsets contains three different numbers $a, b, c$ such that $a + b = c$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048074#p2048074]Problem 37:[/url] $(MOR 1)$ $(MOR 1)$ Denote by $[x]$ the greatest integer not exceeding $x$. For all real $k > 1$, define two sequences: \n\\[a_n(k) = [nk]\\text{ and } b_n(k) =\\left[\\frac{nk}{k - 1}\\right]\\]\nIf $A(k) = \\{a_n(k) : n \\in\\mathbb{N}\\}$ and $B(k) = \\{b_n(k) : n \\in \\mathbb{N}\\}$, prove that $A(k)$ and $B(k)$ form a partition of $\\mathbb{N}$ if and only if $k$ is irrational.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046270#p2046270]Problem 38:[/url] $(MOR 2)$ Determine all continuous functions $f$ such that\n\\[f(x + y)f(x - y) = (f(x)f(y))^2\\:\\:\\:\\: \\forall(x, y) \\in\\mathbb{R}^2\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048083#p2048083]Problem 39:[/url] $(MOR 3)$ Let $ABC$ be an isosceles triangle, $AB = AC, \\angle A = 20^{\\circ}$. Let $D$ be a point on $AB$, and $E$ a point on $AC$ such that $\\angle ACD = 20^{\\circ}$ and $\\angle ABE = 30^{\\circ}$. What is the measure of the angle $\\angle CDE$?\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366605&#p366605]Problem 40:[/url] $(NET 1)$ Find one pair of positive integers $a,b$ such that $ab(a+b)$ is not divisible by $7$, but $(a+b)^7-a^7-b^7$ is divisible by $7^7$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049190#p2049190]Problem 41:[/url] $(NET 2)$ Determine positive integers $p, q$, and $r$ such that the diagonal of a block consisting of $p\\times q\\times r$ unit cubes passes through exactly $1984$ of the unit cubes, while its length is minimal. (The diagonal is said to pass through a unit cube if it has more than one point in common with the unit cube.)\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049194#p2049194]Problem 42:[/url] $(NET 3)$ Triangle $ABC$ is given for which $BC = AC + \\frac{1}{2}AB$. The point $P$ divides $AB$ such that $BP : PA = 1 : 3$. Prove that $\\angle CAP = 2\\angle CPA$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=165811&#p165811]Problem 43:[/url] $(POL 1)$ Let $a,b,c,d$ be odd integers such that $0 1$ and $x_i \\in \\mathbb{R}$ for $i = 1,\\cdots, n$. Set \n\\[S_k = x_1^k+ x^k_2+\\cdots+ x^k_n\\]\nfor $k \\ge 1$. If $S_1 = S_2 =\\cdots= S_{n+1}$, show that $x_i \\in \\{0, 1\\}$ for every $i = 1, 2,\\cdots, n.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366610&#p366610]Problem 50:[/url] $(ROM 5)$ Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049199#p2049199]Problem 51:[/url] $(SPA 1)$ Two cyclists leave simultaneously a point $P$ in a circular runway with constant velocities $v_1, v_2 (v_1 > v_2)$ and in the same sense. A pedestrian leaves $P$ at the same time, moving with velocity $v_3 = \\frac{v_1+v_2}{12}$ . If the pedestrian and the cyclists move in opposite directions, the pedestrian meets the second cyclist $91$ seconds after he meets the first. If the pedestrian moves in the same direction as the cyclists, the first cyclist overtakes him $187$ seconds before the second does. Find the point where the first cyclist overtakes the second cyclist the first time.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046268#p2046268]Problem 52:[/url] $(SPA 2)$ Construct a scalene triangle such that \n\\[a(\\tan B - \\tan C) = b(\\tan A - \\tan C)\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048109#p2048109]Problem 53:[/url] $(SPA 3)$ Find a sequence of natural numbers $a_i$ such that $a_i = \\displaystyle\\sum_{r=1}^{i+4} d_r$, where $d_r \\neq d_s$ for $r \\neq s$ and $d_r$ divides $a_i$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049154#p2049154]Problem 54:[/url] $(SPA 4)$ Let $P$ be a convex planar polygon with equal angles. Let $l_1,\\cdots, l_n$ be its sides. Show that a necessary and sufficient condition for $P$ to be regular is that the sum of the ratios $\\frac{li}{l_{i+1}} (i = 1,\\cdots, n; l_{n+1}= l_1)$ equals the number of sides.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049156#p2049156]Problem 55:[/url] $(SPA 5)$ Let $a, b, c$ be natural numbers such that $a+b+c = 2pq(p^{30}+q^{30}), p > q$ being two given positive integers.\n$(a)$ Prove that $k = a^3 + b^3 + c^3$ is not a prime number.\n$(b)$ Prove that if $a\\cdot b\\cdot c$ is maximum, then $1984$ divides $k$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049160#p2049160]Problem 56:[/url] $(SWE 1)$ Let $a, b, c$ be nonnegative integers such that $a \\le b \\le c, 2b \\neq a + c$ and $\\frac{a+b+c}{3}$ is an integer. Is it possible to find three nonnegative integers $d, e$, and $f$ such that $d \\le e \\le f, f \\neq c$, and such that $a^2+b^2+c^2 = d^2 + e^2 + f^2$?\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043852#p2043852]Problem 57:[/url] $(SWE 2)$ Let $a, b, c, d$ be a permutation of the numbers $1, 9, 8,4$ and let $n = (10a + b)^{10c+d}$. Find the probability that $1984!$ is divisible by $n.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046264#p2046264]Problem 58:[/url] $(SWE 3)$ Let $(a_n)_1^{\\infty}$ be a sequence such that $a_n \\le a_{n+m} \\le a_n + a_m$ for all positive integers $n$ and $m$. Prove that $\\frac{a_n}{n}$ has a limit as $n$ approaches infinity.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046262#p2046262]Problem 59:[/url] $(USA 1)$ Determine the smallest positive integer m such that $529^n+m\\cdot 132^n$ is divisible by $262417$ for all odd positive integers $n$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012335#p2012335]Problem 60:[/url] $(USA 2)$ Determine all pairs $(a, b)$ of positive real numbers with $a \\neq 1$ such that\n\\[\\log_a b < \\log_{a+1} (b + 1).\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043853#p2043853]Problem 61:[/url] $(USA 3)$ A fair coin is tossed repeatedly until there is a run of an odd number of heads followed by a tail. Determine the expected number of tosses.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049209#p2049209]Problem 62:[/url] $(USA 4)$ From a point $P$ exterior to a circle $K$, two rays are drawn intersecting $K$ in the respective pairs of points $A, A'$ and $B,B' $. For any other pair of points $C, C'$ on $K$, let $D$ be the point of intersection of the circumcircles of triangles $PAC$ and $PB'C'$ other than point $P$. Similarly, let $D'$ be the point of intersection of the circumcircles of triangles $PA'C'$ and $PBC$ other than point $P$. Prove that the points $P, D$, and $D'$ are collinear.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012331&#p2012331]Problem 63:[/url] $(USA 5)$ Inside triangle $ABC$ there are three circles $k_1, k_2, k_3$ each of which is tangent to two sides of the triangle and to its incircle $k$. The radii of $k_1, k_2, k_3$ are $1, 4$, and $9$. Determine the radius of $k.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049210#p2049210]Problem 64:[/url] $(USS 1)$ For a matrix $(p_{ij})$ of the format $m\\times n$ with real entries, set\n\\[a_i =\\displaystyle\\sum_{j=1}^n p_{ij}\\text{ for }i = 1,\\cdots,m\\text{ and }b_j =\\displaystyle\\sum_{i=1}^m p_{ij}\\text{ for }j = 1, . . . , n\\longrightarrow(1)\\]\nBy integering a real number, we mean replacing the number with the integer closest to it. Prove that integering the numbers $a_i, b_j, p_{ij}$ can be done in such a way that $(1)$ still holds.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046259#p2046259]Problem 65:[/url] $(USS 2)$ A tetrahedron is inscribed in a sphere of radius $1$ such that the center of the sphere is inside the tetrahedron. Prove that the sum of lengths of all edges of the tetrahedron is greater than 6.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=141613#p141613]Problem 66:[/url] $(USS 3)$ Let $1=d_1\\frac{5}{6}R\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043854#p2043854]Problem 68:[/url] $(USS 5)$ In the Martian language every finite sequence of letters of the Latin alphabet letters is a word. The publisher \u201cMartian Words\u201d makes a collection of all words in many volumes. In the first volume there are only one-letter words, in the second, two-letter words, etc., and the numeration of the words in each of the volumes continues the numeration of the previous volume. Find the word whose numeration is equal to the sum of numerations of the words Prague, Olympiad, Mathematics." } { "Tag": [ "MIT", "college", "Harvard", "geometry" ], "Problem": "I know the most obvious answers - MIT, Harvard, and all of the other ivy league schools in the area. Unfortunately, I'm sure that those are, for the most part, beyond me financially, not to mention I would have a slim chance of being accepted. So I have two questions:\r\n\r\n1) I know that these are the \"dream\" schools. But really, is it that big of a deal if you can't get into them? (in terms of jobs after college, expected salaries, etc). And if the answer to this is \"yes\", then how hard is it to transfer into them for graduate school? :lol: \r\n\r\n2) Going back to my original subject, what are some other schools in the New England area that are good for math / possibly computer science related majors? \r\n\r\nI really am passionate about math, but seeing as how it was somewhat of a late discovery (spring of my junior year), I haven't had much of a chance to build an impressive resume to qualify for the more competitive schools.", "Solution_1": "What are your thoughts on liberal arts schools? They're programs in math and computer science don't match up to the likes of Harvard and MIT, but since the schools are so small, you'll get a lot of close attention in those departments. I was thinking Williams and Amherst.", "Solution_2": "Make sure that these schools really are beyond you financially. Harvard is providing quite generous financial aid which might make it cheaper to go to Harvard than other schools.\r\n\r\nhttp://www.hno.harvard.edu/gazette/2007/12.13/99-finaid.html", "Solution_3": "It's a misconception to think that any school is beyond you financially. Just apply and wait for the aid package. If you get into all those really good schools, presumably at least one will give you a nice deal. And even if you have a pretty nice aid package, you can ask for more. I was offered 32k/year for 4 years, but I said \"I think I need more\" and they gave me 35k/year for 4 years instead. $ \\$$12000 is quite a bit of money.\r\n\r\n1. No, it doesn't matter as much as your abilities matter.\r\n\r\n2. Rensselaer Polytechnic Institute. Many people here are attending because they ot rejected by MIT and CalTech. I'd argue it's the third best tech school." } { "Tag": [ "geometry", "trigonometry", "inequalities", "triangle inequality", "trig identities", "Law of Sines", "geometry unsolved" ], "Problem": "The incircle of $ ABC$ is tangent to the sides $ BC$, $ CA$ and $ AB$ respectively at the points: $ A_1$, $ B_1$ and $ C_1$. It is also known that the line $ A_1B_1$ pass through the middle point of the segment $ CC_1$. Find the angles of the triangle if their $ \\text{sin}$'s form an arithmetic progression.", "Solution_1": "My solution is quite unelegant. I hope there is a better approach.\r\nWrite $ a\\equal{}\\left|BC\\right|$, $ b\\equal{}\\left|CA\\right|$, $ c\\equal{}\\left|AB\\right|$ and $ \\alpha\\equal{}\\angle CAB$, $ \\beta\\equal{}\\angle ABC$, $ \\gamma\\equal{}\\angle BCA$. W.l.o.g. $ \\sin\\alpha\\leq \\sin\\beta$. Now there are two cases:\r\n[b]1:[/b] $ \\sin\\gamma\\leq \\sin\\alpha\\leq \\sin\\beta$. By the law of sines and by the initial conditions $ \\sin\\beta\\minus{}\\sin\\alpha\\equal{}\\sin\\alpha\\minus{}\\sin\\gamma\\Leftrightarrow \\minus{}2a\\plus{}b\\plus{}c\\equal{}0$ holds. Since $ A_{1}B_{1}$ passes through the midpoint of $ CC_{1}$ $ C_{1}A_{1}B_{1}$ and $ A_{1}CB_{1}$ have the same altitudes with respect to $ A_{1}B_{1}$, hence the same area ($ A(\\cdot)$ denotes the area):\r\n$ A\\left(ABC\\right)\\minus{}A\\left(AC_{1}B_{1}\\right)\\minus{}A\\left(C_{1}BA_{1}\\right)\\minus{}A\\left(A_{1}CB_{1}\\right)\\equal{}A\\left(A_{1}CB_{1}\\right)$\r\n$ \\Leftrightarrow bc\\sin\\alpha\\minus{}\\left(\\frac{\\minus{}a\\plus{}b\\plus{}c}{2}\\right)^{2}\\sin\\alpha\\minus{}\\left(\\frac{a\\minus{}b\\plus{}c}{2}\\right)^{2}\\sin\\beta\\minus{}2\\left(\\frac{a\\plus{}b\\minus{}c}{2}\\right)^{2}\\sin\\gamma\\equal{}0$\r\n$ \\Leftrightarrow abc\\minus{}a\\left(\\frac{\\minus{}a\\plus{}b\\plus{}c}{2}\\right)^{2}\\minus{}b\\left(\\frac{a\\minus{}b\\plus{}c}{2}\\right)^{2}\\minus{}2c\\left(\\frac{a\\plus{}b\\minus{}c}{2}\\right)^{2}\\equal{}0$\r\n$ \\Leftrightarrow 0\\equal{}ab\\left(2a\\minus{}b\\right)\\minus{}\\frac{a^{3}}{4}\\minus{}b\\left(\\frac{3a}{2}\\minus{}b\\right)^{2}\\minus{}2\\left(2a\\minus{}b\\right)\\left(b\\minus{}\\frac{a}{2}\\right)^{2}\\equal{}\\left(b\\minus{}a\\right)\\left(2b\\minus{}5a\\right)\\left(2b\\minus{}a\\right)/4$.\r\nBecause of $ a\\leq b$ either $ a\\equal{}b$ or $ 2b\\minus{}5a\\equal{}0$. In the first case the triangle is equilateral, in the second case the triangle inequality is not fulfilled.\r\n[b]2:[/b]$ \\sin\\alpha\\leq \\sin\\gamma\\leq \\sin\\beta$. Analogously to case 1 we get:\r\n$ 0\\equal{}bc\\left(2c\\minus{}b\\right)\\minus{}\\left(2c\\minus{}b\\right)\\left(b\\minus{}\\frac{c}{2}\\right)^{2}\\minus{}b\\left(\\frac{3c}{2}\\minus{}b\\right)^{2}\\minus{}\\frac{c^{3}}{2}\\equal{}\\minus{}c\\left(c\\minus{}b\\right)^{2}$.\r\nAgain $ ABC$ has to be equilateral." } { "Tag": [ "induction" ], "Problem": "A gold shipment consists of $ N > 1$ bars containing a total of $ 2^{100}$ gold atoms. The number of atoms in each bar is also a power of $ 2$ (which can vary from bar to bar). Show the bars can be loaded onto two ships so that each carries the same amount of gold. (What is the value of the shipment?)", "Solution_1": "We will prove this by induction after $ n \\geq 1$, for a total of $ 2^n$ gold atoms. For $ n\\equal{}1$ it is clear, since the only possibility is $ N\\equal{}2$, two bars of one atom each. Assume the hypothesis proven up to $ n > 1$.\r\nLet the $ N$ bars contain $ 2^{a_i}$ atoms, with $ \\sum_{i\\equal{}1}^N 2^{a_i} \\equal{} 2^{n\\plus{}1}$, and $ 0 \\leq a_N \\leq \\cdots \\leq a_1$. \r\n\r\nIf $ a_N > 0$, take $ b_i \\equal{} a_i \\minus{} 1$, so $ \\sum_{i\\equal{}1}^N 2^{b_i} \\equal{} 2^{n}$. We can apply the induction hypothesis to obtain a partition $ I \\cap J \\equal{} \\emptyset$, $ I \\cup J \\equal{} \\{1,2,\\ldots,N\\}$ and $ \\sum_{i \\in I} 2^{b_i} \\equal{} \\sum_{j \\in J} 2^{b_j} \\equal{} 2^n$. Multiplying by $ 2$ proves that this is a valid partition for $ n\\plus{}1$ too.\r\n\r\nIf $ a_N \\equal{} 0$, let $ k$ be the least index such that $ a_k \\equal{} 0$. Then $ N \\minus{} k \\plus{} 1 \\equal{} 2M$ needs be even. Take $ a'_k \\equal{} a'_{k\\plus{}1} \\equal{} \\cdots \\equal{} a'_{N\\minus{}M} \\equal{} 1$ (\"pack\" the one-atom bars in pairs) and $ a'_i \\equal{} a_i$ for $ i \\equal{} 1, 2, \\ldots, k\\minus{}1$, so $ \\sum_{i\\equal{}1}^{N\\minus{}M} 2^{a'_i} \\equal{} 2^{n\\plus{}1}$. We find ourselves now in the previous case (with $ a'_{N\\minus{}M} \\equal{} 1 > 0$ and $ N\\minus{}M > 1$), so we can find a valid partition." } { "Tag": [ "quadratics", "number theory solved", "number theory" ], "Problem": "[color=green]Prove that if (2 to the power n) - 15 = x square, then n = 4 or n = 6.[/color]\r\n\r\nThanks very much guys", "Solution_1": "For large even $n$, $2^{n}$ is a square and it will not be \"close\" to another square.\r\n\r\nFor odd $n$, the LHS is $\\equiv 2 \\bmod 3$, not a quadratic residue." } { "Tag": [ "geometry", "rectangle", "puzzles" ], "Problem": "How many rectangles (of any size and in any position) are there on a standard 8 x 8 checkerboard? \r\n\r\n[hide]you can check ur solution here: http://www.projecteureka.org/problem/question/377[/hide]", "Solution_1": "There are [b]1296[/b] rectangles:\r\nTo choose the locations of the vertical sides, there are 9C2=36 ways to pick them. The same goes for horizontal sides, so the total ways to make a rectangle is 36*36=1296." } { "Tag": [], "Problem": "A girl bought a dog for $\\$10, sold it for $\\$15, bought it back for $\\$20, and finally sold it for $\\$25. Did the girl make or lose money, and how much did she make or lose?", "Solution_1": "[hide=\"da anwer\"] \n$-10+15-20+25=10$\n She made 10 dollars[/hide]", "Solution_2": "[hide]-10+15-20+25=+10\nShe MADE $10.[/hide]", "Solution_3": "[quote=\"MCrawford\"]A girl bought a dog for $\\$10, sold it for $\\$15, bought it back for $\\$20, and finally sold it for $\\$25. Did the girl make or lose money, and how much did she make or lose?[/quote]\r\n\r\n[hide]$-10+15-20+25=10$[/hide]" } { "Tag": [ "ratio", "geometric series" ], "Problem": "16 hours and 160 bottles of beer after the New Year's movie, Bobert decides to walk up the Leaning Tower of Treething with Caddlebot and drop him off. Every time Caddlebot hits the ground, he bounces to 1/4 of his previous height. In total, he moves 350 feet. How tall is the Leaning Tower of Treething?", "Solution_1": "[hide=\"Hmmm\"]The LToT is x feet tall. The amount that Caddlebot moves after his original fall can be represented by $2\\sum_{i=1}^{\\infty} \\left(\\frac 14\\right)^i\\cdot x$, or at least I hope it can because otherwise my solution is wrong. So\n$x+2\\sum_{i=1}^{\\infty} \\left(\\frac 14\\right)^i\\cdot x=350$\nThe first term of the infinite geometric series is x/4, and the common ratio is 1/4. So plugging into the formula for such a series,\n$x+2\\cdot\\frac{x/4}{1-1/4}=350$\n$x+2\\cdot\\frac x3=350$\n$5x/3=350$\n$x=210$\nIs 210 feet the right answer? Probably not.\n[/hide]\r\nP.S. This is more Intermediate than Getting Started I think...", "Solution_2": "Yeah, I think summing an infinite geometric series is just across the line into (easy) Intermediate Forum level, so I'll move this.", "Solution_3": "No offense but does the problem has to involve FBI or CIA actions? (a.k.a. crime :? ) It's fine with me since it's fictional and funny but I'm just wondering. ;)", "Solution_4": "I must resort to chatspeak to express my reaction to what Silverfalcon said.\n\n\n\n\"wtf?\"\n\n\n\nYes, it does involve the CIA. I paid the CIA agent Bobert five hundred dollars plus a cat poison bottle to push Caddlebot off of the Leaning Tower of Treething.\n\n\n\n[hide]\n\nYep, it is 210 feet high.\n\n[/hide]", "Solution_5": "He means, why do all of your problems involve people killing or trying to kill other people and animals for no good reason? You could just have easily had Bobert kick a basketball off the top of the LToT. :P \r\nI solved a problem involving an infinite geometric series correctly?! I'm going to go celebrate! :lol:", "Solution_6": "[hide]\nLet x equal the height of the tree.\n\n2(x/(1-(1/4)) - x = 350\n2x / (3/4) - x = 350\n(8/3)x - x = 350\n(5/3)x = 350\nx = 210\n[/hide]", "Solution_7": "Geez treething. I didn't mean for it to be violent to the point that it's reputation is gonna go down like this. I mean god your next problem gonna be like this:\r\n\r\n\"Frankinfueter shoots 15000000 AK - 47 rounds into Caddlebot, if the AK - 47's rate of fire is 300 rounds per sec how long does it take to shoot the rounds into Caddlebot?\"\r\n\r\nAnd that is EVIL. But I still hope that everyone catches on. So if you're posting a word problem please put in Albatross and friends. I'll soon have a name for all 26 letters of the alphabet. Also if we could get this in the AMC it would be so awesome." } { "Tag": [], "Problem": "Questions is : What would be the amount of compound interest on $8,000. invested for one year at 6%, compounded quarterly? Round your answer to the nearest dollar?????? :what?:", "Solution_1": "In order to do this, you need the compound interest formula.\r\n\r\nThis is: (if i remember it correctly)\r\n\r\nP(1 + r/n) ^ nt\r\n\r\nP = principal, or the initial amount invested\r\nr = rate, in a decimal (e.g. 6% = .06)\r\nn = number of years over which the investment occurrs\r\nt = number of compoundings in one year. Note that quarterly = 4 compoundings/yr.\r\n\r\nKnowing this, plugging the formula into your calculator should solve this.", "Solution_2": "[hide=\"Hint\"]Let me give an example to illustrate the concept of compound interest. Say the initial amount is 1000 dollars, with 10% interest compounded monthly. \n\nSince interest is compounded every month:\nAfter 1 month, there would be $ 1000d + 1000d*10\\% = 1100d$.\nAfter 2 months, there would be $ 1100d + 1100d*10\\% = 1210d$.\n\nBasically, interest is compounded on the total amount of money at each point in time (including whatever interest has been earned so far), rather than in simple interest where only the initial amount counts.[/hide]\n\n[hide=\"Solution\"]There are 4 quarters in a year, so the interest of 6% would be compounded 4 times.\n\n$ 1 + 6\\% = 1.06$\n\\begin{align*}8000d*1.06^4 - 8000d & = 8000d*1.26247696 - 8000d \\\\\n& = 0.26247696*8000d \\\\\n& \\approx\\boxed{2100d}\\end{align*}[/hide]", "Solution_3": "$ 8000 \\cdot (1 \\plus{} \\frac {6}{100\\cdot4})^{4} \\minus{} 8000$", "Solution_4": "ZhangPeijin is correct, I misunderstood what the compound interest of 6% referred to. So since the 6% is yearly interest, if it's compounded quarterly then each quarter there would be 1.5% interest added, for 4 quarters. So, just evaluate $ 8000*1.015^4\\minus{}8000$ to get the solution." } { "Tag": [ "summer program", "PROMYS", "calculus", "calculus computations" ], "Problem": "Can anyone check whether this proof has enough detail and is rigorous enough? \r\n\r\nProve that for any $a, b, c, d \\in R$, $-(a-b) = b-a$. I know this sounds trivial, but the proof has to be detailed as possible. \r\n\r\nProof: $-(a-b) = (-1)(a+(-b)) = (-1)(a)+(-1)(-b) =-a+b = b-a$ by commutative field property. \r\n\r\nIs this a reasonable proof?\r\n\r\n2ndly, If $S_{1}, S_{2}$ are nonempty subsets of $R$ that are bounded from above, prove that $lub \\{x+y: x \\in S_{1}, y \\in S_{2}\\}= lub S_{1}+lub S_{2}$. lub stands for least upper bound. \r\nAgain here's my proof:\r\n\r\nThe conditions given imply that $x \\le lub S_{1}$ and $y \\le lub S_{2}$. Adding the two gives $x+y \\le lub S_{1}+lub S_{2}$.\r\n\r\nThis almost seems correct, BUT, how can I show that $lub S_{1}+lub S_{2}$ is smaller than ANY OTHER upper bound (let's call it $n$)?\r\n\r\nAs always, help is appreciated.", "Solution_1": "[quote=\"bythecliff\"]Proof: $-(a-b) = (-1)(a+(-b))$[/quote]\n\nIf you want a rigorous proof from axioms, this is unacceptable. Prove this.\n\n[quote=\"bythecliff\"]$(-1)(a)+(-1)(-b) =-a+b$[/quote]\r\n\r\nThis also needs to be proven. Namely, that $(-1)(-b) = b$.\r\n\r\n[hide=\"A simpler rigorous proof\"] We want to show that $a-b$ and $b-a$ are additive inverses.\n\n$(a-b)+(b-a) = (a+(-b))+(b+(-a)) = \\left( a+(-b+b) \\right)+(-a) = \\left( a+0 \\right)+(-a) = a+(-a) = 0$\n\nBefore this is acceptable, we need to show that the additive identity is unique, and then that the additive inverse is unique. This is left as an exercise, since it is very simple. [/hide]", "Solution_2": "OK, let me do the first proof again: \r\n\r\nlet's prove a simpler case of $-a = (-1)(a)$ for all $a \\in R$. $(-1)(a)+a = a((-1)+1) = a(0) = 0$, so $(-1)(a)$ is a solution to the equation $x+a = 0$. \r\n\r\nAlso, $x+a = 0 \\Rightarrow x+a+(-a) = 0+(-a) \\Rightarrow x+0 = (-a) \\Rightarrow x =-a$.\r\n\r\nSo both $x = (-1)(a)$ and $x =-a$ are solutions to the equation $x+a = 0$, and because the solution is unique, $(-1)(a) = (-a)$. Hence, $-(a-b) = (-1)(a+(-b))$.", "Solution_3": "[quote=\"bythecliff\"]and because the solution is unique[/quote]\r\n\r\nNeeds to be proven. (You asked for rigor! :P )", "Solution_4": "[quote=\"t0rajir0u\"][quote=\"bythecliff\"]and because the solution is unique[/quote]\n\nNeeds to be proven. (You asked for rigor! :P )[/quote]\r\n\r\nNot so much that it needs to be proven as doesn't need to be there at all. If $x = y$ and $x = z$, $y = z$ by transitivity of equality.", "Solution_5": "[quote=\"JBL\"]Not so much that it needs to be proven as doesn't need to be there at all. If $x = y$ and $x = z$, $y = z$ by transitivity of equality.[/quote]\r\n\r\nNo, no, his statement was that if $t+a = 0$ and $t+b = 0$ it follows that $a = b$. To prove this statement rigorously you need to show that the additive identity is unique. :P \r\n\r\nThis is super PROMYS-level picky, though." } { "Tag": [ "search" ], "Problem": "at twelve oclock the second minut and hour hands all coincide. What is the next time, in hours, minutes and seconds(to the nearest second), the minute and hour hands will coincide", "Solution_1": "The hour hand coicides with the minute hand 11 times in 12 hours, therefore if we say the hour is $ h$, the time would be $ h$ hours $ h$ minutes and $ \\frac{12}{11}h$ seconds. we want to find a time where $ \\frac{12}{11}h$ is equal to $ h$. then we get $ \\frac{1}{11}h=0$ so $ h=0$. that means the only time that the minute, second, and hour hand will coincide, will be 12 hours from 12:00.", "Solution_2": "are you sure, it only asks when hour and minute not second.", "Solution_3": "[quote=\"dingzhou\"]are you sure, it only asks when hour and minute not second.[/quote]\r\nThe question asks for the nearest second.", "Solution_4": "Dude next time you post a question use a search button (or you can just scroll down on this page)\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=157457\r\n\r\nthis post was on [b]this[/b] page (if you scroll down you will find it) and ive seen this question atleast 10 times here.", "Solution_5": "Some people might not have seen the question before and wouldn't know to search it... it's not like he's spamming or anything.\r\n\r\nAnyways, the wording of the question should be improved.. but it asks for when just the minute and hour hands coincide next.\r\n[hide]We can easily see that they will coincide after one o'clock. Let $ x$ equal the number of minutes that will pass before the minute and hour hands coincide. Since 12 minutes will pass before the hour hand moves one mark, we have\n$ x=\\frac{x}{12}+5$\n$ x=\\frac{60}{11}$ minutes $ \\approx 5$ minutes $ 27$ seconds.\nSo the answer is 1 hour 5 minutes 27 seconds.[/hide]", "Solution_6": "isnt there a formula to this. like the $ n$th time past 12 that the hand will coincide will be $ \\frac{12}{11}n$ because there are 12 hours and the hours hand makes 1 full circle when the minute hand makes 12 making it 12 11ths hours the first time they coincide." } { "Tag": [ "radical axis", "geometry unsolved", "geometry" ], "Problem": "Given a circle $(O_{k})$ and a triangle $ABC$. \r\nLet the circle $(O_{k})$ meet the line $BC$ at $A_{1}$ and $A_{2}$, the line $CA$ at $B_{1}$ and $B_{2}$, and the line $AB$ at $C_{1}$ and $C_{2}$. \r\nWe construct the three circles with diameters $A_{1}A_{2}$, $B_{1}B_{2}$, $C_{1}C_{2}$. \r\n\r\nProve that the radical center of these three circles is the isogonal conjugate of the center of $(O_{k})$ with respect to triangle $ABC$.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=16404 ; please tell me if you want the solution to be somewhat more detailed.\r\n\r\n darij" } { "Tag": [ "function", "calculus", "derivative", "inequalities", "algebra unsolved", "algebra" ], "Problem": "$ f: \\mathbb{R^{ \\plus{} }}\\rightarrow\\mathbb{R}$\r\n\r\n$ h(x) \\equal{} f(x) \\minus{} x^3$ and $ g(x) \\equal{} f(x) \\minus{} 3x$ are increasing in $ \\mathbb{R}$\r\n\r\nshow that $ T(x) \\equal{} f(x) \\minus{} x^2 \\minus{} x$ is incrising too\r\n\r\nNB: $ f$ can be note derivative", "Solution_1": "What does 'greated' means?\r\nDo you mean 'increasing'?\r\n\r\nPierre.", "Solution_2": "[hide=\"@Pierre\"]Bsr mr.Pierre et desol\u00e9 pour mon anglais;\nje vx dire par greatest croissante j ai pas su la traduction exacte.\nps: f peux etre non derivable.\nmerci ;)[/hide]", "Solution_3": "Let $ a,b$ be two arbitrary reals with $ a > b$. We are given that\r\n\r\n$ f(a) \\minus{} f(b) > a^3 \\minus{} b^3$\r\n$ f(a) \\minus{} f(b) > 3a \\minus{} 3b$\r\nand we want to show that\r\n$ f(a) \\minus{} f(b) > a^2 \\plus{} a \\minus{} b^2 \\minus{} b$\r\n\r\nLet $ C \\equal{} \\frac{f(a) \\minus{} f(b)}{a \\minus{} b}$ (note $ a \\minus{} b > 0$). The given inequalities become\r\n$ C > a^2 \\plus{} ab \\plus{} b^2$\r\n$ C > 3$\r\nand what we want is\r\n$ C > a \\plus{} b \\plus{} 1$\r\n\r\nIf we show this holds true for $ a > b \\geq 1$ and $ 1 \\geq a > b$, then the result for the remaining cases follows by transitivity ($ f(a) > f(1) > f(b)$).\r\n\r\n$ a > b \\geq 1$: $ C > a^2 \\plus{} ab \\plus{} b^2 \\equal{} (a\\plus{}b)a \\plus{} b^2 > (a\\plus{}b)(1) \\plus{} 1^2 \\equal{} a \\plus{} b \\plus{} 1$\r\n$ 1 \\geq a > b$: $ C > 3 > a \\plus{} b \\plus{} 1$" } { "Tag": [ "pigeonhole principle" ], "Problem": "I found this problem in The Art and Craft of Problem Solving:\r\n\r\nProve that for any positive integer n, there exists a positive multiple of n that only uses the digits 7 and 0.", "Solution_1": "Consider the remainders when $ 7, 77, 777, ....$ are divided by $ n$.", "Solution_2": "Thank you! That helped.\r\n\r\nFull proof:\r\n\r\nConsider the first $ n$ numbers of the sequence 7, 77, 777, 7777...\r\n\r\nWe have two cases:\r\n\r\n1. One of them is a multiple of $ n$.\r\nIn this case we are done.\r\n\r\n2. None of them are multiples of $ n$.\r\nThen there are $ n\\minus{}1$ possible remainders upon division by $ n$. By the pigeonhole principle, two of the numbers must have the same remainder. So their difference is a multiple of n. But their difference only contains the digits 7 and 0, so we are done.", "Solution_3": "very nice soln!" } { "Tag": [], "Problem": "im finishing my freshman year of high school at north ridge high school. already i have been asked about colleges. i know that i want to go to a christian college but i also want to leave home. the only christian colleges i know of are here in indiana or in the surrounding states. does anyone know of good christian colleges away from home?", "Solution_1": "This depends a lot on your definition of Christian in what kind of colleges you are looking for. I assume, because you are posting here on AoPS, that you might be interested in a college with a strong math or science program besides, but I'm not sure how Christian is Christian enough for your requirements. (My point of reference is going to a secular state university for my undergraduate degree, where I met plenty of very devout Christians from all over the world.) What colleges do you have in mind so far? That might give me an idea of colleges in other regions of the country with similar characteristics. Or just tell me in more detail what characteristics you are looking for.", "Solution_2": "Taylor University\n", "Solution_3": "Samford University. Its in Birmingham, Alabama. Where I live." } { "Tag": [ "modular arithmetic" ], "Problem": "In year $ N$, the $ 300^\\text{th}$ day of the year is a Tuesday. In year $ N \\plus{} 1$, the $ 200^{\\text{th}}$ day of the year is also a Tuesday. On what day of the week did the $ 100^\\text{th}$ day of year $ N \\minus{} 1$ occur?\r\n\r\n$ \\textbf{(A)}\\ \\text{Thursday} \\qquad \\textbf{(B)}\\ \\text{Friday} \\qquad \\textbf{(C)}\\ \\text{Saturday} \\qquad \\textbf{(D)}\\ \\text{Sunday} \\qquad \\textbf{(E)}\\ \\text{Monday}$", "Solution_1": "[hide]\nFirst, notice that $ 100\\equiv 2\\pmod 7$. Thus, if the $ 300$th day of year $ N$ was a Tuesday, the $ 200$th day of $ N$ was a Sunday. Since $ 365\\equiv 1\\pmod 7$, we need to have year $ N\\plus{}1$ be a leap year if day $ 200$ of $ N\\plus{}1$ is a Tuesday. Thus $ N$ is not a leap year. The $ 100$th day of $ N$ is a Friday, so the $ 100$th day of $ N\\minus{}1$ is a Thursday. Answer is $ A$.\n[/hide]", "Solution_2": "N has 366 days since the number of days between 300-th day of N year and 200-th day of N+1 year must be divided by 7. Now, the number of days between 100-th day of N-1 year and 300-th day of N year is 565 days. (565+2)%7=0 so since 302-d day of N year is Thursday so the 100-th day of N-1 year is Thursday.", "Solution_3": "Hehehe. This came up on an old AMC, right? I distinctly recall using the TI-89's calendar feature on this one :P", "Solution_4": "this is WAY too easy for a #18", "Solution_5": "[quote=\"apiarist1\"]N has 366 days since the number of days between 300-th day of N year and 200-th day of N+1 year must be divided by 7. Now, the number of days between 100-th day of N-1 year and 300-th day of N year is 565 days. (565+2)%7=0 so since 302-d day of N year is Thursday so the 100-th day of N-1 year is Thursday.[/quote]\r\n\r\nDoesn't N+1 have 366 days?", "Solution_6": "No, I explained 'why'. See above.", "Solution_7": "Alternate Solution:\n\n$300\\equiv6\\mod7$ so January 6 of year $N$ was a Tuesday so January 1 of year $N$ was a Thursday.\n\n$200\\equiv4\\mod7$ so January 4 of year $N+1$ was a Tuesday so January 1 of year $N+1$ is a Saturday. \n\n$366\\equiv2\\mod7$. If year $N$ was a leap year, then January 1 of year $N+1$ would be on a Saturday which satisfies the conditions. Thus, year $N-1$ is not a leap year. So January 1 of the year $N-1$ is a Wednesday. Since $100\\equiv2\\mod7$, then the $100$th day of year $N-1$ is on a Thursday.", "Solution_8": "I just used the general pattern they gave to find that year (N-1)'s 400-365=65th day of Year N which is a Tuesday. You find 65 (mod 7)=2 so Tuesday+2 days=Thursday. Is this a valid solution?" } { "Tag": [ "geometry", "perimeter", "rectangle", "Asymptote" ], "Problem": "A man has a $ 10$ m $ \\times$ $ 10$ m square garden. In the center is a $ 2$ m $ \\times$ $ 2$ m square patch which he cannot use. He divides his usable space into four congruent rectangular patches. What is the number of meters in the perimeter of each rectangle?", "Solution_1": "I don't know how to draw it, but they are 6m x 4m, so the perimeter is 20m.", "Solution_2": "[asy]draw((0,0)--(10,0)--(10,10)--(0,10)--cycle);\ndraw((6,10)--(6,4));\ndraw((0,6)--(6,6));\ndraw((4,0)--(4,6));\ndraw((4,4)--(10,4));[/asy]\r\n\r\nIf we let $ w$ be the width and $ l$ be the length, you get \\[ w\\plus{}l\\equal{}10\\] \\[ w\\plus{}2\\plus{}w\\equal{}10\\]\r\nSolving gets us $ w\\equal{}4$ and $ l\\equal{}6$, so the perimeter is $ 2(4\\plus{}6)\\equal{}\\boxed{20}$", "Solution_3": "[quote=\"5849206328x\"][asy]draw((0,0)--(10,0)--(10,10)--(0,10)--cycle);\ndraw((6,10)--(6,4));\ndraw((0,6)--(6,6));\ndraw((4,0)--(4,6));\ndraw((4,4)--(10,4));[/asy]\n\nIf we let $ w$ be the width and $ l$ be the length, you get \\[ w\\plus{}l\\equal{}10\\] \\[ w\\plus{}2\\plus{}w\\equal{}10\\]\nSolving gets us $ w\\equal{}4$ and $ l\\equal{}6$, so the perimeter is $ 2(4\\plus{}6)\\equal{}\\boxed{20}$[/quote]\n\nYou get upvoted for the awesome asymptote picture!", "Solution_4": "EXCUSE ME PLEASE TELL ME WHERE THIS COMES FROM! ASAP", "Solution_5": "Why do you need to know ASAP", "Solution_6": "Since we get $w+l=10$, and the perimeter is $2(l+w)$, we don't have to solve for them, our answer is just $2(10)=\\boxed{20}$" } { "Tag": [ "probability", "blogs", "geometry", "Support", "calculus", "articles", "induction" ], "Problem": "I'm making this topic because some discussion about this has gotten into the probability of GOD topic, when it obviously does not belong there. This is only for scientific evidence for or against evolution, or for or against any other explanation for the diversity of life on earth. Talk about religious implications in the 'science and religion' thread.", "Solution_1": "okay i'm moving this to round table...", "Solution_2": "I thought I'd also bring in the last thing Lynnelle said on that thread:\r\n\r\n[quote=\"Lynnelle Ye\"]Define 'religion'. Certain interpretations of the Christian religion conflict with evolution. Other interpretations don't--accounting for the theistic evolutionist majority at http://www.christianforums.com (check out the creo/evo debate forum there). I don't know as much about other religions versus science, but my point is that religion conflicts if you think it conflicts, and doesn't conflict if you don't think it conflicts. I mean, there are still a few geocentrists and even flat-earthers around, and they claim heliocentrism and a round earth conflict with their religion, which is fine because that's partially how their individual religion is defined. If you see what I mean.\nThat wasn't very coherent, but I hope you can get some meaning out of it...\nP.S. Maybe we need a separate thread for evolution.[/quote]", "Solution_3": "According to some (eg Aismov), simple things like price of papyrus in old days have much to do with how the theory was documneted. If you think I am not making any sense please read this blog:\r\n\r\n[url=http://www.sumware.com/creation.html]How it Happened[/url]\r\n\r\nEnjoy!", "Solution_4": "It is almost the definition of science that ANY body of ideas will come into conflict with it (or more precisely: enter the contest as one of the contenders) upon making assertions specific enough that they can be tested, compared with other theories' conclusions, etc. In other words, science is a kind of (ruthless) natural selection process within the realm of all possible sufficiently-specific theories.\r\n\r\nThe only way for religion, or any other doctrine (politics, politesse, fashion,....) to avoid conflict with science is to only make assertions nebulous enough to escape any evaluation or comparison. Assertions that are \"not even wrong\", as Pauli put it. \r\n\r\nIf particular religious doctrines or their interpreters choose not to go near the arena, OK --- but once they enter the space of meaningful theories, they are subject to evaluation and comparison like all others.", "Solution_5": "Please define evolution.", "Solution_6": "Religion and evolution are completely different topics that cannot be grouped together. Religions is based on certain assumptions that cannot be argued against, and evolution is based on other assertions of mankind.", "Solution_7": "[quote=\"mcalderbank\"]Religions is based on certain assumptions that cannot be argued against,[/quote]\n\nReligious assumptions/assertions are unarguable only if they are too nebulous to determine; \"not even wrong\", in other words. When they are more specific, such as claims that the world is several thousand years old, they become highly arguable.\n\n\n[quote] and evolution is based on other assertions of mankind.[/quote]\r\n\r\nEvolution is a mathematical phenomenon, such as faster reproducers eventually dominating a population. These theorems and mathematical phenomena are applicable to biology and many other subjects.", "Solution_8": "[quote=\"white_horse_king88\"]Please define evolution.[/quote]\r\nThe tendency for the frequency of different alleles in a population to change over time. (Paraphrasing a definition occasionally posted on http://www.christianforums.com ; I think they got it from http://www.talkorigins.org , though I'm not sure) Common descent is also part of the theory I believe, though it's not included in the above definition.", "Solution_9": "[quote=\"LynnelleYe\"][quote=\"white_horse_king88\"]Please define evolution.[/quote]\nThe tendency for the frequency of different alleles in a population to change over time.[/quote]\r\n\r\nIt's really much broader than that.\r\n\r\nEvolution is more or less the same thing as population dynamics (dynamics in the sense \r\nof \"dynamical systems\"), which involves ideas from many areas of mathematics.\r\nIn the applications to biology one looks at the evolution of molecular, cellular, or organismic\r\npopulations, but the phenomena are generic and visible everywhere. For example, a group of\r\npeople with 5 children per family will eventually dominate a population where everyone else has 2 children per family.\r\n\r\nWhat people normally think of as evolution in Darwin's sense, is the application of such ideas\r\nto situations where heredity, variation, and replication are all present. But such contexts need not be biological.", "Solution_10": "Well, by that definition, evolution obviously occurs. Changes happen all the time. \r\n\r\nThe real question is, has Darwinian evolution (or maybe we should just call it \"macroevolution\" occured? Has one type of organism EVER been seen to turn into another type of organism (eg. whales into cows)?", "Solution_11": "[quote=\"boy_21\"]Well, by that definition, evolution obviously occurs. Changes happen all the time. \n\nThe real question is, has Darwinian evolution (or maybe we should just call it \"macroevolution\" occured? Has one type of organism EVER been seen to turn into another type of organism (eg. whales into cows)?[/quote]\r\n\r\nIt's not exactly one type of organism \"turning\" into another.\r\n\r\nBasically, mutations occur, and the mutants are \"naturally selected\", and over long period of time these mutants no longer breed with their parent population and we have speciation.\r\n\r\nThere are evidences that whales and cows are genetically related (DNA sequence, homologous structures...)", "Solution_12": "Right, so lets say we speed up the mutation rate of fruit flies and see if we get any beneficial mutations. We could simulate millions of years of \"evolution\" in a few years. Can you guess what would happen?\r\n\r\n\r\nPlease, [i][b]PLEASE[/b][/i] do not mistake selection of certian alleles that allow a population to survive better in certain conditions for mutations. I've heard that done WAY to many times. (eg powdered moths)[/i]", "Solution_13": "[quote=\"boy_21\"]Well, by that definition, evolution obviously occurs. Changes happen all the time. [/quote]\n\nAnd mathematical models can be applied to quantify various aspects of the changes. \n\n\n[quote] The real question is, has Darwinian evolution (or maybe we should just call it \"macroevolution\" occured? Has one type of organism EVER been seen to turn into another type of organism (eg. whales into cows)?[/quote]\r\n\r\nFunny you should mention cows. Breeding of domesticated animals, such as the hundreds of types of dogs (originally bred from wolves) and cows (there are no naturally occuring breeds) should answer your question. \r\nThe evolution in vivo (so to speak) of new strains of e.g. HIV virus is a another, current example on a much shorter timescale. It was predicted on theoretical grounds roughly how fast this would happen. \r\nOther examples:\r\nBioengineering of various bacterial strains to acquire desirable traits such as metabolization of toxins.\r\n\"Directed evolution\" of fruit flies to increase lifespan by an order of magnitude.\r\nThese both happened on a lab-experiment timescale.", "Solution_14": "[quote=\"boy_21\"]Right, so lets say we speed up the mutation rate of fruit flies and see if we get any beneficial mutations. We could simulate millions of years of \"evolution\" in a few years. Can you guess what would happen?[/quote]\r\n\r\nWhat is your point, if any? Is the above in answer to some statement (which one?) in this thread? \r\n\r\nCertainly there have been a lot of experiments with flies, bacteria, and other organisms that illustrate \r\nthe evolutionary phenomenon on a short time scale.", "Solution_15": "[quote=\"white_horse_king88\"]two major clashes in evolution. TIME [age of earth] and MEANS [God's creations are perfect and need not evolve]. This can just go on, because both sides made an assumption which cannot be proven by OBSERVATION.[/quote] \r\nThe evolutionary side of those questions is based on observation, not mere assumption. It is observed fact that all species, including humans, evolve (i.e. that heredity and variation are present; for humans it is debatable when and how large population differences would arise, but small ones are very apparent between families, tribes, and national groups). The conclusions one can draw from this are proven by essentially mathematical reasoning (population dynamics of various models), not observation or assumption, and one then can compare them with further observation.", "Solution_16": "[quote=\"LynnelleYe\"] We don't know if evolution is driven by God. It doesn't matter--it still happens. God is outside of science, so whether or not he has anything to do with it is beside the point. [/quote]\r\n\r\nYou are still making the assumption that macroevolution does happen. All that have been listed so far as \"visible\" evolution are cases of adaption, or microevolution, which is far from proving that macroevolution occurs. You have taken 100 small steps along the side of a canyon and claimed that you can therefore walk accross the 100-foot canyon. Each example of microevolution (adaption) is a small step, but that doesn't give any evidence to macroevoultion. Simply put, evolution requires the addition of [i]information[/i], and no one has ever proven that adaption to a new environment or resistance to certain drugs is done by the addidtion of new information. It has been proven that it is possible to create a system that will adapt, but it never been proven that the system can evolve into a completely different system.\r\n\r\nSo can you please stop listing proof of adaption, and try to show me some scientific evidance for evolution?", "Solution_17": "This is the example my teacher uses for macroevolution, the squirrels across the Grand Canyon. Because of isolation for thousands of years, these two types of squirrels are different species from each other because they can no longer reproduce with each other.", "Solution_18": "Macro-evolution does exist. Let's say that evolution can be defined as the change of allele frequencies in a population, which is the accepted definition among the entire scientific community.\r\n\r\nOk.\r\n\r\nKeep in mind that natural selection, which leads to evolution acts on the population, and if we can prove that it acts on the individuals(micro-scale) of a population, then by inference, it acts on the population(macro-scale). \r\n\r\nObservation: Natural populations of all organisms have the potential to increase rapidly because organisms can produce far more offspring than are required merely to replace the parents. A populations consists of all the individuals of one species in a particular area.\r\n\r\nNevertheless, the sizes ofmost natural populations and the resources abailable to maintain them remain relatively constant over time.\r\n\r\nConclusion: There is competition for survival and reproduction. In each generation many individuals must die young, fail to reproduce, produce few offspring, or produce less-fit offspring that fail to survive and reproduce in their turn.\r\n\r\nObservation: Individual members of a populatioin differ from one another in their ability to obtain resources and withstand evnironmental extremes, escape predators, and so on.\r\n\r\nConclusion 2. The most well-adapted, (\"fittest\") indivudals in one generation tend to be the ones who leave the most offspring. The environment \"selects\" for the better adapted traits.\r\n\r\nAt least some of the variation among individuals in traits that affect survivalor reproduction is due to genetic differences that may be passed from parent to offspring.\r\n\r\nConclusion 3. Over generations, differences, or unequal, reproduction among individuals with different genetic makeup changes the overall genetic composition of the population. This process is evolution.\r\n\r\nNotice that as long as allele frequencies change, the population evolves.\r\n\r\nNow consider a fault in transcription or replication of DNA. This occurs one in about every 100,000 genes or so. Now these mutations code for different proteins which affect the genetic genotype and phenotype of certain characteristics of a species. Now, the environment is selective for or against phenotypes, so the advantagious phenotype will survive in the individual and the trait is carried to the next generation. Now, as more random mutations occur, they survive, and there is a large genetic drift in the entire population, cause evolution on the macro-scale. The evidence is enormous. \r\n\r\nYour stance is basically - induction cannot be used as proof, when induction is one of the best forms of proof out there.\r\n\r\nInduction and logic support the macroevolution of populations from observation at the micro-scale.\r\n\r\nThe squirrels Sunny refers to are the Albert and Kaibab species - they are different species because they no longer interbreed. One lives on the north ridge and the other the south ridge. This form of speciation, or isolation, is caused by geography.\r\n\r\nYou say forget about proof and show evidence? Proof is evidence.", "Solution_19": "I am not sure it is even worth it any more. If you made the effort, you would notice that the people on this forum have given you [i]several[/i] instances of macro-evolution, myself included. Without an understanding of the process or the willingness to listen and learn, this information is pointless. I retire from this argument in frustration.", "Solution_20": "[quote=\"boy_21\"][quote=\"LynnelleYe\"] We don't know if evolution is driven by God. It doesn't matter--it still happens. God is outside of science, so whether or not he has anything to do with it is beside the point. [/quote]\n\nYou are still making the assumption that macroevolution does happen. All that have been listed so far as \"visible\" evolution are cases of adaption, or microevolution, which is far from proving that macroevolution occurs. You have taken 100 small steps along the side of a canyon and claimed that you can therefore walk accross the 100-foot canyon. \n\nSo can you please stop listing proof of adaption, and try to show me some scientific evidance for evolution?[/quote]\r\n\r\nBlah Blah.\r\n\r\nI will use my own horrible metaphor since you choose to use one that's completely irrelevant.\r\n\r\nJust because you haven't seen 1 billion dollars in person does not mean that 1 billion dollars does not exist. What does this example prove? Nothing more than yours'. What does your example prove? Nothing much. Settled.", "Solution_21": "I have given you fossils. I have given you evidence from our DNA. I have given you examples of us observing new species evolving--note that 'macroevolution' is defined as speciation. I could give you more if you want. Others in this thread have given you more as well. And you still say that we are 'providing no evidence'. FYI, when someone posts a link, it is usually meant to be clicked on.\r\n\r\nP.S. Your analogy is messed up. It'd be more like walking around your block, and claiming that given enough time you could walk 20 miles. Which is true. You don't have to be able to walk those 20 miles in one step, which is what you're claiming that evolution requires.", "Solution_22": "[quote=\"JBL\"]That's not true at all -- it would be exceedingly easy for me to write a computer program to generate a sonnet. I would simply have it run, in lexicographic order, over all possible strings of words in the English language up to a certain length. It would produce a lot of garbage, which an interested reader would skip over, but it would also produce poetry and, if I let it run long enough, any other written work you like. If I had a database of english words, I could start pumping out Haikus right now. \n\nAlso, I note that you tend to respond with actual challenges only with sarcasm and it makes you a much less convincing person to listen to. HIV yields HIV, yes, but only because we happen to know exactly where it came from. It is demonstratably different, in fundamental ways, from pre-existing strains of HIV. Of course, we have the difficulty of making a distinction here because HIV doesn't reproduce sexually, but HIV today is not \"the same\" HIV that was around thirty years ago, it's an exceedingly different creature, and, if it makes sense to talk about species of viri, it would be safe to call it a distinct species.[/quote]\r\n\r\nYEAY! Hitchhiker's Guide to the Galaxy.\r\n\r\nThose monkeys are going to type up Hamlet sooner or later.\r\n\r\nboy21 - In your example of the CD from ealier, you mention that the CD is still a CD. But that's generalizing. For example, if one species of Squirrel becomes a separate species, we can say that they are organisms. Sure whatever. But, the truth is, as compared to the information stored in a CD, the information which composed the organism, DNA, has changed.\r\n\r\nYou do not need additional information for evolution. Like LynnelleYe said, it's just speciation - which comes from change in allele frequency by random mutation, or comes from isolation. It's all logic layered upon logic if you just flat out disregard scientific proof.", "Solution_23": "[quote=\"boy_21\"] 10 systems that Behe outlines that are absolutely dependant on every major part of the system--for example the human eye. If you take away or change any major part, the system does not work (or all of a bunch of minor parts, like the rods or cones).[/quote]\n\nThat is an expected result of evolution, not contradictory to evolution. It is amazing, and not in a good way, that Behe considers such signatures of evolution to be evidence against it. \n\nEvolution optimizes critical systems, including specialized organs (the eye) or basic biological machinery (the Krebs cycle). This means that needless components will usually be removed over time, leaving an \"irreducible\" set. It also means that minor modifications to a system will render it less efficient, or inoperable.\n\n\n[quote] Therefore the system, as long as it has been in existance, has worked exactly as it does now. [/quote]\r\n\r\nThe much likelier conclusion is that evolution reached a steady (locally optimal) state of that particular system, and since then the system has worked as it does now. If the selective pressure changes so will the system (e.g. if humans move to a dark underground, the structure of the eye would change; note the difference in eye structure of near-surface and deep-water fish).", "Solution_24": "People here may be interested on [url=http://www.cnn.com/2005/EDUCATION/05/02/life.evolution.reut/index.html]this article[/url] about some trial that is taking place in Kansas right now about if evolution should or should not be taught in school!", "Solution_25": "[quote=\"boy_21\"][quote]It would help to define 'new, useful information'[/quote]\n\nGood point. What is information?\n\nI once heard a lecture about information theory by Dr. Werner Gitt (this genius German guy) which I found absolutely fascinating. In fact, maybe we should start a new thread on information theory. Anyways, to make a really long story really short, you cannot randomly generate information. You cannot make a computer program that will write an original sensible sonnet or an ode. \n[/quote]\nHeck yes you can. I made a random name generator the other day. It just takes some linguistics knowledge.\n[quote]\nThe whole point is, you need to add information for macroevolution (think whales to cows), so random mutations aren't going to do it.[/quote]\r\nThe mutations add information by changing some of the other information. Some of this is bad(cystic fibrosis) and some of this is good ([url=http://www.msnbc.msn.com/id/5278028/]more muscles[/url]), but new is really just changing a bunch of things. I can get some fossil records if you need them.", "Solution_26": "I'm new to this site and have read all these posting.\r\n\r\nBoth sides seem to be talking past each other. This reminds me of discussions between members of two different religions. (Perhaps it is.) Many of the examples are clearly not on point. (Similar to the argument that a great warrior Achilles couldn't past a turtle in a race, if the turtle had a head start and kept moving, since when the warrior reached the place the turtle was when the warrior started, the turtle would have moved a bit more. . . .) The problem with this historical logic problem is that it only applies until the point where both are the same distance from the starting point.\r\n\r\nSimilarly the random generation of \"information\" doesn't occur in the stated examples. Random data is selected by an observer, who is the \"generator\" of the new \"information.\" This is also a \"problem\" with the use of fossils that are sorted to \"show\" evolution. Many times the location and dating of the fossil are not the critical aspect in placing a fossil in an evolution line.\r\n\r\nThe examples of micro-evolution are good and well accepted by most people on both sides of this discussion, so references to micro-evolution, such as the relative populations of dark and light moths when London went from a dark background (high coal dust deposition) to a lighter background (lower coal dust deposition) are just \"noise\" in this discussion.\r\n\r\nNext, the Bible says many times that acceptance is based on faith alone, so anyone who thinks that they have a \"proof\" for ID is in direct conflict with the major source of inspiration for ID. Since God says that acceptance is based on faith alone, he could easily have set up the fossil record to support a much older system. No where is there a requirement that when God created the earth, it had to start with any form. If the stars are really million of light years away, why wouldn't one assume that the \"in transit\" light was created well, in transit?\r\n\r\nIf one wants to explore faith, ask where did the material come from for the Big Bang?\r\n\r\nIf I were to find a watch on a beach, what scientific proof would be required to support the idea that there was a watchmaker somewhere? I think none. The simplest organism is so much more complex than the simplest living organism.\r\n\r\nLet's get out of biology and into chemistry. Why do all living things use the same optically active molecules? (Just like a person's two hands are very similar, but can't be placed so the fingers are lined up when both hands face the same way, certain molecules, essential to life, are distinct. All life uses the same orientation, but no process has ever been found that doesn't use a living force (perhaps in selecting an optically active chemical) that doesn't result in a racemic mixture (equal quantities of both the right hand molecule and left hand molecule). First, this is relevant to evolution because it doesn't require that the material under study be alive and be studied for many, many years. Second, even if the initial start was with God or an ID, the fact that subsequent evolution occurred would still destroy the basic idea of evolution: NO GOD.\r\n\r\nFinally, let\u2019s apply the second law of thermodynamics to this theory of evolution. Everything runs down, becoming more disordered. (If you find a way to not have this increase in disorder occur, I\u2019d like to try it out on my room.) The only way a room becomes more ordered and something becomes more complex is as a sub-system of a system becoming more disordered.\r\n\r\nA tree combines the carbon dioxide it brings in through its leaves, with the water and minerals brought in through the roots, catalyzed with chlorophyll, and powered by the sun to make sugars and wood. However, when one considers the sun that generates the energy the system is loosing order some energy is converted into waste heat, that can never be recovered. So if everything is winding down and we don\u2019t want to talk about the watch maker, what about asking about the watch winder? Where did the order that we all live on come from?\r\n\r\nThis is probably enough from me. (Some may say too much.)\r\n\r\nAs we go along, we should all remember that not many are likely to change their views, let's be reasonable in discussing this issue that is important to many people.", "Solution_27": "[quote=\"wizard2378\"] Second, even if the initial start was with God or an ID, the fact that subsequent evolution occurred would still destroy the basic idea of evolution: NO GOD.\n[/quote]\r\nNowhere does evolution say there is no god. Some evolutionist also think there is no god, but evolution itself does not exclude the possibility", "Solution_28": "You're right that the evolution theory itself doesn't preclude a belief in God, but this is the essence of the public and legal debate.", "Solution_29": "[quote=\"wizard2378\"]You're right that the evolution theory itself doesn't preclude a belief in God, but this is the essence of the public and legal debate.[/quote]\r\n\r\nYes, I agree, the evolution debate is mostly a vieled God vs No God debate.\r\n\r\nAlso, the Second Law ONLY applies to closed systems. The Universe is a closed system, not the earth. We have the sun coming in for example. And it does not exclude local flucuations in entropy, resulting in one area actually gaining more energy." } { "Tag": [ "function", "algebra", "domain", "complex numbers", "imaginary numbers" ], "Problem": "If f(x) = x^2 - 3 and g(x) = sqrt(x+2), then find f(g(x))\r\n\r\nthanks", "Solution_1": "in f(x) the x stands not just for an x but can stand for any variable thingy...so f(g(x)) is just replacing all the x's in the f(x) equation by g(x) so we hvae\r\n\r\n\r\nf(g(x))=(g(x))^2-3.\r\n\r\nplug in g(x) we have x+2-3=x-1,\r\n\r\nbut you have to know that x+2>0, so $ x\\geq\\minus{}2$ otherwise it wont work\r\n\r\n\r\nso its $ x\\minus{}1$ with $ x\\geq\\minus{}2$\r\n\r\nwell imactually assuming the question said that x is in its domain because im betting you arent dealing with imaginaries here.", "Solution_2": "f(g(x)) is broken down to two simple steps\r\n1.Given x, do g(x)\r\n2.Take result of g(x) and treat it as x for f(x)\r\n\r\nSo in your question:\r\n\r\nf(g(x))= [Sqrt(x+2)]^2-3=\r\n[b]x+2[/b]-3=\r\n[b]x-1[/b]\r\n \r\nAnd as stated above, x cannot be less than -2, otherwise g(x) would be an imaginary number", "Solution_3": "g(x)=squa(x+2) -2 }{\\|w_l\\|^2}w_l\r\n\\]\r\nCan you fill in the details from here? If you need more help, this is called the Gram-Schmidt process. The idea is to essentially subtract off the non-orthogonal parts (that you have already taken care of) so that all you have left is something orthogonal to the previous ones (and non-zero, do you see why?)." } { "Tag": [ "blogs", "\\/closed" ], "Problem": "1. Why doesn't the text wrap work for for the Shoutbox? [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=681]Example.[/url]\r\n\r\n2. Is there a linebreak system designed for the Custom Block?", "Solution_1": "1. You mean when you have a very long word? Because CSS doesn't know how to cut that in two. You have to avoid those basically.\r\n\r\n2. For now the Custom Block is not that customizable. I will be fully BBcodefiable in the future." } { "Tag": [ "LaTeX", "geometry", "parallelogram", "ratio", "algebra", "polynomial", "analytic geometry" ], "Problem": "We need something challenging/interesting on here... so I'm gonna start posting some problems. Feel free to post your solutions. If you solve one, you can post your own questions too!\r\n\r\nAdded bonus: because I suck at LaTeX, you get my problems in deluxe ascii math style!\r\n\r\nProblem #1)\r\n[code]\nLet n>1 be an integer.\n\nLet S be the set of all complex numbers k such that\n\n n\n -----\n \\ 2i\n ) k = 0\n /\n -----\n i = 0\n\nLet T be the set:\n n+1\n{k : k is an element of S}\n\nProve that T is always {-1,1}, regardless of the choice of n.[/code]\r\n\r\n[color=darkred][Mod LaTeX addition, for those of you who prefer properly typesetted mathematics.][/color]\r\n\r\nLet $n>1$ be an integer. Let $S$ be the set of all complex numbers $k$ such that\r\n\\[ \\sum_{i=0}^{n} k^{2i} = 0 \\]\r\nLet $T$ be the set $\\{ k^{n+1} : k \\text{ is an element of } S \\}$. Prove that $T$ is always $\\{-1, 1\\}$, regardless of the choice of $n$.\r\n\r\nElyot's backlash:\r\nIf you're gonna latex it, at least do it properly :P. I changed the incorrect sum from 1 to n to a correct sum from 0 to n.\r\n\r\nActually I have nothing against latex (it's very nice), but the ascii art makes it much funnier in my opinion. You gotta admit... that ascii sigma is PIMPIN! ;) \r\n\r\nAdditional backlash:\r\nThere is no such word as \"typesetted\". The correct past tense of typeset is, in fact, typeset.", "Solution_1": "Here's a problem proposed by me in the Mathematical Mayhem (Crux v30 n7)\r\n\r\n[size=150]M153[/size].\r\nTwo similar triangles $APB$ and $BQC$ are erected externally on a triangle $ABC$. If $R$ is a point such that $PBQR$ is a parallelogram, show that triangles $ARC$ and $APB$ are similar.", "Solution_2": "[quote=\"billzhao\"]Here's a problem proposed by me in the Mathematical Mayhem (Crux v30 n7)\n\n[size=150]M153[/size].\nTwo similar triangles $APB$ and $BQC$ are erected externally on a triangle $ABC$. If $R$ is a point such that $PBQR$ is a parallelogram, show that triangles $ARC$ and $APB$ are similar.[/quote]\r\n\r\n[hide=\"Click to unhide my solution\"]\nI got:\n- use angle chasing to prove angle ABC = angle APR = angle RQC.\nThen basic similar triangle ratios gives:\n- spiral similarity about B, ratio a/c, angle ABC, maps AB : CB \n- spiral similarity about P, ratio a/c, angle ABC, maps AP : RP\n- spiral similarity about Q, ratio a/c, angle ABC, maps RQ : CQ\nHence:\n- ABC, APR, RQC are similar.\nThen:\n- desired result immediately follows from the ratio AC:AR:RC = AB:AP:RQ (RQ=BP, parallelogram)\n[/hide]\r\n\r\nMaybe some minor additions to the angle chasing step for special cases, but I think that's about it. It originally seemed hard, but not so bad after you figure out what's going on. Nice one, though.\r\n\r\nNote: my solution also works if the triangles are erected internally, so long as P and Q are not the same point (otherwise things get degenerate, but theoretically still work)\r\n\r\nSome teasing: the solution to my problem is about 3 or 4 lines... but hard to see. There's very little complex number stuff involved... just polynomials, roots of unity, and a simple algebraic trick (or three).", "Solution_3": "[quote=\"Elyot\"]\nYou gotta admit... that ascii sigma is PIMPIN!\n[/quote]\r\n\r\nNo, no it's not \"pimpin\". Stick to LaTeX.\r\n--\r\n\r\nNow onto the problem.\r\n\r\nLet $x \\in \\mathbb{C}$, and $K(x) = \\sum_{j=0}^n x^{2j}$. Notice that $(x^2 - 1)K(x) = (x^2)^{n+1} - 1$. (*)\r\n\r\nSuppose $x \\in S$. Then $K(x) = 0$, and it implies $(x^{n+1})^2 = 1$, implying $x^{n+1} = -1,$ or $1$. So $T$ can only contain $-1$ or $1$. To show both values are possible, set $x^{n+1} = \\pm 1$, where $x^2 \\ne 1$. Then (*) implies $K(x) = 0$, as required.", "Solution_4": "[quote=\"billzhao\"]Here's a problem proposed by me in the Mathematical Mayhem (Crux v30 n7)\n\n[size=150]M153[/size].\nTwo similar triangles $APB$ and $BQC$ are erected externally on a triangle $ABC$. If $R$ is a point such that $PBQR$ is a parallelogram, show that triangles $ARC$ and $APB$ are similar.[/quote]\r\n\r\nOooh I like it. Nothing to post though, cuz my solution turned out to be basically the same as Elyot's. Gotta love those roots of unity eh. :P", "Solution_5": "Yes, good job... that's my exact solution. Actually that polynomial has very very interesting properties. You can prove that the set of roots of the polynomial 1 + x^2 + x^4 + ... + x^(2n) are the unique (2n+2)th roots of unity, excluding 1 and -1 (multiplying by n^2-1 puts these roots back into the polynomial). For n=1, you get i and -i... but for n>=2, you will have 2 consecutive roots somewhere in there (if you order them by the angle in polar coordinates), one of which must be an (n+1)th root of 1 and the other which must be an (n+1)th root of -1.\r\n\r\n---------------------------------------\r\n\r\nI have another one... the topic this time is FUNCTIONS, SEQUENCES, AND RECURSION!\r\n\r\nPart a) is easy and worth 1 point. Part b) is hard and worth 6 points.\r\n\r\na) Prove that for any real constant k>0, the polynomial x^3 + k*x - 12 always has exactly one real positive root.\r\n\r\nb) For k>0, Let f(k) be the real root of the polynomial x^3 + k*x - 12. Consider the sequence a[1], a[2], a[3]... defined recursively by a[1]=1, a[n] = f(a[n-1]). Prove that this sequence converges and find the limit of the sequence as n goes to infinity.", "Solution_6": "I also have one\r\n\r\nLet $a_1, a_2, \\dots, a_n$ be distinct reals, where $n \\ge 2$ is an integer. Prove the identity \r\n\r\n$\\displaystyle \\sum_{cyc}^n \\frac{(\\prod_{i=2}^n a_i) (\\sum_{i=2}^n a_i)}{\\prod_{i=2}^n (a_1 - a_i)} = (-1)^{n+1} \\sum_{i=1}^n a_i$", "Solution_7": "Wow. That problem is... well... not too hard... but so ugly. I could write up a solution... but something like that is just not fun to write up.\r\n\r\nDo you have an elegant solution, or is this just a bunch of algebraic manipulation, ugliness, cancelling, etc? I can prove a few lemmas that simplify it a bit... but dude... that question is a monstrosity!", "Solution_8": "Yes, I have elegant solution.\r\n\r\n[hide=\"Hint\"]\nLagrange Interpolation\n[/hide]\n\n\n[hide=\"Answer\"]\n\nLet $s = \\sum_{i=1}^n a_i$. Consider the unique polynomial $f(x)$ (of degree n-1) such that for $i = 1, \\dots, n$, $f(a_i) = s - a_i$.\nBy the Lagrange interpolation formula,\n\n$f(x) = \\displaystyle \\sum_{cyc}^n \\frac{(\\prod_{i=2}^n x -a_i) (\\sum_{i=2}^n a_i)}{\\prod_{i=2}^n (a_1 - a_i)}$ \n\nwhence \n\n$f(0) = (-1)^{n+1} \\displaystyle \\sum_{cyc}^n \\frac{(\\prod_{i=2}^n a_i) (\\sum_{i=2}^n a_i)}{\\prod_{i=2}^n (a_1 - a_i)}$ (*)\n\nConsider $g(x) = f(x) + x - s$. It is also a polynomial of degree $n-1$, that has $n$ zeroes (at $a_i$). This implies that $g(x)$ is the zero function. Thus, $f(x) = s - x$, implying $f(0) = s$. From (*), the result follows.\n\n[/hide]", "Solution_9": "1) CANADIAN team refers to UGLY, brute force solutions as GRUNGING.\r\n\r\n2) CANADIAN national languages are English and FRENCH.\r\n\r\n3) The FRENCH word for GRUNGE is LA GRANGE.\r\n\r\n4) To have any sensible solution, your problem demands the use of LA GRANGE Interpolation.\r\n\r\n5) Therefore, your problem is UGLY!\r\n\r\nYou just got PWNED!\r\n :stretcher: :stretcher:\r\n\r\nSeriously though... most people don't use those kind of techniques in solutions... that problem would never appear on the IMO for that reason... or even on the Putnam. I can easily make difficult problems that rely on very very deep knowledge in specialized areas of mathematics... it's much tougher to make challenging problems that can be solved with high school math.", "Solution_10": "Lagrange interpolation hardly qualifies as 'very very deep knowledge'. It is usually very obvious when you want to use it, and this is certainly one of those cases (just look at the denominator!)", "Solution_11": "Uh... I agree with Elyot.\r\n\r\nIt's no fun for high school people to try and solve a problem that requires a formula that no high school students ever heard of. Not only that, it's EXTREMELY easy to make up math problems that require advanced knowledge and like 2 tricks nobody would ever think of. However, that's not math. That's just showing off...\r\n\r\nYou did get pwned... Lagrange is ugly... :P", "Solution_12": "And I think both of you are wrong. Saying that you pwned someone doesn't equate to actually doing so, whatever you might like to believe.", "Solution_13": "Dude, don't blame your problems on me. You should thank me now that you have a new trick under your belt.\r\n\r\n[size=75]\n[b][u]Anyways, the idea in the solution is not a new idea either.[/u][/b] And the \"formula\" isn't really contrived: you should know that n+1 points (say, $(x_i, y_i)_{i=0}^n$) uniquely determine an $n^{th}$ degree polynomial, and it is obvious to find such a polynomial $p(x) = \\sum_{i=0}^n y_i \\prod_{j=0, j\\ne i}^n \\frac{x-x_j}{x_i-x_j}$ satisfying the n+1 points. Therefore it is the unique one.\n\nAlso, I could've bitched that your roots problem, part (a) was just the Descartes Rule of Signs. [u]But I'm not a whiny bitch.[/u]\n\nBy the way, ascii was never cool... actually... maybe in 1995 but thats it.\n[/size]\r\n\r\nAnother problem:\r\n\r\nFind the largest real $k$ (in terms of m) so that $|xy+yz+zx| \\ge k|x+y+z|$, for all $x,y,z \\in \\mathbb{C}$, each with absolute value $m$.", "Solution_14": "Solved your next problem... it's more fair I think. Too easy for putnam though.\r\n\r\n[hide]\n- Convert x,y,z to m*cis(a),etc form.\n- Multiply stuff together and common factor out m.\n- Lemma: |cis(a) + cis(b) + cis(c)| = |cis(a+b) + cis(b+c) + cis(a+c)|\nProof: \n- by obviousness, |cis(a) + cis(b) + cis(c)| = |cis(-a) + cis(-b) + cis(-c)|\n- rotate the RHS about the origin by an angle of a+b+c. Lemma proven.\nSo in fact, km <= m^2, so k<=m, so largest possible k is m.\n[/hide]", "Solution_15": "Here is another.\r\n\r\na,b,c are reals. Prove $\\sum_{cyc} \\frac{a}{bc-a^2} =0 \\Rightarrow \\sum_{cyc} \\frac{a}{(bc-a^2)^2} =0$. Is it true for complex numbers?", "Solution_16": "i'm sorry but what the hell is 'cyc'?\r\n...maybe i can learn somethin here", "Solution_17": "Solution in spoiler.\r\n[hide]\nLet $P_k = \\sum_{cyc} \\frac{a^k}{bc-a^2}$. Then $P_0P_1 = P_2$.\n[/hide]", "Solution_18": "[quote=\"Elyot\"]\n2) CANADIAN national languages are English and FRENCH.\n[/quote]\nCorrect, this is why I'll permit myself to put a quote in French (tough I've seen Canadians be kind of hostile to that language ...)\n[quote=\"Elyot\"]\nvery very deep knowledge in specialized areas of mathematics...[/quote]\r\n\r\nTo answer in French \"C'est genant de dire\" that this is \"very very deep\" knowledge. This is basically knowledge any first year undergraduate must *at last* have :) Moreoever Lagrange interpolation formula is 18-th century mathematics. Needless to say it cannot be very very deep (something deep must be recent) or specialized (Lagrange had time to work in mathematics and physics so had no time for specialization ;))\r\nFinally I've already seen Putnam question relying on a specific trick, just like here.", "Solution_19": "Did someone call pizza hut?", "Solution_20": "[quote=\"Singular\"]Did someone call pizza hut?[/quote]\r\n\r\nOMG. You inspired me to craft a problem! This will be an easy problem for many of you, and very hard if you're bad at geometric inequalities, but have fun with it. I guarantee that there is a pretty, non-bash solution.\r\n\r\nLet a \"pizza hut roof\" (PHR) be a hexagon ABCDEF such that AF, BE, and CD are all parallel. (see the attached pizza hut logo... note that I used the old pizza hut logo, because all the symmetry is gone in the new, modern, fancy pizza hut logo). We're going to assume all the nice symmetrical properties of the PHR are true (ie ABC reflected through a line perpendicular to AF maps onto FED). Think of it like two isosceles trapezoids glued together. I'm also going to give you two more things:\r\n- AF < BE < CD, to preserve it's \"pizza-hut-roofiness\".\r\n- B and E are inside ACDF (we want a nice, sleek roof... not a bloated, ugly one)\r\nHopefully I didn't forget anything. You can ignore degenerate roofs.\r\n\r\n[img]http://www.248am.com/images/pizzahut.gif[/img]\r\n\r\nOK, now on to the problem. Suppose that AD, FC, and BE are concurrent and meet at Q. Let AE and BF meet at P, and BD and CE meet at R.\r\n\r\nIf p is a point inside the roof, let f(p) be the length of the line segment through p, parallel to AF, with both of its endpoints on the perimeter of the roof. For example, f(Q) is just the length of BE (since Q lies on BE).\r\n\r\nProve f(Q) > 2*f(P)*f(R)/(f(P)+f(R))\r\n\r\nBONUS: I'll also send a cookie to anybody who can make a nicer pizza hut problem (rule: everything before \"OK, now on to the problem.\" must be the same.)\r\n\r\n~ Elyot", "Solution_21": "Lemma: Suppose ABCD is a trapezoid, with E the intersection of AC and BD. Its left to the reader to prove that the length of the line through E parallel to AB and CD is $\\frac{2|AB||CD|}{|AB|+|CD|}$.\r\n\r\n--\r\n\r\nOn to the problem: let $x,y,z$ = lengths $AF, BE, CD$.\r\n\r\nSo we need to show $\\frac{2xz}{x+z} > \\frac{2 \\frac{2xy}{x+y} \\frac{2yz}{y+z} }{ \\frac{2xy}{x+y} + \\frac{2yz}{y+z} }$\r\n\r\nIt becomes $\\frac{xz}{x+z} > \\frac{2y^2xz}{xy(y+z) + yz(x+y)}$, becoming $y < \\frac{2xz}{x+z}$.\r\n\r\nConsider Q on XY, with XY parallel to AD, X on AC, Y on DF. Then $y = BE < XY = \\frac{2xz}{x+z}$ (BE is given inside ACDF) and result follows.", "Solution_22": "That's really a lot of algebra... do you realize 2ab/(a+b) is just the harmonic mean in disguise? :D Do you see the two-line solution now?", "Solution_23": "nope - present two line please", "Solution_24": "OK... First we replace the right hand side of your lemma with HM(AB, CD) where HM is the harmonic mean.\r\n\r\nSo we'll do the same thing as you did with x,y,z. then:\r\n2*f(p)*f(r)/(f(p)+f(r)) = HM(f(p),f(r)) = HM(HM(x,y),HM(y,z)) = HM(x,y,y,z) = HM(y,HM(x,z)). By essentially the argument that you gave in the last line, we have y_<\r\n\r\n\r\nlol... no one in my school got this question -____-", "Solution_31": "For finite sets, the cardinality of the set is just the number of elements.", "Solution_32": "Just curious,[b] blahblahblah [/b],when did you start learning abstract algebra? My teacher told me it's usually in the third year of university. I've started reading a book called Modern Algebra, the one was called \"the most classical algebra textbook\" , what type of textbooks do you use?\r\nPlus, I was told that rational solutions of such equations: $y^2 = F(x) = x^3 + ax +b (a,b \\in Q)$, (typical elliptic curves), forms groups. But why?", "Solution_33": "how do you know that any set of cardinality n has 2^n subsets?\r\n\r\nsorry for all the question. i just really wanna understand what i dont understand :roll: lol", "Solution_34": "I started learning abstract algebra in Grade 12. Nothing too significant, just the basic theory of groups (it was for one of my IB courses). Since then, I've taken an undergraduate course on rings and fields and a graduate course in groups and rings with some representation theory thrown in. The two algebra texts that I have the most experience with are Dummit and Foote's Abstract Algebra, which is a huge text and a good text to start learning from, and Isaacs' Algebra: A Graduate Course.\r\n\r\nIt's hard to say what the most classical algebra textbook is. There's a brief discussion of a lot of them [url=http://www.ocf.berkeley.edu/~abhishek/chicmath.htm#i:general-abstract-algebra]here[/url], not having read most of them I can't really comment. Personally, I'd hesitate to read anything too old; older textbooks are often very good but difficult to get much out of on a first reading.\r\n\r\nAs for your question on elliptic curves, I can't say that I know; I haven't taken a course in algebraic geometry yet.", "Solution_35": "There is $1$ set with $0$ elements, $\\binom{n}{1}$ sets with $1$ element, $\\binom{n}{2}$ sets with $2$ elements, etc.\r\n\r\nHence a set of cardinality $n$ has\r\n\\[ \\sum^n_{j=0}\\binom{n}{j}=(1+1)^n=2^n \\]\r\nsubsets.", "Solution_36": "[quote=\"Chocoboom\"]how do you know that any set of cardinality n has 2^n subsets?[/quote]blahblahblah's answer makes use of the binomial expansion, which is part of the grade 12 curriculum. However, you can also think of it this way: each element of the set can either be in the subset or not in the subset, so there are 2 possibilities, and n elements. Thus $2^n$.", "Solution_37": "lol i havnt learned the gr 12 curriculum yet :P \r\n\r\nanyways, i think i get it now thanks everyone\r\nAoPS rocks XD", "Solution_38": "Now, I'm sure all of you are a big fan of 3D geometry problems that are very hard to visualize but easy to solve. This is one of those. Once you figure out the trick, it's a simple problem.\r\n\r\nLet a \"pigeon\" be a point in 3-space with integer coordinates. Define a \"flock\" F of size n to be an ordered list of n pigeons [p1, p2, p3, ... pn] with the following properties:\r\n\r\n- n is an integer and n > 3.\r\n- No two pigeons have the same coordinates.\r\n- For all integers x and y in {1..n}, if x-y-1 is divisible by n, then the distance between pigeons px and py is 1.\r\n(for all you newbs out there, 0 IS divisibly by n for all n)\r\n- For all integers w,x,y, and z in {1..n} if w-x-1, x-y-1, and y-z-1 are all divisible by n, then pigeons pw, px, py, and pz are NOT coplanar. 4 Pigeons are coplanar if there exists a plane in 3-space such that all 4 of them lie on it.\r\n\r\nFor example, [(0,0,0), (0,0,1), (0,-1,1), (1,-1,1), (1,-1,0), (1,0,0)] is a flock.\r\n\r\nProve:\r\na) If a flock is of size n, then n is divisible by 6.\r\nb) For every integer k, there exists a flock of size 6k." } { "Tag": [ "inequalities" ], "Problem": "Let be $P$ any point inside $\\triangle{ABC}$. Prove or disprove:\r\n\r\na) $AP$, $BP$ and $CP$ can be sides of a new triangle (so the triangle's inequality holds) for an acute-angled $\\triangle{ABC}$.\r\n\r\nb) $AP$, $BP$ and $CP$ can be sides of a new triangle (so the triangle's inequality holds) for any $\\triangle{ABC}$.", "Solution_1": "[quote=\"Jos\u00e9\"]Let be $P$ any point inside $\\triangle{ABC}$. Prove or disprove:\n\na) $AP$, $BP$ and $CP$ can be sides of a new triangle (so the triangle's inequality holds) \nfor an acute-angled $\\triangle{ABC}$.\n\nb) $AP$, $BP$ and $CP$ can be sides of a new triangle (so the triangle's inequality holds) \nfor any $\\triangle{ABC}$.[/quote]\r\n\r\n\r\nWhen you first read this problem, you think that the part (a) is always true but the part (b) is not \r\n\r\nConsider an isosceles acute triangle with $AB=AC$ enough larger than $BC$\r\n\r\nThen take a point $P$ near to the base $BC$\r\n\r\nYou'll get $PB+PC not true for all triangles", "Solution_2": "[quote=\"pontios\"][quote=\"Jos\u00e9\"]Let be $P$ any point inside $\\triangle{ABC}$. Prove or disprove:\n\na) $AP$, $BP$ and $CP$ can be sides of a new triangle (so the triangle's inequality holds) \nfor an acute-angled $\\triangle{ABC}$.\n\nb) $AP$, $BP$ and $CP$ can be sides of a new triangle (so the triangle's inequality holds) \nfor any $\\triangle{ABC}$.[/quote]\n\n\nWhen you first read this problem, you think that the part (a) is always true but the part (b) is not \n\nConsider an isosceles acute triangle with $AB=AC$ enough larger than $BC$\n\nThen take a point $P$ near to the base $BC$\n\nYou'll get $PB+PC not true for all triangles[/quote]\r\n\r\nWhat about equilaterals triangles??? \r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=76616", "Solution_3": "[quote=\"pontios\"]not true for all triangles[/quote]\r\n\r\nHere, I mean that this fact could be true for some triangles, but not for all.\r\n\r\nI don't mean that it is false for all triangles\r\n\r\n\r\nI think that we can find a lot of triangles (not equilateral) such that this fact is true", "Solution_4": "[quote=\"pontios\"][quote=\"pontios\"]not true for all triangles[/quote]\n\nHere, I mean that this fact could be true for some triangles, but not for all.\n\nI don't mean that it is false for all triangles\n\n\nI think that we can find a lot of triangles (not equilateral) such that this fact is true[/quote]\r\n\r\nOhhh, OK then." } { "Tag": [ "algorithm", "number theory unsolved", "number theory" ], "Problem": "There is a keypad entry system on a safe door. There is no 'clear' button, but the safe will open automatically if the last n digits entered match the n digit 'password' required. \r\nIf there are m digits to choose from, what is the fewest number of button presses you would need to type in to guarantee access if the password is n digits long? \r\nThis is a well known problem, usually called the [b]shortest common superstring problem[/b]. It is NP hard. It has very pratical applications in genome mapping, and there's lots of work on generating a common superstring that is no worse than (some limit) x as long as the ideal string.\r\nCan you prove or disprove that the minimal length sequence, is the total number of unique combinations + the length of the code - 1 ?\r\nx(m,n) = m^n + n - 1", "Solution_1": "This is essentially asking for a proof of the existence of a de Bruijn sequence of $m$ alphabets and length $n$. De Bruijn sequences exist for all $m, n \\in \\mathbb{N}$.", "Solution_2": "Indeed. I think this problem was posted a long time ago by alekk in the \"Combinatorics\" section. A nicer way to pose the problem would be considering the letters are arranged around a circle (in a clockwise direction) s.t. all words of $n$ letters are eventually obtained, and the last $n-1$ letters of a code are the first $n-1$ of the next code, and then we would require $m^n$ letters (I say it's nicer because $m^n$ certainly looks nicer than $m^n+n-1$ :)).", "Solution_3": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5624]Here's[/url] the thread that grobber was referring to.\r\n\r\nThe following is an algorithmic construction of a de Bruijn sequence of $m+1$ alphabets $0, 1, \\ldots, m$ and length $n$. (The sequence generated should be read cyclically, the way grobber described it.)\r\n\r\nConsider the following array which is divided into two parts: the left part contains a single column which contains nodes $0, 1, \\ldots, L - 1$ where $L = (m+1)^{n-1}$. The right part contains $m+1$ rows each of which contains the labels $0, 1, \\ldots, m$ in this order.\r\n\r\n[b]Node --- Labels[/b]\r\n$0 \\hspace{6ex} \\mbox{ } 0 \\mbox{ } 1 \\mbox{ } 2 \\ldots m$\r\n$1 \\hspace{7ex} 0 \\mbox{ } 1 \\mbox{ } 2 \\ldots m$\r\n$2 \\hspace{7ex} 0 \\mbox{ } 1 \\mbox{ } 2 \\ldots m$\r\n$\\ldots$\r\n$L-1 \\hspace {2ex} \\mbox{ } 0 \\mbox{ } 1 \\mbox{ } 2 \\ldots m$\r\n \r\n[b]Algorithm:[/b]\r\n[b]Step 1.[/b] Start with the node $i = 0$ and set the sequence to be generated to be empty.\r\n[b]Step 2.[/b] (Iterative) If the row with node $i$ is not empty, then append the rightmost label $s$ in the row with node $i$ to the sequence already generated and remove this label from the row and evaluate $i \\leftarrow (i(m+1) + s) \\mbox{ mod } L$. Otherwise stop.\r\n\r\nFrom Step 2, note if the last $n$ alphabets of the sequence generated are read as a number in base $L$, this is simply the node to go to in the next step.\r\n\r\n[b]Lemma.[/b] When Step 2 of the algorithm reaches the row with node $i$, then, if $i$ is non-zero, the row is non-empty, i.e. Step 2 always succeeds.\r\n\r\n[b]Proof.[/b] Suppose the row with node $i (\\neq 0)$ is empty. That means that Step 2 has reached it $m+2$ times, because there are only $m+1$ labels in the row with node $i$. But this is impossible except when $i = 0$ since there are only $m+1$ numbers in the range from $000\\ldots0$ ($n$ zeros) to $mmm \\ldots m$ ($n$ $m$'s) which have the same non-zero remainder when divided by $L$. Hence, the algorithm must terminate at the row with node $0$ when step 2 reaches this row for the $(m+2)$th time.\r\n\r\nNow, we note that to reach the row with node $0$, we have to first empty a row with node a multiple of $(m+1)^{n-2}$ but not $L$. Hence, if the row with node $0$ is empty, then all the rows with nodes a multiple of $(m+1)^{n-2}$ but not $L$ have been emptied. To empty all those rows, we have to first empty all the rows which are multiples of $(m+1)^{n-3}$ but not $(m+1)^{n-2}$ and so on. This means that when the algorithm terminates, the table is empty and the sequence generated has exactly $(m+1)^n$ alphabets.\r\n\r\nThat no subsequence with length $n$ can occur more than once follows from Step 2. Thus the sequence generated is a de Bruijn sequence.\r\n\r\n\r\n[b]Reference[/b]: Xie, Shenquan. (1987). Notes on de Bruijn Sequences. [i]Discrete Applied Mathematics, 16[/i], pp. 157-177.\r\n\r\nP.S. Thanks to Orlando for helping me fix the array. :)" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "If a,b,c are positive real numbers prove that\r\n(a^7)+(b^7)+(c^7)>=(a+b+c)[(abc)]^2", "Solution_1": "(7,0,0) > (3,2,2)\r\nthen muirhead.\r\nOf course it's the same to use amgm.", "Solution_2": "Also from Chebychev and AM-GM : \r\n[tex]3(a^7+b^7+c^7)\\geq(a+b+c)(a^6+b^6+c^6)\\geq3(a+b+c)(a^2b^2c^2) [/tex]" } { "Tag": [ "greatest common divisor", "algebra", "polynomial", "geometry" ], "Problem": "So, as a warmup to remember Visual Basic (which may be a bad thing), I'm going to make a math program to make my (and others') life easier. The CPU-intensive subroutines will probably be written in C++ for efficiency, and the GUI will be made in the aforementioned VB. So, the reason I'm posting is: I need ideas for useful tools people will need. Currently, I'm going to start with:\r\n\r\nPrime factorisation\r\nLCD\r\nGCF\r\nSolving/Factoring Polynomials, maybe graphing, although there's GraphCalc for that\r\n\r\nSo, I need ideas. Offer them up, and I'll integrate what's practical. Version 0.01 should be available pretty soon.", "Solution_1": "[quote=\"sponge008\"]\nLCD\nGCF[/quote]\r\n\r\nWhat's that?? \r\n\r\nAn idea: If you write a number n, you're shown all divisors of n.", "Solution_2": "[quote=\"Jos\u00e9\"][quote=\"sponge008\"]\nLCD\nGCF[/quote]\n\nWhat's that?? \n\nAn idea: If you write a number n, you're shown all divisors of n.[/quote]\r\n\r\nGood idea, as well as the sum, product, and number of divisors.\r\n\r\nLCD=least common denominator/multiple\r\nGCF=greatest common factor.", "Solution_3": "How about... solving triangles (given 2 sides and an angle, or something like that) or solving linear equations.\r\n\r\nAlso, you should see your local calculator.. it has a lot of things that people find useful :)\r\n\r\nEdit--I'm talking about stuff like TI-83 graphing calculators.\r\nEdit II--you should also check out Mathematica.", "Solution_4": "But if you don't want to pay for Mathematica, try Maxima.\r\n\r\nAren't most of these things already out there, or am I just spoiled by my OS? For instance, a good graphing program, K Interactive Geometry, and Python for raw computation does everything I need.", "Solution_5": "[quote=\"solafidefarms\"]But if you don't want to pay for Mathematica, try Maxima.\n\nAren't most of these things already out there, or am I just spoiled by my OS? For instance, a good graphing program, K Interactive Geometry, and Python for raw computation does everything I need.[/quote]\r\n\r\nTarget users are:\r\n1) Computer noobs or seminoobs who don't know how to do most of this stuff and would be more able to use this than other programs (such as the aforementioned Python, or in my case, C++)\r\n\r\n2) Lazy bastards like me who don't feel like doing each case individually. :D" } { "Tag": [ "LaTeX" ], "Problem": "I'm experiencing trouble with my paragraphs in LaTeX. I use the book class, but after linebreaks (\\\\) I don't get any indents.", "Solution_1": "It might be different with the book class, but I doubt it: Make sure you have a blank line before the new paragraph starts (\\\\ does a line break, NOT a paragraph break), and if you REALLY REALLY REALLY need to put indents in [i]artificially[/i], use \\indent.", "Solution_2": "But isn't standard to get an indent after a linebreak in a paragraph?", "Solution_3": "A line break simply puts you on the next line of the [i]same[/i] paragraph. It doesn't start the next paragraph, and so it doesn't indent. If you want to go to the next paragraph, leave a blank line before you start the paragraph." } { "Tag": [ "search", "AMC", "AIME", "geometry", "ARML" ], "Problem": "Remember [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1472359248&t=148061]Stretch[/url]? It's back.\r\n\r\n[b]Rules:[/b]\r\nYou may post whenever you like, as long as you were not the last poster.\r\nWhen you post, you must add one letter to the current word and anagram it.\r\nThe resulting word must be a real English word by the official game dictionary found [url=http://www.scrabulous.com/twl_dictionary.php]here[/url].\r\nPlease post all previous words as well as your new word [b]in bold[/b].\r\nUse your word in a sentence to assure you know what it means.\r\nYou may not post a word with the same basic root as the previous (LOBSTERS may not follow LOBSTER).\r\nIf you believe that no word can follow the previous word, you may type the word [b]FORFEIT[/b] to end the round.\r\n\r\n[b]Scores:[/b]\r\nWhen you end a round, scores are added as follows:\r\nWhoever posted the last word receives N-5 points, where N is the number of letters in the word. \r\nYou receive 3 points as well, just for ending the round.\r\nIf you end a round when there was another possible word which a player points out, you lose 3 points instead of gaining 3.\r\nThere is also a 3 point penalty for posting a word that does not exist in the official dictionary. Check it before posting.\r\nPlease post the scoreboard after ending a round.\r\n\r\n[b]Example:[/b]\r\nSIT\r\nSUIT\r\nQUITS\r\nSQUINT\r\nINQUEST\r\nANTIQUES\r\nQUAINTEST\r\nQUANTITIES\r\nEQUIDISTANT\r\n\r\n---\r\n\r\n[b]PIG[/b]\r\n\r\nSentence: \"A pig is an animal.\"", "Solution_1": "[b]PIG\nGRIP[/b]\r\n\"The man had a strong grip.\"", "Solution_2": "[b]PIG\nGRIP\nGRIPE[/b]\r\n\r\n\"His tone of voice GRIPES me\"\r\n\r\nScores: \r\n\r\nmiyomiyo: -2\r\ncf249:-1\r\nBOGTRO: 0\r\n\r\nlol...", "Solution_3": "To clear things up: nobody receives points until the round ends. There are no scores currently. Also, only the most recently added word should be bolded. Sorry if that was unclear.\r\n\r\nPIG\r\nGRIP\r\nGRIPE\r\nThe [b]PINGER[/b] continued to ping obnoxiously.\r\n\r\n(I think that's a word.)", "Solution_4": "PIG\r\nGRIP\r\nGRIP\r\nPINGER\r\nI had fun [b]REAPING[/b] the benefits.\r\n\r\nProblems: \r\n-The dictionary we are using does not recognize words that are greater than 8 letters.\r\n-Is there a penalty for saying a word that does not exist?\r\n-Soup?", "Solution_5": "PIG\r\nGRIP\r\nGRIP\r\nPINGER\r\nREAPING\r\nI'm [b]REPAYING[/b] Sam the fourteen fifteenths of my soul that I owe him.\r\n\r\nSolutions:\r\n- I replaced with this [url=http://www.scrabulous.com/twl_dictionary.php]better dictionary[/url].\r\n- 3 point penalty. I'll add that into the rules.\r\n- Absolutely.", "Solution_6": "PIG\r\nGRIP\r\nGRIP\r\nPINGER\r\nREAPING\r\nREPAYING\r\n[b]PREPAYING?[/b] (does that count?)\r\nI am [b]prepaying[/b] 4/5 of the lease on the car", "Solution_7": "Sure, though REPLAYING was more obvious.\r\n\r\nPIG\r\nGRIP\r\nGRIP\r\nPINGER\r\nREAPING\r\nREPAYING\r\nPREPAYING\r\n[b]REAPPLYING[/b]\r\n\r\nThere's no better way to not get sunburned than reapplying sunscreen every fifteen minutes.\r\n(As a challenge, I'm using 15 in every sentence I write.)", "Solution_8": "[b]Forfeit[/b]\r\n\r\n\r\nI hope, I hope, I hope.", "Solution_9": "[b]TEA[/b]", "Solution_10": "TEA\r\nI like the color [b]TEAL[/b].", "Solution_11": "TEA\r\nTEAL\r\n[b]STALE[/b]\r\n\r\n\"The bread was stale.\"", "Solution_12": "TEA\r\nTEAL\r\nSTALE\r\n[b]STEALS[/b]\r\n\r\nmiyomiyo:7", "Solution_13": "TEA \r\nTEAL \r\nSTALE\r\nSTEALS\r\n[b]ATLASES[/b]\r\n\r\ni looked up the location of Kazakhstan on many ATLASES.\r\n\r\nmiyomiyo has 5. read the rules", "Solution_14": "TEA\r\nTEAL\r\nSTALE\r\nSTEALS\r\nATLASES \r\n[b]LACTASES[/b]\r\n\r\nMilk has many lactases (I think).", "Solution_15": "ABY\r\nBABY\r\nTABBY\r\n[b]BATBOY[/b]\r\n\r\nthe guy stunk at hitting, but he was so overpaid they decided to make him a batboy and force him to do all the work in the dugout", "Solution_16": "[b]FORFEIT[/b]\r\n\r\nNew Word:\r\n[b]HIS[/b]\r\nThe boy said the fallen quarter was his.\r\n\r\nLast round, both yongyi and james were given 3 points, when forfeiting and \"flagons\" should both earn just 2 points. Previous to that, 33286 earned a point and somehow I missed that he was already on the list, and put him on again. These should be combined into one score. The scores have been adjusted accordingly.\r\n[hide=\"Score:\"]\n1. mathforlife: 16\n2. BOGTRO: 14 \n3. james4l: 13\n3. 33286: 13\n5. mz94: 12 \n6. Flame: 9 \n6. miyomiyo: 9 \n8. woser456: 7 \n9. Yongyi781: 6 \n10. Quickster94: 5 \n10. Brut3Forc3: 5 \n10. i_like_pie: 5 \n13. lingomaniac88: 4 \n14. distracted523: 3 \n15. infinity4ever: 2 \n15. cf249: 2 \n15. krakola45: 2 \n18. pacman2812: -6[/hide]", "Solution_17": "HIS\r\n[b]THIS[/b]\r\n\r\nI could have formed an inappropriate word, but instead I decided to use this.\r\n\r\n[hide=\"Score:\"]\n1. mathforlife: 16\n2. BOGTRO: 14 \n3. james4l: 13\n3. 33286: 13\n5. mz94: 12 \n6. Flame: 9 \n6. miyomiyo: 9 \n8. woser456: 7 \n9. Yongyi781: 6 \n10. Quickster94: 5 \n10. Brut3Forc3: 5 \n10. i_like_pie: 5 \n13. lingomaniac88: 4 \n14. distracted523: 3 \n15. infinity4ever: 2 \n15. cf249: 2 \n15. krakola45: 2 \n18. pacman2812: -6[/hide]", "Solution_18": "HIS\r\nTHIS\r\n[b]SHIFT[/b]\r\n\r\nThe boy shifted his position on the couch.", "Solution_19": "HIS \r\nTHIS \r\nSHIFT\r\n[b]FIFTHS[/b]\r\n\r\nI will divide the pie into fifths, then eat 20 percent.", "Solution_20": "[b]Forfeit.[/b]\r\n\r\nNew Word:\r\n[b]DAY[/b]\r\n\r\nAn apple a [b]day[/b] keeps the doctor away.\r\n\r\n[hide=\"Scores\"]1. mathforlife: 17 \n2. BOGTRO: 14 \n3. james4l: 13 \n3. 33286: 13 \n5. mz94: 12 \n6. Flame: 9 \n6. miyomiyo: 9 \n8. woser456: 7 \n9. Yongyi781: 6 \n10. Quickster94: 5 \n10. Brut3Forc3: 5 \n10. i_like_pie: 5 \n13. cf249: 4\n13. lingomaniac88: 4 \n15. distracted523: 3 \n16. infinity4ever: 2 \n16. krakola45: 2 \n18. pacman2812: -6[/hide]", "Solution_21": "[b][i][u]DRAY.[/u][/i][/b]\r\nMedieval people used a dray to haul objects.\r\n\r\nmust be hard for you scorekeepers [hide=\":jump:\"][hide=\":jump:\"][hide=\":jump:\"]o.O Those smileys stop idk[/hide][/hide][/hide]", "Solution_22": "DAY\r\nDRAY\r\n[b]HARDY[/b]\r\n\r\nhardy boys.", "Solution_23": "I dispute 1=2's sentence... but w/e.\r\n\r\nDAY \r\nDRAY \r\nHARDY \r\n[b]HARDLY[/b]\r\n\r\nThe poor family [i]hardly[/i] had any food.", "Solution_24": "DAY \r\nDRAY \r\nHARDY \r\nHARDLY \r\n[b]HALYARD[/b]\r\n\r\n\r\nyou use a halyard on a ship to lower the flag", "Solution_25": "DAY \r\nDRAY \r\nHARDY \r\nHARDLY \r\nHALYARD \r\n[b]CHARLADY[/b]\r\n\r\nCharlady is the British equivalent of \"charwoman\", which is a woman who cleans in a large house or office.", "Solution_26": "[b]FORFEIT[/b]\r\n\r\nNew Word:\r\nSAW\r\n\r\nI saw all the ARML teams in their bright shirts.\r\n\r\n[hide=\"Score\"]\n1. mathforlife: 19\n2. BOGTRO: 14 \n3. james4l: 13 \n3. 33286: 13 \n5. mz94: 12 \n6. Flame: 9 \n6. miyomiyo: 9 \n8. woser456: 7 \n9. Yongyi781: 6 \n10. Quickster94: 5 \n10. Brut3Forc3: 5 \n10. i_like_pie: 5 \n13. cf249: 4 \n13. lingomaniac88: 4\n15. panjia123: 3\n15. distracted523: 3 \n17. infinity4ever: 2 \n17. krakola45: 2 \n19. pacman2812: -6 \n[/hide]", "Solution_27": "SAW\r\n[b]PAWS[/b]\r\n\r\nThe dog used its paws to change the channel to a nature documentary", "Solution_28": "SAW\r\nPAWS\r\n[b]WARPS.[/b]\r\nWarps are science fictional thingys to get from one place to another, usually.", "Solution_29": "SAW\r\nPAWS\r\nWARPS\r\n[b]SPRAWL[/b]\r\n\r\nMy cats sprawl on the floor and sleep all day." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "function", "geometry solved" ], "Problem": "Given a regular tetrahedron $ABCD$ with edge length $1$ and a point $P$ inside it.\r\nWhat is the maximum value of $\\left|PA\\right|+\\left|PB\\right|+\\left|PC\\right|+\\left|PD\\right|$.", "Solution_1": "I assume $P$ can also be on the border of $ABCD$, because otherwise the maximum can't be reached (in case the maximum is really what we want; are you sure the problem didn't ask for the minimum?).\r\n\r\nI'll use the fact that the distance between two points in (or on) the regular tetrahedron with side $\\ell$ is at most $\\ell$. Through $P$ draw a plane parallel to $(BCD)$, cutting $AB,AC,AD$ in $B',C',D'$ respectively. In the plane $(B'C'D')$ draw a parallel through $P$ to $C'D'$, cutting $B'C',B'D'$ in $C'',D''$ respectively. \r\n\r\nWe now have $PA\\le AB'\\ (1)$, $PB\\le PB'+BB'\\le B'C''+BB'\\ (2)$, $PC+PD\\le DD'+D'D''+D''C''+C''C'+C'C\\ (3)$. If you draw the picture properly you'll see that the sum of the three RHS's is exactly $3\\ell$ ($3$, in our case, since $\\ell=1$). The equality is reached only when $P$ is one of the vertices of the tetrahedron.", "Solution_2": "Really P can be on the boundary.\r\nNice solution.", "Solution_3": "[quote=\"hardsoul\"]Given a regular tetrahedron ABCD with edge length 1 and a point P inside it.\nWhat is the maximum value of |PA|+|PB|+|PC|+|PD|.[/quote]\r\n\r\nI am just seeing that this is a corollary of the well-known fact that the maximum of a convex function inside a convex polyhedron is reached in a vertex of the polyhedron. This time, the convex function is |PA| + |PB| + |PC| + |PD| (the distance function is convex!). Actually, Grobber's solution was just an elementary formulation of this.\r\n\r\n[b]EDIT:[/b] See http://www.mathlinks.ro/Forum/viewtopic.php?t=19485 for details.\r\n\r\n Darij", "Solution_4": "[color=darkred][size=150]Excuse me .\ni wanted to know why the function |PA| + |PB| + |PC| + |PD| is convex .\nthanks for your help.[/size][/color]", "Solution_5": "[quote=hardsoul]Given a regular tetrahedron $ABCD$ with edge length $1$ and a point $P$ inside it.\nWhat is the maximum value of $\\left|PA\\right|+\\left|PB\\right|+\\left|PC\\right|+\\left|PD\\right|$.[/quote]\n\nWe claim that the maximum value is $3$. Equality occurs at any one of the vertices. Let $f(P) = AP + BP + CP + DP$. Suppose $P$ is the point for which $f$ attains it maximum value.\n\n[b]Claim 1:[/b] $P$ lies on one of the face of the tetrahedron \n\n[i]Proof.[/i] Suppose not then consider the ellipsoid $\\mathcal{C}_1, \\mathcal{C}_2$ with foci $(A,B) ; (C,D)$. with sum of distance from foci $AP + BP$ and $CP +DP$ respectively. Note that there exists another ellipsoid $\\mathcal{C}_3$ with foci $(C,D)$ and sum of focal distance slightly larger than that of $\\mathcal{C}_2$ such that $\\mathcal{C}_3$ meets $\\mathcal{C}_1$ inside the tetrahedron. This leads to a contradiction and we get the desired result. $\\square$\n\n[b]Claim 2:[/b] $P$ lies on the sides of equilateral triangle $\\triangle ABC$ if $P$ maximizes $g(P) = AP + BP + CP$\n\n[i]Proof.[/i] Similar to proof of [b]Claim 1[/b]. $\\square$\n\nBy our [b]Claim 1[/b] and [b]Claim 2[/b], we have $P$ lies on the edges of the tetrahedron let\u2019s say $AB$. Note that $AP + BP + CP + DP = 1 + CP + DP \\leq 3$. $\\blacksquare$" } { "Tag": [ "function" ], "Problem": "Let $ f(x,y) $ be a periodic function satisfying the condititon $ f(x,y) = f ((2x+2y),(2y-2x)) \\forall x,y \\in \\mathbb{R} $. Now define a function $ g(x) = f ( 2^X , 0 ) $ . Then prove that $g$ is periodic.", "Solution_1": "How do u define periodicity of functions of MULTIPLE varaibles ?? :?:", "Solution_2": "[quote=\"santosh_blr\"]How do u define periodicity of functions of MULTIPLE varaibles ?? :?:[/quote]\r\n\r\nI ask it too..." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "I had a question about subsequences and de limit inferrior and superrior\r\n\r\n1. I have a bounded convergent sequence xn\r\n\r\nIs it then always correct to say that lim inferrior=limit superrior=Limit of xn?\r\nIs there a theorem that concludes this?\r\n\r\n2. I have a bounded non-convergent sequence yn\r\n\r\nI want to make a subsequence for the limit superrior, How do I do that without knowing yn? (I only know it is bounded and non-convergent)", "Solution_1": "1 is correct, you should use the fact that if a sequence is convergent then every its subsequences are also convergent to the same limit." } { "Tag": [ "geometry", "cyclic quadrilateral" ], "Problem": "Let $ABCD$ be a cyclic quadrilateral, let $M$ be the midpoint of $AB$, let $N$ be the intersection of $AC$ and $BD$, and let $O$ be the point where the line passing through $M$ and $N$ intersects $CD$.\r\n\r\nProve that $\\angle DON=\\angle CON=90^{\\circ}$.\r\n\r\nedited to correct problem statement.", "Solution_1": "Sorry but I still don't think the problem's statement is correct. Please look at the picture I added.", "Solution_2": "Indeed. Maybe I transcribed the problem wrong; I'll check when I get home. Thanks." } { "Tag": [ "calculus", "integration", "limit", "LaTeX", "trigonometry", "complex analysis", "complex analysis unsolved" ], "Problem": "1. What is the best way of getting $ Res(f,0)$\r\n while $ f(z) \\equal{} \\frac {z}{sinz(1 \\minus{} cosz)}$\r\n I know it cat can be done with $ \\displaystyle\\lim_{z\\to0}\\frac {d}{dz}(z^2f(z))$\r\n but - is there a better way?\r\n\r\n2. what is $ \\int_{|z| \\equal{} 2} zsin(\\frac {z \\plus{} 1}{z \\minus{} 1})$\r\n (not the answer of course but some direction).\r\n\r\nThanks", "Solution_1": "Hey Nick, suppose $ f(z)$ is analytic for $ |z| > R_0$, then for $ R > R_0$:\r\n\r\n$ \\displaystyle\\mathop\\oint\\limits_{|z| \\equal{} R} f(z)dz \\equal{} 2\\pi i\\mathop\\textbf{Res}\\limits_{z \\equal{} 0}\\left[\\frac {1}{z^2}f\\left(\\frac {1}{z}\\right)\\right]$\r\n\r\nIn case you're interested in checking it numerically via Mathematica:\r\n\r\n[code]\nIn[21]:=\ng[z_] := z*Sin[(z + 1)/(z - 1)]\nNIntegrate[g[z]*2*I*Exp[I*t] /. z -> 2*Exp[I*t], {t, 0, 2*Pi}]\nN[4*Pi*I*(Cos[1] - Sin[1])]\n\nOut[22]=\n-7.771561172376096*^-16 - 3.7845972565184955*I\n\nOut[23]=\n0. - 3.7845972369939314*I\n[/code]", "Solution_2": "I would calculate the residue like this (and I think everyone else would too). It takes a long time to type in latex, but it's quick on paper.\r\n\r\n$ \\frac {\\sin z}{z} \\equal{} 1 \\minus{} \\frac {z^2}{6} \\plus{} O(z^4)$ and $ 1 \\minus{} \\cos z \\equal{} \\frac {z^2}{2} \\plus{} O(z^4)$ so their product is $ \\frac {z^2}{2}(1 \\plus{} O(z^2))$. Using $ \\frac {1}{1 \\plus{} z} \\equal{} 1 \\plus{} O(z)$ we get\r\n\r\n$ f(z) \\equal{} \\frac {2}{z^2} \\plus{} O(1)$ so the residue is zero.", "Solution_3": "[quote=\"Kalle\"] (and I think everyone else would too)[/quote]\r\n$ f(z)\\equal{}f(\\minus{}z)$ hence residue at point $ 0$ is zero. :blush:" } { "Tag": [ "number theory", "least common multiple", "modular arithmetic", "number theory unsolved" ], "Problem": "Given $ a,b,n\\in\\mathbb{N}^*$ satisfy $ \\gcd(a,n) \\equal{} \\gcd(b,n) \\equal{} 1$. Prove that:\r\n$ \\mathrm{ord}_nab \\equal{} \\mathrm{lcm}(\\mathrm{ord}_na;\\mathrm{ord}_nb)$", "Solution_1": "That's wrong, you just have $ \\mathrm{ord}_n(ab) | \\mathrm{lcm}(\\mathrm{ord}_n(a),\\mathrm{ord}_n(b))$. Otherwise already $ a\\equiv b\\equiv \\minus{}1 \\mod n$ is a counterexample.", "Solution_2": "Oh, I see, so sorry.\r\n\r\nThe problem first is: If $ \\gcd(\\mathrm{ord}_na;\\mathrm{ord}_nb)\\equal{}1$ then $ \\mathrm{ord}_nab\\equal{}\\mathrm{ord}_na.\\mathrm{ord}_nb$.\r\n\r\nThen if we remove the condition $ \\gcd(\\mathrm{ord}_na;\\mathrm{ord}_nb)\\equal{}1$, could you please tell me how can we express $ \\mathrm{ord}_nab$ in terms of $ \\mathrm{ord}_na$ and $ \\mathrm{ord}_nb$", "Solution_3": "Wtf is $ ord_n$..? :)", "Solution_4": "I think that $ ord_n a$ is the least number such that \r\n\r\n$ a^{ord_n a}\\equiv 1$ $ mod$ $ n$", "Solution_5": "Right.\r\n\r\nMashimaru: In general, you cannot express $ \\mathrm{ord}_n(ab)$ solely with $ \\mathrm{ord}_n(a)$ and $ \\mathrm{ord}_n(b)$. It depends on $ a,b$ itself. The most general thing is the one I mentioned: $ \\mathrm{ord}_n(ab)$ divides $ \\mathrm{lcm}(\\mathrm{ord}_n(a),\\mathrm{ord}_n(b))$.", "Solution_6": "Greatly thanks to you, ZetaX.\r\n\r\nWe can put $ \\mathrm{lcm}(\\mathrm{ord}_na;\\mathrm{ord}_nb)\\equal{}k_a\\mathrm{ord}_na\\equal{}k_b\\mathrm{ord}_nb$ then:\r\n$ (ab)^{\\mathrm{lcm}(\\mathrm{ord}_na;\\mathrm{ord}_nb)}\\equal{}(a^{\\mathrm{ord}_na})^{k_a}.(b^{\\mathrm{ord}_nb})^{k_b}\\equiv 1\\pmod n$ so $ \\mathrm{ord}_nab|\\mathrm{lcm}(\\mathrm{ord}_na;\\mathrm{ord}_nb)$" } { "Tag": [ "floor function", "ratio", "geometric sequence", "algebra proposed", "algebra" ], "Problem": "Prove that among the elements of the sequence $\\left\\{ \\left\\lfloor n\\sqrt{2003} \\right\\rfloor \\right\\}_{n\\geq 1}$ one can find a geometric progression having any number of terms, and having the ratio bigger than $k$, where $k$ can be any positive integer.\r\n\r\n[i]Radu Gologan[/i]", "Solution_1": "Fix two positive integers $k,m>1$. Now take some $n$ for which $\\{n\\sqrt{2003}\\}<\\frac 1{k^m}$. Now, for each $i\\in\\overline{1,m-1}$, we have $\\lfloor k^{i-1}n\\sqrt{2003}\\rfloor\\le\\frac{k^in\\sqrt{2003}}k<\\lfloor k^{i-1}n\\sqrt{2003}\\rfloor+\\frac 1k$, which, after multiplication with $k$, gives $k\\lfloor k^{i-1}n\\sqrt{2003}\\rfloor\\le k^in\\sqrt{2003}\\frac{1}{S(n)}+\\frac{1}{S(n+1)}+\\frac{1}{S(n+2)}+...\\forall n>1$", "Solution_1": "Help me please!!!!" } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "Let a>0 and the sequence xn=a^1/n. Calculate the limit of\r\n(n^3)(x_n-x_(n+1)-x_(n+2)+x_(n+3)).", "Solution_1": "Well, the limit is 4lna.", "Solution_2": "One idea wich conduct to 4lna, an asymptotic expansion of x(n) when n \r\ntends to +\\infty\r\nx(n) = 1 +lna/n + (lna)^2/2n^2 + (lna)^3/(6n^3) + o(1/n^3)\r\ndo the same for x(n+1), x(n+2), x(n+3) and get 4.lna wich probably the coefficient of 1/n^3 in the asymptotic expansion of \r\nx(n)-x(n+1)-x(n+2)-x(n+3)" } { "Tag": [ "IMO", "IMO 2005" ], "Problem": "Ok, as asked by some members, we will try to make a guide in order to defend yourselves in spanish there in Mexico:\r\n\r\n[b]Basic Supervivence[/b]\r\n\r\nHi My name is (your name here)= Hola, me llamo (your name here).\r\n\r\nI am from (your country here)= Yo soy de (your country here)\r\n\r\nwhat time is it?= Que hora es?\r\n\r\nWhats your name?= Como te llamas?\r\n\r\nHow did you do in the test?= Como te fue en el examen?\r\n\r\nWhere is the bathroom?= Donde queda el ba\u00f1o?\r\n\r\nI want some tequilla please!= Quiero Tequila por favor!\r\n\r\nCould you help me please, i am lost= Me puede ayudar, estoy perdido\r\n\r\nWhere is the dinner?= Donde es la comida?\r\n\r\nI am in room (your room number)= Estoy en la habitacion (room number)\r\n\r\nI love Mexico= Amo Mexico\r\n\r\nI am a Mathlinker, my nickname there is (your nickname)= Yo soy un Mathlinker, mi nombre de pila es (nickname).\r\n\r\nWhere is (name of hotel) Hotel?= Donde queda el hotel (nombre del hotel)?\r\n\r\nWe need a wake up servie at 7 am= Necesitamos servicio de despertador a las 7 (siete!) am\r\n\r\nI cant find my sit for the exam= No puedo encontrar mi puesto para el examen\r\n\r\nHow much is it?= Cuanto vale eso?\r\n\r\nPlease give me a (name of product you wanna buy)= Por favor me vende un (name of product)\r\n\r\n[b]More advanced stuff[/b]\r\n\r\nHi girl, you are very good looking= Hola nena, eres muy linda\r\n\r\nCan you give me your phone number?=Me puedes dar tu numero?\r\n\r\nSo we can party tonight?= para que salgamos esta noche?\r\n\r\nI will be going with 600 friends that are staying with me at the hotel!= Voy con 600 amigos que estan conmigo en el hotel :D \r\n\r\nI hope it helps, wanna know something else? let me know! :D", "Solution_1": "[quote=\"Pascual2005\"]what time is it?= Que hora es?\n\nI hope it helps, wanna know something else? let me know! :D[/quote]This is no good as long as we don't know how to count in spanish :P :D", "Solution_2": "[quote=\"Valentin Vornicu\"]This is no good as long as we don't know how to count in spanish :P :D[/quote]\r\nActually, \"Voy con 600 amigos que estan conmigo en el hotel\" would be even more amusing, because I'm sure that anybody who's learning from the above list wouldn't be able to say 600 in Spanish, and I certainly can't imagine saying \"Voy con [i]six hundred[/i] amigos que estan conmigo en el hotel\". :D", "Solution_3": "1=uno\r\n2=dos\r\n3=tres\r\n4=cuatro\r\n5=cinco\r\n6=seis\r\n7=siete\r\n8=ocho\r\n9=nueve\r\n10=diez\r\n11=once\r\n12=doce\r\n13=trece\r\n14=catorce\r\n15=quince\r\n16=dieciseis\r\n17=diecisiete\r\n18=dieciocho\r\n19=diecinueve\r\n20=vainte\r\n30=treinta\r\n40=cuarenta\r\n50=cincuenta\r\n60=sesenta\r\n70=setenta\r\n80=ochenta\r\n90=noventa\r\n100=cien\r\n600= seicientos", "Solution_4": "How do you say 21? Vainte e uno? (just in case someone decides to play some blackjack :D :D).", "Solution_5": "yes", "Solution_6": "We can bet beans! or tequila! :D", "Solution_7": "Well, what will be spanish of:\r\n\r\n1) I don't know spanish much.\r\n\r\n2) Can you speak english?\r\n\r\n3) Thank You\r\n\r\n4) Good morning\r\n\r\n5) Good night\r\n\r\n6) Bye", "Solution_8": "The only sentence I know in Spanish is: \"Ola chica, como estas, tu es muy querida...\"\r\n\"querida\" may be substituted by \"caliente\"... :D", "Solution_9": "[quote=\"mahbub\"]Well, what will be spanish of:\n\n1) I don't know spanish much.\n\n2) Can you speak english?\n\n3) Thank You\n\n4) Good morning\n\n5) Good night\n\n6) Bye[/quote]\r\nWell, \r\n2)\u00bfPuede usted hablar ingl\u00e9s?\r\n3)Gracias. \r\n4)Buenos d\u00edas.\r\n5)Buenas noches. \r\n6)Adios. \r\n (There's several for this one) Some are more formal than others. \r\n\r\n***And remember, if you are talking to someone older than you, someone you don't know, or someone who respect needs to be shown to, use the usted form and not tu. ***\r\n\r\nHope this helps!!! :)", "Solution_10": "1) is \"no se mucho espanol\"\r\n\r\n2) might also be said \"habla usted ingles?\"", "Solution_11": "Well...I know this in Spanish:\r\n\r\nlindos ojos !!\r\n\r\nis that right??\r\n :D", "Solution_12": "[quote=\"matty\"]Well...I know this in Spanish:\n\nlindos ojos !!\n\nis that right??\n :D[/quote]\r\nIf you wanted to say \"Cute Eyes\" yes, it is right.\r\nBy the way, if someone wants to know his/her guide name, just tell me, maybe i already meet him/her.\r\nSee you in M\u00e9rida", "Solution_13": "i am noot in any imo. so what do u mean by guide? will the host country give a guide with each country? if so then who is for our country?", "Solution_14": "[quote=\"mahbub\"]i am noot in any imo. so what do u mean by guide? will the host country give a guide with each country? if so then who is for our country?[/quote]yes, every country has a guide, to help that team familiarize with the hosting country's way of things :)", "Solution_15": "AHA! that's good!\r\n\r\nwell how long they will stay with us daily?", "Solution_16": "[quote=\"Dave Vather\"][quote=\"matty\"]Well...I know this in Spanish:\n\nlindos ojos !!\n\nis that right??\n :D[/quote]\nIf you wanted to say \"Cute Eyes\" yes, it is right.\n[/quote]\r\n\r\nyup!\r\nthis is what i meant!\r\n :D", "Solution_17": "[quote=\"mahbub\"]AHA! that's good!\n\nwell how long they will stay with us daily?[/quote]\r\nI heard that they must wait you patiently to you arrive, and must be whit you everytime, Go to bed after the last one of your country and be awake before the first one.", "Solution_18": "[quote=\"Dave Vather\"][quote=\"mahbub\"]AHA! that's good!\n\nwell how long they will stay with us daily?[/quote]\nI heard that they must wait you patiently to you arrive, and must be whit you everytime, Go to bed after the last one of your country and be awake before the first one.[/quote]That's a tough life :) Especially when we all know some teams don't go to bed to early :D :heli:", "Solution_19": "[quote=\"Valentin Vornicu\"][quote=\"Dave Vather\"][quote=\"mahbub\"]AHA! that's good!\n\nwell how long they will stay with us daily?[/quote]\nI heard that they must wait you patiently to you arrive, and must be whit you everytime, Go to bed after the last one of your country and be awake before the first one.[/quote]That's a tough life :) Especially when we all know some teams don't go to bed to early :D :heli:[/quote]\r\nIts a little prize considerating that you are going to the IMO free. Also most of the guides are math students that make lots of partys, so they are used to going to bed late. (But don\u00b4t tell your guide that i told you this)", "Solution_20": "How do I say: Did you score 21 today? ;)", "Solution_21": "[quote=\"kueh\"]How do I say: Did you score 21 today? ;)[/quote]I think you need to learn how to spell that in Chinese :D :D", "Solution_22": "Dave Vather: Did you meet the french guide :P ?", "Solution_23": "To Kueh: \"Lograste 21 puntos hoy?\" look for how to spell 21 \"veinta ei ouno\"\r\n\r\nTo Valentin: Yes, some teams dont even get to bed!\r\n\r\nTo Igor: Dont worry Igor, Latin girls are hot! :) Ask Orl! :D", "Solution_24": "[quote=\"Igor\"]Dave Vather: Did you meet the french guide :P ?[/quote]\r\nYes, In fact, you have one of the better guides of the contest. He is one of the few guides that know guides and he speaks french very well, since he lived in France. I don\u00b4t know his name, everybody here know him as \"El french\"", "Solution_25": "Dave Vather: \r\nDo you know if Armenia guide is a girl? :P \r\nAnd what's her(his) name?", "Solution_26": "As far as i know there are no bad words in spanish?\r\nis there any.", "Solution_27": "[quote=\"Vahe Musoyan\"]Dave Vather: \nDo you know if Armenia guide is a girl? :P \nAnd what's her(his) name?[/quote]\r\nIt seems to be not yet defined. You should ask ni\u00f1o, maybe he knows.\r\n\r\nBy the way, there are lots of bad words in spanish", "Solution_28": "[quote=\"Dave Vather\"]By the way, there are lots of bad words in spanish[/quote]I would really appreciate if you guys don't start posting them here! ;)", "Solution_29": "Who knows anything about the Albanian guide." } { "Tag": [], "Problem": "to play: Label the letters A-Z with points 1-26, respectively. Then think of a word that begins with A in less than one minute. If you fail, you have to use the letter A. Write your word down. Repeat with the rest of the letters. Add up your total points, and that's your score.\r\n\r\n\r\n[hide=\"mine\"]\n\naerodynamics \nbullion\ncalifornium\ndynamite\neuphemism\nfrancium\ngranite\nhallway\nigneous\njocular\nkiss\nlonely\nmullet\nnovice\noverestimate\npollution\nquotation\nrainbow\nsodium\ntainted\nunderestimate\nvociferous\nwasteful\nxylophone\nyellow\nzebra[/hide]\r\n\r\nMy score: 2357", "Solution_1": "i don't really get how this works o.O", "Solution_2": "So in the first minute, you think of a word that begins with the letter a, and in the second minute you think of a word that begins with the letter b, so on. Then calculate your score.", "Solution_3": "how does the scoringwork?", "Solution_4": "you add the point value of your letters, like the word \"am\" will get 1+13=14 points. So will the letter n.", "Solution_5": "Okay, lemme try:\r\n\r\n[hide=\"An attempt\"]\nAssassinations\nBosses\nCluttering\nDizziness\nExtraction\nFuzzy\nGrosses\nHappiness\nIsosceles\nJazzy\nKisses\nLozenges\nMosses\nNastiness\nOrganizations\nPreppily\nQuestioningly\nRazorblades\nSeizures\nTestosterone\nUnanimous\nVerifyingly\nWastefully\nXylophones\nYellow (Y is a hard one)\nZyzzyva[/hide]\r\n\r\nI think it's better with 10 sec. That's how I did it. I don't feel like adding mine up.", "Solution_6": "I can write a program easily to calculate scores\r\n\r\nIf I ever feel like it", "Solution_7": "[hide=\"blargh\"]\nargumentative\nberyllium\ncondominium\ndysprosium\nectoplasm\nfluorine\ngadolinium\nheterozygous\nindexing\njuxtaposition\nkindling\nluminous\nmendelevium\nneodymium\nosmosis\npraseodymium\nquadrillionth\nrubidium\nstrontium\ntypewriters\nuranium\nvanadium\nwryly\nxylophone\nytterbium\nzyzzyva (yeah I stole this one)\n[/hide]\nscore: 3335\n\nI realized after I was past [hide]c[/hide] that I missed [hide]chlorofluorocarbon[/hide].\r\n\r\n\r\nmiyomiyo: you got 3258", "Solution_8": "[hide]antidisestablishmentarianism\nbystander\ncretaceous\ndisillusionment\nepidemiologist\nfraternity\ngeophysicist\nhydrothermal\nintegumentary\njanissary\nkaleidoscope\nlithification\nmesosphere\nneanderthal\nornithologist\npseudoscience\nquintessential\nreexamination\nsyzygy\ntechnological\nusurper\nviviparious\nwee-wee\nxylophone\nytterbium\nzulu\n\nbeh[/hide]" } { "Tag": [ "geometry", "AMC", "USA(J)MO", "USAMO", "Putnam", "probability", "IMO Shortlist" ], "Problem": "He has already won 3 gold medals in IMO 2000,2001,2002. If he won a gold medal in IMO 2003, then he will have 4 gold medals. He will be the second IMO contestant after Reid Barton to have 4 gold medals.", "Solution_1": "Yes he did. He also had won a bronze medal before his four gold ones.", "Solution_2": "You mean he won 5 medals at the IMO 1 bronze and 4 gold medals?", "Solution_3": "so basically reid is not a record man anylonger :D", "Solution_4": "Yes he is .. He also got a gold medal 2 times in the IOI and as far as I know no one so far has won 2 different international olympiads in the same year :D He won the compsci one too in 2001 with a score of 580 out of 600.", "Solution_5": "and he also got another gold medal in some other olympiad, biology or chemistry - that's a rumor that i coudn't infirm or accept by now", "Solution_6": "IOI(1 gold medal), IMO(4 gold medals) and (chemistry or biology) he must be a genius! :D ( i'm talking about reid barton)", "Solution_7": "well Christian Reiher in maths won more medals than Reid Barton, so in maths C.R. is the record holder!", "Solution_8": ":D interesting debate\r\n\r\nI got to meet both of them very briefly at the IMO 2002 (Reid was in the army of USA leaders )\r\n\r\none more point in favor of Reid : Reid was 18 when he won the last gold medal at the IMO, and won the highest score ever in the IOI, while Christian is older, 19 or even more\r\n\r\nI can't imagine him staying so long in high school because he has study problems, because I heard he even did some research, so it's just a trick to keep him in business", "Solution_9": "Nikolai Dourov is also very strong, his average IMO score is higher than Reid Barton.\r\n\r\nNikolai Dourov:\r\nIMO gold medals 1996,1997,1998\r\nIOI gold medals 1996\r\nIOI silver medals 1995,1997,1998\r\nIMO 1996 7 7 7 7 2 7\r\nIMO 1997 7 7 7 4 7 7\r\nIMO 1998 7 7 7 7 7 4\r\naverage IMO score = 38.33\r\n\r\nReid Barton\r\nIMO 1998 4 7 7 7 0 7\r\nIMO 1999 7 7 7 7 2 4\r\nIMO 2000 7 7 7 7 7 4\r\nIMO 2001 7 7 7 7 7 7\r\naverage IMO score = 36.75", "Solution_10": "Reid has submitted this year one of the most beautiful and hardest problems ever to be found on the IMO ShortList - it's a shame that it didn't got elected - we will have to wait up untill next year to check it out!\r\n\r\nMihai Manea also scored 3 gold medals and next year proposed a problem at the IMO 2002: problem no. 4.", "Solution_11": "What problem is that? Algebra, Geometry, Combinatorics or Number Theory?", "Solution_12": "Talking about the numbers of medals and scores , it seems that every country has his genius! :D", "Solution_13": "hey wait,but isnt ciprian the only one to have three perfect scores(if i'm not mistaken).but i really luv to know which of these math guys were brilliant at physics!!.that wud be the sign of real genius ,in my view :D", "Solution_14": "ciprian manolescu was a genius at almost everything he put his mind into - unfortunately in Romania you CANNOT participate to more than ONE international olympiad ... damn :(", "Solution_15": "well I am from ROmania and I didn't know that... anyway.. the people who came out with this \"rull\" should take the example of other contries! some of them won in the same year gold at IMO and IOI, perhaps we have fellows that can do that too, but they are stopped by this \"rull\" .. :(", "Solution_16": "There are many people who are good in both math and physics. For example, Gregory Price, Timothy Abbott, Daniel Gulotta, Ari Turner, Steven J. Byrnes. All of them are from USA, and they have good results in USAMO and USAPhO. Ari Turner is an IPhO gold medallist, and he is also a Putnam Fellow(top 5). Gregory Price is good in Math, Physics and Infomatics, he represented USA in IPhO and IOI and performed very well in USAMO, being a winner of USAMO 2001 and 2002 (top 12).", "Solution_17": "Since fredbel6 accused Germany and C.Reiher of cheating and no-one expressed a different opinion, let me do it. I do not know Reiher nor R.Barton personally and I have not met them. I just know of them. I believe that both of them are brilliant and I would not like to compare them. How could I?\r\n\r\nHowever, I think that the above accusation for a student like Reiher and a country as Germany is very bad and above all unfair. \r\n\r\nI cannot accept that it is so easy to fool the IMO committees and so many math leaders and have students compete at the Olympiad without being eligible. Especially, if these students have attracted enough publicity, like Reiher, with their achievements. If there was something that violates the rules of the IMO, the other countries would react. Moreover, if he was eligible because they \"kept\" him in high school so that he breaks Barton's record as fredbel6 suggested, how were they so sure that he would get a gold medal in the IMO 2003? And without accusations of this type...? It is like a conspiracy and it does not make any sense to me.\r\n\r\nI checked the website \r\n\r\nhttp://www.bundeswettbewerb-mathematik.de/imo/\r\n\r\nand it says that C.Reiher attends KLASS 13 and that this is his fifth olympiad. If you check the previous years at that site, you will see that it is not at all rare for the German team to have members that attend that klass; i.e. to be so \"old\". In 2002, he attended KLASS 11 and so on. I must say that I am not familiar with the German system though.\r\n\r\nAlso, I do not know how old he is and so I do not base my judgement on rumors. Even if he is 19 (is this true 100%?) the eligible age is 20, as long as you do attend high-school because of your country's system. \r\n\r\nAlso, even if he did some research, this does not imply anything suspicious either. It is not at all rare for students of that level to do some kind of research. You may find many examples, especially here in the USA.\r\n\r\nI hope you all agree!", "Solution_18": "How come a 13-grader could be regarded as cheating if he's under 20? Hong Kong Team has many F.7-ers. (equivalent to grade 13) Without knowing the education systems one just shouldn't accuse another of that.", "Solution_19": "Concerning the German education system, it is not seldom to find 19 year olds at school; we have 13 grades and there is usually no reason for anybody to jump over a grade unless he is very bored and very good as I was when I jumped over the 2 grade. When I jumped over the 6 grade I was not so good any more and I indeed have missed some important stuff so that my grades in biology, chemistry, sports now are not really pleasant :-) And moreover, I have lost two olympiad years - damn!\r\n\r\n dg", "Solution_20": "edit: removed (silly mistake, didn't see this had multiple pages)", "Solution_21": "[quote=\"darij grinberg\"]Concerning the German education system, it is not seldom to find 19 year olds at school; we have 13 grades and there is usually no reason for anybody to jump over a grade unless he is very bored and very good as I was when I jumped over the 2 grade. When I jumped over the 6 grade I was not so good any more and I indeed have missed some important stuff so that my grades in biology, chemistry, sports now are not really pleasant :-) And moreover, I have lost two olympiad years - damn!\n\n dg[/quote]\r\nare you still in high-school? :)", "Solution_22": "Of course I am - in the 11 grade, and I still have 3 years of math olympics before me. I have actually a chance to get into the IMO this year - if I will belong to the 6 best of the 16 training participants.\r\n\r\n Greetings, Darij", "Solution_23": "that's nice, so I will expect you on the contestants list on MathLinks contest? :D (in case I put more geometry questions :D:D).", "Solution_24": "[quote=\"Anonymous\"][...][/quote]\r\n\r\nSorry - I just had forgotten to login. Darij", "Solution_25": "[quote=\"Achilleas Sinefakopoulos\"]Moreover, if he was eligible because they \"kept\" him in high school so that he breaks Barton's record as fredbel6 suggested, how were they so sure that he would get a gold medal in the IMO 2003? And without accusations of this type...? It is like a conspiracy and it does not make any sense to me.[/quote]\r\n\r\n s it illegal to keep them in high school btw :?", "Solution_26": "[quote=\"Valentin Vornicu\"]that's nice, so I will expect you on the contestants list on MathLinks contest? :D (in case I put more geometry questions :D:D).[/quote]\r\n\r\nWell, I'll try... thanks anyway, btw, what do we get as prizes? an upgrade of the user level?\r\n\r\n Darij", "Solution_27": "well I gaved diplomas and in the first edition an insignia, and in the second some money (not much though - some 100 dollars) to the winners.\r\n\r\nIt's called a surprise prize you know :) \r\n\r\nBut pretty much you earn by learning from the contest :P.", "Solution_28": "Aha, clear, it was really a joke question...\r\n\r\n dg\r\n\r\nPS. I have finally decided to take part in the contest after I have solved one problem...", "Solution_29": "Well, I don't really understand why do people speak so mcuh about Reiher who won his first gold medal in the 10th grade and they forget about so many Russians who score higher gold medals in the 9th grade?\r\nIt's a lack of luck that Russia has just 11 grades, or the Russians could smash everybody!", "Solution_30": "well scoring 4 golds is a great achievement, even if you start in 10th grade.\r\n\r\nI see that you are from Moldovia, iura. There is a Moldovian student, Boreico which has scored last year (8th grade!) a silver medal with 28p. He participated at the Romanian NMO and went first on the NMO at 9th grade, and also 1st at the Romanian TST (unofficially). He will probably score 4 gold medals as well at the IMO. It remains to be seen how many full scores will he achieve (I belive he's the only one by now capable of making Ciprian Manolescu's Record :D). Do you know him?", "Solution_31": "Yes, it's me. \r\nAnyway, it's early to speak about 4 gold medals or even 3 perfect scores!\r\nI will do my best, I hope. Time will show what is this \"best\"", "Solution_32": "Hi,\r\n\r\nI doubt your statement that Russia would have many more IMO record if they had 12 grades at school. I agree that it increases their opportunities for sure but not too much.\r\n\r\nContestans usually cannot participate too often in IMO since there is a lot of competition among the participating students. Imagine a huge country with many people, what are the chances, that a student of 9th (or even younger) defeats somebody from 10th or 11th grade ?\r\n\r\nI want to provide an example. Imagine a student from Sankt-Peter(s)burg (formerly Leningrad) which have a superb math tradition. There students have to be the best at 3-4 rounds, then the Russian NMO follows. Finally they still have to be among the TOP6 at the Russian TST(s). It is possible but it requires a unique performance, doesn't it ? \r\n\r\nAnyway I wish you good luck at IMO !!!", "Solution_33": "Maybe you are right, but imagine the the Russian team at 2002 IMO(6 gold medals) would participate at 2003 IMO (that would probably happen if there were 12 grades). Don't you think they would smash China?", "Solution_34": "Of course they would smash China, but they wouldn't have smashed Bulgaria (which distroyed China). China had the worst team in years last year ... so I guess we'll see an incredible come back this year ;) \r\n\r\nanyway, as I see it Moldovia is approaching her first presence in top 10 countries, with her best 3 students better than all Romanian students this year :( \r\n\r\nthere is another side with this 12 grade thing. Think how many of the Russians which went to IMO in 9th grade would go in if 12th grades were allowed ... same number as USA 9th graders, Romania 9th graders, China 9th graders ... at most 1 (USA had an exception in 1999 when they had two 9th graders with gold medals: Reid Barton and Gabriel Caroll -> but that was a big exception). So again the probability of scoring a 4-times-gold medal at the IMO is smaller ...", "Solution_35": "[quote=\"iura\"]Maybe you are right, but imagine the the Russian team at 2002 IMO(6 gold medals) would participate at 2003 IMO (that would probably happen if there were 12 grades). [/quote]\r\n\r\nI agree with Valentin that introducing a further grade, e.g. a 12th grade, does not increase chances too much. And your argument is tricky iura as you suppose that right now the new grade would have been introduced. So if the team would have participated in IMO with students mostly in grade 12 you cannot suppose anymore they also would have been at IMO with mostly the same students being in grade 11 back then. It is much more probable that in this year the elder guys would have made up the team, isn't it ?", "Solution_36": "Just a little correction Valentin..\r\nMoldova, not Moldovia :)", "Solution_37": "upss :blush: I ment the English name of Moldova which is [url=http://www.nationmaster.com/encyclopedia/Moldovia]Moldovia[/url]", "Solution_38": "one thing you forgot is that Reid Barton also had the absolute #1 spot on the 2001 IMO while Christian Reiher never had an absolute #1 score. that's why Reid Barton is my favourite among those two :)" } { "Tag": [], "Problem": "Three people's first names are Josie, Nicki and Sara. Their last\nnames are, not necessarily in this order, Croll, Liu and Menkin.\nTheir favorite colors, not necessarily in this order, are red,\nyellow and blue. The number of letters in a person's favorite\ncolor is the same as the number of letters in either their first\nname or their last name. What is the first name and the last\nname of the person whose favorite color is blue?", "Solution_1": "blue has 4 letters.\r\n\r\nso it's neither Josie or Nicki, so Sara.\r\n\r\nand yellow is Menkin, Red is Liu.\r\n\r\nso the last name of person who likes blue is Croll.\r\n\r\nAnswer : Sara Croll" } { "Tag": [ "function", "trigonometry", "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ A(x) \\equal{} \\frac{1}{\\cos x} \\equal{} \\sum_{n\\ge0}a_nx^n$. Recall that \r\n\r\n$ \\cos x \\equal{} \\sum_{n\\ge0}(\\minus{}1)^n\\frac{x^{2n}}{(2n)!} \\equal{} 1 \\minus{} \\frac{x^2}{2!} \\plus{} \\frac{x^4}{4!} \\minus{} \\frac{x^6}{6!} \\plus{} \\ldots$\r\n\r\nCompute $ a_0$ through $ a_4$. (Hint: $ A(x) \\cos x \\equal{} 1$)", "Solution_1": "We have $ \\cos x \\sum_{n \\ge 0} a_n x^n \\equal{} (a_0\\plus{}a_1 x \\plus{}a_2 x^2\\plus{}a_3x^3\\plus{}a_4 x^4\\plus{}...) \\left(1\\minus{}\\frac{x^2}{2!} \\plus{} \\frac{x^4}{4!} \\minus{} ....\\right) \\equal{} 1$\r\n\r\nComparing Coefficients on both sides\r\n\r\n$ a_0 \\equal{} 1; a_1 \\equal{} 0$\r\n\r\n$ a_2 \\minus{} \\frac{a_0}{2!} \\equal{} 0 \\Rightarrow a_2 \\equal{} \\frac{1}{2}$\r\n\r\n$ a_3 \\minus{} \\frac{a_1}{2!} \\equal{} 0 \\Rightarrow a_3 \\equal{} 0$\r\n\r\n$ a_4 \\minus{} \\frac{a_2}{2!}\\plus{}\\frac{a_0}{4!} \\equal{} 0 \\Rightarrow a_4 \\equal{} \\frac{5}{24}$\r\n\r\nAnd you can check that these are correct here: http://en.wikipedia.org/wiki/Trigonometric_function", "Solution_2": "http://en.wikipedia.org/wiki/Euler_number\r\nhttp://mathworld.wolfram.com/SecantNumber.html\r\nhttp://mathworld.wolfram.com/EulerNumber.html\r\n\r\n darij" } { "Tag": [ "function", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "f:R-->R and f(x) is continuous function such that [tex] f(x)f(-x)=cos^2x[/tex]\r\nFind f(x)?", "Solution_1": "i don't think there is just one continuous function which satisfies that equation. e.g. you might choose $f(x)$ for positive $x$ quite arbitrary(but continuous) and then choose $f(-x)$ s.t. the equation is satisfied. this always gives as a solution(maybe not in some particular cases, but in general...e.g. you have to take care of the positive $x$ where $\\cos^2 x=0$; if $f(x)=0$ there, you have to take some $f(-x)$ s.t. $f$ stays continuous at $-x$...).", "Solution_2": ":( may be...my english is very very bad\r\nmy problem is :\r\nConsider a continuous function f(x):R-->R satisfies that equation:[tex]f(x)f(-x)=cos^2x[/tex] \r\nFind f(x)\r\nI think[tex] f(x)=cosx\\sqrt{\\frac{g(x)}{g(-x)}}[/tex]\r\nwith g(x)>0 and g(x) is a continuous function ;but I haven't proved :( :D", "Solution_3": "This is a really minor issue, but I feel like saying it. The word \"continuous\" in English has two [i]u[/i]'s: [u]contin[b]u[/b]ous[/u]. Around here it sometimes seems that the one-[i]u[/i] misspelling occurs more frequently than the correct spelling.\r\n\r\n(/pedantry)", "Solution_4": "your conjecture can't be true, a counterexample is $f(x)=1+\\sin x$ (the square root would always be nonnegative, therefore the sign of $f(x)$ would be determined just by $\\cos x$ if you're statement was true; however the above function is always nonnegative -> contradiction).", "Solution_5": "[quote=\"Peter Scholze\"]your conjecture can't be true, a counterexample is $f(x)=1+\\sin x$ (the square root would always be nonnegative, therefore the sign of $f(x)$ would be determined just by $\\cos x$ if you're statement was true; however the above function is always nonnegative -> contradiction).[/quote]\r\nthank you\r\nI had a mistake\r\nI thought [tex] 1+sinx=cosx\\sqrt{\\frac{1+sinx}{1-sinx}}=f(x)[/tex](choose g(x)=1+sinx)\r\nBut[tex] f(x) =cosx\\sqrt{\\frac{1+sinx}{1-sinx}}[/tex] isn't continuous function :blush: \r\nOh;can you post your solution? :D \r\nI haven't solved this problem" } { "Tag": [ "trigonometry", "inequalities", "function", "inequalities unsolved" ], "Problem": "Let $A, B, C$ be real numbers in the interval $\\left(0,\\frac{\\pi}{2}\\right)$. Let \\begin{align*} X &= \\frac{\\sin A\\sin (A-B)\\sin (A-C)}{\\sin (B+C)} \\\\ Y &= \\frac{\\sin B\\sin(B-C)\\sin (B-A)}{\\sin (C+A)} \\\\ Z &= \\frac{\\sin C\\sin (C-A)\\sin (C-B)}{\\sin (A+B)} . \\end{align*} Prove that $X+Y+Z \\geq 0$.", "Solution_1": "Hope that the following solution is correct.\r\nSuppose that C>=B>=A,then sinC>= sinB>= sinA,sinAsin(A-B)sin(A-C)/sin(B+C)>=0.\r\nThe inequality reduces to show that \r\nsinCsin(C-B)sin(C-A)/sin(A+B) >=sinBsin(C-B)sin(B-A)/sin(A+C).\r\nNote that sinC>= sinB,sin(C-B)>=0.So it would sufficient to show that\r\nsin(C-A)sin(C+A)>=sin(B-A)sin(B+A)is equivalent to cos2A>=cos2B<=> sin 2 A>=sin 2 B.", "Solution_2": "[quote=\"hardsoul\"]Hope that the following solution is correct.\nSuppose that C>=B>=A,then sinC>= sinB>= sinA,sinAsin(A-B)sin(A-C)/sin(B+C)>=0.\nThe inequality reduces to show that \nsinCsin(C-B)sin(C-A)/sin(A+B) >=sinBsin(C-B)sin(B-A)/sin(A+C).\nNote that sinC>= sinB,sin(C-B)>=0.So it would sufficient to show that\nsin(C-A)sin(C+A)>=sin(B-A)sin(B+A)is equivalent to cos2A>=cos2B<=> sin 2 A>=sin 2 B.[/quote]\r\n\r\nNice! :D Well, I didn't find any mistake except the last part, which should be $\\cos{(2A)} \\geq \\cos{(2B)} \\Leftrightarrow \\sin^2{A} \\leq \\sin^2{B}$, which is true since $A \\leq B$.\r\n\r\nOr, you could have finished the last part with by noting that $\\sin{(C-A)} \\geq \\sin{(B-A)}$, and $\\sin{(C+A)} \\geq \\sin{(B+A)}$. I think it's right... anyways, cool solution.", "Solution_3": "Does anyone notice that this inequality is a lot like Schur's except with trig?", "Solution_4": "as me ,I use 'Shur :lol: '.thank you problem very easy", "Solution_5": ":) ,i too.if interested ask for sol.", "Solution_6": "Denote $a=\\sin A, b=\\sin B, c=\\sin C$\r\n\r\nThen it is equivalent to,\r\n\r\n$\\sum \\sin A\\sin (A-B)\\sin (A-C)\\sin (A+B)\\sin (A+C)\\\\ =\\sum \\sin A(\\sin ^2A-\\sin ^2B)(\\sin ^2A-\\sin ^2C)\\\\ =\\sum a(a^2-b^2)(a^2-c^2)\\\\ \\geq 0$\r\n\r\nThe last inequalitiy refers to Schur's.", "Solution_7": "Assume WLOG that $ A \\ge B \\ge C $. Note that $ \\sin $ is an odd, nonnegative, monotonically increasing function in the interval $ \\left(0, \\frac{\\pi}{2}\\right) $. Moreover, since $ \\frac{\\sin{A}}{\\sin{(B + C)}} + \\frac{\\sin{C}}{\\sin{(A + B)}} \\ge \\frac{\\sin{B}}{\\sin{(C + A)}} $ we have that Theorem $ 3c $ in http://web.mit.edu/~darij/www/VornicuS.pdf immediately solves the problem." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "You are given 3 equations:\r\n$x-y-2z=-3$\r\n$x+3y=30$\r\n$y+4z=10$\r\nWhat is $xyz$?", "Solution_1": "i brute forced it\r\n[hide]\nx=6, y=8, z=1/2\nso \n24[/hide]", "Solution_2": "[quote=\"davidlizeng\"]i brute forced it\n[hide]\nx=6, y=8, z=1/2\nso \n24[/hide][/quote]\r\nThat's right but what equations did you use?", "Solution_3": "He probably used all the equations.", "Solution_4": "i used the three given ones....\r\n[hide]\nx+3y=30\ny+4z=10\nx-y-2z=-3\n\nso first solve for x in terms of y\nyou get x=30-3y from the first equation\nthen solve for y in terms of z\nfrom the second one you get y=10-4z\nplug this one into the x=30-3y to get\nx=12z\nplug this and y=10-4z into x-y-2z=-3\nand solve for z which is 1/2\nx would be 6 and y =8\n[/hide]\r\ni'm sure there's a better solution", "Solution_5": "[hide]Multiply the last equation by $-2$ so you can cross out the z, then using that to the second. So $2x-2y-4z=-6$\n + $y+4z=10$ Solving for this you get $2x-y=4$. Then multiply the middle equation by -2 so we have $2x-y=4$\n +$-2x-6y=-60$.\nSolving this gets you y=8. Putting y into the middle and the last you can get x=8 and z=1/2. So $xyz=24$.[/hide]", "Solution_6": "[quote=\"mathgeniuse^ln(x)\"]He probably used all the equations.[/quote]\r\nNo, I meant how did he get there with the given equations.", "Solution_7": "It's easy to set up an augmented matrix and then row reduce it, perhaps this is not applicable here (but took three steps to find z by elimination method.", "Solution_8": "[quote=\"akech\"]It's easy to set up an augmented matrix and then row reduce it, perhaps this is not applicable here (but took three steps to find z by elimination method.[/quote]\r\n\r\ndude, this is mathcounts...", "Solution_9": "[hide=\"Ok, here is the full solution\"]Subtract the second equation from the first:\n$-4y-2z=-33$\nMultiply this by $2$ and add this to the third equation:\n$-7y=-56$\n$y=8$\nFrom the second and third equations,\n$x=6$\n$z=\\frac{1}{2}$\nSo $xyz=24$[/hide]" } { "Tag": [ "modular arithmetic" ], "Problem": "[b]Problem.[/b] If $ m,n\\in\\mathbb{N}$, $ m\\neq n$, and $ a\\equiv 1\\pmod 2$ then $ \\gcd(a^{2^{m}}\\plus{}2^{2^{m}}, a^{2^{n}}\\plus{}2^{2^{n}}) \\equal{} 1$.", "Solution_1": "[hide]Well $ a^{2^{m}}\\plus{}2^{2^{m}}|a^{2^{n}}\\minus{}2^{2^{n}}$ if $ m0$. Since $f_{t}(x,y)$ is uniformly continuous in both $t$ and $(x,y)$, there is $\\delta>0$ such that $|f_{t}(x,y)-(r,q)|1$ then $4-(y+z)^2$ and $(k-1)(k+3-(y+z)^2)$ must have same sign and this can be done only for $k\\ge6$ and it cannot be smaller because we can choose let say $(y+z)^2=4,01$\r\nif $k<1$ then $4-(y+z)^2$ and $(k-1)(k+3-(y+z)^2)$ must have opposite sign and this can't be done because we can choose let's say $(y+z)^2=3,01$\r\n\r\nIt seems ok but I'm not sure about it....", "Solution_2": "[quote=\"Adalbert\"]\nnow whe must see what are the conditions in which is ${1\\over{k-1}} \\ge{1\\over{k+3-(y+z)^2}}$\n[/quote]We can assume $k\\ge1$. Then your condition is equivalent to $4\\ge(y+z)^2$. But that is not always true as you state yourself. If it were, the originial ineq would be trivial. :)", "Solution_3": "The answer is $k \\geq 7$. For $k=7$, equality occurs again when $x=y=2z$ or any other cyclic permutation.", "Solution_4": "[quote=\"Vasc\"]The answer is $k \\geq 7$. For $k=7$, equality occurs again when $x=y=2z$ or any other cyclic permutation.[/quote]Vasc, this is just incredible... an equality case in the middle of nowhere :fool: Now I would be curious to see how you prove it from there :roll:", "Solution_5": "For $k \\geq 9$ we can use Cauchy-Schwarz Inequality. \r\nFor $k=7$, we have to show that\r\n$x^2+y^2+z^2=3$ implies $\\sum \\frac 1{10-(y+z)^2} \\geq \\frac 1 {2}$.\r\nThe ineq. is equivalent to\r\n$\\sum \\frac 1{10-(s-x)^2} \\geq \\frac 1 {2}$,\r\nwhere $s=x+y+z$. Using (n-1)Equal Variable-Theorem (since $x^2+y^2+z^2=constant$ and $x+y+z=constant$), it suffices to consider $y=z$.\r\nSo, the inequality reduces to $(x-1)^2(3x^2-1)^2 \\geq 0$. :lol:", "Solution_6": "[quote=\"Vasc\"]For $k \\geq 9$ we can use Cauchy-Schwarz Inequality. \n[/quote]Please give a hint, how? (I'd thought of that already, as the ineq occured to me first with $k=9$... but I can't see how :huh:)", "Solution_7": "Write first the inequality as\r\n$\\sum \\frac{3+x^2-2yz}{k+x^2-2yz} \\geq \\frac{6}{k-1}$,\r\nthen apply Cauchy-Schwarz." } { "Tag": [ "geometry", "trapezoid", "trigonometry" ], "Problem": "$ ABCD$ is an isosceles trapezium.\r\nUpper base $ \\overline {AD}\\equal{}2$ ,lower base $ \\overline{BC}\\equal{}8$.If $ M$is the mid-point of $ AB$ and $ MD \\perp CD$.\r\nFind the area of isosceles trapezium $ ABCD$.", "Solution_1": "[quote=\"vinskman\"]$ ABCD$ is an isosceles trapezium.\nUpper base $ \\overline {AD} \\equal{} 2$ ,lower base $ \\overline{BC} \\equal{} 8$.If $ M$is the mid-point of $ AB$ and $ MD \\perp CD$.\nFind the area of isosceles trapezium $ ABCD$.[/quote]\r\nLet $ \\overline{MD}\\equal{}c,\\overline{DA}\\equal{}a,\\overline{DC}\\equal{}b$,so $ c\\equal{}\\frac{1}{2}(5a\\plus{}b)$ and $ b*c\\equal{}0$ so $ (5a\\plus{}b)b\\equal{}0$\r\n$ 10|b|cos\\angle ADC\\plus{}b^2\\equal{}0$ so $ |b|\\equal{}10cos\\angle ABC$ AH-altitude of trapezium so $ AH\\equal{}10sin\\angle ABCcos\\angle ABC$ and \r\n$ BH\\equal{}10cos^2\\angle ABC$, and $ BC\\equal{}AD\\plus{}20cos^2\\angle ABC$ so $ cos\\angle ABC\\equal{}\\frac{1}{\\sqrt{5}}$ and $ AH\\equal{}5$ and finaly\r\n$ S_{ABCD}\\equal{}AH(AD\\plus{}BC)/2\\equal{}25$ :)", "Solution_2": "$ |AM| \\equal{} |BM| \\equal{} \\frac {b}{2}$\r\n\r\n$ \\cos \\angle ABC \\equal{} \\frac {3}{b}$ $ \\implies$ $ \\frac {b^2}{4} \\plus{} 10 \\equal{} |MD|^2 \\equal{} 25 \\minus{} \\frac {b^2}{4}$\r\nThen $ \\boxed{b \\equal{} \\sqrt {30}}$\r\n$ S_{ABCD} \\equal{} (8 \\plus{} 2)/2.\\sqrt {21}\\equal{}5\\sqrt{21}$", "Solution_3": "[quote=\"dima ukraine\"][quote=\"vinskman\"]$ ABCD$ is an isosceles trapezium.\nUpper base $ \\overline {AD} \\equal{} 2$ ,lower base $ \\overline{BC} \\equal{} 8$.If $ M$is the mid-point of $ AB$ and $ MD \\perp CD$.\nFind the area of isosceles trapezium $ ABCD$.[/quote]\nLet $ \\overline{MD} \\equal{} c,\\overline{DA} \\equal{} a,\\overline{DC} \\equal{} b$,so $ c \\equal{} \\frac {1}{2}(5a \\plus{} b)$ and $ b*c \\equal{} 0$ so $ (5a \\plus{} b)b \\equal{} 0$\n$ 10|b|cos\\angle ADC \\plus{} b^2 \\equal{} 0$ so $ |b| \\equal{} 10cos\\angle ABC$ AH-altitude of trapezium so $ AH \\equal{} 10sin\\angle ABCcos\\angle ABC$ and \n$ BH \\equal{} 10cos^2\\angle ABC$, and $ BC \\equal{} AD \\plus{} 20cos^2\\angle ABC$ so $ cos\\angle ABC \\equal{} \\frac {1}{\\sqrt {5}}$ and $ AH \\equal{} 5$ and finaly\n$ S_{ABCD} \\equal{} AH(AD \\plus{} BC)/2 \\equal{} 25$ :)[/quote]\r\n\r\n$ \\boxed{BC \\equal{} AD \\plus{} 20cos^2\\angle ABC}$ so $ cos\\angle ABC$ [color=red]o[/color]$ \\equal{} \\frac {1}{\\sqrt {5}}$ and $ AH \\equal{} 5$\r\nSorry,I think you put wrong data to get the incorrect result!! :juggle:", "Solution_4": "Ptolemy!\r\nPut F on DC so $ MF \\parallel BC$\r\n$ AF\\equal{}MD\\equal{}k$, $ AM\\equal{}DF\\equal{}m$\r\nBy Ptolemy on AMFD,\r\n$ 10\\plus{}m^2\\equal{}f^2$\r\nBy Pythag on AMF,\r\n$ m^2\\plus{}f^2\\equal{}25$\r\n$ m\\equal{} \\frac{\\sqrt{15}}{2}$\r\nBy Pythag, the height of the trapezoid is $ \\frac{\\sqrt{21}}{2}$, the area is $ 5\\sqrt{21}$.", "Solution_5": "[quote=\"dgreenb801\"]Ptolemy!\nPut F on DC so $ MF \\parallel BC$\n$ AF \\equal{} MD \\equal{} k$, $ AM \\equal{} DF \\equal{} m$\nBy Ptolemy on AMFD,\n$ 10 \\plus{} m^2 \\equal{} f^2$\nBy Pythag on AMF,\n$ m^2 \\plus{} f^2 \\equal{} 25$\n$ m \\equal{} \\frac {\\sqrt {15}}{2}$\nBy Pythag, the height of the trapezoid is $ \\frac {\\sqrt {21}}{2}$, the area is $ 5\\sqrt {21}$.[/quote]\r\n\r\nbeautiful and elegant work... :jump:", "Solution_6": "Let $N=BC\\cap MD$; clearly $BN=AD\\ (*)$, hence $CN=AD+BC=10\\ (\\ 1\\ )$. Let $H$ be the projection of $D$ onto $BC$. Easily $CH=(BC-AD)/2=3$, so from the right-angled $\\triangle CDN\\implies DH^2=CH\\cdot NH=3\\cdot 7=21$. From $(*)$ we get $[ABCD]=[CDN]=CN\\cdot DH/2=5\\sqrt 21$.\n\nBest regards,\nsunken rock", "Solution_7": "Let $AB\\cap CD=P$ and $F$ be the midpoint of $CD$. Then, $AP=PD=\\frac{x}{3}$ by similar triangles. The right triangle $PDM$ gives us $\\frac{25x^2}{36}-\\frac{x^2}{9}=\\frac{21x^2}{36}=MD^2$, so $MD=\\frac{\\sqrt{21}x}{6}$. Now, in right triangle $DMF$, $MD^2+DF^2=16$, so $\\frac{21x^2}{36}+\\frac{x^2}{4}=16$ or $\\frac{30x^2}{36}=16$, hence $x=\\frac{24}{\\sqrt{30}}$. Then my calculations get off..." } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "Hello! I have a question about finding where a line crosses it's x-axis.\r\n\r\nI have found an equation of my graphed line, and found the slope of the line also. Now I'm trying to find it's x-axis intercept. I only know how to find the y-axis intercept using the slope-intercept form. Is there another form to find the x-axis? Help. Here are some numbers to work with= (62, 157.5) and (64, 162.6)\r\n\r\nThanks.", "Solution_1": "In your equation of your line, if you make the y equal to 0, you will find the x intercept. Likewise, if you make the x zero, you will make the y intercept. Try to see why, in y intercept form, the last term is your y intercept", "Solution_2": "[quote=\"dogseatcheese\"]In your equation of your line, if you make the y equal to 0, you will find the x intercept. Likewise, if you make the x zero, you will make the y intercept. Try to see why, in y intercept form, the last term is your y intercept[/quote]\r\n\r\n\r\nThank you so much! Now, I think I'm having a mental block...My line equation goes like this:\r\n\r\n-5.1x + 2y = -1.2\r\n\r\nSo, how do I make y=0? Is there anything special I have to do to besides multiply y by 0?\r\n\r\nThis graphing is more confusing than I thought at first.\r\n\r\n :oops:", "Solution_3": "[quote=\"heide.rumble\"][quote=\"dogseatcheese\"]In your equation of your line, if you make the y equal to 0, you will find the x intercept. Likewise, if you make the x zero, you will make the y intercept. Try to see why, in y intercept form, the last term is your y intercept[/quote]\n\n\nThank you so much! Now, I think I'm having a mental block...My line equation goes like this:\n\n-5.1x + 2y = -1.2\n\nSo, how do I make y=0? Is there anything special I have to do to besides multiply y by 0?\n\nThis graphing is more confusing than I thought at first.\n\n :oops:[/quote]\r\n\r\nTo \"make\" y=0, you just substitute 0 for y. So for your equation:\r\n$-5.1x+2\\times0=-1.2, -5.1x=-1.2, \\text{ and } x=\\frac{4}{17}.$ So the x-intercept is at point $(\\frac{4}{17},0)$.", "Solution_4": "Here's another way to think about it:\r\nIf you have a line in the form $Ax+By=C$, Then the slope is $\\frac{-A}{B}$, x intercept is $\\frac{A}{C}$, and the y intercept is $\\frac{B}{C}$.", "Solution_5": "That's basically the same thing except the rearrangement is done in variables and you're not really thinking. :P", "Solution_6": "*drops on couch*too lazy to think, shortcuts are better (if you understand them, so I guess it is thinking, ha). :)", "Solution_7": "I am still confused. I can't see how we got 4/17? I'm getting 4 & 1/4 (or 4/9). I think I'm missing some logical step somewhere along the way. That must be it.\r\n\r\nHere's what I do after hearing the shortcut:\r\n\r\n-5.1/-1.2 = 4.25 = 4 & 1/4 or 4/9.\r\n\r\nWithout hearing the shortcut, I get something completely different. On the original equation, these are my steps:\r\n\r\n-5.1x + (2*0) = -1.2\r\n-5.1x = -1.2\r\ndivide both sides out by -5.1 to get x alone,\r\nx=-1.2/-5.1\r\nwhich equals pretty much nothing, and definitely not 4/17ths.\r\n\r\nHelp, what step am I missing?\r\n\r\nI know, I know, I should be getting this by now. Thanks for being patient with me.", "Solution_8": "The equations for x-intercept and y-intercept should of been $\\frac{C}{A}$ and $\\frac{C}{B}$.\r\n\r\n$x=\\frac{-1.2}{-5.1}$ does equal $\\displaystyle\\frac{4}{17}$ths.\r\n\r\n$\\frac{-1.2}{-5.1}=\\frac{12}{51}=\\frac{4}{17}$", "Solution_9": "Oops. :blush: major typo. sorry heide.rumble.", "Solution_10": "[quote=\"pianoforte\"]The equations for x-intercept and y-intercept should of been $\\frac{C}{A}$ and $\\frac{C}{B}$.\n\n$x=\\frac{-1.2}{-5.1}$ does equal $\\displaystyle\\frac{4}{17}$ths.\n\n$\\frac{-1.2}{-5.1}=\\frac{12}{51}=\\frac{4}{17}$[/quote]\r\n\r\n\r\n :idea: No one had ever told me to take the decimals out of a fraction before. It makes sense (duh what is a fraction, but a decimal in the first place).\r\n\r\nThanks! I knew it must have been something really simple that I was missing. Here I was trying to divide the darn things. LOL (Sometimes I really think I should have been born a blond, cause I don't do justice to my fellow brunettes.) But what I lack in math brains I make up for in other areas. (I saved my house from burning down on Thursday everyone!!! :lol: ) So, I thank you all for explaining me through my weak spots!!!\r\n\r\nYou are all WONDERFUL!!!!!" } { "Tag": [ "logarithms" ], "Problem": "Let $r = log_{6}27$, find $log_{\\sqrt2}({108-2r})^{3-r}$", "Solution_1": "[hide=\"Maybe...\"]\nIs this the answer by any chance? My solution is really long, and I don't want to tex it up until I'm sure this is right.\n\n$6\\log_6 (102+6\\log_6 2)$[/hide]", "Solution_2": "Actually I remember the problem wrong. It should be $log_{\\sqrt2}108^{3-r}-2r$ :D", "Solution_3": ":cursing: :furious: \r\n\r\nyou mean i've got to start all over now? oh well, that happens sometimes. :roll:", "Solution_4": "Is the $-2r$ included in the log? I'll assume not, unless I find that it doesn't work that way.\r\n\r\n[hide]$\\log_627=3\\log_63$\n\n$\\log_{\\sqrt{2}} (108^{3-r})-2r=(3-3\\log_63)\\log_{\\sqrt{2}}(108)-6\\log_63$\n\n$(3-3\\log_63)(\\log_{\\sqrt{2}}4+\\log_{\\sqrt{2}}27)-6\\log_63=(3-3\\log_63)(4+6\\log_23)-6\\log_63$[/hide]" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": ":( :( :(", "Solution_1": "Find H(orthocenter).Line GH cuts the circle at A and B with B in opositive to H .Bissector HB cuts the circle at C and D.Triangle ACD is the answer", "Solution_2": "thanks :lol:" } { "Tag": [ "geometry", "rectangle", "modular arithmetic", "combinatorics unsolved", "combinatorics" ], "Problem": "Given an nXn chessboard with the bottom right hand square removed, let S(n) be the set of numbers k such that this board can be tiled by kx1 rectangles, e.g. S(4) = (1, 3). Prove that x is an element of S(n) if and only if x divides (n-1).", "Solution_1": "We can easily show we can cover it if $ k \\equal{} n \\minus{} 1$, just cover the bottom right most square that isn't deleted horizontally, then the next bottom right most up from that and so on, we will be left with a $ n \\minus{} 1$ by $ 1$ Vertical uncovered block which we can cover. To show this is true for all x such that x divides $ n \\minus{} 1$, just split our kx1 block into m smaller blocks and arrange them to make our larger n-1 * 1 block, which we know we can tile.\r\n\r\nTo show we cannot cover it if x does not divide n-1, we have $ n^2 \\minus{} 1$\r\nsquares = $ (N \\plus{} 1)(N \\minus{} 1)$ so one of k,1 divides N+1 and the other divides N-1. Now we just have to show it doesn't work if k divides only N+1 and not N-1", "Solution_2": "How do we prove that it doesn't work if k divides only n+1 and not n-1?", "Solution_3": "Couldn't figure out that bit, anyone know how to do it?", "Solution_4": "You are wrong about the condition you are trying to prove: that $ k \\mid n^2 \\minus{} 1$ and $ k \\not \\mid n \\minus{} 1$ does [i]not[/i] imply $ k \\mid n \\plus{} 1$. For example, $ 15 \\mid 34^2 \\minus{} 1 \\equal{} 33 \\cdot 35$ and $ 15 \\not \\mid 33$ and $ 15 \\not \\mid 35$.", "Solution_5": "Thanks JBL. \r\n\r\nI can see what's happening by trying to tile when k does not divide n-1, but can't turn it into a proof.\r\n\r\nDoes anyone know how to solve this problem?", "Solution_6": "Arrange the board so that the \"bottom\" = \"first\" row is the one with the corner removed. Let's say we're tiling with $ k$-blocks, and we begin by looking at our tiling from the first row. We have $ n \\minus{} 1 \\equiv r \\not\\equiv 0 \\pmod k$, so in the first row there must be a number of vertical blocks congruent to $ r \\mod k$. Then the second row has one more space, so the number of vertical blocks that start there is congruent to 1 mod $ k$. Then for the 2nd through $ k$th rows, the number of vertical blocks that start there is divisible by $ k$. But then in the $ k \\plus{} 1$th row, we again need a number congruent to $ r$ mod $ k$ many vertical blocks to begin there. Then in the next row we need $ \\equiv 1 \\pmod k$ new vertical blocks. And so on. Because $ r \\not \\equiv 0 \\pmod k$ and $ 1 \\not \\equiv 0 \\pmod k$ (or equivalently because $ k$ doesn't divide $ n \\minus{} 1$), in [i]every[/i] row above the first, we have some vertical block that isn't terminating at that row. This actually proves something a fair bit stronger than the original claim, I believe, assuming no errors on my part :)", "Solution_7": "You can do this in a completely algebraic way. The if part is clear. The only if part can be done by letting $ \\varepsilon\\equal{}e^{i\\frac{2\\pi}{k}}$. By assigning the number $ \\varepsilon^{i\\plus{}j}$ to the square $ (i,j)$, where $ (1,1)$ is the bottom left corner and $ (n,n)$ is the top right corner, we must have $ \\sum_{i,j}\\varepsilon^{i\\plus{}j}\\equal{}\\varepsilon^{n\\plus{}1}\\iff (\\varepsilon^{n\\plus{}1}\\minus{}1)(\\varepsilon^{n\\minus{}1}\\minus{}1)\\equal{}0$. Therefore, $ k|n\\minus{}1$ or $ k|n\\plus{}1$. The latter case is impossible, and so $ k|n\\minus{}1$, as required. $ \\Box$", "Solution_8": "[quote=\"boxedexe\"]You can do this in a completely algebraic way. The if part is clear. The only if part can be done by letting $ \\varepsilon \\equal{} e^{i\\frac {2\\pi}{k}}$. By assigning the number $ \\varepsilon^{i \\plus{} j}$ to the square $ (i,j)$, where $ (1,1)$ is the bottom left corner and $ (n,n)$ is the top right corner, we must have $ \\sum_{i,j}\\varepsilon^{i \\plus{} j} \\equal{} \\varepsilon^{n \\plus{} 1}\\iff (\\varepsilon^{n \\plus{} 1} \\minus{} 1)(\\varepsilon^{n \\minus{} 1} \\minus{} 1) \\equal{} 0$. Therefore, $ k|n \\minus{} 1$ or $ k|n \\plus{} 1$. The latter case is impossible, and so $ k|n \\minus{} 1$, as required. $ \\Box$[/quote]\r\nAnd why $ k|(n\\plus{}1)$ is impossible?", "Solution_9": "Wow. A terribly unjustified statement. I assumed $ n\\plus{}1$ was a prime for some unknown reason and excluded the trivial case $ k\\equal{}1$.", "Solution_10": "Thanks JBL. Now I know what I should have done." } { "Tag": [ "logarithms", "floor function", "ceiling function", "algebra unsolved", "algebra" ], "Problem": "Sequence $ {a_n,n \\equal{} 1,2,3,...}$.\r\nWe know that\r\n$ a_1 \\equal{} 1$,\r\n$ a_{n \\plus{} 1} \\equal{} a_n \\plus{} \\sqrt {a_n} \\plus{} \\frac {1}{2},n \\equal{} 0,1,2,...$.\r\n\r\nFind the formula of $ {a_n}$.", "Solution_1": "I can prove only acimptotic formul $ \\sqrt{a_n}\\equal{}\\frac n2 \\plus{}\\frac{\\ln n}{4}\\plus{}c\\plus{}O(\\frac{\\ln n}{n})$ for some constant c.", "Solution_2": "Do you have any reason at all to suspect that this sequence has a nice closed form?", "Solution_3": "[quote=\"t0rajir0u\"]Do you have any reason at all to suspect that this sequence has a nice closed form?[/quote]\r\n\r\nThe problem is a really contest problem I heard that I can't solve it.", "Solution_4": "Are you sure it isn't $ a_{n \\plus{} 1} \\equal{} \\left\\lfloor a_n \\plus{} \\sqrt {a_n} \\plus{} \\frac {1}{2} \\right\\rfloor$? That one actually has a nice answer and it would explain the half.", "Solution_5": "[quote=\"t0rajir0u\"]Are you sure it isn't $ a_{n \\plus{} 1} \\equal{} \\left\\lfloor a_n \\plus{} \\sqrt {a_n} \\plus{} \\frac {1}{2} \\right\\rfloor$? That one actually has a nice answer and it would explain the half.[/quote]\r\n\r\nI can't be sure.I heard this problem from someone who also only heard from sombody!\r\n\r\nBy the way,how about(how to solve) your sequence?", "Solution_6": "$ a_n$ turns out to be of the form $ \\left\\lceil \\frac {an^2 \\plus{} bn \\plus{} c}{d} \\right\\rceil$ and then it remains only to compute the relevant coefficients." } { "Tag": [ "function", "inequalities", "number theory" ], "Problem": "1) We take the real positivs numbers x,y,z,t such that x+y+z+t=1.Prove that sqrt(x^2y^2+z^2t^2)+sqrt(x^2t^2+y^2z^2)>=(sin45)/4.\r\n2) Find the natural number n such that (2^n+4^n+...+2006^n)/(1^n+3^n+...+2005^n) is also natural.", "Solution_1": "Solution to problem 2: n = 0. Natural since 2005/2005 = 1. Did you mean positive integer solution?", "Solution_2": "The previous posting should read: 1003/1003 = 1 is a natural number.", "Solution_3": "In number theory, natural numbers are usually taken to be ${1,2,3...}$. So natural numbers and positive integers are the same.\r\n\r\n(And yes, it seems silly to me too, so let's not have an arguement over it)\r\n\r\nAnd, in future, you may wish to the use the edit function rather than double posting :)", "Solution_4": "I am really curious about the flaw which either exists in this argument or in the intentions of the problem setter: Feedback appreciated!\r\n\r\nThe numerator can be factorised as 2^n(1^n + 2^n + ...+1003^n).\r\nThere are 1003 odd terms in the denominator (all terms are powers of odd numbers in the sequence 2k-1 for k = 1, 2, ..., 1003), so the denominator itself is odd.\r\nOdd numbers do not have even factors. Therefore the denominator has no proper factor in common with 2^n. All factors of the denominator are factors of the numerator. No proper factor of the denominator is a factor of 2^n. Therefore all factors of the denominator are factors of (1^n + 2^n + ... + 1003^n). Therefore the denominator, 1^n + 3^n + ...2005^n must divide (1^n + 2^n + ...1003^n).\r\nBut that is absurd unless n = 0 since both expressions have the same number of terms, and every term except the first is greater in the denominator, the first term being equal (to 1) in both cases. The denominator is strictly greater than an expression it must divide exactly.\r\nAs this is clearly impossible, no such n > 0 exists.\r\n\r\nIn brief:\r\n(i) The denominator is odd so has nothing in common with the 2^n factor in the numerator.\r\n(ii) But the other factor in the numerator is smaller than the denominator.\r\n(iii) There must therefore be odd primes left in the denominator after all\r\npossible cancellation has been carried out, so the answer whatever it is is not an integer.", "Solution_5": "For problem 1, the inequality is false when t and y are really really tiny.\r\n\r\nHowever, the real reason for my posting here was not to tell you that, but rather to say:\r\n\r\nBRITISH!!!!" } { "Tag": [ "algebra", "function", "domain", "number theory", "prime factorization", "Ring Theory", "Galois Theory" ], "Problem": "Lef $ f$ in $ K[x]$ be irreducible and let $ L/K$ be a normal field extension. Show that if $ g, h$ are normalized prime factors of $ f$ in $ L[x]$, there exist $ \\sigma \\in G(K/L)$ such that $ \\sigma g \\equal{} h$\r\n\r\nDeduce that in the prime factorization of $ f$ in $ L[x]$ all prime factors have the same exponent.", "Solution_1": "Reduce to the case where $ L/K$ is separable. Then $ L/K$ can be assumed to be Galois with group $ G\\equal{}Gal(L/K)$. Hence $ G$ operates $ K[X]$-linearly on the Dedekind domain $ L[X]$ with $ Fix_G(L[X])\\equal{}K[X]$. By assumption, $ gL[X]$ and $ hL[X]$ are prime ideals of $ L[X]$ lying above the prime ideal $ fK[x]$ of $ K[X]$. By Galois theory of Dedekind domains, the prime ideals $ gL[X]$ and $ hL[X]$ are conjugated by some $ \\sigma\\in G$. Because $ g$ and $ h$ are normalized it follows that they are conjugated by $ \\sigma$ as elements in $ L[X]$.\r\n\r\nThe additional claim follows trivially.", "Solution_2": "What is going on with this problem? \r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244275]just 2 days ago[/url]\r\n\r\nInterestingly enough, this wants you to deduce that the prime factors are of the same degree, whereas I wanted to use that fact to prove this.", "Solution_3": "Ouups, I was not aware of the fact of this double posting. \r\n\r\n@ son of Skanderbeg: This should be avoided in any case. :mad:" } { "Tag": [ "LaTeX", "inequalities unsolved", "inequalities" ], "Problem": "For a,b,c,d>0 and abcd=1 we have\r\n$ \\frac{a}{b}$+$ \\frac{b}{c}$+$ \\frac{c}{d}$+$ \\frac{d}{a}$a+b+c+d", "Solution_1": "My Latex in the last version no good :lol: I write new \r\n[b][i]We have[/i][/b] $ a,b,c,d>0$ [i]and[/i] $ abcd\\equal{}1$\r\n[b]Prove that[/b]\r\nIf $ a/b\\plus{}b/c\\plus{}c/d\\plus{}d/aa\\plus{}b\\plus{}c\\plus{}d$" } { "Tag": [ "inequalities", "geometry", "search", "inequalities unsolved" ], "Problem": "$ a,b,c$ are the side lengths of $ \\triangle ABC$, $ S$ is the area of $ \\triangle ABC$\r\n\r\nprove:\r\n\r\n$ \\frac{ab\\plus{}bc\\plus{}ca}{4S}\\geq\\sqrt{3}$", "Solution_1": "Similar with the well known Weiscnbock ineq", "Solution_2": "See here:\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=14388&search_id=1616039182[/url]\r\n[url]http://www.mathlinks.ro/viewtopic.php?search_id=1616039182&t=31598[/url]" } { "Tag": [ "IMO", "IMO 2008" ], "Problem": "I noticed some participants are already registrated, but I can only see the number of participants but not the link that lead to the pics and names etc...\r\n when we can see the participants? we must wait until all register??", "Solution_1": "Click on the country code? (at http://www.imo-official.org/year_reg.aspx )", "Solution_2": "thanks... i was trying to do this on the http://www.imo-2008.es haha" } { "Tag": [], "Problem": "Prove that for every integer $ n\\geq 3$, the equation $ x^{n}+y^{n}=z^{n+1}$ has infinitely many solutions in positive integers $ x,y,z$", "Solution_1": "From the eye you can see that $ (2,2,2)$ is a solution. Afterwards if $ \\alpha \\in \\mathbb N$ then $ (2\\cdot \\alpha^{n+1},2\\cdot \\alpha^{n+1},2\\cdot \\alpha^{n})$ si also a solution." } { "Tag": [ "geometry", "ratio", "circumcircle" ], "Problem": "From a point within a triangle, line segments are drawn to the vertices. A necessary and sufficient condition that the three triangles thus formed have equal areas is that the point be:\r\n\r\n$ \\textbf{(A)}\\ \\text{the center of the inscribed circle} \\qquad \\\\\r\n\\textbf{(B)}\\ \\text{the center of the circumscribed circle}\\qquad \\\\\r\n\\textbf{(C)}\\ \\text{such that the three angles fromed at the point each be }{120^\\circ}\\qquad \\\\\r\n\\textbf{(D)}\\ \\text{the intersection of the altitudes of the triangle}\\qquad \\\\\r\n\\textbf{(E)}\\ \\text{the intersection of the medians of the triangle}$", "Solution_1": "[hide=\"My Solution\"]\nIt is known that medians form equal areas so the answer is $ \\fbox{(E)}$.\n[/hide]", "Solution_2": "Answering the questions you post doesn't really give anyone else a chance to respond.", "Solution_3": "Are all these problems for the resources page or something? :maybe:", "Solution_4": "@grn_trtle: Every ten problems I posted, I went through problems that weren't answered and added the solution. So it's not like I completely didn't give anybody a chance to post.\r\n\r\n@ThinkFlow: Yeah.", "Solution_5": "the rihgt answer is last one. because median bisects an area of triangle.", "Solution_6": "Ok the right answer is letter A because the centroid of a triangle divides the triangle in three triangles with equal areas, but the ''if and only if'' is not that easy, so a rigorous proof to this problem (as geometry requests) has not been posted yet on this topic.", "Solution_7": "I think you mean letter E, because letter A involves the incenter.", "Solution_8": "Let the triangle be ABC, with medians AP,BQ, and CR and centroid N. I can prove the centroid is sufficient.\r\nProof 1: [BRC]=[ABP]=1/2[ABC], so subtracting [BRNP] from both sides, [ARN]=[NPC]. Similarly for the other three. But [BNP]=[NPC] since they have the same base and height, so all six triangles BNP, PNC, PNQ, QNA, ANR, and RNB have the same area. So [BNC]=[CNA]=[ANB].\r\nProof 2: It is well-known that AN/NP=2. This can be proved either by Menelaus or by the fact that the medial triangle, POR, is similar to the original triangle with ratio 1/2. So dropping perpendiculars from A and N to BC, we find the perp from A is three times the perpendicular from N, so [NBC]=1/3[ABC]. Similarly for the other three triangles.\r\n\r\nNow to prove it is necessary.\r\nDrop perpendiculars U and V from N to BC and AC, respectively. We must have $ NU \\cdot BC\\equal{}NV \\cdot AC, or \\frac{NU}{NV}\\equal{}\\frac{BC}{AC}$= a constant for the areas to be equal. From here it follows that N must lie on CR. Similarly it must lie on BQ and AP, so we are done." } { "Tag": [ "linear algebra", "matrix", "function", "linear algebra unsolved" ], "Problem": "$(a(i,j))\\in M(n,R)$ such that $a(i,j)+a(j,k)+a(k,i)=0$ for any i,j,k \r\n\r\nProve that there exist a sequence $t(1),...,t(n)$ such that $a(i,j)=t(i)-t(j)$ for\r\nall i,j", "Solution_1": "I think it is something related to graph theory. For example, if one consider\r\nan electronique circuit, with $n$ vertices, each vertice connected to every others. \r\nAn oriented edge from $i$ to $j$ has a number that represents a difference of potential. All\r\nwe know here is that, on every $3$-cycle, the sum of the differences of potential is null. And what we \r\nwant to prove is that there actually exists a potential function, that is a number $t(i)$ for each vertices, such that\r\nthe number on the edge from $i$ to $j$ is precisely $t(j)-t(i)$.", "Solution_2": "so one can check that:\r\n$t(1)=0$ and $t(i)=a(1,i)$ for $i \\geq 2$ works ..." } { "Tag": [ "AMC", "AIME" ], "Problem": "I can't figure out this one on the 1996 AIME\r\n\r\nFind the average value of |a1 - a2| + |a3 - a4| + |a5 - a6| + |a7 - a8| + |a9 - a10| for all permutations a1, a2, ... , a10 of 1, 2, ... , 10.", "Solution_1": "[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=5026]Go here[/url]" } { "Tag": [ "calculus", "inequalities", "inequalities proposed" ], "Problem": "Given $ a, b, c \\geq\\ 0$. Prove that:\r\n$ \\frac {3}{\\sqrt {2}}.\\sqrt {a^2 \\plus{} b^2 \\plus{} c^2} \\geq\\ \\frac {a^2 \\plus{} b^2}{a \\plus{} b} \\plus{} \\frac {b^2 \\plus{} c^2}{b \\plus{} c} \\plus{} \\frac {c^2 \\plus{} a^2}{c \\plus{} a}$\r\n\r\nPS: Please don't use calculus :)", "Solution_1": "And I think it's harder than the old result: $ \\sum\\ \\frac{a^2\\plus{}b^2}{a\\plus{}b} \\geq\\ \\sqrt{3(a^2\\plus{}b^2\\plus{}c^2)}$\r\n :)", "Solution_2": "this inequality is equivalent to :\r\n$ \\frac{3}{\\sqrt{2}} \\geq \\sum\\frac{cos^2x}{siny\\plus{}sinz}$", "Solution_3": "[quote=\"Evariste-Galois\"]this inequality is equivalent to :\n$ \\frac {3}{\\sqrt {2}} \\geq \\sum\\frac {cos^2x}{siny \\plus{} sinz}$[/quote]\r\nCan you post your full proof? Thank you :)", "Solution_4": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. Prove that:\n$ \\frac {3}{\\sqrt {2}}.\\sqrt {a^2 \\plus{} b^2 \\plus{} c^2} \\geq\\ \\frac {a^2 \\plus{} b^2}{a \\plus{} b} \\plus{} \\frac {b^2 \\plus{} c^2}{b \\plus{} c} \\plus{} \\frac {c^2 \\plus{} a^2}{c \\plus{} a}$\n\nPS: Please don't use calculus :)[/quote]\r\n\r\nMy solution :)\r\nAssume:\r\n$ a \\ge b \\ge c$\r\nWe have:\r\n$ RHS \\le \\frac{a^2\\plus{}b^2}{a\\plus{}b}\\plus{}a\\plus{}b \\le 3\\sqrt{\\frac{a^2\\plus{}b^2}{2}} \\le LHS$\r\nDone! :)", "Solution_5": "[quote=\"caubetoanhoc94\"][quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. Prove that:\n$ \\frac {3}{\\sqrt {2}}.\\sqrt {a^2 \\plus{} b^2 \\plus{} c^2} \\geq\\ \\frac {a^2 \\plus{} b^2}{a \\plus{} b} \\plus{} \\frac {b^2 \\plus{} c^2}{b \\plus{} c} \\plus{} \\frac {c^2 \\plus{} a^2}{c \\plus{} a}$\n\nPS: Please don't use calculus :)[/quote]\n\nMy solution :)\nAssume:\n$ a \\ge b \\ge c$\nWe have:\n$ RHS \\le \\frac {a^2 \\plus{} b^2}{a \\plus{} b} \\plus{} a \\plus{} b \\le 3\\sqrt {\\frac {a^2 \\plus{} b^2}{2}} \\le LHS$\nDone! :)[/quote]\r\nNice proof, caubetoanhoc94 :) \r\nMy proof is for the stronger:\r\n$ \\frac{6(a^2\\plus{}b^2\\plus{}c^2)(a\\plus{}b\\plus{}c)}{3\\sum\\ a^2\\plus{}2\\sum\\ ab} \\geq\\ \\sum\\ \\frac{a^2\\plus{}b^2}{a\\plus{}b}$\r\n :)" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "A circle with center O is tangent to the sides AB, BC, CA and DA of a convex quadrilateral ABCD at K, L, M, and N respectively. The lines KL and MN intersect at point S. Prove that BD is perpendicular to OS.", "Solution_1": "I didn't consider the problem to be so difficult. However, I hope you'll enjoy solving it and hopefully help me out with the solution. Thank you in advance!", "Solution_2": "just see that the line BD is the polar of S in the circle (O) then the result comes . :wink: , You can read here for further questions about polars http://mathworld.wolfram.com/Polar.html", "Solution_3": "[quote=\"Tiger100\"]just see that the line BD is the polar of S in the circle (O) then the result comes . :wink: , You can read here for further questions about polars http://mathworld.wolfram.com/Polar.html[/quote]\r\n\r\nI have heard something about poles and polars but I would never think of using them. The site You recommended seems really useful. Thank you very much!" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let $x ,y$ such that $1 \\le x^{2}+y^{2}-xy \\le 2$\r\n\r\nprove that : $x^{4}+y^{4}\\ge \\frac{2}{9}$ .", "Solution_1": "[quote=\"ramox\"]let $x ,y$ such that $1 \\le x^{2}+y^{2}-xy \\le 2$\n\nprove that : $x^{4}+y^{4}\\ge \\frac{2}{9}$ .[/quote]\r\nWhy $x^{2}+y^{2}-xy \\le 2$ $?$\r\nI think, $x^{4}+y^{4}\\ge \\frac{2}{9}$ is true for all $x$ and $y$ such that $1 \\le x^{2}+y^{2}-xy$ only. :wink:", "Solution_2": "The problem was to prove that $\\frac{2}{9}\\le x^{4}+y^{4}\\le 8$ .\r\n$x, y$ real numbers such that $1\\le x^{2}+y^{2}-xy \\le 2$ .\r\nthe other inequality is easy to prove ...", "Solution_3": "OK. $x^{4}+y^{4}\\geq\\frac{2}{9}\\cdot(x^{2}+y^{2}-xy)^{2}\\Leftrightarrow(x+y)^{2}(7x^{2}-10xy+7y^{2})\\geq0,$ which true.\r\nThus, $x^{4}+y^{4}\\geq\\frac{2}{9}.$" } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "Find the 1000th derivative of f(x) = x*e^(-x). I did the problem but i just want to check my answer.\r\n\r\ni found the 1st two derivatives and i got:\r\n\r\nf '(x) = e^(-x) - [x*e^(-x)]\r\n\r\nf ''(x) = -2e^(-x) + x*e^(-x)\r\n\r\nso i noticed the coefficient of the first term is equal to the order of the derivative and that the signs of both terms alternate and concluded that the 1000th derivative would have 1000 as the coefficient in the first term and that since 1000 is an even number, the signs are the same as what appears in the 2nd derivative and got: \r\n\r\n1000th derivative of f(x) = -1000e^(x) + x*e^(-x)\r\n\r\nIs this correct?", "Solution_1": "Yes. Prove it by induction.\r\n$f^{(n)}(x)=(-1)^{n}(x-n)e^{-x}$", "Solution_2": "There's Leibnitz's rule\r\n\\[(f(x)g(x))^{(n)}=\\sum_{k=0}^{n}\\binom nkf^{(k)}(x)g^{(n-k)}(x)\\]\r\nYou have f(x)=x, hence $f^{(k)}(x)=0$, k>1. This is a general method for such problems." } { "Tag": [ "inequalities" ], "Problem": "Let $ a,b,c \\in R^\\plus{}$ and $ ab\\plus{}bc\\plus{}ca \\equal{} 3$\r\nProve that \\[ \\sum_{cyclic}\\frac{a^3}{bc(b\\plus{}c)^3} \\geq \\frac{3}{8}\\]", "Solution_1": "[hide]Hint:\nUsing the Chebyshev inequality after assuming that a>=b>=c[/hide]", "Solution_2": "Multiplying the LHS by $ ab\\plus{}bc\\plus{}ca$ and the RHS by $ 3$ homogenizes the inequality, right?", "Solution_3": "[quote=\"iwishiwasgoodatmath\"]Multiplying the LHS by $ ab \\plus{} bc \\plus{} ca$ and the RHS by $ 3$ homogenizes the inequality, right?[/quote]\r\n\r\nRight.", "Solution_4": "I would just do the problem with the AM-GM inequality...\r\n\r\n[hide=\"Solution\"]\nBy AM-GM:\n$ \\sum_{cyclic}\\frac {a^3}{bc(b \\plus{} c)^3} \\geq \\frac{3abc}{(a\\plus{}b)(a\\plus{}c)(b\\plus{}c) \\sqrt[3]{a^2b^2c^2}}$\n\nAgain by AM-GM, we have:\n$ 1\\equal{}\\frac{ab\\plus{}ac\\plus{}bc}{3}\\geq \\sqrt[3]{a^2b^2c^2}$\n\nThus,\n$ \\frac{3abc}{(a\\plus{}b)(a\\plus{}c)(b\\plus{}c) \\sqrt[3]{a^2b^2c^2}} \\geq \\frac{3abc}{(a\\plus{}b)(a\\plus{}c)(b\\plus{}c)}$\n\nClearly, we must prove that $ \\frac{abc}{(a\\plus{}b)(a\\plus{}c)(b\\plus{}c)}\\geq \\frac{1}{8}$, which is shown, yet again, through AM-GM...\n\n$ (a\\plus{}b)(a\\plus{}c)(b\\plus{}c)\\equal{}a^2b\\plus{}a^2c\\plus{}b^2a\\plus{}b^2c\\plus{}c^2a\\plus{}c^2b\\plus{}abc\\plus{}abc \\geq 8abc$.\n\nThat completes the proof...[/hide]\r\n\r\nI think this is my first post in the Pre-Olympiad Section :)", "Solution_5": "[quote=\"krsattack\"]I would just do the problem with the AM-GM inequality...\n\n[hide=\"Solution\"]\nBy AM-GM:\n$ \\sum_{cyclic}\\frac {a^3}{bc(b \\plus{} c)^3} \\geq \\frac {3abc}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c) \\sqrt [3]{a^2b^2c^2}}$\n\nAgain by AM-GM, we have:\n$ 1 \\equal{} \\frac {ab \\plus{} ac \\plus{} bc}{3}\\geq \\sqrt [3]{a^2b^2c^2}$\n\nThus,\n$ \\frac {3abc}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c) \\sqrt [3]{a^2b^2c^2}} \\geq \\frac {3abc}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c)}$\n\nClearly, we must prove that $ \\frac {abc}{(a \\plus{} b)(a \\plus{} c)(b \\plus{} c)}\\geq \\frac {1}{8}$, which is shown, yet again, through AM-GM...\n\n$ (a \\plus{} b)(a \\plus{} c)(b \\plus{} c) \\equal{} a^2b \\plus{} a^2c \\plus{} b^2a \\plus{} b^2c \\plus{} c^2a \\plus{} c^2b \\plus{} abc \\plus{} abc \\geq 8abc$.\n\nThat completes the proof...[/hide]\n\nI think this is my first post in the Pre-Olympiad Section :)[/quote]\r\nbut $ \\frac{abc}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)} \\leq \\frac{1}{8}$ :wink:", "Solution_6": "Oh... woops :oops:", "Solution_7": "[quote=\"apollo\"]Let $ a,b,c \\in R^ \\plus{}$ and $ ab \\plus{} bc \\plus{} ca \\equal{} 3$\nProve that\n\\[ \\sum_{cyclic}\\frac {a^3}{bc(b \\plus{} c)^3} \\geq \\frac {3}{8}\n\\]\n[/quote]\r\nI'm surprised that this way works :) \r\n\r\nby AM-GM we have\r\n$ \\frac {a^3}{bc(b \\plus{} c)^3} \\plus{} \\frac {b \\plus{} c}{16abc} \\plus{} \\frac {b \\plus{} c}{16abc} \\plus{} \\frac {b \\plus{} c}{16abc} \\geq \\frac {1}{2bc}$\r\ncreating 2 other similar inequality, we have :\r\n$ LHS \\plus{} \\frac {3}{8}(\\frac {1}{ab} \\plus{} \\frac {1}{bc} \\plus{} \\frac {1}{ac}) \\geq \\frac {1}{2}(\\frac {1}{ab} \\plus{} \\frac {1}{bc} \\plus{} \\frac {1}{ac})$ , so that $ LHS \\geq \\frac{1}{8}(\\frac {1}{ab} \\plus{} \\frac {1}{bc} \\plus{} \\frac {1}{ac})$\r\nand by Cauchy-S we have $ \\frac {1}{8}(\\frac {1}{ab} \\plus{} \\frac {1}{bc} \\plus{} \\frac {1}{ac}) \\geq \\frac {1}{8}\\frac {9}{ab \\plus{} bc \\plus{} ac} \\equal{} \\frac {3}{8}$", "Solution_8": "[quote=\"HTA\"][quote=\"apollo\"]Let $ a,b,c \\in R^ \\plus{}$ and $ ab \\plus{} bc \\plus{} ca \\equal{} 3$\nProve that\n\\[ \\sum_{cyclic}\\frac {a^3}{bc(b \\plus{} c)^3} \\geq \\frac {3}{8}\n\\]\n[/quote]\nI'm surprised that this way works :) \n\nby AM-GM we have\n$ \\frac {a^3}{bc(b \\plus{} c)^3} \\plus{} \\frac {b \\plus{} c}{16abc} \\plus{} \\frac {b \\plus{} c}{16abc} \\plus{} \\frac {b \\plus{} c}{16abc} \\geq \\frac {1}{2bc}$\ncreating 2 other similar inequality, we have :\n$ LHS \\plus{} \\frac {3}{8}(\\frac {1}{ab} \\plus{} \\frac {1}{bc} \\plus{} \\frac {1}{ac}) \\geq \\frac {1}{2}(\\frac {1}{ab} \\plus{} \\frac {1}{bc} \\plus{} \\frac {1}{ac})$ , so that $ LHS \\geq \\frac {1}{8}(\\frac {1}{ab} \\plus{} \\frac {1}{bc} \\plus{} \\frac {1}{ac})$\nand by Cauchy-S we have $ \\frac {1}{8}(\\frac {1}{ab} \\plus{} \\frac {1}{bc} \\plus{} \\frac {1}{ac}) \\geq \\frac {1}{8}\\frac {9}{ab \\plus{} bc \\plus{} ac} \\equal{} \\frac {3}{8}$[/quote]\r\nNice solution :coolspeak: \r\nThere is another Nice solution with Holder's Inequality.", "Solution_9": "Well using Holder was a little more suggested by the problem than AM-GM\r\n\r\n[hide]\n\\[ (1\\plus{}1\\plus{}1)(bc\\plus{}ca\\plus{}ab)\\left( \\sum{\\frac{a^3}{bc(b\\plus{}c)^3}}\\right) \\ge \\left (\\sum{\\frac{a}{b\\plus{}c}}\\right) ^3\\ge \\frac{27}{8}\\]\n[/hide]\r\n\r\nNice inequality by the way\r\n:)" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "here\r\n\r\n[img]http://www.geocities.com/zettz1/alg.jpg[/img]\r\n\r\nthanks...", "Solution_1": "BLANK = prime. Now try to prove it.\r\n\r\nFor the first part, assume that an algebraic $ \\alpha$ is a root of $ a_n x^n \\plus{} ... \\plus{} a_0$ for rational $ a_i$. Multiply this with a common denominator of all $ a_i$, thus we may assume that all $ a_i$ are integers. Then multiply again, this time by $ a_n^{n\\minus{}1}$ and show that $ a_n\\alpha$ is an algebraic integer.", "Solution_2": "Suppose a=bc\r\ni\r\nthen N(a)= N(bc)= N(b) N(c)= prime \r\n\r\nwhich implies that N(b) or N(c) is a 1 which implies that b or c is a unit which implies that a is irreducible.\r\n\r\ndoes that work?", "Solution_3": "Yes :)" } { "Tag": [], "Problem": "Let $f(n)$ be the number of lattice points in the region $2|x|+|y|\\le n$. Find the formula for $f(n)$.", "Solution_1": "my answer...\r\n[hide]\nAssuming that the lattice points on the borders count, $f(n)=n^2+n+1$[/hide]", "Solution_2": "That's correct ;) Post your solutoin.", "Solution_3": "We are to consider $f(n+1)-f(n)$.", "Solution_4": "[quote=\"kunny\"]We are to consider $ f(n+1)-f(n) $.[/quote]\r\nYes, that's the correct approach.\r\nDoes anyone have a complete solution to share?" } { "Tag": [ "algebra", "polynomial", "calculus", "Russia" ], "Problem": "The polynomials $ P(x)$ and $ Q(x)$ are given. It is known that for a certain polynomial $ R(x, y)$ the identity $ P(x) \\minus{} P(y) \\equal{} R(x, y) (Q(x) \\minus{} Q(y))$ applies. Prove that there is a polynomial $ S(x)$ so that $ P(x) \\equal{} S(Q(x)) \\quad \\forall x.$", "Solution_1": "I solve it firstly in case q has no double roots.\r\n(1) Let p(x)=q(x)c(x)+r(x).\r\nThen writing the equation for z[1]..z[deg[q]] solutions of q(x)=0 we get\r\nr(x) is constant.\r\nPerforming (1) for c(x) and so on, we get the desired result\r\nSo suppose q has double root.\r\nNow we substitute q(x)=q'(x)+c to have q'(x) no double roots.\r\nThe condition follows/\r\nBut I can't prove at the moment such q' exists ( though it seems obvious)", "Solution_2": "http://www.bath.ac.uk/~masgcs/ukimo2004/cc2sol.pdf here is a solution", "Solution_3": "Is it really true that for any polynomial Q(x) there is some c such that Q(x)+c doesn't have double roots?\r\nIt sounds natural but how to prove it? (Look above at iura's solution)", "Solution_4": "Of course it's true: every double root is also a root of the derivative. Since the derivative remains unchanged if you add a constant $c$ to the polynomial, simply add a constant $c$ s.t. $P(x)+c$ is non-zero in all the roots of the derivative $P'$.", "Solution_5": "Does this work?\nLet $P(x)=Q(x)T(x)+U(x)$, where $\\deg U < \\deg Q$.\nThen, $Q(x)-Q(y) | Q(x)T(x)+U(x)-Q(y)T(y)-U(y)$ i.e. $Q(x)-Q(y) | Q(y)(T(x)-T(y))+U(x)-U(y)$. Let $r$ be a root of $Q$. Substitute $y=r$, implying that $Q(x)-Q(r) | Q(r)(T(x)-T(r))+U(x)-U(r)$ i.e. $Q(x) | U(x)-U(r)$. However, $\\deg U < \\deg Q$, which means that $U$ is a constant. \nThus, $Q(x)-Q(y) | Q(y)(T(x)-T(y))$. However, since $\\gcd (Q(x)-Q(y), Q(y)) = \\gcd(Q(x), Q(y)) = 1$, $Q(x)-Q(y) | (T(x)-T(y))$, which means that we are back where we started, so we can repeat our process until $T=$ constant, in which case we are done. ", "Solution_6": "Let $P(x)=\\sum_{i=0}^np_ix^i$ ,$Q(x)=\\sum_{i=0}^mq_ix^i$, W.L.O.G $p_n,q_n>0$. Let $\\omega$ be a primitive $m-th$ root of unity. Substitute $y=\\omega x$. The degree of $Q(\\omega x)-Q(x)$ is strictly less than $m$, therefore the degree of $P(\\omega x)-P(x)$ is less than $n$ ( $R(x,\\omega x)$ may be the zero polynomial ), therefore $p_n \\omega^n -p_n=0$, which implies $m|n$. Let $P_1(x)=P(x)-\\frac{p_n}{q_n^{n/m}}(Q(x))^{n/m}$, $R_1(x,y)=\\left(R(x,y)-\\frac{p_n}{q_n^{n/m}}\\frac{(Q(x))^{n/m}-(Q(y))^{n/m}}{Q(x)-Q(y)}\\right)$. We have $P_1(x)-P_1(y)=R_1(x,y)(Q(x)-Q(y))$ and $\\deg(P_1)<\\deg(P)$ and hence by induction hypothesis we are done !\n\n", "Solution_7": "Does anyone know if my solution to this problem is correct? Thank you so much for all your help!", "Solution_8": "Looks correct to me.", "Solution_9": "Cool, thank you very much for the confirmation, r3mark! I really appreciate it!", "Solution_10": "We have $Q(x)-Q(y) \\mid P(x)-P(y)$. Use the Division Algorithm to write $P(x)=A(x)Q(x) + B(x)$ for some $A,B\\in \\mathbb{R}[x]$ with $\\deg B < \\deg Q$. We have\n\\[ Q(x)-Q(y) \\mid \\big[A(x)Q(x)-A(y)Q(y) \\big]+\\big[B(x)-B(y)\\big] \\qquad (\\clubsuit). \\]\nLet $\\{r_1,\\ldots,r_k\\}$ be the set of distinct roots of $Q$, and let $Q(x) = (x-r_1)^{a_1}\\cdots (x-r_k)^{a_k}$ for some $(a_i) \\in \\mathbb{Z}_{\\ge 1}$. If $r$ and $s$ are roots of $Q$, then plugging in $x=r$ and $y=s$ above implies $B(r)=B(s)$. Hence $B(r_1)=\\cdots=B(r_k)$. Therefore, \n\\[ B(x)-c = (x-r_1)^{b_1}\\cdots (x-r_k)^{b_k} \\qquad (\\spadesuit)\\]\nfor some constant $c\\in \\mathbb{R}$ and for some $(b_i)\\in \\mathbb{Z}_{\\ge 1}$. \n-----------------------\nFrom $(\\clubsuit)$, we get\n\\[ Q(x)-Q(y) \\mid \\big[A(x)-A(y) \\big]Q(x) +\\big[B(x)-B(y)\\big]. \\]\nWe know $x-y\\mid P(x)-P(y)$ for any polynomial $P$. Hence we can write -- as a polynomial statement -- the following:\n\\[ \\frac{Q(x)-Q(y)}{x-y} \\Biggr| \\left[ \\frac{A(x)-A(y)}{x-y} \\right ]Q(x) +\\left[\\frac{B(x)-B(y)}{x-y}\\right]. \\]\nTaking the limit as $y$ approaches $x$, the above implies \n\\[ Q'(x) \\mid A'(x)Q(x) + B'(x) \\implies A'(x)Q(x)+B'(x) = Q'(x)S(x,y) \\qquad (\\heartsuit)\\]\nfor some $S(x,y) \\in \\mathbb{R}[x,y]$. \n----------------------\n[color=#00f][b]Lemma: [/b][/color]If $Q$ has a root $r$ with multiplicity $n$, then $r$ is also a root of $B$, and it has multiplicity at least $n$. \n\n[color=#00f][i]Proof:[/i][/color] Replace $n$ with $n+1$. We know $Q(r)=0$ and $Q'(r)=0$, and so on till $Q^{(n)}(r)=0$. Taking $n-1$ derivatives of $(\\heartsuit)$, every term on the LHS (except $B^{(n)}(x)$) and every term on the RHS will have a term $Q^{(m)}(x)$ for some $m\\in \\{0,\\ldots,n\\}$, and is hence $0$. Therefore, $B^{(n)}(x)=0$, and also by the same logic $B^{(m)}(x)=0$ for any $m \\le n$. Therefore, $B$ has a root at $r$ with multiplicity $n$ as well. $\\blacksquare$\n\n[color=#00f][b]Claim: [/b][/color] $B(x)$ is a constant polynomial. \n\n[color=#00f][i]Proof:[/i][/color] Suppose $Q$ has a root $r_i$ with multiplicity $a_i\\ge 2$. By the Lemma, so does $b$, so $(x-r_i)^{a_i} \\mid B(x)$. Hence $B'(x)$ has a root at $r_i$ with multiplicity at least $a_i-1$. By $(\\spadesuit)$, $B'(x)$ has a root at $r_i$ with multiplicity $b_i-1$. Therefore, $b_i-1\\ge a_i-1$, so $b_i\\ge a_i$. Summing over all $i$, we get $\\deg B \\ge \\deg Q$, contradiction. Therefore, we must have $B(x)=c$ in fact, since the $0$ polynomial has any number as a root, and any root has infinite multiplicity. $\\blacksquare$\n--------------------\nNow we are almost done. We claim $A(x)$ also works in the place of $P(x)$ in the original statement. Indeed, since $P(x)=A(x)Q(x)+c$, and $Q(x)-Q(y) \\mid P(x)-P(y)$, we have $Q(x)-Q(y) \\mid A(x)Q(x)-A(y)Q(y)$, which implies $Q(x)-Q(y) \\mid [A(x)-A(y)]Q(x)$. But $\\gcd(Q(x),Q(x)-Q(y)) = \\gcd(Q(x),Q(y))=1$ since they are polynomials purely in different variables. Therefore, $Q(x)-Q(y)\\mid A(x)-A(y)$, as claimed. Now we can continue the process, and backtracking once the process is finished, we get that $P(x)$ is a polynomial in $Q(x)$ (the process essentially builds up the polynomial in $Q$, increasing the degree by $1$ at each step). \n\n[hide=Remarks]\nThe main ideas and motivation of the solution are:\n[list]\n[*] Writing $P=AQ+B$ using Division Algorithm. This is motivated since the final result of the problem is that we want to show $P$ is a polynomial in $Q$, so we want to ``build up'' $P$ in terms of $Q$, motivating using Division Algorithm. We want to show $B$ is a constant. \n[*] After manipulating the above into $(\\clubsuit)$, the idea is to plug in $r$ and $s$ if they are two roots of $Q$ -- this gets rid of everything except $B(r)-B(s)$, so it is also $0$. From here, we have that the set of roots of $B$ (translated by a constant) and $Q$ is the same. This is a pretty key step, and a trick the problem uses a lot -- vanish a lot of stuff so the remaining stuff is also $0$.\n[*] Now that the set of roots is the same, the hard part of the problem is dealing with multiplicity. The best way to deal with multiplicity in general is through derivatives. We get $Q' \\mid A'Q+B'$ after some trickery. We need to use the fact that $\\deg Q > \\deg B$, and using $(\\heartsuit)$, we can prove that the multiplicities of the roots of $B$ and $Q$ are the same; not just that they have the same set of roots. Now we are basically done. \n[/list] \n[/hide]", "Solution_11": "Here is a solution based on looking at coefficients. Please let me know if it doesn't work.\n\nWe induct on $\\deg P$, keeping $Q$ constant. WLOG suppose that both $P$ and $Q$ are monic.\nSuppose that $\\deg Q=d$. The main claim is that $d \\mid \\deg P$. Suppose the terms of $R(x,y)$ with maximum degree are $r(x,y)$. Then the maximum degree terms of $R(x,y)(Q(x)-Q(y))$ are $(x^d-y^d)r(x,y)$, which should also be $x^r-y^r$ for some $r$ by considering $P(x)-P(y)$. Evidently there must be an $x^k$ term in $r(x,y)$, else $y \\mid (x^d-y^d)r(x,y)$ in the polynomial sense\u2014which is untrue. Then there is a $-x^ky^d$ term in the (unsimplified) representation of $P(x)-P(y)$, but this is illegal so it must be cancelled out by an $x^{k-d}y^d$ term in $r(x,y)$ (since $x^d(x^{k-d}y^d)-y^d(x^k)=0$). Then there is a $-x^{k-d}y^{2d}$ term in $P(x)-P(y)$, so it must be cancelled out by an $x^{k-2d}y^{2d}$ term in $r(x,y)$, and so on. This process only terminates when $k-di=0$; for example, if $k-di=d-1>0$ at some point, then we still need an $x^{k-d(i+1)}y^{d(i+1)}$ term in $r(x,y)$, which is illegal. This readily implies that $d \\mid k$. Since $\\deg P = d+k$, our claim is true.\nWe are now ready to induct. Suppose that some (monic) $P_0$ exists with $\\deg P_0=ad$ and $P_0(x)-P_0(y)=R_0(x,y)(Q(x)-Q(y))$. Let $P(x)=P_0(x)-Q(x)^a$, which has degree at most $ad-1<\\deg P_0$. It is clear that $Q(x)-Q(y) \\mid Q(x)^a-Q(y)^a$, so $P(x)-P(y)=R(x,y)(Q(x)-Q(y))$ for some suitable choice of $R$ as well, completing the inductive step.\nIt remains to take care of the base case of $\\deg P = 0$, but this is trivial; thus we are done. $\\blacksquare$", "Solution_12": "Firstly , note that for each arbitrary complex number $\\alpha \\in \\mathbb{C}$ we have $Q(x)-Q(\\alpha)|P(x)-P(\\alpha)$ and for each two polynomials with this property , we'll show that we have $deg(Q)|deg(P)$. So suppose that for polynomial $P_{0}(x) \\in \\mathbb{C}[X]$ we have :\n$$P(x)=P_{0}(x)(Q(x)-Q(\\alpha))+P(\\alpha)$$\nNow we'll show that for each number $\\beta \\in \\mathbb{C}$ that $Q(\\beta) \\neq Q(\\alpha)$ , we have $Q(x)-Q(\\beta)|P_{0}(x)-P_{0}(\\beta)$ and thus , one can see that :\n$$Q(x)-Q(\\beta)|P(x)-P(\\beta)=P_{0}(x)(Q(x)-Q(\\alpha))+P(\\alpha)-P(\\beta) \\implies P_{0}(x)(Q(x)-Q(\\alpha)) \\equiv P_{0}(\\beta)(Q(x)-Q(\\alpha)) \\pmod{Q(x)-Q(\\beta)}$$\nSo while polynomials $Q(x)-Q(\\alpha)$ and $Q(x)-Q(\\beta)$ are coprime , as the result we can get :\n$$P_{0}(x) \\equiv P_{0}(\\beta) \\pmod {Q(x)-Q(\\beta)}$$\nNote that the equality $deg(P_{0})=deg(P)-deg(Q)$ holds , then by defining polynomials $P_1(x) , P_2(x) , ...$ in the same way , for a number $k \\in \\mathbb{N}$ we have $deg(P_k) < deg(Q)$ and while for a number $\\omega \\in \\mathbb{C}$ we have $Q(x)-Q(\\omega)|P_k(x)-P_k(\\omega)$ , thus polynomial $P_k(x)$ is constant which implies $deg(Q)|deg(P)$ obviously. \n\nNow suppose that we have $P(x)=a_{nk}x^{nk}+...+a_1x+a_0$ and $Q(x)=b_nx^n+...+b_1x+b_0$. Then we define polynomial $P_0(x)$ such as below : \n$$P_0(x)=P(x)-\\frac{a_nk}{b_{n}^{k}}Q(x)^{k}$$\nAnd according to this defining , note that for each number $\\alpha \\in \\mathbb{C}$ , polynomial $Q(x)-Q(\\alpha)$ divides $P_0(x)-P_0(\\alpha)$ too , because one can see that :\n$$P(x)-P(\\alpha) \\equiv \\frac{a_{nk}}{b_{n}^{k}}(Q(x)^k-Q(\\alpha)^k) \\equiv 0 \\pmod {Q(x)-Q(\\alpha)}$$\nSo we have $deg(Q)|deg(P_0)$ and by defining polynomials $P_1 , P_2 , ...$ in the same way , while $deg(P_{t+1}) < deg(P_t)$ there exist a number $t \\in \\mathbb{N}$ such that $P_t(x) \\equiv 0$ and for some complex numbers $c_k , ... , c_0 \\in \\mathbb{C}$ one can see that :\n$$P_t(x)=P(x)-c_kQ(x)^k-c_{k-1}Q(x)^{k-1}-...-c_0=0 \\implies S(x)=c_kx^k+c_{k-1}x^{k-1}+...+c_0 , P(x)=S(Q(x))$$\nSo we're done. ", "Solution_13": "Solved with [b]everythingpi3141592[/b]\nBy the divisibility condition we have that when $Q(x) = Q(y) \\implies P(x) = P(y)$, which essentially helps us prove that all turning points of $Q(x)$ are of $P(x)$, now if we construct a polynomial $S(x)$ such that when $P'(x) = 0 \\implies S'(Q(x)) = 0$ (that is the turning points of $P(x)$ which aren't of $Q(x)$ are of $S(Q(x))$), which is just a matter of langrage interpolation, so we're done.\n\nThis tells us so because like from here we get that $P'(x) = k \\cdot S'(Q(x)) \\cdot Q'(x)$, where k is just a constant, now note that if some constant was vanished in taking the derivative of $P(x)$, we can just add that up to $S(x)$ accordingly.", "Solution_14": "We use Lagrange interpolation polynomial to do this.\n\nFirst we take $ m $ points over function $ Q$ such that:$$Q(x_1)=y_1,Q(x_2)=y_2,\\cdots Q(x_m)=y_m.$$\nwhere $m>\\frac{\\deg P}{\\deg Q}$.\nThen by Lagrange interpolation polynomial we deduce that there exists a function $S:\\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n$$P(x_k)=S(Q(x_k))$$\nholds for $k=1,2,\\cdots m$.Now what we need to prove is that actually $P=S \\circ Q$.\nUsing the fact that $Q(x)-Q(y)|P(x)-P(y)$ and $x-y|f(x)-f(y)$ for any $f\\in R[x]$,then we have \n$$Q(x)-y_k \\bigg| P(x)-S(Q(x))$$\nholds for any $k=1,2,\\cdots m$.Then it is easy to get that $\\gcd (Q(x)-y_i,Q(x)-y_j)=1$ for $1\\leq i < j \\leq m $.Thus we get\n$$\\prod_{k=1}^{m}\\bigg( Q(x)-y_k \\bigg) \\bigg| P(x)-S(Q(x))$$.But look at the expotent we must have $\\deg P-S\\circ Q=0$,thus we get $P=S\\circ Q$ holds in $ \\mathbb{R}$.$\\blacksquare$", "Solution_15": "It suffice to show the problem with real replaced by complex, for if we get that to work, then $f(x)=P(g(x))$ for some $P\\in \\mathbb{C}[x]$. Now, taking conjugates imply that $f(x) = \\overline{P}(g(x)) = P(g(x))$ (since $g\\in \\mathbb{R}[x]$), which implies $\\overline{P}=P$, so $P\\in \\mathbb{R}[x]$.\n\nFix $g$; we induct on $\\deg f$. The base case, $f\\equiv 0$, is clear.\n\nIt suffices to show that $\\deg g \\mid \\deg f$, because if that is the case, say $\\deg f = k\\deg g$, then we can apply inductive hypothesis on $f(x)-cg(x)^k$ where $\\deg(f(x)-cg(x)^k) < \\deg (f(x))$.\n\nLet $n=\\deg g$. The main intuition is that if $x\\in \\mathbb{R}$ is a large real, then there is a root around $x\\omega_n$, where $\\omega_n=e^{\\frac{2\\pi i}{n}}.$\n\nLet's formalize that intuition. Let $h(x)=g(x+b)$ for some constant $b$ such that $[x^{n-1}]h(x)=0$. Write $$h(x) = h_nx^n + h_{n-2}x^{n-2} + \\cdots + h_0$$\n\nNow we can see that \n\n$$|h(x\\omega_n) - h(x)| = O(x^{n-2}) $$\n\nLet $z$ be a unit complex number. Then \n\n$$\\frac{\\partial h}{\\partial z}(x) = \\lim_{\\substack{d\\to 0^+ \\\\ d\\in \\mathbb{R}^+}} \\frac{h(x+dz)-h(x)}{d}= zh'(x)$$\n\nWe use a Newton-style approximation method to find $x'$ such that $|x\\omega_n-x'| = O(x^{-1})$ and $h(x')=h(x)$. Consider a sequence of points $x_0,x_1,\\cdots$ defined by $x_0=x\\omega_n$ and\n\n$$x_k-x_{k-1} = \\frac{h(x) - h(x_{k-1})}{h'(x)}$$\n\nWe induct on $k$ to show that $x_k-x_{k-1} = O(x^{-k})$. For the base case $k=1$, check that \n\n$$h(x_1) = h\\left(x_0 + \\underbrace{\\frac{h(x)-h(x_0)}{h'(x)}}_{O(x^{-1})}\\right)=h(x_0)+nx_0^{n-1} \\frac{h(x)-h(x_0)}{h'(x)} + O(x^{n-3})$$\n\nThus, $$h(x)-h(x_1) = (h(x)-h(x_0))\\left( 1-\\frac{nx_0^{n-1}}{h'(x)}\\right) = (h(x)-h(x_0)) O(1/x)$$\n\nInductive step is similar: keep in mind that $x_{k-1}=x\\omega_n + O(1/x)$ is a consequence of inductive hypothesis.\n\n$$h(x_k) = h\\left(x_{k-1} + \\underbrace{\\frac{h(x)-h(x_{k-1})}{h'(x)}}_{O(x^{-k})}\\right)=h(x_{k-1})+nx_{k-1}^{n-1} \\frac{h(x)-h(x_{k-1})}{h'(x)} + O(x^{n-k-2})$$\n\nThus, $$h(x)-h(x_k) = (h(x)-h(x_{k-1}))\\left( 1-\\frac{nx_{k-1}^{n-1}}{h'(x)}\\right) = (h(x)-h(x_{k-1})) O(1/x)$$\n\nSince the sequence of points $x_0,\\cdots,$ is Cauchy and $\\mathbb{C}$ is a complete metric space, it must converge to $x'$. We can see that $h(x)=h(x')$. This means that for any large $x$, there exists a constant $a$ such that $g(x) = g(x\\omega_n + a + O(1/x))$. This means that $f(x) = f(x\\omega_n + a + O(1/x)) \\rightarrow n\\mid \\deg f$", "Solution_16": "If $Q(x)$ is constant then $P(x)$ is constant, so assume $\\deg Q(x)\\ge 1$. Let us fix $Q(x)$. We prove the problem statement by strong induction on $\\deg(P(x))$. We can shift $P(x),Q(x)$ by a constant, hence we can assume that $P(0)=Q(0)=0$. Substitute $y=0$, implying that $Q(x)\\mid P(x)$. Hence $P(x)=T(x)Q(x)$ for a polynomial $T(x)\\in\\mathbb R(x)$. Hence\n$$Q(x)-Q(y)\\mid T(x)Q(x)-T(y)Q(y)$$\nso\n\\begin{align*}\nQ(x)-Q(y)&\\mid T(x)Q(x)-T(y)Q(y)+T(y)(Q(x)-Q(y))\\\\\n&=(T(x)-T(y))Q(x).\n\\end{align*}\nNow \n$$\\gcd(Q(x)-Q(y),Q(x))=\\gcd(Q(x),Q(y))=1$$\nas these are polynomials of different variables, so $Q(x)-Q(y)\\mid T(x)-T(y)$. Since $\\deg T(x)<\\deg P(x)$, we are done by induction hypothesis.", "Solution_17": "Here is another idea. If $Q(x)$ is constant, then $P(x)$ is constant as well, so assume $Q(x)$ is nonconstant. Again, we fix $Q(x)$ and induct on $\\deg P(x)$. WLOG assume $P(x)$ and $Q(x)$ are monic. Assume $\\deg Q(x)=m$ and $\\deg P(x)=n$. By taking terms of makimal degree we see that $x^n-y^n=r(x,y)(x^m-y^m)$ for a polynomial $r(x,y)\\in\\mathbb R[x,y]$. Substituting $y=1$ implies that $x^m-1\\mid x^n-1$ in $\\mathbb R[x]$. It is well-known that $\\gcd(x^m-1,x^n-1)=x^{\\gcd(m,n)}-1$, hence $\\gcd(m,n)=m$ so $m\\mid n$. Now let $d\\doteqdot\\frac{n}{m}$. Since $Q(x)-Q(y)\\mid Q(x)^d-Q(y)^d$, we can replace $P(x)$ by $P(x)-Q(x)^d$ and continue by induction hypothesis.", "Solution_18": "We will strong induct on $\\deg P$. Note obviously $\\deg Q < \\deg P$.\n\nNote that we can shift $P$ and $Q$ by any constant and the condition will still be satisfied; thus, set $P(x) = Q(x)R(x)$ for some polynomial $R$. Then $$Q(x)-Q(y) \\mid (Q(x)R(x) - Q(y)R(y)) - (Q(x)R(x) - Q(y)R(x)) = Q(y)(R(x) - R(y))$$ in $\\mathbb R[x, y]$. Note that as $Q$ is nonconstant, $\\gcd(Q(x) - Q(y), Q(y)) = 1$, hence $$Q(x) - Q(y) \\mid R(x) - R(y).$$ But by the inductive hypothesis, as $\\deg R < \\deg P$, we have $R(x) = S(Q(x))$ for some $S$. Then $$P(x) = S(Q(x))Q(x) = (S+1)(Q(x)).$$", "Solution_19": "blud what is this question\n\nWrite \n\\[P=A_dQ^d+\\dots+A_0Q^0\\]\nfor polynomials $A_0\\dots,A_d$ with degree less than $\\text{deg }Q$. Our aim is to show that all $A_i$ are constant. Define $\\equiv$ for polynomials in the obvious way (dunno if this is legit)\nNotice that \n\\[\\sum A_i(y)Q(y)^i=P(y)\\equiv P(x)=\\sum A_i(x)Q(x)^i\\equiv \\sum A_i(x)Q(y)^i\\pmod {Q(x)-Q(y)}.\\]\nHence $Q(x)-Q(y)$ divides\n\\[\\sum (A_i(x)-A_i(y))Q(y)^i\\]\nSince the maximal degree of $x$ in this is $\\text{deg }Q-1$ and yet $Q(x)-Q(y)$ has a maximal degree in $x$ of $\\text{deg }Q$ it follows that the entire expression is equal to $0$. Now we just induct from the top down. FTSOC suppose at some point $A_i$ isn't constant. Consider the maximal $m$ where this holds, i.e.,\n\\[A_m(x)-A_m(y)\\ne 0.\\]\nClearly then the degree of $y$ in the original expression at least $m\\cdot \\text{deg }Q$. But for all $n= n^4 si prin urmare n <= 2 si am incercat cazurile...\r\n\r\n10x anyway...\r\n\r\n\r\nas mai avea oproblema care mi-a dat bataie de cap ziua asta...si ma voi mai gandi la ea dar cred ca oricum am nevoie de ajutor :blush: :\r\nsa se demonstreze ca ecuatia x 2 +y 2 +z 2 +t 2 =2^2004,\r\nunde 0 <= x <= y <= z <= t,are exact doua solutii in multimea nr intregi :blush: sper sa nu ma fac de ras\r\n\r\n\r\nmerci" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all the functions $f$ such that $f^2(x)\\cdot f(\\frac{1-x}{1+x})=64x$ \r\n\r\nfor each $x\\in R-\\left\\{ 0,-1,1 \\right\\}$", "Solution_1": "Substitute $x$ by $\\frac{1-x}{1+x}$ then:\r\n$f(x).f^2(\\frac{1-x}{1+x})=64\\frac{1-x}{1+x}$ with the original equation we have:\r\n$f^3(x)=64x^2\\frac{1-x}{1+x}$\r\n$f(x)=4\\sqrt[3]{x^2\\frac{1-x}{1+x}}$\r\nEasy to verify its indeed solution.", "Solution_2": "Yes AYMANE nice solution :)" } { "Tag": [], "Problem": "Order the numbers 1, 2, ... , n any way you want. Add up the differences between successive numbers, with all differences counted positively: for example, if you start with the ordering 5, 3, 1, 4, 2, you get 2 + 2 + 3 + 2 = 9. What is the largest number you can obtain in this way?", "Solution_1": "Not sure if this is right but if you order them like this\r\n\r\n$n, 1, n-1, 2, n-2, 3...$ then the sum is\r\n$\\frac{n(n-1)}{2}$", "Solution_2": "Well thats not right here is something that gives a higher sum if you order them like this\r\n\r\n$2,n,1,n-1,3,n-2...$ then the sum is\r\n$\\frac{n(n-1)}{2}+1$...hmm is there something higher??" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c,d,e,f $ be positive real numbers. \r\n\r\n\r\nProve that: \r\n\r\n\r\n(a^3+d^3)(b^3+e^3)(c^3+ f^3 ) \\geq (abc+def)^3", "Solution_1": "Speaking about Holder and Huygens inequality...", "Solution_2": "just by brute force, we may take x = a/d, y = b/e, z = c/f, and obtain\r\nx+y+z + (xy)+(yz)+(zx) >= 3xyz + 3xyz, which is trivial.", "Solution_3": "Or just by Cauchy(Hey I mean AM-GM :D )." } { "Tag": [ "geometry", "trapezoid", "perimeter" ], "Problem": "Just like other forums, this will be a marathon thread that is restricted to ONLY PROBLEMS TAUGHT AT A REGULAR MIDDLE SCHOOL LEVEL! Please do not go overboard with problem difficulty. Thanks!\r\n\r\nSimply answer the problem given by the poster above you, showing all work. Then post a new problem yourself, so as to keep it going.\r\n\r\nI'll start:\r\n\r\nNew Problem:\r\n\r\nWhat is the area of a trapezoid with bases 4 and 6 and height 12?", "Solution_1": "$ A \\equal{} \\frac12 \\times b_1b_2 \\times h$\r\n\r\n$ A \\equal{} \\frac12 \\times 4 \\times 6 \\times 12$\r\n\r\n$ A \\equal{} 5 \\times 12$\r\n\r\n$ A \\equal{} \\boxed{60}$\r\n\r\n[b]Problem 2:[/b]\r\n\r\nFind the number of centimeters in the hypotenuse with legs 9 and 40.", "Solution_2": "Sum of squares of legs is 1600+81=1681, so hypotenuse is square root of 1681 or 41.\r\n\r\nNew problem:\r\n\r\nOlly is 12 years older than Izzy. Two years later, Olly will be twice as old as Izzy. Find their current ages.", "Solution_3": "For some reason, I think this is useless because there are only a few users who cannot solve questions like those, and half of them probably never come online. And, easy problems are too easy.\r\n\r\nOh, BTW the answer to your problem is 41.", "Solution_4": "[quote=\"D3m0n Shad0w\"]Please do not create any more marathons in Classroom Math - they're never going to last and it's really a waste of effort.[/quote]\r\n\r\nyeah...", "Solution_5": "[quote=\"gauss1181\"]Sum of squares of legs is 1600+81=1681, so hypotenuse is square root of 1681 or 41.\n\nNew problem:\n\nOlly is 12 years older than Izzy. Two years later, Olly will be twice as old as Izzy. Find their current ages.[/quote]\r\n\r\nolly is 22, izzy is 10.\r\n\r\nnew: find the area of a square with the same perimeter as the right triangle with legs 11 and 60.", "Solution_6": "waaayyy too hard for classroom math.", "Solution_7": "so evaluating 7*3 is waaayyy to hard for classroom math.\r\nin that case\r\nNEW PROBLEM\r\nevaluate: 3+7", "Solution_8": "[quote=\"qwertythecucumber\"][quote=\"D3m0n Shad0w\"]Please do not create any more marathons in Classroom Math - they're never going to last and it's really a waste of effort.[/quote]\n\nyeah...[/quote]\r\n\r\nDOT DOT DOT", "Solution_9": "$ 2^4\\equal{}\\boxed{16}$ ways.\r\n\r\nNew Problem:\r\n\r\nA truck travels 30 miles per hour. How far does it travel in 150 minutes?", "Solution_10": "[quote=\"AIME15\"][quote=\"qwertythecucumber\"][quote=\"D3m0n Shad0w\"][size=200][b]Please do not create any more marathons in Classroom Math - they're never going to last and it's really a waste of effort.[/b][/size][/quote]\n\nyeah...[/quote]\n\nDOT DOT DOT[/quote]\r\n\r\nIs anyone listening to these posts??", "Solution_11": "[quote=\"athunder\"]uhh define anyone... gauss1181 probably has and she is fine with... soooo we will make it continue :)\n\n\nLOL WHAT AM I SAYING wow... does anyone in aops really find this marathon to be useful? i would say its used to increase post count which is lame.[/quote]\n\nwhy not continue this marathon", "Solution_12": "Also, if you're going to start a marathon, use hide tags.\n\n[code] [hide=Problem # or Solution #] Text [/hide] [/code]\n\n[hide=\" Optional\"]\n\nTo box in your answers use\n\n[code] \\boxed {ANSWER} [/code]\n\n[/hide]", "Solution_13": "Please do not revive marathons after this amount of inactivity. There are many other marathons in this forum that are essentially the same as this one (except this covers all the MS classroom math subjects in one marathon). Locked." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "geometry", "probability", "AIME", "Putnam" ], "Problem": "Hello AoPS users. I am beginning a new game, and in it, people sign up, and the first person to sign up will be the contestant. Anybody else that signs up is a mob member. The contestant is going to go against the mob. The contestant's goal is to beat everybody else. The mob's goal is to beat the contestant. Basically, I will ask everybody questions. The questions will be on this post. You will have to pm me your answers. Anybody that answers the questions wrong will be eliminated. The contestant has life lines that he can use. I will tell them their lifelines later. \r\n\r\nAt the end of each competition there are two possibilities; \r\n\r\nA.) The contestant wins. \r\n\r\nIf this happens, then he can be the contestant has four choices. \r\n \r\n A.A.) He can be the contestant again. \r\n A.B.) He can choose somebody from the mob to be the next contestant. \r\n A.C.) He can leave it to me to decide randomly. \r\n A.D.) He can let the mob compete. \r\n\r\n\r\nB.) The mob wins. \r\n\r\nIf this happens then they will all have a vote. These are the three choices they have. \r\n\r\n A.A.) They can compete against each other. \r\n A.B.) They can let me pick who becomes the new contestant. \r\n A.C.) They vote on who becomes the new contestant. \r\n \r\n\r\nAnybody that chooses to compete after somebody has won will become a mob member. \r\n\r\nSign-ups begin now.", "Solution_1": "Hmmm, this looks like my game from months ago... I join.", "Solution_2": "ok\r\n\r\ni am in", "Solution_3": "umm... wait. math questions. right??\r\n\r\n><\r\ni'm in.", "Solution_4": "I'm in\r\n\r\nIf pythag011 joints I quit.", "Solution_5": "I'll join, but your games never end and take way tooo long.", "Solution_6": "im in\r\n\r\ni guess", "Solution_7": "Is lotsofmath the first contestant?\r\n\r\nI join.\r\n\r\nARe these math questions?", "Solution_8": "I'll join too.\r\n\r\nAnd I don't think they're math questions. If they're math questions this competition will take too long, since most people here are used to answering math questions [b]correctly[/b]. (like in the last 1vs100, it went on for a long time)", "Solution_9": "[quote=\"ProtestanT\"]I'll join too.\n\nAnd I don't think they're math questions. If they're math questions this competition will take too long, since most people here are used to answering math questions [b]correctly[/b]. (like in the last 1vs100, it went on for a long time)[/quote]\r\n\r\n*cough* add USAMO level questions *cough*", "Solution_10": "im in yayyaya", "Solution_11": "why not? /join.", "Solution_12": "idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join!", "Solution_13": "[quote=\"1=2\"]If pythag011 joints I quit.[/quote]\r\n\r\nI quit", "Solution_14": "I join. [b]Mewto55555 chooses to be first contestant[/b] Pwnt. I got there first.", "Solution_15": "wait, are you sure I answered?", "Solution_16": "Yes you did.", "Solution_17": "*yawn* drop out.", "Solution_18": "I'm positive that I sent you a PM with my answer.", "Solution_19": "Oh, yes sorry crazypianist answered the question, but pythag didn't. Only about 5 hours left to pm me. \r\n\r\nStill waiting on; \r\nmewto55555 \r\nBOGTRO \r\npythag011", "Solution_20": "Alright, mewto and BOGTRO are too late, and they are eliminated from this competition. \r\n\r\njames4l, you can either use one of your lifelines, or you can answer the question. If you would like to use a lifeline, then just say so, and I will show you the three lifelines.", "Solution_21": "I drop out.", "Solution_22": "james4l got the question correct, so now there are only five mob members left. Unfortunately, one of these five people got the question incorrect. \r\n\r\nIt was one of these five; \r\n\r\ncrazypianist116 \r\npythag011 \r\nazjps \r\nSMB \r\nnick42 \r\n\r\n[hide]pythag011 got the question correct \n\n[hide]crazypianist1116 got the question correct \n\n[hide]azjps got the question correct \n\nnick42 got the question correct \n\nSuperMathBoy was incorrect. [/hide][/hide][/hide]\r\n\r\n\r\njames4l has 75,000 up for grabs, so he can take the money, or try to earn more by reducing the mob.", "Solution_23": "Go for more (I won't settle for less than 300k!)", "Solution_24": "lol, if you want more than 300k, you'll have to narrow the mob down to two people.\r\n\r\nWhich topic do you pick; \r\n\r\nArithmetic \r\nAlgebra \r\nGeometry \r\nProbability", "Solution_25": "Probability.", "Solution_26": "Okay, the question will be posted tomorrow.", "Solution_27": "Here is the question; \r\n\r\nWhat is the probability that and integer between 1 and 1000 inclusive is divisible by 3, but not 4, and 5, but not 7? \r\n\r\njames4l, please wait to solve this problem until the mob has answered. \r\n\r\n\r\nBTW - The answer to the previous problem was 112, in case I haven't said already.", "Solution_28": "Do you mean, \"is a multiple of 3 and 5 but not a multiple or 4 and/or 7\" or \"is a multiple of 3 and 5 but not 4 and 7?\"", "Solution_29": "It is divisble by 3 and 5, but it is not divisible by 4 or 7." } { "Tag": [], "Problem": "tamame tavabe haghighi $f$ ra tori peida konid ke :\r\n$f(x^{2}+f(f(x)-y))=2f(f(x))+2y^{2}, x,y \\in \\mathbb R$", "Solution_1": "tamame tavabe haghighi $f$ ra tori biabid ke \r\n$(x+y)(f(x)-f(y))=(x-y)f(x+y) ,x,y \\in\\mathbb R$" } { "Tag": [ "geometry", "rectangle", "ratio", "similar triangles", "geometry unsolved" ], "Problem": "Let two circles $ (O; R)$ and $ (O; r)$ with $ R > r$. Let $ P$ is a fixed point on $ (O; r)$ and let $ B$ is a change point on $ (O;R)$. Let $ BP$ meet $ (O;R)$ at $ C$, let $ (d)$ is a line pass point $ P$ and $ (d) \\perp BP$, let $ (d)$ meet $ (O;r)$ at point $ A$. \r\n\r\na) Find $ BC^2\\plus{}CA^2\\plus{}AB^2$ by $ R$ and $ r$.\r\n\r\nb) Let $ I$ is midpoint of $ AB$. Find locus of point $ I$ ?", "Solution_1": "[quote=\"thanhnam2902\"]Let two circles $ (O; R)$ and $ (O; r)$ with $ R > r$. Let $ P$ is a fixed point on $ (O; r)$ and let $ B$ is a change point on $ (O;R)$. Let $ BP$ meet $ (O;R)$ at $ C$, let $ (d)$ is a line pass point $ P$ and $ (d) \\perp BP$, let $ (d)$ meet $ (O;r)$ at point $ A$. \n\na) Find $ BC^2 \\plus{} CA^2 \\plus{} AB^2$ by $ R$ and $ r$.\n\n[/quote]\r\n\r\n[hide=\"Solution a)\"] Let's circle $ (O;r)$ and line that pass through $ PB$ intersects second time in $ K$. Then we have : \n$ CK\\equal{}BP$, $ AK\\equal{}2r$\nDetermine $ CK\\equal{}BP\\equal{}a$, $ KP\\equal{}x$, $ AP\\equal{}b$.\n$ AB^2\\plus{}AC^2\\plus{}BC^2\\equal{}2AP^2\\plus{}BP^2\\plus{}PC^2\\plus{}BC^2\\equal{}2b^2\\plus{}a^2\\plus{}(a\\plus{}x)^2\\plus{}(2a\\plus{}x)^2\\equal{}2(b^2\\plus{}x^2)\\plus{}6a(a\\plus{}x)\\equal{}2\\cdot 4r^2\\plus{}6(R\\minus{}r)(R\\plus{}r)\\equal{}8r^2\\plus{}6R^2\\minus{}6r^2\\equal{}6R^2\\plus{}2r^2$[/hide]", "Solution_2": "[color=darkblue]Let $ M$ be foot of the perpendicular from $ O$ onto $ BC$ and let $ BC$ cuts $ (O,r)$ at $ T$. Then $ T$, $ O$ and $ A$ are collinear and $ OM$ is midline of the triangle $ TPA$. So $ OM \\equal{} \\dfrac{1}{2}PA$. Also note that $ M$ is midpoint of $ BC$. Denote $ \\{G\\} \\equal{} OP \\cap AM$. Then from similar triangles $ OMG$ and $ PGA$ we obtain that\n\n$ \\frac {MG}{GA} \\equal{} \\frac {OG}{GP} \\equal{} \\frac {OM}{AP} \\equal{} \\frac {1}{2}$\n\nThus $ G$ is centroid of the triangle $ BAC$ and also $ OG \\equal{} \\frac {1}{3}OP \\equal{} \\frac {1}{3}r$.\n\nNow apply Leibnitz theorem to the triangle $ BAC$ and point $ O$ and obtain\n\n$ AB^2 \\plus{} BC^2 \\plus{} CA^2 \\equal{} 3OA^2 \\plus{} 3OB^2 \\plus{} 3OC^2 \\minus{} 9OG^2 \\equal{} 3r^2 \\plus{} 6R^2 \\minus{} r^2 \\equal{} 2r^2 \\plus{} 6R^2$.\n\nFor $ b)$ part let us prove the value $ IQ$ is constant where $ Q$ is midpoint of $ OP$. Then, the required locus will be circle with center $ Q$ and radius $ QI$.\n\nBy median theorem we have\n\n$ 8QI^2 \\equal{} 4AQ^2 \\plus{} 4BQ^2 \\minus{} 2AB^2$\n$ 4BQ^2 \\equal{} 2BO^2 \\plus{} 2BP^2 \\minus{} PO^2 \\equal{} 2R^2 \\plus{} 2BP^2 \\minus{} r^2$\n$ 4AQ^2 \\equal{} 2AO^2 \\plus{} 2AP^2 \\minus{} PO^2 \\equal{} 2R^2 \\plus{} 2AP^2 \\minus{} r^2$\n\n$ 8QI^2 \\equal{} 4R^2 \\minus{} 4r^2 \\plus{} 2BP^2 \\plus{} 2AP^2 \\minus{} 2AB^2 \\equal{} 4R^2 \\minus{} 4r^2$[/color]\r\n\r\n[img]http://pic.ipicture.ru/uploads/090112/b7cU3tBq7d.png[/img]", "Solution_3": "[color=darkblue]Your solution is nice. I can find locus of point by use homothetic transformation.\n\n+) Easy we get $ PBMA$ is a rectangle $ \\Longrightarrow$ $ I$ is midpoint of PM.\n\n+) Hence we get $ \\overrightarrow {PI} \\equal{} \\frac{1}{2} \\overrightarrow {PM}$\n\n$ \\Longrightarrow$ point $ I$ be image of point $ M$ of homothetic center $ O$, ratio $ k\\equal{}\\frac{1}{2}$\n\nBut $ M \\in (O) \\Longrightarrow I \\in (O')$ with $ (O')$ be image of $ (O)$ of homothetic center $ O$, ratio $ k\\equal{}\\frac{1}{2}$.[/color]" } { "Tag": [ "group theory", "abstract algebra", "function", "LaTeX", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $G$ be a group with $m$ elements and let $H$ be a proper subgroup of $G$ with $n$ elements. For each $x\\in G$ we denote $H^x = \\{ xhx^{-1} \\mid h \\in H \\}$ and we suppose that $H^x \\cap H = \\{e\\}$, for all $x\\in G - H$ (where by $e$ we denoted the neutral element of the group $G$).\r\n\r\na) Prove that $H^x=H^y$ if and only if $x^{-1}y \\in H$;\r\nb) Find the number of elements of the set $\\bigcup_{x\\in G} H^x$ as a function of $m$ and $n$.\r\n\r\n[i]Calin Popescu[/i]", "Solution_1": "Consider the rel P on G by the following aPb iff a(b^-1) belongs to G.We can show that it is an equivalence relation.\r\nNow work with the equivalence classes generated by P.\r\nI think we can get H_x=H_y or id I do anything wrong. :?", "Solution_2": "a) $H^x=H^y\\Leftrightarrow xHx^{-1}=yHy^{-1}\\Leftrightarrow y^{-1}xHx^{-1}y=H\\Leftrightarrow H^{y^{-1}x}=H\\Leftrightarrow y^{-1}x\\in H$.\r\nThe last equivalence follows from the condition $H^x\\cap H=\\{e\\}$ for $x\\in G-H$.\r\n\r\nb) First we show that $H^x$ and $H^y$ can only have trivial intersection ($\\{e\\}$ or $H^x$). If $x^{-1}y\\in H$ we know with a) that $H^x=H^y$. Now consider the case $x^{-1}y\\not\\in H$. Then we cannot have $h_1,h_2\\in H\\backslash\\{e\\}$ with $xh_1x^{-1}=yh_2y^{-1}$ since otherwise $y^{-1}xh_1x^{-1}y=h_2$ and $H\\cap H^{x^{-1}y}\\neq\\{e\\}$ which is a contradiction to one of the conditions. Now we have exactly $[G:H]=\\frac{m}{n}$ different sets of the form $H^x$ with pairwise intersection $\\{e\\}$. In total we get \r\n$n+\\left(\\frac{m}{n}-1\\right)\\cdot(n-1)=\\frac{mn-m+n}{n}$\r\nelements in $\\cup_{x\\in G} H^x$.\r\n\r\nI hope I didn't miscalculate in the end of b)\r\n\r\nMisha", "Solution_3": "I dont think so because there is a problem in Herstein where one has to give an example of group G such that for a subgroup H of G :cup: H_x x belonging to G where ( :cup: H_x x belonging to G\r\nis defined in the same way)is not the whole of G.I have constructed a similar example.", "Solution_4": "I don't see a contradiction to my solution. :? \r\nCould you please explain what you mean (and use LATEX) ?\r\n\r\nMisha", "Solution_5": "[quote=\"Misha123\"]I don't see a contradiction to my solution. :? \nCould you please explain what you mean (and use LATEX) ?\n\nMisha[/quote]\r\nWhat contradiction!!!I said theres a similar problem in Herstein (the 2nd part) & that two I have done similarly.\r\nSorry I ll try to learn latex.I know its very clumsy& its equally difficult for me to express myself.. :(", "Solution_6": "All right! I just misunderstood you, sorry :blush: \r\n\r\n\r\nMisha", "Solution_7": "O never mind it happens all the time", "Solution_8": "(a) If $x^{-1}y\\in H$, then $\\{(y^{-1}x)h(x^{-1}y)\\mid h\\in H\\}=H$ (note that $(y^{-1}x)^{-1}=x^{-1}y$): for each $h'\\in H$, we can take $h:=(x^{-1}y)h'(y^{-1}x)$ which satisfies $(y^{-1}x)h(x^{-1}y)=h'$. It follows that $H^x=H^y$.\n\nNow assume $H^x=H^y$. Clearly either both $x$ and $y$ are in $H$ or none of them are in $H$. Since $|H^x|=|H^y|=n$, there is a permutation $\\sigma$ of $H$ such that $xhx^{-1}=y\\sigma(h)y^{-1}$ for all $h\\in H$. It follows that $H^{y^{-1}x}=\\{(y^{-1}x)h(x^{-1}y)\\mid h\\in H\\}=H$. But this can only happen when $y^{-1}x\\in H$, which implies $x^{-1}y\\in H$, as desired.\n\n(b) We claim that for all $x,y\\in G,x^{-1}y\\not\\in H$, we have $H^x\\cap H^y=\\{e\\}$. Assume otherwise. Then there exists $h_1,h_2\\in H$ with $h_1,h_2\\neq e$ such that $xh_1x^{-1}=yh_2y^{-1}$ so $(y^{-1}x)h_1(x^{-1}y)=h_2$. But now $H^{y^{-1}x}$ contains $h_2\\neq e$, a contradiction.\n\nIt follows that for all $x,y\\in G$, either $H^x=H^y$ or $H^x\\cap H^y=\\{e\\}$. For each $x\\in G$, the set $G_x:=\\{y\\in G\\mid H^x=H^y\\}$ has cardinality $n$ (there are exactly $n$ $y\\in G$ such that $H^x=H^y$: one for each element in $H$). Thus the $G_x$ define equivalence classes: $G_x\\cap G_y=\\{e\\}$ if and only if $y\\not\\in G_x$. There are $\\frac{m}{n}$ such equivalence classes so $\\left|\\bigcup_{x\\in G}H^x\\right|=1+\\frac{m}{n}(n-1)$. $\\square$" } { "Tag": [ "conics", "ellipse" ], "Problem": "If the major and minor axes of an ellipse are denoted by 2a and 2b respectively, why is the distance between the diagonals $ \\sqrt{a^{2}-b^{2}}$? What other properties of ellipses are worth taking note? Any help/information is appreciated :)", "Solution_1": "when you say diagonal do you mean chord??\r\n\r\n---\r\nyou should be able to prove tangent/normal formula, chord formula\r\n\r\nfoci directrices etc\r\n\r\nalso you should be able to prove that PS+PS'=2a", "Solution_2": "Sorry I meant the distance between the foci, not the diagonals.. :blush: my mistake. I read somewhere that the distance the foci is $ \\sqrt{a^{2}-b^{2}}$..", "Solution_3": "hmm that is an interesting property I have neevr come across\r\n\r\n$ \\sqrt{a^{2}-b^{2}}$\r\n$ \\sqrt{a^{2}-a^{2}(1-e^{2})}$ using $ b^{2}$=$ a^{2}(1-e^{2})$\r\n$ \\sqrt{a^{2}(1-(1-e^{2})}$\r\n$ \\sqrt{a^{2}e^{2}}$\r\n$ ae$\r\n\r\nand distance from origin to foci is ae therefore \r\n$ \\sqrt{a^{2}-b^{2}}$=$ ae$" } { "Tag": [ "calculus", "integration", "limit", "function", "Putnam", "logarithms", "calculus computations" ], "Problem": "If $00$. $ Show$ $ that$\r\n$ \\sum_{cyc}\\sqrt{a^{4}\\plus{}a^{2}b^{2}\\plus{}b^{4}}\\geq\\sum_{cyc}a\\sqrt{2a^{2}\\plus{}bc}$", "Solution_1": "[quote=\"Leonhard Euler\"]$ Let$ $ a,b,c > 0$. $ Show$ $ that$\n$ \\sum_{cyc}\\sqrt{a^{4}\\plus{}a^{2}b^{2}\\plus{}b^{4}}\\geq\\sum_{cyc}a\\sqrt{2a^{2}\\plus{}bc}$[/quote]\r\n$ \\sum_{cyc}\\sqrt{a^{4}\\plus{}a^{2}b^{2}\\plus{}b^{4}}\\geq\\sum_{cyc}a\\sqrt{2a^{2}\\plus{}bc}\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a^{2}b^{2}\\minus{}a^{2}bc)\\plus{}2\\sum_{cyc}\\sqrt{(a^{4}\\plus{}a^{2}b^{2}\\plus{}b^{4})(a^{4}\\plus{}a^{2}c^{2}\\plus{}c^{4})}\\geq2\\sum_{cyc}ab\\sqrt{(2a^{2}\\plus{}bc)(2b^{2}\\plus{}ac)}.$\r\nBut $ (a^{4}\\plus{}a^{2}b^{2}\\plus{}b^{4})(a^{4}\\plus{}a^{2}c^{2}\\plus{}c^{4})\\geq\\left(a^{4}\\plus{}\\frac{a^{2}(b^{2}\\plus{}c^{2})}{2}\\plus{}b^{2}c^{2}\\right)^{2}$ and\r\n$ 2a^{2}\\plus{}bc\\plus{}2b^{2}\\plus{}ac\\geq2\\sqrt{(2a^{2}\\plus{}bc)(2b^{2}\\plus{}ac)}.$\r\nThus, it remains to prove that\r\n$ \\sum_{cyc}(a^{2}b^{2}\\minus{}a^{2}bc\\plus{}2a^{4}\\plus{}4a^{2}b^{2})\\geq\\sum_{cyc}(2a^{3}b\\plus{}2a^{3}c\\plus{}2a^{2}bc),$ which obviously true.\r\nSee here the stronger inequality:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=165319" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Solve the equation\r\n$ max \\limits_{b}(min \\limits_{a}(a^2-b^2+ab+ax-bx))=3$", "Solution_1": "$ max \\limits_{b}(min \\limits_{a}(a^2 - b^2 + ab + ax - bx)) = max_b(-\\frac{(x+b)^2}{4}-b^2-bx)=\\frac 14 max_b(-5b^2-6bx-x^2)=\\frac{x^2}{5}.$\r\nTherefore $ x=\\pm \\sqrt{15}$." } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "If $f(x)=x+sin x$ find $f^-1 (x)$. :)", "Solution_1": "Is this an olympiad problem?............ :|", "Solution_2": "I don't think, but our algebra teacher that teaches us algebra for olympiad said this problem. ;)", "Solution_3": "And he told you that it has a solution?!! :huh:", "Solution_4": "$|\\sin x -\\sin y|\\le |x-y|$. At first, prove $f\\nearrow$. Maybe $f^{-1}(x_0)$, where $x_0$ is known. For example, $f^{-1}(\\pi )=\\pi$. $\\boxed {f^{-1}(y)=x\\Longleftrightarrow f(x)=y}$." } { "Tag": [ "geometric transformation", "geometry", "power of a point", "radical axis", "geometry proposed" ], "Problem": "(A.Zaslavsky, 9--11) Given two circles. Their common external tangent is tangent to them at points $ A$ and $ B$. Points $ X$, $ Y$ on these circles are such that some circle is tangent to the given two circles at these points, and in similar way (external or internal). Determine the locus of intersections of lines $ AX$ and $ BY$.", "Solution_1": "Assume without loss of generality that the third circle $ \\Gamma$ is tangent internally to the first two given ones (the other case can be treated analogously). Denote by $ X^{\\prime}$, and by $ Y^{\\prime}$ the intersections of the line $ XY$ with the first, and second circle respectively, and let $ O_{1}$, $ O_{2}$ be the centers of these first two given circles $ \\gamma_{1}$, and $ \\gamma_{2}$; finally let $ Q$ be the center of $ \\Gamma$. Since $ \\angle{O_{1}XX^{\\prime}}\\equal{}\\angle{O_{1}X^{\\prime}X}\\equal{}\\angle{QXY}\\equal{}\\angle{QYX}\\equal{}\\angle{O_{2}YY^{\\prime}}\\equal{}\\angle{O_{2}Y^{\\prime}Y},$ we have that $ O_{1}X \\| O_{2}Y^{\\prime}$, and $ O_{1}Y \\| O_{1}X^{\\prime}$. Thus, $ \\angle{AXX^{\\prime}}\\equal{}\\angle{BY^{\\prime}Y}$, and since $ AB$ is tangent to $ \\gamma_{2}$, we conclude that $ \\angle{AXX^{\\prime}}\\equal{}\\angle{BY^{\\prime}Y}\\equal{}\\angle{ABY}$. This shows that the points $ A$, $ B$, $ X$, and $ Y$ lie on a same circle.", "Solution_2": "Let $ \\omega_1 ,\\omega_2$ denote the two given circles and $ \\omega$ denotes the third circle internally/externally tangent to $ \\omega_1 ,\\omega_2$ at $ X,Y.$ Let $ O$ be the exsimilicenter of $ \\omega_1 \\sim \\omega_2,$ $ X,Y$ are the exsimilicenters/insimilicenters of $ \\omega \\sim \\omega_1$ and $ \\omega \\sim \\omega_2.$ Therefore, by Monge & d'Alembert theorem $ XY$ passes through $ O$ $\\Longrightarrow$ $ X,Y$ and $ A,B$ are inverse points under the direct inversion through pole $ O$ that takes $ \\omega_1$ and $ \\omega_2$ into each other $ \\Longrightarrow$ $ A,B,X,Y$ are concyclic $ \\Longrightarrow$ $ P \\equiv AX \\cap BY$ moves on the radical axis of $ \\omega_1 ,\\omega_2.$", "Solution_3": "First, let's see that $\\{T \\} \\equiv AX \\cap BY$ belongs to the third circle.\n\n[b]Proof:[/b] if $\\omega_1$ and $\\omega_2$ are the first two circles and $\\Omega$ - the third one, then $X$ is the similicenter (in- or ex-) of $\\omega_1$ and $\\Omega$ and, obviously, if $\\{X, T' \\} \\equiv AX \\cap \\Omega$, then the tangent to $\\Omega$ at $T'$ is parallel to $AB$, the same for $B$ and $T\"$, $T\"$ being the intersection of $BY$ with $\\Omega$, hence $T\" \\equiv T' \\equiv T$.\n\nNext, if $(TZ$ is the direction $(BA$ of this tangent, we have $\\widehat{BAT}=\\widehat{XTZ}$ $(1)$, but from the circle $\\Omega $: $\\widehat{XTZ}=\\widehat{XYT}$, $(2)$, from $(1)$ and $(2)$ getting that $ABYX$ is cyclic, i.e. $T$ belongs to the radical axis of $\\omega_1$ and $\\omega_2$.\n\nBest regards,\nsunken rock" } { "Tag": [ "factorial", "superior algebra", "superior algebra unsolved" ], "Problem": "The problem reads: \r\n\"Prove that if a>1, then there is a prime p with a 0$. Prove the inequality:\r\n\r\n$ abc(a\\plus{}2)(b\\plus{}2)(c\\plus{}2) \\le \\left( 1 \\plus{} \\frac{2}{3}(ab\\plus{}bc\\plus{}ca) \\right)^3$", "Solution_1": "[quote=\"Mashimaru\"]Let $ a,b,c > 0$. Prove the inequality:\n\n$ abc(a \\plus{} 2)(b \\plus{} 2)(c \\plus{} 2) \\le \\left( 1 \\plus{} \\frac {2}{3}(ab \\plus{} bc \\plus{} ca) \\right)^3$[/quote]\r\n\r\nThis inequality is from KMO(Korean Mathematical Olympiad) 2006, Round 2.\r\n\r\n[Proof]\r\nTransform the RHS into this form;\r\n$ RHS \\equal{} ( \\frac {(1 \\plus{} ab \\plus{} bc) \\plus{} (1 \\plus{} bc \\plus{} ca) \\plus{} (1 \\plus{} ca \\plus{} ab)}{3})^3$\r\n$ \\geq (1 \\plus{} ab \\plus{} bc)(1 \\plus{} bc \\plus{} ca)(1 \\plus{} ca \\plus{} ab)$ due to AM-GM.\r\n\r\nMoreover, by Cauchy Inequality, the following holds;\r\n$ (1 \\plus{} ab \\plus{} bc)(bc \\plus{} ca \\plus{} 1) \\geq (\\sqrt {bc} \\plus{} a \\sqrt {bc} \\plus{} \\sqrt {bc})^2 \\equal{} bc(a \\plus{} 2)^2$\r\n$ (1 \\plus{} bc \\plus{} ca)(ca \\plus{} ab \\plus{} 1) \\geq (\\sqrt {ca} \\plus{} b \\sqrt {ca} \\plus{} \\sqrt {ca})^2 \\equal{} ca(b \\plus{} 2)^2$\r\n$ (1 \\plus{} ca \\plus{} ab)(ab \\plus{} bc \\plus{} 1) \\geq (\\sqrt {ab} \\plus{} b \\sqrt {ab} \\plus{} \\sqrt {ab})^2 \\equal{} ab(c \\plus{} 2)^2$\r\n\r\nMultiplying the inequalities and taking square roots of both inequalities kills the problem =)", "Solution_2": "Great solution, [b]qwerty414[/b] :D \r\n\r\n[b]P/S:[/b] non-homogenous inequalities are always as hard as its beauty :blush:" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $E$ be a real normed vectorial space and $H$ an hyperplan of $E$.Show that $E\\setminus H$ is connexe par arcs iff $H$ is not closed.", "Solution_1": "Dear Killer, I remember you that the last RMS issue should not be posted on Mathlinks till later. And moreover, I remember you that the same problem was already posted by you and solved by grobber. ;)", "Solution_2": "Actually ...I posted this problems in order for someone to give me a solution for them,since I dont know how to solve some of them and I find them interesting.And that's the purpose of mathlinks,isn't it Harazi?I dont see why shouldnt I post them on this site(you know very well that they do not belong to that RMS isssue.They are there simply because they were given in last year(s) at the ENS oral exams.So everyone is free to think over them.I dont understand your remark at all ;) )\r\n\r\nAnd sorry if i posted this pb for a second time.I simply forgot.In that case..some moderator should delete it.If not ,enjoy it.", "Solution_3": "Here is this link to problem http://www.mathlinks.ro/Forum/viewtopic.php?highlight=arc&t=60058#" } { "Tag": [ "algebra", "difference of squares", "special factorizations" ], "Problem": "What is the value of $ 513^2 \\minus{} 487^2$?", "Solution_1": "Utilizing difference of squares, $ 513^2\\minus{}487^2\\equal{}(513\\minus{}487)(513\\plus{}487)\\equal{}26(1000)\\equal{}\\boxed{26000}$.", "Solution_2": "Sorry, accidentally pressed \"report error\" on this problem." } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Let $ P$ be a polynomial with integer coefficients.Suppose that \r\n(i)$ P(n)>n$ $ \\forall n>0$\r\n(ii)For each positive integer $ m$,there exists an integer $ k>0$ such that $ P_k(1)$ is divisible by $ m$. \r\nProve that $ P(x)\\equal{}x\\plus{}1$.\r\n\r\nNote: $ P_k(x)\\equal{}P(P(P(...P(x)))$", "Solution_1": "Mathlinkers,could you help me with this.Thanks.", "Solution_2": "because $ P(n)>n$ so we take $ m\\equal{}P(1)\\minus{}1$\r\nthere exists a positive integer k such that $ P_{k}(1)$ is divisible by $ P(1)\\minus{}1$ it follows that $ P(1)$ is divisible $ P(1)\\minus{}1$ and we must have P(1)=2;\r\nwe prove that the degree of P equals 1, if not :\r\nwe take m=$ P_{l}(2)\\minus{}P{l\\minus{}1}(2)$ ,with a value of $ l$ we have a value of k which $ P_{k}(2)$ is divisible by m;\r\n so with a large $ l$ we must have k$ \\ge$l\r\nbut we have with $ k \\ge l$ we have $ P_{k}(2)\\minus{}P_{k\\minus{}1}(2)$ is divisible m so, we must have $ P_{l\\minus{}1}(2)$ is divisible m with a large l; so $ P_{l\\minus{}1}(2) \\ge P_{l}(2)\\minus{}P{l\\minus{}1}(2)$ \r\nbut the degree of P is greater than 1 so |$ P_{l}(2)\\minus{}2P_{l\\minus{}1}(2)$| is very large when l is enough large so the degree of P equals 1;\r\nand, now it is easy to find that P(x)=x+1", "Solution_3": "You got a mistake in typing . I think you mean \"We take $ m = P_l(2) - P_{l - 1}(2)$\" when you wrote $ m = P_l(2) - P{l - 1)(2)}$\"^^ and I don't know the last step .\r\nWhere is the relation between proving \" $ 2P_{l - 1}(2) \\ge P_l(2)$ is imposible when $ l$ large enough\"(1) and \"|$ 2P_{l - 1)(2) - P_l(2)|}$ is very large when $ l$ large enough \" ?\r\n\r\nTo prove (1) ,when can do like this:\r\n* $ P_{l - 1}(2)$ is very large when $ l$ large enough\r\n* if $ degP(x) > 1$ then $ P(x) > 2x$ when $ x$ large enough \r\n=> $ P( P_{l - 1}(2)) > 2P_{l - 1}(2)$ when $ l$ large enough" } { "Tag": [ "factorial", "inequalities", "absolute value", "complex numbers", "number theory unsolved", "number theory" ], "Problem": "Please someone help me with this i got stuck somwhere in the middle:\r\nfind all integer solutions to:\r\n\r\n$ y^{3}\\equal{}x^{2}\\plus{}x\\plus{}1$", "Solution_1": "Only $ x\\equal{}0,x\\equal{}\\minus{}1.$", "Solution_2": "Delete... :blush:", "Solution_3": "Thanks. very nice solution. it turns out to be these are the only solutions $ (x,y)\\equal{}(\\minus{}1,1),(0,1),(18,7),(\\minus{}19,7)$", "Solution_4": "[quote=\"tdl\"]We have: $ (y \\minus{} 1)(y^2 \\plus{} y \\plus{} 1) \\equal{} x(x \\plus{} 1)\\Rightarrow x(x \\plus{} 1)\\vdots (y \\minus{} 1)$\nBecause $ GCM(x,x \\plus{} 1) \\equal{} 1$ then we have two case:\n1) $ x \\equal{} k(y \\minus{} 1)$ and $ y^2 \\plus{} y \\plus{} 1 \\equal{} k(x \\plus{} 1)$, in which $ k$ be a natural number. (Easy to check that $ k > 1$)\n$ y^2 \\plus{} y \\plus{} 1 \\equal{} k(ky \\minus{} k \\plus{} 1)\\Leftrightarrow y^2 \\plus{} (1 \\minus{} k^2)y \\plus{} k^2 \\minus{} k \\plus{} 1 \\equal{} 0$\n2) $ x \\plus{} 1 \\equal{} k(y \\minus{} 1)$ and $ y^2 \\plus{} y \\plus{} 1 \\equal{} kx$: Similarly![/quote]\r\nIt is wrong. For exfmple $ 28|20*21$, but $ 28\\not |20$ and $ 28\\not |21$.", "Solution_5": "well... lol yes its obviously wrong. :blush:\r\nur absolutly right. u can't asume anything like that from $ (y \\minus{} 1)(y^2 \\plus{} y \\plus{} 1) \\equal{} x(x \\plus{} 1)$\r\n\r\nAnyway I approached it simmilar to that fermat problem $ x^2 \\plus{} 2 \\equal{} y^3$:\r\n\r\n$ y^3 \\equal{} x^2 \\plus{} x \\plus{} 1\\Leftrightarrow4y^3 \\equal{} (2x \\plus{} 1 \\plus{} i\\sqrt3)(2x \\plus{} 1 \\minus{} i\\sqrt3)$\r\nfrom what u can deduce $ (1 \\plus{} i\\sqrt3)(a \\plus{} bi\\sqrt3)^3 \\equal{} (2x \\plus{} 1\\pm\\sqrt3i)$ and from this $ a^3 \\plus{} 3a^2b \\minus{} 9ab^2 \\minus{} 3b^3 \\equal{} \\pm1$ which could be said unleast looks harder than the original equasion. At this point i gave up.\r\n\r\nCan anyone prove it? and can please someone tell me if my thinking above is correct as an idea, even if it doesn't lead anywhere at this particular problem, because i don't quite handle this kind of aproach yet.", "Solution_6": "[quote=\"Jure the frEEEk\"]well... lol yes its obviously wrong. :blush:\nur absolutly right. u can't asume anything like that from $ (y \\minus{} 1)(y^2 \\plus{} y \\plus{} 1) \\equal{} x(x \\plus{} 1)$\n\nAnyway I approached it simmilar to that fermat problem $ x^2 \\plus{} 2 \\equal{} y^3$:\n\n$ y^3 \\equal{} x^2 \\plus{} x \\plus{} 1\\Leftrightarrow4y^3 \\equal{} (2x \\plus{} 1 \\plus{} i\\sqrt3)(2x \\plus{} 1 \\minus{} i\\sqrt3)$\nfrom what u can deduce $ (1 \\plus{} i\\sqrt3)(a \\plus{} bi\\sqrt3)^3 \\equal{} (2x \\plus{} 1\\pm\\sqrt3i)$ and from this $ a^3 \\plus{} 3a^2b \\minus{} 9ab^2 \\minus{} 3b^3 \\equal{} \\pm1$ which could be said unleast looks harder than the original equasion. At this point i gave up.\n\nCan anyone prove it? and can please someone tell me if my thinking above is correct as an idea, even if it doesn't lead anywhere at this particular problem, because i don't quite handle this kind of aproach yet.[/quote]\r\n$ Z[i\\sqrt 3 ]$ is not factorial, for example $ 2*2 \\equal{} (1 \\plus{} i\\sqrt 3 )(1 \\minus{} i\\sqrt 3)$, but you can use $ Z[e^{\\frac {2\\pi i}{3}}]$.", "Solution_7": "Here's my solution:(3o minutes for it) :blush: \r\nI easily prove that y is positive integer!\r\nThen I have 2 cases considered:\r\n1st case:x>=2\r\nThen I have:8y^3=8x^2+8x+8\r\nAnd I use this inequality: (2x)^3<8x^2+8x+8<(2x+1)^3\r\nSo this equation doesn't have any roots for this case!\r\nWhen x=1 then y^3=3(unreasonable)\r\nWhen x=0 then y^3=1.Therefore,(0,1) is one root of this equation!\r\nx=-1 then y=0(another root)\r\nx=-2 then y^3=3(unreasonable)\r\n2nd case:x=<-3\r\nSince x(x+1) is divisible by 2, then I have:y must be an odd integer\r\nLet y=(2k+1)(k>=0 and k is integer)\r\nThen replacing it into the equation I have:2k(3+4k^2+6k)=x^2+x\r\nSince x<=-3 then x must divide by k or x must divide by 3+4k^2+6k\r\nLet x=m.k then I have:k(-m^2+8k+12)=-4(m,k is integer and k>=0)\r\nSimilar to the 2nd equation(using 2 variable) I have:3=k(2kl-7-4k)(k, l is integer and k>=0)\r\nFinally I'm done! :lol:", "Solution_8": "[quote=\"ghjk\"]\nAnd I use this inequality: (2x)^3<8x^2+8x+8<(2x+1)^3\n[/quote]\r\n\r\nWhy $ (2x)^3<8x^2\\plus{}8x\\plus{}8<(2x\\plus{}1)^3$, where is root $ x\\equal{}18,y\\equal{}7$?\r\nYou are wrong!", "Solution_9": "[quote=\"tdl\"][quote=\"ghjk\"]\nAnd I use this inequality: (2x)^3<8x^2+8x+8<(2x+1)^3\n[/quote]\n\nWhy $ (2x)^3 < 8x^2 \\plus{} 8x \\plus{} 8 < (2x \\plus{} 1)^3$, where is root $ x \\equal{} 18,y \\equal{} 7$?\nYou are wrong![/quote]\r\n\r\nYeah, you're right :blush: . So i decide to take off the 1st case and againI have to consider the case when: /x/>=3(since x=2 is not the value of this equation).Then you can continue with the rest of my solution! :)", "Solution_10": "[quote=\"ghjk\"]\nSince x<=-3 then x must divide by k or x must divide by 3+4k^2+6k\n[/quote]\r\nsorry but this is not true. same mistake as above. check rust's post\r\n\r\n@Rust thanks for the hint\r\n\r\n$ Z[i\\sqrt3]$ isn't factorial but $ |x\\parallel{}y| \\equal{} |z|$ for comlex $ x,y,z$ limits the number of \"choises for factorisation\". I mean $ 4 \\equal{} 4\\cdot1 \\equal{} 2\\cdot2 \\equal{} (1 \\minus{} i\\sqrt3)(1 \\plus{} i\\sqrt3)...$ but there are only a few of this possible factorisations. now please tell me if this is right or wrong:\r\n\r\nwe're operating with numbers of form $ a \\plus{} bi\\sqrt3$ where $ a,b$ are integers.\r\nnow $ 4y^3 \\equal{} (2x \\plus{} 1 \\minus{} i\\sqrt3)(2x \\plus{} 1 \\plus{} i\\sqrt3)$. if there is such number of form $ a \\plus{} bi\\sqrt3$, other than $ 1$, that devides $ 2x \\plus{} 1 \\minus{} i\\sqrt3$ and $ 2x \\plus{} 1 \\plus{} i\\sqrt3$ than it has to devide $ 2i\\sqrt3$. noticing 2 doesn't devide $ 2x \\plus{} 1\\pm\\sqrt3i$ and with a limited selection of such numbers with absolute value of less or equal to $ 2i\\sqrt3$ we come to the conclusion that the only such number could be $ i\\sqrt3$.\r\n\r\nif that were the case than $ y^3$ would be devisable by $ 27$, but $ x^2 \\plus{} x \\plus{} 1$ can't be devisable by $ 9$. therefore there is no such number other than $ 1$ that would devide both $ 2x \\plus{} 1 \\minus{} i\\sqrt3$ and $ 2x \\plus{} 1 \\plus{} i\\sqrt3$\r\n\r\n$ 4$ has to devide $ (2x \\plus{} 1 \\minus{} i\\sqrt3)(2x \\plus{} 1 \\plus{} i\\sqrt3)$ and since $ 2$ doesn't devide $ 2x \\plus{} 1\\pm\\sqrt3i$ we come to the only possible conclusion that $ (1 \\plus{} i\\sqrt3)$ devides $ 2x \\plus{} 1 \\plus{} i\\sqrt3$ and $ (1 \\minus{} i\\sqrt3)$ devides $ 2x \\plus{} 1 \\minus{} i\\sqrt3$ or the other way around.\r\n\r\ntherefore we come to the conclusion that $ 2x \\plus{} 1 \\plus{} i\\sqrt3$ can be written in te form $ (1\\pm\\sqrt3i)(n \\plus{} mi\\sqrt3)^3$ with $ n,m$ integers.\r\n\r\nFrom this, with regarding the imaginary part, we now get the equasion:\r\n$ n^3 \\plus{} 3n^2m \\minus{} 9nm^2 \\minus{} 3m^3 \\equal{} \\pm1$ which we unfortunatelly don't know how to solve either", "Solution_11": "can anyone solve this one? it's equivalent. i give up :( \r\n\r\n$ y^3\\minus{}3x^2y\\pm1\\equal{}x^3$" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "If p, q are primes then a group G of order $ p^2q$ can not be simple. I have a solution but i feel it contains a mistake...\r\nIf q

p$; the # of p-Sylow subgroup is 1 or q. If it is 1 then we are done, suppose it is $ q$. # of q-Sylow subgroups is 1, $ p$ or $ p^2$. If it is 1 we are done; $ q>p$ so it can not be p, then it is $ p^2$ and $ p^2 \\minus{} 1\\equiv 1 (mod \\ q)$. q is prime, so we must have $ p \\plus{} 1$ divides $ q$ and thus $ q \\equal{} p \\plus{} 1$, so q=3 and p=2, we can check this particular case easily. \r\nAre there any mistakes?", "Solution_3": "Solvability can be proved easier: We already proved that there is a nontrivial normal subgroup. N and G/N are of order p,q or p^2. Therefore N and G/N are solvable and so is G.", "Solution_4": "[quote=\"Gockel\"]Solvability can be proved easier: We already proved that there is a nontrivial normal subgroup. N and G/N are of order p,q or p^2. Therefore N and G/N are solvable and so is G.[/quote]\r\nHow about $ |G/N|\\equal{}pq$?", "Solution_5": "Oh, yeah, i forget this case. But this case is known to be solvable too for it has a normal syslowgroup." } { "Tag": [ "MIT", "college" ], "Problem": "Herzlichen Gl\u00fcckwunsch allen deutschen IMO-Teilnehmern!\r\n;) :first: ;) :first: ;) \r\n\r\nImmerhin habt ihr 2,5 von drei gro\u00dfen Gronau-Zielen erf\u00fcllt (F\u00fcr Nicht-Insider: Die Ziele lauten: mindestens eine Gold-Medaille, eine Medaille f\u00fcr jeden und TOP 10). Hier zum Vergleich die Bilanzen der Vorjahre:\r\n\r\n2004: kein Ziel\r\n2003: zwei Ziele\r\n2002: ein Ziel\r\n2001: ein Ziel\r\n\r\nNochmal herzlichen Gl\u00fcckwunsch!", "Solution_1": "Herzlichen Gl\u00fcckwunsch an Alle!\r\n\r\nwenn die Ziele damals die gleichen waren hast du 2003 mit 2002 vertauscht oder ist ne Anerkennung seit neustem eine Medalie ;-) ? (Friedrich und Alex hatten beide nur ne A.) daf\u00fcr waren wir 2002 aber auf platz 10", "Solution_2": "Strimmt, du hast nat\u00fcrlich Recht. Das gute Jahr 2002 liegt ja schon drei Jahre zur\u00fcck! ;)", "Solution_3": "Und Christians Ziel, vor der Mongolei zu landen, haben wir auch erf\u00fcllt. Landen im w\u00f6rtlichen Sinne...\r\n\r\n dg", "Solution_4": "Mein Gl\u00fcckwunsch an alle IMO-Teilnehmer. :)" } { "Tag": [ "function", "abstract algebra", "group theory", "superior algebra", "superior algebra solved" ], "Problem": "a)(H.B.Mann)Let $G$ be a finite group,and let $S$ and $T$ be(not necessarily distinct) noempty subsets.Prove that either $G=ST$ or $|G|\\geq |S|+|T|$.\r\nb)Prove that every element in a finite field is a sum of two squares.\r\n\r\n[i]Rotman[/i]", "Solution_1": "b/ find the number of squares. The conclusion will follows.", "Solution_2": "a) Assume $|G|<|S|+|T|$.\r\nLet $x\\in G$. The sets $S^{-1}\\cdot x$ and $T$ must intersect because $|S^{-1}\\cdot c|+|T|=|S|+|T|>|G|$. Let $s^{-1}x=t$. Then $x=st$ and we are done. I denoted $S^{-1}\\cdot x=\\{s^{-1}x|s\\in S\\}$. $S^{-1}\\cdot x$ and $S$ have the same number of elements because the application $G\\to G: g\\to g^{-1}\\cdot x$ is bijective (be careful about the distinction bijective vs. isomorphism).\r\n\r\nb)\r\nIf $K$ is of characteristic 2, then all the elements of $K$ are squares and we have nothing to prove. This is because the function $K\\to K: x\\to x^2$ is Frobenius's automorphism, hence surjective.\r\n\r\nIf $K$ has an odd number of elements $n$, then it has odd characteristic. The kernel of the group homomorphism $K^*\\to K^*: x\\to x^2$ is $\\pm 1$, hence it's image has $\\frac{n-1}2$ elements. So $K$ has $\\frac{n+1}2$ squares (the element 0 which is a square has to be counted also in $K$).\r\nLet $S$ be the set of squares. Then $|S|+|S|=n+1>n=|K|$ and we can apply (a) for $S=T$ and $G=(K,+)$.\r\n\r\nThis followed alekk's idea.", "Solution_3": "[quote=\"amfulger\"]\nIf $K$ is of characteristic 2, then all the elements of $K$ are squares and we have nothing to prove. This is because the function $K\\to K: x\\to x^2$ is Frobenius's automorphism, hence surjective.\n\n[/quote]\r\n\r\nI no understand,help me!", "Solution_4": "We have $(x+y)^2=x^2+y^2$ and $(xy)^2=x^2y^2$, so squaring is a (ring) homomorphism. A field has no nontrivial ideals, so a nonzero homomorphism from a field is injective. An injective map from a finite set to itself is surjective, so squaring is an automorphism.\r\n\r\nThe same applies to the $p$th power map in a finite field of characteristic $p$. This is known as the Frobenius automorphism, and it actually generates the automorphism group of the field.\r\n\r\nThese maps do not need to be surjective if the field is infinite.", "Solution_5": "[quote=\"jmerry\"]We have $(x+y)^2=x^2+y^2$ and $(xy)^2=x^2y^2$, so squaring is a (ring) homomorphism. A field has no nontrivial ideals, so a nonzero homomorphism from a field is injective. An injective map from a finite set to itself is surjective, so squaring is an automorphism.\n\nThe same applies to the $p$th power map in a finite field of characteristic $p$. This is known as the Frobenius automorphism, and it actually generates the automorphism group of the field.\n\nThese maps do not need to be surjective if the field is infinite.[/quote]\r\n\r\nThank you :)" } { "Tag": [ "limit", "logarithms", "Euler", "real analysis", "real analysis unsolved" ], "Problem": "1.Find all $\\alpha \\in \\mathbb{R}$ such that $\\exists \\lim_{n\\to+\\infty}x_{n}\\in \\mathbb{R}$\r\n$x_{n}=1^{\\alpha}+2^{\\alpha}+...+n^{\\alpha}+\\alpha .lnn$\r\n2.the same question with:\r\n$y_{n}=1^{\\alpha}+2^{\\alpha}+...+n^{\\alpha}+\\frac{1}{\\alpha .n^{\\alpha}}$", "Solution_1": "I will discuss only the first problem.\r\n\r\nDefinitely that if $\\alpha \\geq 0$ then the limit is $\\infty$.\r\n\r\nlet us consider the case when $\\alpha=-\\frac{1}{\\beta}<0$, where $\\beta >0$.\r\n\r\nThen $x_{n}=1+\\frac{1}{2^{\\beta}}+\\cdots+\\frac{1}{n^{\\beta}}-\\frac{\\ln{n}}{\\beta}$.\r\n\r\nWe distinguish here three cases:\r\n\r\n1) $\\beta =1$, in this case the sequence converges to $\\gamma$.\r\n\r\n2) $If \\beta >1$ then the sequence diverges.\r\n\r\n3) If $\\beta \\in (0, 1)$ then we have that \r\n\r\n$x_{n}=\\left(1+\\frac{1}{2^{\\beta}}+\\cdots+\\frac{1}{n^{\\beta}}-\\frac{n^{1-\\beta}}{1-\\beta}\\right)+\\frac{n^{1-\\beta}}{1-\\beta}-\\frac{\\ln n}{\\beta}$\r\nNow, since the paranthesis converges and $\\lim \\left(\\frac{n^{1-\\beta}}{1-\\beta}-\\frac{\\ln n}{1-\\beta}\\right)=\\infty$ it simply follows that your sequence diverges, hence the only real values for which the initial sequence converges is $\\alpha=1$ for which the limit is nothing else but the famous Euler-Mascheroni constant.", "Solution_2": "[quote=\"didilica\"]I will discuss only the first problem.\n\nDefinitely that if $\\alpha \\geq 0$ then the limit is $\\infty$.\n\nlet us consider the case when $\\alpha=-\\frac{1}{\\beta}<0$, where $\\beta >0$.\n\nThen $x_{n}=1+\\frac{1}{2^{\\beta}}+\\cdots+\\frac{1}{n^{\\beta}}-\\frac{\\ln{n}}{\\beta}$.\n\nWe distinguish here three cases:\n\n1) $\\beta =1$, in this case the sequence converges to $\\gamma$.\n\n2) $If \\beta >1$ then the sequence diverges.\n\n3) If $\\beta \\in (0, 1)$ then we have that \n\n$x_{n}=\\left(1+\\frac{1}{2^{\\beta}}+\\cdots+\\frac{1}{n^{\\beta}}-\\frac{n^{1-\\beta}}{1-\\beta}\\right)+\\frac{n^{1-\\beta}}{1-\\beta}-\\frac{\\ln n}{\\beta}$\nNow, since the paranthesis converges and $\\lim \\left(\\frac{n^{1-\\beta}}{1-\\beta}-\\frac{\\ln n}{1-\\beta}\\right)=\\infty$ it simply follows that your sequence diverges, hence the only real values for which the initial sequence converges is $\\alpha=1$ for which the limit is nothing else but the famous Euler-Mascheroni constant.[/quote]\r\nYes,that's right.I have a similar solution\r\npart 1:$\\alpha =-1$\r\nand part 2:$\\alpha =-\\frac{1}{2}$" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "inequalities", "triangle inequality" ], "Problem": "How many different tetrahedrons are there with whole number-long\r\n\r\nedges, and with 17 as the sum of the lengths of the edges?", "Solution_1": "[quote=\"moldlee\"]How many different tetrahedrons are there with whole number-long\n\nedges, and with 17 as the sum of the lengths of the edges?[/quote]\r\n[hide]$x+y+z+a=17$\nballs walls urns sticks dots barriers laters\n$\\binom{16}{3}=560$[/hide]", "Solution_2": "[quote=\"bpms\"]\n[hide]$x+y+z+a=17$\nballs walls urns sticks dots barriers laters\n$\\binom{16}{3}=560$[/hide][/quote]\n\n[hide]1. Tetrahedrons have more than four edges.\n2. Triangle Inequality[/hide]" } { "Tag": [ "inequalities", "Cauchy Inequality", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be sidelengths of a triangle. If $ A$ and $ B$ are like below, prove that $ AB\\ge9$.\r\n\r\n$ A \\equal{} \\frac {a^2 \\plus{} bc}{b \\plus{} c} \\plus{} \\frac {b^2 \\plus{} ca}{c \\plus{} a} \\plus{} \\frac {c^2 \\plus{} ab}{a \\plus{} b}$\r\n\r\n$ B \\equal{} \\frac {1}{\\sqrt {(a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a)}} \\plus{} \\frac {1}{\\sqrt {(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b)}} \\plus{} \\frac {1}{\\sqrt {(c \\plus{} a \\minus{} b)(a \\plus{} b \\minus{} c)}}$", "Solution_1": "We will prove that \r\n$ A \\ge a \\plus{} b \\plus{} c$ and $ B \\ge \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c}$\r\n\r\nTherefore, $ AB \\ge (a \\plus{} b \\plus{} c)\\left( \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\right) \\ge 9$\r\n\r\n[b][u]First part[/u][/b]:\r\n\r\n$ \\sum_{cyc} \\frac {a^2 \\plus{} bc}{b \\plus{} c} \\plus{} \\sum_{cyc}a \\equal{} \\sum_{cyc}\\frac {(a \\plus{} b)(a \\plus{} c)}{b \\plus{} c}$\r\n\r\nApplying cauchy inequality, we derive\r\n\r\n$ \\frac {(a \\plus{} b)(a \\plus{} c)}{b \\plus{} c} \\plus{} \\frac {(b \\plus{} a)(b \\plus{} c)}{a \\plus{} c} \\ge 2(a \\plus{} b)$\r\n\r\n[b][u]Second part[/u][/b]:\r\n\r\n$ \\sum_{cyc}\\frac {1}{\\sqrt{(a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a)}} \\ge \\sum_{cyc}\\frac {2}{(a \\plus{} b \\minus{} c) \\plus{} (b \\plus{} c \\minus{} a)} \\equal{} \\sum_{cyc}\\frac {1}{b}$ \r\n\r\nWe are done!", "Solution_2": "[quote=\"great math\"]We will prove that \n$ A \\ge a \\plus{} b \\plus{} c$ and $ B \\ge \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c}$\n\nTherefore, $ AB \\ge (a \\plus{} b \\plus{} c)\\left( \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\right) \\ge 9$\n\n[b][u]First part[/u][/b]:\n\n$ \\sum_{cyc} \\frac {a^2 \\plus{} bc}{b \\plus{} c} \\plus{} \\sum_{cyc}a \\equal{} \\sum_{cyc}\\frac {(a \\plus{} b)(a \\plus{} c)}{b \\plus{} c}$\n\nApplying cauchy inequality, we derive\n\n$ \\frac {(a \\plus{} b)(a \\plus{} c)}{b \\plus{} c} \\plus{} \\frac {(b \\plus{} a)(b \\plus{} c)}{a \\plus{} c} \\ge 2(a \\plus{} b)$\n\n[b][u]Second part[/u][/b]:\n\n$ \\sum_{cyc}\\frac {1}{\\sqrt {(a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a)}} \\ge \\sum_{cyc}\\frac {2}{(a \\plus{} b \\minus{} c) \\plus{} (b \\plus{} c \\minus{} a)} \\equal{} \\sum_{cyc}\\frac {1}{b}$ \n\nWe are done![/quote]\r\n\r\nNice solution, great math!" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "There are 100 books on a shelf. We want to select 3 books which are not neighbors. How many selections are there?", "Solution_1": "standard method here...\r\n\r\neach selection corresponds to a sequence of length 100 where three of the terms are 1 and the rest are 0. we are then looking for the number of such sequences where no two 1's are next to each other. call such sequences \"good\". given a good sequence, we may remove exactly one zero from between successive 1's in the sequence (these 0's are available because the sequence is good) for a total removal of two 0's. the result is a sequence of length 98 where three of the terms are 1 and the rest are 0's and on which there are no restrictions. conversely, given any sequence of length 98 where three terms are 1 and the rest 0's, on which there are no restrictions, we may insert a 0 in each of the two blocks between the three 1's (one or more of these blocks may be empty, of course) to get a good sequence. we have a one-to-one onto correspondence, so the answer is $ 98 \\choose 3$.", "Solution_2": "[quote=\"dogcat\"]There are 100 books on a shelf. We want to select 3 books which are not neighbors. How many selections are there?[/quote]\r\nwe can suppose that $ {(a\\ books)\\ (book\\ A)\\ (b\\ books)\\ (book\\ B)\\ (c\\ books)\\ (book\\ C)\\ (100 \\minus{} (a \\minus{} 1 \\minus{} b \\minus{} 1 \\minus{} c \\minus{} 1)\\ books)}$ where $ a\\ge 0$ etnd $ b,c > 0$\r\nso we have to find the numbre of solution of $ (a \\plus{} 1) \\plus{} (b \\plus{} 1) \\plus{} (c \\plus{} 1)\\le 100$ where $ (a,b,c)\\in\\mathbb{N}\\times\\mathbb{N}^*\\times \\mathbb{N}^*$\r\n$ \\equal{} |\\{(a,b,c)\\in \\mathbb{N}\\times\\mathbb{N}^*\\times \\mathbb{N}^*|\\ a \\plus{} b \\plus{} c\\le 97\\}|$\r\n$ \\equal{} |\\{(a,b,c)\\in \\mathbb{N}\\times\\mathbb{N}\\times \\mathbb{N}|\\ a \\plus{} b \\plus{} c\\le 95\\}|$\r\n$ \\equal{} \\sum_{k \\equal{} 0}^{95}C_{k \\plus{} 2}^{2}$\r\n$ \\equal{} \\sum_{k \\equal{} 0}^{95}\\frac {(k \\plus{} 2)(k \\plus{} 1)}{2}$\r\n$ \\equal{} \\frac {\\sum_{k \\equal{} 1}^{96}k^2 \\plus{} \\sum_{k \\equal{} 1}^{96}k}{2}$" } { "Tag": [ "function", "linear algebra", "matrix", "symmetry", "superior algebra", "superior algebra solved" ], "Problem": "[i]F[/i] is a field of numbers. A mapping $f:F\\to F$ satisfy:\r\n(1) for any matrices [b]A[/b], [b]B[/b] of size [i]n[/i], it holds f([b]A[/b]+[b]B[/b])=f([b]A[/b])+f([b]B[/b])\r\n(2) for any $k \\in F$, it holds that f(k[b]A[/b]) = k f([b]A[/b]), where [b]A[/b] is a matrix of size [i]n[/i].\r\n(3) for any matrices [b]A[/b], [b]B[/b] of size [i]n[/i], f([b]A[/b][b]B[/b])=f([b]B[/b][b]A[/b])\r\n(4) f([b]E[/b])=n\r\n\r\nShow that f([b]X[/b]) = tr([b]X[/b])", "Solution_1": "[quote=\"liyi\"][i]F[/i] is a field of numbers. A mapping $f:F\\to F$ satisfy:\n[/quote]\r\n\r\nI bet you ment A mapping $f:M_n(F)\\to F$ satisfy:", "Solution_2": "$F(A)+F(B)=F(A+B) \\Rightarrow F(0_n)=0$.\r\n\r\nLet $E_{i,j}$ be the matrixe $(e_{i,j}^{k,l})_{k,l=\\overline{1,n}}$ given by \r\n$e_{i,j}^{k,l}=\\{\\begin{array}{cc} 1,&if\\ (i,j)=(k,l) \\\\ 0,& otherwise\\end{array}$.\r\nLet $f_{i,j}=F(E_{i,j})$. It is easy to see that $E_{i,j}\\cdot E_{k,l}\\neq 0_n$ and $E_{k,l}\\cdot E_{i,j}\\neq 0_n$ iff i=j=k=l. \r\nLet $L=\\{ f_{i,j} | i\\neq j\\}$ It is easy to see that $ab=0 \\forall a,b \\in F$, not necessarily distinct. This quickly leads to L={0}. So $f_{i,j}=0$ $\\forall i\\neq j$.\r\n\r\n$(E_{i,i}+E_{i,j})(E_{i,i}+E_{j,i})=2E_{i,i}$\r\n$(E_{i,i}+E_{j,i})(E_{i,i}+E_{i,j})=E_{i,i}+E_{i,j}+E_{j,i}+E_{j,j}$\r\nSo $2\\cdot f_{i,i}=2\\cdot F(E_{i,i})=F(2\\cdot E_{i,i})=F(E_{i,i}+E_{i,j}+E_{j,i}+E_{j,j})=f_{i,i}+f_{j,j}$\r\nSo $f_{i,i}=f_{j,j}$. And this is for all $i,j=\\overline{1,n}$.\r\n\r\n$F(E)=n \\Rightarrow \\sum_{k=0}^n f_{k,k}=n \\Rightarrow f_{k,k}=1 \\forall k=\\overline{1,n}$\r\n\r\nThere is no problem now to get F(A)=Tr(A).", "Solution_3": "[quote=\"amfulger\"] It is easy to see that $E_{i,j}\\cdot E_{k,l}\\neq 0_n$ and $E_{k,l}\\cdot E_{i,j}\\neq 0_n$ iff i=j=k=l. \nLet $L=\\{ f_{i,j} | i\\neq j\\}$ It is easy to see that $ab=0 \\forall a,b \\in F$, not necessarily distinct. This quickly leads to L={0}. So $f_{i,j}=0$ $\\forall i\\neq j$.\n[/quote]\r\n\r\nReplace this by:\r\nIt is easy to see that $E_{i,j}\\cdot E_{k,l}\\neq 0_n$ and $E_{k,l}\\cdot E_{i,j}\\neq 0_n$ iff j=k and i=l. \r\nLet $L=\\{ f_{i,j} | i=2 for which the ineq is true for any a,b,c pozitive reals:\r\nsqrt(a+sqrt(b+sqrt(c)))>=(abc)^(1/n).\r\n\r\nbye bye!", "Solution_1": "Contest from Vaslui again. The answer is 14. \r\n\r\nYou can prove that 14 works by repeated applications (2, actually) of AM-GM in such a way that you make the exponents of a, b, c equal. For example, in the first step you apply AM-GM for b, (c^1/2)/2 and (c^1/2)/2. I think you can see what I mean. \r\n\r\nTo prove that 14 is the ONLY n you assume this works for n>14 (it's similar for n<14) and you take some sequences a_k, b_k, c_k which tend to 0 (or 00, I can't remember), in such a way that you have equality in the 2 applications of AM-GM. You'll eventually get a contradiction if you choose 0 or 00 in the right way. I really have no time for posting a soln right now, but this is what I did and I got 10 points for this (that's the max you can get for a prb)." } { "Tag": [ "limit", "linear algebra", "matrix", "real analysis", "real analysis unsolved" ], "Problem": "Let for any $E\\subset R^{2}$ and $\\alpha\\in (0,2]$ consider \r\n\\[\\parallel E\\parallel _\\alpha=\\lim_{\\delta\\to 0}\\inf\\{\\sum_{i=1}^{n}(diam(E_{i}))^\\alpha\\mid\\ E\\subset \\bigcup E_{i},\\ diam(E_{i})<\\delta\\}.\\]\r\nSuppose $T=M_{2}(R)$. Find \r\n\\[\\parallel T\\parallel _\\alpha=\\sup_{\\parallel E\\parallel _\\alpha=1}\\parallel TE\\parallel _\\alpha.\\]", "Solution_1": "what is $M_{2}(\\mathbb{R})$ ?", "Solution_2": "Matrix $2\\times 2$. I though it is a common notation on this forum.", "Solution_3": "$T\\in M_{2}(\\mathbb R)$, then. The [url=http://en.wikipedia.org/wiki/Singular_value_decomposition]singular value decomposition[/url] reduces the problem to diagonal $T$ with nonnegative entries. Let $\\lambda_{1}\\ge \\lambda_{2}$ be the diagonal elements. Naturally, $\\parallel T\\parallel_{\\alpha}\\le \\lambda_{1}^{\\alpha}$. This is attained for $0<\\alpha\\le 1$ by appropriate subsets of $\\mathbb R e_{1}$. For $\\alpha\\in (1,2]$ one should expect $\\parallel T\\parallel_{\\alpha}=\\lambda_{1}\\lambda_{2}^{\\alpha-1}$, but I can't prove either inequality. :(" } { "Tag": [ "calculus", "integration", "function", "real analysis", "real analysis unsolved" ], "Problem": "Suppose that $(E,A,\\phi)$ is a measurable space of finite measure and $f_{n}$ are measurable and converge pointwise to some $f$. Also, assume that for some increasing function $H$ which goes to $\\infty$ at $\\infty$ we have $\\sup_{n\\geq 1}{\\int_{E}{|f_{n}|H(|f_{n}|)d\\phi}}<\\infty$. Then $f_{n}$ converges to $f$ in $L^{1}$.", "Solution_1": "The condition with $H$ implies that $(f_{n})$ is a uniformly integrable family, that is, for any $\\epsilon>0$ there is $M$ such that $\\sup_{n}\\int_{|f_{n}|>M}|f_{n}|<\\epsilon$. Uniform integrability implies convergence in $L^{1}$ as follows: (1) truncate the sequence at level $M$ and apply the Lebesgue domination to the truncated sequence $f_{n}^{M}$; (2) control the contribution of $f_{n}-f_{n}^{M}$ with Fatou's lemma." } { "Tag": [ "inequalities", "geometry", "circumcircle", "trigonometry", "geometry unsolved" ], "Problem": "Let $ABC$ be a triangle, let $H$ be the orthocentre and $L,M,N$ the midpoints of the sides $AB, BC, CA$ respectively. Prove that\n\\[HL^{2} + HM^{2} + HN^{2} < AL^{2} + BM^{2} + CN^{2}\\]\nif and only if $ABC$ is acute-angled.", "Solution_1": "It seems easy. Let $R$ be the circumadius of triangle $LMN$. Of course the point $H$ is it's circumcenter ( a well known fact). Also segments $AL$,$BM$,$CN$ are equal to the sides of triangle $LMN$ thus all we want to do, is to prove the following \\[ LM^2+MN^2+NL^2>3R^2 \\] and this is easy to prove using using formulas concerning triangles.\r\nI think that even the stronger on holds \\[ LM^2+MN^2+NL^2>8R^2 \\]", "Solution_2": "Well, that well known fact is wrong ... the orthocenter isn't the center of the Feuerbach circle. The well known fact is that the orthocenter of $LMN$ is the circumcenter of $ABC$ ...", "Solution_3": "[b][u]Lemma.[/u][/b] $\\boxed {\\ \\mathrm {\\ acute}\\ \\triangle ABC\\Longrightarrow 8R^20\\ .$ [u]Remark.[/u] $a^2+b^2+c^2=8R^2(1+\\cos A\\cos B\\cos C)\\le 9R^2$\r\nbecause it is well-known that in any triangle $ABC$, $\\cos A\\cos B\\cos C\\le \\frac 18\\ .$\r\n\r\n[b][u]The proof of the proposed problem.[/u][/b] It is well-knownn that $AH=2R\\cos A$ a.s.o. Therefore,\r\n$4\\cdot HM^2=2\\cdot(HB^2+HC^2)-a^2=8R^2(\\cos^2B+\\cos^2C)-a^2=$\r\n$8R^2(2-\\sin^2B+\\sin^2C)-a^2=16R^2-2(b^2+c^2)-a^2\\ .$ Thus,\r\n$4\\cdot\\sum HM^2=48R^2-5(a^2+b^2+c^2)\\ .$ From the above lemma results the relation\r\n$\\frac 13 (a^2+b^2+c^2)\\le 4\\cdot\\sum HM^2<6(a^2+b^2+c^2)-5(a^2+b^2+c^2)=a^2+b^2+c^2$, i.e.\r\n$\\boxed {\\ \\frac 13 (AL^2+BM^2+CN^2)\\le HM^2+HN^2+HL^2 existen $ x,y \\in Z^\\plus{}$ talque $ p\\equal{}x^2\\plus{}y^2$, mi duda es q si es conocido q estos $ x,y$ son unicos (porq lo son).", "Solution_8": "Amilcar, esses x e y s\u00e3o de fato \u00fanicos.\r\nN\u00e3o sei provar, mas s\u00e3o! :P", "Solution_9": "[quote=\"aev5peru\"]Bueno hablando de primos q se pueden expresar como la suma de dos cuadrados.\np=1(mod4), => existen $ x,y \\in Z^ \\plus{}$ talque $ p \\equal{} x^2 \\plus{} y^2$, mi duda es q si es conocido q estos $ x,y$ son \u00fanicos (porq lo son).[/quote]\r\nBUeno, conosco dos formas de probar este hecho.\r\n-La primera es usando que si $ n\\equal{}x^2\\plus{}y^2$ , el n\u00famero de soluciones $ (x,y)$ en los n\u00fameros enteros es $ 4(d_1\\minus{}d_3)$, donde $ d_i$ es el n\u00famero de divisores congruentes a $ i$ m\u00f3dulo 4.\r\n-La segunda es suponer por contradicci\u00f3n: $ p\\equal{}x^2\\plus{}y^2\\equal{}a^2\\plus{}b^2$.\r\nchaufas!" } { "Tag": [ "integration", "calculus", "logarithms", "limit", "function", "derivative", "real analysis" ], "Problem": "im trying to calculate the asymptotic estimate of\r\n\r\n$ P_d(z)\\equal{}\\sum_n z^n n^{\\minus{}d}$ as $ z\\uparrow 1$,\r\n\r\neven a link to a nice description on how to do things like this in general would be appreciated,\r\n but then again so would a solution if one exists", "Solution_1": "Start that sum at $ n\\equal{}1$,so we don't have to worry about powers of zero.\r\n\r\nQuick facts: $ P_d(z)$ is bounded for $ d>1$. The series converges uniformly on the closed disk, and the asymptotic estimate is simply the sum $ \\zeta(d)$ of the series. For $ d\\equal{}1$, this is $ \\minus{}\\ln(1\\minus{}z)$. It's $ \\frac{z}{1\\minus{}z}$ for $ d\\equal{}0$, $ \\frac{z}{(1\\minus{}z)^2}$ for $ d\\equal{}\\minus{}1$, and $ \\frac{q_d(z)}{(1\\minus{}z)^{1\\minus{}d}}$ for integer values of $ d$, where $ q_d$ is a polynomial. Since $ P_{d\\minus{}1}(z)\\equal{}zP_d'(z)$, we have $ q_{d\\minus{}1}(z)\\equal{}(1\\minus{}z)q_d'(z)\\plus{}(1\\minus{}d)q_d(z)$, and $ q_{d\\minus{}1}(1)\\equal{}(1\\minus{}d)(\\minus{}d)(\\minus{}d\\minus{}1)\\cdots (1)\\equal{}(1\\minus{}d)!$.\r\n\r\nConjecture: $ P_d(z)\\approx \\begin{cases}\\zeta(d)& d>1\\\\ \\minus{}\\ln(1\\minus{}z)& d\\equal{}1\\\\ \\Gamma(1\\minus{}d)\\cdot(1\\minus{}z)^{d\\minus{}1}& d<1\\end{cases}$\r\n\r\nThe first two lines are already established, so we focus on the third, with $ d<1$. Here's one way to do it:\r\nAs $ z\\to 1^{\\minus{}}$, the sum $ \\sum_{n\\equal{}1}^{\\infty}\\left(z^{n\\minus{}1}\\minus{}z^n\\right)f(z^n)$ converges to $ \\int_0^1 f(t)\\,dt$; this is just the Riemann integral with a geometric partition. It would be nice to match the sum $ \\sum_{n\\equal{}1}^{\\infty}z^n n^{\\minus{}d}$ to this pattern. The $ z^n$ is easy to handle, and leaves us with merely a $ \\frac1z\\minus{}1$ factor to account for in the first term. To get $ n^{\\minus{}d}$, we take $ f(t)\\equal{}(\\minus{}\\ln t)^{\\minus{}d}$ The sum becomes\r\n$ I(d)\\equal{}\\int_0^1 (\\minus{}\\ln t)^{\\minus{}d}\\,dt$\r\n$ \\equal{}\\lim_{z\\to 1^\\minus{}}\\sum_{n\\equal{}1}^{\\infty}\\left(\\frac1z\\minus{}1\\right)\\cdot z^n\\cdot \\left(\\minus{}\\ln z^n\\right)^{\\minus{}d}$\r\n$ \\equal{}\\lim_{z\\to 1^\\minus{}}\\left(\\frac1z\\minus{}1\\right)\\sum_{n\\equal{}1}^{\\infty}z^n\\cdot n^{\\minus{}d}\\cdot (\\minus{}\\ln z)^{\\minus{}d}$\r\n$ \\equal{}\\lim_{z\\to 1^\\minus{}}\\left(\\frac{1\\minus{}z}{z}\\right)\\cdot(\\minus{}\\ln z)^{\\minus{}d}\\cdot\\sum_{n\\equal{}1}^{\\infty}z^n\\cdot n^{\\minus{}d}$\r\nNow, $ \\minus{}\\ln z\\approx 1\\minus{}z$ for $ z$ near 1, so we get that $ P_d(z)\\cdot (1\\minus{}z)^{1\\minus{}d}\\approx I(d)$. Substituting $ t\\equal{}e^{\\minus{}x}, dt\\equal{}\\minus{}e^{\\minus{}x}\\,dx$, $ I(d)\\equal{}\\int_0^{\\infty} x^{\\minus{}d}e^{\\minus{}x}\\,dx\\equal{}\\Gamma(\\minus{}d\\plus{}1)$. This completes the proof of the estimate $ P_z(d)\\approx I(d)\\cdot (1\\minus{}z)^{d\\minus{}1}\\equal{}\\Gamma(1\\minus{}d)(1\\minus{}z)^{d\\minus{}1}$ for $ d<1$.\r\n\r\nThat was a fairly tricky method. There are certainly others, but the key to anything which will get you the constant is to estimate the sum with an integral.", "Solution_2": "thank you very much, but i was actually trying to find an estimate of how $ P_d(z)$ approaches these limits.\r\n\r\nI am actually working with a slightly different sum $ \\hat{P}_d(z)\\equal{}\\sum_{n\\ge 1} z^n L(n) n^{\\minus{}d}$ where $ L(n)$ is a slowly varying function (i.e. $ \\lim_{x \\to \\infty} \\frac{L(cn)}{L(n)}\\equal{}1$ for all c) we can just generalize to $ L(x) \\to 1$ as $ x\\to \\infty$ \r\nand i have a nice relationship between $ z(\\beta), \\beta$ and $ \\hat{P}_d(z(\\beta))$,\r\n\r\nso knowing how $ \\hat{P}_d(z)\\to \\zeta(d)$ would give me something to examine how $ z(\\beta) \\uparrow 1$ as $ \\beta \\downarrow \\beta_0$", "Solution_3": "That's pretty simple. Just differentiate term by term in $ z$:\r\n$ P_d(z)\\approx P_d(1) \\plus{} (z \\minus{} 1)\\sum_{n \\equal{} 1}^{\\infty}nz^{n \\minus{} 1}\\cdot n^{ \\minus{} d}$\r\n$ \\approx P_d(1) \\plus{} \\frac {z \\minus{} 1}{z}\\sum_{n \\equal{} 1}^{\\infty}z^n\\cdot n^{1 \\minus{} d}$\r\n$ \\approx P_d(1) \\minus{} \\frac {1 \\minus{} z}{z}P_{d \\minus{} 1}(z)$\r\n\r\nThe derivative basically just takes you up one power; this also applies to your variant.\r\n\r\nOK, that formula's only really valid when the series for $ P_{d\\minus{}1}$ converges uniformly. When $ d$ is too small for that, we use $ \\frac1z\\int_z^1 P_{d\\minus{}1}(t)\\,dt$. That integral converges and thus produces a meaningful estimate as long as $ d>1$. For larger $ d$, even more derivatives can be taken." } { "Tag": [], "Problem": "James got 75% of the questions on a test correct. If he answered 11 of the first 13 questions correctly, and 2/3 of the remaining questions correctly, how many questions were on the test?", "Solution_1": "Too easy for HSB.\r\n\r\n[hide]$ \\frac{11+2x}{13+3x}=\\frac{3}{4}$\n\n$ 44+8x=39+9x$\n\n$ x=5$\n\n$ 13+3\\cdot5=\\boxed{28}$[/hide]", "Solution_2": "I understand what you did.\r\n\r\nWhy did you use x for 2 and 3?\r\n\r\nI set it up this way:\r\n\r\n11/13 + (2/3)x = 0.75 and got no where.\r\n\r\nYour method is much easier.", "Solution_3": "[quote=\"sharkman\"]I understand what you did.\n\nWhy did you use x for 2 and 3?\n\nI set it up this way:\n\n11/13 + (2/3)x = 0.75 and got no where.\n\nYour method is much easier.[/quote]\r\nI put the number of questions that James got right on top, and the total number on the bottom. The problem states he got $ 11$ of the first $ 13$ right, so I put $ 11$ on top and $ 13$ on the bottom. He also got $ \\tfrac23$ of the remaining questions right, so I added $ 2x$ on top, and $ 3x$ on the bottom, and set the whole thing equal to $ \\tfrac34$. Then I solved for $ x$, and found the total amount of questions once I knew what $ x$ was.", "Solution_4": "post spam in the games and fun factory forum, not here, okay?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "find all natural (x,y,z):\r\n$ x^2\\equal{}2(y\\plus{}z)$\r\n$ x^6\\equal{}y^6\\plus{}z^6\\plus{}31(y^2\\plus{}z^2)$", "Solution_1": "$ x^2 \\equal{} 2(y \\plus{} z)$\r\n$ x^6 \\equal{} y^6 \\plus{} z^6 \\plus{} 31(y^2 \\plus{} z^2)$\r\n\r\n$ 8(y\\plus{}z)^3\\equal{} y^6 \\plus{} z^6 \\plus{} 31(y^2 \\plus{} z^2)$\r\n\r\nWLOG $ z \\ge y$\r\nSuppose that $ z\\ge 4$\r\n$ z (z^3\\minus{}64)\\plus{}31>0 \\iff z^4\\plus{}31>64z \\iff z^6\\plus{}31z^2 >64z^3$ \r\n$ 64z^3 \\ge 8(y\\plus{}z)^3$ contradiction\r\nthen $ y \\le z \\le 3$\r\n\r\nWith only solution for $ (y,z)\\equal{}(1,1)$ adn $ x\\equal{}2$\r\n\r\nis it ok?" } { "Tag": [ "trigonometry", "inequalities", "search", "inequalities unsolved" ], "Problem": "For $a, b,c >0$. Prove that\r\n$a^{2}+b^{2}+c^{2}+2abc+1\\geq 2(ab+bc+ca)$\r\n----------------------------------------------------------------From TTT2 Magazine - Viet Nam", "Solution_1": "$a=x^{3},b=y^{3},c=z^{3}$\r\n$2x^{3}y^{3}z^{3}+1\\ge 3x^{2}y^{2}z^{2}$\r\nNow use: $\\sum_{cyc}x^{6}+3x^{2}y^{2}z^{2}\\ge \\sum_{sym}x^{4}y^{2}$", "Solution_2": "As I know, this problems is belong to Darji's", "Solution_3": "Let $a=\\cos \\alpha ,\\ b=\\cos \\beta ,\\ c=\\cos \\gamma \\left (0<\\alpha ,\\beta , \\gamma <\\frac{\\pi}{2}\\right)$", "Solution_4": "[quote=\"Mr.JN2P\"]For $a, b,c >0$. Prove that\n$a^{2}+b^{2}+c^{2}+2abc+1\\geq 2(ab+bc+ca)$\n----------------------------------------------------------------From TTT2 Magazine - Viet Nam[/quote]\r\n\r\nSet $a=x+1,b=y+1,c=z+1$", "Solution_5": "[quote=\"kunny\"]Let $a=\\cos \\alpha ,\\ b=\\cos \\beta ,\\ c=\\cos \\gamma \\left (0<\\alpha ,\\beta , \\gamma <\\frac{\\pi}{2}\\right)$[/quote]\r\nI think you are right only if $a,b,c\\in (0,1)$ but there is no restriction.", "Solution_6": "Let $b+c=s,\\ bc=t$, since $b>0,\\ c>0\\Longleftrightarrow s>0,\\ t>0,\\ s^{2}\\geq 4t$.\r\nOur inequality will become $a^{2}+2(s-t)a+s^{2}-4t+1\\geq 0$.\r\nProof:$f(a): =a^{2}+2(s-t)a+s^{2}-4t+1\\geq a^{2}-2(s-t)a+1$\r\n$=\\{a-(s-t)\\}^{2}+1-(s-t)^{2}$. We prove that $1-(s-t)^{2}\\geq 0$.\r\nLet $P=\\{(s,\\ t)|s>0,\\ t>0,\\ t\\leq \\frac{1}{4}s^{2}\\}, \\ Q=\\{(s,\\ t)|s>0,\\ t>0,-1\\leq s-t\\leq 1\\}$.\r\nSketching the region of $P$ and $Q$ on $s-t$ plane gives $P\\in{Q}\\Longleftrightarrow f(a)\\geq 0$ which completes the proof.", "Solution_7": "[quote=\"Mr.JN2P\"]For $a, b,c >0$. Prove that\n$a^{2}+b^{2}+c^{2}+2abc+1\\geq 2(ab+bc+ca)$[/quote]\r\n\r\nSee darij's post on\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?search_id=575919256&t=19666\r\n\r\ncarij" } { "Tag": [ "inequalities", "search" ], "Problem": "Given real numbers $ a,b,c$ such that $ \\frac {1}{2} \\le a,b,c \\le 1$\r\nProve that $ 2\\le \\sum \\frac {a \\plus{} b}{1 \\plus{} c} \\le 3$.", "Solution_1": "[quote=\"Agr_94_Math\"]Given real numbers $ a,b,c$ such that $ \\frac {1}{2} \\le a,b,c \\le 1$\nProve that $ 2\\le \\sum \\frac {a \\plus{} b}{1 \\plus{} c} \\le 3$.[/quote]\r\n\r\nSearch for Romania 2006 .", "Solution_2": "$ \\frac {a \\plus{} b}{1 \\plus{} c} < \\equal{} \\frac {2(a \\plus{} b)}{3}, \\frac {b \\plus{} c}{1 \\plus{} a} < \\equal{} \\frac {2(b \\plus{} c)}{3}, \\frac {a \\plus{} c}{1 \\plus{} b} < \\equal{} \\frac {2(a \\plus{} c)}{3}$.on adding we get $ \\sum_cyc \\frac {a \\plus{} b}{1 \\minus{} c} < \\equal{} \\frac {4(a \\plus{} b \\plus{} c)}{3} < \\equal{} 3.$\r\ni am unable to prove the left hand side :(", "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?t=84221\r\nalso,\r\n[quote=\"aadil\"]$ \\frac {a \\plus{} b}{1 \\plus{} c} < \\equal{} \\frac {2(a \\plus{} b)}{3}, \\frac {b \\plus{} c}{1 \\plus{} a} < \\equal{} \\frac {2(b \\plus{} c)}{3}, \\frac {a \\plus{} c}{1 \\plus{} b} < \\equal{} \\frac {2(a \\plus{} c)}{3}$.on adding we get $ \\sum_cyc \\frac {a \\plus{} b}{1 \\minus{} c} < \\equal{} \\frac {4(a \\plus{} b \\plus{} c)}{3} < \\equal{} 3.$\ni am unable to prove the right hand side :([/quote]\r\nam i too dum but how is $ \\frac {4(a \\plus{} b \\plus{} c)}{3} < \\equal{} 3$ that is not true is it?wat abt $ a\\equal{}b\\equal{}c\\equal{}1$" } { "Tag": [ "function", "algebra", "domain", "algebra unsolved" ], "Problem": "I want to find the function solution of this equation:\r\n\r\n$ xf(y)\\plus{}yf(x)\\equal{}(x^2\\plus{}y^2)f(xy)$.\r\n\r\nI found $ f(0)\\equal{}0$, and $ f(x)\\equal{}\\frac{1}{x}$ if $ x\\neq 0$. But I don't know if it is the \r\n\r\nunique solution.\r\n\r\nThanks in advance for some hint.", "Solution_1": "Hint:\r\n[hide]Fill in $ x \\equal{} 1$ (i suppose it's in the domain) and look what happens[/hide]\n\nSolution:\n[hide] Filling in $ x \\equal{} 1$ gives us $ yf(1) \\equal{} y^2f(y)$, so all satisfying functions are of the form\n\\[ f(x) \\equal{} \\frac {c}{x}\\]\nfor some $ c \\in \\mathbb{R}$; for all $ x \\neq 0$. They do all satisfy the conditions. [/hide]" } { "Tag": [ "algebra", "polynomial", "linear algebra", "matrix", "invariant", "linear algebra unsolved" ], "Problem": "Find the class of similitude of those real matrices $A$, $3nx3n$ such that $A^3=0$ and $rank A=2n$.", "Solution_1": "perhaps I'm wrong but cyclic spaces decomposition seems to work:\r\n\r\n1/ it is easy to check that the minimal polynomial of $A$ is $x^3$\r\n2/ we know that a matrix $M$ is similar to $A$ iff their invariant of similitude are the same (this is why it is called invariant of similitude isn't it :D ) So there exists a sequence of polynomial $P_1,..,P_k=x^3$ such that $P_i | P_{i+1}$ that are these invariants of similitude. So $P_i$ is either $x$, $x^2$ or $x^3$. Suppose that $P_1=P_2=..=P_a=x$, $P_{a+1}=P_2=..=P_{a+b}=x^2$, $P_{a+b+1}=P_2=..=P_{k}=x^3$. Then if we use the fact that $Rank(A)=2^n$ we can see that in fact $P_1=..=P_k=x^3$, if I did not made stupid mistakes.\r\n\r\n So I think that the similitude class of $A$ is exactly those matrices such that $A^3=0$ and $rank(A)=2n$. Is it ?", "Solution_2": "there exist $P\\in GL_{3n}(R)$ such that \r\n\r\n$P^{-1}AP=\\left(\\begin{array}{ccc}0&0&0\\\\I_n&0&0\\\\0&I_n&0\\end{array}\\right)$", "Solution_3": "so my answer was correct :)", "Solution_4": "Yes, but the proof is too complicated, Alekk. There is a simple proof in 4 lines. ;) I don't know if at X they will accept all the classical things you presented. Anyway, I like more your solution, since it is more natural." } { "Tag": [ "real analysis", "geometry", "rectangle", "algebra", "function", "domain", "advanced fields" ], "Problem": "Given $n \\in \\mathbb{N}$, let $\\lambda_{n}$ denote Lebesgue measure in $\\mathbb{R}^{n}$. Show that $\\lambda_{m+n}$ = $\\lambda_{m} \\times \\lambda_{n}$.\r\n\r\nIt probably suffices to show $\\lambda^{*}_{m+n}$ = $(\\lambda_{m} \\times \\lambda_{n})^{*}$. (Equality of the outer measures)", "Solution_1": "Check any construction of the product measure: such a measure is entirely characterized by its values on measurable rectangles. In the case of the product of two Lebesgue measures, this product clearly corresponds to the lebesgue measure of the (cartesian) product space on the measurable rectangles whose sides are open intervals, and from here it is easy to show that they correspond on any measurable rectangles.\r\n\r\nP.S: $B_{m} \\times B_{n} = B_{m+n}$ but $L_{m} \\times L_{n} \\subsetneq L_{m+n}$ where $B$ and $L$ denote the Borel and Lebesgue $\\sigma$-algebras respectively. In order to obtain $L_{m+n}$, you need to complete $L_{m} \\times L_{n}$ first. That means what we did above is rigorously true only for Borel $\\sigma$-algebras, otherwise the measures we talked about are different (since their domains are different).", "Solution_2": "I'm still having difficulty with this problem. Guess I need more help." } { "Tag": [ "limit", "function", "real analysis", "real analysis theorems" ], "Problem": "[b]If you read this thread and you're intimidated by all the discussion, please don't care. What I really want to know is: Is it obvious that we are ALLOWED TO REARRANGE THE TERMS IN A POSITIVE SERIES? If there is an easy proof/way to informally understand it please tell me. Even if there isn't one, please tell me so I can stop thinking about this![/b]\r\n\r\n\r\nThis deals with a Mandelbrot problem fron intermediate (http://www.mathlinks.ro/Forum/viewtopic.php?t=67347). We are to show that $\\sum\\limits_{j=1}^{\\infty}\\sum\\limits_{k = 1}^{\\infty}\\frac{1}{(j+k)^3} = \\sum_{n=1}^\\infty \\frac{n-1}{n^3}$. To quote scorpius, the method to see this is: \r\n\r\n[quote=\"scorpius119\"]Because the equation $j+k=n$ has $n-1$ solutions in positive integers $j$ and $k$, the term $\\frac{1}{n^3}$ will appear $n-1$ times in the double sum, so the sum becomes $\\sum_{n=1}^\\infty \\frac{n-1}{n^3}$. [/quote]\r\n\r\nNow, always locking myself up in petty details as I am prone to do (I've spent the past two hours thinking about this. Being me is a curse), I see huge problems with this reasoning. When I establish equality between two series, I want the partial sums to equal each other (at least sometimes), so that the limits are equal to each other. But for this reasoning, the partial sums are NEVER actually equal to each other. So, while the reasoning is easy enough to understand intuitively, I don't see how to formalize the \"rearrangement\" of terms that is done... seeing as we can't cut off at a number of terms and establish the equality. \r\n\r\nDo you understand my problem with this? If you read the original thread you will see that I've reduced the issue to just regrouping one series instead of the original \"double series\"*, but the same kind of problem with the rearrangement of terms appear. \r\n\r\nPlease help!\r\n\r\n\r\n* I've shown that $\\sum_{j=1}^m\\sum_{k=1}^{\\infty}\\frac{1}{(j+k)^3} = \\sum_{n=1}^{\\infty}\\frac{n-1}{n^3} - \\sum_{k=m+2}^{\\infty}\\frac{k - m - 1}{k^3}$, but again, this is not true if you only look at a partial sum instead of infinity.", "Solution_1": "the fact is that series with positive terms commute, and can be rearranged...\r\nit's a matter of absolute convergence, as far as i know..", "Solution_2": "Ok. A lot of strange things happening... Is this true then:\r\nIf both series are convergent, then $\\sum_{k=1}^{\\infty}f(k) + \\sum_{j=1}^{\\infty}f(j) = \\sum_{n=1}^{\\infty}(f(n) + g(n))$\r\n\r\nIt should be true from the definition of limits, but it seems like such strange things happen with series. So, is this result true, or I is there something I did not think of?", "Solution_3": "[i was just stating false and stupid things :p]\r\n\r\nanyway.. your question is slightly different (the original one): is it always true that $\\sum_i \\sum_j a_{i,j} = \\sum_j \\sum_i a_{i,j}$ or $\\sum_i \\sum_j a_{i,j} = \\sum_s \\sum_k a_{k,s-k}$?\r\nin general, answer is no...\r\ni don't know the sharp hypotesis you need on it, though.. maybe someone (kent, fedja, moubinool) could give us some enlightenment :)", "Solution_4": "I am going nuts now. Nothing is what I want it to be... Ok, so what's wrong here:\r\n\r\n$\\sum_{k=1}^{\\infty}(f(k) + g(k)) = \\lim_{n\\to\\infty}\\sum_{k=1}^{n}(f(k) + g(k)) = \\lim_{n\\to\\infty}\\left(\\sum_{k=1}^{n}f(k) + \\sum_{k=1}^{n}g(k)\\right) = \\lim_{n\\to\\infty}\\sum_{k=1}^{n}f(k) + \\lim_{n\\to\\infty}\\sum_{k=1}^{n}g(k) = \\sum_{k=1}^{\\infty}f(k) + \\sum_{k=1}^{\\infty}g(k)$\r\n\r\nFrom this we could make the appropriate changes in variable, and what I said should be true (we are working under the assumption that both series are convergent). Why is this not true? This is upsetting. :P", "Solution_5": "er..\r\nok, i'll edit it :p\r\nthat works, my mistake :D", "Solution_6": "That's the beauty of the statement scorpius made. It is a different procedure for counting.\r\n\r\n$\\sum\\limits_{j=1}^{\\infty}\\sum\\limits_{k = 1}^{\\infty}\\frac{1}{(j+k)^3} = \\frac1{2^3}+\\frac1{3^3}+\\frac1{4^3}+...+\\frac1{3^3}+\\frac1{4^3}+...+\\frac1{4^3}+...$\r\n\r\nFor this problem you are staring at infinite infinitely long series. If you just look at the grouping's I wrote you can easily see the relevance of the single summation.\r\n\r\nThe problem is you can't easily look at a partial sum, unless you change the limit so that it might be like \r\n\r\n$\\sum_{j=1}^{m} \\sum_{k=1}^{n-j} \\frac1{(j+k)^3}$\r\n\r\nIgnoring the ellipsis's above is this function for $m=3$ and $n=4$\r\n\r\nIf you want to more rigorously show that you are allowed to rearrange the terms as shown, although this is similar to the original method, then we can:\r\n\r\nFirst let's make a chart for all $(j+k)$\r\n2 3 4 5 6 7 8 9 10 11 ...\r\n 3 4 5 6 7 8 9 10 11 ...\r\n 4 5 6 7 8 9 10 11 ...\r\n 5 6 7 8 9 10 11 ...\r\n 6 7 8 9 10 11 ...\r\n 7 8 9 10 11 ...\r\n 8 9 10 11 ...\r\n 9 10 11 ...\r\n 10 11 ...\r\n 11 ...\r\n\r\nNow we can obviously see the first approach in action.\r\n\r\nLet $S_j = \\sum_{i=1}^{\\infty} \\frac1{(j+k)^3}$\r\n\r\nThen $S_j - S_{j+1} = \\frac1{(j+1)^3}$\r\n\r\nEach successive $S_j$ is losing it's previous first term. Then you can show that there are a given number of $j+k$ for the entire array of values, in fact $j+k-1$\r\n\r\nSo this can be rewritten as $\\sum_{j+k=2}^{j+k = \\infty} \\frac{j+k-1}{(j+k)^3} = \\sum_{n=2}^{\\infty} \\frac{n-1}{n^3} = \\sum_{n=1}^{\\infty} \\frac{n-1}{n^3}$\r\n\r\nHopefully my explanation helps, but essentially you are just tracking the the sums of the two indices through the primary index.", "Solution_7": "[quote=\"Kalle\"]$\\sum_{k=1}^{\\infty}(f(k) + g(k)) = \\lim_{n\\to\\infty}\\sum_{k=1}^{n}(f(k) + g(k)) = \\lim_{n\\to\\infty}\\left(\\sum_{k=1}^{n}f(k) + \\sum_{k=1}^{n}g(k)\\right) = \\lim_{n\\to\\infty}\\sum_{k=1}^{n}f(k) + \\lim_{n\\to\\infty}\\sum_{k=1}^{n}g(k) = \\sum_{k=1}^{\\infty}f(k) + \\sum_{k=1}^{\\infty}g(k)$[/quote]\r\n\r\nI think you meant that.\r\n\r\nWhere are you having your mistakes?", "Solution_8": "[quote=\"FMako\"]\nWhere are you having your mistakes?[/quote]\r\nNah, apparently I'm not having any. It's just that ma_go said that this statement wasn't true at first, but he then changed his mind. Well, the only thing I really feel unsure about is the change of variable, I guess. If we have the sum of two limits with two different variables, it is always OK to make a substitution so that they depend on the same variable, right? So using the limit laws we can evaluate this as one limit?\r\n\r\nAnd for the original problem, I'm glad you did all the work for me, but I still don't believe you :P I am okay with the fact that we have a rearrangement, I'm just not sure that we have shown the first series to have the same value as the second series. Isn't your more rigorous approach just a more rigorous way of saying that it is a rearrangement? Maybe I'm missing something.", "Solution_9": "I didn't do a really good job of verifying it, mainly because it seems obvious.\r\n\r\nI guess I don't understand what you don't understand.", "Solution_10": "[quote=\"FMako\"]I didn't do a really good job of verifying it, mainly because it seems obvious.\n\nI guess I don't understand what you don't understand.[/quote]\r\nThe value of a series DOES depend on order! See for example http://mathworld.wolfram.com/RiemannSeriesTheorem.html . I don't know if there is some obvious proof that rearrangements don't matter for positive series. I couldn't come up with it.." } { "Tag": [ "function", "algebra", "polynomial", "calculus", "derivative", "calculus computations" ], "Problem": "42. What is the greatest value of $ b$ for which any function $ f$ that satisfies the following properties must also satisfy $ f(1) < 5$?\r\n\r\n(i) $ f$ is infinitely differentiable on the real numbers;\r\n(ii) $ f(0)\\equal{}1$, $ f'(0)\\equal{}1$, and $ f''(0)\\equal{}2$; and\r\n(iii) $ |f'''(x)| 1$).!\r\nabout Giuga's conjecture:\r\n http://en.wikipedia.org/wiki/Giuga%27s_conjecture", "Solution_2": "this looks similar to the giuga's conjecture .\r\nThat is if $ \\displaystyle\\sum_{i \\equal{} 1}^{n \\minus{} 1}\\ i^{n \\minus{} 1} \\plus{} 1 \\equiv 0\\bmod{n}$ then $ n$ is prime :D" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "On aline, take three distinct points A,B,C in this order. Draw the tangnts AD,AE to the circle $ (\\omega)$ whose diameter is BC (D,E are the touching points). H is the orthogonal projection of D on CE. P is the midpoint of DH. CP meets again $ (\\omega)$ at Q. Show that the circumcircle (ADQ) is tangent to the line AC.", "Solution_1": "[quote=\"mr.danh\"]On aline, take three distinct points A,B,C in this order. Draw the tangnts AD,AE to the circle $ (\\omega)$ whose diameter is BC (D,E are the touching points). H is the orthogonal projection of D on CE. P is the midpoint of DH. CP meets again $ (\\omega)$ at Q. Show that the circumcircle (ADQ) is tangent to the line AC.[/quote]\r\n$ DE$ cut $ XY$ at the point X. \r\nBecause $ X,P$ is midpoints of $ DE$ and $ DH$ so $ XP\\parallel{}HE$ \r\nSo $ \\angle{DXP} \\equal{} \\angle {DEC} \\equal{} \\angle{DBP}$\r\nSo $ B,X,P,D$ are cyclic .\r\nBecause $ DH\\perp EC$ so $ XP\\perp DP$\r\nIt means that $ B,X,P,D$ lie on the circle diameter $ DX$ .Call it is $ (C)$\r\nBecause $ AX\\perp DX$ so $ AX$ is a tangent of $ (C)$\r\nSo $ \\angle{BXA} \\equal{} \\angle{BDX} \\equal{} \\angle{AEB}$\r\nIt follows that $ A,B,E,C$ are cyclic on the circle with diameter $ AE$. \r\nBecause $ A,B,E,C$ are cyclic so $ \\angle {XAQ} \\equal{} \\angle {BEX} \\equal{} \\angle {ADB}$\r\nSo $ AC$ is tangent of the circle $ (ADQ)$ at point A.", "Solution_2": "[quote=\"mr.danh\"]On aline, take three distinct points A,B,C in this order. Draw the tangnts AD,AE to the circle $ (\\omega)$ whose diameter is BC (D,E are the touching points). H is the orthogonal projection of D on CE. P is the midpoint of DH. CP meets again $ (\\omega)$ at Q. Show that the circumcircle (ADQ) is tangent to the line AC.[/quote]\r\nLook here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=123012#123012\r\nHere is another solution:\r\n[hide=\"Solution\"]\nLet $ \\angle ACE \\equal{} \\beta$ and $ \\angle QCB \\equal{} \\alpha$. Furthermore, let $ DE$ and $ BC$ to meet at $ X$. Now,\\[ \\angle QDX \\equal{} \\angle QDB \\plus{} \\angle BDE \\equal{} \\angle BCE \\plus{} \\angle QCB \\equal{} \\alpha \\plus{} \\beta\\]Also, since the center of the circle lies on $ BC$, we have that $ D$ and $ E$ are symmetric about $ BC$, so $ DX \\equal{} XE$. This means that $ XP$ is a midline of $ \\triangle DEH$, so $ XP\\parallel EH$. Then, $ \\angle PXC \\equal{} \\angle XCE \\equal{} \\beta$, so $ \\angle QPX \\equal{} \\angle PXC \\plus{} \\angle PCB \\equal{} \\alpha \\plus{} \\beta \\equal{} \\angle QDX$ and so $ QDPX$ is cyclic. Now, $ \\angle XPD \\equal{} \\angle EHD \\equal{} 90$, so $ \\angle QXD \\equal{} \\angle QPD \\equal{} \\angle DPX \\minus{} \\angle QPX \\equal{} 90 \\minus{} \\alpha \\minus{} \\beta$. $ D$ and $ E$ are symmetric about $ BC$, so $ \\angle DXA \\equal{} 90$, which means that $ \\angle QXA \\equal{} 90 \\minus{} \\angle QXD \\equal{} \\alpha \\plus{} \\beta$. Also, $ \\angle QEB \\equal{} \\angle QCB \\equal{} \\alpha$ and $ \\angle BEA \\equal{} \\beta$. Thus, $ \\angle AEQ \\equal{} \\angle AEB \\plus{} \\angle BEQ \\equal{} \\alpha \\plus{} \\beta \\equal{} \\angle QXA$, so $ QXEA$ is cyclic. Since $ D$ and $ E$ are symmetric about $ BC$, we notice that $ \\beta \\equal{} \\angle AEB \\equal{} \\angle ACE \\equal{} \\angle BCD \\equal{} \\angle BEX$, so $ \\angle AEX \\equal{} \\angle AEB \\plus{} \\angle BEX \\equal{} 2\\beta$. Therefore, $ \\angle XQA \\equal{} 180 \\minus{} \\angle AEX \\equal{} 180 \\minus{} 2\\beta$, so $ \\angle QAX \\equal{} 180 \\minus{} \\angle AQX \\minus{} \\angle QXA \\equal{} \\beta \\minus{} \\alpha$. Yet, $ \\angle QCD \\equal{} \\angle BCD \\minus{} \\angle BCQ \\equal{} \\beta \\minus{} \\alpha \\equal{} \\angle QAX$, so $ \\angle ADQ \\equal{} \\angle QCD \\equal{} \\angle QAX$, so the circumcircle of $ \\triangle AQD$ is tangent to $ AC$ at $ A$. [/hide]", "Solution_3": "Let $ T\\equal{}DE\\cdot BC$ then $ DT\\equal{}TE$ this means $ TP\\parallel CE$ $ \\Rightarrow$ $ PT\\perp DH$ because $ P$ is midpoint then, $ \\angle{TDP}90\\minus{}\\angle{DEH}\\equal{}90\\minus{}\\alpha\\equal{}\\angle{THP}$ because, $ \\angle{PQD}\\equal{}\\alpha$ $ \\Rightarrow$ $ QDPT$ is cyclic. Then, $ TQ\\perp QD$ let $ DQ\\cdot AB\\equal{}X$ then, $ X$ is midpoint of $ AT$ $ \\angle{ADX}\\equal{}\\angle{DCP}$ and, $ \\angle{PCH}\\equal{}\\angle{QDT}$ and $ \\triangle DHC \\equiv \\triangle{ATD}$. and because $ TQ\\perp XD$ $ XQ\\cdot XD\\equal{}XT^2\\equal{}XA^2$ $ \\Rightarrow$ $ AX$ is tangent to the circumcircle of $ \\triangle{AQD}$" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Dtermine all x,y such that 2^x+3^y is a perfect square.", "Solution_1": "Suppose $2^x+3^y=z^2$. First we check small cases: for $x=0$ we have $z^2-1=(z+1)(z-1)=3^y$, then $z+1, z-1$ are powers of $3$ that differ by $2$, so $z-1=1, z+1=3, z=2, y=1$. Similarly if $y=0$ we have $(z+1)(z-1)=2^x$, so $z+1, z-1$ are powers of $2$ that differ by $2$. Then $z-1=2, z+1=4, z=3, 2^x=8, x=3$. So we've found $(0, 1)$ and $(3, 0)$. Now suppose $x, y \\geq 1$.\r\n\r\nClearly since $3 \\mid 3^y$, and $3 \\nmid 2^x$ we see we must have $z^2 \\equiv 2^x \\equiv 1 \\pmod{3}$, i.e. $x=2n$. Now since $x \\geq 2$, $4 \\mid 2^x, 4 \\nmid 3^y$, so $z^2 \\equiv 3^y \\equiv 1 \\pmod{4}$, i.e. $y=2k$.\r\n\r\nSo $2^{2n}+3^{2k}=z^2$. We have $3^{2k}=(z+2^n)(z-2^n)$. So $z-2^n=3^a, z+2^n=3^b$ are powers of 3 that differ by $2^{n+1}$, that is, $3^a+2^{n+1}=3^b$. So either $3 \\mid 2^{n+1}$ or $a=0$. Since the first is absurd, we have $a=0$, that is $z-2^n=1, z+2^n=2^{n+1}+1=3^{2k}$. Then we have $2^{n+1}=(3^k+1)(3^k-1)$, so the factors are powers of 2 and differ by 2, which implies $3^k+1=4, 3^k-1=2, k=1, 2^{n+1}=8, n+1=3, n=2, 2n=x=4, y=2k=2$. So we have the solution $(4, 2)$ and there are no other solutions." } { "Tag": [ "puzzles" ], "Problem": "It's more powerful than God.\r\nIt's more evil than the devil.\r\nThe poor have it.\r\nThe rich need it.\r\nIf you eat it, you'll die.\r\nWhat am I?", "Solution_1": "[hide]nothing[/hide]", "Solution_2": "It is [hide]nothing[/hide]\n\nbut when I first found this teaser on braingle <-- place where you can ACTAULLY spam but I don't post, I didn't get the eating part, but now I do.\n\n[hide]nothing is more powerful than God\nnothing is more evil than the devil\nthe poor have nothing\nthe ricch need nothing\nIf you eat nothing, you will die[/hide]" } { "Tag": [ "Support" ], "Problem": "The new topic is: Democracy is best served by strict separation of church and state.\r\n\r\nThis is excellent - though controversial.", "Solution_1": "Hmm...I don't believe it's such a great topic or that it's one of the most controversial issues. It isn't horrible of anything, but definitely could have been much better.", "Solution_2": "Why? I don't think it's all that bad, besides the fact that I can't seem to figure out a good value and VC for my AFF case.", "Solution_3": "History is too one-sided on this. How many successful clergy-controlled governments do you know of?\r\n\r\nBut that's not even the main issue. Where do you plan to find evidence? What arguments can you use? The argument is generally fairly one-sided towards the AFF, unless of course NEG can pull out an argument about the definition of \"strict\" and how a complete seperation of church and state isn't beneficial and etc.\r\n\r\nI mean, just to name a few AFF arguments: freedom of religion, tolerance of different religions, political equality towards all religions, likeliness of discrimination, policy dictated by religious beliefs not necessarily accepted by populace, etc. And that's just what I came up with in the last 10 seconds.\r\n\r\nSo basically, NEG is running an uphill race here. And to be honest, I realli can't think of a good NEG argument off the bat except moral values.\r\n\r\nBut hey...we could discuss possible...uh...arguments here.", "Solution_4": "I agree. The negative has the challenge of proving grounds for which democracy is best served, and that's where the fundamental principles of religion might be handy. After all, where do we get all our moral pithy sayings? From religious beliefs, and religion often dictates what is moral. Thus, applying morality, religion might be essential for that matter. If not a theocracy, at least not utter, absolute, separation of church and state. :?", "Solution_5": "America was founded on religious beliefs.", "Solution_6": "[quote=\"SirErnest\"]America was founded on religious beliefs.[/quote]\r\n\r\nFunny, I thought the settlers came to Jamestown to plant 'baccy.", "Solution_7": "[quote=\"yif man12\"][quote=\"SirErnest\"]America was founded on religious beliefs.[/quote]\n\nFunny, I thought the settlers came to Jamestown to plant 'baccy.[/quote]\r\n\r\nThat was a side effect. :D \r\nThey came to get land for their king/queen/ruler.", "Solution_8": "[quote=\"Danbert\"][quote=\"yif man12\"][quote=\"SirErnest\"]America was founded on religious beliefs.[/quote]\n\nFunny, I thought the settlers came to Jamestown to plant 'baccy.[/quote]\n\nThat was a side effect. :D \nThey came to get land for their king/queen/ruler.[/quote]\r\n\r\nThey came for a lot of reasons. But that has nothing to do with separation of church and state. :-)\r\n\r\nAnd they still believed in randomly finding large piles of gold back then.", "Solution_9": "See, the problem with that, is that it is way too ethnocentric, since the USA isn't even mentioned in the resolution.\r\n\r\nThus, to use it, you have to establish it as a democracy first, and even at that first step, you'll run into huge roadblocks if your opponents knows their stuff. \r\n\r\nThen, you have to prove how that relates to the resolution... which is also difficult.", "Solution_10": "[quote=\"MithsApprentice\"]And they still believed in randomly finding large piles of gold back then.[/quote]\r\n\r\nI wouldn't mind randomly finding a large pile of gold.", "Solution_11": "[quote=\"yif man12\"][quote=\"MithsApprentice\"]And they still believed in randomly finding large piles of gold back then.[/quote]\n\nI wouldn't mind randomly finding a large pile of gold.[/quote]\r\n\r\nNeither would I! :lol:", "Solution_12": "eh stacks of hundred dollar bills work better for me :-)", "Solution_13": "What is the formal procedure for a debate?", "Solution_14": "AFF CASE: 6 MIN\r\nCX: 3 MIN\r\nNEG CASE / 1NR: 7 MIN\r\nCX: 3 MIN\r\n1AR: 4 MIN\r\n2NR: 6 MIN\r\n2AR: 3 MIN\r\n\r\nSo, each must have a value and a value criterion by which to judge the value given. This is a value debate, not policy, so there isn't alot of burdens and specific examples and incidences. Those aren't qualified evidence. It is more important to argue philosophy and use philosophy as a backbone. Then, you can build your case from your contentions.\r\n\r\nSO, a case would be.\r\n\r\nOpenning:\r\n\r\nValue:\r\nValue C:\r\n\r\nDefinitions:\r\n\r\nContention 1:\r\n\r\nSub A:\r\n\r\nSub B:\r\n\r\nContention 2:\r\n\r\nSub A:\r\n\r\nSub B:\r\n\r\nContention 3:\r\n\r\nSub A:\r\n\r\nSub B:", "Solution_15": "What does that mean?", "Solution_16": "Policy Debate > Value Debate\r\n\r\nSorry, I was too tempted. My brother did Value Debate (aka Lincoln-Douglas or LD), but I do Policy Debate (Cross-X or CX). I've seen an LD round before and thought it was so very slow. Then again, if anyone watches a CX round... I'd monitor their heart rate.", "Solution_17": "could just as easily be definitions then value...which is actually how most cases i know of are set up...\r\n\r\nand there can be any number of contentions...although 2/3 is optimal...", "Solution_18": "Public Forum annhilates LD and Policy. What I hate about policy. Anything (peeing in the ocean, butterfly effect) could lead to global nuclear war and ultimate human genocide in policy. What I hate about LD: all the topics suck this year.", "Solution_19": "[quote=\"churchilljrhigh\"]Anything (peeing in the ocean, butterfly effect) could lead to global nuclear war and ultimate human genocide in policy. [/quote]\r\n\r\nGuess you've never heard of a factor we know as \"significance\".", "Solution_20": "Tell me what you all think about using these philosophers for my cases:\r\nAff- Neitzche, Aristotole, Thomas Jefferson, James Madison\r\n\r\nNeg- Spinoza, George Washington, James Madison, St. Thomas Aquinas, Kant's Universiality\r\n\r\nI know what you'll say about using founding fathers for philosophers when the topic's not limited to the U.S. but they're ideas are no less substantial just because they were influences on the U.S.\r\n\r\nAlso, anyone know anything about the Hegelian Dialect? I printed it out, but as of now I'm way to lazy to actually read the thing.", "Solution_21": "[quote=\"MithsApprentice\"][quote=\"churchilljrhigh\"]Anything (peeing in the ocean, butterfly effect) could lead to global nuclear war and ultimate human genocide in policy. [/quote]\n\nGuess you've never heard of a factor we know as \"significance\".[/quote]\r\n\r\nIt doesn't matter how significant nuclear war is or how much the impact will be or how it outweigh all D/As. \r\nWhen you run a Nuclearism Kritik...........that is if you glean my meaning.", "Solution_22": "[quote=\"Katok\"]Tell me what you all think about using these philosophers for my cases:\nAff- Neitzche, Aristotole, Thomas Jefferson, James Madison\n\nNeg- Spinoza, George Washington, James Madison, St. Thomas Aquinas, Kant's Universiality\n\nI know what you'll say about using founding fathers for philosophers when the topic's not limited to the U.S. but they're ideas are no less substantial just because they were influences on the U.S.\n\nAlso, anyone know anything about the Hegelian Dialect? I printed it out, but as of now I'm way to lazy to actually read the thing.[/quote]\r\n\r\nMy two cents:\r\nLD Philosophers\r\nhttp://plato.stanford.edu/index.html\r\nPolicy Forums\r\nhttp://www.cross-x.com", "Solution_23": "[quote=\"churchilljrhigh\"]It doesn't matter how significant nuclear war is or how much the impact will be or how it outweigh all D/As. \nWhen you run a Nuclearism Kritik...........that is if you glean my meaning.[/quote]\r\n\r\nActually I was referring to peeing in the ocean. Anyone suggesting that peeing in the ocean would ultimately lead to nuclear war needs to get their brain checked. It is if they can even find evidence to support their claim in the first place.\r\n\r\nAnd honestly, I don't know what you mean. This significance factor as well as arguments of feasability usually known down outrageous arguments.", "Solution_24": "Thank you, you're wonderful. I'll check it out. Anyone else?", "Solution_25": "[quote=\"MithsApprentice\"][quote=\"churchilljrhigh\"]It doesn't matter how significant nuclear war is or how much the impact will be or how it outweigh all D/As. \nWhen you run a Nuclearism Kritik...........that is if you glean my meaning.[/quote]\n\nActually I was referring to peeing in the ocean. Anyone suggesting that peeing in the ocean would ultimately lead to nuclear war needs to get their brain checked. It is if they can even find evidence to support their claim in the first place.\n\nAnd honestly, I don't know what you mean. This significance factor as well as arguments of feasability usually known down outrageous arguments.[/quote]\r\n\r\nThe negative has a ton of cheap arguments to make if their are some evident problems in the aff case (Extra T and Effects T). Yes, that's the whole point, the negative is saying that topicality is a voter issue (sure, neg is making the judge believe that the aff are crazy, not to sure if that will get you any courtesy points though).", "Solution_26": "I have no idea what that post was supposed to suggest. Care to clarify?\r\n\r\nAnd fyi, policy debate actually requires some argumentation skills. It's not a game of who can make the stupidest remarks.", "Solution_27": "[quote=\"MithsApprentice\"]It's not a game of who can make the stupidest remarks.[/quote]\r\n\r\nDang. Debate's not gonna be my future occupation.", "Solution_28": "If you argue on the neg side that religion in democracy promotes moral actions and decisions, how would you support it, especially if your opponent brings up examples such as abortion rights in the US Supreme Court (and please don't jump on my back about the topic not being about just the US:?)", "Solution_29": "[quote=\"History Genius\"]If you argue on the neg side that religion in democracy promotes moral actions and decisions, how would you support it, especially if your opponent brings up examples such as abortion rights in the US Supreme Court (and please don't jump on my back about the topic not being about just the US:?)[/quote]\r\n\r\nI would stay away from morality, since, unlike our previous cases, this really has nothing to do with it. In the topic states nothing about any moral obligation to protect the people, only to give the best service to [u] the democracy[/u]. We want the system that will best profit the democracy, not necessarily the people that live in it. Are you catching what I'm saying?\r\n\r\nP.S. I love James and Cat!", "Solution_30": "[quote=\"History Genius\"]If you argue on the neg side that religion in democracy promotes moral actions and decisions, how would you support it, especially if your opponent brings up examples such as abortion rights in the US Supreme Court (and please don't jump on my back about the topic not being about just the US:?)[/quote]\r\n\r\nSee, in my opinion, your value should at least be democracy related, since you need to prove that a non-strict separation upholds democracy. Then you would have to tie religion into your VC, since it will most likely be religion in your arguments which influences whether a democracy will be good or bad, or successful or not.", "Solution_31": "why do you guys talk so much about founding fathers, the united states, blalbala it's democracy that counts not united states. you can define democracy in a lot of different ways so the U.S constitution is not importent and btw u.s is a republic not democracy.\r\n\r\ni'l try knowledge v and cultural diverstity for my neg", "Solution_32": "The United States is a democracy - it may have characteristics which define a republic, but it still runs on a democratic system.", "Solution_33": "hmm...that's a very detailed debate folks...i mean it depends on how u define democracy. the US is definitely not a pure democracy...but it's like half and half. some issues everyone votes on and other things we just vote for representatives to vote on", "Solution_34": "It's just because we have a electoral system which throws it out of whack. Otherwise...", "Solution_35": "Not only that -- in a \"pure\" democratic system, every citizen would have an opportunity to vote on every issue that required a societal decision. (Not that this is necessarily a desirable system, but it is the logical extreme of a democracy.) In our system, we have it broken down into all sorts of odd categorizations -- I mean, I have at least 3 major legislative bodies and 4 executive branches which are responsible for running things -- in New York, I hardly ever have the opportunity to vote directly on an issue.", "Solution_36": "What would you debate for a neg side anyway then? Couldn't you say that the reason why other countries have become so poor and war torn is because of neglient leaders and immoral dictators?", "Solution_37": "I dont get what you guys are talking about :blush:" } { "Tag": [ "inequalities", "ratio", "trigonometry", "geometry", "triangle inequality", "inequalities unsolved" ], "Problem": "Prove that $\\frac{1}{m_{a}}+\\frac{1}{m_{b}}+\\frac{1}{m_{c}}>\\frac{5}{s}$,\r\n\r\nwhere $m_{a}$, $m_{b}$, $m_{c}$ are the three medians of a triangle $ABC$, and $s=\\frac{a+b+c}{2}$ is the semiperimeter of this triangle.", "Solution_1": "We have : $m_A < \\frac{b+c}{2}$, $m_B < \\frac{a+c}{2}$, $m_C < \\frac{a+b}{2}$\r\n\r\nThus : $m_A+m_B+m_C < 2s$\r\n\r\nThus $\\frac{1}{m_A}+\\frac{1}{m_B}+\\frac{1}{m_C}>= \\frac{9}{m_A+m_B+m_C} > 9/2$\r\n\r\nThen, I'm not sure your inequality is true.", "Solution_2": "I forgot the s, we have thus\r\n\r\n$\\frac{1}{m_A}+\\frac{1}{m_B}+\\frac{1}{m_C}>= \\frac{9}{m_A+m_B+m_C} > \\frac{4.5}{s}$", "Solution_3": "I agree with you.\r\n\r\n$\\displaystyle \\frac{1}{m_a}+\\frac{1}{m_b}+\\frac{1}{m_c}>\\frac{5}{s}>\\frac{4.5}{s}$ ???", "Solution_4": "Labiliau, your estimation is correct but it is pretty weak. You should try to improve it :) (or disprove the inequality by W. Janous)", "Solution_5": "it seams right, because there is only one \"extreme\" situation, where it is equal, but never smaller...\r\ni mean, when one side is zero ...\r\n\r\nbut i could prove it yet, but cauchy in the way it is used is definitly too weak and also the inequalities $2m_a\\frac{10}{x+y+z}\\]", "Solution_7": "Maybe not the most beautiful proof ever posted on ML, but still a proof:\r\n\r\n[color=blue][b]Theorem 1.[/b] If $m_a$, $m_b$, $m_c$ are the medians and $s =\\frac{a+b+c}{2}$ is the semiperimeter of a triangle ABC, then $\\frac{1}{m_a}+\\frac{1}{m_b}+\\frac{1}{m_c}>\\frac{5}{s}$.[/color]\r\n\r\n[i]Proof of Theorem 1.[/i] We will verify a different inequality:\r\n\r\n$\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}>\\frac{15}{2\\left(m_a+m_b+m_c\\right)}$.\r\n\r\nThis inequality will readily yield Theorem 1 by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59415]the medians duality principle[/url]: In fact, applying this inequality to the medians triangle of triangle ABC, which has sidelengths $m_a$, $m_b$, $m_c$ and medians $\\frac34a$, $\\frac34b$, $\\frac34c$, we obtain $\\frac{1}{m_a}+\\frac{1}{m_b}+\\frac{1}{m_c}>\\frac{15}{2\\left(\\frac34a+\\frac34b+\\frac34c\\right)}$, what simplifies to $\\frac{1}{m_a}+\\frac{1}{m_b}+\\frac{1}{m_c}>\\frac{5}{\\left(\\frac{a+b+c}{2}\\right)}=\\frac{5}{s}$, so that Theorem 1 is proven.\r\n\r\nThus, it remains to prove the inequality\r\n\r\n$\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}>\\frac{15}{2\\left(m_a+m_b+m_c\\right)}$.\r\n\r\nThis rewrites as\r\n\r\n$\\frac23\\left(m_a+m_b+m_c\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)>5$.\r\n\r\nNow, the centroid G of triangle ABC is the point of intersection of its medians $m_a$, $m_b$, $m_c$, and divides these medians in the ratio 2 : 1. Hence, $AG=\\frac23m_a$, $BG=\\frac23m_b$, $CG=\\frac23m_c$, so that $\\frac23\\left(m_a+m_b+m_c\\right)=\\frac23m_a+\\frac23m_b+\\frac23m_c=AG+BG+CG$, and the inequality in question becomes\r\n\r\n$\\left(AG+BG+CG\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)>5$.\r\n\r\nNow, instead of directly attacking this inequality, we will show a more general assertion:\r\n\r\n[color=blue][b]Theorem 2.[/b] If ABC is a triangle and P an arbitrary point in its plane, then\n\n$\\left(AP+BP+CP\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)>5$.[/color]\r\n\r\n[i]Proof of Theorem 2.[/i] We distinguish between two cases:\r\n\r\n[i]Case 1:[/i] The triangle ABC has an angle > 120\u00b0. We can WLOG assume that A > 120\u00b0.\r\n\r\nThen, it is well-known that the point Q in the plane of triangle ABC for which the sum AQ + BQ + CQ is minimal is the vertex A of the triangle ABC. Hence, for every point P in the plane, we have $AP+BP+CP\\geq AA+BA+CA$. This simplifies to $AP+BP+CP\\geq 0+c+b=b+c$. Hence, instead of proving the inequality $\\left(AP+BP+CP\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)>5$, it is enough to verify $\\left(b+c\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)>5$. But this is easy: The triangle inequality for triangle ABC yields b + c > a, and the Cauchy-Schwarz inequality (or AM-HM) yields $\\left(b+c\\right)\\left(\\frac{1}{b}+\\frac{1}{c}\\right)\\geq 4$; thus,\r\n\r\n$\\left(b+c\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)=\\left(b+c\\right)\\frac{1}{a}+\\left(b+c\\right)\\left(\\frac{1}{b}+\\frac{1}{c}\\right)>a\\frac{1}{a}+4=1+4=5$.\r\n\r\nTherefore, the proof of Theorem 2 in Case 1 is complete.\r\n\r\n[i]Case 2:[/i] All angles of triangle ABC are $\\leq 120^{\\circ}$.\r\n\r\nWLOG, we assume that A is the greatest angle of triangle ABC. Then, $A\\leq 120^{\\circ}$ by the condition of Case 2, but on the other hand, $A\\geq 60^{\\circ}$ (in fact, if the angle A would be < 60\u00b0, then the angles B and C would be < 60\u00b0 as well - because A is the greatest angle of triangle ABC -, and thus we would have A + B + C < 60\u00b0 + 60\u00b0 + 60\u00b0 = 180\u00b0, contradicting the sum of the angles in triangle ABC which states that A + B + C = 180\u00b0). From $A\\leq 120^{\\circ}$ and $A\\geq 60^{\\circ}$, it follows that $\\sin A\\geq\\sin 60^{\\circ}=\\sin 120^{\\circ}=\\frac{\\sqrt3}{2}$. On the other hand, the area $\\Delta$ of triangle ABC equals $\\Delta=\\frac12bc\\sin A$. Thus, $\\Delta=\\frac12bc\\sin A\\geq\\frac12bc\\cdot\\frac{\\sqrt3}{2}=\\frac{\\sqrt3}{4}bc$, so that $4\\sqrt3\\Delta\\geq 4\\sqrt3\\frac{\\sqrt3}{4}bc=3bc$.\r\n\r\nNow, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=191721#p191721]http://www.mathlinks.ro/Forum/viewtopic.php?t=19316 post #4[/url] Theorem 2, we have $AP+BP+CP\\geq W$, where $W=\\sqrt{\\frac12\\cdot\\left(a^2+b^2+c^2+4\\sqrt3\\Delta\\right)}$. Thus,\r\n\r\n$AP+BP+CP\\geq\\sqrt{\\frac12\\cdot\\left(a^2+b^2+c^2+4\\sqrt3\\Delta\\right)}\\geq\\sqrt{\\frac12\\cdot\\left(a^2+b^2+c^2+3bc\\right)}$.\r\n\r\nHence, the inequality in question, $\\left(AP+BP+CP\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)>5$, will be evident once the following inequality is shown:\r\n\r\n$\\sqrt{\\frac12\\cdot\\left(a^2+b^2+c^2+3bc\\right)}\\cdot\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)>5$.\r\n\r\nUpon squaring, this inequality becomes\r\n\r\n$\\frac12\\cdot\\left(a^2+b^2+c^2+3bc\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^2>25$,\r\n\r\nand upon multiplication by 2, this becomes\r\n\r\n$\\left(a^2+b^2+c^2+3bc\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^2>50$.\r\n\r\nNow, denote by $h=\\frac{2}{\\frac{1}{b}+\\frac{1}{c}}=\\frac{2bc}{b+c}$ the harmonic mean of the numbers b and c. Then, by the QM-HM inequality, $\\sqrt{\\frac{b^2+c^2}{2}}\\geq h$, so that $\\frac{b^2+c^2}{2}\\geq h^2$ and $b^2+c^2\\geq 2h^2$, and by the GM-HM inequality, $\\sqrt{bc}\\geq h$, so that $bc\\geq h^2$. Finally, $h=\\frac{2}{\\frac{1}{b}+\\frac{1}{c}}$ yields $\\frac{1}{b}+\\frac{1}{c}=\\frac{2}{h}$. Thus,\r\n\r\n$\\left(a^2+b^2+c^2+3bc\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^2\\geq\\left(a^2+2h^2+3h^2\\right)\\left(\\frac{1}{a}+\\frac{2}{h}\\right)^2=\\left(a^2+5h^2\\right)\\left(\\frac{1}{a}+\\frac{2}{h}\\right)^2$,\r\n\r\nand it remains to show that\r\n\r\n$\\left(a^2+5h^2\\right)\\left(\\frac{1}{a}+\\frac{2}{h}\\right)^2>50$.\r\n\r\nThe following nice proof of this inequality is due to Peter Scholze (my own proof was pretty messy): By AM-GM, we have\r\n\r\n$a^2+\\frac53h^2+\\frac53h^2+\\frac53h^2\\geq 4\\sqrt[4]{a^2\\cdot\\frac53h^2\\cdot\\frac53h^2\\cdot\\frac53h^2}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ a^2+5h^2\\geq 4\\sqrt[4]{\\left(\\frac53\\right)^3a^2h^6}$\r\n\r\nand\r\n\r\n$\\frac{1}{a}+\\frac{2}{3h}+\\frac{2}{3h}+\\frac{2}{3h}\\geq 4\\sqrt[4]{\\frac{1}{a}\\cdot\\frac{2}{3h}\\cdot\\frac{2}{3h}\\cdot\\frac{2}{3h}}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ \\frac{1}{a}+\\frac{2}{h}\\geq 4\\sqrt[4]{\\left(\\frac23\\right)^3\\cdot\\frac{1}{ah^3}}$;\r\n\r\nhence,\r\n\r\n$\\left(a^2+5h^2\\right)\\left(\\frac{1}{a}+\\frac{2}{h}\\right)^2\\geq 4\\sqrt[4]{\\left(\\frac53\\right)^3a^2h^6}\\cdot\\left(4\\sqrt[4]{\\left(\\frac23\\right)^3\\cdot\\frac{1}{ah^3}}\\right)^2$\r\n$=64\\sqrt[4]{\\left(\\frac53\\right)^3\\left(\\frac23\\right)^6}>50$\r\n\r\n(in fact, $64\\sqrt[4]{\\left(\\frac53\\right)^3\\left(\\frac23\\right)^6}\\approx 51.101029$). And Theorem 2 is proven in Case 2. This completes the proof of Theorem 1.\r\n\r\n Darij", "Solution_8": "[quote=\"darij grinberg\"]Maybe not the most beautiful proof ever posted on ML, but still a proof[/quote]\r\nwell, i think too, that its a proof ;) ... but it is that easy that i was able to understand it. i didnt expect that, so NICE! ;)\r\njob job darij :)", "Solution_9": "[quote=\"manlio\"]Prove that the triangle inequality\n\\[\\frac{1}{m_{a}}+\\frac{1}{m_{b}}+\\frac{1}{m_{c}}>\\frac{5}{s}\\]\nwhere $m_{a},\\ m_{b},\\ m_{c}$ are medians of a triangle and $s$ is its semi-perimeter.[/quote]\r\nThis is a problem was proposed by Walther Janous, see [url=http://journals.cms.math.ca/CRUX/]CRUX Mathematicorum[/url], Vol. 12 (1986), p. 177, Problem $1137^{*}$. A solution by Ji Chen, see Vol. 18 (1992) , pp. 208 - 209.", "Solution_10": "Can you post a complete solution ? You think all mathlinker can download Crux as you? [/color]", "Solution_11": "I have just merged this topic with another one, so you can find my proof of the inequality above, but I would love to know the one you proposed in CRUX, Ji Chen!\r\n\r\n darij" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Prove that:\r\n$ \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{F_nF_{n \\plus{} 2}^2F_{n \r\n\\plus{} 3}} \\plus{} \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{F_nF_{n \\plus{} 1}^2F_{n \\plus{} 3}} \\equal{} \\frac12$", "Solution_1": "For conbernience, let $ a\\equal{}F_n$, $ b\\equal{}F_{n\\plus{}1}$, $ c\\equal{}F_{n\\plus{}2}$, and $ d\\equal{}F_{n\\plus{}3}$. Then $ a\\plus{}b\\equal{}c$ and $ b\\plus{}c\\equal{}d$.\r\n$ LHS$\r\n$ \\equal{} \\sum_{n\\equal{}1}^\\infty \\frac{1}{ac^2d} \\plus{} \\sum_{n\\equal{}1}^\\infty \\frac{1}{ab^2d}$\r\n$ \\equal{} \\sum_{n\\equal{}1}^\\infty \\left( \\frac{1}{ac^2d} \\plus{} \\frac{1}{ab^2d} \\right)$\r\n$ \\equal{} \\sum_{n\\equal{}1}^\\infty \\left( \\frac{b}{abc^2d} \\plus{} \\frac{c}{ab^2cd} \\right)$\r\n$ \\equal{} \\sum_{n\\equal{}1}^\\infty \\left( \\frac{c\\minus{}a}{abc^2d} \\plus{} \\frac{d\\minus{}b}{ab^2cd} \\right)$\r\n$ \\equal{} \\sum_{n\\equal{}1}^\\infty \\left( \\frac{1}{abcd} \\minus{} \\frac{1}{bc^2d} \\plus{} \\frac{1}{ab^2c} \\minus{} \\frac{1}{abcd} \\right)$\r\n$ \\equal{} \\sum_{n\\equal{}1}^\\infty \\left( \\frac{1}{ab^2c} \\minus{} \\frac{1}{bc^2d} \\right)$\r\n$ \\equal{} \\sum_{n\\equal{}1}^\\infty \\left( \\frac{1}{F_n F_{n\\plus{}1}^2 F_{n\\plus{}2}} \\minus{} \\frac{1}{F_{n\\plus{}1} F_{n\\plus{}2}^2 F_{n\\plus{}3}} \\right) \\ \\ \\leftarrow$ Telescoping sum\r\n$ \\equal{} \\frac{1}{F_1 F_2^2 F_3}$\r\n$ \\equal{} \\frac{1}{2}$" } { "Tag": [ "geometry", "incenter", "search", "circumcircle", "angle bisector", "cyclic quadrilateral", "geometry proposed" ], "Problem": "Let $\\triangle ABC$ be a triangle, $I$ its incentre and $\\gamma _a$ the circle internally tangent to the sides $AB,AC$ and to the circumcircle. \r\nDenote with $M,N$ the points of contact of $\\gamma _a$ with $AB,AC$ respectively. Show that :\r\n\r\n(1) $I, M, N $ are collinear\r\n\r\n(2) $MN \\perp AI$", "Solution_1": "discussed. a search for \"casey's theorem\" / \"tranversal theorem\" would probably pull up the topic as I remember that being used in one of the proofs. I think this might also be IMO78/4. Finally, a recent Belgrade problem was generalized and this was a special case; look for that too.", "Solution_2": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=46418 Soon will appear the solution.", "Solution_3": "Let us to present an elementary proof of the collinearity of $ I,\\ M,\\ N,$ which it is also enough for the proof of the result of $ AI\\perp MN.$\r\n\r\n$ \\bullet$ We denote as $ M,\\ N,$ the intersection points of $ AB,\\ AC$ respectively, from the line through the incenter $ I$ of $ \\bigtriangleup ABC$ and perpendicular to $ AI.$\r\n\r\nThe lines through the points $ M,\\ N$ and perpendicular to $ AB,\\ AC$ respectively, intersect each other at the point so be it $ O',$ lies on the angle bisector of $ \\angle A$ and clearly, the circle $ (O')$ centered at this point with radius $ O'M \\equal{} O'N,$ tangents to $ AB,\\ AC,$ at points $ M,\\ N,$ respectively.\r\n\r\nWe will prove that the circle $ (O'),$ also tangents to the circumcircle $ (O)$ of the given triangle $ \\bigtriangleup ABC.$\r\n\r\n$ \\bullet$ Let be the points $ B'\\equiv (O)\\cap BI$ and $ C'\\equiv (O)\\cap CI.$\r\n\r\nLet also be the points $ P\\equiv OO'\\cap B'N,$ $ P'\\equiv OO'\\cap C'M$ and we will prove trhat $ P'\\equiv P.$\r\n\r\nFrom $ O'N\\parallel OB',$ applying the [b][size=100]Thales theorem[/size][/b], we have that $ \\frac {PO'}{PO} \\equal{} \\frac {O'N}{OB'}$ $ ,(1)$\r\n\r\nSimilarly, from $ O'M\\parallel OC'$ $ \\Longrightarrow$ $ \\frac {P'O'}{P'O} \\equal{} \\frac {O'M}{OC'}$ $ ,(2)$\r\n\r\nFrom $ (1),$ $ (2)$ and because of $ \\frac {O'N}{OB'} \\equal{} \\frac {O'M}{OC'}$ $ ,(3)$ $ \\Longrightarrow$ $ \\frac {PO'}{PO} \\equal{} \\frac {P'O'}{P'O}$ $ \\Longrightarrow$ $ \\frac {PO'}{PO \\minus{} PO'} \\equal{} \\frac {P'O'}{P'O \\minus{} P'O'}$ $ \\Longrightarrow$ $ \\frac {PO}{OO'} \\equal{} \\frac {P'O'}{OO'}$ $ \\Longrightarrow$ $ PO' \\equal{} P'O'$ $ \\Longrightarrow$ $ P'\\equiv P$ $ ,(4)$\r\n\r\n$ \\bullet$ We consider the non-convex hexagon $ ABB'PC'C$ and because of the collinearity of the points $ M\\equiv AB\\cap PC',$ $ I\\equiv BB'\\cap C'C,$ $ N\\equiv B'P\\cap AC,$ based on the [b][size=100]Pascal theorem[/size][/b], we conclude that this hexagon is cyclic.\r\n\r\nSo, the point $ P\\equiv OO'\\cap B'N\\cap C'M,$ lies on the circumcircle of $ \\bigtriangleup ABC.$\r\n\r\nBecause of now, $ OP \\equal{} OB'$ and $ O'N\\parallel OB',$ we conclude that $ O'P \\equal{} O'N$ $ ,(5)$ and similarly we have $ O'P \\equal{} O'M$ $ ,(6)$\r\n\r\nFrom $ (5),$ $ (6)$ and because of the collinearity of the points $ O,\\ O',\\ P,$ we conclude that the circle $ (O'),$ which tangent to $ AB,\\ AC$ at points $ M,\\ N$ respectively, tangents also internally to the circle $ (O)$ at point $ P$ and the proof is completed.\r\n\r\nKostas Vittas.", "Solution_4": "Here is a nice generalization of the collinearity: http://www.mathlinks.ro/viewtopic.php?t=170192 (when cyclic quadrilateral $ ABCD$ degrades to triangle $ ABC$)" } { "Tag": [ "trigonometry", "function", "integration", "calculus", "calculus computations" ], "Problem": "Find\r\n\r\n$\\int{\\sec^{-1}{x}}\\,dx$\r\n\r\n[hide=\"Hint\"]\nUse the substitution $x=\\sec{y}$[/hide]", "Solution_1": "I used $x = \\sec{k}$.\r\nso $dx = \\sec{k}. \\tan{k}dk$\r\n\r\nSubst. inside will get $\\int{k\\sec{k}\\tan{k}}\\,dk$\r\n\r\nThen, use integration by parts, \r\n\r\n$I = k\\sec{k}-\\int{\\sec{k}}\\,dk+C_{1}$\r\n $= k\\sec{k}-ln|\\sec{k}+\\tan{k}|+C$\r\n \r\nSubst. back will get,\r\n$\\int{\\sec^{-1}{x}}\\,dx = x\\sec^{-1}{x}-ln|x+\\sqrt{x^{2}-1}|+C$", "Solution_2": "Correct :lol:", "Solution_3": "would knowing that $\\sec^{-1}(x)=\\cos^{-1}\\left(\\frac{1}{x}\\right)$ help at all?", "Solution_4": "[quote=\"maokid7\"]would knowing that $\\sec^{-1}(x)=\\cos^{-1}\\left(\\frac{1}{x}\\right)$ help at all?[/quote]\r\n\r\nI don't think so. For me, it makes things more complicated. :D", "Solution_5": "for me it was much simpler\r\n[hide=\"here is how it is done\"]\n$\\int\\sec^{-1}(x)dx=\\int\\cos^{-1}\\left(\\frac{1}{x}\\right)dx$\nso integrating by-parts we use\n\n$dv=dx$ $v=x$ $u=\\cos^{-1}\\left(\\frac{1}{x}\\right)$ and $du=\\frac{1}{x\\sqrt{x^{2}-1}}$\n\nso\n\n$\\int\\cos^{-1}\\left(\\frac{1}{x}\\right)dx=x\\cos^{-1}\\left(\\frac{1}{x}\\right)-\\int\\frac{1}{\\sqrt{x^{2}-1}}$\n\nto evaluate the second integral we use $x=\\cosh(\\theta)$\n\nso\n$\\int\\cos^{-1}\\left(\\frac{1}{x}\\right)dx=x\\cos^{-1}\\left(\\frac{1}{x}\\right)-\\cosh^{-1}(x)=x\\sec^{-1}(x)-\\ln(|x+\\sqrt{x^{2}-1}|)+C$\n[/hide]", "Solution_6": "you're right, That's a great solution! :D \r\ndidn't think of using hyperbolic functions. i only learnt them 2days ago :P" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $G$ be a finite group of order $90$ with a non-normal subgroup,$H$,of order $5$.\r\na).Show that $G$ has a subgroup of order $15$.Show this subgroup is abelian.", "Solution_1": "I would not know how to do this without Sylow, you know Sylow theorems?\r\n\r\nThe number of subgroups of order five must be a divisor of ninety, and 1 mod 5\r\nso it is 1 or 6\r\nhowever is also is part of sylow theorems that they are all conjugated\r\nso you said H is not normal, so the number is six\r\n\r\n\r\nnow by this conjugation, we get a permutation representation on G itself, transitively\r\nso the stabilisator of H is 90/6=15 by the orbit formula\r\n\r\n\r\nnow there is a much more general theorem that every group of order p*q with p>>\r\n\r\nChangelog: \r\n> Set R was changed to set Q because of a stupid mistake.\r\n> Roots of any NON-CUBES cannot be rational.\r\n> Added proof of the non-rational statement.\r\n\r\nAssume the contrary:\r\nLet the three primes be $ x,y,z$.\r\nLet (by the definition of an arithmetic progression)\r\n$ \\begin{cases} \\sqrt [3]{y} \\equal{} \\sqrt [3]{x} \\plus{} pr \\\\\r\n\\sqrt [3]{z} \\equal{} \\sqrt [3]{y} \\plus{} qr \\end{cases}$\r\nWhere $ p, q\\in\\mathbb{Z}$, $ p, q\\ne 0$.\r\nThen, multiplying by $ q$ in the first equation and $ p$ in the second, we have\r\n$ \\begin{cases} q\\sqrt [3]{y} \\equal{} q\\sqrt [3]{x} \\plus{} pqr \\\\\r\np\\sqrt [3]{z} \\equal{} p\\sqrt [3]{y} \\plus{} pqr \\\\\r\n\\end{cases}$\r\nThen subtracting the second from the first, we have\r\n$ q\\sqrt [3]{y} \\minus{} p\\sqrt [3]{z} \\equal{} q\\sqrt [3]{x} \\minus{} p\\sqrt [3]{y}$\r\nAnd then, adding $ p\\sqrt [3]{y}$ to both sides we have\r\n$ (p \\plus{} q)\\sqrt [3]{y} \\minus{} p\\sqrt [3]{z} \\equal{} q\\sqrt [3]{x}$\r\nNow, let $ k \\equal{} p \\plus{} q$ for convenience.\r\n$ k\\sqrt [3]{y} \\minus{} p\\sqrt [3]{z} \\equal{} q\\sqrt [3]{x}$ (EQUATION 1)\r\nWe then cube.\r\n$ k^3y \\minus{} 3k^2p\\sqrt [3]{y^2z} \\plus{} 3kp^2\\sqrt [3]{yz^2} \\minus{} p^3z \\equal{} q^3x$\r\nNow, grouping some terms, we have\r\n$ k^3y \\minus{} p^3z \\minus{} 3kp\\sqrt [3]{yz}\\left( k\\sqrt [3]{y} \\minus{} p\\sqrt [3]{z}\\right) \\equal{} q^3x$\r\nSubstituting (1) and adding $ p^3z \\minus{} k^3y$ we have\r\n$ \\minus{} 3kpq\\sqrt [3]{xyz} \\equal{} q^3x \\plus{} p^3z \\minus{} k^3y$\r\nDividing by $ \\minus{} 3kpq$ we have\r\n$ \\sqrt [3]{xyz} \\equal{} \\dfrac{k^3y \\minus{} q^3x \\minus{} p^3z}{3kpq}$\r\nThen, since the numerator and denominator in the RHS are both integer, $ \\text{RHS} \\in\\mathbb{Q}$. But this cannot be rational. Assume $ \\sqrt[3]{xyz}\\equal{}\\dfrac{g}{h}$. Then $ xyz \\equal{} \\dfrac{g^3}{h^3}$, and since $ xyz$ is an integer, let $ n\\equal{}1$. Then $ xyz \\equal{} m^3$, which is impossible since $ x,y,z$ are distinct primes. Therefore, $ \\sqrt [3]{xyz}\\not\\in\\mathbb{Q}$. Contradiction!", "Solution_1": "If I were writing this up, I'd probably start straight from equation 1; that's a well-known property of arithmetic sequences, I would think.\r\n\r\nRoots of any kind cannot be rational is not quite a true statement. $ \\sqrt{4}$? The cube root of the product of three distinct prime numbers on the other hand...", "Solution_2": "For the end, I think you mean Q (rationals), not R (reals). Also, for a bit more rigor, you can go like:\r\nAssume $ \\sqrt[3]{xyz}\\equal{}\\frac{m}{n}$, (a rational with all the usual m,n are integers stuff). Then $ xyz\\equal{}\\frac{m^3}{n^3}$, and since $ xyz$ is an integer, we can let $ n\\equal{}1$. Then we have $ xyz\\equal{}m^3$, which is impossible since $ x,y,z$ are distinct primes.\r\n\r\nAnother method you can use goes as follows:\r\nConsider the polynomial $ f(t)\\equal{}t^3\\minus{}xyz$. Since the polynomial is monic, any rational root of the polynomial must be an integer. Thus, showing that there are no integer roots of the polynomial will prove that the polynomial has no rational roots. Brute force checking with all possible factors of $ xyz$ shows that this is true.\r\n\r\nSince $ \\sqrt[3]{xyz}$ [b]is[/b] a root of f(t), and f(t) has no rational roots, $ \\sqrt[3]{xyz}$ must be irrational.", "Solution_3": "[hide=\"identity\"]Let $ p0$ such that $\\{x|00\\,\\exists\\,\\delta>0$ such that if $00\\,\\exists\\,\\delta>0$ such that if $0 \\angle BUX$. But these angles are easy to compute: they are $ B \\plus{} \\alpha$ and $ C$ respectively. According to our assumption $ B\\ge C$, so we are done.\r\n\r\nSo, let us define the function $ f(t)$ as the unique angle between $ \\minus{} \\pi/2$ and $ \\pi/2$ such that $ \\sin f(t) \\equal{} \\lambda\\sin(t \\plus{} \\alpha)$. Since $ \\lambda\\in(0,1)$, this definition makes sense on the entire real line.\r\n\r\nOur next task will be outrule obtuse angles. So suppose that $ A$ is obtuse. Then $ B \\equal{} f(A)$, $ C \\equal{} f(B)$ and $ A \\equal{} \\pi \\minus{} f(C)$. Note now that $ f(t) < t \\plus{} \\alpha$ if $ 0\\le t\\le\\pi/2$, so $ f(C) < C \\plus{} \\alpha\\le C \\plus{} B$. But then the last equation implies $ A \\plus{} B \\plus{} C > \\pi$ which is impossible.\r\n\r\nNow, when we outruled the possibility of an obtuse angle, we can be sure that $ B \\equal{} f(A)$, $ C \\equal{} f(B)$ and $ A \\equal{} f(C)$, so it is enough to show that $ f$ is a weak contraction. Since it is given by a relation $ \\sin s \\minus{} \\lambda\\sin(t \\plus{} \\alpha)$, it is enough to show that $ |\\cos s| > \\lambda|\\cos(t \\plus{} \\alpha)|$ for $ s,t$ satisfying the relation. But it is obvious since the smaller sine (in absolute value) implies the larger cosine and $ 0<\\lambda<1$. We are done with Problem 2. So, let's go to\r\n\r\n[b]Problem 1[/b]\r\nHere 2 diagrams are possible. \r\n[asy]size(250);\npicture f(real s)\n{\nreal r=1-s;\npair A=D(\"A\",(0,0),SW), B=D(\"B\",(20,0),SE), C=D(\"C\",(12,16),N),\n N=D(\"N\",s*B+r*C,NE), P=D(\"P\",s*C+r*A,NW), M=D(\"M\",s*A+r*B),\n U=IP(A--N,C--M), V=IP(B--P,A--N), W=IP(C--M,B--P);\nfill(U--V--W--cycle,yellow);\nD(A--B--C--A); DPA(A--N^^C--M^^B--P);\nDPA(A--M^^B--N^^C--P, orange+linewidth(1.5));\nreturn CC();\n}\npicture a=f(0.3), b=f(0.6);\nadd(a);\nadd(shift((30,0))*b);[/asy] \r\n(the orange segments are equal and the yellow triangle is equilateral).\r\n\r\nThe argument below applies regardless of which picture is used. We shall denote the length of each orange segment by $ \\ell$, fix the yellow triangle, and consider $ C$ as a function of $ A$ (so, given $ A$, we define $ C$ as the unique point on the ray extending the appropriate side of the yellow triangle such that $ |PC| \\equal{} \\ell$. Of course, we need to show that our definition makes sense (i.e., that such $ C$ exists and is unique). We will do that plus, in addition, we will show that the mapping $ A\\mapsto C$ is a weak contraction. Then, identifying the rays containing $ A,B,C$ by appropriate rotations around the center of the yellow triangle, we will conclude that $ A$, $ B$, and $ C$ lie at the same distance from the corresponding vertices of the triangle, which immediately yields that the triangle $ ABC$ is equilateral.\r\n\r\nLet us look at the following picture:\r\n[asy]size(300);\nfill((0,0)--(2,0)--2*dir(60)--cycle,yellow);\nD((-20,0)--(0,0)--(10,17), black);\nD(L((-8,17),(2,0),0,-0.3),black);\nD(L((-8,17),(2,0),-0.7,0.2),dashed);\npair A=D(\"x\",(-15,0)), B=D(\"y\",(8,13.6),SE), C=D(IP(A--B,(-8,17)--(2,0)));\nD(L(A,B,0,0.2));\nMA(\"60^\\circ\",C,(2,0),A,3);\nMA(\"60^\\circ\",B,(1,sqrt(3)),C,3);\nMC(30,\"M(x,y)\",A--C,N); MC(30,\"L(x,y)\",C--B,N);\nMA(\"\\alpha\",(0,0),A,B,2,2,orange);\nMA(\"\\beta\",B,C,(-8,17),2,2,orange);\nMA(\"\\gamma\",2*B-A,B,(10,17),2,2,orange);[/asy]\r\nHere the position of the point $ A$ is denoted by $ x$ and the position of the point $ C$ by $ y$. The relation connecting $ x$ and $ y$ is $ L(x,y) \\equal{} \\ell$. Note that when $ y$ changes from $ 0$ to $ \\infty$, so does $ L(x,y)$ (on this diagram; on the other diagram $ L(x,y)$ starts at the length of the side of the yellow triangle and ends at $ \\infty$). We need the derivatives $ \\frac {\\partial L}{\\partial x}$ and $ \\frac {\\partial L}{\\partial y}$. The reader needs to be familiar with both calculus and trigonometry to compute them and the computation is a bit boring, but here is the result:\r\n\\[ \\left|\\frac {\\partial L}{\\partial x}\\right| \\equal{} \\frac {L}{L \\plus{} M}\\frac {\\sin\\alpha}{\\sin\\beta}|\\cos\\beta|\r\n\\]\r\nand\r\n\\[ \\left|\\frac {\\partial L}{\\partial y}\\right| \\equal{} \\left|\\cos\\gamma \\minus{} \\frac {M}{L \\plus{} M}\\frac {\\sin\\gamma}{\\sin\\beta}\\cos\\beta\\right|\\ge \\cos\\gamma \\minus{} \\frac {M}{L \\plus{} M}\\frac {\\sin\\gamma}{\\sin\\beta}|\\cos\\beta|\r\n\\]\r\nNote now that $ \\alpha,\\gamma\\in(0,\\pi/3)$ while $ \\beta\\in(\\pi/3,2\\pi/3)$, so the ratios of sines are less than $ 1$ and $ \\cos\\gamma > |\\cos\\beta|$. This allows to conclude that $ \\left|\\frac {\\partial L}{\\partial x}\\right| < \\left|\\frac {\\partial L}{\\partial y}\\right|$, so $ L(x,y)$ is monotone in $ y$ (thus justifying the existence and uniqueness of $ C$, given $ A$) and the mapping defined by the relation $ L(x,y) \\equal{} \\ell$ is a weak contraction. Problem 1 is done as well.\r\n\r\n[b]Final remarks[/b]. I am very far from claiming that the above is the best solution for either of those problems. All I wanted was to introduce a simple (and very well known) idea of how to deal with cyclic identities arising in triangle geometry. Sometimes it works, sometimes it doesn't. Also, the weak contraction property is not the only one that allows to conclude that $ f(f(f(x))) \\equal{} x$ implies $ f(x) \\equal{} x$. Another useful property is $ f' < 0$ (so $ f$ is decreasing). Here is a relatively simple exercise which makes use of this observation:\r\n\r\n[b]Exercise[/b]. On the picture below the green angles are equal and the orange intervals are equal. Prove that the triangle $ ABC$ is equilateral.\r\n[asy]size(200);\npair A=D(\"A\",(0,0),SW), B=D(\"B\",(20,0),SE), C=D(\"C\",(12,16),N),\n X=D(\"X\",(2*B+C)/3,NE), Y=D(\"Y\",(2*C+A)/3,NW), Z=D(\"Z\",(2*A+B)/3);\nDPA(A--Z--C^^B--X--A^^C--Y--B);\nMA(B,A,X,3,9);\nMA(C,B,Y,3,9);\nMA(A,C,Z,3,9);\nDPA(B--Z^^C--X^^A--Y,orange+linewidth(1.5));[/asy]\r\n\r\nTry to solve :).", "Solution_1": "I rode above post it seems very interesting. Maybe I'm wrong but I cannot find any gaps. I also think it is the most universal method I've seen for solving such kind of problems. The only thing I can say - these problems are not very easy at least for me ... even with such universal and convinient method they requires some not routine observations and deep math thinking.\r\n\r\nI cannot solve the last problem. Here is the place to remember again - math is only my hobby. It is interesting to see a solution for the problem - proposed by fedja as an excercise. I think if it is not solved in a week fedja must show us again some beautiful solution on it using calculus or not.", "Solution_2": "OK, let it be one week (starting at the time in borislav_mirchev's post time signature). :)", "Solution_3": ":) it is almost week from my last post.", "Solution_4": "OK, since I'm not certain about my whereabouts at 2PM tomorrow, I'll post the solution today. Let $ x$ be the length of each orange segment and $ \\alpha$ be the size of each green angle. Then, using the Sine theorem, we get\r\n\\[ x \\equal{} |AY| \\equal{} \\frac {\\sin(B \\minus{} \\alpha)}{\\sin(C \\plus{} \\alpha)}2R\\sin C\\,,\r\n\\]\r\nso\r\n\\[ \\sin(B \\minus{} \\alpha) \\equal{} \\lambda\\frac {\\sin(C \\plus{} \\alpha)}{\\sin C}\r\n\\]\r\nwith $ \\lambda \\equal{} \\frac {x}{2R}$.\r\nNow notice that $ |AY| \\equal{} x < |AB|$, so the angle $ B \\minus{} \\alpha$ cannot be obtuse. Thus, $ B \\equal{} f(C)$ where $ f(t) \\equal{} \\alpha \\plus{} \\arcsin\\left(\\lambda\\frac {\\sin(t \\plus{} \\alpha)}{\\sin t}\\right)$. Also, $ C \\equal{} f(A)$ and $ A \\equal{} f(B)$. The function $ f$ is decreasing on the subinterval of $ (0,\\pi)$ on which it is defined (just write $ \\frac {\\sin(t \\plus{} \\alpha)}{\\sin t} \\equal{} \\cos\\alpha \\plus{} \\sin\\alpha\\operatorname{ctg} t$ and recall that $ \\operatorname{ctg}$ is decreasing on $ (0,\\pi)$ to see that). Thus, $ f(f(f(t))) \\equal{} t$ implies $ f(t) \\equal{} t$, so all angles are equal. The end.", "Solution_5": "Short, nice and beautiful.\r\nI'm wondering why no one discovered it.\r\n\r\nIn the configuration given there may be invented many similar problems. They may be probably solved in the manner, proposed by you. I'm wondering if to publish them. If I publish all such problems - I risk to be boring. But sometimes the manner of solving is similar and no so...some creativity is required there. We also have lots of problems ready for math competitions (in mathlinks there are 4 or 5 similar problems) and we may see some different ideas.\r\n\r\nWhat you will advice me - to publish or not to publish more problems of this kind?", "Solution_6": "You mean \"to post or not to post?\". Well, it seems that even my exercise didn't attract any attention (except from you), and, once you've seen the solution (which is almost a verbatim repetition of what I did in Problem 2, only much simpler), you should agree that one week was plenty of time for this question. So I'm not sure that sufficiently many mathlinkers really like this kind of problems. Also, such problems can be created in bundles, so to post all possible variations would be quite boring. On the other hand, you can try to post one more and see if somebody responds. \r\n\r\nIn general, I see it like this: there is a certain culture in almost every subforum and the problems that fit with this culture attract attention from many people, including the best ones, and are solved rather quickly if they are solvable at all, while the posts containing much easier (and sometimes even nicer) problems that do not quite fit with the current \"standards\" are just ignored by those who could really solve them. The dominating culture in the Geometry subforum seems to be what I would call \"Classical Synthetic Geometry\". The folks here are really good at it and the stuff they do is beautiful, indeed, but if you deviate from the mainstream a little bit, they either get confused, or lose interest, or both. There is nothing unusual about it; the people in \"professional mathematics\" are hardly any different with a few exceptions (needless to say, i like the exceptions most :)). \r\n\r\nSo, perhaps, you shouldn't bother posting more than one additional \"prove that the triangle is equilateral\" problem, if any, but if you have other \"non-mainstream\" questions that you consider nice and difficult, by all means post them in this subforum. It is always a good idea to attract the attention of people to something that is not completely foreign to them but also not exactly what they are used to.\r\n\r\nAnyway, I'm just a fleeting guest in this particular subforum, not one of the regulars. If you want a more informed opinion about your question, ask somebody like Darij or Pohoatza ;)." } { "Tag": [ "inequalities", "geometry", "3D geometry", "tetrahedron", "sphere", "LaTeX", "inequalities proposed" ], "Problem": "Can you solve this problem! :( \r\n\r\n\r\nFind a least value of x satisfying this condition: for every tetrahedron with volumes V and radius of the circumsphere and sphere inscribed is R,r respectively\r\nwe have \r\n\\[ V \\leq r \\cdot [x \\cdot R^2+(24\\cdot \\sqrt 3-9x)\\cdot r^2] \\]", "Solution_1": "eum, perhaps can you explain us what this has to do with the IMO in greece? :?", "Solution_2": "might be a problem for shortlist? :maybe:", "Solution_3": "Not be shorlist.\r\n\r\nIdea of problem:\r\nProbable every one knew similar inequalityin plane:\r\n\r\n(1)For every triangle abc with area S and radius of the circum circle and inscribedcircle is R,r, we have \r\nS <= [(3.sqrt3)/2].R.r\r\n\r\nHow to solve (1)?\r\nPut M,N,P are excenter of triangle ABC. Easy to check triangle MNP inscribed in circle with radius R'=2.R and area of MNP =2.R/r. area of ABC. \r\nBy the \"problem circumference\" we have:\r\narea ofMNP<=(3sqrt3)/4.(R'^2) so we have (1)./\r\n\r\nWhat 's the heart of this solution. I thinks it's content of this theorem:\r\n\r\nTheorem in plane: OA=OB=OC=OM and 90degree>#CAM=#MAB then OM perpedicular BC.\r\n\r\nTheorem is heart of problem 1 in IMO 04.\r\nI don't know is there exist similar theorem in spaces. Can you :lol:", "Solution_4": "why don't you solve thispro :D :? it veryvery hard? :D", "Solution_5": "[quote=\"nguyenquockhanh\"]why don't you solve thispro :D :? it veryvery hard? :D[/quote]because it's in the wrong place ... watch carefully where you post. The forum I belive has a clear structure :).", "Solution_6": "sorry for my point, but can you solve it", "Solution_7": "[quote=\"nguyenquockhanh\"]sorry for my point, but can you solve it[/quote]Maybe if I could understand it ... I am editing it to make it $\\LaTeX$ compatible. Please look at it and tell me if it's what you wanted ... :?" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ABC$ a triangle, $N$ the midpoint of $(AC)$, $AD\\bot BC$, $D\\in (BC)$ and $P\\in (AD)$ such that $\\angle ABP \\equiv \\angle NBC$. \r\nProve that $m(\\angle BAC)=90^0$ if and only if $(AP)\\equiv (PD)$.", "Solution_1": "If $\\angle{ABC}=90^o$, then $D=B$, then $P \\in AB$, then $\\angle{ABP}=0^o \\neq \\angle{NBC}$.", "Solution_2": "Sorry for the mistake. Now it's correct." } { "Tag": [ "conics", "ellipse", "function", "symmetry", "calculus", "calculus computations" ], "Problem": "Given an ellipse $ D: \\frac {x^2}{a^2} \\plus{} \\frac {y^2}{b^2} \\equal{} 1\\ (a > 0,\\ b > 0)$ which contains a circle $ C: x^2 \\plus{} y^2 \\equal{} 1.$ Draw a tangent line from the point $ P(0,\\ b)$ on $ D$, let the extented line meets $ D$ at $ Q$. Draw a tangent line from $ Q$, let the extended line meets $ D$ at $ R$, draw a tangent line from $ R$, let the extended line meets $ D$ at $ P$. Express $ a$ as the function of $ b$.\r\n\r\nPlease translate the context into plain English.\r\n\r\nThanks in advance\r\n\r\nkunny", "Solution_1": "Suppose w.l.o.g $Q=(x_0,y_0)$, $x_0>0$, $y_0<0$. Since $PR$ is also a tangent line, by symmetry, $R=(-x_0,y_0)$. Then that $QR$ is tangent to $C$ gives $y_0=-1$. The line $PR$ thus becomes $x_0y+(1+b)x-x_0b=0$, and the distance from the origin to this line is $\\frac{x_0b}{\\sqrt{x_0^2+(1+b)^2}}=1$. Since $Q$ lies on $D$, $\\frac{x_0^2}{a^2}+\\frac{1}{b^2}=1$. Eliminating $x_0$, we get the result:\n\\[a=\\frac{b(1+b)}{b^2-1}.\\]", "Solution_2": "That's a correct answer, that is to say, $a=\\frac{b}{b-1}$.\n\nThe problem is 1990 Kyoto University entarnce exam/Science.", "Solution_3": "Hi, kunny.\n$\\triangle POE\\sim\\triangle PQF\\Longrightarrow\\frac{OE}{PE}=\\frac{QF}{PF}\\Longrightarrow QF=\\frac{b+1}{\\sqrt{b^{2}-1}}$\n$\\frac{QF^{2}}{a^{2}}+\\frac{OF^{2}}{b^{2}}=1\\Longrightarrow a=\\frac{b}{b-1}$", "Solution_4": "Yes, So did I." } { "Tag": [ "email", "\\/closed" ], "Problem": "How do I get those class reminders to stop coming(to my email)?\r\nI actually like the classes, so there's no way that I could forget that there is a class, and I don't need those reminders(and to be honest, it's kind of annoying).", "Solution_1": "set up a message filter in watev email u have and send all email from classes@artofproblemsolving.com directly to Trash\r\n\r\nor u could set up a AoPS folder instead of trash", "Solution_2": "I'd actually prefer if the reminders were closer to the classes, say 19 hours before class instead of 43 hours?", "Solution_3": "[quote=\"junggi\"]I actually like the classes, so there's no way that I could forget that there is a class[/quote]\r\n\r\nErm...not everybody remembers everything they like or care about.", "Solution_4": "But just how can I forget about AoPS classes? :P \r\n\r\n[quote=\"akalra1\"]set up a message filter in watev email u have and send all email from classes@artofproblemsolving.com directly to Trash \n\nor u could set up a AoPS folder instead of trash[/quote]\r\nHow do I do that in yahoo mail?", "Solution_5": "[quote=\"junggi\"]But just how can I forget about AoPS classes? :P \n\n[quote=\"akalra1\"]set up a message filter in watev email u have and send all email from classes@artofproblemsolving.com directly to Trash \n\nor u could set up a AoPS folder instead of trash[/quote]\nHow do I do that in yahoo mail?[/quote]\r\n\r\nGO to options (its on the right) and filters. It might not work for you I have a sbcglobal/yahoo accounts which means I might have some options you don't. Just check and see. \r\n\r\nIf you're using yahoo mail beta it makes you switch back to the old version while you're doing that." } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "Ya se puede discutir y dar a conocer los problemas verdad?", "Solution_1": "creo que no", "Solution_2": "S\u00ed, ya paso el 25 de mayo.", "Solution_3": "bueno yo conosco los problemas ( participe en la olimpiada) alguien quiere discutirlos y comparar soluciones?", "Solution_4": "Por favor, ponga las problemas para discutirlas." } { "Tag": [ "geometry", "3D geometry", "pyramid", "tetrahedron", "LaTeX", "geometry proposed" ], "Problem": "It is a not easy problem.\r\n\r\nThere are four lines (there are no two of them that have two common points or are parallel). Is it possible the angle between [i]any two[/i] of them to be the same?", "Solution_1": "If you have lines $ l_{1}, l_{2}, l_{3}$ which pairwise all form angles $ \\theta$, then either two are parallel, or the triangle they form has all equal angles. No two of the lines in the problem are parallel, so any three of the lines form an equilateral triangle. It then follows that two of the lines are parallel: contradiction, and so it's impossible.", "Solution_2": "But please take a look at the problem with the piramid proposed in Vietnam 1999 :)", "Solution_3": "[quote=\"borislav_mirchev\"]But please take a look at the problem with the piramid proposed in Vietnam 1999 :)[/quote]\r\n\r\nFor this one, i'm thinking if you take this regular tetrahedron, ABCD. WLOG, OX passes through ABC. Consider the circumsphere of ABCO. Extend DO to meet this circumsphere again at P. Then, wLOG, $ X$ also lies on this circumsphere. Perhaps this makes calculations easier...though I don't know...", "Solution_4": "I took in mind the following problem: In space let be given 4 rays: Ox, Oy, Oz, Ot. The angle between any two is the same. Let Or be a ray rotating around O. Let A, B, C, D are the angles forming by it and rays Ox, Oy, Oz, Ot respectively. Prove that: \r\ni) cosA+cosB+cosC+cosD \r\nii) (cosA)^2+(cosB)^2+(cosC)^2+(cosD)^2 \r\ndo not depend on Or.\r\n\r\nIt is ok but ... you proved that the statement is impossible, because it is impossible\r\nthe angle between any two of for lines to be the same ... do you remember? ;)\r\n\r\nP.S. I promised to learn Latex ... but I met very nice girl and now I'm very distracted...", "Solution_5": "[quote=\"borislav_mirchev\"]I took in mind the following problem: In space let be given 4 rays: Ox, Oy, Oz, Ot. The angle between any two is the same. Let Or be a ray rotating around O. Let A, B, C, D are the angles forming by it and rays Ox, Oy, Oz, Ot respectively. Prove that: \ni) cosA+cosB+cosC+cosD \nii) (cosA)^2+(cosB)^2+(cosC)^2+(cosD)^2 \ndo not depend on Or.\n\nIt is ok but ... you proved that the statement is impossible, because it is impossible\nthe angle between any two of for lines to be the same ... do you remember? ;)\n\nP.S. I promised to learn Latex ... but I met very nice girl and now I'm very distracted...[/quote]\r\n\r\nWell in space, there's exactly one configuration where all pairwise angles are the same...namely where the four rays are the rays from the centroid of a regular tetrahedron through the vertices...", "Solution_6": "I've a question. How you prove that in space there is \"exactly one\" configuration?", "Solution_7": "[quote=\"borislav_mirchev\"]I've a question. How you prove that in space there is \"exactly one\" configuration?[/quote]\r\n\r\nWell consider first two rays. Now there are exactly two directions for the third ray such that all three pairwise angles are the same. One direction must be the third ray, and the other the fourth, then. Moreover, the angle between these two rays must be the same. A continuity argument yields the desired result (create a picture in your head where you vary the angle between the first two rays, and as a result, the other two move in and out). This isn't rigorous, but it's a sketch of the complete proof.", "Solution_8": "Another idea is to suppose the angles between the rays are equal... then to take for points M, N, P, Q on the same distance of O. If you take a look at the equal triangles MNO, NPO, PQO, QMO... you can conclude that this is a regular tetrahedron. I remember - similar idea were used somewhere in mathlinks for the Vietnam 1999 problem." } { "Tag": [], "Problem": "CO anh nao` co bai` ve` day so ko\r\npost len cho em ha\r\nthanks", "Solution_1": "Cho so thuc $a$ va day {$x_{n}$} xac dinh boi $x_{1}$= $\\sqrt{2005}$ va $x_{n+1}$=$\\frac{a}{x_{n}^2+1}$.Tim a de day co gioi han huu han khi n dan toi vo cung.", "Solution_2": "bai nay` sach nao` hong co'\r\nve mua sach doc.\r\nanh cho em bai` khac hay hon de", "Solution_3": "ah`\r\nAnh oi; post cho em may bai` luong giac lun nghe\r\nthanks", "Solution_4": "Th\u1eed b\u00e0i n\u00e0y lu\u00f4n \u0111i: Cmr v\u1edbi m\u1ecdi tam gi\u00e1c $ABC$ c\u00f3 $\\sum (tg\\frac{A}{2})^2 +8\\prod sin\\frac{A}{2}$ $\\geq 2$.", "Solution_5": "zozozo123 , sao khoong v\u00f4 http://diendantoanhoc.net\r\n trong do thi b\u00e0i t\u1eadp v\u1ec1 d\u00e3y s\u1ed1 v\u00e0 l\u01b0\u1ee3ng gi\u00e1c b\u1ea1t ng\u00e0n . tha h\u1ed3 l\u00e0m \r\n em \u0111\u1eebng l\u00ean mathlinks \u00edt b\u00e0i v\u1ec1 lo\u1ea1i n\u00e0y l\u1eafm .", "Solution_6": "[color=blue]\u0110\u00fang v\u1eady. [url=http://www.diendantoanhoc.net]Trang nh\u00e0[/url] kh\u00f4ng v\u00e0o m\u00e0 l\u1ea1i v\u00e0o \u0111\u00e2y l\u00e0m b\u00e0i nh\u1ec9?\nK\u1ec3 c\u0169ng l\u1ea1...[/color]" } { "Tag": [ "algebra", "polynomial", "inequalities proposed", "inequalities" ], "Problem": "Suppose that:\r\n \r\n$ H_{n + 1}(x_1,x_2,\\ldots,x_{n + 1}) = x_1^k[A(x_1,x_2,\\ldots,x_{n + 1}) - 2F_n(x_{2},x_{3},\\ldots,x_{n + 1})]$\r\n\r\n${ + x_2^k[A(x_1,x_2,\\ldots,x_{n + 1}) - 2F_n(x_3,x_{4},\\ldots,x_1})]$\r\n\r\n${ + x_3^k[A(x_1,x_2,\\ldots,x_{n + 1}) - 2F_n(x_4,x_{5},\\ldots,x_2})]$\r\n\r\n$ + ......$\r\n\r\n${ + x_{n + 1}^k[A(x_1,x_2,\\ldots,x_{n + 1}) - 2F_n(x_1,x_{2},\\ldots,x_{n }})],$\r\n\r\nwhere\r\n\r\n$ A(x_1,x_2,\\ldots,x_{n + 1}) = F_n(x_1,x_2,\\ldots,x_n) + F_n(x_2,x_3,\\ldots,x_{n + 1}) + \\ldots$\r\n\r\n$ + F_n(x_{n + 1},x_1,\\ldots,x_{n - 1}), (x_1,x_2,\\ldots,x_{n + 1})\\in{R_{ + }^{n + 1},k\\in{N}}$\r\n\r\n\r\nprove (or disprove) that:\r\n\r\n if $ F_n(x_1,x_2,\\ldots,x_n)$ is positive semi-definite homogeneous symmetric \r\n\r\npolynomial,then $ H_{n + 1}(x_1,x_2,\\ldots,x_{n + 1})$ is also positive semi-definite polynomial.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=324338&sid=d9bdd40acb9494da5b76d5542d047e23\r\n16#.\r\npay a attention.\r\nBQ", "Solution_2": "[quote=\"fjwxcsl\"]Suppose that:\n \n$ H_{n + 1}(x_1,x_2,\\ldots,x_{n + 1}) = x_1^k[A(x_1,x_2,\\ldots,x_{n + 1}) - 2F_n(x_{2},x_{3},\\ldots,x_{n + 1})]$\n\n${ + x_2^k[A(x_1,x_2,\\ldots,x_{n + 1}) - 2F_n(x_3,x_{4},\\ldots,x_1})]$\n\n${ + x_3^k[A(x_1,x_2,\\ldots,x_{n + 1}) - 2F_n(x_4,x_{5},\\ldots,x_2})]$\n\n$ + ......$\n\n${ + x_{n + 1}^k[A(x_1,x_2,\\ldots,x_{n + 1}) - 2F_n(x_1,x_{2},\\ldots,x_{n }})],$\n\nwhere\n\n$ A(x_1,x_2,\\ldots,x_{n + 1}) = F_n(x_1,x_2,\\ldots,x_n) + F_n(x_2,x_3,\\ldots,x_{n + 1}) + \\ldots$\n\n$ + F_n(x_{n + 1},x_1,\\ldots,x_{n - 1}), (x_1,x_2,\\ldots,x_{n + 1})\\in{R_{ + }^{n + 1},k\\in{N}}$\n\n\nprove (or disprove) that:\n\n if $ F_n(x_1,x_2,\\ldots,x_n)$ is positive semi-definite homogeneous symmetric \n\npolynomial,then $ H_{n + 1}(x_1,x_2,\\ldots,x_{n + 1})$ is also positive semi-definite polynomial.[/quote]\r\n\r\n[size=200]More generally,we have proved that :[/size]\r\n\r\n[size=150]if $ F_n$ is cyclic symmetric positive semi-definite (homogeneous or \n\ninhomogeneous) ,then $ H_{n + 1}$ is also positive semi-definite .[/size]" } { "Tag": [], "Problem": "Krabby Patties give Sponge Billy Bob bad breath. The number of breath mints he takes varies inversley with the number of minutes he waited after eating a Krabby Patty. On Monday, Sponge Billy Bob ate a Krabby Patty. Exactly four minutes later, he ate five breath mints. On Tuesday, Sponge Billy Bob had to eat twenty breath mints (whoooah!!!!!!!!). How many minutes did Sponge Billy Bob wait?", "Solution_1": "[hide]If x varies inversely with y, then xy will always be the same.\n\nSo on Monday, we get $4 \\cdot 5 = 20$\n\nOn Tuesday, we get $20 \\cdot x = 20$\nSo $x=\\boxed{1}$.[/hide]", "Solution_2": "Please hide your solution. Thank you.", "Solution_3": "[quote=\"Arvind_sn\"]Krabby Patties give Sponge Billy Bob bad breath. The number of breath mints he takes varies inversley with the number of minutes he waited after eating a Krabby Patty. On Monday, Sponge Billy Bob ate a Krabby Patty. Exactly four minutes later, he ate five breath mints. On Tuesday, Sponge Billy Bob had to eat twenty breath mints (whoooah!!!!!!!!). How many minutes did Sponge Billy Bob wait?[/quote]\r\n\r\n[hide]\nmonday- 20\ntuesday 20x=20\nx=1\n[/hide]", "Solution_4": "[quote=\"Arvind_sn\"]Krabby Patties give Sponge Billy Bob bad breath. The number of breath mints he takes varies inversley with the number of minutes he waited after eating a Krabby Patty. On Monday, Sponge Billy Bob ate a Krabby Patty. Exactly four minutes later, he ate five breath mints. On Tuesday, Sponge Billy Bob had to eat twenty breath mints (whoooah!!!!!!!!). How many minutes did Sponge Billy Bob wait?[/quote]\r\n\r\n[hide]20x=20\nx=1[/hide]", "Solution_5": "[hide]xy must always be the same, if x varies inversly with y. \n$20x=20$\n$x=1$[/hide]", "Solution_6": "[hide]because it varies inversely, after just 1 minute Billy must eat 20 mints... [/hide]" } { "Tag": [ "topology", "function", "real analysis", "real analysis unsolved" ], "Problem": "Could you give an example of separable and Hausdorf space that has no countable local base. (there exists at least a point that does not have a countable local basis)", "Solution_1": "The topological product $\\mathbb R^\\mathbb R$ should do the trick.\r\n\r\nIt's Hausdorff because it's a product of Hausdorff spaces. In order to exhibit a countable dense subset, consider the functions $: \\mathbb R\\to\\mathbb R$ which are zero everywhere except for several (i.e. a finite number) disjoint intervals with rational endpoints, where they take rational values (we consider such a function for every combination of intervals and values). Finally, let's prove that this space is not first-countable by showing that $f: \\mathbb R\\to\\mathbb R,\\ f\\equiv 0$ has no countable local basis $\\mathcal B$ of open sets (of course, this works equally well for any element of $\\mathbb R^\\mathbb R$). \r\n\r\nEvery basis element $B_n\\in\\mathcal B$ contains a set of the form $S_n=S(x_1,\\ldots,x_k;\\varepsilon)=\\{g: \\mathbb R\\to\\mathbb R\\ |\\ |g(x_i)|<\\varepsilon,\\ i=\\overline{1,k}\\}$, hence every open set around $f$ must contain some $S_n$. Now consider the open set $S=\\{g: \\mathbb R\\to\\mathbb R\\ |\\ |g(x)|<1\\}$, where $x$ is a real different from all the $x_i$'s appearing in all the $S_n$'s (it's possible to find such a real number $x$ because there are countably many $x_i$). $S$ is an open set containing $f$, but does not contain any of the sets $S_n$, so we have a contradiction.", "Solution_2": "I'm don't understand very well your example. Is it separable ($\\mathbb{R}^{\\mathbb{R}}$ with the product topology ?)", "Solution_3": "Yes, it is.", "Solution_4": "for each $x \\in \\mathbb{R}$, $V_x$ the set of functions that vanish on $\\mathbb{R}-\\{x\\}$ and such that $|f(x)|< 1$ is an open set. So $\\mathbb{R}^{\\mathbb{R}}$ is not separable ?", "Solution_5": "So what's your point? Those $V_x$ are not disjoint. How do you conclude that it's not separable?", "Solution_6": "OK, but for example if $V_x$ is the set of functions $f$ that satisfy:\r\n1/ $f(t)=x$ on $\\mathbb{R}-\\{x\\}$\r\n2/ $|f(x)-x|<1$\r\n\r\nThen the $V_x$ are open disjoint sets, isn't it.", "Solution_7": "They are disjoint, but not open :). Focus man! :) An open neighborhood of $f$ in $\\mathbb R^\\mathbb R$ is a set of the form $\\{g\\in\\mathbb R\\ |\\ |g(x_i)-f(x_i)|<\\varepsilon_i,\\ \\forall i\\in\\overline{1,n}\\}$.", "Solution_8": ":blush: my fault .. Sorry for the stupid things I wrote above :roll:" } { "Tag": [ "function", "integration", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "For $ f \\in L^{p}[a,b]$, $ 1 \\leq p<\\infty$, and a subdivision $ \\Delta\\equal{}\\{ \\xi_{0},\\ldots,\\xi_{m} \\}$ of $ [a,b]$, define the $ \\Delta$-approximant to $ f$ as the step function on $ [a,b]$ that is constant on each subinterval $ [\\xi_{k},\\xi_{k\\plus{}1})$, taking the value $ \\frac{1}{\\xi_{k\\plus{}1}\\minus{}\\xi_{k}}\\int_{\\xi_{k}}^{\\xi_{k\\plus{}1}} f$ there. Show that the ($ L^{p}$) norm of the $ \\Delta$-approximant is at most $ \\parallel{}f\\parallel{}_{p}$.", "Solution_1": "Let a $ \\delta$-approximant be $ g$. \r\n$ \\int_{[a,b]} |g|^p$\r\n$ \\equal{} \\sum \\int_{[a,b]} |g \\chi_{[\\xi_k, \\xi_{k\\plus{}1}]}|^p$\r\n$ \\equal{} \\sum \\int_{[\\xi_k, \\xi_{k\\plus{}1}]} \\frac{1}{(\\xi_{k\\plus{}1} \\minus{} \\xi_k)^p} \\left|\\int_{\\xi_k}^{\\xi_{k\\plus{}1}} f \\right|^p$\r\n$ \\equal{} \\sum \\frac{1}{(\\xi_{k\\plus{}1} \\minus{} \\xi_k)^{p\\minus{}1}} \\left|\\int_{\\xi_k}^{\\xi_{k\\plus{}1}} f \\right|^p$\r\n$ \\leq \\sum \\int_{\\xi_k}^{\\xi_{k\\plus{}1}} |f|^p$ (Holder's inequality)\r\n$ \\equal{} \\| f \\|p$", "Solution_2": "The $ \\Delta$-approximant converges to $ f$ in $ L^{p}$ as the length of the longest subinterval tends to 0. Does it converge to $ f$ in measure?", "Solution_3": "I think yes. For continuous functions, the delta approximant actually converges back to it. We know that L^p functions can be approximated by continuous functions. So what you can do is\r\n$ f \\in L^p$ approximated by $ g \\in C[a,b]$ approximated by $ \\delta g \\in L^p$ approximated by $ \\delta f \\in L^p$. I think that works.", "Solution_4": "In general, does convergence in the $ L^{p}$ norm imply convergence in measure? I think it does by Chebyshev's inequality." } { "Tag": [ "videos", "Support" ], "Problem": "I thought that we get Karma if we answer a problem correctly, even if we've already done the problem before...is this true? Or do we only get Karma the first time around when solving a problem?", "Solution_1": "answering a problem correctly does not get you karma\r\nrating the solution does though\r\n\r\nhere is what gets and loses you karma, taken from the instructions: \r\n\r\n[quote]The karma points are currently as follows, but subject to change: Answering a problem Spend 5 \nRate problem/solution Receive 5 \nSubmit bug report Spend 10 \nBug report determined valid Receive 25 \nWatch video Spend 5 \nRate video Receive 5 \nComment on video Receive 5 \n[/quote]", "Solution_2": "How do you rate/comment on a video?", "Solution_3": "you can rate a video or a problem by scrolling down the solution page and clicking the stars to show how much you liked the solution. You can do the same with the videos by looking where the large comment box is and clicking the stars to rate.", "Solution_4": "[quote=\"vallon22\"]answering a problem correctly does not get you karma\nrating the solution does though\n\nhere is what gets and loses you karma, taken from the instructions: \n\n[quote]The karma points are currently as follows, but subject to change: Answering a problem Spend 5 \nRate problem/solution Receive 5 \nSubmit bug report Spend 10 \nBug report determined valid Receive 25 \nWatch video Spend 5 \nRate video Receive 5 \nComment on video Receive 5\n[/quote][/quote]\r\n\r\n\r\n\r\n\r\n\r\n\r\n bug report????\r\nidk", "Solution_5": "If you discover a bug (error) and send it to them, you get karma.", "Solution_6": "[quote=\"BOGTRO\"]If you discover a bug (error) and send it to them, you get karma.[/quote]\r\n\r\nYou would only get karma if it is a legitimate report about a real problem; this is to prevent spamming of reports." } { "Tag": [ "combinatorial geometry", "advanced fields", "advanced fields unsolved" ], "Problem": "In the following, we denote by $ \\text{conv}\\left(p_1,p_2,...,p_n\\right)$ the convex hull of $ n$ given points $ p_1$, $ p_2$, ..., $ p_n$.\r\n\r\n[b](a)[/b] Let $ a$, $ b$, $ c$ be three non-collinear points in the plane, and let the set $ \\text{conv}\\left(a,b,c\\right)$ be covered by three open sets $ A$, $ B$, $ C$ such that $ \\text{conv}\\left(b,c\\right)\\subseteq A$, $ \\text{conv}\\left(c,a\\right)\\subseteq B$, $ \\text{conv}\\left(a,b\\right)\\subseteq C$. Prove that $ A\\cap B\\cap C\\neq\\emptyset$.\r\n\r\n[b](b)[/b] Let $ a_1$, $ a_2$, ..., $ a_{n \\plus{} 1}$ be $ n \\plus{} 1$ points in $ \\mathbb{R}^n$ which form a non-degenerate $ n$-simplex (this is equivalent to the $ n$ differences $ a_1 \\minus{} a_2$, $ a_1 \\minus{} a_3$, ..., $ a_1 \\minus{} a_n$, $ a_1 \\minus{} a_{n \\plus{} 1}$ being linearly independent). Assume that the set $ \\text{conv}\\left(a_1,a_2,...,a_{n \\plus{} 1}\\right)$ is covered by $ n \\plus{} 1$ open sets $ A_1$, $ A_2$, ..., $ A_{n \\plus{} 1}$ such that $ \\text{conv}\\left(a_1,a_2,...,a_{i \\minus{} 1},a_{i \\plus{} 1},...,a_{n \\plus{} 1}\\right)\\subseteq A_i$ for every $ i\\in\\left\\{1,2,...,n \\plus{} 1\\right\\}$. Prove that $ A_1\\cap A_2\\cap ...\\cap A_{n \\plus{} 1}\\neq\\emptyset$.\r\n\r\n[b](c)[/b] Derive [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105036]the Brouwer fixed point theorem[/url] from problem [b](b)[/b].", "Solution_1": "I have edited the above because I noticed that problem [b](b)[/b] follows from the well-known [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=72491]KKM Theorem[/url] (just set $ F_i \\equal{} \\text{conv}\\left(a_1,a_2,...,a_{n \\plus{} 1}\\right)\\setminus A_i$), what made my post pretty much useless. I have thus added part [b](c)[/b], which seems not to have been discussed here yet.\r\n\r\nIf I remember correctly, [b](a)[/b] was the actual question given to Christian in the 3rd round of the BWM (the west-German mathematical olympiad; note that the 3rd round is a colloquium with a professor to find out whether one is really good at mathematics, so one should expect questions of any possible kind there), and [b](b)[/b] was his generalization. Christian Sattler and me both were pretty proud of our proofs, but it turned out that these proofs are exactly the two ones mentioned by Grobber in http://www.mathlinks.ro/Forum/viewtopic.php?t=72491 , modulo some slight differences in my proof.\r\n\r\nNote that there seems to be a similar fact (also proven using Brouwer) called [url=http://www.springerlink.com/content/p411228841781u81/]Scarf's theorem[/url].\r\n\r\n darij" } { "Tag": [ "geometry", "3D geometry", "sphere", "ratio" ], "Problem": "In a sphere of radius $r$, a \"truncated\" cone (i think this is how it is called in english) of height $r$ is inscribed. The trnc cone has $V_{cone}= \\frac{1}{2} \\cdot V_{sphere}$.\r\nFind out where it is placed, i.e. what are the distances from its bases to the center of the sphere.", "Solution_1": "First I want to find out if this is possible. It should be, if X is the largest [i]cylinder[/i] that can fit in this sphere, if V(X) > 1/2V(sphere) then a truncated cone should work.\r\n\r\nSo the largest such cylinder would have a height equal to its diameter. Let r be the radius of the sphere.\r\nV(sphere) = $ \\frac{4\\pi r^3}{3} $\r\nV(cylinder) = $ \\sqrt{2}r \\vdots \\frac{\\pi r^2}{2} $ = $ \\frac{\\sqrt{2}\\pi r^3}{2} $\r\nThe cylinder to sphere ratio is\r\n$ \\frac{\\sqrt{2}}{2} : \\frac{4}{3} $\r\n$ 3\\sqrt{2} : 8 $\r\n\r\nWhich is more than 1/2, so this problem should be a good one. But I'm still not sure there is a unique solution..." } { "Tag": [ "summer program", "PROMYS", "SuMAC" ], "Problem": "To any people who have gone to either of these camps, approximately how long does it take to be notified if u got in or not? I applied to PROMYS and my friend applied to SUMAC, and we are both interested in finding out (especially me, because if i dont get in, then im going to send off an app to ROSS).", "Solution_1": "I think I got my acceptance letter from SUMaC in early May." } { "Tag": [ "factorial", "function", "integration", "calculus", "trigonometry", "logarithms", "Asymptote" ], "Problem": "I think that the answer to this question involves the gamma function so I posted it here.\r\nHow do you calculate the factorial of a real number that is not an integer? Can the same be done for complex numbers?", "Solution_1": "You can compute the \"factorial\" of any complex number that isn't a negative integer. The accepted conventional defintion of the Gamma function is offset by one from the factorial, so that $x!=\\Gamma(x+1).$\r\n\r\nIf $\\text{Real}(z)>0$ we define $\\Gamma(z)=\\int_0^{\\infty}t^{z-1}e^{-t}dt.$\r\n\r\nThe reason for the restriction to the right hand half plane is so that the improper integral will converge at the $t\\to0^+$ end.\r\n\r\nWe have an extremely important identity: if $\\text{Real}(z)>0,$ then $\\Gamma(z+1)=z\\Gamma(z).$ This can be proved by using the obvious integration by parts. This identity can now be used to define $\\Gamma(z)$ on the rest of the complex plane. That is, since \r\n\r\n$\\Gamma\\left(\\frac12\\right)=\\sqrt{\\pi},$ then $\\Gamma\\left(-\\frac12\\right)= \\frac{\\Gamma\\left(\\frac12\\right)}{-\\frac12}=-2\\sqrt{\\pi}.$\r\n\r\nSimilarly, $\\Gamma\\left(-\\frac32\\right)=\\frac43\\sqrt{\\pi}, \\Gamma\\left(-\\frac52\\right)=-\\frac8{15}\\sqrt{\\pi},$ and so on.\r\n\r\nHowever, as $z\\to0, \\Gamma(z)=\\frac1z+$ a bounded function. That is, the Gamma function has a simple pole of residue 1 at the origin. This in turn propagates through the recursion formula: Gamma has simple poles at all the nonpositive integers, and the residue at $-n$ is $\\frac{(-1)^n}{n!}.$", "Solution_2": "For example,\r\n\r\n$i!=\\Gamma(1+i)=\\int_0^{\\infty}t^ie^{-t}dt =$ \r\n\r\n$\\int_0^{\\infty}\\cos(\\ln t)e^{-t}dt +i\\int_0^{\\infty}\\sin(\\ln t)e^{-t}dt.$\r\n\r\nIf you want to see numerical values for something like this, or graphs, I would suggest a symbolic-manipulation package like Mathematica or Maple.", "Solution_3": "Thanks for the explanation, it really helped. I was surprised that my calculator knew $(-1)!$ was undefined. This made me wonder, is the accepted value of $\\frac{1}{0}$ undefined? My teacher told me that it was $\\infty$, but if you look at the graph of $y=\\frac{1}{x}$, then it could be $\\pm \\infty$. The graph of $y=\\frac{1}{x^2}$ however, shows that $\\frac{1}{0^2}=\\infty$. My calculator (TI-89) says $\\frac{1}{0}=undef$ and $\\frac{1}{0^2}=\\infty$. Can you clarify this since $0=0^2$?", "Solution_4": "Your calculator is making a distinction between $+0$ and $-0$. $0^2=+0$, but it doesn't know which one you meant if you just put in $0$.\r\n\r\nA calculator's number system is not one of the standard ones- there are a lot of funny things at the edges.", "Solution_5": "The 1/0 stuff is a touchy subject. Your calculator and teacher are both talking about limits. I'm just going to presume by your presence in this forum that you know what a limit is :).\r\n\r\nFor 1/x, the limit as x comes from the right and goes to 0 is +inf, while when it comes from the left and goes to zero, is -inf. Therefore, the limit is undefined. For 1/x^2, the limit from both directions is +inf, which is what your calculator is telling you.\r\n\r\nHowever, note that this does not at all mean that 1/0^2=inf, 1/0=inf, or anything of the sort. The value of a function is not in the least determined by the values of the function near it. That's precisely what a limit is (especially from a topological viewpoint). The value of 1/2 has nothing to do with the values of 1/(1.9) or of 1/(2.1), and you would never use those values to calculate it. Likewise, 1/0 has nothing to do with 1/.1 or 1/-.1.", "Solution_6": "Thanks, you both verified what I thought about the calculator.\r\nSo, just to be sure, $\\frac{1}{0}$ and $\\frac{1}{0^2}$ is undefined or not a number but NOT $\\infty$ right?", "Solution_7": "Right.", "Solution_8": "Firstly I want to say that infinity is no real no.Its included in the extended number system to provide a more definitive limit notion.\r\nThe topic about 1/0 is extremely touchy.If you want the tricky limits, its useful to graph the function(not always its easy!).You will get a clear idea about the function,its limits,asymptotes etc..\r\n\r\nGood Luck......." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let $a>2$ be a real number and let $f_{n}(x)=a^{10}x^{n+10}+x^{n}+...+x+1$. Prove that the equation $f_{n}(x)=a$ always has unique root, for all $n\\in\\mathbb{N}$, denote by $x_{n}$ and do prove sequence $x_{n}$ has a finite limit when $n$ run at infinity", "Solution_1": "[quote=\"hien\"]Let $a>2$ be a real number and let $f_{n}(x)=a^{10}x^{n+10}+x^{n}+...+x+1$. Prove that the equation $f_{n}(x)=a$ always has unique root, for all $n\\in\\mathbb{N}$, denote by $x_{n}$ and do prove sequence $x_{n}$ has a finite limit when $n$ run at infinity[/quote]\r\n\r\nAnd http://www.mathlinks.ro/Forum/viewtopic.php?p=750368#750368 (problem 7) are different :maybe:", "Solution_2": "Yes, this is my mistake typing. I have edited:\r\n\r\n[b]Problem 7[/b]\r\nLet $a>2$ be a real number and let $f_{n}(x)=a^{10}x^{n+10}+x^{n}+...+x+1$. Prove that the equation $f_{n}(x)=a$ always has unique root in $(0,+\\infty)$, for all $n\\in\\mathbb{Z}^{+}$, denote by $x_{n}$ and do prove sequence $x_{n}$ has a finite limit when $n$ run at infinity", "Solution_3": "so if, this is my solution \r\n[hide]Fix $n$, we have $f_{n}(x)$ is continuous on $[0,\\infty)$ , $f_{n}(0)=1a$ and $f$ is increasing, therefore first part of problem solved.\n\nNow, for second part, we have $\\frac{1}{a}a,a^{10}q^{9}-a^{10}q^{10}-1>0$ and $f_{n}(x_{n+1})=x_{n+1}^{n+1}(a^{10}x_{n+1}^{9}-1-a^{10}x_{n+1}^{10})+a>a$ for $n$ sufficient large. Therefore $x_{n+1}>x_{n}$ for $n$ sufficient large, and we have done :D[/hide]" } { "Tag": [], "Problem": "A soda manufacturer decided to produce one-liter glass bottles of soda. The retail price of the bottle with the soda is $1.10. If the bottle costs $1.00 more than the soda, how much does the soda cost without the glass bottle?", "Solution_1": "[quote=\"math92\"]A soda manufacturer decided to produce one-liter glass bottles of soda. The retail price of the bottle with the soda is $1.10. If the bottle costs $1.00 more than the soda, how much does the soda cost without the glass bottle?[/quote]\r\n\r\n[hide] soda=x\n\n(100 + x) + x = 110\n\n100 + 2x= 110\n\n2x= 10\n\nx=5\n\nThe soda costs [b] 5 cents [/b] [/hide]", "Solution_2": "[quote=\"math92\"]A soda manufacturer decided to produce one-liter glass bottles of soda. The retail price of the bottle with the soda is $1.10$. If the bottle costs $1.00$ more than the soda, how much does the soda cost without the glass bottle?[/quote]\r\n\r\n[hide=\"And the answer is...\"]Let $b=$ $\\textrm{price of bottle}$ $\\textrm{and}$ $s=$ $\\textrm{price of soda}$:\n\n$b=1+s$\n\n$b+s=1.1$\n\n$b+s=1.1$\n\n$1+2s=1.1$\n\n$2s=0.1$\n\n$s=0.05$\n\n[b]5 cents[/b][/hide]", "Solution_3": "[quote=\"math92\"]A soda manufacturer decided to produce one-liter glass bottles of soda. The retail price of the bottle with the soda is $1.10. If the bottle costs$1.00 more than the soda, how much does the soda cost without the glass bottle?[/quote]\r\n\r\n[hide]\nthe answer is....\n[hide=\"clck ths\"]\n$\\boxed{\\boxed{5\\text{ cents}}}$\n[/hide][/hide]" } { "Tag": [ "geometry", "inequalities", "trigonometry", "function", "inequalities proposed" ], "Problem": "For $m_a,m_b,m_c$ medians of a triangle of area $S$ prove\r\n\r\n\r\n$\\sum \\frac{1}{m_bm_c} \\leq \\frac{\\sqrt{3}}{S}$", "Solution_1": "Using the theorem that the medians $m_a$, $m_b$, $m_c$ form a new triangle $D_m$ with area $\\displaystyle S_m = \\frac34 S$, it is clear that we can apply the inequality\r\n\r\n$\\displaystyle \\frac{1}{bc}+\\frac{1}{ca}+\\frac{1}{ab}\\leq \\frac{3\\sqrt{3}}{4S}$\r\n\r\nto the triangle $D_m$ and get\r\n\r\n$\\displaystyle \\frac{1}{m_{b}m_{c}}+\\frac{1}{m_{c}m_{a}}+\\frac{1}{m_{a}m_{b}}\\leq \\frac{3\\sqrt{3}}{4S_{m}}=\\frac{3\\sqrt{3}}{4\\cdot \\frac{3}{4}S}=\\frac{\\sqrt{3}}{S}$.\r\n\r\nSo it remains to prove the above inequality\r\n\r\n$\\displaystyle \\frac{1}{bc}+\\frac{1}{ca}+\\frac{1}{ab}\\leq \\frac{3\\sqrt{3}}{4S}$.\r\n\r\nIn fact, multiplying it with 2S, you get\r\n\r\n$\\displaystyle \\frac{2S}{bc}+\\frac{2S}{ca}+\\frac{2S}{ab}\\leq \\frac{3\\sqrt{3}}{2}$.\r\n\r\nBut since $S=\\frac{1}{2}bc\\sin A$, we have $\\displaystyle \\frac{2S}{bc}=\\sin A$, and similarly $\\displaystyle \\frac{2S}{ca}=\\sin B$ and $\\displaystyle \\frac{2S}{ab}=\\sin C$, hence it remains to show that $\\displaystyle \\sin A+\\sin B+\\sin C\\leq \\frac{3\\sqrt{3}}{2}$. But this is clear, since the function sin x is concave on the interval $\\left[ 0;\\;\\pi \\right] $, and thus Jensen yields\r\n\r\n$\\displaystyle \\sin A+\\sin B+\\sin C\\leq 3\\cdot \\sin \\frac{A+B+C}{3}=3\\cdot \\sin \\frac{\\pi }{3}=3\\cdot \\frac{\\sqrt{3}}{2}=\\frac{3\\sqrt{3}}{2}$.\r\n\r\nQed.\r\n\r\n Darij", "Solution_2": "Very weak inequality! Almost any approach works!" } { "Tag": [ "inequalities", "geometry solved", "geometry" ], "Problem": "Let ABC be a triangle inscribed in a circle of radius R \r\n\r\nand a, b, c be respectly sides of the triangle ABC.\r\n\r\nProve that:\r\n\r\n $\\frac{R}{r}\\geq\\frac{b}{c}+\\frac{c}{b}$\r\n\r\nPd. r: inradio of the triangle ABC.\r\n\r\n[color=red][Moderator edit: Also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=15807 and at http://www.mathlinks.ro/Forum/viewtopic.php?t=49433 .][/color]", "Solution_1": "Well pretty nice probem.Here is the solution:\r\nGiven inequality is equivalent to the followwing:\r\n[tex]b^{2}+c^{2}\\leq\\frac{2a}{(a+b-c)(b+c-a)(c+a-b)}\\cdot{b^{2}c^{2}} (1)[/tex]\r\nI got it using the square formulas:[tex][ABC]=\\frac{abc}{4R},[ABC]=\\frac{a+b+c}{2}\\cdot{r}[/tex] and [tex][ABC]=\\sqrt{p(p-a)(p-b)(p-c)},[/tex]where [tex]p[/tex] is semiperimeter of the triangle [tex]ABC.[/tex]\r\nNext step is to notice that inequality [tex](1)[/tex] is equivalent to\r\n[tex](\\frac{2ab^{2}}{(a+b-c)(b+c-a)(c+a-b)}-1)(\\frac{2ac^{2}}{(a+b-c)(b+c-a)(c+a-b)}-1)\\geq{1}[/tex]\r\nAnd at last we write\r\n[tex]\\frac{2ab^{2}}{(a+b-c)(b+c-a)(c+a-b)}-1\\geq\\frac{2a}{c+a-b}-1=\\frac{a+b-c}{c+a-b}>0[/tex]\r\nand\r\n[tex]\\frac{2ab^{2}}{(a+b-c)(b+c-a)(c+a-b)}-1\\geq\\frac{2a}{a+b-c}-1=\\frac{c+a-b}{a+b-c}>0\r\n[/tex]\r\nIt remains only to multiple last two inequalities.And equality holds if and only if [tex]a=b=c.[/tex]", "Solution_2": "Here is a simple method:\r\n\r\nPut $x=s-a,y=s-b,z=s-c$ where $s$ is the semiperimeter of the triangle.So it's easy to see that :\r\n\\[\\frac{R}{r}=\\frac{(x+y)(y+z)(z+x)}{4xyz}\\] and\r\n\\[\\frac{b}{c}+\\frac{c}{b}=\\frac{x+z}{x+y}+\\frac{x+y}{x+z}\\]\r\nSo we have to prove \r\n\\[\\frac{y+z}{4xyz}\\ge \\frac{1}{(x+y)^2}+\\frac{1}{(x+z)^2}\\]\r\nfor $x,y,z>0$ . But in fact this follows from that:\r\n \\[\\frac{1}{(u+v)^2}\\le \\frac{1}{4uv}\\]for $u,v>0$\r\nThis complets the proof." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "$ dy/dx \\equal{} y/x \\plus{} 2x \\plus{} 1$\r\n\r\nI solved this using integrating factor but am wondering if there's any other way to solve it.", "Solution_1": "hello, by setting $ u = \\frac{y}{x}$ we have $ y = ux$ and by differentiating with respact to $ x$ we got\r\n$ y^' = u^'x + u$, inserting this in our equation we have $ u^' = \\frac {2x + 1}{x}$ and from here\r\n$ du = \\frac {2x + 1}{x}dx$.\r\nSonnhard." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Prove that \r\n$ (a\\plus{}b\\plus{}c\\plus{}d)^4\\plus{}(a\\plus{}b\\minus{}c\\minus{}d)^4\\plus{}(a\\minus{}b\\plus{}c\\minus{}d)^4\\plus{}(a\\minus{}b\\minus{}c\\plus{}d)^4\\minus{}(a\\plus{}b\\plus{}c\\minus{}d)^4\\minus{}(a\\plus{}b\\minus{}c\\plus{}d)^4\\minus{}(a\\minus{}b\\plus{}c\\plus{}d)^4\\minus{}(\\minus{}a\\plus{}b\\plus{}c\\plus{}d)^4\\equal{}192abcd$", "Solution_1": "Would...there be a point?", "Solution_2": "Let $ T$ be the set of all $ n$-tuples in the form $ (1,\\pm1,\\pm1,\\dots,\\pm1)$. Let $ x : \\equal{} \\left(x_1,x_2,x_3,\\dots,x_n\\right)$. Define\r\n\\[ P(x) : \\equal{} \\sum\\limits_{e \\in T} u_e(x\\cdot{e})^n\\,,\r\n\\]\r\nwhere $ u_e : \\equal{} e_1e_2e_3\\cdots{e_n}$. For each $ n$-tuple $ r : \\equal{} \\left(r_1,r_2,r_3,\\dots,r_n\\right)$ with $ r_1 \\plus{} r_2 \\plus{} r_3 \\plus{} \\cdots \\plus{} r_n \\equal{} n$, the coefficient of $ x_1^{r_1}x_2^{r_2}x_3^{r_3}\\cdots{x_n^{r_n}}$ in $ u_e(x\\cdot{e})^n$ is\r\n\\[ c_r(e) : \\equal{} \\frac {n!}{r_1!r_2!r_3!\\cdots{r_n!}}e_1^{r_1 \\plus{} 1}e_2^{r_2 \\plus{} 1}e_3^{r_3 \\plus{} 1}\\cdots{e_n^{r_n \\plus{} 1}}\\,.\r\n\\]\r\nIf $ r_k \\equal{} 0$ for some $ k$, then $ c_r(e) \\propto e_k$ and hence,\r\n\\[ \\sum\\limits_{e\\in{T}} c_r(e) \\propto \\sum\\limits_{e_k \\equal{} \\pm1} e_k \\equal{} 0\\,.\r\n\\]\r\nTherefore, the only remaining term is $ x_1x_2x_3\\cdots{x_n}$, which has coefficient\r\n\\[ \\sum\\limits_{e\\in{T}}\\frac {n!}{(1!)^n}e_1^{2}e_2^{2}e_3^{2}\\cdots{e_n^{2}} \\equal{} \\sum\\limits_{e\\in{T}}n! \\equal{} n!|T| \\equal{} n!2^{n \\minus{} 1}\\,.\r\n\\]\r\nTherefore, $ P(x) \\equal{} n!2^{n \\minus{} 1}x_1x_2x_3\\cdots{x_n}$.", "Solution_3": "[quote=\"kl2836\"]Prove that \n$ (a \\plus{} b \\plus{} c \\plus{} d)^4 \\plus{} (a \\plus{} b \\minus{} c \\minus{} d)^4 \\plus{} (a \\minus{} b \\plus{} c \\minus{} d)^4 \\plus{} (a \\minus{} b \\minus{} c \\plus{} d)^4 \\minus{} (a \\plus{} b \\plus{} c \\minus{} d)^4 \\minus{} (a \\plus{} b \\minus{} c \\plus{} d)^4 \\minus{} (a \\minus{} b \\plus{} c \\plus{} d)^4 \\minus{} ( \\minus{} a \\plus{} b \\plus{} c \\plus{} d)^4 \\equal{} 192abcd$[/quote]\r\nI have a ideal Let $ LHS\\equal{}P(a,b,c,d)$ try with$ a\\equal{}0,b\\equal{}0,c\\equal{}0.d\\equal{}0$ hence $ P(a,b,c,d)\\equal{}@abcd$ try some values of $ a,b,c,d$ to find $ @\\equal{}192$", "Solution_4": "Batominovski, that is amazing..." } { "Tag": [], "Problem": "Alex can paint $ 25$ sq ft with red paint in $ 5$ minutes, and with blue paint in $ 10$ minutes. Bob can paint $ 36$ sq ft with blue paint in $ 6$ minutes, and with red paint in $ 12$ minutes. A $ 250$ sq ft wall is to be painted half red and half blue. If Alex and Bob work together, how fast can they paint the wall? Give your answer to the nearest minute.", "Solution_1": "[quote=\"i_like_pie\"]Alex can paint $ 25$ sq ft with red paint in $ 5$ minutes, and with blue paint in $ 10$ minutes. Bob can paint $ 36$ sq ft with blue paint in $ 6$ minutes, and with red paint in $ 12$ minutes. A $ 250$ sq ft wall is to be painted half red and half blue. If Alex and Bob work together, how fast can they paint the wall? Give your answer to the nearest minute.[/quote]\r\n\r\n23 minutes", "Solution_2": "Do you have some kind of solution?", "Solution_3": "Does anybody have a solution? And stop rating posts as spam whoever it is.", "Solution_4": "[quote=\"i_like_pie\"]Does anybody have a solution? And stop rating posts as spam whoever it is.[/quote]\r\n\r\nI can try...\r\n[hide]So we have the following information\n\nAlex paints\nRed: $ 5 \\text{ ft}^{2}/\\text{minute}$\nBlue: $ 2.5 \\text{ ft}^{2}/\\text{minute}$\n\nBob paints\nRed: $ 6 \\text{ ft}^{2}/\\text{minute}$\nBlue: $ 3 \\text{ ft}^{2}/\\text{minute}$\n\nSo, we see that together, they paint\nRed: $ 11 \\text{ ft}^{2}/\\text{minute}$\nBlue: $ 5.5 \\text{ ft}^{2}/\\text{minute}$\n\nSo, we need $ 125 \\text{ ft}^{2}$ red and blue paint. Calculating that, we get\n$ \\frac{125}{11}+\\frac{125}{5.5}\\approx 34\\text{ minutes}$[/hide]", "Solution_5": "umm i get 23 minutes too..its actually a little over 23..but you said round..\r\n\r\njust scale everything down to 1 minute.\r\n\r\n[b]vishalarul, your solution does not take into account once someones already done with their part of the job, they can help with the other person, making the other persons job quicker[/b]\r\n\r\nfor example, alex doenst have to paint both blue and red at the same time. alex can paint all red, and bob all blue (sepcializing on their strengths) to make it quicker.\r\n\r\n\r\ni_like_pie, it should be whats the minimum time required i assume..", "Solution_6": "How is this classic? and theres no twist..\r\nI get $ \\frac{375}{16}$ minutes or $ 23.4375$ minutes\r\nYou always want to maximize on a strength for both people and then finish off with the weakness for them together (making it stronger since theres 2 people)", "Solution_7": "The \"classic\" part is that 2 people paint a wall at the same time. The twist is with the different types of paint.\r\n\r\n@ vishal - Each person paints the side that they are fastest at, and whoever finishes first helps the other finish faster. 23 minutes is the correct answer.", "Solution_8": "[quote=\"i_like_pie\"]Alex can paint $ 25$ sq ft with red paint in $ 5$ minutes, and with blue paint in $ 10$ minutes. Bob can paint $ 36$ sq ft with blue paint in $ 6$ minutes, and with red paint in $ 12$ minutes. A $ 250$ sq ft wall is to be painted half red and half blue. If Alex and Bob work together, how fast can they paint the wall? Give your answer to the nearest minute.[/quote]\r\n\r\n[hide=\"fast solution\"]theres a fast way to do these work problems. i think its $ \\frac{1}{\\frac{1}{R_{1}}+\\frac{1}{R_{2}}}$. where $ R_{1}$ refers to Alexs minutes per squ ft and $ R_{2}$ refers to Bob's minutes per squ ft. alex paints a squ ft in $ \\frac{1}{5}$ minutes with red paint and bob does 1 squ ft in $ \\frac{1}{3}$ minutes. pluging the numbers in gets us. $ \\frac{1}{8}$ minutes per squ ft if they work together. since half the wall must be red and the other half blue, they must paint 125 squ ft of red wall, so it will take them $ \\frac{125}{8}$ minutes to do the red side of the wall. the blue side of the wall is the same ting only different numbers: $ \\frac{2}{5}$ and $ \\frac{1}{6}$. plugging the numbers in gets us a rate of 1 squ ft every $ \\frac{2}{17}$ minutes. then multiply it by 125, you get $ \\frac{250}{17}$. add the two together gets us apx. $ 30.33$ minutes.[/hide]\r\n\r\nEDIT: after reading the other posts, i think i_like_pie worded the problem badly because you asked \"how fast CAN they paint the wall.\" i think you meant \"what is the fastest speed in which they can paint the wall\". i just did it the normal way i would do these problems so i didnt get it.", "Solution_9": "[quote=\"abacadaea\"][quote=\"i_like_pie\"]Alex can paint $ 25$ sq ft with red paint in $ 5$ minutes, and with blue paint in $ 10$ minutes. Bob can paint $ 36$ sq ft with blue paint in $ 6$ minutes, and with red paint in $ 12$ minutes. A $ 250$ sq ft wall is to be painted half red and half blue. If Alex and Bob work together, how fast can they paint the wall? Give your answer to the nearest minute.[/quote]\n\n[hide=\"fast solution\"]theres a fast way to do these work problems. i think its $ \\frac{1}{\\frac{1}{R_{1}}+\\frac{1}{R_{2}}}$. where $ R_{1}$ refers to Alexs minutes per squ ft and $ R_{2}$ refers to Bob's minutes per squ ft. alex paints a squ ft in $ \\frac{1}{5}$ minutes with red paint and bob does 1 squ ft in $ \\frac{1}{3}$ minutes. pluging the numbers in gets us. $ \\frac{1}{8}$ minutes per squ ft if they work together. since half the wall must be red and the other half blue, they must paint 125 squ ft of red wall, so it will take them $ \\frac{125}{8}$ minutes to do the red side of the wall. the blue side of the wall is the same ting only different numbers: $ \\frac{2}{5}$ and $ \\frac{1}{6}$. plugging the numbers in gets us a rate of 1 squ ft every $ \\frac{2}{17}$ minutes. then multiply it by 125, you get $ \\frac{250}{17}$. add the two together gets us apx. $ 30.33$ minutes.[/hide]\n\nEDIT: after reading the other posts, i think i_like_pie worded the problem badly because you asked \"how fast CAN they paint the wall.\" i think you meant \"what is the fastest speed in which they can paint the wall\". i just did it the normal way i would do these problems so i didnt get it.[/quote]\r\n\r\nThat works only if you are given two rates. This problem is different, and that method doesn't work.\r\n\r\nThat method works perfectly fine on something like this:\r\n\r\nBob drives at 40 mph to work, and drives at 60 mph back home. What is his average speed for the round trip?\r\n\r\nThen you can use the harmonic mean to get 48 mph.\r\n\r\nI misread the problem at first. I thought that they both specialized in Red. :oops:", "Solution_10": "Actually, that formula isn't the harmonic mean formula, it works when you are given the time, not rate." } { "Tag": [ "geometry", "ratio", "AMC", "AIME", "similar triangles" ], "Problem": "Given a triangle ABC, mark point P on AB such that AP:PB = 3:4, and mark point Q on AC such that AQ:QC = 3:2. Join and extend PQ to meet BC at Z. Prove that BC = CZ.", "Solution_1": "Well, a quick and straightforward solution - not particularly beatiful - is to use vectors. I'll think of a purely geometric solution but it seems difficult to find such a solution.", "Solution_2": "Why is this not just a direct application of Menelaus' Theorem?", "Solution_3": "Most people don't know about Menelaus' theorem, certainly not a grade 12 Algeo class. ;)\r\n\r\nThis problem took me about 30 minutes during the said class, which I'm ashamed about.", "Solution_4": "Can you tell me the Menelaus theorem? It sounds like a good one.", "Solution_5": "Can't explain it better than Mathworld can:\r\n\r\n[url]http://mathworld.wolfram.com/MenelausTheorem.html[/url]\r\n\r\nHowever, this question has a three-line solution that doesn't need any fancy mathematics. 8-)", "Solution_6": "Menelaos theorem is the dual of Ceva's theorem. In a triangle with vertices A, B, and C and points X,Y,Z on the sides AB,BC,CA respectively, X,Y,Z are collinear iff AX/BX*BY/CY*CZ/AZ=1", "Solution_7": "Random comment on Menelaus and Ceva: they are both nice theorems because they're useful and easy to remember. In the same way that we get 16/64 = 1/4 by cancelling 6's, if you cancel letters in Menelaus and Ceva, you get 1 = 1. They are simple, powerful theorems for proving things about ratios and concurrency and colinearity. \r\n\r\nCan someone post a proof of Menelaus? I don't remember how it goes.", "Solution_8": "Wow. Thats a cool thing... Do lots of contest problems use Menalaus' Theorem?", "Solution_9": "yeah. take a look at AIME 2003 A #15.", "Solution_10": "*poke* does anyone have a proof to this question yet? Once again I remind you, it's extremely short and does not involve Menelaus' Theorem (or anything of the sort). ;)", "Solution_11": "i dont know if this is the simple geo soln ur talking about osiris, but heres one with mass points:\n\n[hide] we know that AQ/QC=3/2, so if there is a weight of x at A and y at C, then AQ/QC=3/2=y/x. Also, if there is a weight of z at B, then AP/PB=3/4=z/x. It would be convenient if the weights were all integers, so we can see that x=4, y=6, and z=3. Then, by definition, \n\n3B+nZ=6C (where the number is the weight, the letter is the point). hence n=3, and BC/CZ=3/3=1, so done.[/hide]", "Solution_12": "Not quite. ^^\" My solution involves only parallel lines, similar triangles and proportions of lengths.", "Solution_13": "Eh, since noone seems to have come up with anything, I'll post my solution:\n\n\n\n[hide]Construct CD // PZ, meeting AB at D. Then AP : PD : DB = 3 : 2 : 2. Then PD = DB, therefore BC = CZ.[/hide]", "Solution_14": "wow. how is it that you gained your geometric intuition, like where and when to add lines and stuff? im horrible at geometry and would really like to know.", "Solution_15": "i think that can be considered part of one of the proof of menelaos theorem if you take out the numbers. ill post the proof eventually.", "Solution_16": "I thought of a good way to prove it... If it is 3x:4x, take the 4x and take the midpoint, 2x and call it M. Then extend it to the point C. We can say that this line is parallel to PZ because they have the same ratios of 3:2 on both sides. Triangle BMC and triangle BPZ are similar and they are in the ratio of 2 to 1. QED! :P", "Solution_17": "Arrrg... I just realized that Osiris posted the same solution.", "Solution_18": "In answer to the bit about geometric intuition - the truly talented geometers do indeed appear to be magicians, but you can get there yourself through practice and constantly asking the question 'How did you think to do that?' The answer, you'll find, is often 'I've seen it before', which isn't too satisfying, so just ask again, but ask 'How would you have thought of it the first time?'\r\n\r\nIn this problem, the reason we think 'draw in a parallel' is that this is a problem involving ratios. Ratios mean similar triangles. We get similar triangles from parallel lines. So we just visualize a few parallel lines we could add - the one that solves the problem is easily one of the early candidates to try. \r\n\r\nSo, now you have another couple tools for your toolbox. Doesn't always work, but they're tools good geometers reach for frequently. Specifically: Ratios -> look for similar triangles. Need similar triangles -> draw a line through a point in the diagram parallel to some other line.", "Solution_19": "well the first thing that comes to mind is menelaus... simply because the diagram is all so familiar...\r\n\r\nbut the second thing that comes to mind is a midline... specifically the midline of BPZ parallel to PZ. If u show it hits C ur finished...\r\n\r\nso lets just take the line parallel to PZ through C, since if C is the midpoint of BZ this will be the midline...\r\n\r\nSO how do we SHOW this is the midline... well then BD:DP = 2:2... how can we show this?\r\nWell since AQ:QC = 3:2 and AP:PB = 3:4, then PD: DB = 2:2... well look at that we're finished :)" } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "Prove that the following equation has no solution in integer numbers: \\[ x^2 + 4 = y^5. \\] [i]Bulgaria[/i]", "Solution_1": "We shall prove that $x$ is odd. Indeed, suppose that $x$ is even. Then $y$ is also even, so it's divisible by $16$. It means that $x^2 \\equiv 12 \\mod{16}$ but that's impossible (it can be checked by hand very easily).\r\nNow, we see that the equation is equivalent to the following:\r\n$(x+2i)(x-2i) = y^5$\r\nNow let $d=(x+2i, x-2i)$\r\nIt follows that $d|4i$\r\nBut because $x$ is odd we must have $d|i$. It implies that $x+2i$ and $x-2i$ are coprime, so they have to be the fifth powers.\r\n$x+2i = (a+bi)^5$ and\r\n$x-2i = (a-bi)^5$\r\n(this is because $x+2i$ and $x-2i$ are conjugates)\r\nNow we substract the second equation from the first and, after doing some boring, bot fortunately not too long, computations, we obtain a equation which has no solutions (after comparing the real and imaginary parts or so...).", "Solution_2": "I think looking mod 11 works. We have $y^{10} \\equiv 0, 1 \\pmod{11}$; thus $y^5 \\equiv -1, 0, 1 \\pmod{11}$. Also, $x^2 \\equiv 0, 1, 3, 4, 5, 9 \\pmod{11}$; thus $x^2 + 4 \\equiv 2, 4, 5, 7, 8, 9 \\pmod{11}$. Hence $x^2 + 4$ and $y^{10}$ are different mod 11.", "Solution_3": "Looking $\\mod 16$ gives that $x,y$ are odd (as above) and $y \\equiv 1 \\mod 4$.\r\nNow the equation is equivalent to $x^2+6^2 = y^5+2^5 = (y+2)(y^4-2y^3+4y^2-8y+16)$, and because of $\\gcd(y+2, y^4-2y^3+4y^2-8y+16)|16$ we get that the $\\gcd$ is $1$. Now $y+2 \\equiv 3 \\mod 4$, so there is some prime divisor $q$ of $y+2$ that divides $y$ an odd number of times, then also dividing $y^5+2^5$ an odd number of times, contradicting the fact that this number is sum of two squares.\r\n\r\nTo show that $m=a^2+b^2$ can't be divided by $p^{2s-1}$ and not by $p^{2s}$ for $p \\equiv 3 \\mod 4$, just use inductive reasoning:\r\nLet $m$ be the smallest number for that we haven't proved this. Then let $p$ be as above and look $\\mod p$: $a^2+b^2 \\equiv 0 \\mod p$.\r\n\r\n1. Case: $b \\equiv 0 \\mod p$.\r\nThis gives that $p|a,b$, thus $m=a^2+b^2=p^2 a'^2 + p^2 b'^2 = p^2 (a'^2+b'^2)$ for some integers $a',b'$, giving that the number $m'=\\frac{m}{p^2}=a'^2+b'^2$ is sum of two squares, contradicting the fact that we have shown this as impossible for all smaller $m'$ (note that $p^{2s-3}$ but not $p^{2s-2}$ divides $m'$).\r\n\r\n2. Case: $b \\not \\equiv 0 \\mod p$.\r\nThen multiplication with the inverse $b^{-1} \\mod p$ of $b$ gives for $c=ab^{-1}$ that $c^2 \\equiv a^2 (b^2)^{-1} \\equiv -1 \\mod p$. But that is impossible since it would lead to $1 \\equiv c^{p-1} \\equiv (c^2)^{\\frac{p-1}{2}} \\equiv (-1)^{\\frac{p-1}{2}} \\equiv -1 \\mod p$, clearly impossible.\r\n\r\nThus both cases lead to contradictions, thus done.", "Solution_4": "[quote=\"TomciO\"]\nBut because $x$ is odd we must have $d|i$. It implies that $x+2i$ and $x-2i$ are coprime, so they have to be the fifth powers.\n$x+2i = (a+bi)^5$ and\n$x-2i = (a-bi)^5$\n(this is because $x+2i$ and $x-2i$ are conjugates)\n[/quote]\r\n\r\nCan you explain why \r\n$x+2i = (a+bi)^5$ and\r\n$x-2i = (a-bi)^5$ but not\r\n$x+2i=\\epsilon (a+bi)^5$ and\r\n$x-2i=\\overline{\\epsilon} (a-bi)^5$, where $N(\\epsilon)=1$? You consider the case when $\\epsilon=1$ only.", "Solution_5": "$i = i^5$ :). So for example: $i(a+bi)^5$=$i^5(a+bi)^5=(ai-b^2)^5$\r\nSo there's no problem with writing like that (for odd powers of course).", "Solution_6": "mod 11 :P (standard for integer variables in 5-th power)", "Solution_7": "This problem is too easy for this competition.The idea is using the little theorm of Fermat and we have y^10=1(mod11)=>y^5=-1,0,1(mod11) and x^2+4=2,4,5,7,8,9(mod11) and we discover a contradiction.", "Solution_8": "$y^5=0,\\pm 1\\mod 11$.\nIf $y^5=0\\mod 11$, then $x^2=-4\\mod 11$ - contradition $(\\frac{-4}{11})=-1$.\nIf $y^5=1\\mod 11$, then $11|x^2+36\\to (x/6)^2=-1\\mod 11$ - contradition.\nIf $y^5=-1\\mod 11$, then $x^2=-5=-4^2\\mod 11$ - contradition.", "Solution_9": "Clearly something is wrong with this problem. \nTaking $ \\mod 11$ makes this problem a JBMO P2 rather than a BMO P4. ", "Solution_10": "[quote=sttsmet]Clearly something is wrong with this problem. \nTaking $ \\mod 11$ makes this problem a JBMO P2 rather than a BMO P4.[/quote]\n\nThis is a problem from the 1990s.", "Solution_11": "@above I don't see your point. ", "Solution_12": "[quote=sttsmet]@above I don't see your point.[/quote]\n\nOlder problems tend to be easier ", "Solution_13": "I agree, I just say it should be a bit more complicated... here only a single idea was needed. " } { "Tag": [ "LaTeX" ], "Problem": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=72170]example[/url]\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=69618]doesn't work here[/url]\r\n\r\nWhy is it that the feature when the mouse icon over latex text becomes a hand sometimes, allowing copying to the clipboard, and sometimes it doesn't\r\n\r\nI really don't see a pattern, it would be nice if you could always do that", "Solution_1": "This has got to be an issue with your browser, since it always works fine for me (and in fact LaTeX is rendered the same way -- as a gif file -- everywhere on the forum).", "Solution_2": "I can confirm the hand/no hand difference on two different browsers and systems. On the other hand, I still get a menu appropriate to an image when right-clicking.", "Solution_3": "If my memory doesn't cheat me, the LaTeX image is clickable if and only if the code is over a certain fixed length (so very short codes, like $1+1=2$, are not clickable, while ones like $\\sum_{i=1}^{\\infty}\\frac{1}{2^i}+\\sum_{i=1}^{\\infty}\\frac{1}{2^i}=\\sum_{i=0}^{\\infty}\\frac{1}{2^i}$ are). If you are really desperate to get the code of a short LaTeX string, you can right-click it, go to \"Properties\", and you see the code as \"alternative text\" (well, this is how you do it with Firefox). Or you can quote the message, if you are logged in.\r\n\r\n[b]EDIT:[/b] Of course, Jmerry is right - simply selecting the text and copying works, too. Of course, provided one has a good browser (FF works, IE doesn't - who would expect -, I haven't checked others).\r\n\r\n Darij", "Solution_4": "i use internet explorer a lot and rarely opera\r\n\r\ni am just trying to see the pattern\r\n\r\nso length is the criterium?\r\n\r\ni don't see the logic in that, when you need a specific thing in latex it usually pops up in a short string? \r\n\r\nbut it is true, quoting always works", "Solution_5": "For me, selecting and copying has always worked regardless of length- it cuts out the dollar signs, but you get the text.", "Solution_6": "copying? from the quoted section you mean?\r\n\r\nit's just that i myself need to learn some more latex do be a serious member here, and i wanna help a friend who considers joining and learning latex (well he is a member but has zero posts for now)", "Solution_7": "The point is that each clickable formula is in fact a piece of javascript code. If the formula is less than a certain length limit, there is no need for the code as it is surely well known (so no need to generate further traffic). \r\n\r\nYou can always however over a formula to see how it is written." } { "Tag": [ "trigonometry", "ARML" ], "Problem": "Find the numerical value of\r\n\\[ \\frac{(\\sin 18)(\\cos 12) + (\\cos 162)(\\cos 102)}{(\\sin 22)(\\cos 8) + (\\cos 158)(\\cos 98)}, \\] where all angles are measured in degrees.\r\n\r\n[size=75][color=green]a few modifications by mod[/color][/size]", "Solution_1": "just use the cofunction identities, and angle addition formulas", "Solution_2": "[hide]\nRecall the identities $\\cos (180 - \\alpha) = - \\cos (\\alpha), \\cos (90 + \\alpha) = - \\sin (\\alpha)$. From these we can rewrite the top as $\\sin 18 \\cos 12 + \\cos 18 \\sin 12 = \\sin 30$ and the bottom as $\\sin 22 \\cos 8 + \\cos 22 \\sin 8 = \\sin 30$. Hence the value is $\\boxed{1}$.\n[/hide]", "Solution_3": "[hide][quote=\"t0rajir0u\"][hide]\nRecall the identities $\\cos (180 - \\alpha) = - \\cos (\\alpha), \\cos (90 + \\alpha) = - \\sin (\\alpha)$. From these we can rewrite the top as $\\sin 18 \\cos 12 + \\cos 18 \\sin 12 = \\sin 30$ and the bottom as $\\sin 22 \\cos 8 + \\cos 22 \\sin 8 = \\sin 30$. Hence the value is $\\boxed{1}$.\n[/hide][/quote]\n\nNice solution. I got it right through deciding that the number were ugly, adding or subtracting 2 from each angle, and saying $\\frac{n}{n} = 1$ :P [/hide]" } { "Tag": [ "geometry" ], "Problem": "If the altitude of an equilateral triangle is $ \\sqrt {6}$, then the area is:\r\n\r\n$ \\textbf{(A)}\\ 2\\sqrt {2} \\qquad\\textbf{(B)}\\ 2\\sqrt {3} \\qquad\\textbf{(C)}\\ 3\\sqrt {3} \\qquad\\textbf{(D)}\\ 6\\sqrt {2} \\qquad\\textbf{(E)}\\ 12$", "Solution_1": "[hide=\"Solution\"]The altitude cuts it into two 30-60-90 triangles, so one side of the triangle must be $ 2\\cdot \\frac{\\sqrt{6}}{\\sqrt{3}}\\equal{}2\\sqrt{2}$.\nTherefore, the area is $ \\sqrt{2}\\cdot\\sqrt{6}\\equal{}\\boxed{\\textbf{(B)}\\ 2\\sqrt{3}}$.[/hide]" } { "Tag": [], "Problem": "Marian wishes to buy a new computer that will cost her $ \\$300$. She receives $ \\$5$ per hour for watching her younger brother and $ \\$4$ per hour for helping her mother with chores. Each week she watches her brother for four hours and helps her mother with chores for 10 hours. How many full weeks must she work to earn enough money to buy the computer?", "Solution_1": "Each week she earns $ 5 \\times 4 \\equal{} 20$ dollars for watching her brother and $ 4 \\times 10 \\equal{} 40$ dollars for helping her mother, making a total of 60 dollars a week. Thus it will take her $ 300 \\div 60 \\equal{} 5$ full weeks." } { "Tag": [ "LaTeX" ], "Problem": "I've been including images in my $ \\text{\\LaTeX} $ documents by converting those in [i]JPEG[/i] to [b][i]EPS[/i][/b], and then inserting the appropriate code. This website recommends converting them to [b][i]PDF[/i][/b] instead. Does this approach offer any advantage? It seems to me I would not need to use the [i]epstool[/i] program to adjust the [i]BoundingBox[/i]. Anything else?", "Solution_1": "If your final output is going to be PDF I suggest you plainly use .PNG images (any image editor can convert images to .png) which you can easily insert and compile with pdflatex (won't give you a .dvi output, but it'll be a perfect .pdf). \r\n\r\nOtherwise, I'd go with the EPS, as it's easier to work with when you are also creating PS files.", "Solution_2": "With \\includegraphics*, you can include jpg files directly.", "Solution_3": "Thank you guys for the suggestions.\r\n\r\nUsing [i]\\includegraphics*[/i] I won't need to bother about converting my files to [i]EPS[/i], using [i]epstool[/i], and later having specify the size of picture within the regular [i]\\includegraphics[/i] command. \r\n\r\nInserting images in [i]PDF[/i] format through [i].PNG[/i] images and the [i]pdflatex[/i] package also sounds good." } { "Tag": [ "modular arithmetic" ], "Problem": "Na apodeiksete oti ean $n\\not\\equiv 0 \\pmod{6}$ tote\r\n\r\n$1^{n}+2^{n}+3^{n}+4^{n}+5^{n}+6^{n}\\equiv 0 \\pmod{7}$\r\n\r\nAlexandros\r\n\r\nY.G. Na ipenthimisw (gia tous endiaferomenous) oti exei meinei aliti i deuteri askisi apo ton Arximidi tou 94 pou eixa kanei post.", "Solution_1": "[color=green][b]\u039b\u03ae\u03bc\u03bc\u03b1:[/b][/color]$2^{k}(2^{k}+1) \\equiv 6 \\pmod 7$ \u03b1\u03bd \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b1\u03bd $k \\equiv 1$ \u03ae $2 \\pmod 3$\r\n\r\n[u][b]\u0391\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03bb\u03ae\u03bc\u03bc\u03b1\u03c4\u03bf\u03c2:[/b][/u]\r\n$2^{3m+u}\\equiv 8^{m}2^{u}\\equiv 1^{m}2^{u}\\equiv 2^{u}\\pmod 7$ ,\u03cc\u03c0\u03bf\u03c5 $u=0,1,2$\r\n\u0386\u03c1\u03b1, $2^{k}(2^{k}+1) \\equiv 2(2+1) \\equiv 6 \\pmod 7$ \u03b1\u03bd $k \\equiv 1 \\pmod 3$\r\n$2^{k}(2^{k}+1) \\equiv 4(4+1) \\equiv 20 \\equiv 6 \\pmod 7$ \u03b1\u03bd $k \\equiv 2\\pmod 3$\r\n$2^{k}(2^{k}+1) \\equiv 1(1+1) \\equiv 2 \\pmod 7$ \u03b1\u03bd $k \\equiv 0 \\pmod 3$ \r\n\r\n[size=150][u][color=darkblue][b]\u039b\u03cd\u03c3\u03b7:[/b][/color][/u][/size]\r\n$1^{n}+2^{n}+3^{n}+4^{n}+5^{n}+6^{n}\\equiv 1^{n}+2^{n}+3^{n}+(-1)^{n}+(-2)^{n}+(-3)^{n}\\pmod 7$\r\n\r\n\u0386\u03c1\u03b1, \u03b1\u03bd $n$ \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc\u03c2 \u03c4\u03cc\u03c4\u03b5\r\n$1^{n}+2^{n}+3^{n}+4^{n}+5^{n}+6^{n}\\equiv 1^{n}+2^{n}+3^{n}-1^{n}-2^{n}-3^{n}\\equiv 0 \\pmod 7$\r\n\r\n\u0395\u03bd\u03ce \u03b1\u03bd $n$ \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2 \u03c4\u03cc\u03c4\u03b5\r\n$1^{n}+2^{n}+3^{n}+4^{n}+5^{n}+6^{n}\\equiv 2(1^{n}+2^{n}+3^{n}) \\pmod 7$\r\n\u0395\u03c0\u03b5\u03b9\u03b4\u03ae \u03c4\u03ce\u03c1\u03b1 \u03bf $n$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2 \u03bc\u03b5 $n \\not \\equiv 0 \\pmod 6$ \u03ad\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 $n=2k$ \u03bc\u03b5 $k \\equiv 1$ \u03ae $2 \\pmod 3$, \u03bf\u03c0\u03cc\u03c4\u03b5\r\n$1^{n}+2^{n}+3^{n}+4^{n}+5^{n}+6^{n}\\equiv 2(1^{k}+4^{k}+9^{k}) \\equiv 2(1^{k}+4^{k}+2^{k}) \\equiv$ \r\n$\\equiv 2(2^{k}(2^{k}+1)+1) \\equiv 2(6+1) \\equiv 14 \\equiv 0 \\pmod 7$ \u03c3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03bf \u03bb\u03ae\u03bc\u03bc\u03b1.\r\n\r\nStelios...", "Solution_2": "Mpravo Stelio! Poli wraia proseggisi parolo pou egw exw 2 alles sto mialo mou!\r\n\r\nTin askisi auti tin evala gia ena didaktiko skopo ton opoio tha anaferw se epomeno post giati exw ligi doulitsa simera! Tha mporeite meta apo auto na antimetwpisete polles askiseis tetoiou eidous 'h akomi kai diskoloteres tou tipou \"Na vreite ola ta dinata ipoloipa tis diairesis tou $2\\cdot 3^{n}+3\\cdot 7^{2n+1}+5^{3n+1}-7$ me to $11$\" (Prosoxi! Den dinw poio einai to ipoloipo tis diairesis gia na ginei me epagwgi (pragma eukolo), alla zitw gia tis diafores times tou $n$, ola ta dinata ipoloipa). Episis tha mporeite na kataskeuasete tis dikes sas askiseis (opws auti pou parethesa egw parapanw) kai na kanete tis dikes sas paratiriseis.\r\n\r\nEpisis h askisi auti einai enas kalos tropos (3os tropos ektos apo ekeinon tou Steliou kai ekeinon pou tha anaferw se allo post) gia na thimithoume ena sigekrimeno sxima tis epagwgis me to opoio, se kathe epagwgiko vima den diwxneis mono mia periptwsi alla parapanw apo mia! [Xaraktiristiko einai to paradeigma tou provlimatos \"Na apodeixthei oti gia opoiondipote fisiko arithmo $n$ megalitero tou 6 isxiei to eksis: Kathe tetragwno mporei na diameristhei, me ti voitheia eutheiwn parallilwn pros tis pleures tou, se $n$ to plithos tetragwna (anisa en genei metaksi tous)\"].\r\n\r\nO logos omws gia ton opoio stelnw auto to post einai allos: Paratirisa me vasi tin parapanw askisi pou evala kai me ti voitheia tou ipologisti oti isws i parapanw protasi na isxiei kai genika. Den exw omws kathisei na tin apodeiksw 'h na tin aporripsw. Den kserw an episis anaferetai ws askisi kapou.\r\n\r\n\r\n[size=125]Ean $p$ prwtos $>2$ kai $n$ fisikos me $n\\not\\equiv 0 \\pmod{p-1}$\ntote $1^{n}+2^{n}+\\ldots+(p-1)^{n}\\equiv 0 \\pmod{p}$[/size]\r\n\r\n\r\nTi lete? Elkistikotato etsi? \r\nAlexandros", "Solution_3": "\u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9 \u03b5\u03bb\u03ba\u03c5\u03c3\u03c4\u03b9\u03ba\u03bf\u03c4\u03b1\u03c4\u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7.\u03bc\u03b9\u03b1 \u03b5\u03c1\u03c9\u03c4\u03b7\u03c3\u03b7 \u03bc\u03b7\u03c0\u03c9\u03c2 \u03b7 \u03c3\u03c5\u03bd\u03b8\u03b7\u03ba\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b3\u03b5\u03bd\u03b9\u03ba\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b8\u03b5\u03c4\u03b5\u03b9\u03c2 \u03b1\u03bb\u03b5\u03be\u03b1\u03bd\u03b4\u03c1\u03b5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03c4\u03b9 \u03bf p \u03b4\u03b5\u03bd \u03b4\u03b9\u03b1\u03c1\u03b5\u03b9 \u03c4\u03bf 2^n-1? \u03ba\u03b1\u03b9 \u03ba\u03b1\u03c4\u03b9 \u03b1\u03bb\u03bb\u03bf .\u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03bb\u03c5\u03c3\u03b7 \u03bc\u03b7\u03c0\u03c9\u03c2 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03be\u03b5\u03ba\u03b9\u03bd\u03b7\u03c3\u03c9 \u03bc\u03b5 \u03b5\u03bd\u03b1 \u03c0\u03bb\u03b7\u03c1\u03b5\u03c2 \u03c3\u03c5\u03c3\u03c4\u03c5\u03bc\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b9\u03c0\u03c9\u03bd? :)", "Solution_4": "Kwsta exw elenksei to provlima gia olous tous prwtous tous mikroterous 'h isous tou $29$ kai den exei parousiastei kapoio provlima! Gi'auto pisteuw oti isxiei genika etsi opws to edwsa! \r\n\r\nDen xreiazetai na dwsw tin sinthiki $2^{n}\\not\\equiv 1 \\pmod{p}$ pou anafereis Kwsta. Opws to edwsa me $n\\not\\equiv 0 \\pmod{(p-1)}$ eimaste kalimenoi. Exeis kapoio antiparadeigma mipws?\r\n\r\nOso gia to pws mporei na lithei agnwsto kai se mena pros to paron... Isws na apotelei kali skepsi na pareis ena plires sistima ipoloipwn.\r\n\r\nAlexandros", "Solution_5": "Kwsta molis twra teleiwsa tin dokimi me tous prwtous tou Mersenne pou anafereis mexri ton $2^{13}-1$ (diladi tous $5$ prwtous) kai pali isxiei. Exw ipopsi mou tin idia protasi pou anafereis kai apaitei to $p\\not|2^{n}-1$ (kai tin opoia ap'oti thimamai pragmati douleueis me plires sistima ipoloipwn) alla fainetai na isxiei genika...\r\n\r\nAlexandros", "Solution_6": "\u03bc\u03b1\u03bb\u03bb\u03bf\u03bd \u03b5\u03c7\u03b5\u03b9\u03c2 \u03b4\u03b9\u03ba\u03b9\u03bf", "Solution_7": "\u039a\u03b1\u03bb\u03b7....\r\n\u0391\u03bd \u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03bf\u03c2,\u03c4\u03bf\u03c4\u03b5 $1^{n}+6^{n},2^{n}+5^{n},3^{n}+4^$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb7.......\r\n\u0391\u03bd n=\u03b1\u03c1\u03c4\u03b9\u03bf\u03c2 \u03c4\u03bf\u03c4\u03b5 \u03b8\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03b7\u03c2 6\u03ba+2\u03ae 6\u03ba+4(\u03b8\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03c9 \u03bc\u03bf\u03bd\u03bf \u03c4\u03b7\u03bd \u03bc\u03b9\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03c9\u03c3\u03b7 \u03ba\u03b1\u03b9[b] \u03b5\u03bd\u03c4\u03b5\u03bb\u03c9\u03c2 \u03bf\u03bc\u03bf\u03b9\u03b1 [/b]\u03b1\u03c0\u03bf\u03b4\u03c5\u03ba\u03bd\u03b5\u03b9\u03b5\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03b7 \u03b1\u03bb\u03bb\u03b7...) \r\n\u03c4\u03bf\u03c4\u03b5 \u03c4\u03bf \u03b1\u03bc\u03b5\u03bb\u03bf\u03c2 \u03b3\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 $1^{3k+1}+4^{3k+1}+9^{3k+1}+16^{3k+1}+25^{3k+1}+36^{3k+1}$ \u03c0\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03c5\u03c0\u03bf\u03bb\u03bf\u03b9\u03c0\u03bf \u03bc\u03b5 $2[1^{3k+1}+4^{3k+1}+9^{3k+1}]$\u03ba\u03b1\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03b7$(2,7)=1$ \u03b8\u03b1 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b4\u03bf$1^{3k+1}+4^{3k+1}+9^{3k+1}$ =\u03c0\u03bf\u03bb7\r\n \u0395,\u03ba\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1? \u0395\u03c0\u03b1\u03b3\u03c9\u03b3\u03b9\u03ba\u03b1....\r\n\u0393\u03b9\u03b1 \u03ba=\u03bf, 14=\u03c0\u03bf\u03bb7\r\n\u0395\u03c3\u03c4\u03c9 \u03bf\u03c4\u03b9 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 $1^{3k+1}+4^{3k+1}+9^{3k+1}$=\u03c0\u03bf\u03bb7\r\n\u0393\u03b9\u03b1 \u03ba+1 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5...\u03b1 \u03bc\u03b5\u03bb\u03bf\u03c2=$1^{3k+4}+4^{3k+4}+9^{3k+4}=1^{3k+1}+64*4^{3k+1}+729*9^{3k+1}$\u03c0\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03c5\u03c0\u03bf\u03bb\u03bf\u03b9\u03c0\u03bf \u03bc\u03b5 $1^{3k+1}+4^{3k+1}+9^{3k+1}$\u03c0\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb7 \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03c5\u03c0\u03bf\u03b8\u03b5\u03c3\u03b7...\r\n\u0395\u03bd\u03c4\u03b5\u03bb\u03c9\u03c2 \u03bf\u03bc\u03bf\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03b7 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03c9\u03c3\u03b76\u03ba+4.....\r\nCreatanman,\u03bc\u03b1\u03bb\u03bb\u03bf\u03bd \u03b5\u03c7\u03b5\u03b9\u03c2 \u03b4\u03b9\u03ba\u03b9\u03bf....\u0394\u03bf\u03ba\u03b9\u03bc\u03b1\u03c3\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03bf\u03bd 6\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf \u03c4\u03bf\u03c5 Mersenne,\u03ba\u03b1\u03b9 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9..\r\n\u0391,\u03bd\u03b1 \u03c3\u03b5 \u03c1\u03c9\u03c4\u03b7\u03c3\u03c9 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03c4\u03b9...\u0397 \u03c3\u03ba\u03b7\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b4\u03b9\u03bd\u03b5\u03b9\u03c2 \u03c3\u03c4\u03bf 3\u03bf post \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03b7...(\u03bf\u03c7\u03b9 \u03b7 \u03b3\u03b5\u03bd\u03b9\u03ba\u03b5\u03c5\u03c3\u03b7,\u03b1\u03bb\u03bb\u03b1 \u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7...)\u0394\u03b5\u03bd \u03bb\u03c5\u03bd\u03b5\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03b8\u03b5\u03c9\u03c1\u03b7\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bf\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03bc\u03bf\u03c1\u03c6\u03b5\u03c2 \u03bd=10\u03ba+\u03c5...?[\u03c0\u03b1\u03b9\u03c1\u03bd\u03bd\u03c9 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 10 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b7\u03bc\u03bf\u03c0\u03bf\u03b9\u03b7\u03c3\u03c9 fermat...,(\u03bf\u03c7\u03b9 \u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03b9\u03bf \u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1 :lol: )]\r\n\r\n :D :D :D", "Solution_8": "Vevaiws RAKAR! Auto akrivws prepei na kaneis! Oi askiseis tetoiou eidous exoun na kanoun me to $\\phi(n)$ pou stin prokeimeni periptwsi $\\phi(11)=10$. Tin afinw liges imeres akomi gia na tin dei opoios thelei prin kanw tin geniki paratirisi gia oles autes tis askiseis. Parolo pou tha pareis oles tis periptwseis $10k+u$, me $0\\leq u<9$ einai poli eukolo ean to organwseis se ena omorfo pinakaki. Gia koitakste to kai pali kai kanete tis paratiriseis sas (RAKAR eisai mesa...)\r\n\r\nPame gia tin genikeusi tou allou twra...\r\n\r\nAlexandros", "Solution_9": "[quote=\"cretanman\"]\n[size=125]Ean $p$ prwtos $>2$ kai $n$ fisikos me $n\\not\\equiv 0 \\pmod{p-1}$\ntote $1^{n}+2^{n}+\\ldots+(p-1)^{n}\\equiv 0 \\pmod{p}$[/size]\n[/quote]\r\n\r\n\u039a\u03b1\u03c4'\u03b1\u03c1\u03c7\u03ae\u03bd \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03bf\u03cd\u03bc\u03b5 \u03cc\u03c4\u03b9 $p^{n+2}=p+\\binom{n+2}{1}[1+2+...+(p-1)]+\\binom{n+2}{2}[1^{2}+2^{2}+...+(p-1)^{2}]+...$\r\n$....+\\binom{n+2}{n+1}[1^{n+1}+2^{n+1}+...+(p-1)^{n+1}]$ .\r\n\u0386\u03c1\u03b1 \u03b1\u03bd $p \\mid [1^{k}+2^{k}+...+(p-1)^{k}]$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $k=1,2,...,n$ ,\u03cc\u03c0\u03bf\u03c5 $n \\in \\{1,2,...,(p-3)\\}$, \u03c4\u03cc\u03c4\u03b5 $p \\mid [(n+2)(1^{n+1}+2^{n+1}+...+(p-1)^{n+1})]$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae $n \\in \\{1,2,...,(p-3)\\}$ \u03b8\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03cc\u03c4\u03b9 o $p$ \u03b4\u03b5\u03bd \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03af \u03c4\u03bf $(n+2)$. \u03a3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 $p \\mid [1^{n+1}+2^{n+1}+...+(p-1)^{n+1}]$.\r\n\u0395\u03c0\u03b1\u03b3\u03c9\u03b3\u03b9\u03ba\u03ac \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae $p \\mid [1+2+...+(p-1)]=\\frac{p(p-1)}{2}$ \u03b8\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03cc\u03c4\u03b9 $p \\mid [1^{k}+2^{k}+...+(p-1)^{k}]$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $k=1,2,...,(p-2)$ \r\n\u03ba\u03b1\u03b9 \u03b1\u03c6\u03bf\u03cd $a^{(p-1)m+u}\\equiv (a^{p-1})^{m}a^{u}\\equiv a^{u}\\pmod p$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $a$ \u03bc\u03b5 $(a,p)=1$ (\u03b3\u03b9\u03b1\u03c4\u03af \u03b1\u03c0\u03cc \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 Fermat \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03cc\u03c4\u03b9 $a^{p-1}\\equiv 1 \\pmod p$ \u03b1\u03bd $(a,p)=1$) \r\n\u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 $p \\mid [1^{n}+2^{n}+...+(p-1)^{n}]$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $n \\in \\mathbb N^{*}$ \u03bc\u03b5 $n \\not \\equiv 0 \\pmod{p-1}$, \u03cc\u03c0\u03bf\u03c5 $p$ \u03c0\u03c1\u03ce\u03c4\u03bf\u03c2 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03c2 \u03c4\u03bf\u03c5 $2$.\r\n\r\nStelios...", "Solution_10": "Katapliktiki skepsi Stelio!! Sigxaritiria...\r\n\r\nNomizw oti tha mporouse sigoura na einai ena kalo thema olimpiadas ean i lisi tou einai auti pou parethese o stelios kai den iparxei sintomoteri... Ti lete ki eseis?\r\n\r\nKai pali mpravo!\r\n\r\nAlexandros", "Solution_11": "\u03a9\u03c1\u03b1\u03af\u03b1 \u03a3\u03c4\u03ad\u03bb\u03b9\u03bf :)", "Solution_12": "\u0394\u03b5\u03af\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03b5\u03b4\u03ce \u03b3\u03b9\u03b1 \u03bc\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03bb\u03cd\u03c3\u03b7 . \u038c\u03c0\u03c9\u03c2 \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03b5\u03ba\u03b5\u03af \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03bf\u03bd\u03bf\u03bc\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c9\u03c2 \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 Chevalley-Warning\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=40171", "Solution_13": "The Chevalley-Warning Theorem can be found [url=http://en.wikipedia.org/wiki/Chevalley-Warning_theorem]here[/url].\r\n\r\nNote that the proof given in silouan's link is not completely elementary, since it uses the concept of primitive roots.\r\n\r\nPS: It's very nice to see that the Greek sub-forum is hyper-active." } { "Tag": [ "geometry", "incenter", "geometry unsolved" ], "Problem": "Let G and I be respectively the centroid and the incenter of a give triangle ABC.Prove that if $ AB^2 \\plus{} 2.IC^2 \\equal{} 2.IB^2 \\plus{} AC^2$ then GI is parallel to BC.", "Solution_1": "In order to get $ GI \\parallel{} BC$ we need $ AB \\plus{} AC \\equal{} 2\\cdot BC$ ( * ) but, the given relation leads straight to (*), hence the result.\r\n\r\nBest regards,\r\nsunken rock", "Solution_2": "This is an ongoing problem on The M&Y magazine in Vietnam .\r\n\r\nWell , I dont know whether it is just accident but all two posts by the author mkt are both on going problems on the M&Y magazine ." } { "Tag": [ "calculus", "integration", "LaTeX", "calculus computations" ], "Problem": "HI, can someone help me evaluate this definite integral using the substitution method? \r\n\r\ndefinite integral of 40/(x^2+25) dx from x = 0 to x = 5. I can't figure out what to make for the substitution for du and u (or any other corresponding dummy variables) as to simplify the expression. Could anyone help out? Thanks a bunch. \r\n\r\nSorry, I don't know how to work latex yet.", "Solution_1": "I don't really know where textbook problems like this go in, but I would post them in Pre-Olympiad because they aren't really challenging.\r\n\r\nLet $x=5u, dx=5du$\r\n\r\nThen we have\r\n\r\n$\\int^5_0 f(x)dx={8}\\int^1_0 \\frac {du}{u^2+1}=8(\\arctan 1-\\arctan 0)=2\\pi$\r\n\r\nNote the change of limits on the definite integral, alternately, you get that:\r\n\r\n$\\int \\frac {40}{x^2+25}=8\\arctan \\left(\\frac {x}{5}\\right)+C$", "Solution_2": "blahblahblah, thanks so much for helping. No, the problem did come out of a textbook, was not meant to be challenging. Just that I couldn't figure it out. ;)" } { "Tag": [], "Problem": "I need help!", "Solution_1": "[hide=\"ramblings\"]\nhmmm... always try to get 4 on your turn, so obviously never give the number 4 to your opponent. Number four on your turn guarantees victory. 4 - 1 = 3, 3 - 1 = 2, 2 - 1 = 1.\n\nIf n is prime on your turn, your opponent will have n-1, is that a good thing?\n\n\nIf n is 10 on your turn, you have won (congrats)\n[/hide]\n\n\n[hide=\"definitive\"]\n[code]\n1 <- 2 <- 3\n\n..........^\n..........|\n\n1 <- 2 <- 4 <- 5\n[/code]\n\nThis is the table of forced moves. As you can see, having a four on your turn guarantees a victory if you know what you are doing. However, if your opponent does not know what he is doing, it is possible to win even though the opponent has a four, as shown.\n\nThe number is generally irrelevant (read on) if it is not one of the chart numbers OR a number that can be turned into a number on the chart. There is only one other beneficial number, and it is 10. If you have 10 on your turn, take five away so your opponent is forced to give you a four.\n\nAnother beneficial strategy is leaving your opponent with a prime number. If your opponent is left with a prime number, then they only have one choice, which can be manipulated. Say you have a 22. You take away 11, and your opponent is forced to take away one. You now have 10, and you can turn that 10 into a 5, which all leads to my chart.\n[/hide]", "Solution_2": "Thanks treething for your reply!!! I have some questions: please help me asap\r\n\r\n1)\r\nHow did you came up with 10?\r\nI mean... how do you know that the other numbers will not have the 10 property (that quickly left you or your opponent with 4)?\r\n\r\n2)\r\nDoes leaving your opponent with any other prime number (besides 11) guarantee you the win? or just primes that will help you to 10? and how do you prove that?\r\n\r\n3) \r\nAre there winning strategies for starting players? .... starting number?\r\n\r\n4) \r\nI got to answer this question:\r\nclearly explains whether a longest and a shortest game exist for particular starting numbers :? \r\n\r\nmany thanks :)", "Solution_3": "THANK YOU SO MUCH treething!!! I finished the write-up on time!\r\n:)" } { "Tag": [ "integration", "trigonometry", "function", "calculus", "calculus computations" ], "Problem": "I found this one on a paper, but it doesn't look correct. Can anyone prove me wrong?\r\n$ \\int_0^\\pi\\cos(m\\theta) \\sin(x_{ml}|\\cos(\\theta) \\minus{} p|)\\,d\\theta \\equal{} 2\\int_0^{\\cos^{ \\minus{} 1}p}\\cos(m\\theta) \\sin[x_{ml}(\\cos(\\theta) \\minus{} p)]\\,d\\theta$\r\n\r\n$ p \\in [0,1]$\r\n$ x_{ml} \\in \\Re$\r\n$ m$ is integer\r\n\r\nThanks", "Solution_1": "It doesn't seem to hold for, for example, $ m \\equal{} 1$, $ p \\equal{} 0$, and $ 0 < |x_{ml}| \\leq \\pi$. The left hand side is $ 0$ but the right is strictly positive for $ 0 < x_{ml} \\leq \\pi$ and negative for $ \\minus{} \\pi \\leq x_{ml} < 0$. Given the subscript, I would guess that it is meant that $ x_{ml}$ is not arbitrary, but depends in some very specific way on $ m$.", "Solution_2": "You're right about $ x_{ml}$. It's the $ (l \\plus{} 1)$th zero of the Bessel function of the first kind $ J_m(x)$, where $ l$ is a non-negative integer. $ m$ is also non-negative.\r\n$ x_{ml}$ not arbitrary, but for all purposes in this integral it can be assumed to be.", "Solution_3": "$ \\int_0^\\pi\\cos(m\\theta) \\sin(x_{ml}|\\cos(\\theta)\\minus{}p|)\\,d\\theta \\minus{} \\int_0^{\\cos^{\\minus{}1}p}\\cos(m\\theta) \\sin [x_{ml}(\\cos(\\theta)\\minus{}p)] d\\theta$\r\n\r\n$ \\equal{} \\int_{\\cos^{\\minus{}1}p}^\\pi\\cos(m\\theta) \\sin(x_{ml}(p\\minus{}\\cos(\\theta)))\\,d\\theta \\equal{} 0$ when?", "Solution_4": "when?\r\nIn the paper, it looks as though \"always\" would be the answer. Look at how they make the statement:\r\n[img]http://i191.photobucket.com/albums/z226/pocaracas/eq8_9.gif[/img]\r\n\r\nooops.. I forgot the 2 on the right hand side... it's not too relevant because $ 2*0$ should the $ 0$.", "Solution_5": "$ [0,\\pi] \\equal{} [0,\\cos^{ \\minus{} 1}p] \\cup (\\cos^{ \\minus{} 1}p, \\pi \\minus{} \\cos^{ \\minus{} 1}p) \\cup [\\pi \\minus{} \\cos^{ \\minus{} 1}p,\\pi]$\r\n\r\n$ I \\equal{} \\int_0^\\pi\\cos(m\\theta) \\sin(x_{ml}|\\cos(\\theta) \\minus{} p|)\\,d\\theta \\equal{} I_1 \\plus{} I_2 \\plus{} I_3$\r\n\r\n$ I_1 \\equal{} \\int_0^{\\cos^{ \\minus{} 1}p}\\cos(m\\theta) \\sin [x_{ml}(\\cos(\\theta) \\minus{} p)] d\\theta$\r\n\r\n$ I_2 \\equal{} \\int_{\\cos^{ \\minus{} 1}p}^{\\pi \\minus{} \\cos^{ \\minus{} 1}p} \\cos (m\\theta) \\sin [x_{ml}(p \\minus{} \\cos(\\theta)] d\\theta$\r\n\r\n$ I_3 \\equal{} \\int_{\\pi \\minus{} \\cos^{ \\minus{} 1}p}^\\pi \\cos(m\\theta) \\sin [x_{ml}(\\cos(\\theta) \\minus{} p)] d\\theta$\r\n\r\ni think what we do is show $ I_2 \\equal{} 0, I_1 \\equal{} I_3$", "Solution_6": "$ I_3=\\int_{\\pi-\\cos^{-1}p}^\\pi \\cos(m\\theta) \\sin [x_{ml}(\\cos(\\theta)-p)] d\\theta$\r\n$ \\chi=\\pi-\\theta$\r\n$ =\\int_{\\cos^{-1}p}^0 -\\cos(m(\\pi-\\chi)) \\sin [x_{ml}(\\cos(\\pi-\\chi)-p)] d\\chi$\r\n\r\n$ =\\int_0^{\\cos^{-1}p} -\\cos(m\\pi) \\cos(m\\chi) \\sin [x_{ml} (\\cos \\chi + p )] %Error. \"dchi\" is a bad command.\n$\r\n\r\n$ =(-1)^{m+1} \\int_0^{\\cos^{-1}p} \\cos(m\\chi) \\sin [x_{ml}(\\cos \\chi -p)+2px_{ml}]$\r\n\r\n$ =(-1)^{m+1} \\cos (2px_{ml})\\int_0^{\\cos^{-1}p} \\cos(m\\chi) \\sin [x_{ml}(\\cos \\chi -p)]d\\chi$\r\n$ +(-1)^{m+1} \\sin(2px_{ml}) \\int_0^{\\cos^{-1}p}\\cos[x_{ml}(\\cos \\chi -p)]d\\chi$\r\n\r\nmaybe $ x_{ml}$ matters here?\r\n\r\n$ I_2=\\int_{\\cos^{-1}p}^{\\pi-\\cos^{-1}p}\\sin [x_{ml}(p-\\cos(\\theta)] d\\theta$\r\n\r\n$ \\xi=\\theta - \\frac{\\pi}{2}$, $ \\alpha = \\frac{\\pi}{2}-\\cos^{-1}p$\r\n\r\n$ =\\int_{-\\alpha}^\\alpha \\cos( m\\xi +\\frac{m\\pi}{2}) \\sin [ x_{ml} (cos(\\xi+\\frac{\\pi}{2}) - p)] d\\xi$\r\n\r\n\r\n$ =-\\int_{-\\alpha}^\\alpha [\\cos( m\\xi)\\cos \\frac{m\\pi}{2}-\\sin( m\\xi)\\sin \\frac{m\\pi}{2}]\\sin [ x_{ml} (\\sin(\\xi) + p)] d\\xi$\r\n\r\n$ =\\sin \\frac{m\\pi}{2}\\int_{-\\alpha}^\\alpha \\sin( m\\xi)[\\sin ( x_{ml}\\sin\\xi)\\cos(x_{ml}p)+\\cos (x_{ml}\\sin\\xi )\\sin(x_{ml}p) ] d\\xi$\r\n$ - \\cos \\frac{m\\pi}{2}\\int_{-\\alpha}^\\alpha \\cos( m\\xi)[\\sin ( x_{ml}\\sin\\xi)\\cos(x_{ml}p)+\\cos (x_{ml}\\sin\\xi )\\sin(x_{ml}p) ]d\\xi$\r\n\r\n$ =\\sin \\frac{m\\pi}{2}\\cos(x_{ml}p)\\int_{-\\alpha}^\\alpha \\sin( m\\xi)\\sin ( x_{ml}\\sin\\xi)$\r\n$ + \\sin \\frac{m\\pi}{2}\\sin(x_{ml}p)\\int_{-\\alpha}^\\alpha \\sin( m\\xi)\\cos ( x_{ml}\\sin\\xi)$\r\n$ - \\cos \\frac{m\\pi}{2}\\cos(x_{ml}p)\\int_{-\\alpha}^\\alpha \\cos( m\\xi)\\sin ( x_{ml}\\sin\\xi)$\r\n$ - \\cos \\frac{m\\pi}{2}\\sin(x_{ml}p)\\int_{-\\alpha}^\\alpha \\cos( m\\xi)\\cos ( x_{ml}\\sin\\xi)$\r\n\r\n$ =2\\sin \\frac{m\\pi}{2}\\cos(x_{ml}p)\\int_0^\\alpha \\sin( m\\xi)\\sin ( x_{ml}\\sin\\xi)$\r\n$ -2\\cos \\frac{m\\pi}{2}\\sin(x_{ml}p)\\int_0^\\alpha \\cos( m\\xi)\\cos ( x_{ml}\\sin\\xi)$\r\n\r\nor maybe im not getting anywhere =(", "Solution_7": "Thanks for the effort Ken4000, but I also think it's not leading anywhere... :(" } { "Tag": [ "counting", "derangement", "geometry", "geometric transformation", "rotation" ], "Problem": "JACK, AND, JILL, WENT, UP, THE, and HILL are sitting at a round table. They want to reorder themselves such that JACK isn't in JACK's spot, AND isn't in AND's spot,... (everybody isn't in each other's original spot)How many possible arrangements are there?\r\n\r\nwhat if they were originally just sitting in a row of chairs? thx! :)", "Solution_1": "For the second problem, this becomes the number of derangements of $ 7$ objects. For $ n$ objects this is \\[ \\sum_{i\\equal{}0}^{n}(\\minus{}1)^i \\frac{n!}{i!}\\] Plugging in $ 7$ we get $ \\boxed{1854}$.\r\n\r\nFor the first problem I suppose you can take the derivation of this formula and modify it for the problem?", "Solution_2": "How do you derive that formula? PIE? I keep getting 4320 from epic casework... :maybe:", "Solution_3": "Yeah it's PIE with the sets $ A_i$, $ 1\\leq i\\leq n$, where $ A_i$ is the set of all arrangements where person $ i$ is in the right place.", "Solution_4": "i think for the first problem that it would be 1854*6! cuz for each row of seats, u could put it in a circle and for each circle there are 6! arrangements. am i right? :|", "Solution_5": "I don't know if this problem applies to a case of a round table where rotations are indistinct, since you can't say that any particular seat is Jack's seat. If we do say that seat 1 is Jack's seat, etc. then rotations are distinct and our answer is the same.", "Solution_6": "5...x, what's a sideways, capital \"M\" supposed to mean????", "Solution_7": "??? :maybe:", "Solution_8": "That's a summation symbol.", "Solution_9": "@akhil0422:\r\n\r\n$ \\sum_{k\\equal{}a}^{b}f(k)\\equal{}f(a)\\plus{}f(a\\plus{}1)\\plus{}\\cdots\\plus{}f(b\\minus{}1)\\plus{}f(b)$.", "Solution_10": "So do we have a final answer to the round circle question? To reiterate,\r\n[quote]I don't know if this problem applies to a case of a round table where rotations are indistinct, since you can't say that any particular seat is Jack's seat. If we do say that seat 1 is Jack's seat, etc. then rotations are distinct and our answer is the same.[/quote]" } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "without use of any calculator,determine:\r\nsin(pi/14)+6sin\u00b2(pi/14)-8sin^4(pi/14)", "Solution_1": "Let $z : = e^{i \\frac{\\pi}{14}}$ and $y : = z^{2}$. Then\r\n$\\sin (\\frac{\\pi}{14})+6 \\sin^{2}(\\frac{\\pi}{14})-8 \\sin^{4}(\\frac{\\pi}{14})$ \r\n$= \\frac{1}{2i}(z-z^{-1})-\\frac{3}{2}(z-z^{-1})^{2}-\\frac{1}{2}(z-z^{-1})^{4}$\r\n$= \\frac{1}{2}(-i(z-z^{-1})-3(z^{2}+z^{-2}+2)-(z^{4}-4 z^{2}+6-4z^{-2}+z^{-4}))$\r\n$= \\frac{1}{2}(-z^{8}+z^{6}-z^{4}-z^{-4}+z^{2}+z^{-2})$\r\n$= \\frac{1}{2}(-y^{4}+y^{3}-y^{2}-y^{-2}+y+y^{-1})$\r\n$= \\frac{1}{2}(-y^{6}+y^{5}-y^{4}+y^{3}-y^{2}+y-1+1)$\r\n$=\\frac{1}{2}(-\\frac{y^{7}+1}{y+1}+1) = \\frac{1}{2}$" } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "geometry", "3D geometry", "number theory" ], "Problem": "P1, P2, and P3 are polynomials defined by:\r\n$P1(x) = 1+x+x^{3}+x^{4}+......+x^{96}+x^{97}+x^{99}+x^{100}$\r\n\r\n$P2(x) = 1-x+x^{2}-......-x^{99}+x^{100}$\r\n\r\n$P3(x) = 1+x+x^{2}+......+x^{66}+x^{67}.$\r\n\r\nFind the number of distinct complex roots of P1 * P2 * P3.\r\n\r\n[hide=\"HINT\"]\nGeometric Summation and PIE.[/hide]", "Solution_1": "[hide]$P1(x)=\\frac{x^{101}-1}{x-1}$\n$P2(x)=\\frac{x^{101}+1}{x+1}$\n$P3(x)=\\frac{x^{68}-1}{x-1}$\n\nso $P1(x)$ has 100 roots (don't include 1 because then the fraction's undefined), $P2(x)$ has 100, and $P3(x)$ has 67. i guess none of them are the same, because the ones for $P1(x)$ are in the form $e^{\\frac{2k\\pi}{101}}$, for integral k from 1 to 100, the roots of $P2(x)$ are in the form $-e^{\\frac{2k\\pi}{101}}$, and the roots of $P3(x)$ are in the form $e^{\\frac{2k\\pi}{68}}$.\n\nso there should be 267 distinct roots. [/hide]", "Solution_2": "You didn't look carefully enough at $P_{1}$.", "Solution_3": "[hide]266; the difference in $P1$ still has the same number of roots (100,) but this time, there is a double root (-1.) Therefore, you simply subtract 1 from perfectnumber628's original answer (267) to get the correct answer.[/hide]", "Solution_4": "You've also done something wrong (there are far more overlapping roots than you've listed) but since you didn't actually show any work, I can't be sure what it was.\r\n\r\nP1 is missing every third power, $x^{2}, x^{5}, \\ldots, x^{9}8$.", "Solution_5": "[hide]well look at that, it [b]is[/b] missing every third power. so factor out a $1+x$ to get $P1(x)=(x+1)(1+x^{3}+x^{6}...x^{99})=(x+1)\\left(\\frac{x^{102}-1}{x^{3}-1}\\right)$. so $P1$ has 100 roots, -1 and the 102nd roots of unity, not counting the cube roots of unity. so -1 is a double root, so there are only 99 distinct ones.\n\nnow some of those will be the same as the 68th roots from that other polynomial. the roots of $P1$ are in the form $e^{\\frac{2k\\pi}{102}}$ (k is an integer, $1\\leq k\\leq 101$, k is not 34 or 68), the roots of $P2$ are in the form $e^{\\frac{2k\\pi}{101}}$ (k is an integer from 1 to 100), and the roots of $P3$ are $e^{\\frac{2k\\pi}{68}}$ (k is an integer from 1 to 67).\n\nsince 101 is relatively prime to 102 and 68, none of the roots of $P2$ overlap the other ones. let $k_{1}$ be the k from a root of $P1$ and $k_{3}$ be a k from $P3$. so the roots that aren't distinct are the ones where $\\frac{k_{1}}{102}=\\frac{k_{3}}{68}$\n\nmulitplying by 34, we get $\\frac{k_{1}}{3}=\\frac{k_{3}}{2}$, which means $2k_{1}=3k_{3}$. so $k_{3}$ is divisible by 2, and $k_{1}$ is divisible by 3. i think any even number from 2 to 66 works for $k_{3}$, so there's 33 of them. \n\nso we have $99+100+67-33=233$. so hopefully 233 is the right answer :lol: [/hide]" } { "Tag": [ "calculus", "integration" ], "Problem": "What is the largest integer $x$ for which $\\frac{1}{x}$ is larger than $\\frac{4}{49}$?", "Solution_1": "The answer should be [hide=\"answer\"]$12$[/hide]", "Solution_2": "Yes that is. \r\n[hide=\"brief explanation\"]4/49 is close (but smaller than) 4/48, which is also 1/12. Common sense would tell us that 1/12 would be greater than 4/49. \n1/13 is 4/52, obviously bigger than 4/49. \n$x = 12$[/hide]", "Solution_3": "[quote]1/13 is 4/52, obviously bigger than 4/49[/quote]\r\n\r\nI think you mean smaller. ;)\r\n\r\nA little bit more of a proof... $\\frac{4}{49}=\\frac{1}{x} \\Rightarrow 4x=49 \\Rightarrow x=12.25$\r\nTherefore, for $\\frac{1}{x}>\\frac{4}{49}$, $x<12.25$. And since the problem specifies the largest integer, the answer is $12$.", "Solution_4": "[quote=\"JesusFreak197\"][quote]1/13 is 4/52, obviously bigger than 4/49[/quote]\n\nI think you mean smaller. ;)[/quote]\r\n\r\nWhoops, yes I did. I hope you all realize that I knew that already :blush:", "Solution_5": "[hide]\nBad Method:\n$\\frac{1}{x}>\\frac{4}{49}\\\\49>4x\\\\x<\\frac{49}{4}\\\\x=12\\\\$\n\nGood Method:\n$\\frac{x}{1}<\\frac{49}{4}\\\\x=12$\n^_^\n[/hide]", "Solution_6": "$\\frac{1}{x} = \\frac{4}{4x} < \\frac{4}{49}$\r\n$4x<49 \\rightarrow x<12.25$ so the greatest integral $x$ is 12." } { "Tag": [ "inequalities", "induction" ], "Problem": "How does one get titu's lemma from cauchy's inequality", "Solution_1": "I'll prove this for the 3 variable case. Let $ u \\equal{} <\\sqrt{a}, \\sqrt{b}, \\sqrt{c}>$, and let $ v \\equal{} <\\frac{x}{\\sqrt{a}}, \\frac{y}{\\sqrt{b}}, \\frac{z}{\\sqrt{c}}>$. We have that $ |u|^2|v|^2 \\geq |u \\cdot v|^2$, that is, $ (a \\plus{} b \\plus{} c)(\\frac{x^2}{a} \\plus{} \\frac{y^2}{b} \\plus{} \\frac{z^2}{c}) \\geq (x \\plus{} y \\plus{} z)^2$. Divide both sides by $ (a \\plus{} b \\plus{} c)$ to get the desired result.", "Solution_2": "Or you could just prove it for the two variable case: $ \\frac{a_1^2}{b_1}\\plus{}\\frac{a_2^2}{b_2}\\geq \\frac{(a_1\\plus{}a_2)^2}{b_1\\plus{}b_2}$, which is true if you clear denominators and multiply out (or you can use Cauchy.)\r\nThen, if $ \\sum_{i\\equal{}1}^n\\frac{a_i^2}{b_i}\\geq \\frac{\\left(\\sum_{i\\equal{}1}^n a_i\\right)^2}{\\sum_{i\\equal{}1}^n b_i}$, then $ \\sum_{i\\equal{}1}^{n\\plus{}1}\\frac{a_i^2}{b_i}\\geq \\frac{\\left(\\sum_{i\\equal{}1}^n a_i\\right)^2}{\\sum_{i\\equal{}1}^n b_i}\\plus{}\\frac{a_{n\\plus{}1}^2}{b_{n\\plus{}1}}\\geq \\frac{\\left(\\sum_{i\\equal{}1}^{n\\plus{}1} a_i\\right)^2}{\\sum_{i\\equal{}1}^{n\\plus{}1} b_i}$.", "Solution_3": "He wanted to see how is T2's lemma a special case of CS.\r\n\r\nLike this:\r\n\r\nCS: $ (x_1^2\\plus{}...\\plus{}x_n^2)(y_1^2\\plus{}...\\plus{}y_n^2)\\ge(x_1y_1\\plus{}...\\plus{}x_ny_n)^2$\r\n\r\nTake $ x_i\\equal{}\\frac{a_i^2}{b_i},y_i\\equal{}b_i$, and T2's lemma is there.", "Solution_4": "Simply,\r\nIf there is $ a,b,c$ and real $ x,y,z>0$ then: \\[ \\frac{\\left(a\\plus{}b\\plus{}c\\right)^2}{x\\plus{}y\\plus{}z} \\geq \\frac{a^2}{x}\\plus{} \\frac{b^2}{y}\\plus{}\\frac{c^2}{z}\\]\r\n\r\nProof: All of it is choosing good values for Cauchy then some algebra. Notice:\r\n\\[ \\left(a\\plus{}b\\plus{}c\\right)^2\\equal{}\\left(\\frac{a}{\\sqrt{x}}\\sqrt{x}\\plus{}\\frac{b}{\\sqrt{y}}\\sqrt{y}\\plus{}\\frac{c}{\\sqrt{z}}\\sqrt{z}\\right)^2\\]\r\nNow make: $ \\left(a_1,a_2,a_3\\right)\\equal{}\\left(\\frac{a}{\\sqrt{x}},\\frac{b}{\\sqrt{y}},\\frac{c}{\\sqrt{z}}\\right)$ and $ \\left(b_1,b_2,b_3\\right)\\equal{}\\left(\\sqrt{x},\\sqrt{y},\\sqrt{z}\\right)$ \r\nNow apply CS\r\n\\[ \\left(a\\plus{}b\\plus{}c\\right)^2\\equal{}\\left(\\frac{a}{\\sqrt{x}}\\sqrt{x}\\plus{}\\frac{b}{\\sqrt{y}}\\sqrt{y}\\plus{}\\frac{c}{\\sqrt{z}}\\sqrt{z}\\right)^2 \\geq \\left(\\frac{a^2}{x}\\plus{} \\frac{b^2}{y}\\plus{}\\frac{c^2}{z}\\right)\\left(x\\plus{}y\\plus{}z\\right)\\]\r\nBy dividing we get \\[ \\frac{\\left(a\\plus{}b\\plus{}c\\right)^2}{x\\plus{}y\\plus{}z} \\geq \\frac{a^2}{x}\\plus{} \\frac{b^2}{y}\\plus{}\\frac{c^2}{z}\\]", "Solution_5": "a_1/x_1+a_2/x_2+...a_n/x_n>=(a_1+a_2+....a_n)^2/x_1+x_2+...+x_n.\r\ncross multiply.\r\n(a_1/x_1+a_x/x_2+.....+a_n/x_n)(x_1+x_2....+x_n)>=(a_1+a_2+....+a_n)^2\r\nThis is cauchy!", "Solution_6": "[quote=\"Bugi\"]He wanted to see how is T2's lemma a special case of CS.\n\nLike this:\n\nCS: $ (x_1^2 \\plus{} ... \\plus{} x_n^2)(y_1^2 \\plus{} ... \\plus{} y_n^2)\\ge(x_1y_1 \\plus{} ... \\plus{} x_ny_n)^2$\n\nTake $ x_i \\equal{} \\frac {a_i^2}{b_i},y_i \\equal{} b_i$, and T2's lemma is there.[/quote]\r\n\r\nJust a small typo: $ x_i \\equal{} \\dfrac{a_i}{\\sqrt{b_i}}$, $ y_i\\equal{}\\sqrt{b_i}$", "Solution_7": "[hide=\"proof \"]\nSince \n\n$ \\frac {a_1 ^2}{b_1} \\plus{} \\frac {a_2 ^2 }{b_2} \\geq \\frac {(a_1 \\plus{} a_2) ^2}{b_1 \\plus{} b_2} \\iff (a_1 b_2 \\minus{} a_2 b_1)^2 \\geq 0$\n\nWe have that the inequality is true for $ n \\equal{} 2$ \n\nSuppose it is true for $ n$\n\n$ \\sum_{k \\equal{} 1}^{n} {\\frac {a_k ^2}{b_k}} \\geq \\frac {\\left(\\sum {a_k}\\right)^2}{ {b_k}}$\n\nThen we add $ \\frac {a_{k \\plus{} 1}^2}{b_{k \\plus{} 1}}$ to both sides .\n\n$ \\sum_{k}^{n} {\\frac {a_k ^2}{b_k}} \\plus{} \\frac {a_{k \\plus{} 1}^2}{b_{k \\plus{} 1}} \\geq \\frac {\\left(\\left(\\sum {a_k}\\right)^2 \\plus{} a_{k \\plus{} 1} \\right)^2}{\\sum {{b_k}} \\plus{} b_{k \\plus{} 1}} \\iff \\left((\\sum a_k)^2 \\cdot b_{k \\plus{} 1} \\minus{} a_{k \\plus{} 1} \\sum b_{k} \\right)^2 \\geq 0$ \n\nWe see the inequality is true for $ n \\equal{} k \\plus{} 1$ ,then it's true for every $ n \\in \\mathbb{N}$ ,as desired \n\nThat was the proof of Titu's Lemma by induction :\n\nNow by Cauchy Schwartz inequality \n\n$ \\left(\\frac{a_1 ^2}{b_1} \\plus{} ...\\plus{}\\frac{a_n ^2}{b_n}\\right)\\cdot (b_1 \\plus{}...\\plus{}b_n) \\geq {(a_1 \\plus{} ...\\plus{}a_n)^2}$ ,done [/hide]", "Solution_8": "Please change the sign in my post from $ \\geq$ to $ \\le$ I made a typo but I can't edit it now :|", "Solution_9": "corrections" } { "Tag": [ "\\/closed" ], "Problem": "I think we should keep this forum up-to-date. Does anybody agree?\r\n[img]7129[/img]", "Solution_1": "[quote=\"perfect_radio\"]I think we should keep this forum up-to-date. Does anybody agree?\n[img]7129[/img][/quote]\r\n\r\nROFL its just a minor detail....but if u want Mr. Vornicu can change it :D", "Solution_2": "[quote=\"7h3.D3m0n.117\"]ROFL its just a minor detail...[/quote]\r\n\r\nFor some reason, I have a feeling it is more like a subtle allusion...\r\n\r\n darij", "Solution_3": "It's skin-dependent: the date is 2007 in AOPS style and 2004-2006 in Mathlinks style." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Let $ P(x)$ be a sixth degree polynomial so that the coefficient of $ x^6$ is $ 1$.\r\nIf $ P(1) \\equal{} 1, P(2) \\equal{} 2, P(3) \\equal{} 3, P(4) \\equal{} 4, P(5) \\equal{} 5$ and $ P(6) \\equal{} 6$, what's the value of $ P(7)$?", "Solution_1": "hello, solving the linear system which we get by the conditions we have\r\n$ P(x)\\equal{}x^6\\minus{}21x^5\\plus{}175x^4\\minus{}735x^3\\plus{}1624x^2\\minus{}1763x\\plus{}720$\r\nand we get $ P(7)\\equal{}727$\r\nSonnhard.", "Solution_2": "We have:\r\n$ P(x)\\minus{}x \\equal{} c(x\\minus{}1)(x\\minus{}2)(x\\minus{}3)(x\\minus{}4)(x\\minus{}5)(x\\minus{}6)$ where $ c\\equal{}1$ because P(x) is a monic polynomial, which gives\r\n$ P(7) \\equal{} 6! \\plus{} 7 \\equal{} 727$" } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f: [0,\\infty)\\to\\mathbb{R}$ be a continuous function on $ (0,\\infty)$.Prove that : \r\n\r\n$ f$ has The Intermediate Value Property $ \\Leftrightarrow$ $ \\exists$ a sequence $ (a_n)_n$ , $ a_n>0 \\forall n \\in \\mathbb{N}$ such that $ \\lim_{n\\to\\infty}a_n\\equal{}0$ and $ \\lim_{n\\to\\infty}f(a_n)\\equal{}f(0)$\r\n\r\n[i]author :C.Mortici[/i]", "Solution_1": "The question is not really that interesting. I don't even know why I posted it [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=75435]here[/url] :?:" } { "Tag": [ "function", "trigonometry", "algebra", "domain", "Support", "integration", "calculus" ], "Problem": "I dont really get how to show that a function is of exponential order. Here are some functions from the book:\r\n\r\n$ t^{n}$\r\n\r\n$ \\sin (at)$\r\n\r\n$ e^{\\sqrt}{t}$\r\n\r\nCould someone just explain how its done or do one of these as an example?\r\n\r\nThanks", "Solution_1": "The question is what you mean by exponential order. What I'm going to assume, since you mention differential equations, is that you're looking for function to which you can apply Laplace transform techniques.\r\n\r\nSo assume we have a function $ f$ whose domain is $ [0,\\infty).$ (Sometimes it's useful to consider functions with domain $ \\mathbb{R}$ and support $ [0,\\infty),$ but that doesn't really change anything.\r\n\r\nWhat we need is for $ \\int_{0}^{\\infty}f(t)e^{-st}\\,dt$ to converge for some real $ \\s_{0}.$ Then the integral will converge for all $ s$ such that $ \\text{Re}(s)>s_{0}.$\r\n\r\nA sufficient (but not necessary) condition for this is that $ f$ is locally integrable and that there is some $ s$ such that $ f(t)e^{-st}$ remains bounded as $ s\\to\\infty.$ That's probably what you mean by \"exponential order.\"\r\n\r\n$ t^{n}$ is of exponential order; for every $ s>0,\\ t^{n}e^{-st}\\to0$ as $ t\\to\\infty.$\r\n\r\n$ \\sin(at)$ is of exponential order since $ \\sin(at)e^{0t}$ is bounded. You can say this about any bounded function.\r\n\r\n$ e^{\\sqrt{t}}$ is of exponential order. For any $ s>0,\\ \\sqrt{t}-st\\to-\\infty$ as $ t\\to\\infty,$ therefore $ e^{\\sqrt{t}}e^{-st}\\to0.$ [It doesn't have a \"nice\" Laplace transform, but that's a different question. It does have some Laplace transform.]\r\n\r\n$ t^{m}e^{at}\\cos{bt}$ is of exponential order. And has a nice Laplace transform.\r\n\r\nSo what isn't of exponential order? We need $ e^{b(t)}$ where $ b(t)$ increases at a faster-than-linear rate. For instance, $ e^{t^{2}}.$ Or $ t^{t}.$ Or $ \\Gamma(t).$" } { "Tag": [], "Problem": "Could you,please send me the full solutions of the fourth and sixth Junior Balkan Olympiad.", "Solution_1": "if you post the questions yourself on the forum in the apropriate places you will probably get some answers. otherwise I really doubt whether anyone would write soln to problems whoose statements they don't know :D", "Solution_2": "Try http://www.kalva.demon.co.uk" } { "Tag": [ "calculus", "trigonometry", "derivative" ], "Problem": "Find the maximum value of $ f(x)$, if $ f(x\\plus{}\\pi)\\equal{}12sin(3x)\\plus{}5cos(3x\\plus{}\\frac{\\pi}{2})\\plus{}3$", "Solution_1": "[hide=\"Solution (involves calculus)\"]We know $ f(n\\plus{}\\pi)\\equal{}12\\sin 3n\\plus{}5\\cos\\left(3n\\plus{}\\frac{\\pi}{2}\\right)\\plus{}3$. Plug in $ n\\equal{}x\\minus{}\\pi$ to get $ f(x)\\equal{}12\\sin(3x\\minus{}3\\pi)\\plus{}5\\cos\\left(3x\\minus{}\\frac{5\\pi}{2}\\right)\\plus{}3$.\n\nSo, $ f^\\prime(x)\\equal{}36\\cos(3x\\plus{}3\\pi)\\minus{}15\\sin\\left(3x\\plus{}\\frac{5\\pi}{2}\\right)$.\n\n$ 36\\cos(3x\\minus{}3\\pi)\\minus{}15\\sin\\left(3x\\minus{}\\frac{5\\pi}{2}\\right)\\equal{}0$\n\n$ 36\\cos 3x\\minus{}15\\sin\\left(3x\\plus{}\\frac{\\pi}{2}\\right)\\equal{}0$\n\n$ 36\\cos 3x\\plus{}15\\cos 3x\\equal{}0$\n\n$ \\cos 3x\\equal{}0\\implies x\\equal{}\\frac{\\pi}{6}\\plus{}2\\pi n,\\;n\\in\\mathbb{Z}^\\plus{}$.\n\nNote that half of these values will be minima and the other half maxima. The first derivative test shows that $ x\\equal{}\\frac{\\pi}{6}$ is a minimum, and $ x\\equal{}\\frac{\\pi}{2}$ is a maximum.\n\nFinally, we find $ f\\left(\\frac{\\pi}{2}\\right)\\equal{}12\\sin\\left(\\frac{3\\pi}{2}\\minus{}3\\pi\\right)\\plus{}5\\sin\\left(\\frac{3\\pi}{2}\\right)\\plus{}3\\equal{}12\\minus{}5\\plus{}3\\equal{}\\boxed{10}$[/hide]", "Solution_2": "Non-Calculus solution:\r\n\r\n[hide]$ f(x \\plus{} \\pi) \\equal{} 12\\sin(3x) \\plus{} 5\\cos(3x \\plus{} \\frac {\\pi}{2}) \\plus{} 3$\n\n$ f(x \\plus{} \\pi) \\equal{} 12\\sin(3x) \\plus{} 5\\left(\\cos3x\\cos\\frac {\\pi}{2} \\minus{} \\sin3x\\sin\\frac {\\pi}{2}\\right) \\plus{} 3$\n\n$ f(x \\plus{} \\pi) \\equal{} 7\\sin(3x) \\plus{} 3$\n\nNow we can already see that the maximum will be $ 10$, but we'll proceed formally:\n\n$ x \\plus{} \\pi \\equal{} t$\n\n$ f(t) \\equal{} 7\\sin(3t \\minus{} 3\\pi) \\plus{} 3$\n\n$ f(t) \\equal{} 7(\\sin 3t \\cos 3\\pi \\minus{} \\cos 3t \\sin 3\\pi) \\plus{} 3$\n\n$ f(t) \\equal{} \\minus{} 7\\sin 3t \\plus{} 3$\n\n$ \\sin \\varphi \\geq \\minus{} 1$ so $ f(t)\\geq 7 \\plus{} 3 \\equal{} 10$\n\nwhich is the maximum and it is reached for $ \\sin 3t \\equal{} \\minus{} 1$.\n\n[/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Prove that if $a,b,c$ are positive real then:\r\n\r\n$(a+\\frac{b^2}{c})^2+(b+\\frac{c^2}{a})^2+(c+\\frac{a^2}{b})^2\\ge 12\\frac{a^3+b^3+c^3}{a+b+c}$", "Solution_1": "OK. It seems too hard. My solution is quite long and I wait someone solve it easier.", "Solution_2": "[quote=\"hungkhtn\"]OK. It seems too hard. My solution is quite long and I wait someone solve it easier.[/quote]\r\n\r\nis the question changed?!", "Solution_3": "Yeah, I have strugle with Hung puzzle too. Did you change from\"4\" to \"12\"?\r\nBecause if it is \"4\" then I have no idea where the inequa hold.\r\nAnyway, if it is \"12\" then probably I will try again.", "Solution_4": "Ok. Sorry. I used to post solution in hurry then sometimes It wasn't like of mines at first. :( I change the constant $4$ to $12$.", "Solution_5": "Hi... It's very hard. I like it because $n-1 EV$ can't kill it! :lol: :lol: \r\n\r\n I hope before a month, there's a solution here.", "Solution_6": "[quote=hungkhtn]Prove that if $a,b,c$ are positive real then:\n$(a+\\frac{b^2}{c})^2+(b+\\frac{c^2}{a})^2+(c+\\frac{a^2}{b})^2\\ge 12\\frac{a^3+b^3+c^3}{a+b+c}$[/quote]\nWe have\n\\[\\left(a+\\frac{b^2}{c}\\right)^2+\\left(b+\\frac{c^2}{a}\\right)^2+\\left(c+\\frac{a^2}{b}\\right)^2 - \\frac{12(a^3+b^3+c^3)}{a+b+c} = \\frac{F(a,b,c)}{a^2b^2c^2(a+b+c)},\\]\nfor\n\\[(a+b+c)F(a,b,c) = \\sum abc^2 (2a^2+7ab+2b^2)(a^2+bc-2ab)^2\\]\n\\[\\displaystyle +\\sum c^2a^3\\left[(2a+4b+c)ca+2b(b-2c)^2+2b(a-b)^2\\right](a-b)^2\\]\n\\[+\\sum a^2b^2(b^3+c^3+2abc-b^2c-2bc^2-ca^2)^2 \\geqslant 0.\\]\nThe following inequality is also true\n\\[\\left(a+\\frac{b^2}{c}\\right)^2+\\left(b+\\frac{c^2}{a}\\right)^2+\\left(c+\\frac{a^2}{b}\\right)^2 \\geqslant \\frac{73(a^3+b^3+c^3)-39abc}{5(a+b+c)}\\]" } { "Tag": [], "Problem": "Let be $ A\\equal{}\\{(x,y)\\in \\mathbb{R}^2 \\ |\\ x^2\\plus{}y^2>2\\}$ and $ B\\equal{}\\{(x,y)\\in \\mathbb{R}^2\\ |\\ x\\plus{}y\\plus{}xy<3\\}$ . Determine $ A\\cup B$ .", "Solution_1": "I think you mean A intersect B, not A union B. Anyways, there would still be many solutions, take y=-x, x>10.", "Solution_2": "[quote=\"dgreenb801\"]I think you mean A intersect B, not A union B. Anyways, there would still be many solutions, take y=-x, x>10.[/quote]\r\n\r\nIt is $ A\\cup B$ .", "Solution_3": "Oh I see, you intend for there to be infinitely many solutions, but you have to describe them. Sorry.\r\n\r\nThe answer is all possible ordered pairs but (1,1). Assume an ordered pair doesn't satisfy either. Then\r\n$ 2xy < x^2 \\plus{} y^2 < 2$, so $ xy < 1$\r\nSince $ x\\plus{}y\\plus{}xy>3$, we find $ x\\plus{}y >2$\r\nSquaring, $ x^2 \\plus{} 2xy \\plus{} y^2 > 4$. Adding $ 2 > 2xy$, we get\r\n$ x^2 \\plus{} y^2 > 2$\r\nContradiction\r\n(1,1) works for the equality case, which I left out", "Solution_4": "[quote=\"dgreenb801\"]Oh I see, you intend for there to be infinitely many solutions, but you have to describe them. Sorry.[/quote]\r\n\r\nThe solution is $ \\mathbb{R}\\backslash \\{(1,1)\\}$ .", "Solution_5": "[quote=\"alex2008\"]Let be $ A \\equal{} \\{(x,y)\\in \\mathbb{R}^2 \\ |\\ x^2 \\plus{} y^2 > 2\\}$ and $ B \\equal{} \\{(x,y)\\in \\mathbb{R}^2\\ |\\ x \\plus{} y \\plus{} xy < 3\\}$ . Determine $ A\\cup B$ .[/quote]\r\n\r\n[b][u]Proof[/u].[/b] Observe that $ \\overline {A\\cup B} \\equal{} \\overline A\\cap\\overline B$ and $ (x,y)\\in A\\cup B\\Longleftrightarrow (y,x)\\in A\\cup B$ . Therefore,\r\n\r\n$ (x,y)\\in \\overline {A\\cup B}\\Longleftrightarrow \\left\\|\\begin{array}{c} x^2 \\plus{} y^2 \\le 2 \\\\\r\n \\\\\r\n3\\le x \\plus{} y \\plus{} xy\\end{array}\\right\\|^{(*)}\\ \\bigoplus\\ \\ \\Longrightarrow\\ x^2$ $ \\plus{} y^2 \\plus{} 3\\le x \\plus{} y \\plus{} xy \\plus{} 2\\ \\Longleftrightarrow$\r\n\r\n$ x^2 \\plus{} y^2 \\minus{} xy \\minus{} xy$ $ \\plus{} 1\\le 0\\ \\ \\|\\ \\bigodot\\ 2\\ \\Longleftrightarrow\\ (x \\minus{} y)^2$ $ \\plus{} (x \\minus{} 1)^2 \\plus{} (y \\minus{} 1)^2\\le 0\\ \\Longleftrightarrow\\ x \\equal{} y \\equal{} 1$ .\r\n\r\nSince $ (1,1)$ verify both relations from $ (*)$ obtain $ \\overline {A\\cup B} \\equal{} \\{(1,1)\\}$ . In conclusion, $ A\\cup B \\equal{} \\mathbb R^2 \\minus{} \\{(1,1)\\}$ .", "Solution_6": "To find $ (x,y)$ not belong to $ A \\cup B$.\r\n\r\n$ \\begin{cases} x^2\\plus{}y^2 \\le 2 \\\\\r\n(|x|\\plus{}1)(|y|\\plus{}1|) \\ge (x\\plus{}1)(y\\plus{}1) \\ge 4 \\end{cases}$\r\n\r\nNow $ 2 \\ge \\frac{(x^2\\plus{}1)\\plus{}(y^2\\plus{}1)}{2} \\ge \\frac{(|x|\\plus{}1)^2}{4}\\plus{}\\frac{(|y|\\plus{}1)^2}{4} \\ge \\frac{2(|x|\\plus{}1)(|y|\\plus{}1)}{4} \\ge 2$\r\n\r\nEquality holds $ \\Leftrightarrow x\\equal{}y\\equal{}1$\r\n\r\nSo $ A \\cup B\\equal{}\\mathbb{R}^{2}\\backslash\\{(1,1)\\}$" } { "Tag": [ "algebra solved", "algebra" ], "Problem": "Find all reals $x$ such thaat $x=\\sqrt{2+\\sqrt{2-\\sqrt{2+x}}}$.", "Solution_1": "Should it be : $x = \\sqrt{2+\\sqrt{2-\\sqrt{2+x}}}$? ;)", "Solution_2": "In this case have not real solution.", "Solution_3": ".\n\n\n\n[hide]\n\n\n\n through half angle and double angle formulas.\n\n\n\n[/hide]", "Solution_4": "it's ok Elemennop" } { "Tag": [ "geometry", "inequalities", "trigonometry", "geometry proposed" ], "Problem": "For any quadrilateral with the side lengths $a,$ $b,$ $c,$ $d$ and the area $S,$ prove the inequality$S\\leq \\frac{a+c}{2}\\cdot \\frac{b+d}{2}.$", "Solution_1": "Let $A,B,C,D$ be the vertices of the quadrilateral s.t. $a=AB,b=BC,c=CD,d=DA$. We then have $\\frac{a+c}2\\cdot\\frac{b+d}2=\\frac{ab+cd+ad+bc}4$, which is $\\ge \\frac{ab\\sin\\angle B+cd\\sin\\angle D+ad\\sin\\angle A+bc\\sin\\angle C}4=S$." } { "Tag": [ "MATHCOUNTS" ], "Problem": "I am actually kind of embaressed to ask this question, and I asked similar questions before, but how does one gets friends if I do not know too much about stuff not related to academic subjects, like restaurants, nite clubs, concerts, I am almost clueless on these subjects!!! What do I do? Please help.", "Solution_1": "Bah, I wish there was another way, but- you really must be outgoing. Go talk to people and don't wait for other to talk to you. That's a start.\r\nI am impossibly shy, and thus, I've got no friends. :(", "Solution_2": "[quote=\"ragnarok23\"]Bah, I wish there was another way, but- you really must be outgoing. Go talk to people and don't wait for other to talk to you. That's a start.\nI am impossibly shy, and thus, I've got no friends. :([/quote]\r\n\r\ncould u plz let me be ure friend? ;)", "Solution_3": "I have some friends. I got them by just hanging out with them and eventually, they just started talking to me and thought of me as a friend.", "Solution_4": "Cheer up, everyone! I think the average number of friends an Aopser has is less than 1!\r\nIt's okay to be antisocial! :blush: (just kidding...or maybe not?)\r\n\r\nAnyways, just find somebody with similar interests with you and it will work out fine. ;)", "Solution_5": "Yes, but that is easy to say. I have no idea how to become like everyone else. The thing is, a very large portion of the time, I am home, reading something on an academic subject, I have not even been to the tons of palces most people my age go, like I have never been in bars or nightclubs. I have no clue how to act? By the way Alex, u r Russian right?", "Solution_6": "I have about 5 friends :D . lucky me. of course, there's about 300 kids in my grade, so i don't know if that makes me immensely popular. :( i know other people though, and they know me, but they're not really my friend.", "Solution_7": "[quote=\"hello\"] By the way Alex, u r Russian right?[/quote]\r\nYes I am.", "Solution_8": "[quote=\"hello\"]I am actually kind of embaressed to ask this question, and I asked similar questions before, but how does one gets friends if I do not know too much about stuff not related to academic subjects, like restaurants, nite clubs, concerts, I am almost clueless on these subjects!!! What do I do? Please help.[/quote]\r\n\r\n\r\nXDDDDDD\r\n\r\n\r\n\r\n\r\nacademic enough for you =p ?", "Solution_9": "Find people that love maths and everything will be right\r\nOf course that is really difficult...but that way I had the best meeting of my life just a few days before", "Solution_10": "don't worry about having lots of friends. friendship is not about conforming to someone else's interests, like restaurants and nightclubs. there was this moron at my school who used to be very good at math. but apparently, some of his jealous pseudo-friends told him he'd be more popular if he were less competitive. so this year, he started getting D grades in math and never made the school team, simply because he thought being dumber would get him more friends.\r\n\r\nhowever, true friends would not have demanded that he do worse. true friends try to help their friends be better people and would enjoy your conpany for your own personality, not because you pretend to be interested in restaurants and concerts. as long as you have a few good friends that you can talk to; as long as you have a few real friends that will stand by you when you need help, that's all you need. i myself consider no more than five people in my school \"friends\" at this point. interestingly enough, my best friends are my math team at school and my math rivals across my state.\r\n\r\ndon't conform yourself to the demands of others! :P", "Solution_11": "I have lots of them because everyone is a studyholic and lots of us share interests. Our school is a magnet school and there are lots of math/science people.\r\n\r\nedit: and none of them go to nightclubs or anything like that. Don't worry, when you go to college you will find a ton; all the other people will be in the junior college.", "Solution_12": "You are in college/university already aren't you? Hmm. Find people who share your interests. You say you spend a lot of time studying. Ok, there's probably tons of people who are studying the same thing somewhere, find them, start a conversation about what you're studying.", "Solution_13": "If you are in a math club, you are BOUND to make at least one friend. I made one almost as soon as I got onto the mathcounts team!", "Solution_14": "Well, I am in college now. I am very interested in economics, but I have no clue about who to talk to this on a regular basis, there are barely any people who would want to talk about a subject, it seems as though people just want to get grades rather then study (except at AoPS). But the thing is, I am applying for internships, and because I dont have much friends, I dont know a lot of what today's young adults are interested in and have a hard time keeping up in say conversations at work, cause I dont do these things. So I dont know what to do.", "Solution_15": "just play a sport, teamwork", "Solution_16": "Sign up for facebook, and add as many people as you can.", "Solution_17": "[quote=\"h_s_potter2002\"]Sign up for facebook, and add as many people as you can.[/quote]\r\n\r\n :rotfl: You just made me laugh out loud." } { "Tag": [], "Problem": "If 10^(k) = 1/2, find 10^( k + 3)\r\n\r\nI keep getting 1000, but the math book tells me that the answer is 500.\r\n\r\nHow can that be?", "Solution_1": "10^(k+3)=(10^k)*10^3=1/2*1000=500.", "Solution_2": "We are given $ 10^k\\equal{}\\frac{1}{2}$\r\n\r\nTherefore, multiplying both sides by $ 1000$ gives us\r\n\r\n$ 10^k \\cdot 10^3\\equal{}500 \\implies 10^{k\\plus{}3}\\equal{}500$.\r\n\r\nSo, the answer is $ 500$.", "Solution_3": "In situations like this, it really helps if you post your solution, so we can correct your method. :P", "Solution_4": "Sharkman, your answer of 1000 doesn't make sense. If you let say, $ 10^{k\\plus{}3} \\equal{} 1000$, this means that $ k\\equal{}0$. However, given that $ 10^k \\equal{} \\frac{1}{2}$, the value of $ k$ doesn't make sense. So I guess the real lesson in this would be to [i]check your work[/i]. :P" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "[size=150]For an even-sided bicentric polygon, the diagonals are concurrent at the limiting point of the two circles, whereas for an odd-sided polygon, the lines connecting the vertices to the opposite points of tangency are concurrent at the limiting point. [/size]\r\nhow to prove it??", "Solution_1": "[quote=\"lizhi\"][size=150]For an even-sided bicentric polygon, the diagonals are concurrent at the limiting point of the two circles, whereas for an odd-sided polygon, the lines connecting the vertices to the opposite points of tangency are concurrent at the limiting point. [/size]\nhow to prove it??[/quote]\r\nIt's related to Poncelet's theorem.", "Solution_2": "Hi all,\r\n\r\nI do not think that it is true for bicentric $ n\\minus{}$ gon, when $ n$ is odd.\r\nI have tried to picture as exact as possible and it did not look true.\r\nFor example, draw a diagram scaling with respect to $ R\\equal{}1$, $ r\\equal{}0.75$, $ d\\equal{}0.19094$." } { "Tag": [ "calculus", "inequalities", "inequalities proposed" ], "Problem": "Given:\r\n$a, b, c \\ge 0$\r\n$a+b+c \\ge 6$\r\nProve:\r\n$\\frac{a^{3}}{ab+4}+\\frac{b^{3}}{bc+4}+\\frac{c^{3}}{ca+4}\\ge a+b+c$", "Solution_1": "I solved it for $s\\geq\\frac{1+\\sqrt{145}}{2}$ :maybe:", "Solution_2": "I think there is something wrong with the problem.... \r\n\r\nIf $a=b=c$, then we have \r\n$\\frac{3a^{3}}{a^{2}+4}\\ge 3a \\Rightarrow 3a^{3}\\ge 3a^{3}+12a$ which is obviously false for $a \\ge 0$.\r\n\r\nIf $b=c=0$ then we have\r\n$\\frac{a^{3}}{4}\\ge a$ which is true for $a \\ge 2$. \r\n\r\n:( \r\n\r\n[b]silouan[/b] what did you do? Can we fix this somehow?\r\n\r\nGiven:\r\n$a, b, c ,\\ge 0$\r\n$a+b+c \\ge 6$\r\nwill it hold:\r\n$\\frac{a^{3}}{ab+4}+\\frac{b^{3}}{bc+4}+\\frac{c^{3}}{ca+4}\\ge \\frac{a+b+c}{2}$\r\n :?:", "Solution_3": "[quote=\"Xevarion\"]Given:\n$a, b, c ,\\ge 0$\n$a+b+c \\ge 6$\nwill it hold:\n$\\frac{a^{3}}{ab+4}+\\frac{b^{3}}{bc+4}+\\frac{c^{3}}{ca+4}\\ge \\frac{a+b+c}{2}$\n :?:[/quote]\r\nIt is true. My friend gave a very long proof using calculus. I will post it if someone asks for it. Does anyone have an elementary solution? :D :lol:", "Solution_4": "I think there is no need for calculus here! :) \r\nBy cauchy we have that:\r\n$\\sum_{cyc}\\frac{a^{3}}{ab+4}\\geq \\frac{(\\sum_{cyc}a^{2})^{2}}{\\sum{cyc}(a^{2}b+4a)}$\r\nso all we need to show is that :\r\n$2(\\sum_{cyc}a^{2})^{2}\\geq 4(\\sum_{cyc}a)^{2}+(\\sum_{cyc}a)(\\sum_{cyc}a^{2}b)$\r\nbut the two inequalities :\r\n$(\\sum_{cyc}a^{2})^{2}\\geq 4(\\sum_{cyc}a)^{2}$ and \r\n$(\\sum_{cyc}a^{2})^{2}\\geq (\\sum_{cyc}a)(\\sum_{cyc}a^{2}b)$ are clear!", "Solution_5": "Very nice! If I am not wrong your method even solves the general:\r\n\r\nGiven: $a, b, c \\ge 0$, $a+b+c \\ge 3\\sqrt[n]{t}$\r\n\r\nTo prove: \r\n$\\sum_{cyc}\\frac{a^{n+1}}{a^{n-1}b+t}\\ge \\frac{a+b+c}{2}$\r\n\r\n;) :D\r\n\r\nBy the way, you can use \\left and \\right to make larger parentheses, and your equations will look nicer. :)\r\n\r\n(edited exponent)", "Solution_6": "[quote=\"Albanian Eagle\"] but the two inequalities :\n$\\left(\\sum_{cyc}a^{2}\\right)^{2}\\geq 4\\left(\\sum_{cyc}a\\right)^{2}$ and\n$\\left(\\sum_{cyc}a^{2}\\right)^{2}\\geq \\left(\\sum_{cyc}a\\right)\\left(\\sum_{cyc}a^{2}b\\right)$ are clear! [/quote]\n[quote=\"Xevarion\"]Very nice! If I am not wrong your method even solves the general:\n\nGiven: $a, b, c \\ge 0$, $a+b+c \\ge 3\\sqrt[n]{t}$\n\nTo prove: \n$\\sum_{cyc}\\frac{a^{n+1}}{a^{n-1}b+t}\\ge \\frac{a+b+c}{2}$\n[/quote]\r\nNot so! [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635620#635620]See here![/url]\r\nThe second ineq is not clear when $n$ (as used in this thread) is $\\ge 7$. \r\nCan anyone prove the general ineq in this case? (Note that I had an error as I typed it before and now it is fixed.) I think it is still true despite what happened in the other thread.\r\n\r\n[b]edit[/b] Actually I am wrong: if $t=1, n=49, a=0.95, b=0.95, c=1.1$, then the LHS is about $1.40154$ which is less than $\\frac{3}{2}$. So, I revise my question: what is the largest $n$ for which this holds for all $t$? For a fixed $t$, what is the largest $n$?" } { "Tag": [ "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Hello!\r\n\r\nI have a tiny problem while solving ODE of second order. It goes like this:\r\n\r\nFind two linear independent solutions of differential equation\r\n\r\n$ x^2y''\\plus{}3x(1\\plus{}x)y'\\plus{}(1\\minus{}3x)y\\equal{}0$\r\n\r\nIn the vicinity of $ x\\equal{}0$.\r\n\r\nSo I assume the solution of this form: $ y(x)\\equal{}\\sum_{n\\equal{}0}^{\\infty}x^{n\\plus{}\\lambda}$.\r\n\r\nI put the sum in the ODE, shift the two index so that all sums would have $ x^{n\\plus{}\\lambda}$ part, I solve the indicial equation (paring the terms with the lowest exponent for n=0):\r\n\r\n$ a_0[\\lambda(\\lambda\\minus{}1)\\plus{}3\\lambda\\plus{}1]\\equal{}0$, $ a_0\\ne0$\r\n$ \\lambda_{1,2}\\equal{}\\minus{}1$\r\n\r\nThen I'm trying to get recurrence relation, and I get this for the first solution:\r\n\r\n$ a_n\\equal{}\\minus{}a_{n\\minus{}1}\\frac{3(n\\plus{}\\lambda\\minus{}2)}{(n\\plus{}\\lambda\\plus{}1)^2}$\r\n\r\nSo for $ \\lambda\\equal{}\\minus{}1$ for first solution I have:\r\n\r\n$ a_n\\equal{}\\minus{}a_{n\\minus{}1}\\frac{3(n\\minus{}3)}{n^2}$\r\n\r\nSo my first 3 terms are: \r\n$ a_1\\equal{}\\minus{}a_0\\frac{3(\\minus{}2)}{1^2}$\r\n$ a_2\\equal{}a_0\\frac{3^2(\\minus{}2)(\\minus{}1)}{1^22^2}$\r\n$ a_3\\equal{}0$\r\n\r\nSo all terms from $ a_3$ are zero. Only $ a_0$, $ a_1$ and $ a_2$ are not zero.\r\n\r\nMy first solution is then (straight into sum):\r\n$ y_1(x)\\equal{}a_0x^{\\minus{}1}\\plus{}a_1\\plus{}a_2x$\r\n\r\nNow the second solution should be found by deriving the first solution by $ \\lambda$. But What do I make of recurrence relation?\r\n\r\n$ a_1\\equal{}\\minus{}a_0\\frac{3(\\lambda\\minus{}1)}{(\\lambda\\plus{}2)^2}$\r\n$ a_2\\equal{}a_0\\frac{3^2(\\lambda\\minus{}1)\\lambda}{(\\lambda\\plus{}2)^2(\\lambda\\plus{}3)^2}$\r\n$ a_3\\equal{}\\minus{}a_0\\frac{3^3(\\lambda\\minus{}1)\\lambda(\\lambda\\plus{}1)}{(\\lambda\\plus{}2)^2(\\lambda\\plus{}3)^2(\\lambda\\plus{}4)^2}$\r\n$ a_4\\equal{}a_0\\frac{3^4(\\lambda\\minus{}1)\\lambda(\\lambda\\plus{}1)(\\lambda\\plus{}2)}{(\\lambda\\plus{}2)^2(\\lambda\\plus{}3)^2(\\lambda\\plus{}4)^2(\\lambda\\plus{}5)^2}$\r\nand so on...\r\n\r\nThe thing is: what is the general recurrence relation for the second solution? The term $ a_4$ has $ \\lambda\\plus{}2$ in denominator as well as in numerator so one can cancel the other...\r\n\r\nI would say that it goes sth like this\r\n$ a_n\\equal{}(\\minus{}1)^n a_0 \\frac{3^n\\lambda(\\lambda^2\\minus{}1)}{(\\lambda\\plus{}2)\\cdot\\ldots\\cdot(\\lambda\\plus{}n\\plus{}1)}$\r\n\r\nDoes this makes any sense? \r\nAny help would be appreciated!", "Solution_1": "Quick general principle: if the two values of $ \\lambda$ differ by an integer (which includes the possibility that they are the same), then there may be a solution which is not of the forum $ \\sum_{n \\equal{} 0}^{\\infty}a_nx^{n \\plus{} \\lambda}$ but instead has a leading term of the form $ x^{\\lambda}\\ln x.$", "Solution_2": "hello, try to find a solution of the form\r\n$ y(x)\\equal{}\\frac{Ax^2\\plus{}Bx\\plus{}C}{x}$ with real numbers $ A,B,C$.\r\nSonnhard.", "Solution_3": "[quote=\"Kent Merryfield\"]Quick general principle: if the two values of $ \\lambda$ differ by an integer (which includes the possibility that they are the same), then there may be a solution which is not of the forum $ \\sum_{n \\equal{} 0}^{\\infty}a_nx^{n \\plus{} \\lambda}$ but instead has a leading term of the form $ x^{\\lambda}\\ln x.$[/quote]\n\nYeah, the second solution will be of the following form: $ y_2(x)\\equal{}y_1(x)\\ln(x)\\plus{}\\sum_{n\\equal{}0}^{\\infty}b_n(\\lambda_2)x^{n\\plus{}\\lambda_2}$ but I would still like to make the second solution in the closed form, that is the part with the sum. \n\nFor generating second solution I can derive the first one with the respect to $ \\lambda$ and evaluated in $ \\lambda\\equal{}\\lambda_1$, right?\n\n[quote=\"Dr Sonnhard Graubner\"]hello, try to find a solution of the form\n$ y(x)\\equal{}\\frac{Ax^2\\plus{}Bx\\plus{}C}{x}$ with real numbers $ A,B,C$.\nSonnhard.[/quote]\r\n\r\nI have never heard of that one. How should I find $ A,B,C$?" } { "Tag": [ "ratio", "geometry" ], "Problem": "The trisection points of the sides of an equilateral triangle are connected to each other by drawing segments parallel to the sides of the triangle. Many triangles are formed when this is done. What is the ratio, expressed as a common fraction, of the area of one of the smallest triangles formed to the area of the original triangle? Express your answer as a common fraction.", "Solution_1": "[quote=\"GameBot\"]The trisection points of the sides of an equilateral triangle are connected to each other by drawing segments parallel to the sides of the triangle. Many triangles are formed when this is done. What is the ratio, expressed as a common fraction, of the area of one of the smallest triangles formed to the area of the original triangle? Express your answer as a common fraction.[/quote]\r\n\r\nObviously $ \\frac{1}{3}^2\\equal{}\\frac{1}{9}$.", "Solution_2": "To elaborate on BOGTRO's solution, he is basically saying that since the smallest triangle has height and base $1/3$ of the original triangle, it is going to be $(1/3)^2 = 1/9$ as large." } { "Tag": [ "USAMTS", "\\/closed" ], "Problem": "Can you make a weekly/monthly contest? Like 5 questions and we PM you the answers and whoever gets the most right gets recognized? That would be fun.", "Solution_1": "That would be really hard issue.\r\n\r\nFirst, we have 9829 users. Second, what level are you talking about? We have people down in middle school and we have people up in College Playground.\r\n\r\nSee this is impossible in my opinion. I don't know.. I like the idea though if it's possible as long as it's not too hard. :D", "Solution_2": "I think the way it's working now is fine. There are plenty of contests in the getting started/intermediate (and I've seen a good number in middle school) range which is where the majority of the the website's \"customers\" are in terms of ability. I don't believe there is a way to incorporate everyone on the site in a mathematical ability, because the topics that will be covered will most likely either be way too easy for some and way too hard for others. Like I said, the way it is now is great because the contests are centered on a certain level - and if anyone is worried about those who are higher level not having any contests - then I suggest you write a monthly/weekly contest yourself! I'm sure it would increase your abilities as a mathematician, you would get to know the material much better by being able to view the numerous solutions that are sent to you, and you'd be helping all of your peers. It's a win-win-win situation :)", "Solution_3": "[quote=\"math92\"]Can you make a weekly/monthly contest? Like 5 questions and we PM you the answers and whoever gets the most right gets recognized? That would be fun.[/quote]\r\n\r\nWe already run something like that -- the USAMTS.", "Solution_4": "As Mathew said there are already a few \"official\" contests running, the USAMTS, the MathLinks Contest (IMO Level), and numerous other contests held through the forums by the users (Silverfalcon's COP and others). Try finding the one at your level and ... good luck :P", "Solution_5": "I didn't know there were so many...sorry" } { "Tag": [ "Euler" ], "Problem": "Aratati ca pentru orice numar exista un multiplu care are numai cifre de 5 si 0!", "Solution_1": "Demonstram pentru numarul N.\r\n$N=2^a\\cdot 5^b\\cdot n$ unde $(n, 10)=1$ deci si $(9n, 10)=1$, deci din teorema lui Euler avem $10^{\\phi(9n)}\\equiv 1(mod 9n)$ deci exista $k$ astfel ca $9nk=999...9$ unde in scriere apar $\\phi(9n)$ cifre de 9. Deci $n\\cdot 5k=555....5$ unde apar $\\phi(9n)$ cifre de 5.\r\nAtunci $N\\cdot 2^b\\cdot 5^{a+1}\\cdot k$ are doar cifre de 5 si 0.", "Solution_2": "[quote=\"posabogdan\"]Aratati ca pentru orice numar exista un multiplu care are numai cifre de 5 si 0![/quote]\r\n Consideram numerele a1=50, a2=5050,...an=505050...50(de n ori cate50).Daca gasim n astfel incat n divide pe ak, atunci am rezolvat problema.In caz contrar, doua dintre numerele a1,...,an dau acelasi rest la impartirea cu n.Atunci diferenta lor este divizibila cu n, deci diferenta ap-aq, p>q este multiplul cautat.E foarte usoara problema!" } { "Tag": [ "LaTeX" ], "Problem": "don't get this.\r\n\r\nhow many 8 digit numbers can u make out of the digits 0,0,0,0,1,1,1,1 ???\r\n\r\ncan u explain how u sloved it? thx.", "Solution_1": "you mean digits with 0 and 1? (btw this is not quite olympiad question, it's more like an intermediate question, so I'll move it there). \r\nthe answer is $2^7$ (it's exactly the binary numbers from 0 to 127).", "Solution_2": "No, I believe he means with exactly 4 0's and exactly 4 1's, which gives a different answer. One way to approach the problem is to try and figure out how to solve it in smaller cases and see if you notice a pattern. Another way would be to try and solve it in general -- what if you had n 1's and n 0's? A third method would just be to write them all out -- there just aren't that many of them, Valentin's $2^7 = 128$ setting a good upper bound.", "Solution_3": "first digit has to be 1, then arrange the rest 7!/3!/4!", "Solution_4": "[quote=\"JBL\"]No, I believe he means with exactly 4 0's and exactly 4 1's, which gives a different answer.[/quote]Oh ya, that might make more sense now than what I've thought. Still it's an intermediate topic :D", "Solution_5": "[quote=\"Valentin Vornicu\"][quote=\"JBL\"]No, I believe he means with exactly 4 0's and exactly 4 1's, which gives a different answer.[/quote]Oh ya, that might make more sense now than what I've thought. Still it's an intermediate topic :D[/quote]\r\nHey Valentin, you can still downgrade it to Getting Started, its easy combinatorics anyways :D :P :rotfl:", "Solution_6": "[quote=\"3X.lich\"]Hey Valentin, you can still downgrade it to Getting Started, its easy combinatorics anyways :D :P :rotfl:[/quote]The rule \"if you're not sure where to put this put it into the higher forums\" also applies to us the Admins ;) \r\n\r\nOh, and as of now on, please no off-topic in this topic ...", "Solution_7": "[quote=\"Valentin Vornicu\"][quote=\"3X.lich\"]Hey Valentin, you can still downgrade it to Getting Started, its easy combinatorics anyways :D :P :rotfl:[/quote]The rule \"if you're not sure where to put this put it into the higher forums\" also applies to us the Admins ;) \n\nOh, and as of now on, please no off-topic in this topic ...[/quote]\nYes sir! I won't post off-topic\n\nAnd now, regarding isaacchao's solution:\n[quote=\"isaacchao\"]first digit has to be 1, then arrange the rest 7!/3!/4![/quote]\r\nThe first number has to be a 1 like this: 1 _ _ _ _ _ _ _\r\n\r\nYou have 7 spaces available for the 7 digits remaining: 0,0,0,0,1,1,1\r\n\r\nYou have 3 1's and 4 0's\r\n\r\nSo we have 7! ways to arrange the remaining 7 digits. But there are 3 digits of 1 repeating and so we divide 7! by 3!. But there are also 4 digits of 0 repeating and again we divide 7!/3! by 4!.\r\n\r\nIsaacchao's answer is 7!/3!/4! or in $\\LaTeX$ form ---> $\\frac{7!}{3!4!}=35$" } { "Tag": [ "geometry", "3D geometry", "sphere" ], "Problem": "Can we place or assume point charges(+ve or -ve) inside an initially uncharged [b]conducting[/b] sphere to demonstrate the [b]mirror-image[/b] method in case of a +ve point charge placed in front of the sphere, to show the charge distribution on the sphere because of the point charge?\r\n\r\nAnd if we can do so, how is it possible when a conducting body cannot have any charges inside it?", "Solution_1": "You can represent the sphere with a mirror charge.., however its a bit more tricky.., you have to use spherical inversion to satisfy the boundary condition of zero potential (or other constant potential if not grounded).\r\n\r\nI can be more elaborate regarding the method if you are interested.., if yes, then i will show it tommorrow :D", "Solution_2": "YES, please explain the method quite elaborately because I am a novice at applying this [b]mirror-image[/b] method.", "Solution_3": "The image will be a point charge $ \\;\\; \\minus{} {R \\over r}q\\;\\;$ located at a distance $ \\;\\;{R^2 \\over r}\\;\\;$ from the center of the sphere (of radius $ \\;R\\;$) , in the direction to the point charge $ \\;\\;q\\;\\;$ (that induces the charge on the sphere), located at $ \\;\\; r\\;\\;$ from the spheres center.\r\n\r\nNote that the image is inside the sphere, since $ \\;\\;r > R\\;\\;$ and that it's a virtual position (that's wy it's called image...).\r\n\r\nThe \"real\" charge is distributed (non uniformly) in the spheres surface, but the total value is the same of the image.\r\n\r\nIf the sphere is insulated and it's charge (net) is null there must be a second image at it's center with charge $ \\;\\;{R \\over r}q\\;\\;$.\r\n\r\nBtw, rhe method is better known as Image Method (without \"mirror\") and you can find a very clear discussion with solved problems in this subject, in a Brazilian Book: Problemas de F\u00ecsica - Eletricidade- (Eletrost\u00e1tica) by Eduardo Wilner.", "Solution_4": "That means an [b]image[/b] charge, being a virtual charge, can violate the law that a point charge can stay inside a [b]conducting[/b] sphere?\r\n\r\nPlease correct me if I am wrong.", "Solution_5": "Indeed.\r\n\r\nYou can calculate electrical effects (like field, potential, charge densities, etc.) from the images charge, only outside the sphere, whose surface, in this case, is the boundary that satisfies the condition of an equipotential surface." } { "Tag": [ "LaTeX" ], "Problem": "hi,\r\nWhen I make a figure or table, I need change the format of \\caption{...Ii must be with font size smallest.or different font.\r\nFor example:\r\n---Figure---\r\n[size=75]Figure 1.1: this figure ....[/size]\r\n\r\nCan anyone help me?\r\n\r\nThanks.", "Solution_1": "[code]\\makeatletter\n\\renewcommand{\\fnum@figure}{\\tiny{Figure~\\thefigure}}\n\\makeatother[/code]", "Solution_2": "Hi,\r\nI don't know how I must do it.\r\nI make the figure as:\r\n\r\n\\begin{figure}[..]\r\n\\begin{center}\r\n \\includegraphics[width=...]{..}\r\n \\caption{....}\r\n \\textbf{\\label{fig:...}}\r\n\\end{figure}\r\n\r\nHow I've put it:\r\n\r\n[quote=\"stevem\"][code]\\makeatletter\n\\renewcommand{\\fnum@figure}{\\tiny{Figure~\\thefigure}}\n\\makeatother[/code][/quote]\r\n\r\nI know that I can do tiny the description of the figure, but I don't know how do tiny [b]Figure 1.1[/b]\r\n\r\nThanks", "Solution_3": "Just put the code somewhere before \\begin{figure} and it will make Figure 1.1 tiny as well." } { "Tag": [], "Problem": "Anyone here a fan of orchestral and/or chamber music? What are your favorite works from both? Favorite orchestras? I'm finding myself with an increasing interest in orchestral stuff for some reason. I'm mainly a pianist, but play violin as well, though not seriously at all (like a lesson every other week, don't practice very much). I'm part of a loose string quartet made up of friends from school.\r\n\r\nI don't yet have many preferences in orchestral music, but I like the Egmont and 1812 Overtures of Beethoven and Tchaikovsky (and of course Rossini's William Tell Overture :P ). I should probably listen to a full Wagner opera and a Mahler symphony or two at some point. The Berliner Philharmoniker is, but admittedly I haven't listened to enough of a variety of orchestras to make any kind of judgment.\r\n\r\nAs for chamber music, I like the Shostakovich String Quartets and the well-known Mendelssohn Piano Trio (the first one), particularly the second movement. And to re-emphasize Messiaen's PWNAGE, the Quartet for the End of Time (Piano Trio + Clarinet) is cool.", "Solution_1": "I play the violin pretty seriously and I'm in an orchestra right now. I wasn't a big fan of it before, but once I actually played some, I started liking it a lot. Playing in an actual orchestra is probably the best way to get exposed to it. Dvorak is one of my favorites, especially because of his New World Symphony. Tchaikovsky and Prokofiev are also fun to listen to, though I've only listened to a couple works by each. Prokofiev's Cinderella Ballet was a really cool piece. I also like Mozart and Haydn. Overture to \"The School for Scandal\" by Barber is a really cool piece. So is Histoire du Soldat by Stravinsky.\r\n\r\nI don't really have a preference for orchestra. I just get the one that pops up first in iTunes. They all kinda sound the same to me. :wink: \r\n\r\nI'm not into chamber music as much, although I've played some at my school. From what I've heard though, I like Beethoven string quartets.", "Solution_2": "i have a really sensitive ear towards classical music. i really dont like frederich seitz even tho i play a lot of his concertos. i like beethoven a lot more than bach or mozart. mendelsohn's songs are my favorite. i dislike Vivaldi.", "Solution_3": "Chamber wise, I play in a well noted local chamber group and so far I must say I like the Mozart serenades for 2 violins and cello (which the cello part isn't that bad).\r\n\r\nMy favorite orchestrial music must be Marriage of Figaro by Mozart (which I played) or the Beethoven 9th symphony. Personaly I agree with dreambold on Vivaldi, as he is very repetitve and has the worst cello parts in the history of cello parts. However Elgars great Cello Concerto makes him a favorite for me too.", "Solution_4": "Orchestral:\r\nBeethoven - 9th Symphony (Furtwangler '42), 5th Symphony (Kleiber), 6th Symphony (Walter), 3rd Symphony (Klemperer)\r\nMahler - 9th Symphony (Karajan), 5th Symphony (Barshai), 3rd Symphony (Horenstein), 4th Symphony (Szell), 6th Symphony (Szell), Das Lied von der Erde (Szell)\r\nBrahms - Piano Concerto 2 (Richter)\r\nR. Strauss - Vier letzte Lieder (Norman), Tod und Verklarung (Karajan), Don Quixote (Reiner), Metamorphosen (Kempe)\r\nSchubert - Symphony 8 (Kleiber)\r\n\r\nChamber:\r\nBeethoven - String Quartets (all of them - Alban Berg Quartet), Archduke Trio (Casals), Kreutzer Sonata (Busch and Serkin)\r\nSchubert - Death and the Maiden, String Quintet, Trout Quintet, Piano Trios\r\nBrahms - All of it\r\nBach - Cello Suites (Casals), Sonatas and Partitas for Violin (Milstein)\r\nMozart - Clarinet Quintet\r\n\r\nOpera - \r\nWagner - All of it (especially Tristan (Furtwangler), the Ring (Keilberth), and Parsifal(Knappertsbusch))\r\nR Strauss - Salome, Rosenkavalier, Die Frau ohne Schatten \r\nMozart - Marriage of Figaro, Don Giovanni (both by Jacobs)\r\n\r\nThat should be enough to get you started.", "Solution_5": "I really like the Death and the Maiden by Schubert!", "Solution_6": "Mendelssohn, Dvorak, Schubert, Brahms...there's more, but I don't listen to all that many chamber music.", "Solution_7": "Yeah I just had a chamber music workshop and it was pretty cool. I played a Brahms Piano Quartet and a Dvorak Piano Quintet. \r\nChamber music wise, I prefer Mozart and Mendelssohn (though Dvorak is pretty good)", "Solution_8": "[quote=\"zhentil\"]Orchestral:\nBach - Cello Suites (Casals), Sonatas and Partitas for Violin (Milstein)\n[/quote]\r\nDon't forget The Art of Fugue and the Musical Offering!" } { "Tag": [ "articles" ], "Problem": "Has anybody ever noticed that almost all of Rep123's posts in the round table are about Bush/Republicans/Related?", "Solution_1": "haha yea, I got sortof mad about it here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13956&postdays=0&postorder=asc&&start=0 but then again nobody responded to me. I don't dislike all those people I insulted, but then again their posts about politics bother me a little.", "Solution_2": "I just saw something on Fox news about how the movie was partisan and it was neither fair nor biased.\r\n\r\nI'm surpised they could afford to waste so much time, instead of reporting on real news. Oh, wait.", "Solution_3": "Sorry, JS- didn't read that post, or I would've responded to it.\r\nI agree that you have a point. Sure, I am by no means at an age where I feel comfortable with my own political views. The only things I can go on is what I know. Which is very little. Which is why I try not to say too much either way, because I'm neither pro-Bush [i]or[/i] pro-Kerry, because I think they both have faults that could divide a nation, potentially. But I digress. I hate talking politics.", "Solution_4": "[quote=\"Rep123max\"]Has anybody heard about it? I just ordered the DVD and can't wait for it to come. It's a documentary about how right wing Fox News is and how they spin the news to help their agenda. They interview lots of old employees and analyze memos. Or something like that. Don't expect it to come to a theatre near you. Heres a trailer: outfoxed.org and then go to trailer.[/quote]\r\nThis is really suprising and abnormal, because there is no bias to be found anywhere else in the media, or even in the world. I can't believe Fox News has a bias!!!! OMG WHAT WILL WE DO?!?", "Solution_5": "From the Washington Post:\r\n\r\n\"Let's refer to the US marines we see in the foreground as 'sharpshooters,' not snipers, which carries a negative connotation,\" [Fox]'s senior vice president for news, John Moody, told staffers via e-mail.\r\n\r\nThe entire article is here:[url]http://www.washingtonpost.com/wp-dyn/articles/A58618-2004Jul17.html[/url]" } { "Tag": [ "geometry", "circumcircle", "radical axis", "geometry solved" ], "Problem": "Well yesterday i have created a very nice prob. in geo. Here's it:\r\n\r\n[b]Problem[/b]: Let be given the triangle $ABC$ and a given point $M$ in the interior of the triangle $\\Delta ABC$. We need to find the locus of $N$ in the plane of the triangle $ABC$ so that the powers of $N$ to the circlcumme of triangles $\\Delta AMB$, $\\Delta AMC$ and $\\Delta BMC$ are equals each other!\r\n\r\nany idea ? :)", "Solution_1": "The \"locus\" reduces to the radical center of the three circumcircles. As this circumcircles meet at point $M$, this point is its radical center, so it is the \"locus\".\r\n\r\n[[b]Moderator edit:[/b] Probably the author meant a different problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=57796 .]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "If $a,b,c>0$ then:\r\n$ \\sum \\frac{ab+bc+ca+a^2}{b+c+2a} \\leq \\sum a $.", "Solution_1": "I made de substitucion x=b+c+2a, y=a+c+2b, z=a+b+2c, and i also expressed the inicial numerator as: ab+bc+ca+a^2=b(a+c)+a(a+c)=(a+b)(a+c)=(x+z-y)(x+y-z)/4=x^2/4-(y-z)^2/4, expanding everithing in this way the inequality follows easily!", "Solution_2": "Or we can write the inequality as $ \\displaystyle\\sum\\limits_{\\text{cyc}}\\frac{(a+b)(a+c)}{(a+b)+(a+c)} \\le a+b+c$ and with the substitution $x=a+b,y=b+c,z=c+a$ we must prove $ \\displaystyle\\sum\\limits_{\\text{cyc}}\\frac{xy}{x+y} \\le \\frac{x+y+z}{2}$ but $\\frac{xy}{x+y} \\le \\frac{x+y}{4}$." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that for each positive integer n there exist n consecutive positive integers none of which is a prime or a prime power. \r\n\r\n\r\nHi. I solved that problem and I want to know if my solution is right.\r\n\r\nSolution:Firts see that if $n \\equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ so n is not a prime(obviously) and n is not a power of a prime. Prove: Suposse that $n=p^{a}$ with $n=m.p_{1}^{a_{1}}.p_{2}^{a_{2}}$.Thus $p^{a}=m.p_{1}^{a_{1}}.p_{2}^{a_{2}}$, and since $gcd(p_{1}^{a_{1}}, p)=gcd(p_{2}^{a_{2}}, p)=1$ we must have that $m=p^{a}$, but that implies $n>p^{a}$, contradition!\r\n\r\nSo if we prove that there are n consecutive intengers such that each of them satisfies the condition $n \\equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ the problem's over. So take distinct primes $p_{1},p_{2},p_{3},...,p_{2(n-1)}$, and put: $a \\equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ $a \\equiv-1(mod p_{3}^{a_{3}}.p_{4}^{a_{4}})$ ............. $a \\equiv-(n-1)(mod p_{2n-3}^{a_{2n-3}}.p_{2n-2}^{a_{2n-2}})$ Since $mdc(p_{2i-1}^{a_{2i-1}}.p_{2i}^{a_{2i}}, p_{2j-1}^{a_{2j-1}}.p_{2j}^{a_{2j}})=1$ for each $i \\not=j$,we know by the chinese remainder theorem that this system has a solution.So there are intenger a,a+1,...,a+n-1 such that none of them is prime or power of a prime Q.E.D\r\n\r\n\r\nPlease say me if you note something wrong.\r\nThanks!", "Solution_1": "[quote=\"gabriel ponce\"]Solution:Firts see that if $n \\equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ so n is not a prime(obviously) and n is not a power of a prime. Prove: Suposse that $n=p^{a}$ with $n=m.p_{1}^{a_{1}}.p_{2}^{a_{2}}$.Thus $p^{a}=m.p_{1}^{a_{1}}.p_{2}^{a_{2}}$, and since $gcd(p_{1}^{a_{1}}, p)=gcd(p_{2}^{a_{2}}, p)=1$ we must have that $m=p^{a}$, but that implies $n>p^{a}$, contradition!\n[/quote]\r\nYou can't simply say $gcd(p_{1}^{a_{1}}, p)=gcd(p_{2}^{a_{2}}, p)=1$. But in fact you can make the proof even shorter.\r\n\r\nBut the rest seems good to me, and this small gap can be filled.\r\n\r\nPS: don't use [ tex ] and $\\$$ parallely...", "Solution_2": "Sorry but i didn't understand.Why can't I simply say that $gcd(p_{1}^{a_{1}},p)=gcd(p_{2}^{a_{2}},p)=1$ ?", "Solution_3": "You could have $p=p_{1}$ for example.\r\nBut I would base the whole argument the following way:\r\nBy unique prime factorisation, $n=p^{a}$ has only the prime divisor $p$. Thus we would have $p_{1}=p=p_{2}$ since both are prime divisors of $n$, contradiction.", "Solution_4": "Sorry I forgot to write that $p_{1}\\not= p_{2}$, because with that we can't have $p=p_{1}$ or $p=p_{2}$.", "Solution_5": "This problem has been discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=372271#p372271]here[/url]\r\nYou can have a look at all IMO problems [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16]here[/url] :wink:" } { "Tag": [ "function", "algebra", "domain" ], "Problem": "What is the theorem's name that stated...\r\n\r\nEvery function can be written as the sum of an even function and an odd one.\r\n\r\nthanks!", "Solution_1": "I don't think it has a name. It's just a property of real functions. Wikipedia has the following explanation:\r\n\r\n[quote=\"Wikipedia\"] Any linear combination of even functions is even, and the even functions form a vector space over the reals. Similarly, any linear combination of odd functions is odd, and the odd functions also form a vector space over the reals. In fact, the vector space of all real-valued functions is the direct sum of the subspaces of even and odd functions. In other words, every function can be written uniquely as the sum of an even function and an odd function:[/quote]\r\n\r\nDoes it apply to discontinuous functions, though?", "Solution_2": "Is this possibly an extension of the intermediate value theorem? I know that that one states that for a given domain, continuous functions take on all ranged values within that domain. If you can write any function as a composition of even and odd functions, i.e. for\r\n$f(x)=x^{4}$, define functions g(x) and h(x) such that g(x) is odd and h(x) is even, then write the function composition form\r\n$(g \\circ h)(x) = f(x)$, then you can show with the intermediate value theorem that f(x) takes on all values in the domain of g and h. I'm not sure if that made sense.", "Solution_3": "Compositions aren't really all that related to additions (well, linear combinations), which is the subject of this theorem." } { "Tag": [ "number theory open", "number theory" ], "Problem": "Find all integer n such that (2^(n-1)-1)/n^2 is integer.", "Solution_1": "Obviosly $ n\\equal{}1$ is solution.\r\nIf $ n>1$, let $ p$ is prime divisor n. Let $ T_p$ is period 2 mod p. Because $ p\\not |n\\minus{}1$ and $ p^2|2^{n\\minus{}1}\\minus{}1$ we get p is Weieferich prime and $ T_p|n\\minus{}1$. If $ p^2|n$ we get $ p^4|2^{p\\minus{}1}\\minus{}1$.\r\nAll $ n\\equal{}p$ Wieferich prime is solution. Known only 2 Wieferich primes $ p\\equal{}1093,T_p\\equal{}364,p\\equal{}3511,T_p\\equal{}1755$.\r\nBut $ n\\equal{}1093*3511$ is not solution, because $ lcm(364,1755)\\equal{}49140\\not |1093*3511\\minus{}1$" } { "Tag": [], "Problem": "I am going. The comprehensive tests from previous years are very computationally heavy and time consuming. I haven't been able to finish a single test in under an hour. I need to practice mental computations some this week.\r\n\r\nIf you are going post here. See you there.", "Solution_1": "I am going. but spain park is gonna suck cause all our good people are going to the engineering competition", "Solution_2": "we'll be there :)", "Solution_3": "woah florida people. are you division I?", "Solution_4": "yeah, should be [we might not technically be, but we would register as such for competitive purposes]\r\n\r\nboth the top 2 comprehensive teams were from FL last year, and both are coming again (rickards [us] and buchholz [from gainesville]). the good news for us is that we have the same 3 people as last year + a freshman, except we're juniors this year, and we actually know precalc/trig now :P \r\n\r\nour algebra 2 team should be pretty competitive, but not all of our top people are going; buchholz should have very strong geo, alg2, and comprehensive. fairview middle [our feeder] should have strong teams as usual.", "Solution_5": "lovely.. well i'll be there to help out:) cya there", "Solution_6": "... welp... we won...", "Solution_7": "I went! and got lost!", "Solution_8": "Seriously? how in the world did you get lost?...well I did 2, but that's a matter of being short:) haha", "Solution_9": "[b]I[/b] didn't get lost... it was all jimmy...", "Solution_10": "I was going from the balcony of the auditorium to the floor of the auditorium (which should be simple enough, right? actually, there was another door that I think would have been a lot quicker but it was locked) and it involves at least five turns and weird hallways and a flight of stairs.", "Solution_11": "To think I'm going there next year... *shudders*", "Solution_12": "[quote=\"dynamo729\"]I was going from the balcony of the auditorium to the floor of the auditorium (which should be simple enough, right? actually, there was another door that I think would have been a lot quicker but it was locked) and it involves at least five turns and weird hallways and a flight of stairs.[/quote]\r\n\r\ndid this happen when you were late to the awards?", "Solution_13": "Hahaha I didn't get to go up to the balcony (prbably a good thing). Did you hear when the guy who was calling names out told that kid to get off the rail? :rotfl: Oh btw Cman had to leave after Algebra Awards (well actually, they kicked us out) so I didn't get to see Geo Awards. I heard you won though. Congrats! :D", "Solution_14": "[quote]did this happen when you were late to the awards?[/quote]\n\nYup. I ended up just running to find a door that would lead to the outside and then running around the outside of the school until I got to the front door, because from there I knew how to get to the auditorium. I was actually very surprised to get there before all the awards were called :D\n\n[quote]Hahaha I didn't get to go up to the balcony (prbably a good thing). Did you hear when the guy who was calling names out told that kid to get off the rail? Oh btw Cman had to leave after Algebra Awards (well actually, they kicked us out) so I didn't get to see Geo Awards. I heard you won though. Congrats![/quote] \nYeah, that's Zhijian... he's in my Math Seminar class. It's actually a very fitting thing to his personality to do something like that. \nThanks! How did they kick you out...? \n\n[quote]To think I'm going there next year... *shudders*[/quote]\r\nWhy the shudder? They're a very good school, as much as I'd hate to admit.", "Solution_15": "Well my math teacher said all the 7th and 8th graders had to leave b/c the auditorium wouldn't hold us and high schoolers. :maybe: .... I wish I could've stayed though to see how everyone did." } { "Tag": [ "parameterization", "advanced fields", "advanced fields unsolved" ], "Problem": "Consider the curve $ C: x_1^3 \\plus{} x_2^3 \\plus{} x_0x_1^2 \\equal{} 0$ in $ \\mathbb{C}P^2$.\r\n\r\nGive a rational parameterization $ \\phi: \\mathbb{C} \\cup \\{\\infty\\} \\to \\mathbb{C}P^2$ which induces a bijection between $ \\mathbb{C}$ and the simple points of $ C$.", "Solution_1": "Do you mean something like $ t\\mapsto [t^3\\minus{}1,1,\\minus{}t]$ ?\r\nOr do you want a formula which also works for $ t\\equal{}\\infty$ without any change of chart? (In this one, if you are in a neighbourhood of $ \\infty$, you can divide by $ t^3\\minus{}1$ and then the parametrization works with $ t\\equal{}\\infty$, but gives some problem when $ t^3\\equal{}1$)." } { "Tag": [ "geometry", "rectangle", "power of a point" ], "Problem": "i need help with two of these problems. if there are any useful formulas, please give them to me. thanks:\r\n\r\nwhat is the length, in cm, of the longest line segment that can be contained in a right circular cylinder with a base radius of 6cm and a height of 8cm?\r\n\r\nand\r\n\r\nin circle O, chord AB is 10 cm long and chord CD intersects AB at E such that AE= 6 and CE= 3. what is the length of DE, in cm?", "Solution_1": "[quote=\"mr. math\"]i need help with two of these problems. if there are any useful formulas, please give them to me. thanks:\n\nwhat is the length, in cm, of the longest line segment that can be contained in a right circular cylinder with a base radius of 6cm and a height of 8cm?\n\nand\n\nin circle O, chord AB is 10 cm long and chord CD intersects AB at E such that AE= 6 and CE= 3. what is the length of DE, in cm?[/quote]\r\n\r\n[hide=\"Numero Uno\"]Well it's gonna be diagonal so pretend it's in a rectangle of length 8 and width 2r=2*6=12.\nUsing the pythagorean theorm, we get $8^{2}+12^{2}=64+144=208$\n$\\sqrt{208}=\\boxed{4\\sqrt{13}}$[/hide]\n\n[hide=\"NUmero Dos\"]Using the Power of a Point (AoPS volume 1, chapter 17), we have $6\\cdot4=3x$ where x is DE. We get that $x=\\boxed{8}$[/hide]", "Solution_2": "[quote=\"mr. math\"]in circle O, chord AB is 10 cm long and chord CD intersects AB at E such that AE= 6 and CE= 3. what is the length of DE, in cm?[/quote]\r\n[hide=\"Hint\"] Power of a Point Thereom: For all lines drawn from $P$ to intersect a circle at points $A$ and $B$, the value of $PA\\cdot PB$ is the same. \n\nSo, in this case, $AE\\cdot BE=CE\\cdot DE$.[/hide]", "Solution_3": "[hide=\"hint for 1\"]Where would the longest line drawn be?\n\n[hide=\"another hint\"]Try looking at a right triangle with the diameter as the base and the height of the cylinder as its height.[/hide][/hide]\n\n[hide=\"hint for 2\"]Look at lotrgreengrapes7926's hint :) [/hide]" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Given that \r\n$ a^{1}\\plus{}b^{1}\\plus{}c^{1}\\equal{}3$\r\n$ a^{2}\\plus{}b^{2}\\plus{}c^{2}\\equal{}5$\r\n$ a^{3}\\plus{}b^{3}\\plus{}c^{3}\\equal{}7$,\r\n \r\ncompute\r\n$ a^{4}\\plus{}b^{4}\\plus{}c^{4}$.", "Solution_1": "Let's see if the wiki thing works...\r\n\r\n[hide=\"hint\"]Let $ a,b,c$ be roots of $ P(x)$ and then just use [[Newton's Sums]][/hide]", "Solution_2": "You want \r\n\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_sums\r\n\r\nnot \r\n\r\nhttp://www.artofproblemsolving.com/Wiki/index/Newton's%20Sums", "Solution_3": "[quote=\"xxxyyyy\"]Given that \n$ a^{1}\\plus{}b^{1}\\plus{}c^{1}\\equal{} 3$\n$ a^{2}\\plus{}b^{2}\\plus{}c^{2}\\equal{} 5$\n$ a^{3}\\plus{}b^{3}\\plus{}c^{3}\\equal{} 7$,\n \ncompute\n$ a^{4}\\plus{}b^{4}\\plus{}c^{4}$.[/quote]\r\nHere is a long solution.\r\n[hide]\n$ a\\plus{}b\\plus{}c\\equal{}3$\nSquare this expression,\n$ a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}2(ab\\plus{}ac\\plus{}bc) \\equal{} 9$\nBy equation 2 we got,\n$ 5\\plus{}2(ab\\plus{}ac\\plus{}bc)\\equal{}9$\nSo,\n$ ab\\plus{}ac\\plus{}bc\\equal{}2$.\n\nNow,\n$ a^{3}\\plus{}b^{3}\\plus{}c^{3}\\minus{}3abc \\equal{} (a\\plus{}b\\plus{}c)(a^{2}\\plus{}b^{2}\\plus{}c^{2}\\minus{}ab\\minus{}ac\\minus{}bc)$\nSubsitute what you know,\n$ 7\\minus{}3abc \\equal{} 3(5\\minus{}2)$\nSo,\n$ abc \\equal{}\\frac{16}{3}$.\n\nNow $ ab\\plus{}ac\\plus{}bc\\equal{}2$ so square,\n$ a^{2}b^{2}\\plus{}a^{2}c^{2}\\plus{}b^{2}c^{2}\\plus{}2abc(a\\plus{}b\\plus{}c)\\equal{}4$\nSubstitute what you know,\n$ a^{2}b^{2}\\plus{}a^{2}c^{2}\\plus{}b^{2}c^{2}\\plus{}2\\cdot\\frac{16}{3}\\cdot 3 \\equal{} 4$\nSo,\n$ a^{2}b^{2}\\plus{}a^{2}c^{2}\\plus{}b^{2}c^{2}\\equal{}\\minus{}28$\n\nNow $ a^{2}\\plus{}b^{2}\\plus{}c^{2}\\equal{} 5$ so square,\n$ a^{4}\\plus{}b^{4}\\plus{}c^{4}\\plus{}2(a^{2}b^{2}\\plus{}a^{2}c^{2}\\plus{}b^{2}c^{2})\\equal{}25$\nSubstituee what you know,\n$ a^{4}\\plus{}b^{4}\\plus{}c^{4}\\plus{}2(\\minus{}28) \\equal{} 25$\nNow solve for $ a^{4}\\plus{}b^{4}\\plus{}c^{4}\\equal{} 81$.\n[/hide]", "Solution_4": "[quote=\"sylow_theory\"]\nNow,\n$ a^{3}\\plus{}b^{3}\\plus{}c^{3}\\minus{}3abc \\equal{} (a\\plus{}b\\plus{}c)(a^{2}\\plus{}b^{2}\\plus{}c^{2}\\minus{}ab\\minus{}ac\\minus{}bc)$\nSubsitute what you know,\n$ 7\\minus{}3abc \\equal{} 3(5\\minus{}2)$\nSo,\n$ abc \\equal{}\\frac{16}{3}$.\n[/quote]\r\n\r\n :roll: \r\nIt should be $ abc \\equal{}\\frac{\\minus{}2}{3}$\r\nThen do thing the same as you did, we get $ \\sum(a^{4}) \\equal{} 9$ (Really nice, isn't it?)\r\nBtw, this is not really long. :)", "Solution_5": "[quote=\"archimedes1\"]Let's see if the wiki thing works...\n\n[hide=\"hint\"]Let $ a,b,c$ be roots of $ P(x)$ and then just use [[Newton's Sums]][/hide][/quote]\n\nThis, by the way, is not a bad idea.\n\n[hide=\"Solution\"] Let $ S_{n}\\equal{} a^{n}\\plus{}b^{n}\\plus{}c^{n}$. We know that $ S_{0}\\equal{} 3, S_{1}\\equal{} 3, S_{2}\\equal{} 5, S_{3}\\equal{} 7$ and we want to calculate $ S_{4}$. Well, this is a third-order linear recurrence (in other words, we are using Newton's sums), so it takes the form\n\n$ S_{n}\\equal{} A S_{n\\minus{}1}\\plus{}B S_{n\\minus{}2}\\plus{}C S_{n\\minus{}3}$\n\nIn particular, the corresponding characteristic polynomial\n\n$ x^{3}\\equal{} Ax^{2}\\plus{}Bx\\plus{}C$\n\nHas roots $ a, b, c$, so already we know that $ A \\equal{} a\\plus{}b\\plus{}c \\equal{} 3$. We can easily calculate that $ (a\\plus{}b\\plus{}c)^{2}\\equal{} a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}2(ab\\plus{}bc\\plus{}ca) \\equal{} 9$, hence $ ab\\plus{}bc\\plus{}ca \\equal{}\\minus{}B \\equal{} 2$. Finally, we know that\n\n$ S_{3}\\equal{} A S_{2}\\plus{}B S_{1}\\plus{}C S_{0}\\Leftrightarrow$\n$ 7 \\equal{} 3(5)\\minus{}2(3)\\plus{}3C\\Leftrightarrow$\n$ C \\equal{}\\minus{}\\frac{2}{3}$\n\nTo calculate $ S_{4}$ we simply calculate\n\n$ S_{4}\\equal{} A S_{3}\\plus{}B S_{2}\\plus{}C S_{0}\\equal{} 3(7)\\minus{}2(5)\\minus{}\\frac{2}{3}(3) \\equal{}\\boxed{9}$. [/hide]" } { "Tag": [ "USAMTS" ], "Problem": "I was just curious as to what specific qualities are the graders looking for beyond the simple proof.", "Solution_1": "From the [url=http://www.usamts.org/TipsFAQ/U_FAQ.php#commended]USAMTS FAQ list[/url]:\r\n[quote]A commended score means that a grader reading your solution thought that besides meeting the critera for a score of \"five\" (a concise well-written solution), the grader thought that your solution was unusually well-done, perhaps involving a novel argument or an interesting generalization. We don't have any hard and fast rules on when to check the commended box, leaving that decision to each grader. We estimate that less than 10 per cent of the solutions that get a perfect score of five also get commended. So a commendation means that a solution is among the best of the best, and something for which you should be very proud![/quote]" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "okey, we got favorite books and music here, how 'bout movies\r\nnothing beats Monty Python and the Holy Grail", "Solution_1": "the castle of ....AHHHHHH", "Solution_2": "castle anthrax\r\n\r\n\r\n\r\nhow bout office space!\r\n\r\ni found in walmart yesterday a monty python quest of the holy grail special thingie for 50 bucks or something. it looked cool and im tempted to buy it", "Solution_3": "Huh? :) \r\n\r\nI liked the Beautiful Mind. It's a shame they didn't put that famous speech in the movie, but rather in the deleted scenes. (The one that says something along the line of \"It is important to have a beautiful mind, but it is more important to have a beautiful heart.\" It really moved me :cry:)", "Solution_4": "I [b]REALLY[/b] wanted to see that, but my parents thought it was \"above my head\" even when my sociology teacher told me to see it.\r\n\r\nI really like tho LoTR movies...along with HP (no comments please) and The Sound of Music is pretty good too. There are SOOOOOOOOO many good movies out there tho', I just haven't been allowed to see any of them.", "Solution_5": "beautiful mind was an awesome movie\r\n\r\n\r\nbut has anyone seen office space?", "Solution_6": "I didn't like A Beautiful Mind that much. The movie should have been a lot less romantic. Or a lot more romantic, and it would have made a nice date. :wink:\r\n\r\nI just saw the Italian Job. Now that was a funny movie. \"With speakers so loud...\"", "Solution_7": "Mmm, what romance? (a murmur from the audience: \"Typical guy...\") [hide] Subliminally messaging people...[/hide]\n\nSyntax Error wrote:beautiful mind was an awesome movie\n\n\nbut has anyone seen office space?\n\nI have, on Comedy Central. It was pretty funny, but...er, lets just say I classified that with \"Mars Attack!\"", "Solution_8": "wurg. OFFICE SPACE is in a league of its own. way above any other movay", "Solution_9": "Mmm, I guess so, I mean what other movie is called \"mo[b]vay[/b]\" :lol: \r\n\r\nEnter the Matrix is good I believe, 'course it sucks as a game though :lol:", "Solution_10": "hmm.... I don't really have favorite movies... dunno why I don't either... hehehe... well... I watch Star Wars and LOTR and some comedies and action... but I don't have a fav.", "Solution_11": "Indiana Jones forever! On a more serious note, I really liked the Godfather (no. 1 and 2.)", "Solution_12": "Eh, I also liked Road to Perdition...haven't got a chance to see Godfather in full for obvious reason :)", "Solution_13": "[quote=\"Chinaboy\"]hmm.... I don't really have favorite movies... dunno why I don't either... hehehe... well... I watch Star Wars and LOTR and some comedies and action... but I don't have a fav.[/quote]\r\n\r\nStar Wars = da bomb\r\n\r\nJust had to say that. The old ones (Ep. 4-6) are great! I watched them the other week and I was amazed. Again. And again.", "Solution_14": "The Ring was nice.... but scary... hehehe", "Solution_15": "[quote=\"Chinaboy\"]The Ring was nice.... but scary... hehehe[/quote]\r\nWhich did you like better?" } { "Tag": [ "geometry", "circumcircle", "angle bisector", "geometry proposed" ], "Problem": "In a triangle $ABC$ we have $\\angle A = 120$. On the bisector of angle $A$ we take in this order points $D,E$ such that $BC$ is bisector of angle $ABD$ and $DE=DB$.Prove that $AE = AB+AC$\r\n\r\n[u] Babis [/u]", "Solution_1": "Let $F$ be on $AE$ such that $AC = CF$. Then $\\Delta ACF$ is equilateral. Since $BC$ is the angle bisector of $\\angle ABD$ and $\\angle DBE = \\angle DEB$, then $\\angle ABC = \\angle AEC$ and $ABEC$ is cyclic. So $BC = CE$. Then $ABC \\equiv \\Delta FEC$ which leads to the thesis.", "Solution_2": "[quote=\"Sepp\"]Let $F$ be on $AE$ such that $AC = CF$. Then $\\Delta ACF$ is equilateral. Since $BC$ is the angle bisector of $\\angle ABD$ and $\\angle DBE = \\angle DEB$, then $\\angle ABC = \\angle AEC$ and $ABEC$ is cyclic. So $BC = CE$. Then $ABC \\equiv \\Delta FEC$ which leads to the thesis.[/quote]\r\n\r\n Please , tell me where do you use point $F$ and why $ABEC$ is cyclic ? In fact , my first question would be :'' prove that $ABEC$ is cyclic ''.\r\n ANy way ', I'll try to study your proof more carefully.Probably , I can not see something this moment.\r\n \r\n[u]babis[/u]", "Solution_3": "I am not sure about the correctness of the first proof-however i am sure that it is not complete...\r\n\r\nHere is my solution.\r\n\r\nFirst of all the points $D$ and $E$ are unique!I will prove that $E$ is the midpoint of the bigger arc of $BC$ and $D$ is the intersection of $BE$ with the perpendicular line from $O$ (the center) to $BE$.Suppose that $D, E$ are defined in this way.If both conditions of the problem are true in this case then because of the uniqueness - this is the case!\r\n\r\nIt is obvious after this construction that $BD=DE$.\r\nAlso $\\angle{CBD}=60^{o}-\\angle{DBE}=\\angle{AEC}=\\angle{ABC}$ and so we proved that both conditions are true.\r\n\r\nNow its as well known as obvious to prove that $AB+AC=AE$!\r\n[i]I usually prove this by Ptolemeos Theorem. [/i]", "Solution_4": "[quote=\"Nick Rapanos\"]I am not sure about the correctness of the first proof-however i am sure that it is not complete...\n\nHere is my solution.\n\nFirst of all the points $D$ and $E$ are unique!I will prove that $E$ is the midpoint of the bigger arc of $BC$ and $D$ is the intersection of $BE$ with the perpendicular line from $O$ (the center) to $BE$.Suppose that $D, E$ are defined in this way.If both conditions of the problem are true in this case then because of the uniqueness - this is the case!\n\nIt is obvious after this construction that $BD=DE$.\nAlso $\\angle{CBD}=60^{o}-\\angle{DBE}=\\angle{AEC}=\\angle{ABC}$ and so we proved that both conditions are true.\n\nNow its as well known as obvious to prove that $AB+AC=AE$!\n[i]I usually prove this by Ptolemeos Theorem. [/i][/quote]\r\n\r\n Nick !\r\n\r\n Exactly , this was the way the problem was made.And of course the solution is perfect.\r\n Or , the same , you produce $AD$ to meet the circumcircle of $\\triangle ABC$ and then you prove very easy that $DE = DB$.\r\n\r\n[u] babis [/u]" } { "Tag": [], "Problem": "A railroad track is laid along the arc of a circle of radius of 1800 ft. The circular part of the track subtends a central angle of 40 degrees. How long (in seconds) will it take a point on the front of a train traveling 30 mph to go around this portion of the track?\r\n\r\nThere are several formulas given: \r\nomega = theta/time (for angular speed), \r\nv (speed) = arc length/time, \r\nv = (r*theta)/t, and \r\nv = r*omega (all for linear speed).\r\n\r\nI'm not sure how to work this problem, or which formula(s) to use. Please help", "Solution_1": "[quote=\"RHSLilSweetie07\"]A railroad track is laid along the arc of a circle of radius of 1800 ft. The circular part of the track subtends a central angle of 40 degrees. How long (in seconds) will it take a point on the front of a train traveling 30 mph to go around this portion of the track?\n[/quote]\n\nThe track is an arc with length 2(pi)1800 * (40/360) = 400(pi). The amount of time it takes to go around this track at 30 mph is t = d/v (where t = time, v = velocity, and d = distance). With 400 ft = 5/66 mi, t = (5/66*pi)/30 = (pi/396) hours (approximately .0079).[/quote]", "Solution_2": "In seconds, this is 10pi/66 = (600pi/66) seconds (approximately 28.56)." } { "Tag": [ "function", "calculus", "derivative", "logarithms", "absolute value", "topology", "complex analysis" ], "Problem": "Does there exist an analytic function f =u+iv in D={z : |z|<1} such that |f(z)|=x for all z=x+iy $ \\in D$ ?", "Solution_1": "No, because the absolute value of a complex number can't be negative.\r\n\r\nMore interestingly, can there be an analytic function $ f$ on some open set $ U$ such that $ |f(x\\plus{}iy)|\\equal{}x$ for $ x\\plus{}iy\\in U$?\r\nGiven some analytic $ f$, let's look at the behavior of $ g(x,y)\\equal{}|f(x\\plus{}iy)|^2$. The Laplacian $ \\Delta$ can be written as $ 4\\frac{\\partial^2}{\\partial z\\partial \\overline{z}}$ in complex form (see posts #2 and #6 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88930]here[/url] for the definition of these partial derivative operators), so $ \\Delta g \\equal{}4\\frac{\\partial^2}{\\partial z\\partial \\overline{z}}\\left(f\\cdot \\overline{f}\\right)\\equal{}4f'\\cdot \\overline{f'}\\equal{}4|f'(x\\plus{}iy)|^2$\r\nIf we have $ |f(x\\plus{}iy)|\\equal{}x$, $ g(x,y)\\equal{}x^2$ and $ \\Delta g(x\\plus{}iy)\\equal{}2$. That makes $ |f'(x\\plus{}iy)|\\equal{}\\frac1{\\sqrt{2}}$, a constant. Since $ f'$ is analytic and its absolute value is a constant, $ f'$ is constant and $ f$ is (locally) a linear function. Looking at the possibilities, $ |(a\\plus{}bi)(x\\plus{}iy)\\plus{}(c\\plus{}di)|$ always has circular level sets, and can't be $ x$. Therefore, there is no such $ f$.", "Solution_2": "Jmerry,\r\n\r\nThanks for your solution", "Solution_3": "A different proof that also uses the Laplacian: If $ |f(x\\plus{}iy)| \\equal{} x$ in $ U$, then $ f \\neq 0$ in $ U$, which implies $ \\ln|f|$ is harmonic in $ U$. But $ \\ln|f(x\\plus{}iy)| \\equal{} \\ln x$, and the Laplacian of the latter is $ \\minus{}1/x^2$, contradiction." } { "Tag": [], "Problem": "Find how many solutions has the following system:\r\n\r\n$x^{2004}=x+y^{2004}$\r\n$y^{2004}=y+z^{2004}$\r\n$z^{2004}=z+x^{2004}$", "Solution_1": "[hide=\"Hint\"]Adding all the equations together and canceling out all the terms with an exponent of $2004$, we get:\n\n$x+y+z=0$\n\nIs this helpful?[/hide]", "Solution_2": "I tried to work with that too but got nothing" } { "Tag": [], "Problem": "which is more soluble in water\r\nAgCl or AgBr and why'", "Solution_1": "That would depend on their solubility products\r\n\r\nI would say AgCl because AgCl is much more ionic than AgBr and like dissolves like" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Let $ ABC$ a triangle and $ P$ a point in the same plane, $ A_{1}B_{1}C_{1}$ is the simetric of the triangle $ ABC$ respect to the point $ P$, the point $ A$ belongs to the line $ L_{a}$, and similarly $ L_{b}, L_{c}$, suppose that $ L_{a}, L_{b}$ and $ L_{c}$ concur, define $ A_{2}$ as the intersection of the lines $ L_{a}$ and $ B_{1}C_{1}$, and similarly $ B_{2}, C_{2}$, prove that $ A_{2}, B_{2}$ and $ C_{2}$ are collinear.", "Solution_1": "As Mihai Miculita noticed in [url]http://www.mathlinks.ro/viewtopic.php?p=917325#917325[/url], the problem is [b]wrong[/b].\r\n\r\nSee the attached figure.", "Solution_2": "Yes, sorry; i made a mistake. But i think the problem is correct if the lines $ L_{a}$, $ L_{b}$ and $ L_{c}$ are parallel (I made the graphic with geogebra)." } { "Tag": [], "Problem": "The ordered list of numbers 18, 21, 24, $ a$, 36, 37, $ b$ has median 30 and mean 32. Find the positive difference between $ a$ and $ b$.", "Solution_1": "The median is $ a$ meaning the equation is:\r\n\r\n$ \\frac {18 \\plus{} 21 \\plus{} 24 \\plus{} 30 \\plus{} 36 \\plus{} 37 \\plus{} b}{7} \\equal{} 32$\r\n\r\n$ 166 \\plus{} b \\equal{} 224$\r\n\r\n$ b \\equal{} 58$\r\n\r\n$ b\\minus{}a \\equal{} \\boxed{\\boxed{\\boxed{\\boxed{\\boxed{28}}}}}$" } { "Tag": [ "geometry", "circumcircle", "analytic geometry", "search", "geometry unsolved" ], "Problem": "Given two circles $ S_1,S_2$ with different radii, externally tangent at $ F$. A common external tangent touches $ S_1$ at $ A$ and $ S_2$ at $ B$. Another line is tangent to $ S_2$ at $ C$, parallel to the other line, and intersects $ S_1$ at $ D,E$. The circumcircles of triangles $ ABC$ and $ BDE$ intersect at $ M$.\r\n\r\nProve that $ A,F,C$ are collinear, and so are $ B,F,M$. :)", "Solution_1": "Please check the wording of the problem. It is not clear than A,F,C be collinear. Thanks", "Solution_2": "[color=darkblue]I think he meant that tangent at $ C$ is parallel to common tangent. If so the first part of the problem was given at Russian MO 1994. I'll post my solution.[/color]", "Solution_3": "[color=darkblue]It is clear that points $ O_1, F, O_2$ are collinear. We have $ O_2C \\perp CD, O_1A \\perp AB \\Rightarrow O_2C \\parallel O_1A \\Rightarrow \\angle{AO_1F}\\equal{}\\angle{FO_2C}$. Because triangles $ AO_1F$ and $ FO_2C$ are isosceles, we have $ \\angle{O_1FA}\\equal{}\\angle{CFO_2}$. So points $ A,F,C$ are collinear.[/color]", "Solution_4": "Yes, the first part of the problem cames from Russia MO 1994, Grade 9, problem 2. The second part cames also from Russia 1994, and the second tangent to S_2 is parallel to AB. Probably it be good the posters of the problems must explain where the problems are from....by \"rights of autorship\", at least.", "Solution_5": "Sorry for the mistype, and I've included the source :) \r\n\r\nNow what about the second part?", "Solution_6": "[quote=\"discredit\"]Sorry for the mistype, and I've included the source :) \n\nNow what about the second part?[/quote]\r\nSorry for off-topic,but could you please create a topic in \"National Olympiad\" and post there links to all problems you've posted from ARO 2008...(and it will be better if you will write All-Russian olympiad instead of ARO)", "Solution_7": "I have located the problem in Kvant. I will translate the synthetic solution soon. For another hand, althought can sounds cumbersome, it is possible solve the problem by analytic geometry. In this case probably the best choice is to take A (0,0), and the tangent common to S_1 and S_2 as axis of abscysses. Then, calling r_1 and r_2 the radiuses of S_1 and S_2, respectively, the coordinates of B are (2(r_1r_2)^1/2, 0) and the rest of equations simplify neatly. Of course is need to suppose r_1 greater than r_2 in order the problem have sense. As a hint in the search of the synthetic solution, happend that point A is the circumcenter of the triangle BDE.", "Solution_8": "Supposed solved the part a), that is, A,F,C are collinear, it happends that F is the foot of the altitude from B in the right triangle ABC. In order to prove that F belongs to the common chord of the circles (ABC) and (BDE), we proceed of the following manner:\r\ni) A is the circumcenter of BDE: AO_1 intersects DE at P, say. Then AP = BC = r_2, and by the Pythagoras theorem at APD and O_1PD (O_1 is obviously the center of S_1), we get\r\nPD, and then AD = AB.\r\nii) It suffices to show that point F has the same power with respect to both circumcircles.\r\nAnd this is true; the power of F with respect both circumcircles equals\r\n-4r_1(r_2)^2/(r_1+r_2).", "Solution_9": "The authors of this nice problem are A.Kalinin and V.Dubrovskii. It was published in the journal Kvant with the number M1452" } { "Tag": [ "search", "analytic geometry", "vector", "probability", "combinatorics unsolved", "combinatorics" ], "Problem": "Real numbers are written on the squares of an infinite grid. Two figures consisting of finitely many squares are given. They may be translated anywhere on the grid as long as their squares coincide with those of the grid. It is known that wherever the first figure is translated, the sum of numbers it covers is positive. Prove that the second figure can be translated so that the sum of the numbers it covers is also positive.", "Solution_1": "Actually the original problem was given at All-Russian Olympiad 1994,as the fourth problem,i.e as one of the hardest problem.\r\nSo you could search through the site...", "Solution_2": "[quote=\"Erken\"]Actually the original problem was given at All-Russian Olympiad 1994,as the fourth problem,i.e as one of the hardest problem.\nSo you could search through the site...[/quote]\r\n\r\nYou mean this site? Or where?", "Solution_3": "[quote=\"discredit\"][quote=\"Erken\"]Actually the original problem was given at All-Russian Olympiad 1994,as the fourth problem,i.e as one of the hardest problem.\nSo you could search through the site...[/quote]\n\nYou mean this site? Or where?[/quote]\r\nI was talking about this site...or maybe Google can help you,but if you know russian, then I can scan the page and send it to you by pm. :lol:", "Solution_4": "[quote=\"Erken\"]\nI was talking about this site...or maybe Google can help you,but if you know russian, then I can scan the page and send it to you by pm. :lol:[/quote]\r\n\r\nThe problem is I don't know Russian...\r\n\r\nMaybe you can post at least the outline of the solution? Or the full solution would be good.", "Solution_5": "[quote=\"discredit\"][quote=\"Erken\"]\nI was talking about this site...or maybe Google can help you,but if you know russian, then I can scan the page and send it to you by pm. :lol:[/quote]\n\nThe problem is I don't know Russian...\n\nMaybe you can post at least the outline of the solution? Or the full solution would be good.[/quote]\r\nWell I'll do it later...if not today,then tomorrow for sure. :wink: Or maybe someone else will post a solution", "Solution_6": "It does not seem to me that hard, although perhaps a little awkward to explain. If I didn't miss something, you should be able to assign coordinates to the grid squares (taking some arbitrary square to be (0, 0)), and then consider the sets A and B of coordinates covered by the two figures (placed arbitrarily, as long as they are oriented correctly). Then look at the multiset $ \\{ a \\plus{} b | a \\in A, b \\in B \\}$, where we use vector addition to add $ a$ and $ b$.", "Solution_7": "To [b]discredit[/b]:\r\nI guess now there is no reason for me to post the outline of a solution,because [b]probability 1.01[/b] already done it... :wink:" } { "Tag": [ "LaTeX" ], "Problem": "ok if u don't want a proof by well ordering try it by induction..though both are equivalent\r\nfix $ n$ we will prove that for every $ m$ and any $ n$ there exist $ q$ and $ 0\\leq r < n$ such that $ m \\equal{} nq \\plus{} r$ \r\nfor $ m \\equal{} 1$ this is trivial set $ q \\equal{} 0$ and $ r \\equal{} 1$ \r\nor else if $ n \\equal{} 1$ set $ r \\equal{} 0$ and $ q \\equal{} 1$\r\nnow let this be true for $ m \\equal{} k$ for some natural number $ k$ then\r\nthere exists $ q'$ and $ r'$ such that\r\n$ k \\equal{} nq' \\plus{} r'$ $ 0\\leq r' < n$\r\nnow if $ 0 \\leq r' < n \\minus{} 1$\r\n$ k \\plus{} 1 \\equal{} nq' \\plus{} r' \\plus{} 1 \\equal{} nq \\plus{} r$ where $ q' \\equal{} q$ and $ r' \\plus{} 1 \\equal{} r < n$\r\nif $ r' \\equal{} n \\minus{} 1$\r\nthen $ k \\plus{} 1 \\equal{} n(q' \\plus{} 1) \\equal{} nq \\plus{} 0$\r\nso we are done", "Solution_1": "wth da fool. i fully typed a proof by WOP and then submitted and found it was locked :mad: . Stupid moderator :mad:", "Solution_2": "Nice... :D Can you pm me the full proof rohit? Someone wanted it.. Thats why I posted this question in the first place, sorry about its trivial nature oh master of mathematics!", "Solution_3": "sorry da...when i read the thread there was nothing there u can post it here if u want...", "Solution_4": "dei dumbo ******* i typed full of connotations and things and u say coolly? :mad: :mad: \r\n@shruthi : tell me who wanted it and i shall chose my discretion to tell the place where the proof is :)", "Solution_5": "dei avasarekudeke ....u shud not have posted this\r\n[quote=\"madness\"]Wth go mod n? :huh: \nA remainder can only be less than or equal to its divisor. You want me to prove that :huh: ? . Dont infuriate with such problems :ninja: \n@pardesi: WOP is well ordering principle \nplz look at ur pm[/quote]\r\nthis implied u were disinterested and infuriated :D", "Solution_6": "Well, a person called Arjun Ramesh, whom you would probably not know..", "Solution_7": "posts can be misleading :D And i was infuriated bcos of sth else due to the author which made me show my anger here da :|\r\n\r\nWho is Arjun Ramesh? Where does he study? I need to know these things to tell the source And probably i know his source too :D", "Solution_8": "Aiyo.... He used to study in CV.. Now he is working ok? Internship.. And he was bugging me with random problems, and I got reallllly bugged.. So I just posted the first problem I couldnt take here.. Now can you please?", "Solution_9": "I have a book whose first page is this WOP :D . Std proof. I dont have the patience to type it all again :) . So ask pardesi to type as it involves a lot of tedious symbols :P . i have to go now. have to study for chem pracs", "Solution_10": "dei rohit ore problem ke solution kudeke yen da evelave alatikare :huh:", "Solution_11": "dei naye i have board chem pracs da and as i said the proof involves much of tedious latex da so i cant have to go neways. :D \r\n@shreyas if u see this: Have asked pardesi wat u told me to ask. :D", "Solution_12": "seri let me do it it isn't long\r\nConsider the set $ S\\equal{}[m\\minus{}nq>0: q\\in \\mathbb{N}]$\r\nthat is the set of numbers of the form $ m\\minus{}nq$ is positive and where $ q$ ranges over integers..\r\nby WOP there exists a $ q'$ for which $ m\\minus{}nq'$ is minimum say that value is $ r$ now we prove $ m\\minus{}nq'\\equal{}r n$ then $ m\\minus{}q'r>m\\minus{}(q'\\plus{}1)n\\equal{}r\\minus{}n\\equal{}r'>0$ so a contradiction since we assumed that\r\n$ r$ is the minimum [b]Positive[/b] value of $ m\\minus{}nq$ .\r\nSo done :)", "Solution_13": "Lol.. Thanks anyway.. All the best for chem Rohit...", "Solution_14": "how was ur chem pracs??", "Solution_15": "I didnt have chem pracs today.. had bio.. Went great!", "Solution_16": "chem pracs over and i dont wanna speak much abt it as i am already up in flames :mad: :mad: :mad:", "Solution_17": "The proof goes as follows:\r\nConsider $ S = [b - ak|b - ak \\geq 0,k \\in \\mathbb{Z}]$\r\n$ \\implies b + |ab| \\in S$.\r\n$ \\implies$ $ |S| \\neq 0$. \r\nBy WOP, $ S$ has a least element say $ b - aq = r$ .\r\nIf $ r \\geq |a|$ then $ 0 \\leq r - |a| < r \\in S$ \r\na contradiction.\r\n$ \\implies 0\\leq r < |a|$ .\r\n\r\n[color=blue]MOD EDIT:I $ \\text{\\LaTeX}$ ified it [/color]", "Solution_18": "Thanks Rohit :D And why up in flames about the practs da?", "Solution_19": "viva ques went to my neighbours while i was simply sitting and nodding my head whether they where right as if i was the external examiner :D :huh: Actually i should be paid the fees for the day :P", "Solution_20": "[quote=\"madness\"]viva ques went to my neighbours while i was simply sitting and nodding my head whether they where right as if i was the external examiner :D :huh: Actually i should be paid the fees for the day :P[/quote]\r\n\r\nI know. Same here. I answered most that I got but the fact remains that I got very less. I am very worried. WTH Gandu got the sappest of the quest and he faltered. Wrong NV.", "Solution_21": "same here da!!! :D And i got only 2 ques da :mad:", "Solution_22": "@madness\r\neverything is gud enuf without $ \\text{\\LaTeX}$ da so no need :)", "Solution_23": "i want that belongs to in latex da plz :D :D then only it will look like a professional proof :D", "Solution_24": "$ \\text{\\LaTeX}$ done :)" } { "Tag": [ "pigeonhole principle", "modular arithmetic" ], "Problem": "Prove that every odd integer $ n\\neq1,-1$ Has infinitly many multiples cosisting, in base 10, only of the digits 1and 0, with not 2 consecutive digits 1. ($ n$ not a multiple of 5)\r\n\r\n\r\n :)", "Solution_1": "[hide=\"a generalization\"] For any positive integers $ k$ and $ n$ such that $ (n, 10)=1$, we have that there are infinitely many multiples of $ n$ which are just several versions of $ k$ in a row (for example, if $ k$ were $ 10$, of the form $ 10101010...010$.)\n[/hide]\n\n[hide=\"solution\"] Suppose $ k$ has $ d$ digits. Let $ k_{i}$ just be $ i$ $ k$'s in a row (ie, $ k+10^{d}k+10^{2d}k+...+10^{d(i-1)}k=k\\sum_{m=0}^{i=1}10^{md}$). Then consider the residues modulo $ n$ of\n\\[ k, k_{2}, k_{3}, ... \\]\nClearly, two of these must be congruent modulo $ n$. Then $ n$ divides their difference. We then have something like\n\\[ n|10^{dj}k_{i}\\]\nBut $ (n, 10)=1$, so $ n|k_{i}$. It then follows that $ n$ divides two of $ k_{i}$ in a row, three in a row, etc...[/hide]", "Solution_2": "I don't think that you are done...\r\n\r\nYou showed that there exists $ i$ and $ j$ such that $ n|(k_{i}-k_{j})$...but then you deduce that $ n|k$. \r\n\r\nLike if you think about the way that you proved this, you essentially showed that for any $ k$, $ n|k$.\r\n\r\nLooking at your proof, this would be the different step. If wlog $ i>j$, then $ n|k*10^{j}*\\frac{1}{9}*(10^{i-j}-1)$.\r\n\r\nBasically you proved that there exists $ q$ such that if $ (n,10)=1$, then $ n|(10^{q}-1)$.\r\n\r\nThe core of the problem is still not solved...Of course once you find one $ k$, you can trivially construct more.", "Solution_3": "[quote=\"Altheman\"]I don't think that you are done...\n\nYou showed that there exists $ i$ and $ j$ such that $ n|(k_{i}-k_{j})$...but then you deduce that $ n|k$. \n[/quote]\r\n\r\nThis wasn't the argument. Here is what the logical progression was...\r\n\r\n$ k_{i+j}\\equiv k_{j}\\mod{n}$\r\n$ n|(k_{i+j}-k_{j})$, $ k_{i+j}-k_{j}=10^{dj}k_{i}$\r\n$ \\implies n|10^{dj}k_{i}\\implies n|k_{i}$\r\n\r\nI did not say $ n|k$: rather, I said $ n$ divides [b]some[/b] $ k_{i}$.", "Solution_4": "Haha okay I missed a huge step. I didn't see how you collapsed those terms. \r\n\r\nEssentially you showed that given a positive integer, $ d>1$, there exists $ q$ such that $ n|\\frac{10^{qd}-1}{10^{d}-1}$ where $ (n,10)=1$.\r\n\r\nThis uses like a pidgeonhole argument on partities mod $ n$ and then subtracting, etc.\r\n\r\nThat lemma essentially finishes the problem...\r\n\r\n\r\nI was thinking of doing like $ 1000...0001$ where you extend the number of $ 0$'s, but I didn't think of extending the number of blocks of $ 100000...00$.\r\n\r\nGood solution.", "Solution_5": ":huh: \r\n\r\n\r\nI understand all the prove, but I can't get how does it apply to the problem.\r\n\r\n\r\n???", "Solution_6": "[quote=\"ElChapin\"]:huh: \n\n\nI understand all the prove, but I can't get how does it apply to the problem.\n\n\n???[/quote]\r\n\r\nSo okay...take $ k$ to be any number formed by only 1's and 0's which doesn't have two consecutive 1's in a row and doesn't both begin and end in a 1. $ k=10$, would probably be the simplest example.", "Solution_7": "mmm... I see...\r\nAs you don't even mentioned it, I supose it is trivial to find such $ k$ for all $ n$.\r\nWell nice solution then.\r\n__________________________________________________________________\r\n\r\nMine goes like this:\r\n\r\n\r\n[b]Lemma 1:[/b]\r\n\r\nAll integers $ n$ relativly prime with $ 10$, have a multiple with only ones (a repunit).\r\n[find $ x$ such that $ 9n|10^{x}\\minus{}1$]\r\n\r\n[b]Lemma 2:[/b]\r\n\r\n$ 1\\plus{}x\\plus{}x^{2}\\plus{}...\\plus{}x^{n\\minus{}1}\\equal{}p|q\\equal{}x^{a_{1}}\\plus{}x^{a_{2}}\\plus{}...\\plus{}x^{a_{n}}$ if \r\n$ \\{1, 2,\\cdots,n\\}\\equiv\\{a_{1}, a_{2},\\cdots, a_{n}\\}\\pmod{n}$\r\n[$ n$ roots of the unit]\r\n\r\n\r\n :) \r\n\r\n[hide=\"Solution\"]\nLet $ n|p\\equal{}1\\plus{}10\\plus{}10^{2}\\plus{}...\\plus{}10^{k\\minus{}1}$ with $ k>1$ an integer, and $ q$ as in lemma 2.\nThe problem is reduced then to prove there are infinitely many sets A such that\n$ A\\equal{}\\{a_{1}, a_{2},\\cdots, a_{n}\\}\\equiv\\{1, 2,\\cdots,n\\}\\pmod{n}$ in order to have $ q$ with no two consecutive 1's, wich is plain.\nFor example:\n$ a_{i}\\equal{}b_{i}n\\plus{}i$ for $ i\\equal{}1,2,\\cdots,n$ and $ a_{i\\plus{}1}>a_{i}\\plus{}1$ for $ i\\equal{}1,2,\\cdots,n\\minus{}1$\n\n :) \n[/hide]", "Solution_8": "I guess here is how I would write up a proof.\r\n\r\nLet $ S$ be the set of numbers with the desired property. Define $ f(x)\\equal{}\\frac{100^{x}\\minus{}1}{99}$. Consider the $ n\\plus{}1$ integers: $ f(1)$, $ f(2)$,..., $ f(n\\plus{}1)$ modulo $ n$. Clearly two must be equivalent modulo $ n$ by the pidgeonhole principle. Suppose that $ f(i)\\equiv f(j)\\mod n$ where $ i>j$. Then $ n|[f(i)\\minus{}f(j)]\\equal{}10^{j}*f(i\\minus{}j)$. Since $ (n,10)\\equal{}1$, $ n|f(i\\minus{}j)$. Hence $ S$ is nonempty. \r\n\r\nSuppose for sake of contradiction that $ S$ had a finite number of elements. Since $ S$ is a set of positive integers with a finite number of elements, it has a maximal element. Call it $ x$. If $ x$ has $ d$ digits, then consider $ x*10^{d\\plus{}1}\\plus{}1$. This number is in $ S$ and larger than $ x$, which is a contradiction. Hence $ S$ has an infinite number of elements." } { "Tag": [ "conics", "algorithm", "parameterization", "algebra", "polynomial", "number theory", "Diophantine equation" ], "Problem": "I don't understand how to do them very well. Linear ones are easy, but once they get more advanced I get lost. Could somebody post guidelines for solving them and/or some practice problems?\r\n\r\nThanks in advance.", "Solution_1": "One thing that bites a lot of people is the fact that $ x^2$ represents both a positive and negative solution. In some equations, like with square roots, one must check for extraneous solutions. Also notice that in conics, the constant is also the square of a certain value, not the value itself. That caused me a lot of trouble, and made my Conics test grade ([b][color=red]C-[/color][/b]) by far the worst of my math grades.", "Solution_2": "I don't see the relevance of the post above :huh: What do conics have to do with Diophantine equations?\r\n\r\nLinear Diophantine equations, as you already know, have a standard algorithm for obtaining the solutions. Thanks to Yuri Matiyasevich, we know today that such algorithm doesn't exist for all Diophantine equations (there is a standardized way of solving second-order D. e., and the algorithm for third order is yet to be found). \r\n\r\nWhat should you use for olympiad D.es? Well, the toolbox could include:\r\n\r\n-modular arguments (very helpful when it comes to squares and their residues), which can lead to use of CRT etc.\r\n-factoring and cases-solving\r\n-elliptic curves\r\n-constructing solution families (when asked to prove that an equation has infinitely many solutions)\r\n\r\nhttp://www.mathlinks.ro/index.php?f=466\r\n\r\nPEN Diophantine equations might be too hard, but you could see the application of common techniques, and then return to easier problems on Intermediate and Pre-Olympiad and try to apply the techniques there.\r\n\r\n/just my $ \\frac{1}{50}\\$$/", "Solution_3": "[quote=\"hsiljak\"]I don't see the relevance of the post above :huh: What do conics have to do with Diophantine equations?[/quote]\r\nWell, conics are the next simplest form of Diophantine equation to look at after linear ones. There are a few overlapping perspectives that allow you to solve several classes of them, including the notion of a [url=http://en.wikipedia.org/wiki/Pell%27s_equation]Pell's equation[/url] and some standard factorization arguments (over the integers) allowing you to compute Pythagorean triplets and the solutions to other related equations, as well as more elaborate factorization arguments over other rings. In general, it is known that every conic in two variables has a rational parameterization, hence the solutions to every homogeneous conic in three variables over the integers has a polynomial parameterization.", "Solution_4": "Thank you, Qiaochu: conics were always $ \\mathbb{R}\\to \\mathbb{R}$ in my (algebraic) perspective, since I always related them to my Algebra courses (which never implied searching for its integer solutions). In other words, conics in this context are what I called 'second-order Diophantine equations', right?" } { "Tag": [ "geometry", "incenter", "circumcircle", "projective geometry", "geometry unsolved" ], "Problem": "$ I$ is the incenter of $ ABC$ triangle and $ ABC$ is not isosceles. $ A_1 , B_1 , C_1$ are the points of intersection of incircle to $ BC , AC , AB$ respectively.prove that circumcenters of $ AIA_1 , BIB_1 , CIC_1$ are collinear", "Solution_1": "$ \\mbox{Let } O_a, O_b, O_c \\mbox{circumcenters of }\\triangle{IAA_1}, \\triangle{IBB_1} \\mbox{ and } \\triangle{ICC_1}. \\\\\r\n\\mbox{Let }\r\nA_2\\in BC, AA_1\\perp AA_2; B_2\\in AC, BB_1\\perp BB_2\\\\\r\n \\mbox{ and } C_2\\in AB, CC_1\\perp CC_1\\perp CC_2. \\\\\r\n\\mbox{The points } A_2, B_2, C_2 \\mbox{ are on the trilinear polar line of the point I and the points } \\\\\r\nO_a, O_b \\mbox{ and } O_c \\mbox{ is midpoints of } [IA_2], [IB_2], [IC_2] \\mbox{ respectively}\\Rightarrow \\\\\r\n\\Rightarrow \\mbox{ the points } O_a, O_b, O_c \\mbox{ are colinear.}$", "Solution_2": "your solution is nice\r\nbut is there any other solution?", "Solution_3": "See here: \r\nhttp://www.mathlinks.ro/viewtopic.php?t=283185" } { "Tag": [ "logarithms" ], "Problem": "if log(2)=.3010 and log(3)=.4771 the value of x when 3^(x+3)=135 is approximately \r\nA)5\r\nB)1.47\r\nC)1.67\r\nD)1.78\r\nE)1.63\r\n\r\nno clue how to solve it since i constantly get a log(5) is there any relationship between the log2 the log3 and the log5? I guess this isn't an intermediate problem but please stoop.", "Solution_1": "$ \\log 5\\equal{}\\log \\frac{10}{2}.$", "Solution_2": "sorry i'm a senior in hs and haven't applied myself in math until recently. Could you please state the theorem or rule that you use to find log(10/2)? it would \"make sense\" and i believe you than mean that 10-2 is idk 3log(2)???", "Solution_3": "Could you understand the meaning of $ \\log 2?$", "Solution_4": "if your implying thats theres some generalization that states log(a)=(2a/2) which seems obvious but not with logs then i guess yes; yet i dont know where to go from there. To be honest i don't know how this math is so difficult for me. Maybe i tried to go through aops v1 too fast.", "Solution_5": "[quote=\"adrenalinefix36\"]if log(2)=.3010 and log(3)=.4771 the value of x when 3^(x+3)=135 is approximately \nA)5\nB)1.47\nC)1.67\nD)1.78\nE)1.63\n\nno clue how to solve it since i constantly get a log(5) is there any relationship between the log2 the log3 and the log5? I guess this isn't an intermediate problem but please stoop.[/quote]\r\n\r\n[hide=\"Hint 1\"]$ 135 \\equal{} 27\\cdot5 \\equal{} 3^3\\cdot5$[/hide]\n\n[hide=\"Hint 2\"]We want base 10 here, so note that $ 5\\cdot2 \\equal{} 10$.[/hide]", "Solution_6": "$ \\log 2$ is a number $ x$ such that $ 10^x \\equal{} 2.$ Let $ 10^y \\equal{} 5$ we have $ 10^{x \\plus{} y} \\equal{} 2*5.$ Thus $ x \\plus{} y \\equal{} \\log 2*5\\Longleftrightarrow \\log 2*5 \\equal{} \\log 2 \\plus{} \\log 5.$ \r\nSimilarly $ 10^{x \\minus{} y} \\equal{} \\frac {2}{5}\\Longleftrightarrow x \\minus{} y \\equal{} \\log \\frac {2}{5}\\Longleftrightarrow \\log 2 \\minus{} \\log 5 \\equal{} \\log \\frac {2}{5}$.", "Solution_7": "i'm sorry sorry for my stupidity. After coming up with the two hints, which i had done previously, where do we go? PLease show steps as if i was an alien entirely ignorant of any rule. lol.thanks", "Solution_8": "Did you check my explanation?", "Solution_9": "yes 100%! please show calculation of answer", "Solution_10": "Before showing the answer, complete the following expressions.\r\n\r\nLet $ a,\\ b,\\ c,\\ p,\\ M$ are positive numbers not equal to one.\r\n\r\n1. $ \\log_a M \\plus{} \\log_a N \\equal{} \\boxed{A}.$\r\n\r\n2. $ \\log_a M \\minus{} \\log_a N \\equal{} \\boxed{B}.$\r\n\r\n3. $ \\log_a M^p \\equal{} \\boxed{C}.$\r\n\r\n4. $ \\frac {\\log_c b}{\\log_c a} \\equal{} \\boxed{D}.$\r\n\r\nSorry my first post was useless. :oops: Now you can solve your problem in using formula 4.\r\n\r\nGood luck! :) \r\n\r\nkunny", "Solution_11": "I apologize once more for my ignorance i simply do not understand logs. I guess i understand that some log operations like log(5)=(log(10)/log(2)) but i don't know that value(log(10)). It is it 3(log(2))? by subtraction or am i supposed to know what log(10)?(which i don't understand since im only given log(2) and log(3).", "Solution_12": "What is the meaning of $ \\log_2 2$?", "Solution_13": "yea confusing in reality wouldn't it be log(2)/log(2) but in this case its log(2)-log(2) which is 0? since a-a=0", "Solution_14": "[quote=\"kunny\"]What is the meaning of $ \\log_2 2$?[/quote]\r\n\r\n$ \\log_2 2$ is the number $ x$ such that $ 2^x\\equal{}2$. Thus $ \\log_a a\\equal{}1$ for $ a>0$ and $ a\\neq 1.$", "Solution_15": "[quote=\"kunny\"]$ \\log_2 2$ is the number $ x$ such that $ 2^x \\equal{} 2$. [/quote]\r\n\r\nThis is fundamental. Most times I find that the difficulty students have with logarithms is that they do not understand the basic [b]definition[/b] of a logarithm, so that even if they memorize all of the relevant properties they are constantly confused as to basic calculations like this. Here are some examples, and I will write out the base of the logarithm to avoid all possible confusion.\r\n\r\n$ \\log_{10} 1 \\equal{} 0$ because $ 10^0 \\equal{} 1$.\r\n$ \\log_{10} 10 \\equal{} 1$ because $ 10^1 \\equal{} 10$.\r\n$ \\log_{10} 2 \\equal{} 0.3010...$ because $ 10^{0.3010...} \\equal{} 2$.\r\n$ \\log_{10} 3 \\equal{} 0.4771...$ because $ 10^{0.4771...} \\equal{} 3$.\r\n$ \\log_{10} 1000 \\equal{} 3$ because $ 10^{3} \\equal{} 1000$.\r\n\r\nIf you understand exponents and you understand the definition of a logarithm, then you understand logarithms. Therefore, ask yourself:\r\n\r\n- [i]Do you understand exponents?[/i]\r\n- [i]Do you understand the definition of a logarithm?[/i]\r\n\r\nWhat kunny has been trying to communicate to you is that\r\n\r\n$ \\log_{10} 5 \\equal{} \\log_{10} \\left( \\frac{10}{2} \\right) \\equal{} \\log_{10} 10 \\minus{} \\log_{10} 2 \\equal{} 1 \\minus{} \\log_{10} 2 \\equal{} 1 \\minus{} (0.3010) \\equal{} .6990$.\r\n\r\nThe first equality is arithmetic. The second equality is the logarithm property\r\n\r\n$ \\log_a \\frac{b}{c} \\equal{} \\log_a b \\minus{} \\log_a c$\r\n\r\nwhich follows from the corresponding exponent property\r\n\r\n$ a^{x \\minus{} y} \\equal{} \\frac{a^x}{a^y}$\r\n\r\nwhere $ x \\equal{} \\log_a b, y \\equal{} \\log_a c$. The third equality follows from kunny's discussion that $ \\log_a a \\equal{} 1$.", "Solution_16": "Now i understand so this is process. First you take the equation with exponents and simplify it from 3^(x+3)=135 to 3^x=5. Then seeing the exponents, we convert to logs. We see that log3(5)=x=(log(5))/(log(3)). From there we simplify logs, the log(5) following the common concept of numerical division. Hence, log(10/2)/log(2). The important thing to then realize, is that given the fact that all logs unless noted otherwise are of base 10, and the fundamental relationship of loga(N)/loga(N)=1 by the law of exponents we have (1-log(2))/log(3). From there just use substitution of known quantities. Thanks, i'll try to make it easier next time by learning latex.", "Solution_17": "Congratulations! :)", "Solution_18": "[url=http://www.mathlinks.ro/viewtopic.php?t=82841&search_id=1638154101]Practice[/url]" } { "Tag": [ "limit", "integration", "trigonometry", "logarithms", "calculus", "derivative", "calculus computations" ], "Problem": "Evaluate this limit : \r\n1) $ \\lim_{n \\to + \\infty}{\\frac {\\sqrt {n^{3n}}}{n!}{\\prod_{k = 1}^{n}{sin(\\frac {k}{n\\sqrt {n}})}}} =$\r\n2) $ \\lim_{x \\to + \\infty} {(\\frac {1}{lnx}\\int_{0}^{x}{\\frac {dt}{\\sqrt [5] {1 + t^5}}})} = ?$\r\n3)$ \\lim_{x \\to + 0}(\\frac {1}{x^2}\\int_{0}^{x}{t^{1 + t}dt}) ?$\r\n4)$ \\lim_{n \\to + \\infty}{\\int_{0}^{\\pi}{\\sqrt [n] {x}sin(x) dx}} = ?$\r\n5)$ \\lim_{n \\to + \\infty}{\\int_{0}^{\\frac {\\pi}{2}}{\\frac {sin^n{x}}{\\sqrt {1 + x}}dx} = ?}$\r\n6)$ \\lim_{n \\to + \\infty}{n^4\\int_{n}^{n + 1}{\\frac {xdx}{x^5 + 1}} = ?}$\r\n7)$ \\lim_{n \\to + \\infty}{n^3 \\int_{n}^{2n}{\\frac {xdx}{x^5 + 1}} = ?}$", "Solution_1": "$ \\lim_{x\\to + \\infty}{(\\frac {1}{lnx}\\int_{0}^{x}{\\frac {dt}{\\sqrt [5]{1 + t^{5}}}})} = \\lim_{x\\to\\infty} \\frac {1}{\\sqrt [5]{1 + x^{5}}} \\cdot x = 1$ by Lopitale rule\r\n$ \\lim_{x\\to + 0}(\\frac {1}{x^{2}}\\int_{0}^{x}{t^{1 + t}dt}) = \\lim_{x\\to 0} \\frac {x^{1 + x}}{2x} = \\lim_{x\\to 0} \\frac12 x^x = \\frac12$ . the same idea\r\nsince $ \\sqrt [n]{x}sin(x) \\to \\sin x,\\ n\\to \\infty$ then $ \\lim_{n\\to + \\infty}{\\int_{0}^{\\pi}{\\sqrt [n]{x}sin(x) dx}} \\to \\int_0^{\\pi} \\sin x\\ dx = 2$ by Lebesque convergence theorem\r\nsince $ sin^{n}{x} \\to 0\\ (mod \\lambda)$ then $ \\lim_{n\\to + \\infty}{\\int_{0}^{\\frac {\\pi}{2}}{\\frac {sin^{n}{x}}{\\sqrt {1 + x}}dx} = \\int_{0}^{\\pi/2} 0 \\,dx = 0}$ by Lebesque convergence theorem\r\n\r\n$ \\forall x\\in[n,n + 1]: \\ \\frac {n}{n^5 + 1} \\ge \\frac {x}{x^5 + 1} \\ge \\frac {n + 1}{(n + 1)^5 + 1}$ so $ \\frac {n^5}{n^5 + 1} = n^{4}\\int_{n}^{n + 1}{\\frac {n\\ dx}{n^{5} + 1} \\ge n^{4}\\int_{n}^{n + 1}{\\frac {xdx}{x^{5} + 1} \\ge n^{4}\\int_{n}^{n + 1}{\\frac {n + 1\\ dx}{(n + 1)^{5} + 1} = \\frac {n^5 + n^4}{(n + 1)^5 + 1}}}}$. RHS and LHS tends to 1 when $ n\\to\\infty$ so answer is 1\r\n\r\n$ \\lim_{n\\to + \\infty}{n^{3}\\int_{n}^{2n}{\\frac {xdx}{x^{5} + 1}} = \\lim_{n\\to + \\infty} \\frac {\\frac {2\\cdot 2n}{(2n)^{5} + 1} - \\frac {n}{n^5 + 1} }{ - 3n^{ - 4} } = \\frac13\\frac {n^4(32n^6 - 4n^6 + ....)}{32n^{10} + ...} = \\frac {28}{3\\cdot 32}=\\frac{7}{24}}$ by Lopitale rule\r\n\r\nand 1) \r\n$ \\lim_{n\\to + \\infty}{\\frac {\\sqrt {n^{3n}}}{n!}{\\prod_{k = 1}^{n}{sin(\\frac {k}{n\\sqrt {n}})}}} = \\lim exp( \\sum \\ln\\left(\\frac {\\sin\\frac {k}{n\\sqrt {n}}}{\\frac {k}{n\\sqrt {n}}}\\right))$\r\nnow\r\n\r\n $ \\lim \\sum \\ln\\left(\\frac {\\sin\\frac {k}{n\\sqrt {n}}}{\\frac {k}{n\\sqrt {n}}}\\right) = \\lim \\sum \\ln\\left(\\frac {\\frac {k}{n\\sqrt {n}} - \\frac16 \\frac {k^3}{n^4\\sqrt {n}} + o(\\frac {k^4}{n^6})}{\\frac {k}{n\\sqrt {n}}}\\right) = \\lim \\sum \\ln\\left(1 - \\frac16\\frac {k^2}{n^3} + o(\\frac {1}{n\\sqrt {n}})\\right) = \\lim - \\frac16 \\sum \\frac {k^2}{n^3} + o(\\frac {1}{n\\sqrt {n}}) = - \\frac16 \\frac {n(n + 1)(2n + 1)}{6n^3} + o(\\frac {1}{\\sqrt {n}}) \\ to - 1/36$. one can shown that all $ o(..)$ here is uniformly bounded.. so \r\nanswer is $ e^{ - 1/36}$", "Solution_2": "[quote=\"anik_andrew\"]\n$ \\lim_{n\\to + \\infty}{n^{3}\\int_{n}^{2n}{\\frac {xdx}{x^{5} + 1}} = \\lim_{n\\to + \\infty} \\frac {\\frac {2n}{(2n)^{5} + 1} - \\frac {n}{n^5 + 1} }{ - 3n^{ - 4} } = \\frac13\\frac {n^4(32n^6 - 2n^6 + ....)}{32n^{10} + ...} = \\frac {10}{32}}$ by Lopitale rule\n[/quote]\r\n\r\nThis can't be correct. As $ x \\to + \\infty$ we see that $ f(x) = \\frac {x}{x^5 + 1} \\sim \\frac {1}{x^4}$, so\r\n\r\n$ \\int_{x = n}^{2n} \\frac {x \\, dx}{x^5 + 1} \\sim \\int_{x = n}^{2n} \\frac {dx}{x^4} = \\frac {7}{24n^3}$.\r\n\r\nThus $ \\lim_{n \\to + \\infty} n^3 \\int_{x = n}^{2n} \\frac {x \\, dx}{x^5 + 1} = \\frac {7}{24}$.\r\n\r\nThe error is that you computed the derivative of the integral with respect to $ n$ incorrectly. It should be\r\n\r\n$ \\frac {d}{dn}\\left[\\int_{x = n}^{2n} \\frac {x \\, dx}{x^5 + 1} \\right] = \\frac {d}{dn}[2n]\\frac {2n}{(2n)^5 + 1} - \\frac {n}{n^5 + 1}$.", "Solution_3": "hello, yes for 7) i have also got\r\n$ \\lim_{n \\to \\infty}n^3\\int_{n}^{2n}\\frac{x}{1\\plus{}x^5}\\,dx\\equal{}\\frac{7}{24}$.\r\nSonnhard.", "Solution_4": "atomicwedgie, thanks )) i forgot this derivirtate ) i have edited ) :blush:\r\n\r\nand what do you mean \r\n[quote=\"atomicwedgie\"]\n\n we see that $ f(x) \\equal{} \\frac {x}{x^5 \\plus{} 1} \\sim \\frac {1}{x^4}$, so\n\n$ \\int_{x \\equal{} n}^{2n} \\frac {x \\, dx}{x^5 \\plus{} 1} \\sim \\int_{x \\equal{} n}^{2n} \\frac {dx}{x^4}$.\n\n[/quote] ?", "Solution_5": "hello, for 6)\r\n$ \\lim_{n \\to \\infty}n^4\\int_{n}^{n\\plus{}1}\\frac{x}{x^5\\plus{}1}\\,dx$ i have got $ 1$ as the limit.\r\nSonnhard." } { "Tag": [ "geometry", "incenter", "circumcircle", "projective geometry", "power of a point", "radical axis", "geometry solved" ], "Problem": "Let I be the incentre of triangle ABC. Let the line passing through I perpendicular to AI intersect side BC at A'. Points B', C' are defined similarly. Prove that A', B', C' lie on a line perpendicular to IO, where O is the circumcenter of triangle ABC.", "Solution_1": "First of all we observe that $A'I$ is tangent to the circle $IBC$ (by simple angle chasing)\r\n\r\nDenote center of the circle $IBC$ as $O_1$, its radius - $R_1$ and $M$ -midpoint of $BC$.\r\n\r\nWe will prove that $A'O^2- A'I^2 =R^2$ , same for $B,C$ \r\n(which will give us that this all this points are collinear and $OI\\perp A',B',C'$)\r\n\r\n$A'O^2 = A'M^2+OM^2= A'O_1^2 -O_1M^2+OM^2$\r\n$A'I^2 = A'O_1^2-R_1^2$\r\n\r\n$A'O^2-A'I^2 = R_1^2+OM^2-O_1M^2$\r\n\r\nBut we can see that $OM = Rcos(a), R_1 = 2Rsin(a/2), O_1M = 2Rsin^2(a/2)$\r\n\r\nAnd $A'O^2-A'I^2 = R^2cos^2(a)+4R^2sin^2(a/2)- 4R^2sin^4(a/2)= R^2$ c.t.d.", "Solution_2": "Fantastic! Thanks, prowler.", "Solution_3": "Thank you for a good problem :)", "Solution_4": "Construct polar line of the incenter. Line OI intersects it at X. \r\n Three circles with centers at A',B',C' passing thru I and X are coaxal\r\n with A'-B'-C' the axis - mid-perpendicular to IX, or midline in the\r\n corresponding triangles with vertex at I.\r\n\r\n\r\n \r\n Maj. Pestich", "Solution_5": "I am sorry pestich, but could you explain why $A'X = A'I$ ,\r\n(I mean $A'$ is center of the circle which passes through $I,X$)", "Solution_6": "[quote=\"jhaussmann5\"]Let I be the incentre of triangle ABC. Let the line passing through I perpendicular to AI intersect side BC at A'. Points B', C' are defined similarly. Prove that A', B', C' lie on a line perpendicular to IO, where O is the circumcenter of triangle ABC.[/quote]\r\n\r\nMy proof of this uses more or less the same idea as Prowler's one, but is a bit more general:\r\n\r\nSince the point I is the incenter of triangle ABC, it lies on the angle bisectors of the angles CAB, ABC and BCA. Thus, $\\measuredangle IAB=\\frac{A}{2}$, $\\measuredangle IBA=\\frac{B}{2}$ and $\\measuredangle ICB=\\frac{C}{2}$.\r\n\r\nWLOG assume that the point A' lies on the extension of the segment BC beyound the point B. Then, < A'IB = < AIB - < AIA'. But since $IA^{\\prime}\\perp AI$, we have < AIA' = 90\u00b0, and thus < A'IB = < AIB - 90\u00b0. But by the sum of the angles in triangle AIB, we have $\\measuredangle AIB=180^{\\circ}-\\measuredangle IAB-\\measuredangle IBA=180^{\\circ}-\\frac{A}{2}-\\frac{B}{2}$. Hence,\r\n\r\n$\\measuredangle A^{\\prime}IB=\\measuredangle AIB-90^{\\circ}=\\left(180^{\\circ}-\\frac{A}{2}-\\frac{B}{2}\\right)-90^{\\circ}=90^{\\circ}-\\frac{A}{2}-\\frac{B}{2}$\r\n$=90^{\\circ}-\\frac{A+B}{2}=90^{\\circ}-\\frac{180^{\\circ}-C}{2}$ (since A + B = 180\u00b0 - C by the sum of the angles of triangle ABC)\r\n$=90^{\\circ}-\\left(90^{\\circ}-\\frac{C}{2}\\right)=\\frac{C}{2}=\\measuredangle ICB$.\r\n\r\nHence, by the converse of the tangent-chordal angle theorem, it follows that the line IA' is the tangent to the circumcircle of triangle BIC at the point I. Thus, the point A' is the point where the tangent to the circumcircle of triangle BIC at the point I intersects the line BC. Similarly, the points B' and C' are the points where the tangents to the circumcircles of triangles CIA and AIB at the point I intersect the lines CA and AB, respectively. Now, our assertion that the points A', B', C' lie on one line perpendicular to the line IO becomes a particular case of the following general theorem:\r\n\r\n[color=blue][b]Theorem 1.[/b] Let P be an arbitrary point in the plane of a triangle ABC, and let the tangents to the circumcircles of triangles BPC, CPA, APB at the point P intersect the lines BC, CA, AB at the points A', B', C', respectively. Then, these points A', B', C' lie on one line perpendicular to the line PO, where O is the circumcenter of triangle ABC.[/color]\r\n\r\n[i]Proof of Theorem 1.[/i] Let k be the circumcircle of triangle ABC, and let p be the \"zero circle\" with center P (this is the circle with center P and radius 0). Then, the power of the point A' with respect to the circle k equals $A^{\\prime}B\\cdot A^{\\prime}C$, and the power of the point A' with respect to the circle p equals $A^{\\prime}P^2$. But by the intersecting secant and tangent theorem, we have $A^{\\prime}B\\cdot A^{\\prime}C=A^{\\prime}P^2$ (since the point A' lies on the tangent to the circumcircle of triangle BPC at the point P). Hence, the point A' has equal powers with respect to the circles k and p; thus, the point A' lies on the radical axis of the circles k and p. Similarly, the points B' and C' also lie on this radical axis. Hence, all three points A', B', C' lie on one line, namely on the radical axis of the circles k and p. This radical axis is clearly perpendicular to the line PO, since O and P are the respective centers of the circles k and p, and the radical axis of two circles is always perpendicular to the line joining their centers. This proves Theorem 1.\r\n\r\n Darij" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given$ a,b,c>0$.Prove that:\r\n$ \\frac{20ab}{(a\\plus{}c)(b\\plus{}c)}\\plus{}\\frac{15bc}{(b\\plus{}a)(c\\plus{}a)}\\plus{}\\frac{12ca}{(c\\plus{}b)(a\\plus{}b)} \\ge 11$ :)", "Solution_1": "[hide]Expanding and simplifying, this becomes\n\\[ 9a^2 b + 9ab^2 + a^2 c + ac^2 + 4b^2 c + 4bc^2 \\ge 22abc.\\]\nBut by weighted AM-GM,\n\\begin{align*}\n&9a^2 b + 9ab^2 + a^2 c + ac^2 + 4b^2 c + 4bc^2 \\\\\n&= 36 \\cdot \\frac{a^2 b}{4} + 18 \\cdot \\frac{ab^2}{2} + 12 \\cdot \\frac{a^2 c}{12} + 18 \\cdot \\frac{ac^2}{18} + 12 \\cdot \\frac{b^2 c}{3} + 36 \\cdot \\frac{bc^2}{9} \\\\\n&\\ge 132 \\left( \\frac{a^2 b}{4} \\right)^{36/132} \\left( \\frac{ab^2}{2} \\right)^{18/132} \\left( \\frac{a^2 c}{12} \\right)^{12/132} \\left( \\frac{ac^2}{18} \\right)^{18/132} \\left( \\frac{b^2 c}{3} \\right)^{12/132} \\left( \\frac{bc^2}{9} \\right)^{36/132} \\\\\n&= 22abc.\n\\end{align*}\nEquality occurs if $ a: b: c = 2: 1: 3$.[/hide]", "Solution_2": "[quote=\"nsato\"][hide]Expanding and simplifying, this becomes\n\\[ 9a^2 b + 9ab^2 + a^2 c + ac^2 + 4b^2 c + 4bc^2 \\ge 22abc.\n\\]\nBut by weighted AM-GM,\n\\begin{align*} & 9a^2 b + 9ab^2 + a^2 c + ac^2 + 4b^2 c + 4bc^2 \\\\\n& = 36 \\cdot \\frac {a^2 b}{4} + 18 \\cdot \\frac {ab^2}{2} + 12 \\cdot \\frac {a^2 c}{12} + 18 \\cdot \\frac {ac^2}{18} + 12 \\cdot \\frac {b^2 c}{3} + 36 \\cdot \\frac {bc^2}{9} \\\\\n& \\ge 132 \\left( \\frac {a^2 b}{4} \\right)^{36/132} \\left( \\frac {ab^2}{2} \\right)^{18/132} \\left( \\frac {a^2 c}{12} \\right)^{12/132} \\left( \\frac {ac^2}{18} \\right)^{18/132} \\left( \\frac {b^2 c}{3} \\right)^{12/132} \\left( \\frac {bc^2}{9} \\right)^{36/132} \\\\\n& = 22abc. \\end{align*}\nEquality occurs if $ a: b: c = 2: 1: 3$.[/hide][/quote]\r\nNice solution :)", "Solution_3": "[quote=\"quykhtn-qa1\"]Given$ a,b,c > 0$.Prove that:\n$ \\frac {20ab}{(a \\plus{} c)(b \\plus{} c)} \\plus{} \\frac {15bc}{(b \\plus{} a)(c \\plus{} a)} \\plus{} \\frac {12ca}{(c \\plus{} b)(a \\plus{} b)} \\ge 11$ :)[/quote]\r\nAfter expanding we need to prove that\r\n$ (9b\\plus{}c)a^2\\plus{}(9b^2\\minus{}22bc\\plus{}c^2)a\\plus{}4b^2c\\plus{}4bc^2\\geq0.$\r\nIf $ 9b^2\\minus{}22bc\\plus{}c^2\\geq0$ the inequality is true. \r\nBut for $ 9b^2\\minus{}22bc\\plus{}c^2<0$ enough to prove that\r\n$ (9b^2\\minus{}22bc\\plus{}c^2)^2\\minus{}16bc(b\\plus{}c)(9b\\plus{}c)\\leq0,$ which is equivalent to\r\n$ (3b\\minus{}c)^2(9b^2\\minus{}54bc\\plus{}c^2)\\leq0,$ which is true.", "Solution_4": "[quote=\"arqady\"][quote=\"quykhtn-qa1\"]Given$ a,b,c > 0$.Prove that:\n$ \\frac {20ab}{(a \\plus{} c)(b \\plus{} c)} \\plus{} \\frac {15bc}{(b \\plus{} a)(c \\plus{} a)} \\plus{} \\frac {12ca}{(c \\plus{} b)(a \\plus{} b)} \\ge 11$ :)[/quote]\nAfter expanding we need to prove that\n$ (9b \\plus{} c)a^2 \\plus{} (9b^2 \\minus{} 22bc \\plus{} c^2)a \\plus{} 4b^2c \\plus{} 4bc^2\\geq0.$\nIf $ 9b^2 \\minus{} 22bc \\plus{} c^2\\geq0$ the inequality is true. \nBut for $ 9b^2 \\minus{} 22bc \\plus{} c^2 < 0$ enough to prove that\n$ (9b^2 \\minus{} 22bc \\plus{} c^2)^2 \\minus{} 16bc(b \\plus{} c)(9b \\plus{} c)\\leq0,$ which is equivalent to\n$ (3b \\minus{} c)^2(9b^2 \\minus{} 54bc \\plus{} c^2)\\leq0,$ which is true.[/quote]\r\nNice :)", "Solution_5": "I am very sorry for my doubts..... :?: \r\nI am thinking, that if we replace $ a, b, c$ by $ ka, kb, kc$ respectively then the inequality remains unchanged, then can we assume that : $ abc\\equal{}1$?? :huh: \r\n\r\nCan anyone help me to clear my doubts??\r\nPS: I think I was looking for a proof without Expanding.", "Solution_6": "[quote=\"Potla\"]I am very sorry for my doubts..... :?: \nI am thinking, that if we replace $ a, b, c$ by $ ka, kb, kc$ respectively then the inequality remains unchanged, then can we assume that : $ abc \\equal{} 1$?? :huh: \n\nCan anyone help me to clear my doubts??\nPS: I think I was looking for a proof without Expanding.[/quote]\r\nYes, you can assume any condition,my friend :)" } { "Tag": [], "Problem": "assuming that the inert gas configuration of atoms is their most stable state it follows that it (in. gas. config.)must have the min. energy.(for eg unstable nucleii gain stability by releasing enormous amt.s of energy and are respectively converted to their less energy states which are stable) . It follows therefore that there must be inherent tendency in each particle to release energy and get converted into its most stable state.\r\nNow let us look towards the ionization of metals.All metals of the s block are stabilized by the release of electrons.they are converted into a stabler state and hence energy must be released in the process .WE can say that DEL(H)<0.also there is an increase in the randomness of the system ,thus DEL(S)>0(from a bound electron we get a free one and the no. of particles increases too).thus the ionization process ought to be feasible at all temperatures and must be accompanied with a simultaneous release of energy.\r\nWhat we see in the world is just the reverse.We have to provide heat energy ,light photons etc to the metals for overcoming their ionization energy , ionizatiion potential etc.Why is this so? :?:", "Solution_1": "The ionization process has an activation energy, doesn't it? The [b]net[/b] enthalpy and entropy make it favorable, but an initial energy is necessary to begin the reaction.", "Solution_2": "why at all?? \r\nis this activation energy is used for overcoming the coulombic barrier?\r\n on the other hand,an electron in the outermost shell is nearer to the electron cloud and must experience a greater repulsive force.:-(electrostatic force is inversely proportional to the square of the distance.)\r\n(the nucleus may be assumed to be the size of a tip of a pin paced inside a well of 1 meter diameter :the electro cloud)\r\nthere are reactions which have high act.energy but eventually turn out to be exothermic. Why is the ionization proces not one of them??" } { "Tag": [ "calculus", "integration", "logarithms", "limit", "inequalities", "LaTeX", "induction" ], "Problem": "Let $ (a_n)$ be a sequence of positive real numbers, decreasing and the sum of whose terms is infinite. Prove that the series whose general term is $ min(a_n,\\frac{1}{n}) $ is also divergent. Not a difficult, but a very beautiful problem.", "Solution_1": "The problem came up in the 6th German TST this year and was already posted by Darij.\r\n\r\nMisha", "Solution_2": "It's also from ISL 2003.", "Solution_3": "I'm not really suprised becaused nearly all problems of the German TSTs this year\r\nwere from the ISL 2003.", "Solution_4": "Let's start with a simpler problem: Supose $x_n>0$ and $\\sum x_n=\\infty$. Then we can prove that $\\sum \\min(1,x_n)=\\infty$. This is clear as follows. If $x_n>1$ infinitely many times, then $\\min(1,x_n)=1$ infinitely many times, and the series diverges because its terms don't tend to zero. On the other hand, if $x_n>1$ only finitely many times, then $\\min(1,x_n)=x_n$ for all sufficiently large $n$ and $\\sum \\min(1,x_n)$ diverges by comparison to $\\sum x_n$.\r\n\r\nNow we employ the Cauchy condensation test for convergence: if $(b_n)$ is a nonincreasing sequence of positive numbers, then $\\sum b_n$ converges if and only if $\\sum 2^kb_{2^k}$ converges. (This is merely an extension of Oresme's 14th-century proof of the divergence of the harmonic series).\r\n\r\nThe minimum of two monotone sequences is monotone, so $\\min\\left(a_n,\\frac1n\\right)$ is nonincreasing. By the condensation test, we know that $\\sum 2^ka_{2^k}$ diverges. Applying the condensation test again, we see that $\\sum \\min\\left(a_n,\\frac1n\\right)$ is comparable to $\\sum 2^k\\min\\left(a_{2^k},\\frac1{2^k}\\right)=\\sum\\min\\left(2^ka_{2^k},1\\right)$. The result in the first paragraph shows that this diverges.", "Solution_5": "Your proof is the same as the best proof in the TST. At that time I was not so good in Calculus and I even didn't know the condensation test by Cauchy.\r\n\r\nMisha", "Solution_6": "All tests for the convergence of positive term series are at heart comparison tests. Out of all of those tests, are there any for which \"$a_n$ is decreasing\" is a key hypothesis? There are two: the integral test and the condensation test. Unsurprisingly, these two tests tend to be well suited to the same specific examples, such as $\\sum \\frac1{n(\\ln n)^2}$. It was the hypothesis of decreasing terms that caused me to think of the condensation test.", "Solution_7": "Sorry to revive such an old thread, but here's something related to this. The following problem is sort of a converse to the one posted here:\r\n\r\nGiven a decreasing sequence $(b_n)_n$ of positive reals with the property that for every decreasing sequence of positive reals $(a_n)_n$ with $\\sum a_n=\\infty$ we have $\\sum\\min(a_n,b_n)=\\infty$, prove that $\\inf_n nb_n>0$.\r\n\r\nIt appeared in some AMM from a couple of years ago, I think.", "Solution_8": "That should be $\\liminf$; the long-term behavior is what matters.", "Solution_9": "I'm pretty sure it works with $\\inf$ though.", "Solution_10": "The problem is that it's vacuous with $\\inf$. The $\\inf$ is always positive.", "Solution_11": "Why? Isn't it possible to find large $n$'s for which $nb_n$ approach zero? I mean, no, it's not possible, since that's what the problem states, but why do you find it obvious?", "Solution_12": "i made a nice proof:suppose that Sum(b_n) <+00 where (b_n)=(inf(a_n,1/n))\r\nsince (a_n) and (1/n) are bouth positifs and decreasing it's to for (b_n) , and so\r\n0<=n.b_2n<=Sum(b_k,k=n,...,2n)=o(1) and because (b_n) is decreasing we conclude that b_n=o(1/n) , so for sufficient large n :b_n=a_n witch gaves a contradiction. :P", "Solution_13": "Sorry yassinus,\r\n\r\n\r\nI am probably too stupid, but still can't see why $\\sum_{k=n}^{2n}b_k = o(1)$ (1). For example $\\lim_{n\\to \\infty}\\sum_{k=n}^{2n}\\frac{1}{k} =\\ln 2$, but I assume you use a different argument to get (1)... or do you mean\r\n\r\n$\\sum_{k=n}^{2n}b_k = O(1)$?\r\n\r\n\r\nMany thanks!\r\n\r\n[i][b]Edit[/b][/i] To Kent Merryfield: I don't know the proof that you mean of the so-called ``condensation criterion'', but isn't it just a consequence of the inequalities\r\n\r\n$\\sum_{k=3}^{\\infty}b_k\\leq \\sum_{k=1}^{\\infty}2^k b_{2^k} \\leq 2 \\sum_{k=1}^{\\infty}b_k$?\r\n\r\nMany thanks!", "Solution_14": "jifko777: Yes, that's the proof. Think of it as a generalization inspired by Oresme's ~800 year old proof of the divergence of the harmonic series.", "Solution_15": "Kent, I must confess that that's the easiest proof I have seen so far, which makes it sort of brilliant, doesn't it.\r\n\r\n\r\nCheers.", "Solution_16": "but we know that sum(bk,k=n,+00) ->0 when n->+00 because i suppose the series convergent ( your exemple is not a counter-exemple because harmonic sum is not convergent as we know)\r\n 0<=sum(bk,k=n...2n) <=sum(bk,k=n,+00) and we can conclud that sum(bk,k=n...2n)=o(1) :huh:", "Solution_17": "Thanks [b]yassinus[/b], very nice proof indeed! I presume I wasn't concentrated enough when reading your posting.\r\n\r\n\r\nActually the same technique may be used to prove that the harmonic series diverges.\r\n\r\n\r\nJust a minor request to you: Could you please latex your postings so that they can be read more easily. If you still haven't done latex at all, you can start by holding the mouse pointer on a latex typed formula, then you can see that some symbols appear in a yellow (Windows) box. You can produce the same formula by copying the symbols from the yellow box and putting the dollar sign at the beginning and at the end of your formula.\r\n\r\n\r\nOnce again thanks very much for your reply!", "Solution_18": "Does Cauchy condensation test hold for $ \\sum 3^n a_{3^n}$", "Solution_19": "[quote=\"Kent Merryfield\"]Let's start with a simpler problem: Supose $ x_n>0$ and $ \\sum x_n \\equal{} \\infty$. Then we can prove that $ \\sum \\min(1,x_n) \\equal{} \\infty$. This is clear as follows. If $ x_n>1$ infinitely many times, then $ \\min(1,x_n) \\equal{} 1$ infinitely many times, and the series diverges because its terms don't tend to zero. On the other hand, if $ x_n>1$ only finitely many times, then $ \\min(1,x_n) \\equal{} x_n$ for all sufficiently large $ n$ and $ \\sum \\min(1,x_n)$ diverges by comparison to $ \\sum x_n$.\n\nNow we employ the Cauchy condensation test for convergence: if $ (b_n)$ is a nonincreasing sequence of positive numbers, then $ \\sum b_n$ converges if and only if $ \\sum 2^kb_{2^k}$ converges. (This is merely an extension of Oresme's 14th-century proof of the divergence of the harmonic series).\n\nThe minimum of two monotone sequences is monotone, so $ \\min\\left(a_n,\\frac1n\\right)$ is nonincreasing. By the condensation test, we know that $ \\sum 2^ka_{2^k}$ diverges. Applying the condensation test again, we see that $ \\sum \\min\\left(a_n,\\frac1n\\right)$ is comparable to $ \\sum 2^k\\min\\left(a_{2^k},\\frac1{2^k}\\right) \\equal{} \\sum\\min\\left(2^ka_{2^k},1\\right)$. The result in the first paragraph shows that this diverges.[/quote]\r\n\r\nThe minimum of two monotone sequences is monotone?", "Solution_20": "They have to be monotone in the same direction for that to be true, but yes. (And in this situation, they were monotone in the same direction.)\r\n\r\nSuppose $ (a_n)$ and $ (b_n)$ are nonincreasing sequences and let $ c_n\\equal{}\\min(a_n,b_n).$ Then $ c_n$ is nonincreasing.\r\n\r\nLet $ m 0$.\n\nIt appeared in some AMM from a couple of years ago, I think.[/quote]", "Solution_28": "mlok has done a very close thing in one of his posts. If $ \\liminf nb_n\\equal{}0$, then, for every $ \\varepsilon>0$, we can find arbitrarily large $ N$ with $ b_n<\\varepsilon/N$ for $ n\\ge N$. Now we want to construct $ a_n$. Do it by induction choosing $ \\varepsilon_k\\equal{}2^{\\minus{}2k}$ and, taking $ n_{k\\plus{}1}>2^k n_k$ satisfying the above property for $ \\varepsilon_k$ and putting $ a_n\\equal{}2^{\\minus{}k}/n_k$ for $ 2^{k\\minus{}1} n_{k\\minus{}1}x\u00b04 +y\u00b04 -8 <= -(x\u00b2y\u00b2-4xy+4)=-(xy-4)\u00b2<= 0.\r\nthe second is easy. :wink:", "Solution_2": "yes :lol:" } { "Tag": [ "inequalities" ], "Problem": "Given $ a,b,c \\geq 0\\ $ and $ a+b+c=1 $\r\n\r\nProve:\r\n\r\n$ \\Big(1+\\frac{1}{a}\\ \\Big)\\Big(1+\\frac{1}{b}\\ \\Big)\\Big(1+\\frac{1}{c}\\ \\Big) \\geq 64\\ $", "Solution_1": "I'm just baffled. Are the parenthesis supposed to be that way, and is that 1_b minus, or something else.... like 1/b.", "Solution_2": "[hide] value of a+b+c will be minimized when a+b+c since abc i maximized when this is true while this question involves its inverse\n\n if a=b=c, and a+b+c=1 so they each equal 1/3\n\n(1+1/a)(1+1/b)(1+1/c)>=64\n\n(1+1/a)^3>=64\n\n1+1/a>=4\n\n1/a>=3\n\n1/1/3>=3\n\n3=3\n\nSo the inequality is true as when the values differ they (1+a)(1+b)(1+c) will decrease while the inverse will increase.\n\n[/hide]", "Solution_3": "Chigr wrote:[hide] value of a+b+c will be minimized when a+b+c since abc i maximized when this is true while this question involves its inverse\n if a=b=c, and a+b+c=1 so they each equal 1/3\n(1+1/a)(1+1/b)(1+1/c)>=64\n(1+1/a)^3>=64\n1+1/a>=4\n1/a>=3\n1/1/3>=3\n3=3\nSo the inequality is true as when the values differ they (1+a)(1+b)(1+c) will decrease while the inverse will increase.\n[/hide]\n\n\n\na+b+c can't be minimized, it's a constant.\n\n\n\nAs I said, this is really an AM-GM problem.\n\n\n\nAM-GM for four variables takes the form:\n\na+b+c+d >= 4(abcd)^(1/4). I can prove this if anyone wants.", "Solution_4": "Hi\r\nwe know that\r\n a+1/a>=2,\r\nb+1/b>=2 \r\nand c+1/c>=2.\r\nis that the way to proceed.?\r\n\r\nSriram", "Solution_5": "Hint: \"harmonic mean of 1/a, 1/b and 1/c \" is 3, their AM and GM both should be greater than 3. \n\n\n\nNotice that the LHS is \n\n(1 + (1/a+1/b+1/c) + 2/(abc) ) which is [hide]nothing but 1 + 3*(AM) +2*(GM)^3 \n\n[/hide] \n\n\n\nOr if you like the details: Here they are in (spoiler) \n\nAM >= GM \n\n [hide](a+b+c)/3 = 1/3 >= (abc)^(1/3) or (1/abc) >= 27 [/hide]\n\n\n\nAnd AM >= HM (which is almost the same as AM>=GM done twice) \n\n\n\n [hide]1/3 = (a+b+c)/3 >= 3/(1/a+1/b+1/c) or (1/a+1/b+1/c) >= 9 \n\n(and of course, 1+9+2*27 = 64) \n\n[/hide]", "Solution_6": "Okay a twist on the problem .. Once you solved the original and want a more challange...\r\n\r\nIf (1+1/a)(1+1/b)(1+1/c) = 2 \r\n\r\nobviously here a+b+c is not equal to 1 but supposing a,b,c all are integers what are the possible values of a+b+c ? \r\n\r\n(Problem may be a little harder for general public here so pretend that you are looking at \"pre-olympiad\" forum) :)", "Solution_7": "Since this is going to fall like a stone off the front page of 'intermediate' soon, I will post a very pretty little solution.\r\n\r\n(1+1/a) = (a+1)/a = (a+a+b+c)/c\r\n\r\na+a+b+c >= 4(a)^(1/2)(bc)^(1/4)\r\n\r\nHence (a+a+b+c)/c >= 4(bc)^(1/4)/(a)^(1/2)\r\n\r\nRepeating this for every term of the product, we get:\r\n\r\nLHS >= 4*4*4abc/abc = 64 as desired.", "Solution_8": "Gyan wrote:Hint: \"harmonic mean of 1/a, 1/b and 1/c \" is 3, their AM and GM both should be greater than 3. \n\nNotice that the LHS is \n(1 + (1/a+1/b+1/c) + 2/(abc) ) which is [hide]nothing but 1 + 3*(AM) +2*(GM)^3 \n[/hide] \n\nOr if you like the details: Here they are in (spoiler) \nAM >= GM \n [hide](a+b+c)/3 = 1/3 >= (abc)^(1/3) or (1/abc) >= 27 [/hide]\n\nAnd AM >= HM (which is almost the same as AM>=GM done twice) \n\n [hide]1/3 = (a+b+c)/3 >= 3/(1/a+1/b+1/c) or (1/a+1/b+1/c) >= 9 \n(and of course, 1+9+2*27 = 64) \n[/hide]\n\n\n\nGyan, shouldn't there be 1/ab, etc. terms in there?", "Solution_9": "Sorry should have pointed out that one of 1/abc in '2/abc' came from \r\n\r\n(1/bc+1/ca+1/ab) = (a+b+c)/abc = 1/abc because (a+b+c = 1)\r\n\r\n(Tha'ts how you get 2/abc because in the product you have 1/abc and the three terms which also adds up to 1/abc_", "Solution_10": "[quote=\"Gyan\"]Sorry should have pointed out that one of 1/abc in '2/abc' came from \n\n(1/bc+1/ca+1/ab) = (a+b+c)/abc = 1/abc because (a+b+c = 1)\n\n(Tha'ts how you get 2/abc because in the product you have 1/abc and the three terms which also adds up to 1/abc_[/quote]\r\n\r\nGotcha. I still like my solution better :cool:. I'll try and post a solution to your other problem as soon as I finish my latexing my math homework.", "Solution_11": "[quote=\"sriram_mask\"]Hi\nwe know that\n a+1/a>=2,\nb+1/b>=2 \nand c+1/c>=2.\nis that the way to proceed.?\n\nSriram[/quote]\r\n\r\nNo. That is too weak. Remember, equality holds for a=b=c=1/3, but if we substitute a=1/3 into any of those, we get: 1/3 + 3 = 10/3 >= 2" } { "Tag": [ "Putnam", "combinatorics proposed", "combinatorics" ], "Problem": "In how many ways can one put up $ n$ elements into $ r$ sets $ A_1$, $ A_2$, ..., $ A_r$ such that the intersection of any $ k$ sets is disjoint... ??", "Solution_1": "What is the Putnam problem of which this is a generalization? (Also, I assume you mean that the intersection of any $ k$ of the sets is empty ... we can say that two sets are disjoint if they have empty intersection, but the word is not used for three or more sets (except to say they are [i]pairwise disjoint[/i], which I assume is not what you mean).)", "Solution_2": "This is the generalization of putnam problem A1 1985..\r\n\r\nthis is the link..\r\n\r\nhttp://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1985.pdf\r\n\r\nand when i said intersection of any k sets is disjoint..\r\n\r\ni meant that intersection of any k sets together is empty....not pairwise intersection of sets..", "Solution_3": "For each of the $ n$ elements in our universe, we must choose between 1 and $ k \\minus{} 1$ of the $ r$ sets to place it into. (I'm assuming here that you want the set $ \\bigcup_{i \\equal{} 1}^r A_i$ to be the entire $ n$-set, as in the Putnam problem; otherwise, we can replace \"1\" in the previous sentence with \"0\" and make similar changes in what follows.) This is both necessary (if any element were to be found in $ k$ of our $ r$ sets, they would violate the given condition) and sufficient (no element is in $ k$ sets, so the intersection of any $ k$ sets is empty) to have our desired goal. Thus the answer is just $ \\left(\\sum_{i \\equal{} 1}^{k \\minus{} 1} \\binom{r}{i} \\right)^n$.\r\n\r\nIf we have $ k \\equal{} r$ as in the Putnam problem, this simplifies to $ (2^r \\minus{} 2)^n$.", "Solution_4": "yup correct..:)" } { "Tag": [ "calculus", "integration", "limit", "trigonometry", "calculus computations" ], "Problem": "Let $ a$ be a postive constant number. Given a positive integer $ n$, take an integer $ m$ such that $ m\\leq \\frac{na}{\\pi} NH4 + NO3\r\n\r\nI have tried to balance out the equation, but to no avail.. anyone can help me please?", "Solution_1": "If it's possible, then you can find natural numbers a,b,c and d such that :\r\n$aNH_3N0_3+bH_2O\\rightarrow cNH_4+ dNO_3$. Then you must have the same number of atomes of $N,H$ and $O$. Thus a,b,c and d must satisfy the following system :\r\n$3a+2b=4c$\r\n$2a=c+d$\r\n$3a+b=3d$ that is $(a,b,c,d)=(10\\alpha,3\\alpha,9\\alpha,11\\alpha)$. So you have $10NH_3N0_3+3H_2O\\rightarrow 9NH_4+ 11NO_3$\r\n\r\nEDIT : Oups I misread it is $NH_4$ on the LHS. The system implies then $b=0$. It's impossible", "Solution_2": "There's no H20 in the equation, its just NH4NO3 --> NH4+ + NO3-" } { "Tag": [ "function", "calculus", "derivative", "inequalities", "calculus computations" ], "Problem": "f is continuous on $ [0, \\pi]$ and differentible on $ (0, \\pi)$ satisfying $ f(0)\\equal{}f(\\pi)\\equal{}0, |f'(x)| <1,$ for all $ x \\in (0,\\pi)$ prove that \r\ni) there exists $ \\alpha \\in (0,\\pi) : f'(\\alpha)\\equal{}tanf(\\alpha)$\r\nii)$ |f(x)| < \\frac{\\pi}2$ for all $ x \\in (0, \\pi)$", "Solution_1": "first one is nothing but legrange`s mean value theorem. { a pretty simple application of that!} :lol:", "Solution_2": "can you give the detail of the solution? :lol:", "Solution_3": "2nd is easily proven using the derivative inequality:\r\n\r\n$ \\minus{}1 \\leq f'(x) \\leq 1$\r\n\r\nIntegrating from 0 to x\r\n\r\n$ \\minus{}x \\leq f(x) \\leq x$\r\n\r\nAlso integrating from x to \\pi\r\n\r\n$ \\minus{}\\pi\\minus{}x \\leq \\minus{}f(x) \\leq \\pi\\minus{}x$\r\n\r\n=> $ \\minus{}\\pi\\plus{}x \\leq f(x) \\leq \\pi\\minus{}x$\r\n\r\nNow add both ineq above.\r\n\r\nUsing this result we can now integrate tanx function for any value of f(x) but I cant still prove the first claim.", "Solution_4": "can anyone help me with the first? thanks!" } { "Tag": [ "MATHCOUNTS" ], "Problem": "I never really got the rt=d thing for mathcounts problems i understand that if i drive for 30 minutes at a rate of 60 miles per hour then i will drive 30 miles but not much past that can anyone give some mathcounts level probs that invole rt=d and if possible solutions, u can also pm me or pst useful websites.\r\n\r\nThanks in advance", "Solution_1": "Bob drove for 2 hours at 60 mph and x hours at 75 mph. If he averaged 70 mph for the whole trip, find x.", "Solution_2": "i cross multiplied knowing that the average = total distance/total time\r\nso x=4" } { "Tag": [ "calculus", "FTW" ], "Problem": "At the 1996 grading of Advanced Placement Calculus exams, 530 math teachers graded 126000 tests in 10 hours. Each test had 6 problems. What is the average number of minutes it took one teacher to grade 25 problems? Express your answer to the nearest minute.", "Solution_1": "there were $ 126000$ tests, which means there were $ 126000\\times6$ problems, or $ 756000$ problems\r\n$ 530$ teachers graded $ 756000$ problems in $ 10$ hours, which means that...\r\n$ 1$ teacher graded $ 756000\\div530$ problems in $ 10$ hours, or about $ 1426.415$ problems\r\n\r\nnow, instead of writing it out like this, i will make an equation\r\n$ 1426.415$ problems = $ 10$ hours\r\n\r\nnow i know these equations don't look pretty, since their units are different, but that's not going to matter\r\nwe want to find the time in minutes it took a teacher to grade 25 problems\r\n\r\nwe first convert the hours to minutes\r\n\r\n$ 1426.415$ problems = $ 600$ minutes\r\n\r\ndivide each side by $ 1426.415$ to get the time it took for one teacher to grade on problem\r\n\r\n$ 1$ problem = $ 0.4206$ minutes\r\n\r\nwe want to find the time it took to grade $ 25$ problems, so we multiply both sides by $ 25$\r\n\r\n$ 25$ problems = $ 10.51$ minutes or $ 11$ minutes\r\nour answer is $ \\boxed{11\\text{minutes}}$\r\n\r\ndisclaimer: answer might not be exact due to rounding, but should still be the right answer", "Solution_2": "answer is 9.905 = 10 because 12600/(530*600) is unit time.", "Solution_3": "The answer on FTW is 42.\n\nHere's my method:\n\nEach teacher graded $\\frac{126,000\\times6}{530}=\\frac{75,600}{53}$ problems, each in 10 hours. Thus, each hour, one teacher graded $\\frac{\\frac{75,600}{53}}{10}=\\frac{7,560}{53}$ problems. Each minute, one teacher graded $\\frac{\\frac{7,560}{53}}{60}=\\frac{126}{53}$ problems. Thus, each teacher grades $\\frac{25}{\\frac{126}{53}}=\\frac{1325}{126}$ of $25$ problems per minute, so they need approximately $\\boxed{11}$ minutes to finish 25 problems.\n\n...FTW is wrong?", "Solution_4": "Thats wrong. The actual answer is 42 minutes.\n[quote=\"vallon22\"]there were $ 126000$ tests, which means there were $ 126000\\times6$ problems, or $ 756000$ problems\n$ 530$ teachers graded $ 756000$ problems in $ 10$ hours, which means that...\n$ 1$ teacher graded $ 756000\\div530$ problems in $ 10$ hours, or about $ 1426.415$ problems\n\nnow, instead of writing it out like this, i will make an equation\n$ 1426.415$ problems = $ 10$ hours\n\nnow i know these equations don't look pretty, since their units are different, but that's not going to matter\nwe want to find the time in minutes it took a teacher to grade 25 problems\n\nwe first convert the hours to minutes\n\n$ 1426.415$ problems = $ 600$ minutes\n\ndivide each side by $ 1426.415$ to get the time it took for one teacher to grade on problem\n\n$ 1$ problem = $ 0.4206$ minutes\n\nwe want to find the time it took to grade $ 25$ problems, so we multiply both sides by $ 25$\n\n$ 25$ problems = $ 10.51$ minutes or $ 11$ minutes\nour answer is $ \\boxed{11\\text{minutes}}$\n\ndisclaimer: answer might not be exact due to rounding, but should still be the right answer[/quote]", "Solution_5": "Could someone provide an solution that leads to 42?" } { "Tag": [ "inequalities", "number theory", "relatively prime", "arithmetic series" ], "Problem": "p/q is a proper fraction where p and q are 3 digit numbers.how many pairs of p and q will satisfy the condition (p+1)/(q+3)<=1/2", "Solution_1": "Here's what I got:\r\n\r\n[hide]\n\nWe can cross multiply the inequality without worrying about cases because both the numerator and denominator can't be negative (or it would simplify to a positive fraction). So there are only two possibilities... only the numerator is negative, or both are positive. We can multiply the inequality by 2(q+3) in either case.\n\n(p+1) / (q+3) <= 1/2\n\n2p - q <= 1\n\nBecause p and q are both integers, the difference can't be less than 1.\n\n2p - q = 1\np = (q+1) / 2\n\nSo p and q are always relatively prime (using Euclid's algorithm).\n\nAlso, 100 <= q <= 999, so 51 <= (q+1) / 2 <= 500. \n\nSo there are 500 - 51 + 1 = 450 pairs of (p, q) that satisfy the original conditions.\n\n[/hide]", "Solution_2": "Because p and q are both integers, the difference can't be less than 1. \r\n\r\n\r\nBut I think the difference can be negative?", "Solution_3": "[quote=\"cheeesec4k3\"]\n\n2p - q <= 1\n\nBecause p and q are both integers, the difference can't be less than 1.\n[/quote]\r\n\r\nYou meant integers=naturals. Integer numbers are ...,-3,-2,-1,0,1,2,3,... \r\n\r\nIf you take $ p\\equal{}0,q\\equal{}0$ the difference is 0, so less than 1.", "Solution_4": "Eh, is there such fraction as $ 0/0$? :P \r\nBut I think the difference can be negative (take $ 123/457$ for example)", "Solution_5": "$ 0$ is not a 3-digit number.", "Solution_6": "[quote=\"mathwizarddude\"]Eh, is there such fraction as $ 0/0$? :P \n[/quote]\r\n\r\nThat's just undefined. LOL.", "Solution_7": "You know what I meant... 123/457 is such an example, I showed that p-2q can be less than 1, and someone obviously didn't read that :wink:", "Solution_8": "Wait, are you talking about me...?\r\n\r\nAnyway, whoops, I overlooked that. Hm... I'm not sure what to do with the p - 2q < 1 case... \r\nMaybe I should try a completely different approach.", "Solution_9": "Look at the variable in the numerator, p. Because you're \r\ngoing to add 1 to it, it will vary from 100 to 499. It cannot\r\nequal to 500 (or more), because you're interested in fractions\r\nwhose values are less than or equal to 1/2. \r\nThis is NOT TRUE.\r\n\r\n*** EDIT AT THE BOTTOM OF THE POST ***\r\n\r\nWhen p = 100, the smallest that q can be is 199, so that\r\nwhen 3 is added to it, and 1 added to p, you have the\r\nfraction:\r\n\r\n 101/202.\r\n\r\nKeeping p at 100, q can vary up to and including 996, so that\r\nq + 3 = 999.\r\n\r\nq goes from being 199 to 996.\r\n\r\nTo find the number of numbers of q, subtract and add 1:\r\n\r\n996 - 199 + 1 = 798 (numbers).\r\n\r\n---------------------------------------------------------------\r\n\r\nFor confirmation of a pattern, you might note the next \r\nfew lines for p = 101:\r\n\r\np + 1 becomes 102.\r\n\r\nThe variable in the denominator, q, can vary from 201 up to \r\nand including 996.\r\n\r\n996 - 201 + 1 = 796 (numbers).\r\n\r\n----------------------------------------------------------------\r\n\r\nSkip to the end:\r\n\r\np can be as large as 499 (included).\r\n\r\nThen (q + 3) can be either 998 or 999, so there are \r\ntwo choices for q: 995 or 996, respectively.\r\n\r\n{p = 100} + {p = 101} + ... + {p = 499}\r\n\r\nThe sum of this arithmetic series is the answer to the\r\noriginal problem:\r\n\r\n798 + 796 + . . . + 2 = ?\r\n\r\nLet n = # of terms\r\nLet d = common difference\r\n\r\nn = 499 - 100 + 1 = 400\r\n\r\nd = 2\r\n\r\n$ S_n \\equal{} n(a_1 \\plus{} a_n)/2$\r\n\r\n$ S_{400} \\equal{} 400(798 \\plus{} 2)/2$\r\n\r\n$ S_{400} \\equal{} 160,000$\r\n\r\n\r\nThere are 160,000 pairs p and q that satisfy the condition\r\n(p + 1)/(q + 3) <= 1/2, such that p/q is a proper fraction,\r\nand p and q are each 3-digit numbers.\r\n\r\n*** EDIT REFERENCED FROM ABOVE FOLLOWS *** \r\n\r\nPlease read the post by Mewto55555.\r\n\r\n160,000 + 1 (for that missing case of mine) = \r\n\r\n160,001 . . . the corrected answer", "Solution_10": "[quote=\"Arrange your tan\"]Look at the variable in the numerator, p. Because you're \ngoing to add 1 to it, it will vary from 100 to 499. It cannot\nequal to 500 (or more), because you're interested in fractions\nwhose values are less than or equal to 1/2.\n[/quote]\r\n\r\np=500,q=999" } { "Tag": [ "geometry", "3D geometry", "sphere", "trigonometry" ], "Problem": "How should charges sigma(theta) be distributed on the surface of a sphere of radius R for the field inside the sphere to be uniform and equal to E? Find also the field outside the sphere.[/code]", "Solution_1": "[hide=\"Answer\"]\n$ \\sigma \\equal{} \\sigma_{0} \\cos \\theta$... where $ \\theta$ is the azhimuthal angle\ni know of a non sense meaningless proof...but the actual proof uses heavy mathematics[/hide]", "Solution_2": "Whats Azimuthal Angle :?: \r\nBut i know of a not too difficult proof.May be you shud try it once more :!:", "Solution_3": "azimuthal means with the '$ z$' axis \r\nok is ur proof same as [url=http://www.mathlinks.ro/viewtopic.php?t=168923]here[/url]??" } { "Tag": [ "geometry" ], "Problem": "Taken [i]in verbatim[/i] from the Tournament of the Towns 2003 Fall Senior A-Level Paper:\r\n\r\nSorry, had to be taken down since some people have yet to take it", "Solution_1": "Are you sure that everyone that could have taken this thing took it!?!?\r\n-----------------------------------------------" } { "Tag": [ "quadratics", "number theory unsolved", "number theory" ], "Problem": "Consider number $ F\\equal{}2^{2^{n}}\\plus{}1,n\\in N^*$, satisfy $ (3^{\\frac{F\\minus{}1}{2}}\\plus{}1)\\vdots F$. Prove that F is a prime number", "Solution_1": "Pepins test : $ F$ is a fermat prime iff $ (3^{\\frac{F\\minus{}1}{2}}\\plus{}1)\\vdots F$\r\n\r\nThe if part : We observe that $ F\\minus{}1$ is the order of $ 3$, $ mod p$. Hence, $ F\\minus{}1$ should divide $ phi(F)$ where $ phi$ is the Euler's totient. Since $ phi(F)$ is at most $ F\\minus{}1$ with equality only when $ F$ is prime, we deduce that $ phi(F)\\equal{} F\\minus{}1$ implying F is prime\r\nFor the necessity part, quadratic reciprocity does the job" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$ a,$ $ b$ and $ c$ are non-negative numbers such that only one of them can be zero. Prove that:\r\n\\[ \\frac{a^2\\plus{}b^2\\minus{}2c^2}{\\sqrt{c^2\\plus{}ab}}\\plus{}\\frac{a^2\\plus{}c^2\\minus{}2b^2}{\\sqrt{b^2\\plus{}ac}}\\plus{}\\frac{b^2\\plus{}c^2\\minus{}2a^2}{\\sqrt{a^2\\plus{}bc}}\\geq0\\]", "Solution_1": "it is killed by ineq S.O.S or $ A(a\\minus{}b)^2\\plus{}B(b\\minus{}c)^2\\plus{}C(c\\minus{}a)^2$", "Solution_2": "[quote=\"arqady\"]$ a,$ $ b$ and $ c$ are non-negative numbers such that only one of them can be zero. Prove that:\n\\[ \\frac {a^2 \\plus{} b^2 \\minus{} 2c^2}{\\sqrt {c^2 \\plus{} ab}} \\plus{} \\frac {a^2 \\plus{} c^2 \\minus{} 2b^2}{\\sqrt {b^2 \\plus{} ac}} \\plus{} \\frac {b^2 \\plus{} c^2 \\minus{} 2a^2}{\\sqrt {a^2 \\plus{} bc}}\\geq0\n\\]\n[/quote]\r\nIf $ a,b,c$ are sides of triangle then the inequality is true by Chebysev inequality:\r\n\\[ a^2 \\plus{} b^2 \\minus{} 2c^2 \\ge a^2 \\plus{} c^2 \\minus{} 2b^2 \\ge b^2 \\plus{} c^2 \\minus{} 2a^2 \\\\ c^2 \\plus{} ab \\le b^2 \\plus{} ac \\le a^2 \\plus{} bc\r\n\\]\r\nfor $ a \\ge b \\ge c \\ge 0$.\r\nAssume $ a \\ge b\\plus{}c$, then $ c^2 \\plus{} ab \\le a^2 \\plus{} bc,b^2 \\plus{} ac \\le a^2 \\plus{} bc$ and $ a^2 \\plus{} b^2 \\minus{} 2c^2 \\ge a^2 \\plus{} c^2 \\minus{} 2b^2 >0$, hence\r\n\\[ LHS \\ge \\frac {a^2 \\plus{} b^2 \\minus{} 2c^2}{\\sqrt {a^2 \\plus{} bc}} \\plus{} \\frac {a^2 \\plus{} c^2 \\minus{} 2b^2}{\\sqrt {a^2 \\plus{} bc}} \\plus{} \\frac {b^2 \\plus{} c^2 \\minus{} 2a^2}{\\sqrt {a^2 \\plus{} bc}} \\equal{}0\r\n\\].", "Solution_3": "[quote=\"ehku\"]\nfor $ a \\ge b \\ge c \\ge 0$.\nAssume $ a \\ge b \\plus{} c$, then $ c^2 \\plus{} ab \\le a^2 \\plus{} bc,b^2 \\plus{} ac \\le a^2 \\plus{} bc$ and $ a^2 \\plus{} b^2 \\minus{} 2c^2 \\ge a^2 \\plus{} c^2 \\minus{} 2b^2 > 0$[/quote]\r\nWhy $ a^2 \\plus{} c^2 \\minus{} 2b^2 \\geq 0$ $ ?$ Thank you!", "Solution_4": "I'm sorry, it is a mistake :oops: \r\nThe correct is:\r\nWe needn't consider the case $ a,b,c$ are sides of a triangle.\r\nAssume $ a \\ge b \\ge c \\ge 0$. \r\nIf $ a^2\\plus{}c^2\\minus{}2b^2 \\ge 0$ then \r\n\\[ LHS \\ge \\frac {a^2 \\plus{} b^2 \\minus{} 2c^2}{\\sqrt {a^2 \\plus{} bc}} \\plus{} \\frac {a^2 \\plus{} c^2 \\minus{} 2b^2}{\\sqrt {a^2 \\plus{} bc}} \\plus{} \\frac {b^2 \\plus{} c^2 \\minus{} 2a^2}{\\sqrt {a^2 \\plus{} bc}} \\equal{} 0\r\n\\]\r\nIf $ a^2\\plus{}c^2\\minus{}2b^2 \\le 0$, because $ a^2 \\plus{} b^2 \\minus{} 2c^2 \\ge b^2 \\plus{} c^2 \\minus{} 2a^2$ and $ \\sqrt {c^2 \\plus{} ab} \\le \\sqrt {a^2 \\plus{} bc}$, by Chebysev inequality we get\r\n\\[ LHS \\ge (a^2 \\plus{} b^2 \\minus{} 2c^2\\plus{}b^2 \\plus{} c^2 \\minus{} 2a^2)\\left(\\frac {1}{\\sqrt {c^2 \\plus{} ab}}\\plus{}\\frac {1}{\\sqrt {a^2 \\plus{} bc}}\\right) \\plus{} \\frac {a^2 \\plus{} c^2 \\minus{} 2b^2}{\\sqrt {b^2 \\plus{} ac}} \\\\ \\equal{}(2b^2\\minus{}a^2\\minus{} c^2)\\left(\\frac {1}{\\sqrt {c^2 \\plus{} ab}}\\plus{}\\frac {1}{\\sqrt {a^2 \\plus{} bc}}\\minus{}\\frac {1}{\\sqrt {b^2 \\plus{} ac}}\\right) \\ge 0 \r\n\\]\r\nbecause $ \\frac {1}{\\sqrt {c^2 \\plus{} ab}}\\plus{}\\frac {1}{\\sqrt {a^2 \\plus{} bc}} \\ge \\frac{2}{\\sqrt[4]{(c^2 \\plus{} ab)(a^2 \\plus{} bc)}} \\ge \\frac {1}{\\sqrt {b^2 \\plus{} ac}}$." } { "Tag": [ "algebra", "polynomial", "binomial theorem" ], "Problem": "This is a problem in my algebra 2 book. I understand it but there is a confusion that I have with it. \r\n\r\n$\\sum_{k=0}^{5} \\binom{5}{k} \\cdot (-1)^{k} \\cdot x^{5-k} \\cdot 3^{k} = 32$\r\nthis is an obviously the binomial theorem so it would be\r\n\r\n$(x-3)^5 = 32$\r\n$x-3 = 2$\r\n$x=5$\r\nThis is the answer that they give in the back of the book. The problem is: when you do that, it creates a 5th degree polynomial, so there are 5 solutions for x, not just 1. Does this make the book wrong or what?", "Solution_1": "The other solutions are imaginary. If your Alg II book is like mine it is only looking for solutions in the reals, unless it mentions specifically imaginaries.", "Solution_2": "The imaginary solutions occur when you take the 5th root. \r\n\r\nNote that the other four solutions are:\r\n\r\nx=2(cos 72k+i sin 72k) k=1,2,3,4.\r\n\r\nwhere the sin are in degrees." } { "Tag": [ "AMC" ], "Problem": "How many of the AoPS members are CAMC Mathematics Book Award winners?", "Solution_1": "Nobody?\r\n\r\nCAMC Mathematics Book Award \"is given to the top-scoring student who took the AMC 12 in each region (http://www.unl.edu/amc/f-miscellaneous/regions.html) in grade 9 or less.\"", "Solution_2": "In 10th, 11th, and 12th grades, I got books for doing well on the AMC tests. In 11th grade, I think it was sponsored by CAMC, and in the other years, I think there was a different sponsor. I'm not sure though." } { "Tag": [ "calculus", "algebra unsolved", "algebra" ], "Problem": "$R$ is the reals. $f : R\\rightarrow R$ is continuous and for any $a> 0$, $lim_n\\rightarrow$$ \\infty f(na) = 0$. Prove $lim_x\\rightarrow \\infty f(x) = 0.$", "Solution_1": "This has been discussed [b]a lot[/b]. Try searchoing the \"Calculus\" section" } { "Tag": [ "trigonometry" ], "Problem": "Verify this identity: \r\n\r\ncot (x/2) = (sin x) / (1 - cos x)", "Solution_1": "$ \\cot\\left(\\frac x2\\right)\\equal{}\\frac{\\cos\\left(\\frac x2\\right)}{\\sin\\left(\\frac x2\\right)}$\r\n\r\n$ \\equal{}\\frac{\\frac{\\sqrt{1\\plus{}\\cos x}}2}{\\frac{\\sqrt{1\\minus{}\\cos x}}2}\\equal{}\\sqrt{\\frac{1\\plus{}\\cos x}{1\\minus{}\\cos x}}$\r\n\r\n$ \\equal{}\\sqrt{\\frac{(1\\plus{}\\cos x)(1\\minus{}\\cos x)}{(1\\minus{}\\cos x)^2}}\\equal{}\\sqrt{\\frac{\\sin^2x}{(1\\minus{}\\cos x)^2}}$\r\n\r\n$ \\equal{}\\frac{\\sin x}{1\\minus{}\\cos x}$, as desired (notice that we never had to worry about any sign ($ \\pm$) problems--try it and see).", "Solution_2": "You sort of do. It's nicer to prove the half-angle identities in the other direction: let $ t \\equal{} \\tan \\frac {x}{2}$. Prove the identities\r\n\r\n$ \\cos x \\equal{} \\frac {1 \\minus{} t^2}{1 \\plus{} t^2}$\r\n$ \\sin x \\equal{} \\frac {2t}{1 \\plus{} t^2}$\r\n\r\nThis identity is a straightforward consequence." } { "Tag": [ "ratio", "geometry", "incenter", "rotation" ], "Problem": "An equilateral triangle $ \\bigtriangleup ABC$ is inscribed inside a circle with center $ O$. Circle $ O$ is inscribed in an equilateral triangle $ \\bigtriangleup XYZ$. \r\n\r\nWhat is the ratio between sides $ AB$ to $ XY$ and what is the ratio between the area of $ \\bigtriangleup ABC$ to $ \\bigtriangleup XYZ$?", "Solution_1": "[hide]Since Circle $ O$ is inscribed in $ \\bigtriangleup ABC$, $ O$ is the incenter of the triangle and lies on the incenter of the triangle. Since in an equilateral triangle, all four types of cevians meet in the same place, this is also the centroid of the triangle. If let the side of $ \\bigtriangleup ABC$ equal $ 1$, the height of the triangle is $ \\sqrt 3/2$ and the radius of the circle is $ \\sqrt 3/3$. We now have a triangle circumscribed around this circle. The height of this triangle is $ \\sqrt3$. Then, since the two triangles are similar (they're both equilateral) the ratio of $ AB$ to $ XY$ is the ratio of the heights or $ 1/2$. The ratio of areas is the square of this, $ 1/4$[/hide]", "Solution_2": "Err one thing: $ \\bigtriangleup ABC$ is inscribed circle $ O$, not circle $ O$ inscribed in $ \\bigtriangleup ABC$...", "Solution_3": "[hide=\"Simple solution\"]\nWhat we have is this:\n\n[geogebra]40d4d6ec2ef79020bb96e96a95aca784d631e3e0[/geogebra] \n\nwith ABC being the smaller triangle and XYZ being the larger.\n\nNow, rotate the triangle ABC by 60 degrees with respect to its centroid.\n\nThen, we will have:\n\n [geogebra]1383f4c388fae2a52ec5c0678aff56616fb6a6a5[/geogebra] \n\nObviously, the area of ABC is $ \\frac {1}{4}$ of the area of XYZ.\n\nFurthermore, the ratio between sides AB to XY will be $ \\sqrt {\\frac {1}{4}} \\equal{} \\frac {1}{2}$.[/hide]" } { "Tag": [], "Problem": "Let $ ABCD$ be a square, consider a point $ E$ on segment $ AB$, lets call $ < CEB \\equal{} x$.\r\n$ F$ is a point on segment $ DA$ such that $ < FEC \\equal{} x$.\r\n\r\nProove that $ EF \\equal{} BE \\plus{} DF$\r\n\r\n\r\n :)", "Solution_1": "[quote=\"ElChapin\"]Let $ ABCD$ be a square, consider a point $ E$ on segment $ AB$, lets call $ < CEB \\equal{} x$.\n$ F$ is a point on segment $ DA$ such that $ < FEB \\equal{} x$.\n\nProove that $ EF \\equal{} BE \\plus{} DF$\n\n\n :)[/quote]\r\nHow is this construction possible? $ CEB$ is acute while $ FEB$ is obtuse...", "Solution_2": "Sorry I changed a vertex, it is correct now.\r\n\r\n :)", "Solution_3": "[hide=\"Hint\"]Construct $ G\\in EF$ such that $ EG\\equal{}EB$ and play with triangle congruency.[/hide]" } { "Tag": [ "trigonometry", "calculus", "integration", "function", "calculus computations" ], "Problem": "$ \\sum\\limits_{k \\equal{} 1}^\\infty {\\sin \\frac {1}{{k^2 }}}$\r\n\r\nMy book says convergent and I proved it was convergent but Mathematica 6 does not give me the limit. What does it converge to?\r\n\r\nEDITED", "Solution_1": "Mathematica is wrong.", "Solution_2": "I changed the variable of integration and now as an 'answer' mathematica just writes what I entered.\r\n\r\nIs there a known limit for this series?", "Solution_3": "Doubtful. Using Taylor series, for example, we get a series in terms of $ \\zeta(2k), k \\equal{} 1, 2, ...$, which can be expressed in terms of the Bernoulli numbers, but that's just another series.", "Solution_4": "[quote=\"\u00a7outh\u00a7tar\"]Is there a known limit for this series?[/quote]\r\nI don't know, but I don't see any reason to suspect there is any \"known\" answer.\r\n\r\nConvergence tests for series (such as limit comparison or whatever you used on this one) derive from the theorem that says that a bounded monotone sequence converges. This theorem comes from - in fact is equivalent to - the completeness of the real numbers. I'll give the following crude paraphrase of the completeness of the real numbers: for every limiting process that looks like it ought to converge, there is a real number there to be the limit.\r\n\r\nThis means that by using limits, or sums, or integrals, or solutions of equations, we can express numbers for which we had no previous names, or even create functions (via power series or trigonometric series) for which we had no previous names. I consider this to be a feature, not a bug.\r\n\r\nPut yet another way, there are more real numbers than there are \"nice\" names for real numbers. There is a real number that is the sum of $ \\sum_{k \\equal{} 1}^{\\infty}\\sin\\left(\\frac1{x^2}\\right),$ but it is possible, even likely, that the simplest name for this real number is \"the real number that is the sum of that particular series.\" In effect, that is what Mathematica was telling you. Or, what is the solution to the equation $ x \\equal{} \\cos x$? We can prove there is a solution (Intermediate Value Theorem), but the simplest name for the resulting number is \"the unique real number $ x$ such that $ x \\equal{} \\cos x.$\"\r\n\r\nNote that we should have no trouble getting good numerical approximations to these numbers. A few minutes with a spreadsheet gives me this:\r\n\r\n$ \\sum_{k \\equal{} 1}^{\\infty}\\sin\\left(\\frac1{k^2}\\right)\\approx 1.483522817$\r\n\r\nI'm sure Mathematica could do the same.", "Solution_5": "[quote=\"Kent Merryfield\"]Or, what is the solution to the equation $ x \\equal{} \\sin x$?[/quote]\r\n\r\n$ x \\equal{} 0$? :wink:\r\n\r\n[color=green][Now it says what I meant it to say. OK? - K.M.][/color]", "Solution_6": ":lol: @ KBL\r\n\r\nThanks everyone. Makes it a little harder if you're asked to determine if a series converges to a 'known quantity' because until you find out you won't know that it does." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Find all real $ a_0$ such that the sequence $ a_0, a_1, a_2, ...$ defined by $ a_{n \\plus{} 1} \\equal{} 2^n \\minus{} 3a_n$ has $ a_{n \\plus{} 1} > a_n$ for all $ n\\geq 0$.", "Solution_1": "$ a_{n \\plus{} 1} \\equal{} \\minus{} 3a_n \\plus{} 2^n\\Longleftrightarrow a_{n \\plus{} 1} \\minus{} \\frac {1}{5}\\cdot 2^{n \\plus{} 1} \\equal{} \\minus{} 3\\left(a_n \\minus{} \\frac {1}{5}\\cdot 2^n\\right)$, yielding $ a_n \\equal{} \\frac {1}{5}\\cdot 2^n \\plus{} ( \\minus{} 3)^n\\left(a_0 \\minus{} \\frac {1}{5}\\right)$\r\n\r\n$ \\therefore a_{n \\plus{} 1} \\minus{} a_n \\equal{} \\frac {1}{5}\\cdot 2^n \\minus{} 4( \\minus{} 3)^n\\left(a_0 \\minus{} \\frac {1}{5}\\right) > 0.$", "Solution_2": "$ a_n\\equal{}2^{n\\minus{}1}\\minus{}3a_{n\\minus{}1}\\equal{}2^{n\\minus{}1}\\minus{}3*2^{n\\minus{}2}\\plus{}3^2a_{n\\minus{}2}\\equal{}...\\equal{}2^{n\\minus{}1}\\minus{}3*2^{n\\minus{}2}\\plus{}3^2*2^{n\\minus{}3}\\minus{}...\\plus{}(\\minus{}3)^{n\\minus{}1}\\minus{}(\\minus{}3)^na_0\\equal{}\\frac{2^n\\plus{}3^n}{5}\\plus{}(\\minus{}3)^na_0$.\r\nTherefore from $ a_{n\\plus{}1}>a_n$ we get $ |a_0|\\le \\frac{1}{10}.$" } { "Tag": [ "analytic geometry", "geometry", "geometric transformation", "reflection" ], "Problem": "The point $(4, 3)$ is reflected over the $x$-axis and then over the $y$-axis. What is the sum of the coordinates of the new point?", "Solution_1": "[hide]To reflect over the x-axis, we just multiply the y-coordinate by -1. To do the y-axis, we reverse the formula - we multiply the x by -1. Therefore, the new point have coordinates (-3,-4) for a sum of -7.[/hide]\r\n\r\nBilly", "Solution_2": "[quote=\"solafidefarms\"][hide]To reflect over the x-axis, we just multiply the x-coordinate by -1. To do the y-axis, we do the same. Therefore, the new point have coordinates (-4,-5) for a sum of -9.[/hide]\n\nBilly[/quote]\r\nSlow down a little, and think about the problem before you post. I'm sure you know how to do this problem, but the rule you gave for reflecting over the x-axis isn't correct. And your final answer isn't correct, either.", "Solution_3": "You mean -7, right??? Negate the X and Y values to get -4 and -3, and sum them to get -7. How did someone get -9? Maybe you got the wrong numbers? Sorry if i seem causic... im jw...", "Solution_4": "[hide]it is reflected of the line y=-x\n\nthe point is (-4,-3)\n\n-4-3=-7[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c,p,q$ be positive real numbers. Prove that\r\n\r\n$ \\left.\\frac{a^{p\\minus{}q}b^qc^q}{b^{p\\plus{}q}\\plus{}c^{p\\plus{}q}\\plus{}a^{p\\minus{}q}b^qc^q}\\plus{}\\frac{a^qb^{p\\minus{}q}c^q}{a^{p\\plus{}q}\\plus{}c^{p\\plus{}q}\\plus{}a^qb^{p\\minus{}q}c^q}\\plus{}\\frac{a^qb^qc^{p\\minus{}q}}{a^{p\\plus{}q}\\plus{}b^{p\\plus{}q}\\plus{}a^qb^qc^{p\\minus{}q}}\\leq1\\right.$", "Solution_1": "[quote=\"marin.bancos\"]Let $ a,b,c,p,q$ be positive real numbers. Prove that\n\n$ \\left.\\frac {a^{p \\minus{} q}b^qc^q}{b^{p \\plus{} q} \\plus{} c^{p \\plus{} q} \\plus{} a^{p \\minus{} q}b^qc^q} \\plus{} \\frac {a^qb^{p \\minus{} q}c^q}{a^{p \\plus{} q} \\plus{} c^{p \\plus{} q} \\plus{} a^qb^{p \\minus{} q}c^q} \\plus{} \\frac {a^qb^qc^{p \\minus{} q}}{a^{p \\plus{} q} \\plus{} b^{p \\plus{} q} \\plus{} a^qb^qc^{p \\minus{} q}}\\leq1\\right.$[/quote]\r\n\r\n\r\n\r\nI think it can prove by Muirhead theorem.\r\n\r\n$ \\left.\\frac {a^{p \\minus{} q}b^qc^q}{b^{p \\plus{} q} \\plus{} c^{p \\plus{} q} \\plus{} a^{p \\minus{} q}b^qc^q} \\plus{} \\frac {a^qb^{p \\minus{} q}c^q}{a^{p \\plus{} q} \\plus{} c^{p \\plus{} q} \\plus{} a^qb^{p \\minus{} q}c^q} \\plus{} \\frac {a^qb^qc^{p \\minus{} q}}{a^{p \\plus{} q} \\plus{} b^{p \\plus{} q} \\plus{} a^qb^qc^{p \\minus{} q}}\\leq1\\right.$\r\n\r\n$ \\Longleftrightarrow \\sum_{sym}(a^{2p\\plus{}2q}b^{p\\plus{}q}\\minus{}a^{2p\\plus{}q}b^pc^{2q})\\geq 0$", "Solution_2": "Maybe... My solution doesn't use Muirhead's theorem.", "Solution_3": "I think the solution should use the following inequality\r\n\r\n\\[ b^{p\\plus{}q}\\plus{}c^{p\\plus{}q} \\ge (bc)^{qq}(b^{p\\minus{}q}\\plus{}c^{p\\minus{}q})\\]", "Solution_4": "Dear [b]great math[/b],\r\nThere is a little typo in your inequality. The right form of it is the following\r\n\r\n$ b^{p \\plus{} q} \\plus{} c^{p \\plus{} q}\\ge (bc)^{q}(b^{p \\minus{} q} \\plus{} c^{p \\minus{} q})$\r\n\r\nCould you give a solution to the proposed inequality? \r\n\r\n[b]REMARK[/b]\r\nUsing particular values for $ p$ and $ q$ we get interesting forms for the proposed inequality.\r\nJust an example. For $ p \\equal{} 2, q \\equal{} 1$ this inequality is equivalent to\r\n\r\n$ \\left.\\frac {1}{b^3 \\plus{} c^3 \\plus{} abc} \\plus{} \\frac {1}{c^3 \\plus{} a^3 \\plus{} abc} \\plus{} \\frac {1}{a^3 \\plus{} b^3 \\plus{} abc} \\leq \\frac {1}{abc}\\right.$\r\n\r\nThe problem 2.29 from the excellent book [b]INEQUALITIES WITH BEAUTIFUL SOLUTIONS[/b], written by [b]Vasile Cirtoaje, Vo Quoc Ba Can [/b]and [b]Tran Quoc Anh[/b], has the following statement\r\n\r\nFor $ a,b,c$ positive real numbers, prove that\r\n\r\n$ \\left.\\frac {a^3}{a^3 \\plus{} b^3 \\plus{} abc} \\plus{} \\frac {b^3}{b^3 \\plus{} c^3 \\plus{} abc} \\plus{} \\frac {c^3}{c^3 \\plus{} a^3 \\plus{} abc} \\geq 1\\right.$", "Solution_5": "No ideas ? You can use the [b]great math[/b]'s hint.", "Solution_6": "$ \\left.(b^p\\minus{}c^p)(b^q\\minus{}c^q)\\geq0\\ \\ \\Leftrightarrow\\ \\ b^{p\\plus{}q}\\plus{}c^{p\\plus{}q}\\geq b^pc^q\\plus{}b^qc^p\\ \\ \\Leftrightarrow\\ \\ b^{p\\plus{}q}\\plus{}c^{p\\plus{}q}\\geq b^qc^q(b^{p\\minus{}q}\\plus{}c^{p\\minus{}q})\\ \\ \\Leftrightarrow\\right.$\r\n$ \\left.\\Leftrightarrow\\ \\ b^{p\\plus{}q}\\plus{}c^{p\\plus{}q}\\plus{}a^{p\\minus{}q}b^qc^q\\geq b^qc^q(a^{p\\minus{}q}\\plus{}b^{p\\minus{}q}\\plus{}c^{p\\minus{}q})\\ \\ \\Leftrightarrow\\ \\ \\frac{b^qc^q}{b^{p\\plus{}q}\\plus{}c^{p\\plus{}q}\\plus{}a^{p\\minus{}q}b^qc^q}\\leq \\frac{1}{a^{p\\minus{}q}\\plus{}b^{p\\minus{}q}\\plus{}c^{p\\minus{}q}}\\ \\Leftrightarrow\\ \\right.$\r\n$ \\left.\\Leftrightarrow\\ \\ \\frac{a^{p\\minus{}q}b^qc^q}{b^{p\\plus{}q}\\plus{}c^{p\\plus{}q}\\plus{}a^{p\\minus{}q}b^qc^q}\\leq \\frac{a^{p\\minus{}q}}{a^{p\\minus{}q}\\plus{}b^{p\\minus{}q}\\plus{}c^{p\\minus{}q}}\\right.$\r\n\r\nSimilarly, we get:\r\n \r\n$ \\frac{a^{q}b^{p\\minus{}q}c^{q}}{a^{p\\plus{}q}\\plus{}c^{p\\plus{}q}\\plus{}a^{q}b^{p\\minus{}q}c^{q}}\\leq \\frac{b^{p\\minus{}q}}{a^{p\\minus{}q}\\plus{}b^{p\\minus{}q}\\plus{}c^{p\\minus{}q}}$\r\n\r\n$ \\frac{a^{q}b^{q}c^{p\\minus{}q}}{a^{p\\plus{}q}\\plus{}b^{p\\plus{}q}\\plus{}a^{q}b^{q}c^{p\\minus{}q}}\\leq \\frac{c^{p\\minus{}q}}{a^{p\\minus{}q}\\plus{}b^{p\\minus{}q}\\plus{}c^{p\\minus{}q}}$\r\n\r\nTherefore, adding them up we get the desired inequality.\r\n\r\n[b]What do you think about this problem?[/b]" } { "Tag": [ "function", "counting", "distinguishability", "symmetry", "combinatorics unsolved", "combinatorics" ], "Problem": "Prove that \r\n\r\n\\[ 2^n \\binom{n}{n} + 2^{n-1} \\binom{n+1}{n} + 2^{n-2} \\binom{n+2}{n} + \\ldots + 2 \\binom{2n-1}{n} + \\binom{2n}{n} = 2^{2n} . \\]\r\n\r\n...", "Solution_1": "Here is an algebraic solution. $2^{2n-k}\\binom{k}{n}$ equals to the coefficient of $x^n$ in $2^n (x+\\frac12)^k$. Therefore, the sum in the left hand side of the identity equals the coefficient of $x^n$ in \r\n\\[ 2^n \\sum_{k=0}^{2n} (x + \\frac12)^k = 2^n \\frac{1 - (x+\\frac12)^{2n+1}}{\\frac12 - x} = 2^{n+1} \\left(\\frac{1}{1-2x} - \\frac{(x+\\frac12)^{2n+1}}{1-2x}\\right) \\]\r\n(The summands for $k2003]\n\nn = 2 [p_22003]\n\n..\n\nn = 2003 [p_20032003]\n\n\n\nHas a solution by the Chineese Remainder Theorem.\n\n\n\nThen n-2003,n-2002,..,n-1 are 2003 successive positive integers, each of them is divisible by the 2003th power of a prime number. :mrgreen: [/hide]" } { "Tag": [ "function", "algebra", "polynomial", "logarithms", "calculus", "integration", "inequalities" ], "Problem": "Let $P(n)$ be the number of functions $f: \\mathbb{R} \\to \\mathbb{R}$, $f(x)=a x^2 + b x + c$, with $a,b,c \\in \\{1,2,\\ldots,n\\}$ and that have the property that $f(x)=0$ has only integer solutions. Prove that $n29 (i found this bound using Maple). For lower n u must try it or estimate better." } { "Tag": [ "real analysis", "geometry", "geometric transformation", "invariant", "real analysis unsolved" ], "Problem": "Hi \r\n\r\nI need to give an example of a subset of reals whose length cannot be defined. I was thinking about the set S of rational numbers from zero to one which is not composed of a finite number of intervals, and so no length is defined for it. But as a hint we have that we should use the axiom of choice and that length of (S+a)=length of (S) Could anyone tell me how to apply it to my example if it is correct.\r\n\r\nThank you\r\nRomy", "Solution_1": "http://en.wikipedia.org/wiki/Vitali_set\r\n\r\nEdit: The rationals have Lebesgue measure zero.", "Solution_2": "Thank you for the link. I had seen it but we have not covered measures and Lebesque yet. We merely finished Riemann and gauge integrals.\r\n\r\nDo you have any proof using these concepts.\r\n\r\nThank you", "Solution_3": "The problem with that: what is \"length\"? Trying to define that inevitably leads to Lebesgue measure.", "Solution_4": "Not really, this question can be answered with a much more heuristic approach than that. If we suppose that we to every subset $A$ of the real line can assign a length $l(A) \\ge 0$ in such a manner that:\r\n\r\ni) $l([a,b]) = b-a$\r\nii) $l$ is monotone\r\niii) $l$ is $\\sigma$-additive\r\niv) $l$ is translation-invariant\r\n\r\nThen by the usual construction of a non-measurable set you get a contradiction." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all value n $\\geq$ 3 such that a1,a2,...,an in R: a1+a2+..+an=0 we have:\r\n a1a2+a2a3+...+ana1 $\\leq$ 0", "Solution_1": "For n > 5, if n is even, we can let the first half be 1, the rest be -1\r\nif n is odd, let a2 be 2, the first half, rounded down, exept a2, be 1, the rest be -1\r\nSo the only possible choises are 3,4,5" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove for all $x+y+z=1$ that\r\n\\[\\frac{xy}{\\sqrt{xy+yz}}+\\frac{yz}{\\sqrt{yz+zx}}+\\frac{zx}{\\sqrt{zx+xy}}\\le\\frac{1}{\\sqrt{2}}\\].\r\n\r\nI have worked on it for some time, but I have found that I am... rather bad at inequalities. (very bad. :ninja: )\r\n\r\nI also used Cauchy in more than one way and it didn't work either... maybe I didn't try hard enough?", "Solution_1": "Considering that it's only cyclic and that equality holds for not only (1/3,1/3,1/3) but also (1/2,1/2,0), not getting this problem does not mean you're very bad at inequalities. (And of course, the fact that it was on a Black test... Last summer, I don't know if I would have gotten [b]any[/b] of them right :roll: )\r\n[hide]\nBy Cauchy's Inequality,\n\\[\\left( \\sum_\\text{cyc}\\frac{xy}{\\sqrt{xy+yz}}\\right)^{2}\\leq \\left( \\sum_\\text{cyc}xy(xy+xz)\\right) \\left( \\sum_\\text{cyc}\\frac{xy}{(xy+yz)(xy+xz)}\\right) \\]\nShowing that the right hand side is at most 1/2 is equivalent to\n\\[4(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}+xyz(x+y+z))\\leq (x+y+z)(x+y)(y+z)(z+x) \\]\nwhich follows from\n\\[xy(x-y)^{2}+yz(y-z)^{2}+zx(z-x)^{2}\\geq 0 \\]\n[/hide]", "Solution_2": "Thanks!\r\n\r\nI think the problem was that I replaced $x+y$ with $1-z$ wherever I saw it, kinda confusing the issue :oops: \r\n\r\nAgain, thanks a lot!\r\n\r\n(btw how did you think of using $(xy+yz)^{2}$ in the denominator? is it because it's a square?)\r\n\r\nEDIT: oh, haha, I read $xz$ as $yz$ :oops: \r\n\r\nThanks again!", "Solution_3": "Huh? There isn't $(xy+yz)^{2}$ anywhere in my solution... [hide=\"stuff related to my solution\"]I think that writing\n\\[\\frac{x^{2}y^{2}}{xy+yz}=xy(xy+xz)\\cdot \\frac{xy}{(xy+yz)(xy+xz)}\\]\nin order to use Cauchy might seem mysterious at first, but it has good reasons. The $xy$ appears in both factors to make the application of Cauchy strong enough (if it messes up the (1/2,1/2,0) equality case, it will [b]FAIL[/b]), and the factor of $xy+xz$ appears to turn the cyclic inequality into a symmetric one. (Symmetric inequalities are much easier to deal with.)[/hide]", "Solution_4": "[quote=\"not_trig\"]Prove for all $x+y+z=1$ that\n\\[\\frac{xy}{\\sqrt{xy+yz}}+\\frac{yz}{\\sqrt{yz+zx}}+\\frac{zx}{\\sqrt{zx+xy}}\\le\\frac{1}{\\sqrt{2}}\\]\n[/quote]\r\n$x,$ $y$ and $z$ are positive numbers. :wink: \r\nSee also here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=88439" } { "Tag": [], "Problem": "All worgs are either green, purple, or red. All worgs either have one horn or two horns. Those with one horn are either big or small and always have three eyes. Those with three horns are always small. They either have three eyes or one eye. \r\n\r\nNO GENETIC MUTATIONS!!! \r\n\r\nHow many different types of worgs are there if different types are determined by color, number of horns, size, or number of eyes?", "Solution_1": "[quote=\"game.guy\"]All worgs are either green, purple, or red. All worgs either have one horn or two horns. Those with one horn are either big or small and always have three eyes. Those with three horns are always small. They either have three eyes or one eye. \n\nNO GENETIC MUTATIONS!!! \n\nHow many different types of worgs are there if different types are determined by color, number of horns, size, or number of eyes?[/quote]\r\nWait, you said \"All worgs have either one horn or two horns.\" Then you said \" Those with three horns are always small\". Is it a trick question or a typo?", "Solution_2": "Yeah, i wasnt too sure about that either...\r\nAssuming that it was a typo, i got 12 worgs total.\r\njust get two cases, one case is where the worg has 1 horn, the other is where the worg has 2 horns.\r\nCase 1: the worg has 1 horn\r\nthere are $ 3$ choices for the color of the worg and the worg has two choices for size so the number of worgs with 1 horn is $ 3*2\\equal{}6$\r\nCase 2: the worg has 2 horns\r\nagain, there are $ 3$ choices for the color, and $ 2$ choices for eye number so the total number of worgs with 2 horns is $ 3*2\\equal{}6$\r\n$ 6\\plus{}6\\equal{}12$ worgs total\r\ni think...", "Solution_3": "Those with two horns I meant to say.\r\n\r\nThe answer was twelve by the way. Good job." } { "Tag": [ "trigonometry" ], "Problem": "One leg of a right triangle is 12 inches, and the measure of the\nangle opposite that leg is $ 30^\\circ$. What is the number of inches in\nthe hypotenuse of the triangle?", "Solution_1": "No, it is not a 5,12,13 triangle.\r\n\r\nBy special triangles (and drawing a diagram, or by trig), the hypotenuse of a 30,60,90 triangle is 2 times the length of the leg that is opposite the 30 degree angle. Thus, our answer is $ \\boxed{24}$." } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "Find $u(x,t)$ such that\r\n\\[\\begin{cases}u_{tt}=k(x,t)u_{xx}+f(u,u')+g(x,t), x\\in [0,1], t>0,\\\\ u_{x}(0,t)=h_{1}(u_{t}(0,t)),u_{x}(1,t)=h_{2}(u_{t}(1,t)),\\\\ u(x,0)=\\alpha(x), u_{t}(x,0)=\\beta(x),\\end{cases}\\]\r\nwhere $k,g,h_{1},h_{2},\\alpha,\\beta$ are given and satisfy some conditions which you can estalish by yourself. The weaker conditions, the more interesting your work is.\r\n\r\nThanks", "Solution_1": "The form of this system suggests to me that you need to deepen your understanding of the PDE theory before trying to contribute to it.", "Solution_2": "[quote=\"mlok\"]The form of this system suggests to me that you need to deepen your understanding of the PDE theory before trying to contribute to it.[/quote]\r\n\r\nThanks. I'd like to study the Hadamard well-posedness for this system, but I face some difficulties. Could you help me?", "Solution_3": "Please, use Galerkin's Approximation to deal with this.\r\n\r\nFor this problem, it is unable to give the exact solution.", "Solution_4": "[quote=\"NguyenThanh_545\"]Please, use Galerkin's Approximation to deal with this.\n\nFor this problem, it is unable to give the exact solution.[/quote]\r\n\r\nBut too hard to get the result", "Solution_5": "You should treat when $ h_{i}(z)=z$ first." } { "Tag": [], "Problem": "\u0395\u03c3\u03c4\u03c9 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf $ ABC$ \u03ba\u03b1\u03b9 \u03c4\u03c1\u03b5\u03b9\u03c2 //\u03bb\u03b5\u03c2 \u03b5\u03c5\u03b8\u03b5\u03b9\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03b5\u03c1\u03bd\u03b1\u03bd\u03b5 \u03b1\u03c0\u03bf \u03c4\u03b1 $ A,B,C$. \u039d\u0394\u039f \u03bf\u03b9 \u03b9\u03c3\u03bf\u03b3\u03c9\u03bd\u03b9\u03b5\u03c2 \u03b1\u03c5\u03c4\u03c9\u03bd \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b9\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03b5\u03c2 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf\u03c5 \r\n\u03b4\u03b9\u03b5\u03c1\u03c7\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b1\u03c0\u03bf \u03c4\u03bf \u03b9\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03b9\u03bf.", "Solution_1": "\u03a3\u03c5\u03b3\u03b3\u03bd\u03ce\u03bc\u03b7, \u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae\u03c2 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03ac\u03bb\u03bb\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae; \u0388\u03c7\u03b5\u03b9 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03c4\u03b1 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c3\u03b5 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf;", "Solution_2": "[quote=\"Sakis\"]\u0395\u03c3\u03c4\u03c9 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf $ ABC$ \u03ba\u03b1\u03b9 \u03c4\u03c1\u03b5\u03b9\u03c2 //\u03bb\u03b5\u03c2 \u03b5\u03c5\u03b8\u03b5\u03b9\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03b5\u03c1\u03bd\u03b1\u03bd\u03b5 \u03b1\u03c0\u03bf \u03c4\u03b1 $ A,B,C$. \u039d\u0394\u039f \u03bf\u03b9 \u03b9\u03c3\u03bf\u03b3\u03c9\u03bd\u03b9\u03b5\u03c2 \u03b1\u03c5\u03c4\u03c9\u03bd \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b9\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03b5\u03c2 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf\u03c5 \n\u03b4\u03b9\u03b5\u03c1\u03c7\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b1\u03c0\u03bf \u03c4\u03bf \u03b9\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03b9\u03bf.[/quote]\r\n\r\nK\u03b1\u03b9 \u03bc\u03ac\u03bb\u03b9\u03c3\u03c4\u03b1 \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b1\u03c5\u03c4\u03cc \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf\u03bd \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03ad\u03bd\u03bf \u03ba\u03cd\u03ba\u03bb\u03bf \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 :wink:", "Solution_3": "\u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03b5\u03be\u03b7\u03b3\u03ae\u03c3\u03b5\u03b9 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03ce \u03c4\u03b9 \u03b5\u03bd\u03bd\u03bf\u03bf\u03cd\u03bc\u03b5 \u03b5\u03b4\u03ce \u03bb\u03ad\u03b3\u03bf\u03bd\u03c4\u03b1\u03c2 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b5\u03c2;", "Solution_4": "[quote=\"Spribo\"]\u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03b5\u03be\u03b7\u03b3\u03ae\u03c3\u03b5\u03b9 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03ce \u03c4\u03b9 \u03b5\u03bd\u03bd\u03bf\u03bf\u03cd\u03bc\u03b5 \u03b5\u03b4\u03ce \u03bb\u03ad\u03b3\u03bf\u03bd\u03c4\u03b1\u03c2 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b5\u03c2;[/quote]\r\n\u0394\u03cd\u03bf \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \u03b4\u03b9\u03b1 \u03c4\u03b7\u03c2 \u03ba\u03bf\u03c1\u03c5\u03c6\u03ae\u03c2 \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03b7\u03c2 \u03b3\u03c9\u03bd\u03af\u03b1\u03c2 ( \u03c3\u03c4\u03bf \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03cc \u03ae \u03c3\u03c4\u03bf \u03b5\u03be\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03cc \u03bc\u03ad\u03c1\u03bf\u03c2 \u03b1\u03c5\u03c4\u03ae\u03c2 ) \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ad\u03c2 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b7 \u03b4\u03b9\u03c7\u03bf\u03c4\u03cc\u03bc\u03bf \u03c4\u03b7\u03c2, \u03bf\u03bd\u03bf\u03bc\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 [b]\u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b5\u03c2[/b] \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b7 \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03b7 \u03b3\u03c9\u03bd\u03af\u03b1.\r\n\r\n$ \\bullet$ \u0388\u03bd\u03b1 \u03b3\u03bd\u03c9\u03c3\u03c4\u03cc \u03ba\u03b1\u03b9 \u03c7\u03c1\u03ae\u03c3\u03b9\u03bc\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c3\u03c4\u03b9\u03c2 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b5\u03c2 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2, \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03bb\u03ad\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03ac\u03bd $ P$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c4\u03c5\u03c7\u03b1\u03af\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c3\u03c4\u03bf \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ \\bigtriangleup ABC,$ \u03bf\u03b9 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b5\u03c2 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \u03c4\u03c9\u03bd $ AP,\\ BP,\\ CP,$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b7\u03bd \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 ( \u03b7 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b1 \u03c4\u03b7\u03c2 $ AP,$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 $ \\angle A$ \u03ba\u03bb\u03c0 ), \u03c4\u03ad\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03ad\u03c3\u03c4\u03c9 $ P',$ \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03bf\u03bd\u03bf\u03bc\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03c5 $ P,$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03bf \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf $ \\bigtriangleup ABC.$\r\n\r\n\u03a4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ P,\\ P',$ \u03bf\u03bd\u03bf\u03bc\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 [b]\u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b1 \u03c3\u03c5\u03b6\u03c5\u03b3\u03ae[/b] ( = [b]isogonal conjugates[/b] ) \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf $ \\bigtriangleup ABC.$\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.", "Solution_5": "\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03bb\u03cd!" } { "Tag": [ "trigonometry" ], "Problem": "Suppose that $0^\\circ < \\theta < 90^\\circ$, simplify:\r\n\\[\r\n\\arctan\\left( \\frac{1+\\sin \\theta}{\\cos \\theta} \\right)\r\n\\]", "Solution_1": "Heres a geometric solution:\r\n\r\n[hide]\nSuppose you have a circle center O and radius 1. Let A and B be points on the circumference such that $\\angle\\ AOB=\\theta$. Let $AC$ be the perpendicular from $A$ to line $OB$. Then $OC=\\cos\\ \\theta$ and $AC=\\sin\\ \\theta$. \n\nNow construct a circle with center A and radius $OA=1$. Extend $AC$ to P on the second circle so that $PC=1+\\sin\\ \\theta$. Let $\\angle\\ POA=\\alpha$. \n\nThen triangle $POC$ is right angled at $\\angle\\ PCO$, so $\\angle\\ POC+\\angle\\ OPC=90\\Rightarrow\\ 2\\alpha+\\theta=90\\Rightarrow\\ \\alpha=\\frac{90-\\theta}{2}$\n\nThis means that $\\angle\\ POC=\\alpha+\\theta=\\frac{90+\\theta}{2}$. But notice that $\\tan(\\alpha+\\theta)=\\frac{1+\\sin\\ \\theta}{\\cos\\ \\theta}$ hence $\\frac{90+\\theta}{2}=\\alpha+\\theta=\\arctan\\ (\\frac{1+\\sin\\ \\theta}{\\cos\\ \\theta})$\n\n[/hide]" } { "Tag": [ "algebra", "polynomial", "induction", "number theory", "relatively prime", "topology", "superior algebra" ], "Problem": "Is there an integer $ n>1$ such that $ X$ is reducible over $ \\mathbb{Z}_n[X]$?", "Solution_1": "[quote=\"Fleury\"]Is there an integer $ n > 1$ such that $ X$ is reducible over $ \\mathbb{Z}_n[X]$?[/quote]\r\n\r\nYes. For example in $ (\\mathbb{Z}/6\\mathbb{Z})[X]$ we have: $ X\\equal{}(3X\\plus{}4)(4X\\plus{}3),$ or in $ (\\mathbb{Z}/10\\mathbb{Z})[X]$ we have: $ X\\equal{}(5X\\plus{}4)(4X\\plus{}5)\\equal{}(5X\\plus{}6)(6X\\plus{}5).$", "Solution_2": "Thanks ysharifi,\r\nHow does one find all such integers $ n>1$?", "Solution_3": "[quote=\"Fleury\"]Thanks ysharifi,\nHow does one find all such integers $ n > 1$?[/quote]\r\n\r\nIn general, suppose $ R$ is a commutative ring with identity such that $ R$ has a non-trivial idempotent $ e.$ By non-trivial I mean $ e \\neq 0, 1.$ Then in $ R[X]$ we'll have: \r\n\r\n$ X \\equal{} (eX \\plus{} 1 \\minus{} e)((1 \\minus{} e)X \\plus{} e),$ which proves that $ X$ is reducible in $ R[X].$\r\n\r\nNow if $ n$ has at least two prime divisors, then $ R \\equal{} \\mathbb{Z}/n\\mathbb{Z}$ will have some non-trivial idempotent and thus $ X$ will be reducible in $ R[X].$ So the only case that\r\n\r\nneeds to be investigated is when $ n \\equal{} p^k,$ where $ p$ is a prime. If $ k \\equal{} 1,$ then $ X$ is obviously irreducible. So we may assume that $ k \\geq 2.$", "Solution_4": "[quote=\"Fleury\"]Now if $ n$ has at least two prime divisors, then $ R \\equal{} \\mathbb{Z}/n\\mathbb{Z}$ will have some non-trivial idempotent[/quote]\r\nFor anyone who took as long as I did to figure out why this is true, here's a proof:\r\n\r\nWrite $ n \\equal{} ab$, with $ a$ and $ b$ relatively prime, and $ 1 < a < b$. Let $ k$ be the multiplicative inverse of $ a$ mod $ b$. Then $ ka \\minus{} 1$ is divisible by $ b$, so $ ka(ka \\minus{} 1)$ is divisible by $ n$, so $ ka$ is idempotent mod $ n$. Note also that $ 0 < ka < n$. Since $ 1 < a < b$, we also have $ 1 < k < b$, so $ ka > 1$.", "Solution_5": "As for $ n\\equal{}p^k$, it seems that $ 1\\equal{}(1\\minus{}p^{k\\minus{}1} X)(1\\plus{}p^{k\\minus{}1} X)$, so [i]everything[/i] is reducible...", "Solution_6": "[quote=\"ysharifi\"]...when $ n \\equal{} p^k,$ where $ p$ is a prime. If $ k \\equal{} 1,$ then $ X$ is obviously irreducible...[/quote]\r\nWhat about $ X\\equal{}X^{\\frac{p\\minus{}1}{2}}X^{\\frac{p\\plus{}1}{2}}$?", "Solution_7": "Equality of polynomials is not the same as those polynomials taking the same values.", "Solution_8": "[quote=\"LydianRain\"]As for $ n \\equal{} p^k$, it seems that $ 1 \\equal{} (1 \\minus{} p^{k \\minus{} 1} X)(1 \\plus{} p^{k \\minus{} 1} X)$, so [i]everything[/i] is reducible...[/quote]Irreducible elements are non-units by definition. But your examples are units. For $ k\\ge 2$, a polynomial $ f(X)\\in(\\mathbb Z/p^k)[X]$ which is not a unit is irreducible in $ (\\mathbb Z/p^k)[X]$ iff $ f(X)\\mod p$ is irreducible in $ (\\mathbb Z/p)[X]$.", "Solution_9": "[quote=\"jmerry\"]Equality of polynomials is not the same as those polynomials taking the same values.[/quote]\r\nOK, with that definition then $ X$ is irreducible over $ \\mathbb{Z}/p\\mathbb{Z}$.", "Solution_10": "[quote=\"Third Edition\"]Irreducible elements are non-units by definition. But your examples are units. For $ k\\ge 2$, a polynomial $ f(X)\\in(\\mathbb Z/p^k)[X]$ which is not a unit is irreducible in $ (\\mathbb Z/p^k)[X]$ iff $ f(X)\\mod p$ is irreducible in $ (\\mathbb Z/p)[X]$.[/quote]\r\nYes, you're right. Whoops. Showing that $ X$ irreducible in $ \\mathbb{Z}/p$ implies it is irreducible in $ \\mathbb{Z}/p^k$ can be done with a relatively straightforward induction, (similar to my observation that $ 1\\minus{}p^{k\\minus{}1} X$ is a unit) and this completes the original discussion.\r\n\r\nBut I'm still not seeing why a polynomial that is irreducible in $ \\mathbb{Z}/p^k$ must be irreducible in $ \\mathbb{Z}/p$...", "Solution_11": "In fact, that's only almost true, take for example $ x^2 \\minus{} p$ (didn't check it, but it should work). For nice cases, Hensel's lemma does the job.", "Solution_12": "\"Nice cases\" means anything but a power of an irreducible polynomial. If the two factors are relatively prime, it's pretty easy to build up.", "Solution_13": "[quote=\"LydianRain\"]But I'm still not seeing why a polynomial that is irreducible in $ \\mathbb{Z}/p^k$ must be irreducible in $ \\mathbb{Z}/p$...[/quote]\r\n... this part of my statement is bluntly false in general. Sorry! :( I was led by the nasty idea that irreducibility of elements is preserved by nilpotent quotients (in both directions): This is true for the property of being a unit on the element level and - and this was the tricky thing which trapped me - for the property of a Zariski closed set of being irreducible. But the latter does not properly translate to elements which I overlooked.\r\n\r\n$ X\\minus{}p\\in\\mathbb Z/p^n[X]$ is irreducible since $ p$ is no square in $ \\mathbb Z/p^n$, for $ n>1$." } { "Tag": [ "inequalities" ], "Problem": "Given $a+b+c=1$ and that all are positive reals, prove (or disprove?) that:\r\n\r\n$\\frac{(ab)^{2}+1}{ab}+\\frac{(bc)^{2}+1}{bc}+\\frac{(ac)^{2}+1}{ac}\\leq \\frac{82}{81abc}$", "Solution_1": "It is equivalent to\r\n\\[81a^{2}b^{2}c^{2}\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\leq 1 \\]\r\nNow, the LHS has $256$ terms, and so does $(a+b+c)^{5}$, so the inequality $LHS \\leq (a+b+c)^{5}= 1$ is true by Muirhead's inequality (apply it many times)\r\nIf you don't like Muirhead, call it rearrangement :wink:", "Solution_2": "Pretty much the same solution as @above, and I think you meant $243$ terms not $256$.\n\nMultiply everything by $abc$ and you got\n\\[\\sum_{cyc}((ab)^2c+c)\\leq\\frac{81}{82}\\]\nminus $a+b+c$ from LHS and $1$ from RHS and we have\n\\[81abc\\sum_{cyc}ab\\leq 1=(a+b+c)^5\\]\nwhich can be proven by Muirhead.\n\nremark: this is a 17 year old problem. Fewer and fewer inequalities can be killed by Muirhead now... :( " } { "Tag": [ "logarithms" ], "Problem": "Prove that\r\n\r\n$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{4\\cdot\\textrm{log}_3 \\, 12\\,+\\,\\textrm{log}^2_3\\,4\\,+\\,2\\cdot\\textrm{log}^2_3\\,12\\,-\\,3\\cdot\\textrm{log}_3\\,4\\cdot\\textrm{log}_3\\,12\\,-\\,2\\cdot\\textrm{log}_3\\,4}{\\textrm{log}_3\\,4\\,-\\,2\\cdot\\textrm{log}_3\\,12}\\,=\\,-3$.\r\n\r\nNote: $\\textrm{log}^c_b\\,a\\,=\\,\\left(\\textrm{log}_b\\,a \\right)^c$.", "Solution_1": "Eq: (B + C)/A = -3\r\n\r\nA: log4 - 2log12 = log4 - 2(log3 + log4) = log4 - 2(1 + log4) = -2 - 2log4 (log(c)c = 1)\r\nB: 4log12 - 2log4 = 4(log3*4) - 2log4 = 4(log3 + log4) - 2log4 = 4(1 + log4) - 2log4 = 4 + 2log4\r\nC: (log4)^2 + 2(log12)^2 - 3log4*log12 \r\n= (log4)^2 + 2(1 + 2log4 + (log4)^2) - 3log4*(log3 + log4) // (a+b)^2 = ...\r\n= (log4)^2 + 2(1 + 2log4 + (log4)^2) - 3(log4)^2 - 3log4\r\n= 2 + log4 - 3 = log4 - 1\r\n\r\nB + C = ( 4 + 2log4 ) + (log4 - 1 ) = 3log4 + 3 = 3 (1 + log4)\r\nA = -2(1 + log4)\r\n\r\nThere is some erorr, but it's can be way to solve", "Solution_2": "[quote=\"LoboSolitario\"]Prove that\n\n$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{4\\cdot\\textrm{log}_3 \\, 12\\,+\\,\\textrm{log}^2_3\\,4\\,+\\,2\\cdot\\textrm{log}^2_3\\,12\\,-\\,3\\cdot\\textrm{log}_3\\,4\\cdot\\textrm{log}_3\\,12\\,-\\,2\\cdot\\textrm{log}_3\\,4}{\\textrm{log}_3\\,4\\,-\\,2\\cdot\\textrm{log}_3\\,12}\\,=\\,-3$.\n\nNote: $\\textrm{log}^c_b\\,a\\,=\\,\\left(\\textrm{log}_b\\,a \\right)^c$.[/quote]\r\nJust let $a=\\log_34$, so $\\log_312=\\log_34+\\log_33=a+1$\r\n\r\n$\\frac{4\\log_312+\\log_3^24+2\\log_3^212-3\\log_34\\cdot\\log_312-2\\log_34}{\\log_34-2\\log_312}$\r\n\r\n$=\\frac{4(a+1)+a^2+2(a+1)^2-3a(a+1)-2a}{a-2(a+1)}$\r\n\r\n$=\\frac{4a+4+a^2+2a^2+4a+2-3a^2-3a-2a}{a-2a-2}$\r\n\r\n$=\\frac{3a+6}{-a-2}=-3$ because $a\\neq -2$\r\n\r\nI completely do not understand dragansm's answer, sorry. :roll:", "Solution_3": "[quote=\"dragansm\"]Eq: (B + C)/A = -3\n\nA: log4 - 2log12 = log4 - 2(log3 + log4) = log4 - 2(1 + log4) = -2 - [b]2[/b]log4 (log(c)c = 1)\nB: 4log12 - 2log4 = 4(log3*4) - 2log4 = 4(log3 + log4) - 2log4 = 4(1 + log4) - 2log4 = 4 + 2log4\nC: (log4)^2 + 2(log12)^2 - 3log4*log12 \n= (log4)^2 + 2(1 + 2log4 + (log4)^2) - 3log4*(log3 + log4) // (a+b)^2 = ...\n= (log4)^2 + 2(1 + 2log4 + (log4)^2) - 3(log4)^2 - 3log4\n= 2 + log4 [b]- 3[/b] = log4 - 1\n\nB + C = ( 4 + 2log4 ) + (log4 - 1 ) = 3log4 [b]+ 3[/b] = 3 (1 + log4)\nA = [b]-2(1 + log4)[/b]\n\nThere is some erorr, but it's can be way to solve[/quote]\r\n\r\nthe bold stuff is wrong, little miscalculations nothing big, but should be corrected ...", "Solution_4": "[quote=\"dragansm\"]Eq: (B + C)/A = -3\n\nA: log4 - 2log12 = log4 - 2(log3 + log4) = log4 - 2(1 + log4) = -2 - [color=red][b]2[/b][/color]log4 [b]<-------Wrong[/b] (log(c)c = 1)\nB: 4log12 - 2log4 = 4(log3*4) - 2log4 = 4(log3 + log4) - 2log4 = 4(1 + log4) - 2log4 = 4 + 2log4\nC: (log4)^2 + 2(log12)^2 - 3log4*log12 \n= (log4)^2 + 2(1 + 2log4 + (log4)^2) - 3log4*(log3 + log4) // (a+b)^2 = ...\n= (log4)^2 + 2(1 + 2log4 + (log4)^2) - 3(log4)^2 - 3log4\n= 2 + log4 - [color=red][b]3[/b][/color] = log4 - 1 [b]<--------Wrong[/b]\n\nB + C = ( 4 + 2log4 ) + (log4 - 1 ) = 3log4 + 3 = 3 (1 + log4) [b]B+C is wrong because of above error[/b]\nA = -2(1 + log4) [b]A is wrong because of above error[/b]\n\nThere is some erorr, but it's can be way to solve[/quote]\r\nLook at the arrows and red numbers I put above and that should point out where you made the mistake, it was a hard to read answer...", "Solution_5": "If that is an advise, thank you 10000th User.\r\n\r\nMy english is bad, so forgiveme if i posted this problem in wrong forum.\r\n\r\nAnd about time, a russian friend brought me this problem." } { "Tag": [], "Problem": "Solve the system:\r\n\r\n\\[ \\frac{xy}{x+y}=\\frac{1}{3} \\]\r\n\\[ \\frac{yz}{y+z}=\\frac{1}{4} \\]\r\n\\[ \\frac{xz}{x+z}=\\frac{1}{5} \\]", "Solution_1": "[hide]$\\frac{xy}{x+y} = \\frac13, \\frac{yz}{y+z} = \\frac14, \\frac{xz}{x+z} = \\frac15$\nreciprocating gives:\n\n$\\frac{x+y}{xy} = 3, \\frac{y+z}{yz} = 4, \\frac{x+z}{xz} = 5$\n\n$\\frac 1x + \\frac 1y = 3, \\frac 1y + \\frac 1z = 4, \\frac 1x + \\frac 1z = 5$\n\nLet $a, b, c$ represent $\\frac 1x, \\frac 1y, \\frac 1z$ respectively\n\nThen:\n$a+b=3$ \n$b+c=4$ \n$a+c=5$\nWe have 3 variables and 3 equations, so we can have a potential solution set. Solving gives:\n\n$a=2, b=1, c=3$\n\nWhich means that: $x=\\frac12, y=1, z=\\frac13$\n[/hide]", "Solution_2": "Hmm, probably a better way, but...\r\n\r\n[hide]\nWe can use substitution to find x and z in terms of y (not sure why I decided to solve for y)\n$3xy=x+y \\rightarrow x = \\frac{y}{3y-1}$\n$4yz=y+z \\rightarrow z = \\frac{y}{4y-1}$\n$5zx=z+x \\rightarrow \\frac{5y^2}{(3y-1)(4y-1)} = \\frac{y}{3y-1} + \\frac{y}{4y-1}$\n$5y^2 = 4y^2 - y + 3y^2 - y \\rightarrow y = 1$\nThen simple subsitution gives us the solution $\\left( \\frac 12 , 1 , \\frac 13 \\right)$\n[/hide]", "Solution_3": "Probably an even [i] better [/i] way, but.. :lol: \r\n\r\n[hide] We know:\n\n $\\frac{1}{a} + \\frac{1}{b}=\\frac{a+b}{ab}$\n\n So in each equation we simply flip the fractions to get:\n\n\n$\\frac{x+y}{xy} = 3$\n\n$\\frac{y+z}{yz} = 4$\n\n$\\frac{x+z}{xz} = 5$\n\n Now we can re-write each equation as:\n\n $\\frac{1}{x} + \\frac{1}{y} = 3$\n\n $\\frac{1}{y} + \\frac{1}{z} = 4$\n\n $\\frac{1}{x} + \\frac{1}{z} = 5$\n\n Now we have three equations we can easily solve by substitution, doing so yields:\n\n $( \\frac{1}{2} , 1 , \\frac{1}{3} )$[/hide]", "Solution_4": "Haha, I wasn't being conceited, I meant there probably is a better way than mine :oops:", "Solution_5": "It was easy ... I done it like G-UNIT :)" } { "Tag": [ "inequalities", "search", "function", "calculus", "derivative", "algebra", "polynomial" ], "Problem": "Prove that for positive real numbers $x$, $y$, $z$, \\[ x^3(y^2+z^2)^2 + y^3(z^2+x^2)^2+z^3(x^2+y^2)^2 \\geq xyz\\left[xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2\\right].\\] [i]Zarathustra (Zeb) Brady.[/i]", "Solution_1": "[hide=\"How Zeb intended this problem to die\"]Expand and multiply both sides by $ x\\plus{}y\\plus{}z$ to make loads of stuff cancel.[/hide]", "Solution_2": "[quote=\"MellowMelon\"]Prove that for positive real numbers $ x,y,z$,\n\\[ x^3(y^2 \\plus{} z^2)^2 \\plus{} y^3(z^2 \\plus{} x^2)^2 \\plus{} z^3(x^2 \\plus{} y^2)^2 \\geq xyz[xy(x \\plus{} y)^2 \\plus{} yz(y \\plus{} z)^2 \\plus{} zx(z \\plus{} x)^2].\\]\n[/quote]\r\nIt's easy SOS.", "Solution_3": "[quote=\"MellowMelon\"]Prove that for positive real numbers $ x,y,z$,\n\\[ x^3(y^2 \\plus{} z^2)^2 \\plus{} y^3(z^2 \\plus{} x^2)^2 \\plus{} z^3(x^2 \\plus{} y^2)^2 \\geq xyz[xy(x \\plus{} y)^2 \\plus{} yz(y \\plus{} z)^2 \\plus{} zx(z \\plus{} x)^2].\\]\n[i]Zarathustra (Zeb) Brady.[/i][/quote]\r\nLet $ x\\equal{}1/a,y\\equal{}1/b,z\\equal{}1/c$, the inequality becomes\r\n$ \\sum \\frac{(b^2\\plus{}c^2)^2}{a^3b^4c^4} \\ge\\frac{1}{abc}\\sum \\frac{(b\\plus{}c)^2}{b^3c^3},$\r\nor equivalently,\r\n$ \\sum a(b^2\\plus{}c^2)^2 \\ge \\sum a^3(b\\plus{}c)^2.$ (1)\r\nWe have $ \\sum a(b^2\\plus{}c^2)^2 \\equal{}\\sum a(b^4\\plus{}c^4)\\plus{}2\\sum ab^2c^2 \\equal{}\\sum a^4(b\\plus{}c) \\plus{}\\sum a^2bc(b\\plus{}c),$ so the last inequality can be written as\r\n$ \\sum [a^4(b\\plus{}c)\\plus{}a^2bc(b\\plus{}c)\\minus{}a^3(b\\plus{}c)^2] \\ge 0,$\r\nthat is\r\n$ \\sum a^2(b\\plus{}c)(a\\minus{}b)(a\\minus{}c) \\ge 0.$\r\nThis can be proved easily using Vornicu Schur.\r\n\r\nEspecially, we have a very nice identity which implies (1), it is\r\n$ LHS\\minus{}RHS\\equal{}\\sum a(b\\minus{}c)^2(b\\plus{}c\\minus{}a)^2.$", "Solution_4": "[quote=\"can_hang2007\"] $ \\sum a^2(b \\plus{} c)(a \\minus{} b)(a \\minus{} c) \\ge 0.$\nThis can be proved easily using Vornicu Schur.\n\nEspecially, we have a very nice identity which implies (1), it is\n$ LHS \\minus{} RHS \\equal{} \\sum a(b \\minus{} c)^2(b \\plus{} c \\minus{} a)^2.$[/quote]\r\nI think the following generalization holds for any $ k\\ge 0$ and nonnegative $ a,b,c$:\r\n\r\n$ \\sum (b \\plus{} c)(a \\minus{} b)(a \\minus{} c)(a \\minus{} kb)(a \\minus{} kc)\\ge 0,$\r\n\r\nwith equality for $ b \\equal{} c \\equal{} a$, for $ b \\equal{} c \\equal{} 0$, for $ b \\equal{} c \\equal{} \\frac a{k}$ (or any cyclic permutation).\r\nWe have also a nice identity.", "Solution_5": "[quote=\"can_hang2007\"] $ \\sum a^2(b \\plus{} c)(a \\minus{} b)(a \\minus{} c) \\ge 0.$\nThis can be proved easily using Vornicu Schur.\n\nEspecially, we have a very nice identity which implies (1), it is\n$ LHS \\minus{} RHS \\equal{} \\sum a(b \\minus{} c)^2(b \\plus{} c \\minus{} a)^2.$[/quote]\r\nThis is an interesting stronger inequality\r\n\r\n$ \\sum a^2(b \\plus{} c)(a \\minus{} b)(a \\minus{} c) \\ge \\frac {4(a\\minus{}b)^2(b\\minus{}c)^2(c\\minus{}a)^2}{a\\plus{}b\\plus{}c}.$", "Solution_6": "[quote=\"Vasc\"][quote=\"can_hang2007\"] $ \\sum a^2(b \\plus{} c)(a \\minus{} b)(a \\minus{} c) \\ge 0.$\nThis can be proved easily using Vornicu Schur.\n\nEspecially, we have a very nice identity which implies (1), it is\n$ LHS \\minus{} RHS \\equal{} \\sum a(b \\minus{} c)^2(b \\plus{} c \\minus{} a)^2.$[/quote]\nI think the following generalization holds for any $ k\\ge 0$ and nonnegative $ a,b,c$:\n\n$ \\sum (b \\plus{} c)(a \\minus{} b)(a \\minus{} c)(a \\minus{} kb)(a \\minus{} kc)\\ge 0,$\n\nwith equality for $ b \\equal{} c \\equal{} a$, for $ b \\equal{} c \\equal{} 0$, for $ b \\equal{} c \\equal{} \\frac a{k}$ (or any cyclic permutation).\nWe have also a nice identity.[/quote]\r\nThe inequality is in degree 5th, therefore if we transform it into $ pqr$ form, the degree of $ r$ will be $ 1$, from which follows that we only need to prove it in the case $ abc\\equal{}0$ or $ (a\\minus{}b)(b\\minus{}c)(c\\minus{}a)\\equal{}0$ (according from an theorem which I have posted already on the forum). But in both cases, the inequality is trivial, so we are done.", "Solution_7": "[quote=\"can_hang2007\"][quote=\"Vasc\"][quote=\"can_hang2007\"] $ \\sum a^2(b \\plus{} c)(a \\minus{} b)(a \\minus{} c) \\ge 0.$\nThis can be proved easily using Vornicu Schur.\n\nEspecially, we have a very nice identity which implies (1), it is\n$ LHS \\minus{} RHS \\equal{} \\sum a(b \\minus{} c)^2(b \\plus{} c \\minus{} a)^2.$[/quote]\nI think the following generalization holds for any $ k\\ge 0$ and nonnegative $ a,b,c$:\n\n$ \\sum (b \\plus{} c)(a \\minus{} b)(a \\minus{} c)(a \\minus{} kb)(a \\minus{} kc)\\ge 0,$\n\nwith equality for $ b \\equal{} c \\equal{} a$, for $ b \\equal{} c \\equal{} 0$, for $ b \\equal{} c \\equal{} \\frac a{k}$ (or any cyclic permutation).\nWe have also a nice identity.[/quote]\nThe inequality is in degree 5th, therefore if we transform it into $ pqr$ form, the degree of $ r$ will be $ 1$, from which follows that we only need to prove it in the case $ abc \\equal{} 0$ or $ (a \\minus{} b)(b \\minus{} c)(c \\minus{} a) \\equal{} 0$ (according from an theorem which I have posted already on the forum). But in both cases, the inequality is trivial, so we are done.[/quote]\r\nYes, Can_hang. We have also\r\n\r\n$ \\sum (b \\plus{} c)(a \\minus{} b)(a \\minus{} c)(a \\minus{} kb)(a \\minus{} kc)\\equal{}\\sum a(b\\minus{}c)^2(b\\plus{}c\\minus{}(k\\plus{}1)a)^2$.", "Solution_8": "can anyone help me to prove that\r\n$ \\sum a^r(b\\plus{}c)(a\\minus{}b)(a\\minus{}c) \\ge 0$ where a,b,c,r are positive real.\r\ni cant prove it using vornicu schur because i cant understand vornicu schur", "Solution_9": "[quote=\"ishfaq420haque\"]can anyone help me to prove that\n$ \\sum a^r(b \\plus{} c)(a \\minus{} b)(a \\minus{} c) \\ge 0$ where a,b,c,r are positive real.\ni cant prove it using vornicu schur because i cant understand vornicu schur[/quote]\r\nThe following reasoning is strongest for the Vornicu-Schur's method. \r\nLet $ a\\geq b\\geq c.$ Hence, $ \\sum_{cyc}a^r(b \\plus{} c)(a \\minus{} b)(a \\minus{} c)\\geq$\r\n$ \\geq a^r(b \\plus{} c)(a \\minus{} b)(a \\minus{} c) \\plus{} b^r(a \\plus{} c)(b \\minus{} a)(b \\minus{} c) \\equal{}$\r\n$ \\equal{} (a \\minus{} b)(a^r(b \\plus{} c)(a \\minus{} c) \\minus{} b^r(a \\plus{} c)(b \\minus{} c)) \\equal{}$\r\n$ \\equal{} (a \\minus{} b)((ab \\minus{} c^2)(a^r \\minus{} b^r) \\plus{} c(a \\minus{} b)(a^r \\plus{} b^r))\\geq0.$", "Solution_10": "[quote=\"MellowMelon\"]Prove that for positive real numbers $ x,y,z$,\n\\[ x^3(y^2 \\plus{} z^2)^2 \\plus{} y^3(z^2 \\plus{} x^2)^2 \\plus{} z^3(x^2 \\plus{} y^2)^2 \\geq xyz[xy(x \\plus{} y)^2 \\plus{} yz(y \\plus{} z)^2 \\plus{} zx(z \\plus{} x)^2].\\]\n[i]Zarathustra (Zeb) Brady.[/i][/quote]\r\nHere is my solution\r\n\\[ {{x}^{3}}{{\\left( {{y}^{2}}\\plus{}{{z}^{2}} \\right)}^{2}}\\plus{}{{y}^{3}}{{\\left( {{z}^{2}}\\plus{}{{x}^{2}} \\right)}^{2}}\\plus{}{{z}^{3}}\\left( {{x}^{2}}\\plus{}{{y}^{2}} \\right)\\ge xyz\\left[ xy{{\\left( x\\plus{}y \\right)}^{2}}\\plus{}yz{{\\left( y\\plus{}z \\right)}^{2}}\\plus{}zx{{\\left( z\\plus{}x \\right)}^{2}} \\right]\\]\r\n\\[ \\Leftrightarrow x[{{(x{{y}^{2}}\\plus{}x{{z}^{2}})}^{2}}\\minus{}{{({{y}^{2}}z\\plus{}y{{z}^{2}})}^{2}}]\\plus{}y[{{(y{{z}^{2}}\\plus{}y{{x}^{2}})}^{2}}\\minus{}{{({{x}^{2}}z\\plus{}x{{z}^{2}})}^{2}}]\\plus{}z[{{(z{{x}^{2}}\\plus{}z{{y}^{2}})}^{2}}\\minus{}{{({{x}^{2}}y\\plus{}x{{y}^{2}})}^{2}}]\\ge 0\\]\r\n\\[ \\Leftrightarrow x(x{{y}^{2}}\\plus{}x{{z}^{2}}\\plus{}{{y}^{2}}z\\plus{}y{{z}^{2}})[{{y}^{2}}(x\\minus{}z)\\plus{}{{z}^{2}}(x\\minus{}y)]\\plus{}y(y{{x}^{2}}\\plus{}y{{z}^{2}}\\plus{}{{x}^{2}}z\\plus{}x{{z}^{2}})[{{x}^{2}}(y\\minus{}z)\\plus{}{{z}^{2}}(y\\minus{}x)]\\plus{}z(z{{x}^{2}}\\plus{}z{{y}^{2}}\\plus{}{{x}^{2}}y\\plus{}x{{y}^{2}})[{{x}^{2}}(z\\minus{}y)\\plus{}{{y}^{2}}(z\\minus{}x)]\\ge 0\\]\r\nOr\r\n$ x^2 (y\\minus{}z)(x^2 y^2\\plus{}yz^2\\plus{}x^2 yz\\plus{}xyz^2\\minus{}x^2 z^2\\minus{}z^2 y^2\\minus{}x^2 yz\\minus{}xy^2 z)\\plus{}y^2 (z\\minus{}x)(x^2 z^2\\plus{}z^2 y^2\\plus{}x^2 yz\\plus{}xy^2 z\\minus{}x^2 y^2\\minus{}x^2 z^2\\minus{}xy^2 z\\minus{}xyz^2 )\\plus{}z^2 (x\\minus{}y)(x^2 y^2\\plus{}z^2 x^2\\plus{}xy^2 z\\plus{}xyz^2\\minus{}x^2 y^2\\minus{}y^2 z^2\\minus{}x^2 yz\\minus{}xyz^2 )\\geq 0$\r\n$ \\Leftrightarrow x^3 (y\\minus{}z)^2 (xy\\plus{}xz\\minus{}yz)\\plus{}y^3 (z\\minus{}x)^2 (xy\\plus{}yz\\minus{}zx)\\plus{}z^3 (x\\minus{}y)^2 (yz\\plus{}zx\\minus{}xy)\\geq 0$\r\nAnd we have S.O.S method: $ S\\equal{}S_x (y\\minus{}z)^2\\plus{}S_y (z\\minus{}x)^2\\plus{}S_z (x\\minus{}y)^2\\geq 0$\r\nwith \\[ {{S}_{x}}\\equal{}{{x}^{3}}\\left( xy\\plus{}zx\\minus{}yz \\right),{{S}_{y}}\\equal{}{{y}^{3}}\\left( xy\\plus{}yz\\minus{}zx \\right),{{S}_{z}}\\equal{}{{z}^{3}}(yz\\plus{}zx\\minus{}xy)\\]\r\nAssume \\[ x\\ge y\\ge z\\] \\[ \\Rightarrow {{S}_{y}}\\ge 0\\]\r\nWe have\r\n\\[ {{S}_{x}}\\plus{}{{S}_{y}}\\equal{}{{x}^{4}}y\\plus{}{{x}^{3}}z\\left( x\\minus{}y \\right)\\plus{}x{{y}^{4}}\\plus{}{{y}^{3}}z\\left( y\\minus{}x \\right)\\ge z\\left( x\\minus{}y \\right)\\left( {{x}^{3}}\\minus{}{{y}^{3}} \\right)\\equal{}z{{\\left( x\\minus{}y \\right)}^{2}}\\left( {{x}^{2}}\\plus{}xy\\plus{}{{y}^{2}} \\right)\\ge 0,\\]\r\n\\[ {{S}_{y}}\\plus{}{{S}_{z}}\\equal{}{{y}^{4}}z\\plus{}x{{y}^{3}}\\left( y\\minus{}z \\right)\\plus{}y{{z}^{4}}\\plus{}x{{z}^{3}}\\left( z\\minus{}y \\right)\\ge x\\left( y\\minus{}z \\right)\\left( {{y}^{3}}\\minus{}{{z}^{3}} \\right)\\equal{}x{{\\left( y\\minus{}z \\right)}^{2}}\\left( {{y}^{2}}\\plus{}yz\\plus{}{{z}^{2}} \\right)\\ge 0\\]\r\nThus, \\[ S\\equal{}\\left( {{S}_{x}}\\plus{}{{S}_{y}} \\right){{\\left( y\\minus{}z \\right)}^{2}}\\plus{}\\left( {{S}_{y}}\\plus{}{{S}_{z}} \\right){{\\left( x\\minus{}y \\right)}^{2}}\\plus{}2{{S}_{y}}\\left( x\\minus{}y \\right)\\left( y\\minus{}z \\right)\\ge 0\\]\r\nThe end.", "Solution_11": "[quote=\"Vasc\"][quote=\"can_hang2007\"] $ \\sum a^2(b \\plus{} c)(a \\minus{} b)(a \\minus{} c) \\ge 0.$\nThis can be proved easily using Vornicu Schur.\n\nEspecially, we have a very nice identity which implies (1), it is\n$ LHS \\minus{} RHS \\equal{} \\sum a(b \\minus{} c)^2(b \\plus{} c \\minus{} a)^2.$[/quote]\nThis is an interesting stronger inequality\n\n$ \\sum a^2(b \\plus{} c)(a \\minus{} b)(a \\minus{} c) \\ge \\frac {4(a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2}{a \\plus{} b \\plus{} c}.$[/quote]\r\n\r\nAssume : $ c\\equal{}min\\{a;b;c\\}$\r\n\r\nWith : $ a\\equal{}x\\plus{}c \\ ; \\ b\\equal{}y\\plus{}c \\ ; \\ x,y \\geq 0$\r\n\r\nIt is equivalent to : \r\n\r\n$ 6c^4x^2\\minus{}6c^4xy\\plus{}6c^4y^2\\plus{}14c^3x^3\\minus{}9c^3x^2y\\minus{}9c^3xy^2\\plus{}14c^3y^3\\plus{}10c^2x^4\\plus{}c^2x^3y\\minus{}18c^2x^2y^2\\plus{}c^2xy^3\\plus{}10c^2y^4\\plus{}2cx^5\\plus{}5cx^4y\\minus{}7cx^3y^2\\minus{}7cx^2y^3\\plus{}5cxy^4\\plus{}2cy^5\\plus{}x^5y\\minus{}4x^4y^2\\plus{}6x^3y^3\\minus{}4x^2y^4\\plus{}xy^5 \\geq 0$\r\n\r\nWhich is true :)", "Solution_12": "Is the Muirhead inequality allowed here?\r\n\r\nwell even if its not its very easy to make it with A-G\r\n\r\n(4,3,0)+(3,2,2)$ \\geq$ (4,2,1)+(3,2,2)\r\nand this is obvious?\r\nhave I maybe miss understood the task?(it seems too easy)", "Solution_13": "Standard mistake, I thought also that Muirhead kills it. \r\n\r\nIt's not 4,3,0 + 3,2,2 $ \\ge$ 4,2,1 + 3,2,2\r\n\r\nIt's 4,3,0 + 3,2,2 $ \\ge$ 4,2,1 + 3,3,1 and that isn't possible with Muirhead. \r\n\r\nBut maybe, when multiplying with x+y+z (brilliant idea if it's true, see post #2) it will reduce to something solvable with Schur/Muirhead...", "Solution_14": "you are right my mistake sorry and thanks for showing it to me\r\nthis x+y+z is very very nice idea but I didn't manage to solve it after doing that so\r\ncould someone provide that solution?", "Solution_15": "Hm... I also tried multiplying, but we are left with (5,2,1)+2(4,3,1) $ \\le$ (5,3,0)+(4,4,0)+(3,3,2)\r\n\r\nMurihead can't do it, Schur applies only with (a,0,0)... Maybe it's something else.", "Solution_16": "Does anyone has a link where is explained Vornicu Schur method\r\nI've never heard of it.\r\nMuirhead solves it if you square the ineq\r\nyou have (8,6,0)+(7,7,0)+2(7,4,3)+3(6,4,4)+(8,3,3)+4(5,5,4)+2(7,5,2)+2(6,6,2)+2(6,5,3) >=\r\n(8,4,2)+3(6,6,2)+2(6,5,3)+3(5,5,4)+(8,3,3)+4(6,5,3)+2(7,5,2)+2(7,4,3)\r\nis equal as\r\n(8,6,0)+(7,7,0)+(5,5,4)>=(8,4,2)+(6,6,2)+(6,4,4)\r\nLHS>=(8,4,2)+2(6,6,2)>=RHS\r\nfirst time is muihead + A-G second just muirhead\r\nif I made another error please correct me cause this is done manually by hand so I would be suprised if its correct.", "Solution_17": "Why, and how did you use AM-GM on (x,y,z) ??? \r\n\r\nFor Vornicu-Schur, search\r\nhttp://www.mathlinks.ro/portal.php?t=162684", "Solution_18": "thanks for the link\r\nand Am-GM\r\nis used to prove that (7,7,0)+(5,5,4)>=2(6,6,2)\r\n$ x^7y^7\\plus{}x^5y^5z^4\\geq 2 x^6y^6z^2$\r\n$ y^7z^7\\plus{}y^5z^5x^4\\geq 2 y^6z^6x^2$\r\n$ z^7x^7\\plus{}z^5x^5y^4\\geq 2 z^6x^6y^2$\r\nand summing this 3 inequalities you get this ineq up here.\r\nAnd the question why I don't understand (is there a way to do it without it(I couldn't find it)?)", "Solution_19": "The inequality is equivallent to\r\n\\[ x^{3}(y - z)^{2}(y + z) + y^{3}(z - x)^{2}(z + x) + z^{3}(x - y)^{2}(x + y)\\geq{x^{2}yz(y - z)^{2} + xy^{2}z(z - x)^{2} + xyz^{2}(x - y)^{2}}.\\]\r\n. \r\n Dividing both sides by $ xyz$ we get:\r\n\\[ \\frac {(xy - zx)^{2}(xy + xz - yz)}{yz} + \\frac {(yz - xy)^{2}(yz + xy - zx)}{zx} + \\frac {(xz - yz)^{2}(zx + zy - xy)}{xy}\\geq{0}.\\]\r\n. Substitute $ xy = a, yz = b, zx = c$. The inequality reduces to $ \\frac {(a - b)^{2}(a + b - c)}{c} + \\frac {(b - c)^{2}(b + c - a)}{a} + \\frac {(c - a)^{2}(c + a - b)}{b}\\geq{0}$. Rewrite the inequality as\r\n$ \\frac {a(a - b)^{2}}{c} + \\frac {b(a - b)^{2}}{c} + \\frac {b(b - c)^{2}}{a} + \\frac {c(b - c)^{2}}{a} + \\frac {a(c - a)^{2}}{b} + \\frac {c(c - a)^{2}}{b}\\geq{(a - b)^{2} + (b - c)^{2} + (c - a)^{2}}.$ Substitute $ |a - b| = m$, $ |b - c| = n$, $ |c - a| = k$. \r\n The last inequality is equivallent to $ (m^{2}\\frac {a}{c} + n^{2}\\frac {c}{a}) + (m^{2}\\frac {b}{c} + k^{2}\\frac {c}{b}) + (n^{2}\\frac {b}{a} + k^{2}\\frac {a}{b})\\geq{m^{2} + n^{2} + k^{2}}.$ Using AM-GM, $ LHS\\geq{2mn + 2mk + 2kn}$, but easy to prove that $ m,n,k$ are the lengths of a triangle(maybe degenerate),or all of them is $ 0$. (To prove this, wlog assume $ m$ is maximum between $ (x,y,z)$. Than, by the well known inequality $ |x| + |y|\\geq{|x + y|}$, $ k + n = |b - c| + |c - a|\\geq{|(b - c) + (c - a)| = |a - b| = m}$, q.e.d.)\r\n So,\r\n\\[ LHS\\geq{2mn + 2nk + 2km} = (m(n + k - m) + n(k + m - n) + k(m + n - k)) + (m^{2} + n^{2} + k^{2})\\geq{m^{2} + n^{2} + k^{2}},\\]\r\nQ.E.D.", "Solution_20": "We've to show $\\sum x^3(y^4+z^4)+2\\sum x^3y^2z^2\\geq xyz(\\sum xy(x^2+y^2)+2\\sum(xy)^2)$.\nNow suppose $xyz=1$ and lagrange function, \n$f(x,y,z,\\lambda)=\\sum xy(\\frac {1}{x^4}+\\frac{1}{y^4})+2\\sum x-(\\sum\\frac {x^2+y^2}{z}+2\\sum\\frac{1}{x^2})$.\nAfter differentiating both side respectively WRT $x,y,z$ we obtain, \n$-\\frac{3y+3z}{x^4}+\\frac{y^3+z^3}{y^3z^3}+2-4x^3-2x(\\frac{1}{y}-\\frac{1}{z})+\\frac{y^2+z^2}{x^2}=\\lambda yz$ and similarly others.\nSo, we've like,\n$-\\frac {3a}{x^3}+bx+2x-4x^4-2x^2c+\\frac {d}{x}+\\frac {2}{x^2}+x$ is constant for $x,y,z$ where $a,b,c,d$ are same for $x,yz$.\nNow notable part is, again after derivative WRT $x,y,z$ we obtain, \n$-\\frac {9a}{x^3}+b'x+2x-4x^4-2x^2c'+\\frac {d'}{x}+\\frac {2}{x^2}+x$ is constant for $x,y,z$ where $a',b',c',d'$ are same for $x,yz$ , hence we've like,\n$b''x+2x-4x^4-2x^2c''+\\frac {d''}{x}+\\frac {2}{x^2}+e'x$ is constant for $x,y,z$ where $b'',c'',d'',e'$ are same for $x,yz$.\nAgain doing same with $-\\frac {9a}{x^3}+b'x+2x-4x^4-2x^2c'+\\frac {d'}{x}+\\frac {2}{x^2}+x$ we obtain,\n$b'''x+2x-4x^4-2x^2c'''+\\frac {d'''}{x}+\\frac {2}{x^2}+e''x$ is constant for $x,y,z$ where $b''',c''',d'',e''$ are same for $x,yz$. (well this is easy to show all of $b''\\neq b''', c''\\neq c''',d''\\neq d''', e'\\neq e''$.\nSo now we get $-4x^4-2x^2c''''+\\frac {d''''}{x}+\\frac {2}{x^2}$ , so similarly going on we get a constant polynomial in $x$ for three values with coefficients of $x^k$ are vanishing, for a fixed $k$ (since in each step coefficients of $x^k$ are different for two same degree polynomial in $x$).\nSo keep going on we get $mx$ is constant for $x,y,z$ where $m$ is symmetric in $x,y,z$ and not nonzero either. So $x=y=z$ and so done.", "Solution_21": "After expanding, we wish to show that \\[\\left(\\sum\\limits_{sym} x^3y^4\\right) + \\left(2\\sum\\limits_{cyc} x^3y^2z^2\\right) \\ge \\left(\\sum\\limits_{sym} x^4y^2z\\right) + \\left(2\\sum\\limits_{cyc} xy^3z^3\\right).\\] By putting the difference $\\text{LHS} - \\text{RHS}$ into Chinese Dumbassing Notation, or by being especially observant, we see that this is equivalent to the \"sum\" (even though the signs point in opposite directions) of two sum of squares identities: \\[\\sum\\limits_{cyc} x^4z^3 + x^4y^3 \\ge \\sum\\limits_{cyc} x^4y^2z + x^4yz^2 \\quad \\text{(1)}\\]\\[\\sum\\limits_{cyc} 2x^3y^2z^2 \\le \\sum\\limits_{cyc} x^3y^3z + x^3yz^3. \\quad \\text{(2)}\\] By doing a little bit of algebra, we see that $(1)$ is equivalent to \\[\\sum\\limits_{cyc} x^4\\left(y^3 + z^3 - y^2z - yz^2\\right) \\ge 0\\]\\[\\iff \\sum\\limits_{cyc} x^4\\left(y + z\\right)\\left(y - z\\right)^2 \\ge 0. \\quad \\text{(3)}\\] Meanwhile, $(2)$ is equivalent to \\[0 \\le \\sum\\limits_{cyc} x^3yz\\left(y^2 + z^2 - 2yz\\right)\\]\\[\\iff 0 \\le \\sum\\limits_{cyc} x^3yz\\left(y - z\\right)^2. \\quad \\text{(4)}\\] Now, even though their signs point in different directions, we attempt to \"add\" $(3)$ and $(4).$ We then need to show that \\[\\sum\\limits_{cyc} x^3\\left(zx + xy - yz\\right)\\left(y - z\\right)^2 \\ge 0.\\] Now, by the symmetry of the inequality, we assume WLOG that $z \\ge y \\ge x.$ Then if we let $S_x = x^3\\left(zx + xy - yz\\right)$ and denote $S_y$ and $S_z$ similarly, the inequality is equivalent to \\[S_x\\left(y - z\\right)^2 + S_y\\left(z - x\\right)^2 + S_z\\left(x - y\\right)^2 \\ge 0. \\quad \\text{(5)}\\] We now use a standard sum of squares trick. (See [url]http://www.mit.edu/~evanchen/handouts/SOS_Dumbass/SOS_Dumbass.pdf[/url] for more detail)\n\nBy our ordering assumption, it follows that $y$ is the median of $\\{x, y, z\\}.$ Therefore \\[\\left(y - z\\right)\\left(y - x\\right) \\le 0\\]\\[\\implies \\left(z - x\\right)^2 \\ge \\left(x - y\\right)^2 + \\left(y - z\\right)^2.\\]\nIn addition, clearly $S_y = xy + yz - zx \\ge yz - zx \\ge 0.$ Therefore, we have that \\[S_x\\left(y - z\\right)^2 + S_y\\left(z - x\\right)^2 + S_z\\left(x - y\\right)^2 \\ge\\]\\[\\ge S_x\\left(y - z\\right)^2 + S_y\\left(\\left(x - y\\right)^2 + \\left(y - z\\right)^2\\right) + S_z\\left(x - y\\right)^2\\]\\[= \\left(S_y + S_z\\right)\\left(x - y\\right)^2 + \\left(S_x + S_y\\right)\\left(y - z\\right)^2\\]\\[= yz\\left(x - y\\right)^2 + xy\\left(y - z\\right)^2 \\ge 0.\\]\nTherefore, it follows that $(5)$ is true, which, upon expansion, is equivalent to the desired result.", "Solution_22": "Expand the desired inequality as\n\\[\\sum_{\\mbox{sym}} x^4y^3 + 2xyz[x^2yz+xyz^2+xy^2z]\\ge \\sum_{\\mbox{sym}} x^4y^2z + 2xyz[x^2y^2+y^2z^2+z^2x^2].\\]\nThis rearranges to\n\\[\\sum_{\\mbox{cyc}}x^4[y^3+z^3-y^2z-yz^2] = \\sum_{\\mbox{cyc}}x^4(y-z)^2(y+z) \\ge xyz[(xy-yz)^2+(xy-xz)^2+(xz-yz)^2].\\]\nRearrange again to\n\\[0\\le \\sum_{\\mbox{cyc}}\\left[x^4(y-z)^2(y+z)-x^3yz(y-z)^2\\right] = \\sum_{\\mbox{cyc}}x^3(y-z)^2[xy+xz-yz].\\]\nLet $yz=a, xz=b, xy=c$ to see it is equivalent to prove\n\\[0\\le \\sum_{\\mbox{cyc}}bc\\cdot (b-c)^2[b+c-a].\\]\nWe are done so long as $a,b,c$ satisfy the triangle inequality. Suppose they do not, so WLOG $a=b+c+t$ for some $t>0$. Then the desired inequality is\n\\[tbc(b-c)^2 \\le (b+c+t)c(b+t)^2(2c+t) + (b+c+t)b(c+t)^2(2b+t).\\]\nThis rewrites as\n\\[tbc(b-c)^2 \\le (b+c+t)[4b^2c^2+5bct(b+c)+2(b+c)^2t^2+(b+c)t^3].\\]\nIn fact, this is trivial because\n\\[(b+c+t)[4b^2c^2+5bct(b+c)+2(b+c)^2t^2+(b+c)t^3]\\ge 5(b+c)^2bct > (b+c)^2bct> (b-c)^2bct.\\]" } { "Tag": [ "quadratics", "number theory", "national olympiad" ], "Problem": "The Albanian NMO was held 2 week ago. Here\u2019s the grade 10 paper. If you\u2019re looking for sth challenging this is not stuff for you.\r\n\r\n\r\n\r\n1-Prove that this equation can\u2019t have three real positive roots not all of them equal.\r\n\r\nx^3 - x^2 + px \u2013 p/9=0 (p\u22600)\r\n\r\nFind the value of p such that all three roots are equal.\r\n\r\n\r\n2-Let ABC be an isosceles triangle (AB=BC) and O its circumcenter. M is the midpoint of BO and M\u2019 it\u2019s symmetric point to the midpoint of AB. Let K=AB\u2229MM\u2019. Let L be a point on BC such that angle BLM= angle CLO. Prove that K,L,O,B lie on circle.\r\n\r\n3-Let\u2019s call a \u201cdomino\u201d such a figure\t What\u2019s the maximal number of such dominos that can be put in a 8x8 table in such a way that the sides of the domino concide with the sides of the unit squares of the table?\r\n\r\n4-Prove that if x^2 + y^2 is divisible by p where p is a prime of the form 4m-1 than both x and y are divisible by p.\r\n\r\n5-Find all natural numbers a,b,c such that\r\n(a+b)/(a+c)=(b+c)/(b+a) and ab+bc+ca is a prime.", "Solution_1": "[quote]3-Let\u2019s call a \u201cdomino\u201d such a figure What\u2019s the maximal number of such dominos that can be put in a 8x8 table in such a way that the sides of the domino concide with the sides of the unit squares of the table?[/quote]\r\n\r\nWhat does \"domino\" look like? Can you give a figure for it?", "Solution_2": "[quote=\"Iris Aliaj\"]\nM\u2019 it\u2019s symmetric point to the midpoint of AB. Let K=AB\u2229MM\u2019.[/quote]\r\n\r\nDo you mean [tex]K[/tex] is the midpoint of [tex]AB[/tex]?", "Solution_3": "I had made the figure on word but it didn't apperar and i dont know how to do it. it's kind of a cross made of 5 squares 3 acrossa nd three horizontally.\r\n\r\nyeah K is the midpoint of AB", "Solution_4": "Hmm, but then I don't get why you need [tex]M'[/tex]", "Solution_5": "Thank you for the problems $Iris Aliaj$.Our olympiad also passed these days. :D \r\nAnd the difficulty of our problems were almost same(in my opinion).\r\nBetwean did you participate in the contest?If yes what did you score?", "Solution_6": "yes i did partecipate but we haven't got the results yet. i think i did very well and I'm excepting to be in the top 2 :D", "Solution_7": "hi\r\ni think problem 5 was easy but what about problem 4?\r\nall i know is if\r\n$x^2+y^2=m$ and $p=4k-1$ and $p|x^2+y^2$\r\nthen $m=p^{2k'}n$ , $(p,n)=1$ and so $p^2|x^2+y^2$\r\nbut i couldn't solve this\r\nhow did you solve this?or anyone has idea?", "Solution_8": "Suppose that neither $x$ nor $y$ is divisible by $p$.\r\nBy Fermat's little theorem $x^{p-1}=y^{p-1}(modp)$ also by the given conditions $x^4=y^4(modp)$.\r\nObviously if $k$ is the smallest positive integer for which $x^k=y^k(modp)$ then $k|4$ so it is easy to see that $k=4$.\r\nNext it is easy to see that $4|(p-1)$ but then $p=4k+1$ contradiction. :)", "Solution_9": "My solution of 4,\r\nSuppose x,y are not multiples of p\r\n$x^2=-y^2(mod p)$\r\n$(\\frac{-y^2}{p})=(-y^2)^{\\frac{p-1}{2}}=-1 (mod p)$ (Fermat's little theorem)\r\nThus $-y^2$ is a quadratic nonresidue. QED", "Solution_10": "thanks hardsoul and siuhochung\r\nbut siuhochung i dont get something\r\n[quote]$(\\frac{-y^2}{p})=(-y^2)^{\\frac{p-1}{2}}=-1 (mod p)$ (Fermat's little theorem) [/quote]\r\nfirst we supposed that $y$ is not divisible by $p$,so $(\\frac{-y^2}{p})$ is not an integer and can not be used in mod.\r\ncould you explain your solution again? :?", "Solution_11": "[quote=\"shuresh\"]thanks hardsoul and siuhochung\nbut siuhochung i dont get something\n[quote]$(\\frac{-y^2}{p})=(-y^2)^{\\frac{p-1}{2}}=-1 (mod p)$ (Fermat's little theorem) [/quote]\nfirst we supposed that $y$ is not divisible by $p$,so $(\\frac{-y^2}{p})$ is not an integer and can not be used in mod.\ncould you explain your solution again? :?[/quote]\r\n\r\nit 's the legendre symbol." } { "Tag": [], "Problem": "Propose a synthesis of 4-(carboximethyl)-cyclopent-2-enone (a cyclopentanone ring having a double bond between carbons 2/3 and the group $ \\minus{}CH_2CO_2H$ attached to carbon 4) from cyclopentadiene. You can only use other organic starting materials with 2 carbons maximum, and any necessary inorganic reagents.\r\n\r\nNote: this problem was created by me some years ago, and my solution uses 3 steps.", "Solution_1": "cyclopentadiene + PCC( oxidises allylic methylene group to keto group) foll by reaction with methyl cyanide in presence of base(michael reaction) foll by hydrolysis of CN to COOH.", "Solution_2": "You cannot do that, because cyclopentadienone is a very unstable compound (since it is antiaromatic) and as soon as formed it immediatly undergoes a Diels-Alder reaction with itself.", "Solution_3": "so what, we can generate cyclopentadienone in situ rite. i mean benzyne also dimerises but we definitely have reactions in which benzyne is the intermediate", "Solution_4": "But benzyne is not antiaromatic. Also, even generated in situ it would immediatly undergo Diels-Alder with itself. I have written under the topic title \"Advanced\" for some reason. This type of problems is not supposed to be solved by students of your level (no offense meant). There are plenty of problems out there that I think at this stage would be much more usefull to you (spectroscopy problems, for example)." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Everyone knows that if $ E\\equal{} \\{ x \\in (0,1) \\ni x \\notin \\mathbb{Q} \\}$ then m(E) = 1. That's what we expect. But, I want to go through the mechanics of applying the definition of outer measure to this set. The outer measure will be equal to the measure, provided the set is measurable. Definition of outer measure:\r\n\r\n$ m*E \\equal{} \\inf_{E\\subset U I_n} \\sum l(I_n)$\r\n\r\nThen {In} are countable collections of open intervals that cover E. Now what I'm thinking is that even if we split E so that we exclude every rational (that is, split (0,1) in to (0, 1/2)and(1/2, 1) ...etc. ) our open cover still has a total measure of 1 since we are only removing one point at a time and a point has measure 0. I guess I'm just bothered that this process of splitting is clearly countable, but using any of the countable sequences that included all rationals between (0,1) we will never remove all of the rationals from any given neighborhood. We can always go back and take more out. \r\n\r\n :huh: \r\n\r\nAnd then this is also saying that every non-rational number is in one of the intervals in the set {In}-- but it is never alone-- in fact there must be uncountably many other non-rationals in there with it. And I can't even name one of the intervals in {In} any way. if I say (a/b, c/d) we can always split it *again* --\r\n\r\nI guess that is all pretty cool but also a little disturbing. I guess I'm still not really comfortable with the notion of \"dense\" -- \r\n\r\n :o \r\n\r\nHow would you prove that the measure of this set is 1? Is there another way?", "Solution_1": "You know $ m^*(Q) \\equal{} 0$ and $ m^*([0,1]) \\equal{}1$, right?", "Solution_2": "I think he's looking for a constructive proof using the definition of outer measure. If you enumerate the rationals in $ [0,1]$, say $ (r_{n})$, let $ I_{n}\\equal{}(r_{n}\\minus{}\\frac{\\epsilon}{2^{n\\minus{}1}},r_{n}\\plus{}\\frac{\\epsilon}{2^{n\\minus{}1}})$, where $ \\epsilon>0$ is chosen so that $ \\sum_{n\\equal{}1}^{\\infty}\\frac{\\epsilon}{2^{n}}<1$. You know that the outer measure of an interval is equal to its length. Note that $ [0,1]\\setminus\\bigcup_{n\\equal{}1}^{\\infty}I_{n}\\subseteq\\mathbb{R}\\setminus\\mathbb{Q}$. Also, $ m^*\\left(\\bigcup_{n\\equal{}1}^{\\infty}I_{n}\\right)\\leq\\sum_{n\\equal{}1}^{\\infty}I_{n}<1$ by construction. But there's no reason we can't choose $ \\epsilon$ so that $ \\sum_{n\\equal{}1}^{\\infty}\\frac{\\epsilon}{2^{n}}<\\frac{1}{k}$, where $ k\\in\\mathbb{N}$. In other words, we can shave off smaller and smaller neighborhoods of rational numbers and enlarge our set of irrationals. This isn't exactly what you want (or is it?) but I think it's a little more constructive.", "Solution_3": "Jrav, thanks, that is more along the lines of what I was thinking of!" } { "Tag": [ "ARML", "geometry", "ratio", "AMC", "USA(J)MO", "USAMO", "induction" ], "Problem": "Once the awards have been handed out and people have survived their bus trips back home, I'd like to use this thread here for people to share their thoughts on this year's contest. Too hard? Too easy? Too [i]awesome?[/i] :wink: \r\n\r\nGood luck to all y'all's this weekend, and remember (it's noted elsewhere) that that sites might not be perfectly in sync throughout this year's ARML, so wait until Saturday evening, after the awards, to do any discussin' :)", "Solution_1": "Partial results:\r\n\r\nTeams:\r\n1. Lehigh Fire, 215\r\n2. Exeter Red, 201\r\n3. San Francisco Bay Area A1, 197\r\n4. New York City A1, 196\r\n5. Montgomery A1, 196\r\n5. Exeter Red, 192\r\n6. Southern California A1, 191\r\n7. North Carolina A1, 186\r\n8. San Diego Surf, 182\r\n9. Indiana Gold, 182\r\nGeorgia: South Carolina A1, 173\r\nIowa: Chicago A1, 177\r\nLas Vegas: Washington Gold, 166\r\nPenn: Thomas Jefferson A1, 180\r\n\r\nDivision B:\r\n1. Thomas Jefferson B1, 156\r\n2. Andover B1, 150\r\n3. Lehigh Ice, 144\r\n4. Central Jersey B1, 136\r\n5. San Diego Sun, 130\r\n6. Utah B1, 129\r\n7. Maine B1, 126\r\nGeorgia: Grissom B1, 115\r\nIowa: Michigan Naturals, 112\r\nLas Vegas: Colorado B1, 104\r\nPenn: NYC B3, 124\r\n\r\nInternational:\r\n1. Macau\r\n\r\n(Site awards for teams were not all announced)\r\n\r\nIndividuals:\r\n1. Alex Song, Ontario West\r\n2. Warut Suksimpong, Exeter\r\n3. Qin Xuan Pan, Montgomery\r\n4. Junu Bae, Western Penn\r\n5. Chong Gu, NYC\r\n6. David Yang, SoCal\r\n7. Bayley Wang, SFBA\r\n8. Zhifan (Ivan) Zhang, SoCal\r\n9. Aki Hogge, Montgomery\r\n10. Robert Yang, Utah\r\nGeorgia site: John Berman, NC\r\nIowa site:\r\nLas Vegas site: Sam Keller, Washington Gold\r\nPenn site: Ofir Nachum, Western Mass A1\r\n\r\nThe top three individuals scored 10, with the other winners at 9.\r\n\r\n[color=green][Edited for Exeter/Andover correction.][/color]", "Solution_2": "It was really a great competition this year. The Team Contest was harder than usual which is sometimes enjoyable. The Power Round, I thought, was pretty easy aside from such problems as #10 and, according to my teammates, 5(b). The Individual round I failed completely =.= I got 1, 2, 3, 5, 6, and 9. For 4, I was just being stupid and tried to solve for the actual legs of the triangle rather than the sum of their reciprocals (of course, in the last 30 seconds, I realized my stupidity and got the correct set of ratios, but was just a little too late). #7 I forgot the 00, 00 case, and so got 41 instead of 42. #8 I didn't really try (as I had a [correct] feeling that my answer to #7 was missing something...) and #10 I was close to getting, but the time constraints wouldn't allow for it. So, while I should have gotten an 8 easy, I am instead left with a 6. Disappointing a little, but the problems were really fun and I had a great time.\r\n\r\nBy the way, how do you do Team #9?", "Solution_3": "The Power Round and Individual Questions were far easier this year compared to last year. Our team, Chesapeake A1, got a 44 on the Power Round after a more disappointing score last year and losing our best seniors. As for the individual questions, I got the first eight and should have gotten nine had I not wasted time trying to bash #10 and just trying to check #9, where I had messed up the surface area of a cylinder. :mad: Excluding 10, none of the problems seemed that difficult. I would have liked to make tiebreakers; I think I would have been one of the few non-USAMO qualifiers to do so had I made tiebreakers. I would have gotten the question quickly too; I guessed the correct answer by virtue of having absolutely no idea on how to proceed and dividing the two numbers given. :rotfl:\r\n\r\nThe Team Round seemed harder, but I think it was because our proctor messed up our timing and we had a couple minutes of consternation trying to figure out exactly how much time we had left. We failed badly on Relays.\r\n\r\nI liked this ARML more than last year's ARML. The Song Contest at Penn State never ceases to delight.", "Solution_4": "[quote=\"jmerry\"]Partial results:\n\nTeams:\n1. Lehigh Fire, 215\n2. SFBA A1, 197\n3. New York City A1, 196\n4. Montgomery A1, 196\n5. Exeter Red, 192\n6. Southern California A1, 191\n7. North Carolina A1, 186\n8. San Diego Surf, 182\n9. Indiana Gold, 182\n\nDivision B:\n1. Exeter B1, 159\n2. Thomas Jefferson B1, 156\n3. Lehigh Ice, 144\n4. Central Jersey B1, 136\n5. San Diego Sun, 130\n6. Utah B1, 129\n7. Maine B1, 126\n\nInternational:\n1. Macau\n\nIndividuals:\n1. Alex Song, Ontario West\n2.\n3.\n4.\n5.\n6. David Yang, SoCal A1\n7. Bayley Wang, SFBA A1\n8. Zhifan (Ivan) Zhang, SoCal A1\n9. , Alabama\n10. Robert Yang, Utah B1\nGeorgia site:\nIowa site:\nLas Vegas site: Sam Keller, Washington Gold\nPenn site:\n\nThe top three individuals scored 10, with the other winners at 9.[/quote]\r\nI remember from PA site:\r\n2. someone from Exeter\r\n3. Qinxuan Pan, Montgomery\r\n4. someone from NY?\r\n5. Junu Bae, Western PA\r\n\r\nyeah that didnt really help :P", "Solution_5": "The ones that jmerry is missing are:\r\n\r\n2. Warut Suksimpong, Exeter\r\n3. Qin Xuan Pan, Montgomery\r\n4. Junu Bae, Western Penn\r\n5. Chong Gu, NYC\r\n9. Aki Hogge, Montgomery\r\n\r\nGeorgia site winner: John Berman, NC\r\n\r\nI believe that Montgomery is Montgomery County, Maryland. Not Montgomery Alabama. For the 1st time in a while the normal Alabama team didn't come. There was a Grissom, AL team that was the Georgia B site winner.", "Solution_6": "how did the alabama individuals do?", "Solution_7": "My team (AAST Mu A) answered everything on power, and tied for third. I think we got a few points off for style/handwriting/stuff like that, I dunno. Well, we did use group theory. Or maybe we ACTUALLY FAILED but I do not wish to believe that. \r\n\r\n[quote]The Power Round, I thought, was pretty easy aside from such problems as #10 and, according to my teammates, 5(b). [/quote]\nGuess what the only problem I did the whole time period was! Oddly enough, I recently found [url=http://mathworld.wolfram.com/EulerZigzagNumber.html]this[/url], and the answer was on [url=http://www.research.att.com/~njas/sequences/A000111]this[/url], but of course I forgot it. I made up my own much more complicated formula ($ a_1 \\equal{} 1$ and $ a_{n \\plus{} 2} \\equal{} \\sum_{k \\equal{} 1}^{\\frac {n \\plus{} 1}{2}} {n \\plus{} 1 \\choose 2k \\minus{} 1} a_{2k \\minus{} 1} a_{2n \\minus{} 2k \\plus{} 2}$, or something like that.) I just proved that formula's validity and bashed it out to find $ a_5$ and $ a_9$. Proud that I got it, I assigned it to a minion who wasn't doing anything. And he got it wrong, which resulted in me doing the fastest computations in my life to change the answer in the last minute or so. \n\nIn the team round, I was assigned to work on #9 the whole time, and therefore did not contribute to the team round at all. And since I only worked on 5 for power, I can't tell you anything about those. \n\nI failed individual, I don't really want to talk about it. Same with relay. \n\nThe only thing I'll say is that I used complementary counting once and it screwed me over. And then I didn't use it and I got screwed over again. Individual #4 and relay 1 #3, respectively. \n\nI must say I was fond of #8. It was a fun little problem. I wrote #10 in terms of complex numbers (a tool on recently learned here), and ended up with a cubic, and ran out of time. \n\nOverall, my team failed relay. One member read 12 as 21. \n\nOverall, my team failed to place at all. Not even in the singing contest. \n\nBy the way, I was the brave volunteer who held up the microphone.\n\n[quote]6. David Yang, SoCal A1 [/quote]\r\nSURPRISE!!!!1111111", "Solution_8": "[quote=\"J.Mayers\"]\nBy the way, how do you do Team #9?[/quote]\r\nSay that $ a\\plus{}b\\equal{}c$, then you know that $ b\\equal{}c\\minus{}a$. You can substitute this in, expand, multiply the first equation by 8, and then set the coefficients of each $ a$ term equal to each other. Then you solve for $ c$. Sorry--I don't have a copy of the problem, and nor did I get it in the round, but this is the solution our coach gave us afterwards.", "Solution_9": "Actually, the third place team should read \"New York City A1: Murph and the Magictones\"", "Solution_10": "I used induction for power round #10. How did everyone else do it?", "Solution_11": "[quote=\"jmerry\"]\nIndividuals:\n1. Alex Song, Ontario West\n[/quote]\r\nIs this the same Alex Song from Waterloo, Ontario who qualified for USAMO last year as a 5th grader? I refuse to believe a 6th grader won individuals....", "Solution_12": "[quote=\"buzzer11\"]I used induction for power round #10. How did everyone else do it?[/quote]\r\n\r\nWhat was the problem again? I didn't look at it during the round.\r\n\r\nComments:\r\n\r\nIndividual Round:\r\nThis round seemed a lot easier this year. 10 was hard but a lot of people i know bashed it and got a quadratic in x^2 but got -8 root 3 instead of +8root3.\r\nPower Round:\r\nEhh, I only looked at 2 problems, so I might be wrong, but the problems I looked at weren't very hard and according to my team 10 was the only hard question?\r\nTeam Round:\r\nThis round was really good... it was also harder than normal. My team (SoCal A1) missed 4 problems on this.....\r\nTiebraker #1: I got this by looking at a case where everything seemed to work, but I missed that the triangle was degenerate... arg I didn't solve this legitly but at least I got the right answer...", "Solution_13": "Was there a \"smart\" way to do individual #10?", "Solution_14": "For #10, divide both sides by $ 2$. Now take the tangent of both sides. Tangent of $ \\frac{\\pi}{2}$ is undefined, so the denominator of the tangent of the left side is $ 0$. If we have $ \\tan \\left(a\\plus{}b\\plus{}c\\plus{}d\\plus{}e\\right)$, the denominator is $ e_0\\minus{}e_2\\plus{}e_4$, so set that equal to $ 0$ and you have a quadratic in $ x^2$.", "Solution_15": "#3 in relay round, had the question \"...has T+1 colors of finger polish...\" implying T is an integer. was passed back \"10^(1+rt5)/4\". i was like... uh okay let's just pretend he meant to pass back 5. probably should have guessed power of 10 = 10 but whatever", "Solution_16": "[quote=\"Ravi B\"][quote=\"Team Round #9\"]Let $ a$ and $ b$ be real numbers such that\n\\[ a^3 \\minus{} 15 a^2 \\plus{} 20a \\minus{} 50 \\equal{} 0 \\quad \\textrm{and} \\quad 8 b^3 \\minus{} 60 b^2 \\minus{} 290 b \\plus{} 2575 \\equal{} 0.\n\\]\nCompute $ a \\plus{} b$.[/quote]\nHere's a cheapo method for solving Team Round #9. By the form of the question, it seems likely that the roots of the first equation are some constant $ c$ minus the roots of the second equation. Let's make that assumption. Hence the sum of the roots of the first equation is $ 3c$ minus the sum of the roots of the second equation. By Vieta's formula, the sum of the roots of the first equation is $ 15$ and the sum of the roots of the second equation is $ \\frac {15}{2}$. So we get the equation $ 15 \\equal{} 3c \\minus{} \\frac {15}{2}$. Solving for $ c$ gives $ c \\equal{} \\boxed{\\frac {15}{2}}$..[/quote]\r\nHere's another cheapish method, found by grn_trtle:\r\n\r\nMultiply the first equation by $ 8$ and add it to the second, to obtain $ 8b^3\\minus{}60b^2\\minus{}290b\\plus{}8a^3\\minus{}120a^2\\plus{}160a\\plus{}2175 \\equal{} 0.$ \r\nNow, we guess that this factors as $ (a \\plus{} b \\minus{} c)(Aa^2 \\plus{} Bb^2 \\plus{} Cab \\plus{} Da \\plus{} Eb \\plus{} F) \\equal{} 0$, where $ c$ is the desired answer to this problem. \r\nComparing coefficients of $ a^3$ and $ b^3$, we have $ A \\equal{} B \\equal{} 8$. \r\nLooking at the coefficients of $ a^2$ and $ b^2$, we get $ \\minus{} 8c \\plus{} D \\equal{} \\minus{} 120$ and $ \\minus{} 8c \\plus{} E \\equal{} \\minus{} 60$. Subtracting, we obtain $ D \\minus{} E \\equal{} \\minus{} 60$. \r\nLooking at the coefficients of $ a$ and $ b$, we have $ \\minus{} Dc \\plus{} F \\equal{} 160$ and $ \\minus{} Ec \\plus{} F \\equal{} \\minus{} 290$. Subtracting, we have $ \\minus{} c(D \\minus{} E) \\equal{} \\minus{} 450$. \r\nSubstituting $ D \\minus{} E \\equal{} \\minus{} 60$, we obtain $ c \\equal{} \\frac {15}{2}$.", "Solution_17": "[quote=\"ra5249\"]#3 in relay round, had the question \"...has T+1 colors of finger polish...\" implying T is an integer. was passed back \"10^(1+rt5)/4\". i was like... uh okay let's just pretend he meant to pass back 5. probably should have guessed power of 10 = 10 but whatever[/quote]\r\n\r\nThe final answer is $ (T\\plus{}1)T^4\\minus{}(T\\plus{}1)T$. R1-2 is a log-base-10 equation, so my thinking was that if the person in position #2 messed up or had a wrong answer to begin with, they would pass back a power of 10. This, coupled with the fact that $ (T\\plus{}1)T^4$ is a royal pain to compute for all but a few values of $ T$ (like 2, 3, and 10) leads me to believe that a 2-man relay team would have stood a good chance at arriving at the correct final answer. If position 3 receives some goofy non-integer power of 10, their first (and only) guess for the intended $ T$ should be 10.\r\n\r\nI told this to a fellow coach, who said agreed, but who had heard several people in position #2 say they solved and got $ \\log x\\equal{}1$ and passed back $ x\\equal{}1$ :blush: It's hard to use at least 3 different colors of fingernail polish when you only have one color :P", "Solution_18": "On relay 1-3, I was passed the correct answer, $ 10$, I still didn't get the problem right! I got that there were $ (T \\plus{} 1)T ^ 4 \\equal{} 110000$ ways for Kay to paint her fingernails such that no two consecutive fingernails have the same color, but not necessarily using at least three colors. I then attempted to subtract the number of combinations that used fewer than three colors, but I thought that there were $ 2 ^ 5 \\cdot \\binom{11}{2}$ of those combinations, instead of $ 2 \\cdot \\binom{11}{2}$, because I didn't realize that I had to subtract combinations where no two fingernails are the same color! :blush:", "Solution_19": "It was pretty fun for my first ARML. I got a 5 on my B1 with two mistakes, but it was a very fun experience. In the team round we did pretty bad, with a 2, but I checked and corrected #1 and got #3 for our second problem. The power round was mostly a failure for us. I thought (and many of my teammates) thought it focused to much on terminology and much less on actual mathematical ability. I had to check back multiple times just to see what some of the terms meant. In result, we got a 23/50 there. Our relay and super relays were pretty much disasters. On the first our third person (I was 2nd) couldn't get it for 5 or so minutes. On the 2nd, the first guy gave me a coordinate, not a number. So I guessed. I was wrong.\r\n\r\nOn the super relay, I was at position 7. Unluckily, the lines on both sides stopped at around 3 and 10. We never were close to finishing.", "Solution_20": "I thought Team #9 was ridiculously contrived - as soon as you start solving the cubics (depressing them) the answer just appears out of thin air. Once you remembered your trig and had the insight to see the denominator had to be 0, Individual #10 was an algebra bash. Overall, I felt the problems on Individual were sort of ugly this year.", "Solution_21": "I had a cleanish solution to individual #10 using complex numbers, which I'll post later when I have time.\r\n\r\n@archimedes1: So you finally got that system of coefficient equations to work out :D", "Solution_22": "Here's my solutions: \r\n[hide=\"No Complex Numbers\"]\nLet $ x \\equal{} \\tan(A) \\Rightarrow A \\equal{} \\tan^{ \\minus{} 1}(x)$ and $ 3x \\equal{} \\tan(B) \\Rightarrow B \\equal{} \\tan^{ \\minus{} 1}(3x)$. \n\nWe are trying to solve $ 6A \\plus{} 4B \\equal{} \\pi \\Leftrightarrow 3A \\plus{} 2B \\equal{} \\frac {\\pi}{2} \\Rightarrow \\tan(3A)\\tan(2B) \\equal{} 1$. \n\nUsing triple and double angle formulas (Which aren't too hard to derive during the 10 mins): \n\n$ \\tan(3A) \\equal{} \\dfrac{3\\tan(A) \\minus{} \\tan^3(A)}{1 \\minus{} 3\\tan^2(A)} \\equal{} \\dfrac{3x \\minus{} x^3}{1 \\minus{} 3x^2}$ \n\n$ \\tan(2B) \\equal{} \\dfrac{2\\tan(B)}{1 \\minus{} \\tan^2(B)} \\equal{} \\dfrac{6x}{1 \\minus{} 9x^2}$ \n\nThus, we have: \n$ \\dfrac{3x \\minus{} x^3}{1 \\minus{} 3x^2} \\cdot \\dfrac{6x}{1 \\minus{} 9x^2} \\equal{} 1$ \n\n$ (3x \\minus{} x^3)(6x) \\equal{} (1 \\minus{} 3x^2)(1 \\minus{} 9x^2)$\n\n$ 18x^2 \\minus{} 6x^4 \\equal{} 27x^4 \\minus{} 12x^2 \\plus{} 1$ \n\n$ 33x^4 \\minus{} 30x^2 \\plus{} 1 \\equal{} 0$ \n\nUsing the quadratic formula yields: \n$ x^2 \\equal{} \\dfrac{ \\minus{} ( \\minus{} 30) \\pm \\sqrt {( \\minus{} 30)^2 \\minus{} (4)(33)(1)}}{2(33)} \\equal{} \\dfrac{15 \\pm 8\\sqrt {3}}{33}$ \n\nNow since clearly $ 0 < A < B$, we need $ \\frac {\\pi}{2} \\equal{} 3A \\plus{} 2B > 5A \\Rightarrow 0 < A < \\frac {\\pi}{10}$, and thus, $ 0 < \\tan(A) < \\tan(\\frac {\\pi}{10}) < \\frac {1}{2}$ \n\nThus, the desired solution is $ x^2 \\equal{} \\boxed{\\dfrac{15 \\minus{} 8\\sqrt {3}}{33}}$.[/hide]\n\n[hide=\"Complex Numbers\"]\nNote that $ arg(1\\plus{}xi) \\equal{} \\tan^{\\minus{}1}(x)$, and $ arg(1\\plus{}3xi) \\equal{} \\tan^{\\minus{}1}(3x)$ \n\nThen, the equation becomes: \n$ 6arg(1\\plus{}xi) \\plus{} 4arg(1\\plus{}3xi) \\equal{} \\pi$\n\n$ 3arg(1\\plus{}xi) \\plus{} 2arg(1\\plus{}3xi) \\equal{} \\frac{\\pi}{2}$ \n\n$ arg\\left[(1\\plus{}xi)^3(1\\plus{}3xi)^2\\right] \\equal{} \\frac{\\pi}{2}$ \n\n$ (1\\plus{}xi)^3(1\\plus{}3xi)^2 \\equal{} Ki$ for some $ K \\in \\mathbb{R^\\plus{}}$ \n\n$ Re[(1\\plus{}xi)^3(1\\plus{}3xi)^2] \\equal{} 0$ \n\nAfter some tedious expansion, this becomes: \n$ Re[(33 x^4\\minus{}30x^2\\plus{}1)\\plus{}(9x^5\\minus{}46x^3\\plus{}9x)i] \\equal{} 33 x^4\\minus{}30x^2\\plus{}1 \\equal{} 0$ \n\nThen proceed the same way as in the non-complex numbers method\n[/hide]", "Solution_23": "Yeah, that's the second solution in the official packet.. the complex numbers one is the first.", "Solution_24": "That was my solution as well, though I didn't think to divide by 2 while taking the actual test :blush: Time pressure...\r\n\r\nAt least I saw the complex numbers behind it, so I'm happy with myself...", "Solution_25": "Complex numbers is essentially what I did, except I didn't divide by 2, thus making it even more bashy. I somehow ended up with a cubic in $ x^2$ instead. Probably made an algebra mistake somewhere. \r\n\r\nProbably shouldn't have spent so much time checking 9. \r\n\r\nEDIT: Wow I somehow didn't see grn_turtle's post.", "Solution_26": "hey guys... check the national rankings again!", "Solution_27": "OMG really? That stinks lol.\r\n\r\nYeah, this year there were many embarrassing mistakes made by the ARML staff. Including telling my team that we were one of the top 5 teams...and then thirty minutes later, they tell us that they were kidding lol.", "Solution_28": "Team contest problem $ 9$ is an old idea with different numbers...\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=268928\r\n\r\nThe tiebreaker problem 1 was quite pretty!\r\n\r\nOverall, the test looks great. Too bad my ARML days are over :(", "Solution_29": "the official solution to tiebreaker one was weird....I think it had some typos?\r\n\r\nIt took 5 lines to show that $ \\angle ADE \\equal{} \\angle CBE$ when it could have just said $ \\angle ADE \\equal{} 180 \\minus{} \\angle EDC \\equal{} \\angle CBE$ since $ BCDE$ is cyclic. \r\n\r\nAlso, the angle bisector through me off when I worked this since I think it is irrelevant to the problem?\r\n\r\nI just saw this problem as:\r\n\r\n since $ BPC\\sim EPD$ and $ ABC\\sim ADE$ (since $ BCDE$ is cyclic), \r\n\r\n\\[ \\dfrac{BP}{PE} \\equal{} \\dfrac{BC}{DE} \\equal{} \\dfrac{AC}{AE} \\equal{} 3.\\]" } { "Tag": [], "Problem": "What is the value of $ 1\\minus{}2\\plus{}3\\minus{}4\\plus{}5\\minus{}6\\plus{} \\ldots \\minus{}98\\plus{}99\\minus{}100$?", "Solution_1": "$ 1\\minus{}2 \\plus{} 3\\minus{}4 \\plus{} ... \\plus{} 99\\minus{}100$\r\n\r\n$ \\minus{}1 \\plus{} \\minus{}1 \\plus{} ... \\plus{} \\minus{}1$\r\n\r\n$ \\boxed{\\minus{}50}$", "Solution_2": "Alternatively, its the sum of the first $ 50$ odd integers minu the sum of the first $ 50$ even integers. The sum of the first $ n$ odd integers is $ n^2$. The sum of the first $ n$ even integers is $ n(n\\plus{}1)$. So, $ 50^2\\minus{}50(51)\\equal{}50(\\minus{}1)\\equal{}\\boxed{\\minus{}50}$" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Two circles $C_1$ and $C_2$ intersect at $S$. \r\nThe tangent in $S$ to $C_1$ intersects $C_2$ in $A$ different from $S$.\r\nThe tangent in $S$ to $C_2$ intersects $C_1$ in $B$ different from $S$.\r\nAnother circle $C_3$ goes through $A, B, S$.\r\nThe tangent in $S$ to $C_3$ intersects $C_1$ in $P$ different from $S$ and $C_2$ in $Q$ different from $S$.\r\nProve that the distance $PS$ is equal to the distance $QS$.", "Solution_1": "As an angle between a chord and a tangent in the circle $C_3$, the angle < BSP is equal to the angle < BAS. Also, as an angle between a chord and a tangent in the circle $C_1$, the angle < ASB equals the angle < SPB. In other words, we have < BSP = < BAS ans < SPB = < ASB. Hence, the triangles BSP and BAS are similar, and it follows that PS : BS = SA : BA. In other words, PS : BS = AS : AB. Thus, $PS \\cdot AB = AS \\cdot BS$. Similarly, $QS \\cdot AB = AS \\cdot BS$. Hence, PS = QS as intended to show.\r\n\r\n Darij", "Solution_2": "[b][u]Remark.[/u][/b] A intersting metric relation :\r\n\r\n$C_k=C(O_k,r_k),\\ k\\in \\overline {1,2},\\ \\triangle SAQ\\sim \\triangle BSP\\sim \\triangle BAS\\Longrightarrow \\underline {\\overline {\\left| \\frac {AQ}{BP}=\\left( \\frac {r_2}{r_1}\\right) ^2.\\right| }}$", "Solution_3": "Ancient topic, but it in Contest Collections, so...\nWith $T$ - midpoint of $AB$, $TS$ - middle line of trapezoid $AQPB$. Done.\n", "Solution_4": "Let $\\angle SAB=\\alpha,\\angle SBA=\\beta,\\angle BSA=\\gamma$. Then, by the tangents, we have $$\\angle PSB=\\alpha,\\angle SPB=\\gamma,\\angle QSA=\\beta,\\angle SQA=\\gamma.$$ Hence, we also have $$\\angle PBS=\\beta,\\angle QAS=\\alpha.$$ Hence, by Law of Sines spam, $$SQ=PS\\cdot\\frac{\\sin \\gamma}{\\sin \\beta}\\cdot\\frac{\\sin \\beta}{\\sin \\alpha}\\cdot \\frac{\\sin\\alpha}{\\sin\\gamma}=PS,$$ so we are done." } { "Tag": [ "limit", "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "Let $a_0>0$ and $a_{n+1} = \\arctan a_n$ for $n \\ge 0$. \r\nShow that $\\lim_{n \\to 0} a_n = 0$ ([i]easy[/i]). \r\nThen find numbers $L>0, \\, \\alpha$, possibly depending on $a_0$, such that $\\lim_{n \\to \\infty}\\frac{a_n}{Ln^\\alpha} = 1$, or prove that no such numbers exist.", "Solution_1": "classical method:\r\n$arctan(x)=x(1-2x^2+o(x^2))$ thus $arctan(x)^a=x^a(1-2ax^2+o(x^2))$. So if one chooses $a=-2$,\r\n$arctan(x)^{-2}=x^{-2} +2 +o(1)$.\r\nSo $u_{n+1}^{-2}-u_n^{-2}$ tends towards $2$. This is why $u_n$ is equivalent to $(2n)^{-\\frac{1}{2}}$", "Solution_2": "[quote=\"alekk\"] $arctan(x)=x(1-2x^2+o(x^2))$ thus $arctan(x)^a=x^a(1-2ax^2+o(x^2))$. [/quote]\r\nWhy?\r\n\r\n\r\n\r\nEdit- I figrd it out. Stuid me!", "Solution_3": "[quote=\"alekk\"]classical method:\n$\\arctan(x)=x(1-2x^2+o(x^2))$[/quote]\r\n$\\arctan(x)=x(1-x^2/3+o(x^2))$, giving a different constant. But I'd still like to see a proof that goes beyond formal asymptotics :)", "Solution_4": "what do you mean ? (i.e what is formal asymptotics :) )", "Solution_5": "[quote=\"alekk\"]what do you mean ? (i.e what is formal asymptotics :) )[/quote][i]Formal [/i] asymptotics is asymptotics without proving that the asymptotic statement is valid (\"Expand in powers of $\\epsilon$, compare coefficients, the result is ...\") :huh: \r\n\r\nWhat I was looking for is a proof that makes sense to a first year college student or advanced high school student. The problem can be stated at this level, so there should be a solution not too far from this level.", "Solution_6": "Using derivatives one can show:\r\n$\\forall x > 0, x- \\frac{x^3}{3} < arctan(x) < x - \\frac{x^3}{3} + \\frac{x^5}{5}$\r\nso: $- \\frac{1}{3} < \\frac{a_{n+1} - a_n}{a_n^3} < - \\frac{1}{3} + \\frac{a_n^2}{5}$\r\n\r\n$\\frac{a_{n+1}}{a_n} = \\frac{arctan(a_n)}{a_n} \\to 1$\r\n\r\n$\\frac{1}{a_{n+1}^2} - \\frac{1}{a_n^2} = \\frac{a_n - a_{n+1}}{a_n^3} \\cdot [(\\frac{a_n}{a_{n+1}})^2 + \\frac{a_n}{a_{n+1}}] \\to \\frac{2}{3}$\r\n\r\nUsing Stoltz-Cesaro, $\\frac{1}{n \\cdot a_n^2} \\to \\frac{2}{3}, L=\\sqrt{\\frac{3}{2}}, \\alpha=- \\frac{1}{2}$" } { "Tag": [ "linear algebra", "matrix", "induction", "algebra", "polynomial", "quadratics" ], "Problem": "Consider the matrix $ A\\equal{}\\left(\\begin{array}{cc}6 & 5\\\\ \\ \\minus{}3 & \\minus{}2\\end{array}\\right).$ Compute $ A^n, n \\in \\mathbb{N}^*.$", "Solution_1": "Applying Cayley-Hamilton's theorem, we get $ A^2 \\plus{} 4A \\plus{} 3I \\equal{} O$. ($ I$ is the unit matrix, and $ O$ is all-zero matrix)\r\n\r\nIt can be rewritten as\r\n$ A(A \\plus{} I) \\equal{} \\minus{} 3(A \\plus{} I)\\cdots\\boxed{1}$\r\n$ A(A \\plus{} 3I) \\equal{} ( \\minus{} 1)(A \\plus{} 3I)\\cdots\\boxed{2}$\r\n\r\nBy induction, \r\n\r\n$ A^n(A \\plus{} I) \\equal{} ( \\minus{} 3)^n(A \\plus{} I)\\cdots\\boxed{1}'$\r\n$ A^n(A \\plus{} 3I) \\equal{} ( \\minus{} 1)^n(A \\plus{} 3I)\\cdots\\boxed{2}'$\r\n\r\n$ \\frac {1}{2}\\cdot(\\boxed{2}' \\minus{} \\boxed{1}')$, we get $ A^n \\equal{} \\frac {( \\minus{} 1)^n \\minus{} ( \\minus{} 3)^n}{2}A \\plus{} \\frac {3( \\minus{} 1)^n \\minus{} ( \\minus{} 3)^n}{2}I,\\quad{n\\in{N}}$\r\n\r\nAnother method;\r\n\r\nLet's think about a polynomial identity.\r\n$ x^n \\equal{} (x \\plus{} 1)(x \\plus{} 3)Q_{n}(x) \\plus{} p_{n}x \\plus{} q_{n}$ We can define $ p_{n},q_{n}$ by substitute $ x \\equal{} \\minus{} 1, \\minus{} 3$. \r\nAnd then, substitute $ x$ to matrix $ A$, we get the same answer. (Think $ 1, 3$ to $ I, 3I$.)", "Solution_2": "[quote=\"J.Y.Choi\"]Applying Cayley-Hamilton's theorem, we get $ A^2 \\plus{} 4^A \\plus{} 3I \\equal{} O$. [/quote]\r\n\r\nThat's wrong.\r\n\r\n[color=green][It was a typo. I've fixed it for him. - K.M.][/color]", "Solution_3": "Alternatively, just diagonalize it. We've seen the characteristic polynomial; the eigenvalues are $ 3$ and $ 1.$\r\n\r\nFor the eigenvalue $ 3,$ find the null space of $ \\begin{bmatrix}3 & 5 \\\\\r\n\\minus{} 3 & \\minus{} 5\\end{bmatrix},$ getting $ \\begin{bmatrix}5 \\\\\r\n\\minus{} 3\\end{bmatrix}.$\r\n\r\nFor the eigenvalue $ 1,$ find the null space of $ \\begin{bmatrix}5 & 5 \\\\\r\n\\minus{} 3 & \\minus{} 3\\end{bmatrix},$ getting $ \\begin{bmatrix} \\minus{} 1 \\\\\r\n1\\end{bmatrix}.$\r\n\r\nLet $ P \\equal{} \\begin{bmatrix}5 & \\minus{} 1 \\\\\r\n\\minus{} 3 & 1\\end{bmatrix},$ so that $ P^{ \\minus{} 1} \\equal{} \\frac12\\begin{bmatrix}1 & 1 \\\\\r\n3 & 5\\end{bmatrix}.$\r\n\r\nWe have that $ P^{ \\minus{} 1}AP \\equal{} \\begin{bmatrix} 3 & 0 \\\\\r\n0 & 1\\end{bmatrix} \\equal{} D,$ or $ A \\equal{} PDP^{ \\minus{} 1}.$\r\n\r\nWe can then raise that to any power:\r\n\r\n$ A^n \\equal{} PD^nP^{ \\minus{} 1} \\equal{} \\frac12\\begin{bmatrix}5 & \\minus{} 1 \\\\\r\n\\minus{} 3 & 1\\end{bmatrix} \\begin{bmatrix}3^n & 0 \\\\\r\n0 & 1\\end{bmatrix} \\begin{bmatrix} 1 & 1 \\\\\r\n3 & 5\\end{bmatrix}$\r\n\r\n$ \\equal{} \\frac12\\begin{bmatrix}5 & \\minus{} 1 \\\\\r\n\\minus{} 3 & 1\\end{bmatrix}\\begin{bmatrix}3^n & 3^n \\\\\r\n3 & 5\\end{bmatrix} \\equal{} \\frac12\\begin{bmatrix}5\\cdot3^n \\minus{} 3 & 5\\cdot3^n \\minus{} 5 \\\\\r\n\\minus{} 3\\cdot3^n \\plus{} 3 & \\minus{} 3\\cdot3^n \\plus{} 5\\end{bmatrix}.$", "Solution_4": "[quote=\"kunny\"][quote=\"J.Y.Choi\"]Applying Cayley-Hamilton's theorem, we get $ A^2 \\plus{} 4^A \\plus{} 3I \\equal{} O$. [/quote]\n\nThat's wrong.\n\n[color=green][It was a typo. I've fixed it for him. - K.M.][/color][/quote]\r\n\r\nSorry, yet it is wrong. :(", "Solution_5": "The sum of each column of A is 3, therefore 3 is an eigenvalue with eigenspace spanned by (5,-3). Observe now that 6-5=1 and -3-(-2) = -1, so 1 is the other eigenvalue with eigenspace spanned by (1,-1). Hence,\r\n\r\n$ A^n \\equal{} \\left[\\begin{array}{cc}1 & 5 \\\\\r\n\\minus{} 1 & \\minus{} 3\\end{array}\\right] \\left[\\begin{array}{cc}1 & 0 \\\\\r\n0 & 3^n \\end{array}\\right]\\left[\\begin{array}{cc}1 & 5 \\\\\r\n\\minus{} 1 & \\minus{} 3\\end{array}\\right]^{ \\minus{} 1} \\equal{} \\frac12 \\left[\\begin{array}{cc}5 \\cdot 3^n\\minus{}3 & 5(3^n\\minus{}1) \\\\ 3(1\\minus{}3^n) & 5\\minus{}3^{n\\plus{}1}\\end{array}\\right]$.", "Solution_6": "That's correct. \r\n\r\nWhy don't you examine wheter the result is correct or not by substituting $ n \\equal{} 0,\\ 1$? :wink: \r\n\r\nAnswer:\r\n\\[ \\boxed{\\boxed{A^n \\equal{} \\frac {3^n \\minus{} 1}{2}A \\minus{} \\frac {3^n \\minus{} 3}{2}E}}\\]", "Solution_7": "Yikes! I really screwed that up. The eigenvalues are $ 3$ and $ 1,$ not $ \\minus{}3$ and $ \\minus{}1.$ The characteristic polynomial is $ x^2\\minus{}4x\\plus{}3,$ not $ x^2\\plus{}4x\\plus{}3.$ And my diagonalization post was riddled with errors. I've tried to clean it up, but I won't swear I got everything.", "Solution_8": "Sorry,i made a mistake. (Applying Cayley Hamilton's theorem, we get $ A^2\\minus{}4A\\plus{}3I\\equal{}O$.) But the approach was right, though.... :)", "Solution_9": "That's O.K. but you should edit it. :wink: \r\n\r\nAnyway here is my solution:\r\n\r\nThe equation of the line passing through the points $ (1,\\ 1),\\ (3,\\ 3^n)$ on $ y \\equal{} x^n\\ (n\\geq 2)$ is $ y \\equal{} \\frac {3^n \\minus{} 1}{2}(x \\minus{} 1) \\plus{} 1\\Longleftrightarrow y \\equal{} \\frac {3^n \\minus{} 1}{2}x \\minus{} \\frac {3^n \\minus{} 3}{2}$, we can write $ x^n \\equal{} (x \\minus{} 1)(x \\minus{} 3)Q(x) \\plus{} \\frac {3^n \\minus{} 1}{2}x \\minus{} \\frac {3^n \\minus{} 3}{2}$ where $ Q(x)$ is a monic polynomial with degree $ n\\geq 2.$, thus by Caley-Hamilton theorem, we have $ A^2\\minus{}4A\\plus{}3E\\equal{}\\mathbb{O}$.\r\n\r\n$ \\therefore A^n \\equal{} (A^2 \\minus{} 4A \\plus{} 3E)Q(A) \\plus{} \\frac {3^n \\minus{} 1}{2}A \\minus{} \\frac {3^n \\minus{} 3}{2}E$, yielding $ A^n \\equal{} \\frac {3^n \\minus{} 1}{2}A \\minus{} \\frac {3^n \\minus{} 3}{2}E\\ (n\\geq 0)$", "Solution_10": "By Caley-Hamilton thorem we have $ A^2 - 4A + 3E = \\mathbb{O}\\Longrightarrow A^n - 4A^{n - 1} + 3A^{n - 2} = \\mathbb{O}\\ (n\\geq 2).$\r\nwe can write the equation in two ways as follows:\r\n\r\n$ \\left\\{ \\begin{array}{cc}{ll} A^n - A^{n - 1} = 3(A^{n - 1} - A^{n - 2}) & \\quad \\\r\nA^n - 3A^{n - 1} = A^{n - 1} - 3A^{n - 2} & \\quad \\end{array} \\right.$\r\n\r\n$ \\therefore \\left\\{ \\begin{array}{{cc}{ll} A^n - A^{n - 1} = 3^{n - 1}(A - E})\\ \\cdots [1] & \\quad \\\r\nA^n - 3A^{n - 1} = A - 3E \\ \\ \\ \\ \\ \\ \\ \\cdots [2] & \\quad \\end{array} \\right.$\r\n\r\n$ ([1]*3 - [2])\\div 2$ gives $ A^n = \\frac {3^n - 1}{2}A - \\frac {3^n - 3}{2}E\\ (n\\geq 0)$", "Solution_11": "My penance for screwing this up: I led off my class today with exactly this example. We're studying eigenvalues, eigenvectors, and diagonalization.\r\n\r\nMy second lecture example was a formula for the powers of $ \\begin{bmatrix}0&1\\\\1&1\\end{bmatrix}.$\r\n\r\nThat's always a good example.", "Solution_12": "[quote=\"Kent Merryfield\"]\nMy second lecture example was a formula for the powers of $ \\begin{bmatrix}0 & 1 \\\\\n1 & 1\\end{bmatrix}.$\n\nThat's always a good example.[/quote]\r\n\r\nLet $ \\alpha ,\\ \\beta \\ (\\alpha < \\beta)$ be the roots of the quadratic equation $ \\lambda ^2 \\minus{} \\lambda \\minus{} 1 \\equal{} 0$,\r\n\r\nDoing as my previous post, #10, we obtain $ A^n \\equal{} \\frac {\\beta ^ n \\minus{} \\alpha ^n}{\\beta \\minus{} \\alpha}A \\plus{} \\frac {\\beta ^{n \\minus{} 1} \\minus{} \\alpha ^ {n \\minus{} 1}}{\\beta \\minus{} \\alpha}E$, yielding $ \\boxed{\\boxed{A^n \\equal{} \\left( \\begin{array}{cc} f(n \\minus{} 1) & f(n) \\\\\r\nf(n) & f(n \\plus{} 1) \\end{array} \\right)\\ (n\\geq 0)}}$ for some $ f(n)$.\r\n\r\nWhat's $ f(n)$? That's left for you. :D \r\n\r\nI have more 4 solutions. :lol:\r\n\r\n1.Diagonalization\r\n\r\n2\uff0eEigenvalue, Eigenvector\r\n\r\n3. Reccurence\r\n\r\n4. [color=blue]Spectrum Decomposition[/color]", "Solution_13": "I worked it into an explicit formula for the $ n$th Fibonacci number.", "Solution_14": "Yes, at first sight the problem, it occured to me.\r\n\r\nKent Merryfield, see my previous post, What's ''Spectrum'' in Spectrum Decomposition that appeares in Japanese University entrance examination?" } { "Tag": [ "floor function" ], "Problem": "What is the greatest prime factor of $ 55^{100}\\plus{}55^{101}\\plus{}55^{102}$?", "Solution_1": "We can factor out a $ 55^{100}$ to get $ 55^{100}\\plus{}55^{101}\\plus{}55^{102}\\equal{}55^{100}(55^0\\plus{}55^{1}\\plus{}55^{2})\\equal{}55^{100}(1\\plus{}55\\plus{}3025)\\equal{}55^{100}(3081)$\r\n3+8+1=12 so we can factor out a 3.\r\n\r\n$ 5^{100}\\cdot11^{100}\\cdot3\\cdot1027$\r\n\r\nIs 1027 prime? No it's 13x79 so our largest factor is 79.", "Solution_2": "\\begin{eqnarray*}\r\n1027\r\n& = & 1000 + 27 \\\\\r\n& = & 10^3 + 3^3 \\\\\r\n& = & (10+3)(10^2 - 10\\times 3 + 3^2) \\\\\r\n& = & 13 \\times 79\r\n\\end{eqnarray*}\r\n\r\n$ 79$ is prime.\r\nBecause $ \\left\\lfloor \\sqrt{79}\\right\\rfloor = 8$. So $ 2, 3, 5, 7 \\nmid 79$.\r\n\r\n\r\n\r\nBoth $ \\nmid$ and $ \\not |$ are unshapely.", "Solution_3": "Please excuse the revival, but what does the $ \\nmid$ symbol mean?" } { "Tag": [ "calculus", "derivative", "function", "real analysis", "real analysis unsolved" ], "Problem": "Anyone can demonstrate how to deduce Fa\u00e1 di Bruno's formula for nth derivative of the composition f(g(t)),if f(t) and g(t) are functions for which all necessary derivatives are defined?\r\n\r\nPlease moderators Kent,blahblahblah,fedja,.., May you solve it?\r\nthanks", "Solution_1": "Anyone can deduce it???\r\nthanks" } { "Tag": [ "percent" ], "Problem": "The chart below was created from the results of a radio station survey. What percent\nof the males surveyed listen to the station?\n\\begin{tabular}{| c | c | c | c|}\n\\hline &Listened to Station&Did not Listen to Station & Total\\\\ \\hline\nMale & $ 39\\%$& $ 13\\%$&$ 52\\%$ \\\\ \\hline\nFemale &$ 29\\%$ & $ 19\\%$ & $ 48\\%$ \\\\ \\hline\nTotal & $ 68\\%$ & $ 32\\%$ & $ 100\\%$ \\\\ \\hline\n\\end{tabular}", "Solution_1": "$ \\begin{tabular}{| c | c | c | c|} \r\n\\hline &Listened to Station&Did not Listen to Station & Total\\\\ \\hline \r\nMale & 39\\%&13\\% & 52\\%\\\\ \\hline \r\nFemale & 29\\%& 19\\% & 48\\%\\\\ \\hline \r\nTotal & 68\\% & 32\\%& 100\\% \\\\ \\hline \r\n\\end{tabular}$\r\n\r\n$ \\frac{39}{52} \\equal{} \\boxed{75\\%}$" } { "Tag": [ "geometry", "3D geometry", "sphere", "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Show that the infinite-dimensional sphere $S^{\\infty} = \\cup_{n} S^{n}$ is contractible.", "Solution_1": "[quote=\"chungyc\"]Show that the infinite-dimensional sphere $S^{\\infty} = \\cup_{n} S^{n}$ is contractible.[/quote]\r\n\r\nHow's the book reading going so far?\r\n\r\nYou seem to have lots of questions, do you ever come around to answering them?\r\n\r\nI'm just curious by the way. I plan on studying Algebraic Topology sometime soon, like maybe in the Summer.", "Solution_2": "I would say that the book is suitable for beginners like myself because it has pictures and lots of descriptions to illustrate concepts. My lecturer isn't really helping much with the way he's teaching so I am struggling with the subject. Knowing the theory is one matter; being able to apply it to problems is another. I have been going through the exercise at the end of chapter 0, and so far I've only been able to solve about half of the questions. Those that I have posted are some questions that I've not been able to answer.", "Solution_3": "[quote=\"chungyc\"]I would say that the book is suitable for beginners like myself because it has pictures and lots of descriptions to illustrate concepts. My lecturer isn't really helping much with the way he's teaching so I am struggling with the subject. Knowing the theory is one matter; being able to apply it to problems is another. I have been going through the exercise at the end of chapter 0, and so far I've only been able to solve about half of the questions. Those that I have posted are some questions that I've not been able to answer.[/quote]\r\n\r\nHave you considered reading other Introductions to Algebraic Topology?\r\n\r\nI find it better to have two texts. I follow one, and if it isn't clear at one point, I look it up in the other. That sometimes works, but then again there are plenty of books at the university library, which is where I assume you study.\r\n\r\nIf you are capable of answering half the questions, that's pretty good by my standard. The best thing to do is to just move on to Chapter 1, and come back to it later. Unless of course, you need to answer those questions for the next chapter.\r\n\r\nI wish I can help, but like I mentionned, I'm just not there yet.", "Solution_4": "Evidently every homotopy group is trivial, to see this just map some n-sphere to the infinite-sphere.\r\n\r\nOn the other hand, you may construct a homotopy from the id to constant map. For example, you slide the (n-1)-sphere along the n-sphere continuously to a point in the homotopy map F(x,t) when t belongs to [1-2^-(n-1),1-2^-n].", "Solution_5": "Or, you can just look at Example 1.B.3 in Hatcher.", "Solution_6": "It seems as if everyone uses Munkres for point-set topology, but no one uses the second half of the book for algebraic topology. Is it lacking in some way, or other books just better? I don't know since I haven't read any algebraic topology yet." } { "Tag": [ "geometry", "Pythagorean Theorem" ], "Problem": "In the figure below, ABCD is a square piece of paper 6 cm on each side. Corner C is folded over so that it coincides with E, the midpoint of $ \\overline{AD}$. If $ \\overline{GF}$ represents the crease created by the fold, what is the length of $ \\overline{FD}$? Express your answer as a common fraction.\n[asy]import geometry_dev;\nsize(150);\npair A = (0,0), B = (0,1), C = (1,1), D = (1,0);\npath square = A--B--C--D--cycle;\ndraw(square);\nlabel(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE);\npair E = midpoint(A--D);\nline CE = line(C,E);\npair X = midpoint(C--E); line FG = perpendicular(X,CE); \npair[] intwithsquare = intersectionpoints(FG,square);\npair G = intwithsquare[0];\npair F = intwithsquare[1];\ndraw(F--G,dashed);\nlabel(\"F\",F,E);\nlabel(\"G\",G,W);[/asy]", "Solution_1": "[geogebra]b8443d910511f13fa2910517f147499ae08c0739[/geogebra] \r\n\r\nBy pythagorean theorem\r\n\r\n(6-x)^2+9=x^2\r\nx^2-12x+36+9=x^2\r\n45=12x\r\nx=15/4", "Solution_2": "since 6-x is the hypotenus (i killed that spelling) of the right triangle, shouldn't it be (x-6)^2[b]-[/b]9?", "Solution_3": "Yes, I think it should be.\r\n\r\n$ (6 \\minus{} x)^2 \\minus{} 9 \\equal{} x^2$\r\n\r\n$ 36 \\minus{} 12x \\plus{} x^2 \\minus{} 9 \\equal{} x^2$\r\n\r\n$ 27 \\minus{} 12x \\equal{} 0$\r\n\r\n$ x \\equal{} \\boxed{\\frac{9}{4}}$" } { "Tag": [ "LaTeX", "irrational number" ], "Problem": "The proposition below is true or false? If it's true, then prove.\r\n\r\n$D \\in \\mathbb{N} \\thickspace \\wedge \\thickspace d \\in \\mathbb{N} \\thickspace \\wedge \\thickspace D = d^2 \\sqrt{d} \\quad \\Rightarrow \\quad \\sqrt{d} \\in \\mathbb{N}$", "Solution_1": "What does $\\wedge$ mean? Well I think if $D$ and $d$ are both $\\in N$, then $d^2 \\in N$. Since $D=d\\sqrt{d}$, $D$ is in $N$, so the other side of the equation must be also in $N$, so $\\sqrt{d}$ is also in $N$.", "Solution_2": "[quote=\"shobber\"]What does $\\wedge$ mean?[/quote]It means \"and\".", "Solution_3": "I think it is true, As shobber siad. So Bruno Bonagura Are we right? :huh:", "Solution_4": "I don't know :|\r\nI think shobber afirmation is false. Look...\r\n\r\n$D = d^2\\sqrt{d} \\thickspace \\Rightarrow \\thickspace \\sqrt{d} = \\frac{D}{d^2}$\r\n\r\nBoth $D$ and $d^2$ are natural but their division may not result a natural too. \r\n\r\nI'm really confused! :?", "Solution_5": "$d$ is an integer so $\\sqrt{d}$ [b]can't[/b] be a fraction. It is obvious but I am trying to prove it... This is a very difficult problem for me...", "Solution_6": "[quote=\"Bruno Bonagura\"]I don't know :|\nI think shobber afirmation is false. Look...\n\n$D = d^2\\sqrt{d} \\thickspace \\Rightarrow \\thickspace \\sqrt{d} = \\frac{D}{d^2}$\n\nBoth $D$ and $d^2$ are natural but their division may not result a natural too. \n\nI'm really confused! :?[/quote]\r\n\r\nSince you KNOW $D=d^2\\sqrt d$ and $D\\in \\mathbb{N}$ --> $d^2\\sqrt{d}$ must also be Natural . And if $d^2\\sqrt{d}\\in \\mathbb{N}$ $\\wedge$ $d\\in \\mathbb{N}$ --->$\\sqrt d\\in \\mathbb{N}$ . So i guess shobber isnt wrong ....", "Solution_7": "[quote=\"shyong\"]Since you KNOW $D=d^2\\sqrt d$ and $D\\in \\mathbb{N}$ --> $d^2\\sqrt{d}$ must also be Natural . And if $d^2\\sqrt{d}\\in \\mathbb{N}$ $\\wedge$ $d\\in \\mathbb{N}$ --->$\\sqrt d\\in \\mathbb{N}$ . So i guess shobber isnt wrong ....[/quote]\r\n\r\nI also think it is right but I don't believe in \"guess mathematics\". I'm trying to prove it.\r\n\r\nWe know that for every $a,b \\in \\mathbb{N}$ we have $ab \\in \\mathbb{N}$ (field axiom). But a solution for this problem is to prove that: $a,b \\in \\mathbb{N} \\thickspace \\wedge \\thickspace a = bc \\thickspace \\Rightarrow \\thickspace c \\notin \\mathbb{N}$ It's true? :?", "Solution_8": "[quote=\"Bruno Bonagura\"][quote=\"shyong\"]Since you KNOW $D=d^2\\sqrt d$ and $D\\in \\mathbb{N}$ --> $d^2\\sqrt{d}$ must also be Natural . And if $d^2\\sqrt{d}\\in \\mathbb{N}$ $\\wedge$ $d\\in \\mathbb{N}$ --->$\\sqrt d\\in \\mathbb{N}$ . So i guess shobber isnt wrong ....[/quote]\n\nI also think it is right but I don't believe in \"guess mathematics\". I'm trying to prove it.\n\nWe know that for every $a,b \\in \\mathbb{N}$ we have $ab \\in \\mathbb{N}$ (field axiom). But a solution for this problem is to prove that: $a,b \\in \\mathbb{N} \\thickspace \\wedge \\thickspace a = bc \\thickspace \\Rightarrow \\thickspace c \\notin \\mathbb{N}$ It's true? :?[/quote]\r\n\r\nLets see , if $ab\\in \\mathbb{N}$ , then is it always $a,b\\in \\mathbb{N}$ . Well , this is not true here since it can be $a=\\frac 12$ and $b=10$ . But if $a\\in \\mathbb{N}$ , how would you say about $b$ ? Could $b$ always be natural ? No again since $b$ CAN be a fraction \r\n. \r\nOk back to our question , now since $d^2\\sqrt d\\in \\mathbb{N}$ and $d^2\\in \\mathbb{N}$ , $\\sqrt d$ can be either Natural or Fraction . But you know that $d$ is Natural , and a square root of natural number wont give you a fraction (it can be either an integer or irrational number obvious) . So in conclusion , $\\sqrt d \\in\\mathbb{N}$", "Solution_9": "[quote=\"Bruno Bonagura\"]The proposition below is true or false? If it's true, then prove.\n\n$D \\in \\mathbb{N} \\thickspace \\wedge \\thickspace d \\in \\mathbb{N} \\thickspace \\wedge \\thickspace D = d^2 \\sqrt{d} \\quad \\Rightarrow \\quad \\sqrt{d} \\in \\mathbb{N}$[/quote]$D=d^2\\sqrt{d}\\Longrightarrow\\sqrt{d}=\\frac{D}{d^2}\\in\\mathbf{Q}$. It is well known, as shyong just said, that when $d\\in\\mathbf{N}$, $\\sqrt{d}$ is either irrational or an integer. Since it is in $\\mathbf{Q}$, it's not irrational, so $\\sqrt{d}\\in\\mathbf{N}$. What's all the confusion about?", "Solution_10": "The fact $d\\in \\mathbb{N} \\implies \\sqrt{d}\\notin\\mathbb{Q-N}$. So that means $\\sqrt{d}$ is either natural or irrational. Then $D=d^2\\sqrt{d} \\implies \\sqrt{d}\\in \\mathbb{Q}$. Thus, the intersection of these two implies that it is natural.", "Solution_11": "Now it seems a little bit clearer... Thanks by the help!\r\n\r\nBut somebody can prove that? \r\n\r\n$d \\in \\mathbb{N} \\implies \\sqrt{d} \\in (\\mathbb{R - Q}) \\thickspace \\vee \\thickspace \\sqrt{d} \\in \\mathbb{N}$ \r\n\r\nPlease?", "Solution_12": "[quote=\"Bruno Bonagura\"]Now it seems a little bit clearer... Thanks by the help!\n\nBut somebody can prove that? \n\n$d \\in \\mathbb{N} \\implies \\sqrt{d} \\in (\\mathbb{R - Q}) \\thickspace \\vee \\thickspace \\sqrt{d} \\in \\mathbb{N}$ \n\nPlease?[/quote]\r\n\r\nI don't like the excessive use of symbols - words are nice too. Anyways,\r\n\r\n$d \\in \\mathbb{N} \\implies \\sqrt{d} \\in (\\mathbb{R - Q}) \\thickspace \\vee \\thickspace \\sqrt{d} \\in \\mathbb{Q}$\r\n\r\nis a tautology, as soon as we know that square roots do exist (they do).\r\n\r\nSo we want to show that $\\sqrt{d}\\not\\in(\\mathbb{Q-N})$. Assume otherwise. Then we have $\\sqrt{d}=m/n$ with $(m,n)=1$ and $n>1$, so $d=m^2/n^2$, and $(m^2,n^2)=1$, and $n^2>1$, so that $d\\not\\in \\mathbb{N}$, a contradiction.", "Solution_13": "Well it was me the unblessed who formuled this question ( http://www.somatematica.com.br/forumsm/viewtopic.php?p=18843#18843 ).I based myself on that proposition to solve a very dificult question, but if that proposition is wrong, the solution to that question is wrong too. Bruno Bonagura participates of that forum and indicated me this forum.\r\n\r\nWell, I will use Bruno Bonagura logic:\r\n$\\sqrt{d} = \\frac{D}{d^2} \\ \\ \\ \\{D, d\\} \\in \\mathbb{N}$\r\n\r\n$D=d^2\\sqrt{d} \\ \\Rightarrow \\ d^2$ and $\\sqrt{d}$ are divisors of $D$. If $d^2$ is a divisor of $D$, then $\\frac{D}{d^2}$, which is equal to $\\sqrt{d}$, is a natural number.\r\n\r\n\r\nIs that right??\r\n\r\n\r\n[size=75][i]edit: I liked that Latex code![/i][/size] :D" } { "Tag": [ "algebra", "polynomial", "integration", "algebra unsolved" ], "Problem": "Let be a natural number $ k$ and a polynomial $ P(x)\\in R[x]$ such that $ P(2x) = 2^{k-1}(P(x)+P(x+\\frac{1}{2}))$ $ \\forall x\\in R$ . Prove that :\r\n $ P(3x) = 3^{k-1}( P(x)+P(x+\\frac{1}{3})+P(x+\\frac{2}{3}))$ $ \\forall x\\in R$\r\nComment : This problem looks very nice ! It's the reason why I post it again . \r\nand to N.T.Tuan : I'll try to corect my mistakes . :)", "Solution_1": "[quote=\"Nbach\"]Let be a natural number $ k$ and a polynomial $ P(x)\\in R[x]$ such that $ P(2x) = 2^{k-1}(P(x)+P(x+\\frac{1}{2}))$ $ \\forall x\\in R$ . Prove that :\n $ P(3x) = 3^{k-1}( P(x)+P(x+\\frac{1}{3})+P(x+\\frac{2}{3}))$ $ \\forall x\\in R$\nComment : This problem looks very nice ! It's the reason why I post it again . \nand to N.T.Tuan : I'll try to corect my mistakes . :)[/quote]\r\nObviosly if $ P(x)$ is solution\r\n(1) $ P_{k}(2x) = 2^{k-1}(P_{k}(x)+P_{k}(x+\\frac{1}{2})$,\r\nthen $ aP(x)$ is solution too. Therefore we can consider only monic polinoms (with leader coefficient=1).\r\nFrom (1) we get \r\n(2) $ P_{k}'(2x) = 2^{k-2}(P_{k}'(x)+P_{k}'(x+\\frac{1}{2})$ and \r\n(3) $ P_{k}(0)(2^{1-k}-1) = P_{k}(\\frac{1}{2})$.\r\nFrom (2) and (3) we get unique monic polinom \r\n$ P_{0}\\equiv 1,\\ P_{1}= x-\\frac{1}{2},\\ P_{2}(x) = x^{2}-x+\\frac{1}{6},...P_{k}(x) = k\\int_{0}^{x}P_{k-1}(y)dy-c,c =-\\frac{k}{2-2^{1-k}}\\int_{0}^{1/2}P_{k-1}(y)dy.$ \r\nTherefore $ P_{k}(x)$ is $ k-th$ polinom Bernulli.\r\nProve. Let $ f_{k}(x) = P_{k}(1-x)$, then $ f(2x) = P_{k}(1-2x) = 2^{k-1}(P_{k}(\\frac{1}{2}-x)+P_{k}(1-x)) = 2^{k-1}(f_{k}(x)+f_{k}(\\frac{1}{2}+x)$,\r\ntherefore $ P_{k}(1-x) = (-1)^{k}P_{k}(x)$ and $ P_{k}'(x) = kP_{k-1}(x)$. It give$ P_{k}(x)\\equiv B_{k}(x)$.\r\nBy same method from $ P_{k}(3x) = 3^{k-1}(P(x)+P(x+\\frac{1}{3})+P(x+\\frac{2}{3})$ we get $ P_{k}(x)\\equiv B_{k}(x)$ for monic polinoms." } { "Tag": [ "limit", "integration", "calculus", "calculus computations" ], "Problem": "It's been a couple years since I took Calc 1a, and my friend asked me a question that I couldn't figure out\r\n\r\n\r\n\r\n$ \\lim_{x \\minus{} > \\infty} \\int_{x}^{x^2} e^{ \\minus{} t^2} dt$\r\n\r\n\r\n\r\nThere has to be some sort of trick that I'm missing\r\nI tihnk the answer is 0, but I have no idea if that is right nor how to show it\r\n\r\n\r\nAny help is appreciated\r\n\r\nThanks", "Solution_1": "For $ x>1,$\r\n\r\n$ 0<\\int_x^{x^2}e^{-t^2}\\,dt<\\int_x^{\\infty}e^{-t^2}\\,dt$\r\n\r\n$ \\int_x^{\\infty}e^{-t^2}\\,dt<\\int_x^{\\infty}e^{-t}\\,dt=e^{-x}$\r\n\r\nAnd that goes to zero. We could make the estimate much tighter than this, but we don't need to in order to solve the problem.\r\n\r\nGeneralization: Suppose $ \\int_0^{\\infty}f(t)\\,dt$ converges and suppose that $ g(x) 1$ . Thus $ 1 < x < x^2$ and exists $ c_x\\in \\left(x,x^2\\right)$ so that\r\n\r\n$ \\int_{x}^{x^2} e^{ - t^2}\\ \\mathrm {dt} = \\left(x^2 - x\\right)\\cdot e^{ - c^2_x}\\ .$ Thus, $ x < c_x < x^2$ $ \\implies$ $ \\lim_{x\\to\\infty}c_x = \\infty$ , \r\n\r\n$ 0 < \\frac {x^2 - x}{e^{c^2_x}} < \\frac {x^2 - x}{e^{x^2}}$ , $ \\lim_{x\\to\\infty}\\frac {x^2 - x}{e^{x^2}} =\\lim_{x\\to\\infty}\\frac {x^2-x}{x^2}\\cdot\\frac {x^2}{e^{x^2}}=0$ $ \\implies$ $ L = 0\\ .$", "Solution_4": "I also used mean value theorem of Integral, Virgil Nicula. :)" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "need help in this urgently\r\n\r\nhow do you prove the differentiability of the function a(x,y) if we know b(x,y) and b(x,y)*a(x,y) are both differentiable?\r\n\r\nthanks", "Solution_1": "We don't: take $ b\\equiv 0$ and choose $ a$ to be whatever you want. Of course, you didn't mean this case, so, in other situations, just recall that the quotient of two differentiable functions is differentiable when the denominator does not vanish. \r\n\r\nNow as a moderator, I have to remark that this is [b]not[/b] a homework help hotline, so normally requests for \"urgent help\" with routine questions are ignored at best and result in not answering your posts at all at worst ;) (the fact that I answered now is due to a combination of my good mood and your being new to the forum :))." } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "9) Find the equation of the line which intersects the positive x and positive y-axis, contains the point $(-5/2,3)$ and the sum of whose intercepts is 4.\r\n\r\n10) What is the slope of the line passing though the vertices of the graphs of the equations $y=x^2-4x+3$ and $y=-x^2-4x-1$?", "Solution_1": "[quote=\"cowpi\"]9) Find the equation of the line which intersects the positive x and positive y-axis, contains the point $(-5/2,3)$ and the sum of whose intercepts is 4.\n\n10) What is the slope of the line passing though the vertices of the graphs of the equations $y=x^2-4x+3$ and $y=-x^2-4x-1$?[/quote]\r\n[hide=\"9\"]\nIf y=y intercept and x=xintercept, we know that from the point given $\\frac{3}{\\frac{5}{2}+x}=\\frac{y}{x}$ and $y+x=4$. Substituting $y=4-x$ into the first equation and solving for x, $x=\\frac{5}{2}$ and then $y=\\frac{3}{2}$, then finding the slope and yintercept the equation is $\\boxed{y=-\\frac{3}{5}x+\\frac{3}{2}}$[/hide]\n[hide=\"10\"]\nVertex: 2 and -2 \nfor the first one, $f(2)=-1$ and for the second, $f(-2)=3$. Points are (2,-1) and (-2,3) and the slope is $\\boxed{-1}$. [/hide]", "Solution_2": "[hide=\"9\"]s=slope\nb=y intercept\n$3=-5s/2+b$\n$b-b/s=4$\n$-6/5+2b/5=s$\n$b-\\frac{b}{\\frac{-6}{5}+\\frac{2b}{5}}=4$\n$b-\\frac{5b}{-6+2b}=4$\n$-6b+2b^2-5b=-24+8b$\n$2b^2-19b+24=0$\n$b=3/2,8$\nsince both intercepts are postive, $b=3/2$\n$\\frac{\\frac{3}{2}}{\\frac{-5}{2}}=\\frac{-3}{5}$\n\n$y=\\frac{-3x}{5}+\\frac{3}{2}$[/hide]\n\n[hide=\"10\"]$\\frac{-b}{2a}=x coordinate of vertex$\nreenter in and you get\n$(2,-1) (-2,3)$\n$y-2=-(x+1)$\n\n$y=-x+1$\noops answered wrong question $-1$[/hide]", "Solution_3": "[hide=\"Number 10\"]\nI graphed the lines on my calculator and found the vertices are $(-2,3)$ and $(2,-1)$. Apply slope formula to get $\\boxed {-1}$.[/hide]" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "MATHCOUNTS", "ARML", "percent", "factorial" ], "Problem": "The list is comprised of every state we have received information from with at least a 37 cutoff and the other option. I think Indiana or California will win the team.", "Solution_1": "I'm rooting for NY, although I have a feeling Texas is going to win again...\r\nP.S. Is ESPN airing the countdown round?When?", "Solution_2": "I'm rooting for Indiana, and I have a feeling they will win", "Solution_3": "I doubt IN or TX will win again. I think Indiana's lost a lot of depth compared to last year's team, and TX has also lost its top scorers. When in doubt, vote Cal.", "Solution_4": "Michigan all the way. We have a chance.", "Solution_5": "Yes for Michigan.\r\n\r\nWho else voted for Michigan, I thought Neil and I were the only \"Michiganders\" (I think).\r\n\r\nDefinitely many teams have a chance, but, though the Michigan cutoff wasn't that high, we have NEIL GURRAM, that says it all.\r\n\r\nMostly what matters is the team round, so as long as the team can work together and has good strategy, you should get a 10 on the team round, and with it, the national championship.", "Solution_6": "Thanks Kyyuanmathcount, I couldn't have said it better myself.", "Solution_7": "the maryland team will own in the competition. THey are amazing. :D", "Solution_8": "Kansas is going to win :rotfl: ! ......I voted for Michigan.", "Solution_9": "Ohio has a 37 cutoff, but seeing as I am the only one in the state and I already voted for another, it isn't going to get any votes...", "Solution_10": "[quote=\"kyyuanmathcount\"]Yes for Michigan.\n\nWho else voted for Michigan, I thought Neil and I were the only \"Michiganders\" (I think).\n\nDefinitely many teams have a chance, but, though the Michigan cutoff wasn't that high, we have NEIL GURRAM, that says it all.\n\nMostly what matters is the team round, so as long as the team can work together and has good strategy, you should get a 10 on the team round, and with it, the national championship.[/quote]\r\n\r\nTeam round isn't going to be the determining factor, it's going to be the team's average.\r\n\r\nTop 5 I think will be Texas, California, Indiana, Michigan, and somewhere else.\r\n\r\nI don't think Texas will win this year =\\ California has Kevin Yang and generally high state scores obviously maybe 2nd or 3rd? Indiana I think will win, their team is still pretty good, Michigan I think will also get 2nd or 3rd just missing Indiana :P", "Solution_11": "[quote=\"kyyuanmathcount\"]Who else voted for Michigan, I thought Neil and I were the only \"Michiganders\" (I think).\n[/quote]\r\n\r\nMICHIGAN PRIDE!!!\r\n\r\nYou thought wrong :D", "Solution_12": "I voted for MI because I'm from MI, but I still think the MI team lost everything when Alan Huang left.", "Solution_13": "But we have gained stuff, and don't think that we will not do well just because Alan Huang is gone.", "Solution_14": "Obviously South Carolina will win it all. They've got me...lol.", "Solution_15": "hey, I won kinda fairly...", "Solution_16": "I thought the answer was unfair at the time, but don't they start the timer right a fter you buzz in anyways?", "Solution_17": "[quote=\"Treething\"]hey, I won kinda fairly...[/quote]\r\n\r\ni'm not saying you didn't, from what i heard, you just used it to your advantage.", "Solution_18": "Virginia was 1/4 point ahead of Washington, Indiana was 1/4 point behind Washington, Texas was far back (2-4 points?)", "Solution_19": "[quote=\"janieluvsmusic\"][quote=\"Treething\"]hey, I won kinda fairly...[/quote]\n\ni'm not saying you didn't, from what i heard, you just used it to your advantage.[/quote]\r\n\r\n\r\nsee, loopholes are for winners :D\r\n\r\nneal wu could've beat me 4-0 if he used that to his advantage, Andrew could've beaten me 4-2 probably\r\n\r\nbut nathan benjamin was funny either way", "Solution_20": "[quote=\"Fanatic\"]Virginia was 1/4 point ahead of Washington, Indiana was 1/4 point behind Washington, Texas was far back (2-4 points?)[/quote]\r\n\r\nheh. 2.5", "Solution_21": "what was the cutoff for team awards?", "Solution_22": "Do you mean top ten or top 3?", "Solution_23": "[quote=\"janieluvsmusic\"]what was the cutoff for team awards?[/quote]\r\n\r\nI think you need like top 3 teams, and Pulak definitely could've improved 11 points for TX to get third. Pulak's score, let's just say, was within 3 points of half of Neal Wu's score.", "Solution_24": "Klebian's 24 :ninja:", "Solution_25": "[quote=\"Ignite168\"][quote=\"janieluvsmusic\"]what was the cutoff for team awards?[/quote]\n\nI think you need like top 3 teams, and Pulak definitely could've improved 11 points for TX to get third. Pulak's score, let's just say, was within 3 points of half of Neal Wu's score.[/quote]\r\n\r\n35 would've put me in 14th place, which I wasn't really expecting anyway, so I doubt I could've improved that much myself.", "Solution_26": "Congratulations to the Virginia kids for their showing at Nationals. They did a great job and deserve all their kudos!!!\r\n\r\nAlso, congratualtions to the 4 Texas kids. 4th in the nation is fantastic. I am proud of everything that you did to prepare this year. A special congratulations for Kevin Chen for finishing 3rd in the Nation as a 7th grader. 29th as a 6th grader and 3rd as a 7th grader..WOW! A great improvement!! \r\n\r\nAlso congratualtions to everyone that went to Nationals. No matter how you did you are all STATE CHAMPIONS!!!! No one can ever take that away from you. You are all the best in your state!\r\n\r\n :lol: :lol: :lol: :lol: :lol: \r\n\r\nJeff Boyd\r\nTexas Coach", "Solution_27": "yay\r\n\r\nnow i just have to start preparing for next year", "Solution_28": "I have no next year :( goodbye mathcounts.\r\nI'll probably be back as some sort of assistant coach though ;)", "Solution_29": "[quote=\"samath\"]I have no next year :( goodbye mathcounts.\nI'll probably be back as some sort of assistant coach though ;)[/quote]\r\nBut, you did make countdown, and you did do pretty well." } { "Tag": [ "real analysis", "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Given $ \\varepsilon > 0$, construct an open set $ E\\subset [0,1]$ that is dense in $ [0,1]$ such that $ m(E) \\equal{} \\varepsilon$ ($ m(E)$ is the Lebesgue measure)\r\n\r\nI thought of a set which seems to work but is cumbersome to write down: Take open intervals of lenghts $ \\frac {\\varepsilon}{2}$, $ \\frac {\\varepsilon}{4}$, ...\r\nDivide the first into two and put the parts at the ends of the intervals, then place the second centred around $ \\frac {1}{2}$ and so on.\r\n\r\nThis looks something like: \r\n\r\n$ \\left(0,\\frac {\\varepsilon}{4}\\right)\\cup\\left(1 \\minus{} \\frac {\\varepsilon}{4},1\\right)\\cup\\left(\\frac {1}{2} \\minus{} \\frac {\\varepsilon}{8},\\frac {1}{2} \\plus{} \\frac {\\varepsilon}{8}\\right)\\cup\\left(\\frac {1}{4} \\plus{} \\frac {\\epsilon}{32},\\frac {1}{4} \\plus{} \\frac {3\\varepsilon}{32}\\right)\\cup\\left(\\frac {3}{4} \\minus{} \\frac {3\\epsilon}{32},\\frac {3}{4} \\minus{} \\frac {\\varepsilon}{32}\\right)\\cup\\ldots$\r\n\r\nAny ideas of a more elegant solution?\r\nThanks!", "Solution_1": "choose an enumeration $ q_1,q_2,...$ of the rationals in $ [0,1]$ and choose balls around the $ q_n$ of radius $ \\epsilon 1/2^{n\\plus{}1}$ (or something like that).", "Solution_2": "That gets you an open set with measure strictly less than $ \\epsilon$, since the balls overlap.\r\n\r\nThe construction in the first post works. Another idea is the complement of a Cantor-type set:\r\nLet $ a_n$ be a sequence of numbers in $ (0,1)$ with $ \\prod_n (1\\minus{}a_n)\\equal{}1\\minus{}\\epsilon$. Repeat this procedure: at step $ n$, remove the open middle $ a_n$ part of each remaining interval. After infinitely many steps, we are left with a closed set with empty interior and measure $ 1\\minus{}\\epsilon$. Its complement is the desired dense open set of measure $ \\epsilon$.", "Solution_3": "Thanks jmerry\r\n\r\n[quote=\"jmerry\"]Repeat this procedure: at step $ n$, remove the open middle $ a_n$ part of each remaining interval.[/quote]\r\n\r\nI guess you mean \"remove the open middle parts of all remaining intervals so as total length removed is $ a_n$\"\r\n\r\nAlso, is there a better way to write down my original construction?", "Solution_4": "No. I mean my $ a_n$ to be the proportion removed at each step. The original Cantor set uses $ a_n\\equal{}\\frac13$, so its measure is $ \\prod_{n\\equal{}1}^{\\infty}(1\\minus{}\\frac13)\\equal{}0$." } { "Tag": [ "logarithms", "calculus", "calculus computations" ], "Problem": "(1) Solve $ x\\frac {dy}{dx} \\plus{} y \\equal{} 0$.\r\n\r\n(2) Solve $ x\\frac {dy}{dx} \\plus{} y \\equal{} x\\ln x$ in using (1).", "Solution_1": "hello, let's solve your equation (1), writing your equation in the form\r\n$ \\frac{dy}{y}\\equal{}\\minus{}\\frac{1}{x}dx$ from here we get all solutions in the form\r\n$ y(x)\\equal{}\\frac{C}{x}$ with $ x\\neq0$.\r\nFor $ x\\equal{}0$ we have $ y(0)\\equal{}0$.\r\nSonnhard.", "Solution_2": "hello, to your second problem, $ y(x) = \\frac {C}{x}$ is a solution of the homogenius part of the equation $ y^'+\\frac {y}{x} = \\ln(x)$,by the ansatz $ y(x) = \\frac {C(x)}{x}$ we will find a special solution of the inhomgenius part, we have $ y^'(x) = \\frac {C^'(x)x - C(x)}{x^2}$ inserting this in the equation we get $ C^'(x) = x\\ln(x)$ and from here $ C = \\frac {1}{2}x^2\\ln(x) - \\frac {1}{4}x^2$, finally we get\r\n$ y(x) = \\frac {1}{2}\\ln(x)x - \\frac {1}{4}x + \\frac {C}{x}$.\r\nSonnhard.", "Solution_3": "That's correct." } { "Tag": [ "floor function", "number theory proposed", "number theory" ], "Problem": "Let $ x_{1},x_{2},\\cdots,x_{n}$ be real numbers such that the identity\r\n\r\n$ \\left\\lfloor mx_{1}\\right\\rfloor\\plus{}\\left\\lfloor mx_{2}\\right\\rfloor\\plus{}\\cdots\\plus{}\\left\\lfloor mx_{n}\\right\\rfloor \\equal{}\\left\\lfloor m\\left(x_{1}\\plus{}x_{2}\\plus{}\\cdots\\plus{}x_{n}\\right)\\right\\rfloor$\r\n\r\nholds for all positive integers $ m$. Prove that at most one of the $ x_{k}$'s ($ 1\\leq k\\leq n$) can be non-integer.", "Solution_1": "[hide=\"Hint (zero and one non-integer)\"]Using that $ \\lfloor z\\plus{}r\\rfloor\\equal{}z\\plus{}\\lfloor r\\rfloor$ with $ z\\in\\mathbb Z$ and $ r\\in\\mathbb R$ it is easy to prove that it is true for zero or one non-integer[/hide]\n[hide=\"Hint 2 (Simplification for the two or more non-integer)\"]Using that if $ x_{1},x_{2},\\cdots,x_{n}$ works then $ x_{1}\\minus{}z_{1},x_{2}\\minus{}z_{2},\\cdots,x_{n}\\minus{}z_{n}$ also works with $ z_{i}\\in\\mathbb Z$ we just need to prove that $ f_{1},\\cdots,f_{n}$ does not work for ($ 0< f_{i}< 1$) and ($ 2\\leq n$)[/hide]", "Solution_2": "let ${ S =\\{i\\in[|1,n|]:\\ x_{i}\\not\\in\\mathbb{Z}}$ and we suppose that $ S =\\{a_{1}+r_{1},a_{2}+r_{2},...,a_{h}+r_{h}\\}$ for $ h\\ge 2$ and $ \\forall i\\in\\{1,2,...,h\\}:\\ (r_{i},a_{i})\\in]0,1[\\times\\mathbb{Z}$.\r\nthen $ \\forall m\\in\\mathbb{Z}:\\ \\sum_{i = 1}^{h}[mr_{i}] = [m\\sum_{i = 1}^{h}r_{i}]\\ gives\\ \\sum_{i = 1}^{h}\\{mr_{i}\\}< 1$ where $ (\\{x\\}= x-[x])$\r\nthen $ {\\forall\\epsilon > 0:\\ m =\\frac{1}{r_{2}(1+\\epsilon)}\\ gives\\ \\{\\frac{r_{1}}{r_{2}(1+\\epsilon)}\\}+\\{\\frac{r_{2}}{r_{2}(1+\\epsilon)}\\}<\\sum_{i = 1}^{h}\\{\\frac{r_{i}}{r_{2}(1+\\epsilon)}\\}< 1}$\r\nwe get $ \\forall\\epsilon > 0:\\ \\{\\frac{r_{1}}{r_{2}(1+\\epsilon)}\\}<\\frac{\\epsilon}{1+\\epsilon}$\r\n$ \\epsilon\\to 0$ gives $ \\{\\frac{r_{1}}{r_{2}}\\}=0$ gives $ r_{1}= 0$ (impossible) \r\nso $ |S|\\le 1$" } { "Tag": [ "algebra", "polynomial", "search", "Vieta", "number theory", "prime numbers", "AMC" ], "Problem": "For each positive integer $ n$, let $ f(n)\\equal{}n^4\\minus{}360n^2\\plus{}400$. What is the sum of all values of $ f(n)$ that are prime numbers?\r\n\r\n$ \\textbf{(A)}\\ 794\\qquad \r\n\\textbf{(B)}\\ 796\\qquad \r\n\\textbf{(C)}\\ 798\\qquad \r\n\\textbf{(D)}\\ 800\\qquad \r\n\\textbf{(E)}\\ 802$", "Solution_1": "[hide=\"hint\"]Try to factor $ f(n)$ as a difference of squares. See Problem 1 at [url]http://www.princeton.edu/~ploh/docs/math/b1-polynomials.pdf[/url] [/hide]", "Solution_2": "[hide=\"Solution\"]Factor $ f(n) \\equal{} n^4 \\minus{} 360n^2 \\plus{} 400 \\equal{} (n^2 \\plus{} 20)^2 \\minus{} 400n^2 \\equal{} (n^2 \\minus{} 20n \\plus{} 20)(n^2 \\plus{} 20n \\plus{} 20)$. WLOG take $ n\\geq0$. Then we must have $ n^2 \\minus{} 20n \\plus{} 20 \\equal{} 1$ (or $ (n\\minus{}19) (n\\minus{}1) \\equal{} 0$) and $ n^2 \\plus{} 20n \\plus{} 20$ prime. Both $ n \\equal{} 1$ and $ n \\equal{} 19$ satisfy the primality condition, producing $ f(n) \\equal{} 41$ and $ f(n) \\equal{} 761$, respectively. Adding, we have $ 41 \\plus{} 761 \\equal{} 802$.[/hide]", "Solution_3": "This might be a stupid question, but what's WLOG?", "Solution_4": "[quote=\"360_Fan\"]This might be a stupid question, but what's WLOG?[/quote]\r\n\r\nWithout Loss of Generality.", "Solution_5": "[hide=\"Terribly Stupid Solution\"]Check $ f(1)$. It's prime. $ f(2)$ and a few afterward are negative. Because we've memorized our squares, we see that $ f(n)\\equal{}n^2(n^2\\minus{}360)\\plus{}400$ becomes positive again at or before $ f(19)$. $ f(18)$ is still negative, so we test $ f(19)\\equal{}761$. Since any remaining output is higher than any of the answer choices, there can be no more primes, and $ 41$ alone is too small. So the answer is $ 41\\plus{}761\\equal{}\\boxed{802}$.[/hide]", "Solution_6": "Sorry to revive.\n[hide=\"Solution\"]Let $f(n)=p$ for some prime number $p$. Then, $n^4-360n^2+400-p=0$. It only remains for us to check perfect squares. To make the search easier, note that the sum of the roots must be $0$ by Vieta's Formulas. Thus, if $n$ is a positive integer, the polynomial can be factored as $(n^2-a^2)(n^2-b^2)$ for some positive integers $a$ and $b$. Let $a^2=m$ and $b^2=n$. Then, $m+n=360$ and $mn=400-p$. Moreover, we only need to check $(m,n)=(-1,-359)$ and $(m,n)=(1,-361)$, which result in $p=41$ and $p=761$, respectively. Both $41$ and $761$ are prime, and $41+761=(D) \\boxed{802}$.[/hide]", "Solution_7": "amctrivial grind: This factors as $(n^2-20n+20)(n^2+20n+20)$. This can only be prime when one of these values is $1$. If the first value is $1$, then $n^2-20n+19=0$, so $n = 1, 19$. If the second value is $1$, then $n^2+20n+19=0$, so $n = -1, -19$. Only the first two values are positive, and plugging in gives $41, 761 \\implies \\boxed{802}$." } { "Tag": [ "geometry", "circumcircle", "incenter", "geometry solved" ], "Problem": "Let ABC be a triangle and P a varying point on the arc BC of the circumcircle of ABC. Prove that the circle through P and the incenters of PAB and PAC pass through a fixed point independent of P.", "Solution_1": "I have the feeling this was posted before.\r\n\r\n[color=red][Moderator edit: Yes, http://www.mathlinks.ro/Forum/viewtopic.php?t=555 .][/color]", "Solution_2": "I don't know but somehow I have the feeling that paul_mathematics likes to repost problems which are already on the forum." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "find all function f:$R\\to R$ and $f$ continuous such that : \r\n $f(f(f(x))) +f(x)=2x$ for all x in real numbers.", "Solution_1": "Look around on the first page of this section and you'll find the answer.", "Solution_2": "Here, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88009" } { "Tag": [ "geometry", "incenter", "angle bisector", "geometry unsolved" ], "Problem": "ABC a triangle, I the incenter of ABC, U, V, W the symmetrics of I wrt BC, CA, AB, A'B'C' the orthic triangle of ABC,\r\nY the point of intersection of BC and B'C', Z\tthe point of intersection of AB and A'B', By, Bz the internal bisectors wrt $ A'R$ is the angle bisector of $ \\angle{C'A'B}$.\r\nSimilarly, $ C'R$ is the angle bisector of $ \\angle{A'C'B}$\r\nThen $ A'R$ and $ C'R$ are 2 angle bisectors of triangle $ BA'C'$.\r\n\r\nSince, $ AI$ is the line bisector of $ VW$, after a few deductions, we have $ \\angle{EIR'} \\equal{} \\angle{EIA}$\r\nTherefore, $ ER'.EA \\equal{} DP'.DC \\equal{} r^2$\r\n\r\nFrom this, we can position the points $ P'$ and $ R'$. Then I use the algebraic transformation to deduce : $ \\frac{BP}{BP'} \\equal{} \\frac{BR}{BR'}$\r\n(I 'm just lazy to write all of this here :D )\r\n\r\n(QED)", "Solution_2": "Can you explain your \"algebraic transformation\"?\r\nThank you\r\nJean-Louis", "Solution_3": "\"Algebraic transformation\" means calculating all the segments.\r\n\r\nWe can calculate $ BP, BP', BR, BR'$ (based on $ a,b,c$ - the 3 sides of triangle $ ABC$) and then all that we have to do is proving an identity.", "Solution_4": "I have found a proof based on \"Two parallels (2) and (3)\".\r\nI will put it on my web site which I will open soon.\r\nJean-Louis" } { "Tag": [ "geometry" ], "Problem": "How do you think you did?", "Solution_1": "Um, there's already a topic about this. :maybe:", "Solution_2": "Oh well. \r\n\r\n\r\nAre you Eric Yang the guy who scored an 85 last year?\r\n\r\nThe results come up tomorrow morning. :lol:", "Solution_3": "I guess results won't be up till later today. :maybe:", "Solution_4": "oops i won\r\n\r\nedit> roflrofl\r\n\r\nVestavia won\r\n\r\nand we only had people compete in three divisions even tho all five counted\r\n\r\nthat's hilarious", "Solution_5": "haha actually only four counted.\r\n\r\n\r\nA 12th grader at our school got a 50. 11th grader highest was 60, 10th grade highest at our school 60, and 9th grade highest at our school 60. \r\n\r\nSo those four add up to 230, the final score. Yeah I got a 34 lol bet you can't guess which one I am. \r\n\r\nVestavia only entered in 10th 11th and 12th. \r\nScores for the three added up to 240. :rotfl: \r\n\r\nThe scores were a bit down this year...", "Solution_6": "ARGH HOW DID I GET A 52 MAYBE THESE PEOPLE CAN'T GRADE \r\nor maybe:\r\n1) Wrong answer key given\r\n2) Teachers can't grade\r\n\r\nMAN I AM SO MAAAAAAAAAAAAAAAADDDD! :cursing:", "Solution_7": "lol that happened to me on the fall startup. \r\n\r\nI was told I had a higher score than the posted score that came up a few weeks later. \r\n\r\n\r\noh well I need to speed up lolz in non geometry ciphering", "Solution_8": "two of our top people's grades dropped from what our teacher expected (wayne zhu and evan zhao both went from 88 to 84)\r\n\r\nsurprisingly mine didn't drop so I guess I missed different problems", "Solution_9": "you probably did harder problems that had a definite answer. maybe the answer key given had a few typos or something. :lol: \r\n\r\n\r\nmine didn't drop either. Ahhhh Hopefully AoPS Vol. 2 will help me do more of these hard problems, or at least learn some stuff.", "Solution_10": "MINE DROPPED BECAUSE MY TEACHER DIDN'T SUBMIT MY ANSWER FORM!!!!!!!!!!!!!!!!!!!!" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a$, $b$, $c\\ge-\\frac{7}{15}$ such that $a+b+c=1$. Prove that\r\n\\[\\frac{1}{1+a+a^2}+\\frac{1}{1+b+b^2}+\\frac{1}{1+c+c^2}\\ge\\frac{27}{13}\\]\r\n\r\n[edit]\r\n :blush: Sorry, I messed up the constant. It should now work.", "Solution_1": "We have \\[\\frac{1}{1 + a + a^2} \\geq \\frac{162 - 135a}{169}\\] and the similar inequalities, since the above inequality is equivalent to \\[(15a + 7)(3a - 1)^2 \\geq 0.\\]\r\nAdding these gives the result." } { "Tag": [ "geometry", "ratio", "geometry unsolved" ], "Problem": "1) Let ABC be a fixed triangle . With compass and ruler draw a line which divides the area of the triangle into two equal parts.\r\n\r\n2) With compass and ruler draw a line wich divides the area of the triangle into two parts such that the ratio of the areas of the parts be equal to k (k is the length of a given line segment).", "Solution_1": "Maybe you want some extra restrictions? Like the line passing through some given point? Because otherwise it's easy.. :?", "Solution_2": "Indeed, there is a fixed point. Not so many time ago I wrote a solution for a problem, where cases $k=1$ and $k=2$ asked (actually we were asked to divide triangle into three parts of equal area). But the idea can be easily applied for your problem.\r\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=28947 .", "Solution_3": "hi \r\n\r\nas you have said , i have checked the question agian \r\n\r\nyes it needs a fixed point \r\n\r\nsorry for the mistake \r\n\r\nand mr.myth thank you for your link \r\nit was intersting" } { "Tag": [ "function", "complex analysis", "complex analysis unsolved" ], "Problem": "If a function $ f$ statisfies $ f(z^2)\\equal{}f(z)\\minus{}z$, can you conclude that $ f$ has a singularity in the point $ z\\equal{}1$ and $ z\\equal{}\\minus{}1$. Justify your answer ! Thanks in advance.", "Solution_1": "Assuming you mean a [i]meromorphic[/i] function $ f$ on the whole complex plane--you really need to be specific about these things--then it seems that no such function exists. So you can conclude whatever you want.\r\n\r\nSpecifically, repeated application of the equation for $ z\\in \\mathbb{C}$, $ |z| < 1$, gives $ f(z^{2^n})\\equal{}f(z) \\minus{} \\sum_{i\\equal{}1}^n z^i$, so $ f(0)\\equal{}f(z) \\minus{} \\frac{z}{1\\minus{}z}$. Then $ f(z^2)$ would have a pole at $ \\minus{}1$, while $ f(z)\\minus{}z$ does not, contradiction.", "Solution_2": "Somehow I saw the complicated answer to this question first. :roll: \r\n\r\nYou can also just set $ z\\equal{}1$ if $ f$ has no singularity at $ 1$, then you get $ 0\\equal{}\\minus{}1$, a contradiction. Then, if $ f$ has no singularity at $ \\minus{}1$, the right hand side has a finite limit as $ z\\to \\minus{}1$, but the left hand side does not, contradiction." } { "Tag": [ "number theory", "prime factorization" ], "Problem": "Let $ A$, $ M$, and $ C$ be digits with\r\n\\[ (100A \\plus{} 10M \\plus{} C )(A \\plus{} M \\plus{} C ) \\equal{} 2005.\r\n\\]What is $ A$?\r\n\r\n$ \\textbf{(A)}\\ 1 \\qquad\r\n\\textbf{(B)}\\ 2 \\qquad\r\n\\textbf{(C)}\\ 3 \\qquad\r\n\\textbf{(D)}\\ 4 \\qquad\r\n\\textbf{(E)}\\ 5$", "Solution_1": "[hide=\"Click for solution\"]\nThe prime factorization of $ 2005$ is $ 5 \\times 401$ so $ 100A\\plus{}10M\\plus{}C\\equal{}401$ and $ A\\plus{}M\\plus{}C\\equal{}5$. Because they are digits, $ A\\equal{}4$, $ M\\equal{}0$, $ C\\equal{}1$ or $ \\boxed{\\textbf{(D)}}$.\n[/hide]", "Solution_2": "Solution: [hide]Seeing the multiplication shows that 2005 = 5 * 401. The LHS is essentially a 3-digit number times the sum of the digits. We see that 401 * 5 works. Therefore A = 4 and the answer is D[/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let G be a graph with n vertices with no three cycles, yet contains an x-cycle where x is odd and >3. Prove that there exists a vertex in G with degree at most 2n/5.\r\n\r\nBomb", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=5763" } { "Tag": [ "calculus", "integration", "trigonometry", "geometry", "3D geometry", "sphere", "analytic geometry" ], "Problem": "$ \\int\\int(3x\\plus{}4y^2)dxdy$ where R is bounded by $ y\\equal{}\\sqrt{4\\minus{}x^2}$,$ y\\equal{}\\sqrt{1\\minus{}x^2}$,$ y\\equal{}0$\r\n\r\n$ \\int\\int\\int(x^2)dxdydz$ where Q is above $ z\\equal{}\\sqrt{3x^2\\plus{}3y^2}$ and inside $ z\\equal{}\\sqrt{25\\minus{}x^2\\minus{}y^2}$\r\n\r\nVolume of a cylinder $ (x\\minus{}1)^2\\plus{}(y\\minus{}1)^2\\equal{}2$ between $ z\\equal{}0$ and $ z\\equal{}10\\minus{}x^2\\minus{}y^2$\r\n\r\nThe first one I found the limits of x to be -4 to -1, and y from 0 to ?\r\nThe second and third I'm not really sure about. Thanks in advance.", "Solution_1": "[quote=\"xerosmythe\"]$ \\int\\int(3x + 4y^2)dxdy$ where R is bounded by $ y = \\sqrt {4 - x^2}$,$ y = \\sqrt {1 - x^2}$,$ y = 0$\n\n$ \\int\\int\\int(x^2)dxdydz$ where Q is above $ z = \\sqrt {3x^2 + 3y^2}$ and inside $ z = \\sqrt {25 - x^2 - y^2}$\n\nVolume of a cylinder $ (x - 1)^2 + (y - 1)^2 = 2$ between $ z = 0$ and $ z = 10 - x^2 - y^2$\n\nThe first one I found the limits of x to be -4 to -1, and y from 0 to ?\nThe second and third I'm not really sure about. Thanks in advance.[/quote]\r\n\r\n[hide]$ 1)$ $ \\int\\int_R (3x + 4y^2) ~dA$ is bounded by the two semicircles above the x-axis with radii $ r = 1$ and $ r = 2$.\n\n$ R = \\{ (x,y) \\in \\mathbb{R}^2 | ~~ - 2 < = x < = 2, ~~\\sqrt {1 - x^2} < = y < = \\sqrt {4 - x^2} \\}$. But we also can convert this to polar so:\n$ R = \\{ (r,\\theta) \\in \\mathbb{R}^2 | ~~0 < = \\theta < = \\pi, ~~1 < = r < = 2 \\}$\n\n$ \\int\\int_R (3x + 4y^2) ~dA = \\int_{ - 2}^{2}\\int_{\\sqrt {1 - x^2}}^{\\sqrt {4 - x^2}}(3x + 4y^2)~dy~dx$\n$ = \\int_0^{\\pi}\\int_1^2 (3r\\cos \\theta + 4r^2 \\sin^2 \\theta) r~dr~d\\theta$\n\n\n$ 2)$ $ \\int\\int\\int_Q x^2 ~dV$ where $ Q$ is bounded by the cone $ z = \\sqrt {3x^2 + 3y^2}$ and the hemisphere $ z = \\sqrt {25 - x^2 - y^2}$.\n\nThe easiest way to do this one is using spherical coordinates. Obviously the radius of the sphere is $ 5$ so, $ \\rho = 5$. Obviously $ 0 < = \\theta < = 2\\pi$. Using $ z = \\sqrt {3x^2 + 3y^2}$, and substituting what we know $ x,y,$ and $ z$ to be in spherical coordinates, we see $ 0 < = \\phi < = \\frac {\\pi}{6}$.\n\n$ Q=\\{ (x,y,z) \\in \\mathbb{R}^3 | ~~ -\\frac{5}{2}<=x<=\\frac{5}{2}, ~~-\\sqrt{\\frac{25}{4}-x^2}<=y<=\\sqrt{\\frac{25}{4}-x^2}, ~~\\sqrt{3x^2+3y^2}<=z<=\\sqrt{25-x^2-y^2} \\}$\n$ Q=\\{ (\\rho,\\phi,\\theta) \\in \\mathbb{R}^3 | ~~0<=\\rho<=5, ~~0<=\\phi<=\\frac{\\pi}{6}, ~~0<=\\theta<=2\\pi \\}$\n\nSo $ \\int\\int\\int_Q x^2 ~dV = \\int_0^{2\\pi}\\int_0^{\\frac {\\pi}{6}}\\int_0^5 \\rho ^2 \\sin ^2 \\phi \\cos ^2 \\theta ~\\rho ^2 \\sin \\phi ~d\\rho ~d\\phi ~d\\theta$\n\n\n$ 3)$ $ V=\\int\\int\\int_E 1 ~dV$ where\n\n$ E=\\{ (x,y,z) \\in \\mathbb{R}^3 | ~~1-\\sqrt{2}<=x<=1+\\sqrt{2}, ~~1-\\sqrt{2-(x-1)^2}<=y<=1+\\sqrt{2-(x-1)^2}, ~~0<=z<=10-x^2-y^2 \\}$\n\n$ E=\\{ (x,y,z) \\in \\mathbb{R}^3 | ~~-\\sqrt{2}<=x<=\\sqrt{2}, ~~\\sqrt{2-x^2}<=y<=\\sqrt{2-x^2}, ~~0<=z<=10-(x-1)^2-(y-1)^2 \\}$\n$ E=\\{ (r,\\theta,z) \\in \\mathbb{R}^3 | ~~0<=\\theta<=2\\pi, ~~0<=r<=\\sqrt{2}, ~~0<=z<=8-r^2+2r(\\cos \\theta + \\sin \\theta) \\}$\n\n$ V=\\int\\int\\int_E 1 ~dV=\\int_0^{2\\pi}\\int_0^{\\sqrt{2}}\\int_0^{8-r^2+2r(\\cos\\theta+\\sin\\theta)} r~dz~dr~d\\theta$[/hide]", "Solution_2": "1. Draw a picture of the region. After that, you'll see that if you change to polar coordinate system, 1 <= r <= 2 and 0 <= theta <= Pi. The calculation is much easier using polar system.\r\n2. Q is bounded by a sphere(top) and an elliptic cone(bottom). Draw a graph. Do the integration in this order instead: dz, dy, dx. For z, the limit is from sqrt(3x^2+3y^2) to sqrt(25-x^2-y^2). For y, the limit is from -sqrt(25/4-x^2) to sqrt(25/4-x^2). For x, the limit is from -5/2 to 5/2.\r\n You can also change to spherical coordinate system to simplify your calculation. In this case, r is from 0 to 5. theta is from 0 to 2PI, and phi is from 0 to Pi/6.\r\n3. The Region is bounded by the cylinder, z=0 and the elliptic paraboloid. This is pretty much the same as 2. Do the integration in the order of dz, dy, dx. The limits will be clear if you draw a graph." } { "Tag": [ "trigonometry", "function", "quadratics", "vector", "algebra", "real analysis", "real analysis unsolved" ], "Problem": "I am just beginning differential equations, and wanted to find the solution to $y'' + y = 0$. I know the solution is $A\\cos{t}+B\\sin{t}$, but I had trouble in the following (after setting up the characteristic equation).\r\n\r\nI found roots of the characteristic equation to be $i$ and $-i$. So I said the solutions to the differential equation were $e^{it}$ and $e^{-it}$. The real valued solution basis would then be $\\cos{t}$ for $e^{it}$. When I try to find the real solution basis for $e^{-it}$, I come up with the same. However, it seems it should be sine instead of cosine (judging by the actual solution that I found online).\r\n\r\n$z=e^{it}=\\cos{t}+i\\sin{t}\\implies\\textrm{Re}(z)=\\cos{t}$\r\n\r\n$z^{-1}=e^{-it}=\\cos{t}-i\\sin{t}\\implies\\textrm{Re}(z)=\\cos{t}$\r\n\r\nWhy does $e^{-it}$ coorespond to the real valued solution $\\sin{t}$?\r\n\r\nThanks for your help.\r\n\r\nAlex", "Solution_1": "Since $e^{it} = \\cos t + i\\sin t$, and $e^{-it} = \\cos t - i\\sin t$, we have $\\cos t = \\frac{e^{it}+e^{-it}}{2}$ and $\\sin t = \\frac{e^{it}-e^{-it}}{2i}$. So each of the pairs $\\{\\cos t,\\sin t\\}$ and $\\{e^{it},e^{-it}\\}$ can be expressed in terms of the other.", "Solution_2": "Suppose $L$ is a homogeneous linear differential operator with real coefficients, and suppose that $w(t)$ is a complex-valued function and $I$ an interval with no singular points of $L$ such that for $t\\in I,$ $L(w)=0.$ Then (in all cases, for $t\\in I$):\r\n\r\n1. $L(\\overline{w})=0.$\r\n\r\n2. If $\\text{Re}(w)=u$ and $\\text{Im}(w)=v$, then $L(u)=0$ and $L(v)=0.$\r\n\r\n3. If there exists a $t\\in I$ such that $\\frac{w'(t)}{w(t)}\\in\\mathbb{C}\\setminus\\mathbb{R},$ then $u$ and $v$ are linearly independent. If we now assume that $L$ is a second order linear differential operator, the complete general solution of $L(y)=0$ is $y=c_1u+c_2v.$\r\n\r\nAs for the proof:\r\n\r\n1. Since the coefficients of $L$ are real, $\\overline{L(w)}=L(\\overline{w}).$\r\n\r\n2. From (1), we have that $w$ and $\\overline{w}$ are solutions of the homogeneous linear differential equation. Hence, so also are any linear combinations of $w$ and $\\overline{w},$ notably $u=\\text{Re}(w)=\\frac{w+\\overline{w}}2$ and $v=\\text{Im}(w)=\\frac{w-\\overline{w}}{2i}.$ Note that this is exactly what rgep did above.\r\n\r\n3. We have $w=u+iv$ where $u$ and $v$ are real. Then\r\n\r\n$\\frac{w'}w=\\frac{u'+iv'}{u+iv}=\\frac{(u'+iv')(u-iv)}{(u+iv)(u-iv)} = \\frac{(uu'+vv')+i(uv'-vu')}{u^2+v^2}.$\r\n\r\nThe condition that this take on a non-real value is precisely the condition that $uv'-vu'\\ne0$ at some point. But this is simply the Wronskian of $u$ and $v$, and if the Wronskian is ever nonzero on $I,$, then $u$ and $v$ are independent and, for second order equations, span the solution space.\r\n\r\n---\r\n\r\nPutting this to practical use:\r\n\r\nConsider the differential equation $y''+2y'+5y=0.$ Is we assume solutions of the form $y=e^{rt},$ we get the quadratic equation $r^2+2r+5=0,$ which has roots $r=-1\\pm2i.$\r\n\r\nLet's just focus on the solution $w=e^{(-1+2i)t}=e^{-t}e^{2it}=e^{-t}(\\cos 2t+i\\sin 2t).$\r\n\r\nThen $u=\\text{Re}(w)=e^{-t}\\cos 2t$ and $v=\\text{Im}(w)=e^{-t}\\sin 2t.$\r\n\r\nThese are independent, and the general solution can be written as\r\n\r\n$y(t)=c_1e^{-t}\\cos 2t+c_2e^{-t}\\sin 2t.$\r\n\r\nYou may worried about justifying the line \"assume solutions of the form ....\" The justification comes after the fact. However we came up with that, we did find solutions, and having found a two-dimensional vector space of solutions (in particular, we can solve for all possible initial conditions), the existence-uniqueness theorem tells us we have found all solutions." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $ A$ be a $ n\\times n$ matrix with real entries.\r\n\r\nProve that we cannot have\r\n\\[ A^2 \\equal{} \\text{diag}(\\alpha_1,\\ldots,\\alpha_n)\r\n\\]\r\nwhere $ \\alpha_i<0$ for $ 1\\leq i\\leq n$.", "Solution_1": "$ A\\equal{}\\begin{bmatrix}0&\\minus{}1\\\\1&0\\end{bmatrix}.$ $ A^2\\equal{}\\begin{bmatrix}\\minus{}1&0\\\\0&\\minus{}1\\end{bmatrix}.$\r\n\r\nNow, what's the real question supposed to be?", "Solution_2": "How about \"$ n$ is odd\" as an additional hypothesis? Then it would be true, since $ A$ would have to have at least one real eigenvalue and hence $ A^2$ would have at least one nonnegative eigenvalue.", "Solution_3": "actually I copied the question correctly; there must have been an error. That is a nice argument for the odd case though.", "Solution_4": "[quote=\"blahblahblah\"]actually I copied the question correctly; there must have been an error. That is a nice argument for the odd case though.[/quote]Transfer to a different university.", "Solution_5": "[quote=\"blahblahblah\"]That is a nice argument for the odd case though.[/quote]\r\n\r\nA better one would be $ \\alpha_1\\cdot\\alpha_2\\cdot ...\\cdot\\alpha_n\\equal{}\\det\\left(A^2\\right) \\equal{} \\left(\\det A\\right)^2\\geq 0$. ;)\r\n\r\n darij" } { "Tag": [ "superior algebra", "superior algebra theorems" ], "Problem": "In a ring with identity and with proper ideals, there always exist maximal ideals. Is the statement true for rings with a nontrivial multiplication and with no identity?", "Solution_1": "http://sierra.nmsu.edu/morandi/notes/NoMaxIdeals.pdf, Theorem 4", "Solution_2": "[quote=\"-oo-\"]http://sierra.nmsu.edu/morandi/notes/NoMaxIdeals.pdf, Theorem 4[/quote]\r\nGood link, thank you very much! :wink:" } { "Tag": [ "inequalities", "calculus", "inequalities proposed" ], "Problem": "I solve this problem today , and I think the inequality is very strict :D \r\n\r\n$ \\sqrt {\\frac {(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{8abc}} \\geq 1 \\plus{} \\frac {3}{5} \\left( 1 \\minus{} \\sqrt {\\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}} \\right)$\r\n\r\nwith the condition $ a,b,c$ are positive.\r\n\r\n\r\nP.S CAS solution is welcome. :)", "Solution_1": "It's obviously beacause: $ (a\\plus{}b)(b\\plus{}c)(c\\plus{}a)(a\\plus{}b\\plus{}c)^2 \\geq\\ 24abc(a^2\\plus{}b^2\\plus{}c^2)$\r\n(Next, assume $ a^2\\plus{}b^2\\plus{}c^2\\equal{}3.$ Let $ a\\plus{}b\\plus{}c\\equal{}t$, we are easily to complete the proof)\r\n :)", "Solution_2": "[quote=\"nguoivn\"]It's obviously beacause: $ (a \\plus{} b)(b \\plus{} c)(c \\plus{} a)(a \\plus{} b \\plus{} c)^2 \\geq\\ 24abc(a^2 \\plus{} b^2 \\plus{} c^2)$\n(Next, assume $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3.$ Let $ a \\plus{} b \\plus{} c \\equal{} t$, we are easily to complete the proof)\n :)[/quote]\r\n\r\nThen I will change it to \r\n\r\n$ \\sqrt {\\frac {(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{8abc}} \\geq 1 \\plus{} \\frac {17}{25} \\left( 1 \\minus{} \\sqrt {\\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}} \\right)$\r\n\r\nand your method does not work . Just check $ t \\equal{} 2.9993$ :D", "Solution_3": "$ \\sqrt {\\frac {(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{8abc}} \\geq 1 \\plus{} \\frac {11}{16} \\left( 1 \\minus{} \\sqrt {\\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}} \\right)$ is stronger.", "Solution_4": "[quote=\"arqady\"]$ \\sqrt {\\frac {(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{8abc}} \\geq 1 \\plus{} \\frac {11}{16} \\left( 1 \\minus{} \\sqrt {\\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}} \\right)$ is stronger.[/quote]\r\n\r\nYeah Arqady! \r\nI think the best constant k is k = 0.6883 But I guess you have solved it already.", "Solution_5": "[quote=\"Gibbenergy\"]\nI think the best constant k is k = 0.6883 But I guess you have solved it already.[/quote]\r\nI have got $ k_{max}\\equal{}0.688300972...$", "Solution_6": "[quote=\"arqady\"][quote=\"Gibbenergy\"]\nI think the best constant k is k = 0.6883 But I guess you have solved it already.[/quote]\nI have got $ k_{max} \\equal{} 0.688300972...$[/quote]\r\n\r\nArqady, Can you please show us your solution?\r\n\r\nThanks :)", "Solution_7": "For $ k\\equal{}\\frac{11}{16}$, using my hint above and the old result:\r\n$ \\frac{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}{abc} \\geq\\ 8\\plus{}4\\sqrt[]{2}\\minus{} \\frac{4\\sqrt[]{2}(ab\\plus{}bc\\plus{}ca)}{a^2\\plus{}b^2\\plus{}c^2}$\r\n$ (a\\plus{}b)(b\\plus{}c)(c\\plus{}a)(a\\plus{}b\\plus{}c)^2 \\geq\\ 24abc(a^2\\plus{}b^2\\plus{}c^2)$\r\nConsider 2 cases of t, we have done.\r\n :)", "Solution_8": "[quote=\"nguoivn\"]For $ k \\equal{} \\frac {11}{16}$, using my hint above and the old result:\n$ \\frac {(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{abc} \\geq\\ 8 \\plus{} 4\\sqrt []{2} \\minus{} \\frac {4\\sqrt []{2}(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2}$\n$ (a \\plus{} b)(b \\plus{} c)(c \\plus{} a)(a \\plus{} b \\plus{} c)^2 \\geq\\ 24abc(a^2 \\plus{} b^2 \\plus{} c^2)$\nConsider 2 cases of t, we have done.\n :)[/quote]\r\n\r\n[I edited]\r\n\r\n\r\nI think you did try to solve the problem. But there is a little mistake if you use Hungkhtn's inequality ? :) \r\nConsider Arqady's case in which $ k \\equal{} \\frac {11}{16}$. Let $ t \\equal{} \\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}$\r\n\r\nFor the first inequality, Hung's inequality is applicable if $ t > 2.933899575$\r\n\r\nFor the second inequality, it will fail if $ t > 2.877890984$", "Solution_9": "[quote=\"Gibbenergy\"][quote=\"nguoivn\"]For $ k \\equal{} \\frac {11}{16}$, using my hint above and the old result:\n$ \\frac {(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{abc} \\geq\\ 8 \\plus{} 4\\sqrt []{2} \\minus{} \\frac {4\\sqrt []{2}(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2}$\n$ (a \\plus{} b)(b \\plus{} c)(c \\plus{} a)(a \\plus{} b \\plus{} c)^2 \\geq\\ 24abc(a^2 \\plus{} b^2 \\plus{} c^2)$\nConsider 2 cases of t, we have done.\n :)[/quote]\n\n[I edited]\n\n\nI think you did try to solve the problem. But there is a little mistake if you use Hungkhtn's inequality ? :) \nConsider Arqady's case in which $ k \\equal{} \\frac {11}{16}$. Let $ t \\equal{} \\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}$\n\nFor the first inequality, Hung's inequality is applicable if $ t > 2.933899575$\n\nFor the second inequality, it will fail if $ t > 2.877890984$[/quote]\r\n\r\nWhat is the name of CAS program that you are using to determine such accuracies?", "Solution_10": "Very sorry. I was wrong in calulus :oops:", "Solution_11": "[quote=\"shaaam\"][quote=\"arqady\"][quote=\"Gibbenergy\"]\nI think the best constant k is k = 0.6883 But I guess you have solved it already.[/quote]\nI have got $ k_{max} = 0.688300972...$[/quote]\n\nArqady, Can you please show us your solution?\n\nThanks :)[/quote]\r\nIt's a very ugly.\r\nLet $ f(a,b,c) = \\sqrt {\\frac {(a + b)(b + c)(c + a)}{8abc}} - 1 - k \\left( 1 - \\sqrt {\\frac {ab + bc + ca}{a^2 + b^2 + c^2}} \\right),$ $ abc = 1$ and $ a = \\max\\{a,b,c\\}.$\r\nHence, $ f(a,b,c) - f\\left(a,\\sqrt {bc},\\sqrt {bc}\\right)\\geq$\r\n$ \\geq(\\sqrt b - \\sqrt c)^2\\left(\\frac {k(a^3 - 4a - 4)a^2}{2(a^3 + 2)^{1.5}\\sqrt {2a\\sqrt a + 1}} + \\frac {a^2\\sqrt a + 4\\sqrt a + 1}{4(\\sqrt {2(a^4 + a)} + \\frac {a^{1.5} + 1}{\\sqrt [4]a})}\\right)\\geq0.$ \r\nThe last inequality is true for all $ 0 < k\\leq0.7$ and $ a\\geq1.$\r\nThus, we need to check when $ f(a,1,1)\\geq0,$ which gives\r\n$ k\\leq\\min{\\frac {\\sqrt {a^2 + 2}(\\sqrt {a^2 + 2} + \\sqrt {2a + 1})}{2\\sqrt a(\\sqrt a + 1)^2} = 0.688300972...}$", "Solution_12": "[quote=\"arqady\"][/quote][quote=\"shaaam\"][quote=\"arqady\"][quote=\"Gibbenergy\"]\nI think the best constant k is k = 0.6883 But I guess you have solved it already.[/quote]\nI have got $ k_{max} = 0.688300972...$[/quote]\n\nArqady, Can you please show us your solution?\n\nThanks :)[/quote][quote=\"arqady\"]\nIt's a very ugly.\nLet $ f(a,b,c) = \\sqrt {\\frac {(a + b)(b + c)(c + a)}{8abc}} - 1 - k \\left( 1 - \\sqrt {\\frac {ab + bc + ca}{a^2 + b^2 + c^2}} \\right),$ $ abc = 1$ and $ a = \\max\\{a,b,c\\}.$\nHence, $ f(a,b,c) - f\\left(a,\\sqrt {bc},\\sqrt {bc}\\right)\\geq$\n$ \\geq(\\sqrt b - \\sqrt c)^2\\left(\\frac {k(a^3 - 4a - 4)a^2}{2(a^3 + 2)^{1.5}\\sqrt {2a\\sqrt a + 1}} + \\frac {a^2\\sqrt a + 4\\sqrt a + 1}{4(\\sqrt {2(a^4 + a)} + \\frac {a^{1.5} + 1}{\\sqrt [4]a})}\\right)\\geq0.$ \nThe last inequality is true for all $ 0 < k\\leq0.7$ and $ a\\geq1.$\nThus, we need to check when $ f(a,1,1)\\geq0,$ which gives\n$ k\\leq\\min{\\frac {\\sqrt {a^2 + 2}(\\sqrt {a^2 + 2} + \\sqrt {2a + 1})}{2\\sqrt a(\\sqrt a + 1)^2} = 0.688300972...}$[/quote]\r\n\r\nI appreciate :) \r\n\r\n(P.S why do you assume $ abc = 1$)", "Solution_13": "[quote=\"shaaam\"]\n\n why do you assume $ abc \\equal{} 1$[/quote]\r\nBecause the original inequality is homogeneous and $ abc\\equal{}1$ simplifies this inequality.", "Solution_14": "[quote=\"arqady\"][/quote][quote=\"shaaam\"][quote=\"arqady\"][quote=\"Gibbenergy\"]\nI think the best constant k is k = 0.6883 But I guess you have solved it already.[/quote]\nI have got $ k_{max} = 0.688300972...$[/quote]\n\nArqady, Can you please show us your solution?\n\nThanks :)[/quote][quote=\"arqady\"]\nIt's a very ugly.\nLet $ f(a,b,c) = \\sqrt {\\frac {(a + b)(b + c)(c + a)}{8abc}} - 1 - k \\left( 1 - \\sqrt {\\frac {ab + bc + ca}{a^2 + b^2 + c^2}} \\right),$ $ abc = 1$ and $ a = \\max\\{a,b,c\\}.$\nHence, $ f(a,b,c) - f\\left(a,\\sqrt {bc},\\sqrt {bc}\\right)\\geq$\n$ \\geq(\\sqrt b - \\sqrt c)^2\\left(\\frac {k(a^3 - 4a - 4)a^2}{2(a^3 + 2)^{1.5}\\sqrt {2a\\sqrt a + 1}} + \\frac {a^2\\sqrt a + 4\\sqrt a + 1}{4(\\sqrt {2(a^4 + a)} + \\frac {a^{1.5} + 1}{\\sqrt [4]a})}\\right)\\geq0.$ \nThe last inequality is true for all $ 0 < k\\leq0.7$ and $ a\\geq1.$\nThus, we need to check when $ f(a,1,1)\\geq0,$ which gives\n$ k\\leq\\min{\\frac {\\sqrt {a^2 + 2}(\\sqrt {a^2 + 2} + \\sqrt {2a + 1})}{2\\sqrt a(\\sqrt a + 1)^2} = 0.688300972...}$[/quote]\r\n\r\nThis is a nice and natural approach and I also had the same result. \r\n@nguoivn: I actually like your sols, but it will be more convenient if you can write your sols more detail. \r\n@Shaaam : I use Ti to find the root of equation. \r\n If you ask for CAS, I guess MatLab is #1, following by Mathcad, Maple, Mathematica, PolyMath,... etc" } { "Tag": [ "limit", "function", "integration", "real analysis", "real analysis solved" ], "Problem": "Let $\\{a_i\\},\\{b_i\\}$ be decreasing sequences of positive numbers s.t. $\\displaystyle \\sum_{n\\ge 1}a_n=\\sum_{n\\ge 1}b_n=\\infty$. Does it follow that $\\displaystyle \\sum_{n\\ge 1}\\sqrt{a_ib_i}=\\infty$?", "Solution_1": "Note that sayng that $\\min(a_i,b_i)\\le \\sqrt{a_ib_i}\\le \\max (a_i,b_i)$ doesn't suffice, because we can find sequences $\\{a_i\\},\\ \\{b_i\\}$ which have the mentioned properties, but $\\displaystyle \\sum_{n\\ge 1}\\min(a_n,b_n)<\\infty$. \r\n\r\nHere's a concrete case: what if $b_n=\\frac 1n$? Does it follow that $\\displaystyle \\sum_{n\\ge 1}\\sqrt{\\frac {a_n}n}=\\infty$?", "Solution_2": "I don't have the answer for the first question, though I think it is no, but for the second question the answer is yes, it follows:\r\n $ (\\sqrt{\\frac{a_1}{1}}+...+\\sqrt{\\frac{a_n}{n}})^2\\geq\\sum_{i=1}^n \\sqrt{\\frac{a_i}{i}}\\cdot\\sum_{j=1}^i {\\sqrt{\\frac{a_j}{j}}\\geq\\sum_{i=1}^n a_i }$", "Solution_3": "Yeah, I realized it wasn't that hard. I think that if $\\{x_n\\}$ is a decreasing sequence of positive numbers s.t. $\\displaystyle \\sum_{n\\ge 1}x_n<\\infty$ then $\\displaystyle \\lim_{n\\to \\infty}nx_n=0$. It's related to the following problem (posted and solved before on the forum: if $f : (0,\\infty)\\to (0,\\infty)$ is a strictly decreasing function s.t $\\displaystyle \\int^{\\infty}_0 f(x)dx<\\infty$ then $\\displaystyle \\lim_{n\\to \\infty}xf(x)=0$. We can define a continuous function $f$ s.t. $f(n)=x_n$ and we're done.\r\n\r\nWe apply this result to $x_n=\\sqrt{\\frac {a_n}n}$. We get $\\displaystyle \\lim_{n\\to \\infty}n\\sqrt{\\frac {a_n}n}=0$. On the other hand, $a_n=\\sqrt{\\frac {a_n}n}\\cdot n\\sqrt{\\frac {a_n}n}$, which gets really small when compared to $\\sqrt{\\frac {a_n}n}$, and if the latter has a convergent series, then so must $a_n$. If we look at it backwards, if $a_n$ has a divergent series, then so must $\\sqrt{\\frac {a_n}n}$.\r\n\r\n[b] Moderator edit: Edited for readability (there was a problem with : in math mode in the original post)[/b]", "Solution_4": "if i'm right the lemma which is used in your solution come from vojtec could you refer me to proof Grobber?", "Solution_5": "Sorry, I can't seem to find it, although I distinctly remember me posting it and Moubinool giving a really nice solution. The one I know isn't that nice, and I don't really feel like posting it :). It was indeed proposed for Voytch Yarnk (if that's the correct spelling :D).", "Solution_6": "I can't forgot that magnific solution! Let $ F(x)=\\int_0^x f(t) dt $. Then \r\n $ F(2x)-F(x) $ goes to zero, but $ F(2x)-F(x)\\geq xf(2x)$." } { "Tag": [ "geometry" ], "Problem": "Square $ABCD$ is inscribed in a circle. Point $X$ lies on minor arc $AB$ such that $[XCD]=993$ and $[XAB]=1$. Find $[XAD]+[XBC]$.", "Solution_1": "[hide]\n\nLet r = radius of the circle\nThen $\\sqrt{2}r$ = the length of one side of the square\nLet h = the height of $\\triangle{XAB}$ = distance from X to AB\n$[XCD] = \\frac{1}{2}(\\sqrt{2}r)(\\sqrt{2}r+h) = 993$\n$[XAB] = \\frac{1}{2}(\\sqrt{2}r)(h) = 1$\n$r^2=992$\nBut since the height of XAD and XBC sum to $\\sqrt{2}r$,\n$[XAD]+[XBC] = \\frac{1}{2}r^2 = 496$\n\n\n[/hide]", "Solution_2": "[hide=\"Unfortunately, that is incorrect.\"]Your answer is off by a factor of 2...[/hide]", "Solution_3": "I think minsoens made a small error at the end. $[XAD]+[XBC] = \\frac{1}{2}r^2 = 496$ should be $[XAD]+[XBC]=\\frac{1}{2}\\cdot(r\\sqrt{2})^2=r^2=992$.", "Solution_4": "992 is correct. \r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?&t=55093 :D", "Solution_5": "I may sound stupid, but what's [XAD] etc mean?", "Solution_6": "[quote=\"JavaMan\"]I may sound stupid, but what's [XAD] etc mean?[/quote]I think that [XAD] indicates the area of the triangle XAD.", "Solution_7": "[hide=\"another sol\"]\nlet side be $s$,\nand perpendicular distance from $X$ to nearest side be $x$,\n$\\frac{1}{2}*s*(s+x)=993$\n$\\frac{1}{2}*s*x=1$\nSo,\n$\\frac{1}{2}*s^2=992$\nWe need to find:\n$\\frac{1}{2}*s*y+\\frac{1}{2}*s*z$\nbut $y+z=s$,\nSo,\n$\\frac{1}{2}*s^2$\nwhich is:\n$\\boxed{992}$\n[/hide]" } { "Tag": [ "\\/closed" ], "Problem": "If one uses \" in a description, it always gets converted to \\\". I haven't tested, but maybe other characters are influenced as well.", "Solution_1": "[quote=\"ZetaX\"]If one uses \" in a description, it always gets converted to \". I haven't tested, but maybe other characters are influenced as well.[/quote]\r\n\r\nYeah, and it is kind of annoying if you edit the description of a competition several times, e.g. I got \\^{10}\" several times. As I understand that is kind of escaping mechanism due to the coding but then there should be a post-running script trimming them off again.", "Solution_2": "In fact, the number of \\ grows exponentially, causing a lot of fun :D" } { "Tag": [ "function", "college", "integration", "real analysis", "real analysis solved" ], "Problem": "if f{x} is a real valued differentiable function with f{1} =1.If f{x} prime is equal to\r\n1/{x^2+ f{x}^2},then proove that f{x}<1+pi/4 for all x.\r\n\r\ncan u please tell me a book in which these r found in plenty.", "Solution_1": "Berkeley Problems... :?", "Solution_2": "well I come from India and don't know what you r talking about.can u specifically tell me the source and please solve it also", "Solution_3": "Hmm. Maybe you can find it at [url]http://math.berkeley.edu/~desouza/pb.html[/url].\r\nI've got a book of that.\r\n\r\nI think the conclusion is $f(x)<1+\\frac{\\pi}{4}$ for $x\\geq 1$.\r\n\r\n$f'(x)>0$ so $f$ is strictly increasing, and $f(x)>1$ for $x> 1$.\r\n\r\n$\r\nf(x)-f(1) = \\int_1^x f'(t)dt = \\int_1^x \\frac{dt}{t^2+f(t)^2} <\\int_1^x \\frac{dt}{t^2+1} < \\int_1^\\infty \\frac{dt}{t^2+1} = \\frac{\\pi}{4}\r\n$\r\n\r\nThe conclusion follows easily.", "Solution_4": "yup that is the answer\r\nthanks lily for the link" } { "Tag": [ "AMC", "AMC 12", "AIME" ], "Problem": "Five students $ A, B, C, D, E$ took part in a contest. One prediction was that the contestants would finish in the order $ ABCDE$. This prediction was very poor. In fact, no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so. A second prediction had the contestants finishing in the order $ DAECB$. This prediction was better. Exactly two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so. Determine the order in which the contestants finished.", "Solution_1": "I don't know if this is correct or not, this is my first time doing this kind of logic problem... \r\n\r\nFrom \"two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so,\" we see that\r\n\r\n1) The two that finished in the correct place must be consecutive.\r\n\r\nProof:\r\nSuppose they are not consecutive. Then they form at least a block of 3 (e.g. D_E..) which means that the other two numbers must satisfy the finishing consecutively condition. However, the number between them cannot be correct, which means a second consecutive pair cannot be formed. Contradiction.\r\n\r\n2) The two that are correct and consecutive must be at the ends.\r\n\r\nProof:\r\nIf they are in the middle, then the other pair is fixed, making it correct also, leaving 4 correct places. Contradiction.\r\n\r\nThen we test:\r\n\r\nIf DA is in the correct place, we have:\r\nDACBE-- WRONG, E, C is in the correct place from the very first condition\r\nDABEC-- WRONG, AB is a correct consecutive pair, violates first condition\r\n\r\nIf CB is in the correct place, we have:\r\nAEDCB-- WRONG, A is in the correct place\r\nEDACB-- Correct. Solution.", "Solution_2": "Very hard problem, might i ask what is the correct solution, i'm not discounting iniquitious' solution only i don't necessarily understand it all )=, i'm not used to seeing logic problems outside of the mensa workbooks and this seems 20x harder.", "Solution_3": "[hide=\"correct answer\"]EDACB, nice job iniquitus[/hide]\n[hide=\"lightentropy\"]btw this was an IMO question, yea it was 1963, but it was still IMO[/hide]", "Solution_4": "[quote]btw this was an IMO question, yea it was 1963, but it was still IMO[/quote]\r\nYeah the level of difficulty has increased a lot since then :(", "Solution_5": "that makes sense still, i'm more at the AMC12/ and can do maybe 2-3 problems on AIME so it would make sense that i was unaware how to do this. I need to do more logic problems, they are interesting and fun.\r\ncool post" } { "Tag": [ "geometry", "analytic geometry", "geometric transformation", "reflection" ], "Problem": "got this from a middle school chinese mathbook.\r\n\r\nTwo points on a Cartesian plane, A and B, are located at (5,4) and (1,1) respectively. There lies a point P that is located on either the X or the Y axis. What are the coordinates of P so that AP + BP is minimized.", "Solution_1": "[hide=\"hint\"]\nreflections\n[/hide]", "Solution_2": "yeah, reflections are definitely the way because if you can place AP and a reflected point on a single line, it minimizes the value of AP + BP." } { "Tag": [ "probability", "percent", "probability and stats" ], "Problem": "Seventy percent of the kids who visit a doctor have high temperature and 30% of the kids who have high temperature, also have smallpox. What's the probability that a kid who goes to the doctor's has a temperature and smallpox?", "Solution_1": "Just multiply. 21%.\r\n\r\nAs a bit of advice, please don't post multiple threads with the same title. We like to be able to tell threads apart. \"Probability\" is a very bad title for this forum, since it says almost nothing about the content.", "Solution_2": "Strictly speaking, unsufficient data: $ P(H|D)\\equal{}0.7$, $ P(S|H)\\equal{}0.3$, $ P(S\\cap H|D)\\equal{}?$. Some independence assumption is used.", "Solution_3": "Actually, it's an implicit assumption that \"who visit the doctor\" is the whole universe here." } { "Tag": [ "quadratics" ], "Problem": "Hi.\r\n\r\nHexagonal number nth term: $ f_6(n)\\equal{}2n^2\\minus{}n$\r\ntriangular number nth term: $ f_3(n)\\equal{}0.5(n^2\\plus{}n)$\r\n\r\n\r\nLooking at a proof here to prove every hexagonal number is also triangular they did:\r\n\r\n$ f_6(n) \\equal{} f_3(2n\\minus{}1)$ and continued from that point.\r\n\r\nCan someone explain where did 2n-1 come from?\r\n\r\nThanks alot.", "Solution_1": "What this says is that the $ n^{th}$ hexagonal number is the same thing as the $ (2n \\minus{} 1)^{th}$ triangular number. Now, there are a couple of ways to understand this statement.\r\n\r\n1. If you look at those algebraic expressions, they're pretty similar; a linear and a quadratic term, no constant term. There might be a way to turn one into the other. Now, in $ f_3$ the linear and quadratic term are the same whereas in $ f_6$ one is twice the other, so perhaps we should look at $ f_3(2n) \\equal{} 2n^2 \\plus{} 2n$. But the linear term is wrong, so we keep tweaking this until we arrive at the right answer.\r\n\r\n2. Can you figure out a way to transform hexagons into triangles? (I think this is much more fun. Play around with small cases, although make sure you understand the definition of a hexagonal number.)", "Solution_2": "I dont understand the goal nor the way to do it :(\r\n\r\nAre we trying to guess terms to put into $ f_3(n)$ so it becomes like the form of $ f_6(n)$ ?", "Solution_3": "Yes, pretty much, but I wouldn't advise you to think about the problem that way since it provides no motivation: there's no reason to suspect that there is a way to do this at all, so I would prefer to see what happens if you try to do it geometrically (although I don't know if it works).", "Solution_4": "find m(n) such that \r\n$ 2n^2 \\minus{} n \\equal{} 0.5(m^2 \\plus{} m)\r\n\\\\4n^2 \\minus{} 2n \\equal{} m^2 \\plus{} m \r\n\\\\m^2 \\plus{} m \\plus{} \\dfrac{1}{4} \\equal{} 4n^2 \\minus{} 2n \\plus{} \\dfrac{1}{4}\r\n\\\\(m\\plus{}\\frac{1}{2})^2 \\equal{} (2n \\minus{} \\frac{1}{2})^2 \r\n\\\\m \\plus{} \\frac{1}{2} \\equal{} (2n \\minus{} \\frac{1}{2})\r\n\\\\m \\equal{} 2n \\minus{} 1$\r\n\r\nwe take the positive square root on the secondlast line because we require $ m \\geq 0$" } { "Tag": [], "Problem": "If $t^23,-3>3^1/2\n\nso t=1/3 [/hide]" } { "Tag": [], "Problem": "What is the smallest integer which is larger than $ \\sqrt[3]{26,991}$?", "Solution_1": "we see that 30^3=27000. we are in a countdown round, which means no time to test 29^3. we assume the answer is 30. and we are correct :D" } { "Tag": [ "function", "topology", "complex analysis", "complex analysis unsolved" ], "Problem": "Fix $ a \\in D$ (open unit disc) and let $ T_{a}(z) \\equal{} \\frac{z\\minus{}a}{1\\minus{}\\overline{a}z}$. Let $ f$ and $ g$ analtytic functions in an open set containing $ \\overline{D}$, such that $ |f(z)| > |g(z)|$ for all $ |z| \\equal{} 1$. Prove the function $ T_{a}f \\plus{} g$ has at least one zero in $ D$.\r\n\r\nThis is how I find the problem, i guess $ T_{a}f$ is the composite function, first $ f$ then $ T_{a}$.\r\n\r\nCan someone give me a hint for this one.\r\n\r\nThx", "Solution_1": "Suppose the contrary and conclude by [url=http://en.wikipedia.org/wiki/Rouche's_theorem]Rouch\u00e9's Theorem[/url] that $ T_a f (z)$ has no zeros in $ D$, clearly a contradiction since $ T_a f(a) \\equal{} 0$.", "Solution_2": "Thank you for your answer, I dont know why $ T_{a}f(a) \\equal{} 0$, since\r\n\r\n\\[ T_{a}f(a)\\equal{}\\frac{f(a)\\minus{}a}{1\\minus{}\\overline{a}f(a)} \\equal{} 0,\\] only when $ f(a)\\equal{}a$ and we dont know this.\r\nI was trying to prove that there exists some $ z_{0}$ such that $ f(z_{0}) \\equal{} a$ and this is clearly a zero for $ T_{a}f$, but i think this cant be done.\r\n\r\nI repeat i found this problem and dont know if its composite or the product, if its product then clearly\r\n$ T_{a}f(a) \\equal{} T_{a}(a)f(a) \\equal{} 0$, but I dont see how $ |T_{a}f(z)| > |g(z)|$ for all $ |z|\\equal{}1$ so I can use\r\nRouche Theorem.", "Solution_3": "My apologies, I misread your problem.\r\n\r\nIf it's composition, the statement doesn't seem to be true. What if both $ f$ and $ g$ are constant, for example?" } { "Tag": [ "function", "algebra", "domain", "real analysis", "calculus", "calculus computations" ], "Problem": "Hi,\r\nWhile I was reading about uniform continuity of real functions, I visited this post and read the #4 on it:\r\n[url]http://www.physicsforums.com/showthread.php?t=52795[/url]\r\n\r\nThe description is clear, but I could not figure it out, how to prove that f(x)=1/x is not uniformly continuous. Do you have recommendations please?", "Solution_1": "Continuity really means to be able to draw the graph of the function with having to take the hand of ur paper. that is it means there are no breaks in the graph in a particular interval . For example consider the function $ f(x) = x$ in any interval for instance in $ ( 0, \\infty )$ we say this function is continuous throughout the interval because when we draw its graph it has no breaks .\r\n\r\nBasically continuity needs to be checked in each and every point inorder to say its uniformly continuous in the interval.\r\n\r\nSo a function $ f(x)$ can be said as continuous at x belonging to the intervalv $ [ X , Y ]$ if there exists a $ x_{0}$ in the neighborhood of of x such that $ |x - x_{0} | < \\epsilon$ where $ \\epsilon > 0$ is a very small real number and $ lim_{x \\rightarrow x_{0^ - } } f(x) = lim_{x \\rightarrow x_{0^ + } } = f( x_{0} )$ \r\n\r\nHence we see that for the function $ f(x) = \\frac {1}{x}$ it is not contionuous at 0 because $ lim_{x \\rightarrow{0^ - } } f(x) \\neq lim_{x \\rightarrow{0^ + } } f(x)$and $ f( 0)$ is not defined . hence it is not uniformly continuous.", "Solution_2": "Continuity of a function is only defined at points. Continuity of a function over a certain domain means it is continuous at each point of that domain. Uniform continuity is a stronger statement, saying that in each point of the domain you can use the same distance between arguments to get the images within some distance $ \\epsilon$.\r\n\r\nvijaymenon's argument is flawed.", "Solution_3": "[quote]vijaymeno: ...is not defined . hence it is not uniformly continuous. [/quote]\r\nThank you for the fast answer. \r\nMeanwhile, please consider as it is mentioned at #4 of the post in:\r\n[url]http://www.physicsforums.com/showthread.php?t=52795[/url]\r\nWe are considering $ (0,\\plus{}\\infty)$, not [X,Y] as you mentioned. \r\nThe writer of the #4 post there said \"f(x)=1/x IS contiuous, but not uniformly continuous on the interval $ (0,\\plus{}\\infty)$. \r\nHence I can't agree with the proof construction provided in your post:\r\n\"it is not uniformly continuous, because it is not continuous at all\" (if I understand it right). The proof you provide, is around $ x_{0} \\equal{} 0$, which is not our consideration, as \"uniformly continuous\" says that it is \"absolutely independent of any $ x_{0}$\". \r\nThe problem here is, to find an $ \\epsilon>0$, for which we prove there is NO $ \\delta > 0$ which breaks the uniformity on the interval. This is what I tried to do but failed. It must be possible, as in some intervals, say, between (0,1) the function itself is indefinitely big, and it may cause problem to define $ \\epsilon>0$ and $ \\delta > 0$. What I am wondering about, is how to formally write down this idea! let's continue to think together around it :wink:", "Solution_4": "For any $ \\delta > 0$, consider the values of the function $ f(x)$ at $ x_0$ and $ x_0 \\plus{} \\delta$, where $ x_0$ is the positive root of $ x^2 \\plus{} \\delta x \\minus{} \\delta \\equal{} 0$. Now, figure out how I came up with that :wink:", "Solution_5": "[color=green]Redundant quote deleted by moderator.[/color]\r\n\r\nI didn't figure it out. please clarify.", "Solution_6": "First, do you understand the significance of the points $ x_0$ and $ x_0 \\plus{} \\delta$ that I chose? That is, have you done what I said in my first sentence?\r\n\r\n\r\nThere's no need to quote the entire post immediately preceding yours -- you're responding to it by default.", "Solution_7": "Consider two points, $ x=\\frac{1}{n}$ and $ y=\\frac{1}{n+1}$, with $ n\\in\\mathbb{Z}_{+}$. Then $ |x-y|=\\frac{1}{n}-\\frac{1}{n+1}|=|\\frac{n+1}{n(n+1)}-\\frac{n}{n(n+1)}|=|\\frac{1}{n(n+1)}|$. As $ n\\rightarrow\\infty$, this goes to 0. However, $ |f(x)-f(y)|=|n-n-1|=1$, which does not go to 0." } { "Tag": [ "quadratics", "analytic geometry", "calculus", "conics", "parabola", "symmetry", "special factorizations" ], "Problem": "What is the smallest value of $x^2+10x$ where $x$ is a real number?", "Solution_1": "[hide]Using the formula to find the minimum and maximum we find it is -10/2 so the answer is -5.[/hide]", "Solution_2": "Read the question more carefully...", "Solution_3": "[hide] The minimum is -5, so plug that in to get:\n\n(-5)^2 + 10(-5)\n\n25 - 50\n\n= -25\n\n[/hide]", "Solution_4": "-b/2a \r\n-(10)/2(1)\r\nx=-5 where sqr(x) + 10x is least so\r\nsqr(-5)+10(-5)\r\n25+-50\r\n-25", "Solution_5": "[hide]\n\nf(x) = x^2 + 10x\n\n-b/2a = -10/2 = -5\n\nx = -5, so f(x) = -25\n\n\n\n\n\n[/hide]", "Solution_6": "Where does this -b/2a come from? Why does that work?", "Solution_7": "Given a quadratic in standard form $ax^2+bx+c=y$, where a>0, the lowest point on the graph lies at the vertex. The x-coordinate of the vertex point is -b/2a, and so plugging that value into the original equation for x yields the smallest value.\r\n\r\nThis can be seen by completing the square, which gives the vertex-form of the equation, $y=a(x-h)^2+k$,where (h,k) is the vertex of the function.", "Solution_8": "-b/2a is the formula used to find the x value of the lowest, or highest, value of y depending on whether it opens up or down. Once you get the x value you plug it into the original equation and out pops the minimum y value", "Solution_9": "[quote=\"#include\"]-b/2a is the formula used to find the x value of the lowest, or highest, value of y depending on whether it opens up or down. Once you get the x value you plug it into the original equation and out pops the minimum y value[/quote]\r\n\r\nWe know what the formula IS but how would you prove it? A formula is practically useless unless you know where it comes from. nolachrymose has the right idea but hasn't really shown that -b/2a is the x-coordinate of the vertex.", "Solution_10": "$ax^2+bx+c=$\r\n\r\n$a\\left(x^2+\\frac{bx}{a}\\right)+c=$\r\n\r\n$a\\left(x^2+\\frac{bx}{a}+\\frac{b^2}{4a^2}\\right)-\\frac{b^2}{4a^2}+c=$\r\n\r\n$a\\left(x+\\frac{b}{2a}\\right)^2-\\frac{b^2}{4a^2}+c$\r\n\r\nWe want the square to be zero to either maximize or minimize the expression, so $x=-\\frac{b}{2a}$", "Solution_11": "$\\frac{-b}{a}$ is the sum of the roots.$ \\frac{-b+\\sqrt{b^2-4ac}}{2a}+\\frac{-b-\\sqrt{b^2-4ac}}{2a}=\\frac{-b}{a}$\r\nThe average of these two roots is then the x coordinate of the vertex or $\\frac{-b}{2a}$", "Solution_12": "[hide=\"Short Answer\"]\nThe minimum occurs at $\\frac{-b}{2a}=\\frac{-10}{2}=-5$ Therefore $x=-5$ and the smallest y value is -25\n :D \n[/hide]", "Solution_13": "um diophantus, don't post calculus in the beginner forum.", "Solution_14": "The arithemetic mean of the two roots will be the vertex, which is the sum of the roots divided by 2. Hence, -b/(2a).", "Solution_15": "To those who have explained their solution by simply claiming that $x=-\\frac{b}{2a}$ at the min/max of a quadratic equation, how do you know that? You saw it written in a box in a textbook?\r\n\r\nThe algebraic solution to this problem that is not based on such assumptions is:\r\n\r\n$(x+5)^2\\ge0\\Longrightarrow(x+5)^2-25\\ge-25\\Longrightarrow{x}^2+10x+25-25\\ge-25\\Longrightarrow{x}^2+10x\\ge-25,$\r\nand equality is attained for $x=-5$, so $-25$ is the minimum value.\r\n\r\nThere was a similar discussion about this in a different thread.. this is what I said:\r\n\r\n[quote=\"AntonioMainenti\"]... If you don't know how to use calculus or what it means, you can also do this algebraically. $y=ax^2+bx+c$ can be manipulated to $a\\left(x+\\frac{b}{2a}\\right)^2+\\frac{4ac-b^2}{4a}$, so $\\left(x+\\frac{b}{2a}\\right)^2\\ge0\\Rightarrow{a}\\left(x+\\frac{b}{2a}\\right)^2+\\frac{4ac-b^2}{4a}\\ge\\frac{4ac-b^2}{4a}$, and equality is attained for $x=\\frac{-b}{2a}$, so $\\frac{4ac-b^2}{4a}$ is the extreme (min or max) value in all cases. ...[/quote]", "Solution_16": "[quote=\"Magnara\"]The arithemetic mean of the two roots will be the vertex, which is the sum of the roots divided by 2. Hence, -b/(2a).[/quote]\r\nSorry, I was under the impression you were trying to determine roots from this equation rather than the vertex", "Solution_17": "[quote=\"Magnara\"]\nThe arithemetic mean of the two roots will be the vertex, which is the sum of the roots divided by 2. Hence, -b/(2a).[/quote]\r\n\r\nYep :D This works because a parabola has a line of symmetry through the vertex.", "Solution_18": "Does that argument assume that the quadratic has two real roots?", "Solution_19": "[quote=\"Ravi B\"]Does that argument assume that the quadratic has two real roots?[/quote]Yeah, you can't really interpret it geometrically unless it has real roots, or a double root, but still all quadratics have two roots (by the Fundamental Theorem of Something-er-rather).", "Solution_20": "I agree that a quadratic always has two roots. But when the roots are complex, I'm not sure how Magnara was going to prove that the arithmetic mean of the roots is the x-coordinate of the vertex. It's still true algebraically, but I'm not sure how to prove it geometrically.", "Solution_21": ":( Hi.\r\nAre these problems solvable? I have tried setting them up in standard form Ax^2+Bx+C=0 but not getting any where. Thanks\r\n\r\nProblem 1: (4x+1)(2x-3)=3(x+4)\r\n\r\n\r\nProblem 2: (x-11)^2=(3x+6)^2\r\n\r\n", "Solution_22": "For the 1st one. Multiply out (or distribute) both sides and rearrange terms.\r\n[hide]You'll end up with $8x^2-13x-15=0$[/hide]\n\nThe same for the second one. Multiply them all out and rearrange terms.\n[hide]You'll end up with $8x^2+58x-85=0$[/hide]\r\n\r\nThen, you can always solve with the quadratic formula.", "Solution_23": "Refering to AntonioMainenti's post, here's the other thread discussed a few weeks ago on quadratic optimization.\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26049[/url]", "Solution_24": "We can use -b/2a to get -10/2=-5 for the smallest value.\r\nf(-5)=(-5)2+(10)*-5= -25.\r\nI do not know how to do a rigorous proof, but It is the vertex of the parabola, so that is the minimum." } { "Tag": [ "geometry", "trapezoid", "geometric transformation", "homothety", "Pythagorean Theorem" ], "Problem": "a trapezium with four lengths of sides are 1,2,3,4. Find its area?", "Solution_1": "Letting the sides of length $ 1$ and $ 4$ be the bases, we can split the longer base into $ x$, $ 1$, and $ y$, so that $ 4 \\equal{} x \\plus{} y \\plus{} 1 \\implies x \\plus{} y \\equal{} 3$.\r\nApplying the Pythagorean Theorem on two right triangles, we have\r\n\\[ 2^2 \\minus{} x^2 \\equal{} h^2 \\equal{} 3^2 \\minus{} y^2 \\implies y^2 \\minus{} x^2 \\equal{} \\left(y \\plus{} x\\right)\\left(y \\minus{} x\\right) \\equal{} 5.\r\n\\]\r\nPlugging in $ x \\plus{} y \\equal{} 3$ and solving for $ y \\minus{} x$, we have\r\n\\[ y \\minus{} x \\equal{} \\frac {5}{3}.\r\n\\]\r\nThen\r\n\\[ 2y \\equal{} 3 \\plus{} \\frac {5}{3} \\implies y \\equal{} \\frac {7}{3}.\r\n\\]\r\nThen we find that\r\n\\[ h \\equal{} \\sqrt {9 \\minus{} y^2} \\equal{} \\sqrt {9 \\minus{} \\frac {49}{9}} \\equal{} \\sqrt {\\frac {32}{9}} \\equal{} \\frac {4\\sqrt {2}}{3}.\r\n\\]\r\nThen the area is\r\n\\[ A \\equal{} \\frac {1}{2}\\left(b_1 \\plus{} b_2\\right)h \\equal{} \\boxed{\\frac {10\\sqrt {2}}{3}}.\r\n\\]\r\n\r\nYou can also use a homothety centered at the intersection of the two slants.", "Solution_2": "[quote=\"123456789\"]Letting the sides of length $ 1$ and $ 4$ be the bases, we can split the longer base into $ x$, $ 1$, and $ y$, so that $ 4 \\equal{} x \\plus{} y \\plus{} 1 \\implies x \\plus{} y \\equal{} 3$.\nApplying the Pythagorean Theorem on two right triangles, we have\n\\[ 2^2 \\minus{} x^2 \\equal{} h^2 \\equal{} 3^2 \\minus{} y^2 \\implies y^2 \\minus{} x^2 \\equal{} \\left(y \\plus{} x\\right)\\left(y \\minus{} x\\right) \\equal{} 5.\n\\]\nPlugging in $ x \\plus{} y \\equal{} 3$ and solving for $ y \\minus{} x$, we have\n\\[ y \\minus{} x \\equal{} \\frac {5}{3}.\n\\]\nThen\n\\[ 2y \\equal{} 3 \\plus{} \\frac {5}{3} \\implies y \\equal{} \\frac {7}{3}.\n\\]\nThen we find that\n\\[ h \\equal{} \\sqrt {9 \\minus{} y^2} \\equal{} \\sqrt {9 \\minus{} \\frac {49}{9}} \\equal{} \\sqrt {\\frac {32}{9}} \\equal{} \\frac {4\\sqrt {2}}{3}.\n\\]\nThen the area is\n\\[ A \\equal{} \\frac {1}{2}\\left(b_1 \\plus{} b_2\\right)h \\equal{} \\boxed{\\frac {10\\sqrt {2}}{3}}.\n\\]\nYou can also use a homothety centered at the intersection of the two slants.[/quote]\r\n\r\n\r\nwould you have another idea in which the smaller base of trapezium is not 1,eg.2 or 3? Then ,is it any possible solution? :blush:" } { "Tag": [], "Problem": "Given any integer $n\\ne 0$ can you find an integer $x$ for which $1+nx$ is composite integer?", "Solution_1": "[quote=\"digger\"]Given any integer $n\\ne 0$ can you find an integer $x$ for which $1+nx$ is composite integer?[/quote]\r\n\r\nwhat does the word \"composite\" mean?", "Solution_2": "not prime. :wink:", "Solution_3": "[hide]\nYes. $1+n(n+2)=n^{2}+2n+1=(n+1)^{2}$, which is obviously composite. \n[/hide]", "Solution_4": "[hide=\"Another approach\"] By Bezout's Lemma we can find $y$ such that\n\n$py-nx = 1$\n\nFor an arbitrary $p$ such that $gcd(n, p) = 1$. By a slight extension of Bezout's Lemma, we can find infinite such $y$. [/hide]", "Solution_5": "[quote=\"amirhtlusa\"]not prime. :wink:[/quote]\r\n1 is not prime. :wink:", "Solution_6": "A composite number, according to mathworld is:\r\n\r\n[quote]A composite number n is a positive integer n>1 which is not prime (i.e., which has factors other than 1 and itself)[/quote]\r\n\r\nGoogle is your friend. :)", "Solution_7": "[quote=\"anirudh\"]A composite number, according to mathworld is:\n\n[quote]A composite number n is a positive integer n>1 which is not prime (i.e., which has factors other than 1 and itself)[/quote]\n\nGoogle is your friend. :)[/quote]\r\nNo, it's not.\r\nWikipedia is your only online friend. :D\r\nedit: sorry that is not true. Anything AoPS is your best online friend", "Solution_8": "What about the AoPS Wiki?" } { "Tag": [ "MIT", "college", "blogs" ], "Problem": "You probably think that somewhere in the net there are some great websites you never had the opportunity to discover, and in this topic we can exchange a looooooot by just asking : what's your favorite websites?\r\n(of course it includes (http://www.mathlinks.ro));)", "Solution_1": "http://www.artofproblemsolving.com\r\nhttp://www.gamedev.net\r\nhttp://www.slashdot.com\r\nhttp://video.google.com/videoplay?docid=1491516901670441597", "Solution_2": "http://www.myspace.com\r\nhttp://www.myyearbook.com\r\nhttp://www.dogpile.com\r\nhttp://www.blogtop100.com\r\nhttp://www.miniclip.com\r\nhttp://www.artistdirect.com\r\nhttp://www.abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijk.com\r\nhttp://www.funnyjunk.com\r\netc etc....i dunt really know what wud interest nerds tho\r\nim just on this site as an xcuse 4 studying cuz of my mom lol...bet some others r too", "Solution_3": "[url=http://artofproblemsolving.com]artofproblemsolving.com[/url]\r\n[url=http://youtube.com]youtube.com[/url]\r\n[url=http://google.com]google.com[/url]", "Solution_4": "[url]http://www.woot.com[/url]", "Solution_5": "[url=http://www.skyscrapercity.com]www.skyscrapercity.com[/url]\r\nvery nice, not only have they pictures of all skyscrapers, you can ask questions, show your own pictures, and they often show pictures of parks, bridges, rivers etc, you also get to see a lot of cities like in eastern europe or africa or north korea, where you rarely catch a glimpse of\r\n\r\n\r\n[url=http://www.imdb.com]imdb[/url]\r\nmother of all movie and tv sites, pages about everything, links to all actors and directors, trivia, fora, etc\r\n\r\n[url=http://www.space.com]space.com[/url]\r\ncovers everything about space : unmanned, manned, astronomy\r\n\r\n\r\n[url=http://www.nasa.gov]www.nasa.gov[/url]\r\nvery nice site\r\n\r\n[url=http://en.wikipedia.org/wiki/Main_Page]wikipedia[/url] well you all know this encyclopedia right", "Solution_6": "Some websites I like a lot:\r\n\r\nhttp://www.scienceblogs.com/pharyngula\r\n\r\nA weblog written by a developmental biologist.\r\n\r\nhttp://www.nws.noaa.gov/\r\n\r\nThe National Weather Service. (If you don't like Senator Rick Santorum of Pennsylvania, you should use their website instead of commercial weather companies to check the forecast. Besides, weather is cool.)\r\n\r\nhttp://zenoferox.blogspot.com/\r\n\r\nA weblog written by a math instructor at the college level.\r\n\r\nhttp://www.pestiside.hu\r\n\r\nA news/weblog site for English-speakers in Budapest.", "Solution_7": "http://mathworld.wolfram.com/", "Solution_8": "[url=http://www.langorigami.com/]http://www.langorigami.com/[/url]\r\n[url=http://www.origami.vancouver.bc.ca/home.html]http://www.origami.vancouver.bc.ca/home.html[/url]\r\n[url=http://www.asahi-net.or.jp/~qr7s-kmy/g/]http://www.asahi-net.or.jp/~qr7s-kmy/g/[/url]\r\n[url=http://www11.ocn.ne.jp/~origami/]http://www11.ocn.ne.jp/~origami/[/url]\r\n\r\n[url=http://www.discovery.com/]http://www.discovery.com/[/url]", "Solution_9": "[url=http://en.wikipedia.org/wiki/Main_Page]wikipedia[/url]\r\n[url=http://google.com]google[/url] (random googling and going on the sites i get)\r\n[url=http://www.artofproblemsolving.com/Forum/index.php]AoPS[/url]\r\n[url=http://facebook.com]facebook[/url]\r\n[url=http://http://ocw.mit.edu/index.html]MIT OpenCourseWare[/url]\r\ni mostly surf the net for time wasting purposes, not much interest in knowledge :(\r\nits amazing how many AoPSers have blog sites/myspace-like-sites as their favorites, is that true of most people?", "Solution_10": "[url=http://www.gamefaqs.com]www.gamefaqs.com[/url] >_>\r\n[url=http://www.wikipedia.com]www.wikipedia.com[/url]", "Solution_11": "[url=http://www.mugglenet.com]Mugglenet[/url]\r\n[url=http://www.jkrowling.com]JK Rowling's site[/url]\r\nand ofcourse this site.\r\n\r\nI'll post more once I think of them.", "Solution_12": "[url]http://www.goodquotes.com[/url]", "Solution_13": "http://www.oma.org.ar\r\nwww26.brinkster.com/omaregionpilar", "Solution_14": "I can't believe nobody has mentioned it yet....\r\n\r\nhttp://www.theonion.com\r\n\r\nI [i]can't[/i] be the only AoPS'er that loves that site :D", "Solution_15": "homestarrunner.com is the best", "Solution_16": "[url=http://www.mathlinks.ro]www.mathlinks.ro[/url] : for math things obiouvsly :D\r\n[url=http://www.hi5.com]www.hi5.com[/url] : keep in touch with friends, collegues, etc. I check it regularily but I don't use it as a mail box at all.\r\n[url=http://www.imdb.com]www.imdb.com[/url] : i watch a movie only if is well rated ($\\geq 6.0$). This is very useful and true in the same time.\r\n[url=http://www.explodingdog.com/drawonpaper/]www.explodingdog.com/drawonpaper/[/url] : nice drawings. I like them very much. I used on as an avatar :P.\r\n\r\nI didn't mentioned but it's a popular thing to use Google or Y! Mail :)", "Solution_17": "The most hilarious clip:\r\nhttp://www.endofworld.net", "Solution_18": "What was the adress of that page, whose owner won 1 million dollars????", "Solution_19": "[url=http://Wikipedia.org]Wikipedia, of course[/url]\r\n\r\n[url=http://myspace.com]And Myspace[/url]\r\n\r\n[url=http://www.research.att.com/~njas/sequences/]OEIS[/url] is pretty fun, too.\r\n\r\nAnd, to round it off, [url=http://rinkworks.com]Rinkworks[/url]", "Solution_20": "[url=http://digg.com]digg.com[/url]", "Solution_21": "Post number thousand!\r\n\r\n\r\nif you just want a quick look\r\n\r\n[url]http://www.skyscraperpicture.com/index1.htm[/url]\r\n\r\n\r\nfor closer inspection :\r\n\r\n[url]http://www.skyscrapercity.com[/url]\r\n[url]http://www.emporis.com[/url]", "Solution_22": "http://www.kurzweilai.net is fascinating. Among other things, it contains links to a lot of really interesting science news stories. (Mostly related to biotech, nanotech, and AI/robotics.)", "Solution_23": "Wolfram (mathworld)!" } { "Tag": [ "algebra", "polynomial", "geometry", "analytic geometry" ], "Problem": "Any guesses?\r\n\r\n... I've always screwed up on the last problem.. I'm placing my bet on either polynomial problem, or messy analytic geometry! :D", "Solution_1": "yucky number theory? :D :D :D", "Solution_2": "I think it'll be a polynomial this year.\r\nLast year I got the last problem. but screwed up the one before it. lol" } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Find all functions such that |f(x)-f(y)|<=|x-y| for all x,y>0 and there exist positive constants a, b, c satisfying \r\n\r\n f(ax)+f(b/x)=f(cx) for all x. \r\n\r\n(Edg\u00e1r Dobrib\u00e1n, Cluj-Napoca)", "Solution_1": "Lipschitz functions are uniformly continuous and can be extended to the limit points of their domains. Hence, $ \\lim_{x\\to0^{\\plus{}}}f(x)$ exists. Just to have a notation, call this limit $ f(0).$\r\n\r\n$ \\lim_{x\\to0^{\\plus{}}}f\\left(\\frac{b}{x}\\right)\\equal{}\\lim_{x\\to0^{}}(f(cx)\\minus{}f(ax))\\equal{}f(0)\\minus{}f(0)\\equal{}0.$\r\n\r\nHence, $ \\lim_{x\\to\\infty}f(x)\\equal{}0.$\r\n\r\nBy taking the limit of the same expression as $ x\\to\\infty,$ we conclude that $ f(0)\\equal{}0.$\r\n\r\n---\r\n\r\nJust getting started; I don't have a whole argument yet.", "Solution_2": "We have that $ f(y) \\equal{} f(uy)\\plus{}f\\left(\\frac{v}{y}\\right)$.\r\nHence, for $ z : \\equal{}\\frac{v}{y}$, $ f\\left(\\frac{v}{z}\\right) \\equal{} f\\left(\\frac{uv}{z}\\right)\\plus{}f\\left( z\\right)$, so $ f(uz) \\equal{}\\minus{}f\\left(\\frac{uv}{z}\\right)$.\r\nPutting $ \\frac{v}{t}: \\equal{} uz$ we get that $ f\\left(\\frac{v}{t}\\right) \\equal{}\\minus{}f\\left( u^{2}t\\right)$.\r\nTherefore, $ f(y) \\equal{} f(uy)\\minus{}f\\left( u^{2}y\\right)$, which implies $ f\\left( uy\\right) \\equal{} f\\left( u^{2}y\\right)\\minus{}f\\left( u^{3}y\\right)$, i.e. $ f(y) \\equal{}\\minus{}f\\left( u^{3}y\\right)$.\r\n\r\nIf $ u \\equal{} 1$, then $ f(y) \\equal{} f(y)\\plus{}f\\left(\\frac{v}{y}\\right)$, so $ f\\equiv 0$.\r\nElse, $ f(y) \\equal{}\\minus{}f\\left( u^{3}y\\right) \\equal{} f\\left( u^{6}y\\right) \\equal{}\\minus{}f\\left( u^{9}y\\right) \\equal{}\\ldots$.\r\nAs Kent Merryfield showed, that has limit zero, so that's all.\r\n\r\nThe only solution is $ f\\equiv 0$.\r\n\r\nPS: If I made any mistakes, you'll have to excuse me. It's really late here. :blush:" } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "He,\r\nI have a problem with L-series and Dirichlet characters:\r\n\r\nLet $ \\chi$ be a non-principal charcter modulo $ q$ with $ \\chi(\\minus{}1)\\equal{}1$. Express $ L(2,\\chi)$ as a finite sum.\r\n\r\nI have shown that $ L(2, \\chi)\\equal{}L(2,\\chi^{'})\\prod_{p|q} ((1\\minus{}\\chi^{'}(p)p^{\\minus{}2})$ where $ \\chi^{'}$ is a primitive \r\ncharacter that induces $ \\chi$.\r\n\r\nNow I have to show that $ L(2,\\chi^{'})$ is a finite product/sum... so I can put things together and solve the problem.\r\n\r\nAnyone?", "Solution_1": "I found a solution. Thanks...!" } { "Tag": [ "calculus", "integration", "function", "algebra open", "algebra" ], "Problem": "Is there any (interesting) relation between two functions [tex]f[/tex] and [tex]g[/tex] such that [tex]f, g \\in C^{\\infty}[/tex] and [tex]f, g : R \\rightarrow R[/tex] and:\r\n[tex]\\begin{eqnarray*}\r\n \\int_{g ( a )}^{g ( b )} f ( x ) \\tmop{dx} = \\int_{f ( a )}^{f ( b )} g ( x\r\n ) \\tmop{dx} & & \r\n\\end{eqnarray*}[/tex]\r\nCan we ''find'' [tex]f[/tex] and [tex]g[/tex] ?\r\n\r\nThanks !", "Solution_1": ":blush: I just realized I made an awfull mistake (the post has been edited)\r\nBy differentiating both terms i got the relation [tex]f(g(b))=g(f(b))[/tex] for any real [tex]b[/tex] (I suppose the [tex]b[/tex] in the statement is any real [tex]b[/tex]). That's all !", "Solution_2": "f is g^(-1)" } { "Tag": [ "geometry" ], "Problem": "We have a triangle $ABC$ such that $AB=13$, $BC=14$, and $CA=15$. Let $M$ be the midpoint of $BC$, and let $N$ be the midpoint of $AB$. What is the value of $\\frac{[AMN]}{[ABC]}$? Express your answer as a common fraction (in lowest terms).", "Solution_1": "what do you mean by $[]$? is it area? is it angle :maybe:", "Solution_2": "[quote=\"anirudh\"]what do you mean by $[]$? is it area? is it angle :maybe:[/quote] It's common notation for area. :D", "Solution_3": "Is there a common notation for volume?\r\n\r\n[hide]$[AMN]=\\frac12 [ABM]=\\frac14 [ABC]$\n\n$\\boxed{\\frac{1}{4}}$[/hide]", "Solution_4": "i have attached the sol", "Solution_5": "[quote=\"anirudh\"]i have attached the sol[/quote]\r\nUm, that isn't a sol, that is just a picture..." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c>0$ . Prove that :\r\n\r\n\\[ \\frac{a\\plus{}b}{c}\\plus{}\\frac{b\\plus{}c}{a}\\plus{}\\frac{c\\plus{}a}{b}\\plus{}\\frac{9abc}{(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)}\\ge 7\\]", "Solution_1": "[quote=\"Pain rinnegan\"]Let $ a,b,c > 0$ . Prove that :\n\\[ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {9abc}{(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}\\ge 7\\]\n[/quote]\r\nLet $ a \\plus{} b \\plus{} c \\equal{} 3u,$ $ ab \\plus{} ac \\plus{} bc \\equal{} v^2$ and $ abc \\equal{} w^3.$\r\nHence, $ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {9abc}{(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}\\ge 7\\Leftrightarrow$\r\n$ \\Leftrightarrow\\frac {9uv^2 \\minus{} 3w^3}{w^3} \\plus{} \\frac {9w^3}{9uv^2}\\geq7\\Leftrightarrow$\r\n$ \\Leftrightarrow 9u^2v^4 \\minus{} 10uv^2w^3 \\plus{} w^6\\geq0\\Leftrightarrow(uv^2\\minus{}w^3)(9uv^2\\minus{}w^3)\\geq0,$ which is obvious.", "Solution_2": "[quote=\"Pain rinnegan\"]Let $ a,b,c > 0$ . Prove that :\n\\[ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {9abc}{(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}\\ge 7\\]\n[/quote]\r\nNot really strong ... \r\n\r\nJust use AM-GM's inequality : \r\n\r\n$ LHS \\equal{}\\left( \\frac {a \\plus{} b}{2c} \\plus{} \\frac {b \\plus{} c}{2a} \\plus{} \\frac {c \\plus{} a}{2b} \\plus{} \\frac {9abc}{(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)} \\right)$ $ \\plus{}\\left( \\frac {a \\plus{} b}{2c} \\plus{} \\frac {b \\plus{} c}{2a} \\plus{} \\frac {c \\plus{} a}{2b} \\right)$ $ \\geq 4\\sqrt [4]{\\frac {9(a \\plus{} b)(b \\plus{} c)(c \\plus{} a) }{8(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)} } \\plus{} 3 \\geq 7$", "Solution_3": "The following inequality is also true.\r\nLet $ a,$ $ b$ and $ c$ are positive numbers. Prove that\r\n\r\n\\[ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {81abc}{(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}\\ge 15\\]", "Solution_4": "[quote=\"arqady\"]The following inequality is also true.\nLet $ a,$ $ b$ and $ c$ are positive numbers. Prove that\n\\[ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {81abc}{(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}\\ge 15\\]\n[/quote]\r\nRewrite the inequality as \r\n$ \\frac{(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)}{abc}\\plus{}\\frac {81abc}{(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)} \\ge 18$\r\nWhich is clearly true according to the AM-GM ineq for 18 numbers.", "Solution_5": "Nice proof, leedt26!\r\nThe following inequality is also true.\r\nLet $ a,$ $ b$ and $ c$ are positive numbers, $ 0\\leq k\\leq6\\plus{}6\\sqrt3\\cos10^{\\circ}.$ Prove that:\r\n\\[ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b}\\ge2\\sqrt{9\\plus{}\\frac{k(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}ac\\minus{}bc)}{ab\\plus{}ac\\plus{}bc}}\\]\r\nwhere $ k\\equal{}6\\plus{}6\\sqrt3\\cos10^{\\circ}$ is the best value, for which this inequality is true for all positives $ a,$ $ b$ and $ c.$", "Solution_6": "The following is also true :\r\n\r\n\\[ \\frac{a\\plus{}b}{c}\\plus{}\\frac{b\\plus{}c}{a}\\plus{}\\frac{c\\plus{}a}{b}\\plus{}\\frac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2}\\ge 7\\]\r\n\r\nwith $ a,b,c> 0$", "Solution_7": "[quote=\"Pain rinnegan\"]The following is also true :\n\\[ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}\\ge 7\\]\nwith $ a,b,c > 0$[/quote]\r\n\r\nwe have\r\n\r\n$ \\sum_{cyc}\\frac{a\\plus{}b}{c}\\plus{}\\frac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2}\\minus{}6\\equal{}\\sum_{cyc}\\left(\\frac{a}{b}\\plus{}\\frac{b}{a}\\minus{}2\\right)\\plus{}\\frac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2}$\r\n$ \\equal{}\\sum_{cyc}\\frac{(a\\minus{}b)^2}{ab}\\plus{}\\frac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2} \\ge\\sum_{cyc}\\frac{(a\\minus{}b)^2}{a^2\\plus{}b^2\\plus{}c^2}\\plus{}\\frac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2}$\r\n$ \\equal{}\\frac{a^2\\plus{}b^2\\plus{}c^2\\plus{}(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\plus{}bc\\plus{}ca)}{a^2\\plus{}b^2\\plus{}c^2} \\ge 1$", "Solution_8": "[quote=\"Pain rinnegan\"]The following is also true :\n\\[ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}\\ge 7\\]\nwith $ a,b,c > 0$[/quote]\r\nHere is stronger:\r\nLet $ a,$ $ b$ and $ c$ are positive numbers. Prove that:\r\n\\[ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {4\\sqrt2(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2}\\ge 6\\plus{}4\\sqrt2\\]", "Solution_9": "[quote=\"arqady\"][quote=\"Pain rinnegan\"]The following is also true :\n\\[ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}\\ge 7\\]\nwith $ a,b,c > 0$[/quote]\nHere is stronger:\nLet $ a,$ $ b$ and $ c$ are positive numbers. Prove that:\n\\[ \\frac {a \\plus{} b}{c} \\plus{} \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {4\\sqrt2(ab \\plus{} bc \\plus{} ca)}{a^2 \\plus{} b^2 \\plus{} c^2}\\ge 6 \\plus{} 4\\sqrt2\\]\n[/quote]\r\n\r\nI think $ k\\equal{}4\\sqrt{2}$ is the best and it's easy by S.O.S :lol:" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that $ S_3$ is not the direct product of any family of its proper subgroups. The same is true of $ \\mathbf{Z}_{p^n}$ (p prime, n $ \\geq 1$) and $ \\mathbb{Z}$. (Hungerford 1.8.1)\r\n\r\nFirst, let me make sure I understand the question. They want me to prove that there is no isomorphism between any direct product of proper subgroups and the original group right?\r\n\r\nSecond, here is my attempt for $ S_3$. Let me know if I am on the right track. There are 6 elements in $ S_3$ so we need a subgroup of size two and a subgroup of size three. The only subgroups of size two are the transpositions and the only subgroups of size three are the derangements. So, WLOG any candidate for such a direct product would be isomorphic to $ < (2,1,3) > \\times < (3,1,2) >$. However, this direct product has only one element of order two namely $ (2,1,3) \\times (1,2,3)$ unlike $ S_3$ therefore it cannot be isomorphic.", "Solution_1": "Yes you understood the question.\r\nYour argument is ok. Note that the group you obtained is $ \\mathbb Z_6$ (as this one is cyclic and $ S_3$ is not, this gives another proof).\r\nBy the way, for the second part, you may want to show that no cyclic group is a product of proper subgroups (assuming that proper means: not the trivial group; if you only exclude the whole group, you would have $ \\mathbb Z \\cong (2 \\mathbb Z) \\times \\{0\\}$).", "Solution_2": "[quote=\"ZetaX\"] you may want to show that no cyclic group is a product of proper subgroups.[/quote]\r\n\r\nWait, that's not true: $ Z_2 \\times Z_3 \\equal{} Z_6$.", "Solution_3": "Ups :blush: :wink:", "Solution_4": "for the $ \\mathbb{Z}_{p^n}$ part, i would use the fundamental theorem of finite abelian groups, but this may be a bigger hammer than is needed, or has been handed to you yet.", "Solution_5": "It is easy to prove that if finite cyclic group is a direct product of its subgroups then orders of that subgroups are pairwise relatively prime.\r\n \r\nFor the third part - every nontrivial subgroup of $ \\mathbb{Z}$ is isomorphic to $ \\mathbb{Z}$, and for every $ k \\equal{} 2,3,\\ldots$, $ \\ \\mathbb{Z}$ is not isomorphic to $ \\mathbb{Z}^{k}$, since $ \\mathbb{Z}^{k}$ is not cyclic." } { "Tag": [ "LaTeX", "induction", "graph theory", "combinatorics solved", "combinatorics" ], "Problem": "Let's color each of the edges of the complete graph with $n$ vertices either in red or blue.\r\nDetermine a close formula for the least number $f(n)$ of monochromatic triangles that appear in the figure.\r\n\r\nPierre.", "Solution_1": "Don't you use the classical theorem about the number of edges, I can't remember it now :(, but it's the formula for f(k) and f(n(n-1)/2 - k) minimum of them ?", "Solution_2": "???Sorry but I don't see what you are talking about...\r\n\r\nPierre.", "Solution_3": "hmm... what about $[\\frac {n-3} 3]$? It is a lower bound, but probably not very close.", "Solution_4": "Far from best possible... :cool:\r\n\r\nPierre.", "Solution_5": "That's what I thought...the result has been found in 1959 :( \r\nIn aonther hand I've less than 50 years late, thus maybe with some extra efforts...\r\n\r\nPierre.", "Solution_6": "A much better lower bound is $n(n-1)(n-2)/120$. It is obtained just by the fact that in a $K_6$ there is always a monochromatic triangle. now that we may choose $n(n-1)(n-2)(n-3)(n-4)(n-5)/720$ (what is the Latex-command for a binomial coefficient?) 6-element subsets of our graph, we get that much triangles, and every triangle is counted at most $(n-3)(n-4)(n-5)/6$ times, so we get that lower bound. I don't know if it's the best one, but it looks really good I think.\r\n\r\nPeter", "Solution_7": "Ok, I just give the answer :\r\n\r\n$f(2a+2) = \\frac {a(a-1)(a+1)} {3}$,\r\n$f(4a+1) = \\frac {2a(4a+1)(a-1)} {3}$,\r\n$f(4a+3) = \\frac {(2a+1)(4a+3)(2a-1) + 3} {6}$\r\n\r\nPierre.", "Solution_8": "can you prove it please?\r\n\r\nThanx", "Solution_9": "Yes I can, since I proposed it :D \r\nI'll give a solution later, if no one solved it :(\r\n\r\nPierre.", "Solution_10": "Ok Pascual, since I've got a little time, I can give you my solution now. As always it won't be a short post so I will divide it into two or three parts to avoid computer bugs....\r\n\r\nLet $A_1,...,A_n$ be the vertices.\r\nFor $i,j,k$ pairwise distinct in $ \\{1,...,n \\}$, let $p_i(j,k) = 2$ if the edges $A_iA_j$ and $A_iA_k$ have the same color, and $p_i(j,k)=-1$ otherwise.\r\n\r\nNote that if the triangle $A_iA_jA_k$ is monochromatic then $p_i(j,k)+p_j(k,i)+p_k(i,j)=6$ and $p_i(j,k)+p_j(k,i)+p_k(i,j)=0$ otherwise.\r\n\r\nLet $p_i= \\sum p_i(j,k)$ (the sum is over $1 \\leq j < k \\leq n$ and $j,k \\neq i$\r\nand let $p= \\sum_{i=1}^n p_i$.\r\n\r\nLet $b$ (resp. $r$) denotes the number of blue (resp. red) triangles.\r\nSince each pair of adjacent edges belongs to one and only one triangle, we then have :\r\n$p=6(b+r)$. (1)\r\n\r\nIn another hand, for a given $i$, let $k$ be the number or red edges with endpoint $A_i$. Thus, there are $n-1-k$ blue edges with endpoint $A_i$. We then have :\r\n${p_i = 2C_k^2 + 2C_{n-1-k}^2 - C_k^1C_{n-1-k}^1 = 3(k - \\frac {n-1} 2} )^2 + \\frac {(n-5)(n-1)} {4}$. (2)\r\n\r\nFirst case : If $n=2a+2$ is even.\r\nThen, from (2), we deduce that $p_i \\geq \\frac {3} {4} + \\frac {(n-5)(n-1)} {4} = a(a-1)$ for each $i$.\r\nUsing (1), it follows that $b+r \\geq \\frac {1} {6} na(a-1) = \\frac {a(a+1)(a-1)} {3}$.\r\nSince both $f(2a+2)$ and $ \\frac {a(a+1)(a-1)} {3}$ are integers, we have $f(2a+2) \\geq \\frac {a(a+1)(a-1)} {3}$.\r\n\r\nConversely, let's consider the complete graph whose edges $A_iA_j$ are colored red if $i=j$ mod[2], and blue otherwise.\r\nIt is easy to verify that this graph has exactly $2C_{a+1}^3$ monochromatic triangles, which provs that $f(2a+2) \\leq \\frac {a(a+1)(a-1)} {3}$.\r\nThus $f(2a+2) = \\frac {a(a+1)(a-1)} {3}$ as claimed.\r\n\r\nPierre.", "Solution_11": "Now, let's turn to the second case, $n =2a+1$ is odd :\r\n\r\nUsing (2), we deduce that $p_i \\geq \\frac {(n-5)(n-1)} {4} = a(a-2)$ for each $i$.\r\nUsing (1), it follows that $b+r \\geq \\frac {1} {6} na(a-2) = \\frac {a(2a+1)(a-2)} {6}.$\r\nSince $f(2a+1)$ is an integer, we deduce that $f(2a+1) \\geq [ \\frac {a(2a+1)(a-2)} {6} ]$ where $[x]$ denotes the least integer greater than or equal to $x$ (and not the integer part....).\r\nThis is equivalent to \r\n$f(4a+1) \\geq \\frac {2a(4a+1)(a-1)}{3}$ (which is an integer), and\r\n$f(4a+3) \\geq \\frac {(2a+1)(4a+3)(2a-1)+3} {6}$ (because $\\frac {(2a+1)(4a+3)(2a-1)} {6}$ is the half an odd integer).\r\n\r\nConversely :\r\n- If $n = 4a+1$.\r\nFirst, for $a=1$, we consider the coloring of the edges of the complete graph $K_5$ seen as all the line segments joining two of the vertices of a regular convex pentagon, where the sides of the pentagon are blues and the diagonals are red. In that case, there is no monochromatic triangle, so that $f(5)=0$.\r\nNote that in this graph, from each vertex there are the same number of red edges than blue ones. Such a graph will be called a balanced graph.\r\n\r\nNow, assume that for a given $a \\geq 2$, we have a $K_{4(a-1)+1}$ balanced edge bicolored complete graph.\r\nLet's give a complete graph $K_{4a+1}$. And color its edges as follows :\r\nFirst we construct a balanced $K_{4(a-1)+1}$ for the subgraph whose vertices are $A_5,A_6,...,A_{4a+1}$, from the induction hypothesis.\r\nThen, color the edges of the subgraph whose vertices are $A_1,A_2,...,A_5$ as in the regular pentagon above.\r\nAt last, for each $i \\geq 6$, give color of $A_5A_i$ to $A_1A_i$ and to $A_2A_i$, and the opposite ne to $A_3A_i$ and $A_4A_i$.\r\nIt is easy to verify that such a bicoloring gives us a balanced $K_{4a+1}$.\r\n\r\nFor such a complete graph, for each $i$ the number of red edges from $A_i$ is equal to $k=2a$, from which we deduce (using (1)) that $p_i = \\frac {(n-5)(n-1)} {4}$. It easily follows from above that $b+r= \\frac {n(n-5)(n-1)} {4}$ which gives $f(4a+1) \\leq \\frac {(2a(4a+1)(a-1)} {3}$, and we are done.\r\n\r\n- If $n=4a+3$ :\r\nLet's color the edges of $K_{4a+3}$ as follows :\r\nFirst color the edges of the subgraph $K_{4a+1}$ whose vertices are $A_3,...,A_{4a+3}$ to obtain a balanced graph as above.\r\nThen for each $i \\geq 4$ give the color $A_3A_i$ to $A_1A_i$ and the opposite one to $A_2A_i$.\r\nNow give the color blue to $A_1A_2$ and red to $A_1A_3$ and $A_2A_3$.\r\nIt is easy to verify that for each $i \\neq 3$, we have exactly $k=2a+1$ red edges from $A_i$, so that $p_i = \\frac {(n-5)(n-1)} {4}$.\r\nMoreover the number of red edges from $A_3$ is $2a$, so that $p_3 = 3+ \\frac {(n-5)(n-1)} {4}$.\r\nIt follows that $p = 3 + \\frac {n(n-5)(n-1)} {4}$ and $f(4a+3) \\geq b+r = \\frac\r\n{(2a+1)(4a+3)(2a-1)} {6} + \\frac {1} {2}$, and we are done.\r\n\r\nPierre.", "Solution_12": "I never said thanks.... :? \r\n\r\nSo it is time: Thanks Pierre for such a wonderful solution!", "Solution_13": ":) \r\n\r\nPierre.", "Solution_14": "I think it is a theorem called Goodman.What else,I wonder what Pierre means in French,could you tell me? :blush:", "Solution_15": "It is his name ;)" } { "Tag": [ "trigonometry" ], "Problem": "Let be given a [u][b]acute-triangle[/b][/u] $ABC$. Prove that: $\\left(\\sum\\cos A\\right)^{2}\\le\\sum\\sin^{2}A$", "Solution_1": "[hide]\nWriting $\\sin^{2}A=1-\\cos^{2}A$ etc. and using the formula\n\\[\\cos^{2}A+\\cos^{2}B+\\cos^{2}C+2\\cos A\\cos B\\cos C=1\\]\nit is equivalent to\n\\[2\\cos A\\cos B+2\\cos B\\cos C+2\\cos C\\cos A-4\\cos A\\cos B\\cos C\\leq 1\\]\nWLOG let $\\angle A\\leq \\angle B\\leq \\angle C$, so that $\\angle C\\geq 60^\\circ$ and $\\cos C\\leq \\frac{1}{2}$. The left hand side of the above becomes\n\\[2\\cos A\\cos B(1-2\\cos C)+2\\cos C(\\cos A+\\cos B)\\]\nAs $2\\cos A\\cos B=\\cos (A+B)+\\cos (A-B)=\\cos (A-B)-\\cos C\\leq 1-\\cos C$, it remains to prove that\n\\[(1-\\cos C)(1-2\\cos C)+2\\cos C(\\cos A+\\cos B)\\leq 1\\]\nBut the left hand side equals $1+\\cos C(2\\cos A+2\\cos B+2\\cos C-3)$, and $\\cos A+\\cos B+\\cos C\\leq 3\\cos \\left( \\frac{A+B+C}{2}\\right) =\\frac{3}{2}$ by Jensen.\n[/hide]", "Solution_2": "We have: $2\\cos B\\cos C=\\sqrt{\\sin 2A\\cot A\\sin 2B\\cot B}\\le\\frac{1}{2}(\\sin 2A\\cot B+\\sin 2B\\cot A)$\r\nHence: \r\n\\begin{eqnarray*}2\\sum\\cos B\\cos C&\\leq&\\frac{1}{2}\\sum\\sin 2A\\cot B+\\sin 2B\\cot A\\\\ &=&\\frac{1}{2}\\sum\\cot A(\\sin 2B+\\sin 2C)\\\\ &=&\\frac{1}{2}\\sum\\cot A.2\\sin A\\cos(B-C)\\\\ &=&-\\frac{1}{2}\\sum 2\\cos(B+C)\\cos(B-C)\\\\ &=&-\\frac{1}{2}\\sum\\cos 2B+\\cos 2C\\\\ &=&3-2\\sum\\cos^{2}A \\end{eqnarray*}\r\n\\[\\Longleftrightarrow\\left(\\sum\\cos A\\right)^{2}\\le 3-\\sum\\cos^{2}A=\\sum\\sin^{2}A \\]\r\n[b]Remark.[/b] Use: $\\sum\\cos^{2}A=1-2\\prod\\cos A$ we have:\r\n\\[\\boxed{\\left(\\sum\\cos A\\right)^{2}\\le\\sum\\sin^{2}A\\Longleftrightarrow\\sum(\\cos B+\\cos C)^{2}\\le 3\\Longleftrightarrow 2\\sum\\cos B\\cos C\\le 1+4\\prod\\cos A}\\]", "Solution_3": "[b] Remark 2.[/b] Use: $\\sum\\cos A=1+\\frac{r}{R}$ we have:\r\n\\[\\boxed{\\left(\\sum\\cos A\\right)^{2}\\le\\sum\\sin^{2}A\\Longleftrightarrow 4(R+r)^{2}\\le a^{2}+b^{2}+c^{2}}\\]", "Solution_4": "I've just found it at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14567]trig question[/url]. And I'm very interesting with [b]treegoner's solution[/b], therefore I'll rewrite [b]treegoner's solution[/b] in particular $m=n=p=1$ at here.\r\n[b]Lemma 1.[/b] $2\\cos A.\\sin B.\\sin C=\\sin^{2}B+\\sin^{2}C-\\sin^{2}A$ \r\n[b]Lemma 2.[/b] $\\sum\\frac{\\cos A}{\\sin B\\sin C}=2$\r\nCome back with our problem:\r\nWe have:\r\n\\begin{eqnarray*}\\left(\\sum\\cos A\\right)^{2}&=&\\left(\\sum\\sqrt{\\cos A\\sin B\\sin C}\\cdot\\sqrt{\\frac{\\cos A}{\\sin B\\sin C}}\\right)^{2}\\\\ &\\leq& \\left(\\sum \\cos A\\sin B\\sin C\\right)\\left(\\sum\\frac{\\cos A}{\\sin B\\sin C}\\right)\\\\ &=&\\sum\\sin^{2}B+\\sin^{2}C-\\sin^{2}A\\\\ &=&\\sum\\sin^{2}A \\end{eqnarray*}" } { "Tag": [ "inequalities" ], "Problem": "Prove that if $ a,b,c>0$ then \r\n\r\n$ \\sqrt{abc}(\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c})\\plus{}(a\\plus{}b\\plus{}c)^{2}\\geq 4\\sqrt{3abc(a\\plus{}b\\plus{}c)}$", "Solution_1": "From Schur's inequality and AM-GM we get \r\n\\[ a^2\\plus{}b^2\\plus{}c^2\\plus{}\\sqrt{abc}(\\sqrt a\\plus{}\\sqrt b\\plus{}\\sqrt c)\\ge 2(ab\\plus{}bc\\plus{}ca)\\]\r\nTherefore it remains to show that \r\n\\[ ab\\plus{}bc\\plus{}ca\\ge\\sqrt{3abc(a\\plus{}b\\plus{}c)}\\]\r\nwhich is just AM-GM." } { "Tag": [], "Problem": "x = y\r\nx\u00b2 = xy\r\n2x\u00b2-2xy = x\u00b2-xy\r\n2(x\u00b2-xy) = x\u00b2-xy\r\n2 = 1", "Solution_1": "The third step doesnt make sense", "Solution_2": "[quote=\"jimli\"]x = y\nx\u00b2 = xy\n2x\u00b2-2xy = x\u00b2-xy\n2(x\u00b2-xy) = x\u00b2-xy\n2 = 1[/quote]\r\n\r\nStep 4, i.e. $2(x^2-xy)=x^2-xy$ implies division by zero.\r\n\r\nIt's not a puzzle, it's a mistake.", "Solution_3": "[quote=\"Dr. No\"][quote=\"jimli\"]x = y\nx\u00b2 = xy\n2x\u00b2-2xy = x\u00b2-xy\n2(x\u00b2-xy) = x\u00b2-xy\n2 = 1[/quote]\n\nStep 4, i.e. $2(x^2-xy)=x^2-xy$ implies division by zero.\n\nIt's not a puzzle, it's a mistake.[/quote]\r\n\r\nI knew it is a mistake. However, division by zero can be \"done\"(cough).\r\n\r\n1/1=1/2+1/2^2+1/2^3...\r\n1/2=1/3+1/3^2+1/3^3...\r\n1/3=1/4+1/4^2+1/4^3...\r\n\r\n1/0=1/1+1/1^2+1/1^3...= :inf:", "Solution_4": "there is post in the High School--Other Problem Solving forum talking about $1=2$, there are many proofs. You may find it interesting. :D", "Solution_5": ":lol: You're proof basically proves that Infinity = Infinity.", "Solution_6": "[quote=\"shobber\"]there is post in the High School--Other Problem Solving forum talking about $1=2$, there are many proofs. You may find it interesting. :D[/quote]\r\n\r\nI take exception to the use of the word [i]proof[/i] in this context. The possibility of division by zero is itself founded upon a contradiction. Therefore, any result that follows from division by zero will be logically inconsistent by definition. Such so called proofs are not clever, they're just ignorant.", "Solution_7": "jimli,\r\n\r\nI moved this from Classroom Math to the Basics as we try to leave Classroom Math for ordinary middle school curriculum while the Basics serves as the forum for learning the mechanics of math that eventually leads to advanced high school curriculum.", "Solution_8": "if x=y, then x 2 -xy is 0. dividing by 0 is not allowed", "Solution_9": "Your proof is invalid" } { "Tag": [ "geometry", "incenter", "circumcircle", "analytic geometry", "trigonometry", "geometry proposed" ], "Problem": "Let $ABC$ a triangle and his cirmumcircle. Let $I$ the incenter of the triangle and let \r\n$D,E,F$ the of intersection of the lines $AI,BI,CI$ with the circumcircle.\r\n\r\nProve that $\\frac{AI}{ID}+\\frac{BI}{IE}+\\frac{CI}{IF}\\geq 3$", "Solution_1": "$l_a$ - the length of the bisector of the angle $\\widehat{A}\\ ,\\ X\\in BC\\cap AD\\Longrightarrow$\r\n$\\frac{IA}{b+c}=\\frac{IX}{a}=\\frac{l_a}{2p}$. It is well-known the relation $l_a=\\frac{2}{b+c}\\sqrt{bcp(p-a)}$.\r\nThus, $IA=\\frac{b+c}{2p}\\cdot l_a\\Longrightarrow IA^2=\\frac{bc(p-a)}{p}=\\frac{abc(p-a)}{ap}=\\frac{4Rrp(p-a)}{ap}\\Longrightarrow$\r\n$IA^2=2Rr\\cdot \\frac{b+c-a}{a}\\Longrightarrow \\sum IA^2=2Rr\\cdot \\left(-3+\\sum \\frac{b+c}{a}\\right)\\ .$\r\n$IA\\cdot ID=IB\\cdot IE=IC\\cdot IF=2Rr\\ ;\\ \\sum \\frac{b+c}{a}\\ge 6\\ .$\r\n$==================================$\r\n$\\sum \\frac{IA}{ID}=\\sum \\frac{IA^2}{IA\\cdot ID}=\\frac{1}{2Rr}\\cdot\\sum IA^2=$\r\n$-3+\\sum \\frac{b+c}{a}\\ge -3+6=3\\ .$Therefore, $\\sum \\frac{IA}{ID}\\ge 3\\ .$\r\n\r\n[b]Remark.[/b] For a inner point $M$ denote the second intersections $D$, $E$, $F$\r\nof the circumcircle with the lines $AM$, $BM$, $CM$ respectively. For $M: =O$,\r\nis evidently $\\sum \\frac{OA}{OD}=3$; also, for $M: =G$, $\\sum \\frac{GA}{GD}=3$.\r\nFor any inner point $M$ is true the relation $\\sum \\frac{MA}{MD}\\ge 3\\ ?$\r\n[u]Silouan, now you have a research them ![/u]\r\n[hide=\"Here is a extension.\"]Denote the intersections $X\\in AM\\cap BC$, $Y\\in BM\\cap CA$, $Z\\in MC\\cap AB$. Let $ABC$ be a triangle and let $M$ be a inner point with\nthe barycentrical coordinates $(x,y,z)$ w.r.t. the triangle $ABC$, i.e. $x>0$, $y>0$, $z>0$, $x+y+z=1$ and $\\frac{x}{[BMC]}=$ $\\frac{y}{[CMA]}=$ $\\frac{z}{[AMB]}$. Denote $p=a^2yz+b^2zx+c^2xy$. Then $MA\\cdot MD=MB\\cdot ME=MC\\cdot MF=p$, $(y+z)^2AX^2=(b^2z+c^2y)(y+z)-a^2yz$, $MA=(y+z)AX$ $\\Longrightarrow$ $AM^2=c^2y+b^2z-p$. Therefore, $\\sum \\frac{MA}{MD}=$ $\\sum \\frac{MA^2}{MA\\cdot MD}=$ $\\frac 1p\\left[\\sum (b^2z+c^2y)-3p\\right]=$ $-3+\\frac 1p\\sum a^2(y+z)$ a.s.o.[/hide]", "Solution_2": "Nice solution mr Virgil .My solution was ungly with trigonometry but this is qute and small. Thanks", "Solution_3": "I have another solution:\r\nWe have $AI=4Rsin\\frac{B}{2}sin\\frac{c}{2}$.\r\n$DI=BD=2Rsin\\frac{A}{2}$.\r\nThus $\\frac{AI}{DI}=\\frac{2sin\\frac{B}{2}sin\\frac{C}{2}}{sin\\frac{A}{2}}$ $=$ $2(\\frac{1}{1-tg\\frac{B}{2}tg\\frac{C}{2}} -1)$.\r\nIt suffice to prove that $\\sum\\frac{1}{1-tg\\frac{B}{2}tg\\frac{C}{2}}-3 \\geq \\frac{3}{2}$\r\n$\\leftrightarrow$ $\\sum\\frac{1}{1-tg\\frac{B}{2}tg\\frac{C}{2}} \\geq \\frac{9}{2}$,which is true because $\\sum tg\\frac{B}{2}tg\\frac{C}{2} =1$.", "Solution_4": "Moreover,we have $IA^2+IB^2+IC^2 \\leq ID^2+IE^2+IF^2$." } { "Tag": [ "ceiling function", "logarithms", "floor function", "number theory", "least common multiple", "number theory proposed" ], "Problem": "I'm finding difficult to understand the following (for reasons i will explain), which is part\r\nof the \"PRIMES is in P\" paper by AKS, \r\n[url]http://www.math.princeton.edu/~annals/issues/2004/Sept2004/Agrawal.pdf[/url]. \r\nMaybe someone can help me pointing out where my misunderstanding is.\r\n\r\nLemma1: if $ n \\geq 7$, $ lcm(1,2,\\ldots,n)\\geq 2^n$\r\nDefine $ o_r(n)$ to be the order of $ n$ in $ (\\mathbb{Z}/r\\mathbb{Z})^*$, for $ (n,r)=1$.\r\nClaim: Suppose n>2. Then there exist an $ r\\leq \\lceil \\log_2^5 n \\rceil$ s.t. $ o_r(n)>\\log_2^2n$\r\nProof: Let $ r_1,\\ldots,r_t$ be all numbers s.t. either $ o_{r_i}(n)\\leq \\log_2^2 n$ or $ r_i|n$. Then each $ r_i$\r\ndivide the product \r\n$ n\\cdot \\prod_{i=1}^{\\lfloor log_2^2 n\\rfloor}(n^i-1)1$, then since $ s$ does not divide $ n$ and $ (s,n)\\in \\{r_1, \\ldots, r_t\\}$, $ r=\\frac{s}{(s,n)}\\not\\in \\{r_1,\\ldots,r_t\\}$ and so $ o_r(n)>\n\\log_2^2 n$[/b]\r\n\r\nMy problem is on [b]bolded[/b] text. Consider $ n=6$ and $ s=4$. We have $ s\\leq\\lceil \\log_2^5 n\\rceil$\r\nand $ s\\not | n$, and $ o_r(n)$ not defined as $ (s,n)>1$, hence $ s \\not\\in\\{r_1, \\ldots, r_t\\}$. \r\nBut ${ r=\\frac{s}{(s,n)}=2 | n}$, so it is not possible that $ o_r(n)>log_2^2 n$, nor $ r\\not\\in\\{r_1, \\ldots, r_t\\}$. ?!?", "Solution_1": "I think you are right, the proof seems to be wrong... :huh:\r\n\r\nHowever, it can be fixed, by choosing $ r$ that does not divide\r\n\\[ P \\equiv n^2\\cdot \\prod_{i \\equal{} 1}^{\\lfloor \\log_2^2 n\\rfloor}(n^i \\minus{} 1).\r\n\\]\r\nThen the original argument works. Assume that $ r \\equal{} s/(s,n) \\in \\{r_1,\\dots,r_t\\}$. Then either (i) $ r$ divides $ n$ or (ii) $ r$ is coprime to $ n$. In the first case, $ s$ divides $ n^2$; in the second case, $ r$ divides $ \\prod_{i \\equal{} 1}^{\\lfloor \\log_2^2 n\\rfloor}(n^i \\minus{} 1)$, and so $ s$ divides $ P$. We get a contradiction.\r\n\r\nUpdate: in fact, what this proves is that $ r \\equal{} s/(s,n)$ is not in $ \\{r_1,\\dots, r_t\\}$. But $ r$ is not necessarily coprime to $ n$. :(", "Solution_2": "[quote]In the first case, $ s$ divides $ n^2$ ;[/quote]\r\n\r\nUhm, can you please explain this a little more ? Is the claim:\r\n$ r\\in \\{r_1, \\ldots, r_t\\}\\wedge (r,n) > 1 \\Rightarrow r \\equal{} \\frac {s}{(s, n)}|n^2 \\Rightarrow r|n \\Rightarrow s|n^2$ ?\r\nI suspect the $ 2$ in $ n^2$ is not enough... :( \r\n\r\nAnyway, thank you for your reply :)", "Solution_3": "Yes!!!!! I got it! They've fixed it in this new version, I've just found on agrawal homepage:\r\n[url]http://www.cse.iitk.ac.in/~manindra/algebra/primality_v6.pdf[/url]\r\n\r\nJust replace the \"big product\", $ P$, by \r\n$ n^{\\lfloor \\log_2 \\lceil \\log_2^5 n\\rceil\\rfloor}\\cdot\\prod_{i \\equal{} 1}^{\\lfloor \\log_2^2 n\\rfloor}(n^i \\minus{} 1)$\r\n\r\nSo the journal version has definitely this minor mistake... good to know :D\r\nAnyway thank you Yuri for having pointed out that we must somehow take a bigger power of $ n$." } { "Tag": [ "Columbia" ], "Problem": "Bueno, llego la hora de las apuestas!, esta comunidad es supuestamente iberoamericana, entonces pues me gustaria que en este tema apostemos los resultados de las olimpiadas venideras, Centro, Imo e Ibero, en cuanto aa pais iberoamericano que ganara. Y tambien pues para que vallamos posteando los equipos que se vallan escogiendo:\r\n\r\nMis predicciones son: \r\nCentro: Mexico\r\nIMO: Colombia\r\nIbero: Colombia\r\n\r\nEn cuanto al equipo de colombia les cuento que ya empezo la olimpiada colombiana, de la gente que ya ha viajado este a\u00f1o parece repetiran 3 ( Jose Gabriel Acevedo (imo2004), Esteban Gonzales (Imo 2003, ibero 2004), Y yo (imo 2003, 2004, ibero 2003). De resto parece ser gente nueva, entonces no creo que el equipo este tan fuerte como el a\u00f1o pasado que iban 4 ex imos.\r\n\r\nEn cuanto a otros paises, supongo Manuel Y Leonardo van a ir...(leo Chavez=chang?), pues hay un argentino nuevo en el foro con el nombre Severius o algo asi, y creo que esta trabajando bien, supongo que tambien ira, se tambien que Sherry (como se escriba) parece que va a participar con los EEUU, de resto ni idea.\r\n\r\nOtra cosa es que creo que venezuela este a\u00f1o va a tener un equipo mas fuerte por lo visto en el entrenamiento de enero aqui en colombia donde participaron, y tambien parece que muchos paises de centroamerica van a ir a imo, que bien!\r\n\r\nEspero pongan sus predicciones y vallamos discutiendo los equipos y todas estas cosas!\r\n\r\nSuerte", "Solution_1": "I also noticed new strong member Severius. :) I am very happy, because my IMO was held in Argentina and I like this country very much :)\r\n\r\nI think Colombia will have at least one gold medal ;)\r\n\r\nGood luck to all participants!", "Solution_2": "Bueno lo lamento Pascual pero la ibero este a\u00f1o la va a ganar Argentina!! a pesar de ke sean locales, si si si... con respecto a la IMO weno, este a\u00f1o no se te tiene ke escapar el oro eh! \r\n Centroamericana no tengo idea kienes participan ni nada, pero supongo ke debe ser algo parecido a la del Cono Sur... \r\n\r\nbtw talking about Severius, i have to say he surprised everyone here (and hope he still doing that!) because he first participated in the Argentina National Olympiad in 2004, and he won in his level, then got a silver medal at the Rioplatense Math Olympiad :lol: \r\n\r\nAsi ke supongo ke el ira a la IMO, acompa\u00f1ado de Gabriel Carvajal (imo2004,ibero2004), Maxi Camporino(Cono sur 2004), Ignacio Rossi(imo2004), y weno la verdad no estoy muy seguro de los demas, es muy incierto el selectivo de estte a\u00f1o... \r\n \r\n Lo bueno de ke la IMO sea en Mexico este a\u00f1o, es ke van a invitar seguramente a muchos paises de latinoamerica ke usualmente no van a la IMO... \r\n lo malo es ke yo ya no puedo ir :S\r\n\r\nSuerte a todos! \r\n\r\nRaMlaF...", "Solution_3": "Pascual le tienes a ganar a sherry por mas de 5 puntos pa bajarle el guille! (asi se dice alla en colombia jaj?)", "Solution_4": "Jaja, Ramiro: yo crei que usted iba! ademas la ibero se queda en casa, esperamos llevar este a\u00f1o un equipo IMO a la Ibero, jaja, aunque nada es oficial, vamos a ver si por lo menos me dejan ir a mi!\r\n\r\nManuel: jaja, claro, ademas esa vieja ya no va a representar su pais!, supongo que toco ganarle, no?\r\n\r\nResto de gente: hagan sus pollas!\r\n\r\nMyth: I see you know some spanish! Great! :D :D", "Solution_5": "[quote=\"RaMlaF\"] Lo bueno de ke la IMO sea en Mexico este a\u00f1o, es ke van a invitar seguramente a muchos paises de latinoamerica ke usualmente no van a la IMO... \n[/quote]\r\n\r\nLo malo es que segun parece, todos los paises centroamericanos que van este anio, no van a ir en los anios posteriores, o al menos eso es lo que he escuchado de varios de los entrenadores. El asunto es que, como en todo lado, para estos eventos escatiman la plata, y pues, ni modo, van este anio por ser en Mexico y quedar bien cerca, pero el otro, no les dan plata ni para los pasajes. Que triste la historia del tercer mundo :( \r\n\r\nBueno, nos vemos,", "Solution_6": "[quote=\"Pascual2005\"]Jaja, Ramiro: yo crei que usted iba! cagada que nO!, ademas la ibero se queda en casa, esperamos llevar este a\u00f1o un equipo IMO a la Ibero, jaja, aunque nada es oficial, vamos a ver si por lo menos me dejan ir a mi!\n[/quote]\r\n\r\njaja weno.. no es ke sea tan viejo (17), pero termine el colegio en diciembre y no me dejan ir al selectivo :( ... desde ya ke de todas formas los ke vayan van a tener ke traerme un tekila de regalo :D", "Solution_7": "si es que no no lo tomamos todos! jaja!\r\n\r\nDjimenez: han tratado esos paises de buscar patrocinio con las empresas locales? a nosotros nos ha ayudado!", "Solution_8": "I think that brazilian time is the best.", "Solution_9": "[quote=\"Pascual2005\"]Djimenez: han tratado esos paises de buscar patrocinio con las empresas locales? a nosotros nos ha ayudado![/quote]\r\n\r\nLo ignoro. A quienes conozco trabajan mas que todo en la parte de entrenamiento, no tanto en la parte administrativa.", "Solution_10": "jaja manuel no sea despectivo orale! jaja! solo esperemos que no nos partan como siempre haciendo problemas!", "Solution_11": "Hey, kiubo!, como andan todos. De los equipos de Venezuela para las olimpiadas venideras solo tengo informacion de la IMO, al parecer los candidatos mas fuertes somos Roland Hablutzel (Ibero 2004), Andres Guzman (Ibero 2004), y yo (Ibero 2003-IMO 2004). Al parecer puede que Rodrigo (IMO 2003-IMO 2004) no participe este a\u00f1o, pero nada se sabe aun. Ya comenzaron las clases, yo me incorporare pronto, porque ahora ando en US", "Solution_12": "Por cierto, quien es chavez????? no es chang o si??", "Solution_13": "Are you also going to Clay Academy, Leonardo?", "Solution_14": "No Myth, he is not going, he is living in the USA right now, because his father is there i think, and he is going back to Venezuela soon.", "Solution_15": "Leo, le mande un PM a Chavez porque me imagino que con ese nick seguramente es Chang (por lo que he oido, no?), sin embargo no ha respondido! suerte", "Solution_16": "Mis predicciones:\r\nIMO: M\u00e9xico\r\nCentro: M\u00e9xico\r\nIbero: M\u00e9xico\r\n\r\ncon suerte, los veo en la IMO", "Solution_17": "[quote=\"Dave Vather\"]Mis predicciones:\nIMO: M\u00e9xico\nCentro: M\u00e9xico\nIbero: M\u00e9xico\n\ncon suerte, los veo en la IMO[/quote]\r\n\r\nDejame adivinar... Sos de Mexico... Yujuuu, pegue! (Digo, la verdad es que tanta seguridad, pues sorprende). Entre los paises latinoamericanos que van a la IMO, pues yo apostaria hacia Argentina, Brasil o Colombia antes que a Mexico. Con referencia a la Ibero, pues igual, apostaria a Argentina, Brasil o Mexico antes que a Mexico. Con respecto a la Centroamericana, pues talvez Colombia, talvez si, Mexico!!", "Solution_18": "Si los mexicanos no apost\u00e1ramos por M\u00e9xico, \u00bfQui\u00e9n lo har\u00eda?\r\nCreo que si hay un a\u00f1o para ser el primer pa\u00eds Iberoamericano en la IMO deber\u00eda ser este. La competencia por llegar a la IMO est\u00e1 m\u00e1s re\u00f1ida que en a\u00f1os pasados. La Ibero todav\u00eda est\u00e1 lejos pero creo que la podemos ganar, y la centro la llevamos ganando por 3 a\u00f1os seguidos, la pasada por m\u00e1s de 20 puntos y los tres primeros lugares. No la podemos perder.\r\nSi consigo ir a la IMO estoy dispuesto a apostar (al menos, por m\u00ed)", "Solution_19": "[quote=\"Dave Vather\"]Si consigo ir a la IMO estoy dispuesto a apostar (al menos, por m\u00ed)[/quote]\r\n\r\nBuena Suerte!!", "Solution_20": "jajajajajaja pascual q malas predicciones\u00a1\u00a1\u00a1\u00a1 mexiko por todos lados ser\u00e1 kien gane. yo konozco a los integrantes del equipo tenemos a varios repetidores. y eso q en los a\u00f1os pr\u00f3ximos a\u00f1os el equipo mexikano quedar\u00e9 mucho mejor xk tendr\u00e1 a integrantes komo yo. todav\u00eda no me konocen a nivel nacional jajaja aunque he ganado 2 bronces en los 2 nacionales q he ido peo las kosas este a\u00f1o jejejeje seran mucho mejor tenemos en mi estado a varios integrantes del equipo q son mui buenos jejejeje nimodo pascual pa la proxima\u00a1\u00a1\u00a1\u00a1", "Solution_21": "[quote=\"marko avila\"]jajajajajaja pascual q malas predicciones\u00a1\u00a1\u00a1\u00a1 mexiko por todos lados ser\u00e1 kien gane. yo konozco a los integrantes del equipo tenemos a varios repetidores. y eso q en los a\u00f1os pr\u00f3ximos a\u00f1os el equipo mexikano quedar\u00e9 mucho mejor xk tendr\u00e1 a integrantes komo yo. todav\u00eda no me konocen a nivel nacional jajaja aunque he ganado 2 bronces en los 2 nacionales q he ido peo las kosas este a\u00f1o jejejeje seran mucho mejor tenemos en mi estado a varios integrantes del equipo q son mui buenos jejejeje nimodo pascual pa la proxima\u00a1\u00a1\u00a1\u00a1[/quote]\r\n\r\nEspero no sonar mala gente o grocero (aunque probablemente lo suene) pero si quer\u00e9s hacer cosas bien en Olimpiadas, por qu\u00e9 no empez\u00e1s con tu ortograf\u00eda y tu redacci\u00f3n. Tanto aqu\u00ed como en [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=200957#p200957]este otro post[/url] dejan muchisimo que desear. \r\n\r\nPor otro lado, dependiendo de la edad que teng\u00e1s, dos bronces en una olimpiada nacional no son un tan buen augurio, en especial por que de ir a la IMO o a la Ibero, te encontrar\u00e1s con gente que gano oros en sus respectivas olimpiadas nacionales durante cuatro o cinco anios consecutivos.\r\n\r\nDicen que una persona debe dejar que sus hechos hablen por \u00e9l y no \u00e9l por sus hechos, en especial si estos aun no han ocurrido!", "Solution_22": "Jaja parece que los de mexico se tienen mucha fe, pero no se acuerdan quien los voltio en atenas :D :D :D !! jaja mentiras, aun asi marko, lo que dice david es verdad, lo que uno halla hecho en olimpiadas no le asegura una futura buena marca en imo...por ejemplo...yo nunca gane la olimpaida nacional colombiana pero si fui el mejor en imo, y el que gano la olimpiada no clasifico al equipo ;) ;) De todas maneras la confianza es buena y me alegra esten tan motivados para las futuras olimpiadas...Dave, enserio creo que este a\u00f1o les vamos a ganar, usted dice que apuesta por usted...en ese caso yo apuesto por mi...jaja solo por seguirle la corriente pues la verdad la imo no lo es todo...hay objetivos mas grandes...aunque tengo que reconocer en la imo se pasa muy rico! :D :D", "Solution_23": "El topic ha sido partido, solamente he movido los posts que estaban mas o menos fuera de lugar en este tema y las he colocado en otro. Sigamos haciendo predicciones sobre las distintas olimpiadas del 2005. :D :P", "Solution_24": "He borrado lo que escribi, para no generar otra discucion en este foro :)", "Solution_25": "Ahhh!!!\r\n\r\nOlvide dejar aqui una indicacion. Toda la discusion sobre ortografia fue puesta en un post aparte. Lo que pasa es que parece que mientras yo hice esto y Pascual le dio submit a su mensaje solo unos segundos despues. Estoy tratando de ver como pasar un solo mensaje a otro post.\r\n\r\nBueno, sigamos aqui solo con olimpiadas.", "Solution_26": "[quote=\"djimenez\"][quote=\"RaMlaF\"] Lo bueno de ke la IMO sea en Mexico este a\u00f1o, es ke van a invitar seguramente a muchos paises de latinoamerica ke usualmente no van a la IMO... \n[/quote]\n\nLo malo es que segun parece, todos los paises centroamericanos que van este anio, no van a ir en los anios posteriores, o al menos eso es lo que he escuchado de varios de los entrenadores. El asunto es que, como en todo lado, para estos eventos escatiman la plata, y pues, ni modo, van este anio por ser en Mexico y quedar bien cerca, pero el otro, no les dan plata ni para los pasajes. Que triste la historia del tercer mundo :( \n\nBueno, nos vemos,[/quote]\r\n\r\nHey!!!!! Eso no suena bien, de hecho, en teor\u00eda, una vez que hemos sido invitados, podemos seguir participando en la IMO :D ademas, yo opino que si ha habido dinero (que no s\u00e9 de d\u00f3nde lo sacaron) para mandar delegaci\u00f3n a Argentina en el 2003, tambi\u00e9n les alcanzar\u00e1 para mandar a Eslovenia el otro a\u00f1o, pues creo Argentina esta igual de lejos que Eslovenia de aqu\u00ed jajajaja :)", "Solution_27": "Pues mi amigo, te recomiendo que revises tu mapa, por que estas bastante equivocado, puesto que El Salvador y Argentina estan a algo asi como 5500 kilometros de distancia, mientras que Eslovenia esta a mas de 10000. Y por otro lado, a Argentina se vol\u00f3 en Setiembre, temporada baja, y por el continente, que es muchisimo mas barato. Haciendo una busqueda por algunas agencias de viajes por Internet, mandar a un grupo de 6 personas (4 participantes + 1 tutor + 1 delegado) de San Salvador a Buenos Aires, costaria, en setiembre, al rededor de $950 por cabeza, osea, al rededor de $5700 por todos. Ahora, mandar a 8 personas a Eslovenia, pues hay que hacer un vuelo interatlantico, que por lo general no son baratos, ademas, en temporada alta. Haciendo la misma cuenta, pues el viaje de San Salvador, El Salvador a Ljubljana, Slovenia saldria al rededor de $2850 por cabeza, y dado que el equipo puede ser de 6, mas el tutor y el delegado, pues estamos hablando de ocho tiquetes, pues es casi $23000 por todos. No mi amigo, no es lo mismo mandar a un equipo a la Ibero en Argentina, que a la IMO en Slovenia.", "Solution_28": ":( :( :( qu\u00e9 triste...................... bueno, con un poco de suerte quiz\u00e1 podamos participar........... en fin, prefiero dejar que el destino siga su curso....................\r\n\r\nPor cierto, mi nombre es Gabriel, y como puedes ver (c\u00f3mo no, no se mira la banderita :mad:) soy de El Salvador (si te preguntas c\u00f3mo es que llegu\u00e9 hasta aqu\u00ed, es una historia muy larga y triste, jajajajajajaja :) )", "Solution_29": "Gabriel lo unico que necesitamos los optimistas son ganas, nada de cuentas! asi que espero verlos en emxico y en las olimpiadas venideras", "Solution_30": "Pues si, lo que hace falta es solamente ganas de hacer mate. Lastimosamente muchas veces los paises no participan en este tipo de actividades por que el boleto es muy caro, aun cuando todo lo demas es gratis, pero pueden mandar a un equipo completo de Futbol al otro lado del mundo en primera clase y pagarles los mejores hoteles... a veces eso me da rabia!! :mad: \r\n\r\nPor cierto, bienvenido Gabriel. Si queres hacer una presentacion algo mas extensa, pues podes ir a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30603]este thread[/url] y contarnos alli todo lo que queras de vos.\r\n\r\nPor cierto, lo de la banderita lo podemos arreglar... Tenes alguna imagen de la bandera de el Salvador en las dimensiones apropiadas (usualmente son 32x20 pixeles, pero mientras tenga ese radio de 8/5 entre el largo y el ancho, se puede cambiar las dimensiones). Si la tenes, mandamela en un PM y trato de que la cambien!!\r\n\r\nCiao,", "Solution_31": "Bueno, todos sabemos que la inversion que se tiene que hacer para mandar un equipo a una olimpiada por lo general no es barata. Puede pasarle como a Venezuela, que comenzo participando en la IMO de 1981 (un momento de auge economico aca), pero mandaban en general gente que no estaba bien preparada y por lo general salian mal. Para el 83 (un golpe muy fuerte en la economia venezolana, comienzo de una crisis que se ha sentido hasta ahora :( ), dejamos de participar, y no fue sino hasta 1998 (creo) que volvimos a ir. Desde entonces, han habido a\u00f1os en donde por la falta de dinero se manda solo 1 participante....cosa que es medio triste, pero a la vez, es de cierto modo una manera de mantenernos llendo. Me imagino que eso lo podrian hacer los paises latinoamericanos que van por primera vez a esta IMO, y que les parece que mandar un equipo completo a Slovenia es demasiado caro. Seria interesante ver mas de nuestra gente en olimpiadas del mundo :)", "Solution_32": "Por cierto, todavia no tengo noticias de la seleccion del equipo de mi pais, pero tengo noticias del de Argentina. Por si a alguien le interesa\r\n\r\nEquipo Argentino - 46\u00b0 Olimp\u00edada Matem\u00e1tica Internacional - M\u00e9xico\r\nGabriel Carvajal - Z\u00e1rate - Buenos Aires \r\nGerm\u00e1n Gieczewski - San Isidro - Buenos Aires \r\nMaximiliano Camporino - Cdad. de Buenos Aires \r\nNicol\u00e1s Romero - Posadas -Misiones \r\nIgnacio Rossi - Garin - Buenos Aires \r\nTom\u00e1s Ibarlucia - Tandil - Buenos Aires \r\n\r\nAlguien mas ya sabe quienes van de su pais?", "Solution_33": "Hola a todos, soy Lucas Andisco (IMO 2003, IMO 2004, Ibero 2004), para los que no saben qui\u00e9n es.... Germ\u00e1n Gieczewski=Severius (el apodo se lo puse yo!, jaja, ahora nadie sabe que se llama Germ\u00e1n, todo el mundo le dice Severius, jaja). Gaby y Nacho fueron a la IMO 2004 y Gabriel tambi\u00e9n fue a la Ibero 2004. Maxi y Nacho fueron a la Cono 2004. Tom\u00e1s hab\u00eda quedado suplente para el equipo Imo del a\u00f1o anterior y Nico para el de Cono, creo.\r\nSeguro que alguien ya lo dijo antes, el equipo de Espa\u00f1a se conoce, de los que me acuerdo est\u00e1n Elisa (IMO e Ibero 2004) y Teixido (IMO 2004).\r\nCreo que el de Brasil ya debe estar tambi\u00e9n, pero no estoy seguro. El que pod\u00eda participar es Gaby Boujokas, que le fue muuuuy bien en IMO 2004, no sac\u00f3 oro por el problema 1, y tal vez Fabio Diaz Moreira, que (no estoy seguro) por lo que dijeron andaba con ganas de hacer un a\u00f1o m\u00e1s de olimp\u00edadas.\r\n\u00bfC\u00f3mo le va Pascual? Cuidado usted con proposiciones extra\u00f1as para las damas de esta IMO.... Ahhh, \u00bfvieron que la linda suizita tambi\u00e9n va a la IMO?\r\nPascual y Leo p\u00f3nganse las pilas para traer excelentes medallas en esta IMO... pero para la ibero guarda eh, el equipo de Argentina pinta bien. Bueno, un saludo a todos.", "Solution_34": "Lucas, como sabe que la suizita va??? jaja....", "Solution_35": "\u00bfPor qu\u00e9 casi todos los que van de Argentina son de Buenos Aires?\r\n\u00bfEst\u00e1 muy centralizada (la olimpiada) all\u00e1 o es simplemente que los de Buenos Aires son mejores?.\r\nPor cierto M\u00e9xico hace estas dos semanas sus \u00faltimos 5 ex\u00e1menes que deciden quien va y quien no. Les informo yo antes que nadie", "Solution_36": "Pascual, los que estamos en la 'hermandad del salamitrosky' tenemos 'contactos', jajaja\r\nEn la prueba no se descuide con albanitas y esas cosas, usted ya tiene experiencia amigo, jeje.\r\nCon respecto a lo de Bs As, hay muchos factores, sin embargo el principal, para la olimp\u00edada y las dem\u00e1s cosas, es que la provincia de Bs As es por muuuuuuucho la provincia con m\u00e1s habitantes del pa\u00eds. Despu\u00e9s hay otros motivos, pero \u00e9se es en l\u00edneas grales el m\u00e1s importante.", "Solution_37": "Uy como asi, es que ya montaron hermandad y todo?\r\n\r\nFresco que este a\u00f1o ya me volvi muy juicioso con las muejres :)", "Solution_38": "M\u00e9xico tiene ya a 5/6 partes de su selecci\u00f3n:\r\nPablo Sober\u00f3n de Morelos (6a OMCC)\r\nJoshua Hern\u00e1ndez de Coahuila (5a OMCC)\r\nGuevara Guevara (si, su nombre es igual a su ape\u00eddo) de Zacatecas\r\nIsaac Buenrostro de Jalisco (6a OMCC)\r\nDavid Torres de Guanajuato (yo :lol: ) (6a OMCC)\r\ny el sexto es alguno de:\r\nMario Huicochea (o algo por el estilo) del DF\r\nDaniel Garc\u00eda (45 IMO) de Chihuahua\r\nGonzalo Montealv\u00e1n (45 IMO y 5a OMCC) de Puebla.\r\nPor cierto, los 5 que estamos seguros todos repetimos y respecto a lo de Buenos Aires,\r\nel estado m\u00e1s poblado de M\u00e9xico es el estado de M\u00e9xico (no confundir con DF) y no\r\nes muy bueno que digamos.", "Solution_39": "a toda la comunidad olimpica .... un gran saludo desde tierras mexicanas y aun mas especifico desde merida....... ciudad donde ene stos dias se lleva a cabo la 46 IMO .... es para mi un placer estar con ustedes amantes de la matematica y del razonamiento para resolver problemas ... porque ... \u00bf que otra cosa es las matematicas sino resolver problemas?....... .....\r\nreciban desde aqui felicitaciones por participar o ganar medallas para sus respectivos paises........ felicidades para los mexicanos.... que ni moda tendremos que esperar otro a\u00f1o para ganar la primer medalla de oro en estas competencias.......... a prepararse para eslovenia......... \r\naun asi felicidades por ser los mejores del pais.......... listod para campeche?...... si alguien sabe quien repite aparte del harry potter joshua ( quien estuvo en ixtapa , sabe a que me refiero)........ \r\nseguimos en contacto por este fabuloso foro. y a prepararse para las otras olimpiadas", "Solution_40": "En primer lugar , felicitar a la selecci\u00f3n colombiana que ha mostrando un buen nivel en esta 41a IMO - Mexico, y quisiera que los miembros de esta comunidad, comparta y comente los resultados de los equipso de sus paises... en vista de las proximas ediciones...\r\n\r\n\r\nPascual... podrias decirme cual fue el equipo que represento a colombia este a\u00f1o?..\r\nsegun la codificacion que se maneja en el certamen (COL1, COL2, ETC.)\r\n\r\nBueno .. y hablando de Peru... ha sido la primera vez que se ha enviado un equipo completo en los ultimos 5 a\u00f1os ......\r\nQue pena que ....se nos escaparon dos platas este a\u00f1o.. :| \r\n\r\nSim embargo... hay muchas esperanzas para las proximas ediciones puesto que este equipo es relativamente joven y varios de los chicos tienes para dos o tres a\u00f1os m\u00e1s en estas competiciones... \r\n\r\nSalu2.\r\n\r\n\r\nCarlos Bravo... :) \r\nLima - Peru", "Solution_41": "Bueno al parecer Pascual va ganando las apuestas, s\u00f3lo falta ver como queda la ibero, yo tambi\u00e9n le apuesto a Colombia aunque no se sabe todav\u00eda el equipo. El equipo de Colombia para la imo este a\u00f1o fue:\r\n\r\nCOL1 Jos\u00e9 Gabriel Acevedo (centro 03, imo 04) (Yo!)\r\nCOL2 Daniel Bland\u00f3n (centro 04)\r\nCOL3 Esteban Gonzalez (centro 02, ibero 02, imo 03, ibero 04)\r\nCOL4 Miguel Moreno\r\nCOL5 David Quinche (centro 04)\r\nCOL6 Pascual Restrepo (imo 03, ibero 03, imo 04)", "Solution_42": "felicidades a todos los colombianos y a los demas latinos!!\r\n\r\npor si acaso yo soy manuel de puerto rico imo 2004 pero no fui a esta imo, pero seguramente imo 2006!!!", "Solution_43": "[quote=\"matemario\"]\naun asi felicidades por ser los mejores del pais.......... listod para campeche?...... si alguien sabe quien repite aparte del harry potter joshua ( quien estuvo en ixtapa , sabe a que me refiero)........ \nseguimos en contacto por este fabuloso foro. y a prepararse para las otras olimpiadas[/quote]\r\nPues de hecho es m\u00e1s f\u00e1cil decir qui\u00e9n no repite. Unicamente Washa (H\u00e9ctor) de Chihuahua no repite. Los dem\u00e1s podemos repetir.\r\nPor cierto, parece que Colombia y M\u00e9xico vamos empatados pues M\u00e9xico gan\u00f3 la centro pero Colombia la IMO (aunque claro cuenta m\u00e1s haber ganado la IMO :mad: ) Pero espero que en la Ibero nos desquitemos. Por cierto, felicidades a Pascual por su plata, a los peruanos (se ve\u00edan chiquitos pero todos me ganaron :( ) y a todos los latinos que fueron a la IMO", "Solution_44": "he he Gracias y por supuesto estare en cartagena defendiendo mi territorio! :D" } { "Tag": [ "videos" ], "Problem": "Post Youtube video on this thread, but please, do not put pointless ones, and try to put funny videos, and don't put ones that will probably get this thread locked.", "Solution_1": "[youtube]zV2-zhIDqOA[/youtube]", "Solution_2": "[youtube]FT060JGp9sQ[/youtube]", "Solution_3": "[youtube]7JX-Bx0BETQ[/youtube]", "Solution_4": "This Italian band Rhapsody of Fire's hilariously terrible music video for the song \"Unholy Warcry\"\r\n[i]Playing time is 10:32... watch it all or don't watch it. Also, let it time to load first, pauses kill it[/i]\r\n\r\n[youtube]ZYpxQEPzpJk[/youtube]", "Solution_5": "More Weird Al.\r\n\r\n[youtube]uKrCAZsJsAI[/youtube]", "Solution_6": "[youtube]ILHlrx_oRfI[/youtube]\r\n\r\nThere are more on YouTube...", "Solution_7": "[youtube]y2YR80hTRmw[/youtube]", "Solution_8": "lol, \"wierd\"" } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "hello!\r\nprove that not exist $ (a,b,c)\\in{IN^{3}}$ / $ a^{2} + b^{2} + c^{2}\\equiv7[8]$.", "Solution_1": "any square is $ 0$ or $ 1$ or $ 4$ modulo $ 8$, therefore the sume of three squares is never $ 7$ modulo $ 8$.\r\n\r\nPierre.", "Solution_2": "Squares are equal to $ 0,1,4\\pmod{8}$ so looking at the different cases you can conclude the claim\r\nIn fact it is easy to see than the numbers of the form $ 4k$, when squared are of the form $ 8l$, the number of the form $ 4k \\plus{} 2$ are of the form $ 8l \\plus{} 4$ and finally any odd number squared:\r\n$ (2x \\plus{} 1)^2 \\equal{} 4x^2 \\plus{} 4x \\plus{} 1 \\equal{} 4(x(x \\plus{} 1) \\plus{} 1$ The first summand is clearly of the form $ 8l$ so the odd numbers are of the form $ 8l \\plus{} 1$ when squared\r\n\r\nDaniel\r\n\r\n\r\nEDIT:Pierre was faster :D" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "prove that if the equation \\[ x^3 \\plus{} ax{}^2 \\plus{} bx \\plus{} c \\equal{} 0\r\n\\]\r\nhas three distinet real roots, then so is\r\n\\[ x^3 \\plus{} ax{}^2 \\plus{} ( \\minus{} a^2 \\plus{} 4b)x \\plus{} a^3 \\minus{} 4ab \\plus{} 8c \\equal{} 0\r\n\\]", "Solution_1": "why don't you heelp me?" } { "Tag": [], "Problem": "If 74 hens lay 74 dozen eggs in 74 days and if 37 hens eat 37 kilograms of wheat in 37 days, how many kilograms of wheat are needed to produce 1 dozen eggs?\r\n[hide=\"Answer\"]\n2 kilograms[/hide]", "Solution_1": "[quote=\"DonkeyKong\"]If 74 hens lay 74 dozen eggs in 73 days and if 37 hens eat 37 kilograms of wheat in 37 days, how many kilograms of wheat are needed to produce 1 dozen eggs?\n[hide=\"Answer\"]\n2 kilograms[/hide][/quote]\r\nI'll assume you meant 74 days because it doesn't work out that way.\r\n1 hen eats 1 kg of wheat in 37 days.\r\n1 hen lays 1 dozen eggs in 74 day.\r\n1 hen eats 2 kg of wheat in 74 days.", "Solution_2": "[quote=\"ZhangPeijin\"][quote=\"DonkeyKong\"]If 74 hens lay 74 dozen eggs in 73 days and if 37 hens eat 37 kilograms of wheat in 37 days, how many kilograms of wheat are needed to produce 1 dozen eggs?\n[hide=\"Answer\"]\n2 kilograms[/hide][/quote]\nI'll assume you meant 74 days because it doesn't work out that way.\n1 hen eats 1 kg of wheat in 37 days.\n1 hen lays 1 dozen eggs in 74 day.\n1 hen eats 2 kg of wheat in 74 days.[/quote]\r\nI am sorry for the typo. I couldn't see the simple logical explaination behind this problem. Thanks.", "Solution_3": "Why don't you divide the days as well?", "Solution_4": "by your logic:\n\n37 hens => 37kg in 37 days\n\n1 hen => 1kg in 1 day\n\n1 hen => 37 kg in 37 days\n\n37 hens => 37*37kg in 37 days\n\nlol..." } { "Tag": [ "geometry", "trapezoid" ], "Problem": "Find the area of the isosceles trapezoid in square feet. Express your answer in simplest radical form.\n[asy]draw((0,0)--(18,0)--(15,6.3)--(3,6.3)--cycle);\nlabel(\"7'\",(1.5,3.15),WNW);\nlabel(\"12'\",(9,6.3),N);\nlabel(\"18'\",(9,0),S);[/asy]", "Solution_1": "First, drop a \"convenient altitude\".\r\n\r\n[asy]draw((0,0)--(18,0)--(15,6.3)--(3,6.3)--cycle);\nlabel(\"7'\",(1.5,3.15),WNW);\nlabel(\"12'\",(9,6.3),N);\nlabel(\"12'\",(9,0),S);\ndraw((3,6.3)--(3,0));\ndraw((15,6.3)--(15,0));\nlabel(\"3'\",(1.5,0),S);\nlabel(\"3'\",(16.5,0),S);[/asy]\r\n\r\nThe height is clearly $ 2\\sqrt{10}$ by Pythagorean theorem.\r\n\r\n[asy]draw((0,0)--(18,0)--(15,6.3)--(3,6.3)--cycle);\nlabel(\"7'\",(1.5,3.15),WNW);\nlabel(\"12'\",(9,6.3),N);\nlabel(\"12'\",(9,0),S);\ndraw((3,6.3)--(3,0));\ndraw((15,6.3)--(15,0));\nlabel(\"3'\",(1.5,0),S);\nlabel(\"3'\",(16.5,0),S);\nlabel(\"$2\\sqrt{10}$'\",(3,3.15),E);[/asy]\r\n\r\nThus, the area is $ 30 \\times 2 \\sqrt{10} \\div 2$ or $ \\boxed{30\\sqrt{10}}$.", "Solution_2": "we know that these are isoceles so we know can drop 12 \n\nand the \"remaining side is 6 so we divide that by 2 and get 3 on each side\n\n\nhypotenuse theorem: a^2+b^2=c^2\n\nwe get 2sqrt 10 for the height \n\nsolving for area =(A1+A2)HEIGHT/2\n\nwe get the answer of 30sqrt10\n\n\n\n\n\nP.S. can someone tell me how to write exponents like 5^2=25 \n\nif you can thank you so much!!" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all primes p for which (2^(p-1)-1)/p is a square", "Solution_1": "That has been solved on http://www.mathlinks.ro/Forum/viewtopic.php?t=1055 .\r\n\r\n darij", "Solution_2": "As I know it's for Iran Olympiad . :huh: ;)" } { "Tag": [], "Problem": "Let x,y,z is positive integer number, Find all x,y,z\r\n\\[ xyz \\equal{} 4(x \\plus{} y \\plus{} z)\r\n\\]", "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=232404]See this thread.[/url]" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "I'm given 2 matrices, 4*4 and 4*2. If I'm asked to find whether the span of A equals the span of B, could I just create one 4*6 matrix and solve for linear independence?\r\n\r\nSorry if this is vague. If needed, I'll put an example into mathtype and post it.", "Solution_1": "What exactly do you mean by the span of $ A?$ That's the part that is unclear about your question.\r\n\r\nOne possible meaning: consider $ A$ as the operator $ x\\mapsto Ax,$ and ask about its range. This is the same as the column space of $ A$ or span of the columns of $ A.$ It's fairly likely that this is what you're talking about.\r\n\r\nNow, what would you do with that one $ 4\\times 6$ matrix? Its column space is the sum (as subspaces) of the column spaces of $ A$ and $ B$. If those two are equal, then the column space of the combined matrix is exactly the same. \r\n\r\nThis will work. We do three row reductions, just for the purpose of counting pivots (that is, determining rank). If $ \\text{rank}(A)\\equal{}\\text{rank}(B)\\equal{}\\text{rank}[A\\mid B],$ then $ \\text{col}(A)\\equal{}\\text{col}(B).$", "Solution_2": "Thanks for the quick response.\r\n\r\nI was given 2 matrices, A and B, and asked if Span[A] = Span[B]. I think I see what you mean, even though a lot of this stuff is still confusing to me.[/code]" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Tha points $ A_1$,$ B_1$,$ C_1$ are chosen on the sides $ AB$,$ BC$,$ CA$ of the triangle $ ABC$ s.t the lines $ AA_1$,$ BB_1$ and $ CC_1$ are concurrent at the point $ D$.The lines $ A_1C_1$ and $ BB_1$ are intersecting at the point $ E$.Prove that if $ BD\\equal{}2B_1D$ then $ BE\\equal{}B_1E$.", "Solution_1": "the problem is a particolar case of this:\r\n\r\nLemma: Let a triangle ABC, a line $ r$ and a point $ P \\in r$, call $ E: BP \\cap AC$, $ F: CP \\cap AB$ and $ D: AP \\cap EF$. So the locus of the point D when P move on the line $ r$ is a line.", "Solution_2": "Sorry, but how to prove your lemma? :wallbash_red:", "Solution_3": "Call $ Q: PA \\cap BC$, so you can use that D is the harmonic cojugate of Q w.r.t. AP :wink:", "Solution_4": "Are there any other solutions,please help! :(" } { "Tag": [ "MATHCOUNTS", "number theory", "\\/closed" ], "Problem": "It's possible that you already realised or are on fixing it, but:\r\n\r\nFarenhajt today told me that he wasn't allowed to edit one of his posts.\r\nI first checked in Number Theory on some posts and had no problem. By that I thought it's a temporary bug only hitting him.\r\nBut I tried in the test forum now, and I've the same problem (I don't want to try through all forums ;) ).\r\n\r\nIn fact, I think it affects everyone not being a moderator of that special forum.", "Solution_1": "This is what I got: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=115798\r\n\r\nI was just about to post it now. \r\n\r\nBig picture\r\n[img]\r\nhttp://i96.photobucket.com/albums/l167/adasarathy/edit.jpg[/img]\r\n\r\nZoom of problem:\r\n\r\n[img]http://i96.photobucket.com/albums/l167/adasarathy/editbig.jpg[/img]\r\n\r\nThis is a problem.", "Solution_2": "I can't edit the above post, so I will post again.\r\n\r\nERROR\r\n[img]http://i96.photobucket.com/albums/l167/adasarathy/edit.jpg[/img]\r\n\r\nI can't edit anything now!!! :o", "Solution_3": "I get the selfsame problem. Wonder if it was somebody hacking?", "Solution_4": "[quote=\"ZetaX\"]It's possible that you already realised or are on fixing it, but:\n\nFarenhajt today told me that he wasn't allowed to edit one of his posts.\nI first checked in Number Theory on some posts and had no problem. By that I thought it's a temporary bug only hitting him.\nBut I tried in the test forum now, and I've the same problem (I don't want to try through all forums ;) ).\n\nIn fact, I think it affects everyone not being a moderator of that special forum.[/quote]\r\nWow, the same thing is happening to me (anywhere except MathCounts, where I am a moderator). I can't edit any of my posts at all! :o \r\n\r\nBut this is likely (and hopefully) just a temporary setting the admins have turned on for some reason (either that or a really weird bug :wink: ).", "Solution_5": "Same problem.", "Solution_6": "No bug guys. Editing is now limited to $x$ hours ([i]the amount is open for debate[/i] - I think 24 hours is enough). After that only moderators can edit the posts in the specific forums (ie you have to be a moderator of the forum where the post lies so that you can edit it).\r\n\r\nThis is to stop people from going back and deleting their posts using edit method (people have done and are doing that for different reasons).\r\n\r\n[POST EDIT :D] Okay there's a bug in the code :) There should be 24 hours available for free editing (after that you can ask the moderator of the forum if it's a really important edit, or post a new post :) ).", "Solution_7": "Should be fixed now: time for editing 24 hours.", "Solution_8": "It doesn't work. My latest post (excluding this one) is 4 hours old and I still can't access it.", "Solution_9": "Same here, even for newly posted posts (but I can delete posts everywhere).", "Solution_10": "EXACTLY THE SAME HERE! :( \r\n\r\nWon't let me edit, says I can't edit my posts anymore. oh well", "Solution_11": "Hm, I can't edit that post above either.", "Solution_12": "Now it's finally fixed :) Small undefined variable, computed the ellapsed time starting 1970 :D Obviously all posts were more than 1 day old :P", "Solution_13": "thanks a lot lol :wink:", "Solution_14": "[quote=\"Valentin Vornicu\"]No bug guys. Editing is now limited to $x$ hours ([i]the amount is open for debate[/i] - I think 24 hours is enough). After that only moderators can edit the posts in the specific forums (ie you have to be a moderator of the forum where the post lies so that you can edit it).\n\nThis is to stop people from going back and deleting their posts using edit method (people have done and are doing that for different reasons).\n\n[POST EDIT :D] Okay there's a bug in the code :) There should be 24 hours available for free editing (after that you can ask the moderator of the forum if it's a really important edit, or post a new post :) ).[/quote]\r\n\r\nBut why do we need this?", "Solution_15": "I said why we need it right in the middle of the post. Read it carefully ;)", "Solution_16": "It seems to be working properly now. I can edit a post from 16 hours ago, but not one from 30 hours ago.\r\n\r\nI agree with 24 hours as the time limit. After that, a new post is usually a better idea.", "Solution_17": "[quote=\"Valentin Vornicu\"]I said why we need it right in the middle of the post. Read it carefully ;)[/quote]\r\n\r\nYes, I just did. :( So, if we were going to keep a scoreboard in a competition, we can't edit it though. :(\r\n\r\nEDIT: Okay. I'll just post a new list. :)", "Solution_18": "[quote=\"anirudh\"][quote=\"Valentin Vornicu\"]I said why we need it right in the middle of the post. Read it carefully ;)[/quote]\n\nYes, I just did. :( So, if we were going to keep a scoreboard in a competition, we can't edit it though. :([/quote]\r\nok, now read the last part of that post" } { "Tag": [ "LaTeX" ], "Problem": "I'm in a high school honors algebra class & have a question about 2 problems I am stuck on! This stuff's really really hard. Any help would be appreciated! \r\n\r\nz-7/z-8 = 1/z-8. Solve for z.\r\n\r\nAlso, \r\n\r\n5x+3/x-4 + 3x/4-x.", "Solution_1": "$ \\text{\\LaTeX}$ would make this a lot easier to understand. Is the first problem\r\n$ \\frac{z-7}{z-8}=\\frac{1}{z-8}$ or $ z-\\frac{7}{z-8}=\\frac{1}{z-8}$ or something else entirely?\r\n\r\nI don't get what you're asking in the second problem. That's not an equation.", "Solution_2": "It's the first setup you mentioned. I am so lost on that one. Any help would be appreciated.", "Solution_3": "hello, multipying for $ z \\ne 8$ both sides of the equation by $ z\\minus{}8$ we get\r\n$ z\\minus{}7\\equal{}1$ and adding $ 7$ on both sides we get the solution $ z\\equal{}8$ which is impossible, so your equation has no solutions.\r\nSonnhard.", "Solution_4": "Thank you!", "Solution_5": "hello, is your second term $ \\frac{3\\plus{}5x}{x\\minus{}4}\\plus{}\\frac{3x}{4\\minus{}x}$?\r\nSonnhard.", "Solution_6": "[quote=\"gal220\"] \n\nz-7/z-8 = 1/z-8. Solve for z.\n\n[/quote]\r\n\r\ngal220,\r\n\r\nas long as you are typing horizontally, place I) parentheses**\r\naround the numerators that are more than one monomial,\r\nand place II) parentheses around the denominators that are \r\na) more than a monomial, b) are the product of at least\r\ntwo constants, c) product of at least two variables, and/or \r\nd) are the product of at least a variable and a constant##\r\nbecause of the Order of Operations.\r\n\r\nYou could have had this for a proper equation:\r\n\r\n(z - 7)/(z - 8) = 1/(z - 8)\r\n \r\n** or other grouping symbols\r\n\r\n##\r\n(Specific example) . . . classification(s) from above\r\n\r\nEx. 1/(z - 8) . . . IIa\r\n\r\nEx. (x - 2)/(5e) . . . I, IIb\r\n\r\nEx. 7/(3*4) . . . IIb\r\n\r\nEx. 10y/(abc) . . . IIc\r\n\r\nEx. (ab + 1)(x^4 + 4x)/(-3w) . . . I, IId", "Solution_7": "I think that \"+\" in the second one is supposed to be an \"=\" (the one that's surrounded by spaces).\r\n\r\nAssuming that's true, $ 5x\\plus{}\\frac{3}{x}\\minus{}4\\equal{}\\frac{3x}{4}\\minus{}x\\implies 20x^2\\plus{}12\\minus{}16x\\equal{}3x^2\\minus{}4x^2$.\r\n\r\n$ 21x^2\\minus{}16x\\plus{}12\\equal{}0$ has no real solutions.", "Solution_8": "I think he was trying to say (5x+3)/(x-4)=(3x)/(4-x)", "Solution_9": "Oh yeah, maybe. In that case the only answer is $ \\minus{}\\frac{3}{8}$.", "Solution_10": "hello, why don't you use $ \\text{\\LaTeX}$? it looks much better\r\nFind all real solutions of the equation $ \\frac{5x+3}{x-4}=\\frac{3x}{4-x}$.\r\nSonnhard." } { "Tag": [ "limit", "trigonometry", "calculus", "calculus computations" ], "Problem": "Calculate :\r\n\r\n1.$ \\lim_{x\\rightarrow 0}\\frac{1\\minus{}\\cos^5x\\cdot \\cos^3 2x\\cdot \\cos^33x}{x^2}$\r\n\r\n2.$ \\lim_{n\\rightarrow \\infty} n(\\sqrt[3]{n^3\\plus{}3n^2\\plus{}2n\\plus{}1}\\plus{}\\sqrt{n^2\\minus{}2n\\plus{}3}\\minus{}2n)$\r\n\r\n3.$ \\lim_{x\\rightarrow 0}\\frac{1\\minus{}(1\\minus{}\\sin \\ x)(1\\minus{}\\sin \\ 2x)^2...(1\\minus{}\\sin \\ nx)^n}{x}\\ , \\ n\\in \\mathbb{N}^*$", "Solution_1": "1.\r\n\t\r\n$ \\cos ^{5}x\\cos ^{3}2x\\cos ^{3}3x$\r\n\t\r\n$ \\equal{}\\left( 1\\minus{}\\frac{x^{2}}{2}\\plus{}...\\right) ^{5}\\left( 1\\minus{}2x^{2}\\plus{}...\\right) ^{3}\\left( 1\\minus{}\\frac{9x^{2}}{2}\\plus{}...\\right) ^{3}$\r\n\t\r\n$ \\equal{}\\left( 1\\minus{}\\frac{5x^{2}}{2}\\plus{}...\\right) \\left( 1\\minus{}6x^{2}\\plus{}...\\right) \\left( 1\\minus{}\\frac{27x^{2}}{2}\\plus{}...\\right)$\r\n\t\r\n$ \\equal{}1\\minus{}22x^{2}\\plus{}...$\r\n\t\r\n$ \\lim_{x\\rightarrow 0}\\frac{1\\minus{}\\cos ^{5}x\\cos ^{3}2x\\cos ^{3}3x}{x^{2}}\\equal{}\\lim_{x\\rightarrow 0}\\frac{1\\minus{}(1\\minus{}22x^{2}\\plus{}....)}{x^{2}}\\equal{}22$", "Solution_2": "3.\r\n\t\r\n$ (1\\minus{}\\sin x)(1\\minus{}\\sin 2x)^{2}(1\\minus{}\\sin 3x)^{3}....(1\\minus{}\\sin nx)^{n}$\r\n\t\r\n$ \\equal{}(1\\minus{}x\\plus{}...)(1\\minus{}2x\\plus{}...)^{2}(1\\minus{}3x\\plus{}...)^{3}....(1\\minus{}nx\\plus{}...)^{n}$\r\n\t\r\n$ \\equal{}(1\\minus{}x\\plus{}...)(1\\minus{}2^{2}x\\plus{}...)(1\\minus{}3^{2}x\\plus{}...)....(1\\minus{}n^{2}x\\plus{}...)$\r\n\t\r\n$ \\equal{}1\\minus{}\\left( 1\\plus{}2^{2}\\plus{}3^{2}\\plus{}...\\plus{}n^{2}\\right) x\\plus{}...$\r\n\t\r\n$ \\equal{}1\\minus{}\\frac{n}{6}(2n\\plus{}1)(n\\plus{}1)x\\plus{}...$\r\n\t\r\n$ \\lim_{x\\rightarrow 0}\\frac{1\\minus{}(1\\minus{}\\sin x)(1\\minus{}\\sin 2x)^{2}(1\\minus{}\\sin 3x)^{3}....(1\\minus{}\\sin nx)^{n}}{x}\\equal{}\\frac{n}{6}(2n\\plus{}1)(n\\plus{}1)$", "Solution_3": "2)\r\n\t\r\n$ \\lim_{n\\rightarrow \\infty }n\\left( \\root{3}\\of{n^{3}\\plus{}3n^{2}\\plus{}2n\\plus{}1}\\plus{}\\sqrt{n^{2}\\minus{}2n\\plus{}3}\\minus{}2n\\right)$\r\n\t\r\n$ \\equal{}\\lim_{n\\rightarrow \\infty }n^{2}\\left( \\root{3}\\of{1\\plus{}\\frac{3}{n}\\plus{}\\frac{2}{n^{2}}\\plus{}\\frac{1}{n^{3}}}\\plus{}\\sqrt{1\\minus{}\\frac{2}{n}\\plus{}\\frac{3}{n^{2}}}\\minus{}2\\right)$\r\n\t\r\n$ \\equal{}\\lim_{x\\rightarrow 0\\plus{}}\\frac{1}{x^{2}}\\left( \\root{3}\\of{1\\plus{}3x\\plus{}2x^{2}\\plus{}x^{3}}\\plus{}\\sqrt{1\\minus{}2x\\plus{}3x^{2}}\\minus{}2\\right)$\r\n\t\r\n$ \\equal{}\\lim_{x\\rightarrow 0\\plus{}}\\frac{\\root{3}\\of{1\\plus{}3x\\plus{}2x^{2}\\plus{}x^{3}}\\plus{}\\sqrt{1\\minus{}2x\\plus{}3x^{2}}\\minus{}2}{x^{2}}$\r\n\t\r\n$ \\equal{}\\lim_{x\\rightarrow 0\\plus{}}\\frac{\\frac{2(\\minus{}3\\plus{}3x\\plus{}5x^{2})}{9\\left( 1\\plus{}3x\\plus{}2x^{2}\\plus{}x^{3}\\right) ^{5/3}}\\allowbreak \\plus{}\\frac{2}{\\left( \\left( 1\\minus{}2x\\plus{}3x^{2}\\right) \\right) ^{3/2}}}{2}$\r\n\t\r\n$ \\equal{}\\frac{2}{3}$" } { "Tag": [ "algebra", "polynomial", "quadratics", "algebra unsolved" ], "Problem": "factorize:\r\nx^3-9xy^2-10y^3", "Solution_1": "It's a homogeneous bivariate polynomial, so you can use the change $x=zy$. Luckily, the resulting univariate polynomial in $z$ can be factored with the usual trick to find a rational root ($z=-2$) and then with the quadratic formula." } { "Tag": [ "probability", "probability and stats" ], "Problem": "the probability of A speaking the truth is 3/4 and that of B speaking the truth is 7/10. they both assert that a white ball has been drawn out of a bag containing 6 balls all of different colours. what is the probability of the truth of their statement?\r\n\r\nwhat I did:\r\n\r\nlet W denote the event of drawing a white ball and A the event that both A & B state that a white ball has been drawn. we need to find P(W|A).\r\n\r\n$ P(W|A)$ = $ \\frac {P(WA)}{P(A)}$\r\n\r\n =$ \\frac {P(A|W) P(W)}{P(A|W) P(W) \\plus{} P(A|W^C) P(W^C)}$\r\n\r\n = $ \\frac {\\frac {3}{4}\\frac {7}{10} \\frac {1}{6}}{\\frac {3}{4}\\frac {7}{10} \\frac {1}{6} \\plus{} \\frac {1}{4} \\frac {3}{10} \\frac {5}{6}}$\r\n\r\n = $ \\frac {21}{36}$ = $ \\frac {7}{12}$.\r\n\r\nbut the answer in the book is $ \\frac {35}{36}$\r\n\r\nwhere did I go wrong? :maybe:", "Solution_1": "[quote=\"Attila the Hun\"]where did I go wrong?[/quote] In assuming the answer in your book must be correct ;)", "Solution_2": "[quote=\"JBL\"][quote=\"Attila the Hun\"]where did I go wrong?[/quote] In assuming the answer in your book must be correct ;)[/quote]\r\n\r\njust what I was thinking.", "Solution_3": "I agree that you should not always believe what is written in a book, but in this case you should. \r\nIn fact I claim that both your answer and the answer in the book are correct, it is just that you have \r\ndifferent interpretations of the problem. I also claim that the answer in the book makes more sense, \r\nsince intuitively you should expect that if both A and B come up with the same color, \r\nit is more likely that they speak the truth. So what goes wrong? \r\n\r\nThe problem is that it is not clearly specified what A and B say when they lie, \r\nand there are several possible interpretations of that. \r\n\r\nOne possibility is that the lie is to say \"white\" if the ball is not white and one of the other colors if the ball is white. \r\nThis is what your are assuming in your calculation, but somehow this assumption is very artificial since it \r\nimplies that A and B have agreed on a certain type of lie before the experiment, and the color white is treated \r\ndifferently from all the other colors. \r\n\r\nA more natural assumption would be that if sombedy lies about the drawn color, he chooses one of the other 5 colors \r\nat random. With this assumption you have $ P(A|W^c) \\equal{} \\frac 1 4 \\frac 1 5 \\frac 3 {10} \\frac 1 5 \\frac 5 6$, which gives \r\n$ P(W|A) \\equal{} \\frac{35}{36}$. \r\n\r\nThis kind of problem of interpretation occurs frequently when you are dealing with conditional probabilities; \r\nthe most famous example is probably the monty hall problem, where the different interpretations concern \r\nthe strategy of the game show host.", "Solution_4": "thank you solyaris! I got it now." } { "Tag": [ "integration" ], "Problem": "Sa se determine toate functiile polinomiale $f: R\\rightarrow R$ astfel incat $\\int_{x}^{x+1}f(t)\\: dt=x^{2},x\\in\\: R$", "Solution_1": "Fie $F\\in \\int f(x)dx$. Atunci conditia se scrie $F(x+1)-F(x)=x^{2}$. Pentru $n\\in N$ obtinem $F(n)=F(0)+\\frac{(n-1)n(2n-1)}{6}$. Fie $G(x)=F(0)+\\frac{(x-1)x(2x-1)}{6}$ un polinom. Cum $F(x)=G(x)$ pentru o infinitate de valori (pe multimea numerelor naturale) rezulta ca $F(x)=G(x),\\forall x \\in R$. Deci $f(x)=\\frac{1}{6}(6x^{2}-6x+1)$" } { "Tag": [ "inequalities" ], "Problem": "Let $a,$ $b,$ $c,$ $d$ and $e$ are positive numbers. Prove that\r\n$\\frac{a}{a+2b+2c}+\\frac{b}{b+2c+2d}+\\frac{c}{c+2d+2e}+\\frac{d}{d+2e+2a}+\\frac{e}{e+2a+2b}\\geq1.$", "Solution_1": "[hide=\"Try it.\"] Use Cauchy inequality[/hide]", "Solution_2": "Intersting, I think I've seen this inequality before also:\r\n\\[\\sum_{cyc}\\frac{a}{a+2b+c}\\geq 1 \\]\r\nfor 4 varialbes $a,b,c,d$.\r\n\r\nEdit: ---------------------------------- The following is for arqady's problem :) \r\n\r\nUsing Cauchy we obtain:\r\n\\[LHS \\geq \\frac{(\\sum a)^{2}}{\\sum a^{2}+2\\sum ab}\\]\r\nbut the latter equals one.", "Solution_3": "I don't believe it!!!\r\nI'm so sorry, look what I just found (I hadn't check the inequlities forum in quite a while) http://www.mathlinks.ro/Forum/viewtopic.php?t=134422\r\n\r\nThe ineq. I remembered was from an old Mildorf problems packet.... didn't realize that it was just discussed!!", "Solution_4": "hikaru123, me@home, see here my first post:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=134422\r\nI think, it's harder. :wink:" } { "Tag": [ "probability" ], "Problem": "In probability, do you only directly multiply P's together when each P's sample space is not equal to the other P's sample space?\r\n\r\nI have noticed that when multiplying probabilities together in which sample spaces are the same set of elements, one must multiply the product of the P's by the number of permutations that the events can occur in.\r\n\r\nHowever, when the sample spaces are not exactly the same, multiplication by permutation of P's is not needed. Is this idea correct?", "Solution_1": "Does anyone know or have a correction to this?" } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "rotation", "geometry proposed" ], "Problem": "[b][size=100][color=DarkBlue]A triangle $ \\bigtriangleup ABC$ is given, with circumcircle $ (O)$ and $ \\angle A \\equal{} 60^{o}$ and let be the points $ D\\equiv AC\\cap BO$ and $ E\\equiv AB\\cap CO,$ where $ O$ is the center of $ (O).$ We denote as $ M,$ the midpoint of the side-segment $ BC$ and let be the point $ N\\equiv (O)\\cap MO,$ in to the same side of $ BC,$ as vertex $ A.$ Prove that $ PM \\equal{} PN,$ where $ P$ is the midpoint of the segment $ DE.$[/color][/size][/b]\r\n\r\nKostas Vittas.", "Solution_1": "Obviously $ NBC$ is an equilateral triangle. Let's call $ P \\equiv CN \\cap BO$ and $ Q \\equiv BN \\cap CO$; $ P$ and $ Q$ are midpoints of $ CN$ and $ BN$. As the angles $ \\angle ABN \\equal{} \\angle ACN$, the isosceles triangles $ CDN$ and $ BEN$ are equal. From these triangles we get $ RD \\equal{} QE$; as from the equilateral triangle $ BCN$ we have $ OR \\equal{} OQ$.\r\nIf $ P \\equiv DE \\cap RQ$, applying Ceva to the triangle $ DEO$ with the transversal $ RPQ$, we get $ PE \\equal{} PD$, i.e. $ P$ belongs to $ QR$, the midline of the triangle $ BCN$, hence $ PN \\equal{} PM$.\r\n\r\n[i] Remark: In my figure the points $ P$ and $ R$ are interchanged[/i]\r\n\r\nBest regards,\r\nsunken rock", "Solution_2": "In $ \\bigtriangleup ABC$ with $ \\angle A=60^\\circ$, circumcenter $ O$, let $ D\\equiv AC\\cap BO$ and $ E\\equiv AB\\cap CO$. Let $ M$ and $ P$ be the midpoints of $ BC$ and $ DE$ respectively. Denote circumcircles of triangles $ AED$ and $ ABC$ by $ C_1$ and $ C_2$ respectively\r\n\r\nWe observe that $ AEOD$ is cyclic. The center of the spiral similarity ($ T$) that carries $ ED$ into $ BC$ is found to be the intersection of $ C_1$ and $ C_2$ (different from $ A$, and in case the two circles are tangent, $ A$ is obviously the center of central similarity) . Let $ S%Error. \"neqA\" is a bad command.\n$ be the intersections of these two circles.\r\n\r\n$ S$ is also the center of spiral similarity($ T'$) that carries $ EB$ into $ DC$. But observe that $ EB=DC$ . Hence, $ T'$ is simply a rotation and we get that $ SE=SD$ and $ SB=SC$\r\n\r\nDenote by $ N$ the center of $ C_1$. Since $ SNP$ is carried into $ SOM$ by $ T$, $ \\bigtriangleup SNO$ is similar to $ \\bigtriangleup SPM$ and since $ SN=NO$, $ SP=PM$", "Solution_3": "Here is my solution for this problem \n[b]Solution[/b] \nWe have: $\\widehat{DOE}$ = $\\widehat{BOC}$ = $120^0$ = $180^0$ $-$ $\\widehat{BAC}$ \nSo: $A$, $D$, $O$, $E$ lie on a circle \nThen: $\\widehat{ODE}$ = $\\widehat{OAE}$ = $\\widehat{OBE}$ \nHence: $\\triangle$ $BDE$ is $E$ - isosceles \nSo: $BE$ = $DE$ \nSimilarly: $DE$ = $CD$ \nThen: $BE$ = $CD$ = $DE$ \nHence: ($ADE$) passes through $N$ is midpoint of $\\stackrel\\frown{BAC}$ of ($O$) \nThen: there exist a spiral similar: $S^{k; \\alpha}_N$: $E$ $\\mapsto$ $B$; $F$ $\\mapsto$ $C$ (With $k$ = $\\dfrac{NB}{NE}$; $\\alpha$ = $\\widehat{BNE}$) \nBut: $P$, $M$ are midpoint of $DE$, $BC$ \nSo: $S^{k; \\alpha}_N$: $P$ $\\mapsto$ $M$ \nHence: $\\triangle$ $NBE$ $\\sim$ $\\triangle$ $NMP$ \nWe also have: $\\widehat{EBN}$ + $\\widehat{BNE}$ = $\\widehat{AEN}$ = $\\widehat{AON}$ = 2 . $\\widehat{EBN}$ \nSo: $\\widehat{EBN}$ = $\\widehat{BNE}$ \nThen: $\\triangle$ $BNE$ is $E$ - isosceles \nHence: $\\triangle$ $MNP$ is $P$ - isosceles \nTherefore: $PM$ = $PN$ " } { "Tag": [ "function", "geometry", "rectangle", "algorithm" ], "Problem": "I know only one language fluently: Java. I have seen many java games but i don't even know where to start with making one. can someone please help me? like attach some source code or something?", "Solution_1": "I'm pretty good with Java (currently taking the AP class), and I've written a couple of games in it. For example, I wrote a decent, albeit very simple, version of Tetris from scratch in two days, and created a wheel of fortune program with a partner for last year's final. I still have the code for both of these in case you're interested (although the Tetris has been heavily modified and is now much better).\r\n\r\nIf you give at least a broad idea of what kind of game you were thinking of, I can try to provide an example to serve as a reference. :)", "Solution_2": "It's not hard to code simple games in Java (which I mostly use as well), but often you need other languages like C++, Pascal, or Python to run more complex games. I am currently self-studying C/C++; it's much harder to code a functional C program, but it is DEFINITELY faster.", "Solution_3": "Well there are two main parts to making a game- the thing itself and then representing it. If you've been programming for a little bit you probably already know how to make the game itself. Take a simple game like pong. You would start by creating objects for the paddles and the ball and the playing field. You would run a loop which changes the ball's position over time, and looks for collisions. You would look for input from the keyboard and change the position of the paddle. You might have some AI to control the other paddle, yadda yadda.\r\n\r\nThis would technically be a game, but it wouldn't be very fun. For graphics in Java, you'd need to use applets. There are tons of resources and documentations that will tell you how to create applets and what functions are available to you. You can start messing around creating shapes and seeing what you can do; this part is pretty simple.\r\n\r\nOnce you're comfortable with that you can have your program get the positions of the paddles and the ball, throw some rectangles and a circle up on the screen, have it keep updating and you're good to go.\r\n\r\nthat's a simple example, but it has many elements common to any game. Accepting user input, manipulating variables inside objects, having an AI algorithm, displaying your information as graphics, etc.\r\n\r\nI may be able to show you some old examples if you like. Good luck!" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Find the intergral of ln(cosx)dx.", "Solution_1": "This is a horribly ugly indefinite integral; are there any limits you left out?", "Solution_2": "You know, we do have a calculus forum. A moderator might consider moving this.", "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?highlight=solution&t=120089" } { "Tag": [ "AMC" ], "Problem": "Which year's AMC/AHSME's should I take as practice?\r\n\r\nSince calculators are not allowed, do you think that this year's tests will be more like the previous few years' tests or older AHSME's?", "Solution_1": "All of them. :wink: \r\n\r\nThey all have neat problems.", "Solution_2": "[quote=\"boo\"]Which year's AMC/AHSME's should I take as practice?\n\nSince calculators are not allowed, do you think that this year's tests will be more like the previous few years' tests or older AHSME's?[/quote]\r\nIf you've noticed, very few problems on AMC10/12 need a calculator, and for very few of them does a calculator actually speed it up significantly enough to help...", "Solution_3": "I don't recommend taking any test more than couple, namely from recent years (2007-2005).\r\n\r\nInstead, I say definitely do the ones that require you to add/subtract because that's where most stupid errors are going to occur. You have a week left so don't expect to suddenly have an epiphany and learn everything. That's just not possible. Relax and just sharpen up your add/subtract skills so you can do that correctly and if you can, that saves time as well.", "Solution_4": "That was my problem last year, lol. I was like i'll just learn everything the day before >_>" } { "Tag": [ "calculus", "integration", "trigonometry", "geometry", "geometric transformation", "rotation", "calculus computations" ], "Problem": "Let $ C$ be the parameterized curve for a given positive number $ r$ and $ 0\\leq t\\leq \\pi$, \r\n\r\n$ C: \\left\\{\\begin{array}{ll} x \\equal{} 2r(t \\minus{} \\sin t\\cos t) & \\quad \\\\\r\ny \\equal{} 2r\\sin ^ 2 t & \\quad \\end{array} \\right.$\r\n\r\nWhen the point $ P$ moves on the curve $ C$, \r\n\r\n(1) Find the magnitude of acceleralation of the point $ P$ at time $ t$.\r\n\r\n(2) Find the length of the locus by which the point $ P$ sweeps for $ 0\\leq t\\leq \\pi$.\r\n\r\n(3) Find the volume of the solid by rotation of the region bounded by the curve $ C$ and the $ x$-axis about the $ x$-axis.\r\n\r\nEdited.", "Solution_1": "[quote=\"kunny\"]Let $ C$ be the parameterized curve for a given positive number $ r$ and $ 0\\leq t\\leq \\pi$, \n\n$ C: \\left\\{\\begin{array}{ll} x = 2r(t - \\sin t\\cos t) & \\quad \\\\\ny = 2r\\sin ^ 2 t & \\quad \\end{array} \\right.$\n\nWhen the point $ P$ moves on the curve $ C$, \n\n(1) Find the acceleralation of the point $ P$ at time $ t$.\n\n(2) Find the length of the locus by which the point $ P$ sweeps for $ 0\\leq t\\leq \\pi$.\n\n(3) Find the volume of the solid by rotation of the region bounded by the curve $ C$ and the $ x$-axis about the $ x$-axis.[/quote]\r\n\r\nsecond problem,note that the length of the locus\r\n $ L=\\int_{0}^{\\pi}\\sqrt{x'^2+y'^2}dt=\\int_{0}^{\\pi}\\sqrt{(2r)^2[1-(\\cos^2 t-\\sin^2 t)]^2+(2r2\\sin t\\cos t)^2}dt\n=2r\\int_{0}^{\\pi}\\sqrt{(1-\\cos^2 t+\\sin^2 t)^2+4\\sin^2 t(1-\\sin^2 t)}dt=2r\\int_{0}^{\\pi}2\\sin tdt=8r$", "Solution_2": "[quote=\"kunny\"]Let $ C$ be the parameterized curve for a given positive number $ r$ and $ 0\\leq t\\leq \\pi$, \n\n$ C: \\left\\{\\begin{array}{ll} x = 2r(t - \\sin t\\cos t) & \\quad \\\\\ny = 2r\\sin ^ 2 t & \\quad \\end{array} \\right.$\n\nWhen the point $ P$ moves on the curve $ C$, \n\n(1) Find the acceleralation of the point $ P$ at time $ t$.\n\n(2) Find the length of the locus by which the point $ P$ sweeps for $ 0\\leq t\\leq \\pi$.\n\n(3) Find the volume of the solid by rotation of the region bounded by the curve $ C$ and the $ x$-axis about the $ x$-axis.[/quote]\r\nthe first problem,you need calculate$ \\frac{d^2 y}{d x^2},\n\\frac{dy}{d x}=\\frac{dy/dt}{dx/dt}=\\frac{4r\\sin t\\cos t}{2r[1-(\\cos^2t-\\sin^2t)]}=\\cot t\n\\frac{d^2 y}{d x^2}=\\frac{d\\cot t}{dx}=\\frac{d\\cot t}{dt}\\frac{1}{\\frac{dx}{dt}}\n=(-1-\\cot^2 t)\\frac{1}{4r\\sin^2 t}$", "Solution_3": "[quote=\"kunny\"]Let $ C$ be the parameterized curve for a given positive number $ r$ and $ 0\\leq t\\leq \\pi$, \n\n$ C: \\left\\{\\begin{array}{ll} x \\equal{} 2r(t \\minus{} \\sin t\\cos t) & \\quad \\\\\ny \\equal{} 2r\\sin ^ 2 t & \\quad \\end{array} \\right.$\n\nWhen the point $ P$ moves on the curve $ C$, \n\n(1) Find the acceleralation of the point $ P$ at time $ t$.\n\n(2) Find the length of the locus by which the point $ P$ sweeps for $ 0\\leq t\\leq \\pi$.\n\n(3) Find the volume of the solid by rotation of the region bounded by the curve $ C$ and the $ x$-axis about the $ x$-axis.[/quote]\r\nthe third problem volume $ V\\equal{}\\int_{0}^{\\pi}\\pi y^2 \\frac{d x}{dt}dt\\equal{}\\int_{0}^{\\pi}\\pi(2r\\sin^2 t)^2 4r\\sin^2 t\\equal{}16\\pi r^3\\int_{0}^{\\pi}\\sin^6 t dt\\equal{}5\\pi^2 r^3$", "Solution_4": "Yes, those answers are correct, sorry I have just edited context of (1)." } { "Tag": [], "Problem": "In how many ways can 7 people be seated in a row of chairs if two of the people, Wilma and Paul, refuse to sit next to each other?", "Solution_1": "there are $ 7!$ ways to arrange the chairs. now, lets subtract when the two are sitting next to eachother. making them one person, there are $ 6$ ways to arrange them, and then $ 5!$ to arrange the rest. however, they can be arranged in $ 2$ ways themselves. thus, the answer is $ 7!\\minus{}6\\cdot2\\cdot5!\\equal{}5040\\minus{}12\\cdot120\\equal{}5040\\minus{}1440\\equal{}\\boxed{3600}$." } { "Tag": [ "inequalities", "probability", "induction", "inequalities unsolved" ], "Problem": "Let m,n $\\in N, x \\in$ [0,1]. Show that\r\n\r\n$(1-x^n)^m + (1-(1-x)^m)^n \\geq 1$", "Solution_1": "Let $f(x) = (1-x^n)^m+(1-(1-x)^m)^n$, $\\forall x \\in [0,1]$.\r\nWe have : $f(0) = 2$ and $f(1) = 1$.\r\nTo show that $f(x) \\ge 1$ in $[0,1]$, we just have to show that $f$ decreases on this interval.\r\n$f'(x)= -n \\cdot m \\cdot x^{n-1} \\cdot (1-x^n)^{m-1}+n \\cdot m \\cdot (1-x)^{m-1} \\cdot (1-(1-x)^m)^{n-1} < 0$\r\n$\\iff \\left(\\frac{1-(1-x)^m}x \\right)^{n-1} < \\left(\\frac{1-x^n}{1-x} \\right)^{m-1}$,\r\nor, if we develop :\r\n$(1-x+..+m(-x)^{n-1})^{m-1} < (1+x+..+x^{m-1})^{n-1}$.\r\n\r\n:)", "Solution_2": "[quote=\"mathmanman\"]Let $f(x) = (1-x^n)^m+(1-(1-x)^m)^n$, $\\forall x \\in [0,1]$.\nWe have : $f(0) = 2$ and $f(1) = 1$.\nTo show that $f(x) \\ge 1$ in $[0,1]$, we just have to show that $f$ decreases on this interval.\n$f'(x)= -n \\cdot m \\cdot x^{n-1} \\cdot (1-x^n)^{m-1}+n \\cdot m \\cdot (1-x)^{m-1} \\cdot (1-(1-x)^m)^{n-1} < 0$\n$\\iff \\left(\\frac{1-(1-x)^m}x \\right)^{n-1} < \\left(\\frac{1-x^n}{1-x} \\right)^{m-1}$,\nor, if we develop :\n$(1-x+..+m(-x)^{n-1})^{m-1} < (1+x+..+x^{m-1})^{n-1}$.\n\n:)[/quote]\r\n\r\n.. and why is the last inequaity true?? :?", "Solution_3": "That is an easy one inequality solving in not very famous strategy.\r\n\r\nAnyway it was not proposed in Czech MO... the source you wrote is incorrect.", "Solution_4": "Probabilistic interpretation:\r\n\r\nLet $x$ be the probability of flipping a coin heads, so $1-x$ is the chance of flipping tails. Note $1-x^n$ is the probability that at least one tail will appear so $(1-x^n)^m$ is the probability that after $mn$ flips, at least one tail will appear in each of the intervals flips $1$ to $n$, $n+1$ to $2n$, etc $(m-1)n+1$ to $mn$. Similarly, $(1-(1-x)^m)^n$ can be interpreted as the probability that at least one heads will appear in each of set of flip #s $\\{1,n+1,\\dots (m-1)n+1\\}$, $\\{2,n+2,\\dots,(m-1)n+2\\}$ etc. At least one of these must occur -- for instance, if the first doesn't occur, there exists a $k$ so that flips $kn+1, \\dots , (k+1)n$ are all heads and the second event occurs. If the second event doesn't occur, one of the sets of flips $\\{\\ell,n+\\ell,\\dots (m-1)n+\\ell\\}$ contains all tails... so we're done!!", "Solution_5": "[quote=\"M4RI0\"]Let m,n $ \\in N, x \\in$ [0,1]. Show that\n\n$ (1 \\minus{} x^n)^m \\plus{} (1 \\minus{} (1 \\minus{} x)^m)^n \\geq 1$[/quote]\r\n\r\nif x=1, 0: the inequality is true. Let's assume x is not equal to 0 or 1.\r\ndefine: $ f(x)\\equal{}1\\minus{}(1\\minus{}x)^m$. The inequality is then equivalent to $ f(x)^{n}\\geq f(x^n)$. I will show this by induction:\r\nn=1: trivial\r\nn=2: \r\n$ (1\\minus{}(1\\minus{}x)^m)^2 \\geq 1\\minus{}(1\\minus{}x^2)^m \\equal{}> \\frac{(1\\minus{}x^2)^m\\plus{}((1\\minus{}x)^2)^m}{2} \\geq (1\\minus{}x)^m \\equal{}> \\frac{(1\\plus{}x)^m\\plus{}(1\\minus{}x)^m}{2} \\geq 1$\r\nand this is true by Power Mean (for every $ m \\geq 1$).\r\nLet's assume that it's true for some integer n=k>1. for n=k+1:\r\n$ f(x)^{k\\plus{}1}\\equal{}f(x)f(x)^k \\geq f(x)f(x^{k})$\r\nit is sufficient to show that $ f(x)f(x^k)\\geq f(x^{k\\plus{}1})$, which is equivalent to:\r\n$ (1\\minus{}x^{k\\plus{}1})^m\\plus{}(1\\minus{}x\\minus{}x^k\\plus{}x^{k\\plus{}1})^m \\geq (1\\minus{}x^k)^m\\plus{}(1\\minus{}x)^m$\r\nwhich is true by Karamata\\Majorization.\r\nSo it is true for every $ m \\geq 1$, and natural n.\r\n \r\nThAzN1 - how did you come up with this interpretation? :o\r\n\r\nxirti: you are wrong, because f(0)=1 and so it doesn't decrease in the interval.", "Solution_6": "[color=green]Hnm, the general problem was posted by me. :) [/color]" } { "Tag": [ "search" ], "Problem": "did any of you know that 0.9999999..... (re-occuring 9) equals 1. So far I've seen 2 proofs of this. 1 of them:\r\n\r\n1/3=.3333333\r\nmultiply each side by 3\r\n3/3 or 1 = .999999\r\n\r\n Ive known this for sometime now, have you? :huh:", "Solution_1": "Doesn't work. That requires knowing that 1/3 = .33333333333333333333333.\r\n\r\nA better way is this:\r\n\r\n[hide]x = .999999...\n10x = 9.99999...\nSubtract\n9x = 9.000000000000000...\nx = 1.000000000000.....[/hide]", "Solution_2": "[quote=\"A+MATH\"]Doesn't work. That requires knowing that 1/3 = .33333333333333333333333.\n\nA better way is this:\n\n[hide]x = .999999...\n10x = 9.99999...\nSubtract\n9x = 9.000000000000000...\nx = 1.000000000000.....[/hide][/quote]\r\n\r\nThats the one I learned in my math class. That's pretty cool.", "Solution_3": "Yea i also knew about the substarction method but i didnt bother postign it.", "Solution_4": "[quote=\"A+MATH\"]Doesn't work. That requires knowing that 1/3 = .33333333333333333333333.\n\nA better way is this:\n\n[hide]x = .999999...\n10x = 9.99999...\nSubtract\n9x = 9.000000000000000...\nx = 1.000000000000.....[/hide][/quote]\r\n\r\nits just the numbers repeating thingy for like .432432432=432/999 and like .222222=2/9 so then .99999 must equal 9/9 or just 1", "Solution_5": "Yeah, I learned that 2 years ago in algebra 1.", "Solution_6": "Yeah, I learned that 2 years ago in algebra 1." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let x_i, y_i >0 and $ \\sum^n x_i \\equal{}\\sum^n y_i \\equal{}n$.\r\nProve that\r\n\r\n$ 2n \\sum^n \\frac{1}{x_i\\plus{}y_i} \\ge \\frac{n^2}{2} \\plus{} \\frac{1}{n} \\sum^n \\frac{x_i y_i}{x_i\\plus{}y_i}$", "Solution_1": "[quote=\"4865550150\"]Let x_i, y_i >0 and $ \\sum^n x_i \\equal{} \\sum^n y_i \\equal{} n$.\nProve that\n\n$ 2n \\sum^n \\frac {1}{x_i \\plus{} y_i} \\ge \\frac {n^2}{2} \\plus{} \\frac {1}{n} \\sum^n \\frac {x_i y_i}{x_i \\plus{} y_i}$[/quote]\r\n\r\n\r\n$ 2n\\sum^{n}\\frac{1}{x_i\\plus{}y_i}\\equal{}\\sum^{n}(x_i\\plus{}y_i)*\\sum^{n}\\frac{1}{x_i\\plus{}y_i}\\geq n^{2}$\r\n\r\n$ 4ab\\leq (a\\plus{}b)^{2}$ <--> $ \\frac{ab}{a\\plus{}b}\\leq\\frac{a\\plus{}b}{4}$\r\n\r\n$ \\frac{n^2}{2} \\plus{} \\frac {1}{n} \\sum^n \\frac {x_i y_i}{x_i \\plus{} y_i}\\leq\\frac {n^2}{2} \\plus{} \\frac {1}{n} \\sum^{n}\\frac{x_i\\plus{}y_i}{4}\\equal{}\\frac {n^2}{2} \\plus{} \\frac {1}{n}*\\frac{2n}{4}\\equal{}\\frac{n^{2}\\plus{}1}{2}$\r\n\r\n$ 2n \\sum^n \\frac {1}{x_i \\plus{} y_i}\\geq n^{2}\\geq\\frac{n^{2}\\plus{}1}{2}\\geq\\frac{n^2}{2} \\plus{} \\frac {1}{n} \\sum^n \\frac {x_i y_i}{x_i \\plus{} y_i}$", "Solution_2": "Is this inequality true?\r\n\r\n$ \\sum^n \\frac {2 \\minus{} x_i y_i}{x_i \\plus{} y_i} \\ge \\frac {n}{2}$", "Solution_3": "Yes of course, as the above solution shows.\r\n\r\n$ \\sum^{n}\\frac{2}{x_{i}\\plus{}y_{i}}\\ge n$ and\r\n\r\n$ \\sum^{n}\\frac{x_{i}y_{i}}{x_{i}\\plus{}y_{i}}\\leq\\sum^{n}\\frac{x_{i}\\plus{}y_{i}}{4}\\equal{}\\frac{n}{2}$\r\n\r\nI suppose this is what you meant in the original (i.e. putting $ n$ instead of $ \\frac 1n$ in the RHS) :)", "Solution_4": "[quote=\"spanferkel\"]Yes of course, as the above solution shows.\n\n$ \\sum^{n}\\frac {2}{x_{i} \\plus{} y_{i}}\\ge n$ and\n\n$ \\sum^{n}\\frac {x_{i}y_{i}}{x_{i} \\plus{} y_{i}}\\leq\\sum^{n}\\frac {x_{i} \\plus{} y_{i}}{4} \\equal{} \\frac {n}{2}$\n\nI suppose this is what you meant in the original (i.e. putting $ n$ instead of $ \\frac 1n$ in the RHS) :)[/quote]\r\n\r\nyes, thx. :)" } { "Tag": [], "Problem": "Angie walks and jogs as a part of her normal workout. One day she walked for 20 minutes at 2 mph, jogged for 30 minutes at 4 mph, and cooled down by walking for 10 minutes at 2 mph. How many miles did she travel during the hour workout?", "Solution_1": "Using $ d\\equal{}rt$, we have $ \\frac{1}{3}\\times{2}\\plus{}\\frac{1}{2}\\times{4}\\plus{}\\frac{1}{6}\\times{2}\\equal{}\\frac{2}{3}\\plus{}2\\plus{}\\frac{1}{3}\\equal{}\\boxed{3}$." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "we say that a sequences $ {a}_{0},{a}_{1},{a}_{2}....{a}_{n}$ of real numbers is p-balanced for some positive integer p if\r\n$ {a}_{0}\\plus{}{a}_{p}\\plus{}{a}_{2p}\\plus{}...\\equal{}{a}_{1}\\plus{}{a}_{p\\plus{}1}\\plus{}...\\equal{}.....\\equal{}{a}_{p\\minus{}1}\\plus{}{a}_{2p\\minus{}1}\\plus{}....$\r\nLet the sequences $ {a}_{0},{a}_{1},...{a}_{49}$ be p-balanced for p=3,5,7,11, 13,17.PROVE THAT $ {a}_{0}\\equal{}{a}_{1}\\equal{}{a}_{2}\\equal{}...{a}_{49}\\equal{}0$.", "Solution_1": "The finite sums involved do not contain the same number of terms; is that intentional?" } { "Tag": [], "Problem": "very easy!\r\n1) What three numbers between o and 10 can be multiplied together to make a product that equals their sum?\r\nits so easy\r\nreally easy!\r\nmean it\r\n2) The average of five numbers is 20. The largest number is 18. What is the sum of the other numbers?\r\nlol this one is easy too\r\nreally really easy!", "Solution_1": "1.[hide] eh.. i guessed, since you said it was easy, i guess 1, 2, 3 and guess what? lucky me... hehe[/hide]\n\n2. [hide]5*20-18 = 82... but i have a question, if my answer is correct, then 18 can't be the largest number.... unless i'm totally mis-interrupting the problem... again[/hide]", "Solution_2": "[quote=\"LuCy4EvA\"]1.[hide] eh.. i guessed, since you said it was easy, i guess 1, 2, 3 and guess what? lucky me... hehe[/hide]\n\n2. [hide]5*20-18 = 82... but i have a question, if my answer is correct, then 18 can't be the largest number.... unless i'm totally mis-interrupting the problem... again[/hide][/quote]\r\n\r\nI think ur right too. 18 cant be the largest number.", "Solution_3": "Yo! My sister just make a mistake. The highest number is 28. :D", "Solution_4": "1.) 1x2x3=6 and\r\n1+2+3=6\r\nSo the answer is 1,2,3\r\n\r\n2.) Yeah, the largest number can't be 18. :huh:", "Solution_5": "o ya sorry about the mistake!\r\nas my brother said ealier the \r\nlargest number is 28!\r\nsry bout tat\r\nit was a typin mistake\r\n :blush:" } { "Tag": [ "modular arithmetic", "search", "number theory unsolved", "number theory" ], "Problem": "(IMO-1990) Determine all integers n>1 such that\r\n\r\n$ \\frac{2^n\\plus{}1}{n^2}$ is an integer.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?p=366466#366466", "Solution_2": "what notation is this: $ r\\equal{}ord_q{2}$", "Solution_3": "That would be the smallest integer $ r$ such that $ 2^r\\equiv 1\\mod q$.\r\n\r\nFor any integer $ k$ with $ 2^k\\equiv 1\\mod q$, we must have $ r|k$.", "Solution_4": "I didn't understand why $ 9\\not|n$.\r\nWhy does r must be even?\r\nwhy $ 2^n\\equiv1 \\pmod{q}$ is a contradiction? shouldn't it be a contradiciton if it was like this $ 2^n\\equiv1 \\pmod{n^2}$", "Solution_5": "Why $ 9\\not|n$.\r\n\r\nLemma. If $ 2^k\\plus{}1$ is divisible by $ 3^t$, then $ 3^{t\\minus{}1}$ divides $ k$. I'll show this by induction. The base case $ t\\equal{}1$ is easy; clearly $ 3^{0}\\equal{}1$ divides anything. Now suppose $ t\\ge 2$. Then $ k$ is divisible by $ 3$ (just look at $ 2^k\\plus{}1\\mod 9$). So let $ k\\equal{}3k'$.\r\n\r\n$ 2^{3k'}\\plus{}1\\equal{}(2^{k'}\\plus{}1)(2^{2k'}\\minus{}2^{k'}\\plus{}1)$\r\n\r\nWe can quickly check that $ a^2\\minus{}a\\plus{}1\\equiv 0\\mod 9$ has no solutions, so $ 2^{2k'}\\minus{}2^{k'}\\plus{}1$ is divisible by at most $ 3$. Since $ 3^t|2^{3k'}\\plus{}1$, we know that $ 3^{t\\minus{}1}|2^{k'}\\plus{}1$. Thus $ 3^{t\\minus{}2}|k'$ by the inductive hypothesis, so $ 3^{t\\minus{}1}|k$, so the lemma is proved.\r\n\r\nHey I just realized this also follows from [url=http://www.mathlinks.ro/viewtopic.php?search_id=345802527&t=205928]lifting the exponent[/url].\r\n\r\nSo suppose $ 3^{t}$ is the largest power of $ 3$ which divides $ n$, and assume $ t\\ge 2$. Then if $ \\frac{2^n \\plus{} 1}{n^2}$ is an integer, we must have $ 3^{2t}$ dividing $ 2^n \\plus{} 1$. By the lemma, $ 3^{2t\\minus{}1}$ divides $ n$. But if $ t\\ge 2$ then $ 2t\\minus{}1>t$, and we assumed that $ 3^t$ was the largest power of $ 3$ dividing $ n$, a contradiction.\r\n\r\n\r\n\r\nWhy $ r$ is even.\r\n\r\nSo he says $ 2^{2n}\\equiv 1\\mod q$. Now we know $ r|2n$, but if $ r$ is odd, then we would have $ r|n$. Then we would have $ 2^{n}\\equiv 1\\mod q$.\r\n\r\nBut we know that $ 2^{n}\\equiv \\minus{}1\\mod n^2$, which means $ 2^{n}\\equiv \\minus{}1\\mod q$, since $ q$ is a divisor of $ n^2$. This contradicts $ 2^{n}\\equiv 1$, so we must have $ r$ is even.", "Solution_6": "So you proved that for any n that is a power of three with $ t>\\equal{}2$ there will be no solution. But how do I assure that there will be no solutions for $ n\\equal{}5,7,11...$ ? I mean a number that is not a power of three.", "Solution_7": "[quote=\"danilorj\"]So you proved that for any n that is a power of three with $ t > \\equal{} 2$ there will be no solution. But how do I assure that there will be no solutions for $ n \\equal{} 5,7,11...$ ? I mean a number that is not a power of three.[/quote]\r\n\r\nWell I showed that $ 9$ cannot divide $ n$, and then the remaining cases are handled later in that proof in Fermat's Little Turtle's link...", "Solution_8": "[quote=\"Fermat's Little Turtle\"] http://www.mathlinks.ro/Forum/viewtopic.php?p=366466#366466 [/quote]\r\nHave any of you understood what [url=http://www.mathlinks.ro/viewtopic.php?p=1239228#1239228]#3[/url] in this link mean?" } { "Tag": [ "ratio" ], "Problem": "If $ 1 \\minus{} y$ is used as an approximation to the value of $ \\frac {1}{1 \\plus{} y}$, $ |y| < 1$, the ratio of the error made to the correct value is:\r\n\r\n$ \\textbf{(A)}\\ y \\qquad \\textbf{(B)}\\ y^2 \\qquad \\textbf{(C)}\\ \\frac {1}{1 \\plus{} y} \\qquad \\textbf{(D)}\\ \\frac {y}{1 \\plus{} y} \\qquad \\textbf{(E)}\\ \\frac {y^2}{1 \\plus{} y}\\qquad$", "Solution_1": "[hide=\"Solution\"]\n$ \\frac{1}{1 + y} = \\frac{1 - y}{1 - y^2}$\n\nNow we need to find the ratio of the error make to the correct value.\n\n$ \\frac{1 - y}{\\frac{1 - y}{1 - y^2}} = 1 - y^2$\n\nThe value I got isn't one of the choices so I'll go with $ \\boxed{(B) y^2}$.\n[/hide]\r\n\r\nCan anyone tell me what I did wrong or what I still need to do?", "Solution_2": "[quote=TachyonPulse][hide=\"Solution\"]\n$ \\frac{1}{1 + y} = \\frac{1 - y}{1 - y^2}$\n\nNow we need to find the ratio of the error make to the correct value.\n\n$ \\frac{1 - y}{\\frac{1 - y}{1 - y^2}} = 1 - y^2$\n\nThe value I got isn't one of the choices so I'll go with $ \\boxed{(B) y^2}$.\n[/hide]\n\nCan anyone tell me what I did wrong or what I still need to do?[/quote]\n\n$$\\frac{\\frac{1}{1+y}-{1+y}}{\\frac{1}{1+y}}$$ is the ratio of the error made to the correct value. But otherwise, your answer is correct. :D" } { "Tag": [ "USAMTS" ], "Problem": "Any idea when the Round 1 Solutions will be up?", "Solution_1": "They are up. Go to the transcripts of old Math Jams, and one of the very recent ones is \"USAMTS Round 1\". Each problem is explained.", "Solution_2": "I know, but I mean the official solutions on the USAMTS site.", "Solution_3": "I hope to have them up early next week." } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "Show that\r\n\r\n$\\int^{1}_{0}(1-t^{2})^{n}\\,dt = \\frac{(2^{2}n)(n!)^{2}}{(2n+1)!}$\r\n\r\nSo far, all I know is that\r\n\r\n$\\int(1-t^{2})^{n}\\,dt = \\frac{(n+1)(1-t^{2})^{n+1}}{(-2t)!}$ \r\n\r\nCan somebody help me get $\\frac{(n+1)(1-t^{2})^{n+1}}{(-2t)!}$ to $\\frac{(2^{2}n)(n!)^{2}}{(2n+1)!}$", "Solution_1": "[quote=\"problem_solver\"]Can somebody help me get $\\frac{(n+1)(1-t^{2})^{n+1}}{(-2t)!}$ to $\\frac{(2^{2}n)(n!)^{2}}{(2n+1)!}$[/quote]\r\nNo, we can't help you do that - because the first expression makes no sense. Whatever you think you've done already can't possibly be correct.\r\n\r\nI can think of several ways to do this. Here's one:\r\n\r\nUse that this is an even function to write the integral as\r\n\r\n$\\frac12\\int_{-1}^{1}(1-t^{2})^{n}\\,dt= \\frac12\\int_{-1}^{1}(1+t)^{n}(1-t)^{n}\\,dt$\r\n\r\nNow integrate that by parts $n$ times.", "Solution_2": "Thank you Kent Merryfield, that was very helpfull :) \r\n\r\nI actually wrote this out and it took a lot of paper...and it eventually worked..\r\n\r\nI see this sub forum has many interesting integrals for me to solve... :idea:" } { "Tag": [ "analytic geometry", "modular arithmetic", "geometry", "rectangle", "AMC", "counting" ], "Problem": "The set $ G$ is defined by the points $ (x,y)$ with integer coordinates, $ 3\\le|x|\\le7$, $ 3\\le|y|\\le7$. How many squares of side at least $ 6$ have their four vertices in $ G$?\r\n[asy]defaultpen(black+0.75bp+fontsize(8pt));\nsize(5cm);\npath p = scale(.15)*unitcircle;\ndraw((-8,0)--(8.5,0),Arrow(HookHead,1mm));\ndraw((0,-8)--(0,8.5),Arrow(HookHead,1mm));\nint i,j;\nfor (i=-7;i<8;++i) {\nfor (j=-7;j<8;++j) {\nif (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp);\ndraw((-3,-.2)--(-3,.2),black+0.5bp);\ndraw((3,-.2)--(3,.2),black+0.5bp);\ndraw((7,-.2)--(7,.2),black+0.5bp);\ndraw((-.2,-7)--(.2,-7),black+0.5bp);\ndraw((-.2,-3)--(.2,-3),black+0.5bp);\ndraw((-.2,3)--(.2,3),black+0.5bp);\ndraw((-.2,7)--(.2,7),black+0.5bp);\nlabel(\"$-7$\",(-7,0),S);\nlabel(\"$-3$\",(-3,0),S);\nlabel(\"$3$\",(3,0),S);\nlabel(\"$7$\",(7,0),S);\nlabel(\"$-7$\",(0,-7),W);\nlabel(\"$-3$\",(0,-3),W);\nlabel(\"$3$\",(0,3),W);\nlabel(\"$7$\",(0,7),W);[/asy]$ \\textbf{(A)}\\ 125\\qquad \\textbf{(B)}\\ 150\\qquad \\textbf{(C)}\\ 175\\qquad \\textbf{(D)}\\ 200\\qquad \\textbf{(E)}\\ 225$", "Solution_1": "I went ahead and added the diagram DPatrick made.", "Solution_2": "Cool. I edited it a little because I'm crazy and want it to look as much like the diagram on the page.", "Solution_3": "According to someone that I know, the answer can be obtained by computing $ (1^2 \\plus{} 2^2 \\plus{} 3^2 \\plus{} 4^2 \\plus{} 5^2 \\plus{} 4^2 \\plus{} 3^2 \\plus{} 2^2 \\plus{} 1^2) \\plus{} (1^2 \\plus{} 2^2 \\plus{} 3^2 \\plus{} 4^2 \\plus{} 5^2 \\plus{} 6^2 \\plus{} 7^2)$. I understand how the stuff in the first parentheses is gotten but don't understand the second part...\r\n\r\nAnother person I know used an inccorect method that should have gotten him answer of 325 but didn't add correctly and got 225 which is the correct answer!!! :o", "Solution_4": "[quote=\"tinytim\"]According to someone that I know, the answer can be obtained by computing $ (1^2 \\plus{} 2^2 \\plus{} 3^2 \\plus{} 4^2 \\plus{} 5^2 \\plus{} 4^2 \\plus{} 3^2 \\plus{} 2^2 \\plus{} 1^2) \\plus{} (1^2 \\plus{} 2^2 \\plus{} 3^2 \\plus{} 4^2 \\plus{} 5^2 \\plus{} 6^2 \\plus{} 7^2)$. I understand how the stuff in the first parentheses is gotten but don't understand the second part...\n\nAnother person I know used an inccorect method that should have gotten him answer of 325 but didn't add correctly and got 225 which is the correct answer!!! :o[/quote]\r\n\r\nI didn't solve this question correctly, but I'm going to guess that the second part comes from squares that do not have integer lengths. I'm probably wrong, though.\r\n\r\nA person I know brute-forced the answer and got it right. o.O", "Solution_5": "[quote=\"Math Geek\"][quote=\"tinytim\"]According to someone that I know, the answer can be obtained by computing $ (1^2 \\plus{} 2^2 \\plus{} 3^2 \\plus{} 4^2 \\plus{} 5^2 \\plus{} 4^2 \\plus{} 3^2 \\plus{} 2^2 \\plus{} 1^2) \\plus{} (1^2 \\plus{} 2^2 \\plus{} 3^2 \\plus{} 4^2 \\plus{} 5^2 \\plus{} 6^2 \\plus{} 7^2)$. I understand how the stuff in the first parentheses is gotten but don't understand the second part...\n\nAnother person I know used an inccorect method that should have gotten him answer of 325 but didn't add correctly and got 225 which is the correct answer!!! :o[/quote]\n\nI didn't solve this question correctly, but I'm going to guess that the second part comes from squares that do not have integer lengths. I'm probably wrong, though.\n\nA person I know brute-forced the answer and got it right. o.O[/quote]\r\n\r\nYour guess is correct. :) The solution is misrepresented there, though.", "Solution_6": "Can somebody post the official solution?", "Solution_7": "This has to be the fastest solution:\r\n\r\n[hide=\"Beastly Solution\"]Clearly each square must have one vertex in each quadrant. Let $ A, B, C, D$ be the vertices in quadrants 1, 2, 3, 4, respectively. Let $ B', C', D'$ be the points in quadrant 1 corresponding to $ B, C, D$, respectively. It should be intuitively clear that $ AB'C'D'$ is also a square (we can prove this using transformations or vectors). Now if $ A, B', C', D'$ are all the same point, this point corresponds to one square, which gives us 25 squares. If $ A, B', C', D'$ are not all the same point, then they form a small square. This small square corresponds to eight squares, since we can choose a vertex of the square to be A and then an order to label the other vertices of the small square, clockwise or counterclockwise. Therefore, the number of squares is equivalent to $ 25\\equiv 1\\pmod{8}$. The only answer choice that satisfies this is E, so we can bubble E and move on. Or turn in the test. :) [/hide]", "Solution_8": "For some reason I get 220. \r\nThe method I use is to add up all the squares from one to five, , reasoning that this corresponds to side lengths from 6-10. Then I multiply by 4 to fall short 5 from the answer. Could someone tell me if this is just serendipitous or how my method should count the next 5 squares?", "Solution_9": "[quote=\"matt276eagles\"] This small square corresponds to eight squares [/hide][/quote]\r\n\r\nit's 4, not 8", "Solution_10": "[hide=\"Solution\"]Note that the set of all congruent squares in the same orientation form a rectangle in the first quadrant with their first point. Every different rectangle forms a different set of congruent squares so we have\n\\[1\\cdot1 + 1\\cdot2 + \\cdots + 1\\cdot5 + 2\\cdot1 + \\cdots + 5\\cdot 5 = (1+2+3+4+5)^2 = \\boxed{225}.\\][/hide]", "Solution_11": "[quote=thkim1011][hide=\"Solution\"]Note that the set of all congruent squares in the same orientation form a rectangle in the first quadrant with their first point. Every different rectangle forms a different set of congruent squares so we have\n\\[1\\cdot1 + 1\\cdot2 + \\cdots + 1\\cdot5 + 2\\cdot1 + \\cdots + 5\\cdot 5 = (1+2+3+4+5)^2 = \\boxed{225}.\\][/hide][/quote]\nThis seems cool but I don't entirely get it. Can somebody elaborate what they mean by and how they got the final value\n", "Solution_12": "Math1331Math, please draw all possible squares of side length $10$. Now look at the first quadrant, take a red marker, and mark each vertex red. You will end up with a rectangle. In other words, if we can draw in a square in this grid, see how much we can shift it horizontally ($x$) and vertically ($y$), this tells us there are $xy$ congruent squares. We can uniquely define a square by the pair $(x, y)$.", "Solution_13": "Hey, I made a graph on Desmos that could help! The solution 2 on AoPS (site: https://www.artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_25 ) may seem confusing, and I didn't understand the part that said all squares in $G_1$ could be mapped to four squares in $G$. Here is a graph for everyone that needs help and doesn't understand that part either:\n\n[hide=Click to reveal hidden link... if you dare!]https://www.desmos.com/calculator/tmyuq5zxru[/hide]\n\n(Make sure that you show the black dots. This is simple to do:\n\nOn the graph that has $x_2$ and $y_2$:\nclick anywhere on the $y_2$ column except for the circle\n..........|\n..........|\n.........\\/\n$x_2$| O $y_2$\n$3$| $6$\n$6$| $7$\n$7$| $4$\netc.| etc.\n\nThat will make the black dots visible.)", "Solution_14": "[quote=Math1331Math][quote=thkim1011][hide=\"Solution\"]Note that the set of all congruent squares in the same orientation form a rectangle in the first quadrant with their first point. Every different rectangle forms a different set of congruent squares so we have\n\\[1\\cdot1 + 1\\cdot2 + \\cdots + 1\\cdot5 + 2\\cdot1 + \\cdots + 5\\cdot 5 = (1+2+3+4+5)^2 = \\boxed{225}.\\][/hide][/quote]\nThis seems cool but I don't entirely get it. Can somebody elaborate what they mean by and how they got the final value[/quote]\n\nEach congruent squares first vertex can be shifted around among certain points which end up forming a rectangle, and each shift forms another square. Each rectangle corresponds to one set of congruent squares, and all the rectangles have a set of squares so we need to count how many points in each of all the different rectangles. This is the first long sum, which is equivilant to $(1+2+3+4+5)^2$ due to the expansion of $(v+w+x+y+z)^2$.\n\nHope that cleared things up\n\n~Mudkipswims42" } { "Tag": [ "LaTeX" ], "Problem": "Find two four digit numbers whose product is $4^{8}+6^{8}+9^{8}$.\r\n \r\nCan anyone give me a hint on this problem? I got \r\n\r\n$4^{8}+6^{8}+9^{8}=2^{16}+3^{8}2^{8}+3^{16}$ and $(2^{8}+3^{8})^{2}=2^{16}+2\\times2^{8}3^{8}+3^{16}$. please don't tell me how to solve it, just give an example or a formula or something. \r\n\r\nthanks :)\r\nn how do you delete your own post?\r\n\r\n[size=59][color=red]b-flat[/color][/size]", "Solution_1": "[hide=\"try\"]$a^{2}-b^{2}$ with what you have written over there.[/hide]", "Solution_2": "[hide=\"hint\"]\n$2^{16}+2^{8}3^{8}+3^{16}=(2^{8}+3^{8})^{2}-2^{8}3^{8}$. \n\nthen use what shady said above. [/hide]", "Solution_3": "On these types of problems, try to add zero creatively :D", "Solution_4": "oh i was so close, thanks for the hint guys. n thanks for correcting my latex, b-flat." } { "Tag": [ "number theory", "relatively prime", "number theory unsolved" ], "Problem": "$n, m$ are positive integers of different parity, and $n > m$. Find all integers $x$ such that $\\frac{x^{2^n} - 1}{x^{2^m} - 1}$ is a perfect square.", "Solution_1": "It's not hard to see that\r\n$\\frac{x^{2^n}-1}{x^{2^m}-1}=(x^{2^m}+1)(x^{2^{m+1}}+1)\\cdots (x^{2^{n-1}}+1)=A$.\r\n\r\nFirst case, if $x$ is odd then $v_2(A)=n-m\\equiv_2 1$ so $A$ can't\r\nbe perfect square. If $x$ is even, then we will prove that\r\n$(x^{2^a}+1,x^{2^b}+1)=1$.\r\n\r\nLet $(x^{2^a}+1,x^{2^b}+1)=d$ then\r\n$d|x^{2^b}+1-(x^{2^a}+1)=x^{2^a}(x^{2^a(2^{b-a}-1)}-1)$. How\r\n$(x^{2^a}+1,x^{2^a})=1$ then it's $d|x^{2^a(2^{b-a}-1)}-1$.\r\n\r\nNumber $2^{b-a}-1$ is odd so\r\n$d|x^{2^a}+1|(x^{2^a})^{2^{b-a}-1}+1=x^{2^a(2^{b-a}-1)}+1$.\r\n\r\nSo $d|x^{2^a(2^{b-a}-1)}+1 - (x^{2^a(2^{b-a}-1)}-1) =2$ but number\r\n$x^{2^a}+1$ is odd, because $x$ is even, so it has to be $d=1$.\r\n\r\nNow we proved that numbers $x^{2^m}+1, x^{2^{m+1}}+1, \\cdots x^{2^{n-1}}+1$ are relatively prime in pairs, and their product is\r\nsuppose to be perfect square, so they are all squares, and that is\r\nnot possible because $(x^{2^{m-1}})^290^{\\circ}$; hence, the vectors $\\overrightarrow{IH}$ and $\\overrightarrow{IS}$ form an obtuse angle with each other, i. e. their scalar product $\\overrightarrow{IH}\\cdot\\overrightarrow{IS}$ is $\\leq 0$. Hence,\r\n\r\n$AH\\cdot AI+BH\\cdot BI+CH\\cdot CI\\geq\\left(AI^2+BI^2+CI^2\\right)+\\overrightarrow{IH}\\cdot\\left(3\\cdot\\overrightarrow{SI}\\right)$\r\n$=\\left(AI^2+BI^2+CI^2\\right)+\\overrightarrow{IH}\\cdot\\left(-3\\cdot\\overrightarrow{IS}\\right)=\\left(AI^2+BI^2+CI^2\\right)-3\\cdot\\underbrace{\\overrightarrow{IH}\\cdot\\overrightarrow{IS}}_{\\leq 0}$\r\n$\\geq AI^2+BI^2+CI^2$.\r\n\r\nNow, we will show the inequality\r\n\r\n[color=green][b](1)[/b][/color] $AI^2+BI^2+CI^2\\geq a^2+b^2+c^2-6R^2$.\r\n\r\nOnce this inequality will be proven, we will have\r\n\r\n$4\\left(AH\\cdot AI+BH\\cdot BI+CH\\cdot CI\\right)\\geq 4\\left(AI^2+BI^2+CI^2\\right)\\geq 4\\left(a^2+b^2+c^2-6R^2\\right)$\r\n$=4\\left(3R^2-\\left(9R^2-\\left(a^2+b^2+c^2\\right)\\right)\\right)=3\\cdot\\left(2R\\right)^2-4\\left(9R^2-\\left(a^2+b^2+c^2\\right)\\right)$\r\n$=3\\cdot\\left(2R\\right)^2-9\\cdot HS^2$,\r\n\r\nand thus the problem will be solved. Thus, in order to solve the problem, it remains to prove the inequality [color=green][b](1)[/b][/color].\r\n\r\nIf the incircle of triangle ABC touches the side CA at a point Y, then the triangle AYI is right-angled at Y (since the tangent to a circle at a point is perpendicular to the radius through this point), and its catets are IY = r (the radius of the incircle) and AY = s - a, where $s=\\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC (this follows from the well-known formulas for the distances from the vertices of a triangle to the points where the incircle touches the adjacent sides). Hence, by the Pythagoras theorem, applied to the triangle AYI, we get $AI^2=AY^2+IY^2=\\left(s-a\\right)^2+r^2$. Thus,\r\n\r\n$\\left(AI^2+BI^2+CI^2\\right)-\\left(a^2+b^2+c^2\\right)=\\sum\\limits_{\\text{cyclic}} AI^2-\\sum\\limits_{\\text{cyclic}} a^2$\r\n$=\\sum\\limits_{\\text{cyclic}}\\left(\\left(s-a\\right)^2+r^2\\right)-\\sum\\limits_{\\text{cyclic}} a^2=3r^2+\\sum\\limits_{\\text{cyclic}}\\left(\\left(s-a\\right)^2-a^2\\right)$\r\n$=3r^2+\\sum\\limits_{\\text{cyclic}}\\left(\\left(s-a\\right)-a\\right)\\left(\\left(s-a\\right)+a\\right)=3r^2+\\sum\\limits_{\\text{cyclic}}\\left(s-2a\\right)s$\r\n$=3r^2+s\\sum\\limits_{\\text{cyclic}}\\left(s-2a\\right)=3r^2+s\\left(3s-2\\left(a+b+c\\right)\\right)=3r^2+s\\left(3s-2\\cdot 2s\\right)$\r\n$=3r^2-s^2$.\r\n\r\nHence, we can equivalently transform the inequality [color=green][b](1)[/b][/color] as follows:\r\n$AI^2+BI^2+CI^2\\geq a^2+b^2+c^2-6R^2$\r\n$\\Longleftrightarrow$ $\\left(AI^2+BI^2+CI^2\\right)-\\left(a^2+b^2+c^2\\right)\\geq -6R^2$\r\n$\\Longleftrightarrow$ $3r^2-s^2\\geq -6R^2$ $\\Longleftrightarrow$ $3r^2+6R^2\\geq s^2$.\r\n\r\nNow, since $\\left(3r^2+6R^2\\right)-\\dfrac13\\left(4R+r\\right)^2=\\dfrac23\\left(R-2r\\right)^2\\geq 0$, we have $3r^2+6R^2\\geq\\dfrac13\\left(4R+r\\right)^2$, so it will be enough to prove that $\\dfrac13\\left(4R+r\\right)^2\\geq s^2$. This is equivalent to $\\left(4R+r\\right)^2\\geq 3s^2$, or, after taking the square root, to $4R+r\\geq\\sqrt3 s$. Now, I am sure this inequality appeared somewhere on ML, but the proof is so easy that it would be probably more appropriate than a link: It is known that $4R+r=r_a+r_b+r_c$, where $r_a$, $r_b$, $r_c$ are the exradii of triangle ABC (a proof was given at http://www.mathlinks.ro/Forum/viewtopic.php?t=50373 , another one in [url=http://www.math.fau.edu/yiu/Geometry.html]Paul Yiu, Notes on Euclidean Geometry[/url], \u00a72.4.1). Hence, the inequality in question becomes $r_a+r_b+r_c\\geq\\sqrt3 s$. Now use the formulas $r_a=s\\tan\\frac{A}{2}$, $r_b=s\\tan\\frac{B}{2}$, $r_c=s\\tan\\frac{C}{2}$ to rewrite this as $s\\tan\\frac{A}{2}+s\\tan\\frac{B}{2}+s\\tan\\frac{C}{2}\\geq\\sqrt3 s$, what simplifies to $\\tan\\frac{A}{2}+\\tan\\frac{B}{2}+\\tan\\frac{C}{2}\\geq\\sqrt3$. Now, this follows from Jensen's inequality, applied to the function $x\\to\\tan x$, which is convex in the interval [0\u00b0; 90\u00b0[ (all the angles $\\frac{A}{2}$, $\\frac{B}{2}$, $\\frac{C}{2}$ lie in this interval):\r\n\r\n$\\tan\\frac{A}{2}+\\tan\\frac{B}{2}+\\tan\\frac{C}{2}\\geq 3\\tan\\frac{\\frac{A}{2}+\\frac{B}{2}+\\frac{C}{2}}{3}=3\\tan\\frac{A+B+C}{6}$\r\n$=3\\tan\\frac{180^{\\circ}}{6}$ (since A + B + C = 180\u00b0 by the sum of angles in triangle ABC)\r\n$=3\\tan 30^{\\circ}=3\\frac{1}{\\sqrt3}=\\sqrt3$.\r\n\r\nThis completes the solution.\r\n\r\n Darij", "Solution_3": "Very good but big solution darij :) . I am wondering if there is a manner to solve fast this ineq. Like campos theknik.We will see", "Solution_4": "[u]I am sorry, Darij, why offered you such quikly the your solution ?[/u]\r\n\r\nI can affirm that I were certainly that must to use the position of the incenter $I$ w.r.t. the circle with the diameter $HO$.\r\nThe main inequality is $\\sum AH\\cdot AI\\ge \\sum AI^2$ The strong inequality $\\overrightarrow {IH}\\cdot \\overrightarrow {IG}\\le 0$ is the basis of the previous inequality. The proposed inequality is equivalently with the following inequality:\\[ \\underline {\\overline {\\left| \\ \\blacktriangle \\ HI^2+\\sum AH\\cdot AI\\ge 3R^2+OI^2\\ \\blacktriangle\\ \\right| }} \\]", "Solution_5": "[quote=\"Virgil Nicula\"]:\\[ \\underline {\\overline {\\left| \\ \\blacktriangle \\ HI^2+\\sum AH\\cdot AI\\ge 3R^2+OI^2\\ \\blacktriangle\\ \\right| }} \\][/quote]\r\n\r\nWhich can be proved... how?", "Solution_6": "I think we can prove that \r\nwith any M $MA^2+MB^2+MC^2 \\ge a^2+b^2+c^2-6R^2$ quite simple by using Leibnitz fomula:\r\n$\\sum MA^2 =3MG^2 + \\sum GA^2 \\ge \\sum GA^2 = \\frac{a^2+b^2+c^2}{3}$\r\nand if we use $a^2+b^2+c^2 \\le 9R^2$ then we have \r\n$\\frac{a^2+b^2+c^2}{3} \\ge a^2+b^2+c^2-6R^2$\r\nmay be I have some mistakes \r\nbest regard" } { "Tag": [ "floor function", "induction", "function", "logarithms", "inequalities", "calculus", "derivative" ], "Problem": "Let $a_1,a_2,a_3,\\ldots,a_n$ be a sequence of non-negative integers, where $n$ is a positive integer. Let\r\n\\[ A_n={a_1+a_2+\\cdots+a_n\\over n}\\ . \\] \r\nProve that\r\n\\[ a_1!a_2!\\ldots a_n!\\ge\\left(\\lfloor A_n\\rfloor !\\right)^n \\]\r\nwhere $\\lfloor A_n\\rfloor$ is the greatest integer less than or equal to $A_n$, and $a!=1\\times 2\\times\\cdots\\times a$ for $a\\ge 1$(and $0!=1$). When does equality hold?", "Solution_1": "[quote=\"shobber\"]Let $a_1,a_2,a_3,\\ldots,a_n$ be a sequence of non-negative integers, where $n$ is a positive integer. Let\n\\[ A_n={a_1+a_2+\\cdots+a_n\\over n}\\ . \\] \nProve that\n\\[ a_1!a_2!\\ldots a_n!\\ge\\left(\\lfloor A_n\\rfloor !\\right)^n \\]\nwhere $\\lfloor A_n\\rfloor$ is the greatest integer less than or equal to $A_n$, and $a!=1\\times 2\\times\\cdots\\times a$ for $a\\ge 1$(and $0!=1$). When does equality hold?[/quote]\r\nhttp://www.kalva.demon.co.uk/apmo/asoln/asol021.html", "Solution_2": "I don't know why I can't open all the links from this website.", "Solution_3": "try use proxy.", "Solution_4": "Kalva is not working anymore. Can someone post a solution? I'm stucked. :P \r\n\r\nI have tried induction but failed.", "Solution_5": "There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \\equal{} \\log x!$ \r\nIt is understandable because $ \\log x! \\equal{} \\log 1 \\plus{} \\log 2 \\plus{} \\cdots \\log x$ and $ \\log x$ is an increasing function.\r\nHowever I dont see a rigorous proof now.\r\nAlso it can be made monotonically increasing(not strictly) by making $ f(x) \\equal{} 0\\ \\forall x\\geq 0$ and $ \\leq 1$\r\n\r\n$ \\sum_{i \\equal{} 1}^{n}f(a_n)\\geq nf(A_n)$ by Jensen's inequality\r\n$ f(A_n)\\geq f(\\lfloor A_n \\rfloor)$ because $ f(x)$ is monotonically increasing.\r\n$ \\sum_{i \\equal{} 1}^{n}f(a_n) \\equal{} \\sum_{i \\equal{} 1}^{n}\\log a_n! \\equal{} \\log\\prod a_n!$\r\n$ nf(\\lfloor A_n \\rfloor) \\equal{} n\\log \\lfloor A_n \\rfloor ! \\equal{} \\log (\\lfloor A_n \\rfloor !)^n$\r\nThus proved. Equality holds for $ a_1 \\equal{} a_2 \\equal{} \\cdots \\equal{} a_n$ or when some of $ a_i$ are $ 1$ and some are $ 0$.", "Solution_6": "Kalva still lives in our hearts... and someplace else (Webarchive). So here's a link to the solution:\r\n\r\n[url]http://web.archive.org/web/20040313130703/www.kalva.demon.co.uk/apmo/asoln/asol021.html[/url]", "Solution_7": "Great solution in Kalva! But I can't help saying that the equality case given there is not truly an \"iff\" statement :P", "Solution_8": "I think those archive is a little bit different from the real kalva. For example, I didn't see tournament of the towns in kalva before (am I blind?) :P", "Solution_9": "[quote=\"Akashnil\"]There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \\equal{} \\log x!$ \n[/quote]\r\n$ f(x)\\equal{}\\log x!$ is concave!", "Solution_10": "[quote=\"Sunjee\"][quote=\"Akashnil\"]There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \\equal{} \\log x!$ \n[/quote]\n$ f(x) \\equal{} \\log x!$ is concave![/quote]\r\n\r\nHonestly, terms convex and concave always seem to be so confusing: in different literature, I used to find diametrically opposite definitions: in some books I've used, what is concerned concave is convex in others: so some use the terms concave up and concave down- now that's a little more unambiguous. :wink:", "Solution_11": "[quote=\"Sunjee\"][quote=\"Akashnil\"]There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \\equal{} \\log x!$ \n[/quote]\n$ f(x) \\equal{} \\log x!$ is concave![/quote]\r\nFor some function defined on reals, I mean it to be convex iff the 2nd derivative is always non-negetive. And I'm pretty sure there is a convex function $ f(x)$ which has the required property.\r\n\r\nEdit: Can we call a sequence convex? In that case the definition would be $ a_{n \\minus{} 1} \\plus{} a_{n \\plus{} 1}\\geq 2a_n$. Jensen possibly holds there.", "Solution_12": "[quote=\"Akashnil\"]There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \\equal{} \\log x!$ \nIt is understandable because $ \\log x! \\equal{} \\log 1 \\plus{} \\log 2 \\plus{} \\cdots \\log x$ and $ \\log x$ is an increasing function.\nHowever I dont see a rigorous proof now.[/quote]\r\nCorrect! In fact, there is exactly one such function; it is the logarithm of the [url=http://en.wikipedia.org/wiki/Gamma_function]gamma function[/url] (shifted by one). \r\n\r\nNow: I actually think that the right-hand-side [i]divides[/i] the left-hand-side under suitable conditions, but I'm not sure what they are.", "Solution_13": "I don't understand in kalva web !\r\nCould anyone explain it? Thanks.", "Solution_14": "Just start with $ a_1\\equal{}a_2...\\equal{}a$\r\nThen clearly we have equality. Now think about what happens when we increase one of the $ a_i$ by one, and decrease another by $ 1$. Then we are dividing the $ LHS$ by $ a$, and multiplying by $ a\\plus{}1$. In this way we see any change preserves the inequality.", "Solution_15": "[quote=\"Akashnil\"]There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \\equal{} \\log x!$ \nIt is understandable because $ \\log x! \\equal{} \\log 1 \\plus{} \\log 2 \\plus{} \\cdots \\log x$ and $ \\log x$ is an increasing function.\nHowever I dont see a rigorous proof now.\n.[/quote] \r\n\r\nSince\r\n$ \\log (x\\plus{}1)! \\minus{} \\log x! \\equal{} log (x\\plus{}1) \\geq 0$\r\n\r\nso $ f(x)\\equal{} \\log x!$ is increasing!", "Solution_16": "I came up with a solution. Does it work?\r\n\r\n[b]Solution[/b]\r\n\r\nLet $ \\lfloor A_n \\rfloor \\equal{} p$. Let $ b_m \\equal{} a_m \\minus{} p$ for all $ a_m \\ge p$ and let $ c_k \\equal{} p \\minus{} a_k$ for all $ a_k \\le p$. Now note that since $ \\sum a_i \\ge np$, it follows that $ \\sum b_m \\ge \\sum c_k$. Now consider the following product. Define that,\r\n\r\n$ P \\equal{} \\frac{(p\\plus{}1)(p\\plus{}2) \\dots (p\\plus{}b_1)(p\\plus{}1)(p\\plus{}2) \\dots (a\\plus{}b_m)}{(p\\minus{}c_1\\plus{}1)(p\\minus{}c_1\\plus{}2) \\dots (p)(p\\minus{}c_2\\plus{}1)(p\\minus{}c_2\\plus{}2) \\dots (p\\minus{}c_k\\plus{}1) \\dots (p)}$\r\n\r\nThere are a total of $ \\sum b_m$ factors in the numerator each of which is greater than or equal to $ p$ and there are a total of $ \\sum c_k$ factors in the denominator each of which is less than or equal to $ p$ (since it is possible that for some $ i$, $ b_i$ or $ c_i$ is $ 0$). Since $ \\sum b_m \\ge \\sum c_k$, it follows that $ P \\ge 1$.\r\n\r\nNow note that\r\n\r\n$ P \\equal{} \\frac{a_1!}{p!} \\cdot \\frac{a_2!}{p!} \\cdot \\dots \\cdot \\frac{a_n!}{p!}$\r\n\r\nHence since $ P \\ge 1$, it follows that $ a_1!a_2! \\dots a_n! \\ge (p!)^n$.\r\n\r\nEquality holds when $ P \\equal{} 1$ which can only occur if all $ c_i$ and $ b_i$ are $ 0$ which occurs when $ a_1 \\equal{} a_2 \\equal{} \\dots \\equal{} p$.", "Solution_17": "For all $a_i$ being equal the equality case is evident. Now assume that all $a_i$ are not equal, and without loss of generality let $a_1 \\leqslant a_2 \\leqslant ... \\leqslant a_n$. Then for some $x$ such that $a_n \\geqslant a_1 +x$, the transformation $(a_1, a_2,...,a_n)\\mapsto (a_1 +x, a_2, ...,a_n-x)$ decreases the left hand side. Clearly, the right hand side remains invariant under this transformation. The value of the left hand side decreases by every such transformation. This transformation cannot be performed when all the numbers are equal, hence the minimum value of the left hand side is when all $a_i$ are equal. That, as explained above, gives the equality case. Hence we're done.", "Solution_18": "Notice that by taking $\\ln$ of both sides, the inequality becomes: $\\sum_{i=1}^{n}\\ln {a_i!}\\ge\\ln {(\\lfloor A_n\\rfloor)^n}$\nFurthermore let $f(x)=\\ln {x!}\\Longrightarrow f'(x)=\\psi(x+1)\\Longrightarrow f''(x)=\\psi_1(x+1)>0$ thus the function is strictly convex on the non-negative integers.\nThus $\\sum_{i=1}^{n}\\ln{a_i!}=\\sum_{i=1}^{n}f(a_i)\\overset{\\text{Jensen}}{\\ge}n f\\bigg(\\frac{\\sum_{i=1}^{n}a_i}{n}\\bigg)=nf(A_n)=n\\ln {A_n!}$\nFurthermore: $n\\ln {A_n!}\\ge\\ln {(\\lfloor A_n\\rfloor)^n}\\Longleftrightarrow (e^{\\ln {A_n!}})^n\\ge e^{\\ln {(\\lfloor A_n\\rfloor)^n}}\\Longrightarrow (A_n)^n\\ge(\\lfloor A_n\\rfloor)^n\\Longrightarrow A_n!\\ge\\lfloor A_n\\rfloor$ which is clearly true $\\blacksquare$.\nAnd we are done! :bye: ", "Solution_19": "let a_1>=a_2>=....a_n Now we will prove this by induction\nA_1=a_1 so p(1) is true\nnow let p(n-1) be true and note p(n) is\nrewarding\n(a_n)!/([A_n-1]!)>=(([A_n])/([A_n-1]))^n\nand this is true because\n(a_n)-([A_n-1])=n(([A_n])-([A_n-1])) and a_n>=[A_n]\nand equality hold iff (a_i)!=(a_j)! for all 1<=(i,j)<=n" } { "Tag": [ "limit", "geometry", "3D geometry", "calculus", "calculus computations" ], "Problem": "Find:\r\n\\[ \\lim_{n \\to \\infty} \\frac{\\sqrt{n^2+1}+\\sqrt{n}}{\\sqrt[3]{n^2+n}-\\sqrt{n}} \\]", "Solution_1": "This appears to diverge, as the numerator is $O(n)$ and the denominator is $O(n^{2/3})$. Perhaps the first term in the numerator is also meant to be a cube root?", "Solution_2": "I agree with Xevarion's answer, just divide both numerator and denominator by $n$ and it shows that the limit diverges..." } { "Tag": [ "function", "algebra solved", "algebra" ], "Problem": "A function f is defined on the set D. The set D is the set of rationals being different from 0 and 1. The function f satisfies for all x \\in D the following equation:\r\n\r\n f(x) + f(1 - 1/x) = x \r\n\r\nDetermine f !", "Solution_1": "Put in the equation x, (x-1)/x and 1/(1-x) than\r\nf(x)+f((x-1)/x)=x\r\nf((x-1)/x)+f(1/(1-x))=(x-1)/x\r\nf(1/(1-x))+f(x)=1/(1-x)\r\n\r\nthan f(x)=1/2(x+1/(x-1)-(x-1)/x)" } { "Tag": [ "inequalities" ], "Problem": "If [i]x[/i],[i]y[/i],[i]z[/i] are positive integers and the sum of any two greater than the third,prove tht..\\\r\n\r\n $ [1\\plus{}(y\\minus{}z)/x]^{x}[1\\plus{} (z\\minus{}x)/y]^{y}[1\\plus{}(x\\minus{}y)/z]^{z}$ < 1.\r\n :lol: :!:", "Solution_1": "as $ x,y,z$ are sides of a triangle\r\nWe can put $ x \\equal{} a \\plus{} b,y \\plus{} b \\plus{} c,z \\equal{} c \\plus{} a$\r\nPlugging this into the inequality we get \r\n$ LHS \\equal{} \\prod_{cyc}({{1 \\plus{} \\frac {b \\minus{} a}{a \\plus{} b}})^{a\\plus{}b}} \\\\\r\n\\equal{} \\prod_{cyc}{({\\frac {2a}{a \\plus{} b})^{a\\plus{}b}}} \\\\\r\n\\le \\frac {2a \\plus{} 2b \\plus{} 2c}{a \\plus{} b \\plus{} b \\plus{} c \\plus{} c \\plus{} a} \\equal{} 1$\r\nThe Inequality applied here is the weighted AM-GM inequality......." } { "Tag": [ "limit", "integration", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Prove that \r\n\r\n$ \\displaystyle\\lim_{n\\to\\infty}\\frac{1}{n}\\int^n_1 \\|\\dfrac{n}{x}\\|\\,dx\\equal{}log_3(4/\\pi)$", "Solution_1": "I don't quite see how this is number theory...", "Solution_2": "NO ONe???????", "Solution_3": "What does $ \\left \\Vert \\dfrac{n}{x}\\right \\Vert$ stand for? Is that the fractional part of $ \\frac {n}{x}$? If that's the case then\r\n\\[ \\lim_{n\\to \\infty}\\frac {1}{n}\\int_1^n{\\left \\Vert \\dfrac{n}{x}\\right \\Vert dx} \\equal{} 1 \\minus{} \\gamma \\neq \\log_3{\\frac {4}{\\pi}}\r\n\\]\r\nIt must be something else then?" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "We choose a subset of $ \\{1, 2, \\ldots, 2n\\plus{}1 \\}$ with maximal cardinality so that none of its elements is equal to the sum of others.\r\nHow many elements does it have?", "Solution_1": "I interpreted [quote=\"mathmanman\"]the sum of others[/quote] to mean \"the sum of some subset of the others,\" not \"the sum of some pair of the others.\" In that case, $ n \\plus{} 1$ is too large when $ n \\geq 4$ ($ 9 \\equal{} 5 \\plus{} 3 \\plus{} 1$). mathmanman, which problem was intended?", "Solution_2": "Indeed, your interpretation was the intended one [b]JBL[/b].", "Solution_3": "I have moved Tdl's solution for the \"the sum of some pair of others\" version into the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=1491 . Let's use this particular topic to discuss the \"the sum of some subset of the others\" version.\r\n\r\n darij", "Solution_4": "We can take subset $ \\{n\\plus{}1,n\\plus{}2,...,2n\\plus{}1\\}$ with property described because sum of every subset with at least 2 elements is $ \\geq 2n\\plus{}3$.\r\n\r\nSuppose we can take elements $ a_1,a_2,...,a_{n\\plus{}2}$ with required property which are sorted in increasing order. Consider set $ \\{a_{n\\plus{}2}\\minus{}a_1,a_{n\\plus{}2}\\minus{}a_2,...,a_{n\\plus{}2}\\minus{}a_{n\\plus{}1}\\} \\cup \\{a_1,a_2,...,a_{n\\plus{}1}\\}$. Every element of this set is $ \\leq 2n$ and $ \\geq 1$ hence we have $ a_{n\\plus{}2}\\equal{}a_{i}\\plus{}a_{j}$ at least two times for some $ 1\\leq i,j\\leq n\\plus{}1$. It can't be $ i\\equal{}j$ both times so we have contradiction, which show that maximal cardinality is $ n\\plus{}1$." } { "Tag": [ "calculus", "Harvard", "college", "Princeton", "search", "linear algebra", "graph theory" ], "Problem": "I'm going through calculus currently and I'd also like to get started in some higher level math. I would like to learn some more advanced combinatorics and linear algebra. Any book recommendations?", "Solution_1": "How advanced do you want your linear algebra? Are you looking for computationally-based or proof-based? I like \"Linear Algebra\" by Friedberg, Insel, and Spence (nice, hard book). I know they use Axler's \"Linear Algebra Done Right\" at Harvard but I don't know much about the book.", "Solution_2": "At this point, I don't really know which one I'd like to go with (the two types you mentioned), because I know nothing about linear algebra and I don't know how this choice will affect me.", "Solution_3": "As for linear algebra, someone posted [url=http://joshua.smcvt.edu/linearalgebra/]this online book[/url] a while back, which is a fairly comprehensive introductory textbook; though really, mostly any (non-proof based) introductory textbook should do. Princeton uses Hoffman/Kunz, and JRav mentioned another couple of commonly used authors, though the material may be (comparatively) advanced. \r\n\r\nAdvanced combinatorics is a bit more vague; while there may be comprehensive texts on combinatorics, you'll probably have to refine your search a bit more; perhaps enumerative combinatorics? graph theory?", "Solution_4": "graph theory.\r\n\r\nEDIT: Is proof-based really hard to deal with? Cause, I feel sort of guilty when I just \"learn' a theorem without knowing where it came from .", "Solution_5": "I should have been more clear with what I meant with \"non-proof based\"; it does not mean that the basic theorems are not proved. The link I provided does include proofs of the theorems and applications. However, there's a difference (and not just for linear algebra) between covering introductory fully-rigorous linear algebra and introductory computation-based linear algebra; it is rather difficult to do the former without any intuition from the latter. For example, a rigorous linear algebra text will use a much more general context (eg, work in any fields and not just the real numbers) and discuss less of the applications, so it will be more difficult to gain hands-on intuition.", "Solution_6": "[quote=\"azjps\"]I should have been more clear with what I meant with \"non-proof based\"; it does not mean that the basic theorems are not proved. The link I provided does include proofs of the theorems and applications. However, there's a difference (and not just for linear algebra) between covering introductory fully-rigorous linear algebra and introductory computation-based linear algebra; it is rather difficult to do the former without any intuition from the latter. For example, a rigorous linear algebra text will use a much more general context (eg, work in any fields and not just the real numbers) and discuss less of the applications, so it will be more difficult to gain hands-on intuition.[/quote]\r\n\r\nOh god. No. Not multiple fields. Not again after writing that script for sagemath.org. Yeah, I think i'll with go with computational-based.", "Solution_7": "All that means is that you choose scalars that don't have to be real numbers. It's not bad. Besides, you'll have to get used to it for abstract algebra, where $ \\mathbb{R}$ is only one example." } { "Tag": [ "integration", "calculus", "analytic geometry", "real analysis", "real analysis unsolved" ], "Problem": "Let $p(y(t)) = \\frac{1}{\\sigma\\sqrt{2\\pi}}e^{-\\frac{1}{2}\\frac{(y(t)-m)^2}{\\sigma^2}}$ be gaussian distribution of aleator variable $y(t)$, where $m$ is the mean value of $y(t)$. Show that $\\int_{m-3\\sigma}^{m+3\\sigma}p(y(t))dy(t) \\geq 0.95$.", "Solution_1": "[quote=\"jianuovidiu\"]Let $p(y(t)) = \\frac{1}{\\sigma\\sqrt{2\\pi}}e^{-\\frac{1}{2}\\frac{(y(t)-m)^2}{\\sigma^2}}$ be gaussian distribution of aleator variable $y(t)$, where $m$ is the mean value of $y(t)$. Show that $\\int_{m-3\\sigma}^{m+3\\sigma}p(y(t))dy(t) \\geq 0.95$.[/quote]\r\nSubstitute $z = ((y(t)-m)/\\sigma$ to reduce the problem to the case $m = 0, \\, \\sigma = 1$. Then consider the [b]square[/b] of the integral:\r\n$\\frac{1}{2\\pi}\\left( \\int_{-3}^3 e^{-x^2/2} dx \\right) = \\frac{1}{2\\pi}\\int_{-3}^3 \\int_{-3}^3 e^{-\\frac{x^2 +y^2}{2}} dx dy \\ge \\frac{1}{2\\pi} \\iint_{D_3} e^{-\\frac{x^2 +y^2}{2}} dx dy$\r\nwhere $D_3$ is the circle with radius 3 about the origin. Use polar coordinates to compute the last integral as\r\n$\\int_0^3 r e^{-r^2/2}dr = 1 - e^{-9/2}$ and conclude that the integral in question is at least $\\sqrt{1-e^{-9/5}} \\approx 0.994$. The exact value is more like 0.997. \r\n\r\n[i] This is the same trick that one uses to compute $\\int_{-\\infty}^\\infty e^{-x^2/2}dx$.[/i]" } { "Tag": [ "geometry", "rectangle" ], "Problem": "Is there any way to determine the largest rectangle that can be inscribed in a triangle? What a bout the largest square?\r\n\r\nthanks :)", "Solution_1": "[hide]\nLet $ M\\in AB, \\ N\\in AC$ such that $ MN\\parallel BC$. Let $ K,L$ be points on $ BC$ such that $ KLMN$ be a rectangle\n\nOf course the angles $ B,C$ shouldn't be obtuse. Let $ AD$ be the altitude from $ A$\n\nLet $ E\\in MN \\cap AD$\n\nLet $ \\lambda \\equal{} \\frac {AM}{AB}$. Then\n\n$ AM \\equal{} \\lambda AB$\n$ AN \\equal{} \\lambda AC$\n$ LD \\equal{} \\lambda BD$\n$ DK \\equal{} \\lambda DC$\n$ AE \\equal{} \\lambda AD$\n\nThen $ DE \\equal{} (1 \\minus{} \\lambda)AD$\n$ LK \\equal{} LD \\plus{} DK \\equal{} \\lambda(BD \\plus{} DC) \\equal{} \\lambda BC$\n\n$ (KLMN) \\equal{} LK \\cdot DE \\equal{} \\lambda(1 \\minus{} \\lambda)BC\\cdot AD \\equal{} \\lambda(1 \\minus{} \\lambda)\\cdot 2(ABC)$\n\nSo we obtain maximum value for $ (KLMN)$ when $ \\lambda \\equal{} \\frac {1}{2}$. This maximum value is equal to $ \\frac {(ABC)}{2}$\n\n$ \\star$ The rectangle $ KLMN$ is square only if $ KL \\equal{} ED \\iff \\lambda BC \\equal{} (1 \\minus{} \\lambda)AD \\iff \\lambda \\equal{} \\frac {AD}{AD \\plus{} BC}$\n\n$ \\star$ The largest rectangle is a square if and only if $ AD \\equal{} BC$\n\n[color=brown]Note \n$ (X)$ denotes the area of $ X$[/color][/hide]" } { "Tag": [], "Problem": "How many digits are in the number $25^{21}\\cdot2^{48}?$", "Solution_1": "[hide]$5^{42}\\cdot 2^{48}= 5^{42}\\cdot 2^{42}\\cdot 2^{6}=10^{42}\\cdot 2^{6}= 64\\cdot 10^{42}$. There are 42 zeroes, plus 2 digits, total of 44 digits.[/hide]", "Solution_2": "That is correct! :)", "Solution_3": "[quote=\"i_like_pie\"]How many digits are in the number $25^{21}\\cdot2^{48}?$[/quote]\r\n\r\n[hide=\"Solution\"]Firstly, you have to orgainze the data (unless you are willing to compute it, which I am not). The number of zeros must be found.\n\n$5^{42}\\times 2^{48}= 5^{42}\\times 2^{42}\\times 2^{6}$. That is equal to: $10^{42}\\times 2^{6}= 64\\times 10^{42}$. THerefore, there are 42 zeroes. Then, the number of other digits must be found. There are two of these. Therefore, is it $42+2=\\boxed{44\\text{ digits}}$.[/hide]", "Solution_4": "[hide=\"Content Hidden Reveal to Click\"] Prime factoring, we have $2^{48}*5^{42}$. We note that this is equal to $10^{42}*2^{6}$, which is 64 with 42 zeroes after it, a total of 44 digits.[/hide]", "Solution_5": "[hide]$25^{21}\\cdot 2^{48}=5^{42}\\cdot 2^{48}= 10^{42}\\cdot 2^{6}$\n\n$2^{6}$ has 2 digits. $10^{42}$ has 42 zeroes. 2+42=44 digits[/hide]", "Solution_6": "whats the point of posting the same solution 4 times?\r\n-jorian", "Solution_7": "Because other people want to try to solve the problem...\r\n\r\nAnd when posting a solution, the poster doesn't know that it's the same solution as the person above them. You might have a different solution that may be helpful to people reading the thread.", "Solution_8": "And it's fantastic that you read them without trying the problem...", "Solution_9": "[quote=\"perfect628\"]And it's fantastic that you read them without trying the problem...[/quote]\n\nThat's why we hide our answers. :wink:\n\n[quote=\"jhredsox\"]whats the point of posting the same solution 4 times?[/quote]\r\n\r\nBecause others solve it, and don't look at the hidden answers below them, so they don't know really if it has been posted. Also, more answers can clarify small holes in other's logic.", "Solution_10": "I personally don't think these are adequate reasons. Solve the problem, then read the other solutions.\r\n\r\n#1: You can solve the problem and not post it. :wink: \r\n#2: The small chance that you have a different solution is overwhelmed by the benefit of reading the previous solutions, and thus not adding to unnecessary clutter.\r\n#3: Reading previous solutions may reveal errors (both on your part and others') , also helping readers of the forum. :P", "Solution_11": "[hide]\n5*2=10\nand every 0 is a digit.\n$5^{42}\\times2^{48}$\n$10^{4}2\\times2^{6}$\n$2^{6}=64$\n2+42=44 digits[/hide]", "Solution_12": "[hide=\"solution\"]$25^{21}\\times 2^{48}= 5^{42}\\times 2^{48}= 10^{42}\\times 64$\n\nThat can be written as\n\n[u]64000...000[/u]\n42 zeros\n\nTherefore, there are 44 digits.[/hide]" } { "Tag": [ "AMC", "AIME", "geometry", "perpendicular bisector", "cyclic quadrilateral" ], "Problem": "10. Triangle ABC has AC = BC and \u2220ACB = 106 degrees. M is a point inside the triangle such that \u2220MAC = 7 degrees and \u2220MCA = 23 degrees. Find \u2220CMB. \r\n\r\nOkay this one took me 50 minutes probably because I'm stupid, but still the solution I came up with wasn't obvious and was kind of nice.", "Solution_1": "Hmm... I'll take a shot at it.\r\n\r\n[hide]Ok... darn. I can't make a diagram but notice that you can extend AM to a point, lets say D on the perpendicular bisector of AB extended from C.\n\nAngle CMD is 30 degrees, and so is angle MCD. So, MD $\\cong$ CD. Notice that AD $\\cong$ BD, so $\\angle$DAB $\\cong$ $\\angle$DBA. So, $\\angle$ADC $\\cong$ $\\angle$BDC. And BD = BD. So, the two triangles BMD and BCD are congruent. Wow... So, by, I thought I'd never use this on AIME, CPCTC, $\\angle$BMD $\\cong$ $\\angle$BCD, and $\\angle$BCD = $53$. (106/2). So, that's $\\angle$BMD. Yay.[/hide]", "Solution_2": "I'm a retarded idiot. It took me 20 minutes to draw perpendicular bisector of AB, another 30 minutes to miss the obvious stuff. \r\nHere's what I did:\r\n\r\nDraw perpendicular bisector of AB and let D be the midpoint of AB. Let AM intersect CD at I. $\\angle EAI= \\angle ECI = 30$. So ACIE is a cyclic quadrilateral, so $\\angle CAI= \\angle CEI = 7$. BEI= 60-7=53. since MEB=60, MIB=120. MIBE is cyclic quadrilateral. IEB=IMB=53. CMI=30, so CMI+IMB=83=CMB.", "Solution_3": "[hide]Draw point N inside ABC such that triangle CNM is equilateral. triangle NCB :cong: MCA by SAS. angle MNB = 360-150-60 = 150, so triangle MNB :cong: triangle CNB by SAS. Triangle BCM is isosceles, angle CMB = angle MCB = 83[/hide]", "Solution_4": "[hide]Angle ceva:\nsin( XG; thus, if X' is the reflection of the point X in the point G, so that X'G = XG, then AG > X'G, and thus the point X' lies inside the segment AG. Hence, the segment UX' intersects the segment GM at an interior point R. Of course, this implies $MG\\geq RG$. Now, since UG = VG (because G is the midpoint of the segment VU) and X'G = XG, we have $\\frac{UG}{X^{\\prime}G}=\\frac{VG}{XG}$, so that Thales yields UX' || VX, what, again, yields by Thales $\\frac{UG}{RG}=\\frac{VG}{NG}$. Since UG = VG, this becomes RG = NG. But $MG\\geq RG$; thus, $MG\\geq NG$.\r\n\r\nNow, since the area of a triangle equals $\\frac12$ of the product of two sides and the sine of the angle between them, we have $\\left[MGU\\right]=\\frac12\\cdot UG\\cdot MG\\cdot\\sin\\measuredangle UGM$ and $\\left[NGV\\right]=\\frac12\\cdot VG\\cdot NG\\cdot\\sin\\measuredangle VGN$, so that UG = VG, $MG\\geq NG$ and < UGM = < VGN yield $\\left[MGU\\right]\\geq\\left[NGV\\right]$, and thus\r\n\r\n$\\left[MBN\\right]=\\left[BUGN\\right]+\\left[MGU\\right]\\geq\\left[BUGN\\right]+\\left[NGV\\right]=\\left[UBV\\right]=\\frac49S$.\r\n\r\nOn the other hand, again since the area of a triangle equals $\\frac12$ of the product of two sides and the sine of the angle between them, we have $\\left[AGM\\right]=\\frac12\\cdot AG\\cdot MG\\cdot\\sin\\measuredangle AGM$ and $\\left[XGN\\right]=\\frac12\\cdot XG\\cdot NG\\cdot\\sin\\measuredangle XGN$, so that AG > XG, $MG\\geq NG$ and < AGM = < XGN yield $\\left[AGM\\right]\\geq\\left[XGN\\right]$, and thus\r\n\r\n$\\left[MBN\\right]=\\left[BAGN\\right]-\\left[AGM\\right]\\leq\\left[BAGN\\right]-\\left[XGN\\right]=\\left[ABX\\right]=\\frac12S$.\r\n\r\nSo, altogether, we have found that $\\frac49S\\leq\\left[MBN\\right]\\leq\\frac12S$, and the problem is solved.\r\n\r\n[b]EDIT:[/b] More can be said: Since [MBN] + [AMNC] = S, we have [AMNC] = S - [MBN], so $\\left[MBN\\right]\\leq\\frac12S$ yields $S\\geq 2\\cdot\\left[MBN\\right]$ and thus $\\left[AMNC\\right]=S-\\left[MBN\\right]\\geq 2\\cdot\\left[MBN\\right]-\\left[MBN\\right]=\\left[MBN\\right]$. Hence, $\\frac{\\left[AMNC\\right]}{\\left[MBN\\right]}\\geq 1$. On the other hand, from $\\frac49S\\leq\\left[MBN\\right]$ we see that $\\left[AMNC\\right]=S-\\left[MBN\\right]=\\frac94\\cdot\\frac49S-\\left[MBN\\right]\\leq\\frac94\\cdot\\left[MBN\\right]-\\left[MBN\\right]=\\frac54\\cdot\\left[MBN\\right]$, so that $\\frac{\\left[AMNC\\right]}{\\left[MBN\\right]}\\leq\\frac54$. Hence, $1\\leq\\frac{\\left[AMNC\\right]}{\\left[MBN\\right]}\\leq\\frac54$.\r\n\r\n Darij", "Solution_5": "I can suppose $M\\in (AB), N\\in (AC)$. I note $AM=xc, BN=yb$, where $x,y\\in {\\left[ \\frac 12 ,1\\right]},\\ [MAN]=xy\\cdot S\\ $.\r\nIs wellknown the relation $\\ \\frac {MB}{MA}+\\frac{NC}{NA}=1,\\ i.e.\\ \\frac 1x +\\frac 1y =3$. Consequently, $[MAN]-min.\\Longleftrightarrow xy-min.$\r\n$\\Longleftrightarrow \\frac 1x\\cdot \\frac 1y -max.\\ (\\ \\frac 1x +\\frac 1y =3-constant\\ )\\Longleftrightarrow \\frac 1x =\\frac 1y =\\frac32\\Longleftrightarrow x=y=\\frac 23 $. Thus,\\[\\overline{\\underline{\\left| [MAN]\\ge \\frac 49 \\cdot S\\right| }}\\ \\ (1)\\].\r\nI note $\\ u=\\frac 1x,\\ v=\\frac 1y\\Longrightarrow u,v\\in [1,2],\\ u+v=3$. Consequently, $[MAN]-max\\Longleftrightarrow xy-max.$\r\n$\\Longleftrightarrow uv-min.\\Longleftrightarrow u(1-u)-min.$, where $u\\in [1,2] \\Longleftrightarrow u\\in \\{ 1,2\\}$. Thus,\r\n\\[\\overline{\\underline{\\left| [MAN]\\le \\frac 12 \\cdot S\\right| }}\\ \\ (2)\\] In conclusion, from the relations (1) and (2) results that $|[BCNM]-[MAN]|=|S-2\\cdot [MAN]|\\in {\\left[ 0,\\frac 19\\right].}$\r\n\r\n[b]Remark.[/b] If$\\ I\\in MN$, Then $\\ b\\cdot \\frac{MB}{MA}+c\\cdot \\frac{NC}{NA}=a$. You can propose a alike problem !\r\n\r\nSoon I will present a generalization of this problem for a any point $P\\in (ABC)$\r\n without use barycentric coordinates.", "Solution_6": "[quote=\"levi\"]I can suppose $M\\in (AB), N\\in (AC)$. I note $AM=xc, BN=yb$, where $x,y\\in {\\left[ \\frac 12 ,1\\right]},\\ [MAN]=xy\\cdot S\\ $.\nIs wellknown the relation $\\ \\frac {MB}{MA}+\\frac{NC}{NA}=1,\\ i.e.\\ \\frac 1x +\\frac 1y =3$. Consequently, $[MAN]-min.\\Longleftrightarrow xy-min.$\n$\\Longleftrightarrow \\frac 1x\\cdot \\frac 1y -max.\\ (\\ \\frac 1x +\\frac 1y =3-constant\\ )\\Longleftrightarrow \\frac 1x =\\frac 1y =\\frac32\\Longleftrightarrow x=y=\\frac 23 $. Thus,\\[\\overline{\\underline{\\left| [MAN]\\ge \\frac 49 \\cdot S\\right| }}\\ \\ (1)\\].\nI note $\\ u=\\frac 1x,\\ v=\\frac 1y\\Longrightarrow u,v\\in [1,2],\\ u+v=3$. Consequently, $[MAN]-max\\Longleftrightarrow xy-max.$\n$\\Longleftrightarrow uv-min.\\Longleftrightarrow u(1-u)-min.$, where $u\\in [1,2] \\Longleftrightarrow u\\in \\{ 1,2\\}$. Thus,\n\\[\\overline{\\underline{\\left| [MAN]\\le \\frac 12 \\cdot S\\right| }}\\ \\ (2)\\] In conclusion, from the relations (1) and (2) results that $|[BCNM]-[MAN]|=|S-2\\cdot [MAN]|\\in {\\left[ 0,\\frac 19\\right].}$\n\n[b]Remark.[/b] If$\\ I\\in MN$, Then $\\ b\\cdot \\frac{MB}{MA}+c\\cdot \\frac{NC}{NA}=a$. You can propose a alike problem !\n\nSoon I will present a generalization of this problem for a any point $P\\in (ABC)$\n without use barycentric coordinates.[/quote]\r\n[u][b]Generalization[/b]:[/u] Let $\\ \\triangle ABC$, a point $P(x,y,z),\\ x,y,z\\ge o,\\ x+y+z=1$ (barycentric coordinates)\r\n\r\nand the points $M\\in (AB),\\ N\\in (AC)$ such that $\\ P\\in MN$. Prove that:\r\n\r\n\\[\\overline{\\underline{\\left| 4yz\\cdot S\\le [MAN]\\le \\frac{max\\ \\{ y,z \\} }{x+max\\ \\{ y,z \\} }\\cdot S\\right|}}\\ \\ \\ (*)\\]\r\n [b]Remarks.[/b] The point ${\\ P\\in [ABC]}$, because $x,y,z\\ge 0$. In the relation (*) we have equalities iff:\r\n\r\n$\\ \\nwarrow \\frac{AM}{yc}=\\frac{AN}{zb}=2;\\ \\nearrow \\ M=B\\ (for\\ y\\le z)\\ or\\ N=C\\ (for\\ y>z)$.\r\n\r\n[u][i]Some particular cases:[/i][/u]\r\n\r\n1. $\\ P:=G\\Longrightarrow |||\\nwarrow MN\\parallel BC\\ \\nearrow M=B\\vee N=C.$\r\n2. $\\ P:=I\\Longrightarrow |||\\nwarrow AM=AN=\\frac{bc}{s}\\ \\nearrow M=B\\vee N=C$. (For $\\ \\nwarrow $, the points M,N belong to the circle which is tangent to the lines AB,AC and to the circumcircle)." } { "Tag": [ "inequalities", "function", "inequalities proposed" ], "Problem": "Prove that for any three positive real numbers $ a,b,c, \\frac{1}{a\\plus{}b}\\plus{}\\frac{1}{b\\plus{}c}\\plus{}\\frac{1}{c\\plus{}a} \\ge \\frac{9}{2} \\cdot \\frac{1}{a\\plus{}b\\plus{}c}$.", "Solution_1": "$ a\\plus{}b\\equal{}x$ , $ b\\plus{}c\\equal{}y$ ,$ c\\plus{}a\\equal{}z$ after cauchy-schwarz inequality", "Solution_2": "We know that ,for all $ x,y,z,m,n,p$ positive reals it's true that: \r\n$ \\frac{x^2}{m}\\plus{}\\frac{y^2}{n}\\plus{}\\frac{z^2}{p}\\ge\\frac{(x\\plus{}y\\plus{}z)^2}{m\\plus{}n\\plus{}p}$\r\nwith this inequality : $ \\frac{1}{a\\plus{}b}\\plus{}\\frac{1}{b\\plus{}c}\\plus{}\\frac{1}{c\\plus{}a}\\ge\\frac{(1\\plus{}1\\plus{}1)^2}{(a\\plus{}b)\\plus{}(b\\plus{}c)\\plus{}(c\\plus{}a)}\\equal{}\\frac{9}{2} \\frac{1}{a\\plus{}b\\plus{}c}$", "Solution_3": "We can also assume $ a\\plus{}b\\plus{}c\\equal{}1$ since the inequality is homogeneous. Now the function $ f(x)\\equal{}\\frac{1}{1\\minus{}x}$ is convex and by Jensen's inequality it is done.", "Solution_4": "that's how to derive the overrated Nesbitt's Inequality :lol:" } { "Tag": [ "geometry", "geometric transformation", "reflection" ], "Problem": "A disabled tanker leaks kerosene (n=1.2) into the Persian Gulf, creating a large slick on the top of water (n=1.3). If you are scuba diving directly under the region of the slick, whose thickness is 460nm, while the Sun is overhead, for which wavelength(s) of visible light is the transmitted intensity strongest?", "Solution_1": "i'll probably run out of oxigen waiting for a interference pattern to show.but i decide to take a helicopter and of course since i am in persian gulf i'll stay low.but wait...something is happening i see some colours from the poluted sea...the kerosene must be it so i try to explain myself what's going on.i know the path difference is 2*n1*d+lambda/2 and the intensity is stronghest if the path is k*lambda.since the wavelength of the visible light ranges from 400 to 700 nanometers then i have to solve this system:\r\n2*n1*d=(k-1\\2)/lambda and 400 nm2$ and $y>2$, compute the minimum value of $2x+y$.", "Solution_1": "[hide=\"Solution\"] $\\frac{1}{2}(2x+y) \\ge \\left( \\frac{1}{x}+\\frac{1}{y}\\right) (2x+y) \\ge \\left( \\sqrt{2}+1 \\right)^{2}= 3+2 \\sqrt{2}\\implies 2x+y \\ge \\boxed{6+4 \\sqrt{2}}$\n\nThis minimum is obtained when $x = 2+\\sqrt{2}, y = 2 \\sqrt{2}+2$. [/hide]" } { "Tag": [ "trigonometry" ], "Problem": "An incline of weight $\\ M$ moves rests on a frictionless surface. On the incline is a block of mass $\\ m$ and there is a coefficient of friction $\\ \\mu$ between the incline and the block. A horizontal force of magnitude $\\ F$ acts horizontally on the incline. Find the minimum and maximum values of $\\ F$ such that the block does not slip.", "Solution_1": "This is an interesting problem, and it's too bad it's been overlooked. But could you please clarify wheather the incline rests or not (it's much easier if it rests), and is the angle of inclination given?", "Solution_2": "The angle is given to be $\\theta$. I don't know what you mean by does the incline rest. It says in the problem that the incline is moving.", "Solution_3": "Consider in two dimensional view,-a right triangle ABC with right angle at B.The base is BC and the hypotenuse is BC. Then as per your problem , the mass $m$ is placed on the hypotenuse AC.Assume that the force is applied on face AC.Now the force would required would be maximum if the direction of $F$ has the effect of PUSHing side AC and the force would be minimmum if $F$ has the effect of PULLing AC. \r\n\r\nLet $N$ be the normal reaction force between the mass $m$ and $M$\r\nand let a be the common horizontal acceleration of the masses.\r\nThe equations of motion are-\r\n\r\n$F+{\\mu}N{\\cos{\\theta}}-N{\\sin{\\theta}}= Ma$(for minimum force) \r\n $F-{\\mu}N{\\cos{\\theta}}+N{\\sin{\\theta}}= Ma$ 9for force to be maximum \r\n\r\nThe other two equations aarre common for both \r\n$N{\\cos{\\theta}}+{\\mu}N{\\sin{\\theta}}= mg$[for vertical equilibrium since $M$ and $m$ both should have equal vertical accelerations and the vertical acceleration of M is zero obviously]\r\n\r\n$N{\\sin{\\theta}}-{\\mu}N{\\cos{\\theta}}= ma$[M and m should have equal horizontal accelerations a] \r\n\r\nSolving these three equations we get the minimum and maximum values of the required force as - \r\n$F_{min}$ $=$ $(M-m)g$ $\\frac{1-{\\mu}cot{\\theta}}{1+{\\mu}tan{\\theta}}$ and \r\n$F_{max}$ $=$ $(M+m)g$ $\\frac{1-{\\mu}cot{\\theta}}{1+{\\mu}tan{\\theta}}$" } { "Tag": [ "geometry", "circumcircle", "analytic geometry", "Euler", "graphing lines", "slope", "geometry proposed" ], "Problem": "Find the coordinates of the center of the circumcenter of the triangle with vertices at (0,0), (a,0) and (b,c) .", "Solution_1": "If the coordinates are $ \\left(x_{0},y_{0}\\right)$, it is clear that $ x_{0}\\equal{}\\frac{a}{2}$. Then, $ R^{2}\\equal{}\\frac{a^{2}}{4}\\plus{}y_{0}^{2}\\equal{}\\left(\\frac{a}{2}\\minus{}b\\right)^{2}\\plus{}\\left(y_{0}\\minus{}c\\right)^{2}$, or $ y_{0}\\equal{}\\frac{b^{2}\\plus{}c^{2}\\minus{}ab}{2c}$, and we are done...", "Solution_2": "I've just worked on this problem which asked to show the different steps:\r\n\r\nEuler was correct that the centroid, orthocenter, and circumcenter are collinear. \r\n\r\nFollow these steps :\r\n\r\nDraw a \u201cgeneric\u201d triangle with vertices at (0,0), (a,0) and (b,c). \r\n\r\nFind the equations of any two medians. Each will be in terms of a, b, and c. \r\n\r\nNow solve those two equations simultaneously to find the centroid. \r\n\r\nIn similar fashion, find the orthocenter and circumcenter. \r\n\r\nYour results will again be in terms of a, b, and c. \r\n\r\nNow show that the slopes from the centroid to each of the other centers are equal. \r\n\r\nThere, you\u2019ve proved that the three centers are collinear! \r\n\r\nBonus :\r\n\r\ntry to prove that the orthocenter lies twice as far from the centroid as the circumcenter does.", "Solution_3": "[color=darkblue]B.t.w. $ \\frac{\\overline{GH}}{\\overline{GO}}\\equal{}\\frac{\\overline{GN}}{\\overline{GI}}\\equal{}\\minus{}2$ (where $ N$ is the [b]Nagel's point[/b]) for [b]Leo Yard[/b] a little theory :\n\n$ \\left\\{\\begin{array}{c}A(x_{1},y_{1})\\ ,\\ B(x_{2},y_{2})\\\\\\\\ d\\ :\\ \\Delta (x,y)\\equiv ax\\plus{}by\\plus{}c\\equal{}0\\\\\\\\ A\\not\\in d\\ ,\\ B\\not\\in d\\ ,\\ AB\\not\\parallel d\\\\\\\\ M\\in AB\\cap d\\end{array}\\right\\|$ $ \\implies$ $ \\frac{\\overline{MA}}{\\overline{MB}}\\equal{}\\frac{\\Delta (x_{1},y_{1})}{\\Delta (x_{2},y_{2})}\\equal{}\\frac{ax_{1}\\plus{}by_{1}\\plus{}c}{ax_{2}\\plus{}by_{2}\\plus{}c}$.[/color]" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME" ], "Problem": "Ah I want to calm down a little but I can't, I can't concentrate on homework or do anything else.. help!!!!", "Solution_1": "Digging up time capsules seems to work...", "Solution_2": "Knock yourself out.", "Solution_3": "breathe deeply\r\n\r\ntake a nap, i took a nap but when i woke up... still no call :( not a good feeling..", "Solution_4": "hahah, i didn't even make USAMO. i'm so lucky. wait...no i'm not.", "Solution_5": "taking a nap is a good idea...convincing yourself that you have a relatively important physics test tomorrow that has been neglected so far is also a good idea.", "Solution_6": "Play me a game of chess. :P\r\n\r\nI need my points back.", "Solution_7": "I made MOP in 2002 and 2003 but last year I just had a total meltdown and didn't make it. Believe me when I say that I was stunned at how poorly I had done (I knew I'd done poorly, but it was just stunning)\r\n\r\nThis year I'm a Winner.\r\n\r\nSo if you didn't make MOP its not the end of the world, regardless of your expectations.", "Solution_8": "[quote=\"beta\"]Ah I want to calm down a little but I can't, I can't concentrate on homework or do anything else.. help!!!![/quote]\r\n\r\nhope homework is all you have, i've got a test tomorrow. . .", "Solution_9": "i have 3 tests and a journal to do... ew..", "Solution_10": "Write your own obituary. :D \r\n\r\nSeriously though, I am writing my own obituary. But only because that's my english writing assignment due Thursday. :)", "Solution_11": "Still no call for me!! While a lot of the freshmen I know already got it.. sigh it's over :(", "Solution_12": "haha i didn't do any homework monday or tuesday, then missed half my classes, so i'm still behind now.\r\n\r\nif any of you get \"f\"s, file stress as an illness and see if you get excused :D", "Solution_13": "I feel your pain, Beta...", "Solution_14": "Wait did any 9th grader on the East Coast got the call?", "Solution_15": "*Hopefully* the scores list is soon to come.\r\n\r\nBy the way I want to thank the MOP/AMC people for all of their hard work. In all the confusion of today and yesterday, it seems we have forgotten all the important effort they put in to get these graded and then decide who to invite, so thanks very much for all of that.\r\n\r\nJB", "Solution_16": "Ditto to that. I went a little insane over the last few days. Let that insanity be a compliment to the AMC! :D", "Solution_17": "[quote=\"jackb1115\"]*Hopefully* the scores list is soon to come.[/quote]\r\n\r\n\r\nhttp://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/05usamo-results.html", "Solution_18": "so can you explain the meaning of this \"hidden link\"? or are all of us to be left hanging on the edge of our seat? (or maybe i should know already sorry)", "Solution_19": "It's not a hidden link, you just need to refresh the page with your browser.", "Solution_20": "[quote] It's not a hidden link, you just need to refresh the page with your browser.\n\n[/quote]\r\n\r\nHaha", "Solution_21": "Wow. Average is 13.74!!?", "Solution_22": "Wow, I'm like way below average. :(", "Solution_23": "I feel like puking.", "Solution_24": "Yeah, me too. I totally overestimated... well not by that much, but by a fair amount.", "Solution_25": "Yea... actually i did overestimate too... but o well.", "Solution_26": "Do I want to ask?", "Solution_27": "No. You don't.", "Solution_28": "Since I'm sure you guys are wondering, (from an unofficial but reliable source)\r\n\r\nBrian Lawrence, 10th grader, got the 41.\r\n\r\nEric Price with 35\r\n\r\nI got a 30 btw, so I guess they liked my stuff for the LHS of #6 better than I did.", "Solution_29": "Oh well, I overestimated as well. I think we all do." } { "Tag": [ "calculus", "integration", "floor function" ], "Problem": "There is an expression\r\n\r\n\r\n1000!/10 to the power n (only 10 the the power n)\r\n\r\nwhat is the value of n so that the expression has the least integral value.???\r\n\r\nDo solve it", "Solution_1": "In other words, how many zeros are at the end of $1000!$?", "Solution_2": "[quote=\"t0rajir0u\"]In other words, how many zeros are at the end of $1000!$?[/quote]\r\n\r\nThere are 200 multiples of 5 from 1 to 1000, among these multiples, $25=5^{2}$, $125=5^{3}$, $625=5^{4}$, therefore $1000!$ is divisible by $5^{206}$, hence there are 206 zeros at the end of $1000!$, n = 206.", "Solution_3": "[quote=\"schoolgirl\"][quote=\"t0rajir0u\"]In other words, how many zeros are at the end of $1000!$?[/quote]\n\nThere are 200 multiples of 5 from 1 to 1000, among these multiples, $25=5^{2}$, $125=5^{3}$, $625=5^{4}$, therefore $1000!$ is divisible by $5^{206}$, hence there are 206 zeros at the end of $1000!$, n = 206.[/quote]\r\n\r\nYou undercounted. Look at multiples.", "Solution_4": "isnt it\r\n\r\n$\\text{Total}=\\lfloor\\frac{1000}{5}\\rfloor+\\lfloor\\frac{1000}{25}\\rfloor+\\lfloor\\frac{1000}{125}\\rfloor+\\lfloor\\frac{1000}{625}\\rfloor=200+40+8+1=\\boxed{249}$", "Solution_5": "ya that looks correct", "Solution_6": "[quote=\"SM4RT\"]isnt it\n\n$\\text{Total}=\\lfloor\\frac{1000}{5}\\rfloor+\\lfloor\\frac{1000}{25}\\rfloor+\\lfloor\\frac{1000}{125}\\rfloor+\\lfloor\\frac{1000}{625}\\rfloor=200+40+8+1=\\boxed{249}$[/quote]\r\n\r\nThis is also what I got." } { "Tag": [ "probability", "factorial", "geometry", "3D geometry", "logarithms", "puzzles" ], "Problem": "How can you make 24 with 10,16,9 and 4 and JUST division, subtraction, multiplication and/or addition?", "Solution_1": "9*16/(10-4)\r\n\r\nHow about try:\r\n\r\n10, 10, 4, 4\r\n8, 8, 3, 3\r\n1, 3, 4, 6\r\n\r\n:D I also have a program that can generate these if you want more.", "Solution_2": "Hehe nice. I got the last two first, since I'm well acquainted with 1,3,4,6. Spent two weeks at camp trying to figure it out freshmen year. Heck, it's even on my shirt. And knowing that helped with 3,3,8,8. Beta told me 4,4,10,10. ^_^\r\n\r\nAre there multiple ways to do any one of them?", "Solution_3": "When my cousin and I are on a long car ride and one of us has cards handy, we always spend the whole time dealing random combinations and trying to make them equal to 24. Once I argued with her for so long over whether or not we should be allowed to use exponentiation...", "Solution_4": "Hehe, a lot of people at our school actually play 24 with cards. Keep going until loser runs out. That's usually me. :oops:\r\n\r\nI always get 1,3,4,6 though! :)", "Solution_5": "How about my personal favorite, 7, 7, 3, 3?", "Solution_6": "I actually made a list that shows all the solvable cases.\r\n\r\nI also like this one\r\n1, 4, 5, 6\r\nThere are two solutions.", "Solution_7": "One is \"obvious\" from 1 3 4 6 :D", "Solution_8": "[quote=\"lightrhee\"]I actually made a list that shows all the solvable cases.\n[/quote]\r\nSome people might be interested to look at this list(even though it might be wrong, I'm quite sure it is accurate).\r\nHere is my list of non-solvable cases of 24.\r\n 1, 1, 1, 1\r\n 1, 1, 1, 2\r\n 1, 1, 1, 3\r\n 1, 1, 1, 4\r\n 1, 1, 1, 5\r\n 1, 1, 1, 6\r\n 1, 1, 1, 7\r\n 1, 1, 1, 9\r\n 1, 1, 1, 10\r\n 1, 1, 2, 2\r\n 1, 1, 2, 3\r\n 1, 1, 2, 4\r\n 1, 1, 2, 5\r\n 1, 1, 3, 3\r\n 1, 1, 4, 11\r\n 1, 1, 4, 13\r\n 1, 1, 5, 9\r\n 1, 1, 5, 10\r\n 1, 1, 5, 11\r\n 1, 1, 5, 12\r\n 1, 1, 5, 13\r\n 1, 1, 6, 7\r\n 1, 1, 6, 10\r\n 1, 1, 6, 11\r\n 1, 1, 6, 13\r\n 1, 1, 7, 7\r\n 1, 1, 7, 8\r\n 1, 1, 7, 9\r\n 1, 1, 7, 11\r\n 1, 1, 7, 12\r\n 1, 1, 7, 13\r\n 1, 1, 8, 9\r\n 1, 1, 8, 10\r\n 1, 1, 8, 11\r\n 1, 1, 8, 12\r\n 1, 1, 8, 13\r\n 1, 1, 9, 9\r\n 1, 1, 9, 10\r\n 1, 1, 9, 11\r\n 1, 1, 9, 12\r\n 1, 1, 10, 10\r\n 1, 1, 10, 11\r\n 1, 2, 2, 2\r\n 1, 2, 2, 3\r\n 1, 2, 5, 11\r\n 1, 2, 7, 13\r\n 1, 2, 8, 11\r\n 1, 2, 8, 12\r\n 1, 2, 9, 9\r\n 1, 2, 9, 10\r\n 1, 2, 10, 10\r\n 1, 3, 3, 13\r\n 1, 3, 5, 5\r\n 1, 3, 7, 11\r\n 1, 3, 10, 13\r\n 1, 3, 11, 13\r\n 1, 4, 4, 13\r\n 1, 4, 7, 10\r\n 1, 4, 8, 10\r\n 1, 4, 9, 9\r\n 1, 4, 10, 13\r\n 1, 4, 11, 11\r\n 1, 4, 11, 12\r\n 1, 4, 11, 13\r\n 1, 4, 12, 13\r\n 1, 4, 13, 13\r\n 1, 5, 5, 7\r\n 1, 5, 5, 8\r\n 1, 5, 7, 7\r\n 1, 5, 11, 13\r\n 1, 5, 12, 13\r\n 1, 5, 13, 13\r\n 1, 6, 6, 7\r\n 1, 6, 7, 7\r\n 1, 6, 7, 8\r\n 1, 6, 7, 13\r\n 1, 6, 9, 11\r\n 1, 6, 10, 10\r\n 1, 6, 10, 11\r\n 1, 6, 11, 11\r\n 1, 6, 13, 13\r\n 1, 7, 7, 7\r\n 1, 7, 7, 8\r\n 1, 7, 7, 13\r\n 1, 7, 8, 13\r\n 1, 7, 10, 10\r\n 1, 7, 10, 11\r\n 1, 7, 11, 11\r\n 1, 7, 11, 12\r\n 1, 7, 11, 13\r\n 1, 8, 8, 13\r\n 1, 8, 9, 9\r\n 1, 8, 9, 10\r\n 1, 8, 10, 10\r\n 1, 8, 11, 11\r\n 1, 8, 12, 13\r\n 1, 8, 13, 13\r\n 1, 9, 9, 9\r\n 1, 9, 9, 10\r\n 1, 9, 9, 11\r\n 1, 9, 9, 13\r\n 1, 9, 10, 10\r\n 1, 9, 10, 11\r\n 1, 9, 12, 13\r\n 1, 9, 13, 13\r\n 1, 10, 10, 10\r\n 1, 10, 10, 11\r\n 1, 10, 10, 13\r\n 1, 10, 11, 11\r\n 1, 10, 11, 13\r\n 1, 10, 13, 13\r\n 1, 11, 11, 11\r\n 1, 13, 13, 13\r\n 2, 2, 2, 2\r\n 2, 2, 2, 6\r\n 2, 2, 5, 13\r\n 2, 2, 7, 9\r\n 2, 2, 7, 11\r\n 2, 2, 8, 11\r\n 2, 2, 8, 13\r\n 2, 2, 9, 9\r\n 2, 2, 9, 13\r\n 2, 2, 10, 12\r\n 2, 3, 3, 4\r\n 2, 3, 9, 11\r\n 2, 3, 10, 11\r\n 2, 4, 7, 13\r\n 2, 4, 9, 11\r\n 2, 4, 11, 13\r\n 2, 4, 12, 13\r\n 2, 5, 5, 5\r\n 2, 5, 5, 6\r\n 2, 5, 7, 12\r\n 2, 5, 9, 9\r\n 2, 5, 9, 13\r\n 2, 5, 11, 11\r\n 2, 5, 11, 13\r\n 2, 5, 13, 13\r\n 2, 6, 7, 7\r\n 2, 6, 9, 13\r\n 2, 6, 11, 11\r\n 2, 6, 13, 13\r\n 2, 7, 7, 7\r\n 2, 7, 7, 9\r\n 2, 7, 8, 10\r\n 2, 7, 9, 9\r\n 2, 7, 9, 12\r\n 2, 7, 10, 13\r\n 2, 7, 11, 11\r\n 2, 7, 11, 13\r\n 2, 7, 13, 13\r\n 2, 8, 11, 13\r\n 2, 9, 9, 9\r\n 2, 9, 9, 10\r\n 2, 9, 11, 12\r\n 2, 9, 12, 12\r\n 2, 10, 10, 10\r\n 2, 10, 12, 12\r\n 2, 10, 13, 13\r\n 3, 3, 3, 13\r\n 3, 3, 4, 10\r\n 3, 3, 5, 8\r\n 3, 3, 5, 11\r\n 3, 3, 7, 10\r\n 3, 3, 8, 11\r\n 3, 3, 10, 10\r\n 3, 3, 10, 11\r\n 3, 3, 10, 12\r\n 3, 3, 11, 11\r\n 3, 3, 13, 13\r\n 3, 4, 6, 7\r\n 3, 4, 7, 13\r\n 3, 4, 8, 8\r\n 3, 4, 9, 10\r\n 3, 4, 10, 11\r\n 3, 4, 11, 11\r\n 3, 4, 13, 13\r\n 3, 5, 5, 5\r\n 3, 5, 5, 10\r\n 3, 5, 5, 13\r\n 3, 5, 7, 7\r\n 3, 5, 8, 10\r\n 3, 5, 9, 11\r\n 3, 5, 11, 13\r\n 3, 6, 7, 11\r\n 3, 6, 8, 11\r\n 3, 6, 10, 13\r\n 3, 7, 7, 11\r\n 3, 7, 8, 10\r\n 3, 7, 10, 12\r\n 3, 7, 11, 13\r\n 3, 8, 8, 13\r\n 3, 8, 10, 13\r\n 3, 8, 11, 13\r\n 3, 10, 10, 10\r\n 3, 10, 10, 11\r\n 3, 10, 10, 13\r\n 3, 10, 11, 11\r\n 3, 10, 12, 12\r\n 3, 10, 12, 13\r\n 3, 10, 13, 13\r\n 3, 11, 11, 11\r\n 3, 11, 11, 13\r\n 3, 11, 12, 13\r\n 3, 11, 13, 13\r\n 3, 13, 13, 13\r\n 4, 4, 4, 13\r\n 4, 4, 5, 9\r\n 4, 4, 6, 6\r\n 4, 4, 6, 7\r\n 4, 4, 7, 11\r\n 4, 4, 9, 9\r\n 4, 4, 9, 10\r\n 4, 4, 9, 13\r\n 4, 4, 10, 11\r\n 4, 4, 11, 11\r\n 4, 4, 13, 13\r\n 4, 5, 5, 11\r\n 4, 5, 5, 12\r\n 4, 5, 5, 13\r\n 4, 5, 9, 11\r\n 4, 6, 6, 11\r\n 4, 6, 6, 13\r\n 4, 6, 7, 11\r\n 4, 6, 7, 13\r\n 4, 6, 8, 11\r\n 4, 6, 9, 11\r\n 4, 6, 10, 13\r\n 4, 6, 11, 13\r\n 4, 7, 7, 9\r\n 4, 7, 7, 10\r\n 4, 7, 7, 12\r\n 4, 7, 7, 13\r\n 4, 7, 10, 13\r\n 4, 8, 10, 13\r\n 4, 9, 9, 9\r\n 4, 9, 9, 11\r\n 4, 9, 9, 13\r\n 4, 9, 10, 10\r\n 4, 9, 11, 13\r\n 4, 9, 12, 13\r\n 4, 9, 13, 13\r\n 4, 10, 10, 10\r\n 4, 10, 10, 13\r\n 4, 10, 11, 11\r\n 4, 10, 13, 13\r\n 4, 11, 11, 11\r\n 4, 11, 11, 12\r\n 4, 11, 11, 13\r\n 4, 11, 12, 12\r\n 4, 11, 13, 13\r\n 4, 12, 12, 13\r\n 4, 12, 13, 13\r\n 4, 13, 13, 13\r\n 5, 5, 5, 7\r\n 5, 5, 5, 8\r\n 5, 5, 5, 10\r\n 5, 5, 5, 11\r\n 5, 5, 5, 13\r\n 5, 5, 6, 9\r\n 5, 5, 6, 10\r\n 5, 5, 6, 12\r\n 5, 5, 6, 13\r\n 5, 5, 7, 9\r\n 5, 5, 7, 12\r\n 5, 5, 7, 13\r\n 5, 5, 9, 12\r\n 5, 5, 9, 13\r\n 5, 5, 10, 12\r\n 5, 6, 6, 11\r\n 5, 6, 6, 13\r\n 5, 6, 7, 10\r\n 5, 6, 7, 11\r\n 5, 6, 8, 11\r\n 5, 7, 7, 7\r\n 5, 7, 7, 8\r\n 5, 7, 7, 12\r\n 5, 7, 7, 13\r\n 5, 7, 8, 11\r\n 5, 7, 8, 12\r\n 5, 7, 8, 13\r\n 5, 7, 9, 9\r\n 5, 7, 11, 12\r\n 5, 7, 12, 13\r\n 5, 8, 8, 11\r\n 5, 8, 8, 12\r\n 5, 8, 9, 9\r\n 5, 8, 9, 10\r\n 5, 8, 10, 10\r\n 5, 8, 10, 13\r\n 5, 8, 11, 11\r\n 5, 8, 12, 13\r\n 5, 8, 13, 13\r\n 5, 9, 9, 9\r\n 5, 9, 9, 10\r\n 5, 9, 9, 13\r\n 5, 9, 10, 12\r\n 5, 9, 11, 11\r\n 5, 9, 11, 12\r\n 5, 9, 13, 13\r\n 5, 10, 10, 10\r\n 5, 10, 11, 12\r\n 5, 10, 11, 13\r\n 5, 10, 12, 12\r\n 5, 11, 11, 11\r\n 5, 11, 11, 12\r\n 5, 11, 11, 13\r\n 5, 11, 12, 13\r\n 5, 11, 13, 13\r\n 5, 12, 12, 12\r\n 5, 12, 12, 13\r\n 5, 12, 13, 13\r\n 5, 13, 13, 13\r\n 6, 6, 6, 7\r\n 6, 6, 6, 13\r\n 6, 6, 7, 7\r\n 6, 6, 7, 8\r\n 6, 6, 7, 13\r\n 6, 6, 9, 9\r\n 6, 6, 10, 10\r\n 6, 6, 10, 11\r\n 6, 6, 11, 11\r\n 6, 6, 13, 13\r\n 6, 7, 7, 7\r\n 6, 7, 7, 8\r\n 6, 7, 7, 9\r\n 6, 7, 7, 12\r\n 6, 7, 7, 13\r\n 6, 7, 8, 8\r\n 6, 7, 8, 13\r\n 6, 7, 9, 10\r\n 6, 7, 9, 11\r\n 6, 7, 9, 13\r\n 6, 7, 10, 11\r\n 6, 7, 13, 13\r\n 6, 8, 8, 13\r\n 6, 8, 10, 10\r\n 6, 8, 12, 13\r\n 6, 9, 9, 9\r\n 6, 9, 9, 13\r\n 6, 9, 10, 10\r\n 6, 9, 10, 13\r\n 6, 9, 11, 11\r\n 6, 9, 13, 13\r\n 6, 10, 10, 11\r\n 6, 10, 10, 12\r\n 6, 10, 11, 11\r\n 6, 10, 11, 13\r\n 6, 10, 13, 13\r\n 6, 11, 11, 11\r\n 6, 11, 11, 13\r\n 6, 11, 13, 13\r\n 6, 13, 13, 13\r\n 7, 7, 7, 7\r\n 7, 7, 7, 8\r\n 7, 7, 7, 9\r\n 7, 7, 7, 10\r\n 7, 7, 7, 11\r\n 7, 7, 7, 13\r\n 7, 7, 8, 8\r\n 7, 7, 8, 9\r\n 7, 7, 8, 10\r\n 7, 7, 8, 12\r\n 7, 7, 8, 13\r\n 7, 7, 9, 9\r\n 7, 7, 9, 11\r\n 7, 7, 9, 12\r\n 7, 7, 9, 13\r\n 7, 7, 10, 10\r\n 7, 7, 10, 11\r\n 7, 7, 10, 12\r\n 7, 7, 11, 11\r\n 7, 7, 13, 13\r\n 7, 8, 8, 8\r\n 7, 8, 9, 9\r\n 7, 8, 9, 11\r\n 7, 8, 10, 12\r\n 7, 8, 11, 11\r\n 7, 8, 13, 13\r\n 7, 9, 9, 9\r\n 7, 9, 9, 10\r\n 7, 9, 9, 11\r\n 7, 9, 9, 12\r\n 7, 9, 10, 10\r\n 7, 9, 10, 13\r\n 7, 9, 11, 13\r\n 7, 9, 12, 13\r\n 7, 10, 10, 10\r\n 7, 10, 10, 13\r\n 7, 10, 11, 11\r\n 7, 10, 11, 12\r\n 7, 10, 13, 13\r\n 7, 11, 11, 11\r\n 7, 11, 11, 12\r\n 7, 11, 11, 13\r\n 7, 11, 12, 12\r\n 7, 11, 12, 13\r\n 7, 11, 13, 13\r\n 7, 12, 12, 12\r\n 7, 12, 13, 13\r\n 7, 13, 13, 13\r\n 8, 8, 8, 8\r\n 8, 8, 8, 9\r\n 8, 8, 9, 9\r\n 8, 8, 9, 10\r\n 8, 8, 10, 10\r\n 8, 8, 10, 11\r\n 8, 8, 11, 11\r\n 8, 8, 13, 13\r\n 8, 9, 9, 9\r\n 8, 9, 9, 10\r\n 8, 9, 9, 11\r\n 8, 9, 9, 13\r\n 8, 9, 10, 10\r\n 8, 9, 10, 11\r\n 8, 9, 13, 13\r\n 8, 10, 10, 10\r\n 8, 10, 10, 11\r\n 8, 10, 10, 13\r\n 8, 10, 11, 12\r\n 8, 10, 11, 13\r\n 8, 11, 11, 11\r\n 8, 11, 11, 12\r\n 8, 11, 11, 13\r\n 8, 11, 12, 13\r\n 8, 11, 13, 13\r\n 8, 12, 12, 12\r\n 8, 12, 12, 13\r\n 8, 12, 13, 13\r\n 8, 13, 13, 13\r\n 9, 9, 9, 9\r\n 9, 9, 9, 10\r\n 9, 9, 9, 11\r\n 9, 9, 9, 13\r\n 9, 9, 10, 10\r\n 9, 9, 10, 11\r\n 9, 9, 10, 12\r\n 9, 9, 11, 11\r\n 9, 9, 13, 13\r\n 9, 10, 10, 10\r\n 9, 10, 10, 11\r\n 9, 10, 10, 12\r\n 9, 10, 11, 11\r\n 9, 10, 13, 13\r\n 9, 11, 11, 12\r\n 9, 11, 11, 13\r\n 9, 12, 12, 13\r\n 9, 12, 13, 13\r\n 9, 13, 13, 13\r\n 10, 10, 10, 10\r\n 10, 10, 10, 11\r\n 10, 10, 11, 11\r\n 10, 10, 13, 13\r\n 10, 11, 11, 11\r\n 10, 11, 13, 13\r\n 11, 11, 11, 11\r\n 11, 11, 13, 13\r\n 13, 13, 13, 13", "Solution_9": "have alotta time on your hands lol?? :P \r\n\r\nirt JBL's 3,3,7,7\r\n[hide]\n(3+3/7)*7\nyea that ones pretty cool \n[/hide]", "Solution_10": "Hello! lightrhee\r\nDo you know how many cases are solvable?\r\nIf possible, would you post them here also?", "Solution_11": "Well, every case that he didn't list would be solvable...", "Solution_12": "Out of $_{16}C_4$=1820, 458 cases are not solvable. However, probability of non-solvable case to occur is not $\\frac{458}{1820}$. Some cases are \"heavier\" then other cases.", "Solution_13": "Try 1, 4, 5, 6", "Solution_14": "Hey! I'm back from Waterloo Seminar.\r\nSome people wanted to see my program for generating the list. My code in Maple is shown below. If anyone finds a bug or a way to improve, please let me know(since I'm a horrible programmer).\r\n\r\n[color=red]> restart;\n> cmake_24:=proc(am,bm,cm,dm)\n> check:=false:\n> fn:=[am,bm,cm,dm]:\n> fn:=ListTools[Reverse](sort(fn)):\n> for a from 1 by 1 to 3 do\n> for b from a+1 by 1 to 4 do\n> tn:=fn: tn[a]:=0: tn[b]:=0: tn:=ListTools[Reverse](sort(tn)): x:=fn[a]: y:=fn[b]: oo:=[x+y,x-y,x*y,x/y,y/x]:\n> for c from 1 by 1 to 5 do\n> tn[3]:=oo[c]:\n> for d from 1 by 1 to 2 do\n> for e from d+1 by 1 to 3 do\n> sn:=tn: sn[d]:=0: sn[e]:=0: sn:=ListTools[Reverse](sort(sn)): x:=tn[d]: y:=tn[e]: os:=[x+y,x-y,y-x,x*y,0,0]:\n> if y<>0 then os[5]:=x/y end if:\n> if x<>0 then os[6]:=y/x end if:\n> for f from 1 by 1 to 6 do\n> if os[f]>0 then\n> sn[2]:=os[f]: x:=sn[1]: y:=sn[2]: ot:=[x+y,x-y,y-x,x*y,0,0]:\n> if y<>0 then ot[5]:=x/y end if:\n> if x<>0 then ot[6]:=y/x end if:\n> for g from 1 by 1 to 6 do\n> if ot[g]=24 then check:=true end if:\n> end do: end if: end do: end do: end do: end do: end do: end do:\n> check\n> end proc: \n\n> count:=0:\n> for x from 1 by 1 to 13 do\n> for y from x by 1 to 13 do\n> for z from y by 1 to 13 do\n> for w from z by 1 to 13 do\n> if cmake_24(x,y,z,w)=false then\n> print(x,y,z,w): count:=count+1\n> end if:\n> end do:\n> end do:\n> end do:\n> end do:[/color]", "Solution_15": "I'm assuming we're only using *,/,-. + because with stuff like ! things like 1,1,1,1 become possible as (1+1+1+1)! = 24, see?", "Solution_16": "[quote=\"8th Army\"]Try 1, 4, 5, 6[/quote]\r\n\r\n6/(5/4-1)", "Solution_17": "[quote=\"opple\"][quote=\"8th Army\"]Try 1, 4, 5, 6[/quote]\n\n6/(5/4-1)[/quote]\r\nThere is one more solution. Can you find it? :D", "Solution_18": "Well it's essentially the same but...\r\n4/(1-5/6)\r\n\r\nHere are some easy ones :D ...\r\n6, 6, 6, 6\r\n5, 5, 5, 5\r\n4, 4, 4, 4\r\n3, 3, 3, 3\r\n12, 12, 12, 12\r\n\r\nCHALLENGE PROBLEM :D \r\n1, 1, 1, 24", "Solution_19": "[quote=\"K81o7\"]CHALLENGE PROBLEM :D \n1, 1, 1, 24[/quote]\r\n1*1*1*24?", "Solution_20": "[quote=\"K81o7\"]Here are some easy ones :D ...\n[/quote]\r\n[hide]6+6+6+6\n5*5 - 5/5\n4*4+4+4\n3*3*3 - 3\n(12+12)*12/12[/hide]", "Solution_21": "lightrhee: The answer to 1,4,5,6 that I saw is 4/(1-5/6)=24 (which is similar to what opple said). Is that what you were thinking of?\r\n\r\n[hide=\"I remember 24...\"]\nI remember playing the 24 game. I always did the \"Platinum\" cards (Decimals, Fractions, Exponents, Algebra... though they changed Decimals to Integers recently). To win (or get anywhere) in the tournaments, you had to memorize every card to the point where you could quickly glance at four numbers and immediately spit out the final solution. That took a lot of the fun out of it, but at least the cards they have at the final rounds required a little more than pure memorization.\n\nI always liked the four cards they had on the sides of the box. As I recall, some of them required really weird solutions.\n\nSomebody should find the probability of \"eight times three\" being a solution to a randomly selected Single Digit card. I proctored that one year and all I heard was \"Eight times three... Eight times three... Eight times three\" :P[/hide]", "Solution_22": "[quote=\"no_one_2000\"]lightrhee: The answer to 1,4,5,6 that I saw is 4/(1-5/6)=24 (which is similar to what opple said). Is that what you were thinking of?\n\n[hide=\"I remember 24...\"]\nI remember playing the 24 game. I always did the \"Platinum\" cards (Decimals, Fractions, Exponents, Algebra... though they changed Decimals to Integers recently). To win (or get anywhere) in the tournaments, you had to memorize every card to the point where you could quickly glance at four numbers and immediately spit out the final solution. That took a lot of the fun out of it, but at least the cards they have at the final rounds required a little more than pure memorization.\n\nI always liked the four cards they had on the sides of the box. As I recall, some of them required really weird solutions.\n\nSomebody should find the probability of \"eight times three\" being a solution to a randomly selected Single Digit card. I proctored that one year and all I heard was \"Eight times three... Eight times three... Eight times three\" :P[/hide][/quote]\r\nFor 1, 4, 5, 6, there are only two solutions. $\\frac4{1-\\frac56}$ and $\\frac6{\\frac54-1}$", "Solution_23": "Yeah...the challenge problem was a joke, as you probably guessed...\r\nI just found all of the ones below 15 which I could think of which were 4 of 1 number to make 24.\r\nHere are some more...\r\n1, 2, 5, 10\r\n3, 6, 6, 6\r\n2, 4, 6, 10 (find two ways)", "Solution_24": "here's a challenge one for yall. 2,5,5,5. It's on the \"impossible\" list but i found a way to solve it.\r\n\r\nYou can also do it with 2,5,5,6 which is also on the impossible list.\r\n\r\nsame with 2,9,9,9, but with the rules that I play with, the solution is banned. The other two aren't though.\r\n\r\n4,4,6,6 has a solution that is also banned.\r\n\r\nI'll stop with these but think about them.", "Solution_25": "[quote=\"Hamster1800\"]here's a challenge one for yall. 2,5,5,5. It's on the \"impossible\" list but i found a way to solve it.\n\nYou can also do it with 2,5,5,6 which is also on the impossible list.\n\nsame with 2,9,9,9, but with the rules that I play with, the solution is banned. The other two aren't though.\n\n4,4,6,6 has a solution that is also banned.\n\nI'll stop with these but think about them.[/quote]\r\nAre $25-\\frac55$ and $25-6+5$ the ones you have in your mind? Or did you actually find legal solutions(making 25 with 2 and 5 is not allowed in normal 24 game)?\r\nWhat do you mean by banned? You mean like $(\\frac{9-\\frac99}2)!$?", "Solution_26": "Those aren't the solutions I had in mind. Keep trying!\r\n\r\nThe rules that I play with require:\r\n\r\nno factorials\r\n\r\nany radical must be associated with a number, for example, to take a cube root, you need a 3.\r\n\r\nLogarithms must have a whole number base which counts as a card (however any time i try to apply logarithms, it can be solved anyway lol)", "Solution_27": "Is it $5^2-\\frac55$ and $5^2-6+5$?(powers are also not allowed in normal 24 game)", "Solution_28": "yep, but I play with powers so i guess i don't play by the rules.\r\n\r\nTry the other ones now." } { "Tag": [ "calculus", "integration", "function", "limit", "geometry", "rectangle", "analytic geometry" ], "Problem": "Let $f:[0,1]\\to\\mathbb{R}$ be a continuous function and let $\\{a_n\\}_n$, $\\{b_n\\}_n$ be sequences of reals such that\r\n\\[ \\lim_{n\\to\\infty} \\int^1_0 | f(x) - a_nx - b_n | dx = 0 . \\]\r\n\r\nProve that:\r\n\r\na) The sequences $\\{a_n\\}_n$, $\\{b_n\\}_n$ are convergent;\r\n\r\nb) The function $f$ is linear.", "Solution_1": "Assuming it's not linear, take three non-collinear points $(a,f(a)),(b,f(b)),(c,f(c))$ on its graph. We can then choose rectangles around the three points with sides parallel to the axes, so that there's no line intersecting all three rectangles. If, for example, the rectangle around $a$ has its vertical sides on the lines of $x$-coordinate $a-\\varepsilon$ and $a+\\varepsilon$, then for $a_nx+b_n$ which does not contain any point in this rectangle, the quantity $\\int_{x-\\varepsilon}^{x+\\varepsilon}|f(x)-a_nx-b_n|dx$ is always larger than some $\\delta_a>0$. We can find $\\delta_b,\\delta_c$ in a similar way, and then we will always have $\\int^1_0|f(x)-a_nx-b_n|dx>\\min(\\delta_a,\\delta_b,\\delta_c)$, which contradicts the hypothesis.", "Solution_2": "Assume that $ \\int_{0}^{1} |ax+b| \\leqslant \\varepsilon $ .Then\r\n$ \\int_{0}^{1} ax+b \\leqslant \\int_{0}^{1} |ax+b| \\leqslant \\varepsilon $\r\nand\r\n\r\n$ \\int_{0}^{1} ax^{2}+bx \\leqslant \\int_{0}^{1} |x||ax+b| \\leqslant\r\n\\int_{0}^{1} |ax+b| \\leqslant \\varepsilon $.\r\n\r\nTherefore $ \\frac{1}{2}a + b \\leqslant \\varepsilon $ and $ \\frac{1}{3}a + \\frac{1}{2}b\r\n\\leqslant \\varepsilon $ . Changing $ a $ to $ -a $ and $ b $ to $ -b $ we get that\r\n$ | a + 2b | \\leqslant 2 \\varepsilon $ and $ | 2a + 3b | \\leqslant 6 \\varepsilon $.\r\nTherefore $ |a| = | 2(2a+3b) - 3(a+2b) | \\leqslant 3 | a + 2b | + 2 | 2a + 3b |\r\n\\leqslant 18 \\varepsilon $ and $ |b| = | 2(a+2b) - (2a+3b) | \\leqslant 2 | a + 2b | +\r\n| 2a + 3b | \\leqslant 10 \\varepsilon $.\r\n\r\nNow if $ \\int_{0}^{1} | f(x) - a_{n}x - b_{n} | \\leqslant \\varepsilon $ and\r\n$ \\int_{0}^{1} | f(x) - a_{m}x - b_{m} | \\leqslant \\varepsilon $\r\nthen $ \\int_{0}^{1} | (a_{n}-a_{m})x + (b_{n} - b_{m}) | \\leqslant 2 \\varepsilon $\r\nso $ |a_{n} - a_{m}| \\leqslant 36 \\varepsilon $ and $ | b_{n} - b_{m} | \\leqslant\r\n20 \\varepsilon $. So if $ \\lim_{n \\rightarrow \\infty} \\int_{0}^{1} | f(x) - a_{n}x - b_{n} |\r\n = 0 $ then $ \\{ a_{n} \\} $ and $ \\{ b_{n} \\} $ are Cauchy sequences, hence they are convergent.\r\nSuppose $ a = \\lim_{n \\rightarrow \\infty} a_{n} $ and\r\n$ b = \\lim_{n \\rightarrow \\infty} b_{n} $. Then we have\r\n$ \\int_{0}^{1} | f(x) - ax - b | = 0 $, so because $ f $ is continuous, we get that\r\n$ f(x) = ax + b $." } { "Tag": [ "search" ], "Problem": "i was just wondering how die hard u harry fans are. so submit your answer to the poll. i certainly will go at midnight. but its not like its life or death to me. anyways. nuff said", "Solution_1": "oops.\r\nthe poll didnt work\r\njust say yes or no or sumat like that. since im lazy i wont make another poll or anything. but the options open to anyone else. nuff said", "Solution_2": "Erm, I'm not a [i]fanatic[/i] per se, but I'll probably read it it a couple of hours anyway so why bother.\r\nOh, btw, search for it on Kazaa. There's a bunch of funny fakes...\r\n\r\n\r\n\r\n-Isaac", "Solution_3": "What exactly is the yes/no question? But yep I definitely love harry potter. Will be reading the 5th book as soon as I can get my hands on it, but then I'll probably read it far too fast and then I finish :(", "Solution_4": "oops again. the question was are u gonna go get harry potter 5 at midnight when it comes out. but the poll didnt work. nuff said", "Solution_5": "No...I live in a small town, we'll probably get it in a few months once the wagons come 'round 8-) . Anyways, yesterday on CNBC they said at the Amazon warehouse there were 8 ft barricades and armed guards [b]just[/b] for the Ord of Phoenix shipment.", "Solution_6": "HAHAHAHA... just read on Yahoo... 7680 copies were stolen... lol.\r\n2 1/2 inches thick and 896 pages... can't wait...\r\n\r\n\r\n\r\n-Isaac", "Solution_7": "I'm not getting it because I don't especially like Harry Potter.", "Solution_8": "I'm probably going to get it the day after.", "Solution_9": "Well I'm not going at midnight, but I definitely want to get it. :D", "Solution_10": "even though its a few days after the book came out. call the store first to make sure they have it. cause at my borders they said that the next shipment would be on tuesday. which is today. if any of you guys went to a local book store i commend you. cause not that many people think of those places. they automatically go to borders or barnes and nobel. or something like that.", "Solution_11": "Have any of you read the 5th book yet? I finished it two days ago and I was IMMENSELY disappointed. I was waiting for an action packed and suspenseful book... I got a elongated and boring book with about only 200 pages of action. The first 600 pages were total gibberish. Rowling like purposely elongated it to make it 896 pages! It was generally a good story... if you can over look a half-a-dozen hundred pages of pointless stuff that could be deleted from the story to make it better. If it was only shorter... like 500 pages or even 400 pages, it would have been much more interesting... did I mention that it ended really abruptly too?... not to mention how HARRY SCREAMS AT THE READER EVERY OTHER PAGE! Thats about my criticism... I'm writing a review for the book... not done yet...\r\n\r\n-Taylor", "Solution_12": "it was pretty good. but i agree wit dragonslayerxp. i got rather annoyed with how many \"...\"s they had and how many \"-\"s they had. every other line. sheesh. she totally changed her writing style. the last 100 pages were the best. it took a whole 200 pages just to get to hogwash. c'mon give me a break. goblet of fire was much better.", "Solution_13": "Ah well. Think teen angst.", "Solution_14": "Teen Angst sums it up fairly well though... but I think its still an understatement. I doubt teen angst causes you to go around yelling your head off at insignificant things every other line. I also agree with syntax error about how the last 100 pages were the best... well, actually I would say the last 100 pages that were 20 pages from the end. The last 20 pages I thought were dull. She talks about how Harry was mad and didn't want to go to the feast. Then she spends about 2 pages talking about how he looks at Cho and how she's going out with another guy... I thought that was lame.\r\n\r\nA very disappointed reader,\r\n\r\nTaylor", "Solution_15": "now when i think about it, i dont mind the book with pages that could be cut out. this is so cause weve waited (well, most of us) 3 years for this book. and i dont really mind as long as it continues the series along.....", "Solution_16": "[color=indigo]I got the book the day it came out, started reading it a week later and finished in a week.[/color]", "Solution_17": "it took you a whole week to read it. wow", "Solution_18": "[color=violet][i]LESS[/i] than a week. I was busy visiting my cousin and didn't have much time to read. I could read it in a few days, except I didn't have enough time.\n\nP.S. Mom says [/color][color=indigo]THIS COLOR[/color][color=violet] is hard to read.[/color]", "Solution_19": "sure is, but if you highlight it, it doesnt matter. highlighting is funnnn zilla", "Solution_20": "woo, syntax is gonna love spoiler text huh?\r\nI just read Jean Aule's book [u]The Clan of the Cave Bear[/u] in a few days. It was a lot better than I thought it would be.\r\n\r\n\r\n\r\n-Isaac", "Solution_21": "Read that sometime last year - it was interesting.", "Solution_22": "Anybody read James Herriot?\r\n\r\n\r\n\r\n-Isaac", "Solution_23": "Herriot? Alfred Wight? Heard of him...have yet to read though. How are his books?", "Solution_24": "Alfred Wight? No clue who he is. James Herriot was a veterinarian in Britain around WW2. He wrote a series of really good books about his daily life. They're really funny :lol:. Try them sometime...\r\n\r\n\r\n\r\n-Isaac", "Solution_25": "James Herriot is the pen name of James Alfred Wight, if I remember correctly.", "Solution_26": "James Herriot ugh. whatever floats your boat i guess. he was a little redundant with his titles..." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Assume that $ S\\equal{}(a_1, a_2, \\cdots, a_n)$ consists of $ 0$ and $ 1$ and is the longest sequence of number, which satisfies the following condition: Every two sections of successive $ 5$ terms in the sequence of numbers $ S$ are different, i.e., for arbitrary $ 1\\le i1\n\nIf a=1 then the equation becomes 1=2, a contradiction. If a<1, then a^nb if a^2-a+1>b. Let a^2-a+1=c and c>a\n\nac\n\ncb$ so a>b\n\n\n[/hide]\n[hide][/hide][/hide]", "Solution_2": "I get this strange feeling I'm missing something, since this seemed really easy... Oh well, I'll post my solution anyway.\r\n\r\n[hide]\nWe have that $a^{n}= a+1 > 1$, so we have that $a > 1$. The second equation tells us that $b^{2n}= b+3a > 3+b > 3$. This tells us that $b>1$. Now we can show that $b^{2n}-b$ is increasing at least when $b > 1$. Suppose that $x$ and $y$ are real numbers such that $x>y>1$. This means that $x^{2n-1}-1>y^{2n-1}-1 \\implies x^{2n}-x > x(y^{2n-1}-1) > y^{2n}-y$ (true since $x^{2n-1}-1,y^{2n-1}-1 > 0$). Suppose that $b \\ge a > 1$. Then we have \\[b^{2n}-b-3a \\ge a^{2n}-a-3a = (a+1)^{2}-4a = (a-1)^{2}> 0\\] So we have $b^{2n}> b+3a$ if $b \\ge a$, which is impossible. Thus we have $a > b$ and we're done.[/hide]", "Solution_3": "Mine solution is:\r\na^n=a+1 <=> \r\n1) a^2n=a^2+2a+1.\r\n2) b^2n=b+3a\r\n1)-2) gets: \r\na^2n-b^2n=a^2+1-(a-b)\r\na^2n-b^2n+(a-b)=a^2+1\r\n(a-b)(a+b)(a^2+a^2)(a^4+b^4)...(a^n+b^n)+(a-b)=a^2+1\r\n(a-b)((a+b)(a^2+b^2)(a^4+b^4)...(a^n+b^n)+1)=a^2+1 <=> (a,b - positive)\r\n(a-b)(POSITIVE NUMBER)=(POSITIVE NUMBER) .... so\r\na-b>0\r\na>b => a is larger than b.", "Solution_4": "We square and rearrange the first equation and also rearrange the second.\n\\[ a^{2n}\\minus{}a\\equal{}a^2\\plus{}a\\plus{}1\\]\n\\[ b^{2n}\\minus{}b\\equal{}3a\\]\nIt is trivial that $a>1$, $b>1$, thus\n\\[ a^2\\plus{}a\\plus{}1>3a\\]\n\\[ a^{2n}\\minus{}a>b^{2n}\\minus{}b\\]\nwhere we substituted to achieve the result. This writes $(a-b)(a^{2n-1} + a^{2n-2}b + \\cdots + ab^{2n-2} + b^{2n-1} - 1) > 0$, and as the second factor is positive, it follows $a>b$.", "Solution_5": "Posted before at least 3 times.\r\n\\[{ a-b=\\frac{(a-1)^2}{a^{2n-1}+a^{2n-2}b+...+b^{2n-1}-1}}>0.\\]", "Solution_6": "@limac, I know this is a really late post, but you can show that:\n$f(x)=x^(2n)-x$ is an increasing function when x>1. Wouldn't that do the trick.\nAre there any other ways of proving this?", "Solution_7": "Let assume for sake of contradiction that $b>a$, and as above $a,b>1$ so, $b^{2n}>a^{2n}$, $b+3a>a^2+2a+1$, $b+a>a^2+1$, since we assume that $b>a$, $b/a>1$, so $2>b/a+1>a+1/a$, so $2>a+1/a$, contradiction, since $a>1$", "Solution_8": "I claim that $a>b$ for all positive integers $n$.\n\nFirst note that from the equality $a^n=a+1$ we have that raising $a$ to an integer power increases its value, implying that $a>1$. Similarly, we may deduce $b>1$. Now note that we can write $a^{2n}=a^2+2a+1$, so\\[a^{2n}-b^{2n}=(a^2+2a+1)-(b+3a)=(a^2-a+1)-b.\\]In order to prove our claim, it suffices to show that $a^2-a+1>b$ for all $n$. If we can do this, then we have $a^{2n}-b^{2n}>0$, so $a^{2n}>b^{2n}\\implies a>b$.\n\nSuppose the contrary, that $b\\geq a^2-a+1$. As $b$ is positive, increasing $b$ increases the discrepancy between $b^{2n}$ and $b$. Therefore, if we can show that $b^{2n}>b+3a$ for only the case $b=a^2-a+1$, then our desired follows.\n\nNote that from $a>1$ we have $a-1>0$, and squaring this gives $a^2-2a+1>0\\implies a^2-a+1>a$. Now raising both sides to the $(2n)^{\\text{th}}$ power gives\\[(a^2-a+1)^{2n}>(a^n)^2=(a+1)^2=(a^2-a+1)+3a\\implies b^{2n}>b+3a,\\] contradiction. Therefore $a^2-a+1>b$ for all positive integers $n$, and we are done. $\\blacksquare$", "Solution_9": "Note that $a^n=a+1>1\\Rightarrow a>1$ and, thus, $b^{2n}=b+3a>3\\Rightarrow b>1$. Now noting that $(a-1)^2>0\\Rightarrow a^2+1>2a$, \n\\begin{align*}\na^{2n}-a&=a^2+a+1\\\\\n&>3a\\\\\n&=b^{2n}-b\n\\end{align*}\nNow, note that $f(x)=x^{2n}-x$ is increasing for all $x>1$ (seen easily by differentiating). Thus, we may conclude that $f(a)>f(b)\\Rightarrow a>b$.$\\blacksquare$", "Solution_10": "Another solution using Calculus:\nFirst its easy to notice using the expressions that $a>1,b>1$\nConsider the function $f(x)=x^{2n}-x-3a$ for $x\\in [1,\\infty)$ clearly $x=b$ is a root of it. We see that $f'(x)>0$ for all $x\\in [1,\\infty)$, thus $f$ is strictly increasing on this interval. We know that such a function can intersect $x-$axis at most once thus $b$ is the only possible root. Now we have $f(a)=(a-1)^2>0$ and $f(1)<0$ thus using Intermediate Value Theorem $1 0$ show\r\n\r\n$ \\begin{aligned}&\\sum_{k\\in\\Bbb{Z}} \\frac {\\cos 2kx}{k^4 + \\frac {\\alpha^4}{4}} = \\frac {2\\pi}{\\alpha^3} \\biggl(\\sin \\alpha x \\,\\cosh \\alpha x - \\cos\\alpha x\\,\\sinh\\alpha x + \\biggr.\r\n\\\\\r\n&\\left.\\frac {(\\sinh\\alpha\\pi + \\sin\\alpha\\pi)\\cos\\alpha x\\,\\cosh\\alpha x - (\\sinh\\alpha\\pi - \\sin\\alpha \\pi)\\sin\\alpha x\\,\\sinh\\alpha x}{\\cosh\\alpha\\pi - \\cos\\alpha\\pi}\\right)\\end{aligned}$", "Solution_1": "it's easy.\r\nconsider :\r\n$ \\sum \\frac {e^{2ikx}}{k^{4} \\plus{} \\frac {\\alpha^{4}}{4}} \\equal{} \\sum \\frac {( \\minus{} 1)^{k}e^{ik(2x \\minus{} \\pi)}}{k^{4} \\plus{} \\frac {\\alpha^{4}}{4}}$\r\nthus:\r\n$ 2x \\minus{} \\pi \\in ( \\minus{} \\pi,\\pi)$\r\nhence consider\r\n$ \\int_{C}\\frac {e^{iz(2x \\minus{} \\pi)}}{\\sin(\\pi z)(z^{4} \\plus{} \\frac {\\alpha^{4}}{4})}dz$\r\nprove that:\r\nif $ C$ is big rectangle then \r\n$ \\sum_{k} \\frac {( \\minus{} 1)^{k}e^{ik(2x \\minus{} \\pi)}}{k^{4} \\plus{} \\frac {\\alpha^{4}}{4}} \\equal{} \\minus{} \\pi\\sum_{res (z^{4} \\plus{} \\frac {\\alpha^{4}}{4})}\\frac {e^{iz(2x \\minus{} \\pi)}}{\\sin(\\pi z)(z^{4} \\plus{} \\frac {\\alpha^{4}}{4})}$\r\njust take $ Re \\sum \\frac {e^{2ikx}}{k^{4} \\plus{} \\frac {\\alpha^{4}}{4}}$ .Done", "Solution_2": "More general we can write $ \\frac {1}{k^n \\plus{} \\alpha^n}$ as $ \\frac {1}{n\\, \\alpha^{n \\minus{} 1}} \\sum_{j \\equal{} 0}^{n \\minus{} 1} \\frac {e^{\\frac {i\\pi}{n}(2j \\plus{} 1) }}{\\alpha\\, e^{\\frac {i\\pi}{n}(2j \\plus{} 1)} \\minus{} k}$\r\n\r\nand use the formula $ \\sum_{k\\in\\Bbb{Z}} \\frac {( \\minus{} 1)^k \\cos kx}{\\beta \\minus{} k} \\equal{} \\pi \\, \\frac {\\cos\\beta x}{\\sin\\beta \\pi}$ to obtain\r\n\r\n$ \\sum_{k\\in\\Bbb{Z}} \\frac {( \\minus{} 1)^k \\cos kx}{k^n \\plus{} \\alpha^n} \\equal{} \\frac {\\pi}{n\\, \\alpha^{n \\minus{} 1}} \\sum_{j \\equal{} 0}^{n \\minus{} 1} \\frac {\\cos\\left(\\alpha\\, e^{\\frac {i\\pi}{n}(2j \\plus{} 1)}x\\right) \\, e^{\\frac {i\\pi}{n}(2j \\plus{} 1)}}{\\sin\\left(\\alpha\\, e^{\\frac {i\\pi}{n}(2j \\plus{} 1)}\\pi\\right) }$", "Solution_3": "[quote=\"Doe John\"]More general we can write ...[/quote]\r\n.. by same idea just write :\r\n$ \\sum_{k} \\frac {( \\minus{} 1)^{k}e^{ik(x \\minus{} \\pi)}}{k^{n} \\plus{} \\alpha^{n}} \\equal{} \\minus{} \\pi\\sum_{res (z^{n} \\plus{} \\alpha^{n})}\\frac {e^{iz(x \\minus{} \\pi)}}{\\sin(\\pi z)(z^{n} \\plus{} \\alpha^{n})}$" } { "Tag": [], "Problem": "Find positive integers $ a,b$ if for every $ x,y\\in[a,b]$, $ \\frac1x\\plus{}\\frac1y\\in[a,b]$.", "Solution_1": "[hide=\"solution\"]\ncondition:$ a\\le b$.\nit's necessary that $ \\frac{1}{x}\\plus{}\\frac{1}{y}\\ge a$,then let $ x\\equal{}y\\equal{}a$,we have $ a^2\\le 2$ i.e.$ a\\equal{}1$.\nLet $ x\\equal{}y\\equal{}b$,we have $ b\\le 2$ and we get $ b\\equal{}1$or$ b\\equal{}2$.\n\n1)when $ b\\equal{}1$\nwhen $ x\\equal{}y\\equal{}1$, $ \\frac1x \\plus{} \\frac1y\\equal{}2>1\\equal{}b$,ruled out\n\n2)when $ b\\equal{}2$\nif $ x,y\\in[1,2]$ then $ \\frac1x \\plus{} \\frac1y\\ge \\frac12\\plus{}\\frac12\\equal{}1$ and $ \\frac1x \\plus{} \\frac1y\\le \\frac11\\plus{}\\frac11\\equal{}2$ works.\n\nso $ (a,b)\\equal{}(1,2)$.\n\n[/hide]" } { "Tag": [], "Problem": "I'm having a hard time with the simplifying issues. Can someone please tell how to simplify the following expression:\r\n\r\n2(x^2-xy)-2(2x^2+1)+xy+3x\r\n\r\nThank you, any help would be very much appreciated.", "Solution_1": "[hide]$2(x^2-xy)-2(2x^2+1)+xy+3x\\\\\n=2x^2-2xy-4x^2-2+xy+3x\\\\\n=-2x^2-xy+3x-2$[/hide]", "Solution_2": "okay, well i have the answer:\r\n-2x^2 +3x-xy-2\r\nnow, you can factor that also. Do you want me to?\r\nAll you have to do is take it one number at a time. Do distributive property and then simplify. Don't forget to put your answer in descending order. Hope this helped. If you have any more questions, just ask me. Good luck!!", "Solution_3": "Click Mr. Green for the answer!\r\n[hide=\" :D \"]2(x^2-xy)-2(2x^2+1)+xy+3x \n2x^2-2xy-4x^2-2+xy+3x\n-2x^2-xy+3x+xy[/hide]", "Solution_4": "[hide]2(x^2-xy)-2(2x^2+1)+xy+3x \n\n2x^2-2xy-4x^2-2+xy+3x\n\n[b]-2x^2-xy+3x-2[/b][/hide]", "Solution_5": "Wow jeanna, I bet that really helped him he knew what to do =p\r\n\r\n[hide]\n2(x^2-xy)-2(2x^2+1)+xy+3x \n2(x^2-xy)-2(2x^2+1)+xy+3x\n2(x-y)-2(2x^2+1)+xy+3x\n(x-y)-(2x^2+1)+xy+3x/4\n(-y)-(2x+1)+xy+3x/4\n(-y)+-(2x+1)+xy+3x/4\n(-y)+-(x+1)+xy+3x/8\n(-y)+(-x)+xy+3x/7\n(-y+-x)+(xy)+3x/7\n3x/7\nx/2.3 repeating\n\nWow I really think I messed this up. If mine is different from anyone else that has a lot more stars than me, completely ignore me![/hide]", "Solution_6": "[hide]\n\\[ \\\\2(x^2-xy)-2(2x^2+1)+xy+3x \\\\=2x^2-2xy-4x^2-2+xy+3x \\\\=-2x^2-xy+3x-2 \\][/hide]", "Solution_7": "Please don't revive topics from a long time ago (meaning more than one month old and outdated). Thanks. :)" } { "Tag": [ "algebra", "polynomial", "real analysis", "real analysis unsolved" ], "Problem": "For $n\\in{\\mathbb N}$ consider the series $\\sigma(n,x): =\\sum_{k=0}^{\\infty}\\frac{k^{n}}{k!}x^{k}.$ If $x^{n}=\\sum_{k=0}^{n}S(n,k)x(x-1)\\cdots (x-k+1)\\; ,\\; \\forall x \\in{\\mathbb R},$ let \r\n $T(n,m;x): =\\sigma(n,x)\\sum_{k=0}^{m}S(m,k)x^{k}.$ [b] Prove that for any positive integers $n,m$ and all ${x > 0}$ \n\\[T(n,m;x)=T(m,n;x)\\; . \\]\n\n\n\n[/b]", "Solution_1": "[quote=\"flip2004\"]For $n\\in{\\mathbb N}$ consider the series $\\sigma(n,x): =\\sum_{k=0}^{\\infty}\\frac{k^{n}}{k!}x^{k}.$ If $x^{n}=\\sum_{k=0}^{n}S(n,k)x(x-1)\\cdots (x-k+1)\\; ,\\; \\forall x \\in{\\mathbb R},$ let \n $T(n,m;x): =\\sigma(n,x)\\sum_{k=0}^{m}S(m,k)x^{k}.$ [b] Prove that for any positive integers $n,m$ and all ${x > 0}$\n\\[T(n,m;x)=T(m,n;x)\\; . \\]\n[/b][/quote]\r\nYes, but what is $S(n,k)$?", "Solution_2": "[quote=\"Extremal\"] Yes, but what is $S(n,k)$?[/quote]\r\nThe polynomials $\\phi_{0}(x): =1\\; ,\\; \\phi_{k}(x): =x(x-1)\\cdots (x-k+1) \\; ,(k\\ge 1),$ is a basis in the linear space $\\Pi_{n}$ of all real polynomials of degree $\\le n .$\r\n The reason is that [b]degree[/b]$(\\phi_{k}) = k\\; , k\\in \\{0,1,\\ldots ,n\\}\\; .$ \r\n Therefore each polynomial from $\\Pi_{n}$ (e.g. $x^{n}$) may be written in an uniques way as a linear combinations of $\\{\\phi_{0}(x),...,\\phi_{n}(x)\\}.$\r\n As you may observe, $\\parallel S(n,k)\\parallel $ are the coefficients of this linear combinations . \r\n\r\nIn fact these numbers are well-known: i.e. are the [b]Stirling numbers of second kind[/b]." } { "Tag": [ "logarithms", "arithmetic sequence", "number theory open", "number theory" ], "Problem": "A Niven number is a natural number divisible by the sum of its digits. Does there exists an infinite arithmetic progresion free of Niven numbers?\r\n I have no idea, but I think the problem is very beatiful. Don't hesitate to think about it!", "Solution_1": "Nobody after two years! :o", "Solution_2": "Sorry, mathmanman, I promised you to publish my solution one year ago, but since I have no time to develope my forum I will publish it here :(\r\n\r\nConsider an arbitrary arithmetic progression of natural numbers $an+b$. \r\n\r\n1) There is a natural number of form $99...90..0$ which is divisible by $a$. So we may WLOG assume $a=99...90..0$ (just consider subsequence). \r\nLet $A$ be the decimal representation of $a$ and $B$ be the decimal representation of $b$. Let $s(m)$ is the sum of digits of natural number $m$.\r\n\r\n2) We will find $k$ s.t. $(s(ak+b),s(a))=q$, where $q=1$ if $3\\nmid b$, $q=3$ if $3\\mid b$ and $9\\nmid b$, $q=9$ if $9\\mid b$. Indeed, if $(s(a),s(b+a\\cdot 0))\\neq q$, then consider $11...1\\cdot a$ (number of '1' excedes number of '9' in $A$ by one). It is easy to show $s(11...1a)=9+s(a)$. So if we choose $k=11...1\\cdot 10^{h}$, where $h$ is greater than number of digits in $B$, then $(s(ak+b),s(a))=(s(b)+s(a)+9,s(a))=(s(b)+9,s(a))$. It means we may achive $(s(ak+b),s(a))=(s(b)+9r,s(a))$ for appropriate $k$. Thus just take $s$ satisfying $(s(b)+9r,s(a))=q$ (note that $s(a)=9m$). Denote $ak+b$ as $b$. \r\n\r\n3) Numbers $\\frac{s(b)}{q}$ and $\\frac{s(a)}{q}$ are coprime. It follows there is sufficient large prime $p$ (e.g. $p>10a$) s.t. $pq=d\\cdot s(a)+s(b)$ (by Dirichlet theorem).\r\n\r\n4) We will show there are a number in progression of form \\[A0..0A0..0A...A0..0B=X\\] ($d$ symbols $A$) which is divisible by $p$ (it clear it is divisible by $q$). Note that remainder of $A\\cdot 10^{s}$ modulo $p$ depends only on remainder of $s$ modulo $p-1$. Moreover $10^{s}$ modulo $p$ gives at least $\\log_{10}p$ distinct remainders. So each $A$ in $X$ gives us a freedom of at least $\\log_{10}p$ remainders. By Cauchy-Devenport theorem we conclude $X$ can take at least \\[\\min(p,d\\cdot (\\log_{10}p-1)+1)\\] remainders. But $d \\approx \\frac{pq}{s(a)}$ for large $p$, therefore $d\\cdot (\\log_{10}p-1)+1>p$ for large $p$. It means $X$ takes remainder 0 modulo $p$ and $s(X)=pq$.\r\n\r\nHope it is correct." } { "Tag": [ "trigonometry" ], "Problem": "Hey I was wondering if there's an approach to finding nontrivial solutions to problems such as $ x\\minus{}\\cos{x}\\equal{}0$ without a calculator? I really hate having to use the thing. :maybe: \r\n\r\n[hide=\"example\"]Let $ f(x)\\equal{}\\frac{\\cos{x}}{x}$. Find the equation of the line that is tangent to $ f$ at exactly two distinct points.[/hide]\r\nThanks!", "Solution_1": "You mean an algebraic one? No, not really. Most \"transcendental equations\" don't have closed-form solutions, and there's no reason to expect one." } { "Tag": [ "articles", "Harvard", "college", "calculus", "geometry" ], "Problem": "My next article for my college paper will be on exchange rates. I am trying to do a little research on both fixed and flexible exchange rates. Any opinions here?", "Solution_1": "Well, why r u people not answering?", "Solution_2": "I guess that there aren't a whole lot of AoPSers interested in Economics yet. It has always seemed to me that math people got into Economics mostly after getting to college if they ever do and most of the users on the site are not yet in college.\r\n\r\nI think the exchange rates question is fairly obvious, but I'm sure some would disagree. Pegging one currency to another masks the opinions of economic participants on the relative value of each. One of three things must eventually happen -- (1) demand for the overvalued currency will dry up completely resulting in an inability to carry on international trade (and even domestic trade with partners who have international interests), (2) an arbitrager like George Soros will arbitrage the relative value so dramatically that it will cost the taxpayers from the nation with the currency peg billions of dollars, or (3) the peg will be removed resulting in an economic shock to the holders of the currency that is instantly devalued by the supply/demand effects of the marketplace.\r\n\r\nI keep holding my breath that the Chinese currency peg will be lucky enough to escape international disaster only because its true value will move back into the direction of the current peg. I guess I should have mentioned this as a fourth possibility except that exactly hitting a peg is like throwing a dart from the moon and hitting a dartboard in Cleveland. There will almost surely be some economic fallout.", "Solution_3": "I guess I am always showing that I am illogical but can you explain to me the following in even more straightforward logic? I was trying to find a lot of literature on exchange rate regimes and a bit on the depreciation on the dollar, but I could not understand too much of the articles, I still do not have a great grasp of why they are good or bad.\r\n\r\n\r\n(1) demand for the overvalued currency will dry up completely resulting in an inability to carry on international trade (and even domestic trade with partners who have international interests), (2) an arbitrager like George Soros will arbitrage the relative value so dramatically that it will cost the taxpayers from the nation with the currency peg billions of dollars, or (3) the peg will be removed resulting in an economic shock to the holders", "Solution_4": "[quote=\"hello\"]\n(1) demand for the overvalued currency will dry up completely resulting in an inability to carry on international trade (and even domestic trade with partners who have international interests), (2) an arbitrager like George Soros will arbitrage the relative value so dramatically that it will cost the taxpayers from the nation with the currency peg billions of dollars, or (3) the peg will be removed resulting in an economic shock to the holders[/quote]\r\n\r\nA peg requires that the government continually take one side of a currency trade. This does not actually deter the market from believing the relative value is different, so the government simply trades and trades and trades, but the capacity to take one side of a trade is finite and risk levels grow very high with such a one-sided trade. Meanwhile, holders of the currency that should be more highly valued are wary about trading away their currency and holders of the currency that is overvalued find fewer and fewer people will to take their currency in the marketplace because people educated in finance don't believe in the value of the currency.\r\n\r\nI don't want to go into Soros' arbitrage of the peg of the pound, but I'm sure you can Google it.\r\n\r\nAny time a currency is dramatically devalued all at once, economic instability results for a number of reasons. I don't want to write a paper about this right now, but Google on the results of the devaluation of the Russian ruble in 1998 and you'll see what I'm talking about. \r\n\r\nFor that matter, the Soros attack on a currency peg I mentioned above is a more moderate example that was paid for by taxpayers. Taking a billion or two out of an economic in one fell swoop doesn't make life easier for a lot of people. That said, it was better sooner than later.", "Solution_5": "I am not sure I follow the following, sorry that I dont understand anything.\r\n\r\nA peg requires that the government continually take one side of a currency trade. This does not actually deter the market from believing the relative value is different, so the government simply trades and trades and trades, but the capacity to take one side of a trade is finite and risk levels grow very high with such a one-sided trade. Meanwhile, holders of the currency that should be more highly valued are wary about trading away their currency and holders of the currency that is overvalued find fewer and fewer people will to take their currency in the marketplace because people educated in finance don't believe in the value of the currency.", "Solution_6": "hello,\r\n\r\nIf you asked more specific questions, MCrawford would probably be better able to answer in a manner that would help you understand, for instance...\r\n\r\n\"Could you explain what sides of a currency trade are?\"\r\n\r\n\"What is relative value?\"\r\n\r\n\"Why is 'one sided trading' finite and why does the risk grow?\"\r\n\r\nor maybe something like...\r\n\r\n\"Could you maybe provide a live example of this following situation that might help me understand this, 'Meanwhile, holders of the currency that should be more highly valued are wary about trading away their currency and holders of the currency that is overvalued find fewer and fewer people will to take their currency in the marketplace because people educated in finance don't believe in the value of the currency.'\"\r\n\r\nAnd then of course... maybe a thank you? :)", "Solution_7": "OK, I want to thank Mathew for all the help he has provided in all my econ forums. Thanks for the suggesstion to you Spoon also. Well, I am not familiar with all the sentences that you copying from Mathew, so I guess I might need an explanation with that. Sorry about not specifying what I need help with exactly, that is because I understand like nothing from the whole paragraph, so the only way I could have written it was how I did.\r\n\r\nWhat I understand about exchange rates (maybe that will help explain what I dont understand) is that flexible exchange rates have a market in which they are traded, like stocks and bonds, so people can buy them, right? In fixed exchange rates, the currency is fixed to another value, for example 8.28 rambini (what is that by the way, Chinese currency or what?) equals $1. So that the Chinese currency is fixed and is not exchanged on the market. I know that a flexible exchange rate causes inflation, because monetary policy has to monitor it or something like that (I would be glad for a more thorough explanation) and that is why it is not always adopted, yet a fixed exchange rate slows economic growth but manages to hold down inflation. I know that most countries are in the flexible exchange system these days but some like China are not, the US government want them to become flexible. I looked at some academic literature to understand the topic but I understood very little I think. I saw that some economists were not for completely flexible exchange rates like this harvard economist Frankel suggested pegging the dollar to the export price or somthing like that. So that is all I understood, or at least most of what I understood, I am trying to find the logic of the rest of what is going on.\r\n\r\nAlso, I am interested in the value of the dollar, if it is good or bad that its declining and how much longer will it decline? I looked at an interview with economists that had different opinions. I understand that it is good from the perspective that we can get more exports but obviously there is a negative side like prices going up. So I wanted to understand the logic of this and understand it in furthur detail.\r\n\r\nI would like to thank Mathew once again for all his help and anyone else who replies to my questions.", "Solution_8": "Unfortunately I do not have the time to write complete explanations of my thoughts on all these topics. My hope was to provide some ideas to think about and research.\r\n\r\nIn regards to the falling relative value of the dollar, I don't regard the changing of the relative values of currencies as good or bad overall unless it's dramatic because volatility of value eats wealth (a la declining Sharpe ratios). Otherwise, the shifting of values appropriately rewards successful economic activity.", "Solution_9": "ok, thanks. I will try todo research even though I did some I will try again. I tried to find most of my info in academic papers, maybe that is why I did not find a lot of success? Is it too advanced to be reading academic papers when completing only 2 introductory econ classes? Should I just read news type articles? Thanks", "Solution_10": "I personally never favored reading a lot of academic articles (at least for the first two or three years) in order to learn economics. I think a better education comes from practical game theory, playing economics-based strategy games (like poker), working in an environment that involves economic decision making, and reading just enough to enable you to gain academic context for everything you've learned.\r\n\r\nIt doesn't hurt to study enough mathematics such that you're always one step ahead of your economics education in terms of understanding the kinds of processes you are examining and exploring.", "Solution_11": "I think in my whole undergraduate program at my college, there is no calculus based courses for econ which is really strange considering that top colleges have them almost since the beginning of their undergrad career, of course they are more advanced, but nevertheless.", "Solution_12": "I never really understood why a trade deficit means debt. Doesnt it just mean that there is much more coming into the country? What if all the money that is being paid for the products coming in is paid right away? Isnt that not a problem of debt?", "Solution_13": "my article regarding this is published, u can take a look and comment on it, nobody comnes to this forum, darn:\r\n\r\nhttp://www.thetriangle.org/media/paper689/news/2005/07/22/EdOp/U.Dollar.Threatened.In.Global.Market.Bad.Policies-963860.shtml", "Solution_14": "Give it time, and get the word out on other forums and this place might take off, hopefully. There aren't many alternatives. \r\n\r\nThe Mundell-Fleming is useful in determining the relative merits of exchange rate regimes. In this model, monetary policy (interest rate/ money suppy changes) effect national output under flexible exchange rates but not fixed. The opposite is true of fiscal policy (taxes, government spending). Output does not rise with a flexible x-rate regime due to 'crowding out'.\r\n\r\nThere are many scenarios aside from purely flexible and fixed x-rate regimes. I live in the 'ultimate', a common currency union of european nations. This evolved from a target-zone regime of semi-fixed rates. \r\n\r\nRegarding national debt and trade deficits, consider:\r\n(S)aving = (I)nvestment in a closed economy (eg the World)\r\nIf a country saves more than it invests, it is borrowing to the rest of the world (net capital outflows). This is accompanied by a trade surplus (exports > imports) as foreign countries have more of your money than you of theirs.\r\nIt is an identity (necessarily true) that\r\nS-I = X-M\r\nPaul Krumgan's 'pop internationalism' is highly readable", "Solution_15": "I really cannot think of any \"advantage\" for more than one currency.\r\nCurrency trade is nothing more than shear gamble: we are gambling the on the exchange rate, which has no contribution to the production output.\r\nI demand if anyone have better idea on it :?:", "Solution_16": "The nice thing about competing currencies is that they give us a very objective measure of the relative health between economies and with that a gauge of the relative strength of various economic policies.\r\n\r\nCurrency trading could be viewed as gambling, but so could any capital investment. However, that description is a little misleading. I do believe that some people have better insight into relative values between forms of capital than others. To the common investor, it might be throwing darts at boards. But there are people with great skill at playing darts.", "Solution_17": "Bchuchi, so you think that exchange rates on markets are determined randomly, is that what you mean?", "Solution_18": "[quote=\"hello\"]Bchuchi, so you think that exchange rates on markets are determined randomly, is that what you mean?[/quote]\r\n\r\nI don't think that's what he's saying. There is a standard criticism about many markets, currencies in particular, that the relative value movements are random.\r\n\r\nThis is true in some sense as any knowledge of future moves that are well understood by the marketplace should be priced in and therefore movements should be stochastic (not exactly, but close enough -- market crashes tend to be more one-way than booms, so downward moves are on average larger, though less frequent, for any security that is evaluated against a \"high standard\" currency).\r\n\r\nHowever, we can't assume that the relative values of currencies are well understood by the market. Indeed, if they were, we would not see discrete shifts of value -- and we do.", "Solution_19": "Mathew, I am not sure what you mean by understood by relative currencies being understood by the market?", "Solution_20": "I believe what Matthew means is that the rates fluctuate showing that the relative value of the currencies is understood by the participants of the marketplace. \r\nCurrencies are harder to trade and to forecast primarily due to the sheer amount of geopolitical factors that feed into the market and influence the participants. It is an area which is very interconnected in nature which is the reason why you have actions from one currency precipitating down to others. For example a good example of this is last week when the chinese officials raised the range of traded between the RMB and the dollar. This resulted in a series of currency moves which caused sudden spikes and then reverted back to where they were. USDJPY had a spike down (RMB strength is favorable to the Japanese a big trading partner) while the EURUSD had a spike upwards. In both these cases prices reverted back to the normal range after 5 minutes or so.", "Solution_21": "Whether \"the exchange rate is a Brownian motion\" requires a large sample set of data to be verfiied. The problem is: \"what is the good of having more than one currencies in one unified globalized market?\", possibly could we look consider the problem in terms of \"transaction cost\"? Could currency be merely consuming \r\ntransaction cost but with no added value to the total production of the world economy? :idea:", "Solution_22": "It might also be mentioned that something could appear to behave with Brownian motion and still be arbitragable.\r\n\r\nThe notion of randomness has been much better understood in recent years, especially in regards to market behavior. It's often best to view randomness as compression of information to the point where we can't predict it's behavior. Of course, information is different for everybody, so what might be random to one person might not be random to another who has a more complete information set." } { "Tag": [], "Problem": "I have a question about problem No.456 in the September Issue.\r\nI'm not quite sure what is counted as one move. Am i supposed to count the two that moves Token k to cup m and the other which moves\r\nToken k+1 to cup k? or just regard both as one move? \r\n\r\nPS: I'm not discussing the solution but if it's not supposed to talk about, please delete it. Thanks", "Solution_1": "This is from a problem set that is not due until October 31, so it should not be discussed until then. The topic is locked." } { "Tag": [ "trigonometry", "inequalities", "inequalities proposed" ], "Problem": "If $ ab+bc+ac+2abc=1$ and $ a,b,c>0$ Prove that \r\n$ \\sqrt{ab}+\\sqrt{bc}+\\sqrt{ac}\\ge{\\frac{3}{2}}$", "Solution_1": "In every triangle there is a relation: $ \\cos^{2}\\alpha+\\cos^{2}\\beta+\\cos^{2}\\gamma+2\\cos\\alpha\\cos\\beta\\cos\\gamma = 1$ and it is also true that if $ x, y, z$ satisfy the condition $ x^{2}+y^{2}+z^{2}+2xyz=1$ then there are cosines in some triangle.\r\nUsing this for $ \\sqrt{ab}, \\sqrt{bc}, \\sqrt{ca}$ we are left with proving that $ cos\\alpha+\\cos\\beta+cos\\gamma \\geq \\frac{3}{2}$ but it is well known and easy with Jensen.", "Solution_2": "Just a little note: The sign of the inequality is reversed ;)\r\n(it should be $ \\cos\\alpha+\\cos\\beta+\\cos\\gamma\\leq \\frac32$)", "Solution_3": "[quote=\"Litlle 1000t\"]If $ ab+bc+ac+2abc=1$ and $ a,b,c>0$ Prove that \n$ \\sqrt{ab}+\\sqrt{bc}+\\sqrt{ac}\\ge{\\frac{3}{2}}$[/quote]\r\nIt is wrong. Counterexample, $ a\\to 0$, $ b=c=1$.", "Solution_4": "for the edited problem you can also use $ x=\\sin{\\frac{A}{2}},y=\\sin{\\frac{B}{2}},z={\\frac{C}{2}}$\r\ntake a look at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=149752[/url] posts #$ 11,14$", "Solution_5": "At first a clarification: meaning inequality must be reversed!\nNow my solution:\nWith the substitutions $ a=\\ \\frac {a} {2} $, $ b=\\ \\frac {b} {2} $, $ c=\\ \\frac {c} {2} $ inequality is rewritten as follows:\n1) $ \\sqrt(ab)+\\sqrt(bc)+\\sqrt(ca)\\le 3 $ \nif $ a $,$ b $\u200b\u200b,$ c $ are positive numbers such that $ ab + bc + ac + abc = 4 $\nThe substitutions $ a=\\ \\frac{yz}{x} $, $ b=\\ \\frac{zx}{y} $, $ c=\\ \\frac{xy}{z} $ transform (1) in the following equivalent relationship:\n1')$ x+y+z \\le 3 $ \nvalid if $ x $, $ y $,$ z $are positive numbers such that $ x ^ 2 + y ^ 2 + z ^ 2 + xyz = 4 $\nThe result of (1 ') is well-known! Not insist!\n______________________________________\nSandu Marin, Bucuresti, Romania" } { "Tag": [ "geometry", "parallelogram", "ratio", "analytic geometry" ], "Problem": "A parallelogram $ ABCD$ is given where $ P,Q,R,S$ are mid points of $ AB,CD,DA,BC$ respectively.Find the ratio of areas of the shaded region to that of the parallelogram.", "Solution_1": "For those who don't want to download:\r\n[asy]size(1.5inch); pair A,B,C,D,P,Q,R,S; A=(0,0);B=(3,0);C=(4,3);D=(1,3);P=A/2+B/2;Q=C/2+D/2;R=D/2+A/2;S=B/2+C/2; draw(A--B--C--D--cycle);draw(A--S);draw(B--Q);draw(C--R);draw(D--P); label(\"$A$\",A,SW);label(\"$B$\",B,SE);label(\"$C$\",C,NE);label(\"$D$\",D,NW); label(\"$P$\",P,SE);label(\"$Q$\",Q,NE);label(\"$R$\",R,SW);label(\"$S$\",S,NE);[/asy]\r\nThe shaded region to which the question refers are the 4 smallest triangles. I would shade them, but I don't want to compute the coordinates of the vertices not on the parallelogram boundary.", "Solution_2": "[hide=\"extremely non-rigorous\"]WLOG, ABCD is a square. After that, it's area chasing time.[/hide]", "Solution_3": "Just skew the diagram so it is a square. This transformation preserves length ratios [and hence area ratios]\r\n\r\nI think it is 1 to 5.", "Solution_4": "[hide=\"hint\"]$ RACS$ is a parallelogram because one pair of opposite sides are $ \\cong$ and $ \\parallel $ so $ \\angle{RAS}\\cong\\angle{RCS}$. So similarly $ \\angle{DPA}\\cong\\angle{BQC}$ So by ASA we get that the two triangles are congruent. So from there on by vertical angles and we are done. [/hide]" } { "Tag": [ "Support", "geometry" ], "Problem": "Hey everyone,\r\nI seem to get distracted by a lot of things (fantasy baseball news, NBA boxscores, checking collegeconfidential, AIM, facebook, checking AOPS and other message boards, etc). Throughout my last four years of high school, I often procrastinated and waited until the absolute last minute to do everything. Many times, I would not get certain tasks done, and I have missed out on many wonderful opportunities as a result. I was wondering what methods fellow AOPSers use to keep their focus on the tasks at hand? There are so many smart and hardworking people on this board, so I was hoping that I can learn from you guys! Thanks in advance :)\r\n\r\nPS. I definitely have OCD, but I'm not sure if I have ADD.", "Solution_1": "lol definately do not procrastinate, especially in high school. In middle school it might be allowed but in high school grades actually count in your future. \r\n\r\n2.) always plan your \"homework\" time ahead of time...for example if you have an after school activity and lots of homework, force yourself to not get on the computer until your homeowkr is done", "Solution_2": "I'm actually already a senior in high school, unfortunately :oops: . I was wondering though, do most people just eventually develop the habit of not procrastinating if they stick to a set schedule for a long enough time (such as going home, doing all of their homework, then doing anything else)? I'm hoping to erase my procrastinating ways before I go off to college next year, where the classes will be harder, and the consequences for waiting until the last minute to do everything will probably be much more severe.", "Solution_3": "I have quite a strict anti-procrastination policy, and it's always worked well for me. Whenever I do start to slack, I always regret it.", "Solution_4": "Be mildly autistic. :P \r\n\r\nI have a lot of trouble with this, too. I haven't figured out a good solution yet. The real problem is that none of my homework is interesting.\r\n\r\n[quote=\"7h3.D3m0n.117\"]force yourself to not get on the computer until your homeowkr is done[/quote]\r\n\r\nWhen you have to type all of your homework because your teachers refuse to read your handwriting, this could get tricky :oops:", "Solution_5": "[quote=\"t0rajir0u\"][quote=\"7h3.D3m0n.117\"]force yourself to not get on the computer until your homeowkr is done[/quote]\n\nWhen you have to type all of your homework because your teachers refuse to read your handwriting, this could get tricky :oops:[/quote]\r\n\r\nlol touche =)\r\n\r\nwell don't get on homework time consuming sites (like AoPS, yes this has happened before and i had to do an english essay the last minute....darn you AoPS! lol)...also try to avoid using AIM because of annoying ppl that might IM you but sometimes if im relaly busy I just turn it to away.", "Solution_6": "get a friend on AIM that has the same hw with you.\r\nrace against him to finish it with quality or better, set a time where you guys have to confirm that each other finished the work.\r\n\r\n\r\nanother suggestion to slowly cut down is to pick a favorite activity, say visiting aops. say you have a 6 paragraph essay for hw and each paragraph takes you 20 minutes. then after each paragraph, allow yourself a 5 minute break on aops or every other paragrpah. depnding on ure liking. might take longer to type the essay (3 hours instead of 2) but its better than procrastinating.", "Solution_7": "how about:\r\n\r\n\r\n\r\nPRocrastinate! \r\n\r\n\r\nSome people do well under pressure, possibly even better.", "Solution_8": "i would start a new topic, but this is similar.\r\n\r\nmy problem is, if theres something i dont want to do, i just dont.\r\n\r\nlike if i have hw, and its boring, i dont do it. and i cant help it, and my grades suck because of that.\r\n\r\nhelp?", "Solution_9": "[quote=\"neelnal\"]get a friend on AIM that has the same hw with you.\nrace against him to finish it with quality or better, set a time where you guys have to confirm that each other finished the work.\n\n\nanother suggestion to slowly cut down is to pick a favorite activity, say visiting aops. say you have a 6 paragraph essay for hw and each paragraph takes you 20 minutes. then after each paragraph, allow yourself a 5 minute break on aops or every other paragrpah. depnding on ure liking. might take longer to type the essay (3 hours instead of 2) but its better than procrastinating.[/quote]\r\n\r\nYes, these are very good suggestions. The second idea might even improve efficiency, because sometimes you run into blocks when working full-time on a single task (for essay-type stuff).", "Solution_10": "Okay here is as tip I would rarely give:\r\n\r\nsometimes if there is absolutely no way of getting something done just do it the period before it is due...i do that alot but you have to careful not to get caught...also beware of teachers that walk around too much and can see you doing it...that would suck =)", "Solution_11": "[quote=\"7h3.D3m0n.117\"]Okay here is as tip I would rarely give:\n\nsometimes if there is absolutely no way of getting something done just do it the period before it is due...[/quote]\r\n\r\nHeh... There is a rule among Murphy's laws stating something like\r\n\\[(\\text{time left for the project}) \\cdot (\\text{effort invested in completing the project}) = \\text{const.}\\]\r\nAnd a corollary: if there wasn't for the last moment, nobody would have gotten anything done ;)\r\n\r\nFunny how everyday stuff supports Murphy's laws...\r\n\r\nBut on topic, I think that you just have to make yourself do everything on time. No tricks -- just don't let yourself get up and do anything other than homework.", "Solution_12": "[quote=\"7h3.D3m0n.117\"]Okay here is as tip I would rarely give:\n\nsometimes if there is absolutely no way of getting something done just do it the period before it is due...i do that alot but you have to careful not to get caught...also beware of teachers that walk around too much and can see you doing it...that would suck =)[/quote]\r\n\r\nThis isn't a tip, this is my life. I have elaborate systems where I do homework for every class in the class immediately preceding it; it's high pressure, but it works and I haven't done homework at home in a couple months.", "Solution_13": "what if there is a test that takes up the whole period for everyone in the period before your homework is due one day ?", "Solution_14": "[quote=\"now a ranger\"]what if there is a test that takes up the whole period for everyone in the period before your homework is due one day ?[/quote]\r\n\r\nyou rush the test and rush the homework...\r\n\r\n idont' really ever recall a situation where a test required a full hour", "Solution_15": "[quote=\"Elemennop\"][quote=\"7h3.D3m0n.117\"]Okay here is as tip I would rarely give:\n\nsometimes if there is absolutely no way of getting something done just do it the period before it is due...i do that alot but you have to careful not to get caught...also beware of teachers that walk around too much and can see you doing it...that would suck =)[/quote]\n\nThis isn't a tip, this is my life. I have elaborate systems where I do homework for every class in the class immediately preceding it; it's high pressure, but it works and I haven't done homework at home in a couple months.[/quote]\r\nI have a different system -- I do the homework during the class and day that it is assigned, and if I am feeling tired I do it during the next period as well.", "Solution_16": "This is one area in which having dial-up internet is really handy. If I don't need the Internet for homework, I stay disconnected. I also play music to distract me from all of the other distractions. It sounds sort of weird and illogical, but it seems to work.", "Solution_17": "[quote=\"randomdragoon\"][quote=\"Elemennop\"][quote=\"7h3.D3m0n.117\"]Okay here is as tip I would rarely give:\n\nsometimes if there is absolutely no way of getting something done just do it the period before it is due...i do that alot but you have to careful not to get caught...also beware of teachers that walk around too much and can see you doing it...that would suck =)[/quote]\n\nThis isn't a tip, this is my life. I have elaborate systems where I do homework for every class in the class immediately preceding it; it's high pressure, but it works and I haven't done homework at home in a couple months.[/quote]\nI have a different system -- I do the homework during the class and day that it is assigned, and if I am feeling tired I do it during the next period as well.[/quote]\r\n\r\ni do that in math but my teacher caught me...now its harder to do it as she walks around too much and i have to hide the math book every other problem :(", "Solution_18": "[quote=\"7h3.D3m0n.117\"]\ni do that in math but my teacher caught me...now its harder to do it as she walks around too much and i have to hide the math book every other problem :([/quote]\r\n\r\nHold up, your math teacher stops you from working more advanced topics in class even though you're clearly bored with the work being done? Blasphemous!", "Solution_19": "I agree and I would love to skip that class into Geometry but my parents wont let me take the Algebra II test.", "Solution_20": "To maintain one's concentration, I leave everything off to the near end (like 7h3.d3m0n.117 said) and read a book at the same time. It always helps. For me of course. Only problem is that you might write something like, \"He did that,\" on ur math homework." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "This is a very beautiful problem:\r\n\r\nGiven $n>3$ prove that there exist $2^{n-1}$ points in the plane, no \r\nthree collinear, s.t. no $2n$ form a convex polygon.\r\n\r\nI already know two solutions: one on http://www.kalva.demon.co.uk/short/soln/sh94c7.html and another my own.\r\nCan you find another solutions?\r\nCan you prove an upper bound? (this must be harder)", "Solution_1": "It is a famous Erdos-Szekeres problem.\r\n\r\nI know very nice article devoted to this problem in the Bulletin of the AMS, October 2000, volume 37, number 4.\r\n\r\nIn this article following result is proved:\r\n[i]there are $2^{n-2}$ points, no three collinear, s.t. no $n$ form convex polygon.[/i]\r\n\r\nThe famous Erdos-Szekeres conjecture that for any $2^{n-2}+1$ points we can find $n$, which form convex polygon.", "Solution_2": "myth, is there are version of this theorem for polytopes?", "Solution_3": "Rrrr... [b]Myth[/b]\r\n\r\nYes, there is high-dimensional version of the problem.\r\n\r\nJust little problem:\r\n\r\n[b]Statement.[/b]\r\n[i]Let $N_d(n)$ be minimal number s.t. it is always possible to choose convex $n$-top from given $N_d(n)$ points in general position in $\\mathbb{R}^n$.\nProve that $N_d(n)\\leq N_{d-1}(n)\\leq...\\leq N_2(n)\\leq {2n-5 \\choose n-3}+2$.[/i]\r\n\r\nNote there are many different estimates for $N_d(n)$.", "Solution_4": "Actually this is Klein-Erdos-Szekeres Problem or Happy end problem. And you can find something romance about it.\r\nYou can see at :\r\n http://mathworld.wolfram.com/HappyEndProblem.html\r\nThe reason why it was called \"Happy end problem\" ... Go and see \r\nhttp://martin.winkler.net/happy_end.html\r\n\r\n\r\n\r\nSorry :D :P", "Solution_5": "'Happy ending problem'....\r\n\r\nPierre.", "Solution_6": "Hi all,\r\n\r\njust to remember you guys I am working on the ISL/ILL project and I kindly requested you to tell me that you post any of these problems, either per PM or per posting in the ISL year you took the problem from. :) Otherwise I don't hesitate to repost problems and if I find yours by chance guess what will happen to it... :no:" } { "Tag": [ "function", "quadratics", "algebra", "special factorizations", "algebra proposed" ], "Problem": "$ a)$ Determine the functions $ f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that:\r\n\r\n$ 3 \\cdot f(x) \\cdot f(y) \\minus{} 5 \\cdot f(x)\\plus{}7 \\cdot f(y)\\plus{}4\\equal{}xy, \\forall x,y \\in \\mathbb{R}$\r\n\r\n$ b)$ Consider $ f: \\mathbb{R} \\rightarrow \\mathbb{R}$, $ f(x) \\equal{} \r\n\\begin{cases} \r\n x\\minus{}4, & x<2m \\\\\r\n 3x\\minus{}2, & x \\ge 2m\r\n\\end{cases}$, $ m \\in \\mathbb{R}$.\r\n\r\nDetermine $ m \\in \\mathbb{R}$ such that:\r\n\r\n$ f(2m\\minus{}1)\\plus{}f(m^2\\plus{}m\\plus{}1)\\plus{}f(2m\\plus{}2)\\equal{}m f(2m\\minus{}2)$", "Solution_1": "a) Can it be for all $ x,y \\in \\mathbb{R}$? Just put $ x\\equal{}y\\equal{}0$ we get a quadratic equation that has no real solution.\r\n\r\nb)We have $ m^2\\minus{}m\\plus{}1>0$ by completing the square \r\n$ \\Leftrightarrow m^2\\plus{}m\\plus{}1>2m$\r\n\r\nSo the equation becomes $ (2m\\minus{}5)\\plus{}(3m^2\\plus{}3m\\plus{}1)\\plus{}(6m\\plus{}4)\\equal{}m(2m\\minus{}6)$\r\n$ \\Leftrightarrow m^2\\plus{}17m\\equal{}0$, so $ m\\equal{}0$ or $ \\minus{}17$" } { "Tag": [ "AMC", "AIME" ], "Problem": "$m,n$ are integers such that $m^2+3m^2n^2 = 30n^2 + 517$. Find $3m^2n^2$. (AIME 1987)", "Solution_1": "[hide]$m^2(1+3n^2)=10(1+3n^2)+507$\n$(m^2-10)(1+3n^2)=507=3*13^2$\ncycle through the factors: $1*507, 3*169, 13*39$\n$m^2-10=39$ and $1+3n^2=13$ is the only pair with integer $m$ and $n$, so\n$3m^2n^2=12*49=\\boxed{588}$[/hide]", "Solution_2": "[hide]Let $m^2=a$ and $n^2=b$. Then we want to find $3ab$. We know $a+3ab=30b+517$. Using Simon's factoring trick (subtract $10+30b$ on both sides, factor out the $a$, then factor out $10$, and then factor out $1+3b$) we have $(3b+1)(a-10)=507$. Factoring $507=13^2\\cdot 3$. Since both $a$ and $b$ are perfect squares we find $a-10=39$, so $a=49$ and $3b+1=13$, so $b=4$. We want the value $3ab=3\\cdot 49\\cdot 4=588$.[/hide]" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $a_n$ the number of digits greater than or equal to 5 in $n$. Prove that $\\sum_{n\\geq 1}{\\frac{a_{2^n}}{2^n}}=\\frac{2}{9}$.", "Solution_1": "This is another cute problem by harazi I would like to give a second life to. I should confess that, at first, I did not believe it and even tried to disprove the statement using a computer. But when I got $0.222222222...$ for the sum of the first $40$ terms, I decided that it might be true after all, though it took me some time after that to figure the problem out. The only hint I can give without ruining the problem is that, unlike many other problems by harazi, it is [i]very simple[/i].\r\n\r\nAs usual, if nobody figures it out in a couple of weeks, I'll post a solution. :)", "Solution_2": "We can use the relation:\r\n\r\n$s(2^n) = 2s(2^{n-1}) - 9a_{2^{n-1}}$\r\n[try to prove this :)]\r\n\r\nwhere $s(x)$ is the sum of digits of x\r\n\r\nDivide by $2^n$, sum and get : $s(2^n)/2^n - 1 = -(9/2)x_n$ where $x_n$ is the partial sum of our serie.\r\n\r\nFrom here the result is clear because $s(2^n)/2^n \\to 0$ [$s(2^n) \\leq l(2^n) = O(n)$] where $l(x)$ = nr of digits of x", "Solution_3": "do you know the proof of the relation you used ?\r\n\r\n[Edit] ok, I've got it. ;)", "Solution_4": "nice solution,mareleG;\r\nwas you solution the same, fedja ?", "Solution_5": "[quote=\"eugene\"]\nwas you solution the same, fedja ?[/quote]\r\nYes. In a sense it is [b]the[/b] solution. At least, I'll be extremely amazed if somebody finds a really different approach.", "Solution_6": "i see these $O(n)$ these days looking on mathlinks\r\nwhat does it mean ?\r\nis it related to asymptotic notation with big tetha that stuff introductory in cormen ?", "Solution_7": "[quote=\"spx2\"]i see these $O(n)$ these days looking on mathlinks\nwhat does it mean ?[/quote]\r\nRead [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31517]Kent's stickie[/url] in Computations and Tutorials." } { "Tag": [ "LaTeX" ], "Problem": "let n is nature and n>2. Prove that $(n^{n^{n^{n}}}-n^{n^{n}})\\vdots 9$ :lol:", "Solution_1": "What is it exactly that we have to prove?", "Solution_2": "maybe he wants us to prove that $9|n^{n^{n^{n}}}-n^{n^{n}}$", "Solution_3": "I suppose that you could let $n{}^n{}^n=x$, where you know that $n|x$. From there, you have\r\n$n^x-x$\r\nAnd then...mods?\r\n :o \r\nAnd...\r\nUh...\r\n :huh: \r\n:? \r\nnooooooooo\r\n :(", "Solution_4": "clearly 2| n^n - n\r\n\r\nso n^(n^n - n)=n^2k=1 (mod 6), or\r\n\r\nn^(n^n - n) - 1=0 (mod 6), and\r\n\r\n[n^n][n^(n^n - n) - 1]=0 (mod 6), thus\r\n\r\nn^{[n^n][n^(n^n - n) - 1]}=n^6j=1 (mod 9), by Euler's Theorem, or\r\n\r\nn^{[n^n][n^(n^n - n) - 1]} - 1=0 (mod 9), hence\r\n\r\n[n^(n^n)][n^{[n^n][n^(n^n - n) - 1]} - 1]=0 (mod 9), or\r\n\r\n[n^(n^n)][n^[n^(n^n) - n^n] - 1]=0 (mod 9), or\r\n\r\nn^(n^(n^n)) - n^(n^n) = 0 (mod 9)\r\n\r\ni really need to learn Latex :( :oops:" } { "Tag": [ "ARML", "geometry", "geometric transformation", "reflection", "trigonometry", "algebra", "polynomial" ], "Problem": "Hello to any participants of the ARML forum... You may or may not know about the previous ARML Practice Series (located at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30264]here[/url] (Solutions/Results [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32273]here[/url]) and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86168]Here[/url] (Solutions/Results [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86171]here[/url])), but I (along with my partner in crime, pythag011) am ready and willing to revive another! That is, if there is interest. Please PM me or post here if you are interested!\r\n\r\nIf we get enough responses, we'll start up a ARML Practice Series 3 thread, along with a companion Scoreboard/Results thread. There will be a new problem each day, and a format likely similar to the previous Practice Series (different point values, cumulative contest, and possible mini-\"reward\" (mmm, food) for winners who go to the same site as me. (Sorry, people not going to Las Vegas...))\r\nIf not, pythag and I will probably still post some problems, but in either the AMC or Pre-Olympiad forum, and probably not as many...\r\n\r\nFor now, here is a sample of problems that at least somewhat reflect the problems that will be present in the series...\r\n\r\n1. Let $ x,y$ be two reals such that $ x^3 \\plus{} y^3 \\equal{} 2xy \\minus{} \\frac {8}{27}$. Let $ a$ and $ b$ be the maximum and minimum possible values of $ x \\plus{} y$, respectively. Find $ a \\minus{} b$. [Source: Brut3Forc3]\r\n\r\n2. For positive integer $ k$, let $ f_k(x) \\equal{} \\frac {1}{k}(\\sin^k x \\plus{} \\cos^k x)$ Find the minimum possible value of $ f_4(x) \\minus{} f_6(x)$. [Source: 103 Trigonometry Problems]\r\n\r\n3. Prove that there are infinitely many triples of positive integers $ (a,b,c)$ such that $ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca \\plus{} a \\plus{} b \\plus{} c}$ is a positive integer. [Source: pythag011]\r\n\r\n(Note: As anyone with prior knowledge of myself and pythag011 would know, my problems will tend to the easier side, whereas pythag's will tend to the harder side. Most likely, this problem will be dealt with by adjusting the point value of corresponding problems.)\r\n\r\nOnce again, please post here or PM me if you are interested in participating! (Please? :P)", "Solution_1": "Hey, I'm trying out for ARML chicago, but I'm not sure if I'm ready. I think this would be great practice! Please start this series. Thanks!", "Solution_2": "Okay, I've received several responses, so this is a go! I think I'll start it on Friday, at around 8 PM PST, with problems going up at around that time every day after.\r\n\r\n(But signups are still open!)", "Solution_3": "[quote=\"appillai\"]Hey, I'm trying out for ARML chicago, but I'm not sure if I'm ready. I think this would be great practice! Please start this series. Thanks![/quote]\r\nTry solving these 3 problems above first! They are all nice problems!\r\nHere's my solution for the 1st one:\r\nWe have: $ (x\\plus{}y)^3\\plus{}8/27\\minus{}3xy(x\\plus{}y\\minus{}2/3)\\equal{}0$.\r\nFactorize this one we get: $ (x\\plus{}y\\plus{}2/3)[(x\\plus{}y)^2\\minus{}2(x\\plus{}y)/3\\plus{}4/9\\minus{}3xy]\\equal{}0$\r\nWe either have $ x\\plus{}y\\equal{}\\minus{}2/3 or (x\\plus{}y)^2\\minus{}2(x\\plus{}y)/3\\plus{}4/9\\equal{}3xy$\r\nSince $ (x\\plus{}y)^2/4 \\geq xy$, we have: $ 0 \\geq (x\\plus{}y)^2\\minus{}3(x\\plus{}y)^2/4\\plus{}4/9\\minus{}2(x\\plus{}y)/3$, which means $ 0 \\geq [(x\\plus{}y)/2\\minus{} \\minus{}2/3]^2$.\r\nThe equality occurs when $ x\\plus{}y\\equal{}4/3$. \r\nTherefore, summing from both cases, we get max$ (x\\plus{}y)\\equal{}4/3$. The equality occurs at $ x\\equal{}y\\equal{}2/3$", "Solution_4": "[quote=\"ghjk\"][quote=\"appillai\"]Hey, I'm trying out for ARML chicago, but I'm not sure if I'm ready. I think this would be great practice! Please start this series. Thanks![/quote]\nTry solving these 3 problems above first! They are all nice problems!\nHere's my solution for the 1st one:\nWe have: $ (x \\plus{} y)^3 \\plus{} 8/27 \\minus{} 3xy(x \\plus{} y \\minus{} 2/3) \\equal{} 0$.\nFactorize this one we get: $ (x \\plus{} y \\plus{} 2/3)[(x \\plus{} y)^2 \\minus{} 2(x \\plus{} y)/3 \\plus{} 4/9 \\minus{} 3xy] \\equal{} 0$\nWe either have $ x \\plus{} y \\equal{} \\minus{} 2/3 or (x \\plus{} y)^2 \\minus{} 2(x \\plus{} y)/3 \\plus{} 4/9 \\equal{} 3xy$\nSince $ (x \\plus{} y)^2/4 \\geq xy$, we have: $ 0 \\geq (x \\plus{} y)^2 \\minus{} 3(x \\plus{} y)^2/4 \\plus{} 4/9 \\minus{} 2(x \\plus{} y)/3$, which means $ 0 \\geq [(x \\plus{} y)/2 \\minus{} \\minus{} 2/3]^2$.\nThe equality occurs when $ x \\plus{} y \\equal{} 4/3$. \nTherefore, summing from both cases, we get max$ (x \\plus{} y) \\equal{} 4/3$. The equality occurs at $ x \\equal{} y \\equal{} 2/3$[/quote]\r\nI have changed the problem a bit since you saw it, and it now includes finding the min. (But it's basically the same.) Good solution!\r\n\r\nFor the other two, 2 is fairly easy, whereas 3 is much harder, and not likely to be seen on an ARML.", "Solution_5": "I think even 1 is too hard for an ARML problem.", "Solution_6": "I'll post solutions to the other 2.\r\n\r\n[hide=\"3\"]\nThe problem is just a simple application of backwards Vieta jumping.\n\nNote that (1,2,4) is a solution to $ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca \\plus{} a \\plus{} b \\plus{} c} \\equal{} 1.$ We claim there are infinitely many solution. Assume that (a,b,c) is a root, with $ a \\le b \\le c$ and a,b,c positive. Multiply out to get $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} ab \\plus{} bc \\plus{} ca \\plus{} a \\plus{} b \\plus{} c,$ or $ a^2 \\minus{} (b \\plus{} c)a \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} b \\minus{} c \\equal{} 0.$ Consider the polynomial $ x^2 \\minus{} (b \\plus{} c)x \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} b \\minus{} c.$ By the above, one of its roots is a. Vieta gives us that b+c-a is also a root, and that it is greater than c, so (b+c-a,b,c) and (b,c,b+c-a) are solutions. We can keep doing this to give us infinitely many solutions.[/hide]\n\n[hide=\"2\"]Note that $ sin^6 \\plus{} cos^6 \\equal{} (sin^2 \\plus{} cos^2)(sin^4 \\minus{} sin^2cos^2 \\plus{} cos^4) \\equal{} sin^4 \\minus{} sin^2cos^2 \\plus{} cos^4.$ Thus, the expression equals $ \\frac {sin^4 \\plus{} cos^4}{4} \\minus{} \\frac {sin^4 \\minus{} sin^2cos^2 \\plus{} cos^4}{6} \\equal{} \\frac {sin^4}{12} \\plus{} \\frac {sin^2cos^2}{6} \\plus{} \\frac {cos^4}{12} \\equal{} \\frac {sin^4 \\plus{} 2sin^2cos^2 \\plus{} cos^4}{12} \\equal{} \\frac {(sin^2 \\plus{} cos^2)^2}{12} \\equal{} \\frac {1}{12},$ so it always equals 1/12.[/hide]", "Solution_7": "Free problems are free problems. Who doesn't like free problems? :P\r\n\r\nI'd definitely be interested.", "Solution_8": "[quote=\"Differ\"]Free problems are free problems. Who doesn't like free problems? :P\n\nI'd definitely be interested.[/quote]\r\n\r\nThat is why I'll do it and i need that practice in general.", "Solution_9": "I'm interested, although I probably won't do much because difficulty/schoolwork. I promise I'll look at them though :P", "Solution_10": "These are some lovely problems. I will definitely be doing this.", "Solution_11": "I'm interested. I need some practice.", "Solution_12": "I would like to participate.", "Solution_13": "[quote=\"pythag011\"]I'll post solutions to the other 2.\n\n[hide=\"3\"]\nThe problem is just a simple application of backwards Vieta jumping.\n\nNote that (1,2,4) is a solution to $ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca \\plus{} a \\plus{} b \\plus{} c} \\equal{} 1.$ We claim there are infinitely many solution. Assume that (a,b,c) is a root, with $ a \\le b \\le c$ and a,b,c positive. Multiply out to get $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} ab \\plus{} bc \\plus{} ca \\plus{} a \\plus{} b \\plus{} c,$ or $ a^2 \\minus{} (b \\plus{} c)a \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} b \\minus{} c \\equal{} 0.$ Consider the polynomial $ x^2 \\minus{} (b \\plus{} c)x \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} b \\minus{} c.$ By the above, one of its roots is a. Vieta gives us that b+c-a is also a root, and that it is greater than c, so (b+c-a,b,c) and (b,c,b+c-a) are solutions. We can keep doing this to give us infinitely many solutions.[/hide][/hide][/quote]\nJust checked over this, there's a slight error:\n[hide]You said \"Multiply out to get $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} ab \\plus{} bc \\plus{} ca \\plus{} a \\plus{} b \\plus{} c,$ or $ a^2 \\minus{} (b \\plus{} c)a \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} b \\minus{} c \\equal{} 0.$\", but the polynomial is actually $ a^2 \\minus{} (b \\plus{} c\\plus{}1)a \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} b \\minus{} c \\equal{} 0$.\nThis means that $ (a,b,c)$ being a root actually implies that $ (b,c,b\\plus{}c\\minus{}a\\plus{}1)$ is a root. However, that doesn't invalidate the proof at all.[/hide]", "Solution_14": "I will look at them but i dont think i will have the time to do all of them... srry... :(" } { "Tag": [], "Problem": "compute $ (\\sqrt\\frac {1}{2} \\plus{} \\sqrt\\frac {i}{2})^ {12}$ without using a calculator.", "Solution_1": "[hide]\nWhat is so nice about this question anyway? It's tedious and not very interesting, in my opinion. Looks like a homework problem.\n[hide]\n$ \\left(\\sqrt{\\frac{1}{2}} \\plus{} \\sqrt{\\frac{i}{2}}\\right)^{12} \\equal{} \\left(\\frac{1}{2} \\plus{} 2\\sqrt{\\frac{i}{4}} \\plus{} \\frac{i}{2}\\right)^6 \\equal{} \\left(\\frac{1\\plus{}i}{2} \\plus{} \\sqrt{i}\\right)^6$.\n\nSince\n\n$ \\sqrt{i} \\equal{} \\left(e^{i \\pi/2}\\right)^{1/2} \\equal{} e^{i \\pi/4} \\equal{} \\frac{1\\plus{}i}{\\sqrt{2}}$,\n\nwe have\n\n$ \\left(\\frac{1\\plus{}i}{2} \\plus{} \\sqrt{i}\\right)^6 \\equal{} \\left(\\frac{1}{\\sqrt{2}} \\plus{} 1 \\right)^6 (e^{i\\pi/4})^6 \\equal{} \\left(\\frac{1}{\\sqrt{2}} \\plus{} 1\\right)^6 (\\minus{}i) \\equal{} \\frac{\\minus{}i}{8} (1\\plus{}\\sqrt{2})^6$.\n\nNow all that remains is to expand the real factor in parentheses:\n\n$ (1\\plus{}\\sqrt{2})^6 \\equal{} (3 \\plus{} 2\\sqrt{2})^3 \\equal{} 3^3 \\plus{} 3(3^2)(2\\sqrt{2}) \\plus{} 3(3)(2\\sqrt{2})^2 \\plus{} (2\\sqrt{2})^3 \\equal{} 99\\plus{}70\\sqrt{2}$.\n\nHence the answer is $ \\frac{\\minus{}i}{8}(99 \\plus{} 70 \\sqrt{2})$.\n[/hide][/hide]", "Solution_2": "I kind of had a more interesting solution in mind but oh well.....", "Solution_3": "Sorry to double post but why do all my posts get moved here when i post them in HSB?", "Solution_4": "These problems are not HSB level." } { "Tag": [ "limit", "algebra unsolved", "algebra" ], "Problem": "The sequence $(x_{n})$ is defined by $x_{1}= a \\in R$ and\r\n$x_{n}+1 = 3x_{n}^{3}-7x_{n}^{2}+5x_{n}$for n = 1, 2, 3,\r\n Find all $a$ for which $(x_{n})$ has a finite limit and determine\r\nthat limit.", "Solution_1": "I think your problem is as follows.\r\nPlease edit your post.\r\n$x_{n+1}=3x_{n}^{3}-7x_{n}^{2}+5x_{n}\\ (n=1,2,3,\\cdots).$", "Solution_2": "1) If $x_{1}\\le 0$ then $\\lim_{n\\to \\infty }x_{n}=0.$ \r\n2)If $x_{1}>4/3$ then $\\lim_{n\\to \\infty }x_{n}=\\infty .$\r\n3) There are sequence $y_{1}=1, 3y_{n+1}^{3}-7y_{n+1}^{2}+5y_{n+1}=y_{n}$. If $03 /p is not an integer, since p is prime.\r\n\r\nLet S(p) be the LHS of the desired identity.\r\nClearly, S(2) = 0, so that the identity holds in this case thus, from now, we assume that p = 2q+1 is odd.\r\nFor k = 1,...,q, we have :\r\n(p-k) 3 / p = p 2 - 3pk + 3k 2 - k 3 / p.\r\nThus, from (1), we deduce that :\r\n[(p-k) 3 / p ] + [ k 3 / p] = p 2 - 3pk + 3k 2 - 1.\r\nThen :\r\nS(p) = \\sum {1 \\leq k \\leq q} ( [(p-k) 3 / p ] + [ k 3 / p] )\r\n= \\sum {1 \\leq k \\leq q} (p 2 - 3pk + 3k 2 - 1)\r\n= (p 2 - 1)q - 3pq(q+1) + q(q+1)(2q+1)/2\r\n= (p 2 - 1)(p-2)/4, as desired.\r\n\r\nPierre." } { "Tag": [ "LaTeX", "search" ], "Problem": "Hi all, I have a \"problem\" of sorts that I do not know how to solve. On my laptop, if I build a PDF document in TexnicCenter, then view it, then make some changes and recompile and attempt to build again, I get 3 errors along the lines of:\r\n\r\n- I can't write on file 'test.pdf'\r\n\r\n- Emergency Stop!\r\n\r\n- Fatal error occurred, the output PDF file is not finished!\r\n\r\nConsequently, I must always close the currently opened PDF file (created from a previous build) before I can recompile my source code. On my other computer, this is not necessary as it always just overwrites the current PDF file, so I'm wondering if there is some setting or something else that is causing this to happen, or something I can do?\r\n\r\nThanks.", "Solution_1": "In TeXnicCenter, go to the \"Define Output Profiles...\" option of the Build menu.\r\n\r\nMake sure that \"Latex => PDF\" is selected, and click on the Viewer tab.\r\n\r\nThen the one at the bottom -- \"Close document before running (La)TeX\" -- the \"DDE command\" button needs to be selected, and the command is \r\n\r\n[DocClose(\"%bm.pdf\")]\r\n\r\nCompare the settings on this page on your laptop with those on your desktop, and see what's different. If it still doesn't work, click on the \"Wizard\" button and try to reset the output profiles.\r\n\r\nOn the other hand, every once in a while TeXnicCenter will no longer talk to Adobe Reader in the way that it's suppose to. This often necessitates closing both TeXnicCenter and Reader and then restarting them, or in some cases rebooting your PC.", "Solution_2": "Ah ok I got it now. I did have those close document commands previously which was confusing me. However, for some reason under \"View Project's Output\" and \"Forward Search\", it was using Command line arguments instead of DDE Commands. \r\n\r\nFor both \"View project's output\" and \"Forward search\", DDE command must be selected, the command should be \r\n\r\n[DocOpen(\"%bm.pdf\")][FileOpen(\"%bm.pdf\")]\r\n\r\nand for Server, enter \"Acroview\" (without the quotes), and for Topic, enter \"Control\" (without the quotes). Also make sure those are the entries for Server and Topic under the close document part. \r\n\r\nWow, that was annoying.\r\n\r\nEDIT: Oh wow, check this out:\r\n\r\nhttp://en.wikipedia.org/wiki/TeXnicCenter\r\n\r\nIt has all the information you want as well as more goodies, such as making it so your PDF documents go back to the page you were originally on!", "Solution_3": "hmm, i tried to set texniccenter so that its reopens the file after a recompilation at the same location as it was just before by copying the lines from wikipedia\r\n\r\nbut when i try to compile, its gives an error that says it cannot excecute the commands...\r\neven after a restart...\r\n\r\n :(", "Solution_4": "Note that if you follow the Wikipedia instructions for sending a pdf back to where it was previously opened, the pdf must have already been opened, or you will get an error message. The error message does not actually affect the compilation, and you'll get it the first time you compile and open a pdf after starting TexnicCenter. Afterward, every build and recompile will send you to whatever page you had previously been looking at.", "Solution_5": "nope, even after a pc reboot it still doesn't work.\r\n\r\n(i already mentioned that i first needed to open it but that didn't help either", "Solution_6": "That's strange, are you sure you have Server: Acroview, and Topic: Control?", "Solution_7": "hmm, now i know that it can be done i begon to miss it, so here's what i have\r\n\r\ni'm using texniccenter 1 beta 6,31 'Firenze'\r\nand acrobat reader Versie 7.0.5.2005092300\r\n\r\nin attachment you can find how i set the viewer part in texniccenter\r\nthe command i used were these i copied from \r\nany solution on this?? [url=http://en.wikipedia.org/wiki/Texniccenter]wikipedia[/url]", "Solution_8": "Hmm, that is exactly identical to what I have. I have no idea what the problem is. I believe I am using beta version 7.00, but otherwise there is no difference.", "Solution_9": "hm, even after an update to texniccenter 7 it still doesn't work :(", "Solution_10": "Windows can't handle files properly. Only one process at a time can open them. So if you have Adobe Acrobat open, and LaTeX attempts to write, you have an error.", "Solution_11": "[quote=\"Watson\"]Windows can't handle files properly. Only one process at a time can open them. So if you have Adobe Acrobat open, and LaTeX attempts to write, you have an error.[/quote]Not quite. I have a file opened with Acrobat, and I can re-write it using WinEdT." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "search", "group theory", "linear algebra", "superior algebra" ], "Problem": "Hi,\r\n\r\nDoes anyone know of any book/website which has examples of combinatorics or other problems which can be solved by elementary group theory? Can anyone give an example of a combinatoric problems which can be solved by group theory?\r\n\r\nThanks", "Solution_1": "See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1116182#1116182 for an application of group theory to USAMO 2008 #6.\r\n\r\nI've actually also been interested in exactly the same thing, i.e. finding more examples of applications of group theory to olympiad problems.", "Solution_2": "There are, to my mind, two big ways group theory comes up in Olympiad-level (especially combinatorics) problems: [url=http://en.wikipedia.org/wiki/Burnside%27s_lemma]Burnside's lemma[/url] and linear algebra over finite fields (especially $ \\mathbb{F}_2$). See, for example, Gowers' discussion of the latter [url=http://gowers.wordpress.com/2008/07/31/dimension-arguments-in-combinatorics/]here[/url].\r\n\r\nEdit: I also love citing [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=272516508&t=215429]IMO 2008 #5[/url]; the solution I gave implicitly uses the discrete Fourier transform on $ (\\mathbb{Z}/2\\mathbb{Z})^n$, although I didn't recognize it at the time, but I guess I wouldn't call this purely a \"group-theoretic\" technique: you know it as the roots of unity filter." } { "Tag": [ "integration", "calculus", "logarithms", "function", "trigonometry", "calculus computations" ], "Problem": "$ \\int e^x ln(x) dx$\r\n\r\nIts sufficent to compute :$ \\int ln(ln(x)) dx$", "Solution_1": "The indefinite integral is not elementary.", "Solution_2": "do you mean it can not be computed?", "Solution_3": "[quote=\"FOURRIER\"]$ \\int e^x ln(x) dx$\n\nIts sufficent to compute :$ \\int ln(ln(x)) dx$[/quote]\r\n\r\n$ \\int e^x \\ln x dx \\& = \\int \\ln x d \\left( e^x \\right) = e^x \\ln x - \\int \\frac {e^x}{x} dx$\r\n\r\nRecall : $ e^x = \\sum_{\\tiny{k = 0}}^{\\tiny \\infty} \\frac {x^k}{k!} \\;\\; \\Rightarrow \\;\\; \\frac {e^x}{x} = \\sum_{\\tiny{k = 0}}^{\\tiny \\infty} \\frac {x^{\\tiny{k - 1}}}{k!}$\r\n\r\nSo : \r\n\r\n\\begin{align} \\int e^x \\ln x dx & = e^x \\ln x - \\int \\frac {e^x}{x} dx \\\\\r\n& = e^x \\ln x - \\int \\left( \\frac{1}{x} + \\sum_{\\tiny{k = 1}}^{\\tiny \\infty} \\frac {x^{\\tiny{k - 1}}}{k!} \\right) dx \\\\\r\n& = e^x \\ln x - \\ln x + \\sum_{\\tiny{k = 1}}^{\\tiny \\infty} \\int \\frac {x^{\\tiny{k - 1}}}{k!} dx \\\\\r\n& = e^x \\ln x - \\ln x + \\sum_{\\tiny{k = 1}}^{\\tiny \\infty} \\frac {x^{\\tiny k}}{k.k!} \\end{align}", "Solution_4": "You can do certain definite integrals, but the indefinite integral is impossible. (I suppose you could express it in terms of Ei(x) if you want)\r\n$ \\int_{ \\minus{} \\infty}^{0} e^{x}ln|x|dx \\equal{} \\minus{} \\gamma$", "Solution_5": "[quote=\"FOURRIER\"]do you mean it can not be computed?[/quote]\r\n\r\nI mean it cannot be expressed in terms of \"elementary functions\" (unless you consider the [url=http://en.wikipedia.org/wiki/Exponential_integral]exponential integral[/url] an elementary function).", "Solution_6": "[quote=\"FOURRIER\"]$ \\int e^x ln(x) dx$[/quote]\n\nrather than just staring at the ceiling and coming up with these integrals and posting them on the forum, do some preliminary work yourself . for example, when you see a definite integral, first convince yourself that the integral converges ( and don't blindly use a software for that). if you find that it converges, proceed to find out a closed form....... failing that, post it on the forum saying what you actually did in your work and where you may have gotten stuck so others can try to point you in the right direction.\n\ndon't just blurt out some indefinite integral like :D \n\n$ \\int \\; \\ln\\,\\left(\\dfrac{\\tan^{ \\minus{} 1}\\;\\sqrt {\\; 1 \\plus{} x^3\\; }}{e^x \\minus{} 2\\cos^{ \\minus{} 1}\\,x^3 \\plus{} \\sin\\, \\sqrt x}\\right) \\;\\textbf dx$\n\nand expect someone to solve it. it doesn't work that way, unfortunately.\n\n\n[quote=\"FOURRIER\"]\nIts sufficent to compute :$ \\int ln(ln(x)) dx$[/quote]\r\n\r\nsame thing with your vague hints... :| care to explain what you mean by \"it's sufficient\" ... i am interested to know...", "Solution_7": "[quote=\"misan\"]\n[quote=\"FOURRIER\"]\nIts sufficent to compute :$ \\int ln(ln(x)) dx$[/quote]\n\nsame thing with your vague hints... :| care to explain what you mean by \"it's sufficient\" ... i am interested to know...[/quote]\n\nPut $ t \\equal{} e^x$\n\n\n\n[quote=\"misan\"][quote=\"FOURRIER\"]$ \\int e^x ln(x) dx$[/quote]\n\nrather than just staring at the ceiling and coming up with these integrals and posting them on the forum, do some preliminary work yourself . for example, when you see a definite integral, first convince yourself that the integral converges ( and don't blindly use a software for that). if you find that it converges, proceed to find out a closed form....... failing that, post it on the forum saying what you actually did in your work and where you may have gotten stuck so others can try to point you in the right direction.\n[/quote]\r\n\r\n\r\nwhat do you mean by convergence of an integral??", "Solution_8": "misan is talking about improper definite integrals. Often a problem like this (where the integrand does not have an elementary antiderivative) is phrased as a definite rather than indefinite integral, and the answer turns out to be expressible in terms of well-known constants (rather than functions)." } { "Tag": [ "inequalities", "geometry proposed", "geometry" ], "Problem": "Prove that\r\n\r\n$ h_a\\plus{}h_b\\plus{}h_c\\geq\\frac{9[ABC]}{s}$", "Solution_1": "Rewrite the inequality as : $ \\frac{2[ABC]}{a}\\plus{}\\frac{2[ABC]}{b}\\plus{}\\frac{2[ABC]}{c}\\ge \\frac{9[ABC]}{s}$\r\n\r\n$ \\Longleftrightarrow\\ \\frac 1a\\plus{}\\frac 1b\\plus{}\\frac 1c\\ge\\frac{9}{a\\plus{}b\\plus{}c}$ , which is true by [b]Cauchy-Schwarz[/b] ." } { "Tag": [ "inequalities", "algebra", "binomial theorem", "number theory unsolved", "number theory" ], "Problem": "For every natural number $ n$ prove the inequality\r\n\\[ (n\\minus{}1)^n\\plus{}2n^n\\leq (n\\plus{}1)^n\\leq 2(n\\minus{}1)^n\\plus{}2n^n.\\]", "Solution_1": "Dividing by $ n^n$ yields:\r\n\r\n$ (1\\minus{}\\frac{1}{n})^n\\plus{}2 \\le (1\\plus{}\\frac{1}{n})^n \\le 2(1\\minus{}\\frac{1}{n})^n\\plus{}2$\r\n\r\nJust looking at the right side and binomial theorem you only have to show that\r\n\r\n$ \\binom{n}{2k}\\frac{1}{n^{2k}}\\minus{}3\\binom{n}{2k\\plus{}1}\\frac{1}{n^{2k\\plus{}1}} \\ge 0$\r\n\r\nthat should be straight forward...\r\n\r\nlhs should solve analogously", "Solution_2": "[quote=\"wauwau\"]\n\nJust looking at the right side and binomial theorem you only have to show that\n\n$ \\binom{n}{2k}\\frac {1}{n^{2k}} \\minus{} 3\\binom{n}{2k \\plus{} 1}\\frac {1}{n^{2k \\plus{} 1}} \\ge 0$\n\n[/quote]\r\n\r\nI don't think it's just this..." } { "Tag": [ "LaTeX", "calculus", "calculus computations" ], "Problem": "Find the exact values of:\r\n\r\n$\\sum_{n=1}^{\\infty}\\frac n{2^n}$\r\n\r\n$\\sum_{n=1}^{\\infty}\\frac{n^2}{2^n}$\r\n\r\n$\\sum_{n=1}^{\\infty}\\frac{n^4}{2^n}$", "Solution_1": "First one is equal with $2$.hope I'm not mistaken. :?", "Solution_2": "a_1 = 1/2, r=2\r\n\r\nsum=(1/2)/(1-1/2)^2\r\n=(1/2)/(1/4)\r\n=2\r\n\r\nso that confirms your answer", "Solution_3": "#1\r\nderivation:\r\n\r\nsum = 1*1/2 + 2*1/4 + 3*1/8+ 4*1/16 + ...\r\n= 1*1/2 + (1+1)*1/4 + (1+1+1)*1/8 + (1+1+1+1)*1/16 + ....\r\n\r\ndefine P = 1/2 + 1/4 + 1/8 + 1/16 + ... = 1\r\n\r\nsum = P + 1/2P + 1/4P + 1/8P + ....\r\n= P(1+1/2+1/4+....)\r\n= 1*2\r\n=2\r\n\r\n#2\r\n\r\n i get 6", "Solution_4": "How did you get $6$? ;)", "Solution_5": "#2\r\n\r\nlet P be the original sum\r\n\r\n(1/2)*P shifts all the term to the right\r\n\r\nP - 1/2P =1/2P gives the first term from the original summation + sum n=1 to inf [ ((n+1)^2-n^2)/2^(n+1)]\r\n\r\nthe new summation can be calculated using the technique used in #1 . we get 1/2P = 1/2 + 5/2\r\n\r\nP = our desired sum = 6\r\n\r\ni believe #3 can be solved using the same method. i'll try to use latex", "Solution_6": "Very nice! :D ;)", "Solution_7": "$P = \\sum_{n=1}^{\\infty}\\frac {n^4}{2^n} = \\frac 1{2} + \\sum_{n=1}^{\\infty}\\frac {(n+1)^4}{2^{n+1}}$\r\n$\\frac 1{2}P = \\sum_{n=1}^{\\infty}\\frac {(n)^4}{2^{n+1}}$\r\n$\\frac 1{2}P = P - \\frac 1{2}P = \\frac 1{2} + \\sum_{n=1}^{\\infty}\\frac {(n+1)^4-n^4}{2^{n+1}}$\r\n$ = \\frac 1{2} + \\sum_{n=1}^{\\infty}\\frac {4n^3 + 6n^2 + 4n + 1}{2^{n+1}}$\r\n$ = \\frac 1{2}+ 2\\sum_{n=1}^{\\infty}\\frac {n^3}{2^n} + 3\\sum_{n=1}^{\\infty}\\frac {n^2}{2^n} + 2\\sum_{n=1}^{\\infty}\\frac n{2^n} + \\frac 1{2}$\r\n$ = 23 + 2\\sum_{n=1}^{\\infty}\\frac {n^3}{2^n}$\r\n\r\n$Q = \\sum_{n=1}^{\\infty}\\frac {n^3}{2^n}$\r\n$\\frac 1{2}Q = \\frac 1{2} + \\sum_{n=1}^{\\infty}\\frac {(n+1)^3 - n^3}{2^{n+1}}$\r\n$Q = 1 +\\sum_{n=1}^{\\infty}\\frac {3n^2 + 3n + 1}{2^n}$\r\n$Q = 26$\r\n\r\n$P = 2*(23 + 2*26)$\r\n$P = 2*75$\r\n$\\sum_{n=1}^{\\infty}\\frac {n^4}{2^n} = P = 150$" } { "Tag": [], "Problem": "Does anyone have gregx007's old USNCO pdf notes? \r\nCould you post them by any chance? The wayBackMachine (internet archive) hath failed me, and, as far as I can tell gregx007's account is inactive.", "Solution_1": "here it is. saved it from a while ago.\r\n\r\nenjoy :)", "Solution_2": "Thanks, adchia :D" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "$ \\sum_{n} \\frac{(\\minus{}1)^{[\\sqrt{n}]}}{n}$\r\n\r\n[color=darkblue]where $ [x]$ denotes the biggest integer less than $ x$.[/color]", "Solution_1": "hello, your sum does not converge.\r\nSonnhard.", "Solution_2": "The series converges: http://www.mathlinks.ro/Forum/viewtopic.php?p=708340#708340", "Solution_3": "$\\frac{1}{2}$", "Solution_4": "Dr. Graubner is right.", "Solution_5": "No, Sonnhard's not right. Like him, you make the mistake of putting too much trust in computer algebra systems. This depends on putting the terms into blocks of the same sign and estimating their size, which computer tools are not clever enough to notice." } { "Tag": [ "geometry", "projective geometry", "linear algebra" ], "Problem": "I must now learn a subject affine geometry but I haven't any good ebook for this , so if you have it , up it here for me . Thanks :)", "Solution_1": "[quote=\"mita2002\"]I must now learn a subject affine geometry but I haven't any good ebook for this , so if you have it , up it here for me . Thanks :)[/quote]\r\nPlease reply me , I don't wanna this but I haven't got any good book for this , more it's hard to find one", "Solution_2": "Nobody replied this topic , so Can you let me know the name and author of the book about Affine geometry you think it's best", "Solution_3": "I don't know if it's the \"best\" (as it's the only book on the subject I've ever read), but I like [url=http://www.amazon.com/Affine-Projective-Geometry-M-Bennett/dp/0471113158/ref=sr_1_1?ie=UTF8&s=books&qid=1242330269&sr=8-1]Affine and Projective Geometry[/url] by M. K. Bennett. It's approach is very much tied to the subjects of Linear Algebra and Abstract Algebra (particularly Division Rings and Fields), so a good knowledge of those subjects is necessary for a thorough understanding of this book." } { "Tag": [ "calculus", "probability", "linear algebra" ], "Problem": "when i come here i see all these topics about quants but there are hardly any (if any at all) about actuaries. I was wondering if someone could provide me with more information on this field of work.", "Solution_1": "What are actuaries?", "Solution_2": "Think of actuaries as the \"engineers\" of the insurance business (which is broad enough to include pension and benefit programs). They work for insurance companies, pension funds, and as consultants. Some also work for the government agencies that regulate insurance companies or oversee government pension programs. Twenty years ago, most of the top officers of insurance companies were trained as actuaries (and most of the top officers of oil companies were trained as engineers or geologists). That may not be true any more in either case.\r\n\r\nActuaries have a very defined career path and structure, with a series of nationally-set exams being the centerpiece. The first exam involve calculus, linear algebra, probability and statistics. Some of the later exams are mathematical in nature, but others involve finance, law, or regulations. Becuase of the mathematical nature of some of the early exams, it's easer for college math majors or minors to break into the field than for others, but it's not an absolute requirement.\r\n\r\nThe descriptions of \"quants\" that have appeared in this forum make it appear to be a high-risk, high-reward, high-stress job with an emphasis on creativity and dynamic problem-solving. The actuarial career path is nearly the opposite - a stable-income, regular-hours job in which knowledge and experience count for a lot.\r\n\r\nThe actuarial societies' own propaganda site would be a place to start: http://www.beanactuary.org .", "Solution_3": "Interesting.\r\nSounds familiar but I haven' heard of the term actuaries before.", "Solution_4": "I have been an actuary for 30 plus years.\r\n\r\nWe are applied mathematicians specializing in insurance. Generally we determine the insurance premium to charge for some sort of insurance deal (life, health, pension, property, casualty), or we determine the amount of funds to set aside for reserves for the eventual claims for such deals. Actuaries do many other things, and some are president of their companies.\r\n\r\nWe have a series of actuarial exams to pass to be fully credentialed. The first exam has changed from what is described above. We probably do not have the same kind of work hours and pressures of financial engineers, but we are not necessarily 9 to 5 ers either. We have the solvency of our firms in our hands. We also have a code of ethics, and can potentially lose our credentials, like doctors or lawyers - many other professionals have nothing like that.\r\n\r\nA friend of mine has a good intro page to actuaries called Actuarial Advice: http://users.aol.com/fcas/advice.html" } { "Tag": [ "inequalities", "logarithms", "function", "integration", "calculus", "calculus computations" ], "Problem": "Show that $ \\frac{1}{x} > \\ln (1\\plus{}x) \\minus{} \\ln x > \\frac{1}{1\\plus{}x}$ for $ x>0$\r\n\r\nDeduce that, for $ x>0$, $ (1\\plus{} \\frac{1}{x})^{x}$ increases and $ (1\\plus{}\\frac{1}{x})^{x\\plus{}1}$ decreases as $ x$ increases.", "Solution_1": "[quote=\"AndrewTom\"]Show that $ \\frac {1}{x} > \\ln (1 \\plus{} x) \\minus{} \\ln x > \\frac {1}{1 \\plus{} x}$ for $ x > 0$\n\nDeduce that, for $ x > 0$, $ (1 \\plus{} \\frac {1}{x})^{x}$ increases and $ (1 \\plus{} \\frac {1}{x})^{x \\plus{} 1}$ decreases as $ x$ increases.[/quote]\r\nUse the mean value theorem .. :cool:", "Solution_2": "Since $ y \\equal{} \\frac {1}{t}\\ (t > 0)$ is an decreasing and convex function, \r\n\r\n$ x\\leq t\\leq x \\plus{} 1\\ (x > 0)\\Longrightarrow \\frac {1}{x \\plus{} 1}\\leq t\\leq \\frac {1}{x}$\r\n\r\n$ \\therefore \\int_x^{x \\plus{} 1} \\frac {1}{x \\plus{} 1}\\ dt < \\int_x^{x \\plus{} 1} \\frac {1}{t}\\ dt < \\int_x^{x \\plus{} 1} \\frac {1}{x}\\ dt$\r\n\r\n$ \\therefore \\frac {1}{x \\plus{} 1} < \\ln (1 \\plus{} x) \\minus{} \\ln x < \\frac {1}{x}$", "Solution_3": "A nice proof of the given inequality(often called Napier's ineq) is in the book [Proof without Words] by American Mathematical Association." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "a,b,c,d are real numbers\r\nproof\r\n[code](1/a+1/b+1/c+1/d)^2 >= 1/a^2 + 4(a^2+b^2) + 12/(a^2+b^2+c^2) + 18/(a^2+b^2+c^2+d^2)[/code]", "Solution_1": "[quote=\"colorfuldreams\"]a,b,c,d are real numbers\nproof\n[code](1/a+1/b+1/c+1/d)^2 >= 1/a^2 + 4(a^2+b^2) + 12/(a^2+b^2+c^2) + 18/(a^2+b^2+c^2+d^2)[/code][/quote]\r\n\r\n\r\n\r\n$ (\\frac{1}{a} \\plus{} \\frac{1}{b} \\plus{} \\frac{1}{c} \\plus{} \\frac{1}{d})^2\\geq \\frac{1}{a^2} \\plus{}$ $ \\frac{4}{a^2 \\plus{} b^2} \\plus{} \\frac{12}{a^2 \\plus{} b^2 \\plus{} c^2} \\plus{} \\frac{18}{a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2}$\r\n\r\n??", "Solution_2": "[quote=\"hedeng123\"][quote=\"colorfuldreams\"]a,b,c,d are real numbers\nproof\n[code](1/a+1/b+1/c+1/d)^2 >= 1/a^2 + 4(a^2+b^2) + 12/(a^2+b^2+c^2) + 18/(a^2+b^2+c^2+d^2)[/code][/quote]\n\n\n\n$ (\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\plus{} \\frac {1}{d})^2\\geq \\frac {1}{a^2} \\plus{}$ $ \\frac {4}{a^2 \\plus{} b^2} \\plus{} \\frac {12}{a^2 \\plus{} b^2 \\plus{} c^2} \\plus{} \\frac {18}{a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2}$\n\n??[/quote]\r\n\r\nYES", "Solution_3": "take $ a\\equal{}\\minus{}b ,c\\equal{}\\minus{}d$, $ abcd>0$\r\nthen $ LHS\\equal{}0$,$ RHS>0$\r\nso your statement is wrong :wink:", "Solution_4": "[quote=\"\u222bFaiLurE\u222e\"]take $ a \\equal{} \\minus{} b ,c \\equal{} \\minus{} d$, $ abcd > 0$\nthen $ LHS \\equal{} 0$,$ RHS > 0$\nso your statement is wrong :wink:[/quote]\r\n\r\nsorry\r\nthe question is\r\n$ a, b,c ,d$are real positive numbers" } { "Tag": [ "puzzles" ], "Problem": "Guess the word palindrome.\r\nPronounced as one letter but written with three, only two different letters are used to make me.\r\nI'm double, I'm single I'm black, blue, and gray. \r\nI'm read from both ends and the same either way.", "Solution_1": "[hide=\"answer\"]Eye[/hide]\r\n\r\nJust go through the letters...", "Solution_2": "An eye is the answer", "Solution_3": "There is no reason to revive a 5-year old thread, especially when you are not adding anything to the discussion." } { "Tag": [ "Asymptote", "LaTeX", "function", "geometry", "rectangle", "analytic geometry", "conics" ], "Problem": "how do you draw semi-figures with asy?\r\nI know how to draw regular figures, but couldn't figure out how to draw semi-figures..", "Solution_1": "Uh?\r\n\r\ndraw((x,y)--(x2,y2)-- [etc.] );\r\n\r\nwill draw about anything. Consult asymptote.pdf inside your asymptote download for a full tutorial.\r\n\r\nTo use asymptote on the forums:\r\n\r\n[asy]size(200); pair A,B,C,D,P; A=(0,0); B=(200,0); C=(200,150); D=(0,150); P=(124,85); draw(A--B--C--D--cycle); label(\"A\",A,(-1,0)); dot(A); label(\"B\",B,(0,-1)); dot(B); label(\"C\",C,(1,0)); dot(C); label(\"D\",D,(0,1)); dot(D); dot(P); label(\"P\",P,(1,1)); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label(\"w\",(124,0),(0,-1)); label(\"x\",(200,85),(1,0)); label(\"y\",(124,150),(0,1)); label(\"z\",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85));[/asy]\r\n\r\nfor example.\r\n\r\n is for the wiki.\r\n\r\nAnd there's a specific forum from asymptote and latex questions: http://www.artofproblemsolving.com/Forum/index.php?f=123\r\n\r\nEDIT: Okay, moved.", "Solution_2": "[color=red]Unnecessary quote removed by moderator[/color]\r\n\r\nI know how to do those... \r\nWhat I want to know is how to draw semi-circles and other semi-stuffs.", "Solution_3": "Semi-circles can be drawn with arc command. For instance, \r\ndraw(arc((0,0),2,0,180));\r\nproduces\r\n[asy]draw(arc((0,0),2,0,180));[/asy]\r\nAs to semi-stuff, that depends on what is that stuff. There is no command \"semi()\" with picture argument in asy (for obvious reasons: not everything can be divided into two equal halves). If you tell exactly what other semi-stuff you want to draw, somebody will show you how to do it :)", "Solution_4": "Oh, semi?\r\n\r\nWell, half of a square is just a rectangle. Unless you're splitting it down the diagonal?\r\n\r\nWell, anyways, you could write a function to draw a rectangle given it's square;'s coordinates, double, I suppose. You could repeat the process for any other polygon, you wouldn't need anything except draw commands and a few conditionals.\r\n\r\nSemi circles and ellipses are trickier...", "Solution_5": "Well, how do you close the bottom?\r\n\r\nEDIT: Never mind.. I figured it out.\r\n[asy]draw((-2,0)--(2,0)); draw(arc((0,0),2,0,180));[/asy]" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Hi, do you know if there is even an official Maryland Mathcounts website? I believe there should be since 3 Marylanders placed in the 2009 Nationals, but I can't find it, so I can't register.", "Solution_1": "sorry I counldn't find a website but try going to mathcounts.org and seeing if they have anything\r\nalso, i think the registration deadline was sometime in the past few days, so you might want to hurry up", "Solution_2": "The deadline is over, and I couldn't find a Maryland website either :(...", "Solution_3": "[quote=\"TheDeity\"]Hi, do you know if there is even an official Maryland Mathcounts website? I believe there should be since 3 Marylanders placed in the 2009 Nationals, but I can't find it, so I can't register.[/quote]\r\n\r\nCompletely unrelated, but if you are referring to like national countdown, that would be 3 Marylanders in 2008 Denver.", "Solution_4": "umm, i'm pretty sure that there is a maryland mathcounts website, but it hasn't been updated for several years. anyway, I think you're supposed to register on the official MathCounts website.", "Solution_5": "yea. But I think the deadline's already over. Our teacher proabably already registered or something", "Solution_6": "oh, ok. thank you all.", "Solution_7": "no problem. Did you get a chance to register though?" } { "Tag": [], "Problem": "Find the eighteen digit in decimal of numbers $ \\sqrt[3]{37}$", "Solution_1": "Super title... Use titles that have more content, see http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 .\r\nAnd I don't see the fun of calculating digits ;)", "Solution_2": "Help me,please!", "Solution_3": "Your calculator can do it if you're tricky. :)" } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "Hello, I have the following problem:\r\n\r\nYou have the numbers from 1 to k and you take n elements of this set.\r\nNow I want to prove that there are two distinct subsets of this set which have the same sum of their elements.\r\nInteresting for me are the following cases:\r\n\r\na) k=2^(n-1)\r\nb) k=2^n-1\r\n\r\nI would be pleased if you could find a solution (positive or negative) for this problem.\r\n\r\nNaphthalin", "Solution_1": "ok assume that $X=\\{1,...,k\\},A=\\{x_1,x_2,...,x_i\\},B=\\{y_1,y_2,...,y_j\\}$\r\nand also $A \\bigcup B=X$ such that $\\sum_{r=1}^i x_r=\\sum_{s=1}^j y_s$\r\nwe have:$1+2+...+k=\\frac{k(k+1)}{2}$\r\nbut if we can decompose the set into two equal subsets then the sum of the elements of $X$ must be an even number\r\nthere for $4|k(k+1)$ there for $k \\equiv 0,-1 (\\bmod 4)$\r\nnow we prove that for every $k$ such that $k \\equiv 0,-1 (\\bmod 4)$ we can decompose $X$ into two equal subsets\r\nassume that $k=4n$ or $k=4n-1$\r\nnow use induction on $n$ for $n=1$ the problem is solved\r\nnow for $k=4n$ assume that we have solved the problem for $k=4n$ now we want to solve it for $k=4n+4$\r\nput $A'=A \\bigcup \\{4n+1,4n+4\\},B'=B \\bigcup \\{4n+2,4n+3\\}$\r\nby the induction hypothesis we have $|A|=|B|$ but $|A'|=|A|+4n+1+4n+4,|B'|=|B|+4n+2+4n+3$ there for $|A'|=|B'|$\r\nnow for $k=4n-1$ assume that we have solved the problem for $k=4n-1$ now we want tpo solve it for $k=4n+3$\r\nwe know that $A \\bigcup B=X$ there for number $1$ should be in $A$ or $B$ assume that $1 \\in B$ now move $1$ from $B$ to $A$\r\nnow put $A'=A \\bigcup \\{4n-1,4n+3\\},B'=B \\bigcup \\{4n+1,4n+2\\}$\r\nby the induction hypothesis we have $|A|=|B|$ but $|A'|=|A|+1+4n-1+4n+3,|B'|=|B|-1+4n+1+4n+2$\r\nthere for $|A'|=|B'|$\r\nand the problem is solved for every $k \\equiv 0,-1 (\\bmod 4)$\r\nnow for $n>2$ we know that $k=2^{n-1} \\equiv 0 (\\bmod 4)$\r\nand for $n\\geq 2$ we know that $k=2^n-1 \\equiv -1 (\\bmod 4)$\r\nthere for we can do it for part $a,b$ with the above conditions\r\nand the problem goes false for part $a$ for $n\\leq 2$ and it goes false for part $b$ for $n<2$", "Solution_2": "Thanks for your answer, but I think you understood it in a wrong way because I didn't describe my problem exactly enough.\r\nMy problem is: I have the set $Y=\\{1,\\dots ,k\\}$ and I take a subset $X$ with $n$ different elements of the set $Y$. Now I want to take two distinct subsets $M$ and $N$ of $X$ which have the same sum of their elements.\r\nInteresting for me are the two following cases:\r\nCan I find two distinct subsets $M$ and $N$ of $X$ with the same sum of their elements for every subset $X$ of $Y$ for\r\na) $k=2^{n-1}$\r\nb) $k=2^{n}-1$ ?\r\n\r\nNaphthalin", "Solution_3": "ooooooooooooooooooooopps :oops: \r\nsorry\r\nIm really sorry I`ll try this one thanks for the advice", "Solution_4": "no problem bcause i calcuted it and the maximum $k$ for different $n$ is:\r\n$k=13$ for $n=4$ instead of $k=15$, $k=24$ for $n=5$ instead of $k=31$, $k=46$ for $n=6$ instead of $k=65$ and $k=85$ for $n=7$ instead of $k=127$.\r\nnow i want to have the maximum $k$ for each $n$.\r\n\r\nNaphthalin" } { "Tag": [ "induction" ], "Problem": "we call a L-tromino a figure with 3 squares joined together in an \"L\" shape.\r\n\r\na square courtyard with side length $2^n$ units is divided into unit squares by lines parallel to its sides. one of the unit quares is reserved for a drainage outlet.\r\n\r\nprove that the rest of the courtyard can be tiled with unborken, non-overlapping l-tromino shaped tiles.", "Solution_1": "A bit hard to explain without a figure.\r\n\r\nCut the courtyard in 4 parts. The part (let's consider it is the upper right) with the missing square is coverable by induction hypothesis.\r\n\r\nLets remove the bottom right square of the upper left part and cover it with 'L'\r\nLets remove the upper right square of the bottom left part and cover it with 'L'\r\nLets remove the bottom left square of the botom rigth part and cover it with 'L'\r\n\r\n.........X|...........\r\n---------l---------\r\n.........X|X..........\r\n\r\n\r\n\r\nThere is a small 'L'-shaped hole that can be covered with an 'L' between the 3 parts.\r\n\r\nQED :P" } { "Tag": [], "Problem": "$1$ The sum of several consecutive positive integers is $171966464$. what are the numbers?\r\n\r\n$2$ In how many ways can $608608$ be expressed as the sum of several consecutive numbers?\r\n\r\n$3$ Given $N$, in how many ways can $N$ be expressed as the sum of consecutive positive integers?", "Solution_1": "[hide=\"Hint\"] A sum of $2k-1$ consecutive integers is divisible by $2k-1$. [/hide]", "Solution_2": "[quote=\"t0rajir0u\"][hide=\"Hint\"] A sum of $2k-1$ consecutive integers is divisible by $2k-1$. [/hide][/quote]\r\n\r\nCan you give some more details?", "Solution_3": "[hide=\"Explanation of hint\"] The $2k-1$ numbers\n\n$(a-(k-1))+(a-(k-2))+...+(a-1)+a+(a+1)+...+(a+(k-2)+(a+(k-1))$\n\nSum by Gaussian pairing to\n\n$(2k-1)a$\n\nBecause $a-(k-1) = b$ must be positive this is better written as\n\n$(2k-1)(b+(k-1))$ [/hide]\n[hide=\"Second hint\"] A sum of $2k$ integers is congruent to $k \\bmod 2k$. [/hide]", "Solution_4": "My homework over break was something a lot like this. It says \"Call T(n) the number of ways to represent a positive integer n as the sum of consecutive positive integers.\" Then it asks you to prove all of the following:\r\na. T(prime) = 2\r\nb. T(power of 2) = 1\r\nc. T(prime raised to the power of a) = a+1\r\nd. T((p^a)(q^b)) = (T(p^a))(T(q^b)) if p and q are distinct odd primes and a and b are positive integers.\r\n\r\n(Any time it says \"prime\", 2 is NOT included.)" } { "Tag": [ "integration", "vector", "linear algebra", "matrix", "algebra", "polynomial", "real analysis" ], "Problem": "Prove that there exist real coefficients not all zero, such that:\r\n\r\n$ (a_0 \\plus{} ... \\plus{} a_{n\\minus{}1})^2 \\equal{} n^2 \\int_0^1{(a_0 \\plus{} a_1 x \\plus{} ... \\plus{}a_{n\\minus{}1}x^{n\\minus{}1})^2 dx}$\r\n\r\nwhere n is integer at least 2 ( for example in case n = 2 we can find vector $ a_0,a_1 \\equal{} \\minus{}3a_0$ )", "Solution_1": "Let's rephrase this as a linear algebra problem.\r\n\r\nIndex vectors in $ \\mathbb{R}^n$ and $ n\\times n$ matrices so that the indices run from $ 0$ to $ n\\minus{}1.$ Let $ H$ be the $ n\\times n$ matrix whose $ (i,j)$ entry is $ \\frac1{i\\plus{}j\\minus{}1}.$ Let $ J$ be the $ n\\times n$ matrix with every entry equal to $ 1.$ We're asked to show that for some nonzero vector $ a,$\r\n\\[ a^T\\left(n^2H\\minus{}J\\right)a\\equal{}0\\]\r\nThis requires precisely that the matrix $ n^2H\\minus{}J$ be singular.\r\n\r\nFor the $ n\\equal{}2$ example that mareleG gave, we have that\r\n\\[ n^2H\\minus{}J\\equal{}\\begin{bmatrix}3&1\\\\1&\\frac13\\end{bmatrix}\\]\r\nand it is clear that that is a singular matrix and that $ a\\equal{}\\begin{bmatrix}1&\\minus{}3\\end{bmatrix}^T$ lies in its null space.\r\n\r\nFor $ n\\equal{}3,$ $ n^2H\\minus{}J\\equal{}\\begin{bmatrix}8&\\frac72&2\\\\ \\frac72&2&\\frac54\\\\ 2&\\frac54&\\frac45\\end{bmatrix}.$\r\n\r\nA brute force calculation shows that this is also a singular matrix.\r\n\r\nNow, as for the general case ...", "Solution_2": "this was my approach too, the next step is to prove singularity of that matrix.\r\nhint: same technique used for Cauchy matrix, with elements $ 1/(i \\plus{} j)$ or $ 1/(x_i \\plus{} y_j)$\r\nusing this problem one can find a solution for problem 4 - SEEMOUS 2008, different from official solution", "Solution_3": "That $ n^2$ really reminds me of the result [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=146373]here[/url] that $ j^T H^{ \\minus{} 1}j \\equal{} n^2$, where $ j$ is the $ n\\times 1$ vector of all ones.\r\n\r\nOK, what if $ a \\equal{} H^{ \\minus{} 1}j$? We have\r\n$ a^T(n^2H \\minus{} J)a \\equal{} j^T(H^{ \\minus{} 1})^T(n^2H \\minus{} jj^T)H^{ \\minus{} 1}j$\r\n$ \\equal{} n^2 j^T H^{ \\minus{} 1} H H^{ \\minus{} 1}j \\minus{} j^T H^{ \\minus{} 1}j j^T H^{ \\minus{} 1}j$\r\n$ \\equal{} n^2 j^T H^{ \\minus{} 1}j \\minus{} (j^T H^{ \\minus{} 1}j)^2 \\equal{} n^4 \\minus{} n^4 \\equal{} 0$\r\n\r\nSecond thought: the connection is even closer than I thought at first. The appropriate polynomial $ A(x)\\equal{}a_0\\plus{}a_1x\\plus{}\\cdots\\plus{}a_{n\\minus{}1}x^{n\\minus{}1}$ has the property that $ \\langle A(x),x^k\\rangle\\equal{}1$ for each $ k\\in \\{0,1,\\cdots,n\\minus{}1\\}$ (or it's a multiple of that one)", "Solution_4": "It seems that Colombian Olympiad 06 problem is the same as SEEMOUS problem :)\r\nHilbert matrix is a particular case of Cauchy matrix $ 1/(x_i \\plus{} y_j)$\r\nMy solution relies on elementary proving that matrix $ 1/(x_i \\plus{} y_j) \\minus{} 1$ is singular when $ x_1 \\plus{} ... \\plus{} x_n \\plus{} y_1 \\plus{} ... \\plus{} y_n \\equal{} 1$ and choose $ x_i \\equal{} i/n^2, y_j \\equal{} (j \\minus{} 1)/n^2, i,j \\equal{} 1,...,n$" } { "Tag": [ "modular arithmetic" ], "Problem": "Prove that if x^2 + xy + y^2 is divisible by 10 it's divisible by 100.", "Solution_1": "$10|x^{2}\\implies 10|x$.\r\n$10|y^{2}\\implies 10|y$.\r\n\r\nThus $100|xy$.\r\n\r\nSo yeah...", "Solution_2": "How does this prove it?", "Solution_3": "[quote=\"miyomiyo\"]How does this prove it?[/quote]\r\nSince $10|x$, $100|x^{2}$.\r\n\r\nAlso, $100|y^{2}$.\r\n\r\nSo $x^{2}+xy+y^{2}=100(\\text{blah})$.", "Solution_4": "Why is $x$ divisible by $10$?", "Solution_5": "[quote=\"miyomiyo\"]Why is $x$ divisible by $10$?[/quote]\r\nSince $10|(x^{2}+xy+y^{2})$, it divides each term of the sum.", "Solution_6": "[quote=\"rnwang2\"][quote=\"miyomiyo\"]Why is $x$ divisible by $10$?[/quote]\nSince $10|(x^{2}+xy+y^{2})$, it divides each term of the sum.[/quote]\r\n\r\num, that's not true with 3,7,13,19,31,37,43,61... (primes that are 1 mod 6). You have to show that it's actually true for 2 and 5.", "Solution_7": "Here's a proof with two cases.\r\n\r\nCase 1: $x=y$. Then $3x^{2}\\equiv 0 \\pmod{10}$, so we conclude that $x \\equiv 0 \\pmod{100}$ and the result follows from there.\r\n\r\nCase 2: $x \\neq y$. Rewrite the expresion as $\\frac{x^{3}-y^{3}}{x-y}\\equiv 0 \\pmod{10}$. Thus $\\frac{x^{3}-y^{3}}{x-y}=10k$ for some integer $k$, or $x^{3}-y^{3}=(x-y)10k$, from which it follows that $x^{3}\\equiv y^{3}\\pmod{10}$.\r\nThis last congruence implies $x \\equiv y \\pmod{10}$, so our original equation becomes $3x^{2}\\equiv 0 \\pmod{10}$, from which we get $x^{2}\\equiv 0 \\pmod{10}$ and then $x \\equiv 0 \\pmod{10}$.\r\nThus we actually have $x^{2}\\equiv 0 \\pmod{100}$, so that we can write $x^{2}+xy+y^{2}\\equiv 3x^{2}\\equiv 0 \\pmod{100}$.", "Solution_8": "I must have been hallucinating when I wrote those posts. :blush:", "Solution_9": "I cant find it anywhere, but what does the | mean when you post things such as $10|x$ etc..\r\n\r\nthanks :oops:", "Solution_10": "[quote=\"GlenthemanN\"]I cant find it anywhere, but what does the | mean when you post things such as $10|x$ etc..\n\nthanks :oops:[/quote]\r\n\r\nIt means that 10 divides x, or x is divisible by 10." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "This is to my friend, pvthuan, but all are welcome to submit solution.\r\nFor any positive real numbers $a,b$ and $c$,\r\n\\[\\frac{1}{a^{2}+bc}+\\frac{1}{b^{2}+ca}+\\frac{1}{c^{2}+ab}\\le\\frac{a+b+c}{ab+bc+ca}\\left(\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a}\\right) \\]", "Solution_1": "Here's my proof. \r\n\r\nUsing Jensen's ineq on $f(x) = \\frac{1}{x}$.\r\n\r\nWe get $ \\sum_{cyclic}\\frac{1}{a^{2}+bc}\\leq \\frac{9}{\\sum_{cyclic}a^{2}+\\sum_{cyclic}ab}$.\r\n\r\nand $ (a+b+c)\\sum_{cyclic}\\frac{1}{a+b}= 3+\\sum_{cyclic}\\frac{c}{a+b}\\geq \\frac{9}{2}$ (nebit's ineq)\r\n\r\nSo, it is enought to prove that\r\n\r\n$\\displaystyle ab+bc+ca \\leq a^{2}+b^{2}+c^{2}$.\r\n\r\nThis is self-evident.\r\n\r\nQ.E.D.", "Solution_2": "[quote=\"Chang Woo-JIn\"]Here's my proof. \n\nUsing Jensen's ineq on $f(x) = \\frac{1}{x}$.\n\nWe get $\\sum_{cyclic}\\frac{1}{a^{2}+bc}\\leq \\frac{9}{\\sum_{cyclic}a^{2}+\\sum_{cyclic}ab}$.\n\n[/quote]\r\n\r\nI think it's invalid. the sign whould be reversed.", "Solution_3": "Oh.... \r\n\r\nI did stupid works..", "Solution_4": "[quote=\"Chang Woo-JIn\"]\nWe get $\\sum_{cyclic}\\frac{1}{a^{2}+bc}\\leq \\frac{9}{\\sum_{cyclic}a^{2}+\\sum_{cyclic}ab}$.\n[/quote]\r\nYou have mistake here. ;)", "Solution_5": "[quote=\"Chang Woo-JIn\"]\nWe get $\\sum_{cyclic}\\frac{1}{a^{2}+bc}\\leq \\frac{9}{\\sum_{cyclic}a^{2}+\\sum_{cyclic}ab}$.[/quote]\r\n\r\nYes, i did the same mistake, but i saw it when i corrected.Then i find an ugly solution.Here is my solution.\r\n$\\frac{1}{a^{2}+bc}+\\frac{1}{b^{2}+ac}+\\frac{1}{c^{2}+ac}\\leq \\frac{a+b+c}{ab+bc+ca}(\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a})$\r\nEqualizing denominators,opening brackets and simplifying :D we get \r\n$3a^{2}b^{2}c^{2}\\sum_{cyclic}(a^{2}b+b^{2}a)+3a^{3}b^{3}c^{3}+a^{2}b^{2}c^{2}(a^{3}+b^{3}+c^{3}) \\leq abc(a^{6}+b^{6}+c^{6})+abc(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3})+2\\sum_{cyclic}(a^{5}b^{4}+b^{5}a^{4})+abc\\sum_{cyclic}(a^{5}b+b^{5}a)$\r\nUsing following inequalities \r\n$1) abc(a^{6}+b^{6}+c^{6}) \\geq \\frac{abc}{3}(a^{3}+b^{3}+c^{3})^{2}\\geq a^{2}b^{2}c^{2}(a^{3}+b^{3}+c^{3})$\r\n$2) 2\\sum_{cyclic}(a^{5}b^{4}+b^{5}a^{4}) \\geq 12a^{3}b^{3}c^{3}$\r\n$3) abc\\sum_{cyclic}(a^{5}b+b^{5}a) \\geq 2abc(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3})$\r\nWe must prove that\r\n$3abc(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3})+9a^{3}b^{3}c^{3}\\geq 3a^{2}b^{2}c^{2}\\sum_{cyclic}(a^{2}b+b^{2}a)$\r\n$\\Longrightarrow a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}+3a^{2}b^{2}c^{2}\\geq abc\\sum_{cyclic}(a^{2}b+b^{2}a)$\r\nWhich is Schur inequality.", "Solution_6": "My proof involves very little computation and no expansion ;-)\r\nNote that this inequality is stronger than \r\n\\[\\frac{1}{a^{2}+bc}+\\frac{1}{b^{2}+ca}+\\frac{1}{c^{2}+ab}\\le \\frac{3(a+b+c)^{2}}{2(ab+bc+ca)^{2}}\\]\r\nwhich is mentioned at http://www.mathlinks.ro/Forum/viewtopic.php?t=116853", "Solution_7": "Thank you Duc for this nice gift." } { "Tag": [ "LaTeX" ], "Problem": "Hi guys, I've got a project to write on a maths topic, and am looking to LaTeX to write it. It's not my first encounter with it, but I'm no pro by any stretch of the imagination :s\r\nFirstly, I have a title page - fine. Before moving onto the contents page, I want a page with just a couple of quotes, to set the scene a bit, you know. So what I want is a page with a couple of quotes which are roughly near the middle of the page vertically. I tried using the command \\vspace(2.5in) and then the quote environments etc, but the space doesn't appear. I tried preceding it with \\text{ } which worked, so I guess the vspace needs something (in my case at the top of the page) away from which to set my quote. Ok so what I did works, but it hardly seems \"correct\" or neat. Is there a less clumsy way to do this?\r\nThe next trouble I have (which I haven't yet been able to solve at all) is in page numbering. Ideally I want no page numbering on the title page, then start counting in Roman numerals starting with said quotes page as page i, and then when I get to the \"meat\" of the project start counting in Arabic from 1. Does that make sense, what I'm trying to do? Can anyone help me do this?\r\n\r\nMany thanks, and if it helps, you're all such lovely people ;)", "Solution_1": "Hi, I found this post from Google because I was looking for a solution to the vspace not showing up at the top of a page. It's been a few weeks since you posted, so you probably found what you need, or it's too late now, but I can help you with page numbering if you haven't found it yet.\r\n\r\nTo avoid numbering on your title page, try enclosing it in a \\begin{titlepage} ... \\end{titlepage} environment.\r\n\r\nTo get roman numerals starting at i on the page after this, you just need \\pagenumbering{roman}.\r\n\r\nThen what you want is \\pagenumbering{arabic} \\setcounter{page}{1}.\r\n\r\nI hope this helps." } { "Tag": [ "Alcumus", "videos", "Support" ], "Problem": "I noticed that a user named [u]GAMEBOT[/u] is posting all the Alcumus Problems and Video Discussions. Why? Is anyone else allowed to post some? ? ? ?", "Solution_1": "GameBot is a bot that was made to do exactly that, when someone hits \"Discuss Problem on Forum\" or \"Discuss Video on Forum\".", "Solution_2": "Then why is it not on the Bot list?" } { "Tag": [ "algebra", "polynomial", "Vieta", "quadratics", "quadratic formula" ], "Problem": "hello. this is an easy problem.\r\n\r\nLet $ P(x)\\equal{}x^3\\minus{}8x^2\\plus{}mx\\plus{}n$ for some $ m,n\\in\\mathbb{Z}$. It is known that $ P(x)$ has three distinct positive roots and that exactly one is an integer, which is the sum of the other two roots. How many values of $ n$ are possible ?", "Solution_1": "[hide]\nLet the roots be $ a,b,c$ with $ a,b \\not\\in \\mathbb{Z}$ and $ c, ab, (a\\plus{}b) \\in \\mathbb{Z}$ and $ a\\plus{}b\\equal{}c$\n\nUsing Vieta's, and the facts that integers are closed under addition\n\n$ a\\plus{}b\\plus{}c \\equal{} 8 \\Longrightarrow a\\plus{}b\\equal{}c\\equal{} 4$\n\n$ b\\equal{}4\\minus{}a \\Rightarrow a(4\\minus{}a) \\in \\mathbb{Z}$\n\nSince $ a(4\\minus{}a) \\equal{} 4\\minus{}(a\\minus{}2)^2$ we know $ (a\\minus{}2)^2$ is an integer, but $ (a\\minus{}2)$ is not\n\nSo $ a\\equal{} 2 \\plus{} \\sqrt{k}$ where $ k \\in \\mathbb{N}$ and is not a surd.\n\nHence $ n \\equal{}\\minus{}abc\\equal{} \\minus{}4(2\\plus{}\\sqrt{k})(2\\minus{}\\sqrt{k}) \\equal{} 4(k\\minus{}4)$ which has infinite solutions[/hide]", "Solution_2": "[hide=\"solution\"]\nWLOG let $ x_1>x_2>x_3>0$.\nFrom Vieta's we have:\n\n$ x_1\\plus{}x_2\\plus{}x_3\\equal{}8$\n\n$ x_1\\equal{}x_2\\plus{}x_3 \\implies x_1\\equal{}4$\n\n$ x_1x_2x_3\\equal{}\\minus{}n \\implies 4x_2x_3\\equal{}\\minus{}n$\n\nSubstituting $ x_3\\equal{}4\\minus{}x_2$ gives us $ 4x_2^2\\minus{}16x_2\\minus{}n\\equal{}0$ (we would get same quadratic if we substituted $ x_2$) thus\n\n$ x_{2,3}\\equal{}\\frac{4\\pm \\sqrt{16\\plus{}n}}{2}$\n\nSince roots are positive $ 4\\minus{}\\sqrt{16\\plus{}n}>0 \\implies n<0$.\n\nSimilarly $ 16\\plus{}n \\geq 0 \\implies \\minus{}16 \\leq n<0$.\n\nSince $ x_2$ and $ x_3$ can't be integers we have to take out $ n\\equal{}\\minus{}12$ and $ n\\equal{}\\minus{}16$ (which would give $ x_2\\equal{}\\{3,2\\}, x_3\\equal{}\\{1,2\\}$).\n\nIf we return to starting polynomial and plug in $ x\\equal{}4$ we get $ 4m\\plus{}n\\equal{}64$ or $ m\\equal{}16\\minus{}\\frac{n}{4}$ so $ n$ must be multiple of $ 4$ which gives us $ n\\equal{}\\{\\minus{}4,\\minus{}8\\}$ for total of $ \\boxed{2}$ values.[/hide]", "Solution_3": "[quote=\"maths486\"]hello. \n\nLet $ P(x) \\equal{} x^3 \\minus{} 8x^2 \\plus{} mx \\plus{} n$ for some $ m,n\\in\\mathbb{Z}$. It is known that $ P(x)$ has three distinct positive roots and that exactly one is an integer, which is the sum of the other two roots. How many values of $ n$ are possible ?[/quote]\r\n\r\nEDIT: Solution not consistent with known solutions.\r\n\r\nMy answer doesn't agree with the other conclusions.\r\n\r\n[hide]Let a, b be integers\nLet n = ab\nBecause all the roots are positive, $ n \\ne 0$\nThen a $ \\ne$ 0, b $ \\ne 0.$\n\n\nLet x = a be the only positive integer root of P(x) = 0.\n\nDivide P(x) by the corresponding factor (x - a) and get the quotient\n$ x^2 \\plus{} (a \\minus{} 8)x \\plus{} m \\plus{} a(a \\minus{} 8)$ and a remainder of $ ab \\plus{} am \\plus{} a^2(a \\minus{} 8)$.\n\nSet the remainder equal to zero for this to have its three factors and solve for m:\n$ m \\equal{} \\minus{} a^2 \\plus{} 8a \\minus{} b$.\n\nSuppose you want to factor the quadratic in x, and solve for its roots:\n\n$ x \\equal{} \\frac {(8 \\minus{} a) \\pm \\sqrt {(a \\minus{} 8)^2 \\minus{} 4(1)[m \\plus{} a^2 \\minus{} 8a]}}{2a}$\n\n$ x \\equal{} \\frac {(8 \\minus{} a) \\pm \\sqrt {a^2 \\minus{} 16a \\plus{} 64 \\minus{} 4m \\minus{} 4a^2 \\plus{} 32a}}{2a}$\n\n$ x \\equal{} \\frac {(8 \\minus{} a) \\pm \\sqrt { \\minus{} 3a^2 \\minus{} 16a \\plus{} 64 \\minus{} 4m}}{2a}$\n\nSubstitute $ m \\equal{} \\minus{} a^2 \\plus{} 8a \\minus{} b$ and simplify:\n\n$ x \\equal{} \\frac {(8 \\minus{} a) \\pm \\sqrt {(a \\minus{} 8)^2 \\plus{} 4b}}{2a}$ . . . (**)\n\nBesides x = a (the integer root), these second and third roots from\nthis quadratic formula (**) add up to\n\n$ \\frac {8 \\minus{} a}{a} \\equal{} \\minus{} 1 \\plus{} \\frac {8}{a}$\n\nWhat integer values may we choose for a?\n\nPotentially, the variable a = 1, 2, 4, 8, as it is a positive integer, and it \ndivides 8.\n\nLook at their placements into (**).\n\nWhen a = 8, one of the x-values becomes $ \\frac { \\minus{} \\sqrt {4b}}{16}$, and so \nit violates having three positive roots.\n\nIn (**), if a = 1, 2, or 4, then with suitable negative b-values, both x-values\nare positive along with their respective a-values.\n\n\nI am getting three n values, then, for my answer.\n\n\nEDIT: This solution is not consistent with the known soution.\n[/hide]\r\n\r\nPlease check my work.", "Solution_4": "[quote=\"Arrange your tan\"]\nPlease check my work.[/quote]\r\nBut you didn't bother to check ours..?\r\n\r\nAnyway I missed that the roots were positive,\r\n\r\n$ (x\\minus{}4)(x\\minus{}2\\plus{}\\sqrt{k})(x\\minus{}2\\minus{}\\sqrt{k}) \\equal{} x^3 \\minus{} 8x^2 \\plus{} (20\\minus{}k)x \\minus{} 4(4\\minus{}k)$\r\n\r\nSo $ n\\equal{}\\minus{}4(4\\minus{}k)$ with $ k$ an integer, but not a surd\r\n\r\nSince roots are positive $ 2\\minus{}\\sqrt{k}$ is positive, therefore $ k\\le 4$,\r\n but $ 1, 4$ are a perfect square, and $ k\\equal{}0$ implies two roots aren't distinct, so $ k\\equal{}2,3$\r\n\r\nAnd $ n\\equal{}\\minus{}4, \\minus{}8$", "Solution_5": "hello \r\n\r\n[hide]\n\nI think we have the correct answer by Flame. We get $ 2$ possible values for $ n$ since the problem also restricts $ m$ to be an integer, and also since all the roots should be positive. We will get $ a\\equal{}4$ for the integer root and we must therefore have $ 2\\plus{}\\sqrt{b}$ and $ 2\\minus{}\\sqrt{b}$ as roots also, where $ b\\in\\mathbb{Z}$ \n\n$ 2\\minus{}\\sqrt{b}>0$ only when $ b\\equal{}2,3$. We can't have $ b\\equal{}0,1$ since we only have one rational (integer) root. There are $ 2$ values of $ b$ which will correspond to two values for $ n$. \n\n\n\n[/hide]", "Solution_6": "[quote=\"maths486\"] this is an easy problem.\n\n[/quote]\r\nIn the future, leave off subjective descriptions such as these\r\nto try to entice problem solvers." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $A,B$ two real matrix of order 3, such that $rank(A)=rank(B)=1$.\r\n\r\n Prove that $det(A+B)=0$. ;)", "Solution_1": "$Ker(A)$ and $Ker B$ are two planes. So $Ker(A) \\cap Ker(B) \\neq \\{0\\}$. Hence there exists $v$ such that $Av=Bv=0$. Conclusion follows ?", "Solution_2": "$\\text{rank}(A+B)\\le\\text{rank}(A)+\\text{rank}(B).$\r\n\r\nThis is easy to see: the union of a basis for the columns of $A$ and a basis for the columns of $B$ must span the column space of $A+B.$\r\n\r\nSo $\\text{rank}(A+B)\\le2$ and hence $A+B$ is singular.\r\n\r\nThis is basically the same argument that alekk gave, just stated a little differently.", "Solution_3": "[quote=\"alekk\"]$Ker(A)$ and $Ker B$ are two planes. So $Ker(A) \\cap Ker(B) \\neq \\{0\\}$[/quote]\r\n\r\nwhy these two plane are not parallel ?", "Solution_4": "Since $ 2+2>3$, Moubi! Just apply Grassman theorem." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c\\ge 0$ s.t. $ ab \\plus{} bc \\plus{} ca \\equal{} 3$. Prove that:\r\n\r\n$ (a \\plus{} b \\plus{} c)(a^2b \\plus{} b^2c \\plus{} c^2a)\\ge 9$\r\n\r\nEdit: oops, a bit inattentive. I see it's too easy. Please don't write more replies in this topic.", "Solution_1": "[quote=\"Inequalities Master\"]Let $ a,b,c\\ge 0$ s.t. $ ab \\plus{} bc \\plus{} ca \\equal{} 3$. Prove that:\n\n$ (a \\plus{} b \\plus{} c)(a^2b \\plus{} b^2c \\plus{} c^2a)\\ge 9$[/quote]\r\nBut it's Cauchy-Schwartz. :wink:" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "$ a_{0} \\equal{}5000$, $ a_{n\\plus{}1} \\equal{} \\frac {a_{n}^2 } {a_{n} \\plus{}1}$\r\n\r\nThen find the value of $ [a_{2007}]$.\r\n\r\nI think it is easy for you. Please solve it!", "Solution_1": "from the condition we have $ a_{n \\plus{} 1} \\equal{} \\frac {1}{a_n \\plus{} \\frac {1}{a_n}}$ which means $ \\frac {1}{a_{n \\plus{} 1}} \\equal{} \\frac {1}{a_n} \\plus{} a_n$\r\nnow let {$ u_k$}={$ \\frac {1}{a_k}$} we get the same problem as the problem you 've posted http://www.mathlinks.ro/viewtopic.php?t=205618" } { "Tag": [], "Problem": "\u0391\u03bd\u03bf\u03af\u03b3\u03c9 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf topic,\u03ce\u03c3\u03c4\u03b5 \u03bf\u03b9 \u03c0\u03b1\u03bb\u03b1\u03b9\u03cc\u03c4\u03b5\u03c1\u03bf\u03b9 \u03bd\u03b1 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03b5\u03c4\u03b5 \u03c3\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03ac\u03c4\u03c9\u03bd \u03b4\u03b9\u03b1\u03c7\u03b5\u03af\u03c1\u03b7\u03c3\u03b7\u03c2 \u03c4\u03c9\u03bd \u03bb\u03bf\u03b3\u03b1\u03c1\u03b9\u03b1\u03c3\u03bc\u03ce\u03bd \u03bc\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03cc,\u03c4\u03b9 \u03ac\u03bb\u03bb\u03bf \u03c0\u03c1\u03bf\u03ba\u03cd\u03c8\u03b5\u03b9.\u0399\u03b4\u03af\u03c9\u03c2 \u03b3\u03b9\u03b1 \u03bc\u03b1\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b3\u03b9\u03bf\u03c5\u03c2...\r\n\u039a\u03b1\u03b9 \u03be\u03b5\u03ba\u03b9\u03bd\u03ac\u03c9 \u03bc\u03b5 \u03c4\u03b1 \u03c0\u03c1\u03ce\u03c4\u03b1 \u03b5\u03c1\u03c9\u03c4\u03ae\u03bc\u03b1\u03c4\u03b1:\r\n1)\u03c0\u03ce\u03c2 \u03b8\u03b1 \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b7 \u03c5\u03c0\u03bf\u03b3\u03c1\u03b1\u03c6\u03ae \u03bc\u03bf\u03c5;\r\n2)\u03b3\u03b9\u03b1\u03c4\u03af \u03cc\u03c4\u03b1\u03bd \u03c3\u03c4\u03ad\u03bb\u03bd\u03c9 pm \u03b5\u03b4\u03ce \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03bc\u03ad\u03c1\u03b5\u03c2 \u03bc\u03bf\u03c5 \u03c4\u03b1 \u03b2\u03b3\u03ac\u03b6\u03b5\u03b9 \u03c3\u03c4o outbox \u03ba\u03b1\u03b9 \u03cc\u03c7\u03b9 \u03c3\u03c4o sentbox;\u03a0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc \u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03b9\u03ce\u03c3\u03c9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03c0\u03ac\u03c1\u03b5\u03b9 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b7 \u03c3\u03b5 \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1!(\u03a3\u03b7\u03bc.:\u03c3\u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03ae \u03bc\u03bf\u03c5 \u03c4\u03b1 \u03ad\u03b2\u03b3\u03b1\u03b6\u03b5 \u03c3\u03c4\u03bf sentbox \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b9\u03ba\u03bf\u03b9\u03bd\u03c9\u03bd\u03bf\u03cd\u03c3\u03b1 \u03ba\u03b1\u03bd\u03bf\u03bd\u03b9\u03ba\u03ac)", "Solution_1": "\u03a4\u03b1 \u03bc\u03b7\u03bd\u03cd\u03bc\u03b1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf outbox \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bc\u03b7\u03bd\u03cd\u03bc\u03b1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03c6\u03c4\u03ac\u03c3\u03b5\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b7 \u03c3\u03c4\u03bf\u03bd \u03c0\u03c1\u03bf\u03bf\u03c1\u03b9\u03c3\u03bc\u03cc \u03c4\u03bf\u03c5\u03c2, \u03b5\u03af\u03c4\u03b5 \u03b3\u03b9\u03b1\u03c4\u03af \u03bf \u03bc\u03b5\u03bb\u03bb\u03bf\u03bd\u03c4\u03b9\u03ba\u03cc\u03c2 \u03c0\u03b1\u03c1\u03b1\u03bb\u03ae\u03c0\u03c4\u03b7\u03c2 \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b7 \u03c3\u03c5\u03bd\u03b4\u03b5\u03b8\u03b5\u03af \u03b5\u03af\u03c4\u03b5 \u03b3\u03b9\u03b1\u03c4\u03af \u03b3\u03b9\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03bb\u03cc\u03b3\u03bf \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b7 \u03b4\u03b9\u03b1\u03b2\u03ac\u03c3\u03b5\u03b9 \u03c4\u03bf \u03bc\u03ae\u03bd\u03c5\u03bc\u03b1. \u039c\u03cc\u03bb\u03b9\u03c2 \u03c0\u03b1\u03c1\u03b1\u03bb\u03ac\u03b2\u03b5\u03b9 \u03c4\u03bf \u03bc\u03ae\u03bd\u03c5\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b4\u03b9\u03b1\u03b2\u03ac\u03c3\u03b5\u03b9 \u03b8\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c6\u03b5\u03c1\u03b8\u03b5\u03af \u03b1\u03c5\u03c4\u03cc\u03bc\u03b1\u03c4\u03b1 \u03c3\u03b5 \u03c3\u03ad\u03bd\u03b1 \u03c3\u03c4\u03bf\u03bd \u03c6\u03ac\u03ba\u03b5\u03bb\u03bf sentbox.\r\n\u0391\u03bd \u03c4\u03ce\u03c1\u03b1 \u03b5\u03af\u03c3\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03cc\u03c4\u03b9 \u03bf \u03c0\u03b1\u03c1\u03b1\u03bb\u03ae\u03c0\u03c4\u03b7\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03bb\u03ac\u03b2\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03b4\u03b9\u03b1\u03b2\u03ac\u03c3\u03b5\u03b9 \u03c4\u03bf \u03bc\u03ae\u03bd\u03c5\u03bc\u03ac \u03c3\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1'\u03cc\u03bb\u03b1 \u03b1\u03c5\u03c4\u03ac \u03c3\u03c5\u03bd\u03b5\u03c7\u03af\u03b6\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf outbox, \u03c4\u03cc\u03c4\u03b5 \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03bc\u03b5 \u03c4\u03bf \u03bb\u03bf\u03b3\u03b1\u03c1\u03b9\u03b1\u03c3\u03bc\u03cc \u03c3\u03bf\u03c5.\r\n\r\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03bc\u03c6\u03b1\u03bd\u03b9\u03c3\u03c4\u03b5\u03af \u03c4\u03ce\u03c1\u03b1 \u03b7 \u03c5\u03c0\u03bf\u03b3\u03c1\u03b1\u03c6\u03ae \u03c3\u03bf\u03c5 \u03c0\u03b7\u03b3\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 \u03c0\u03ac\u03bd\u03c9 \u03b4\u03b5\u03be\u03b9\u03ac \u03b5\u03ba\u03b5\u03af \u03c0\u03bf\u03c5 \u03bb\u03ad\u03b5\u03b9 profile. \u0391\u03c0\u03cc \u03c4\u03b9\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ad\u03c2 \u03c3\u03c4\u03b1 \u03b1\u03c1\u03b9\u03c3\u03c4\u03b5\u03c1\u03ac \u03c0\u03b7\u03b3\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 Preferences-->Posting a message \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf Always attach my signature \u03c4\u03b9\u03ba\u03ac\u03c1\u03b5\u03b9\u03c2 \u03c4\u03bf Yes (...\u03b1\u03c5\u03c4\u03ac \u03b5\u03c6\u03cc\u03c3\u03bf\u03bd \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b2\u03ac\u03bb\u03b5\u03b9 \u03bc\u03b9\u03b1 \u03c5\u03c0\u03bf\u03b3\u03c1\u03b1\u03c6\u03ae \u03b1\u03c0\u03cc \u03c4\u03bf Profile-->Signature control panel) .\r\n\r\n\u038c\u03bb\u03b1 \u03b8\u03ad\u03bc\u03b1 \u03c3\u03c5\u03bd\u03ae\u03b8\u03b5\u03b9\u03b1\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03bb\u03af\u03b3\u03bf \u03c8\u03b1\u03be\u03af\u03bc\u03b1\u03c4\u03bf\u03c2. :)", "Solution_2": "[quote]\u0393\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03bc\u03c6\u03b1\u03bd\u03b9\u03c3\u03c4\u03b5\u03af \u03c4\u03ce\u03c1\u03b1 \u03b7 \u03c5\u03c0\u03bf\u03b3\u03c1\u03b1\u03c6\u03ae \u03c3\u03bf\u03c5 \u03c0\u03b7\u03b3\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 \u03c0\u03ac\u03bd\u03c9 \u03b4\u03b5\u03be\u03b9\u03ac \u03b5\u03ba\u03b5\u03af \u03c0\u03bf\u03c5 \u03bb\u03ad\u03b5\u03b9 profile. \u0391\u03c0\u03cc \u03c4\u03b9\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ad\u03c2 \u03c3\u03c4\u03b1 \u03b1\u03c1\u03b9\u03c3\u03c4\u03b5\u03c1\u03ac \u03c0\u03b7\u03b3\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 Preferences-->Posting a message \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf Always attach my signature \u03c4\u03b9\u03ba\u03ac\u03c1\u03b5\u03b9\u03c2 \u03c4\u03bf Yes (...\u03b1\u03c5\u03c4\u03ac \u03b5\u03c6\u03cc\u03c3\u03bf\u03bd \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b2\u03ac\u03bb\u03b5\u03b9 \u03bc\u03b9\u03b1 \u03c5\u03c0\u03bf\u03b3\u03c1\u03b1\u03c6\u03ae \u03b1\u03c0\u03cc \u03c4\u03bf Profile-->Signature control panel) .[/quote]\r\n\r\n :blush: \u03a4\u03bf \u03b5\u03af\u03c7\u03b1 \u03c8\u03ac\u03be\u03b5\u03b9,\u03c4\u03bf \u03b5\u03af\u03c7\u03b1 \u03c6\u03c4\u03b9\u03ac\u03be\u03b5\u03b9 \u03b1\u03bb\u03bb\u03ac \u03ad\u03c6\u03c5\u03b3\u03b1 \u03b1\u03c0' \u03c4\u03bf \u03c0\u03b1\u03c1\u03ac\u03b8\u03c5\u03c1\u03bf \u03c7\u03c9\u03c1\u03af\u03c2 submit!\u03a4hx,anyway(\u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b1 pm).\r\n\u03a3\u03b7\u03bc.:\u039f kostas \u03ad\u03c6\u03c4\u03b1\u03b9\u03b3\u03b5 \u03c0\u03bf\u03c5 \u03c4\u03b1 \u03b4\u03b9\u03ac\u03b2\u03b1\u03b6\u03b5 \u03c4\u03cc\u03c3\u03bf \u03b3\u03c1\u03ae\u03b3\u03bf\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03bd\u03cc\u03bc\u03b9\u03b6\u03b1 \u03cc\u03c4\u03b9 \u03c0\u03ae\u03b3\u03b1\u03b9\u03bd\u03b1\u03bd \u03b1\u03c0\u03b5\u03c5\u03b8\u03b5\u03af\u03b1\u03c2 \u03c3\u03c4\u03bf sendbox :lol:", "Solution_3": "\u03b5\u03c4\u03c3\u03b9 \u03b5??? :D" } { "Tag": [ "function", "algebra", "domain" ], "Problem": "This is problem E2 from the AMY packet. It should be \"Easy,\" but I'm not seeing it:\r\nIs there a function from the real numbers, excluding zero, to the real numbers excluding zero such that:\r\nf(f(x)) = 1/x\r\n\r\nfor all x in the domain?", "Solution_1": "I don't have said packet and I didn't attend the program, but\r\n\r\n[hide]Yes. Define $ f(1) \\equal{} \\minus{}1, f(\\minus{}1) \\equal{} 1$ and for all $ x > 1$,\n$ f(x) \\equal{} \\minus{}x$\n$ f(\\minus{}x) \\equal{} 1/x$\n$ f(1/x) \\equal{} \\minus{}1/x$\n$ f(\\minus{}1/x) \\equal{} x$[/hide]", "Solution_2": "you might want to think of it as making a piecewise function that breaks up the intervals like this:\r\n\r\n$ ( \\minus{} \\infty, \\minus{} 1)$\r\n$ \\{ \\minus{} 1\\}$\r\n$ ( \\minus{} 1,0)$\r\n$ (0,1)$\r\n$ \\{1\\}$\r\n$ (1, \\infty)$" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME" ], "Problem": "so guys, what do yo think this year's usamo cutoff and floor values are going to be?\r\n\r\nI would start off by saying 215 and 7.", "Solution_1": "My guess is 233/9", "Solution_2": "223/8 is my guess.", "Solution_3": "eh you just copied last year's cutoff. I think the usamo expansion will make a diff:\r\n\r\n225/8 or 9.", "Solution_4": "I guess, 216.5/7 or 8", "Solution_5": "mid 220's/8 (seems to be a common guess)", "Solution_6": "probably 215/7 because of the expansion and I don't think it was as easy as last year and we are better this year than last year and last year's seniors are gone so..", "Solution_7": "234/9 for me.", "Solution_8": "I'd say 216.5 and 7, though that might be wishful thinking since I got a 7.", "Solution_9": "270/13, we're all screwed.", "Solution_10": "225/8... just because I got a 224.5", "Solution_11": "i don't see how some people think that the Expansion will affect the index/floor that much!\r\ni mean think about it, the expansion of 150 students means increase of 2~3 students for each state on average!\r\nnow consider this, last year there were 9/10 graders in each state whom had the floor value, but did not qualify because of tie-brakers, also in each state there were [i]way more than 2~3[/i] students whom were within one point of the floor so they the expansion does not mean a drop of one point. \r\nnow that is only for 10- graders, as for the 11/12 graders, there are again people who were very close to the cutoff and did not qualify, because somebody else in their state was closer to the cutoff than them, and had qualified through state index!\r\n\r\nanyways, my guess:\r\n230/8+ (by 8+ i mean that not all 10- graders will qualify and they will have tie-brakers)", "Solution_12": "People should keep in mine that Index-Floor*10 <= 150... some of them are pretty borderline...\r\n\r\nIn any case, I think 226.5/8 (for those silly 146.5'ers).", "Solution_13": "Looking at last year's and they year before's statistics, about half the people who got the floor value or more got one less than the floor. So the expansion should drop the floor by about one.\r\n\r\nThe floor's for previous years has been 6, 8, 7, 9 in that order. There's a pattern there, so without the expansion, this year's floor should have been an 8. But because of the expansion, it's a 7.\r\n\r\nAnd I think that this AIME was a bit harder than last year's, for the people who are borderline.\r\n\r\nSo looking at all that, the USAMO cutoff should be 216/7.\r\n\r\n\r\nGahh, I feel like I'm just making excuzes for me to get in.", "Solution_14": "the cutoff is definitely 6, there's no question about it. :harhar:", "Solution_15": "It was 5 once, it can be 5 again! :lol:", "Solution_16": "Hmm...gonna go with 555.5/5.\r\n\r\nSeriously though, 222.5/8. Same thing really. \r\n\r\n\r\nIf we get it right, what do we win? :P", "Solution_17": "You automatically qualify for USAMO if you guess :D \r\n\r\nI would say 7 or 8 floor, seeing as it was slightly harder than last year (which had a 9 floor), and the expansion will probably drop it by a point.", "Solution_18": "[quote=\"Fermatprime\"]Hmm...gonna go with 555.5/5.\n\nSeriously though, 222.5/8. Same thing really. \n\n\nIf we get it right, what do we win? :P[/quote]\r\n\r\nHow do you get a 142.5 on the AMC?", "Solution_19": "What are the tiebreakers for USAMO?", "Solution_20": "Let's say the floor is 8, but there are more than 120 10- with 8. They compare the AMC+AIME scores to determine which ones get it. Question though, would they just compare among the people who scored 8 or will an 8er with 10 points higher than a 9er on the AMC get preference?", "Solution_21": "im saying 223/8", "Solution_22": "Im guessing 215/7 :(", "Solution_23": "Floor of 8 and an index of low 220's.", "Solution_24": "Index: 215\r\nCutoff: 8 for both", "Solution_25": "Hmm... I'll say a 242 index, yay i'll barely qualify :rotfl: \r\n\r\nNo, seriously, my guess is 221/8, possibly a few 7's with tiebreakers.", "Solution_26": "[quote=\"scorpius119\"]Hmm... I'll say a 242 index, yay i'll barely qualify :rotfl: \n\nNo, seriously, my guess is 221/8, possibly a few 7's with tiebreakers.[/quote]\r\nHow do they do tiebraekers?", "Solution_27": "explained in post #21 ;) \r\n\r\nbased on AMC score", "Solution_28": "222/8\r\n\r\n222.5 is my index. ;)", "Solution_29": "Umm 250/12.\r\n\r\nJust kdding.\r\n\r\nI agree with 223/ 8 ish. 220 lets some 7's in so a little over 220 makes sense.", "Solution_30": "[quote=\"paladin8\"][quote=\"Fermatprime\"]Hmm...gonna go with 555.5/5.\n\nSeriously though, 222.5/8. Same thing really. \n\n\nIf we get it right, what do we win? :P[/quote]\n\nHow do you get a 142.5 on the AMC?[/quote]\r\n\r\nThe person who sets the floor doesn't have to have the lowest index, it's possible that someone scores a 143/8.", "Solution_31": "Ah, good point. That's why you get the really low floors. I still think one of those 146.5 scorers got an 8. Just to spite them for leaving a single question blank.", "Solution_32": "300/15. Duh!\r\n\r\nBut seriously, I'm thinking 220ish/hopefully 7 or 8.", "Solution_33": "The floor is definitely going to be 9.\r\nThe index is going to be 233.0 give or take 1.5", "Solution_34": "The floor is definitely going to be 9.\r\nThe index is going to be at most 233, the 13th Fibonacci Number.", "Solution_35": "Neal, just for that, the floor will be a 10. :rotfl:", "Solution_36": "I will guess 217/8. :D", "Solution_37": "[quote=\"h_s_potter2002\"]So looking at all that, the USAMO cutoff should be 216/7.[/quote]\r\n\r\nKind of funny that I was off by one for each thing.", "Solution_38": "[quote=\"h_s_potter2002\"][quote=\"h_s_potter2002\"]So looking at all that, the USAMO cutoff should be 216/7.[/quote]\n\nKind of funny that I was off by one for each thing.[/quote]\r\n\r\nI guessed 216.5/8 the day before yesterday.\r\nSee, I missed only 0.5!!!" } { "Tag": [], "Problem": "What's the easiest way to find how to say/write something in a different language?\r\n\r\nfor example: how do you ask \"what supplies do we need for Chinese 2\" in Chinese?\r\n\r\nIt might be possible to find individual words easily through a dictionary but I'm not sure how to piece the words together unless I assume that the grammar is the same as English's", "Solution_1": "If the phrase you want to figure out is pretty standard, then a phrasebook or an introductory textbook would be your best bet. Otherwise, a grammar of the language will tell you the syntactic and inflectional rules, but you may have to spend a while figuring out how to correctly apply those rules. \r\n\r\nFinding a native speaker to help you is always great, but you have to trust that they didn't misunderstand what you want or are pulling a prank on you (cf. \"My Big Fat Greek Wedding\")." } { "Tag": [ "trigonometry", "inequalities solved", "inequalities" ], "Problem": "Let $a,b,c$ be positive reals satisfying $a+b+c=abc$\r\nShow that \r\n\\[ \\sum a(1-b^2)(1-c^2) \\leq \\frac{4}{27}(a+b+c)^3 \\]", "Solution_1": "By using trigonometry ( puting a = tanA ,b= tanB, c=tanC ...) we can show that:\r\n \\[ \\sum a(1-b^2)(1-c^2) = 4abc \\]\r\nAfter that, apply AM-GM and get the answer. :lol:", "Solution_2": "[quote=\"darkmaster\"]By using trigonometry ( puting a = tanA ,b= tanB, c=tanC ...) we can show that:\n \\[ \\sum a(1-b^2)(1-c^2) = 4abc \\]\nAfter that, apply AM-GM and get the answer. :lol:[/quote]\r\n\r\n \\[ \\sum a(1-b^2)(1-c^2) = 4abc \\]\r\n\r\n\r\nWHY??" } { "Tag": [], "Problem": "A gum ball machine contains 9 red, 7 white and 8 blue gum balls. What is the least number of gum balls a person must buy to ensure getting four balls of each color?", "Solution_1": "If we take 17 balls, it could be 9 red, 8 blue. So the answer is 18.", "Solution_2": "Unfortuantely, the problem asks for 4 balls of each color. Let's take the worst case scenario. We get 9 red and 8 blue. We still need 4 white so its 9+8+4=21.", "Solution_3": "Oooh I see! I thought 4 balls with balls in every color. Sorry all!" } { "Tag": [ "algebra", "polynomial", "system of equations", "algebra unsolved" ], "Problem": "Hopefully I am not totally off-base on how to use these forums (if I am, I'm sorry), but I have a couple of problems that I would really like to see the solutions to:\r\n\r\n1: First show that if \r\np(z) = z^4 + 3z - 2 = (z^2 - az + b)(z^2 + az + c)\r\nthen a^6 + 8a^2 - 9 = 0. Then solve the equation p(z) = 0\r\n\r\n2: Solve the equation z^10 + z^8 + 2z^6 + 2z^4 + 2z^2 + 2 = 0\r\n\r\n\r\n\r\nIf you expand (z^2 - az + b)(z^2 + az + c) and compare coefficients you get a non-linear system of equations:\r\nc - a^2 + b = 0\r\na(b-c) = 3\r\nbc = -2\r\n\r\nI would guess that you're supposed to show that a^6 + 8a^2 - 9 = 0 from this, but I simply can't do it. I can show some other things that let you solve the equation, like \r\n(b^2 - 4/b^2)*(b + 2/b) = 9\r\nbut I still can't show what's asked for.\r\n\r\nFor the second problem I'm pretty much clueless.\r\n\r\nAll help appreciated (these are from past tests, http://www.matematik.lu.se/matematiklu/kurser/MAT131/material/ttu.pdf 73 and 74)!", "Solution_1": "From $b+c=a^2$ and $b-c=\\frac{3}{a}$, you can get $2b=a^2+\\frac{3}{a}$ and $2c=a^2-\\frac{3}{a}$.\r\n\r\nPlugging these into the last one gives $\\left(a^2+\\frac{3}{a}\\right)\\left(a^2-\\frac{3}{a}\\right)=a^4-\\frac{9}{a^2}=-8$, which rearranges to the equation in $a$ you're looking for.", "Solution_2": "Thanks!\r\n\r\nThe other one can be solved like this: Let w = z^2. \r\nw^5 + w^4 + 2w^3 + 2w^2 + 2w + 2 = 0\r\nw^4(w + 1) + 2w^2(w + 1) + 2(w + 1) = 0\r\n(w + 1)(w^4 + 2w^2 + 2) = 0\r\n\r\nThen you factorize and solve it." } { "Tag": [], "Problem": "This is from: [url=http://www.icoool.com/index.asp]http://www.icoool.com/index.asp[/url]\r\n\r\nMaybe this is for liyi! :D \r\n\r\n\u4e00\u3001\u9009\u62e9\u9898(\u6bcf\u989810\u5206):\r\n\r\n1.\u4e0a\u6d77\u8bdd\u91cc\u901a\u5e38\u7ba1\u201c\u4eca\u5929\u201d\u53eb\r\nA.\u4eca\u65e5 B.\u4eca\u591c C.\u4eca\u671d D.\u4eca\u5929 \r\n\r\n2.\u4e0a\u6d77\u8bdd\u91cc\u901a\u5e38\u7ba1\u201c\u6628\u5929\u201d\u53eb\r\nA.\u6628\u65e5 B.\u6628\u591c C.\u6628\u671d D.\u6628\u5929 \r\n\r\n3.\u4e0a\u6d77\u8bdd\u91cc\u201c\u4f60\u201d\u7684\u8bfb\u97f3\u4e3a \r\n\uff08[]\u4e2d\u4e3a\u56fd\u9645\u97f3\u6807\uff09\r\nA.nong B.ou C.yi D.[ne] \r\n\r\n4.\u4e0a\u6d77\u8bdd\u91cc\u201c\u4e8c\u5341\u4e8c\u201d\u7684\u8bfb\u97f3\u548c\u4e0a\u6d77\u8bdd\u4e2d ____ \u76f8\u540c\r\nA.\u5c3c\u77f3\u800c B.\u800c\u5c3c C.\u96be\u77f3\u800c D.\u96be\u5c3c \r\n\r\n5.\u73b0\u4ee3\u4e0a\u6d77\u8bdd\u4e2d\uff0c\u4e0b\u5217\u54ea\u4e2a\u5b57\u7684\u53d1\u97f3\u548c\u5176\u4ed6\u4e0d\u540c\uff1f\r\nA.\u9e7f B.\u516d C.\u9646 D.\u9732 \r\n\r\n6.\u540c\u4e0a,\u4e0b\u5217\u54ea\u4e2a\u5b57\u7684\u53d1\u97f3\u548c\u5176\u4ed6\u4e0d\u540c\uff1f\r\nA.\u738b B.\u671b C.\u9ec4 D.\u6a2a \r\n\r\n\u4e8c\u3001\u9009\u62e9\u586b\u7a7a\u9898(\u6bcf\u98985\u5206):\r\n\r\n1.\u201c\u4fac\u53ea\u5c0f\u8d64\u4f6c\uff01\u201d\u5176\u4e2d\u201c\u8d64\u4f6c\u201d\u53ef\u4ee5\u7528 ___ \u4ee3\u66ff\u3002\r\nA.\u5783\u4e09 B.\u762a\u4e09 C.\u732a\u5934\u4e09 D.\u80ae\u4e09 \r\n\r\n2.\u201c\u963f\u62c9\u591c\u996d\u5403\u683c\u662f\u6cb9\u7116\u843d\u82cf\u3002\u201d\u843d\u82cf\u5c31\u662f ___ \u3002\r\nA.\u756a\u8304 B.\u897f\u7ea2\u67ff C.\u8304\u5b50 D.\u841d\u535c \r\n\r\n3.\u201c\u771f\u5012\u9709\uff0c\u53c8\u88ab\u9ec4\u725b\u65a9\u4e86\u4e00\u5200\u3002\u201d\u8fd9\u91cc\uff0c\u9ec4\u725b\u53ef\u80fd\u662f ___ \u3002\r\nA.\u5c60\u592b B.\u9ec4\u8272\u7684\u725b C.\u7968\u8d29\u5b50 D.\u7a7f\u7740\u9ec4\u8272\u725b\u4ed4\u670d\u7684\u4eba \r\n\r\n4.\u201c\u8fb0\u5149\u4e0d\u591a\u4e86\uff0c\u53eb\u90e8\u53c9\u5934\u5427\uff01\u201d\u53c9\u5934\u662f\u6307 ___ \u3002\r\nA.\u51fa\u79df\u8f66 B.\u9762\u5305\u8f66 C.\u76f4\u5347\u98de\u673a D.\u6469\u6258\u8f66 \r\n\r\n5.\u201c\u4f0a\u8001\u574d\u677f\u683c\u3002\u201d\u574d\u677f\u53ef\u4ee5\u7528 ____ \u4ee3\u66ff\u3002\r\nA.\u4e18 B.\u6d0b\u76d8 C.\u5341\u4e09\u70b9 D.\u8d3c\u5fd2\u516e\u516e \r\n\r\n6.\u54ea\u53e5\u662f\u9519\u8bef\u7684\uff1f\r\nA.\u4fac\u963f\u662f\u642d\u9519\u4e86\uff1f B.\u4fac\u642d\u9519\u662f\u4f10\uff1f C.\u963f\u642d\u9519\u5566\u4fac\uff1f D.\u4fac\u662f\u642d\u9519\u4f10\uff1f \r\n\r\n7.\u54ea\u79cd\u8bf4\u6cd5\u4e0d\u5408\u4e4e\u901a\u5e38\u7528\u6cd5\uff1f\r\nA.\u4fac\u5f53\u6211\u6d0b\u76d8\u554a\uff1f B.\u4fac\u5f53\u6211\u51b2\u5934\u554a\uff1f C.\u4fac\u5f53\u6211\u5c48\u6b7b\u554a\uff1f D.\u4fac\u5f53\u6211\u6206\u5927\u554a\uff1f \r\n\r\n8.\u201c\u4f0a\u62c9\u4e13\u95e8\u65a9\u6d0b\u8471\u5934\u3002\u201d\u8fd9\u53e5\u8bdd\u7684\u610f\u601d\u53ef\u80fd\u662f ___\r\nA.\u4ed6\u4eec\u662f\u52a0\u5de5\u6d0b\u8471\u7684\u4e13\u5bb6\u3002 B.\u4ed6\u4eec\u6572\u4e86\u4e00\u4e2a\u53eb\u6d0b\u8471\u5934\u7684\u4eba\u4e00\u7b14\u7af9\u6760\u3002 C.\u4ed6\u4eec\u65e2\u662f\u6d0b\u76d8\u4e5f\u662f\u51b2\u5934\u3002 D.\u4ed6\u4eec\u4e13\u9a97\u5916\u56fd\u4eba\u7684\u94b1\u3002", "Solution_1": "\u6211\u4e0d\u662f\u4e0a\u6d77\u4eba...\u5728\u4e0a\u6d77\u4e0a\u5b66...\u7531\u4e8e\u6211\u4eec\u73ed\u4e0a\u6ca1\u4ec0\u4e48\u4e0a\u6d77\u4eba...\u6211\u8fd8\u662f\u4e0d\u4f1a\u4e0a\u6d77\u8bdd", "Solution_2": "[quote=\"shobber\"]This is from: [url=http://www.icoool.com/index.asp]http://www.icoool.com/index.asp[/url]\n\nMaybe this is for liyi! :D \n\n\u4e00\u3001\u9009\u62e9\u9898(\u6bcf\u989810\u5206):\n\n1.\u4e0a\u6d77\u8bdd\u91cc\u901a\u5e38\u7ba1\u201c\u4eca\u5929\u201d\u53eb\nA.\u4eca\u65e5 B.\u4eca\u591c C.\u4eca\u671d D.\u4eca\u5929 \n\n2.\u4e0a\u6d77\u8bdd\u91cc\u901a\u5e38\u7ba1\u201c\u6628\u5929\u201d\u53eb\nA.\u6628\u65e5 B.\u6628\u591c C.\u6628\u671d D.\u6628\u5929 \n\n3.\u4e0a\u6d77\u8bdd\u91cc\u201c\u4f60\u201d\u7684\u8bfb\u97f3\u4e3a \n\uff08[]\u4e2d\u4e3a\u56fd\u9645\u97f3\u6807\uff09\nA.nong B.ou C.yi D.[ne] \n\n4.\u4e0a\u6d77\u8bdd\u91cc\u201c\u4e8c\u5341\u4e8c\u201d\u7684\u8bfb\u97f3\u548c\u4e0a\u6d77\u8bdd\u4e2d ____ \u76f8\u540c\nA.\u5c3c\u77f3\u800c B.\u800c\u5c3c C.\u96be\u77f3\u800c D.\u96be\u5c3c \n\n5.\u73b0\u4ee3\u4e0a\u6d77\u8bdd\u4e2d\uff0c\u4e0b\u5217\u54ea\u4e2a\u5b57\u7684\u53d1\u97f3\u548c\u5176\u4ed6\u4e0d\u540c\uff1f\nA.\u9e7f B.\u516d C.\u9646 D.\u9732 \n\n6.\u540c\u4e0a,\u4e0b\u5217\u54ea\u4e2a\u5b57\u7684\u53d1\u97f3\u548c\u5176\u4ed6\u4e0d\u540c\uff1f\nA.\u738b B.\u671b C.\u9ec4 D.\u6a2a \n\n\u4e8c\u3001\u9009\u62e9\u586b\u7a7a\u9898(\u6bcf\u98985\u5206):\n\n1.\u201c\u4fac\u53ea\u5c0f\u8d64\u4f6c\uff01\u201d\u5176\u4e2d\u201c\u8d64\u4f6c\u201d\u53ef\u4ee5\u7528 ___ \u4ee3\u66ff\u3002\nA.\u5783\u4e09 B.\u762a\u4e09 C.\u732a\u5934\u4e09 D.\u80ae\u4e09 \n\n2.\u201c\u963f\u62c9\u591c\u996d\u5403\u683c\u662f\u6cb9\u7116\u843d\u82cf\u3002\u201d\u843d\u82cf\u5c31\u662f ___ \u3002\nA.\u756a\u8304 B.\u897f\u7ea2\u67ff C.\u8304\u5b50 D.\u841d\u535c \n\n3.\u201c\u771f\u5012\u9709\uff0c\u53c8\u88ab\u9ec4\u725b\u65a9\u4e86\u4e00\u5200\u3002\u201d\u8fd9\u91cc\uff0c\u9ec4\u725b\u53ef\u80fd\u662f ___ \u3002\nA.\u5c60\u592b B.\u9ec4\u8272\u7684\u725b C.\u7968\u8d29\u5b50 D.\u7a7f\u7740\u9ec4\u8272\u725b\u4ed4\u670d\u7684\u4eba \n\n4.\u201c\u8fb0\u5149\u4e0d\u591a\u4e86\uff0c\u53eb\u90e8\u53c9\u5934\u5427\uff01\u201d\u53c9\u5934\u662f\u6307 ___ \u3002\nA.\u51fa\u79df\u8f66 B.\u9762\u5305\u8f66 C.\u76f4\u5347\u98de\u673a D.\u6469\u6258\u8f66 \n\n5.\u201c\u4f0a\u8001\u574d\u677f\u683c\u3002\u201d\u574d\u677f\u53ef\u4ee5\u7528 ____ \u4ee3\u66ff\u3002\nA.\u4e18 B.\u6d0b\u76d8 C.\u5341\u4e09\u70b9 D.\u8d3c\u5fd2\u516e\u516e \n\n6.\u54ea\u53e5\u662f\u9519\u8bef\u7684\uff1f\nA.\u4fac\u963f\u662f\u642d\u9519\u4e86\uff1f B.\u4fac\u642d\u9519\u662f\u4f10\uff1f C.\u963f\u642d\u9519\u5566\u4fac\uff1f D.\u4fac\u662f\u642d\u9519\u4f10\uff1f \n\n7.\u54ea\u79cd\u8bf4\u6cd5\u4e0d\u5408\u4e4e\u901a\u5e38\u7528\u6cd5\uff1f\nA.\u4fac\u5f53\u6211\u6d0b\u76d8\u554a\uff1f B.\u4fac\u5f53\u6211\u51b2\u5934\u554a\uff1f C.\u4fac\u5f53\u6211\u5c48\u6b7b\u554a\uff1f D.\u4fac\u5f53\u6211\u6206\u5927\u554a\uff1f \n\n8.\u201c\u4f0a\u62c9\u4e13\u95e8\u65a9\u6d0b\u8471\u5934\u3002\u201d\u8fd9\u53e5\u8bdd\u7684\u610f\u601d\u53ef\u80fd\u662f ___\nA.\u4ed6\u4eec\u662f\u52a0\u5de5\u6d0b\u8471\u7684\u4e13\u5bb6\u3002 B.\u4ed6\u4eec\u6572\u4e86\u4e00\u4e2a\u53eb\u6d0b\u8471\u5934\u7684\u4eba\u4e00\u7b14\u7af9\u6760\u3002 C.\u4ed6\u4eec\u65e2\u662f\u6d0b\u76d8\u4e5f\u662f\u51b2\u5934\u3002 D.\u4ed6\u4eec\u4e13\u9a97\u5916\u56fd\u4eba\u7684\u94b1\u3002[/quote]\r\n\r\nOu Xuo Sang Zei Wu, Gu Jiae Dong Sang Hae Wu Cu Fe Dou!\r\n\u6211 \u8bf4 \u5e38 \u5dde \u8bdd, \u4f30 \u8ba1 \u548c \u4e0a \u6d77 \u8bdd \u5dee \u4e0d \u591a!\r\n\u9009\u51e0\u4e2a\u6ce8\u97f3\u78b0\u78b0\u8fd0\u6c14 1.C 2.C 3.A 4.A 5.D 6.B", "Solution_3": "\u6700\u540e\u4e00\u9898\u5e94\u8be5\u662f D \u5427! (\u6d0b\u8471\u7684\u90a3\u4e2a)." } { "Tag": [ "inequalities", "trigonometry", "quadratics", "algebra", "polynomial", "AMC", "Elmo" ], "Problem": "So... I'm going to shamelessly take advantage of the mathematical excitement generated by IMO season and release the 2009 Entirely Legitimate Junior Mathematical Olympiad for your mid-summer delight! \r\n\r\nTraditionally, the veterans at MOP divide themselves into team leaders and coordinators, prepare an ELMO for the rookies, and then mark the exam in IMO fashion. The most agonizing decision of the ELMO Grand Jury, perhaps even more so than the choice of the problems themselves, is what the acronym \"ELMO\" shall stand for. In past years, ELMO has been short for \"Extremely Loquacious Mathematical Olympiad\" and \"Eric Larson Mathematical Olympiad.\" In honor of the new USA Jr. Math Olympiad, however, we opted to pose an ELMO for Krishanu Roy Sankar and an ELJMO for everyone else, and we selected a title for the exam that we thought would underscore its legitimacy. Not being exactly masters of subtlety, we wrote what was indeed an entirely legitimate junior math olympiad.\r\n\r\nThe six problems below represent the best of the 2009 ELJMO shortlist, as voted by the team leaders, who comprised the enfranchised portion of the jury. Being one of the seniors in black who did not secure a position on the IMO team, I faithfully agreed to serve as Supreme Grand Ayatollah and facilitate the problem selection and coordination processes. Afterward, Sherry Gong and I worked to compile a LaTeX solution manual, which will be posted after everyone here has had the opportunity to wrestle with the problems. All of us at MOP invested a great deal of effort into this contest, whether we helped put together the exam or took it as competitors, so I hope you will all enjoy the fruits of that labor as best you can. If you would like to take the ELJMO as a mock test, you may feel free to send your solutions to me via private message, and I will grade them in accordance with the official rubrics (also voted by the team leaders). Happy problem-solving!\r\n\r\n--DBR\r\n\r\nDAY I\r\n\r\n1. (Evan O'Dorney) Let $ a,b,c$ be positive integers such that $ a^2 \\minus{} bc$ is a square. Prove\r\nthat $ 2a \\plus{} b \\plus{} c$ is not prime.\r\n\r\n2. (David B Rush) Let $ ABC$ be a triangle such that $ AB < AC$. Let $ P$ lie on a line through $ A$ parallel to line $ BC$ such that $ C$ and $ P$ are on the same side of line $ AB$. Let $ M$ be the midpoint of segment $ BC$. Define $ D$ on segment $ BC$ such that $ \\angle BAD \\equal{} \\angle CAM$, and define $ T$ on the extension of ray $ CB$ beyond $ B$ so that $ \\angle BAT \\equal{} \\angle CAP$. Given that lines $ PC$ and $ AD$ intersect at $ Q$, that lines $ PD$ and $ AB$ intersect at $ R$, and that $ S$ is the midpoint of segment $ DT$, prove that if $ A,P,Q,$ and $ R$ lie on a circle, then $ Q,R,$ and $ S$ are collinear.\r\n\r\n3. (Wenyu Cao) Let $ a,b,c$ be nonnegative real numbers. Prove that\r\n\\[ a(a \\minus{} b)(a \\minus{} 2b) \\plus{} b(b \\minus{} c)(b \\minus{} 2c) \\plus{} c(c \\minus{} a)(c \\minus{} 2a) \\geq 0.\\]\r\nDAY II\r\n\r\n4. (Allen Yuan) Let $ n$ be a positive integer. Given $ n^2$ points in a unit square, prove that there exists a broken line of length $ 2n \\plus{} 1$ that passes through all the points.\r\n\r\n5. (Evan O'Dorney) Let $ ABCDEFG$ be a regular heptagon with center $ O$. Let $ M$ be the centroid of $ \\triangle ABD$. Prove that $ \\cos^2(\\angle GOM)$ is rational and determine its value.\r\n\r\n6. (John Berman) Let $ p$ be an odd prime and $ x$ be an integer such that $ p \\mid x^3 \\minus{} 1$ but $ p \\nmid x \\minus{} 1$. Prove that\r\n\\[ p \\mid (p \\minus{} 1)!\\left(x \\minus{} \\frac {x^2}{2} \\plus{} \\frac {x^3}{3} \\minus{} \\cdots \\minus{} \\frac {x^{p \\minus{} 1}}{p \\minus{} 1}\\right).\\]", "Solution_1": "[hide]1. Consider the quadratic $ p(x)\\equal{}bx^2\\plus{}2ax\\plus{}c$. Note that this is a polynomial over $ \\mathbb{Z}$. Since its discriminant is a perfect square, it has rational roots, i.e. it can be factored over the rationals. Therefore it can be factored over the integers. Now just note that $ p(1)\\equal{}2a\\plus{}b\\plus{}c$. [/hide]", "Solution_2": "Is it necessary to send in a specific way, i.e. do I have to send in everything at the same time, or can I send a problem at a time?", "Solution_3": "[quote=\"xpmath\"]Is it necessary to send in a specific way, i.e. do I have to send in everything at the same time, or can I send a problem at a time?[/quote]\r\n\r\nWhatever you like. :)", "Solution_4": "My dear Grand Toyota (did I get that right?),\r\n\r\nCould the ELMO problems also be released? I have only seen the problems 1 and 4 on the ELMO as they were also on the ELJMO.", "Solution_5": "Hmm...is there a way to somehow get hold of ELMO problems from past years?", "Solution_6": "[quote=Phelpedo][hide]1. Consider the quadratic $ p(x)\\equal{}bx^2\\plus{}2ax\\plus{}c$. Note that this is a polynomial over $ \\mathbb{Z}$. Since its discriminant is a perfect square, it has rational roots, i.e. it can be factored over the rationals. Therefore it can be factored over the integers. Now just note that $ p(1)\\equal{}2a\\plus{}b\\plus{}c$. [/hide][/quote]\n\nReally great solution here is my idea. \nLet $x^2=a^2-bc$ and $t=2a+b+c$\nSo $tb=(a+b-x)(a+b+x)$\nLet's take the contrary. $t$ is prime. So we have two cases $t|a+b-x$ or $t|a+b+x$\n$1)$ $t|a+b-x$ which means $a+b-x\\ge 2a+b+c$ $\\Longrightarrow $ $a+c+x\\le 0$\n$2)$ $t|a+b+x$ which means $\\sqrt{a^2-bc}\\geq a+c$ $\\Longrightarrow $ $t\\le 0$.", "Solution_7": "[quote=gighiuhui]\n3. (Wenyu Cao) Let $ a,b,c$ be nonnegative real numbers. Prove that\n\\[ a(a \\minus{} b)(a \\minus{} 2b) \\plus{} b(b \\minus{} c)(b \\minus{} 2c) \\plus{} c(c \\minus{} a)(c \\minus{} 2a) \\geq 0.\\]\n[/quote]\nLet $a,b,c$ are positive real numbers . [url=https://artofproblemsolving.com/community/c6h1660340p10538050]Prove that [/url]$$a^3+b^3+c^3+2(ab^2+bc^2+ca^2)\\geq3(a^2b+b^2c+c^2a).$$\n\n", "Solution_8": "For [hide=1, you get get]\n$bc=a^2 -k^2 \n=> c= \\frac{a^2 -k^2}{b}\n=> 2a +b +c = 2a + \\frac{b^2 +a^2 -k^2}{b}= \\frac{a^2 +b^2 +2ab -k^2}{b}= \\frac{(a+b)^2 -k^2}{b}= \\frac{(a+b+k)(a+b-k)}{b}$, which we know is an integer.\nAs $a > k$, we are done." } { "Tag": [ "probability", "probability and stats" ], "Problem": "Hello everyone,\r\n\r\nI am working on my political stats homework, and I am having trouble interpreting the data. I understand all the computations and I believe I have done them all correctly and they are shown in the PDF. However, I don't know how to interpret the data, I don't know exactly what the data is telling me. If someone could help in explaining what the data means and what its telling me that would be great. According to the instructor, the exam will be very similar to the following problems I have in the PDF. a I understand the math, but I don't understand exactly what he wants me to interpret and see in the data. There are two problems and everything I don't understand is marked in Red text in the PDF. I have checked my answers a couple of times and they seem to be all correct so its not so much the math, its the interpretation of the data.\r\n\r\nhere is the link to my PDF.\r\n\r\nhttp://www.geocities.com/terry.green6/StatHW.pdf\r\n\r\nif someone has enough time to check my math that would be great!! Thank you all!!!!", "Solution_1": "I can help you only with the mathematical part:\r\n\r\n1a) Imagine that you have just 2 voting districts $ A$ and $ B$ with populations $ 10$ and $ 90$ respectively. Imagine that in $ A$ 0% voted for the candidate and in $ B$ 100% voted for him. Then the mean of 0% and 100% is 50% but the portion of the total population voting for the candidate is 90%\r\n\r\n2) You set up the equation for $ Z$ correctly but $ \\frac{0.52}{1}$ is meaningless. Instead you need to determine the value of $ Z$ from the table looking at where the probability becomes $ 0.02$.\r\n\r\nAs to the interpretation part, you can restate the definitions, of course, but that's all that comes into my head. Ask your instructor what he wants there.", "Solution_2": "Thanks you, I think I got all thw stuff down too after massive amounts of googeling. Apreciate the help." } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Let R=F[x] with F field. Describe the field of fractions of R.\r\n\r\nThanks", "Solution_1": "$F(x)$.\r\n\r\nYou're welcome." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c$ are real number such that $a+b+c=0$ and $a^{2}+b^{2}+c^{2}=6$. Find the maximum of following expression:\r\n$N=a^{2}b+b^{2}c+c^{2}a$", "Solution_1": "Nobody solve it :roll: welcome and waiting a proof to it ... General case is true ." } { "Tag": [ "function", "analytic geometry", "calculus", "derivative", "Asymptote", "limit", "calculus computations" ], "Problem": "Hi,\r\n\r\nI need to analyze this function $ y\\equal{}\\frac{1\\minus{}x^3}{x^2}$ by these steps:\r\n1. Find function's definition of area.\r\n2. Find crosspoints with the coordinate axes.\r\n3. Find 1st derivative, determine intervals where function is ascending/decreasing. Find extreme points.\r\n4. Find 2nd derivative, determine intervals where function is concave upward /concave downward. Find points of inflection\r\n5. Find asymptotes.\r\n\r\nThis is my work so far:\r\n\r\n1. $ Dom y\\equal{}\\mathbb{R}$ \\ $ {0}$\r\n[b][color=red]2. [/color][/b]No crosspoints with 0y axis.\r\n with 0x axis: $ \\frac{1\\minus{}x^3}{x^2}\\equal{}0$\r\n\r\n\r\n[b][color=red]3.[/color] [/b]\r\n\r\n$ y'\\equal{}\\frac{(1\\minus{}x^3)'x^2\\minus{}(1\\minus{}x^3)(x^2)'}{(x^4}\\equal{}\\frac{\\minus{}3x^2(x^2)\\minus{}(1\\minus{}x^3)2x}{x^4}\\equal{}\\frac{\\minus{}3x^4\\minus{}2x\\plus{}2x^4}{x^4}\\equal{}\\frac{\\minus{}2\\minus{}x^3}{x^3}\\equal{}\\frac{\\minus{}2}{x^3}\\minus{}1$\r\n\r\nNow, I have to find critical points, but I trapped here.\r\n\r\n[b][color=red]4. [/color][/b] \r\n\r\n$ y''\\equal{} (\\frac{\\minus{}2}{x^3}\\minus{}1)'\\equal{}(\\frac{\\minus{}2}{x^3})'\\minus{}(1)'\\equal{}\\frac{(\\minus{}2)'(x^3)\\plus{}(\\minus{}2)(x^3)'}{x^6}\\equal{}\\frac{\\minus{}6}{x^4}$\r\n\r\n$ \\frac{\\minus{}6}{x^4}\\equal{}0$ There is no inflection points.\r\n\r\n5. Vertical asymptote is $ x\\equal{}0$ because $ \\lim_{x\\to 0}\\frac{1\\minus{}x^3}{x^2}\\equal{}\\infty$\r\n\r\nThere is no horizontal asymptotes because because the degree of the numerator is larger than the denominator.\r\n\r\nSlant asymptote is $ y\\equal{}\\minus{}x$ because $ (1\\minus{}x^3): x^2\\equal{}\\minus{}x\\plus{}\\frac{1}{x^2}$\r\n\r\n\r\nCoud you check me and help with missing points?", "Solution_1": "To find the critical points, you just set the derivative equal to 0: $ \\minus{}\\frac{2}{x^{3}}\\minus{}1\\equal{}0\\Rightarrow \\minus{}\\frac{2}{x^{3}}\\equal{}1\\Rightarrow \\minus{}2\\equal{}x^{3}\\Rightarrow x\\equal{}\\sqrt[3]{\\minus{}2}$ (only one of these is real).", "Solution_2": "hello, the only real solution of the equation $ x^3\\equal{}1$ is $ 1$ so you have only one intersection point with the $ x$ axes.\r\nSonnhard." } { "Tag": [], "Problem": "Alex takes $t$ days to complete a job and Bob takes $(t + 5)$ days to complete the same job. Suppose Bob works alone on the job for $3$ days. After this, Alex works together with Bob on the job and they take a further $4.8$ days to complete the job. \r\nWhat is the value of $t$?\r\n\r\nSource: Singapore ASEAN Scholarship Math Entrance Examination", "Solution_1": "C'mon, this isn't that hard.. application of rates.", "Solution_2": "[hide]\nIf Alex does $\\frac{1}{t}$ jobs per day, and he works for a total of 4.8 days, he does $\\frac{4.8}{t}$ jobs in the time he works. Bob works for 7.8 days at a rate of $\\frac{1}{t+5}$ jobs per day, so he does $\\frac{7.8}{t+5}$ jobs in the time he works.\n\nTogether, they complete 1 job.\n$\\frac{7.8}{t+5} + \\frac{4.8}{t} = 1$\n$t = 10 \\textbf{ days}$\n\n[/hide]", "Solution_3": "[quote=\"aidan\"]Alex takes $t$ days to complete a job and Bob takes $(t + 5)$ days to complete the same job. Suppose Bob works alone on the job for $3$ days. After this, Alex works together with Bob on the job and they take a further $4.8$ days to complete the job. \nWhat is the value of $t$?\n\nSource: Singapore ASEAN Scholarship Math Entrance Examination[/quote]\n\n[quote=\"aidan\"]C'mon, this isn't that hard.. application of rates.[/quote]\r\n\r\nMaybe you should wait a couple of days before saying that ;) \r\n\r\n[hide=\"Solution\"]Bob can do $\\frac{1}{t+5}$ in one day, so after working for 3 days he did $\\frac{3}{t+5}$ of the job. \n\nAlex can do $\\frac{1}{t}$ of the job in one day. So he working together with Alex can do $\\frac{1}{t+5}+\\frac{1}{t}=\\frac{2t+5}{t(t+5)}$ of the job in one day. \n\nSo, our equation is $\\frac{4.8(2t+5)}{t(t+5)}+\\frac{3}{t+5}=1$ and solving gives \n$t=\\boxed{10}$.[/hide]", "Solution_4": "[quote=\"robinhe\"][quote=\"aidan\"]Alex takes $t$ days to complete a job and Bob takes $(t + 5)$ days to complete the same job. Suppose Bob works alone on the job for $3$ days. After this, Alex works together with Bob on the job and they take a further $4.8$ days to complete the job. \nWhat is the value of $t$?\n\nSource: Singapore ASEAN Scholarship Math Entrance Examination[/quote]\n\n[quote=\"aidan\"]C'mon, this isn't that hard.. application of rates.[/quote]\n\nMaybe you should wait a couple of days before saying that ;) \n\n[hide=\"Solution\"]Bob can do $\\frac{1}{t+5}$ in one day, so after working for 3 days he did $\\frac{3}{t+5}$ of the job. \n\nAlex can do $\\frac{1}{t}$ of the job in one day. So he working together with Alex can do $\\frac{1}{t+5}+\\frac{1}{t}=\\frac{2t+5}{t(t+5)}$ of the job in one day. \n\nSo, our equation is $\\frac{4.8(2t+5)}{t(t+5)}+\\frac{3}{t+5}=1$ and solving gives \n$t=\\boxed{10}$.[/hide][/quote]\n\nYeah, instead of a couple of minutes at least wait a couple of hours.\n\nPeople can't always be online when you are.\n\nI also got the same and hopefully correct answer of\n\n[hide]$10$[/hide]" } { "Tag": [ "geometry", "parallelogram", "perpendicular bisector", "geometry proposed" ], "Problem": "[url=http://mathworld.wolfram.com/vanLamoenCircle.html]vanLamoen circle[/url] can be rewrite as\r\n\r\nThree medians $ AA',BB',CC'$ of triangle $ ABC$ are concurrent at $ G$, then perpendicular bisector of $ GA,GA',GB,GB',GC,GC'$ intersect base a cyclic hexagon.\r\n\r\nTwo my extensions\r\n\r\nThree medians $ AA',BB',CC'$ of triangle $ ABC$ are concurrent at $ G$\r\n\r\n1. Take $ A_1,A_2\\in AA'$ such that $ \\overrightarrow{A_1A_2} \\equal{} \\frac {1}{2}\\overrightarrow{AA'}$ similarly we have $ B_1,B_2,C_1,C_2$. Draw two lines through $ A_1,A_2$ and perpendicular to $ AA'$, similarly for cyclically $ B,C$, we get six lines. Prove that six lines intersect, respectively (cyclically $ A,B,C$), base a cyclic hexagon.\r\n\r\n2. Let $ A_1,A_2$ be midpoint of $ GA,GA'$ similarly we have $ B_1,B_2,C_1,C_2$. Draw through $ A_1,A_2$ two parallel lines, similarly for cyclically $ B,C$, we get six lines. Prove that six lines intersect, respectively (cyclically $ A,B,C$), base a hexagon inscribed a conic.", "Solution_1": "See the figure. $ A'$ is the midpoint of $ SV$\r\n$ TMUQ$ is a parallelogram $ \\Rightarrow S_{TMUQ}\\equal{}B_1B_2.TQ\\equal{}A_1A_2.TM$\r\n$ \\Rightarrow \\frac{TM}{TQ}\\equal{}\\frac{B_1B_2}{A_1A_2}\\equal{}\\frac{BB'}{AA'}$\r\n$ \\angle TNR\\equal{}\\angle ASC'\\equal{}\\angle CSV, \\angle TRN\\equal{}\\angle C'SB\\equal{}\\angle SCV$\r\nTherefore $ \\Delta TRN\\sim \\Delta VCS \\Rightarrow \\frac{TR}{TN}\\equal{}\\frac{VC}{VS}\\equal{}\\frac{BB'}{AA'}$\r\n$ \\Rightarrow \\frac{TM}{TQ}\\equal{}\\frac{TR}{TN} \\Rightarrow TM.TN\\equal{}TR.TQ$\r\n$ \\Rightarrow M,N,Q,R$ are concylic. Similarly we get the result (1)" } { "Tag": [ "geometry" ], "Problem": "OK so this isn't reaaaally maths, its more psychology or something, no idea. But has anyone does any sort of study on how people can't pick random numbers? Like just picking numbers from 1 to 10, most people would pick 3 or 7, and in the game rock/paper/scissors or janken or whatever its called, I think people have worked out strategies based on the fact that humans cannot pick a random option of out three.. has anyone thought about this much? I thought it could make a good project or something.", "Solution_1": "Well holmes, in theory you are correct no one can pick random numbers. There is a specific area of computer science which deals with this for there are many uses in programming that have to do with picking random numbers. Gah, if only I could find that windows 2k magazine article... \r\n\r\nGoogle it, you'll find loads of stuff..", "Solution_2": "[quote=\"TripleM\"] Like just picking numbers from 1 to 10, most people would pick 3 or 7[/quote]\r\n\r\nOr they might NOT pick 3 or 7 just because most people DO. Or they might because some people wouldn't pick them. Or not. Or maybe. Or not...I think this would be WAY too complicated, since you'd have to study everything that could possibly influence a human's decision (for instance, my birthday might be on the 3rd, or my best friend's dog's mom's birthday might be on the 7th, or...\r\n\r\nI think it's easier to just say they're close enough to be random. Or maybe \"random with slightly different probabilities.\" But nothing is truly random. Except for my e-mails.", "Solution_3": "Yeah...they're so random that forget to send them. :wink: \r\n\r\n-interesting_move", "Solution_4": "If you're interested in the learning aspects of this, I would suggest Fudenberg and Levine's unpublished paper \"Conditional Universal Consistency\"", "Solution_5": "Where would I get an unpublished paper from :)", "Solution_6": "You can go to the Mathematics ArXiV (pronounced Archive I guess) at [url]http://front.math.ucdavis.edu/[/url].", "Solution_7": "Oooh... this is enough material to keep me busy for a lifetime. Especially considering that I don't understand a lot of the math at this point.", "Solution_8": "It's even more definitely enough to keep you busy for a lifetime considering that it gets updated frequently." } { "Tag": [ "function", "algebra", "domain", "geometry", "geometric transformation", "reflection", "trigonometry" ], "Problem": "Suppose you tell a student that $ f(x) \\equal{} \\frac{1}{x}$ is not continuous at $ x \\equal{} 0$ because it is not defined at $ x \\equal{}0$. Is $ g(x) \\equal{} \\frac{x^2 \\minus{} 1}{x \\minus{} 1}$ continuous at $ x \\equal{} 1$?", "Solution_1": "So, this obviously goes back to the issue raised here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=304721\r\n\r\nMy personal opinion is that we may consider both of these functions as mapping from $ \\bf R$ to $ \\bf R$ plus a new value \"is not defined here,\" and that in this case both functions are discontinuous. Both functions are continuous on the set of values at which they don't take the value \"is not defined here,\" (which in a more serious mathematical context we would just call \"their domains\"), but the latter function has the convenient property that it can be easily extended to a continuous function from $ \\bf R$ to $ \\bf R$.\r\n\r\nThe other way to answer this question (which, on further reflection, may be both more honest and more useful) is to say, \"the function isn't defined at $ x \\equal{} 1$, so it surely can't be continuous there.\" Again, this particular function is interesting because there's an obvious way to extend it to fill the gap, but this extension is not actually the same function.", "Solution_2": "For a significant fraction of the time, I tend to take the attitude that \"removable singularities deserve to be removed.\" If you ask me what the value of $ \\frac{\\sin x}x$ is at $ x\\equal{}0,$ I'll tell you $ 1.$ If we're talking about $ \\int_0^1x\\ln x\\,dx,$ I'll say that that's not even an improper integral - it's the integral of a continuous function on $ [0,1].$\r\n\r\nAs a Calculus I issue: we say that the function is not defined at a certain point, but it's just that point. (In other words, an isolated singularity.) But if I want to have any understanding of how the function behave, I really need to know not just that the function may not be defined there but also how it behaves as $ x$ tends to the troublesome value. And it's in that - the \"how it behaves in the neighborhood\" - that distinguishes $ \\frac1x$ from $ \\frac{x^2\\minus{}1}{x\\minus{}1}.$ Or, for that matter, $ \\frac{\\sin x}{x}.$", "Solution_3": "Yes in a lecture, both explanations given by Kent and JBL will set things straight. But if in addition the answer in the back of the book says that g(x) is continuous at the point in question, then offering a correct view as noted by JBL: \"the function isn't defined at x = 1, so it surely can't be continuous there\" to a student who has already looked up the solution in the back of the book will create a confusion. I am convinced that Kent's suggestion that \"removable singularities deserve to be removed\" before going to the question of continuity will help reduce contradictions in the mind of a beginner.", "Solution_4": "I disagree: in particular, I think people should be introduced to the notion that textbooks can be wrong as early as possible :)" } { "Tag": [ "geometry", "3D geometry", "symmetry", "real analysis", "real analysis unsolved" ], "Problem": "Three ants sit in the three centers of adjacent faces of unit cube 1x1x1.\r\nAt time t=0 all three ants begin to walk towards \r\neach other 1->2->3->1 at equal constant speeds v=1.\r\n\r\nHow much time before the ants meet?", "Solution_1": "The time seems to be exactly $ 1$ -- neat problem!\r\n\r\nOh, a proof, you say?\r\n\r\nLet $ C$ be the corner point that the ants are converging towards. (by symmetry, they must meet there) Let $ A$ be one of the ants, and $ B$ another.\r\n\r\nWe claim that at any time $ t$ prior to the collision, $ \\angle CAB \\equal{} \\frac{\\pi}{4}$.\r\n\r\nProof: Let $ \\theta$ be the angle between line $ CA$ and the adjoining side, and $ D$ be the projection of $ B$ onto the other edge of its face. Then angle $ \\angle BCD \\equal{} \\theta$, by symmetry, so $ \\angle ACB$ is a right angle. Since $ \\triangle ACB$ is isoceles, we're done.\r\n\r\nNote that we do not need the assumption that each ant remains on their own face.\r\n\r\nFinally, we conclude that each ant has a constant radial velocity towards the corner of $ \\frac{\\sqrt{2}}{2}$, and the starting distance is $ \\frac{\\sqrt{2}}{2}$.", "Solution_2": "Ah, anyone reading this should note that I am obviously working on the unfolded cube, and take that into account when I am talking about line segments and such." } { "Tag": [ "geometry" ], "Problem": "if there is a square with side length of r, and there are four circles, each with a radius of r, centered at the four vertices of the square, what is the area of the intersection of the four circles?", "Solution_1": "Let me give you an overview of the solution:\r\nFind the area of the region enclosed by 2 circles. Let this be y\r\nFind the area of the region that looks like an equilateral triangle (the area is actually a sector). Let this be x\r\nAfter you've done that, the area is y-2x.\r\n\r\nIf you are still helpless, download the solution.", "Solution_2": "in your solution, the funny section that looks like a triangle with arcs for sides, is the sector you defined as x, right?\r\n\"notice that x is just an ordinary sector with an angle of 60 degrees.\" \r\nhow do you know its 60 degrees? and isn't a sector a slice of a circle (and must include the circle's center, from which the radii are measured)?" } { "Tag": [ "geometry", "perimeter", "calculus", "integration" ], "Problem": "The primeter of an equalateral 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 centemeters. How many integers are not possible values of d?", "Solution_1": "Are you sure you didn't mean how many integers d there are? :?", "Solution_2": "Please [b]show your solutions[/b]! Posts without solutions have been deleted. :mad:", "Solution_3": "Heyyyyyyy u deleted my post!!!1I was askin him whether i was goin right or not...only then would i know whether my solution was goin right or not and then i would post it.....not fair.", "Solution_4": "[hide=\"Solution\"]\nLet the sides of the triangle be $x+d$,\nSo the sides of the square is $x$,\n$3x+3d-4x=1989$\n$3d-1989=x$\n$3(d-663)=x$\nNow,\n$d$ has to be a positive no. so all integral values of $d$ lesser than $663$ will make $x$ a negative no. or $0$ which is not possible.\nSo $663$ values of $d$ are not possible.\nAnswer=$\\boxed{663}$\n[/hide]\r\nCorrect?" } { "Tag": [ "complex analysis", "complex analysis unsolved" ], "Problem": "Find all M\u00f6bius transformation $ \\tau(z) \\equal{} \\frac {az \\plus{} b}{cz \\plus{} d}$, with $ ad \\minus{} bc \\equal{} 1$ such that $ \\tau(D) \\equal{} D$ where $ D$ is the unit disk $ D \\equal{} \\{z\\in\\mathbb{C}: |z| < 1\\}$.", "Solution_1": "hint[hide]$ d\\equal{}\\bar{a},c\\equal{}\\bar{d}$[/hide]" } { "Tag": [ "calculus", "integration" ], "Problem": "Would someone either give me a hint or explain how to solve one/both of the problems below?\r\n\r\n\"How many integers less than 500 have an odd number of distinct, positive, integer factors? Note: One is a always a factor of a positive integer.\"\r\n\r\n\"Arrange the digits 0 through 9, using each exactly once, to form five two-digit numbers whose product is as large as possible. For example, $10*23*45*67*89=61717050$.\"\r\n\r\nThanks for any help provided.", "Solution_1": "[hide]For first given any $n$, if $a$ is a divisor, so would be $\\dfrac n a$ making number of divisors even...unless it so happens that $a=\\dfrac n a$ ... hmm.. hint ..[b] big hint .[/b].... when is that foing to happen? :) [/hide]", "Solution_2": "2\r\n[hide]\nThis probably isn't the greatest explanation, but you want the largest amount of 10s in the product possible and to create the greatest numbers possible at each step which should occur when you have\n$90 * 81 * 72 * 63 * 54 = 1785641760$\n\nI think this is right, but not 100% sure.[/hide]", "Solution_3": "Thanks for the help Gyan & SnowStorm. Snow, would it be accurate if I said that you're trying to say we want each term to be as big as possible?", "Solution_4": "[hide=\"hint\"]if a number is a perfect square, then it has an odd number of factors...[/hide]", "Solution_5": "Alright, thanks guys:\r\n\r\n[hide=\"Number 1\"]\nSince perfect squares are the only numbers that have an odd number of distinct postive integral factors, we want the number of squares less 500. $\\fbox{22}$\n[/hide]", "Solution_6": "For #2,\r\n[hide]Wouldn't you try to make the numbers as close possible to each other? :?[/hide]", "Solution_7": "[quote=\"SnowStorm\"]2\n[hide]\nThis probably isn't the greatest explanation, but you want the largest amount of 10s in the product possible and to create the greatest numbers possible at each step which should occur when you have\n$90 * 81 * 72 * 63 * 54 = 1785641760$\n\nI think this is right, but not 100% sure.[/hide][/quote]\r\nMaybe this resolves your doubt:\r\nConsider the following mini-problem: Which one is larger 42x31 or 41x32 without using calculator? The main thing is not to solve this mini-problem but to learn how numbers are compared.\r\n\r\nSo let's think about the numbers 31 and 41. If we rewrite them, 42x31=(41+1)x31 and 41x32=41x(31+1). Now lets expand a bit\r\n\r\n42x31=(41+1)x31=41x31+31\r\n41x32=41x(31+1)=41x31+41\r\n\r\nSo, is your solution giving the greatest possible product? :D", "Solution_8": "So we compare every two numbers and such like that? that's a pain....", "Solution_9": "No, I think he was just using an example to prove something. :)", "Solution_10": "ok, so what's the answer and solution? :? I've seen this many times...", "Solution_11": "SnowStorm said it best when he said we want the greatest number of tens in the product as possible. To achieve this we want the bigger digits in the tens place of each term. So the solution is $(90)(81)(72)(63)(54)$" } { "Tag": [], "Problem": "Rosalee plans to open a savings account and a checking account. She has decided to\ndeposit a total of $ \\$45$ per week, such that $ \\$20$ goes into the checking account each\nweek and the remaining money goes into the savings account. When she has deposited\na total of $ \\$200$ into her savings account, how much will she have deposited into her\nchecking account?", "Solution_1": "if $ \\$20$ goes in the checking account, $ \\$25$ goes in the savings account. so after 200/25=8 deposits, the savings will have $ \\$200$, and the checking account will ahve $ \\$(20*8)\\equal{}\\$160$" } { "Tag": [ "geometry", "perpendicular bisector", "angle bisector" ], "Problem": "I am looking for someone to solve the questions below using steps as you go through the lesson. I would like to learn how to solve similar questions on my own. Ready?\r\n\r\n1) How many points are equidistant from points A and B and also 6 inches from line segment AB?\r\n\r\n2) How many points are equidistant from two intersecting lines and also 4 inches from the point of intersection of the lines?", "Solution_1": "#1 Both points must lie on the perpendicular bisector of $AB$. On each side of the line $AB$ there must be exactly one point 6 inches from the line $AB$. (As a point slides along the perpendicular bisector, its distance from $AB$ increases continuously, at some point it passes 6 inches.)\r\n\r\n#2 The locus of points equidistant from two lines is the [url=http://www.artofproblemsolving.com/Wiki/index.php/Angle_bisector]angle bisector[/url] of the two lines: this is a line and passes through the point where they meet. By the same logic as before, there is one point either side of their intersection which is exactly 4 inches from their intersection.", "Solution_2": "So, for question 2, the answer is one point?\r\n\r\nAre not intersecting lines the same as vertical angles?\r\n\r\nVertical angles form 4 angles, right?", "Solution_3": "For question 2 there are [b]2[/b] points: both on the same line that passes through the point of intersection, but one either side of it.\r\n\r\nThe fact that this line is the angle bisector is not really important, so there is no need to worry about vertical angles, we just have a point (where the lines meet) and a light that goes through it.", "Solution_4": "I got it now. Thanks" } { "Tag": [ "function", "limit", "inequalities", "algebra proposed", "algebra" ], "Problem": "Find all functions $ f: \\mathbb{R^\\plus{}} \\rightarrow \\mathbb{R^\\plus{}}$ satisfying $ f(f(x)\\minus{}x)\\equal{}2x$ for all $ x>0.$", "Solution_1": "The equation immediately yields $ f(x)>x$ and $ f$ is surjective. But this requires $ 2x>f(x)\\minus{}x\\Longrightarrow f(x)<3x$. Define two positive sequences via $ a_0\\equal{}1,b_0\\equal{}3$ and $ a_{n\\plus{}1}\\equal{}\\frac{2\\plus{}b_n}{b_n},b_{n\\plus{}1}\\equal{}\\frac{2\\plus{}a_n}{a_n}$ and assume we have shown $ a_nx3$.\r\n(2) We observe for $ n>1$ the inequalities $ a_{n\\plus{}1}\\minus{}a_n\\equal{}(b_{n\\minus{}1}\\minus{}b_n)\\frac{2}{b_nb_{n\\minus{}1}}1$ we note the inequality $ b_{n\\plus{}1}\\minus{}a_{n\\plus{}1}\\equal{}(b_n\\minus{}a_n)\\frac{2}{a_nb_n}<\\frac{2}{3}(b_n\\minus{}a_n)$.\r\n\r\n(1),(2) imply that $ a_n,b_n$ converge to limits $ a\\leq 2\\leq b$. From (3) we conclude $ a\\equal{}b\\equal{}2$." } { "Tag": [ "email" ], "Problem": "If you play on FICS what's your handle??\r\n\r\nmine is [b]andonehotsauce[/b]. Maybe we can play each other", "Solution_1": "Handle: DemonBlade", "Solution_2": "Handle: davidyko\r\nWarning: I absolutely suck at anything except for standard. I can't think right in faster games.", "Solution_3": "Are are you supposed to acquire an account? I have no non-free emails.", "Solution_4": "Meh, chess64 gave me an email address on his site so that's the only way I got my handle.", "Solution_5": "Hm... does school email count as free?" } { "Tag": [ "parameterization", "function", "topology", "geometry", "algebra", "domain", "real analysis" ], "Problem": "1) Show that if S is a (piecewise) smooth k-dimensional surface in $ R^n$ then up to an at most countable number of piecewise smooth surfaces, S can be written as the union of at most countably many parametrized k-dimensional surfaces.\r\n\r\n2) Show that the partition in 1) can be made to be locally finite.", "Solution_1": "Definitions? \r\n\r\nMy guess is that you are asked to eliminate the parameters locally, using the implicit function theorem. For example, if $ x \\equal{} x(t)$, $ y \\equal{} y(t)$ is a parametrized curve, then $ y$ can be written as a function of $ x$ at a neighborhood of any point where $ x'(t)\\ne 0$.", "Solution_2": "Here are the definitions:\r\n\r\nWe call a point a zero-dimensional surface of any smoothness class.\r\nA [i]piecewise smooth one-dimensional surface[/i] is a curve in $ R^n$ which breaks into smooth one-dimensional surfaces when a finite or countable number of zero-dimensional surfaces are removed from it.\r\nA surface S in $ R^n$ of dimension k is [i]piecewise smooth[/i] if a finite or countable number of piecewise smooth surfaces of dimension at most k-1 can be removed from it in such a way that the remainder decomposes into smooth k-dimensional surfaces $ S_i$ (with boundary or without).\r\n\r\nAs you can see, the way the definition is given makes it almost impossible to do any useful mathematics. In the statement of the problem it's not clear exactly what \"partition\" means since it seems likely that the sets in the partition might have some (negligible) overlap. I'm inferring from comments the author makes elsewhere that the intersection of any $ S_i$ and $ S_j$ is a piecewise smooth surface of dimension at most k-1 contained in the intersection of their (nontopological) boundaries. :| \r\n\r\n---\r\n\r\nThe partition of the surface S is locally finite if any compact set K in S can intersect only a finite number of the surfaces $ S_i$.\r\n\r\nI considered something similar to what you said with local charts but I don't see how to show the collection $ S_i$ is countable. In what I interpret as a hint, the author made some offhand comment about each of the pieces $ S_i$ being contained in the range of some local chart of the surface S. Hopefully you see something I don't.", "Solution_3": "Since the surface is embedded into a Euclidean space, its topology has a countable basis. I think this should imply countability of s_i", "Solution_4": "But we don't know that the countable basis has the negligible overlap property, for one. The point of the exercise is to show that these Si can always be constructed and the area of S can be defined invariantly as the sum of the areas of the Si.", "Solution_5": "On further thinking, I think you may be right about the first part. By a similar reasoning, S has a countable atlas and the domains of action of the local charts of that atlas might give the partition we are looking for. However it's still hard to see how to show that they abut on piecewise smooth surfaces of lower dimension. It seems more likely that we should use the connected components of S but is it true that a connected smooth manifold is parametrizable?\r\n\r\nIf it's any help, the author makes an offhand comment which I interpret as a hint: he says that each of the smooth pieces Si is contained in the range of some local chart of the surface S.\r\n\r\nAnd the local finiteness condition?", "Solution_6": "I made made a few errors in my explanation so I'll start over:\r\n\r\nWe call a point a zero-dimensional surface of any smoothness class.\r\nA [i]piecewise smooth one-dimensional surface[/i] is a curve in $ R^n$ which breaks into smooth one-dimensional surfaces when a finite or countable number of zero-dimensional surfaces are removed from it.\r\nA surface S in $ R^n$ of dimension k is [i]piecewise smooth[/i] if a finite or countable number of piecewise smooth surfaces of dimension at most k-1 can be removed from it in such a way that the remainder decomposes into smooth k-dimensional surfaces $ S_i$ (with boundary or without).\r\n\r\nI presume the union is supposed to be pairwise disjoint.\r\n\r\nWe say $ S \\subset R^n$ is a k-dimensional parametrized surface if there is an [i]open connected set[/i] $ D \\subset R^k$and a $ C^1$\r\nhomeomorphism (basically a local chart)$ r: D \\to S$.\r\n\r\nLet's break the problem into smaller steps.\r\n\r\nIf M is a smooth k-dimensional surface in $ R^n$, can we always remove at most countably many piecewise smooth surfaces of dimension at most k-1 so that M breaks up into a union of smooth [i]parametrized[/i] surfaces?\r\n\r\nMy idea was to consider the connected components of M and show that once an atlas for a connected component A of M is fixed, M can be represented as a smooth parametrized surface. This seems plausible but I don't see how to do it.\r\n\r\n[url=http://books.google.com/books?id=q118Hw_7wKgC&pg=PA189&lpg=PA189&dq=%22the+partition+be+locally+finite%22&source=web&ots=Ui4wmQyxMZ&sig=0WXM-Q4YoQOjpDjlvA6NGCnKAtw&hl=en]See here also[/url]" } { "Tag": [ "linear algebra", "matrix", "induction" ], "Problem": "If $ A \\equal{} \\left[\\begin{array}{cc} 7 & 4 \\\\ \\minus{}9 & \\minus{}5 \\end{array}\\right]$ compute $ A^n$.", "Solution_1": "Quite easy \r\n\r\nWell known that for n = 2 matrices $ \\ A^2 \\minus{} TraceA*A \\plus{} det(A)I_2 \\equal{} 0$\r\n$ \\ A^2 \\equal{} 2A \\minus{} I_2$ by induction $ \\ \\minus{} >$ $ \\ A^n \\equal{} nA \\minus{} (n \\minus{} 1)I_2$", "Solution_2": "Yes, that's correct." } { "Tag": [], "Problem": "I have worked on this all day. I need to solve for L, which is in the radicand. \r\n\r\n2 = 2PI(sqrt{L/32})", "Solution_1": "I might be wrong :? \r\n\r\n$2=2\\pi\\sqrt{\\frac{L}{32}}$\r\n\r\nsquare both sides\r\n\r\n$4=4(\\pi^2)\\frac{L}{32}=\\frac{4(\\pi^2)L}{32}$\r\n\r\n$32=(\\pi^2)L$\r\n\r\n$L=\\frac{32}{\\pi^2}$", "Solution_2": "Yes, you got it.\r\n\r\nI needed to square both sides because it is a radical equation.\r\nGood job.", "Solution_3": "[quote=\"symmetry\"]I have worked on this all day. I need to solve for L, which is in the radicand. \n\n2 = 2PI(sqrt{L/32})[/quote]\r\n\r\nAnother way to solve it:\r\n\r\n[hide]\n$2=2\\pi\\sqrt{\\frac{L}{32}}$\n\n$1=\\pi\\frac{\\sqrt{L}}{\\sqrt{32}}$\n\n$1=\\frac{\\pi4\\sqrt{2L}}{32}$\n\n$1=\\frac{\\pi\\sqrt{2L}}{8}$\n\n$\\frac{8}{\\pi}=\\sqrt{2L}$\n\n$\\frac{64}{\\pi^2}=2L$\n\n$L=\\frac{32}{\\pi^2}$\n\nIt may be a little bit longer, but... \n[/hide]\r\n\r\n :?" } { "Tag": [ "blogs", "geometry", "geometric transformation" ], "Problem": "As the subject description already says, What do you think your rank at nats will be?", "Solution_1": "as i can not vouch for myself, i will do so for my team member, dan.\r\n\r\n53rd\r\n\r\nthat's how we do it!\r\n\r\njorian", "Solution_2": "According to my blog 6th written, 1st countdown. :rotfl:\r\nBut I really don't know, I'm hoping I'll get into countdown, but I probably won't crack the top 4 written to get a bye. Anyhow, I consider myself to be pretty strong at countdown, so I have a good shot at going far as long as I don't get matched up against someone really good. But I've countdown maybe 100 times this year, I'm 99-1.", "Solution_3": "Probably (hopefully) in the top 100 written... not in countdown for sure. I'm pretty bad. :blush:", "Solution_4": "most likely in the top 228 spots... unless i get 229 ... which would be kinda sad -_-...", "Solution_5": "[quote=\"13375P34K43V312\"]According to my blog 6th written, 1st countdown. :rotfl:\nBut I really don't know, I'm hoping I'll get into countdown, but I probably won't crack the top 4 written to get a bye. Anyhow, I consider myself to be pretty strong at countdown, so I have a good shot at going far as long as I don't get matched up against someone really good. But I've countdown maybe 100 times this year, I'm 99-1.[/quote]\r\n\r\nnote that i almost won, but he got the win, 3-5\r\n\r\nwho won against you? ni?\r\n\r\njorian", "Solution_6": "I predict 212th place for me.", "Solution_7": "[quote=\"13375P34K43V312\"]According to my blog 6th written, 1st countdown. :rotfl:\nBut I really don't know, I'm hoping I'll get into countdown, but I probably won't crack the top 4 written to get a bye. Anyhow, I consider myself to be pretty strong at countdown, so I have a good shot at going far as long as I don't get matched up against someone really good. But I've countdown maybe 100 times this year, I'm 99-1.[/quote]\r\n\r\nWhat did you get at states?", "Solution_8": "he got 1st the last 2 years, and i'm randomly guessing he got 112th at nats last year...\r\n\r\njorian", "Solution_9": "I meant as a score.", "Solution_10": "its just a guess, but based on my not so awesome past nats scores, id say between 120-140, but dqmot (dont quote me on that)\r\n\r\nEDIT: As a score, i think anywhere between 20-25 maybe, if im lucky, a 27 or 28", "Solution_11": "a 28 gets you around 70th :wink:\r\n\r\njorian", "Solution_12": "ok then, maybe not that high... im gonna get owned at nats. but my goal is to get in the top 100. if not, oh well...", "Solution_13": "This year I predict the Countdown cutoff to be around 30, and the Top 57 cutoff to be around 25.\r\n\r\nIt should be harder this year", "Solution_14": "Isn't the cutoff countdown a little low, even if the test is hard?", "Solution_15": "scores at nationals -_-... the cutoff for countdown will be high this year, considering the fact that most of us can get like 35+ on old nats test... and 30+ on recent ones. or maybe im just assuming things ...", "Solution_16": "i think the cutoff's gonna be like 32 or 33. 30 sounds a bit low.", "Solution_17": "uh,.. i think its gonna be like 35 -_-... really high", "Solution_18": "I agree with Mathcrazed, I think its going to be high, too.", "Solution_19": "Winner: 43\r\nTop four cutoff: 41\r\nTop 12: 37\r\n\r\nMy predictions, even if it's a \"hard\" year.", "Solution_20": "But not that high...", "Solution_21": "I can't say for me but here is some of the others:\r\n\r\nKevin Chen: 1st (or 2nd if by countdown he messes up... :( )\r\nBobby Shen: In the top 15\r\nOther people: In the top 30\r\n\r\nTeam: 1st! w00t", "Solution_22": "Um: I think that people are getting smarter as the years go by so I think that even if the contest is hard, everyone is just going to own it and so the cutoff is 42 to make it to countdown and 45 is the winner.", "Solution_23": "I got 33 at States, what does that normally translate to at nationals?\r\n\r\nProbably not accurate though considering I got 29 at Chapters\r\n\r\nChapter:21/8\r\nState:19/14\r\n\r\nNationals?:17/20.....", "Solution_24": "you can't really translate scores\r\n\r\ni know everyone hates my dt examples, but here's one\r\n\r\n'06: dt got 3rd at school, dt got 12th at chapter, dt got 2nd at state, dt got 92nd at national\r\n'07: dt got 1st at school (should be 2nd, but w/e), 3rd at chapter, dt got 1st at state, dt got x at national\r\n\r\n\r\nim not even going to try to guess any of dan's placings...\r\n\r\njorian", "Solution_25": "For nationals I want a 15+ on the sprint and a double digit on target (harder than you think)", "Solution_26": "Last year I got 105.\r\nThis year, I'm hoping for the top 50+~. ( :o That's a jump of more than 50 places!)", "Solution_27": "I want at least 14 on target and at least 20 on sprint", "Solution_28": "I wat at least a 36.\r\n\r\nI'm hoping for top 25.", "Solution_29": "[quote=\"zserf\"]I got 33 at States, what does that normally translate to at nationals?\n\nProbably not accurate though considering I got 29 at Chapters\n\nChapter:21/8\nState:19/14\n\nNationals?:17/20.....[/quote]\r\nTranslations usually don't work.\r\nLast year I went from a 35 at states to a 33 at nats.\r\nNeal Wu stayed constant with a 42.\r\nSomeone I know went from a 40 to a 24.\r\nSomeone else I know went from a 41 to a 25.\r\nSomeone else I know went from a 39 to a 28.\r\nSo it all depends on the person.", "Solution_30": "Someone I knew went from 30 to 34 :maybe: \r\n\r\nBut yeah, it is weird.", "Solution_31": "[quote=\"diophantient\"]Someone else I know went from a 41 to a 25.\n[/quote]\r\n\r\nYeah that would have been me.\r\nThere are, as I discovered the hard way last year, much more factors than a first-time student my think there are. This is where experience helps, whether you are a coach or a competitor.\r\n\r\nBasically, sleep, quality/quantity of food, type of problems, bathroom issues, and even stuff like clothing and size of paper (they give you like 6*9 sheets) play a role. And the pressure that proctors put on you...my proctor last year thought I was cheating and threatened she would disqualify me so yeah.", "Solution_32": "I hated that moment Carl, it got me distracted.\r\n\r\nJKLOL\r\nBut yeah, I was like, who would cheat at a Math Competition.", "Solution_33": "my prediction: top score = 46 (k.chen i wonder who that could be?)", "Solution_34": "and the 40 to 24 was me.\r\nCarl and I had horrid days, we easily could've been much much higher, so I don't think you should try translating scores.", "Solution_35": "i dont know what ill get but ill predict ill make it into countdown. My parents will probably get mad if i dont make top 5, cuz they've \"invested so much money in mathcounts\". jeez, i wonder whatll happen if i drop. i go t 23rd last yr. i think high score will be 45(dunno who probably kevin. Do i even need to say a last name?). cd cutoff :35, top 4:40" } { "Tag": [ "AMC", "AIME" ], "Problem": "Our free AIME Math Jam will be on Thursday, March 19, starting at 7 PM Eastern / 4 PM Pacific (please note early start time). We will discuss solutions to all 15 problems on the contest. Click on the \"Math Jam\" tab at the top of the page to get more information.", "Solution_1": "About how long will it be?", "Solution_2": "As long as it takes to get through all 15 problems. Probably about 2 hours, give or take a half-hour. Of course, you don't have to be there for the whole thing, and a complete transcript will be posted afterwords.", "Solution_3": "Where will be able to find the transcript?", "Solution_4": "if you hover your mouse over the \"math jams\" option above, you will find a drop down menu that gives two options\r\none says transcripts\r\n\r\nclick on that and you should find it\r\n\r\n*note* you may have a to wait a while for the transcript to be uploaded onto the site", "Solution_5": "thanks! :lol:", "Solution_6": "If you missed the Math Jam, or want to review some of the solutions, the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=251]transcript[/url] is now posted.", "Solution_7": "The AIME II Math Jam is on Friday, April 3 at 7 PM Eastern / 4 PM Pacific. We will discuss the solutions to all 15 problems." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let $ a,b\\in R$ and $ f: R\\rightarrow R$ is an increasing function.Find the functions $ g: [a;b]\\rightarrow [f(a);f(b)]$ such that:\r\n$ g(x) \\minus{} g(y)\\geq|f(x) \\minus{} f(y)|$ for any $ x,y\\in [a;b]$", "Solution_1": "[quote=\"toanIneq\"]Let $ a,b\\in R$ and $ f: R\\rightarrow R$ is an increasing function.Find the functions $ g: [a;b]\\rightarrow [f(a);f(b)]$ such that:\n$ g(x) \\minus{} g(y)\\geq|f(x) \\minus{} f(y)|$ for any $ x,y\\in [a;b]$[/quote]\r\nI think you have a typo .\r\n$ g(x)\\minus{}g(y)\\geq |f(x)\\minus{}f(y)|$ \r\nLet $ x\\to y,y\\to x$ then \r\n$ g(y)\\minus{}g(x)\\geq |f(x)\\minus{}f(y)|$\r\nTherefore $ |f(x)\\minus{}f(y)|\\leq 0$ so $ f(x)\\equal{}c$" } { "Tag": [ "algebra", "polynomial" ], "Problem": "$n$ and $l$ are positive integers.\r\n\r\nFind all possible pairs of $n$ and $l$ such that\r\n\r\n\\[{n\\choose 0}+{n\\choose 1}+{n\\choose 2}+{n\\choose 3}=2^{l}.\\]", "Solution_1": "Well the one solution I know immediately is [hide] $(n,l)=(3,3)$.[/hide] But I don't know if there exist more solutions. If this is useful, I'll save everyone a bit of arithmetic and tell you that it reduces to $n^{3}+5n+6=6 \\cdot 2^{l}$. The usual mods don't give much, other than $4 \\not | \\ n$.", "Solution_2": "I was thinking Lucas' Theorem myself.", "Solution_3": "The polynomial on the left factors as $(n+1)(n^{2}-n+6)$. Thus, you have to have that $n+1 | 3\\cdot 2^{l+1}$, so either $n = 2^{k}-1$ or $n = 3\\cdot 2^{k}-1$. From there, some calculations will give you bounds on the possible size of $k$, from which you can find all values of $n$. (There are 6 of them, I believe.) Unfortunately, this is kind of ugly; there may be a nicer way.", "Solution_4": "jbl, there are only three. but your approach sounds sound. any elaboration on how u obtain your bounds?", "Solution_5": "[quote=\"mao_zai\"]jbl, there are only three. but your approach sounds sound. any elaboration on how u obtain your bounds?[/quote]\r\n\r\nNo, there are 5. (I counted 0 in my previous response, which made 6.)\r\n\r\n${1\\choose 0}+{1\\choose 1}+{1\\choose 2}+{1\\choose 3}=2 = 2^{1}$, ${2\\choose 0}+{2\\choose 1}+{2\\choose 2}+{2\\choose 3}=4 = 2^{2}$, ${3\\choose 0}+{3\\choose 1}+{3\\choose 2}+{3\\choose 3}=8 = 2^{3}$, ${7\\choose 0}+{7\\choose 1}+{7\\choose 2}+{7\\choose 3}=64 = 2^{6}$, and ${23\\choose 0}+{23\\choose 1}+{23\\choose 2}+{23\\choose 3}=2048 = 2^{11}$.\r\n\r\nAs for the actual process, it's fairly ugly. Let me outline the calculations necessary for the case $n = 2^{k}-1$.\r\n\r\nIf $n = 2^{k}-1$, we have $2^{k}\\cdot (2^{2k}-3\\cdot2^{k}+8) = 6\\cdot 2^{l}$. The left hand side is dominated by the $2^{3k}$ term of that product, so one can show that for sufficiently large $k$, it gets bounded strictly between $6 \\cdot 2^{3k-3}$ and $6 \\cdot 2^{3k-2}$. (Basically, you subtract and look at the differences: one term will grow much faster than the others, so eventually the difference will have the same sign as that term.) When that happens, we know there can't be any solutions, so we only need to check the values below the bound. Something similar happens in the case $n = 3\\cdot 2^{k}-1$.", "Solution_6": "thanks for the solution. i omitted the first two cases n=1,2." } { "Tag": [ "limit", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "Let $a,b,A,B$ are positive real numbers. Let sequence $x_{1},x_{2},x_{3},...$ by $x_{1}=a,x_{2}=b,x_{n+1}=A\\sqrt[3]{x_{n}^{2}}+B\\sqrt[3]{x_{n-1}^{2}}(n=2,3,...)$. Prove that exists $lim_{n\\to\\infty}x_{n}$ and find it.", "Solution_1": "Tried the standard trick, but something doesn't seem quite right. Suppose $\\lim_{n\\to\\infty}x_{n}=L$, then $\\lim_{n\\to\\infty}x_{n+1}=L$ and $\\lim_{n\\to\\infty}x_{n-1}=L$ as well. Letting both sides of the equation go to infinity (and using the standard limit laws) we get: $L=A\\cdot L^{2/3}+B\\cdot L^{2/3}\\Rightarrow L=L^{2/3}(A+B)\\Rightarrow L^{5/2}=A+B\\Rightarrow L=(A+B)^{2/5}$. As for proving the limit exists, I'll come back later today, don't have much time right now.", "Solution_2": "[quote=\"HilbertThm90\"]Tried the standard trick, but something doesn't seem quite right. Suppose $\\lim_{n\\to\\infty}x_{n}=L$, then $\\lim_{n\\to\\infty}x_{n+1}=L$ and $\\lim_{n\\to\\infty}x_{n-1}=L$ as well. Letting both sides of the equation go to infinity (and using the standard limit laws) we get: $L=A\\cdot L^{2/3}+B\\cdot L^{2/3}\\Rightarrow L=L^{2/3}(A+B)\\Rightarrow L^{5/2}=A+B\\Rightarrow L=(A+B)^{2/5}$. As for proving the limit exists, I'll come back later today, don't have much time right now.[/quote]\r\n\r\nI think you want $L = L^{\\frac{2}{3}}(A+B) \\Rightarrow L = (A+B)^{3}$.", "Solution_3": "Ah yes. I knew something wasn't right when I did a quick check.", "Solution_4": "And $L=0$???????\r\nWe show that $\\lim_{n\\to\\infty}u_{n}=(A+B)^{3}$.\r\nRemark: if $a<11$; then there exists $\\epsilon$ such that $(\\alpha+\\epsilon)^\\frac{2}{3}<\\alpha$. For all sufficiently large $n$ $u_{n-1}\\le\\alpha+\\epsilon$ and $u_{n-2}\\le\\alpha+\\epsilon$; thus $u_{n}\\le (\\alpha+\\epsilon)^\\frac{2}{3}$ , $u_{n+1}\\le (\\alpha+\\epsilon)^\\frac{2}{3}$ and $z_{n+1}\\le (\\alpha+\\epsilon)^\\frac{2}{3}$; contradiction and $\\alpha=1$ therefore $\\limsup{z_{n}}\\le1$. We show by same method that $\\liminf{z_{n}}\\ge1$. QED", "Solution_5": "First , it is to be prove that the limit exist.\n\nFirst we have to prove $x_n$ do not oscillate. We rewrite the eq. in this way:\n \n$\\frac{A\\sqrt[3]{x^2_n}}{x_{n+1}}+\\frac{B\\sqrt[3]{x^2_{n-1}}}{x_{n+1}}=1$, let $\\sin^2 \\theta_n=\\frac{A\\sqrt[3]{x^2_n}}{x_{n+1}}$ and $\\cos^2 \\theta_n=\\frac{B\\sqrt[3]{x^2_{n-1}}}{x_{n+1}}$.\n\nWe know $sin^2 y$ & $\\cos^2 y$ both are periodic. So we consider $\\frac{A\\sqrt[3]{x^2_2}}{x_3}=\\frac{A\\sqrt[3]{x^2_m}}{x_{m+1}}\\Rightarrow{x^\\frac{2}{3}_{m}}=c_{(2,3)}x_{m+1}$, where $c_{(2,3)}$ be a constant. Again $\\frac{B\\sqrt[3]{x^2_1}}{x_3}=\\frac{B\\sqrt[3]{x^2_{m-1}}{x_{m+1}}\\Rightarrow{x^\\frac{2}{3}_{m-1}}=c_{(1,3)}x_{m+1}$, where $c_{(1,3)}$ be a constant. Then $(\\frac{x_m}{x{m-1}})^{\\frac{2}{3}}=\\frac{c_{(2,3)}}{c_{(1,3)}}=d_m$ ,or, $x_m=d_mx_{m-1}$ .\n Since $d_m\\neq{1}$ for any value of $m$, then $x_n$ do not oscillates. \n\nNow we have two cases. \n[b]case 1:[/b]$x_n>x_{n+1}$, then it is need not to prove that $x_n$ has a limit as $x_n>0$.\n[b]case 2:[/b]$x_nB\\sqrt[3]{x^2_{n-1}}$, then \nwe know that $x_{n+1}<2A\\sqrt[3]{x^2_n}\\Rightarrow{x^3_{n+1}<8A^3x^2_n}\\Rightarrow{x_{n+1}<8A^3}$, this gives $x_n$ is bounded as $A$ is finite.\n\n If we consider that $A\\sqrt[3]{x^2_n}1$\r\n\r\n(the fact that $0<\\ln{x}1$\r\n\r\nHow's this?\r\n[hide]$0<\\frac{\\ln{x}}{x} = \\frac{2\\ln{\\sqrt{x}}}{x}$\n\nSince $x>1$, $\\sqrt{x}>1$ so ${\\frac{\\sqrt{x}}{\\ln{\\sqrt{x}}}}>1$\n\nThen multiplying the upper bound by this factor will increase it, so\n\n${0<\\frac{\\ln{x}}{x} < \\frac{2\\ln{\\sqrt{x}}}{x} ( \\frac{\\sqrt{x}}{\\ln{\\sqrt{x}}}} ) = \\frac{2}{\\sqrt{x}}$[/hide]", "Solution_3": "Well, I'm assuming its $0\\leq\\frac{\\ln x}{x}<\\frac{2}{\\sqrt{x}}\\ \\forall x\\geq 1$. Now, multiplying both sides by $x$, we have ${0\\leq\\ln x<2\\sqrt{x}}$. Raising $e$ to all of them, we have $1\\leq x20$ by bounding the number between two consequently squares (little bit attention for $n$ - odd)and then check similar for $n < 12$ and finally for $ 12 \\leq n \\leq 19$ and get also $n=16$.", "Solution_2": "Can you explain your solution analitically? :o", "Solution_3": "[quote=\"heartwork\"]$n=16$ and $n=20$.\nEasily check for $n>20$ by bounding the number between two consequently squares (little bit attention for $n$ - odd)and then check similar for $n < 12$ and finally for $ 12 \\leq n \\leq 19$ and get also $n=16$.[/quote]\r\n\r\nFor $n=16$ , $2^{12}(2^5+17+2^4)=2^{12}(65)$ which is not a perfect square\r\nFor $n=20$ , $2^{12}(2^5+17+2^8)=2^{12}(305)$ also not a perfect square .... :?", "Solution_4": "In my opinion , theres not solution \r\n\r\nobserve that $2^{17}+17(2^{12})+2^n=2^{12}(2^5+17+2^{n-12})=2^{12}(49+2^{n-12})$\r\n\r\nIf $n> 12$ ,then letting $49+2^{n-12}=(2k+1)^2$ :arrow: $12+2^{n-14}=k(k+1)$ :arrow: $2^{n-14}=(k+4)(k-3)$ . Since one of $k+1$ or $k-3$ must be odd , it follows that $k-3=1$ :arrow: $k=4$ . When $k=4$ it gives us $n=17$ . \r\n\r\nFor $n\\leq 12$ since any even prefect square modulo 8 $\\equiv 0$ . So $n$ must be multiple of $3$ . By trivial for $n=3,6,9,12$ , there are no solution also .\r\nSo the only answer is $n=17$ :P \r\n\r\nEDIT : Thanks for pointing out where is wrong :D", "Solution_5": "I asked someone and he told me that there exist solution, but i didnot find it. Please help me :?", "Solution_6": "There is one solution: $n=17$.\r\n\r\n$2^{17}+2^{16}+2^{12}+2^{17}=2^{12}+2\\cdot2^{15}+2^{18}=(2^6+2^9)^2$\r\n\r\nFor $n>12$, shyong's reasoning works except that the $k=4$ case should not have been ruled out. For $n\\le12$, we can directly check $n=0,2,4,6,8,10,12$ and verify that none are solutions.\r\n\r\n\r\n\r\nA question from an oral round of a local competition:\r\nFind a positive integer value of $n$ such that $2^8+2^{12}+2^n$ is a perfect square.", "Solution_7": "[quote=\"shyong\"]\n\nIf $n\\geq 12$ ,then letting $49+2^{n-12}=(2k+1)^2$ :arrow: $12+2^{n-16}=k(k+1)$ [/quote]\r\nYou were wrong this step shyong :D \r\nIt's $12+2^{n-14}=k(k+1)$\r\n$\\rightarrow 2^{n-14}=(k-3)(k+4)$\r\nThen $k-3=1\\rightarrow k=4\\rightarrow k+4=8\\rightarrow n=17$ and we've done now :D", "Solution_8": "Opps , I didnt notice i made a mistake there :blush: I will edit it .Thanks for pointing it out badboy :P", "Solution_9": "I was rushing and i skipped the \"17\" 1st time... definitely this is not hard. :(" } { "Tag": [], "Problem": "If \r\n\r\n$1$ , $x$ , $y$ , $\\frac{8}{27}$ , $\\frac{16}{81}$\r\n\r\nis a geometric series, find $x$ and $y$", "Solution_1": "[hide]x= 2/3 y= 4/9 multiply by 2/3[/hide]", "Solution_2": "[hide]2/3 and 4/9[/hide]", "Solution_3": "[hide]\n$\\frac{16}{81}\\cdot\\frac{27}{8} =\\frac{2}{3}$\n$\\frac{2}{3}$ and ${\\frac{4}(9}$\n[/hide]", "Solution_4": "[hide]\n$\\frac{8}{27}=\\frac{2^3}{3^3}\\\\\n\\\\\n\\frac{16}{81}=\\frac{2^4}{3^4}\\\\\n\\\\\nx=\\frac{2^1}{3^1}=\\frac{2}{3}\\\\\n\\\\\ny=\\frac{2^2}{3^2}=\\frac{4}{9}$[/hide]", "Solution_5": "2/3 and 4/9. :D" } { "Tag": [ "pigeonhole principle" ], "Problem": "A square of side $n$ is formed from $n^2$ unit squares, each colored in red, yellow or green. Find minimal $n$, such that for each coloring, there exists a line and a column with at least 3 unit squares of the same color (on the same line or column).", "Solution_1": "[hide]Is it just $n=7$? $n=7$ works because of the pigeonhole principle. For any $n$ smaller than this you can just make each row go RYG and cycle through that.[/hide]", "Solution_2": "What if it's the minimal n for which you will always have either 3 consecutive of the same color in a row or column or 3 consecutive of different colors?\r\n\r\nI haven't actually thought about this one (is it even finite?) but seems like an interesting question.\r\n\r\nEdit--OK, it doesn't work... but what if you have the condition that every 2x2 subsquare has all 3 colors?", "Solution_3": "[img]http://www.jia-han.net/images/math.png[/img]\r\n\r\n6 doesn't work, 7 does through pigeonhole principle." } { "Tag": [ "probability", "absolute value" ], "Problem": "Why is standard deviation more commonly used than mean deviation, when in fact, the formula for mean deviation is easier to use than standard deviation?", "Solution_1": "This is essentially because while mean deviation has a natural intuitive definition as the \"mean deviation from the mean,\" the introduction of the absolute value makes analytical calculations using this statistic much more complicated than the standard deviation.", "Solution_2": "This is wrong. The real reason is the appearance of the standard deviation (and not the mean deviation) in the [url=http://en.wikipedia.org/wiki/Central_limit_theorem]central limit theorem[/url], which is one of the two most important theorems in probability theory, and also that the standard deviation is better when sampling from normally distributed populations. There is no corresponding important result which uses the mean absolute deviation." } { "Tag": [ "function", "calculus", "integration", "algebra", "domain", "real analysis", "real analysis unsolved" ], "Problem": "I need some examples of functions which are differentiable but not integrable..", "Solution_1": "If it's differentiable, it's continuous. If it's continuous, it's locally integrable. The only sense in which integrability can fail is that the integral on an infinite domain might not converge.\r\n\r\nIn other words, $ f(x)\\equal{}1$ is the example you're looking for. If that isn't it, you're looking for omething that doesn't exist.", "Solution_2": "If you are searching for a differentiable function whose 'derivative' is not (Riemann) integrable, then refer to [url=http://en.wikipedia.org/wiki/Volterra_function]here[/url].", "Solution_3": "[quote=\"sos440\"]If you are searching for a differentiable function whose 'derivative' is not (Riemann) integrable, then refer to [url=http://en.wikipedia.org/wiki/Volterra_function]here[/url].[/quote]\r\n\r\nThank u very much//SOS 440// Volterra's function is what is was searching for!!!" } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "Hola amigos de este foro, les propongo el siguiente problema de geo,..\r\n\r\nSean P y C sobre el semicirculo con diametro AB, tal que los arcos BC y CD son iguales.\r\nSi AC y BP se intersectan en E y AD y CP se intersectan en F, pruebe que EF es perpendicular a AD, haciendo uso de numeros complejos. :) \r\nBueno, suerte y espero sus respuestas...", "Solution_1": "qu\u00e9 es D?", "Solution_2": "D es el punto que hace BC y CD igual y por eso C es mas cerca de B. P es variable y creo que necesitamos considerar 3 casos: si los puntos en orden son BCDPA, BCPDA, y BPCDA porque en el primer caso, E y F son adentros del semicirculo, en caso 2 E es adentro, y en caso 3, E y F son afueras.\r\n\r\nNo estoy seguro, pero creo que mi interpretacion es correcto porque parece verdadero en mis dibujos. Pero ya no lo he probado :oops:\r\n\r\nEDIT: ay, lo siento, no es necesario que C es mas cerca de B. Si es mas cerca de A, necesitamos extender el circulo, pero todavia es cierto.", "Solution_3": "[quote=\"nicoopm\"]Sean P y C sobre el semicirculo con diametro AB, tal que los arcos BC y CD son iguales. Si AC y BP se intersectan en E y AD y CP se intersectan en F, pruebe que EF es perpendicular a AD.[/quote]\nSolo basta ver que como los arcos $ BC$ y $ DC$ son iguales, entonces los \u00e1ngulos inscritos $ \\angle CPB$ y $ \\angle DAC$ son iguales, esto es $ \\angle EPF \\equal{} \\angle EAF$ $ \\Longrightarrow$ el cuadril\u00e1tero $ PFEA$ es c\u00edclico $ \\Longrightarrow$ $ \\angle AFE \\equal{} \\angle APE \\equal{} 90^{\\circ},$ i.e. $ EF \\perp AD.$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Given p>5 is a prime, prove: \r\nthere exists no less than two positive integers $ 1\\leq a,b\\leq p-1$ such that \r\n$ a^{p-1}\\equiv 1(\\mod p^{2}),b^{p-1}\\equiv 1(\\mod p^{2})$", "Solution_1": "It fails for p=7, where the only possibility is 1." } { "Tag": [ "function", "algebra", "domain", "MATHCOUNTS" ], "Problem": "I have Mathematica at home and can use a school graphing calculator at school. Should I still buy a graphing calculator? If so, which model?", "Solution_1": "Depends what it's for. If you're permitted to use a graphing calculator in your classes, you might as well get one since the teacher will probably design exams that utilize a graphing calculator. If it is strictly optional, it's your call. In high school, we were all required to get a graphing calculator but in college, almost no classes allow calculators on exams, and they are usually not much help. I currently use a graphing calculator only to perform quick arithmetic, graph functions, and check work-I don't use it as a problem solving technique.\r\n\r\nSo in short, it's up to you. I like the TI series I think they start somewhere around $ \\$$80-90, although I could be wrong. The 89 and the 92 are extremely powerful, but their costs are quite high. If you just need simple graphing and algebra capabilities, an 80, 83+, or 84+ will all work fine (I think those are the model numbers).", "Solution_2": "How often would you be away from your computer and need to graph something with greater accuracy than a sketch by hand?\r\n\r\nI feel like graphing calculators are only useful inside the classroom, for tests, class exercises, and the AP exams. It's your call -- if you can get by at school with one and can find one to use for the AP exam, don't bother! Just get a cheap scientific calculator for a few bucks.\r\n\r\nIf you must buy one, peers and eBay sell TI-83+'s cheap. Some people like the HP graphers that use Reverse Polish Notation.", "Solution_3": "[quote=\"JRav\"]The 89 and the 92 are extremely powerful, but their costs are quite high.[/quote]\r\n\r\nAre you [i]shunning[/i] the $ n$spire??? :P\r\n\r\nGraphing calculators are useful for, well, graphing. I have no idea if they can program or not. I have a TI $ n$spire (scientific and graphing, even though I only use the TI-84 mode) so I wouldn't know.\r\nNow tat the AMC 10/12 don't allow calculators I have a sinking feeling that my calculator is going to collect a lot of dust. :P\r\n\r\nI don't think you should get one, you already have Mathematica.", "Solution_4": "Most graphing calculators are $ \\$$100 bricks with 10-30 year old technology and public domain code, imo.\r\n\r\nBut they're $ \\$$100 bricks that you can use on tests! Yay!\r\n\r\n :|", "Solution_5": "As someone has said, Ebay is great. I bought an 84 silver a long time ago, when I didn't know anything. Then I lost it, so I went on Ebay and ordered the cheapest calculator I could find, a TI-86, 35 dollars shipped. Works great for annoying computation in physics and chem. That is the only use I have found for my calculator, as well as when the teacher tells you to do something with the calculator in class.\r\n\r\nSidenote, a pitch for the 86: I believe it's the ideal calculator for Mathcounts because functions can be accessed with one button press from menus." } { "Tag": [], "Problem": "If a number $ X$ consists of 66 digits of threes, and a number $ Y$ consists of 66 digits of sixes, name all the digits that will appear in the product of $ X$ and $ Y$.", "Solution_1": "[hide]3*6 = 18\n33*66 = 2178\n333*666 = 221788\nWe can accurately predict that for an equal number of 3's and 6's above 1, the only digits that appear will be 1,2,7, and 8.[/hide]" } { "Tag": [ "inequalities", "probability and stats" ], "Problem": "Let $ g: \\mathbb{R}\\rightarrow \\mathbb{R}_ \\plus{}$, nondecreasing on $ \\mathbb{R}_\\plus{}$, $ g(x) \\equal{} g( \\minus{} x)$. Let $ X$ be a random variable such that $ |X(\\omega)|\\leq C,\\forall \\omega \\in \\Omega, C > 0$. Then for $ \\varepsilon > 0$ \r\n\r\n$ \\frac {\\mathbb{E}[g(X)] \\minus{} g(\\varepsilon)}{g(C)}\\leq \\mathbb{P}(|X \\minus{} \\mathbb{E}[X]|\\geq \\varepsilon)\\leq \\frac {\\mathbb{E}[g(X \\minus{} \\mathbb{E}[X])]}{g(\\varepsilon)}$\r\n\r\nI have proved right inequality and I would be grateful for some tips who to prove the left one.", "Solution_1": "[quote=\"Mr.Doe\"]Let $ g: \\mathbb{R}\\rightarrow \\mathbb{R}_ \\plus{}$, nondecreasing, $ g(x) \\equal{} g( \\minus{} x)$ .[/quote]\r\nstrange isn't it :maybe: \r\n\r\nAnd what is $ C$ ? And what if $ P(|X \\minus{} E[X]| > \\epsilon) \\equal{} 0$ ?", "Solution_2": "Thank you for pointing out some of the errors, I have corrected them. The problem doesn't say anything in case X is a degenerate variable, and I didn't even thought about it :blush: . I hope everything is ok now." } { "Tag": [], "Problem": "after i registered for SATII test, can i change my mind, and take another additional test on the test day?", "Solution_1": "Yes. The College Board website is a good place to go for questions like this." } { "Tag": [ "ratio" ], "Problem": "Find the ten-thousandths digit of the following series:\r\n\r\n$ 1,4,8,\\frac{32}{3},...$\r\n\r\n(A) 8\r\n(B) 9\r\n(C) 0\r\n(D) 1\r\n(E) 2\r\n\r\nI assume this is asking to find the ten-thousands digit of the [b]sum[/b] of the given series. I have two questions: 1. Is this a divergent series (because the ratios are greater than 1) that wouldn't have a sum? 2. How do I find the next terms of this series (what is the pattern)?", "Solution_1": "The question is just awful -- poorly conceived and poorly implemented. It is, however, possible to guess the pattern that the sequence follows: look at the ratios between consecutive terms." } { "Tag": [], "Problem": "A math teacher requires Noelle to do one homework assignment for each of the first five homework points she wants to earn; for each of the next five homework points, she needs to do two homework assignments; and so on, so that to earn the $ n^{\\text{th}}$ homework point, she has to do $ n\\div5$ (rounded up) homework assignments. For example, when she has 11 points, it will take $ 12\\div5\\equal{}2.4\\rightarrow3$ homework assignments to earn her $ 12^{\\text{th}}$ point. What is the smallest number of homework assignments necessary to earn a total of 25 homework points?", "Solution_1": "This is $ 1\\plus{}1\\plus{}1\\plus{}1\\plus{}1\\plus{}2\\plus{}2\\plus{}2\\plus{}2\\plus{}2\\plus{}3\\plus{}3\\plus{}3\\plus{}3\\plus{}3\\plus{}4\\plus{}4\\plus{}4\\plus{}4\\plus{}4\\plus{}5\\plus{}5\\plus{}5\\plus{}5\\plus{}5\\equal{}5(1\\plus{}2\\plus{}3\\plus{}4\\plus{}5)\\equal{}5(15)\\equal{}\\boxed{75}$." } { "Tag": [], "Problem": "Well , I just had a doubt with the IUPAC nomenclature....Is the name of the compound in the slide that I've attached \r\n$4,4-diethyl-3-methyl-5-Heptyne$ or $4,4-Diethyl-5-methyl-2-Heptyne$ ??", "Solution_1": "It's the 2nd one.", "Solution_2": "I can't see any compound... the first name you listed is incorrect. The proper name for that compound is 3,3-diethyl-4-methyl-1-heptyne (or, using the revised IUPAC rules, 3,3-diethyl-4-methyl-hept-1-yne, although that correct form is less common in most current textbooks and classes).\r\n\r\nSince this gives a completely different compound than the other name, I can't really tell which is correct. Here are the chemical formulas for each one:\r\n\r\n3,3-diethyl-4-methyl-1-heptyne: $\\text{CH}\\equiv\\text{C}-\\text{C(C}_2\\text{H}_5\\text{)}_2-\\text{CH(CH}_3\\text{)}-\\text{CH}_2-\\text{CH}_3$\r\n\r\n4,4-diethyl-5-methyl-2-heptyne: $\\text{CH}_3-\\text{C}\\equiv\\text{C}-\\text{C(C}_2\\text{H}_5\\text{)}_2-\\text{CH(CH}_3\\text{)}-\\text{CH}_3$" } { "Tag": [], "Problem": "Simplify: $ \\frac{\\sqrt{2.5^2\\minus{}0.7^2}}{2.7\\minus{}2.5}$.", "Solution_1": "$ \\frac{\\sqrt{2.5^2\\minus{}.7^2}}{2.7\\minus{}2.5}\\equal{}\\frac{2.4}{.2}\\equal{}\\boxed{12}$." } { "Tag": [ "algebra", "polynomial", "Vieta", "geometric sequence" ], "Problem": "Source: math olympiad treasure\r\nFind $m$ and solve the following equation, knowing that its roots form a geometric sequence:\r\n$x^{4}-15x^{3}+70x^{2}-120x+m=0$.", "Solution_1": "If the roots are $a, aq, aq^{2}, aq^{3}$, then by Vieta we have\r\n\r\n\\[\\begin{eqnarray}a+aq+aq^{2}+aq^{3}&=& 15\\\\ a^{2}q+a^{2}q^{2}+a^{2}q^{3}+a^{2}q^{3}+a^{2}q^{4}+a^{2}q^{5}&=& 70\\\\ a^{3}q^{3}+a^{3}q^{4}+a^{3}q^{5}+a^{3}q^{6}&=& 120\\\\ a^{4}q^{6}&=& m\\\\ \\end{eqnarray*}\\]\r\n\r\n$(1)$ simplifies to $a(1+q+q^{2}+q^{3})=15$, and $(3)$ to $a^{3}q^{3}(1+q+q^{2}+q^{3})=120$. Dividing those two we get $a^{2}q^{3}=8$, from which we obtain $m=a^{4}q^{6}=(a^{2}q^{3})^{2}=64$.\r\n\r\nThe equation becomes\r\n\r\n\\begin{eqnarray*}x^{4}-15x^{3}+70x^{2}-120x+64=0 &\\iff& x^{4}-3x^{3}+2x^{2}-12x^{3}+36x^{2}-24x+32x^{2}-96x+64=0\\\\ &\\iff& x^{2}(x^{2}-3x+2)-12x(x^{2}-3x+2)+32(x^{2}-3x+2)=0\\\\ &\\iff& (x^{2}-12x+32)(x^{2}-3x+2)=0\\end{eqnarray*}\r\n\r\nThe solutions are $1,2,4,8$", "Solution_2": "nice! \\begin{eqnarray*}x^{4}-15x^{3}+70x^{2}-120x+64=0 &\\iff& x^{4}-3x^{3}+2x^{2}-12x^{3}+36x^{2}-24x+32x^{2}-96x+64=0\\\\ &\\iff& x^{2}(x^{2}-3x+2)-12x(x^{2}-3x+2)+32(x^{2}-3x+2)=0\\\\ &\\iff& (x^{2}-12x+32)(x^{2}-3x+2)=0\\end{eqnarray*}\r\nhow did you factor it? is there any method or just trial and error?", "Solution_3": "Well, 64 only has only so many factors, and they form a geometric sequence... I was thinking of just synthetic-dividing the whole thing", "Solution_4": "[quote=\"srulikbd\"]nice! \\begin{eqnarray*}x^{4}-15x^{3}+70x^{2}-120x+64=0 &\\iff& x^{4}-3x^{3}+2x^{2}-12x^{3}+36x^{2}-24x+32x^{2}-96x+64=0\\\\ &\\iff& x^{2}(x^{2}-3x+2)-12x(x^{2}-3x+2)+32(x^{2}-3x+2)=0\\\\ &\\iff& (x^{2}-12x+32)(x^{2}-3x+2)=0\\end{eqnarray*}\nhow did you factor it? is there any method or just trial and error?[/quote]\r\n\r\nBasically, it was kind of obvious what the roots were: $a\\cdot aq^{3}=aq\\cdot aq^{2}$, and $a\\cdot aq\\cdot aq^{2}\\cdot aq^{3}=64$. Therefore the product of the smallest and the largest one is $8$, as well as the product of the other two. From there you very easily see that the roots must be $1,2,4,8$, and the \"factorization\" is just writing that out." } { "Tag": [ "function", "modular arithmetic", "algorithm", "number theory", "greatest common divisor", "floor function" ], "Problem": "Prove that there is no solution to $72x+23y=845$ in $\\mathbb{N}_+^2$.", "Solution_1": "Notice that $845=21^2\\cdot 2+23$\r\ny must be odd\r\nsolve the equation in function of x", "Solution_2": "Or with some modular trickery:\r\n\r\n$845 \\equiv 53 \\pmod {72}$\r\n\r\nSo it must be true that $23y \\equiv 53 \\pmod {72}$\r\n\r\nUsing the euclidian algoritm, we find the modular inverse: $47$.\r\nSo: \\[ y \\equiv 43 \\pmod {72} \\]\r\n\r\nSince we are dealing with $\\mathbb{N}^2_+$, we don't have to consider negative $y$.\r\nThe smallest value for $y$ is $43$, giving \\[ 72x+989=845 \\]\r\nSince $x \\geq 0$, this equation has no solutions.", "Solution_3": "The same method as Jan, except take modulo 23. You get 3x = 17 mod 23, where it is easier to find $x\\ge 21$. Contradiction.", "Solution_4": "Is there any method without using mod?", "Solution_5": "here is a method without mod but i think the concept are still the same ...\r\n\r\nrewrite it as\r\n\r\n$23y=845-72x$\r\n\r\n$\\implies y=36-3x+\\frac{17-3x}{23}$\r\n\r\nSince $y$ is integer , $\\frac{17-3x}{23}$ must also be an integer . By inspection we know that $x=-2$ gives an integer value . And for this value of $x$ , we can get $y=43$\r\n\r\nNow we know that one of the solution is $(x,y)=(-2,43)$\r\n\r\nand we also know that for $ax+by=c$ , if $x_0,y_0$ is the solution , then for any integer $k$ , $x_0+bk,y_0-ak$ is also the solution . \r\n\r\nFrom here you can see that $43-72=-29<0$ or $-2-23=-25<0$ . Both give you negative value . Hence there are no solution for potive integer $x,y$", "Solution_6": "We know that a diofantic equation ax+by = c have intigers solutons if and only if the c divides the gcd(a,b) . The proof simple . Let d= gcd ( a, b) then there are p,q in N with gcd (p,q ) = 1 for which it holds that a = p*d and b = q * d so dxp + bqd =c so c /d and if c/d then the eqyation has intiger solution . For the eqyation 72 x + 23 y = 845 we obtain that gcd(72,23)=1 so because 845 dont divide 1 the equation has no solution in N ^2", "Solution_7": "[quote=\"gdeskos\"]We know that a diofantic equation ax+by = c have intigers solutons if and only if the c divides the gcd(a,b) . The proof simple . Let d= gcd ( a, b) then there are p,q in N with gcd (p,q ) = 1 for which it holds that a = p*d and b = q * d so dxp + bqd =c so c /d and if c/d then the eqyation has intiger solution . For the eqyation 72 x + 23 y = 845 we obtain that gcd(72,23)=1 so because 845 dont divide 1 the equation has no solution in N ^2[/quote]\r\n\r\nNot true ..example \r\n\r\n$3x+4y=7$ . Although $(3,4)=1$ and $7$ doesnt divide $1$ , we still have solution $(x,y)=(1,1)$", "Solution_8": "In the other hand, given the equation $ax+by=c$, where $a, b$ and $c$ are positive integers and $\\gcd(a,b)=1$, the number of positive integer solutions of the equation is $\\left\\lfloor\\frac{c}{ab}\\right\\rfloor$ (Prove that).\r\n\r\nHence, $\\left\\lfloor\\frac{845}{72\\cdot23}\\right\\rfloor=0$ and we are donne.", "Solution_9": "Err what about 3x + 4y = 7 again?", "Solution_10": "[quote=\"probability1.01\"]Err what about 3x + 4y = 7 again?[/quote]\r\n\r\nForget my last post. :blush:", "Solution_11": ":blush: Forgot my last reply . The theorem is like that .The equation ax + by = c has solutions in Z if and only if gcd(a,b) / c . But this situation holds because gcd( 72 , 23) =1 and c = 845 then 1/ 845 . So the equation has intiger solutions and not solution in N^2 .", "Solution_12": "Let ( p,q ) is a solution of the equation ax + by = c . Then the solution of the equation is in the form of x = p + (b/d)* t and y = q - (a/d)* t where d=gcd(a,b) and t is in Z .", "Solution_13": "@Jan:\r\n\r\n[quote]Using the euclidian algoritm, we find the modular inverse: 47.\nSo:[/quote]\r\n\r\nWhat's the Euclidian Algorithm :?", "Solution_14": "http://mathworld.wolfram.com/EuclideanAlgorithm.html :)" } { "Tag": [ "vector", "algebra", "system of equations" ], "Problem": "My Algebra II/Trig teacher gave us a problem set. Here is the \"example\" problem:\r\n\r\nA factory manufactures chairs and tables, each requiring the use of three operations: Cutting, Assembly, and Finishing. The first operation can be used at most $40$ hours, the second at most $42$ hours, and the third at most $25$ hours. A chair requires $1$ hr. of cutting, $2$ hrs. of assembly, and $1$ hr. of finishing. A table needs $2$ hrs. of cutting, $1$ hr. of assembly, and $1$ hr. of finishing. If the profit is $\\$20$ per unit for a chair and $\\$30$ for a table, how many units of each should be manufactured to maximize revenue?\r\n\r\nHere was the objective equation ($C$=chair, $T$=table): $20C+30T$.\r\n\r\nHere are the constraints:\r\n$C+2T\\leq{40}$\r\n$2C+T\\leq{42}$\r\n$C+T\\leq{25}$\r\n$C\\geq{0}$\r\n$T\\geq{0}$\r\n\r\nThe teacher told us to graph the region, locate the vertices of the feasibility region, and figure out which vertex maximizes profit.\r\n\r\nI tried \"solving\" it as a system of equations, but my method did not work. Why?", "Solution_1": "Well I think your main issue is that given $A\\leq B$ and $C\\leq D$, you [b]can't[/b] say that $A-C\\leq B-D$. All you can say is that $A+C\\leq B+D \\implies A-B\\leq D-C$\r\n\r\nWhat I'm trying to say is that you probably saw that:\r\n$2C+T\\leq 40$\r\n$C+T\\leq 25$ and then tried to say:\r\n$C\\leq 15$\r\nbut that isn't true; $C=16$ works perfectly...", "Solution_2": "This is the problem of Linear Programming Method.", "Solution_3": "[quote=\"vishalarul\"]I tried \"solving\" it as a system of [b]equations[/b], but my method did not work. Why?[/quote]\r\n\r\nSimply because they are not [b]equations![/b] That is, $\\le, \\ge$ are not equivalence relations.", "Solution_4": "Hi !\r\n\r\nWe teach that in France in 11th class. Do you want some good courses ? :maybe:", "Solution_5": "I got the vertices as (0,0) (21,0) (0,20) nd (44/3,38/3)..Hence the maximum value gets obtained at C=0 T=20 which 600..Was that the answer???", "Solution_6": "See this one :\r\n[url]http://people.hofstra.edu/faculty/Stefan_waner/RealWorld/Summary4.html[/url]", "Solution_7": "I learned the linear programming method in algebra 2... which is about 9th-10th grade for the U.S.", "Solution_8": "Did you learn the inner product of vector?\r\n\r\nThe answer is $650$ units when $(C,\\ T)=(10,\\ 15).$", "Solution_9": "inner product of vector? i'm not exactly sure what you mean...\r\nanyway thats a great idea to use vectors in this kind of problem.. i haven't thought of that before... let me work it out now!", "Solution_10": "[quote=\"me@home\"]inner product of vector? i'm not exactly sure what you mean...\nanyway thats a great idea to use vectors in this kind of problem.. i haven't thought of that before... let me work it out now![/quote]\r\n\r\nIt was dot product or scalar product. :)" } { "Tag": [ "complex numbers", "algebra unsolved", "algebra" ], "Problem": "P is an odd prime. The integers a, b, c, d are not multiples of p and for any integer n not a multiple of p, we have {na/p} + {nb/p} + {nc/p} + {nd/p} = 2, where { } denotes the fractional part. Show that we can find at least two pairs from a, b, c, d whose sum is divisible by p. \r\n\r\nBut Don't use complex to solve", "Solution_1": "It has a solution with complex numbers? Show it!", "Solution_2": "Take a look on kalva site.", "Solution_3": "I never do it ;) This a problem solving forum, but NOT \"read it on kalva\" forum!", "Solution_4": "It's your problem. I find it completely senseless to ask for a solution using complex numbers and to refuse to look at some other source. Anyway, you are right, this is your forum, so why am I bothering?", "Solution_5": "You will never understand my pedagogical reasons. Sadly.", "Solution_6": "I would rather say fortunately. It means I would spend my life and energy on math forums, which is the least thing I want. Anyway, please delete this message after that, I start spaming.", "Solution_7": "[quote=\"harazi\"]It means I would spend my life and energy on math forums, which is the least thing I want.[/quote]\n\nDont say that harazi , many people have learned countless things from you , .this\n\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=22364&highlight=integrate&start=20 is a way good enough example of your contributions.\n\n\n[quote]I never do it This a problem solving forum, but NOT \"read it on kalva\" forum\n\n you are right, this is your forum, so why am I bothering?\n\nYou will never understand my pedagogical reasons. Sadly[/quote]\r\n\r\nCut it guys . you are just speaking about REFERING TO ANOTHER SOLUTION and see how you are reacting .\r\nAnyone who likes , go look at the other source and who doesnt like, dont . OK now start agian ,Say \" YOU just cant understand My reasons\"" } { "Tag": [], "Problem": "Given any positive integer $n$ , show that there are two positive rational numbers $a$ and $b$ , $a \\not= b$, which are not integers and which are such that $a - b, a^2 - b^2 , \\ldots a^n - b^n$ are all integers.", "Solution_1": "[hide]$a - b | a^n - b^n$ for every positive integer $n$.\n\nSo we can put $a = \\frac{x}{z}$ and $b = \\frac{y}{z}$.\n\nHence $a^n - b^n = \\frac{x^n - y^n}{z^n}$.\n\nNow we can find a number $z^n$ that divides $x - y$.\n\nAn example can be $x = 2005^n + 1$, $y = 1$ and $z^n = 2005$.\n\nSo $a = \\frac{2005^n + 1}{2005}$ and $b = \\frac{1}{2005}$.[/hide]", "Solution_2": "Shouldnt it be something like:\r\n[hide]$x=2^n+1,y=1,z=2$, or anything else in the place of 2 (3,4,5,...)?[/hide]", "Solution_3": "Take $a=\\frac{m}{n}$ and $b=\\frac{1}{n}$.\n\nWe have to find suitable $m$ such that $\\begin{cases} n\\mid m-1\\\\ n^2\\mid m^2-1\\\\ \\cdots \\\\ n^n\\mid m^n-1\\end{cases}$\n\nTaking $m=n^n+1$ suffices as $n^n\\mid m-1\\mid m^i-1$ for any $1\\le i\\le n$.\n\nSo, $(a,b)=\\left(\\frac{n^n+1}{n}, \\frac{1}{n}\\right)$.", "Solution_4": "taking $a=2^n+\\frac{1}{2}$ and $b=1/2$ does the job! :D", "Solution_5": "a=(x^n+1)/x and b=1/x does the job in fact (x is an integer)\nBye" } { "Tag": [], "Problem": "Can you send your test in English!", "Solution_1": "[quote=\"yangml\"]Can you send your test in English![/quote]\r\nWho are you asking? What are you asking for? :huh:" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "\\[ a,b,c \\in [0,1] \\Rightarrow a^2+b^2+c^2 \\le 1+a^2b+b^2c+c^2a \\]", "Solution_1": "[quote=\"Lovasz\"]\\[ a,b,c \\in [0,1] \\Rightarrow a^2+b^2+c^2 \\ge 1+a^2b+b^2c+c^2a \\][/quote]\r\nMaybe $a^2+b^2+c^2 \\le 1+a^2b+b^2c+c^2a$ $?$ ;)\r\nThis is a very old inequality.", "Solution_2": "What about these ones:\r\n$a,b,c \\in [0,1]$\r\n\\[ \\Rightarrow 2(a^3+b^3+c^3) \\le 3+a^2b+b^2c+c^2a \\]\r\n\\[ \\Rightarrow \\frac{a}{bc+1}+\\frac{b}{ca+1}+\\frac{c}{ab+1} \\le 2 \\]", "Solution_3": "Can you post solution for these ones, i don't know what to do with the problems like these. \r\nBy the way, could you show me some methods to kill them?\r\nMerci.", "Solution_4": "[quote=\"Lovasz\"]\\[ \\Rightarrow \\frac{a}{bc+1}+\\frac{b}{ca+1}+\\frac{c}{ab+1} \\le 2 \\][/quote]\r\n\r\nThis one is also well-known.\r\n\r\n$LHS\\leq \\frac{a+b+c}{abc+1}$, so it remains to prove:\r\n$a+b+c-2abc\\leq 2$, and it is enough to check $(0,0,0), (0,0,1),(0,1,1),(1,1,1)$ and all its permutations (why?)\r\n\r\nThe other one I don't know about...", "Solution_5": "[quote=\"Lovasz\"]\\[ a,b,c \\in [0,1] \\Rightarrow a^2+b^2+c^2 \\le 1+a^2b+b^2c+c^2a \\][/quote]\r\n\r\n$f(a)=a^2(b-1)+ac^2+(b^2c+1-b^2-c^2)$\r\n\r\nbecause b-1<0 We have minimum of $f(a)$ on 0 or 1. By the analogy $b,c=0 or 1$ We check all the combination and we will see that this inequality is true :).", "Solution_6": "And the rest :!:" } { "Tag": [ "algebra", "polynomial" ], "Problem": "To solve the equations\r\n\\begin{align*}\r\n2a+3b+c &= 11\\\\\r\n6ab+2ac+3bc &= 24\\\\\r\nabc &= -6\r\n\\end{align*}\r\nVolume 2 says that I can create a polynomial and solve for its roots. However, I don't get how to force these equations to be symmetric roots. Could someone please explain this part?", "Solution_1": "Let $ 2a \\equal{} k$ and $ 3b \\equal{} m$\r\nThen you have\r\n$ k\\plus{}m\\plus{}c \\equal{} 11$\r\n$ km \\plus{} kc \\plus{} mc \\equal{} 24$\r\n$ kmc \\equal{} \\minus{}1$", "Solution_2": "Thanks so much!" } { "Tag": [ "geometry", "3D geometry", "calculus", "calculus computations" ], "Problem": "A cone is made for a circular sheet of paper having radius r, by cutting out a sector and gluing the cut edges of the remaining piece together. What is the maximum volume attainable for the cone.", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138284[/url]" } { "Tag": [ "inequalities", "logarithms", "inequalities unsolved" ], "Problem": "For $x,y,z$ positive real numbers such that $x+y+z=3$ prove that\r\n\r\n$\\displaystyle \r\nx^ry^r + y^rz^r + z^rx^r \\leq \\max(3, 3^{2r} / 2^{2r})$", "Solution_1": "On CRUX Vasc proved this one:\r\n\r\nFor $x,y,z$ positive real numbers and $m=\\frac{ln9-ln4}{ln3}$ we have\r\n\r\n$(\\frac{x^m+y^m+z^m}{3})^2 \\geq (\\frac{xy+yz+zx}{3})^m$\r\n\r\n\r\nand he said this one is the same as our ineq.\r\n\r\nI really don't see how the two inequalities are related :? :?", "Solution_2": "take $a=x^r,b=y^4,c=z^r$. then $x^ry^r+x^rz^r+y^rz^r=ab+ac+bc\\leq 3\\left(\\frac{a^{1/r}+b^{1/r}+c^{1/r}}{3}\\right)^{2r}=3$ if $r\\leq \\frac{\\ln 3}{\\ln 9-\\ln 4}$ by the other one(by power means, we see that it works for $m\\geq \\frac{\\ln 9-\\ln 4}{\\ln 3}$. i don't get the time to post what happens the other way round, wait a minute :D", "Solution_3": "Peter, are you running at the speed of light ?\r\n\r\nI say that as your minute is very dilated in time according to Einstein Relativity Theory, in fact it is again passing.\r\n\r\nNaturally, I am joking ;)", "Solution_4": "uhm yeah, the other way round is just power means as well. to see why we explicitly calculate for $r=\\frac{\\ln 3}{\\ln 9 - \\ln 4}$ observe that for this we obtain equality $3=\\frac{3^{2r}}{2^{2r}}$.", "Solution_5": "Dear Peter, can you please detail better your post.\r\n\r\nI think you mean that:\r\n\r\n[quote]If we put $a=x^r$,$b=y^r$and$c=z^r$ we have two cases.[/quote]\n\n[quote]First case: $mr \\leq 1$\n\n\n$\\displaystyle x^ry^r+x^rz^r+y^rz^r=ab+ac+bc\\leq 3\\left(\\frac{a^{rm}+b^{mr}+c^{mr}}{3}\\right)^{2/m} \\leq 3\\left(\\frac{a+b+c}{3}\\right)^{2r} = 3$ [/quote]\n\n[quote]Second case: $mr > 1$\n\n $\\displaystyle x^ry^r+x^rz^r+y^rz^r=ab+ac+bc\\leq 3\\left(\\frac{a^{rm}+b^{mr}+c^{mr}}{3}\\right)^{2/m} \\leq = ?????$[/quote]\r\n\r\nThank you very much. :)", "Solution_6": "is the original ineq. correct? it doesnt seem to work for x=y=z=1;\r\n\r\nmaybe the RHS should read MIN(3,3^2r/2^2r)?", "Solution_7": "No,it is MAX.", "Solution_8": "[quote=\"chakding\"]is the original ineq. correct? it doesnt seem to work for x=y=z=1;\n\nmaybe the RHS should read MIN(3,3^2r/2^2r)?[/quote]\r\n\r\nit works for x=y=z=1 - try to check it once more", "Solution_9": "for those who didn't understand my previous posts :lol:\r\n\r\nit is easily seen that vasc's inequality is equivalent to the case of $r=\\frac 1m$ of our inequality - and the case $r=\\frac 1m$ is the case where $3=\\frac{3^{2r}}{2^{2r}}$.\r\n\r\nnow, for arbritary $r$ one can use one the one hand that if $r\\leq \\frac 1m$, then\r\n\r\n$x^ry^r+x^rz^r+y^rz^r\\leq 3(\\frac{x^{\\frac 1m}+y^{\\frac 1m}+z^{\\frac 1m}}{3})^{rm}\\leq 3$\r\n\r\nand for $r\\geq \\frac 1m$ we have\r\n\r\n$x^ry^r+x^rz^r+y^rz^r\\leq (x^{\\frac 1m}+y^{\\frac 1m}+z^{\\frac 1m})^{rm}\\leq frac{3^{2r}}{2^{2r}}$.\r\n\r\neach by the case of $r=\\frac 1m$. therefore it suffices to prove vasc's inequality.", "Solution_10": "More detailed, for $r\\geq \\frac{1}{m}\\approx1.35$, we have\r\n\r\n$x^ry^r+y^rz^r+z^rx^r= (x^{1/m}y^{1/m})^{rm}+(y^{1/m}z^{1/m})^{rm}+(z^{1/m}x^{1/m})^{rm}\\leq\r\n(x^{1/m}y^{1/m}+y^{1/m}z^{1/m}+z^{1/m}x^{1/m})^{rm} \\leq 3^{rm}=\\frac{3^{2r}}{2^{2r}}$; ;)" } { "Tag": [ "geometry", "MATHCOUNTS", "\\/closed" ], "Problem": "Is there a way to disregard new posts in certain sections? Specifically, I participate in the Mathcounts forum but don't participate in the Middle School Tournaments, and I don't want new posts in the Middle School Tournaments section to make the general Mathcounts forum say that there are new posts. Any help is greatly appreciated.\r\n\r\nThinkFlow\r\n\r\nP.S. How do you post screenshots? Can you only post images from a webpage?", "Solution_1": "I understand what you want, but at this point you cannot do that, unless you want to completely hide the contents of the Middle School Tournaments section.", "Solution_2": "[quote=\"ThinkFlow\"]Is there a way to disregard new posts in certain sections? Specifically, I participate in the Mathcounts forum but don't participate in the Middle School Tournaments, and I don't want new posts in the Middle School Tournaments section to make the general Mathcounts forum say that there are new posts. Any help is greatly appreciated.\n\nThinkFlow\n\nP.S. How do you post screenshots? Can you only post images from a webpage?[/quote]\r\nSceenshot (I have windows; may be different on macs)\r\nHit Print Screen and hit Ctrl+C\r\nOpen Paint and copy your picture there\r\nedit it however you'd like\r\nSAVE!\r\nPost it by looking at the \"add an attachment\" section, hit browse, select the image, and there you go!", "Solution_3": "On mac its command-shift-3, saves to desktop", "Solution_4": "adding on to my other post...\r\ni don't think you always need to hit Ctrl+C but I do it anyways...just in case.", "Solution_5": "No, PrintScreen automatically saves the image to the clipboard for you. You should [b]NOT[/b] hit Ctrl+C; if you were happening to have selected some text, the image will be ERASED and replaced with the text that you just copied." } { "Tag": [ "probability", "calculus", "integration", "logarithms", "expected value", "geometric series" ], "Problem": "Continuous Probability (might require calculus, I'm not entirely sure on the solution.)\r\n\r\nA number x is chosen from the set of all real numbers between 0 and 1, inclusive.\r\nA real number y is also chosen from the same set. If $y \\ge x$ then we are done. If $y < x$, then another y value is chosen. Find the expected number of y values we have to choose if we are to get $y \\ge x$.", "Solution_1": "Ignore the fact that exactly 0 or 1 is picked or same x and y are picked (Probabilty is 0). There are only two outcomes that can happen x>y or y x in only one pick?\r\nI suppose that makes sense, because if half the time X is less that 0.5, and the EV for that is 1, and half the time x > 0.5, so EV is more than than one, but with lower P.\r\n\r\nCan anyone corroborate this solution?", "Solution_3": "Only y changes; x remains fixed instead of them both being chosen randomly each time. if y 0$ then $ LHS >1$ :)" } { "Tag": [ "probability", "MATHCOUNTS", "AMC", "AMC 10", "AIME" ], "Problem": "it seems like AoPS is filled with way too much sarcastic people, especially in the AMC forum. (yifman, probability, mysticterminator,bubala, among others). Personally, I think its annoying, and tiring. Sick of never seeing straight replies or anything. How about y'all?", "Solution_1": "I find it uttely aggravating these inconsiderate and ill-mannered group of individuals plaugue our poor and sincere forums with satirical wit that is the cause for extensive pain and suffering. May their souls find eternal restlessness in the quintic equations that await us on the other side.", "Solution_2": "[quote=\"G-UNIT\"]I find it uttely aggravating these inconsiderate and ill-mannered group of individuals plaugue our poor and sincere forums with satirical wit that is the cause for extensive pain and suffering. May their souls find eternal restlessness in the quintic equations that await us on the other side.[/quote]\r\n\r\nI'd add G-Unit to your list. :lol:", "Solution_3": "samus - if their rudeness becomes a personal concern, notify the appropriate moderators for post removal. Remember to provide links when you do so. Also, if certain individuals have a continuous and notable trend of ridicule, I'm sure Valentin would like to hear about it.\r\n\r\nAnd G - UNIT, be nice. -.-", "Solution_4": "I don't think anyone in AMC forums have actually posted anything offensive though\r\n\r\nAHHH I AM NOT INCLUDED ON THE LIST I AM SO SAD", "Solution_5": "All right, first! :winner_first: \r\n\r\nI'd like to thank the academy, my 7th grade mathcounts coach for introducing me to AoPS, all the sarcastic people in my life, and everybody who has supported me in this journey, thank you so much!", "Solution_6": "[quote=\"samus\"]it seems like AoPS is filled with way too much sarcastic people, especially in the AMC forum. (yifman, probability, mysticterminator,bubala, among others). Personally, I think its annoying, and tiring. Sick of never seeing straight replies or anything. How about y'all?[/quote]\r\n\r\nI didn't find any sarcasm on the AMC forum (although some of them do plenty of wierd things in the games and fun factory).\r\n\r\nEDIT: nvm, I just found plenty of unnecessary speech", "Solution_7": "[quote=\"yif man12\"]All right, first! :winner_first: \n\nI'd like to thank the academy, my 7th grade mathcounts coach for introducing me to AoPS, all the sarcastic people in my life, and everybody who has supported me in this journey, thank you so much![/quote]\r\n\r\nYou probably need to specifically thank your friend who improved from 75 on AMC10 in December to 150 AMC12/14AIME/42USAMO :lol: . \r\n\r\nAlright still yifman you need to teach me how to be more sarcastic so I can be on the list in future as being formally recognized to be the \"champion\" in AoPS premiere sarcasm competition.", "Solution_8": "[quote=\"beta\"]I don't think anyone in AMC forums have actually posted anything offensive though[/quote]\r\n\r\nsamus did not bring this up because people were saying things that were blatantly offensive to certain people. Instead, he is saying that all the sarcasm is getting annoying. Especially the sarcasm where only the people who know the poster personally will understand. I can usually figure out when someone's being facetious. But I imagine it would annoy other people. I don't think all the sarcasm is necessarily that bad just as long as at some point someone explains the joke.\r\n\r\nYeah, just try to be respectful of others and everything will be fine :)", "Solution_9": "Darn you yifman.. I guess I must settle for a mere :winner_second: \r\n\r\nSeriously though, I usually try to make sure my posts have some explicit content if sarcasm is there (I can only think of one recent instance where I slipped here). I don't think I should have to mark off what is sarcasm and what isn't. And also, of all the people you listed, I don't think any of them really post sarcastically that often, especially the bubala. But anyways very sorry if I've made your AoPS experience miserable.", "Solution_10": "The only topic that I find bubala post sarcastic comment is \r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65275[/url], where almost everyone else on the list except yifman (and not on the list =>me) has posted some thing sarcastic. \r\n\r\nYifman's friend won him the title though.", "Solution_11": "If you want to find some sarcastic comments by bubala look in the AHSIMC forum :D or probably in his blog.\r\n\r\nDon't cut out the sarcasm completely (we need some unintelligeble mathy humor every so often :P ) but be mindful when it appears that other people seem to be left out.", "Solution_12": "[quote=\"joml88\"]If you want to find some sarcastic comments by bubala look in the AHSIMC forum :D or probably in his blog.\n[/quote]\r\n\r\nhahaha AHSIMC forum... how many recent posts there are not sarcastic nor offensive?\r\n\r\nSarcasm about AHSIMC is perfectly okay.", "Solution_13": "[quote=\"beta\"][quote=\"joml88\"]If you want to find some sarcastic comments by bubala look in the AHSIMC forum :D or probably in his blog.\n[/quote]\n\nhahaha AHSIMC forum... how many recent posts there are not sarcastic nor offensive?\n\nSarcasm about AHSIMC is perfectly okay.[/quote]\r\n\r\nYes, sarcasm THERE is definitely allowed. :D In fact AHSIMC = :spam: \r\n\r\nBut oh wait...I'm supposed to be nice. :spider: *cough*", "Solution_14": "[quote=\"MithsApprentice\"]\n\nYes, sarcasm THERE is definitely allowed. :D In fact AHSIMC = :spam: \n\nBut oh wait...I'm supposed to be nice. :spider: *cough*[/quote]\r\n\r\nAh spamming right now is unnecessary since that Metrock guy is gone for awhile(hopefully he will never be back), but if he ever comes back, then we'll go spam /make sarcastic comments. \r\n\r\n ;) spamming/sarcasm/ is not only allowed in AHSIMC but encouraged lol.", "Solution_15": "Why are we still bitter about the iTest? It was free and it was how long ago? Who cares anymore? And do we really not want him to go away and not come back?", "Solution_16": "If he takes our advice seriously, there wouldn't be so many protests. :)\r\n\r\nYes but um...be nice for now guys.", "Solution_17": "[quote=\"solafidefarms\"]And do we really not want him to go away and not come back?[/quote]\r\n\r\nyes that leaves us more time to spam fagjr", "Solution_18": "I want to say that some threads and posts are a whole bunch funnier when you know what they're talking about. As an example, the mock aime iii thread isn't that funny if you don't know about AHSIMC. Then it's hilarious!\r\n\r\nWhile we're on the topic, check out the conflicting quotes at [url]http://en.wikiquote.org/wiki/Sarcasm[/url].\r\n\r\nIMO, sarcasm is only appropriate if the audience understands it. It's up to the poster to determine that. And a note to anyone that feels confused: just ask a question. I found out about AHSIMC when I asked, and then it was funny to read about winning an ipod in the Mock USAMO. ;)" } { "Tag": [ "linear algebra", "linear algebra theorems" ], "Problem": "Hello!\r\n I was wondering if anybody happened to know why the term trace is used for the intersection of a space and a subspace? If not, would you have any ideas on where I could look to find this info?\r\nThank you,\r\nNICK", "Solution_1": "A lot of such information can be found [url=http://members.aol.com/jeff570/mathword.html]here[/url], but not what you are looking for. To me, this usage of \"trace\" seems self-explanatory: when a surface intersects a plane, it leaves a trace in it, i.e. the curve of intersection." } { "Tag": [ "Gauss" ], "Problem": "What was everyone wearing? Here are the colors from Las Vegas:\r\n\r\nColorado: purple\r\nDSW: red (with collars)\r\nMacau: pink\r\nPhilippines: black\r\nNorCal: lime green\r\nOregon: white with black sleeves\r\nRoseburg: black\r\nSan Diego: black\r\nSFBA: purple\r\nSoCal: blue\r\nUtah: light blue\r\nWashington: purple", "Solution_1": "Some of the ones I remember from Penn State:\r\n\r\nAAST: Black\r\nChesapeake: Pink\r\nFairfax Math Circle: Red\r\nHoward County: Pink\r\nLehigh Valley: Black\r\nMaine: White\r\nNYC: Purple\r\nOntario West: Light blue\r\nPhillips Exeter: Red\r\nUpstate New York: Pink", "Solution_2": "SFBA was [b]supposed[/b] to be \"dark blue\"...but they quite failed that.", "Solution_3": "Iowa site : [the ones I remember - possibly wrong]:\r\n\r\nChicago: Yellow\r\nIndiana: Yellow\r\nWisconsin: Red\r\nTexas: Dusty Pink -ish? \r\nIowa : Dark Green\r\nMichigan: Blue\r\nMichigan ICAE : White", "Solution_4": "Connecticut: Purple/Orange Tie-Dye\r\nTJ: Dark Blue", "Solution_5": "Georgia site:\r\n\r\nFlorida - Red\r\nNorth Carolina - Pink\r\nSouth Carolina - Also pink\r\nTennessee - Also red\r\nKentucky - Orange (?)\r\nGeorgia - White", "Solution_6": "From Georgia:\r\n\r\nNorth Carolina: Pink\r\nSouth Carolina: Pink (way too much pink!)\r\nKentucky: gold\r\nFlorida: red (I think)\r\nGeorgia: white\r\n\r\nI'm attaching the pdf for the Georgia shirt. I rather like it this year. The basic design was created by Steve Sigur. Steve passed away last summer after many years of service to the Georgia team.\r\n\r\nTom", "Solution_7": "Iowa:\r\n\r\nMinnesota: light blue\r\nMissouri: green", "Solution_8": "Penn:\r\nVermont: Dark Green", "Solution_9": "PSU:\r\nWest Virginia: White\r\n\r\nAlthough we weren't the only ones wearing white if I remember correctly.", "Solution_10": "Langley (Penn state) - blue", "Solution_11": "eastern mass was in light blue", "Solution_12": "To pianoforte:\r\nMinnesota would be [i]cornflower[/i].\r\n\r\nWe did that since we didn't want wayzata colors (blue and gold).\r\n\r\nbtw, wayzata=26+ of our team (inside joke?)", "Solution_13": "[quote=\"AIME15\"]SFBA was [b]supposed[/b] to be \"dark blue\"...but they quite failed that.[/quote]\r\n\r\n:( Dark blue would have been nice. I would have liked something cooler than Gauss' Line on the back too (but I was never one for knowing random collinearities).", "Solution_14": "My deepest\r\napologies to MN\r\n\r\nI have soiled your shirt with my\r\ninaccuracies.", "Solution_15": "You MEANIE!\r\n\r\nAre you from Missouri? (the other one you said?)", "Solution_16": "yeah (technically kansas, but whatev)" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Prove that for any acute triangle $ ABC$,we have:\r\n$ cosA \\plus{} cosB \\plus{} cosC \\geq 1 \\plus{} 4cosAcosBcosC$", "Solution_1": "[quote=\"SUPERMAN2\"]Prove that for any acute triangle $ ABC$,we have:\n$ cosA + cosB + cosC \\geq 1 + 4cosAcosBcosC$[/quote]\r\n\r\nIt is true for all triangles :\r\n\r\n$ cos A +cos B +cos C = 1 +\\frac{r}{R}$ \r\n\r\n$ 4cosAcosBcosC= \\frac{p^2-(2R+r)^2}{R^2}$\r\n\r\n$ \\Leftrightarrow p^2 \\le 4R^2+5Rr +r^2$ which is true $ ( p^2 \\le 4R^2+4Rr+3r^2$ and $ 2r \\le R )$", "Solution_2": "I think we also can substitute $ cosA\\equal{}\\frac{b^2\\plus{}c^2\\minus{}a^2}{2bc},cos B\\equal{}\\frac{c^2\\plus{}a^2\\minus{}b^2}{2ca},cos C\\equal{}\\frac{a^2\\plus{}b^2\\minus{}c^2}{2ab}$\r\nThen expand the ineq.", "Solution_3": "[quote=\"SUPERMAN2\"]Prove that for any acute triangle $ ABC$,we have:\n$ cosA \\plus{} cosB \\plus{} cosC \\geq 1 \\plus{} 4cosAcosBcosC$[/quote]\r\nI think this ineq is old and very easy! :P\r\nWe have $ cosA\\plus{}cosB\\plus{}cosC\\equal{}1\\plus{}4sin{A/2}sin{B/2}sin{C/2}$\r\nAnd $ cosAcosBcosC \\leq sin{A/2}sin{B/2}sin{C/2}$", "Solution_4": "[quote=\"SUPERMAN2\"]Prove that for any acute triangle $ ABC$,we have:\n$ cosA \\plus{} cosB \\plus{} cosC \\geq 1 \\plus{} 4cosAcosBcosC$[/quote]\r\nIt's true for all triangle. :wink:", "Solution_5": "[quote=\"arqady\"][quote=\"SUPERMAN2\"]Prove that for any acute triangle $ ABC$,we have:\n$ cosA \\plus{} cosB \\plus{} cosC \\geq 1 \\plus{} 4cosAcosBcosC$[/quote]\nIt's true for all triangle. :wink:[/quote]\r\nyes,my sulotion is true for all triangle! :P" } { "Tag": [ "MATHCOUNTS" ], "Problem": "The dates for the MATHCOUNTS school, chapter, state, and national competitions can be found [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=185&z=53]here[/url].\r\n\r\nUntil each competition is completely passed, students should not share any information about the competitions beyond their place. No score information should be discussed and problems should not be posted.", "Solution_1": "According to the pdf file, the last chapters were 2/25.\r\nDoes that mean we can discuss them?", "Solution_2": "[quote=\"biffanddoc\"]According to the pdf file, the last chapters were 2/25.\nDoes that mean we can discuss them?[/quote]\r\n\r\nNo! Not until MATHCOUNTS says you can. The national office emailed us and said that it's still possible chapters could be making up contests late due to weather conditions. This is why the go ahead to post must come from MATHCOUNTS.", "Solution_3": "Ok. I see now", "Solution_4": "I just noticed that the tests are posted online at mathcounts.org\r\nCan we discuss it now?", "Solution_5": "[quote=\"biffanddoc\"]I just noticed that the tests are posted online at mathcounts.org\nCan we discuss it now?[/quote]\r\n\r\nIf they've posted, there's not reason you can't discuss it. Have fun." } { "Tag": [ "geometry" ], "Problem": "block and tackle problem", "Solution_1": "Here's the diagram and problem. Good Luck." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "This shape is built from six unit cubes. How many square units are in the surface area of the shape, where the solid lines indicate the front face of the cubes?\n[asy]size(150);\ndefaultpen(linewidth(0.9));\nvoid drawCube(pair A){\n pair B = (A.x + 1,A.y), C = (A.x + 1, A.y + 1), D = (A.x,A.y + 1);\n draw(A--B--C--D--cycle);\n pair A2 = (A.x + 0.6,A.y + 0.3), B2 = (B.x + 0.6, B.y + 0.3), C2 = (C.x + 0.6, C.y + 0.3), D2 = (D.x + 0.6, D.y + 0.3);\n draw(A2--B2--C2--D2--cycle,dashed);\n draw(A--A2,dashed); draw(B--B2,dashed); draw(C--C2,dashed); draw(D--D2,dashed);\n}\ndrawCube((0,0));\ndrawCube((1,0));\ndrawCube((2,0));\ndrawCube((0,-1));\ndrawCube((2,-1));\ndrawCube((1,1));[/asy]", "Solution_1": "If they didn't touch this would be $ 6 \\times 6\\equal{}36$. Each time they touch we subtract $ 2$. They touch $ 5$ times, so the answer is $ 36\\minus{}2(5)\\equal{}\\boxed{26}$.", "Solution_2": "Or you can just count the number of faces. :)", "Solution_3": "[quote= when u see the thing it is 6x6=36 and when u touch it u subtract 2 and u move it 5 times so the equation is 36 - 2(5)" } { "Tag": [ "inequalities", "geometry", "complex numbers", "graph theory", "\\/closed" ], "Problem": "I've noticed that WOOT is the main program to practice for olympiads. However, I'm wondering whether I should consider taking olympiad courses (ie inequalities, problem solving, geometry) first so that I have some of the helpful background for olympiads. So, are the olympiad courses highly recommended when preparing for olympiads? Or is just simply WOOT a better idea?\r\n\r\nI've also been thinking that taking olympiad courses would teach some aspects of classical mathematics that would be useful later on in life as well as helping in olympiads, whereas WOOT would just simply give proof-writing/olympiad practice.", "Solution_1": "I believe that the olympiad classes are being discontinued due to WOOT.", "Solution_2": "[quote=\"ffdbzathf\"]I believe that the olympiad classes are being discontinued due to WOOT.[/quote]I'm not sure where you got this information -- we are still planning to have Olympiad classes in the fall. We're not offering them in the summer because, in the past, summer Olympiad enrollment has been very low.\r\n\r\nThe Olympiad classes are probably at a higher difficulty level than WOOT (especially the Olympiad subject classes), so if anything I would recommend taking WOOT first, then taking an Olympiad-level class. The main advantage of WOOT is you get to see lots more subjects (though not at the depth that a 12-week subject class can give) and you get several practice tests.", "Solution_3": "[quote=\"DPatrick\"][quote=\"ffdbzathf\"]I believe that the olympiad classes are being discontinued due to WOOT.[/quote]I'm not sure where you got this information -- we are still planning to have Olympiad classes in the fall. We're not offering them in the summer because, in the past, summer Olympiad enrollment has been very low.\n[/quote]\n\nI think ffdbzathf may have saw that in this post (in one of the threads in this subforum): \n\n[quote=\"rrusczyk\"][quote=\"1234567890\"]Is AoPS going to shut down the Olympiad classes and replace them with WOOT? I think you were mentioning that as a possibility before.[/quote]\n\nThis is highly likely. At this point, we expect to end the Olympiad classes and expand/alter WOOT. The olympiad classes will in part be folded into WOOT.[/quote]\r\n\r\nThat'd be great if the Olympiad classes are still going to be held. I was really hoping to take one of them when I saw the mention of the Olympiad classes likely being shut down.", "Solution_4": "Is there any chance that classes, especially WOOT, might be held on weekends? I want to take some courses, but none of the times work for me.", "Solution_5": "[quote=\"Phelpedo\"]Is there any chance that classes, especially WOOT, might be held on weekends? I want to take some courses, but none of the times work for me.[/quote]\r\n\r\nYou can review transcripts.\r\n\r\nEDIT: Read Joe's comments.", "Solution_6": "I want to take WOOT and or Olympiad Problem Solving.\r\n\r\nWell my entire goal is to improve in proof writing.\r\n\r\nIs Olympiad Problem Solving more about like, different ways to write proofs and less about how to solve different subject proofs or what?\r\n\r\nI think WOOT does like alot of subjects and how to as well.. right?\r\n\r\nI'm just wondering if the like 550 dollar difference is worth it lol.", "Solution_7": "WOOT is not designed to teach about proofs. Through test feedback you can gain experience with proofs though. And if you ask on the WOOT message board I'm sure you will get detailed help with proofs. Proving your results is emphasized in WOOT but I wouldn't say that is the only pupose of the class. Neither would I say that is the major difference (or even a difference at all!) between WOOT and the other Olympiad classes since everything I said above could equally describe them as well (except replace \"tests\" with \"midterm\" and \"final\").\r\n\r\nPhelpedo, WOOT classes are always offered twice a week. Both times are weekdays though. Are you certain you can't make at least one of these days? If you are [i]that[/i] busy that you cannot spare a few hours during the week I doubt you would be able to devote enough time to the class to get the full benefit (though you should still gain a lot).\r\n\r\nSaiem, [url=http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php#ps]here's[/url] the description of Olympiad Problem Solving. As you can see each class emphasizes a different proof method and then there are a few classes which are just entitled \"Challenging Problems'. These have a bunch of problems which can be proved using one or more of the proof methods discussed in the class. So they serve to solidify your knowledge and give you some experience into how to determine what method you might want to try. The class doesn't focus on any particular areas such as Algebra, NT, Geometry, etc. but you will get a little bit of each throughout the course. If you want to get good at geometry, Olympiad Geometry is unequaled around AoPS.\r\n\r\nWOOT has a lecture on a different topic every couple weeks. The topics vary and usually are on specific subjects such as collinearity/concurrency, inequalities, NT, complex numbers, graph theory, and so forth. The lecture schedule is very likely to change depending on who lectures next year and the preferences of the students enrolled in the course.", "Solution_8": "BTW, when will the syllabus for next year's WOOT be released?", "Solution_9": "Okay thankyou, it seems like I'm going to take olympiad classes rather than WOOT. Although I may want to practice proof-writing at some point, my main focus right now is problem solving. This year I'll take olympiad courses so as to master those subjects and learn skills I need for olympiads. Then, once I've learned about inequalities, olympiad problem solving, and geometry, and I start to get more serious about olympiads, I'll probably try WOOT. This way I do get the in-depth knowledge of these courses.\r\n\r\nAlso, WOOT is more expensive and takes more time ;) (busy sophmore year)", "Solution_10": "[quote=\"calc rulz\"]Okay thankyou, it seems like I'm going to take olympiad classes rather than WOOT. Although I may want to practice proof-writing at some point, my main focus right now is problem solving. This year I'll take olympiad courses so as to master those subjects and learn skills I need for olympiads. Then, once I've learned about inequalities, olympiad problem solving, and geometry, and I start to get more serious about olympiads, I'll probably try WOOT. This way I do get the in-depth knowledge of these courses.\n\nAlso, WOOT is more expensive and takes more time ;) (busy sophmore year)[/quote]\r\n\r\n[edit: clarified my post a little]\r\n\r\nNo, WOOT isn't all about proof writing. There weren't any lectures dedicated to it. The tests are the only things that will make you better at writing solutions to problems but they also provide you with the ability to have practice with taking the test under real (or close to real) conditions. WOOT is definitely not about just proof writing though by all means you should become better at writing proofs since the graders of the tests will be critiquing you and your peers along with the instructors will be giving you feedback on your solutions to board problems (as long as you post solutions). I've never been to MOP but I'd have to say that WOOT can probably offer a somewhat similar experience (though there are obviously some things that MOP offers that WOOT cannot and vice versa).", "Solution_11": "Yeah Silverfalcon's comment was kind of misleading...", "Solution_12": "[quote=\"chess64\"]Yeah Silverfalcon's comment was kind of misleading...[/quote]\r\n\r\nEh, that's what I thought WOOT was (proof only) heh. BTW, if you knew that I was misleading, why didn't you say that I was misleading (instead of waiting for Joe to do all the talking)? ;)", "Solution_13": "I haven't taken WOOT, so I think somebody else could give a better answer... (BTW, you haven't taken WOOT either, have you? :?)", "Solution_14": "For the olympiad classes, are there any recommended/required books or materials? If so, what are they?", "Solution_15": "[quote=\"joml88\"]\n\n[edit: clarified my post a little]\n\nNo, WOOT isn't all about proof writing. There weren't any lectures dedicated to it. The tests are the only things that will make you better at writing solutions to problems but they also provide you with the ability to have practice with taking the test under real (or close to real) conditions. WOOT is definitely not about just proof writing though by all means you should become better at writing proofs since the graders of the tests will be critiquing you and your peers along with the instructors will be giving you feedback on your solutions to board problems (as long as you post solutions). I've never been to MOP but I'd have to say that WOOT can probably offer a somewhat similar experience (though there are obviously some things that MOP offers that WOOT cannot and vice versa).[/quote]\r\n\r\nHmm though I'd still like to get learn some more in-depth material first and do something like WOOT in another year or something. Then I can use that background while doing general olympiad stuff, in particular get a lot of practice in junior year when I'll probably have my best shot at MOP. It also works better regarding time and money (ie busy sophmore year).\r\n\r\nIn response to DPatrick's comment, aren't olympiad courses more of a 1-time thing while WOOT is a general training program that one would do again and again? So what do you mean by \"go take WOOT\" before doing olympiad courses? (maybe take WOOT for a year or something?) Anyway, I seem to be ready for olympiad courses so that shouldn't be a problem.", "Solution_16": "huh why would you take WOOT twice?", "Solution_17": "[quote=\"chess64\"]huh why would you take WOOT twice?[/quote]\r\n\r\nThe classes, tests, etc., are nearly all different every year (this is part of the reason WOOT is more expensive than our other classes - we have to recreate the course nearly from scratch each year).", "Solution_18": "[quote=\"rrusczyk\"][quote=\"chess64\"]huh why would you take WOOT twice?[/quote]\n\nThe classes, tests, etc., are nearly all different every year (this is part of the reason WOOT is more expensive than our other classes - we have to recreate the course nearly from scratch each year).[/quote]\r\n\r\noh, i didn't know that", "Solution_19": "[quote=\"chess64\"]I haven't taken WOOT, so I think somebody else could give a better answer... (BTW, you haven't taken WOOT either, have you? :?)[/quote]\r\n\r\nNope and I probably won't. Money is a big reason for me and other thing is that I need to study hardcore for AP exams." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "A number of men enter a disreputable establishment and each one leaves a coat and an umbrella at the door. When a message is received saying that the establishment is about to be raided by the police, the men leave hurriedly, and no man gets both the right coat and the umbrella. If there are $ n$ men show that the number of ways in which this can happen is $ n!\\left( n!\\minus{} \\frac{(n\\minus{}1)!}{1!}\\plus{}\\frac{(n\\minus{}2)!}{2!}\\minus{}...\\plus{}(\\minus{}1)^n\\frac{1}{n!}\\right)$.", "Solution_1": "This is direct from inclusion-exclusion: the number of ways that a given $ k$-set of the men get their things is precisely ${ (n - k)!^2}$, and there are $ \\binom{n}{k}$ $ k$-sets of men, so our answer is\r\n\\[ \\sum_{k = 0}^n (-1)^{n - k} \\binom{n}{k} (n - k)! ^2 = n! \\sum_{k = 0}^n (-1)^{n - k} \\frac{(n - k)!}{k!},\\]\r\nas desired.", "Solution_2": "I will try to give a more explicit answer to this problem..\r\n\r\nI understand this problem as gangsters entering a building, and the police raid the place and they take the wrong umbrellas and hats. Let:\r\n$ A_{0}$ - be the set with the collections of all sets in which ALL the gangsters fail to pick up both their items. So $ |A_{0}|$ is what we are looking for.\r\n$ A_{1}$ - the collection of sets in which gangster 1 picks up both his items\r\n$ A_{2}$ - the collection of sets in which gangster 2 picks up both his items\r\n...\r\n$ A_{n}$ - the collection of sets in which gangster n picks up both his items.\r\n\r\nBecause $ A_{0}$ and ($ {A_{1}\\cup A_{2}\\cup...A_{n}}$) are disjoint, we can write:\r\n$ |A_{0}| = |{A_{1}\\cup A_{2}\\cup...A_{n}\\cup A_{0}|}$ \\ $ |{A_{1}\\cup A_{2}\\cup...A_{n}}$| (1)\r\n\r\n[b]1)[/b]The cardinal of the first expression from the right can be deduced directly. It is the total posibilities of picking the hats & umbrellas:\r\ngangster 1->n*n\r\ngangster 2->(n-1)*(n-1)\r\ngangster 3->(n-2)*(n-2)\r\n...\r\ngangster n->1*1 ways, which is\r\n$ |{A_{1}\\cup A_{2}\\cup...A_{n}\\cup A_{0}| = [n!]^2}$\r\n\r\n[b]2)[/b]The cardinal of the second expression from the right will be deduced from the inclusion-exclusion principle:\r\n$ |{A_{1}\\cup A_{2}\\cup...A_{n}| = \\sum |A_{i}| - \\sum |A_{i}\\cap A_{j}| + \\sum |A_{i}\\cap A_{j}\\cap A_{k}| - ... + ( - 1)^{n + 1}|A_{1}\\cap A_{2} \\cap... \\cap A_{n}|}$\r\nEach of the cardinals on the right can be deduced like in 1) , such that:\r\n$ |A_{1}| = |A_{2}| = ... = |A_{n}| = [(n - 1)!]^2$\r\n$ |A_{1}\\cap A_{2}| = |A_{1}\\cap A_{3}| = ... = [(n - 2)!]^2$\r\n...\r\n$ |A_{1}\\cap A_{2} \\cap... \\cap A_{n}| = [1!]^2$, \r\nand the sums have \r\n$ \\binom{n}{1}, \\binom{n}{2},...\\binom{n}{n - 1}$ terms. So we have\r\n$ |{A_{1}\\cup A_{2}\\cup...A_{n}| = \\binom{n}{1}[(n - 1)!]^2 - \\binom{n}{2}[(n - 2)!]^2 + ... + ( - 1)^{n + 1}\\binom{n}{n - 1}[1!]^2}$\r\n\r\nNow we can rewrite (1):\r\n$ |A_{0}| = [n!]^2 - (\\binom{n}{1}[(n - 1)!]^2 - \\binom{n}{2}[(n - 2)!]^2 + ... - ( - 1)^{n + 1}\\binom{n}{n - 1}[1!]^2)$\r\n$ |A_{0}| = [n!]^2 - \\frac {n!(n - 1)!(n - 1)!}{(n - 1)!1!} + \\frac {n!(n - 2)!(n - 2)!}{(n - 2)!2!} - ... + ( - 1)^{n}\\frac {n!1!1!}{1!(n - 1)!}$\r\n$ |A_{0}| = n!\\left( n! - \\frac {(n - 1)!}{1!} + \\frac {(n - 2)!}{2!} - ... + ( - 1)^{n}\\frac {1}{n!}\\right)$\r\nDone." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Could somebody solve this over $\\mathbb{R^+}$? I have a solution but it is pretty complicated, and I hope there is an easier one.\r\n\r\n$x_1+x_2^2+x_3^3=3$\r\n$x_2+x_3^2+x_4^3=3$\r\n$x_3+x_4^2+x_1^3=3$\r\n$x_4+x_1^2+x_2^3=3$", "Solution_1": "hello, i have only found $ x_1\\equal{}x_2\\equal{}x_3\\equal{}x_4\\equal{}1$.\r\nSonnhard.", "Solution_2": "Can you explain why aren't there other solutions?" } { "Tag": [ "vector", "calculus", "integration", "geometry", "3D geometry", "algebra", "polynomial" ], "Problem": "Two problems from the final exam I just gave:\r\n\r\n2. Let $C$ be the closed triangular curve from $(0,0,6)$ to $(0,2,0)$ to $(1,1,2)$ back to $(0,0,6).$ Let $\\vec{F}(x,y,z)=(y+z)\\mathbf{i}+(2x+z)\\mathbf{j}+(2x+2z)\\mathbf{k}.$\r\n\r\nCompute the line integral $\\oint_{C}\\vec{F}\\cdot \\vec{dr}.$\r\n\r\n4. Let $\\vec{G}(x,y,z)$ be defined as the vector field whose direction at $(x,y,z)$ is always directly away from the origin and whose magnitude at $(x,y,z)$ is always the cube of the distance from the origin to $(x,y,z).$\r\n\r\n(a) Write an explicit formula for $\\vec{G}(x,y,z).$\r\n\r\n(b) Let $E=\\{(x,y,z)\\mid x^{2}+y^{2}+z^{2}\\le a^{2}\\}.$ Let $dV$ be $dx\\,dy\\,dz.$ Compute:\r\n\\[\\iiint_{E}\\nabla\\cdot\\vec{G}\\,dV \\]", "Solution_1": "[quote=\"Kent Merryfield\"]4. Let $\\vec{G}(x,y,z)$ be defined as the vector field whose direction at $(x,y,z)$ is always directly away from the origin and whose magnitude at $(x,y,z)$ is always the cube of the distance from the origin to $(x,y,z).$\n\n(a) Write an explicit formula for $\\vec{G}(x,y,z).$\n\n(b) Let $E=\\{(x,y,z)\\mid x^{2}+y^{2}+z^{2}\\le a^{2}\\}.$ Let $dV$ be $dx\\,dy\\,dz.$ Compute:\n\\[\\iiint_{E}\\nabla\\cdot\\vec{G}\\,dV \\]\n[/quote]\r\n\r\nFor this problem, it seems that the [i]n[/i]-dimensional version is just as easy (except perhaps the integral): \r\nLet $\\vec{r}$ denote the usual position vector $\\left< x_{1}, x_{2}, \\ldots, x_{n}\\right>$. Let $E^{n}=\\{( x_{1}, x_{2}, \\ldots, x_{n})\\mid x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2}\\le a^{2}\\}.$ Let $dV^{n}$ be $dx_{1}\\,dx_{2}\\,\\ldots\\, dx_{n}$. Then\r\n\r\n(a) $\\vec{G}(\\vec{x})=\\frac{\\vec{r}}{\\left|\\vec{r}\\right|}\\left|\\vec{r}\\right|^{3}=\\left|\\vec{r}\\right|^{2}\\vec{r}$ \r\n\r\n(b) $\\nabla\\cdot\\vec{G}=\\nabla\\cdot\\left(\\left|\\vec{r}\\right|^{2}\\vec{r}\\right)= \\left|\\vec{r}\\right|^{2}\\nabla\\cdot\\vec{r}+\\vec{r}\\cdot\\nabla\\left(\\left|\\vec{r}\\right|^{2}\\right)=n\\left|\\vec{r}\\right|^{2}+\\vec{r}\\cdot\\left(2\\vec{r}\\right) =(n+2)\\left|\\vec{r}\\right|^{2}$\r\n\r\nLet $E_{+}^{n}=\\{( x_{1}, x_{2}, \\ldots, x_{n})\\mid x_{1}, x_{2}, \\ldots, x_{n}\\ge 0, x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2}\\le a^{2}\\}.$ Then\r\n\r\n\\[\\iint\\cdots\\int_{E^{n}}\\nabla\\cdot\\vec{G}\\,dV^{n}= 2^{n}\\iint\\cdots\\int_{E_{+}^{n}}\\nabla\\cdot\\vec{G}\\,dV^{n}\\]\r\n\\[=2^{n}\\iint\\cdots\\int_{E_{+}^{n}}(n+2)\\left|\\vec{r}\\right|^{2}\\,dV^{n}\\]\r\n\\[=2^{n}(n+2)\\iint\\cdots\\int_{E_{+}^{n}}\\left(x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2}\\right)\\,dV^{n}\\]\r\n\r\nI'll use this theorem:\r\n\r\n[url=http://www.imagehoop.com/view_image/340c92053/DirichletIntegralTheorem.JPG][img]http://www.imagehoop.com/view_image/340c92053/DirichletIntegralTheorem.JPG[/img][/url]\r\n\r\nwith $F(u)=u,t=a,\\alpha_{i}=1,\\beta_{i}=2,\\gamma_{i}=1$ to get\r\n\r\n\\[2^{n}(n+2)\\iint\\cdots\\int_{E_{+}^{n}}\\left(x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2}\\right)\\,dV^{n}= 2^{n}(n+2)\\frac{\\prod_{k=1}^{n}\\left[\\frac{1}{2}\\Gamma\\left(\\frac{1}{2}\\right)\\right]}{\\Gamma\\left(\\sum_{i=1}^{n}\\frac{1}{2}\\right)}\\int_{u=0}^{a}u^{\\sum_{i=1}^{n}\\frac{1}{2}-1}u\\,du \\]\r\n\\[= 2^{n}(n+2)\\frac{\\frac{1}{2^{n}}\\Gamma^{n}\\left(\\frac{1}{2}\\right)}{\\Gamma\\left(\\frac{n}{2}\\right)}\\int_{u=0}^{a}u^{\\frac{n}{2}}\\,du \\]\r\n\\[= \\frac{(n+2)\\pi^\\frac{n}{2}}{\\sqrt{\\pi}\\frac{(n-2)!!}{2^{\\frac{n-1}{2}}}\\right)}\\cdot\\frac{2}{n+2}a^{\\frac{n}{2}+1}\\]\r\n\\[= \\frac{a^\\frac{n+3}{2}\\pi^\\frac{n-1}{2}2^\\frac{n+1}{2}}{(n-2)!!}= \\boxed{\\frac{a}{\\pi}\\cdot\\frac{\\left( 2a\\pi\\right)^\\frac{n+1}{2}}{(n-2)!!}}\\]\r\n\r\nso that for $n=3$ we have \r\n\r\n\\[\\iiint_{E}\\nabla\\cdot\\vec{G}\\,dV = \\frac{a}{\\pi}\\cdot\\frac{\\left( 2a\\pi\\right)^\\frac{3+1}{2}}{(3-2)!!}=\\boxed{ 4\\pi a^{3}}\\]", "Solution_2": "Suffice it to say that I only [i]expected[/i] to see the three dimensional version on my exam. \r\n\r\nThere were two ways to do it. The first is to compute the divergence and integrate that over the volume - which you've done. (Note that the vector field is polynomial, hence smooth everywhere.) The second is to quote the divergence theorem and calculate the integral of $\\vec{G}$ over the sphere. I was hoping to see the second method, but either method would have worked.\r\n\r\nFull credit for the problem was quite rare in my class.", "Solution_3": "[quote=\"Kent Merryfield\"]Two problems from the final exam I just gave:\n\n2. Let $C$ be the closed triangular curve from $(0,0,6)$ to $(0,2,0)$ to $(1,1,2)$ back to $(0,0,6).$ Let $\\vec{F}(x,y,z)=(y+z)\\mathbf{i}+(2x+z)\\mathbf{j}+(2x+2z)\\mathbf{k}.$\n\nCompute the line integral $\\oint_{C}\\vec{F}\\cdot \\vec{dr}.$[/quote]\r\n\r\nLabel the verticies: $A(0,0,6),\\,B(0,2,0),$ and $C(1,1,2)$ so that $\\vec{AB}=<0,2,-6>$ and $\\vec{BC}=<1,-1,2>$ and hence a normal to the surface $S$ (the triangle) is given by $\\vec{AB}\\times \\vec{BC}= <0,2,-6>\\times <1,-1,2> = <-2,-6,-2>$, and since this normal gives positive orientation for $C$ we choose $\\vec{n}=<-2,-6,-2>$. Also,\r\n\\[\\vec{\\nabla}\\times \\vec{F}= \\left|\\begin{array}{ccc}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ \\frac{\\partial}{\\partial x}& \\frac{\\partial}{\\partial y}& \\frac{\\partial}{\\partial x}\\\\ y+z&2x+z&2x+2z \\end{array}\\right| =<-1,-1,1> \\]\r\nNow, by the Curl Theorem (a.k.a. Stokes' Theorem) with $C=\\partial S$ we have\r\n\\[\\oint_{C}\\vec{F}\\cdot \\vec{dr}= \\iint_{S}\\mbox{Curl}\\,\\vec{F}\\cdot d\\vec{S}=\\iint_{S}\\left(\\vec{\\nabla}\\times \\vec{F}\\right)\\cdot \\frac{\\vec{n}}{|\\vec{n}|}\\, dS \\]\r\n\r\n\\[=\\frac{1}{|\\vec{n}|}\\iint_{S}<-1,-1,1>\\cdot <-2,-6,-2>\\, dS \\]\r\n\r\n\\[= \\frac{1}{|\\vec{n}|}(2+6-2)A(S) = \\frac{6}{|\\vec{n}|}\\cdot \\frac{1}{2}|\\vec{n}| =\\boxed{3}\\]", "Solution_4": "About a quarter of my class got full or near-full credit. What I didn't mention was that question 1 was to find an equation for the plane through exactly those three points - that was meant to function as a hint. At least one of the the correct answers was from someone who did not use Stokes but simply paramaterized the three line segments and computed the line integral directly. That was possible.\r\n\r\nThere are some small choices that could be made along the way that would make things look slightly different than what benorin said. You could divide his (non-unit) normal vector by $2.$ You could explicitly talk about the projection on the triangle onto the $xy$ plane.\r\n\r\n$\\LaTeX$ choice: Using < and > gives $<-1,-1,1>.$ Using \\langle and \\rangle gives $\\langle-1,-1,1\\rangle.$ I prefer the latter." } { "Tag": [ "Euler", "algebra", "polynomial", "calculus", "trigonometry", "search" ], "Problem": "Prove that\r\n\r\n$\\frac 1{1^2}+\\frac 1{2^2}+\\frac 1{3^2}+...=\\frac{\\pi^2}{6}$\r\n(my stupidity has been edited out)\r\n\r\nI was hoping there was a way to do this without calculus.", "Solution_1": "thats not true\r\n\r\n1/1 +1/2 +1/3 + 1/4 + ... increases without bound\r\napproximalty logrithmically\r\n\r\nPerhaps you meant\r\n$\\frac 1{1^2}+\\frac 1{2^2}+\\frac 1{3^2}+...=\\frac{\\pi^2}{6}$?", "Solution_2": "[quote=\"mathnerd314\"]Prove that\n\n$\\frac 1{1^2}+\\frac 2{2^2}+\\frac 3{3^2}+...=\\frac{\\pi^2}{6}$\n\nI was hoping there was a way to do this without calculus.[/quote]Well, the way you have written is equivalent to the harmonic series and it diverges! :omighty: \r\n\r\nFor the series $\\frac 1{1^2}+\\frac 1{2^2}+\\frac 1{3^2}+...=\\frac{\\pi^2}{6}$, even mighty Euler used calculus...\r\n\r\n[EDIT]Ack! [b]0714446459923[/b] beat me to it :yankchain:", "Solution_3": "There is definitely no elementary proof. It is not an elementary result.", "Solution_4": "amirhtlusa gives a nice prove using power series :)\r\n\r\n[quote=\"amirhtlusa\"]let $P(x)=1-\\frac{x^2}{3!}+\\frac{x^4}{5!}-\\frac{x^6}{7!}.....$\nclearly $P(0)=1$. to find the roots of $P(x)$, note that for $x\\not=0$:\n$P(x)=x[\\frac{1-\\frac{x^2}{3!}+\\frac{x^4}{5!}-\\frac{x^6}{7!}...}{x}]=\\frac{x-\\frac{x^3}{3!}+\\frac{x^5}{5!}-\\frac{x^7}{7!}...}{x}=\\frac{\\sin{x}}{x}$\nso $P(x)=0 \\Rightarrow \\sin{x}=0 \\Rightarrow x= \\pm k\\pi$ for $k=1,2,3,...$ note that $x=0$ is not a solution. because $P(0)=1$\nnow if you factor $p(x)$ you get:\n$P(x)= (1-\\frac{x}{\\pi})(1-\\frac{x}{-\\pi})(1-\\frac{x}{2\\pi})(1-\\frac{x}{-2\\pi})= [1-\\frac{x^2}{\\pi^2}][1-\\frac{x^2}{4\\pi^2}][1-\\frac{x^2}{9\\pi^2}]+....$\nnow look what happens when you expand the above equation:\nyou get\n$1-[ (\\frac{1}{\\pi^2})+(\\frac{1}{4\\pi^2})+(\\frac{1}{9\\pi^2})+....]x^2+....$now set this equal to the original function for $P(x)$ you get \n$P(x)=1-\\frac{x^2}{3!}+\\frac{x^4}{5!}-\\frac{x^6}{7!}.....= 1-[ (\\frac{1}{\\pi^2})+(\\frac{1}{4\\pi^2})+(\\frac{1}{9\\pi^2})+....]x^2+....$ now take a look at coefficient for the $x^2$ then set them equal to each other, you get:\n$\\frac{-1}{3!}=-(\\frac{1}{\\pi^2}+\\frac{1}{4\\pi^2}+\\frac{1}{9\\pi^2}+....)=\\frac{-1}{\\pi^2}(1+\\frac{1}{4}+\\frac{1}{9}+\\frac{1}{16}+....)$\ntherefore:\n$1+\\frac{1}{4}+\\frac{1}{9}+\\frac{1}{16}+...=\\frac{\\pi^2}{6}$ \n$\\mathbb{QED}$\nso we are done :D[/quote]", "Solution_5": "Oops--duh. That's what I meant. :roll:\r\n\r\nNice proof :thumbup:", "Solution_6": "[quote=\"shyong\"]amirhtlusa gives a nice prove using power series :)\n\n[quote=\"amirhtlusa\"]let $P(x)=1-\\frac{x^2}{3!}+\\frac{x^4}{5!}-\\frac{x^6}{7!}.....$\nclearly $P(0)=1$. to find the roots of $P(x)$, note that for $x\\not=0$:\n$P(x)=x[\\frac{1-\\frac{x^2}{3!}+\\frac{x^4}{5!}-\\frac{x^6}{7!}...}{x}]=\\frac{x-\\frac{x^3}{3!}+\\frac{x^5}{5!}-\\frac{x^7}{7!}...}{x}=\\frac{\\sin{x}}{x}$\nso $P(x)=0 \\Rightarrow \\sin{x}=0 \\Rightarrow x= \\pm k\\pi$ for $k=1,2,3,...$ note that $x=0$ is not a solution. because $P(0)=1$\nnow if you factor $p(x)$ you get:\n$P(x)= (1-\\frac{x}{\\pi})(1-\\frac{x}{-\\pi})(1-\\frac{x}{2\\pi})(1-\\frac{x}{-2\\pi})= [1-\\frac{x^2}{\\pi^2}][1-\\frac{x^2}{4\\pi^2}][1-\\frac{x^2}{9\\pi^2}]+....$\nnow look what happens when you expand the above equation:\nyou get\n$1-[ (\\frac{1}{\\pi^2})+(\\frac{1}{4\\pi^2})+(\\frac{1}{9\\pi^2})+....]x^2+....$now set this equal to the original function for $P(x)$ you get \n$P(x)=1-\\frac{x^2}{3!}+\\frac{x^4}{5!}-\\frac{x^6}{7!}.....= 1-[ (\\frac{1}{\\pi^2})+(\\frac{1}{4\\pi^2})+(\\frac{1}{9\\pi^2})+....]x^2+....$ now take a look at coefficient for the $x^2$ then set them equal to each other, you get:\n$\\frac{-1}{3!}=-(\\frac{1}{\\pi^2}+\\frac{1}{4\\pi^2}+\\frac{1}{9\\pi^2}+....)=\\frac{-1}{\\pi^2}(1+\\frac{1}{4}+\\frac{1}{9}+\\frac{1}{16}+....)$\ntherefore:\n$1+\\frac{1}{4}+\\frac{1}{9}+\\frac{1}{16}+...=\\frac{\\pi^2}{6}$ \n$\\mathbb{QED}$\nso we are done :D[/quote][/quote]\r\nThat proof is nice but not really rigorous as written. You can't just wave your hands and say that an infinite polynomial can be factored in the same way that a finite one can be - I believe to rigorously do so requires concepts from complex analysis. It's the most \"accessible\" proof (and the one printed in aoPS V2), but unfortunately even it reduces one problem (summing the series) to another (proving the factorability of an infinite polynomial).", "Solution_7": "Not only that, look at that step where recalled the Taylor expansion of sin x at x=0. Taylor expansion is a topic within analysis and it requires calculus.", "Solution_8": "Wikipedia has a noncalculus solution\r\n\r\nhttp://en.wikipedia.org/wiki/Basel_problem\r\n\r\nIs it ok to post links like these?", "Solution_9": "***EDIT*** Nevermind, I was thinking of Euler's formula instead of DeMoivre's formula. But, does taking the limit count as part of calculus? :P", "Solution_10": "[quote=\"0714446459923\"]Wikipedia has a noncalculus solution\n\nhttp://en.wikipedia.org/wiki/Basel_problem\n\nIs it ok to post links like these?[/quote]\r\n\r\nThe expansion of the Taylor series of $\\sin x$ is a calculus step :P", "Solution_11": "That's the Euler part. There is also another proof below it that claims to have no calculus but I have doubts to two things:\r\n\r\n1) It has a limit used in the proof\r\n\r\n2) It has Squeeze Theorem, if you read carefully towards the bottom. Squeeze theorem doesn't look like a calculus theorem but I first heard of it earlier this year in my calculus class and never heard of it before when I was taking Trigonometry during summer.", "Solution_12": "[quote=\"10000th User\"]That's the Euler part. There is also another proof below it that claims to have no calculus but I have doubts to two things:\n\n1) It has a limit used in the proof\n\n2) It has Squeeze Theorem, if you read carefully towards the bottom. Squeeze theorem doesn't look like a calculus theorem but I first heard of it earlier this year in my calculus class and never heard of it before when I was taking Trigonometry during summer.[/quote]\r\n\r\nIt uses de Moivre's, for heaven's sakes. de Moivre's is a result that can't really be put on any firm foundation without calculus.", "Solution_13": "I recall someone using this result to prove that pi is irrational.", "Solution_14": "You need calculus for sure, but not for the prove but already to define the infinite sum!\r\n\r\nBut some simple search would give a lot of threads already concerning this!\r\nE.g., most links to further ones are contained in http://www.mathlinks.ro/Forum/viewtopic.php?t=81439 .", "Solution_15": "I was just hoping I didn't need calculus because:\r\n\r\n[quote=\"DPatrick\"] \nPlease do not post calculus problems in this forum. \n[/quote]\r\n\r\nShould I try again somewhere else?", "Solution_16": "[quote=\"mathnerd314\"]Should I try again somewhere else?[/quote]\r\nNo, because it was already posted ;)", "Solution_17": "[quote=\"t0rajir0u\"][quote=\"10000th User\"]That's the Euler part. There is also another proof below it that claims to have no calculus but I have doubts to two things:\n\n1) It has a limit used in the proof\n\n2) It has Squeeze Theorem, if you read carefully towards the bottom. Squeeze theorem doesn't look like a calculus theorem but I first heard of it earlier this year in my calculus class and never heard of it before when I was taking Trigonometry during summer.[/quote]\n\nIt uses de Moivre's, [b]for heaven's sakes[/b]. de Moivre's is a result that can't really be put on any firm foundation without calculus.[/quote]??? What do you mean by that? I hope you are not implying anything derogative there... :o", "Solution_18": "Does the squeeze therom really require calculus?\r\nI thought it would be self evident for any continous, elementary function." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Find minimum of $A=\\frac{a^{n}}{(b+c)^{n}}+\\frac{b^{n}}{(c+a)^{n}}+\\frac{c^{n}}{(a+b)^{n}}$ where $n\\geq 2$ is positive integer and $a,b,c$ are positive reals", "Solution_1": "Jensen's+Nesbitt's = solution. \\[\\frac{a^{n}+b^{n}+c^{n}}{3}\\ge \\left(\\frac{a+b+c}{3}\\right)^{n},\\ n \\ge 1.\\]", "Solution_2": "Why did you think so? This problem don't sovle by your way? It's very difficult? I know there is a rusult that is special case of this problem:\r\n$\\frac{a^{3}}{(b+3)^{3}}+\\frac{b^{3}}{(c+a)^{3}}+\\frac{c^{3}}{(a+b)^{3}}\\geq\\frac{3}{8}$", "Solution_3": "Since you posted the unreal version of Vietnam TST 2005, \\[\\sum_{cyc}{a^{3}\\over (a+b)^{3}}\\ge{3 \\over 8}.\\]" } { "Tag": [ "function", "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "The function f is differentiable on R and the limits $lim_{x\\to \\pm \\infty} f'(x)$ both exist finitely. Show that f is uniformly continuous on R.\r\n\r\nIs this as easy I think it is? You can cut off a large interval compact $[-a,a]$ where we know f is uniformly continuous, since it is continuous. Then for two points outside $[-a,a]$ we can use the mean value theorem $f(x) - f(y) = f'(\\xi)|x-y|$. We just require $|x-y|$ to be small. Now we have a problem if we have one point outisde the interval and one inside it, but if we just consider the larger interval $[-a-d,a+d]$ then f will still be uniformly continuous inside the interval and if we choose two points outside of it we still require $|x-y|$ to be small. Finally if we choose a point outside it and one inside it $f(x) - f(y)$ will still be small enough because it is covered by the first case without the extended interval.\r\n\r\nOf course, I didn't do any details here, but is this how easy an argument could be made? I still feel a little unsure.", "Solution_1": "$f$ satisfies a Lipschitz condition (bounded derivative) for large enough $x$ and small enough $x$ and hence is uniformly continuous there; the rest is the usual bit about continuity being uniform continuity on compact sets. The idea is that $f$ is basically linear at infinity; such functions are uniformly continuous." } { "Tag": [ "floor function", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Prove or disprove:\r\n\r\nIf $ b_{n}$ is a convergent sequence in $ R$ and $ c_{n} \\equal{} n(b_{n} \\minus{} b_{n \\minus{} 1})$ then $ c_{n}$ is a bounded sequence.", "Solution_1": "False.\r\n\r\nLet $ b_n\\equal{}\\sum_{k\\equal{}1}^n\\frac{(\\minus{}1)^k}{\\sqrt{k}}.$\r\n\r\nNow: what happens if we add as an additional hypothesis that $ b_n$ is monotone?", "Solution_2": "If $ b_n$ is a partial sum of a convergent series, then the statement reads: every convergent series (say, with nonnegative terms) is $ O(n^{\\minus{}1})$. Which is clearly false: all we know about the terms of a convergent series is that they are $ o(1)$.", "Solution_3": "Here's an explicit example:\r\n\r\nLet $ b_n\\equal{}\\frac1{\\lfloor \\log_2n\\rfloor}.$\r\n\r\n$ b_n\\to0,$ so it converges, but\r\n\r\n$ c_{2^k}\\equal{}2^k\\left(b_{2^k}\\minus{}b_{2^k\\minus{}1}\\right)\\equal{}\r\n2^k\\left(\\frac1k\\minus{}\\frac1{k\\minus{}1}\\right)$ \r\n\r\nis certainly not bounded." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Prove that in every convex polyhedron,if concurrently it doesn't have quadrilateral plane,pentagonal plane,then have at least four planes are triangle!", "Solution_1": "For a convex polyhedron \r\n\r\ndenote by $ v$ the number of its vertices, respectively $ v_k$ the number of vertices where $ k$ edges meet, for $ k \\geq 3$, hence $ v \\equal{} v_3 \\plus{} v_4 \\plus{} v_5 \\plus{} v_6 \\plus{} \\cdots$;\r\n\r\ndenote by $ f$ the number of its facets, respectively $ f_k$ the number of facets with $ k$ sides, for $ k \\geq 3$, hence $ f \\equal{} f_3 \\plus{} f_4 \\plus{} f_5 \\plus{} f_6 \\plus{} \\cdots \\equal{} f_3 \\plus{} f_6 \\plus{} \\cdots$ (we are given $ f_4 \\equal{} f_5 \\equal{} 0$);\r\n\r\ndenote by $ e$ the number of its edges, so by Euler's formula $ v \\plus{} f \\equal{} e \\plus{} 2$.\r\n\r\nNow a little bit of double counting yields $ 2e \\equal{} 3v_3 \\plus{} 4v_4 \\plus{} 5v_5 \\plus{} 6v_6 \\plus{} \\cdots$, hence $ 2e \\minus{} 3v \\geq 0$. On the other hand, another bit of double counting yields $ 2e \\equal{} 3f_3 \\plus{} 4f_4 \\plus{} 5f_5 \\plus{} 6f_6 \\plus{} \\cdots \\equal{} 3f_3 \\plus{} 6f_6 \\plus{} \\cdots$, hence $ 2e \\minus{} 6f \\geq \\minus{}3f_3$.\r\n\r\nFinally, $ \\minus{}12 \\equal{} 6(e \\minus{} v \\minus{} f) \\equal{} (2e \\minus{} 6f) \\plus{} 2(2e \\minus{} 3v) \\geq \\minus{}3f_3$, therefore $ 3f_3 \\geq 12$, i.e. $ f_3 \\geq 4$, qed.", "Solution_2": "Hi [b]mavropnevma[/b],i'm very glad and thanks so much for your solution!" } { "Tag": [ "\\/closed" ], "Problem": "The way the flags work in Opera (I have version 7.53 with Java) makes no sense.. There are new posts since I've last been here, but they're not marked with orange flags, and some messages which I've viewed a hundred times or so refuse to show a blank flag (it stays orange). Moreover, no flags go blank if I view the topic by pressing previous/next topic from another topic.\r\n\r\nPlease don't tell me something like \"use Internet Explorer\", because that's not really a solution :D:D (and besides, there were problems with that too, and I don't want to go through all existing browsers :)).", "Solution_1": "I think that thing you did (changing the cookie management :)) works: flags are Ok now.", "Solution_2": "I use IE, and as of this morning, every topic cannot be marked read. Even if I read a topic, or press \"mark all read\" I still get the orange flag.", "Solution_3": "I have a lot of topics still showing that bug...", "Solution_4": "It's all in the cookie and session management of phpBB2. Session management is known to have certain problems, and the view-topic-unread mod does not address those problems. \r\n\r\nWe will try to correct this problems in the future, but please continue to report abnormalities in the working of this feature.", "Solution_5": "I think it may be a good start finding patterns in which topic do and don't work, as well as updating phpBB if possible with the many hacks installed.", "Solution_6": "[quote=\"Valentin Vornicu\"]I think that we are better off waiting for the next major release of phpBB. :)[/quote]\r\n\r\nNow that's 2 things you shouldn't do to get an errorless board:\r\n- installing new software as soon as it's released\r\n- installing hacks on newly released software \r\n:D\r\n\r\nNah, I guess we're best off just sitting here and telling people they dreamt about the errors :cool:", "Solution_7": "[quote=\"Peter VDD\"][quote=\"Valentin Vornicu\"]I think that we are better off waiting for the next major release of phpBB. :)[/quote]\n\nNow that's 2 things you shouldn't do to get an errorless board:\n- installing new software as soon as it's released\n- installing hacks on newly released software \n:D\n\nNah, I guess we're best off just sitting here and telling people they dreamt about the errors :cool:[/quote]The last upgrades of phpbb2 do not addess the session handling issue, but rather some security issues. They have no connection with the problem here.", "Solution_8": "[quote=\"Valentin Vornicu\"][quote=\"Peter VDD\"]Now that's 2 things you shouldn't do to get an errorless board:\n- installing new software as soon as it's released\n- installing hacks on newly released software \n:D\n\nNah, I guess we're best off just sitting here and telling people they dreamt about the errors :cool:[/quote]The last upgrades of phpbb2 do not addess the session handling issue, but rather some security issues. They have no connection with the problem here.[/quote]\r\nI don't get the relevance? :?", "Solution_9": "I am saying that upgrading to phpBB 2.0.10 will not solve the problem in this thread. In fact, the version of phpBB displayed at the bottom of each page is a variable that I input in some field, so I might make it phpBB 2.2, but it wouldn't change a thing ;) :)", "Solution_10": "I know that :) But I was replying to this: [quote=\"Valentin Vornicu\"]I think that we are better off waiting for the next major release of phpBB. :)[/quote]", "Solution_11": "[quote=\"Peter VDD\"]I know that :) But I was replying to this: [quote=\"Valentin Vornicu\"]I think that we are better off waiting for the next major release of phpBB. :)[/quote][/quote]The next major release of phpBB2, which is 2.2, will also address these issues, that we are talking about here.", "Solution_12": "Yes, and that's what I replied to: \r\n[quote=\"Peter VDD\"]Now that's 2 things you shouldn't do to get an errorless board:\n- installing new software as soon as it's released\n- installing old hacks on newly released software \n:D[/quote]\r\n\r\nThe first one is usually not too worse for phpBB (only a couple fatal errors)\r\n\r\nBut the second one is a bit worse: Ok, phpBB 2.2 has drastic changes on sessions, but also on a lot of other stuff. And that may cause problems with certain hacks on this board... Honestly, I don't think this version of the unread flags mod will work with phpBB 2.2 :cool:\r\n\r\nAnd until the all the mods are converted to phpBB2.2, that can take a while... possibly even till 2006.", "Solution_13": "The 2.2 release already contains 90% of the mods installed on this board, including the unread topics, categories subforums, global annoucements ;) \r\n\r\nMost likely the only mods that will have to be adapted are the template mods, the latexrender (although I doubt there will be much to adapt to), and a couple of other minor look mods.", "Solution_14": "Oh? In that case much has changed since the last time I checked phpBB2.2, will have to update my experience :) Last time I checked they were polling about keeping unread flags but there was like 70% against :?" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $ A,B \\in M_{3}(C)$ s.t $ AB\\equal{}BA$ and $ det(A\\plus{} B)\\equal{}det(A)\\plus{}det(B)$.\r\nProvet that $ det(A^3\\plus{}B^3)\\equal{}(det(A\\plus{}B))^3$", "Solution_1": "Counterexample: $ A\\equal{}I$, $ B\\equal{}diag(1,\\minus{}3,\\minus{}5)$.\r\n\r\nBut if we assume $ \\det(A\\minus{}B)\\equal{}\\det(A)\\minus{}\\det(B)$ also, then it is true." } { "Tag": [], "Problem": "Hey all, I had no idea where to put this...\r\n\r\nI have a table used to show the rate at which water flows through tubing given the tubing diameter and water column hight. Is there anyway to extract the formula used to calculate this from the infor mation given in the said table?", "Solution_1": "you can try this i guess :\r\n $((pie*d^{2})*((2gh)^{1}/2))/4$" } { "Tag": [ "Gauss", "geometry", "geometry unsolved" ], "Problem": "Is there anyone who knows Gauss theorem?", "Solution_1": "which particular one dya mean?\r\n :: Divergence Theorem, Gauss's Digamma Theorem, Gauss's Double Point Theorem, Gauss's Hypergeometric Theorem, Gauss's Theorema Egregium :lol: :lol: \r\n try to look for what you want \r\n at [url=http://mathworld.wolfram.com/GausssTheorem.html]mathworld.wolfram.com/GausssTheorem.html[/url]\r\nthe triple 's' is correct ,because it's supposed to be Gauss's theorem .", "Solution_2": "I think you mean the following theorem\r\n\r\n[i]Gauss' theorem[/i] The midpoints of a complete quadrilateral's three diagonals are collinear. [/i]", "Solution_3": "@treego: what do you mean by saying the [i]midpoints of a complete quadrilateral [/i]?", "Solution_4": "let $ABCD$ be a quadrilateral ,\r\n\r\nwe call it complete, if we draw its diagoals , and also draw \r\n\r\nlines $AB,CD$and,$AD,BC$. and let them intersect each other.", "Solution_5": "Guys this is wrong place to discusse this kind of quetions :!:", "Solution_6": "[quote=\"Tiks\"]Guys this is wrong place to discusse this kind of quetions :!:[/quote]\r\n\r\ndear, these disscusions have relations to each other.\r\n\r\nthen why shouldnt us disscus them? ;)", "Solution_7": "[quote=\"Ashegh\"][quote=\"Tiks\"]Guys this is wrong place to discusse this kind of quetions :!:[/quote]\n\ndear, these disscusions have relations to each other.\n\nthen why shouldnt us disscus them? ;)[/quote]\r\nYou should disscus them in the section \r\n[color=blue]Geometry theorems and Formuls[/color]! :P", "Solution_8": "[quote=\"Tiks\"][quote=\"Ashegh\"][quote=\"Tiks\"]Guys this is wrong place to discusse this kind of quetions :!:[/quote]\n\ndear, these disscusions have relations to each other.\n\nthen why shouldnt us disscus them? ;)[/quote]\nYou should disscus them in the section \n[color=blue]Geometry theorems and Formuls[/color]! :P[/quote]\r\n\r\nis (the meaning of completequadrilaterall),a theorem?\r\n\r\ndisscussions are available in every where. :wink: :D", "Solution_9": "Maybe you forget what was the purpose of this discution :roll: [quote=\"abdurashidjon\"]Is there anyone who knows Gauss theorem?[/quote]\r\nI think he was asking for Gauss' theorem,isn't he?", "Solution_10": "[quote=\"Tiks\"]Guys this is wrong place to discusse this kind of quetions :!:[/quote]\r\nu said, THIS KIND OF QUESTIONS\r\nofcourse my explanation about perfect quadrilateral,is one of them.\r\n\r\nmay be the other questions, should be disscuss in theorems and formulas \r\n\r\nsection. but mine is only an explanation,\r\n\r\nand explanations dont need any places to say.\r\n\r\nand i can post them every where,\r\n\r\nfor the sake of god plz stop :!: \r\n\r\nNow,the most important thing is to help abdurashidjan.not any thing more.\r\n\r\n :wink:" } { "Tag": [ "algebra", "polynomial", "AMC" ], "Problem": "Rationalize the denominators:\r\n\r\n1) 1/(:rt2: + :rt3: + :rt5:)\r\n\r\n2) 1/(:rt2: + 3:rt2:)\r\n\r\n(Sorry if you've seen these before.)\r\n\r\n\r\n\r\nA harder question that just occurred to me -- those are both algebraic numbers, I believe. Can you find a polynomial that has integer coefficients with either of those numbers as a root? (I can definitely do it for the first one, but the second one will take more work for sure.)", "Solution_1": "JBL wrote:Rationalize the denominators:\n\n1) 1/(:rt2: + :rt3: + :rt5:)\n\n2) 1/(:rt2: + 3:rt2:)\n\n(Sorry if you've seen these before.)\n\n\n\nA harder question that just occurred to me -- those are both algebraic numbers, I believe. Can you find a polynomial that has integer coefficients with either of those numbers as a root? (I can definitely do it for the first one, but the second one will take more work for sure.)\n\n[hide]\n\n1) Multiply by (:rt:2 - :rt:3 - :rt:5)/(:rt:2 - :rt:3 - :rt:5) and then multiply by (-6+2:rt:15)/(-6+2:rt:15) (there was probably an easier way) and simplify to get (3:rt2: + 2:rt3: - :sqrt:30)/12, assuming I didn't blunder.[/hide]", "Solution_2": "2) 1/( 2^(1/2) + 2^(1/3)) => \r\n1/(2^(1/3)(2^(1/6)+1) => 2^(2/3)/(2*(2^(1/6)+1) =>\r\n(2^(2/3)*(2^(1/6)-1))/(2*(2^(1/3)-1). By multiplying by (2^(2/3) + 2^(1/3) + 1) on top and bottom, the expression becomes\r\n\r\n(2^(2/3)*(2^(1/6)-1)*(2^(2/3) + 2^(1/3) + 1))/2", "Solution_3": "Confuted did not blunder, and Tetrahedr0n could do us all a favor by multiplying that out so I know whether or not he's right.\r\n\r\nAnyone have any thoughts on the harder question? (Instead of giving a specific polynomial, you might just give a method for finding it.)", "Solution_4": "This exact problem appeared on our 2003 Collaborative Problem-Solving Contest. If your school participated in the event, your coach received complete solutions to all problems, including this one. Essentially, you can always generate a polynomial by multiplying terms of the form (x - rn) where rn is the nth root of the desired polynomial. For a number like :rt2: + :rt3: + :rt5: you just have to include all of the \"conjugates\" as the other roots. In this case there are eight roots total (all possible combinations of +/- for each of the irrational components).\r\n\r\nIf your school does not currently participate in our Collaborative Problem-Solving Contests, there is still time to register for this year. The contest will occur from April 14th to April 21st, and requires your team to collaborate for one week on fifteen difficult-yet-accessible problems. More information is available on our website." } { "Tag": [], "Problem": "Write $ 1,000,000$ as a sum of a prime number and a perfect square.", "Solution_1": "$ 1,000,000\\equal{}10^6\\equal{}k^2\\plus{}p$\r\n$ (10^3\\minus{}k)(10^3\\plus{}k)\\equal{}p$\r\n\r\nBecause $ p$ is prime, $ 10^3\\minus{}k\\equal{}1$ which implies that $ k\\equal{}10^3\\minus{}1$ and $ p\\equal{}1999$.\r\n\r\nYou can go check to see that $ 1999$ is prime.\r\n\r\n$ 1,000,000\\equal{}(999)^2\\plus{}1999$." } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "conjecture: every uncountable set has a bounded uncontable subset.", "Solution_1": "of course, otherwise every ball of radius $n$ would be countable and thus the union of these balls would be countable.", "Solution_2": "\"Bounded\"?" } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "suppose that {a_k} k=1 to infinity is a sequence with a_k >=0 and a_k >= a_k+1 for all k>=1. prove that if sum_{k=1} to infinity a_k converges then ka_k goes to 0 as k goes to infinity.", "Solution_1": "Condensation test: If $a_n$ is a sequence which decreases to 0, $\\sum a_n$ converges if and only if $\\sum 2^na_{2^n}$ converges.", "Solution_2": "[quote=\"jmerry\"]Condensation test: If $a_n$ is a sequence which decreases to 0, $\\sum a_n$ converges if and only if $\\sum 2^na_{2^n}$ converges.[/quote]\r\n\r\nCan you elaborate, jmerry? Can someone help.... Hw is due this evening. Thanks in advance...", "Solution_3": "Suppose the sum $ \\sum_{k=1}^{\\infty} a_{k} $ converges. Then\r\nclearly $ T_{n} = \\sum_{k=n}^{2n} a_{k} \\leqslant \\sum_{k=n}^{\\infty} a_{k} \\rightarrow 0 $,\r\nwhen $ n \\rightarrow \\infty $. But clearly $ T_{n} \\geqslant (n+1)a_{2n} , na_{2n-1} $,\r\nso $ 2T_{n} \\geqslant 2na_{2n} , (2n-1)a_{2n-1} $ , so subsequence of elements of odd indexes\r\nand subsequence of elements of even indexes of a sequence $ b_{n} = na_{n} $ converge to $ 0 $,\r\nso the sequence $ b_{n} = na_{n} $ converges to $ 0 $." } { "Tag": [ "inequalities", "group theory", "abstract algebra", "vector", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ \\mathcal G$ be the set of all finite groups with at least two elements. \r\n\r\na) Prove that if $ G\\in \\mathcal G$, then the number of morphisms $ f: G\\to G$ is at most $ \\sqrt [p]{n^n}$, where $ p$ is the largest prime divisor of $ n$, and $ n$ is the number of elements in $ G$.\r\n\r\nb) Find all the groups in $ \\mathcal G$ for which the inequality at point a) is an equality.", "Solution_1": "[quote=\"Valentin Vornicu\"]... where $ p$ is the largest divisor of $ n$[/quote]\r\nI assume: prime divisor\u00bf", "Solution_2": "You assume correctly :D", "Solution_3": "Finally, something nice and decent. Take $ a$ an element of order $ p$ and let $ d\\equal{}n/p$, $ x_2,...,x_d$ a system of representatives for the left cosets except the coset $ H$ itself, where $ H\\equal{}$ is the subgroup generated by $ a$. Any morphism is determined by the image of $ a$ and of $ x_2,...,x_d$, thus there are at most $ n^d$ such morphisms. This settles $ i)$. Now, for $ ii)$ assume that we have equality, so for any choice of values of $ f(a),f(x_i)$ one gets a morphism. Let's assume $ |G|>2$ and $ d\\ne 1$. First, I claim that $ ax_2$ must have the form $ x_2a^k$. If not, then $ ax_2\\equal{}x_ja^k$ for some $ j>2$ and some $ k$. But then for any $ g,h,l$ one may find a morphism such that $ f(a)\\equal{}g, f(x_2)\\equal{}h, f(x_j)\\equal{}l$ and so $ gh\\equal{}lg^k$ for all $ g,h,l$, certainly not possible. So, $ ax_2\\equal{}x_2a^k$ and repeating this argument we have $ gh\\equal{}hg^k$ for all $ g,h$. Thus $ g\\equal{}g^k$ for all $ G$ and $ G$ is abelian. Also, recall that for any choice $ g$ one can find a morphism $ f$ such that $ f(a)\\equal{}g$ and so $ g$ has order $ 1$ or $ p$. Thus $ G$ is abelian of exponent $ p$ and so a $ F_p$ vector space, thus it is isomorphic to some $ F_p^k$. No, it is trivial to conclude.", "Solution_4": "Probably Marian Andronache. I can bet. And probably problem 3 from 11th grade is Ion Savu. I've seen similar stuff before from him.", "Solution_5": "Valentin, problem 3 from 11-th grade is proposed by Bogdan Enescu.", "Solution_6": "Does it really matter who proposed them? I mean, when the problem is original and interesting, it matters, but for such problems absolutely anybody can be considered an author. :D", "Solution_7": "[quote=\"harazi\"]Does it really matter who proposed them? I mean, when the problem is original and interesting, it matters, but for such problems absolutely anybody can be considered an author. :D[/quote]\r\n\r\n???\r\n\r\nAnyway, when I saw this problem, I thought, \"Oh, how original and interesting,\" though it's nothing I'm going to tell my future grandchildren about or anything." } { "Tag": [ "trigonometry" ], "Problem": "In right triangle $ABC$ with $\\angle B=90^\\circ$, $\\angle CAB=60^\\circ$. Extend $BC$ and let $D$ be the point on ray $BC$ such that $BC: CD=2: 1$, find the value of $\\tan x$ where $x=\\angle DAC$.", "Solution_1": "Well there is absolutely no trig in this problem other than the definition of tangent in a right triangle...\r\n\r\n[hide]\nLet $CD=a$, then we have $BC=2a$. Since $ABC$ is 30-60-90, we have $AB=\\frac{2}{\\sqrt{3}}a$ and $AC=\\frac{4}{\\sqrt{3}}a$. Hence $\\tan{x}=BC/AB=\\frac{3\\sqrt{3}}{2}$.[/hide]", "Solution_2": "[quote=\"13375P34K43V312\"]Well there is absolutely no trig in this problem other than the definition of tangent in a right triangle...\n\n\nLet $CD=a$, then we have $BC=2a$. Since $ABC$ is 30-60-90, we have $AB=\\frac{2}{\\sqrt{3}}a$ and $AC=\\frac{4}{\\sqrt{3}}a$. Hence $\\tan{x}=BC/AB=\\frac{3\\sqrt{3}}{2}$.[/quote]\r\n\r\n\r\nI get $\\sqrt{3}/11$", "Solution_3": "[hide]\numm.. drawing an image would help, but from the information given, one can conclude that $\\tan{x+60) = \\frac{3\\sqrt{3}}{2}}$.\n\nwell since $\\tan{(a+b)}= \\frac{\\tan{a}+\\tan{b}}{1-\\tan{a}\\tan{b}}$,\n\n$\\frac{\\tan{x}+\\tan{60}}{1-\\tan{x}\\tan{60}}= \\frac{3\\sqrt{3}}{2}$\n\ncross-multiply to get\n\n$2\\tan{x}+2\\tan{60}= 3\\sqrt{3}-3\\sqrt{3}\\tan{x}\\tan{60}$\n\n$2\\tan{x}+2\\sqrt{3}= 3\\sqrt{3}-3\\sqrt{3}\\tan{x}\\cdot\\sqrt{3}$\n\n$11\\tan{x}= \\sqrt{3}$\n\n$\\tan{x}= \\frac{\\sqrt{3}}{11}$\n[/hide]", "Solution_4": "[hide] Let CD=x so then BC=2x and AB=$\\frac{2x\\sqrt{3}}{3}$.\n\nTherefore ${tan(60+x)=\\frac{BD}{AB}=\\frac{3\\sqrt{3}}{2}}$.\n\nUsing Tangent Addition Theorem we get:\n\n$\\frac{tan(60)+tan(x)}{1-tan(60)tan(x)}$\n\nSimplifying:\n\n$\\frac{\\sqrt{3}+tanx}{1-\\sqrt{3}tanx}=\\frac{3\\sqrt{3}}{2}$\n\nSolving for $tanx$: $\\boxed{tanx=\\frac{\\sqrt{3}}{11}}$[/hide]\r\n\r\nEdit: I messed up with the curly brackets." } { "Tag": [ "quadratics", "function", "inequalities" ], "Problem": "Z is a complex number\r\nProve that I Z-2-i I <1 only if I Z +3 +2i I >4", "Solution_1": "i think this works....\r\n\r\nset $Z$=$a+bi$, to make the question $(a-2)^2+(b-1)^2<1$ (i) $\\rightarrow (a+3)^2+(b+2)^2>16$ (ii). \r\n\r\nok so from (i) we know $10) then (ii) is obviously true. But for (i) to be true $a$ must be in the range of $14$.\r\n\r\nLet $Z=a+bi$ for real $a,b$.\r\nThen $0<(a-2)^2+(b-1)^2<1$, so $116$ and $(b+2)^2>4$, we have $|Z+3+2i|>4$.", "Solution_3": "yeah but why is the question to show it's >4 when it's trivially true (we can even say >sqrt(20) so i dont know why it said 4). the 2nd inequality is always true so the conclusion would follow immediately, without even analyzing the first inequality..... its just so weird", "Solution_4": "This question is best seen from a geometric standpoint.Working algebraically is not the most efficient way to proceed in problems like this. \r\nIn the complex plane: \r\nIZ-2-i I=1 is the equation of a circle centre (2,1) radius 1 . Call it C(1)\r\n\r\nIZ+3+2iI =4 is the equation of a circle (-3,-2) radius 4.Call it C(2)\r\n\r\nThe question asks us to show that all points lying inside C(1) are exterior to to C(2).This can only happen if the circles dont intersect and we dont have the situation of C(1) lying entirely within C(2)..\r\nThe distance between the centres of the circles is 6 while the sum of their radii is 5.On the strength of this observation we can conclude that C(1) must lie entirely outside C(2)." } { "Tag": [ "function", "integration", "search", "trigonometry", "calculus", "derivative", "real analysis" ], "Problem": "let $f$ be a continuous function such that $\\int_0^{\\infty}|f(t)|dt < +\\infty$.\r\nshow that the following differential equation $y''+f(t)y=0$\r\nhas at least one unbouded solution. \r\n(if you know the official solution, let others search :-) )", "Solution_1": "[quote=\"alekk\"]\nif you know the official solution, let others search [/quote]\r\n\r\nok", "Solution_2": "I think it is quite simple. Suppose all solutions are bounded and take $f_1, f_2$ two solutions linearly independent. Then the wronskian is a constant function, non-zero. On the other hand, it is clear that $ f_i'$ converge at infinity, since $ f_i$ are bounded and $ f$ is absoutely integrable. Thus $ f_i\\\"$ are integrable on the set of positive reals and thus $f_i'$ convege at infinity. But sine $f_i$ are also bounded, the derivative must in fact converge towars 0. But then the wronskian is a constant non-zero function that tends to 0 in infinity, false. Am I missing something?", "Solution_3": "good :)" } { "Tag": [ "quadratics", "number theory unsolved", "number theory" ], "Problem": "Let $F_{1}=F_{2}=1$ and $F_{n+2}=F_{n+1}+F_{n}.$ If $239|F_{n}$, prove that $n$ is even .", "Solution_1": "$F_{n}=\\frac{1}{\\sqrt 5 }(q_{1}^{n}-q_{2}^{n})$, were $q_{1}=\\frac{1+\\sqrt 5 }{2}, \\ q_{2}=\\frac{1-\\sqrt 5 }{2}$. \r\np=239 is prime number and $(\\frac{5}{239})=1, \\ \\sqrt 5 =\\pm 31 (mod 239)$. It give $F_{n}=\\frac{16^{n}-(-15)^{n}}{31}(mod 239)=\\frac{16^{n}}{31}(1-14^{n})(mod 239)$. Period 14 mod 239 is 238. Therefore $239|F_{n}$ if and only if $238|n$.", "Solution_2": "I wonder how can u find out that $5$ is a quadratic residue mod $239$ and the smallest $n$ so that $14^{n}\\equiv 1 (mod 239)$ is $238$ (because maybe $n=119$ also works...)... \r\nDo you have any formula to do those tasks ? Thx a lot :)", "Solution_3": "More general:\r\nIf $p$ is a prime of the form $4k+3$ and $p|F_{n},$ then $n$ is even. This is easily consequence of the fact that $F_{2n+1}=F_{n}^{2}+F_{n+1}^{2}.$" } { "Tag": [ "induction" ], "Problem": "Does there exist uncountable family $F$ of subset of countable set with finite $x\\cap y$ for all distinct $x,y\\in F?$", "Solution_1": "Without loss of generality, we can assume that our countable set is $E=\\mathbb{N}$. Let $\\Sigma=(A_{i})$ be our uncountable family of subsets of $E$. We assume the result does not hold.\r\nLet $a_{0}$ be the smallest nonnegative integer which is contained in uncountably many $A_{i}$ (the existence of $a_{0}$ is obvious). Denote by $\\Sigma_{0}$ those $A_{i}$ containing $a_{0}$.\r\nBy induction, define $a_{n+1}$ to be the smallest nonnegative integer greater than $a_{n}$ such that $a_{n+1}\\in A_{i}$ for uncountably many $A_{i}\\in \\Sigma_{n}$. Denote by $\\Sigma_{n+1}$ this subset of $\\Sigma_{n}$. \r\nNow write $\\Sigma^{*}=\\cap{\\Sigma_{n}}$. Since by construction: $\\forall n, \\Sigma_{n}-\\Sigma_{n+1}$ is countable, it follows that $\\Sigma^{*}$ is uncountable, and in particular nonempty. For any two elements $X,Y$ of $\\Sigma^{*}$, $X \\cap Y=\\cup \\{a_{n}\\}\\neq \\emptyset$, which is absurd.", "Solution_2": "[quote=\"julien_santini\"]Since by construction: $\\forall n, \\Sigma_{n}-\\Sigma_{n+1}$ is countable, ... [/quote]\r\nHey, why?", "Solution_3": "Because by construction, $\\forall k \\in ]a_{n},a_{n+1}[$, $a_{k}$ is contained in at most a countable number of elements of $\\Sigma_{n}$.", "Solution_4": "Sorry, when writing \"why?\" I havn't mean you will axplain the point. \r\nThat was only to show the first incorrect point of your proof. \r\n\r\nIndeed, what the matter forbids us to assume $a_{0}=0, \\ a_{1}=1$ (just for simplesity) and both $\\Sigma_{1}$ and $\\Sigma_{0}-\\Sigma_{1}$ to be uncountable?\r\n\r\nIt would take me some time to remember my own solution of the problem and those of my friends all rather simple and short but I could never forget the other one because it is perfectly evident and perfectly nice.", "Solution_5": "Well, you are perfectly true.", "Solution_6": "The answer to the question is yes.\r\n\r\nHere is one construction. Let the countable set be the set of finite binary strings. Given an infinite binary string $x$, let $S_{x}$ be the set of finite prefixes of $x$. If $x$ and $y$ are two different infinite binary strings, then $S_{x}\\cap S_{y}$ is finite. Let $F$ be the set of $S_{x}$ for all infinite binary strings $x$. Because there are an uncountable number of infinite binary strings, $F$ is uncountable.", "Solution_7": "Yes, this is almost the same I mensioned above:\r\n\r\nFor a given real $x$ fix any rational consequence which converges to $x$ and let $F_{x}$ be the all their members. It is obvious now that $\\{F_{x}\\}$ is desired family." } { "Tag": [], "Problem": "is it true that bigger head are smarter? but seriousely which one looks better?", "Solution_1": "they both look good if you are confident", "Solution_2": "I don't think either has anything to do with how smart you are.", "Solution_3": "neither looks good\r\n(i now counterexamples for both :roll: )", "Solution_4": "looking good is relative so yeah." } { "Tag": [ "real analysis", "real analysis theorems" ], "Problem": "Let $ A : X \\to U$ be a bounded linear map between Banach spaces. Let $ x_n$ be a weakly convergent sequence in $ X$. Then it seems to me like $ Ax_n$ is weakly convergent also, because if $ l \\in U'$\r\n\r\n$ l(Ax_n) \\equal{} (A'l)(x_n) \\to (A'l)(x) \\equal{} l(Ax)$\r\n\r\nwhere $ A'$ is the transpose of $ A$.\r\n\r\nThis short proof or observation with the transpose seems so strange to me. What is the gist of it? I.e. what is it that really ensures that weak convergence is preserved?", "Solution_1": "I think your way of proving the result is the easiest one can give. Maybe you can write down explicitly what the transpose operator does to reveal the simplicity: $ l(A x_n) \\equal{} (l\\circ A)(x_n) \\to (l\\circ A)(x) \\equal{} l(Ax)$. So what you need is the observation that the sequence $ (l(A x_n))_{n\\in {\\mathbb N}}$ obtained by applying any $ l\\in U^{\\prime}$ to $ (A x_n)$ gives a sequence that is obtained from $ (x_n)$ by applying $ l\\circ A$, which is a bounded functional $ X\\to {\\mathbb K}$, hence an element of $ X^{\\prime}$, to which you can apply the assumption that $ x_n$ converges weakly to $ x$. That's all, I think. \r\nHanno", "Solution_2": "Ohps yeah, I am being dumb. I get confused by notation. Thanks" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "$A,B \\in M(n,C)$ such that \r\n$AB=A$ \r\n$BA=B$\r\n\r\nProve that $(A-B)^2=0$", "Solution_1": "One more example that in Romanian magazines linear algebra= multiplications+additions. Here we have $A^2=(AB)A=A(BA)=AB$ and $B^2=(BA)B=B(AB)=BA$. Conclusion." } { "Tag": [ "quadratics", "real analysis", "real analysis unsolved" ], "Problem": "[color=darkblue]Let $ (u_n)$ is a number sequence and let $ \\begin {cases} u_1 \\equal{} \\alpha \\\\\nu_{n \\plus{} 1} \\equal{} (u_n)^2 \\plus{} \\beta \\end {cases}$\nFind formula of $ u_n$ by $ \\alpha$, $ \\beta$, $ n$.[/color]", "Solution_1": "$ u_1 \\equal{} \\alpha$\r\n$ u_2 \\equal{} \\alpha^2 \\plus{} \\beta$\r\n$ u_3 \\equal{} \\alpha^4 \\plus{}2\\alpha^2 \\beta \\plus{} \\beta^2 \\plus{} \\beta$\r\n$ u_4 \\equal{} \\alpha^8 \\plus{}4\\alpha^6 \\beta \\plus{} 2\\alpha^4(3\\beta^2 \\plus{}\\beta) \\plus{}4 \\alpha^2 (\\beta^3 \\plus{}\\beta^2) \\plus{} [\\beta^4 \\plus{} 2 \\beta^3 \\plus{} \\beta^2 \\plus{}\\beta]$\r\n\r\n\r\n$ u_n \\equal{} \\sum_{k\\equal{}0}^{2^{n\\minus{}1}} \\alpha^{2k} p_{k,n} (\\beta)$\r\n\r\n$ u_{n\\plus{}1} \\equal{} [\\sum_{k\\equal{}0}^{2^{n\\minus{}1}} \\alpha^{2k} p_{k,n} (\\beta)][\\sum_{j\\equal{}0}^{2^{n\\minus{}1}} \\alpha^{2j} p_{j,n} (\\beta)] \\plus{} \\beta$\r\n\r\n$ \\equal{}\\beta\\plus{}\\sum_{k\\equal{}0}^{2^{n\\minus{}1}} \\sum_{j\\equal{}0}^{2^{n\\minus{}1}} \\alpha^{2(k\\plus{}j)} p_{k,n}(\\beta) p_{j,n} (\\beta)$\r\n\r\n\r\n$ p_{0,n\\plus{}1}(\\beta) \\equal{} p_{0,n}^2(\\beta) \\plus{} \\beta$\r\nand for $ 0< l <2^n, p_{l,n\\plus{}1}(\\beta) \\equal{} \\sum_{k\\plus{}j\\equal{}l} p_{k,n}(\\beta) p_{j,n}(\\beta)\\equal{} 2\\sum_{k\\equal{}0}^l p_{k,n} (\\beta) p_{l\\minus{}k,n} (\\beta)$\r\n\r\n\r\nhaving problems solving for these p_k,n explicitly \r\ni am not sure this is really what you wanted though.. =(", "Solution_2": "A generic [url=http://mathworld.wolfram.com/QuadraticMap.html]quadratic map[/url] doesn't have a closed-form solution.", "Solution_3": "[quote=\"t0rajir0u\"]A generic [url=http://mathworld.wolfram.com/QuadraticMap.html]quadratic map[/url] doesn't have a closed-form solution.[/quote]\r\n\r\nThanks t0rajir0u and kenn4000 very much.", "Solution_4": "If $ \\beta \\equal{} \\minus{}2$ then the closed-form solution exists.", "Solution_5": "[color=darkblue]Yes, If $ \\beta \\equal{}\\minus{}2$ then the closed-form solution exists.[/color]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$a,b,c > 0$ and $a+b+c=3$\r\n\r\nProve that $(3-2a)(3-2b)(3-2c) \\geq a^2b^2c^2$", "Solution_1": "What if $a > \\frac{3}{2}$?", "Solution_2": "This has been posted before.. http://www.mathlinks.ro/Forum/viewtopic.php?p=457344#p457344\r\nAnd the sign is reversed" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "(1) Find all function $ f: [0;1] \\rightarrow [0;1]$ which satisfies:\n1/ $ f(x_1) \\neq f(x_2) \\forall x_1 \\neq x_2$\n2/ $ 2x \\minus{} f(x) \\in [0;1] \\forall x \\in [0;1]$\n3/ $ f(2x \\minus{} f(x)) \\equal{} x$\n\n(2) Does there exist a function $ f: \\mathbb R \\rightarrow \\mathbb R$ satisfying these below conditions:\n1/ There exists $ M \\ge 0$ such that $ \\minus{}M \\le f(x) \\le M$\n2/ $ f(1) \\equal{} 1$\n3/ If $ x \\neq 0$ then $ f(x \\plus{} \\frac{1}{x^2}) \\equal{} f(x) \\plus{} f^2(\\frac{1}{x})$.\n\n(3) Find all function $ f,g: \\mathbb R \\rightarrow \\mathbb R$ satisfying\n\\[ f(x \\plus{} g(y)) \\equal{} xf(y) \\minus{} yf(x) \\plus{} g(x) , \\quad \\forall x,y \\in \\mathbb R.\\]\n\n(4) Find all functions $ f: \\mathbb R \\rightarrow \\mathbb R$ satisfying\n1/ $ f(x_1) \\neq f(x_2) \\forall x_1 \\neq x_2$, and\n2/ For all real numbers $x$ and $y$ with $x \\neq y$,\n\\[ f\\left(\\frac{x \\plus{} y}{x \\minus{} y}\\right) \\equal{} \\frac{f(x) \\plus{} f(y)}{f(x) \\minus{} f(y)}.\\]", "Solution_1": "[quote=\"ll931110\"]Problem 1:\nFind all function $ f: [0;1] \\rightarrow [0;1]$ which satisfies:\n1/ $ f(x_1) \\neq f(x_2) \\forall x_1 \\neq x_2$\n2/ $ 2x \\minus{} f(x) \\in [0;1] \\forall x \\in [0;1]$\n3/ $ f(2x \\minus{} f(x)) \\equal{} x$\n[/quote]\r\n\r\nConditions 2. + 3. are enough to conclude $ f(x) \\equal{} x$ :\r\n\r\nLet $ f(a) \\equal{} b$ and the following sequence :\r\n$ u_0 \\equal{} a$\r\n$ u_1 \\equal{} 2a \\minus{} b$\r\n$ u_{n \\plus{} 2} \\equal{} 2u_{n \\plus{} 1} \\minus{} u_n$\r\n\r\nWe obviously have $ f(u_{n \\plus{} 1}) \\equal{} u_n$ $ \\forall n$ and $ u_n\\in[0,1]$ $ \\forall n$\r\n\r\nBut clearly $ u_n \\equal{} n(a \\minus{} b) \\plus{} a$ and so $ a \\equal{} b$, else $ u_n\\in[0,1]$ $ \\forall n$ would be wrong.", "Solution_2": "Thanks pco. I thought it when I was solving this problem, but at that time I wasn't sure my solution is correct.\r\n\r\nWhat about other problems? Who can solve them?", "Solution_3": "[quote=\"ll931110\"]Problem 2:\nDoes there exist a function $ f: R \\rightarrow R$ satisfying these below conditions:\n1/ There exists $ M \\ge 0$ such that $ \\minus{} M \\le f(x) \\le M$\n2/ $ f(1) \\equal{} 1$\n3/ If $ x \\neq 0$ then $ f(x \\plus{} \\frac {1}{x^2}) \\equal{} f(x) \\plus{} f^2(\\frac {1}{x})$[/quote]\r\n\r\n$ f(1) \\equal{} 1$ $ \\implies$ $ f(2) \\equal{} f(1 \\plus{} \\frac {1}{1^2}) \\equal{} f(1) \\plus{} f^2(\\frac {1}{1}) \\equal{} 2$\r\n\r\nLet then $ u$ such that $ f(u)\\geq 2$\r\n\r\nIf $ f(\\frac {1}{u})\\geq 1 \\minus{} f(u)$, let $ v \\equal{} \\frac {1}{u} \\plus{} u^2$.\r\n$ f(v) \\equal{} f(\\frac {1}{u}) \\plus{} f^2(u)$ $ \\geq 1 \\minus{} f(u) \\plus{} f^2(u)\\geq f(u) \\plus{} 1$\r\n\r\nIf $ f(\\frac {1}{u}) < 1 \\minus{} f(u)$, let $ v \\equal{} u \\plus{} \\frac {1}{u^2}$.\r\n$ f(\\frac {1}{u}) < 1 \\minus{} f(u) < 0$ $ \\implies$ $ f^2(\\frac {1}{u}) > f^2(u) \\minus{} 2f(u) \\plus{} 1$\r\n$ f(v) \\equal{} f(u) \\plus{} f^2(\\frac {1}{u})$ $ > f^2(u) \\minus{} f(u) \\plus{} 1\\geq f(u) \\plus{} 1$\r\n\r\nSo, if $ \\exists u$ such that $ f(u)\\geq 2$, then $ \\exists v$ such that $ f(v)\\geq f(u) \\plus{} 1$\r\n\r\nSo, since $ f(2) \\equal{} 2$, $ f(x)$ may be as great as we want, which is impossible, since $ f(x)\\leq M$ $ \\forall x$\r\n\r\nSo, no such function exists.", "Solution_4": "[quote=\"ll931110\"]Problem 4:\nFind all functionj $ f: R \\rightarrow R$ satisfying\n1/ $ f(x_1) \\neq f(x_2) \\forall x_1 \\neq x_2$\n2/ $ f(\\frac {x \\plus{} y}{x \\minus{} y}) \\equal{} \\frac {f(x) \\plus{} f(y)}{f(x) \\minus{} f(y)} \\forall x \\neq y$[/quote]\r\nLet $ P(x,y)$ be the assertion $ f(\\frac{x\\plus{}y}{x\\minus{}y})\\equal{}\\frac{f(x)\\plus{}f(y)}{f(x)\\minus{}f(y)}$\r\n\r\n$ P(x,0)$ $ \\implies$ $ f(1)\\equal{}\\frac{f(x)\\plus{}f(0)}{f(x)\\minus{}f(0)}$ and so $ f(x)(f(1)\\minus{}1)\\equal{}f(0)(f(1)\\plus{}1)$ and so, since $ (f(x)$ cant be a constant, since injective) :\r\n$ f(0)\\equal{}0$\r\n$ f(1)\\equal{}1$\r\n$ P(0,1)$ $ \\implies$ $ f(\\minus{}1)\\equal{}\\minus{}1$\r\n\r\nCompare now $ P(x,1)$ and $ P(xy,y)$ for $ x\\neq 1$ and $ y\\neq 0$. The two LHS are equal, so are the two RHS : $ \\frac{f(x)\\plus{}1}{f(x)\\minus{}1}\\equal{}\\frac{f(xy)\\plus{}f(y)}{f(xy)\\minus{}f(y)}$ And so :\r\n\r\n$ f(x)f(xy)\\minus{}f(x)f(y)\\plus{}f(xy)\\minus{}f(y)$ $ \\equal{}f(xy)f(x)\\plus{}f(y)f(x)\\minus{}f(xy)\\minus{}f(y)$ and so $ \\boxed{f(xy)\\equal{}f(x)f(y)}$ and this is still true for $ x\\equal{}1$ or $ y\\equal{}0$\r\n\r\nLet then $ f(2)\\equal{}a$. $ P(x,1)$ may then be written (using $ f(xy)\\equal{}f(x)f(y)$) : $ f(x\\plus{}1)\\equal{}f(x\\minus{}1)\\frac{f(x)\\plus{}1}{f(x)\\minus{}1}$ and so allow us to compute :\r\n$ f(2)\\equal{}a$\r\n$ f(3)\\equal{}\\frac{a\\plus{}1}{a\\minus{}1}$\r\n$ f(4)\\equal{}a^2$\r\n$ f(5)\\equal{}\\frac{a^2\\plus{}1}{(a\\minus{}1)^2}$\r\n$ f(6)\\equal{}a(a^2\\minus{}a\\plus{}1)$\r\nWriting then $ f(6)\\equal{}f(2)f(3)$ and using the fact that $ f(x)$ is injective (so $ a\\neq 0$), we get $ a\\equal{}2$.\r\n\r\nLet now $ A\\equal{}\\frac{x^2\\plus{}y^2\\plus{}2xy}{x^2\\plus{}y^2\\minus{}2xy}$ for $ x\\neq 0$, $ y\\neq 0$ and $ x\\neq y$\r\n\r\nUsing $ P(x^2\\plus{}y^2,2xy)$ we obtain : $ f(A)\\equal{}\\frac{f(x^2\\plus{}y^2)\\plus{}2f(x)f(y)}{f(x^2\\plus{}y^2)\\minus{}2f(x)f(y)}$ (we needed $ f(2)\\equal{}2$ in order to write this).\r\n\r\nBut $ A\\equal{}(\\frac{x\\plus{}y}{x\\minus{}y})^2$ and so $ f(A)\\equal{}f^2(\\frac{x\\plus{}y}{x\\minus{}y})$ $ \\equal{}\\frac{f^2(x)\\plus{}2f(x)f(y)\\plus{}f^2(y)}{f^2(x)\\minus{}2f(x)f(y)\\plus{}f^2(y)}$\r\n\r\nEquating these two expressions give us $ f(x^2\\plus{}y^2)\\equal{}f(x^2)\\plus{}f(y^2)$, which is obviously also true if $ x\\equal{}0$, or $ y\\equal{}0$, or $ x\\equal{}y$\r\n\r\nSo $ f(x\\plus{}y)\\equal{}f(x)\\plus{}f(y)$ $ \\forall x,y\\geq 0$ and, writing $ f(x)\\equal{}f(x\\plus{}y)\\minus{}f(y)$, we have $ \\boxed{f(x\\plus{}y)\\equal{}f(x)\\plus{}f(y)\\forall x,y}$\r\n\r\nSo we have a classical Cauchy equation with the complementary condition $ f(xy)\\equal{}f(x)f(y)$ which gives, for any $ a>0$ $ f(x\\plus{}a)\\equal{}f(x)\\plus{}f((\\sqrt a)^2)\\equal{}f(x)\\plus{}f^2(\\sqrt a)>f(x)$. So $ f(x)$ is strictly increasing and so $ f(x)\\equal{}xf(1)$\r\n\r\nHence the solution $ f(x)\\equal{}x$", "Solution_5": "@pco: I think I have another way for Problem 4 (at first, thanks for your help)\r\n\r\nFrom $ f(xy) \\equal{} f(x).f(y)$, replacing $ y$ by $ \\frac{1}{x}$ into 2/ gives\r\n$ f(\\frac{x^2 \\plus{} 1}{x^2 \\minus{} 1}) \\equal{} \\frac{(f(x))^2 \\plus{} 1}{(f(x))^2 \\minus{} 1}$ (1)\r\n\r\nHowever, we have $ f(\\frac{x^2 \\plus{} 1}{x^2 \\minus{} 1}) \\equal{} \\frac{(f(x^2) \\plus{} 1}{(f(x^2) \\minus{} 1}$ (2)\r\nCombining (1) and (2) gives $ f(x^2) \\equal{} (f(x))^2 \\forall x$, which means $ f(x) > 0 \\forall x > 0$, which yields f is strictly increasing.\r\n\r\nSo we have $ f(x) \\equal{} x^a$, and return 2/ gives $ a \\equal{} 1$. So $ f(x) \\equal{} x$", "Solution_6": "[quote=\"ll931110\"] From $ f(xy) \\equal{} f(x).f(y)$, replacing $ y$ by $ \\frac {1}{x}$ into 2/ gives\n$ f(\\frac {x^2 \\plus{} 1}{x^2 \\minus{} 1}) \\equal{} \\frac {(f(x))^2 \\plus{} 1}{(f(x))^2 \\minus{} 1}$ (1)\n\nHowever, we have $ f(\\frac {x^2 \\plus{} 1}{x^2 \\minus{} 1}) \\equal{} \\frac {(f(x^2) \\plus{} 1}{(f(x^2) \\minus{} 1}$ (2)\nCombining (1) and (2) gives $ f(x^2) \\equal{} (f(x))^2 \\forall x$, [/quote]\n\nYou dont need this step (using $ y\\equal{}\\frac{1}{x}$) : $ f(xy)\\equal{}f(x)f(y)$ immediately implies $ f(x^2)\\equal{}f^2(x)$\n\n[quote=\"ll931110\"] $ f(x^2) \\equal{} (f(x))^2 \\forall x$, which means $ f(x) > 0 \\forall x > 0$, which yields f is strictly increasing.\n\nSo we have $ f(x) \\equal{} x^a$, and return 2/ gives $ a \\equal{} 1$. So $ f(x) \\equal{} x$ [/quote]\r\n\r\nAnd you're right. Well done :)\r\n\r\nBe careful about the distinction $ x>0$ and $ x<0$", "Solution_7": "Sorry, I'm so silly :D.\r\n\r\nIn fact, we only need to consider whether $ x > 0$, because we have $ f(\\minus{}x) \\equal{} \\minus{}f(x) \\forall x$.\r\n\r\nCan you help me to solve Problem 3, pco? Thanks a lot" } { "Tag": [ "search", "inequalities theorems", "inequalities" ], "Problem": "Is there any substitution for $ x$, $ y$ and $ z$ if if we have condition $ xy\\plus{}yz\\plus{}zx\\plus{}xyz\\equal{}4$ :?:", "Solution_1": "$ x\\equal{}\\frac{2a}{b\\plus{}c},y\\equal{}\\frac{2b}{a\\plus{}c},z\\equal{}\\frac{2c}{a\\plus{}b}$.\r\n :wink:", "Solution_2": "If the range is positive integer,the solution will be only $ (1,1,1)$ :(", "Solution_3": "2 zero\u03b5integer:\r\nI think that he meant to use such substitution in inequalities. :wink:", "Solution_4": "[quote=\"Erken\"]2 zero\u03b5integer:\nI think that he meant to use such substitution in inequalities. :wink:[/quote]\r\n\r\nSorry. :ninja:", "Solution_5": "[quote=\"stephen38\"]$ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{a \\plus{} c}$[/quote]\r\n\r\nIf $ x \\equal{} \\frac {2a}{b \\plus{} c}$ Then $ y$ does not necessary satisfies $ y \\equal{} \\frac {2b}{a \\plus{} c}$\r\n\r\nIm I right?", "Solution_6": "[quote=\"FOURRIER\"][quote=\"stephen38\"]$ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{a \\plus{} c}$[/quote]\n\nIf $ x \\equal{} \\frac {2a}{b \\plus{} c}$ Then $ y$ does not necessary satisfies $ y \\equal{} \\frac {2b}{a \\plus{} c}$\n\nIm I right?[/quote]\r\nyou're wrong:\r\nfirst of all WLOG we can assume that $ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{c \\plus{} a}$ then calculate $ z$ we will have:\r\n\r\n$ \\frac {4ab}{(b \\plus{} c)(c \\plus{} a)} \\plus{} z\\left(\\frac {2a}{b \\plus{} c} \\plus{} \\frac {2b}{a \\plus{} c} \\plus{} \\frac {4ab}{(b \\plus{} c)(c \\plus{} a)}\\right) \\equal{} 4$\r\n\r\n$ \\Rightarrow z\\left(\\frac {2(a \\plus{} b)(a \\plus{} b \\plus{} c)}{(b \\plus{} c)(c \\plus{} a)}\\right) \\equal{} \\frac {4(b \\plus{} c)(c \\plus{} a) \\minus{} 4ab}{(b \\plus{} c)(c \\plus{} a)}$\r\n\r\n$ \\Rightarrow z \\equal{} \\frac {2(c^2 \\plus{} ac \\plus{} bc)}{(a \\plus{} b)(a \\plus{} b \\plus{} c)} \\equal{} \\frac {2c}{a \\plus{} b}$\r\n\r\nso if $ xy \\plus{} yz \\plus{} zx \\plus{} xyz \\equal{} 4$ then we can substitute $ x,y,z$ by:\r\n\r\n$ \\frac {2a}{b \\plus{} c},\\frac {2b}{c \\plus{} a},\\frac {2c}{a \\plus{} b}$", "Solution_7": "[quote=\"BaBaK Ghalebi\"][quote=\"FOURRIER\"][quote=\"stephen38\"]$ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{a \\plus{} c}$[/quote]\n\nIf $ x \\equal{} \\frac {2a}{b \\plus{} c}$ Then $ y$ does not necessary satisfies $ y \\equal{} \\frac {2b}{a \\plus{} c}$\n\nIm I right?[/quote]\nyou're wrong:\nfirst of all WLOG we can assume that $ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{c \\plus{} a}$\n[/quote]\r\n\r\n :coolspeak: \r\n\r\nI was asking if we can assume $ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{c \\plus{} a}$ the rest I know is easy, you search $ z$ :wink: \r\nThank you anyway :yup:", "Solution_8": "[quote=\"FOURRIER\"]\nI was asking if we can assume $ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{c \\plus{} a}$\n[/quote]\r\nWe can assume it because,if we fix the number $ a$ and let $ p,q\\in R$,such that $ p\\plus{}q\\equal{}r$=const and $ x \\equal{} \\frac {2a}{p \\plus{} q}$.\r\nNow let $ p\\equal{}\\frac{y(\\frac{2a}{x}\\minus{}p\\plus{}a)}{2}$.so we can find $ p$ and also $ q$.Now let $ p\\equal{}b,q\\equal{}c$.Thus we are done.", "Solution_9": "[quote=\"Erken\"][quote=\"FOURRIER\"]\nI was asking if we can assume $ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{c \\plus{} a}$\n[/quote]\nWe can assume it because,if we fix the number $ a$ and let $ p,q\\in R$,such that $ p \\plus{} q \\equal{} r$=const and $ x \\equal{} \\frac {2a}{p \\plus{} q}$.\nNow let $ p \\equal{} \\frac {y(\\frac {2a}{x} \\minus{} p \\plus{} a)}{2}$.so we can find $ p$ and also $ q$.Now let $ p \\equal{} b,q \\equal{} c$.Thus we are done.[/quote]\r\n\r\nI see it now :wink:", "Solution_10": "triangle substitution!!!!!\r\nnotice that $ x \\equal{} 2a , y \\equal{} 2b , z \\equal{} 2c$\r\nthen we have $ ab \\plus{} bc \\plus{} ca \\plus{} 2abc \\equal{} 1$\r\nlet $ p \\equal{} \\sqrt {ab} ,q \\equal{} \\sqrt {bc} , r \\equal{} \\sqrt {ca}$\r\nthen we have $ p^{2} \\plus{} q^{2} \\plus{} r^{2} \\plus{} 2pqr \\equal{} 1$\r\nso we can make the substitution:\r\n$ A,B,C \\in [0,\\frac{\\pi}{2}]$\r\nwe have:\r\n$ p \\equal{} cosA ,q \\equal{} cosB , r \\equal{} cosC$\r\nyes!", "Solution_11": "[quote=\"canada\"]Is there any substitution for $ x$, $ y$ and $ z$ if we have condition $ xyz \\plus{} yz \\plus{} zx \\plus{} xy \\equal{} 4$ :?:[/quote]\n[quote=\"stephen38\"]$ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{c \\plus{} a},z \\equal{} \\frac {2c}{a \\plus{} b}$.:wink:[/quote]\r\n$ a: b: c \\equal{} \\frac {x}{x \\plus{} 2}: \\frac {y}{y \\plus{} 2}: \\frac {z}{z \\plus{} 2}$. :yup:", "Solution_12": "Can I somewhere find all usefull substitutions?", "Solution_13": "[quote=\"canada\"]Is there any substitution for $ x$, $ y$ and $ z$ if if we have condition $ xy \\plus{} yz \\plus{} zx \\plus{} xyz \\equal{} 4$ :?:[/quote]\r\nwhat about this one\r\n\\[ a\\plus{}b\\plus{}c\\plus{}abc\\equal{}4\\]\r\n???", "Solution_14": "[quote=\"Wolfbrother\"]Can I somewhere find all usefull substitutions?[/quote]\r\n\r\nHojo lee book", "Solution_15": "you can find more at:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=404216\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=404216&start=20\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=404216&start=40\nand that link:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=551821", "Solution_16": "[quote=\"Wolfbrother\"]Can I somewhere find all usefull substitutions?[/quote]\nYou also can find in Problem from the book of Titu.", "Solution_17": "We have $ \\frac{x}{x+2} +\\frac{y}{y+2} + \\frac{z}{z+2} = 1 $ or $ \\frac{1}{x+2} +\\frac{1}{y+2} + \\frac{1}{z+2} = 1 $" } { "Tag": [ "summer program", "Mathcamp", "MathZoom", "email" ], "Problem": "Who went to the JR thing at Google? What were some of your likes and dislikes? What prizes sis you get? How was it compared to last year? What were your favorite problems?", "Solution_1": "I heard that Ccy owned a person there, and won a book for it.", "Solution_2": "Problems were sorta meh, but I liked the free food and all :D", "Solution_3": "[quote=\"serialk11r\"]Problems were sorta meh, but I liked the free food and all :D[/quote]\r\n\r\nthat's not what I heard", "Solution_4": "Basically I think there were way too many things going on. Some stations were pretty nice, a lot were too elementary. I didn't really like how they just had so many tables everywhere. The descriptions on the sheet they gave told basically nothing about what each table was about.", "Solution_5": "Some of the upstairs tables were nice, except the Pythagoras ones could be easily pwned using some Trig...\r\n\r\nI never got to Folding Fractals (my school people were obsessing over the Its-It, so I had to defend mine...), does anyone have a scanned copy of that? I heard it was interesting...\r\n\r\nI spent[wasted] all my time at the Zome place.", "Solution_6": "I was there with a few other AoPS kids at one table for lunch.. I'll try and summarize the CCY story.\r\n\r\nSo anyways, this lady (fibonacci combinatorist, I think) was doing interesting proofs with Fibonacci identities by actually using combinatorics (i.e. there was a REALLY cool determinant proof done by counting non-intersecting lattice paths and showing a bijection to something. I didn't really get the whole thing, but it was cool)\r\n\r\nAnyways, to start off her talk, she did a \"cool math trick\" which was pretty obvious if you've ever done anything with the Fibonacci #s.. anyways the idea was that each student generated a lucas sequence, and she would sum the first ten terms. (or something like that)\r\n\r\nSo she called on three people from the audience to do the trick on.\r\nNaturally, CCY raised his hand and screamed like a mad man. So he was selected to go up.\r\nThe first two kids dutifully gave here their nice, integer lucas sequences.\r\n\r\nThen, the best part of the whole event happened.\r\nCcy gave her his list, and she tells him Point-Blank to go sit down and not be such a nuisance.\r\nWhy?\r\nHe had started with $ \\pi$ and $ \\phi$, and her mental trick relied upon a quick multiplication of the sum $ 5\\pi \\plus{} 8\\phi$ by 11.\r\nWell, needless to say, he RickRolled her SOOO hard. The look of incredulity on her face was worth the drive.\r\nI was thinking to myself \r\n* You just got Pwnt, ms. Fibonacci lady. Welcome to sparta.*", "Solution_7": "I was watching that from upstairs...it was hilarious.", "Solution_8": "What is a Fibonacci combinatorist? Are they related to Pascal's trianglers?", "Solution_9": "She messes around with Fibonacci Numbers and uses Combinatorics to write proofs", "Solution_10": "wait... could this lady be by any chance\r\njennifer quinn who wrote proofs that really count: the art of combinatorical proof?\r\nor was she a different lady", "Solution_11": "yar", "Solution_12": "um does that mean \"yar\" to teh former or \"yar\" to the latter\r\nand how the * can you post such short a message", "Solution_13": "[quote=\"ccy\"]yar[/quote]\r\n\r\nkudos to you. i would have loved to have been there.", "Solution_14": "Dang congyue is pro. Oh well I guess the trick was pretty obvious then", "Solution_15": "Yea it was Jennifer Quinn. How come I didn't get a free book :|", "Solution_16": "I went to the Pixar one.", "Solution_17": "[quote=\"ccy\"]yar[/quote]\r\nhmm it's been 2 days dude", "Solution_18": "man this thread is still on-topic\r\n\r\nthat is ... kinda boring", "Solution_19": "Haha Jennifer Quinn came to Mathcamp and tried to explain to us why C(n,k)=n!/[(n-k)!k!]. We were like, \"Yeah, uh, you're kind of underestimating us. Just a little bit.\"", "Solution_20": "That's so pro. I never knew anything about combinations!", "Solution_21": "Well, that may be the case, but:\r\nI talked to her after the event (after everyone had cleared out) and she said that she was torn.\r\nShe knew the audience had a number of really smart high schoolers, but she also knew that there were a bunch of newbie middle schoolers who didn't have that strong of a background in combinatorics.\r\n\r\nSo she had difficulty deciding upon the content of the talk, and she went more towards the lower end of the spectrum. She said that she'd rather bore some older kids than confuse the younger ones and possibly drive them away from problem solving. Which is an entirely admirable viewpoint, in my opinion.\r\nSo yes, I guess I'm defending her.", "Solution_22": "Hmmm...yea I thought that was the case too.\r\nBut doing that at Mathcamp? :huh:", "Solution_23": "[quote=\"serialk11r\"]Hmmm...yea I thought that was the case too.\nBut doing that at Mathcamp? :huh:[/quote]\r\nwait till you get a week of dr kevin's classes", "Solution_24": "[quote=\"junggi\"][quote=\"serialk11r\"]Hmmm...yea I thought that was the case too.\nBut doing that at Mathcamp? :huh:[/quote]\nwait till you get a week of dr kevin's classes[/quote]\r\nooh, please elaborate . And isn't he from Mathzoom or something?", "Solution_25": "I wasn't at Julia Robinson, so I'm not arguing with whatever she did there. But there are no newbie middle schoolers at Mathcamp. Well, there are middle schoolers. But definitely no newbies.", "Solution_26": "For the guys who went to Google, I won the Google tour but I haven't been contacted yet. I lost the contact card that the lady gave out. Does anyone know the email of the person to contact?", "Solution_27": "Does anybody know how to contact with Mr. Joshua Zucker besides throughout email? I tried to email him 3 times but he hasn't replied me anything! :( I'm curious about the result since I couldn't come! Does anybody know the specific result of this event? Pleased show me the practical way to contact to him if you know(as phone, through someone else,...).He also promised that he would let me know the result after the events and I wait for a long time since then.", "Solution_28": "there is no \"result\"; it's not a competition\r\nthe only thing you could win was from the raffle and i doubt they kept a record who won what in a convenient place", "Solution_29": "[quote=\"junggi\"]there is no \"result\"; it's not a competition\nthe only thing you could win was from the raffle and i doubt they kept a record who won what in a convenient place[/quote]\r\nHow about those recipients for the Problem Solving Fermat-Pythagore, which is known as the 2nd round? Nobody got anything about it in that event? :o", "Solution_30": "[quote=\"ghjk\"][quote=\"junggi\"]there is no \"result\"; it's not a competition\nthe only thing you could win was from the raffle and i doubt they kept a record who won what in a convenient place[/quote]\nHow about those recipients for the Problem Solving Fermat-Pythagore, which is known as the 2nd round? Nobody got anything about it in that event? :o[/quote]\r\nWhat? What's that? There were [b]no[/b] competitive events. Period.", "Solution_31": "[quote=\"serialk11r\"][quote=\"ghjk\"][quote=\"junggi\"]there is no \"result\"; it's not a competition\nthe only thing you could win was from the raffle and i doubt they kept a record who won what in a convenient place[/quote]\nHow about those recipients for the Problem Solving Fermat-Pythagore, which is known as the 2nd round? Nobody got anything about it in that event? :o[/quote]\nWhat? What's that? There were [b]no[/b] competitive events. Period.[/quote]\r\nHere's what I meant! Prizes for the Problem solving challenge Round 2:\r\nhttp://www.msri.org/specials/festival/index_html\r\nI seem to be lost! Really confused about the prizes for this Problem Solving Challenge Round 2.Maybe they don't talk about it in the event, so is it faked? I put greatly effort on this Problem Solving. :(", "Solution_32": "Hmmm...I seem to vaguely remember a bunch of middle schoolers going up to the stage and getting awards for this..." } { "Tag": [ "function" ], "Problem": "find the inverse function for the following equation:\r\n\r\nf(x)= 3x+2f(x)= 3x+2", "Solution_1": "[hide]\n$f(x) = 3x+2 \\implies y = 3x+2$\n\ntake inverse\n\n$x = 3y+2$\n\n$x-2 = 3y$\n\n$y = \\frac{x-2}{3}$\n\n$f^{-1}(x) = \\frac{x-2}{3}$\n[/hide]" } { "Tag": [], "Problem": "Prove that, if $ F_0\\equal{}F_1\\equal{}1$, and $ F_n\\equal{}F_{n\\minus{}1}\\plus{}F_{n\\minus{}2}$, that $ (F_n)! \\geq n(F_0)!(F_1)!...(F_{n\\minus{}1})!$", "Solution_1": "Edit: IT SHOULD BE $ F_{n\\minus{}2}$ instead of $ F_{n\\minus{}1}$ :blush:", "Solution_2": "[hide=\"hint\"] first prove that if $ a\\plus{}b\\plus{}c\\plus{}...\\plus{}x\\plus{}y \\leq z$, then $ z! \\geq a!b!c!...x!y!$[/hide]" } { "Tag": [ "real analysis", "search", "real analysis theorems" ], "Problem": "let M be a lebesgue measurable subset of the real line with positive measure =a. let 03$ and $ a_{1},...,a_{n}\\geq 0$, whose sum of squares is 1. Solve that:\r\n$ \\frac{a_{1}}{a^{2}_{2}\\plus{}1}\\plus{}...\\plus{}\\frac{a_{n}}{a^{2}_{1}\\plus{}1}\\geq\\frac{4(a_{1}\\sqrt{a_{1}}\\plus{}...\\plus{}a_{n}\\sqrt{a_{n}})^{2}}{5}$", "Solution_1": "$ (a_{1}^{2}(a_{2}^{2}+1)+a_{2}^{2}(a_{3}^{2}+1)+...+a_{n}^{2}(a_{1}^{2}+1))(\\frac{a_{1}}{a_{2}^{2}+1}+...+\\frac{a_{n}}{a_{1}^{2}+1})\\geq (a_{1}\\sqrt{a_{1}}+...+a_{n}\\sqrt{a_{n}})^{2}$ ----> \r\n$ \\frac{a_{1}}{a_{2}^{2}+1}+...+\\frac{a_{n}}{a_{1}^{2}+1}\\geq\\frac{(a_{1}\\sqrt{a_{1}}+...+a_{n}\\sqrt{a_{n}})^{2}}{a_{1}^{2}(a_{2}^{2}+1)+a_{2}^{2}(a_{3}^{2}+1)+...+a_{n}^{2}(a_{1}^{2}+1)}=\\frac{(a_{1}\\sqrt{a_{1}}+...+a_{n}\\sqrt{a_{n}})^{2}}{a_{1}^{2}a_{2}^{2}+a_{2}^{2}a_{3}^{2}+...+a_{n}^{2}a_{1}^{2}+a_{1}^{2}+...+a_{n}^{2}}=$ $ \\frac{(a_{1}\\sqrt{a_{1}}+...+a_{n}\\sqrt{a_{n}})^{2}}{a_{1}^{2}a_{2}^{2}+a_{2}^{2}a_{3}^{2}+...+a_{n}^{2}a_{1}^{2}+1}\\ged [*]\\geq\\frac{(a_{1}\\sqrt{a_{1}}+...+a_{n}\\sqrt{a_{n}})^{2}}{\\frac{1}{4}+1}=\\frac{4}{5}(a_{1}\\sqrt{a_{1}}+...+a_{n}\\sqrt{a_{n}})^{2}$\r\n $ [*]= a_{1}^{2}a_{2}^{2}+a_{2}^{2}a_{3}^{2}+...+a_{n}^{2}a_{1}^{2}\\leq (a_{1}^{2}+a_{3}^{2}+...)(a_{2}^{2}+a_{4}^{2}+...)\\leq (\\frac{a_{1}^{2}+...+a_{n}^{2}}{2})^{2}=(\\frac{1}{2})^{2}=\\frac{1}{4}$ \r\n\r\n\r\ni think n=4 $ a_{1}=a_{2}=a_{3}=a_{4}=\\frac{1}{2}$", "Solution_2": "already posted http://www.mathlinks.ro/viewtopic.php?p=1040#p1040\r\nand zaya_yc why did you assume n even?" } { "Tag": [ "induction", "inequalities" ], "Problem": "Show that $ (1\\minus{}x_{1})(1\\minus{}x_{2})...(1\\minus{}x_{n}) \\ge 1\\minus{}x_{1}\\minus{}x_{2} \\minus{} ... \\minus{} x_{n}$, where $ 0 \\le x_{k} <1, k \\equal{} 1, 2, ..., n$.", "Solution_1": "seeing as how there is \"induction\" in your title, I will try to write an induction solution.\r\n[hide=\"induction solution\"]\nOur base case is $ n\\equal{}1$ which is just $ 1\\minus{}x_1 \\ge 1\\minus{}x_1$ which is true.\n\nok so suppose that it is true for n=a-1. Now we want to prove that the statement is true for n=a.\n\nSo we have our a numbers $ x_1, x_2, \\ldots x_a$\nand it is true that $ \\prod_{k\\equal{}1}^{a\\minus{}1} 1\\minus{}x_k \\ge 1\\minus{}\\sum_{k\\equal{}1}^{a\\minus{}1} x_k$ by our induction hypothesis. We want to prove that \n$ \\left(1\\minus{}x_a\\right)\\prod_{k\\equal{}1}^{a\\minus{}1} 1\\minus{}x_k \\ge 1\\minus{}\\sum_{k\\equal{}1}^{a\\minus{}1} x_k\\minus{}x_a (1)$\nNotice that both sides (especially the LHS) are less than or equal to 1. This is true for the LHS because each product is positive and less than or equal to 1.\nTherefore, $ x_a \\ge x_a \\cdot \\prod_{k\\equal{}1}^{a\\minus{}1} 1\\minus{}x_k \\implies \\minus{}x_a \\cdot \\prod_{k\\equal{}1}^{a\\minus{}1} 1\\minus{}x_k \\ge \\minus{}x_a (2)$\nadding our two inequalities (1) and (2) gives the desired result.\n\nhmm\nequality is achieved when n or n-1 of the $ x_i$ are equal to 0.\nThis can be seen from setting both (1) and (2) as equalities.\n[/hide]\n[/hide]", "Solution_2": "seeing as how there is \"induction\" in your title, I will try to write an induction solution.\r\n[hide=\"induction solution\"]\nOur base case is $ n\\equal{}1$ which is just $ 1\\minus{}x_1 \\ge 1\\minus{}x_1$ which is true.\n\nok so suppose that it is true for n=a-1. Now we want to prove that the statement is true for n=a.\n\nSo we have our a numbers $ x_1, x_2, \\ldots x_a$\nand it is true that $ \\prod_{k\\equal{}1}^{a\\minus{}1} 1\\minus{}x_k \\ge 1\\minus{}\\sum_{k\\equal{}1}^{a\\minus{}1} x_k$ by our induction hypothesis. We want to prove that \n$ \\left(1\\minus{}x_a\\right)\\prod_{k\\equal{}1}^{a\\minus{}1} 1\\minus{}x_k \\ge 1\\minus{}\\sum_{k\\equal{}1}^{a\\minus{}1} x_k\\minus{}x_a (1)$\nNotice that both sides (especially the LHS) are less than or equal to 1. This is true for the LHS because each product is positive and less than or equal to 1.\nTherefore, $ x_a \\ge x_a \\cdot \\prod_{k\\equal{}1}^{a\\minus{}1} 1\\minus{}x_k \\implies \\minus{}x_a \\cdot \\prod_{k\\equal{}1}^{a\\minus{}1} 1\\minus{}x_k \\ge \\minus{}x_a (2)$\nadding our two inequalities (1) and (2) gives the desired result.\n\nhmm\nequality is achieved when n or n-1 of the $ x_i$ are equal to 0.\nThis can be seen from setting both (1) and (2) as equalities.\n[/hide]" } { "Tag": [ "inequalities", "linear algebra", "matrix", "algebra", "polynomial" ], "Problem": "Let $ A \\in \\mathcal{M}_n \\left( \\mathbb{R} \\right)$ with $ \\det{A} \\equal{} 0$ and $ X \\in \\mathcal{M}_{n} \\mathbb{R}$ the matrix with all elements $ 1$, $ x_{ij} \\equal{} 1 , \\forall 1 \\leq i,j \\leq n$.\r\n\r\nProve that: $ \\det{(A \\plus{} X)} \\cdot \\det{(A \\minus{} X)} \\leq 0$.", "Solution_1": "$ X$ is similar over $ \\mathbb{R}$ to $ diag(0,\\cdots,0,n)$. Then we may assume $ X\\equal{}diag(0,\\cdots,0,n)$. If $ A\\equal{}[a_{ij}]$ then $ det(A\\plus{}X)\\equal{}det(A)\\plus{}n\\alpha\\equal{}n\\alpha$ where $ \\alpha$ is the cofactor of $ a_{nn}$. By the same way $ det(A\\minus{}X)\\equal{}\\minus{}n\\alpha$ and $ det(A\\plus{}X)det(A\\minus{}X)\\equal{}\\minus{}n^2\\alpha^2$.", "Solution_2": "hi, i am not good at matrix so can you show me that why X is similar to diag(0,0,...,n) loup blanc, please? thank you very much", "Solution_3": "$ X$ is a real symmetric then it is diagonalizable over $ \\mathbb{R}$. $ rank(X)\\equal{}1$ then $ X$ has at least $ n\\minus{}1$ eigenvalues equal to $ 0$. The last one is $ trace(X)\\equal{}n$.", "Solution_4": "Maybe this will sound stupid, but I`m not sure if we have $ A,X \\in \\mathcal{M}_n \\left( \\mathbb{R} \\right)$, and $ X$ is similar over $ \\mathbb{R}$ with another matrix $ Y \\in \\mathcal{M}_n \\left( \\mathbb{R}\\right)$ then:\r\n\r\n\\[ \\det(A\\plus{}X)\\equal{} \\det(A\\plus{}Y)\\]\r\n\r\nIf this is true, please post a proof. Thank you\r\nAnd yes I know that $ X$ and $ Y$ share the same characteristic polynomial, but this gives only that:\r\n$ X\\equal{}PYP^{\\minus{}1}$ where $ P$ is invertible, and:\r\n$ \\det(A\\plus{}X)\\equal{} \\det(P^{\\minus{}1}AP\\plus{}Y)$, and this is not sufficient..", "Solution_5": "Hi linker,\r\n\"if we have $ A,X \\in \\mathcal{M}_n \\left( \\mathbb{R} \\right)$, and is similar over $ \\mathbb{R}$ with another matrix $ Y \\in \\mathcal{M}_n \\left( \\mathbb{R}\\right)$ then:\r\n$ det(A\\plus{}X)\\equal{}det(A\\plus{}Y)$.\"\r\nI didn't say that. When I write \"we may assume $ X\\equal{}Y$\", then I hide the following:\r\n$ X\\equal{}PYP^{\\minus{}1},A\\equal{}PBP^{\\minus{}1}$ where $ P$ is a real invertible matrix. Then $ B$ is a real matrix s.t. $ det(B)\\equal{}0$. Thus we replace $ X,A$ with $ Y,B$.", "Solution_6": "we have one more general problem:\r\n\" $ A,B \\in \\mathcal{M}_n(R)$ with rank(B)=1. prove that $ det((A\\minus{}B)(A\\plus{}B)) \\leq (detA)^2$ \" :)", "Solution_7": "Use the same method but now there exist 2 cases:\r\ncase 1: we may assume $ B\\equal{}diag(0,\\cdots,0,\\lambda)$ where $ \\lambda$ is a non zero real.\r\ncase 2: we may assume $ B\\equal{}diag(0_{n\\minus{}2},\\begin{pmatrix}0&1\\\\0&0\\end{pmatrix})$." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "You have a 17 x 17 chess table, and on every square there is a coin. Every coin has 2 sides: the face and the number. Innitially, all the coins ar with their faces up, but at each step, you take 5 consecutive coins from any line, column or one of the 2 diagonals and you turn them upside down. Is it possible that after a limited number of steps, all the coins will be with their number sides up?", "Solution_1": "Color the grid as follows:\r\n\r\n[code]\nABBAAABBAAABBAAAB\nAABBAAABBAAABBAAA\nAAABBAAABBAAABBAA\nBAAABBAAABBAAABBA\nBBAAABBAAABBAAABB\nABBAAABBAAABBAAAB\nAABBAAABBAAABBAAA\nAAABBAAABBAAABBAA\nBAAABBAAABBAAABBA\nBBAAABBAAABBAAABB\nABBAAABBAAABBAAAB\nAABBAAABBAAABBAAA\nAAABBAAABBAAABBAA\nBAAABBAAABBAAABBA\nBBAAABBAAABBAAABB\nABBAAABBAAABBAAAB\nAABBAAABBAAABBAAA\n[/code]\r\n\r\nEvery strip contains either 2 or 0 $ B $ squares. Therefore after every flip there will be an even number of $ B $ squares number side up. But by my count there are 115 $ B $ squares in the coloring, so one is not number side up" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,$ $ b,$ $ c$ be positive real numbers such that $ a\\plus{}b\\plus{}c\\equal{}ab\\plus{}bc\\plus{}ca.$ Prove that\r\n$ a^{\\frac{a}{b}}b^{\\frac{b}{c}} c^c \\ge 1.$", "Solution_1": "[quote=\"can_hang2007\"]Let $ a,$ $ b,$ $ c$ be positive real numbers such that $ a \\plus{} b \\plus{} c \\equal{} ab \\plus{} bc \\plus{} ca.$ Prove that\n$ a^{\\frac {a}{b}}b^{\\frac {b}{c}} c^c \\ge 1.$[/quote]\r\nIt looks like sadism. :D", "Solution_2": "See here: http://can-hang2007.blogspot.com/2010/01/inequality-61-v-q-b-can.html" } { "Tag": [ "quadratics", "number theory unsolved", "number theory" ], "Problem": "How many squares be in $ (mod.2^n)$ ??", "Solution_1": "I'll assume that you wish to find the number of quadratic residues $ mod 2^n$.\r\n\r\nLet the number of quadratic residues $ mod 2^n$ be $ f(n)$.\r\n\r\nIt is easy to verify that $ f(2)=2, f(3)=3$.\r\n\r\nClaim: The number of odd quadratic residues are $ 2^{n-3}$.\r\n\r\nProof) We can verify that for all odd k, $ k^2 == (2^{n-1} + k)^2$ $ (mod 2^n )$.\r\n\r\nTherefore, we only need to consider the integers $ 1,3,...,2^{n-1} - 1$.\r\n\r\nWe have that $ k^2 == (2^{n-1} - k)^2$ $ (mod 2^n )$ : we need only to consider the integers $ 1,3, ... , 2^{n-2}-1$.\r\n\r\nHowever, if $ 010$ and belonging to any sequence $ \\{84k+1\\}$, $ \\{84k+13\\}$, $ \\{84k+37\\}$, $ \\{84k+61\\}$ or $ \\{84k+73\\}$.\r\n\r\nFor example : $ 13,37,61,73,97,157,181,...$\r\n\r\nWhich gives for $ c$ : $ 130,370,610,730,970,1570,1810,...$\r\n\r\nFor example, with $ c=130$, we have the three solutions $ (2,11)$, $ (17,3)$ and $ (44,2)$" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "For $ a,b,c$ positive reals prove that\r\n\r\n$ \\sum \\frac{ab}{3a\\plus{}4b\\plus{}2c}\\leq \\frac{a\\plus{}b\\plus{}c}{9}$", "Solution_1": "It's obviously trues by CS because:\r\n$ \\sum\\ \\frac{ab}{a\\plus{}b\\plus{}2c} \\leq\\ \\frac{3}{4} and \\sum\\ \\frac{ab}{2a\\plus{}3b} \\leq\\ \\frac{3}{5}$\r\nPS: $ 3a\\plus{}4b\\plus{}2c\\equal{}a\\plus{}b\\plus{}2c\\plus{}2a\\plus{}3b$ :)" } { "Tag": [ "function", "integration", "trigonometry", "calculus", "calculus computations" ], "Problem": "I am reading a text called \"A guide to distribution theory and Fourier transforms\" by Strichartz and one of the problems asks one to use the Fourier inversion formula to compute $ \\int_{ \\minus{} \\infty}^{\\infty}\\frac {\\sin x}{x}\\,dx.$ I know how to do this integral via elementary real variable approaches and contour integration, but does anyone know the solution suggested by the text?\r\n\r\nAlso, is it permissible to treat $ \\int_{ \\minus{} \\infty}^{\\infty}\\frac {\\sin{\\alpha\\cdot x}}{x}\\,dx$ as a generalized function and then differentiate under the integral with respect to $ \\alpha$ (I am wondering since rewriting $ \\sin{\\alpha\\cdot x}$ as an exponential seems to be the only way I can get Fourier phases to appear in the integrand)?", "Solution_1": "I catch a whiff of possible circularity here, in that I prefer to have this specific integral already computed so that I can use it in a proof of Fourier inversion.\r\n\r\nBut let's say that somehow we already have Fourier inversion. I'm guessing that Strichartz puts his $ 2\\pi$s in a different place than I'm used to; let me go back to my favorite notation.\r\n\r\nGiven appropriate $ f,$ define $ \\widehat{f}(\\xi) \\equal{} \\int_{ \\minus{} \\infty}^{\\infty}f(x)e^{ \\minus{} 2\\pi i\\xi x}\\,dx.$\r\n\r\nFourier inversion would be the statement that under certain fairly mild conditions on $ f$, we can recover $ f(x) \\equal{} \\int_{ \\minus{} \\infty}^{\\infty}\\widehat{f}(\\xi)e^{2\\pi i \\xi x}\\,d\\xi,$ where that integral may be interpreted as an improper integral or even as a principal value (that is, the limit of the integral from $ \\minus{} N$ to $ N.$)\r\n\r\nWe define the \"box function\": $ \\sqcap(x) \\equal{} \\begin{cases}1 & \\text{if }|x| < \\frac12 \\\\\r\n0 & \\text{otherwise}\\end{cases}.$ (Under some circumstances, I may even want to specify the value of the box function at $ \\pm\\frac12$ as $ \\frac12$ rather than as either $ 0$ or $ 1$ - but that's not particularly important right now.)\r\n\r\nBy direct computation, $ \\widehat{\\sqcap}(\\xi) \\equal{} \\frac {\\sin\\pi \\xi}{\\pi\\xi} \\equal{} \\text{sinc}\\,(\\xi)$\r\n\r\nThat was the definition of the function $ ``\\text{sinc}\"$ except that we also have $ \\text{sinc}\\,(0) \\equal{} 1,$ making $ \\text{sinc}$ a smooth function.\r\n\r\nNow if we believe Fourier inversion, we have $ \\sqcap(x) \\equal{} \\int_{ \\minus{} \\infty}^{\\infty}\\text{sinc}\\,(\\xi)e^{2\\pi i\\xi x}\\,d\\xi$ or in particular,\r\n\r\n$ \\sqcap(0) \\equal{} 1 \\equal{} \\int_{ \\minus{} \\infty}^{\\infty}\\text{sinc}\\,(\\xi)\\,d\\xi \\equal{} \\int_{ \\minus{} \\infty}^{\\infty}\\frac {\\sin\\pi x}{\\pi x}\\,dx$\r\n\r\n which is (essentially) the integral you're looking for.\r\n\r\nYou could make the same argument under different \"where are the $ 2\\pi$s\" conventions - just make sure to have those in the right places, and to use the appropriate version of the box function." } { "Tag": [ "MATHCOUNTS", "probability" ], "Problem": "The following question is from the 2006 Mathcounts chapter competiton: \r\n\r\nFive balls are numbered 1 through 5 and placed in a bowl. Josh will randomly choose a ball from the bowl, look at its number and then put it back into the bowl. Then Josh will again randomly choose a ball from the bowl and look at its number. What is the probability that the prodcut of the two numbers will be even and greater than 10? Express your answer as a common fraction.\r\n\r\nCan I also have an explanation on how you came to your answer, please.", "Solution_1": "[hide=\"hint\"]\nLook at products of numbers between 1,5 and then see which ones are even\n[/hide]", "Solution_2": "You can make a table. That would simplify this problem a lot.", "Solution_3": "[hide] There aren't many cases so we can count them. \n\nConsider the first one drawn. \n\nFor example, if the first ball is:\n\n1, the successful outcomes for the second ball are 1-5\n\nSo all the successful outcomes are\n\n1, (none), but 1, 3, and 5 are not even so\n0 successful outcomes\n\n2, (none)\n0\n\n3, (4 through 5) \n1\n\n4, (3 though 5)\n3\n\n5, (3 though 5)\n1\n\n0+0+1+3+1=5\n\n5*5=25 possible outcomes\n\n5/25=[b]1/5[/b]\n[/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If $a,b,c>0$ and $a^{2}+b^{2}+c^{2}=3$ prove that:\r\n \\[\\frac{a^{4}+b^{2}c^{2}+1}{(a^{3}+2bc)^{2}}+\\frac{b^{4}+c^{2}a^{2}+1}{(b^{3}+2ca)^{2}}+\\frac{c^{4}+a^{2}b^{2}+1}{(c^{3}+2ab)^{2}}\\geq 1\\]", "Solution_1": "Nice problem Svejk :) \r\nBy Cauchy-Swartz and AM-GM we have \r\n$3(a^{4}+b^{2}c^{2}+1)=(a^{2}+b^{2}+c^{2})(a^{4}+b^{2}c^{2}+1)\\geq (a^{3}+b^{2}c+c)^{2}\\geq (a^{3}+2bc)^{2}$ \r\n\r\nso $\\frac{a^{4}+b^{2}c^{2}+1}{(a^{3}+2bc)^{2}}\\geq\\frac{1}{3}$ \r\n\r\nSimilarly for the others and add them and we are done :wink:", "Solution_2": "[quote=\"silouan\"]Nice problem Svejk :) \nBy Cauchy-Swartz and AM-GM we have \n$3(a^{4}+b^{2}c^{2}+1)=(a^{2}+b^{2}+c^{2})(a^{4}+b^{2}c^{2}+1)\\geq (a^{3}+b^{2}c+c)^{2}\\geq (a^{2}+2bc)^{2}$ \n\nso $\\frac{a^{4}+b^{2}c^{2}+1}{(a^{2}+2bc)^{2}}\\geq\\frac{1}{3}$ \n\nSimilarly for the others and add them and we are done :wink:[/quote]\r\nCorrection: $(a^{3}+b^{2}c+c)^{2}\\geq (a^{3}+2bc)^{2}\\implies \\frac{a^{4}+b^{2}c^{2}+1}{(a^{3}+2bc)^{2}}\\geq\\frac{1}{3}$.\r\nNice problem." } { "Tag": [], "Problem": "If $ r$, $ s$, and $ t$, are the complex roots of the equation:\r\n$ x^3 \\plus{} 2x^2 \\plus{}5x \\plus{}7$,\r\nwhat is $ r^2 \\plus{}s^2 \\plus{}t^2$?", "Solution_1": "\\[ \\left\\{\\begin{array}{lcl}\r\nr\\plus{}s\\plus{}t & \\equal{} & \\minus{}2, \\\\\r\nrs\\plus{}st\\plus{}tr & \\equal{} & 5, \\\\\r\nrst & \\equal{} & \\minus{}7.\r\n\\end{array}\\right.\\]\r\nAnd\r\n\\[ r^2\\plus{}s^2\\plus{}t^2 \\equal{} (r\\plus{}s\\plus{}t)^2\\minus{}2(rs\\plus{}st\\plus{}tr).\\]\r\n\r\nThus\r\n\\[ r^2\\plus{}s^2\\plus{}t^2 \\equal{} (\\minus{}2)^2 \\minus{} 2 \\cdot 5 \\equal{} 4 \\minus{} 10 \\equal{} \\boxed{\\minus{}6}.\\]", "Solution_2": "A general way of doing this is [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_sums]Newton's Sums[/url]." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all function $f:Z\\longrightarrow Z$ such that $f(-1)=f(1)$ and for any two integer number like $m,n$ we have: \r\n$f(m)+f(n)=f(m+2mn)+f(n-2mn)$", "Solution_1": "substituting m=1, gives: f(1)+f(n)=f(2n+1)+f(-n)\r\nsubstituting m=n, n=-1, gives: f(n)+f(-1)=f(-n)+f(2n-1)\r\nSo f(2n-1)=f(2n+1) for all n, or f(n) = c for all odd integers n, where c is a constant.\r\nIt follows that f(n)=f(-n) for all n.\r\ncan't finish for some reason..." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "No one solved this in the Pre-Olympiad forum, so I've decided to post it here too. \r\n\r\n$f(x)+f\\left(\\frac{1-x}{x}\\right) = 1-x$.", "Solution_1": ":(\r\n\r\n[hide]Why is this message considered too small?[/hide]", "Solution_2": "i would have answered 7 hours ago if i had arrived at the answer. Try i little more right now!\r\nand dude, calm down!", "Solution_3": "[quote=\"rnwang2\"]No one solved this in the Pre-Olympiad forum, so I've decided to post it here too. \n\n$f(x)+f\\left(\\frac{1-x}{x}\\right) = 1-x$.[/quote]\r\n\r\nfor $y=\\frac{1-x}{x}$ we have $f(x)+f(y)=xy$.", "Solution_4": "posted before. and there is a generalisation.you can use surch. :ninja:" } { "Tag": [ "inequalities" ], "Problem": "Prove that\r\n$a^3b+a^3c+b^3a+b^3c+c^3a+c^3b\\geq 2(a^2bc+b^2ac+c^2ab)$", "Solution_1": "$ab(a^2 + b^2) + ac(a^2 + c^2) + bc(b^2 + c^2) \\geq 2abc(a + b + c)$\r\n\r\n$\\sqrt{\\frac{ab(a^2 + b^2)}{2}} + \\sqrt{\\frac{ac(a^2 + c^2)}{2}} + \\sqrt{\\frac{bc(b^2 + c^2)}{2}} \\geq \\sqrt{abc(a+ b+ c)}$\r\n\r\nFor AM-GM it is \r\n\r\n$\\sqrt{\\frac{ab(a^2 + b^2)}{2}} + \\sqrt{\\frac{ac(a^2 + c^2)}{2}} + \\sqrt{\\frac{bc(b^2 + c^2)}{2}} \\geq ab + ac + bc$\r\n\r\nAlso it is $\\sqrt{abc(a + b+ c)} \\leq ab + ac +bc$ because \r\n$a^2bc + b^2ac + c^2ab \\leq a^2b^2 + a^2c^2 + b^2c^2 + 2a^2bc + 2b^2ac + 2c^2ab$\r\nSo the inequality is proved.", "Solution_2": "From Muirhead imediatelly we have that ${3,1,0}\\succ{2,11}$\r\nSo $a^3b\\geq a^2bc$ \r\nWe add them and we are done.", "Solution_3": "Yes, I know it's easy using Muirhead but I wanted to see how other solutions would go :)", "Solution_4": "[quote=\"Andreas\"]$ab(a^2 + b^2) + ac(a^2 + c^2) + bc(b^2 + c^2) \\geq 2abc(a + b + c)$\n\n$\\sqrt{\\frac{ab(a^2 + b^2)}{2}} + \\sqrt{\\frac{ac(a^2 + c^2)}{2}} + \\sqrt{\\frac{bc(b^2 + c^2)}{2}} \\geq \\sqrt{abc(a+ b+ c)}$\n[/quote]\r\nHow do you get from the first step to the second?", "Solution_5": "Oops that's a stupid mistake :blush:", "Solution_6": "well, first of all i think its rearangment as well, but those can be done by am-gm in most cases:\r\n$\\frac{a^3b+a^3c+ab^3+ac^3}{2}\\ge2\\cdot\\sqrt[4]{a^8b^4c^4}=2a^2bc$\r\n\r\ndo that cyclic with $a,b,c$ and add them up u get the needed inequality", "Solution_7": "[quote=\"peeta\"]well, first of all i think its rearangment as well, but those can be done by am-gm in most cases:\n$\\frac{a^3b+a^3c+ab^3+ac^3}{2}\\ge2\\cdot\\sqrt[4]{a^8b^4c^4}=2a^2bc$\n\ndo that cyclic with $a,b,c$ and add them up u get the needed inequality[/quote]\r\n\r\nShall i assume a,b,c to be +ve reals???", "Solution_8": "yeah it should be a,b,c positive reals. otherwise muirhead and am-gm doesnt work and also try $a=-100;b=1;c=0$ ..." } { "Tag": [ "topology", "linear algebra", "matrix", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ k$ to be an odd integer. Let $ M$ to be a $ 2k$-dimensional compact, oriented, smooth manifold without boundary. Prove that the k-th Betti number, i.e. the dimension of the k-th de Rham cohomology $ H^k (M)$, is even.", "Solution_1": "Use Poincare duality; the bilinear pairing defined in it on $ H^k (M)$ is antisymmetric, so the matrix of the pairing has purely imaginary eigenvalues in pairs with their complex conjugate." } { "Tag": [ "geometry" ], "Problem": "points $A,B,C$are given on circle, find $D$ on circle,inwhich:\r\n\r\n\r\n$\\frac{1}{AD}=\\frac{1}{AB}+\\frac{1}{AC}$", "Solution_1": "$AD=\\frac{AB\\dot{AC}}{AB+AC}$\r\n\r\nMultiplication, addition, and division can all be done with a compass and a straightedge, if that's what your looking for.", "Solution_2": "[quote=\"draperp\"]$AD=\\frac{AB\\dot{AC}}{AB+AC}$\n\nMultiplication, addition, and division can all be done with a compass and a straightedge, if that's what your looking for.[/quote]\r\n\r\nim looking for a cleverly solution ;) .", "Solution_3": "let $I$ be an inversion about a circle radius 1, centered at $A$\r\n\r\n$I(B)=B'$ and $I(C)=C'$, have $D'$ on $AC'$, with $C'$ between $A$ and $D'$ such that $C'D'=B'A$, then $I(D')=D$ using inversion", "Solution_4": "[quote=\"Altheman\"]let $I$ be an inversion about a circle radius 1, centered at $A$\n\n$I(B)=B'$ and $I(C)=C'$, have $D'$ on $AC'$, with $C'$ between $A$ and $D'$ such that $C'D'=B'A$, then $I(D')=D$ using inversion[/quote]\r\n\r\ni designed this problem by inversion.\r\n\r\n\r\ni really enjoyed, ure solution.because i love inversive solutions.\r\n\r\n\r\n\r\nu are very clever. ;)", "Solution_5": "[quote=\"Ashegh\"][quote=\"Altheman\"]let $I$ be an inversion about a circle radius 1, centered at $A$\n\n$I(B)=B'$ and $I(C)=C'$, have $D'$ on $AC'$, with $C'$ between $A$ and $D'$ such that $C'D'=B'A$, then $I(D')=D$ using inversion[/quote]\n\ni designed this problem by inversion.\n\n\ni really enjoyed, ure solution.because i love inversive solutions.\n\n\n\nu are very clever. ;)[/quote]\r\n\r\nha, when i saw you post a problem, i knew immediately to think inversion since inversion is your favorite technique :)" } { "Tag": [ "floor function", "number theory", "prime factorization" ], "Problem": "How many $0$'s does $2006!$ end with?", "Solution_1": "[hide=\"hint\"]\nWhat primes in the prime factorization of 2006! would add zeros to the end of the number, and how many of them are there?\n[/hide]", "Solution_2": "[quote=\"Franz Joseph\"]How many $0$'s does $2006!$ end with?[/quote]\r\n[hide=\"solution\"]\ncounting the #of zeros is same as counting the # of 5's in $2006!$.\nso $2006!$ has 401 5's, 80 of 25's, 16 of 125's, and 3 of 625's.\nTotal $401+80+16+3=500$ zeros.\n[/hide]", "Solution_3": "[hide=\"Generalization\"] The number of zeroes at the end of $n!$:\n\n$\\sum_{i=1}^{\\infty}\\left\\lfloor\\frac{n}{5^{i}}\\right\\rfloor$[/hide]", "Solution_4": "[quote=\"lotrgreengrapes7926\"][hide=\"Generalization\"] The number of zeroes at the end of $n!$:\n\n$\\sum_{i=1}^{\\infty}\\left\\lfloor\\frac{n}{5^{i}}\\right\\rfloor$[/hide][/quote]\r\n :coolspeak: Thanks a lot." } { "Tag": [ "LaTeX" ], "Problem": "2006 Nats sprint #26 & #30\r\n\r\n26. What is the product of the two smallest prime factors of $ 2^{1024} \\minus{} 1$ ?\r\n\r\n[hide=\"my answer\"] 21 [/hide]\n\n[hide=\"their answer\"] 15 [/hide]\n\n30. The number $ \\sqrt (17 \\plus{} 12\\sqrt {2})$ is an example of a number expressed in \"embedded radical form\" because of the $ \\sqrt {2}$ under the radical. What is the equivalent value of this number expressed in simplest radical form $ a \\plus{} b\\sqrt {c}$ where $ a$ and $ b$ are integers and $ c$ is a positive integer containing no perfect square factors greater than one?\n\n[hide=\"my answer\"] guh....[/hide]\n\n[hide=\"their answer\"] $ 3 \\plus{} 2\\sqrt {2}$ [/hide]\r\n\r\nNotes: The first square root goes over all that junk. That was a doozy for me Latex-wise. It took me 8 minutes to type! ( :rotfl: )", "Solution_1": "solution for 26\r\n\r\n \r\n[hide]\nfor number 26 factor as the difference of perfect squares.\nso it is (2^512+1)(2^512-1)\nyou can factor the (2^512-1) as the difference of perfect squares\nyou can continue until as (2^512+1)(2^256+1)(2^128+1).........(2^2+1)(2^1+1)(2^1-1)\nthe last 3 terms are 5, 3, and 1\nso 5*3 is 15\n[/hide]", "Solution_2": "[quote=\"dynamo729\"]2006 Nats sprint #26 & #30\n\n26. What is the product of the two smallest prime factors of $ 2^{1024} - 1$ ?\n\n[hide=\"my answer\"] 21 [/hide]\n\n[hide=\"their answer\"] 15 [/hide]\n\n30. The number $ \\sqrt (17 + 12\\sqrt {2})$ is an example of a number expressed in \"embedded radical form\" because of the $ \\sqrt {2}$ under the radical. What is the equivalent value of this number expressed in simplest radical form $ a + b\\sqrt {c}$ where $ a$ and $ b$ are integers and $ c$ is a positive integer containing no perfect square factors greater than one?\n\n[hide=\"my answer\"] guh....[/hide]\n\n[hide=\"their answer\"] $ 3 + 2\\sqrt {2}$ [/hide]\n\nNotes: The first square root goes over all that junk. That was a doozy for me Latex-wise. It took me 8 minutes to type! ( :rotfl: )[/quote]\n[hide=\"26\"]$ 2^{1024}-1=(2^{512}-1)(2^{512+1)}$. If you see why this works, you can keep breaking it down until you get your answer. \n$ (2^{256}-1)(2^{256}+1)(2^{512}+1)$\n$ (2^{128}-1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$\n$ (2^{64}-1)(2^{64}+1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$\n$ (2^{32}-1)(2^{32}+1)(2^{64}+1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$\n$ (2^{16}-1)(2^{16}+1)(2^{32}+1)(2^{64}+1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$\n$ (2^{8}-1)(2^{8}+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$\n$ (2^{4}-1)(2^{4}+1)(2^{8}+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)(2^{128}+1)(2^{256}+1)(2^{512}+1)$\n$ (2^{2}-1)(2^2+1)=15$[/hide]\n\n[hide=\"30\"]First some reasoning. There is only one possibility for $ c$, which is 2. So now we have $ \\sqrt{17+12\\sqrt{2}}=a+b\\sqrt{2}$. Square both sides and match up. \n$ 17+12\\sqrt{2}=a^{2}+2ab\\sqrt{2}+2b^{2}$\nSo now we have two equations, $ a^{2}+2b^{2}=17$ and $ ab=6$. Quick guessing results in $ a=3$ and $ b=2$. So our answer is $ 3+2\\sqrt{2}$[/hide]", "Solution_3": "Some are trickier, such as $ \\sqrt{5\\plus{}2\\sqrt{6}}\\equal{}\\frac{\\sqrt{10\\plus{}4\\sqrt{6}}}{\\sqrt{2}}\\equal{}\\frac{\\sqrt{6}\\plus{}2}{2}$. \r\nTry this one for practice:\r\n\r\n$ \\sqrt{8\\plus{}4\\sqrt{3}}\\equal{}?$", "Solution_4": "how do you know to divide by $ \\sqrt{2}$", "Solution_5": "You multiply by $ \\sqrt{2}/\\sqrt{2}$. It's intuitive. Or you could set up the equation as $ a\\plus{}b\\sqrt{6}$\r\n$ a^2\\plus{}6b^2\\equal{}5$ and $ ab\\equal{}1$.", "Solution_6": "[quote=\"mathcrazed\"]Some are trickier, such as $ \\sqrt {5 \\plus{} 2\\sqrt {6}} \\equal{} \\frac {\\sqrt {10 \\plus{} 4\\sqrt {6}}}{\\sqrt {2}} \\equal{} \\frac {\\sqrt {6} \\plus{} 2}{2}$. \nTry this one for practice:\n\n$ \\sqrt {8 \\plus{} 4\\sqrt {3}} \\equal{} ?$[/quote]\r\n\r\nI got.... $ \\sqrt {5} \\plus{} \\sqrt {3}$, but I don't know how I got it.....", "Solution_7": "[quote=\"mathcrazed\"]Some are trickier, such as $ \\sqrt {5 + 2\\sqrt {6}} = \\frac {\\sqrt {10 + 4\\sqrt {6}}}{\\sqrt {2}} = \\frac {\\sqrt {6} + 2}{2}$. \nTry this one for practice:\n\n$ \\sqrt {8 + 4\\sqrt {3}} = ?$[/quote]\r\n\r\nis the answer $ \\sqrt{2} +\\sqrt{6}$?\r\n\r\nsrry if i didnt use $ \\text{\\LaTeX}$ correctly....", "Solution_8": "[quote=\"mathking123\"][quote=\"mathcrazed\"]Some are trickier, such as $ \\sqrt {5 + 2\\sqrt {6}} = \\frac {\\sqrt {10 + 4\\sqrt {6}}}{\\sqrt {2}} = \\frac {\\sqrt {6} + 2}{2}$. \nTry this one for practice:\n\n$ \\sqrt {8 + 4\\sqrt {3}} = ?$[/quote]\n\nis the answer $ \\sqrt {2} + \\sqrt {6}$?\n\nsrry if i didnt use $ \\text{\\LaTeX}$ correctly....[/quote]\r\n\r\nyep\r\n\r\n[hide=\"method 1\"]\nPlug it into the TI-89 and it will spit out the simplified answer\n[/hide]\n[hide=\"method 2\"]\n$ \\sqrt{8+4\\sqrt{3}}=a+b\\sqrt{3}$\n$ 8+4\\sqrt{3}=a^2+3b^2+2ab\\sqrt{3}$\nThus, $ a^2+3b^2=8$ and $ ab=2$.\nSolve and you find that $ a=b=\\sqrt{2}$.\nNote that you will also find extraneous solutions, so make sure your answer is positive, since that's what the square root sign implies\n[/hide]" } { "Tag": [], "Problem": "Charlene considers herself to be a great bargain shopper because she found a prom dress that cost her only 22 dollars before tax. The dress was on a rack labeled \u201c50% off lowest marked price,\u201d and the lowest marked price was already a 75% reduction from the original price. What was the original price of the dress?", "Solution_1": "[hide] If the dress was already $ 50\\%$ off the marked off price was $ 22(1+\\frac12)==33\\$$. Then, since it was reduced by $ 75\\%$, we have $ x(1-\\frac34)=33\\implies x=\\boxed{132\\$}$.[/hide]", "Solution_2": "[hide]Once again, my solution is quite similar to gaussintraining's. So the dress was already 50% off the marked off pricece was 33 dollars. Since it was reduced by 75%, we have x(1/4)=33 so the answer is 132 dollars.[/hide]" } { "Tag": [ "function", "search", "algebra proposed", "algebra" ], "Problem": "Suppose $ f : R^\\plus{}\\rightarrow R^\\plus{}$ is a decreasing continuous function such that for all $ x; y\\in R^\\plus{}$ , $ f(x \\plus{} y) \\plus{} f(f(x) \\plus{} f(y)) \\equal{} f(f(x \\plus{} f(y))) \\plus{} f(y \\plus{} f(x))$. Prove that $ f(f(x)) \\equal{} x$.", "Solution_1": "[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=371819146&t=137765[/url]" } { "Tag": [], "Problem": "Simplify $ (x\\minus{}a)(y\\minus{}b) \\equal{} ab$. Hence solve $ \\frac{2}{x} \\plus{} \\frac{3}{y} \\equal{} 1$ for integer values of $ x$ and $ y$.", "Solution_1": "this simplifies to a/x+b/y=1 and second results in same diophentine equation", "Solution_2": "[hide=\"Solution\"]\n$ (x\\minus{}a)(y\\minus{}b) \\equal{} ab \\Rightarrow ay\\plus{}bx\\equal{}xy$\n\n$ \\frac{2}{x}\\plus{}\\frac{3}{y}\\equal{} 1 \\Rightarrow 2y\\plus{}3x\\equal{}xy \\Rightarrow a\\equal{}2,b\\equal{}3$\n\n$ (x\\minus{}2)(y\\minus{}3)\\equal{}6\\equal{}6*1\\equal{}3*2 \\Rightarrow (a,b): (8,4),(3,9),(5,5),(4,6)$[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ a/b$ be a fraction with $ a,b$ naturals and $ b>\\equal{}2 gcd(a,b)\\equal{}1$ and $ gcd(b,3)\\equal{}1$ prove that there exists a fraction of the form $ n/3^k\\minus{}1$ equal to the first one", "Solution_1": "is that supposed to be $ \\frac{n}{3^k\\minus{}1}$?", "Solution_2": "$ k \\equal{} \\text{ord}_b(3), n \\equal{} \\frac {3^{\\text{ord}_b(3)} \\minus{} 1}{b} \\cdot a$ which is an integer by definition.", "Solution_3": "yeah, that's too easy a problem to be russian :), that's why i ask" } { "Tag": [ "probability", "expected value" ], "Problem": "A fair 6-sided die is rolled. If I roll $ n$, then I win $ n^2$ dollars. What is the expected value of my win? Express your answer as a dollar value rounded to the nearest cent.", "Solution_1": "$ \\frac{1}{6} \\times 1 \\plus{} \\frac{1}{6} \\times 4 \\plus{} \\frac{1}{6} \\times 9 \\plus{} \\frac{1}{6} \\times 16 \\plus{} \\frac{1}{6} \\times 25 \\plus{} \\frac{1}{6} \\times 36\\approx \\boxed{\\$15.17}$.", "Solution_2": "if only I could win n^2 dollars every time I rolled a die..." } { "Tag": [ "AMC" ], "Problem": "I HEARD THAT MCRAWFORD IS NOT AN ADMIN ANYMORE AND IS MOVING AWAY FROM AOPS!\r\n\r\nWHAT AM I GOING TO DO NOW?\r\n\r\nOH MY GOD! s\r\n\r\n[hide]As arnav would say... :P [/hide]", "Solution_1": "you're:\r\n\r\na) so not funny\r\n\r\nb) late\r\n\r\nc) hes still teaching woot\r\n\r\nd) hes got other things he wants to pursue, which is good for him, and we wish him good luck on his way.", "Solution_2": "Sorry, it's only funny when it's Arnav. :D", "Solution_3": "[quote=\"Pakman2012\"]you're:\n\na) so not funny\n\nb) late\n\nc) hes still teaching woot\n\nd) hes got other things he wants to pursue, which is good for him, and we wish him good luck on his way.[/quote]\r\n\r\n :| \r\n\r\nPart a)\r\n\r\nThat was not really the purpose of this post. I added the z at the end because post titles have to contain at least one lower case letter.\r\n\r\nPart b)\r\n\r\nI came to know of this only recently.\r\n\r\nPart c)\r\n\r\nI didn't know this\r\n\r\nPart d)\r\n\r\nI completely agree with.", "Solution_4": "your guys' arguments are so organized", "Solution_5": "[quote=\"Karth\"][quote=\"Pakman2012\"]you're:\n\na) so not funny\n\nb) late\n\nc) hes still teaching woot\n\nd) hes got other things he wants to pursue, which is good for him, and we wish him good luck on his way.[/quote]\n\n :| \n\nPart a)\n\nThat was not really the purpose of this post. I added the z at the end because post titles have to contain at least one lower case letter.\n\nPart b)\n\nI came to know of this only recently.\n\nPart c)\n\nI didn't know this\n\nPart d)\n\nI completely agree with.[/quote]\r\n\r\nkarth you're in woot aren't you : |", "Solution_6": "yeah Mr. Crawford taught the first class\r\n\r\nremember\r\n\r\nbuilding bridges...\r\n\r\n:ninja:", "Solution_7": "seriously guys\r\n\r\ncome on now\r\n\r\nlet's stay ON TOP OF THE BALL ok" } { "Tag": [], "Problem": "John attended a three-country multicultural festival. At the first exhibit, John paid a 3 dollar admission fee. He then spent half the money he had left on souvenirs. Finally, he spent 3 dollars on food and left. At each of the second and third exhibits, he again paid a 3 dollar admission fee, then spent half of the money he had left for souvenirs, then spent 3 dollars on food, and left. When he left the third exhibit, he had no money remaining. How many dollars did he spend at the festival?", "Solution_1": "[hide]admission=3\nfood=3\nsoveniers=half the money he had left\nmultiply all that by 3 to get what he had at the begining.\n3*3=9\n3*3=9\n\n18\nif theres 30 dollars originally, then the 18 for admission and soveniers then $12 left half that is 6 half that is 3 and then you leave with nothing. There for the answer is he spent $30 at the festival.\n[/hide]", "Solution_2": "[quote=\"MCrawford\"]John attended a three-country multicultural festival. At the first exhibit, John paid a 3 dollar admission fee. He then spent half the money he had left on souvenirs. Finally, he spent 3 dollars on food and left. At each of the second and third exhibits, he again paid a 3 dollar admission fee, then spent half of the money he had left for souvenirs, then spent 3 dollars on food, and left. When he left the third exhibit, he had no money remaining. How many dollars did he spend at the festival?[/quote]\r\n\r\n[hide=\"working backwards\"]\nAt the beginning of the third exhibit, he had 9 dollars.\nAt the beginning of the second exhibit, he had 27 dollars.\nAt the beginning of the first exhibit, he had [b]63 dollars[/b].\n[/hide]", "Solution_3": "[quote=\"MCrawford\"]John attended a three-country multicultural festival. At the first exhibit, John paid a 3 dollar admission fee. He then spent half the money he had left on souvenirs. Finally, he spent 3 dollars on food and left. At each of the second and third exhibits, he again paid a 3 dollar admission fee, then spent half of the money he had left for souvenirs, then spent 3 dollars on food, and left. When he left the third exhibit, he had no money remaining. How many dollars did he spend at the festival?[/quote]\r\n\r\n[hide=\"how i did it\"]3*2+3=9\n12*2+3=27\n30*2+3=[hide=\"the answer\"]63[/hide][/hide]" } { "Tag": [ "geometry", "cyclic quadrilateral", "geometry solved" ], "Problem": "Let $ABCD$ be a cyclic quadrilateral with circumcentre $O$. Let $AC$ and $BD$ meet at $X$, $AB$ and $CD$ meet at $Y$ and $BC$ and $AD$ meet at $Z$. Show that $OY$ and $XZ$ are perpendicular.", "Solution_1": "Read up something about pole and polar/reciprocation or inversion and I'm sure this problem will become a easy task for you.", "Solution_2": "[quote=\"Saul\"]Let $ABCD$ be a cyclic quadrilateral with circumcentre $O$. Let $AC$ and $BD$ meet at $X$, $AB$ and $CD$ meet at $Y$ and $BC$ and $AD$ meet at $Z$. Show that $OY$ and $XZ$ are perpendicular.[/quote]\r\n\r\nThis is just http://www.mathlinks.ro/Forum/viewtopic.php?t=5326 , applied to the cyclic quadrilateral ACBD instead of ABCD.\r\n\r\n darij" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b>0$ and $ x\\in [0,\\frac{1}{2}]$. Prove or disprove $ (a\\minus{}b)^2\\le 2(a^xb^{1\\minus{}x})^2(lna\\minus{}lnb)^2$.", "Solution_1": "The inequality does not hold when x=0. Simply choose x=0, a=2b, LHS = $ b^{2}$, RHS =$ 2b^{2}(ln2)^{2} < b^{2}$" } { "Tag": [], "Problem": "When writing all the integers from 1 to 2007 inclusive, how many 0's are used?", "Solution_1": "[hide]\n\nIn the ones place, a zero appears $ \\lfloor{\\frac{2007}{10}}= 200$ times\nIn the tens place, a zero appears 10 times in every hundred, or $ 190+8 = 198$\nIn the hundreds place, a zero appears 100 times in every thousand or $ 100+8 = 108$\n\nadd it up and you get $ 506$\n[/hide]", "Solution_2": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=894513#894513[/url]\r\n..?" } { "Tag": [], "Problem": "You given 25 different positive numbers. Prove that You can choose two of them such that none of the other numbers equals to the sum or to the difference of the chosen numbers.", "Solution_1": "[quote=\"Inspired By Nature\"]You given 25 different positive numbers. Prove that You can choose two of them such that none of the other numbers equals to the sum or to the difference of the chosen numbers.[/quote]\r\n\r\nWhat about choosing $1, 2, 3, ... , 24, 25$?", "Solution_2": "You can choose the 24,12 for example.", "Solution_3": "[quote=\"SplashD\"]You can choose the 24,12 for example.[/quote]\r\n\r\nOhh obvious. Thanks", "Solution_4": "you can always choose the greaters and none of the other will have the same sum.. \r\n\r\nsorry for my english :oops:", "Solution_5": "The question uses or, not and, so you can't just choose the greater ones", "Solution_6": "[quote=\"SplashD\"]The question uses or, not and, so you can't just choose the greater ones[/quote]\r\nexactly. :) Its a very nice problem.", "Solution_7": "Do you mean positive difference? \r\n\r\nYou can do 29 and 30, 30+29=59 and 29-30=-1 :huh:", "Solution_8": "[quote=\"ragnarok23\"]Do you mean positive difference? \n\nYou can do 29 and 30, 30+29=59 and 29-30=-1 :huh:[/quote]\r\n\r\nobviously because all of the integers are positive", "Solution_9": "Anyone wish to try? :P", "Solution_10": "i can see the delete button now, only for my latest post though. :huh:" } { "Tag": [], "Problem": "I am making a personality test about patience there are four possible choices and 9 questions.\r\n\r\nOne question gives 2 points\r\n\r\nOne gives 1\r\n\r\nOne gives 1\r\n\r\nOne gives 0\r\n\r\nOn every page except the last one, becuase I want to be able to tell if they lied. I am wondering how to make it so there are 4 possibilities\r\n\r\nLiar \r\n\r\nPatient\r\n\r\nImpatient\r\n\r\nVery Impatient\r\n\r\nThe test purposely lies to them at the end and says they wont get a score and I wasted their time, but they do. So if they answer this question incorrectly it means they lied. But I can't figure out how to do that becuase I suck at math lol.\r\n\r\nand the score goes from 1 to 100 points but I can't even figure out the proper range between score types..", "Solution_1": "I think you need to be more specific... I'm not quite sure what you are doing.\r\n\r\nAnyway, from what I can tell, you have 8 question which count from 0-2 each for a max of 16 points. To make the score out of 100, compute 100x/16 where x is the number of points earned.\r\n\r\nAs the lie detector, I don't know what you have in mind. Perhaps if their score lie in the range of the answer they choose, then the test should decide they lied.\r\n\r\nFor the score ranges, my best suggestion is just to experiment or set them arbitrarily - perhaps 0-25, 25-50, 50-75, and 75-100. But be sure to specify the exact boundaries (that is, include every possible score in one of the ranges), because it will be possible to get a score of 25, 50 or 75, and these cannot be between the categories - they have to belong to one or the other. I'd say just count it in the higher category or the lower for every time.\r\n\r\nNote that if you do the scoring as you described, there will only be 17 possible scores - not a very specific scale." } { "Tag": [], "Problem": "n cirles are given in the plane. They divide the plane into parts. Show that you can color the plane with two colors so that no colors with a common boundary line are covered the same way. Such a coloring is called a proper coloring", "Solution_1": "Induction. For n=0 this is obvious. Let we have a plane which is divided by n circles and was 2-colored by +1 and -1. Put n+1-th plane circle colored by +1 inside and by -1 outside and then multiply signs." } { "Tag": [ "geometry", "incenter", "trigonometry", "circumcircle", "parallelogram", "angle bisector", "geometry proposed" ], "Problem": "Let ABC be a rectangular triangle with 0$, there is $ \\delta > 0$ such that $ 0 < |x \\minus{} c| < \\delta$ implies $ |f(x) \\minus{} g(c)| < \\epsilon$. Also, choose any $ d$ such that $ 0 < |d \\minus{} c| < \\delta$. Taking limit $ x \\to d$, we have $ |g(d) \\minus{} g(c)| \\leq \\epsilon$. This shows that $ g$ is continuous.\r\n\r\nFor each $ n$, define $ A_n \\equal{} \\{ x \\in I \\, | \\, |f(x) \\minus{} g(x)| \\geq 1/n \\}$. Since $ \\bigcup_{n\\equal{}1}^{\\infty} A_n \\equal{} I$, some of $ A_n$ must be uncountable. Then $ A_n$ must have at least one limit point, so there is a convergent sequence $ (x_k)_{k\\equal{}1}^{\\infty} \\subset A_n$. Let $ x \\equal{} \\lim_{k\\to\\infty} x_k$. Then $ g(x) \\equal{} \\lim_{k\\to\\infty} f(x_k)$. But taking $ k \\to \\infty$ to the inequality $ |g(x_k) \\minus{} f(x_k)| \\geq 1/n$, we have $ 0 \\equal{} |g(x) \\minus{} g(x)| \\geq 1/n > 0$, a contradiction.\r\n\r\nTherefore, there is no such $ f$.\r\n\r\n\r\nSlightly modifying the proof, we find that if $ f : [a, b] \\to \\mathbb{R}$ has only removable discontinuity, then $ f$ has at most countable discontinuity." } { "Tag": [], "Problem": "we have a insulating ring of mass $ M$ and radius $ r$ and with very light spokes.The spokes join the outer region of the ring to an inner circle of radius $ a$ concentric with the the ring's circumference.we have a magnetic field $ \\vec B$ in this region(circle of radius $ a$ which contains no material) only perpendicular to the plane of the ring,a charge density $ \\lambda$ is given to the circumference of the ring.the magnetic field was instantaneously switched off.find the final angular speed of the ring", "Solution_1": "By Faraday's law, $ E*2 \\pi r = \\pi a^{2}B / t$.\r\n\r\nThus, $ F*t = E*q * t = E* 2 \\pi r * \\lambda * t = \\pi a^{2}B \\lambda$.\r\n\r\n$ \\Delta v = r \\Delta \\omega = \\frac{F*t}{M}$.\r\n\r\nTherefore, $ \\omega = \\frac{F*t}{M*r}= \\frac{\\pi a^{2}B \\lambda}{M r}$.", "Solution_2": "u r right :) \r\nwell the direction is of course $ -\\vec k$", "Solution_3": "If the magnetic field was instantaneously switched off, what is t ?", "Solution_4": "yes well that $ t$ should actually have been $ dt$.just like impulse", "Solution_5": "Faraday's Law : the electromotive force (in this case could be $ \\; E.2\\pi.r\\;$) is given by \r\n\r\n\\[ \\varepsilon\\,=\\;-{{\\partial \\Phi}\\over{\\partial t}\\;}\\] that in this case is infinity ?!\r\nAnd what about the self-inductance ? It's negligible?", "Solution_6": "instantaneously means fastly and not in an 'instant'", "Solution_7": "We need to know \" how fast\". That is; we need the law that govern the flux behavior.", "Solution_8": "we;; it's given by faraday's law", "Solution_9": "Sorry, but I think that in order to aplly Faraday's Law, you must know how the magnetic field varies with time ( see above), an d not the opposite", "Solution_10": "well here u know in a given short time $ \\Delta t$ the magnetic field was reduced to 0 and $ \\frac{d \\phi}{dt}\\approx \\frac{\\Delta \\phi}{\\Delta t}$" } { "Tag": [ "limit", "trigonometry", "function", "calculus", "calculus computations" ], "Problem": "find the limit:\r\nlim (sinx)/x as x goes to 0 if x in degrees", "Solution_1": "[quote=\"bgbgbgbg\"]find the limit:\nlim (sinx)/x as x goes to 0 if x in degrees[/quote]\r\nIsn't it $ 1$?", "Solution_2": "notice that x in degrees not in radian", "Solution_3": "[quote=\"bgbgbgbg\"]notice that x in degrees not in radian[/quote]\r\n\r\nMake the substitution u = x*2Pi/360. x tends to 0+ iff u tends to 0+. Replace all x's with u.", "Solution_4": "it doesnt even count much in hyperreals.", "Solution_5": "$ \\lim_{x \\to 0} \\frac {\\sin x} {x} \\equal{} 1$. It does not matter the unit of measure for $ x$, be it radians, degrees (or maybe $ \\textrm{Fahrenheit}^{\\circ}$ :) ).", "Solution_6": "it would makes sense if you could tell what is the meaning of a real number divided by a degree ?", "Solution_7": "Given the level of understanding this question was asked at, the last few posts really don't fit.\r\n\r\nThere is a perfectly sensible $ \\sin$ function defined on angles, which can be measured in degrees. Before calculus, degrees are as good as any angle measurement system, and cive us nice numbers for nice angles.\r\nSo, now we want to develop the calculus of trigonometric functions. The first crucial limit that comes up is this one: what is $ \\lim_{x\\to 0}\\frac{\\sin x}{x}$. Since we don't know any better, that angle is marked in degrees for now. What is the value of this limit? Some ugly number; if we look closer, it turns out to be $ \\frac{\\pi}{180}$.\r\nWe don't like that, so [i]we invent an entirely new angle measurement[/i], called the radian. Scaling the angle measurement changes the limit, and for the radian's arclength scaling it turns out to be $ 1$.\r\nThat is [b]the[/b] point of radians. From there on, we never use degrees in a calculus setting again." } { "Tag": [ "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "Let V and W be finite-dimensional vector spaces, and let T:V-->W be a linear transformation. Suppose that B is a basis for V. Prove that T is an isomorphism if and only if T(B) is a basis for W.\r\n\r\nPlease help!", "Solution_1": "It suffices to prove that isomorphism preserves linear independence. Let $A={\\alpha_{1}, \\dots, \\alpha_{n}}$ be a basis of V. Suppose that $T(A)=T(\\alpha_{1}), \\dots, T(\\alpha_{n})$ is linearly dependent, i.e. $a_{1}T(\\alpha_{1})+\\dots+a_{n}T(\\alpha_{n})=0$ for some nonzero $a_{1}, \\dots, a_{n}$. Then $0=a_{1}T(\\alpha_{1})+\\dots+a_{n}T(\\alpha_{n})=T(a_{1}\\alpha_{1}+\\dots+a_{n}\\alpha_{n})$ and since T is an isomorphism, $T(x)=0$ implies $x=0$ (kerT={0}) - a contradiction, since A is linearly independent.\r\n\r\nThe inverse implication is similar.", "Solution_2": "I am with you so far in that direction of the proof. I am still having trouble with the reverse direction. Would you mind elaborating please?\r\n\r\nThanks for your help!", "Solution_3": "Let $A=\\alpha_{1}, \\dots, \\alpha_{n}$ be a basis of V and $T(A)=\\beta_{1}, \\dots, \\beta_{n}$ $(\\beta_{i}= T(\\alpha_{i}))$ be a basis of W and suppose that there is some $\\alpha \\neq 0, \\alpha \\in kerT$. Let $\\alpha = a_{1}\\alpha_{1}+\\dots+a_{n}\\alpha_{n}$ for some $a_{1}, \\dots, a_{n}$. $T(\\alpha)=0 \\rightarrow T(a_{1}\\alpha_{1}+\\dots+a_{n}\\alpha_{n})=0 \\rightarrow a_{1}T(\\alpha_{1})+\\dots+a_{n}T(\\alpha_{n}) =0 \\rightarrow a_{1}\\beta_{1}+\\dots+a_{n}\\beta_{n}=0$ - a contradiction, since $\\beta_{1}, \\dots, \\beta_{n}$ were supposed to be linearly independent." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "inequalities", "geometry", "USAMTS", "algebra" ], "Problem": "With the USAMO's coming up in less than two weeks, and me never having had a proof formally graded, I was curious as to what USAMO proofs earned high scores, and which did not. \r\n\r\nIf anyone has links to a site with proofs given for particular problems, and their respective scores, I'd appreciate it a lot. (The USAMO invitation letter said that people taking the test often overestimate their scores; I'm erring on the side of caution to make sure I write clear, concise proofs.) If anyone's kind enough to write their own proofs for problems from previous USAMO's and the like and tell what the score was, I'd much appreciate that too. \r\n\r\nOn that note, I've got a few questions: \r\n1. What does it mean for a solution to be well-written? \r\n2. When are 7s awared? Are they awarded when the solution is completely correct? Do they have to be well-written? Can they be brute force? \r\n3. When and how is partial credit awarded? \r\n4. How is brute force graded? (E.g., massive casework, Muirhead, analytic bashing, and worse...)\r\n5. In geometry problems, is it mandatory for a diagram to be included? Preferred? \r\n6. Which facts can we use in the proof without citation? Which facts can we use with citation? How precise must the citations be? (I once recall solving a problem by using another USAMO solution as a solution; I clearly cannot recall this off the top of my head. Am I simply not allowed to cite this, or can I just say \"on a previous USAMO, [...]\"?)\r\n\r\nSorry if pieces of this post seem incoherent/jumbled up/out of order. I'm quite tired now. Thanks for any responses.", "Solution_1": "1. \"Well-written\" means \"easy to understand.\" I think a good way to explain this is that the grader shouldn't have to think too hard to understand your solution, and the grader shouldn't have to pull out a piece of paper to check any of your steps.\r\n\r\n2. 7's are given for any completely correct solution. Score doesn't depend on elegance at all, so an extreme brute-force solution will get a 7 if it is correct. You can sometimes get a 7 even with a small arithmetic mistake.\r\n\r\n3. In the documentary \"Hard Problems\" about the 2006 IMO team, the concept of \"grading up\" vs. \"grading down\" is discussed. If your solution is basically right but with a few minor errors, it will be graded down, meaning you start with a 7 and lose points for errors. In this case you'll probably end up with a 5-7 depending on the errors. If your solution has a big hole or is just a list of facts or ideas, it will be graded up, meaning you start with a 0 and get points for good stuff. In this case you'll probably get a 0-2 depending again on how useful the stuff you wrote actually is. On some problems it's possible to get a 3-4, but those scores are more unusual. I think the graders have a rubric that tells them how many points to award for each problem.\r\n\r\n4. It's graded the same way as any other solution. If the problem is fundamentally not brute-forceable, like if you tried to reduce a problem with infinitely many cases to finite casework, you'll get nothing. Many people have said that graders aren't as lenient with brute force solutions, but I'm not sure if this is really true.\r\n\r\n5. It's not mandatory as far as I know, but not including a diagram is a good way to get your grader annoyed from the start.\r\n\r\n6. This has been asked a million times. Don't cite a problem from another contest. You don't need to cite really obvious stuff like the associative property of addition, but you need to cite most things. Use common sense.\r\n\r\nIt's extremely true that people overestimate their scores. The first reason for this is because people don't understand the scoring system. If you have a half-complete solution, you don't get a 3 or 4, you get a 1. A significant hole in your solution doesn't cost you 1 point, it costs you 6. That's a big difference. The second reason is because people don't know what constitutes a rigorous proof. You need to consider every detail, and even if your proof ignores just one aspect of the problem or one small case, it's not complete. A lot of people will think they turned in a complete solution and end up losing anywhere from 1 to 6 points.\r\n\r\nIronically, the official USAMO solutions are generally not a good resource for the quality of solution that the USAMO requires. The reason for this is that the official USAMO solutions seem to often gloss over significant details (it's taken me 30 minutes to figure out a single line of an official solution), and that the official solutions will use a method that's easy to rigorize. Often the method you will come up with will not be so easy to rigorize and will require a much longer proof. My USAMO solutions average about 2-3 pages. The official USAMTS solutions might be a better resource for good solutions.", "Solution_2": "Can you cite any theorem that has a name?\r\n\r\n(I tried to post this message for 10 minutes but each time it said \"No post mode specified.\")", "Solution_3": "If you cite a theorem which trivializes the problem, don't expect points. However, as a general rule, if you can name it then you can cite it.", "Solution_4": "If a well known theorem kills the problem, I don't think they would put the problem on the test in the first place. But if that was the case, I think it is only fair that citing that theorem would earn you full credit, since it becomes the contest writers' fault for not putting up a good problem. I think that is what Richard Rusczyk said in the WOOT USAMO session.", "Solution_5": "[quote=\"serialk11r\"]But if that was the case, I think it is only fair that citing that theorem would earn you full credit, since it becomes the contest writers' fault for not putting up a good problem.[/quote]\r\nRegardless of whether it's fair or not, you're probably going to get hammered in practice. As an example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=875055#875055]this ridiculously easy IMO shortlist problem[/url] was on a MOP test once. During the test review session, the graders said that you could not cite that $ 2, 2^2, 2^3, \\ldots$ was periodic modulo $ k$ for any $ k$ because it trivialized the problem; you had to prove it yourself. (much groaning ensued) I would presume there is little difference between how MOP tests and USAMOs are graded.", "Solution_6": "[quote=\"MellowMelon\"][quote=\"serialk11r\"]But if that was the case, I think it is only fair that citing that theorem would earn you full credit, since it becomes the contest writers' fault for not putting up a good problem.[/quote]\nRegardless of whether it's fair or not, you're probably going to get hammered in practice. As an example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=875055#875055]this ridiculously easy IMO shortlist problem[/url] was on a MOP test once. During the test review session, the graders said that you could not cite that $ 2, 2^2, 2^3, \\ldots$ was periodic modulo $ k$ for any $ k$ because it trivialized the problem; you had to prove it yourself. (much groaning ensued) I would presume there is little difference between how MOP tests and USAMOs are graded.[/quote]\r\n\r\nIf you didn't prove the periodicity by yourself, would you still get 2/7 because the proof was correct otherwise?\r\n\r\nAnd to prove the $ {2^n}$ is periodic thing: note that each modulus, after being multiplied by two, has exactly one modulus that can result. Since there are a finite number of moduli (moduluses) that can result (to be exact, at most k-1 because 0 x 2 = 0), the sequence $ {2^n}$ is eventually periodic.\r\n\r\nHere's another problem you might like if you're practising to write proofs.\r\n\r\nProve that, for any odd integer $ k \\ge 3$, the sequence $ 2^0, 2^1, 2^2, 2^3, \\ldots (\\text{mod } k)$ has a period that is a factor of $ k \\minus{} 1$.", "Solution_7": "Um, no. the powers of 2 mod 9 go $ 2, 4, 8, 7, 5, 1$ with period 6. This is not a factor of 9. I think the correct statement, instead of integers, is primes: this problem is pwned by Fermat's little theorem. :)", "Solution_8": "[quote=\"sdkudrgn88\"][quote=\"MellowMelon\"][quote=\"serialk11r\"]But if that was the case, I think it is only fair that citing that theorem would earn you full credit, since it becomes the contest writers' fault for not putting up a good problem.[/quote]\nRegardless of whether it's fair or not, you're probably going to get hammered in practice. As an example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=875055#875055]this ridiculously easy IMO shortlist problem[/url] was on a MOP test once. During the test review session, the graders said that you could not cite that $ 2, 2^2, 2^3, \\ldots$ was periodic modulo $ k$ for any $ k$ because it trivialized the problem; you had to prove it yourself. (much groaning ensued) I would presume there is little difference between how MOP tests and USAMOs are graded.[/quote]\n\nIf you didn't prove the periodicity by yourself, would you still get 2/7 because the proof was correct otherwise?\n\nAnd to prove the $ {2^n}$ is periodic thing: note that each modulus, after being multiplied by two, has exactly one modulus that can result. Since there are a finite number of moduli (moduluses) that can result (to be exact, at most k-1 because 0 x 2 = 0), the sequence $ {2^n}$ is eventually periodic.\n\nHere's another problem you might like if you're practising to write proofs.\n\nProve that, for any odd integer $ k \\ge 3$, the sequence $ 2^0, 2^1, 2^2, 2^3, \\ldots (\\text{mod } k)$ has a period that is a factor of $ k \\minus{} 1$.[/quote]\r\n\r\nIts not k-1; its phi(k), which is k-1 if k is prime.\r\n\r\nIf we prove the theorem as a lemma though, would we get full points?", "Solution_9": "I think so, since that problem is pretty much the lemma.", "Solution_10": "You guys were going a bit off-topic, but ah well. \r\n\r\nWhat qualifies as \"a theorem with a name?\" Would, say, obscure theorems like Ravi's substitution work? \r\n\r\nTo all of you who responded to my question, thank you.\r\n\r\nEDIT: One more thing. How much work do we have to show? For instance, if we're expanding a polynomial expression (perhaps even an inequality =P), do we have to show the work for the expansion?", "Solution_11": "Show enough work that the grader can easily follow your solution without pencil and paper.", "Solution_12": "You can get a 7 without showing work, but it will piss off the grader, so you should be sure that it's right or else you run a huge risk.", "Solution_13": "[quote=\"Zhero\"]What qualifies as \"a theorem with a name?\" Would, say, obscure theorems like Ravi's substitution work?[/quote]\r\n\r\nIf this is what I think you're talking about, with something so simple you might as well prove it (you are speaking of the a=x+y,b=y+z,c=z+x thing with triangles, correct?). In this case, you would just be like, construct the incircle, and by equal tangents AE=AF=x, etc. or whatever. That takes two lines.\r\n\r\nAlso if you feel like you're spitting out the name of an obscure theorem, you should state it. That being said, make sure you actually have the statement correct. If you're using a wrong statement...you will probably not get more than negative zero.", "Solution_14": "I think a lot of people giving advice about USAMO grading are overstating the deductions that can result from small mistakes. If you really only made a calculation error, if your method is what they had in mind, I'm pretty sure you will still get full credit. If you missed an easy case, you're still pretty safe on getting at least a 6. At the same time, though, it is important to understand what constitutes a \"small mistake.\" Just because you think a case you missed looks easy to deal with doesn't mean it is. You can really only judge partial work in the context of a complete solution that relates to it. No matter how interesting and non-trivial your work is, it's only worth points if it can be used to solve the problem.", "Solution_15": "[quote]If you're using a wrong statement...you will probably not get more than negative zero.[/quote]\nWell, I suppose I could raise the bounds to positive zero to make me feel slightly better. \n\nAnd yeah, thanks for the rest of that post too. :P And yes, that was the Ravi's substitution I was referring to. \n\n[quote]You can get a 7 without showing work, but it will piss off the grader, so you should be sure that it's right or else you run a huge risk.[/quote]\r\nThanks. I'll keep that in mind. \r\n\r\nprobability 1.01, that makes me feel slightly better. =D", "Solution_16": "what about geometric transformations like dilation, inversion, etc? do you need to prove there propertises as lemmas (like what implies dilation and what dilation implies etc)...", "Solution_17": "Eh, I think those properties are sufficiently well-known, but in certain cases, you might want to remind the reader of certain facts (e.g., \"Since inversion maps circles to lines, [...]\".)\r\n\r\nOn that note, I've had some discrepancies when informed of the necessary amount of work. If, say, we are doing a Muirhead expansion, what would be optimal: \r\n\r\n1. We actually show our expanding it on the proof sheet. \r\n2. We just show the final expanded sum of symmetric sums. \r\n\r\nIn the case we do have to show the expansion, would you suppose that they would be knowledgeable of the technique I described in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=272496]this thread[/url] and just show the work described in that? Or should I describe that shortcut first? (I'm still not clear on how well-know it is eaxctly, so a response to that thread would be appreciated as well.)\r\n\r\nSimilarly, with an analytic bash, should we show the algebra? \r\n\r\nAlso, I take it that the official solutions are model solutions? In my opinion, they could be a bit clearer at certain areas, as they assume some facts that aren't ENTIRELY obvious (though still somewhat obvious), which I would feel slightly uncomfortable doing in my proofs...", "Solution_18": "I think you should show it, and NOT skip steps. I doubt graders would appreciate solutions by \"magic.\" (Also, it is probably a way for them to discourage brute force, because they don't want to read your long super-computational equations...)", "Solution_19": "Uhh yeah as I said, you can get a 7 without showing work. The example I have is during a MOP test. I did a muirhead bash and my solution went like this.\r\n\r\nWe want .\r\nBy clearing denominators, expanding, and collecting like terms, we get .\r\n\r\nYeah, Yi was pretty pissed at me for that, but I got 7/0.6!", "Solution_20": "MOP grading in that respect may be slightly more lenient though? I still think, to be safe: show all work.", "Solution_21": "[quote=\"not_trig\"]I think you should show it, and NOT skip steps. I doubt graders would appreciate solutions by \"magic.\" (Also, it is probably a way for them to discourage brute force, because they don't want to read your long super-computational equations...)[/quote]\r\n\r\nThen would the method in my topic be sufficiently well-known for that to suffice as the work shown for inequality expansion? Or would I have to give a rigorous explanation as to why that is true? \r\n\r\nAnd thanks, Hamster1800, but I'd rather err on the side of caution. Unless you mean to tell me that MOP grading is identical to USAMO grading, in which case I'd be... very pleased.", "Solution_22": "I heard the mop test had a style score between 0 and 1, which means the probability that you'd get the full solution score on a real competition. So that means Hamster1800's bash solution has 60% chance of earning a perfect 7 on the usamo, which is pretty risky.", "Solution_23": "Thank you, timwu. \r\n\r\nJust in case a bashable inequality shows up: Does anyone know exactly how well-known [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=272496]this technique[/url] is for expanding them? (Just tell if you've seen it before or not.) I do intend to show my work, and this would be far more efficient (and safe, for that matter) than just expanding it out. I just want to know if I have to justify this reasoning in this proof (since if it's sufficiently well-known, I don't.)\r\n\r\nAnd thanks to everyone for all the responses. It helped a lot. :lol:", "Solution_24": "[quote=\"worthawholebean\"]Show enough work that the grader can easily follow your solution without pencil and paper.[/quote]\r\n\r\nI'm just quoting this. \r\n\r\nUsing symmetric notation (correctly!) could probably simplify your technique, but I'm not very familiar with inequalities. I highly doubt usamo would put on such bashable problems anyways.", "Solution_25": "[quote=\"timwu\"][quote=\"worthawholebean\"]Show enough work that the grader can easily follow your solution without pencil and paper.[/quote]\n\nI'm just quoting this. \n\nUsing symmetric notation (correctly!) could probably simplify your technique, but I'm not very familiar with inequalities. I highly doubt usamo would put on such bashable problems anyways.[/quote]\r\n\r\nlol there was a certain #2 last year that elicited quite... a lot of bashing. Coordinates and trig lolol", "Solution_26": "Do we need to give proof before we use Pythagorean Theorem ?", "Solution_27": "I would assume that the Pythagorean theorem can be used without proof.", "Solution_28": "You can use theorems a lot more heavy than the Pythagorean Theorem without proof. Even stuff like Dirichlet's Theorem with no known elementary proof can be used. In general named theorems are okay, although a lot of caveats have already been given in the topic (e.g. write the statement of the theorem if it is obscure).", "Solution_29": "I still don't understand why Muirhead isn't allowed.", "Solution_30": "It's allowed.", "Solution_31": "[quote=\"worthawholebean\"]It's allowed.[/quote]\r\n\r\nI still think it is [i]preferable[/i] to write the AM-GMs though? I think all this stems from Mildorf's inequality packet which says something to the effect of \"graders frown upon Muirhead.\"", "Solution_32": "I don't understand why. Muirhead is a perfectly valid mathematical tool.", "Solution_33": "Do you have to prove these on the USAMO, or are they \"well-known\"?\r\n\r\n1. |x| + |1-x| [u]>[/u] 1 for all real x.\r\n\r\n2. The sum of the first k multiples of n is k(k+1)n/2.\r\n\r\n3. The sum of the first k squares is n(n+1)(2n+1)/6.\r\n\r\n4a. For a 2-D figure with fixed area, the one with the smallest perimeter is the circle.\r\n\r\nAnd its converse,\r\n\r\n 4b. For a 2-D figure with fixed perimeter the one with the max area is the circle.", "Solution_34": "1. A specific case of the triangle inequality, which states that $ |x|\\plus{}|y|\\ge |x\\plus{}y|$ for $ x,y\\in \\mathbb{C}$.\r\n\r\n2. Well-known, don't think it has a name.\r\n\r\n3. Well-known, don't think it has a name.\r\n\r\n4. Isoperimetric principle.", "Solution_35": "[quote=\"Temperal\"]\n\n4. Isoperimetric principle.[/quote]\r\nor calculus?", "Solution_36": "[quote=\"mathemonster\"][quote=\"Temperal\"]\n\n4. Isoperimetric principle.[/quote]\nor calculus?[/quote]\r\nuh, what? calculus of variations is what the bernoullis used to prove the general result, but the name of the theorem is the isoperimetric principle." } { "Tag": [ "USAMTS" ], "Problem": "i know that the round 1 due date is october 14th.\r\n\r\nbut when does the round 1 scores come out? is it october 14th?", "Solution_1": "It certainly takes some time for the graders to grade the papers, upload the grades, etc. However, due to the change in the USAMTS official solutions format, official solutions will be released earlier, so you'll be able to predict your own score sooner.", "Solution_2": "The Round 1 scores and comments will hopefully be released during the first week of November.\r\n\r\nWe are hoping to release both Round 1 solutions and the Round 2 problems within a week of the Round 1 submission deadline.", "Solution_3": "should i submit if i only got 3.5 problems right???", "Solution_4": "You may submit whatever you have.", "Solution_5": "are the official solutions 5 solutions?", "Solution_6": "[quote=\"Lawrence Wu\"]are the official solutions 5 solutions?[/quote]\r\nummm..yeah, just by common sense?", "Solution_7": "[quote=\"gauss1181\"][quote=\"Lawrence Wu\"]are the official solutions 5 solutions?[/quote]\nummm..yeah, just by common sense?[/quote]\r\n\r\ni think he means, are there going to be 1 solution for each problem, or will there be more than 1 solutionfor some problems.\r\n\r\nand i dont know the answer to this", "Solution_8": "varies by problem. There is always at least one solution; occasionally a second or third solution will be given too.", "Solution_9": "Most problems will have one solution. If there's a interesting second (or third) solution method we might present that too.\r\n\r\nFYI -- they should be posted later today (hopefully).", "Solution_10": "the solutions will be posted today?\r\n\r\nYAY!!!!", "Solution_11": "How are commended solutions decided?", "Solution_12": "Probably if the grader thinks its outstanding....or unique.", "Solution_13": "There are no more commended solutions anymore though, I believe.", "Solution_14": "[quote=\"USAMTS site\"]\n\nWe will no longer be issuing \"Commended\" scores. Some issues with doing this in the past is that there were no clear criteria for the graders to follow in determining which solutions would be commended, and also commended solutions were not \"worth anything\" in terms of the point total. Removing commended scores should make the grading process a bit more streamlined.\n[/quote]\r\nQED :D" } { "Tag": [ "number theory", "prime factorization" ], "Problem": "Compute the positive integer $x$ such that\r\n\\[4x^3 - 41x^2 + 10x = 1989.\\]", "Solution_1": "[hide]$1989 = 3^2*13*17$\n\nuse synthetic division to test the factors\n\n$13$ works[/hide]", "Solution_2": "Yep. This question shows the value of knowing the prime factorization of the current year. Does everyone know that the prime factorization of 2005 is $5 \\cdot 401$?" } { "Tag": [ "function", "calculus", "limit", "trigonometry" ], "Problem": "Im unsure of where this goes, so I just put it here.\r\n\r\nMy question is about a proof that he gives (chapter 8) for the intermediate value theorem, defined as following: If $f(x)$ is continuous on $[a,b]$ and $f(a) < 0 < f(b)$, then there is some $x$ in $[a,b]$ such that $f(x) = 0$.\r\n\r\n\"$A = ${$x: a \\le x \\le b$, and $f(x) < 0$ on the interval $[a,x]$}\r\n\r\nClearly $A \\neq \\emptyset$, since $a \\supset A$. In fact, there is some $\\epsilon > 0$ such that $A$ contains all points x satisfying $a \\le x < a + \\epsilon$ . Similarly, b is an upper bound for A and, in fact, there is an $\\epsilon >0$ such that all x satisfying $b - \\epsilon < x \\le b$ are upper bounds for A.\"\r\n\r\nThis is my first gripe. Consider a function that monotonically increases from $b - \\epsilon $ to $b$. Then the only upper bound for the set is b, not all of those other points from $b - \\epsilon $ to $b$, because in the set, there is a value larger than them and bounding.\r\n\r\nThe next part goes on to say: \r\n\r\n\"From these remarks, it follows that A has a least upper bound $\\alpha$ and that $a < \\alpha < b$. We now wish to show that $f(\\alpha) = 0$....\"\r\n\r\nHow can this be, though? How can the least upper bound always be between a and b and not necessarily be b? Plus, why does this least upper bound have to have value = 0? Why cant we then consider negative values to be least upper bounds? \r\n\r\nThe definition provided for least upper bounds at the beginning of the chapter was:\r\n\r\n\"A set A of realy numbers is bounded above if there is a number x such that $x \\ge a$ for ever a in A. Such a number x is called an upper bound for A.\"\r\n\r\nBy this definition, I dont see how his claims are founded.", "Solution_1": "[quote=\"jelyman\"]\n\nThis is my first gripe. Consider a function that monotonically increases from $b - \\epsilon $ to $b$. Then the only upper bound for the set is b, not all of those other points from $b - \\epsilon $ to $b$, because in the set, there is a value larger than them and bounding.[/quote]\r\n\r\nAbout your first question:\r\n\r\nPick any point x from $b - \\epsilon $ to $b$.\r\nIf x were not an upper bound to A that would mean\r\nthat f must be negative for all points between point $a$\r\nand point $x$. In particular, f(x) would be negative \r\nwhich contradicts the right continuity of f at b...\r\nthat is to say it would contradict the fact that f(x) is positive.\r\n\r\nIt follows then, that b can not be a least upper\r\nbound to A, because any point from $b - \\epsilon $ to $b$\r\nis an upper bound for the set A.", "Solution_2": "I'm going to move this to Other Problem Solving Topics. In general, calculus questions should be posted there (unless you think they belong in one of the College forums).", "Solution_3": "BUMP\r\n\r\n(someone pls answer it completely. I really need to understand this)", "Solution_4": "But..let me ask you something...\r\n\r\nDid you understand why the least upper bound of $A$ \r\nmust be strictly less than $b$?\r\n\r\nNotice that any value $x$ for which $f(x)>0$ \r\nmust be an upper bound of $A$. \r\nAnother way to look at it, would be:\r\nLet's say $f(x)>0$ for some $x>a$.\r\nThis means that $x$ can not belong to $A$,\r\nand since $x>a$ then $x$ must be an upper bound of $A$.\r\n(This is due to the fact that $a$ belongs to $A$.)\r\nNow, by the right continuity of $f$ at $b$ there must\r\nexist some interval $(b-\\epsilon,b]$ where\r\n$f$ is stricly positive. Thus, any point belonging \r\nto this interval will be an upper bound to $A$.\r\n\r\nThis shows that $\\alpha0$ and $f(\\alpha)<0$. \r\nBoth cases lead to contradiction so $f(\\alpha)=0$", "Solution_5": "Note that Spivak isn't claiming that $\\sup\\{x:a\\le x\\le b\\}$ isn't $b$. Instead he's claiming that $\\sup\\{x:a\\le x\\le b, f(x)<0\\}$ isn't $b$. I think that's what you're misunderstanding.", "Solution_6": "In fact, Simon, the set Spivak is taking the least upper bound of isn't the set where $f(x)<0$ but a more complicated (and smaller set): the set of all $x$ for which $f(y)<0\\amp\\forall y\\in [a,x]$.\r\n\r\nI don't particularly like this proof, partly because the description of the set $A$ is so complicated that it's too easy for students like jelyman to get tied into knots trying to cope with the definition. When I teach the course, I tend to rely on a different and more direct proof of the IVT: the successive bisection algorithm.\r\n\r\nAssume that we have been introduced to the completeness of the real numbers, and that we have already proceded from that completeness to the all-important Monotone Sequence Theorem: a monotone bounded sequence converges.\r\n\r\nThe plan: Let $a_0=a, b_0=b$. Set $c=(a_0+b_0)/2$. Evaluate $f(c)$. If $f(c)=0$ we are done. If $f(c)<0$ let $a_1 = a_0$ and $b_1=c$. If $f(c)>0$ let $a_1=c$ and $b_1=b_0$. Repeat this process recursively, at each step setting $c=(a_n+b_n)/2$ and evaluating $f(c)$. At every stage of the computation, unless the process terminates, $a_n0$\r\n\r\nIf the process doesn't terminate, we have $(a_n)$ as a bounded nondecreasing sequence and $(b_n)$ as a bounded nonincreasing sequence. By the Monotone Sequence Theorem, both sequences converge. Since $b_n-a_n = (b-a)/2^n \\to 0$ as $n\\to\\infty$, both sequences converge to the same number, which we'll call $x$.\r\nOn the one hand, $f(x)=f(\\lim a_n)=\\lim f(a_n) \\le 0$.\r\nOn the other hand, $f(x)=f(\\lim b_n)=\\lim f(b_n) \\ge 0$.\r\n\r\nIt's both a complete proof and a moderately useful numerical algorithm.\r\n\r\nThe essence of anything that is equivalent to the completeness of the real numbers (the existence of least upper bounds, the Monotone Sequence Theorem, the statement that Cauchy sequences converge, the Bolzano-Weierstrass Theorem) is this: if you have a process that ought to produce a number, then [b]there is[/b] such a number. This allows us to talk about real numbers that we have no conventional names for, there being more real numbers than such names. A good example of such a real number with no conventional name (since we are talking about the IVT): that number $x$ such that $\\cos x = x$.", "Solution_7": "[quote=\"Kent Merryfield\"]I don't particularly like this proof, partly because the description of the set $A$ is so complicated that it's too easy for students like jelyman to get tied into knots trying to cope with the definition. When I teach the course, I tend to rely on a different and more direct proof of the IVT: the successive bisection algorithm.[/quote]\r\n\r\nThe main virtue of Spivak's proof is that it only relies on\r\nthe completeness of the real numbers. The bisection method\r\nrequires many concepts about sequences, convergence, etc..\r\n\r\nThe bisection method is in fact easier to understand\r\nbut I think Spivak's proof is much more elementary\r\nsince it doesn't rely on any previous results (which\r\nwould require proofs on their own)\r\n\r\nThis is of course, only a matter of opinions.", "Solution_8": "Ok I think I understand it now. b can be an upper bound, but the point is that it cant be the least upper bound. Somehow I skipped over that when analysing it.", "Solution_9": "[quote=\"jelyman\"]b can be an upper bound, but the point is that it cant be the least upper bound. [/quote]\r\n\r\nExactly....You got it now!!!! :cool:" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Suppose [tex]x[/tex] is an algebraic number on the unit circle. Does this imply, that [tex]x[/tex] is a root of unity?", "Solution_1": "I believe there are counterexamples in the \"Algebra\" section posted by vess. I think there's a general construction there, and a particular counterexample. \r\n\r\nThe number is a root of unity if all its conjugates also have modulus $1$ though, if I remember correctly.", "Solution_2": "I think every nontrivial rational point on the unit circle will do:\r\nLet be $\\left( \\frac{a}{c} \\right)^2 + \\left( \\frac{b}{c} \\right)^2 = 1$ with integers $a,b,c$.\r\nThen assume there is some $n$ for that you have $\\left( \\frac{a}{c} + \\frac{b}{c} i \\right)^n = 1$ or $ (a+bi)^n = c^n$. By unique prime factorisation in gaussian integers (I think there is a simpler argument here, but I don't see it) you get $a+bi = i^k \\cdot c $ or equivalently $\\pm a=c, b=0$ or $a=0, \\pm b=c$", "Solution_3": "[quote=\"grobber\"]\nThe number is a root of unity if all its conjugates also have modulus $1$ though, if I remember correctly.[/quote]\r\n\r\nSorry, when I said this I was thinking about algebraic integers, not merely algebraic numbers.", "Solution_4": "$\\frac{1+2i}{\\sqrt 5}$" } { "Tag": [ "integration", "trigonometry", "function", "limit", "calculus", "calculus computations" ], "Problem": "suppose we have $ \\int_0^1\\frac{\\sin x}x\\,dx.$\r\n\r\nwhat does say to me that the integrand has limit as $ x\\to0$ ? can I guarantee that converges?\r\n\r\n----------\r\n\r\ni know a theorem that it says when having a bounded function on a compact interval and if it has a finite number of discontinuities is integrable.", "Solution_1": "$ \\lim_{x\\to 0}\\frac{\\sin x}{x}$ is one of those limits you should just [i]know[/i]. You need it for all of the calculus of trigonometric functions anyway.\r\nSo it's a removable discontinuity, and this is basically just the integral of a continuous function on a bounded interval. It can't be calculated exactly, so use numerical methods if you want something quantitative.", "Solution_2": "I do know that limit, and yes, you're recalling the theorem I said before, right?" } { "Tag": [], "Problem": "What is the least value of n such that every subset of {1, 2, ..., 100} with n elements has the property that we can choose 4 distinct elements a, b, c, d so that a+b=c+d ?", "Solution_1": "We will prove that for any subset of at least $ 14$ numbers, the condition holds :\r\nSupposed the subset is {$ a_1,a_2,\\cdots,{a}_{14}$} and the condition doesn't holds, then : ${ ({a}_{14}-{a}_{13})+(a}_{13}-{a}_{12})+\\cdots+({a}_{2}\r\n-{a}_{1})\\leq (1+2+3+\\cdots+14)=101$ contradiction\r\nWe will show that there exist a subset of less than $ 13$ numbers that are not statisfying :\r\n{$ 1,2,3,4,5,6,7,8,9,10,11,12,13$}.", "Solution_2": "It seems as if you had misuderstood the problem.", "Solution_3": "I think the answer is $ 16$ and I'm quite sure it was posted before..", "Solution_4": "I think 12. (No proof yet, stay tuned)", "Solution_5": "This is a proof for n is no greater than 16, but I'm not sure a set with 15 elements that doesn't fulfil the criteria can be constructed.\r\n\r\nSince $ a \\minus{} c \\equal{} d \\minus{} b$ we wish to add as many elements into our set as possible without repeating a difference between two elements. Note that an element cannot appear on both sides of this equation, therefore each element that we add is allowed to have a difference between itself and one element only the same as a diffference already included. Hence the first element $ k_1$ does not create any difference, $ k_2$ creates 1 difference, and any other $ k_m$ creates a further $ m \\minus{} 2$ differences. If we use the smallest possible difference, we have $ 1 \\plus{} 2 \\plus{} 3\\dots \\plus{} 13 \\equal{} 91$, Therefore the maximum number of different differences is 13. Therefore the maximum number of elements without having too many differences is that of $ k_{15}$, that is $ n \\equal{} 15$. I do not think that this optimum can be constructed though.", "Solution_6": "I think it is 21.", "Solution_7": "[quote=\"surshu\"]I think it is 21.[/quote]\r\n\r\nAny justification? Have you found a counterexample set for n=20?" } { "Tag": [ "function", "Divisibility Theory" ], "Problem": "Determine the highest power of $1980$ which divides \\[\\frac{(1980n)!}{(n!)^{1980}}.\\]", "Solution_1": "[quote=\"Peter\"][quote=\"baz\"]I think I can show that this tends to infinity. Does this sound correct to anyone else? See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153926\nfor more information. Thanks[/quote]It certainly tends to infinity. But I think the problem wants an explicit formula in function of $ n$ (and hopefully a more elegant one than the immediate form t0rajir0u gives). At least I think that, I don't have a solution myself, but from the statement I think that is what the problem authors intended.\n\nWelcome to this forum by the way. :)[/quote]", "Solution_2": "It definitely does not tend to infinity.\r\n\r\nThe largest number k such that 1980^k is a factor of ((1980n)!)/((n!)^1980) is no larger than the largest number k such that 11^k is a factor of ((1980n)!)/((n!)^1980). Denote this number by k(n)\r\n\r\nNow, let f(n)=(11n)! / (11^n) (n!). I claim that f(n) is an integer, but is not divisible by 11. For f(0)=1 and f(n+1)/f(n)=(11n+11)(11n+10)(11n+9)(11n+8)...(11n+1)/(11(n+1))=(11n+10)...(11n+1).\r\n\r\nThus the largest power of 11 that divides into (11n)!/n! is n.\r\n\r\nLet T(n)=((1980n)!)/((n!)^1980). Then T(11n)/T(n)=f(1980n)/(f(n)^1980). The right hand side is expressed as a fraction where neither top nor bottom is divisible by 11. Thus T(11n) and T(n) are divisible by exactly the same powers of 11. Thus k(11n)=k(n), so k(n) does not tend to infinity.", "Solution_3": "Interesting. Baz proved there is a subsequence $ (n_k)$ that converges to infinity, and you proved that not every sequence converges to infinity.\r\n\r\nBoth do, however, not answer the original question, do they?", "Solution_4": "[quote=\"Peter\"]Determine the highest power of $ 1980$ which divides\n\\[ \\frac {(1980n)!}{(n!)^{1980}}.\n\\]\n[/quote]\r\n\\[ \\frac {(1980n)!}{(n!)^{1980}}\\equal{}\\prod_{i\\equal{}1}^{1980}{in \\choose n}\r\n\\]\r\nTrying to extend this to something more useful but it just keeps going back into what t0r said", "Solution_5": "Let \\[ or(p)\\equal{}ord_p(\\frac{(1980n)!}{(n!)^{1980}}\\equal{}\\sum_{k\\equal{}1}^{\\infty}([\\frac{1980n}{p^k}]\\minus{}1980[\\frac{n}{p^k}]).\\]\r\nBecause $ 1980\\equal{}2^2*3^2*5*11$ we get highest power is $ x(n)\\equal{}min([\\frac{or(2)}{2}],[\\frac{or(3)}{2}],or(5),or(11)).$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c$ be positive real numbers and $abc=1$. Prove that the inequality\r\n\\[\\frac{1}{1+a^{2}}+\\frac{1}{1+b^{2}}+\\frac{1}{1+c^{2}}\\ge \\frac{1}{1+a}+\\frac{1}{1+b}+\\frac{1}{1+c}\\]\r\n[b]can not be true.[/b]", "Solution_1": "What do u mean? Just give an counter example? :maybe: \r\nCounterexample: just substitute x,y,z with 0.5, 0.5 and 4", "Solution_2": "Don't check any detailed examples? Could you prove that it is false?", "Solution_3": "Sorry, I didn't understand and I don't think you can because for some a,b,c the inequality holds whereas for the others it fails" } { "Tag": [ "AMC", "USA(J)MO", "USAJMO", "ARML", "geometry", "geometric transformation", "dilation" ], "Problem": "I am curious, what are your opinions of the USAJMO. As for me, I am pretty skeptical currently, but am withholding judgement for a bit. Please discuss here (although I don't expect much serious discussion until the ARML people get back).\r\n\r\nEDIT: If a mod could fix the poll I would appreciate it.", "Solution_1": "I agree, the USAJMO will simply dilute the overall quality of the participants associated with USAJMO/USAMO, as top underclassmen will continue to attempt to qualify for USAMO by taking the AMC 12, whereas, the remaining qualifiers for USAJMO will be significantly less talented. Furthermore, this will adversely affect upperclassmen wishing to secure USAMO qualification.", "Solution_2": "I believe if you look at the results and statistics, you will notice that the number of upperclassmen (11-12) vs underclassmen (10 and below) is about what the split for the qualification for the new USAMO/USAJMO. Therefore there will be little to no dilution, nor disadvantage to anyone. You will also notice that a great majority of the younger students score almost nothing on the USAMO making the possibility of scoring positively on the USAJMO a better use of their time and skills.", "Solution_3": "(probably one of the first of said 'ARML people'; I can't sleep)\r\n\r\nAbout the MOP selection process I have no comment.\r\n\r\nA bit troubled by the 270 qualifiers, though. There were 60 more than that this current year, only counting those that came from AMC12, and my score qualified by a margin of 'only' ~30. Getting rid of the 60 might reduce that to around 20-15, and bringing in AMC10 people narrows it even further.\r\n\r\nI also have a feeling that this is a more dire concern for a lot of people who perform much better than I ever could on USAMO.", "Solution_4": "I don't like the idea of USAJMO at all for multiple reasons. I feel like they're pretty much saying that people in 10th grade and below do not score high on USAMO and this discourages them so we should give them an easier test so they don't get discouraged. Well i personally find failure more as a bigger incentive for me to study then I find from success. When i came into seventh grade i couldn't even divide fractions but i made the first cut for the mathcounts team (8/30 was cutoff and i got 9/30) and when i went to the meetings i was like the dumbest person there. Since i was the dumbest person there i worked hard so i could get better and not be the dumbest person there whereas no one else on the team really worked that hard because they were already the top kids in the school (or maybe they had some other reason i don't know about). Due to my hard work the teacher let me past the next cutoff even though my score wasn't too high (10/30 I think it was). However i continued to work hard and i became the best in my middle school at the end of seventh grade. Now in eighth grade I have gotten a perfect on amc 10b a 4 on AIME (should've been 6) and went to various other math competitions. You can see how my performance in math has developed in the past two years. I think that if i was already the best in seventh grade when i first walked in i would not have the motivation to continue to work as hard as i did and so i wouldn't have improved as much. The moral of the story is that failure should motivate students more than success and if it doesn't motivate them then do you really think that being successful will? When you fail you should be thinking that i should work harder to improve my score but when you succeed are you always thinking that?", "Solution_5": "I completely agree with peregrinefalcon, but the smaller number of seats for USAMO might encourage some of the younger students to work even harder. (At least that's what it's making me do.)", "Solution_6": "W8 what is USAJMO. Is it a typo error for USAMO?", "Solution_7": "You should try to...read the stickies, and the many other threads around the AMC forum related to USAJMO?\r\n\r\nSee [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279691]here[/url].", "Solution_8": "I read that whole thread before i made a post so i could make an educated opinion.", "Solution_9": "[quote=\"peregrinefalcon88\"]I read that whole thread before i made a post so i could make an educated opinion.[/quote]\r\nI dislike the USAJMO idea because it makes everything more complicated." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let a,b,c be positive numbers such that a+b+c=ab+bc+ca. Prove that: a+b+c+abc \\ge 4", "Solution_1": "[quote=\"spring_sotana\"]Let a,b,c be positive numbers such that a+b+c=ab+bc+ca. Prove that: a+b+c+abc \\ge 4[/quote]\r\n$a+b+c+abc\\geq4\\Leftrightarrow$\r\n$\\Leftrightarrow(a+b+c)^2(ab+ac+bc)^2+abc(a+b+c)^3\\geq4(ab+ac+bc)^3\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(a^4b^2+b^4a^2-2a^3b^3+3a^4bc-a^3b^2c-a^3c^2b-a^2b^2c^2)\\geq0.$\r\nThe last inequality is obviously true.", "Solution_2": "Why the last one is true?", "Solution_3": "[quote=\"jin\"]Why the last one is true?[/quote]\r\nI haven't check the last step into the last one, but if it was true, you can see from the highest power.\r\n$a^4b^2$ (or $b^4a^2$)can beat all negative term, while $a^4bc$ can beat all negative but $a^3b^3$", "Solution_4": "[quote=\"jin\"]Why the last one is true?[/quote]\r\n$\\sum_{cyc}(a^4b^2+b^4a^2-2a^3b^3+3a^4bc-a^3b^2c-a^3c^2b-a^2b^2c^2)=$\r\n$=\\sum_{cyc}(a^4b^2-2a^3b^3+b^4a^2)+\\sum_{cyc}(2a^4bc-a^3b^2c-a^3c^2b)+\\sum_{cyc}(a^4bc-a^2b^2c^2)=$\r\n$=\\sum_{cyc}a^2b^2(a-b)^2+abc\\sum_{cyc}(a-b)^2(a+b)+\\frac{1}{2}\\cdot abc(a+b+c)\\sum_{cyc}(a-b)^2.$", "Solution_5": "Oh,I see,thank you. I don't care $\\sum(2a^3-a^2b-a^2c)=\\sum(a^3+b^3-a^2b-b^2a)$", "Solution_6": "[quote=spring_sotana]Let $a,b,c$ be positive numbers such that $a+b+c=ab+bc+ca$. Prove that: $$a+b+c+abc \\ge 4$$[/quote]\n[b]Strengthening[/b]\nLet $a,b,c$ be positive numbers such that $a+b+c=ab+bc+ca$. Prove that: $$a+b+c+\\frac{3abc}{a+b+c} \\ge 4$$" } { "Tag": [], "Problem": "are they supposed to send the scores to your coach, or do you not get them?", "Solution_1": "In my state (MA), the coach gets the scores during the ceremony... but it's different for every state. Your coach should get the scores, however, within one month.", "Solution_2": "Some states get their scores at the competition, but some states (like mine) don't give you scores, and just mail them to your coaches. Also, in some states, you have an option to protest and they give you the answers.", "Solution_3": "yeah we got to do that. i wanted to write one about something i cant discuss b/c it would give away diffuculty" } { "Tag": [ "Mafia", "0s", "Game 7" ], "Problem": "chesspro [color=green]confirmed[/color]\nMathWhIz2004 [color=green]confirmed[/color]\n-Terminator- [color=green]confirmed[/color]\nWindSlicer [color=green]confirmed[/color]\nGoBraves [color=green]confirmed[/color]\nMithsApprentice [color=green]confirmed[/color]\njmadsen [color=green]confirmed[/color]\nricebelly \nSudonym \nLynelleYe [color=green]confirmed[/color]\nLucky707 [color=green]confirmed[/color]\nbubala [color=green]confirmed[/color]\nChigr [color=green]confirmed[/color]\nh_s_potter2002 [color=green]confirmed[/color]\nSly Si [color=green]confirmed[/color]\npaladin8 [color=green]confirmed[/color]\n\n16 people, 4 mafia, 2 inspector, 1 doctor, 9 citizens. The night will begin. The PM's will be sent out in about 5-10 minutes, so hold tight. Choose your victim, Mafia!\n\nI forgot to say... the game won't start until i get the responses.", "Solution_1": "Ooh, hello everybody! \r\n\r\nFinally the time has come. :)", "Solution_2": "Ok. I've gotten my PM and I'm ready.", "Solution_3": "I'm here!", "Solution_4": "im alive, unless sum1 didnt tell me i already got lynched", "Solution_5": "lol ... we'll try not to lynch you (first)", "Solution_6": "[quote=\"Chigr\"]im alive, unless sum1 didnt tell me i already got lynched[/quote]\r\n\r\nwell the game officially doesnt start until i get everyone's response back, telling me that they will be playing. When that is all done, i will simply post, \"the game begins.\" Since the night comes first, though, i dunno what will happen.", "Solution_7": "I'm here and ready to go!", "Solution_8": "I'm here... hi everyone :)", "Solution_9": "I think they can choose a person to save during the night. And if the mafia happens to choose that exact same person - the person doesn't die but stays alive, therefore preventing the mafia from succeeding in a killing.", "Solution_10": "Im Here", "Solution_11": "I'm here.", "Solution_12": "Sodonym and MithsApprentice are still yet to confirm (i think) and where is our host?", "Solution_13": "er yeah...I'm good. Okay let's go whenever we're ready.", "Solution_14": "uglyman621 hasnt replied either....this is taking way too long..can we just start?", "Solution_15": "hmm..... well let's make this short. chesspro is dead. \r\n\r\nplease accuse someone.", "Solution_16": "Darn... I died. At least it was fast. \r\n\r\nI know 3 of the 4 mafia...\r\n\r\n[EDIT]: I'm a citizen, btw.", "Solution_17": "[quote=\"chesspro\"]I know 3 of the 4 mafia...[/quote]\r\nErr...don't we all?\r\nSo we should get either mathwhiz or jmadsen. I don't know who's more likely but if one's innocent the other is probably mafia. We have a fair amount of room for error at the moment which is good.", "Solution_18": "I [color=red]accuse sly si[/color]. He's definitely mafia. Unless LynelleYe is. I'm inspector. I know Chigr, mathwhiz, and jmadsen are all innocent. Therefore either sly si or lynelleye is mafia. But I think it's more likely to be sly si.", "Solution_19": "i [color=red]second sly si[/color] and if its not him then we can deal with that tomorow", "Solution_20": "Sly Si on trial!!!\r\n\r\nChigr [color=green]aye[/color]\r\nchesspro\r\nlynelleye\r\nsly si\r\nmathwhiz\r\njmadsen \r\npaladin8 [color=green]aye[/color]\r\nmithsapprentice \r\n\r\nneed three more votes to kill!", "Solution_21": "Well, I vote NAY, naturally. I'm an ordinary citizen, which means either paladin isn't inspector, or LynnelleYe is mafia.\r\n\r\nIs the other inspector still around to confirm or refute paladin's statement? Or was chesspro the other inspector?\r\n\r\nAnyway, I hate to leave in the middle of my trial, but I leave town in about half an hour. So this will be my last post for this game. If you don't kill me, I'll need to be replaced, as I will have no internet access for the next week.\r\n\r\nBye, and good luck!", "Solution_22": "[quote=\"xxreddevilzxx\"]Sly Si on trial!!!\n\nChigr [color=green]aye[/color]\nchesspro\nlynelleye\nsly si\nmathwhiz\njmadsen \npaladin8 [color=green]aye[/color]\nmithsapprentice \n\nneed three more votes to kill![/quote]\r\n\r\nLol, reddevil. I'm dead already. I can't vote.\r\n\r\nSly Si, I am (was?) a citizen.\r\n\r\nLynelleYe, I know everyone knows 3 of the 4 mafia. I just wanted to say that for no reason.", "Solution_23": "As the other inspector, I confirm that what paladin8 said is true. I vote [color=red]aye[/color] on sly si.", "Solution_24": "[quote=\"Sly Si\"]Well, I vote NAY, naturally. I'm an ordinary citizen, which means either paladin isn't inspector, or LynnelleYe is mafia.\n\nIs the other inspector still around to confirm or refute paladin's statement? Or was chesspro the other inspector?\n\nAnyway, I hate to leave in the middle of my trial, but I leave town in about half an hour. So this will be my last post for this game. If you don't kill me, I'll need to be replaced, as I will have no internet access for the next week.\n\nBye, and good luck![/quote]\r\n\r\nWell we aren't SURE you are mafia; there's just a high chance. Anyway, later man.", "Solution_25": "Actually, since chesspro was dead, that will do......and the citizens win!!!!!!!", "Solution_26": "Good job you guys.\r\n\r\nGlad I could help out by saving a few of you :)", "Solution_27": "Awesome! Now to wait for game 9 to start...", "Solution_28": "Nice job, citizens.", "Solution_29": "last post" } { "Tag": [ "integration", "calculus", "logarithms", "calculus computations" ], "Problem": "F(x)=$\\int \\frac{dx}{1+x^{4}}$\r\n find F(x)", "Solution_1": "1. Factor the denominator: $x^{4}+1=(x^{2}+\\sqrt{2}x+1)(x^{2}-\\sqrt{2}x+1).$\r\n\r\n2. Use partial fractions.\r\n\r\n3. Be very patient, and have a substantial quantity of scratch paper available.", "Solution_2": "And I think we should consider a generalization of this problem:\r\n\r\nFind all $\\alpha \\in \\mathbb{R}$ such that the following integration can be solved:\r\n\r\n$\\int \\frac{dx}{1+x^{\\alpha}}$", "Solution_3": "$F(x) = \\frac{1}{2}\\int \\frac{\\left( x^{2}+1 \\right)-\\left( x^{2}-1 \\right)}{x^{4}+1}\\ dx = \\frac{1}{2}\\int \\left( \\frac{1+1/x^{2}}{x^{2}+1/x^{2}}-\\frac{1-1/x^{2}}{x^{2}+1/x^{2}}\\right) \\ dx =$\r\n\r\n$= \\frac{1}{2}\\int \\frac{d(x-1/x)}{(x-1/x)^{2}+2}-\\frac{1}{2}\\int \\frac{d(x+1/x)}{(x+1/x)^{2}-2}$\r\n\r\n$= \\frac{1}{2 \\sqrt 2}\\arctan \\frac{x-1/x}{\\sqrt 2}-\\frac{1}{4 \\sqrt 2}\\ln \\left| \\frac{(x+1/x)-\\sqrt 2}{(x+1/x)+\\sqrt 2}\\right|+C=$\r\n\r\n$= \\frac{1}{2 \\sqrt 2}\\arctan \\frac{x^{2}-1}{x \\sqrt 2}-\\frac{1}{4 \\sqrt 2}\\left| \\frac{x^{2}+x \\sqrt 2+1}{x^{2}+x \\sqrt 2-1}\\right|+C$", "Solution_4": "[quote=\"marshell\"]Find all $\\alpha \\in \\mathbb{R}$ such that the following integration can be solved:\n$\\int \\frac{dx}{1+x^{\\alpha}}$[/quote]\r\nIs $x\\,\\,{}_{2}F_{1}(1/\\alpha,1,1+1/\\alpha,-x^{\\alpha})$ acceptable?", "Solution_5": "[quote=\"greengrass\"]$F(x) = \\frac{1}{2}\\int \\frac{\\left( x^{2}+1 \\right)-\\left( x^{2}-1 \\right)}{x^{4}+1}\\ dx = \\frac{1}{2}\\int \\left( \\frac{1+1/x^{2}}{x^{2}+1/x^{2}}-\\frac{1-1/x^{2}}{x^{2}+1/x^{2}}\\right) \\ dx =$\n\n$= \\frac{1}{2}\\int \\frac{d(x-1/x)}{(x-1/x)^{2}+2}-\\frac{1}{2}\\int \\frac{d(x+1/x)}{(x+1/x)^{2}+2}$\n\n$= \\frac{1}{2 \\sqrt 2}\\arctan \\frac{x-1/x}{\\sqrt 2}-\\frac{1}{4 \\sqrt 2}\\ln \\left| \\frac{(x+1/x)-\\sqrt 2}{(x+1/x)+\\sqrt 2}\\right|+C=$\n\n$= \\frac{1}{2 \\sqrt 2}\\arctan \\frac{x^{2}-1}{x \\sqrt 2}-\\frac{1}{4 \\sqrt 2}\\left| \\frac{x^{2}+x \\sqrt 2+1}{x^{2}+x \\sqrt 2-1}\\right|+C$[/quote]\r\n$= \\frac{1}{2}\\int \\frac{d(x-1/x)}{(x-1/x)^{2}+2}-\\frac{1}{2}\\int$ [b]$\\frac{d(x+1/x)}{(x+1/x)^{2}-2}$[/b]", "Solution_6": "[quote=\"tranthanhnam\"][quote=\"greengrass\"]$F(x) = \\frac{1}{2}\\int \\frac{\\left( x^{2}+1 \\right)-\\left( x^{2}-1 \\right)}{x^{4}+1}\\ dx = \\frac{1}{2}\\int \\left( \\frac{1+1/x^{2}}{x^{2}+1/x^{2}}-\\frac{1-1/x^{2}}{x^{2}+1/x^{2}}\\right) \\ dx =$\n\n$= \\frac{1}{2}\\int \\frac{d(x-1/x)}{(x-1/x)^{2}+2}-\\frac{1}{2}\\int \\frac{d(x+1/x)}{(x+1/x)^{2}+2}$\n\n$= \\frac{1}{2 \\sqrt 2}\\arctan \\frac{x-1/x}{\\sqrt 2}-\\frac{1}{4 \\sqrt 2}\\ln \\left| \\frac{(x+1/x)-\\sqrt 2}{(x+1/x)+\\sqrt 2}\\right|+C=$\n\n$= \\frac{1}{2 \\sqrt 2}\\arctan \\frac{x^{2}-1}{x \\sqrt 2}-\\frac{1}{4 \\sqrt 2}\\left| \\frac{x^{2}+x \\sqrt 2+1}{x^{2}+x \\sqrt 2-1}\\right|+C$[/quote]\n$= \\frac{1}{2}\\int \\frac{d(x-1/x)}{(x-1/x)^{2}+2}-\\frac{1}{2}\\int$ [b]$\\frac{d(x+1/x)}{(x+1/x)^{2}-2}$[/b][/quote]\r\n\r\nOh yes, thank you. I changed it already" } { "Tag": [ "MATHCOUNTS", "modular arithmetic", "number theory", "least common multiple" ], "Problem": "I was doing Mathcounts Chapter Sprint 2002, and could understand how to solve all questions but this one.\r\n\r\n28. The length of a year on planet Mars is exactly 697 days. If Mars has a calendar with a 12-day week, and year 0 begins on the first day of the week, what is the next year that will begin on the first day of the week?\r\n\r\nThanks in advance.", "Solution_1": "[hide]Since $ 697\\equiv 1 \\pmod {12}$, the first day of every year will be the day after the day that the previous year started on.\nSo, the pattern is like this.\nYear 0-Day 1\nYear 1-Day 2\nYear 2-Day 3\netc.\nSo, the year that you will get the 12th day of the week first is Year 11. The next year, Year 12, will be the year where the first day of the week is the first day of the year, so the answer is Year 12. [/hide]", "Solution_2": "I believe you need to find the lcm. Let $ n$ be a non-negative integer. Then $ 697n$ represents the number of days that have passed after $ n$ years. In order for a year to start on the first day of the week, some multiple of 12 days must have passed. So find the lcm, which is $ 8364$. This gives a value of $ n\\equal{}12$, which would mean 12 years.", "Solution_3": "697/12= blah remainder 1\r\nEvery year the day the year starts off with will be moved forward one. Since there are 12 days in a week, that takes 12 years until you can come back on the same day. The year 12 would be the answer" } { "Tag": [ "geometry", "rectangle", "integration", "vector", "calculus", "function", "LaTeX" ], "Problem": "If $ S_1$ is a rectangle with area 1 and $ S_2$ is a rectangle with area 2, then \r\n\r\n$ 2 \\int\\limits_{S_1} \\vec{F} \\cdot d\\vec{A} \\equal{} \\int\\limits_{S_2} \\vec{F} \\cdot d\\vec{A}$\r\n\r\nwhy is this not true?", "Solution_1": "The orientation of the rectangles in the vector field matters.\r\n\r\nThe flux through a rectangle is defined as the integral over the region in question of the DOT PRODUCT of F and dA.\r\n\r\nYou're using a normal product, and so you miss the orientation aspect.", "Solution_2": "What does the \"normal\" product of two vector-valued functions even mean? I think it's reasonably clear that the OP meant dot products and just didn't know how to $ \\text{\\LaTeX}$ them.\r\n\r\nAnyway, I don't understand the question. If $ \\vec{F}$ is nonconstant then there's no reason the two integrals should be related at all.", "Solution_3": "[quote=\"t0rajir0u\"]What does the \"normal\" product of two vector-valued functions even mean? I think it's reasonably clear that the OP meant dot products and just didn't know how to $ \\text{\\LaTeX}$ them.\n\nAnyway, I don't understand the question. If $ \\vec{F}$ is nonconstant then there's no reason the two integrals should be related at all.[/quote]\r\n\r\nI've modified the original question so it clearly says a dot product, hope it makes more sense" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "\\/closed" ], "Problem": "Just like Volumes 1 and 2 prepare students for like AMCs, AIMEs, is there a book that prepares students for USAMO? Is this book the Olympiad-equivalent of Volumes 1 and 2?\r\n\r\nhttp://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=5003\r\n\r\nI'm not sure if I can take WOOT, and if not, I'd like to have a kinda textbook-like book.", "Solution_1": "Also, if anybody has this book, could you give some examples of how this book is like?", "Solution_2": "There is no such book. Mainly because it would be almost impossible to make one. The topics for USAMO/IMO are so widely diversificated that not even a collection of books would be enough to thoroughly cover them. The best you can do is check the Olympiad forums and the resources section (contests) for problems and try to solve them.", "Solution_3": "Problem solving strategies by Engel is close though" } { "Tag": [], "Problem": "A $ 12^{\\prime \\prime}$-diameter pizza and a $ 16^{\\prime \\prime}$-diameter pizza are each cut into eight congruent slices. Jane ate three slices of the $ 12^{\\prime \\prime}$ pizza. Mark ate three slices of the $ 16^{\\prime \\prime}$ pizza. How many more square inches of pizza than Jane did Mark eat? Express your answer as a common fraction in terms of $ \\pi$.", "Solution_1": "So in this problem, we want to find $ \\frac{3}{8} \\cdot 256\\pi \\minus{} \\frac{3}{8} \\cdot 144\\pi$\r\n\r\nWe can easily use the distributive property:\r\n\r\n$ \\frac{3\\pi}{8}(256 \\minus{} 144) \\equal{} \\frac{3\\pi}{8}(112) \\equal{} \\boxed{42\\pi}$.", "Solution_2": "Actually, it says diameters of 12 and 16, so your answer is off by a factor of 4. The real answer should be 10.5pi.", "Solution_3": "No, it shouldn't be $ \\frac{21}{2}\\pi$; $ 42\\pi$ is correct. Why do you need to divide by 4 anyways?", "Solution_4": "Because the question gave diameters, not radii. Therefore, the radii, of each pizza are $ 6$ and $ 8$, respectively. Hence,\r\n\r\n$ \\frac{3\\pi}{8}(8^2\\minus{}6^2)\\equal{}\\frac{3\\pi}{8}(28)\\equal{}\\boxed{10.5\\pi}$.", "Solution_5": "the alcumus correct answer is 21pi/2\n", "Solution_6": "Alcumus solution:They both ate $\\frac{3}{8}$ of a pizza. Therefore, the quantity that Mark ate in excess of Jane is simply $\\frac{3}{8} \\times$ the difference in total area of the pizzas. The 16'' pizza has $64\\pi$ area, and the 12'' pizza has $36\\pi$ area, making for a difference of $28\\pi$. $\\frac{3}{8} \\times 28\\pi = \\boxed{\\frac{21}{2}\\pi}$" } { "Tag": [], "Problem": "The Enigma Is Simple :wink: :\r\n\r\nSome one said: \r\n[quote=\"X\"]\nThere are [b][u][color=red]3[/color][/u][/b] kinds of people : people who [b]know how to count[/b] and people who [b]does not[/b] .\n[/quote]\r\nWhy did he said that there are 3 :gathering: kinds, and he just listed 2 :ddr:", "Solution_1": "There are two kinds of people: Those who like to group people into two different groups, and those that do not.", "Solution_2": "[quote=\"randomdragoon\"]There are two kinds of people: Those who like to group people into two different groups, and those that do not.[/quote]\r\nIt's not the solution :D \r\nThe solution is veeeeery simple :D", "Solution_3": "obvios: He cant count", "Solution_4": "[quote=\"mz94\"]obvios: He cant count[/quote]\r\n\r\nobvious: you have already seen my enigma somewhere :wink:", "Solution_5": "There are $ 10$ types of people, those who can count in binary and those who can't.\r\n\r\n$ 4$ out of $ 3$ people can't do fractions.", "Solution_6": "[quote=\"BanishedTraitor\"]There are $ 10$ types of people, those who can count in binary and those who can't.\n\n$ 4$ out of $ 3$ people can't do fractions.[/quote]\r\nmz94's answer is right :wink:", "Solution_7": "Um,\r\n(1) They weren't posting solutions, they were posting things like the one you said.\r\n(2) Your engima is obvious because it references the reason that he can't count in the enigma.\r\n\r\nStop trying to turn bumper stickers into AoPS threads.", "Solution_8": "[quote=\"miyomiyo\"]Um,\n(1) They weren't posting solutions, they were posting things like the one you said.\n(2) Your engima is obvious because it references the reason that he can't count in the enigma.\n\nStop trying to turn bumper stickers into AoPS threads.[/quote]\n\n[quote=\"i_like_pie\"]Quoted for truth.[/quote]\r\n\r\n\r\n\r\nthough thisi s G&FF, so I suppose a certain amount of leniency should be exercised.", "Solution_9": "Lol, I remember when my older brother told me the binary one when I was in like 1st grade. I'm all like, what, but that's only 2 people!", "Solution_10": "[quote=\"FOURRIER\"]The Enigma Is Simple :wink: :\n\nSome one said: \n[quote=\"X\"]\nThere are [b][u][color=red]3[/color][/u][/b] kinds of people : people who [b]know how to count[/b] and people who [b]does not[/b] .\n[/quote]\nWhy did he said that there are 3 :gathering: kinds, and he just listed 2 :ddr:[/quote]\r\n\r\nHe used bad grammer: is should have been\r\n\r\n\"There are [b][u][color=red]3[/color][/u][/b] kinds of people : people who [b]know how to count[/b] and the person who [b]does not[/b] .\"\r\n\r\nIt's obvious from there.", "Solution_11": "Btw, who's rating all our posts one?", "Solution_12": "[quote=\"1=2\"]He used bad grammer[/quote]\r\n\r\nHe should have said \"There are 3 kinds of people: those who know how to count and those who don't.\"", "Solution_13": "[quote=\"Nerd_of_the_Ages\"]Btw, who's rating all our posts one?[/quote]\r\n\r\nIt doesn't really matter here, at least. :)", "Solution_14": "[quote=\"FOURRIER\"]The Enigma Is Simple :wink: :\n\nSome one said: \n[quote=\"X\"]\nThere are [b][u][color=red]3[/color][/u][/b] kinds of people : people who [b]know how to count[/b] and people who [b][u]does[/u] not[/b] .\n[/quote]\nWhy did he [u]said[/u] that there are 3 :gathering: kinds, and [u]he just[/u] list[u]ed[/u] 2 :ddr:[/quote]\r\n\r\n1. He cant count\r\n2. I underlined the places with spelling/grammer mistakes\r\n3. \"There are 10 kinds of people in the world, the ones who understand binary and the ones who use base 9\" is better", "Solution_15": "I believe grammar is spelled \"grammar\"." } { "Tag": [ "geometry", "trigonometry", "inequalities open", "inequalities" ], "Problem": "Let $ \\triangle ABC$ be a triangle with area $ \\frac{1}{2}$ in which $ a$ is the length of the side opposite to vertex $ A$. Prove that: \r\n\r\n$ a^{2}\\plus{}\\csc (A) \\geq \\sqrt {5}$", "Solution_1": "[quote=\"delegat\"]Let $ \\triangle ABC$ be a triangle with area $ \\frac {1}{2}$ in which $ a$ is the length of the side opposite to vertex $ A$. Prove that: \n\n$ a^{2} \\plus{} \\csc (A) \\geq \\sqrt {5}$[/quote]\r\ndo you mean $ cos(A)$ :?: :|", "Solution_2": "No $ \\csc (A) \\equal{} \\frac{1}{\\sin (A)}$", "Solution_3": "[quote=\"delegat\"]Let $ \\triangle ABC$ be a triangle with area $ \\frac {1}{2}$ in which $ a$ is the length of the side opposite to vertex $ A$. Prove that: \n\n$ a^{2} \\plus{} \\csc (A) \\geq \\sqrt {5}$[/quote]\r\n\r\nsorry i am late i had some problemes with this old pC :oops: \r\n\r\nwe know that $ S(ABC) \\equal{} \\frac {1}{2}bc*sin(A) \\equal{} \\frac {1}{2}$ so $ bc \\equal{} csc(A)$\r\n\r\nlet prouve :$ a^2 \\plus{} bc > \\sqrt(5)$\r\n\r\nusing some triangle formulas $ a^2 \\equal{} b^2 \\plus{} c^2 \\minus{} 2bc*cos(A)$ \r\n\r\nand $ cos(A) \\equal{} \\sqrt(1 \\minus{} (sin(A)^2))$ so $ a^2 \\equal{} b^2 \\plus{} c^2 \\minus{} 2bc*(\\sqrt(1 \\minus{} \\frac {1}{(bc)^2}))$\r\n\r\n$ a^2 \\plus{} bc \\equal{} b^2 \\plus{} c^2 \\minus{} 2\\sqrt((bc)^2 \\minus{} 1) \\plus{} bc$\r\n\r\nand we know tha $ b^2 \\plus{} c^2 > 2bc$ so $ a^2 \\plus{} bc > 3bc \\minus{} 2\\sqrt((bc)^2 \\minus{} 1)$\r\n\r\nthe inq <=> $ 3bc \\minus{} 2\\sqrt((bc)^2 \\minus{} 1) > \\sqrt(5)$ (NB:$ bc > 1$)\r\n\r\nsupposing $ bc \\equal{} x$ $ f(x) \\equal{} 3x \\minus{} 2\\sqrt(x^2 \\minus{} 1) \\minus{} \\sqrt(5)$\r\n\r\n$ f(x) > 0$ <=> $ 5(x^2) \\minus{} 6\\sqrt(5)x \\plus{} 9 \\equal{} (\\sqrt(5)x \\minus{} 3)^2 >0$ \r\n\r\nso $ a^2 \\plus{} bc > \\sqrt(5)$ that include $ a^{2} \\plus{} \\csc (A) \\geq \\sqrt {5}$\r\n\r\nso what do you think :wink:", "Solution_4": "[quote=\"delegat\"]Let $ \\triangle ABC$ be a triangle with area $ \\frac {1}{2}$ in which $ a$ is the length of the side opposite to vertex $ A$. Prove that: \n\n$ a^{2} + \\csc (A) \\geq \\sqrt {5}$[/quote]\r\n$ h_a = 1/a$;\r\n1. if $ h_a \\le a/2$ $ ( a \\ge \\sqrt {2} )$ then : $ \\frac {1}{sin(A)} \\ge \\frac {1}{Sin({\\pi/2})} = 1$ , so for $ a \\ge \\sqrt {2}$ : $ a^{2} + \\csc (A) \\geq 3$;\r\n2. if $ h_a \\ge a/2$ $ ( a \\le \\sqrt {2} )$ then : $ \\frac {1}{sin(A)} \\ge \\frac {a^2}{4} + \\frac {1}{a^2}$ , so for $ a \\le \\sqrt {2}$ : $ a^{2} + \\csc (A) \\geq a^2 + \\frac {a^2}{4} + \\frac {1}{a^2} = \\frac {5}{4}a^2 + \\frac {1}{a^2} \\ge \\sqrt {5}$;\r\nequality is for $ \\triangle$ with sides : $ a = \\sqrt [4]{\\frac {4}{5}}, b = c = \\sqrt {\\frac {3\\sqrt {5}}{5}}$; :)", "Solution_5": "as \"a\" is dimensional and csc(A) is dimensionless the proposition is incorrect", "Solution_6": "[quote=\"castigioni\"]as \"a\" is dimensional and csc(A) is dimensionless the proposition is incorrect[/quote]\r\n\r\nthe proposition is incorrecrt !!!!!!!!! :D so show me something wrong in my solution :evilgrin:", "Solution_7": "Thank you very much for these nice proofs Problem is of course correct and I do not understand this comment posted by [b]castigioni[/b]" } { "Tag": [], "Problem": "Fie a,b,c reale pozitive...sa se demonstreze ineq:\r\n\r\n$\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{9}{(a+b+c)}\\geq 4 (\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a})$", "Solution_1": "Se aplica inegalitatea lui Popoviciu, \\[f(a)+f(b)+f(c)+3 f \\left( \\frac{a+b+c}3 \\right) \\geq 2 f \\left( \\frac{b+c}2 \\right)+2 f \\left( \\frac{c+a}2 \\right)+2 \\left( \\frac{a+b}2 \\right) ,\\] pentru functia convexa $f(x) = \\frac1{x}$.", "Solution_2": "aaa...mijto teorema :| eu o facusem altfel mai \"muncitoreste\" :) dar mijto asta...:)", "Solution_3": "Putem presupune ca avem ordonarea $\\ a\\geq\\ b\\geq\\ c$ sau $\\ c\\geq\\ a\\geq\\ b$.Se stiu inegalitatile $\\frac{1}{a}+\\frac{1}{b}\\geq\\frac{4}{a+b}$ si $\\frac{1}{c}+\\frac{9}{a+b+c}\\geq\\frac{4}{b+c}+\\frac{4}{c+a}$, ultima dintre ele fiind echivalenta cu $\\ 2c((c-a)(c-b)-a^{2}+2ab-b^{2})+(a+b)(c-a)(c-b)\\geq\\ 0$, care e adevarata in ipotezele neristrictive de mai sus.Dar e mai usor cu Popoviciu...", "Solution_4": "[quote=\"stancioiu sorin\"]$\\frac{1}{c}+\\frac{9}{a+b+c}\\geq\\frac{4}{b+c}+\\frac{4}{c+a}$, ultima dintre ele fiind echivalenta cu\n\n$\\mathbf{2c((c-a)(c-b)-a^{2}+2ab-b^{2})+(a+b)(c-a)(c-b) \\geq 0}$,\n\ncare e adevarata in ipotezele neristrictive de mai sus[/quote]\r\n\r\nPoti sa explici mai bine de ce e adevarata inegalitatea in [b]bold[/b]?", "Solution_5": "Cred ca a uitat si el! Desi solutia cu Popoviciu este la indemana iata si o alta solutie!\r\n\r\n Problema.\r\n\r\n Fie $a, b, c$ trei numere reale strict pozitive. Aratati ca \\[\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{9}{a+b+c}\\geq 4\\left(\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a}\\right).\\] Solutie.\r\n\r\n Inmultind cu $a+b+c$ inegalitatea devine: \\[\\frac{a+b+c}{a}+\\frac{a+b+c}{b}+\\frac{a+b+c}{c}+9\\geq 4\\left(\\frac{a+b+c}{a+b}+\\frac{a+b+c}{b+c}+\\frac{a+b+c}{c+a}\\right)\\Leftrightarrow\\] \\[\\frac{b+c}{a}+\\frac{c+a}{b}+\\frac{a+b}{c}+12\\geq 12+4\\left(\\frac{a}{b+c}+\\frac{b}{c+a}+\\frac{c}{a+b}\\right)\\Leftrightarrow\\] \\[\\frac{b+c}{a}+\\frac{c+a}{b}+\\frac{a+b}{c}\\geq4\\left(\\frac{a}{b+c}+\\frac{b}{c+a}+\\frac{c}{a+b}\\right).\\] Aasta ultima e usurica rau avand in vedere ca iese imediat din \\[\\frac{1}{x}+\\frac{1}{y}\\geq\\frac{4}{x+y}.\\]", "Solution_6": "Foarte frumoasa solutia ta Cezar :) . Felicitari!", "Solution_7": "[quote=\"Cezar Lupu\"]\n\n\\[ \\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {a \\plus{} b}{c}\\geq4\\left(\\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{c \\plus{} a} \\plus{} \\frac {c}{a \\plus{} b}\\right).\n\\]\nAasta ultima e usurica rau avand in vedere ca iese imediat din\n\\[ \\frac {1}{x} \\plus{} \\frac {1}{y}\\geq\\frac {4}{x \\plus{} y}.\n\\]\n[/quote]\r\n\r\nBuna !\r\n\r\neu am inlocuit $ x\\equal{}\\frac{a}{b}$ si $ y\\equal{}\\frac{a}{c}$ si imi da ca $ \\frac{b\\plus{}c}{a}\\geq\\frac{4bc}{ab\\plus{}ac}$ :oops:", "Solution_8": "Let $ a,b,c>0$ and $ a\\plus{}b\\plus{}c\\equal{}3$ .Prove that\n$$\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a}+\\frac{2}{9}(bc+ca+ab)\\geq\\frac{13}{6} .$$" } { "Tag": [], "Problem": "\u0388\u03c3\u03c4\u03c9 $T$ \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c4\u03c9\u03bd \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03c9\u03bd \u03b5\u03bd\u03cc\u03c2 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5. \u0398\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $f: T\\rightarrow(0,+\\infty)$ \u03bc\u03b5 \u03c4\u03cd\u03c0\u03bf \r\n$f(ABC)=min\\left(\\frac{b}{a},\\frac{c}{b}\\right),$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf ABC \u03bc\u03b5 $a\\leq b\\leq c.$ \u039d\u03b1 \u03b2\u03c1\u03b5\u03af\u03c4\u03b5 \u03c4\u03bf \u03c0\u03b5\u03b4\u03af\u03bf \u03c4\u03b9\u03bc\u03ce\u03bd \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7\u03c2 \u03b1\u03c5\u03c4\u03ae\u03c2.", "Solution_1": "\u039b\u03cd\u03c3\u03b7:\r\n\r\n \u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 min(\u03b2/\u03b1 , \u03b3/\u03b2)>=1.\r\n\r\n \u0391\u03c0\u03cc \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c3\u03c5\u03bd\u03b7\u03bc\u03b9\u03c4\u03cc\u03bd\u03c9\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b1^2 = \u03b2^2 + \u03b3^2 -2\u03b2\u03b3cosA =>\r\n \r\n (\u03b1/\u03b2)^2 = 1 + (\u03b3/\u03b2)^2 - 2(\u03b3/\u03b2)cosA. \u03bc\u03b5 \u03b3/\u03b2>=1.\r\n\r\n \u0398\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03bf \u03bc\u03ad\u03b3\u03b9\u03c3\u03c4\u03bf \u03c4\u03b7\u03c2 min(y,x) \u03cc\u03c0\u03bf\u03c5 y=\u03b2/\u03b1, x=\u03b3/\u03b2.\r\n =>\r\n (1/y)^2 = 1 + x^2 - 2xcosA. \u0397 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 g(x)=1 + x^2 - 2xcosA (\u03b3\u03b9\u03b1 \u03bf\u03c0\u03bf\u03b9\u03bf\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u0391,\u03c4\u03bf \u0391 \u03b4\u03b5\u03bd \u03b5\u03be\u03b1\u03c1\u03c4\u03ac\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf\r\n x) \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03bd.\u03b1\u03cd\u03be\u03bf\u03c5\u03c3\u03b1 (x>=1 \u03ba\u03b1\u03b9 cosA<=1)).E\u03c0\u03af\u03c3\u03b7\u03c2 \u03b7 (1/y)^2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03bd.\u03c6\u03b8\u03af\u03bd\u03bf\u03c5\u03c3\u03b1. \u0386\u03c1\u03b1 \u03c4\u03bf \u03bc\u03ad\u03b3\u03b9\u03c3\u03c4\u03bf \u03c4\u03b7\u03c2 min(x,y)\r\n \u03b5\u03c0\u03b9\u03c4\u03c5\u03b3\u03c7\u03ac\u03bd\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c4\u03b1\u03bd x=y , \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae 1= x^2 + x^4 - 2(x^3)cosA. => cosA = (x^4 + x^2 - 1) /(2(x^3)) = f(x) \u03b5\u03af\u03bd\u03b1\u03b9\r\n\u03b3\u03bd. \u03b1\u03cd\u03be\u03bf\u03c5\u03c3\u03b1 \u03b3\u03b9\u03b1 x>=1 (\u03c0\u03c1\u03ce\u03c4\u03b7 \u03c0\u03b1\u03c1\u03ac\u03b3\u03c9\u03b3\u03bf\u03c2 \u03b8\u03b5\u03c4\u03b9\u03ba\u03ae) \u03ba\u03b1\u03b9 f(x)<1 , \u03bb\u03cd\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd x^4 + x^2 -1 - 2x^3 = 0 =>\r\n \r\n (x^2 - x - 1)*(x^2 - x + 1)=0 => x0=(1+sqrt(5))/2 . \u0386\u03c1\u03b1 \u03c4\u03bf \u03c0\u03b5\u03b4\u03af\u03bf \u03c4\u03b9\u03bc\u03ce\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf [1,x0) . (E\u03af\u03bd\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03bf \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03cc\u03c4\u03b9 \u03b8\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03b9\u03c2 \u03b5\u03bd\u03b4\u03b9\u03ac\u03bc\u03b5\u03c3\u03b5\u03c2 \u03c4\u03b9\u03bc\u03ad\u03c2 \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b7 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c7\u03b5\u03af\u03c2...)" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "since $Z \\triangleleft Q$, then quotient group $Q/Z$ contains cosets $q+Z$ for $q \\in Q$,if i have $(p+Z)(q+Z)=(pq)+Z$ for $p,q \\in Q$ how would i show that this is not well defined, do i just show that $(p+Z)(q+Z)\\neq(pq)+Z$ ?", "Solution_1": "well-defined means that for different ways of writing the same cosets... the resulting product should come out to the same coset either way...\r\n\r\nhowever, $(1/2+\\mathbb{Z})(1/3+\\mathbb{Z}) = 1/6+\\mathbb{Z}$, whereas $(3/2+\\mathbb{Z})(4/3+\\mathbb{Z}) = 2+\\mathbb{Z} = \\mathbb{Z}$\r\n\r\n(the coset $1/2+\\mathbb{Z}$ is the same as the coset $3/2+\\mathbb{Z}$ and the coset $1/3+\\mathbb{Z}$ is the same as the coset $4/3+\\mathbb{Z}$)", "Solution_2": "Of course, $\\mathbb{Q}/\\mathbb{Z}$ makes perfect sense as an additive structure.\r\n\r\nIt doesn't work under multiplication because $\\mathbb{Z}$ is merely an additive subgroup and not an ideal." } { "Tag": [ "geometry", "ratio", "similar triangles" ], "Problem": "compute the area of the largest square inscribed in a triangle which has sides $5,6,7$\r\n[with proof]", "Solution_1": "WOw this problem is so computational heavy so Im not going to bother solving all those radicals.\r\n\r\nThe basic idea is to find the height and then use similar triangles (use the triangle above the square to the whole triangle)\r\nThe ratio should be\r\n\r\n$\\frac{h}{s}=\\frac{h-d}{d}$\r\n\r\nwhere the height you can find (h)\r\nthe side length (s) is 5,6,or 7 (you have to decide which gives biggest area) :wink: \r\nand d is the side length of square .\r\n\r\nyou only hvae 1 variable (which is d) but the height is a messy radical so im not really going to do it..too lazy\r\n\r\nEDIT: huh what do you mean prove (referring to bottom post)...isnt prove basically doing computational work since this isnt really a proof?", "Solution_2": "[quote=\"da ban man\"]The basic idea is to find the height and then use similar triangles (use the triangle above the square to the whole triangle)[/quote]\r\nWell, the main problem (once you see which side will give the largest square of the three) is to [b]prove[/b] that it's the largest square possible (if it is). :)", "Solution_3": "as he has got the idea i dont think that its hard to prove.... :wink: \r\n[this can be done by showing that no area can be larger than this if two vertices of the square isnt on the base]", "Solution_4": "wait im confused..how cna you have an inscribed square without two of its vertices touching the base?", "Solution_5": "oh i forgot to say...when u use the idea of similar triange u'll find that all the 4 vertices are not on the sides of the triangle so u need to enlarge that and use the idea of limit....and a bit more computation.", "Solution_6": "[quote=\"Moonmathpi496\"]oh i forgot to say...when u use the idea of similar triange u'll find that all the 4 vertices are not on the sides of the triangle so u need to enlarge that and use the idea of limit....and a bit more computation.[/quote]\r\n\r\nim kinda confused here...i thought it was inscribed so all 4 vertices must be on the sides of the triangle...?", "Solution_7": "Here u were asked to find the largest area so not necessary that all the vertices are on the sides of the triangle...\r\nand note that:here inscribed means any square inside the triangle....[probably i mistook to write]\r\n\r\nand about ur last post...if u check u'll find that all the vertices are not on the sides but u have to make the square which has 4 vertices on the sides if u use the fomula u mentioned\r\n\r\n[quote]The basic idea is to find the height and then use similar triangles (use the triangle above the square to the whole triangle)\nThe ratio should be\n\n\\frac{h}{s}=\\frac{h-d}{d}\n\nwhere the height you can find (h) [/quote]", "Solution_8": "[quote=\"Moonmathpi496\"]\n\nand about ur last post...if u check u'll find that all the vertices are not on the sides but u have to make the square which has 4 vertices on the sides if u use the fomula u mentioned\n\n[/quote]\r\n\r\nyes tahts correct but I only mentioned that post because I thought inscribed means have to be touching the sides (which my way would work)..but I guess I was mistaken (line inscribed circle..)" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "a,b,c>0\r\nprove that:\r\n(a^2 - ab + b^2)^0.5 + (b^2 - bc + c^2)^0.5 >= (c^2 + ca + a^2)^0.5", "Solution_1": "you can see that :\r\n http://www.mathlinks.ro/Forum/viewtopic.php?t=78481", "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?t=21405\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=78481\r\n\r\n dg", "Solution_3": "..........", "Solution_4": "b is not appearing on the RHS, so we can write :\r\nLHS>(a^2)^(1/2)+(c^2)^(1/2)=a+c>(a^2+ac+c^2)^(1/2)\r\nwe just put that b=0 and that's it :D" } { "Tag": [ "geometry", "puzzles" ], "Problem": "What comes next in the following sequence:\r\n\r\n15,23,18,6,9,24...\r\n\r\nSee, I can do it too!! :P \r\n\r\n[hide=\"Hint\"]No math required. At all.[/hide]", "Solution_1": "[hide=\"My answer\"]19, 9, ... [/hide]", "Solution_2": "That's correct!", "Solution_3": "Can people who do not live in your area understand the sequence?\r\n\r\nAnd how about...\r\n\r\n31, 41, 59, 26, 53, 58, 97, .... ? :D", "Solution_4": "My sequence is very simple. I can't believe only one person got it." } { "Tag": [ "function", "algebra", "domain", "real analysis", "real analysis solved" ], "Problem": "I found this [url=http://www.math.columbia.edu/~zare/prob_all.html]here[/url]. Check out problem $22$. I think it can also be found in Yaglom, generalized a little bit (if I'm not mistaken, the problem there concerns continuous curves, not necessarily functions, but I can't remember the exact statement).\r\n\r\nIf $g(x)=g(x+c)$ for some $x$, call $c$ a chord of $g$. Which real numbers are chords of all continuous functions $f$ with domain [$0,1$] such that $f(0)=f(1)$?", "Solution_1": "Levi's theorem states that only the reciprocals of positive integers are solutions. This seems to be a classical problem given at oral de ENS and X.", "Solution_2": "Wow! I had no idea there was a theorem stating this. Indeed, these are the solutions. Once you realize this, a proof shouldn't be too difficult to find.", "Solution_3": "Indeed, the proof is not that difficult. Anyway, I do not find it absolutely trivial! ;)", "Solution_4": "harazi, could you please give some more explanation either link concerning Levi's theorem-i haven't heard about this one" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "a) let $1\\le a<2$ be a real number.\r\nProve that if $f: R^+\\to R^+$ is a function satisfying the conditions:\r\ni) $2f(x)\\le af(\\frac{x}{a})+x \\forall x>0$.\r\nii) $f(x)<2005 \\forall x<\\frac{1}{2005}$.\r\nthen $f(x)\\le x \\forall x>0$\r\nb) let $a>1$ prove that exists a function $f: R^+\\to R^+$ satisfying condition (i) and $f(x)>x \\forall x>0$.", "Solution_1": "I think the problem is nice. No one interested in proving it ? :?:", "Solution_2": "I think the problem is nice. No one is interested in proving it ? :?:", "Solution_3": "yes,it is nice. :) \r\nfor part a) if there exist $x_0$ such that $f(x_0) >x_0$\r\nfirst we can prove $a>1$\r\nbecause if $a=1$ then we have $f(x_0) \\geq 2f(x_0) -x_0$ with i) ,contradiction.\r\nlet $f(x_0) -x_0=b,b>0$\r\nwe have $f(\\frac{x_0}{a}- \\frac{x_0}{a}\\ge \\frac{2b}{a}$\r\n\r\n$f(\\frac{x_0}{a^2} - \\frac{x_0}{a^2}\\ge \\frac{4b}{a}$\r\nso ....\r\n$f(\\frac{x_0}{a^n} - \\frac{x_0}{a^2}\\ge \\frac{2^n b}{n a}$\r\nso when $x \\to 0$,$f(x) \\to \\infty$,contradiction with ii).\r\nfor part b)\r\nlet $f(x)=\\frac{1}{x^n}$ n such that $a^n >2$", "Solution_4": "[quote=\"zhaobin\"]yes,it is nice. :) \nfor part b)\nlet $f(x)=\\frac{1}{x^n}$ n such that $a^n >2$[/quote]\r\n\r\nI think you are wrong because $\\frac{1}{x^n}1$", "Solution_5": "[quote=\"Hong Quy\"][quote=\"zhaobin\"]yes,it is nice. :) \nfor part b)\nlet $f(x)=\\frac{1}{x^n}$ n such that $a^n >2$[/quote]\n\nI think you are wrong because $\\frac{1}{x^n}1$[/quote]\r\nwhy, I can't see...\r\nbut I think it is right ;) \r\nwith $f(x)=\\frac{1}{x^n}$ then $af(\\frac{x}{a})=\\frac{a^{n+1}}{x^n}$\r\nand $2f(x)=\\frac{2}{x^n}$\r\nand we can easily see $2f(x)\\le af(\\frac{x}{a})+x \\forall x>0$" } { "Tag": [ "trigonometry", "induction" ], "Problem": "Let $ a, b \\in R$ and $ \\sin{\\frac{a}{2}} \\neq 0$.\r\n\r\nFind $ S\\equal{}\\cos{b}\\plus{}\\cos{(a\\plus{}b)}\\plus{}\\cos{(2a\\plus{}b)}\\plus{}...\\plus{}\\cos{(na\\plus{}b)}$", "Solution_1": "We use\r\n\\[ \\sum_{k \\equal{} 0}^n \\cos(\\alpha \\plus{} kh) \\equal{} \\left\\{\\begin{array}{ll}\\frac {\\cos\\left(\\alpha \\plus{} \\frac {n}{2}h\\right)\\sin\\frac {n \\plus{} 1}{2}h}{\\sin\\frac {h}{2}} & (h \\neq 2m\\pi) \\\\\r\n(n \\plus{} 1) \\cos \\alpha & (h \\equal{} 2m\\pi) \\end{array}\\right.\r\n\\]", "Solution_2": "Thank you very muck. Can you give me other solution?", "Solution_3": "other solution? as in what?\r\n\r\nanyway, Kouichi's formula can be proved simply by induction, with a few sum to product and product to sum identities." } { "Tag": [ "ratio" ], "Problem": "What is $ \\frac{x}{y}$ in the following equation?\r\n\r\n$ x^3\\minus{}4x^2y\\minus{}11xy^2\\plus{}30y^3 \\equal{} 0$", "Solution_1": "[hide]\nFactor $ x^3 \\minus{} 4x^2y \\minus{} 11xy^2 \\plus{} 30y^3 \\equal{} 0$ into\n$ (x \\minus{} 2y)(x \\minus{} 5y)(x \\plus{} 3y) \\equal{} 0$\nThen setting each part equal to zero gives\n$ x \\equal{} 2y$\n$ x \\equal{} 5y$\n$ x \\equal{} \\minus{} 3y$\n\nThus, $ \\frac {x}{y} \\equal{} 2, \\frac {x}{y} \\equal{} 5,$ and $ \\frac {x}{y} \\equal{} \\minus{}3$.\n\nBlarg, sorry, corrected\n[/hide]", "Solution_2": "If $ y \\not\\equal{} 0$, then denote by $ t\\equal{}\\frac{x}{y} \\Rightarrow t^3\\minus{}4t^2\\minus{}11t\\plus{}30\\equal{}0 \\Rightarrow (t\\minus{}2)(t\\minus{}5)(t\\plus{}3)\\equal{}0$. The rest of problem can be deduced easily.", "Solution_3": "[quote=\"27r\"][hide]\nFactor $ x^2 \\minus{} 4x^2y \\minus{} 11xy^2 \\plus{} 30y^3 \\equal{} 0$ [b]<----Typo[/b]: $ x^3$\ninto\n$ (x \\minus{} 2y)(x \\minus{} 5y)(x \\plus{} 3y) \\equal{} 0$\nThen setting each part equal to zero gives\n$ x \\equal{} 2y$\n$ x \\equal{} 5y$\n$ x \\equal{} \\minus{} 3y$\n\nLet $ y \\equal{} 1$ in each and plug the values of $ x$ and $ y$ back into the original equation.\nThe only ratio of $ x/y$ that works is $ x \\equal{} 5y$, or $ \\frac {x}{y} \\equal{} 5$.\n[/hide][/quote]Are you sure that's the only ratio? I'm pretty sure the typo messed up your calculations." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "[color=red]Trying to clear a broblem , I have formulated these questions:[/color]\r\n\r\nLet $ a ,b$ positive integers with $ (a,b) = 1$.\r\n[b]a)[/b]\tProve that for any positive integer $ c$ which is greater than $ ab-a-b$ we can find natural numbers $ x ,y$ such that $ c= xa+yb$\r\n[b]b)[/b]\tCan we find a general formula (with proof)which gives the number of the numbers $ c$ between \r\n\r\n$ 0,ab-a-b$, which can not be expressed in the form $ xa+yb$ ?\r\n\r\n [color=red]Edited [/color].\r\n\r\nSorry :( .I misunderstood the original english text where it was said : the product minus the sum of these numbers is the maximum....\r\n\r\n Thanks for the correction and the solutions.\r\n\r\n [u]Babis[/u]", "Solution_1": "[quote=\"stergiu\"][color=red]Trying to clear a broblem , I have formulated these questions:[/color]\n\nLet $ a ,b$ positive integers with $ (a,b) = 1$.\n[b]a)[/b]\tProve that for any positive integer $ c$ which is greater or equal to $ ab-a-b$ we can find natural numbers $ x ,y$ such that $ c= xa+yb$\n[b]b)[/b]\tCan we find a general formula (with proof)which gives the number of the numbers $ c$ between \n\n$ 0,ab-a-b$, which can not be expressed in the form $ xa+yb$ ?[/quote]\r\n\r\nThat's wrong :\r\n\r\nLet $ a=2$, $ b=3$, $ c=1\\geq ab-a-b=1$, then it does not exist any natural numbers $ x$ and $ y$ such that $ 1=2x+3y$", "Solution_2": "[quote=\"pco\"]That's wrong :\n\nLet $ a=2$, $ b=3$, $ c=1\\geq ab-a-b=1$, then it does not exist any natural numbers $ x$ and $ y$ such that $ 1=2x+3y$[/quote]\r\nExactly for all $ c>ab-a-b$ exist $ x\\ge 0,y\\ge 0$, suth that $ c=ax+by, \\ (a,b)=1$.\r\nFrom $ 1=ax_{0}-by_{0}$ we get a).\r\nb) Let $ S_{0}=ab-a-b$, then $ N=S_{0}+1-\\sum_{y=0}^{[S_{0}/b]}([1+\\frac{S_{0}-yb}{a}])$/", "Solution_3": "Follow the link - it's just the [url=http://mathworld.wolfram.com/CoinProblem.html]coin problem[/url] in the case $ n = 2$. :wink:" } { "Tag": [ "floor function" ], "Problem": "How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not\n12?", "Solution_1": "This question is too hard.\r\nHow are we supposed to know exactly how many multiples of 3 and 4 are between 1 and 2005?!", "Solution_2": "From $ 1$ to $ n$, there are $ \\left \\lfloor \\frac{n}{k} \\right \\rfloor$ multiples of $ k$. Thus:\r\n\r\nThere are $ 668$ multiples of $ 3$, $ 501$ multiples of $ 4$, and $ 167$ multiples of $ 12$ less than $ 2005$. Our desired answer is thus $ (668\\minus{}167)\\plus{}(501\\minus{}167)\\equal{}835$.", "Solution_3": "Ah. I get it.", "Solution_4": "First: How many numbers between 1 and 12 are integer multiples of 3 and 4?\r\n Answer: 5 ( They are 3, 6, 9, 4, and 8).\r\n Therefore, there are 5 integers that are multiples of 3 and 4, but not 12 for every cycle of 12.\r\nSecond: How many cycles of 12 are in 2005? \r\n Well, 2004/12 =167\r\n So, we will get 5 integers 167 times.\r\nAnswer: 167 * 5 = 835", "Solution_5": "Why do you subtract the 167 twice?", "Solution_6": "[hide=\"Explanation\"]When we found the number of multiples of 3, we also included those that were multiples of 12. This is where the first subtracting of the multiples of 12 occurs. Similarly, when we found the multiples of 4, we included the multiples of 12 as well, so we must subtract those again.\n\nAlternately, using the Principle of Inclusion-Exclusion, we find the number of numbers that are either multiples of 3 or 4. From PIE, we have to subtract the multiples of 12 to prevent overcounting. However, since this sum includes the multiples of 12 once, we must subtract yet again.\n\nHope this made sense.[/hide]", "Solution_7": "[hide]It took me a few minutes to figure out this one...\n\n1. 2005 is not divisible by 3, but 2004 is. 2004/3=668\n\n2.1/4 of the numbers that are multiples of 3 are also multiples of 4. (3*1 is not divisible by 4, nor is 3*2, or 3*3, but 3*4 is)\n\n3. so there are 668*3/4=501 numbers that are multiple of 3, but not 12\n\n4. let's now look at the multiples of 4, there are 2004/4=501\n\n5.1/3 of the numbers that are multiples of 4 are also multiples of 3 \n\n6. so there are 501*2/3=334 that are multiple of 4, but not 12\n\n7. so we add them together: 334+501=835\n\n\nso 835 is your answer!\n :lol: :lol: :lol: :lol:[/hide]", "Solution_8": "[hide=Solution] Since $\\frac{2005}{3} = 668\\frac13$, there are 668 multiples of 3 between 1 and 2005. Since $\\frac{2005}{4} = 501\\frac14$, there are 501 multiples of 4 between 1 and 2005. Since $\\frac{2005}{12} = 167\\frac{1}{12}$, there are 167 multiples of 12 between 1 and 2005.\n\nEvery multiple of 12 is also a multiple of 3 and of 4, so there are $668-167 = 501$ multiples of 3 that are not multiples of 12 and $501-167 = 334$ multiples of 4 that are not multiples of 12. That leaves $501 + 334 = \\boxed{835}$ numbers that are multiples of 3 or 4 but not 12. (Note: no number can be a multiple of 3 and 4 without also being a multiple of 12. So, no number is included twice in our count $501+334$.)[/hide]", "Solution_9": "2005 AMC 10B Problems/Problem 13", "Solution_10": "[hide=ALTERNATIVE SOLUTION]\nBy counting, there are $5$ multiples of $3$ and $4$ in every multiple of $12$ (excluding the multiple of $12$). And there are $2005\\div12\\approx167$ multiples of $12$ in the range of $1\\text-2005$. So that's $5\\times167=\\boxed{835}$ multiples of $3$ or $4$ but not $12$ that are between $1$ and $2005$. \n[/hide]" } { "Tag": [ "induction" ], "Problem": "Prove that the sum of squares\r\n$1^2+2^2+...+x^2=\\displaystyle \\frac{x(x+1)(2x+1)}{6}$\r\nDo it algebraicly!\r\n[hide=\"this is my proof\"]\nLet \n$s=\\sum_{a=1}^{x} {a^2}$\n\n$s=1^2+2^2+...+x^2$\n\nThis has x terms. aNother way to write it is:\n\n$s=(1)+(1+3)+(1+3+5)+...+(1+3+...+2x-3+2x-1)$\n\nThere are $x$ $1$'s. There are $x-1$ $3$'s because all terms except $1^2$ have $3$. Then contiue that until there are $2$ $2x-3$'s and $1$ $2x-1$. So \n\n$s=\\sum_{a=1}^{x} {(2a-1)(x-a+1)}$\n\n$s=\\sum_{a=1}^{x} {(-2a^2+2ax+2a+a-x-1)}$\n\n$s=\\sum_{a=1}^{x} {((-2)(a^2)+(2x+3)(a)-(x+1)(1))}$\n\n$s=\\sum_{a=1}^{x} {(-2)(a^2)}+\\sum_{a=1}^{x} {(2x+3)(a)}-\\sum_{a=1}^{x} {(x+1)(1)}$\n\n$s={(-2)}\\sum_{a=1}^{x} {a^2}+{(2x+3)}\\sum_{a=1}^{x} {a}-{(x+1)}\\sum_{a=1}^{x} {1}$\n\n$s=-2s+{(2x+3)}\\sum_{a=1}^{x} {a}-{(x+1)}\\sum_{a=1}^{x} {1}$\n\n$3s={(2x+3)}\\sum_{a=1}^{x} {a}-{(x+1)}\\sum_{a=1}^{x} {1}$\n\n$3s={(2x+3)}\\displaystyle \\frac{(x)(x+1)}{2}-(x+1)(x)$\n\n$3s={(2x+3)}\\displaystyle \\frac{(x)(x+1)}{2}-{(2)}\\displaystyle \\frac{(x)(x+1)}{2}$\n\n$3s={(2x+3-2)}\\displaystyle \\frac{(x)(x+1)}{2}$\n\n$3s=\\displaystyle \\frac{(x)(x+1)(2x+1)}{2}$\n\n$s=\\displaystyle \\frac{(x)(x+1)(2x+1)}{6}$\n[/hide]", "Solution_1": "Hmmm...are we allowed to use induction?", "Solution_2": "I don't know, it becomes almost trivial after applying induction.", "Solution_3": "Using telescoping method,\r\n\r\n$\\sum_{k=1}^n k^2=\\sum_{k=1}^n \\left[\\frac{1}{3}k^3+\\frac{1}{2}k^2+\\frac{1}{6}k-\\left\\{\\frac{1}{3}(k-1)^3+\\frac{1}{2}(k-1)^2+\\frac{1}{6}(k-1)\\right\\}\\right]$\r\n\r\n$=\\frac{1}{3}n^3+\\frac{1}{2}n^2+\\frac{1}{6}n=\\frac{1}{6}n(n+1)(2n+1)$", "Solution_4": "What is induction? And I don't see why you would use it, there are already two good proofs out there", "Solution_5": "induction is just proving that if it works for some k, it also works for some k+1. you also need a base case.\r\n\r\nso suppose your base case is 1. then you prove k works => k+1 works.\r\n\r\nso then 1 works => 2 works => 3 works => 4 works => etc. so it works for all natural numbers.\r\n\r\nit does really trivialize the problem if you know what you're doing.", "Solution_6": "Well, I don't see the good of it here if it can trivialize it because this question isn't that trivial just long algebra or thinking algebra", "Solution_7": "Yeah, the proofs provided are good. Induction is just an easy way to prove it without understanding very much :P. You don't learn much from using it on this proof.", "Solution_8": "Induction proofs are important, and this formula is a great application of the inductive principle. Sure, it doesn't tell you where this formula came from - but it does tell you why it works.", "Solution_9": "So if you use induction, would you just prrove that x=1 works by plugging 1 in, and then proove that $\\frac{x(x+1)(2x+1)}{6} + {(x+1)}^2 = \\frac{(x+1)(1(x+1)+1)(2(x+1)+1)}{6}$?\r\n\r\nBecause it that's right, then inducton is really easy. You can probably tell that I'm not really familiar with induction :blush:", "Solution_10": "Yep, that's it.", "Solution_11": "[quote=\"h_s_potter2002\"]Because it that's right, then inducton is really easy.[/quote]\r\n\r\nThe concept of induction is simple, but that inductive step can be very difficult in some problems. And there's all kinds of cool stuff you can do with induction, such as backwards induction.\r\n\r\nNow, my challenge: Prove, using induction:\r\n\r\nGiven $a_1 \\leq a_2 \\leq ... \\leq a_n$ with $a_1,...,a_n$ positive integers and $1/a_1 + 1/a_2 + ... + 1/a_n=1$ that $a_n \\leq 2^{n!}$" } { "Tag": [], "Problem": "Look around the forum, its everywhere.\r\n\r\n20", "Solution_1": "[b]17[/b]\r\n\r\ni remmebr jongao started at 45 once and the girls still won!! XD", "Solution_2": "21 ", "Solution_3": "18[b][/b]", "Solution_4": "[b]16[/b]", "Solution_5": "20 ", "Solution_6": "[b]17[/b]", "Solution_7": " 21", "Solution_8": "[b]19[/b]", "Solution_9": " 23", "Solution_10": "[b]18[/b]", "Solution_11": "fcawefcewfjwejf 120", "Solution_12": "[b]16[/b]\r\nhmm this is good for us so i'll do it", "Solution_13": "[b]13[/b]", "Solution_14": "[b]11[/b]", "Solution_15": "[b]5[/b]", "Solution_16": "UMMM\r\n\r\n13", "Solution_17": "There are too many people posting at the same time so we don't see each others' posts.", "Solution_18": "umm\r\n\r\nyou added 5 like every single time!!", "Solution_19": "[b][/b]14", "Solution_20": "[b]3[/b]", "Solution_21": "[b]5[/b] 0000000", "Solution_22": "[b]3[/b]", "Solution_23": "000000000000000", "Solution_24": "PowerOfPi is only about 1000 posts from 5.", "Solution_25": "[b]-42[/b]", "Solution_26": "yes, but he only has 4 stars ;)", "Solution_27": "[b]50!!!![/b] Guys win!!!!", "Solution_28": "What? I got an error and couldn't post. Darn :wallbash_red: :wallbash_red:", "Solution_29": "haha nice try\r\n\r\nthe score is like 8 1/2 to 2 1/2 girls winning" } { "Tag": [ "algebra", "polynomial", "ARML" ], "Problem": "If x is a positive integer and x(x+1)(x+2)(x+3)+1=3792, compute x. \r\n\r\nAny ideas?\r\n\r\nFierytycoon", "Solution_1": "There is no such x because x(x+1)(x+2)(x+3)+1=3792 is the same as x(x+1)(x+2)(x+3)=3791 and if x is an odd integer then x+1 is clearly even, the converse gives that x is even, then 3791 must be even, which it's not. Also, x(x+1)(x+2)(x+3) is one less than a perfect square since x(x+1)(x+2)(x+3) + 1 = (x^2 + 3x + 1)^2 an odd perfect square. But 3792 is not odd and neither a perfect square.", "Solution_2": "oh, whoops: i copied the problem incorrectly. It is actually:\r\n\r\nIf x is a positive integer and x(x+1)(x+2)(x+3)+1=379^2, compute x. \r\n\r\nThe exponent didn't get copied correctly. Sorry!\r\n\r\nBut now, I think the problem is not bad. now x(x+1)(x+2)(x+3) = (379-1)(379+1)=(378)(380), so x must be very near Sqrt(379). A bit of guess-and-check should yield the result.\r\n\r\nbtw, I found the problem at http://incolor.inebraska.com/sabalka/arml.html , which has some nice short-answer problems.\r\n\r\nFierytycoon", "Solution_3": "[quote=\"Fierytycoon\"]But now, I think the problem is not bad. now x(x+1)(x+2)(x+3) = (379-1)(379+1)=(378)(380), so x must be very near Sqrt(379). A bit of guess-and-check should yield the result.[/quote]\r\n\r\nSince sqrt(379) :approx: 19.5, we need (x+1) and (x+2) to be around 19 and 20. Indeed, if x is 18, x(x+1)(x+2)(x+3) = 18*19*20*21 = 378*380, which is the desired result. Hence x = 18.\r\n\r\nAlso, factoring (378)(380) gives (2*3^3*7)(2^2*5*19) = (2^3*3^3*5*7*19). Since there's one 19, I'd guess that 19 must be one of the numbers (since 38, 57, etc. would be way too big). Now we want to find numbers around 19. 2*3^2 makes 18. 2^2*5 makes 20. 3*7, the last two prime factors left, make 21. So we have (18)(19)(20)(21), which works. Hence x = 18.", "Solution_4": "I assume this is from ARML, but I forget which year. x(x+1)(x+2)(x+3)+1 is something that comes up from time to time. The thing to remember is that x(x+1)(x+2)(x+3)+1=(x^2+3x+1)^2. You can prove this by reordering the terms to x(x+3)(x+1)(x+2)+1=(x^2+3x)(x^2+3x+2)+1=(x^2+3x+1)^2.", "Solution_5": "[quote=\"ComplexZeta\"]I assume this is from ARML, but I forget which year. x(x+1)(x+2)(x+3)+1 is something that comes up from time to time. The thing to remember is that x(x+1)(x+2)(x+3)+1=(x^2+3x+1)^2. You can prove this by reordering the terms to x(x+3)(x+1)(x+2)+1=(x^2+3x)(x^2+3x+2)+1=(x^2+3x+1)^2.[/quote]\r\n\r\nOh-ho! So in that case, (x^2+3x+1)^2 = 379^2\r\nso x^2+3x+1 = 379\r\nx^2 + 3x - 378 = 0\r\n(x-18)(x+21) = 0\r\nx = 18\r\n\r\nNice...I didn't realize that factored so nicely. Must remember that one.", "Solution_6": "[color=cyan]You got rather lucky there, because you might have missed out on roots.\n\n(x^2+3x+1)^2 = 379^2\nso\n\nx^2+3x+1 = 379 [u][i][b]or[/b][/i][/u] x^2+3x+1 = -379. The second equation just happens not to have any positive whole number solutions. Or real solutions. But that's not important![/color]", "Solution_7": "Well, true, but I didn't bother writing out the -21, either. You're right, I need to be more careful." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $p>5$ be prime number. Prove: \\[ \\binom{qp}{p} \\equiv q\\ (mod \\ p^3) \\]", "Solution_1": "It is Luka theorem \\[ ord_p( \\binom{ap}{bp}-\\binom{a}{b})=ord_p(\\binom{a}{b})+ord_p(a)+ord_p(b)+ord_p(a-b)+v(p),p>2 \\] were $v(3)=2,v(p)\\ge 3,p>3$. Primes with $v(p)>3$ called Wolstenholm's primes. There are only 2 Wolstenholm's primes p<1 000 000 000. There are p=16483,2124679 v(p)=4.", "Solution_2": "Is $ord_p(a)$ the gratest power of $p$ that divides $a$. What is $v(p)$.", "Solution_3": "\\[ v(p)=2+ord_p(\\sum_{k=1}^{p-1}\\frac{1}{k^2}). \\]" } { "Tag": [ "geometric series" ], "Problem": "A ball is dropped from a height of 10 feet. At each bounce it rises to two-thirds of its previous height. How far will it have traveled if it bounces \"forver\"?\r\n\r\nThanks", "Solution_1": "Ahah!\r\n[hide=\"Solution\"]\nGeometric Series Time.\nThe first drop is 10 feet.\nIt then comes up 6.6666 and back down 6.666\nUp 40/9 feet, down 40/9 feet.\nWe wish to evaluate the infinite sum:\n$10+2\\sum_{n=0}^{\\infty}\\frac{20}{3}\\cdot(2/3)^{n}$\n\nBy the formula for solving an infinite series, \nwe get\n$\\frac{a}{1-r}\\longrightarrow 10+2 \\cdot\\frac{20/3}{1-\\frac{2}{3}}\\longrightarrow 10+2\\cdot \\frac{20/3}{1/3}\\longrightarrow \\boxed{50}$ feet.\n[/hide]" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all function $ f: R\\rightarrow R$ so that:\r\ni) $ f(x)\\leq 4\\plus{}2009x$\r\nii) $ f(x\\plus{}y)\\leq f(x)\\plus{}f(y)\\minus{}4$", "Solution_1": "[quote=\"tdl\"]Find all function $ f: R\\rightarrow R$ so that:\ni) $ f(x)\\leq 4 \\plus{} 2009x$\nii) $ f(x \\plus{} y)\\leq f(x) \\plus{} f(y) \\minus{} 4$[/quote]\r\n\r\ni) $ \\implies$ $ f(0)\\leq 4$\r\nii) $ \\implies$ $ f(x\\plus{}0)\\leq f(x) \\plus{} f(0) \\minus{}4$ and so $ f(0)\\geq 4$\r\n\r\nSo $ f(0)\\equal{}4$\r\n\r\nii) $ \\implies$ $ f(x\\plus{}(\\minus{}x)) \\leq f(x) \\plus{} f(\\minus{}x) \\minus{} 4$ and so $ f(x) \\plus{} f(\\minus{}x) \\geq 8$\r\n\r\ni) $ \\implies$ $ f(x)\\leq 4 \\plus{} 2009x$\r\ni) $ \\implies$ $ f(\\minus{}x)\\leq 4 \\minus{} 2009x$\r\nAdding these two lines, we get $ f(x) \\plus{} f(\\minus{}x) \\leq 8$, so (since $ f(x) \\plus{} f(\\minus{}x) \\geq 8$) : $ f(x) \\plus{} f(\\minus{}x) \\equal{} 8$ and the two lines are equalities.\r\n\r\nSo $ f(x)\\equal{}2009x \\plus{} 4$ and it is easy to verify that this solution fit the initial inequations" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Study the convergence of the sequence (x_n)n>=0 where x_0 is a real number and x_(n+1)=[1+2^p*(x_n)^p]^(1/p)-x_n, for every n natural, p is an even natural number.", "Solution_1": "Manifestly there is an increasing sequence : x(n+1) = (1+(2*|x(n)|)^p)^(1/p) - x(n) > x(n) \r\nSuppose a finite limit L for x(n) ===> L = (1+(2*|L|)^p)^(1/p) - L > L (contradiction)\r\nIncreasing sequence and no finite limit ===> lim(n-->+oo) = +oo" } { "Tag": [ "inequalities" ], "Problem": "Farz konid $ a,b,c>0$ va $ ab+bc+ca=3$ sabet konid :\r\n$ a^{3}+b^{3}+c^{3}+7abc \\geq 10$ .", "Solution_1": "in ye rahe hal dare ke ziad ghashango khosh das nis,vali be har hal:\r\n\r\naval az hame midoonim $ (a+b+c)^{2}\\geq 3(ab+bc+ca)=9$ pas $ a+b+c\\geq 3$\r\n----------------------------\r\ntebghe namosavie shur midoonim:\r\n\r\n$ a^{3}+b^{3}+c^{3}+3abc\\geq a^{2}b+ab^{2}+b^{2}c+bc^{2}+c^{2}a+ca^{2}$\r\n\r\nya be ebarati:\r\n\r\n$ a^{3}+b^{3}+c^{3}+3abc\\geq (a+b+c)(ab+bc+ca)-3abc$\r\n\r\npas darim:\r\n\r\n$ a^{3}+b^{3}+c^{3}+6abc\\geq (a+b+c)(ab+bc+ca)$\r\n\r\nke mishe intori neveshtesh:\r\n\r\n$ (a+b+c)^{3}+9abc\\geq 4(a+b+c)(ab+bc+ca)$\r\n\r\nhala ba tavajoh be sharte $ ab+bc+ca=3$ in rabete be shekle zir dar miad:\r\n\r\n$ (a+b+c)^{3}+9abc\\geq 12(a+b+c)$\r\n\r\npas darim:\r\n\r\n$ abc\\geq \\frac{12(a+b+c)-(a+b+c)^{3}}{9}$\r\n\r\nhala age gharar bedim $ S=a+b+c$ mishe:\r\n\r\n$ abc\\geq \\frac{12S-S^{3}}{9}$\r\n-----------------------------------\r\nkhob hala dobare be masale bargardim,soorate hokmemoon moadele ine:\r\n\r\n$ (a+b+c)^{3}+10abc-3(a+b+c)(ab+bc+ca)\\geq 10$\r\n\r\nke ba tavajoh be sharte $ ab+bc+ca=3$ moadele ine:\r\n\r\n$ \\boxed{S^{3}+10abc-9S-10\\geq 0}$\r\n\r\n-------------------------------------------\r\n\r\nkhob hala age az namosavi ke dar avale esbatemoon be das ovordim estefade konim mishe:\r\n\r\n$ S^{3}+10abc-9S-10\\geq S^{3}+10\\frac{12S-S^{3}}{9}-9S-10\\geq 0$\r\n\r\nke age sadash konim mishe:\r\n\r\n$ -S^{3}+39S-90\\geq 0$\r\n\r\nhala ino mishe injoori neveshtesh:\r\n\r\n$ (S-3)(30-S^{2}-3S)\\geq 0$\r\n\r\nhala hamantor ke dar avale esbat gofte shod darim $ S\\geq 3$ pas age paranteze dovomo taine alamat konim mibinim ke age\r\n\r\n$ S\\leq \\frac{-3+\\sqrt{129}}{2}$ bashe masale hal shodast...\r\n\r\npas mimoone baghie halata...\r\n\r\namma baghie halata azin kheili asoontare chera ke darim:\r\n\r\n$ S^{3}+10abc-9S-10\\geq S^{3}-9S-10=(S+2)(S^{2}-2S-5)$\r\n\r\npas kafie sabet konim:\r\n\r\n$ (S+2)(S^{2}-2S-5)\\geq 0$\r\n\r\nke age ino taine alamt konim mibinim ke vaseye $ S\\geq \\frac{2+\\sqrt{24}}{2}$ in namosavi bargharare...\r\n\r\nhala chon $ \\frac{2+\\sqrt{24}}{2}\\leq \\frac{-3+\\sqrt{129}}{2}$ pas hokmemoon be tore kamel esbat shode...", "Solution_2": "Can you solve this problem in English. :lol: \r\nI think you have just proved in this case:$ 3\\leq S\\leq \\frac{\\minus{}3\\plus{}\\sqrt{129}}{2}$ then the inequality is true,and in the case:$ S\\geq\\frac{2\\plus{}\\sqrt{24}}{2}$ it's true too but in the case $ S>\\frac{\\minus{}3\\plus{}\\sqrt{129}}{2}$,you haven't done it", "Solution_3": "read the post completely:\r\nwe want to prove that:\r\n\r\n$ S^3 \\plus{} 10abc \\minus{} 9S \\minus{} 10\\geq 0$ (*)\r\n\r\nnow as we said before,we know that $ S\\geq 3$ now note that if $ S\\leq \\frac { \\minus{} 3 \\plus{} \\sqrt {129}}2$ then we have:\r\n\r\n$ S^3 \\plus{} 10abc \\minus{} 9S \\minus{} 10\\geq \\minus{} S^3 \\plus{} 39S \\minus{} 90 \\equal{} (\\underbrace{S \\minus{} 3}_{\\geq 0})(\\underbrace{30 \\minus{} S^2 \\minus{} 3S}_{\\geq0})\\geq 0$\r\n----------------------------\r\nso it remains to prove (*) for $ S > \\frac { \\minus{} 3 \\plus{} \\sqrt {129}}2$\r\n\r\nwhich is pretty easy because we have:\r\n\r\n$ (*) \\equal{} S^3 \\plus{} 10abc \\minus{} 9S \\minus{} 10\\geq S^3 \\minus{} 9S \\minus{} 10 \\equal{} (S \\plus{} 2)(S^2 \\minus{} 2S \\minus{} 5)$\r\n\r\nso it's sufficient to show that:\r\n\r\n$ (S \\plus{} 2)(S^2 \\minus{} 2S \\minus{} 5)\\geq 0$ for $ S > \\frac { \\minus{} 3 \\plus{} \\sqrt {129}}2$\r\n\r\nbut note that $ S\\geq 3\\Rightarrow S \\plus{} 2\\geq 0$ so we have to show that:\r\n\r\n$ S^2 \\minus{} 2S \\minus{} 5\\geq 0$ (**)\r\n\r\nbut the roots of (**) are $ \\frac {2\\pm\\sqrt {24}}{2}$ so for $ S\\geq \\frac {2 \\plus{} \\sqrt {24}}2$ (**) holds...\r\nbut in the other hand we have:\r\n\r\n$ S\\geq\\frac { \\minus{} 3 \\plus{} \\sqrt {129}}2\\geq\\frac {2 \\plus{} \\sqrt {24}}2$\r\n\r\nthus:\r\n\r\n$ S^2 \\minus{} 2S \\minus{} 5\\geq 0$\r\n\r\nQED" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Sequences {U_n} {V_n} satisfy the following conditions\r\n\r\nU_(n+2) = a.U_(n+1) + b.U_(n)\r\n\r\nV_(n+2) = c.V_(n+1) + d.V_(n) where a, b, c and d are constant.\r\n\r\n\r\nU-(0) , U_(1) , V_(0) , V_(1) are arbitrary numbers subject to V_(0) = U_(0) and V_(1) = U_(2) =a.U_(1) + b.U_(0)\r\n\r\nFind necessary and sufficient conditions on a, b, c and d to ensure that U_(2n) = V_(n) for all n.", "Solution_1": "Solve the characteristic equation... and equate the two numbers?" } { "Tag": [ "modular arithmetic", "puzzles" ], "Problem": "Solve this cryptarithm, where each of the letters represents a different decimal digit from 0 to 9, and S is not zero.\r\n\r\nSAME + SEAM = 2961", "Solution_1": "[b]SAME + SEAM = 2961[/b]\r\n\r\n$ 2000S \\plus{} 110A \\plus{} 11M \\plus{} 101E \\equal{} 2961$.\r\n\r\nClearly $ S \\equal{} 1$. Therefore:\r\n\r\n$ 110A \\plus{} 11M \\plus{} 101E \\equal{} 961$. Factor 11, we get:\r\n\r\n$ 11(10A \\plus{} M) \\plus{} 101E \\equal{} 961$\r\n\r\nTherefore, $ 961 \\minus{} 101E$ must be a multiple of 11. Note that $ 961\\equiv 4 \\pmod {11}$ and $ 101\\equiv 2\\pmod {11}$. Therefore $ E \\equal{} 2$.\r\n\r\nThus, $ 11(10A \\plus{} M) \\equal{} 759 \\implies 10A \\plus{} M \\equal{} 69 \\implies A \\equal{} 6,M \\equal{} 9$.\r\n\r\n[b]Answer:[/b]\r\n[b]A[/b]=6\r\n[b]E[/b]=2\r\n[b]M[/b]=9\r\n[b]S[/b]=1\r\n[b]SAME[/b]=1692\r\n[b]SEAM[/b]=1269\r\n[b]SAME+SEAM[/b]=2961", "Solution_2": "Obviously S is 1. Assume there are carries for the other three place values in the addition (you can check it is not possible otherwise). Then,\r\ne+m=11\r\nm+a=15\r\na+e=8\r\nAdding them and cutting in half, a+e+m=17, so we find a=6, e=2, m=9", "Solution_3": "Hm why are my methods so rigorous... and complicated... :blush:", "Solution_4": "@Yongyi781: Maybe that's just your way. Professionalism is, in most cases, a good thing. However in the practice, shortcuts are always nice, as long as they work." } { "Tag": [], "Problem": "The marked price of a book was $30%\n$ less than the suggested retail price. Alice purchased the book for half the marked price at a Fiftieth Anniversary sale.", "Solution_1": "[quote=\"4everwise\"]The marked price of a book was [30%] less than the suggested retail price. Alice purchased the book for half the marked price at a Fiftieth Anniversary sale.[/quote]\r\n\r\nErr.. is there a question to this? \r\n :lol:" } { "Tag": [ "absolute value" ], "Problem": "If $ x\\plus{} |y| \\minus{} y \\equal{} 12$ and $ |x| \\plus{} x \\plus{}y \\equal{} 10$, find $ x\\plus{}y$.\r\n\r\n[hide]I have tried isolating $ x\\minus{}y$ of one equation and $ x\\plus{}y$ for the other, and get $ 2x \\equal{} 22 \\minus{} |x| \\minus{} |y|$. Nothing comes to mind then. [/hide]", "Solution_1": "[quote=\"jeez123\"]If $ x \\plus{} |y| \\minus{} y \\equal{} 12$ and $ |x| \\plus{} x \\plus{} y \\equal{} 10$, find $ x \\plus{} y$.[/quote]\r\n[hide=\"Solution\"]Adding the two equations yields $ 2x \\plus{} |x| \\plus{} |y| \\equal{} 22$.\n\n[color=darkred][b]Case 1:[/b][/color] $ y\\geq0$\n\n$ x \\equal{} 12\\Longrightarrow |x| \\plus{} |y| \\equal{} \\minus{} 2$\n\nObviously, this is not possible.\n\n[color=darkred][b]Case 2a:[/b][/color] $ y < 0,\\,x\\geq0$\n\n$ x \\plus{} 2y \\equal{} 12,\\,2x \\plus{} y \\equal{} 10\\Longrightarrow\\boxed{x \\plus{} y \\equal{} \\tfrac{22}{3}}$\n\n[color=darkred][b]Case 2b:[/b][/color] $ y < 0,\\,x < 0$\n\n$ x \\plus{} 2y \\equal{} 12,\\,y \\equal{} 10\\Longrightarrow\\boxed{x \\plus{} y \\equal{} 2}$[/hide]", "Solution_2": "[hide=\"Hint\"]Split into four cases - both x,y non-negative; x negative, y non-negative; x non-negative, y negative; both x,y negative - and discuss the system (it's really simple enough for casework).[/hide]\n\n[hide=\"Correct answer\"]$ x \\equal{} {32\\over 5}, y \\equal{} \\minus{} {14\\over 5}, x \\plus{} y \\equal{} {18\\over 5}$[/hide]\n\n[hide=\"Where i_like_pie is wrong\"]Case 2a should be $ x\\boxed{ \\minus{} }2y \\equal{} 12$\nCase 2b can't have both $ y<0$ and $ y\\equal{}10$[/hide]", "Solution_3": "[quote=\"Farenhajt\"]Case 2a should be $ x\\boxed{ \\minus{} }2y \\equal{} 12$\nCase 2b can't have both $ y < 0$ and $ y \\equal{} 10$[/quote]\r\nI definitely agree with your second correction, but how can $ x \\plus{} |y| \\minus{} y \\equal{} 12$ become $ x \\minus{} 2y \\equal{} 12$? The way I see it, you're suggesting the absolute value becomes negative. :huh:", "Solution_4": "Remember the definition of absolute value? $ |y| \\equal{} y$ for non-negative $ y$ and $ |y| \\equal{} \\minus{} y$ for negative $ y$ (notice that $ \\minus{}y$ is actually positive if $ y$ is negative). Since you assumed $ y < 0$ in case 2a, then $ |y| \\equal{} \\minus{} y$", "Solution_5": "Ah, I see. Thanks for correcting me. :)\r\n\r\nI forgot that $ y$ would be negative and it would result in a positive outcome. :oops:", "Solution_6": "[quote=\"i_like_pie\"][quote=\"jeez123\"]If $ x \\plus{} |y| \\minus{} y \\equal{} 12$ and $ |x| \\plus{} x \\plus{} y \\equal{} 10$, find $ x \\plus{} y$.[/quote]\n[hide=Solution]Adding the two equations yields $ 2x \\plus{} |x| \\plus{} |y| \\equal{} 22$.\n\n[color=darkred][b]Case 1:[/b][/color] $ y\\geq0$\n\n$ x \\equal{} 12\\Longrightarrow |x| \\plus{} |y| \\equal{} \\minus{} 2$\n\nObviously, this is not possible.\n\n[color=darkred][b]Case 2a:[/b][/color] $ y < 0,\\,x\\geq0$\n\n$ x \\minus{} 2y \\equal{} 12,\\,2x \\plus{} y \\equal{} 10\\Longrightarrow\\boxed{x \\plus{} y \\equal{} ?}$\n\n[color=darkred][b]Case 2b:[/b][/color][/quote]\r\n\r\nI understood up to those corrected parts above. How do you calculate for $ x\\plus{}y$ in Case 2a? I added both of the equations and got $ 3x\\minus{}y\\equal{}22$ and cannot find the $ x\\plus{}y$ part. Could you finish up this solution? Thanks!", "Solution_7": "You're on the right track. You should add the second equation again:\r\n$ (3x \\minus{} y) \\plus{} (2x \\plus{} y) \\equal{} 22 \\plus{} 10 \\implies 5x \\equal{} 32 \\implies x \\equal{} \\frac {32}{5}$\r\n\r\nWe can plug this into a known equation: $ x \\plus{} y \\equal{} (2x \\plus{} y) \\minus{} x \\equal{} 10 \\minus{} \\frac {32}{5} \\equal{} \\frac {18}{5}$" } { "Tag": [], "Problem": "I couldnt find this one in forum\r\nfactor:\r\n$ {(a\\plus{}b\\plus{}c)}^3\\minus{}(a^3\\plus{}b^3\\plus{}c^3)$", "Solution_1": "[hide]Simplifying yields $ 3\\sum_{cyc}^{}a^2b \\plus{} 6abc$. Noticing the cyclic terms, we try cyclic factors. $ (a\\plus{}b)(a\\plus{}c)(b\\plus{}c)\\equal{} \\sum_{cyc}^{}a^2b \\plus{}2abc$. Multiply by three and we are done.[/hide]" } { "Tag": [ "conics", "ellipse", "LaTeX", "trigonometry" ], "Problem": "In the context of elliptical polarization,every book writes out this formula:\r\n\r\n(E_y/E_0y)^2 + (E_x/E_0x)^2 -2(E_y/E_0y)(E_x/E_0x) cos \u03b5= (sin\u03b5)^2\r\n\r\nwhere \u03b5 is the initial phase difference.\r\n\r\nAfter that they write this is the equation of an ellipse making an angle \u03b1 with the E_x-E_y co-ordinate system such that \r\n\r\ntan 2\u03b1=[[2E_0x*E_0y]cos \u03b5]/[(E_0x)^2 - (E_0y)^2]\r\n\r\nThis is NOt a homework problem;but I want to clarify how does the 2nd relation concering tan 2\u03b1 comes.\r\n\r\nI tried to understand by expanding tan 2\u03b1 and by dividing RHS by (E_0x)^2 throught.\r\nBut the factor cos \u03b5 is annoying.\r\n\r\nCan anyone please tell how the relation comes?", "Solution_1": "Come on kolahalb, use LaTeX please. :wink: its almost impossible to figure out what you write. :huh:", "Solution_2": "In $\\text{\\LaTeX}$: (I think :) )\r\n$\\hline$\r\n\r\nIn the context of elliptical polarization,every book writes out this formula:\r\n\r\n$\\left(\\frac{E_{y}}{E_{0y}}\\right)^{2}+\\left(\\frac{E_{x}}{E_{0x}}\\right)^{2}-2\\left(\\frac{E_{y}}{E_{0y}}\\right)\\left(\\frac{E_{x}}{E_{0x}}\\right) \\cos \\epsilon= (\\sin\\epsilon)^{2}$\r\n\r\nwhere $\\epsilon$ is the initial phase difference.\r\n\r\nAfter that they write this is the equation of an ellipse making an angle $\\alpha$ with the $E_{x}$-$E_{y}$ co-ordinate system such that\r\n\r\n$\\tan 2\\alpha=\\frac{\\left[2E_{0x}\\cdot E_{0y}\\right]\\cos \\epsilon}{[(E_{0x})^{2}-(E_{0y})^{2}]}$\r\n\r\nThis is NOt a homework problem;but I want to clarify how does the 2nd relation concering $\\tan 2\\alpha$ comes.\r\n\r\nI tried to understand by expanding $\\tan 2\\alpha$ and by dividing $RHS$ by $(E_{0x})^{2}$ throught.\r\nBut the factor $\\cos \\epsilon$ is annoying.\r\n\r\nCan anyone please tell how the relation comes?", "Solution_3": "Thanks color,I exactly meant that.\r\n\r\nNow can you hep something about it?" } { "Tag": [ "geometry", "rectangle" ], "Problem": "Hi, I can't solve the following problems, please help.\r\n\r\n1. A rectangle with its dimensions 5m by 12m is surrounded by a path of uniform width. How wide will this path be if the area of the rectangle is the same as the area of the path?\r\n\r\n2. A number is the quotient of 5.52 and 11.5, decrease by 2.79. What is the number?", "Solution_1": "For 1. draw a diagram. Call the width w and split up the path into rectangles, and write the areas of the rectangles in terms of 5, 12, and w. (if this doesn't make sense I can try explaining more)\r\n\r\n2 seems straightforward...\r\n[hide=\"2\"]\n$ \\frac{5.52}{11.5}\\minus{}2.79$\n[/hide]", "Solution_2": "[quote=\"chuchen\"]1. A rectangle with its dimensions 5m by 12m is surrounded by a path of uniform width. How wide will this path be if the area of the rectangle is the same as the area of the path?[/quote]\r\n[hide=\"Hint\"]Set up an equation and solve for the width.[/hide]\n[hide=\"Solution\"]Let the width of the path be $ w$.\n\nThe area of the rectangle is $ 5\\cdot12$, and the area of the path is $ (5 \\plus{} w)(12 \\plus{} w) \\minus{} 5\\cdot12$.\n\nWe know that the two areas are equal, so we can solve for $ w$ by setting those two expressions equal to each other.\n\nExpanding the RHS, we get $ 60 \\equal{} w^2 \\plus{} 17w \\plus{} 60 \\minus{} 60$.\n\nSimplifying and putting everything over on one side gives $ w^2 \\plus{} 17w \\minus{} 60 \\equal{} 0$.\n\nFactoring, $ (w \\minus{} 3)(w \\plus{} 20) \\equal{} 0$.\n\nThe width cannot be negative, so $ \\boxed{w \\equal{} 3\\ \\text{m}}$.[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "I work with three problem,and found nice solution(I think).\r\nProblem 1:(Vasile cirtoaje)\r\nGive $ a,b,c \\ge 0$ and no two which of them are zero.Prove\r\n$ \\frac{1}{a^2\\plus{}ab\\plus{}b^2}\\plus{} \\frac{1}{b^2\\plus{}bc\\plus{}c^2}\\plus{} \\frac{1}{c^2\\plus{}ca\\plus{}a^2} \\ge \\frac{9}{(a\\plus{}b\\plus{}c)^2}$\r\nFirst solution:(by Bui Tuan Vu):\r\nThis ineqlity can be written :\r\n$ \\frac{(a\\plus{}b\\plus{}c)^2}{a^2\\plus{}ab\\plus{}b^2}\\plus{} \\frac{(a\\plus{}b\\plus{}c)^2}{b^2\\plus{}bc\\plus{}c^2}\\plus{} \\frac{(a\\plus{}b\\plus{}c)^2}{a^2\\plus{}ac\\plus{}c^2} \\ge 9$\r\nor :\r\n$ \\sum\\frac{(c\\minus{}a)(c\\minus{}b)}{a^2\\plus{}ab\\plus{}b^2}\\plus{}3\\plus{} \\sum\\frac{3a(b\\plus{}c)}{b^2\\plus{}bc\\plus{}c^2} \\ge 9$\r\nWhich is true by schur's inequality and \r\n$ \\sum\\frac{a(b\\plus{}c)}{b^2\\plus{}bc\\plus{}c^2} \\ge 2$(vasile cirtoaje)\r\nSecond solution(by Bui Ngoc Anh) we use AM-GM and Iran TST 96\r\n$ \\sum\\frac{1}{a^2\\plus{}ab\\plus{}b^2}\\equal{} \\sum\\frac{ab\\plus{}bc\\plus{}ca}{(ab\\plus{}bc\\plus{}ca)(a^2\\plus{}ab\\plus{}b^2)} \\ge \\sum\\frac{4(ab\\plus{}bc\\plus{}ca)}{(a\\plus{}b)^2(a\\plus{}b\\plus{}c)^2} \\ge \\frac{9}{(a\\plus{}b\\plus{}c)^2}$\r\nProblem 2:\r\n$ \\frac{a^2\\plus{}bc}{b^2\\plus{}bc\\plus{}c^2}\\plus{} \\frac{b^2\\plus{}ac}{a^2\\plus{}ac\\plus{}c^2}\\plus{} \\frac{c^2\\plus{}ab}{a^2\\plus{}ab\\plus{}b^2} \\ge 2$\r\nSolution(by Bui Ngoc Anh)\r\nWe use :\r\n$ \\sum\\frac{a(b\\plus{}c)}{b^2\\plus{}bc\\plus{}c^2} \\ge 2$(vasile cirtoaje)\r\nand we can prove the stronger:\r\n$ \\frac{a^2\\plus{}bc}{b^2\\plus{}bc\\plus{}c^2}\\plus{} \\frac{b^2\\plus{}ac}{a^2\\plus{}ac\\plus{}c^2}\\plus{} \\frac{c^2\\plus{}ab}{a^2\\plus{}ab\\plus{}b^2} \\ge \\frac{a(b\\plus{}c)}{b^2\\plus{}bc\\plus{}c^2}\\plus{} \\frac{b(a\\plus{}c)}{a^2\\plus{}ac\\plus{}c^2}\\plus{} \\frac{c(a\\plus{}b)}{a^2\\plus{}ab\\plus{}b^2}$\r\nor \r\n$ \\sum\\frac{(c\\minus{}a)(c\\minus{}b)}{a^2\\plus{}ab\\plus{}b^2} \\ge 0$\r\nWhich is true by schur's inequality.\r\nproblem 3:\r\nGive $ x\\plus{}y\\plus{}z\\plus{}xyz\\equal{}4$ and $ x,y,z \\ge 0$.Prove that:\r\n$ x\\plus{}y\\plus{}z \\ge xy\\plus{}yz\\plus{}zx$\r\nSolution(by Bui Ngoc Anh)\r\nif $ xyz >1$ and by AM-GM we have:$ x\\plus{}y\\plus{}z >3$.\r\nand $ x\\plus{}y\\plus{}z\\plus{}xyz>4$.\r\nSo we have $ xyz \\le 1$ and from $ x\\plus{}y\\plus{}z\\plus{}xyz\\equal{}4$ we get $ x\\plus{}y\\plus{}z \\ge 3$\r\nIf $ xy\\plus{}yz\\plus{}zx \\le 3$,this inequality is obivious true,because $ x\\plus{}y\\plus{}z \\ge 3 \\ge xy\\plus{}yz\\plus{}zx$\r\nand we only need to consider $ xy\\plus{}yz\\plus{}zx \\ge 3$\r\n$ x\\plus{}y\\plus{}z \\ge xy\\plus{}yz\\plus{}zx$ or $ (x\\plus{}y\\plus{}z)^2 \\ge (x\\plus{}y\\plus{}z)(xy\\plus{}yz\\plus{}zx)\\equal{}(4\\minus{}xyz)(xy\\plus{}yz\\plus{}zx)$\r\nand we will to show that\r\n$ (x\\plus{}y\\plus{}z)^2\\plus{}xyz(xy\\plus{}yz\\plus{}zx) \\ge 4(xy\\plus{}yz\\plus{}zx)$\r\nor $ x^2\\plus{}y^2\\plus{}z^2\\plus{}xyz(xy\\plus{}yz\\plus{}zx) \\ge 2(xy\\plus{}yz\\plus{}zx)$\r\nfrom $ xy\\plus{}yz\\plus{}zx \\ge 3$ and $ x\\plus{}y\\plus{}z \\ge 3$,we have\r\n$ xyz(xy\\plus{}yz\\plus{}zx) \\ge 3xyz \\ge \\frac{9xyz}{x\\plus{}y\\plus{}z}$\r\nand finally,we will to show that:\r\n$ x^2\\plus{}y^2\\plus{}z^2\\plus{} \\frac{9xyz}{x\\plus{}y\\plus{}z} \\ge 2(xy\\plus{}yz\\plus{}zx)$\r\nBut it is schur's inequality.\r\nOur proof is completed,equality occur if and only a=b=c.", "Solution_1": "problem #1 can be done easily by Titu's Lemma \r\n[hide=\"#1\"]\n\n$ LHS \\geq \\frac{(1\\plus{}1\\plus{}1)^2}{2a^2 \\plus{}2b^2 \\plus{}2c^2 \\plus{}ab \\plus{}ac\\plus{}bc}$\n\nBy AM-GM\n $ {2a^2 \\plus{}2b^2 \\plus{}2c^2 \\plus{}ab \\plus{}bc\\plus{}ca } \\geq a^2 \\plus{}b^2 \\plus{}c^2 \\plus{}2ab\\plus{}2bc \\plus{}2ca\\equal{}(a\\plus{}b\\plus{}c)^2 \\implies LHS \\geq \\frac{9}{(a\\plus{}b\\plus{}c)^2}$\n\n[/hide]", "Solution_2": "[quote=\"enndb0x\"]problem #1 can be done easily by Titu's Lemma \n[hide=\"#1\"]\n\n$ LHS \\geq \\frac {(1 \\plus{} 1 \\plus{} 1)^2}{2a^2 \\plus{} 2b^2 \\plus{} 2c^2 \\plus{} ab \\plus{} ac \\plus{} bc}$\n\nBy AM-GM\n $ {2a^2 \\plus{} 2b^2 \\plus{} 2c^2 \\plus{} ab \\plus{} bc \\plus{} ca } \\geq a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 2ab \\plus{} 2bc \\plus{} 2ca \\equal{} (a \\plus{} b \\plus{} c)^2 \\implies LHS \\geq \\frac {9}{(a \\plus{} b \\plus{} c)^2}$\n\n[/hide][/quote]\r\n\r\nYou 're wrong :) :P", "Solution_3": "Can you become kind and tell us where is my mistake,instead of just saying \"you're wrong\" ?", "Solution_4": "[quote=\"enndb0x\"]Can you become kind and tell us where is my mistake,instead of just saying \"you're wrong\" ?[/quote]\r\n\r\nyour denominator decreased with AM-GM, so the value increased :D", "Solution_5": "[quote=\"enndb0x\"]Can you become kind and tell us where is my mistake,instead of just saying \"you're wrong\" ?[/quote]\r\n$ \\frac {9}{2(a^2 \\plus{} b^2 \\plus{} c^2)\\plus{}ab\\plus{}bc\\plus{}ca} \\le \\frac {9}{(a \\plus{} b \\plus{} c)^2}$ :)", "Solution_6": "By AM-GM\r\n$ 4(a^2\\plus{}ab\\plus{}b^2)(ab\\plus{}bc\\plus{}ca) \\le (a^2\\plus{}ab\\plus{}b^2\\plus{}ab\\plus{}bc\\plus{}ca)^2\\equal{}(a\\plus{}b)^2(a\\plus{}b\\plus{}c)^2$:)", "Solution_7": "Are you sure the second solution is yours, Bui Ngoc Anh? I think you had read here before ;) (see #11): http://www.maths.vn/forums/showthread.php?t=17775&page=2\r\n\r\n@secrets: Anh nghi la chung ta nen ton trong nhung nguoi giai. Chu neu lay loi giai cua nguoi khac roi nhan lai la cua minh la khong hay! Ok? Anh thay cung co nhieu ban o VN minh nhu the, that tiec! Hay tu tim loi giai cua chinh minh nhe! (Anh noi nhu the vi anh thay sau bai post cua anh o mathsvn co bai post cua em, nen anh nghi em doc loi giai cua anh truoc!)", "Solution_8": "Indeed, it's an old proof of can_hang, secrets.\r\n@can_hang: [hide]Chac la` do secrers vo tinh` + ko de? y' thoi anh a, nhung dung' la` em thay' gan` day co nhieu` truong` hop nhu anh noi' that :-<. Du` sao cung~ ko fai? viec cua minh` nen em ko quan tam lam'. \nHiz, ma` may' tinh' nha` em bi hong? may hom nay nen ko go~ dc them gi`, luc nao` anh ol nhay' may' cho em nhe', em lo qua', giua~ thang' 6 roi`ma` cong viec van~ chua dau vao` dau :( [/hide]", "Solution_9": "@canhang_2007 and nguoivn.\r\nEm xin l\u1ed7i,th\u1ef1c s\u1ef1 l\u00e0 em kh\u00f4ng c\u1ed1 \u00fd,t\u1ea7m hi\u1ec3u bi\u1ebft c\u1ee7a em th\u1ea5p,em c\u0169ng kh\u00f4ng r\u00f5 l\u00e0 l\u1eddi gi\u1ea3i n\u00e0y xu\u1ea5t hi\u1ec7n \u1edf \u0111\u00e2u,em h\u1ee9a l\u00e0 em s\u1ebd kh\u00f4ng bao gi\u1edd post nh\u1eefng topic nh\u01b0 th\u1ebf n\u00e0y n\u1eefa.C\u00e1m \u01a1n anh \u0111\u00e3 nh\u1eafc nh\u1edf", "Solution_10": "Sorry,but I don't read your solution,can hang.In the future,I don't learn inequalities.Thank you.:)\r\n@ canhang_2007 \r\nsr anh nha,em kh\u00f4ng c\u1ed1 \u00fd \u0111\u00e2u,v\u00ec tr\u00ed nh\u1edb em r\u1ea5t t\u1ec7.Em c\u0169ng bi\u1ebft mathlinks l\u00e0 1 trang web r\u1ea5t n\u1ed5i ti\u1ebfng,v\u00e0 ch\u1eafc ch\u1eafn r\u1eb1ng ng\u01b0\u1eddi bi\u1ebft l\u1eddi gi\u1ea3i c\u1ee7a anh r\u1ea5t nh\u00ecu,v\u00ec v\u1eady em \u0111\u00e2u d\u00e1m c\u00f4ng khai l\u1ea5y l\u1eddi gi\u1ea3i c\u1ee7a anh:D.Anh n\u1eb7ng l\u1eddi m\u1ea5t r\u00f9i,\u0111\u00e2y l\u00e0 l\u1ea7n th\u1ee9 hai anh nh\u1eafc nh\u1edf em.C\u00e1m \u01a1n anh nh\u00ecu.Anh \u0111\u1eebng cho r\u1eb1ng em l\u1ea5y l\u1eddi gi\u1ea3i c\u1ee7a anh nha.:) do em kh\u00f4ng bi\u1ebft m\u00e0 thui.\r\n\r\nTrong t\u01b0\u01a1ng lai,em c\u0169ng s\u1ebd kh\u00f4ng h\u1ecdc \u0111\u01b0\u1ee3c b\u1ea5t \u0111\u1eb3ng th\u1ee9c n\u1eefa,c\u00f3 l\u1ebd em ch\u1ec9 c\u1ed1 g\u1eafng n\u1ed1t th\u00e1ng n\u00e0y thui,em s\u1ebd ph\u1ea3i t\u1eadp trung v\u00e0o c\u00e1c m\u00f4n kh\u00e1c,n\u1ebfu kh\u00f4ng pa m\u00e1 em s\u1ebd la em.:D", "Solution_11": "Anh nh\u1edb reply nhanh nha,em kh\u00f4ng mu\u1ed1n b\u1ecb ai gi\u1eadn c\u1ea3.:(H\u00ec,anh nh\u1edb \u0111\u1ecdc k\u0129 hai b\u00e0i post c\u1ee7a em r\u00f9i h\u00e3y reply,r\u00f9i em s\u1ebd nh\u1edd mod kho\u00e1 topic n\u00e0y l\u1ea1i.Co\u00e1 l\u1ebd t\u1eeb nay em s\u1ebd kh\u00f4ng l\u00e0m b\u1ea5t \u0111\u1eb3ng th\u1ee9c n\u1eefa,bi quan ch\u00fat :D:).", "Solution_12": "\u0110\u1ec3 gi\u1ea3i th\u00edch t\u1ea1i sao em l\u1ea1i t\u00ecm ra l\u1eddi gi\u1ea3i tr\u00f9ng v\u1edbi l\u1eddi gi\u1ea3i c\u1ee7a anh.\u0110\u00f3 l\u00e0 do em \"h\u1ecdc\" \u0111\u01b0\u1ee3c AM-GM t\u1eeb anh D\u01b0\u01a1ng \u0110\u1ee9c L\u00e2m\r\n$ 4(a^2\\plus{}ab\\plus{}b^2)(ab\\plus{}bc\\plus{}ca) \\le (a\\plus{}b)^2(a\\plus{}b\\plus{}c)^2$(anh c\u00f3 th\u1ec3 xem t\u1ea1i problem of day tr\u00ean vnineqmath) r\u00f9i em c\u1ed9ng l\u1ea1i.Th\u1eadt tuy\u1ec7t,em th\u1ea5y \u0111\u00f3 l\u00e0 b\u1ea5t \u0111\u1eb3ng th\u1ee9c Iran96 c\u00f3 trong \u0111\u00f3.D\u00f9ng Iran 96,ra b\u00e0i c\u1ee7a Vasc.M\u1ed9t c\u1eadu nh\u00f3c ch\u01b0a bi\u1ebft g\u00ec v\u1ec1 b\u1ea5t \u0111\u1eb3ng th\u1ee9c l\u1ea1i c\u00f3 th\u1ec3 t\u00ecm ra 1 l\u1eddi gi\u1ea3i kh\u00e1 tuy\u1ec7t nh\u01b0 v\u1eady,em c\u1ea3m th\u1ea5y r\u1ea5t vui.:)V\u00ec d\u00f9 sao \u0111\u00f3 c\u0169ng l\u00e0 1 ch\u00fat th\u00e0nh qu\u1ea3 nh\u1ecf.\r\nL\u00fac anh post l\u1eddi gi\u1ea3i c\u1ee7a anh tr\u00ean maths.vn,th\u00fa th\u1ef1c,em kh\u00f4ng quan t\u00e2m,v\u00ec l\u00fac \u0111\u00f3 \u0111\u1ebfn m\u1ed9t ch\u00fat v\u1ec1 AM-GM em c\u00f2n ch\u01b0a v\u1eefng,b\u00e0i Iran96 l\u00e0 1 b\u00e0i m\u00e0 em r\u1ea5t s\u1ee3.V\u00ec v\u1eady,l\u1eddi gi\u1ea3i c\u1ee7a anh em kh\u00f4ng \u0111\u1ecdc k\u0129,v\u00e0 kh\u00f4ng l\u01b0u nh\u00ecu \u1ea5n t\u01b0\u1ee3ng.\r\nD\u00f9 sao,em xin l\u1ed7i anh,v\u00e0 mong anh \u0111\u1eebng k\u1ebft lu\u1eadn 1 r\u1eb1ng em \"l\u1ea5y\" l\u1eddi gi\u1ea3i c\u1ee7a anh.:)C\u00e1m \u01a1n anh.", "Solution_13": "[quote]\n(a+b+c)^2(a+b)^2-4(ab+bc+ca)(a^2+ab+b^2)=(a^2+b^2-ac-bc)^2\n[/quote]\r\n\r\nKids, I know of such solution - not mine, just in case you want to know - long before you get your ears dry. You're all pathetic !!", "Solution_14": "Em ngh\u0129 anh c\u1ea9n n\u1eb7ng l\u1eddi r\u00f9i,kh\u00f4ng th\u1ec3 t\u1eeb l\u1eddi gi\u1ea3i gi\u1ed1ng nhau m\u00e0 suy ra ch\u00e9p l\u1eddi gi\u1ea3i \u0111\u01b0\u1ee3c.C\u00f2n serec c\u0169ng k\u00f4 c\u1ea7n gi\u1ea3i th\u00edch nhi\u1ec1u,ch\u1ec9 c\u1ea7n m\u00ecnh bi\u1ebft l\u1eddi gi\u1ea3i l\u00e0 c\u1ee7a m\u00ecnh,th\u1ebf l\u00e0 \u1ed5n kh\u00f4ng c\u1ea7n ph\u00e0i vi\u1ebft nh\u1eefng 3 b\u00e0i \u0111\u1ec3 l\u00e0m vi\u1ec7c ko \u0111\u00e2u(n\u1ebfu c\u00f3 th\u1eeba th\u1eddi gian th\u00ec h\u00e3y ch\u0103m ch\u1ec9 h\u1ecdc tip B\u0110T \u0111\u1ec3 ng\u01b0\u1eddi kh\u00e1c c\u00f4ng nh\u1eadn m\u00ecnh :D )" } { "Tag": [ "number theory", "prime factorization" ], "Problem": "Here's a fun problem I thought of last night:\r\n\r\nFind all positive integers $n$ such that the proper factors of $n$ have sum and product $n$.", "Solution_1": "Cute!\r\n\r\n[hide]For the sum of proper factors to be n, n must be a perfect number.\nFor the product of proper factors to be n, n must have 4 factors (1, itself, and 2 others).\nThe only number that satisfies both conditions is 6.[/hide]", "Solution_2": "sky9073 - proof for both claims? (and especially that no other numbers exist?)", "Solution_3": "Ok. Proofs..... I'll try.\r\n\r\n[hide]For the sum of proper factors to be n, n must be a perfect number. \n- definition of perfect number.. i think.\n\nFor the product of proper factors to be n, n must have 4 factors (1, itself, and 2 others). \n- The product of proper factors of n = n^(number of factors/2 - 1).[/hide]\r\n\r\nI'll have to think about how to prove the last bit.", "Solution_4": "what's a proper factor? :blush:", "Solution_5": "Factors other than the number itself and 1.", "Solution_6": "Proof of the original (less precise) version of sky's second claim (i.e. for the product of n's proper factors to be n, n must have four factors):\r\n\r\n[hide]Say n has at least three distinct proper factors. Then if the prime factorization of n is n=(p_1^a)(p_2^b). . . then a+b+. . .>=3. For if it is equal to 2, then n=pq or p^2, p and q prime, then n only has two proper factors, a contradiction. So, other than p_1^a, p_2^b, etc, (p_1)*(p_2) must also be a proper factor of n because the sum of the exponents of primes in its factorization is >=3 so that p_1*p_2n, because the product contains all the prime factors of n with their respective exponents, as well as extra repeats of at least two of those primes.[/hide]\r\n\r\nI can prove that version of sky's claim, but I'm still thinking about how to prove this more precise version of her statement:\r\n[quote=\"sky9073\"]\nThe product of proper factors of n = n^(number of factors/2 - 1).\n[/quote]", "Solution_7": "[quote=\"tetrahedr0n\"]Here's a fun problem I thought of last night:\n\nFind all positive integers $n$ such that the proper factors of $n$ have sum and product $n$.[/quote]\r\na1+a2+a3+a4+... none of these are 1, and its equal to a1a2a3a4... and then, we cant have more than 3 factors, because a1a2a3... increases exponentially while a1+a2+... increases arithmetically. Therefore (since 2*2*2>2+2+2) it can only have 2 factors. Since 2*3>2+3, both terms are 2. so 4 is the only number", "Solution_8": "Good proof scrambled. But I still think we should work on proving sky's claim that (product of proper factors of n) = n^((number of factors)/2-1), which would be a much stronger result than anything we've proven in this discussion so far.", "Solution_9": "I think I sort of got the last bit about there being no other numbers..\r\n\r\n[hide]We could start by the fact that all the known perfect numbers are even. (I saw this in Intro NT... someone would have to prove there ARE odd perfect numbers before we can work on those)\n\nSo all known perfect numbers are divisible by 2.\n\nDenoting a number that satisfies both conditions (k is perfect and has 4 factors) k, the proper factors of k are 1, 2, and k/2.\nFrom \"the sum of proper factors of k = k,\" 1 + 2 + k/2 = 3 + k/2 = k. k/2 = 3, k = 6\nTherefore k must equal 6 and no other number....[/hide]", "Solution_10": "[quote=\"goodyfresh741\"]Good proof scrambled. But I still think we should work on proving sky's claim that (product of proper factors of n) = n^((number of factors)/2-1), which would be a much stronger result than anything we've proven in this discussion so far.[/quote]isnt that rather obvious because two proper factors of n multiply to get n. so we have n^[(n-2)/2]=n^[(n)/2-1]", "Solution_11": "[quote=\"Scrambled\"][quote=\"goodyfresh741\"]Good proof scrambled. But I still think we should work on proving sky's claim that (product of proper factors of n) = n^((number of factors)/2-1), which would be a much stronger result than anything we've proven in this discussion so far.[/quote]isnt that rather obvious because two proper factors of n multiply to get n. so we have n^[(n-2)/2]=n^[(n)/2-1][/quote]\r\n\r\nBut doesn't that only work if the number of proper factors is even? Because if the number of factors is odd, then they may not necessarily be able to be paired up the way you stated. Well, the way to do it, I just realized, is that n has an even number of proper factors if and only if n is not a perfect square. Because, any factor a|n such that a 3(q+1) = 4q --> q = 3\n\nn = pq = 6[/hide]", "Solution_14": "probability1.01, I hereby sentence you to....\r\n\r\nActually, I'm glad you posted, especially since its my question.", "Solution_15": "But if proper factors are defined as factors other than 1 and the number, then the proper factors of 6 are 2 and 3, which don't sum to 6...", "Solution_16": "According to the definition I am familiar with, the proper factors of n are all the (positive) factors except n (but not excluding 1). I have sometimes seen the term nontrivial factors used to describe the (positive) factors other than 1 and n.", "Solution_17": "but couldn't n have only the factors p and p^2 (where p is prime) so that it also satisfies the condition that the product of the proper divisors equals n (which is p^3)?", "Solution_18": "Then n would be p^3, and the proper factors 1, p, and p^2.\r\nRemember that the proper factors also have to ADD UP to n..\r\np^2 + p + 1 = p^3 doesn't give integer values for p.", "Solution_19": "oh, yes, now I remember!" } { "Tag": [], "Problem": "Express $ \\frac{1}{1\\plus{}\\frac{1}{1\\plus{}\\frac{3}{4}}}$ as a common fraction.", "Solution_1": "$ \\frac{1}{1\\plus{}\\frac{1}{1\\plus{}\\frac{3}{4}}}$\r\n\r\n$ \\frac{1}{1\\plus{}\\frac{1}{7/4}}$\r\n\r\n$ \\frac{1}{1\\plus{}\\frac{4}{7}}$\r\n\r\n$ \\frac{1}{\\frac{11}{7}}$\r\n\r\n$ \\frac{7}{11}$" } { "Tag": [ "Functional Analysis", "real analysis", "real analysis unsolved" ], "Problem": "suppose $ H$ is Hilbert space. \r\nLet $ L$ is belonging $ H$ and is linear.suppose we have $ f: L \\to R$ linear continuous map. then exist $ !$ extpanded $ F$ on $ H$, such that $ F|_{L} \\equal{} f$ and $ \\parallel{}F\\parallel{} \\equal{} \\parallel{}f\\parallel{}$", "Solution_1": "Is $ L$ a subspace or just any subset?", "Solution_2": "it's subspace", "Solution_3": "This is Hahn-Banach, only quite a bit easier because of the Hilbert space structure. If $ p$ is orthogonal projection onto $ H$, define $ F\\equal{}f\\circ p$." } { "Tag": [ "calculus", "integration", "function", "limit", "derivative", "algebra", "functional equation" ], "Problem": "For arbiterary integers $n,$ find the continuous function $f(x)$ which satisfies the following equation. \\[\\lim_{h\\rightarrow 0}\\frac{1}{h}\\int_{x-nh}^{x+nh}f(t) dt=2f(nx).\\] Note that $x$ can range all real numbers and $f(1)=1.$", "Solution_1": "Note that the limit on the LHS of the given equation has $0/0$ form; naturally the first thing I thought was we should probably use L'Hospital's rule and see where that leads us to.\r\n\r\nSo, using L'Hospital's rule and differentiation under the integral sign, we get \r\n\r\n$\\lim_{h \\to 0}\\left( f(x+nh) \\cdot n-f(x-nh) \\cdot (-n)\\right) = 2f(nx)$ ... [Note we are differentiating w.r.t. $h$]\r\n\r\n$\\Rightarrow 2nf(x) = 2f(nx)$\r\n\r\n$\\Rightarrow nf(x) = f(nx)$ ... $(*)$\r\n\r\nThe above is a nice functional equation, which resembles a lot like Cauchy's equation, viz. $f(x+y) = f(x)+f(y)$ (if I remember it correctly.) We suspect that $f(x) = x$ for all $x \\in \\mathbb{R}.$ Now, there is of course a well-known technique to handle such equations, and this is shown below.\r\n\r\n[b]Claim 1[/b]: $f(x) = x$ for all $x \\in \\mathbb{Z}.$\r\n\r\nJust put $x=1$ in $(*)$ to get $nf(1) = f(n) \\Rightarrow f(n) = n\\cdot 1 = n$ for all $n \\in \\mathbb{Z}.$\r\nAlso, for any integer $m > 1, f(0) = f(m \\cdot 0) = m\\cdot f(0) \\Rightarrow f(0) = 0.$\r\n\r\nThus, our claim is verified.[b]\n\nClaim 2[/b]: $f(x) = x$ for all $x \\in \\mathbb{Q}.$\r\n\r\nLet $x = \\frac{p}{q}$ where $p, q \\in \\mathbb{Z}$ and $q \\not =0.$ \r\nThen, $p = p \\cdot 1 = pf(1) = f(p\\cdot 1) = f(qx) = qf(x) \\Rightarrow f(x) = \\frac{p}{q}= x.$\r\n\r\nThus, our claim is again verified.\r\n\r\n[b]Claim (3)[/b]: $f(x) = x$ for all $x \\in \\mathbb{R}.$\r\nNow for any $x \\in \\mathbb{R}$, let ${x_{n}}$ be a sequence of rational numbers such that $x_{n}\\to x.$ Then, by continuity of $f,$ we have $f(x) = f(\\lim_{x_{n}\\to x}x_{n}) = \\lim_{x_{n}\\to x}f(x_{n}) = \\lim_{x_{n}\\to x}x_{n}\\cdot f(1) = \\lim_{x_{n}\\to x}x_{n}\\cdot 1 = x.$\r\n\r\nThe last claim gives us our answer, viz. $f(x) = x$ for all $x \\in \\mathbb{R}.$ And, we are done.", "Solution_2": "I rather imagined\r\n$F(x) = \\int f(x) dx$, so the LHS becomes\r\n$\\frac{F(x+nh)-F(x-nh)}{h}$\r\nsetting $x' = x-nh$ (and $dx' = dx$)\r\n$2n \\frac{F(x'+2nh)-F(x')}{2nh}= 2n F'(x') = 2n f(x')$\r\nsince we're ultimately thinking of $h \\to 0$, $2n f(x') = \\boxed{2n f(x)}$\r\n\r\n[quote=\"FieryHydra\"]So, using L'Hospital's rule and differentiation under the integral sign, we get \n\n$\\lim_{h \\to 0}\\left( f(x+nh) \\cdot n-f(x-nh) \\cdot (-n)\\right) = 2f(nx)$ ... [Note we are differentiating w.r.t. $h$]\n[/quote]\r\n\r\nWell, I didn't get how you did it with L'Hospital's. Can you do it explicitly?" } { "Tag": [ "trigonometry" ], "Problem": "Find all solutions to this equation:\r\n\r\n$ \\sin 3x \\cos x - \\cos 3x \\sin x = \\displaystyle \\frac{\\sqrt2}{2} $\r\n\r\nany quick or efficient solutions? :D thanks", "Solution_1": "[hide=\"Hint\"]\nThe LHS is equal to $\\sin(3x-x)$\n[/hide]", "Solution_2": "I suppose I'll finish it off.\n\n[hide]\n\nsin(2x) = sqrt(2)/2\n\nsqrt(2)/2 = 45 degrees and 135 degrees and 405 degrees and 495 degrees\n\nAssuming 0 <= x <= 2pi...\n\nx = 22.5 degrees, 67.5 degrees, 202.5 degrees, and 247.5 degrees.\n\n[/hide]", "Solution_3": "ahhhhhh! i see! very slick, i didnt catch the sin(a-b) formula until you pointed it out. thanks a bunch. :lol:" } { "Tag": [ "calculus", "algorithm", "videos", "linear algebra" ], "Problem": "Is calculus important to the people who specialize in computer science and technology? If your answer is \"yes\", please explain why.\r\n\r\nIn my mind, I think other math subjects with \"discreteness\", such as discrete math, linear algebra, and algorithm, are more important than calculus, which mainly focus on continuous object. However, the students majoring in computer science and technology should have a strong background on calculus. I would like to know the reason. (I major in math)", "Solution_1": "I would say it is important to 'many' undergraduates who specialize in computer science. Insofar as there is no way to tell who will or will not wind up needing it, it's reasonable to give all CS students at least some background in calculus. For graduate students, however, I would change 'many' to 'most'.\r\n\r\nA few examples of computer science fields where calculus, or at least principles similar to those underlying calculus, are used extensively: Computer graphics, robotic motion/vision, image/voice recognition, audio and video data compression, digital signal processing.\r\n\r\nBut I think the biggest one, and one that deserves its own category, is numerical computing. Essentially every field of applied science or engineering eventually needs computer algorithms to numerically simulate all sorts of models of continuous (or approximately continuous) phenomena. An understanding of calculus and differential equations in this context is absolutely necessary to develop stable and efficient algorithms (though of course a solid understanding of linear algebra is also crucial). A few examples: [url=http://en.wikipedia.org/wiki/Computational_chemistry]computational chemistry[/url], [url=http://en.wikipedia.org/wiki/Computational_physics]computational physics[/url], [url=http://en.wikipedia.org/wiki/Computational_neuroscience]computational neuroscience[/url], [url=http://en.wikipedia.org/wiki/Bioinformatics]bioinformatics[/url], [url=http://en.wikipedia.org/wiki/Neuroimaging]neuroimaging[/url]; though I'm sure others could mention many more." } { "Tag": [ "geometry", "3D geometry", "rectangle" ], "Problem": "An ice cube tray is in the form of a 2 by 8 rectangle, as shown. In how many different ways can you remove half of the ice cubes in a full tray such that none of the ice cubes remaining in the tray are next to each other horizontally or vertically?\n\n[asy]draw((0,0)--(8,0), linewidth(1));\n\ndraw((0,1)--(8,1), linewidth(1));\n\ndraw((0,2)--(8,2), linewidth(1));\n\n\ndraw((0,0)--(0,2), linewidth(1));\ndraw((1,0)--(1,2), linewidth(1));\ndraw((2,0)--(2,2), linewidth(1));\ndraw((3,0)--(3,2), linewidth(1));\ndraw((4,0)--(4,2), linewidth(1));\ndraw((5,0)--(5,2), linewidth(1));\ndraw((6,0)--(6,2), linewidth(1));\ndraw((7,0)--(7,2), linewidth(1));\ndraw((8,0)--(8,2), linewidth(1));[/asy]", "Solution_1": "the only way to do that is if the icecubes are removed diagonally\r\n\r\ntherefore there are only $ 2$ ways", "Solution_2": "Sort of \"diagonally\", if diagonally means picking cubes out in a checkerboard pattern. Basically you must pull one cube out of each column, but that determines how you pick the adjacent column all the way across. So, depending on how you pick the first cube (top or bottom) you'll get one of the two patterns." } { "Tag": [ "ARML", "email", "geometry", "calculus", "AMC", "USA(J)MO", "USAMO" ], "Problem": "yeah so basically i enjoyed it, the competition and the 7-hour bus rides\r\n\r\nsomeone threatened to do me in my sleep but i don't think he did\r\n\r\nand fanatic/mihika kinda owned me/jeffrey in kemps alsdjfim yeah we're gonna beat you next year\r\n\r\nand uhh i got a cold from playing ultimate frisbee in the rain for an hour so bye and see you all next year", "Solution_1": "to be fair, mihika and I owned SEVERAL people in kemps, not just you and jeffrey :p. Although I didn't keep track all the way, at one point our record was 25 wins, 7 losses. In the end it was probably like 54-15 or so... not counting the free wins we gave you guys when playing to 10 or 15.\r\n\r\n/bragging rights. lol... taking pride in kemps.\r\n\r\nERS... cut underneath my fingernail!! nasty... stung all day for two straight days because I cut it again the second day.\r\n\r\nExcept lyndon ignoring me and running away when I walked by, the three days were great (even playing ultimate frisbee in the rain and stepping in puddles)!\r\n\r\nOf the people that live in Indiana, I think I know like everyone's score... asked all around.", "Solution_2": "This is pretty rediculous. I was gone for 3 days and I havent even come to this site but I cant log on to my other account. I havent even logegd on so I couldnt have done anything, so I find this really weird. Also, Richard, pm me on everything rebirth said.", "Solution_3": "First off I gotta know who you are.", "Solution_4": "[quote=\"cxywe\"]This is pretty rediculous. I was gone for 3 days and I havent even come to this site but I cant log on to my other account. I havent even logegd on so I couldnt have done anything, so I find this really weird. Also, Richard, pm me on everything rebirth said.[/quote]\r\n\r\nSend me an email, as I explained in a different thread. There is a great deal of evidence that you have formed many, many accounts on the site. This is very clearly against our rules. Furthermore, some of the conduct from some of these accounts has been very inappropriate. Perhaps some of this conduct is coming from a sibling or someone else in your household, but at this point, I will continue banning the accounts I see you form (including this newest one) until I have a explanation for the behavior we have seen, and an assurance that it will not continue.", "Solution_5": "[quote=\"Fanatic\"]to be fair, mihika and I owned SEVERAL people in kemps, not just you and jeffrey :p. Although I didn't keep track all the way, at one point our record was 25 wins, 7 losses. In the end it was probably like 54-15 or so... not counting the free wins we gave you guys when playing to 10 or 15.\n\n/bragging rights. lol... taking pride in kemps.[/quote]\r\n\r\nhaha, that was great. I still don't think anyone found out our signal (or lack thereof). remember our strategy for next year so we can beat everyone all over again.\r\n\r\nI thought the trip was great. the competition, however...not so much. I did really badly. It was surprising cuz i was getting around 4's consistantly on the practice tests i took (95-03) and then i bombed the actual thing. i sucked. it was shameful\r\n\r\nDoes anyone know how they pick the teams? And does it have anything to do with going to the practices (which i conveniently didn't figure out about till they were all over)? Thanks.", "Solution_6": "They pick the teams based off scores on other contests and... drumroll... practices!\r\n\r\nYou had no shot for being on gold if you didn't do practices =/. the coaches decide touchy-feely, I don't think there's some special formula for deciding who gets on which team.", "Solution_7": "thanks. when do practices start? and they are in terre haute right?", "Solution_8": "did anyone get number 2 on the first relay and if anyone did, how did they do it. it was the one which was like \"what is the sum of all the values of x if sin x = -1/4\"", "Solution_9": "assuming $x$ is measured in radians and is defined as $0 \\leq x < 2\\pi$\r\n\r\nthe terminal sides of the two possible values of $x$ ends up in quadrants III and IV.\r\n\r\nthink of the two angles separately (drawing would help), and subtract $\\pi$ from both angles. It becomes obvious the remaining angles add up to $\\pi$, so $2\\pi+\\pi = \\boxed{3\\pi}$", "Solution_10": "omg the bus was like oxygen-deprived.\r\n\r\ni had a 7-hour headache on the bus each way. the bus was nice, but the bus ride was torture for me. i felt like throwing up so bad.\r\n\r\nso....anybody agree with me?", "Solution_11": "[quote=\"indianamath\"]omg the bus was like oxygen-deprived.\n\ni had a 7-hour headache on the bus each way. the bus was nice, but the bus ride was torture for me. i felt like throwing up so bad.\n\nso....anybody agree with me?[/quote]\r\nuh\r\ndid you go through 5 hours of INTENSE HEAT IN THE DESERT WITH NO AIR CONDITIONING?\r\nI THINK NOT\r\nYOU \r\nGOT\r\nPWNED", "Solution_12": "did you play ultimate frisbee ina thunderstorm for 40 minutes?\r\n\r\n[hide=\" :lol: \"]I \nTHINK\nNOT\nOWNT.[/hide]", "Solution_13": "Well, technically, there wasn't any thunder...\r\n\r\nWest Lafayette people, how long do you think it'll take for you guys to drive to IUPUI? We have math practices there at times, and it's free (not for ARML, just really hard math in general). I'll talk to Dr. Ji later?", "Solution_14": "[quote=\"indianamath\"]omg the bus was like oxygen-deprived. \n\ni had a 7-hour headache on the bus each way. the bus was nice, but the bus ride was torture for me. i felt like throwing up so bad. \n\nso....anybody agree with me?[/quote]\r\n\r\nit was really hot on the way back. u were lucky u slept through most of it. i swear, temperature must've broken 90 somewhere along the way. the bus was air-conditioned on the way there, so it wasn't that bad.", "Solution_15": "i had a headache the whole ride both ways.", "Solution_16": "[quote=\"indianamath\"]did you play ultimate frisbee ina thunderstorm for 40 minutes?\n\n[hide=\" :lol: \"]I \nTHINK\nNOT\nOWNT.[/hide][/quote]\r\nwell we did play ultimate frisbee in the sprinklers\r\ndoes that count\r\nNOT TO MENTION THE INTENSE HEAT OF LAS VEGAS\r\n\r\nPWNED", "Solution_17": "Just wondering -- why're you visiting indiana's forum?\r\n\r\nBecause we rock? \r\n\r\nlol\r\n\r\nwe're a bunch of 7th, 8th, and 9th graders =/", "Solution_18": "dudeeee.. playing ultimate in the storm was like the highlight of my week\r\n\r\nand yeah the bus ride was torture and i also had a stomach ache.\r\n\r\nwell the last time i've been to IUPUI (science fair asldifjlke) i think it took a little more than an hour, but i could be wrong\r\n\r\nuhh by \"hard math\" do you mean beyond ARML level? cause i was hopeless at last thursday's practice.. :blush:", "Solution_19": "[quote=\"Fanatic\"]Just wondering -- why're you visiting indiana's forum?\n\nBecause we rock? \n\nlol\n\nwe're a bunch of 7th, 8th, and 9th graders =/[/quote]\r\nNo 6th graders? Come on :( If I had gone to ARML, I would have gotten 0, so I didn't go. I don't think any 6th graders went from MI this year, so I'm fine. Maybe next year :ninja:", "Solution_20": "is it just me, or did the ARML individual have WAY more geometry than normal?", "Solution_21": "geometry is cool. ish.\r\n\r\nsorry if that sounds hypocritical, but it's how i feel right now.\r\n\r\naaaand i just took the 2001 AMC 10 without a calculator and got a 138 alsidjflkefmasdf NOT GOOD ENOUGH", "Solution_22": "let's go exeter!\r\n\r\nhaha sorry. i was on the indiana team last year but then changed schools to exeter. who was on the gold team this year?", "Solution_23": "Like, no one...\r\n\r\nNathan Benjamin (9th)\r\nPhil Thomas (10th)\r\nSachin Shinde (9th)\r\nSamuel Dittmer (10th? 11th?)\r\nE.G. Wright (11th, I think)\r\nAshley Neagle (9th? 10th? 11th?)\r\nTyler Zhou (10th? 11th?)\r\nLauren Stephens (9th)\r\nJeffrey Shen (8th)\r\nRichard Ni (8th)\r\nSoo-Hyun Yoo (8th)\r\nYuqing Zhang (7th)\r\nLyndon Ji (7th)\r\nShawn Qian (9th)\r\n??? can't think of last person\r\n\r\ntwo 7th\r\nthree 8th\r\nfour 9th", "Solution_24": "so what are all the subjects you need to know to do well in ARML? I saw some of mathcrazed's posts that you need to know all of high school math (like up to calculus), but then i also heard (from what i consider a pretty reliable source) that most of it is under precalc level, with the occasional precalc problem.\r\n\r\nso i'm planning to finish the geometry and hopefully algebra II honors (precalc) texbooks this summer. will that be enough to have a good knowledge base for ARML?", "Solution_25": "the last person was yixin.\r\n\r\nand sam is going into 11th.", "Solution_26": "i just took some amc 12 without a calc and got a 103.5(ouch ouch ouch ouch)\r\n\r\ni think i got like 15 correct, 1 wrong, and 9 ommited.", "Solution_27": "You guys sent a recruitment letter to me via my school. That was pretty shady... :ninja:", "Solution_28": "huh? what are you ta lkinga bout?", "Solution_29": "[quote=\"Fanatic\"]Just wondering -- why're you visiting indiana's forum?\n\nBecause we rock? \n\nlol\n\nwe're a bunch of 7th, 8th, and 9th graders =/[/quote]\r\nbecause its spamtastic", "Solution_30": "[quote=\"Fanatic\"]Just wondering -- why're you visiting indiana's forum?\n\nBecause we rock? \n\nlol\n\nwe're a bunch of 7th, 8th, and 9th graders =/[/quote]\r\nNo 6th graders?? :o", "Solution_31": "i think we had one.\r\n\r\nthe guy from honey creek named Alex.", "Solution_32": "Mark Botros is a 7th grader (going into 8th).\r\n\r\nAlex is a 6th grader, was on the silver team.", "Solution_33": "[quote=\"indianamath\"]the last person was yixin.\n\nand sam is going into 11th.[/quote]\r\nim pretty sure sam is going into 12th", "Solution_34": "ah i was just glad i was on the gold team.", "Solution_35": "[quote=\"Fanatic\"]Well, technically, there wasn't any thunder...\n\nWest Lafayette people, how long do you think it'll take for you guys to drive to IUPUI? We have math practices there at times, and it's free (not for ARML, just really hard math in general). I'll talk to Dr. Ji later?[/quote]\r\n\r\nit takes about an hour to get to indianapolis. when do these practices start? is there some sort of schedule that they follow?\r\n\r\nand when do the arml practices for this year start (if anyone knows)?", "Solution_36": "[quote=\"mwpl11\"][quote=\"Fanatic\"]Well, technically, there wasn't any thunder...\n\nWest Lafayette people, how long do you think it'll take for you guys to drive to IUPUI? We have math practices there at times, and it's free (not for ARML, just really hard math in general). I'll talk to Dr. Ji later?[/quote]\n\nit takes about an hour to get to indianapolis. when do these practices start? is there some sort of schedule that they follow?\n\nand when do the arml practices for this year start (if anyone knows)?[/quote]\r\n\r\nhmm ill just bump this back up.\r\n\r\nafter USAMO is where it starts. I think first couple of weeks is at Dr. Ji's class (IUPUI) and then its at Terre Haute at Rose Hulman.\r\nI dont think ill be able to go to most of them (only the ones after like May 15ish) so I hope they dont use your practice scores for placement.\r\n\r\nAnd btw, did any of you guys sign up for ICTM. I dont think I did, but i was wondering can you just show up and take the test (i probably wont go-and i forgot to sign up anyways-but just in case im not busy, can i just show up to it or something?)", "Solution_37": "who's going to ARML this year?", "Solution_38": "i think i will, if i am allowed to go by the rest of indiana.", "Solution_39": "I will, except I live in Michigan.", "Solution_40": "somemod should make this 2008 so no one is confused.", "Solution_41": "I'm going. I jokingly mentioned to Dr. Ji that I wanted to bring a couple non-mathnerd friends and he was like \"yeah! the more the merrier!\"... so I'm pretty sure that there'll be a spot for you and your (I forget what it was, but it was pretty good) score on the AIME." } { "Tag": [ "algebra", "polynomial", "Rational Root Theorem", "complex numbers", "number theory" ], "Problem": "hi !!\r\nfind the number of roots$ \\in{\\mathbb{Q}}$ for this equation :\r\n\r\n$ x^{4}+x^{3}+x^{2}-6=0$\r\n\r\nthanx.", "Solution_1": "Just use this ([url=http://en.wikipedia.org/wiki/Rational_root_theorem]Rational root theorem[/url]): if polynomial $ a_nx^n \\plus{} a_{n \\minus{} 1}x^{n \\minus{} 1} \\plus{} \\ldots \\plus{} a_1x \\plus{} a_0$ with integer coefficients has a rational root $ \\frac {k}{l}$ ($ gcd(k,l) \\equal{} 1$), then $ k|a_0$ and $ l|a_n$. In your case, it will be particularily easy to check all the possible roots - there will not be many of them and they will all be integers.\r\n\r\n[size=75](Just a note: I don't think this question belongs in this subforum.)[/size]", "Solution_2": "Checking the possibilities:$ \\pm 1, \\pm2, \\pm3, \\pm6$, we find that there are no integer roots and therefore, since the polynomial is monic, no rational roots. \r\n\r\nDifferentiating, we get $ 4x^{3} \\plus{} 3x^{2} \\plus{}2x$. Setting this equal to zero, we see that there is only one turning point: a local minimum at $ (0, \\minus{}6)$. \r\n\r\nAlso since, for $ f(x) \\equal{} x^{4} \\plus{} x^{3} \\plus{} x^{2} \\minus{}6$, $ f(\\minus{}2) \\equal{} 6, f(\\minus{}1) \\equal{} \\minus{}5; f(1) \\equal{} \\minus{}3, f(2)\\equal{} 28$, we find that there are two real roots, one between $ \\minus{}2$ and $ \\minus{}1$ and one between $ 1$ and $ 2$ and therefore two complex (conjugate) roots.", "Solution_3": "[hide=\"Number Theory solution\"]Assume that there is indeed a rational solution. Since it isn't 0, we know that it can be expressed as a fraction $ \\frac{p}{q}$, where $ \\gcd(p,q)\\equal{}1$, and $ \\frac{p^4}{q^4}\\plus{}\\frac{p^3q}{q^4}\\plus{}\\frac{p^2q^2}{q^4}\\minus{}\\frac{6q^4}{q^4}\\equal{}0$, which means $ p^4\\plus{}p^3q\\plus{}p^2q^2\\minus{}6q^4\\equal{}0$. Since $ \\frac{p}{q}\\neq \\pm 1$, we know that one of $ |p|$ or $ |q|$ is greater than 1.\n\n[b]Case 1: $ |p|>1$[/b]\nThen we know that there exists a prime $ p_1$ such that $ p_1|p$. Note that $ p_1^2|p^4\\plus{}p^3q\\plus{}p^2q^2$. Since $ p^4\\plus{}p^3q\\plus{}p^2q^2\\minus{}6q^4\\equal{}0$, $ p_1^2|6q^4$. This means that $ p_1|q$. But this violates the assumption that $ \\gcd(p,q)\\equal{}1$, and hence there is no rational root such that $ |p|>1$.\n\n[b]Case 2: $ |q|>1$[/b]\nThen we know that there exists a prime $ q_1$ such that $ q_1|q$. Note that $ q_1|p^3q\\plus{}p^2q^26q^4$. Since $ p^4\\plus{}p^3q\\plus{}p^2q^2\\minus{}6q^4\\equal{}0$, $ q_1|p^4$. This means that $ q_1|p$. But this violates the assumption that $ \\gcd(p,q)\\equal{}1$, and hence there is no rational root such that $ |q|>1$.\n\nThis means that there are no rational roots period.[/hide]", "Solution_4": "Er, 1=2, I think that's essentially how you prove the rational root theorem..." } { "Tag": [ "algebra", "polynomial", "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "$p$ is prime.\r\nIs it true that $\\forall A \\in \\mathcal{M}_{n}(\\mathbb{Z}),\\ \\mu_{A^{p}}\\equiv\\mu_{A}~[p] \\textnormal{,}$ where $\\mu_{A}$ is the minimal polnomial of $A$?\r\n\r\nNote: It is true for the characteristic polynomial $\\chi_{A}$.", "Solution_1": "I don't recognize the notation- is that meant to be the minimal polynomial of $A$ considered as a matrix mod $p$, compared to $\\mu_{A}$ considered mod $p$?\r\n\r\nIf so, it's trivially false; let $M$ be any nonzero nilpotent matrix, and let $A=pM$. The two minimal polynomials have different degrees.", "Solution_2": "jmerry,\r\n\r\nYou're right, it was obvious, I should have thought about it first:)\r\nTry to prove it for the characteristic polynomial, it's less obvious (and true this time).", "Solution_3": "one can trigonalize $A$ in an algebraic closure $K$ of $\\mathbb{F}_{p}$ and use the fact that $(a+b)^{p}= a^{p}+b^{p}$ for $a,b \\in K$. \r\n$\\chi_{A}(x)=\\prod (x-\\alpha_{i}) = \\sum a_{k}x^{k}$ and $\\chi_{A^{p}}= \\prod (x-\\alpha_{i}^{p}) = \\sum b_{k}x^{k}$. \r\n\r\nThere are symetric polynomials $P_{k}$ such that $P_{k}(\\alpha_{1}, \\ldots, \\alpha_{n}) = a_{k}$ and $P_{k}(\\alpha_{1}^{p}, \\ldots, \\alpha_{n}^{p}) = b_{k}$. But $b_{k}=P_{k}(\\alpha_{1}^{p}, \\ldots, \\alpha_{n}^{p}) = P_{k}(\\alpha_{1}, \\ldots, \\alpha_{n})^{p}= a_{k}^{p}$. And because $a_{k}\\in \\mathbb{F}_{p}$, $a_{k}^{p}= a_{k}$. This shows $\\chi_{A}= \\chi_{A^{p}}$ modulo $p$.", "Solution_4": "I prefer this one-liner:\r\nThe coefficients of the characteristic polynomial are polynomial (with integer coefficients) in the entries of $A$. These polynomials commute with ring homomorphisms.", "Solution_5": "I don't realy get it :D but maybe, if you elaborate too much it will achieve 4 lines :)" } { "Tag": [ "spam" ], "Problem": "I am the moderator for the group. I started it to increase communication in the state, and to give the national team member somewhere to ask each other questions.", "Solution_1": "Gee, can't Oklahoma be in the spotlight for just a smidgeon?", "Solution_2": "oh dude I remember oklahoma! that's with you and neal and the crazy indian guy OH MAN IT'S OKLAHOMA hmm was the school jenkins? I believe it was jenkins? perhaps?", "Solution_3": "lol, it's close enough to Jenks", "Solution_4": "hey i know neal and his family really well", "Solution_5": "Dis is probably a dead topic, but there is like no one here. I MUST CREATE POPULARITY FOR THE OKLAHOMA FORUM!!!! :D" } { "Tag": [ "function", "analytic geometry", "graphing lines", "slope", "calculus", "derivative" ], "Problem": "If a fucntion is x^3 - x^2 -4x+4, and there is a point (a,b) on the graph on which the tangent line passes through the point (0,-8), what are the values of a and b.\r\n\r\nAnother one:\r\nWith the graph of x^2-xy+y^2=9, at the point (0,3) find the rate of change of the slope of the curve with respect to x", "Solution_1": "i didn't quite understand your first problem but the second problem seems very easy:\r\njust figure out the second derivative of the curve wtith respect to x, i got:\r\nsecond derivative: [((y-2x)/(2y-x))-2]/[((2y-4x)/(2y-x))-1]\r\nplugging in the values of x and y, you get:\r\nsecond derivative equals: -1.5/0. which means that, the rate of change of slope of the curve is infinity.\r\nin orhter vords, the slope of the curve is not differentiable at that point.", "Solution_2": "Anyway I dont really understand the question but shouldn't it be moved to the Calculus section.", "Solution_3": "For the first, b = f'(a)*a - 8. $f'(x) = 3x^2 - 2x - 4$. Thus $b = 3a^3 - 2a^2 - 4a - 8$. Now we also have b = f(a) $\\implies b = a^3 - a^2 - 4a + 4$. Now we have $2a^3 - a^2 - 12 = 0$. Go solve the cubic on your own :P.\r\n\r\nEdit--\r\n\r\n$2a^3 - a^2 - 12 = (a - 2)(2a^2 + 3a + 6)$\r\n\r\nSo a = 2.", "Solution_4": "$2a^3-a^2-12 = (a-2)(2a^2+3a+6) \\Rightarrow a=2$ is the only real solution.\r\n\r\nEDIT - probability's edit beat me to it." } { "Tag": [ "floor function", "modular arithmetic" ], "Problem": "Given a positive integer $a_{0}$, we construct a sequence as follows: If the unit digit of $a_{i}$ does not exceed $5$, then $a_{i+1}$ is obtained by deleting this digit (If nothing remains upon this deletion, the sequence ends). Otherwise, $a_{i+1}=9a_{i}$. Can the sequence be infinite?", "Solution_1": "[hide]\nThe sequence cannot be infinite. Suppose there exists such an infinite sequence. Let $L$ be the least positive integer such that $a_{0}=L$ produces infinite sequence. The units digit of $L$ must exceed 5, since the sequence ends automatically if $1\\leq L\\leq 5$, and if $L>5$ but has units digit at most 5, then\n\\[a_{1}=\\left\\lfloor \\frac{L}{10}\\right\\rfloor \\leq \\frac{L}{10}2$. So $ |z_1|\\neq 0$ and this is equivalent to :\r\n\r\n$ |z_1^2\\minus{}3|\\equal{}1$ and $ |z_1|>2$. So $ z_1^2$ is on the circle centered on $ (3,0)$ and whose radius is $ 1$. \r\n\r\nBut for all these points, we have $ |z_1|^2\\leq 4$ and so $ |z_1|\\leq 2$\r\n\r\nSo no solution." } { "Tag": [ "topology", "real analysis", "real analysis unsolved" ], "Problem": "Let $(X,d)$ be a metric space, and $f:X\\to X$ a map s.t. $d(f(x),f(y))0$. Arguing as before we have that\r\n\\[d(f(l),f(f((l)))\\leq \r\nd(f(l),a_{n_{k}+1})+d(a_{n_{k}+1},a_{n_{k}+2})+d(a_{n_{k}+2},f(f(l))),\\]\r\nand\r\n\\[d(f(l),f(f((l)))\\geq d(a_{n_{k}+1},a_{n_{k}+2})-d(f(l),a_{n_{k}+1})-\r\nd(a_{n_{k}+2},f(f(l))),\\]\r\nand letting $k\\to +\\infty$ we deduce that $d(f(l),f(f(l)))= \\lambda=d(l,f(l))$, and this is absurd if $f$ is a weak contraction.", "Solution_2": "Let me add two more problems inspired by grobber's post.\r\n\r\n1 - Assume that $f$ is a weak contraction with a fixed point $l$. Is it true that Grobber's sequence has a subsequence converging to $l$ for every $a$?\r\n\r\n2 - Assume that $f$ is a weak contraction and that, for a given $a$, Grobber's sequence has a subsequence converging to $l$. Is it true that the whole sequence converges to $l$?", "Solution_3": "The answer to the second one is 'yes', since we know that $\\ell$ is a fixed point, so we have $d(f^{(n+1)}(a),\\ell)=d(f^{(n+1)}(a),f(\\ell))1,\\ f(x)=\\frac{-1+x}2,\\ \\forall x<-1$.", "Solution_4": "[quote=\"grobber\"] i.e. the distance from $f^{(n)}(a)$ to $\\ell$ decreases as $n\\to\\infty$.[/quote]\n\nBut this doesn't mean that the distance tends to 0 ...\n\n[quote=\"grobber\"]As for the first one, I don't think it's true. We can take $X$ to be the subspace of $\\mathbb R$ consisting of $0$ and $[-2,-1)\\cup(1,2]$. We can now put $f(0)=0$ and $f(x)=\\frac{1+x}2,\\ \\forall x>1,\\ f(x)=\\frac{-1+x}2,\\ \\forall x<-1$.[/quote]\r\n\r\nRight.", "Solution_5": "[quote=\"Xamog\"][quote=\"grobber\"] i.e. the distance from $f^{(n)}(a)$ to $\\ell$ decreases as $n\\to\\infty$.[/quote]\n\nBut this doesn't mean that the distance tends to 0 ...\n\n[/quote]\r\n\r\nIt does here, since it tends to $0$ for a subsequence (remember, we're talking about your second problem here, and you said there was a subsequence converging to $\\ell$).", "Solution_6": "Oh, you're right :blush:" } { "Tag": [ "quadratics", "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "can we choose some numbers from n, n+1, n+2, n+3 ,..., n+27, product of chosen numbers equals to product of other numbers?\r\nn is natural number.", "Solution_1": "Very easy!\r\nIf 29 divide not one of the terms\r\nn(n+1).......(n+27)=-1mod29 or -1 isn't a quadratic residue mod 29.\r\nIf 29 divide one of the terms , 29 is a factor only for one of the terms! so 29 can't be a factor of the other hand of the equality. \r\nSo there is no solution.", "Solution_2": "[quote=\"Evariste-Galois\"]Very easy!\nIf 29 divide not one of the terms\nn(n+1).......(n+27)=-1mod29 or -1 isn't a quadratic residue mod 29.\nIf 29 divide one of the terms , 29 is a factor only for one of the terms! so 29 can't be a factor of the other hand of the equality. \nSo there is no solution.[/quote]\r\n\r\nNice :)", "Solution_3": "[quote=\"Evariste-Galois\"]Very easy!\nIf 29 divide not one of the terms\nn(n+1).......(n+27)=-1mod29 or -1 isn't a quadratic residue mod 29.\nIf 29 divide one of the terms , 29 is a factor only for one of the terms! so 29 can't be a factor of the other hand of the equality. \nSo there is no solution.[/quote]\r\n\r\n$ \\minus{}1$ is a quadratic residue $ \\pmod{29}$ since $ 12^2 \\equiv \\minus{}1 \\pmod{29}$", "Solution_4": "that's right!! I did a mistake in calculation. :blush: \r\nI will check this later!", "Solution_5": "[quote=\"pco\"][quote=\"Evariste-Galois\"]Very easy!\nIf 29 divide not one of the terms\nn(n+1).......(n+27)=-1mod29 or -1 isn't a quadratic residue mod 29.\nIf 29 divide one of the terms , 29 is a factor only for one of the terms! so 29 can't be a factor of the other hand of the equality. \nSo there is no solution.[/quote]\n\nNice :)[/quote]\r\n\r\nNot so nice, ... :blush:" } { "Tag": [], "Problem": "Do the following:\r\n/ignore LanguageFilter\r\n\r\nYou actually don't receive the warning \"Hey! No...language\", but sadly, you still get kicked.", "Solution_1": "Um.. That doesn't work for me! :(" } { "Tag": [ "calculus", "derivative", "function", "calculus computations" ], "Problem": "How do you find the derivative of an inverse?\r\nFor example, if given that [i]f'(-2)=-7[/i], how do you find the original function?\r\n\r\nThanks[/i]", "Solution_1": "[quote=\"iamhe\"]How do you find the derivative of an inverse?\nFor example, if given that [i]f'(-2)=-7[/i], how do you find the original function?\n\nThanks[/i][/quote] Do you know about the inverse rule theorem?\r\n\r\n[hide]I think it is something like this. Let $ g$ be the inverse of $ f$ i.e if $ y\\equal{}f(x)$, then $ x\\equal{}g(y)$. Suppose that $ f$ is differentiable at $ g(a)$ and the derivative is non zero. Then $ g$ is differentiable at $ a$ and $ \\frac{dg}{dy}(a)\\equal{}\\frac{1}{\\frac{df}{dx}(g(a))}$. In other words: $ g'(y)\\equal{}\\frac{1}{f'(g(y))}$. Can you use this?[/hide]", "Solution_2": "Quite simply. We go by the rule $ f(f^{\\minus{}1}(x)) \\equal{} x$ and use implicit differentiation (don't forget the chain rule). The result is pretty clear from there.", "Solution_3": "Uhhh,\r\nCould you please elaborate on the implicit differentiation and the chain rule? I am a beginner at calculus and don't understand what you are talking about.\r\nI think I know a little about the chain rule. \r\nIsn't that the derivative of a composition?\r\n\r\nThanks for the help. :)", "Solution_4": "Yes, the chain rule is how you differentiate a composition. It's all you need to calculate the derivative of an inverse- if you assume that inverse is differentiable.\r\nThere are theorems that imply such functions are differentiable when they \"should\" be- those theorems belong to multivariable calculus." } { "Tag": [ "inequalities" ], "Problem": "\u0388\u03c3\u03c4\u03c9 $ n>2$ \u03ad\u03bd\u03b1\u03c2 \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $ x_{1},x_{2},...,x_{n}$ \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03af \u03bc\u03b5 \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 $ 1$. \u039d\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 $ \\prod^{n}_{i\\equal{}1}\\left(1\\plus{}\\frac{1}{x_{i}}\\right)\\geq \\prod^{n}_{i\\equal{}1}\\left(\\frac{n\\minus{}x_{i}}{1\\minus{}x_{i}}\\right)$.", "Solution_1": "Prepei na ginei xrisi tis anisotitas tou Huygens...otan exo xrono tha valo tin lysi..nomizo tin exo dei sto Crux tin askisi", "Solution_2": "\u0392\u03b1\u03b6\u03c9 \u03bc\u03b9\u03b1 \u03bb\u03c5\u03c3\u03b7 \u03c3\u03c4\u03b1 \u03b3\u03c1\u03b7\u03b3\u03bf\u03c1\u03b1 \u03bf\u03c0\u03bf\u03c4\u03b5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03b5\u03c7\u03b5\u03b9 \u03be\u03b5\u03c6\u03c5\u03b3\u03b5\u03b9 \u03ba\u03b1\u03c4\u03b9\r\n\r\n[hide]\u0391\u03c0\u03bb\u03b1 $ n \\minus{} 1$ \u03c6\u03bf\u03c1\u03b5\u03c2 Jensen \u03c3\u03c4\u03b7\u03bd $ f(x) \\equal{} ln\\left(1 \\plus{} \\frac {1}{x}\\right)$ \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 $ n$ \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2 \u03b1\u03bd\u03b1 $ n \\minus{} 1$ \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03b5\u03c3\u03b7 \u03ba\u03b1\u03c4\u03b1 \u03bc\u03b5\u03bb\u03b7.[/hide]" } { "Tag": [ "number theory", "greatest common divisor", "number theory solved" ], "Problem": "Prove that $x^4-y^4=z^2\\ ,\\ (x,y)=1$ has no Not obvious answer in $\\mathbb{Z}.$", "Solution_1": "Consider k^2 - 2x^2 k + y^4. It must have integer roots. Now, the product of the roots is y^4 while the sum is 2x^2. Put the roots as $x^2 + d, x^2 - d$. Then the product is $x^4 - d^2$ so that $x^4 + y^4 = d^2$. (*)\r\n\r\nLemma. If there exists a solution to (*) (x,y,d); then there exists another solution (a,b,c) with c < d.\r\n\r\n$(x^2, y^2, d)$ is a pythagorean triple. If they have a common factor, dividing by this factor yields a smaller pythag triple $(a^2,b^2,c) c \\sqrt{xy}$\r\n\r\n(We know that they are not equal since x and y are not equal)\r\n\r\nSince the digits of the AM can be reversed to make the GM let's let the left side equal AB and the right side equal BA.\r\n\r\n$ AB > BA$\r\n\r\nThis implies that A > B.\r\n\r\nMaybe you can go on from here?", "Solution_3": "Since $ \\sqrt {xy} \\equal{} G$, $ xy \\equal{} G^2$. Similarly, since $ \\frac {x \\plus{} y}{2} \\equal{} A$, $ x \\plus{} y \\equal{} 2A$. Solving for $ x$ and $ y$, we get $ A\\pm{\\sqrt {A^2 \\minus{} G^2}}$. Hence if $ (a,b,c)$ are pythagorean triples such that $ a$, $ b$, and $ c$ are all two digit numbers, there are two solutions for $ x$ and $ y$, namely $ (c \\minus{} a,c \\plus{} a)$ and $ (c \\minus{} b,c \\plus{} b)$. So $ A \\equal{} 65$, and $ G \\equal{} 56$, and $ x \\equal{} 32$, $ y \\equal{} 98$.", "Solution_4": "[hide=\"wrong solution :D \"]\nLet's say that $ \\frac {x \\plus{} y}{2} \\equal{} 10a \\plus{} b \\cdots \\alpha$ and $ \\sqrt {xy} \\equal{} a \\plus{} 10b \\cdots \\beta$\n\nThen, we have $ 10\\alpha \\minus{} \\beta \\equal{} 5(x \\plus{} y) \\minus{} \\sqrt {xy} \\equal{} 100a \\plus{} 10b \\minus{} a \\minus{} 10b \\equal{} 99a$\n\nTake a look at $ \\sqrt {xy}.$ Since $ x$ and $ y$ are both two-digit integers, maximum value of $ \\sqrt {xy}$ is 99.\n\nFirst, we have to find how many value can $ x \\plus{} y$ have.\n\nObtimizing, we have $ 5(x \\plus{} y) \\equal{} 99(1 \\plus{} a) \\implies a \\equal{} 4, 9$\n\nIf $ a \\equal{} 4$, $ x \\plus{} y \\equal{} 99$, which can't be true because $ \\frac {x \\plus{} y}{2}$ is an integer.\n\nIf $ a \\equal{} 9$, $ x \\equal{} y \\equal{} 99$. Then, we have $ \\sqrt {xy}$=99, which also can't be true because $ a \\equal{} b \\equal{} 9$.\n\nThus, there is no answer.[/hide]\r\n\r\nTwin Prime Conjecture: $ 32\\times 98\\neq 33^2$", "Solution_5": "[quote=\"FantasyLover\"][hide=\"solution\"]\nLet's say that $ \\frac {x \\plus{} y}{2} \\equal{} 10a \\plus{} b \\cdots \\alpha$ and $ \\sqrt {xy} \\equal{} a \\plus{} 10b \\cdots \\beta$\n\nThen, we have $ 10\\alpha \\minus{} \\beta \\equal{} 5(x \\plus{} y) \\minus{} \\sqrt {xy} \\equal{} 100a \\plus{} 10b \\minus{} a \\minus{} 10b \\equal{} 99a$\n\nTake a look at $ \\sqrt {xy}.$ Since $ x$ and $ y$ are both two-digit integers, maximum value of $ \\sqrt {xy}$ is 99.\n\nFirst, we have to find how many value can $ x \\plus{} y$ have.\n\nObtimizing, we have $ 5(x \\plus{} y) \\equal{} 99(1 \\plus{} a) \\implies a \\equal{} 4, 9$\n\nIf $ a \\equal{} 4$, $ x \\plus{} y \\equal{} 99$, which can't be true because $ \\frac {x \\plus{} y}{2}$ is an integer.\n\nIf $ a \\equal{} 9$, $ x \\equal{} y \\equal{} 99$. Then, we have $ \\sqrt {xy}$=99, which also can't be true because $ a \\equal{} b \\equal{} 9$.\n\nThus, there is no answer.[/hide]\n\nTwin Prime Conjecture: $ 32\\times 98\\neq 33^2$[/quote]\r\n\r\nmy bad, $ 33$ is supposed to be $ 56$, I edited it, thanks for pointing out.", "Solution_6": "The following is a detailed explanation of what would otherwise be a fast guess-and check method.\r\n\r\n[hide=\"Systematic Search\"]AM-GM must be a multiple of 9 here. Set up the equation:\n\n$ \\frac {x \\plus{} y}{2} \\minus{} \\sqrt {xy} \\equal{} 9 N \\implies \\left( \\sqrt {x} \\minus{} \\sqrt {y} \\right) ^2 \\equal{} 18 N$\n\nThe easiest case to treat is $ N \\equal{} 1$. We have $ \\sqrt {x} \\minus{} \\sqrt {y} \\equal{} 3 \\sqrt {2}$, so we decide that $ x \\equal{} 2 (v \\plus{} 3)^2$ and $ y \\equal{} 2 v^2$. Their respective AM and GM are $ v^2 \\plus{} (v \\plus{} 3)^2$ and $ 2v(v \\plus{} 3)$. We easily find that $ v \\equal{} 4$ generates the only valid integer solution, so $ (x,y) \\equal{} (98,32)$, and their AM/GM's are $ 65, \\; 56$\n\nAnother really easy case to treat is $ N \\equal{} 2$. We have $ \\sqrt {x} \\minus{} \\sqrt {y} \\equal{} 6$, an integer, so both $ x$ and $ y$ must be perfect squares. We check pairs of squares whose squareroots differ by 6 and find no solutions.\n\nFor $ N \\equal{} 3$, we find $ \\sqrt {x} \\minus{} \\sqrt {y} \\equal{} 3 \\sqrt {6}$. Thus, we conclude that $ x \\equal{} 6 (v \\plus{} 3)^2$ and $ y \\equal{} 6 v^2$, whose AM and GM are $ 6v^2 \\plus{} 18v \\plus{} 27$ and $ 6v^2 \\plus{} 18v$. $ AM \\leq 100$, so $ v \\leq \\approx 2.2 \\implies v \\equal{} 1,2$. Neither generates an appropriate solution.\n\nIt's easy to see that we're pretty much hopeless now. If you wish to continue, it looks like this:\n\n[hide]For $ N \\equal{} 4$, we find $ \\sqrt {x} \\minus{} \\sqrt {y} \\equal{} 6 \\sqrt {2}$. Thus, we conclude that $ x \\equal{} 2 (v \\plus{} 6)^2$ and $ y \\equal{} 2 v^2$, whose AM and GM are $ 2v^2 \\plus{} 12v \\plus{} 36$ and $ 2v^2 \\plus{} 12v$. $ AM \\leq 100$, so $ v \\leq \\approx 3.4 \\implies v \\equal{} 1,2,3$. None generate an appropriate solution.\n\n$ N \\equal{} 5$ gives us the following expression:\n$ \\sqrt {x} \\minus{} \\sqrt {y} \\equal{} 3 \\sqrt {10} \\; \\implies \\; x \\equal{} 10 (v \\plus{} 3)^2, \\; \\; y \\equal{} 10 v^2$, whose AM is $ 10 v^2 \\plus{} 30 \\plus{} 45$. $ AM \\leq 100 \\implies v \\leq \\approx 1.3 \\implies v \\equal{} 1$, which doesn't work. We realize that for any more odd numbers, we'll just be in a worse situation (next up is $ 3 \\sqrt {14}$ for $ N \\equal{} 7$), so we give up with odd numbers. Let's see how the next even number fares:\n\nFor $ N \\equal{} 6$, we find $ \\sqrt {x} \\minus{} \\sqrt {y} \\equal{} 6 \\sqrt {3} \\; \\; \\implies x \\equal{} 3 (v \\plus{} 6)^2, \\; \\; y \\equal{} 3 v^2$, whose AM is $ 3 v^2 \\plus{} 18v \\plus{} 54$. $ AM \\leq 100$, so $ v \\leq \\approx 1.9 \\implies v \\equal{} 1$, which doesn't work. We realize that we're in the same dunghole for the remaining even numbers, so we end our search.[/hide]\nThe unique solution is $ (x,y) \\equal{} (98,32)$, with $ AM \\equal{} 65$ and $ GM \\equal{} 56$[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "let $ a$,$ b$ and $ c$ positives reals numbers such that $ a\\plus{}b\\plus{}c\\equal{}1$.\r\nshow that:\r\n$ \\frac{a^3}{a^2\\plus{}b^2}\\plus{}\\frac{b^3}{b^2\\plus{}c^2}\\plus{}\\frac{c^3}{c^2\\plus{}b^2}\\geq\\frac{1}{2}$", "Solution_1": "\\[ \\frac{a^3}{a^2\\plus{}b^2}\\plus{}\\frac{b^3}{b^2\\plus{}c^2}\\plus{}\\frac{c^3}{c^2\\plus{}b^2}\\geq\\frac{1}{2} \\leftrightarrow \\sum_{cyc}\\frac{ab^2}{a^2\\plus{}b^2} \\le \\frac{1}{2}\\] \r\nwhich is obviously true by am-gm; $ a^2\\plus{}b^2 \\ge 2ab$", "Solution_2": "$ \\frac{a^3}{a^2\\plus{}b^2}\\geq a\\minus{}\\frac{b}{2}\\Leftrightarrow b(a\\minus{}b)^2\\geq 0$\r\n$ \\Rightarrow\\sum\\frac{a^3}{a^2\\plus{}b^2}\\geq\\frac{\\sum a}{2}\\equal{}\\frac{1}{2}$" } { "Tag": [ "analytic geometry" ], "Problem": "Given the points K = (0,0), L = (7,\u22121), M = (9,3), P = (6,7), Q = (10,5), and R = (1, 2), triangles KLM and RP Q are congruent. Find coordinates for the point in triangle PQR that corresponds to (6,2) in triangle KLM.", "Solution_1": "It seems that triange $ PQR$ can be got from triangle $ LKM$ by translating $ 1$ to the right and up $ 2$ and then reflecting in the line $ RQ$. This takes $ (6, 2)$ to $ (7, 4)$, which is on the line $ y\\equal{}\\frac{1}{3}x \\plus{} \\frac{5}{3}$, that is, the line $ RQ$.", "Solution_2": "[quote=\"AndrewTom\"]It seems that triange $ PQR$ can be got from triangle $ LKM$ by translating $ 1$ to the right and up $ 2$ and then reflecting in the line $ RQ$. This takes $ (6, 2)$ to $ (7, 4)$, which is on the line $ y \\equal{} \\frac {1}{3}x \\plus{} \\frac {5}{3}$, that is, the line $ RQ$.[/quote]\r\n\r\nI can't tell if that's right, but it might be.\r\n\r\nHow does one set up and lay out a problem like this -- in which one is given a pre-image and an image and is asked to \"reverse engineer\" the transformation -- in a way that's both clear and eliminates doubt as to its correctness?", "Solution_3": "Hi Peterhi. \r\n\r\nI just sketched the two triangles on squared paper and then it was easy to see. You can check by taking other points (forwards or backwards). \r\n\r\nHope this helps; if not, I'll try again.", "Solution_4": "Thanks for the offer -- you've been more than helpful.", "Solution_5": "You're more than welcome!" } { "Tag": [], "Problem": "Do any of you know Forth? Can you recommend any good books/sites for learning Forth?", "Solution_1": "i learned forth some years ago but i think now this was an error: it's an old langage, it's quite long to write efficient program and it's not that readable ... i don,t remember why i learnt that ? and you ?", "Solution_2": "I know that it's an old language, and that it's not very popular (that's why there aren't that many books on it) but I just wanted to try it out.. Maybe it's the RPN that I'm attracted to, I don't know, but anyway, if you have any good tips/pointers/books/websites/etc then that'd be great." } { "Tag": [], "Problem": "What do you do to make other people laugh? What do they do to make you laugh?\r\n\r\nWhenever I recite 'All your base', I never forget the 'Ha ha ha' right after the 'All your base are belong to us'.", "Solution_1": "i tell jokes that are so stupid that u have to laugh", "Solution_2": "Like: Yo mama so fat that she has to wear two watches cause she takes up two time zones? :lol: \r\n\r\nNo offense. Anyone." } { "Tag": [ "blogs" ], "Problem": "hAPPY NEW YEAR!!!!!!!!!!!!!\r\n\r\n\r\n\r\n\r\n\r\n :rotfl: \r\n\r\n\r\nSO WHAT ARE YOUR NEW YEARS RESOLUTIONS?\r\n\r\n\r\nOne of mine is to always check if the Caps Lock is on...", "Solution_1": "How about you make it to scan the first 10 topics or so before posting a double post?", "Solution_2": "[quote=\"Ignite168\"]How about you make it to scan the first 10 topics or so before posting a double post?[/quote]\r\n\r\n\r\nTHE MORE HAPPY WISHES THE MORE MERRIER!!!\r\n\r\n\r\nHAPPY NEW YEAR!!", "Solution_3": "I think restated, we would have:\r\n\r\n\"The more spam the merrier\"\r\n\r\nSeriously this is called games and FUN factory for a reason, not G&spam factory.", "Solution_4": "i was still officially first (look at my blog)\r\n\r\n-jorian" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Is there a decreasing bijection $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ such that $ \\forall x\\in\\mathbb{R}$\r\n\\[ f^{\\minus{}1}(x^3)\\leq f(f(x)^2\\plus{}f(x))\\]\r\nIf there is, find all such functions.", "Solution_1": "As $ x \\to \\minus{} \\infty$, $ x^3 \\to \\minus{} \\infty$ and so $ f^{ \\minus{} 1}(x^3) \\to \\infty$. Similarly, $ f(x) \\to \\infty$ and so $ f(x)^2 \\plus{} f(x) \\to \\infty$ and thus $ f(f(x)^2 \\plus{} f(x)) \\to \\minus{} \\infty$.\r\n\r\n(What an odd question. Where did it come from?)" } { "Tag": [ "real analysis", "superior algebra", "superior algebra unsolved" ], "Problem": "Do there exist a subset $ A$ of $ \\mathbb R^2$ such that $ 0 < \\mu_L(A)$ , and every line intersect $ A$ at finitely \r\nmany points? \r\n\r\n\r\nAllah does not accept any act without knowledge, there is no knowledge without act, so whoever knows, knowledge leads him to act, and whoever does not act gets no knowledge, but belief is a little of a little.[b][/b]", "Solution_1": "Do you mean lebesgue measure \"u_L\" ?", "Solution_2": "[quote=\"christ\"]Do you mean lebesgue measure \"u_L\" ?[/quote]\r\n\r\nyes :wink:", "Solution_3": "A brief solution :\r\nThere is at least one measurable set with the first condition :\r\nA={(x,x^3) belong to R^2| x belong to R} but u_L(A) =0 which not satisfy the other condition .So we can talk about measurable sets such as A .\r\nIf m1 be Lebesgue measure space of R then m2 (Lebesgue measure space of R^2 )will be completion of m1 X m1 (Real and Complex Analasis ,Walter Rudin ,the third Edition, theorem 11.8).\r\nDefining product measure shows that u_1,1(A) =0 (measure associated to m1 X m1) .Then m2(A)=u_L(A)=0 ,so there is not A with these conditions ." } { "Tag": [ "geometry", "trapezoid" ], "Problem": "The diagonals of a trapezoid are perpendicular with lengths 7 and 9. Find the length of the median of the trapezoid.", "Solution_1": "Please explain what you are referring to as the median of the trapezoid... from what I understand, it is nonconstant.", "Solution_2": "The median of a trapezoid is usually defined as the segment connecting the midpoints of each leg\r\n\r\nIt is parallel to each base and is the average of their lengths" } { "Tag": [ "calculus" ], "Problem": "get x by itself.\r\n\r\na^2 x + (a-1) = (a +1)x", "Solution_1": "Your goal is to try to get x by itself. \r\n\r\n$a^2x+a-1=x(a-1)$\r\n\r\n$a^2x+a-1=ax+x$\r\n\r\n$a^2x-ax-x=1-a$\r\n\r\n$x(a^2-a-1)=1-a$\r\n\r\n$x=\\frac{1-a}{a^2-a-1}$", "Solution_2": "I don't get a thank you? :( \r\n\r\nJust kidding. But I would like a thank you for such a fast reply. :(", "Solution_3": "he posted this in classroom and here. :?", "Solution_4": "He posted this in every forum. It was posted in Getting started, classroom, basics, and in the advanced caclculus section. :mad: \r\n\r\nPlease don't do this again. :noo:", "Solution_5": "[quote=\"4everwise\"]He posted this in every forum. It was posted in Getting started, classroom, basics, and in the advanced caclculus section. :mad: \n\nPlease don't do this again. :noo:[/quote] :ninja: sry.. im a newb", "Solution_6": "Oh, sorry for not noticing that. I try to be a good moderator. I will now give a better explanation.\r\n\r\nYou want to get x by itself in some way. In this case, the best way to do that will be to take all the terms with x in them to one side and all the other terms to the other side.\r\n\r\nFirst, multiply everything else so you can do that. \r\n$a^2x+a-1=x(a-1)$\r\nturns into\r\n$a^2x+a-1=ax+x$\r\n\r\nNow comes the relocation. All the stuff with x is on one side, everything else on the other.\r\n$a^2x-ax-x=1-a$\r\n\r\nThis is probably the part where I lost you earlier. Because everything on the left side has an \"x,\" you factor out the x.\r\n$x(a^2-a-1)=1-a$\r\n\r\nBecause this is now multiplication, divide both sides by $(a^2-a-1)$ to get the answer\r\n$x=\\frac{1-a}{a^2-a-1}$\r\n\r\nIf you still don't understand, notify me and I'll try to explain in further detail." } { "Tag": [ "number theory open", "number theory" ], "Problem": "Find all primes numbers $p,q,r$ such that $p+q=r+1$.", "Solution_1": "Sure this is not open\u00bf ;)\r\nBut there probably is a solution for all $r$ ;)", "Solution_2": "yeah, Goldbah should work ;)" } { "Tag": [ "trigonometry", "algebra", "polynomial", "induction", "Irrational numbers" ], "Problem": "Show that $\\tan \\left( \\frac{\\pi}{m} \\right)$ is irrational for all positive integers $m \\ge 5$.", "Solution_1": "We needs the following \r\nLemma: If the polynomial\r\n$ P(x) = a_0 x^n +a_1 x^{n-1}+ ...+ a_{n-1}x + a_n \\in Z[x], a_0 \\ne 0$\r\nadmits as a root an irreducible fraction $ \\frac {a}{b} \\ne 0$,\r\nthen $ a$ divides $ a_n$ and $ b$ divides $ a_0$.\r\nNow let us consider two cases.\r\n\r\n1\u00b0 $ m = 2k+1, k\\in Z$,\r\n\r\n$ \\tan(m \\alpha )= \\frac { \\pm x ^{m} + P_m (x)}{1 + Q_m (x)}$ where \r\n$ x =\\tan \\alpha$, and $ P_m (x)$, $ Q_m (x)$ are polynomials of integer coefficients, of order at most $ m -1$.\r\nBy induction,\r\nfor $ m=1$ it is obvious. Let us suppose that is true for some $ m$, and consider\r\n$ \\tan (m+2) \\alpha =\\tan ( m\\alpha + 2 \\alpha) = \\frac{\\tan m\\alpha +\\tan 2 \\alpha}{1 - ( \\tan m\\alpha)( \\tan 2 \\alpha)} =\r\n(\\frac{\\pm x + P_m (x)}{1 + Q_m(x)} +\\frac{2x}{1 - x^2}) / W(x)=\r\n\\frac {\\pm x^{m+2} \\pm x^m + (1 - x^2) p_m (x) + 2x (1 + Q_m(x))}{1+ Q(x) - x^2(1+Q_m(x))-2x(\\pm x\\m + P_m(x))}$,\r\n(where $ W(x) = 1 - 2x ( \\frac{ \\pm x^m +P_m(x)}{(1-x^2) (1+Q_m(x))} )$ ),\r\nwhich imply that the representation is valid for $ m+2$.\r\nNow, let $ \\alpha = \\frac{ \\pi}{m}$. It follows that $ \\pm x_0^{m} + P_m (x_0) = 0$.\r\n\r\nThe lemme implies that $ x_0 = tan \\alpha$ must be integer.\r\nBut we know, that $ 0 < \\tan \\alpha <1$, so contradiction.\r\n2\u00b0 $ m= 2^k \\cdot n$, with $ n$ is odd. Since\r\n$ \\tan (2 \\alpha) = 2 ( \\frac {\\tan \\alpha}{1- (\\tan \\alpha)^2})$,\r\nand supposing $ \\ tan (\\frac{ \\pi}{ 2^k n} ) \\in Q$ we have that all the terms of the following sequence\r\n $ \\ tan (\\frac{ \\pi}{ 2^{k-1} n} )$, $ \\tan (\\frac { \\pi}{ 2^{k-2} n} )$,.... $ \\ tan (\\frac{ \\pi}{ 2^0 n} )$= $ \\ tan (\\frac{ \\pi}{ n}$ )\r\nare all rational. But that is contradiction.\r\nSo that is done" } { "Tag": [ "geometry", "inequalities", "geometry solved" ], "Problem": "The area of a convex quadrilateral is $64\\ m^{2}$, the sum of the lengths of a diagonal and a pair of opposite sides is $16\\sqrt2$ m. \r\nFind the length of the other diagonal.", "Solution_1": "[quote=\"ice_age\"]The area of a convex quadrilateral is $64\\ m^{2}$, the sum of the lengths of a diagonal and a pair of opposite sides is $16\\sqrt2$ m. \nFind the length of the other diagonal.[/quote]\r\n\r\nmlok, thanks!\r\n\r\nTo summarize, I have the following final solution:\r\n\r\nRecall this inequality: $a\\cdot b\\leq(\\frac{a+b}{2})^{2}$.\r\n\r\nSuppose the length of the known diagonal is $y$, then the sum of the two opposite sides is $(16\\sqrt2-y)$.\r\n\r\nWhen the two opposite sides are perpendicular to the diagonal, the area of the quadrilateral is the largest.\r\n$S_{quadrilateral}\\leq\\frac{1}{2}y(16\\sqrt2-y)\\leq\\frac{1}{2}\\cdot(\\frac{y+(16\\sqrt2-y)}{2})^{2}=64$.\r\n iff $y=16\\sqrt2-y$, i.e., $y=8\\sqrt2$, the equality holds. And only under this condition, the area of the quadrilateral is 64.\r\n\r\nTherefore, the length of the known diagonal is $8\\sqrt2$, and the sum of the two opposite sides is also $8\\sqrt2$.\r\n\r\nFinally, we know that the other diagonal is $16 cm$ in length.", "Solution_2": "[hide=\"An idea\"] Let $a$ and $c$ be two opposite sides and let $p$ be the diagonal such that \\[a+c+p=16\\sqrt{2}\\] The diagonal $p$ splits the quadrilateral into two triangles. These trianges have base $p$ and heights at most $a$ and $c$. Therefore, their areas are at most $pa/2$ and $pc/2$. But the areas add up to $64$, hence \\[p(a+c)\\ge 128\\] You still don't have enough information to find $a$ or $c$, but you can find $a+c$ and $p$, and then the other diagonal.\n\nHint: $x+1/x\\ge 2$ for any $x>0$, with equality only for $x=1$.[/hide]", "Solution_3": "[quote=\"mlok\"][hide=\"An idea\"] Let $a$ and $c$ be two opposite sides and let $p$ be the diagonal such that \\[a+c+p=16\\sqrt{2}\\] The diagonal $p$ splits the quadrilateral into two triangles. These trianges have base $p$ and heights at most $a$ and $c$. Therefore, their areas are at most $pa/2$ and $pc/2$. But the areas add up to $64$, hence \\[p(a+c)\\ge 128\\] You still don't have enough information to find $a$ or $c$, but you can find $a+c$ and $p$, and then the other diagonal.\n\nHint: $x+1/x\\ge 2$ for any $x>0$, with equality only for $x=1$.[/hide][/quote]\r\n\r\nthanks. \r\nand after further searching, I found that actually this question is similar to an IMO 1976 question.\r\nsee url below:\r\nhttp://www.kalva.demon.co.uk/imo/imo76.html \r\n\r\nAlso copied here:\r\nIMO 1976\r\n\r\nProblem A1 \r\nA plane convex quadrilateral has area 32, and the sum of two opposite sides and a diagonal is 16. Determine all possible lengths for the other diagonal. \r\n \r\nSolution \r\nAt first sight, the length of the other diagonal appears unlikely to be significantly constrained. However, a little experimentation shows that it is hard to get such a low value as 16. This suggests that 16 may be the smallest possible value. \r\nIf the diagonal which is part of the 16 has length x, then the area is the sum of the areas of two triangles base x, which is xy/2, where y is the sum of the altitudes of the two triangles. y must be at most (16 - x), with equality only if the two triangles are right-angled. But x(16 - x)/2 = (64 - (x - 8)2)/2 \u2264 32 with equality only iff x = 8. Thus the only way we can achieve the values given is with one diagonal length 8 and two sides perpendicular to this diagonal with lengths totalling 8. But in this case the other diagonal has length 8\u221a2.", "Solution_4": "This is not just similar to, but [b]equivalent to[/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61042]IMO 1976 problem 1[/url] after the quadrilateral is stretched (homothetically) with factor $\\sqrt2$.\r\n\r\n Darij" } { "Tag": [], "Problem": "\\[cos(cos(cos(cos~x)))=sin(sin(sin(sin~x))) \\]", "Solution_1": "it hase $no$ solution :rotfl:", "Solution_2": "It appears that this is a popular question.\r\nPosted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=91594]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105367]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107345]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=113745]here[/url]." } { "Tag": [ "function" ], "Problem": "The function $f(x)$ satisfies $f(2+x) = f(2-x)$ for all real numbers $x$. If the equation $f(x) = 0$ has exactly four distinct real roots, then the sum of these roots is\r\n\r\nA. 0\r\nB. 2\r\nC. 4\r\nD. 6\r\nE. 8", "Solution_1": "$f(2+x)=f(-x+2)$\r\n$f(2+(x+2)=f(-(x+2)+2)$\r\n$f(x+4)=f(-x)$\r\n$f(a)=0$\r\n$f(b)=0$\r\n$f(c)=0$\r\n$f(d)=0$\r\n\r\nIf we substitute $-a, -b, -c, -d$ fo $x$ in our last equation, then we get:\r\n$f(-a+4)=0$\r\n$f(-b+4)=0$\r\n$f(-c+4)=0$\r\n$f(-d+4)=0$\r\nFunction has exactly four distinct roots, so set ${[a,b,c,d]}$is equal to set ${[-a+4, -b+4, -c+4, -d+4]}$\r\n\r\nAfter considering a few cases we get without loosing generality:\r\n\r\n$a=-b+4$\r\n$b=-c+4$\r\n$c=-d+4$\r\n$d=-a+4$\r\nSo the sum of the roots is equal to $\\boxed{8}$", "Solution_2": "it's really possible someone would choose 0, or more possibly 2.\r\nwhy 2? because if it's f(x)=f(-x), aka f(0+x)=f(0-x), the result would have been 0. now we replace the 0 with 2, wouldn't the result be 2?! well, it's actually __(as mentioned above).\r\nfor a single problem, the mistake can be easily detected. But if this problem is only a small segment of a complicated big problem, and you get everything right except here, it's gonna be big trouble(which is what happened to me).", "Solution_3": "Lets take x1, a solution for f(x)=0. There is a real number y so that x1=y+2. so we have f(x1)=f(y+2)=f(2-y)=0, so 2-y is also a solution for f(x)=0, so x2=2-y, so x1+x2=4.\r\nNow we can choose from one of the 2 roots left another root, x3. we follow the same steps as in the previous case and we prove x3+x4=4.\r\nSo x1+x2+x3+x4=8.", "Solution_4": "This function is symmetric by the point 2. And so, the roots are separeted by this center of symmetry. Suppose that $x_1,x_2$ are the \"distances\" from 2 that these roots are. And so, we have as roots $2-x_1,2-x_2$, since they're symmetric by 2, we know where the other roots are $2+x_1,2+x_2$. Summing them we find 8.", "Solution_5": "Isn't this fairly obvious? If $f(x) = 0$, let $x = 2 + y$ for some $y$; now let $x_0 = 2 - y$. We could do this for every solution to $f(x) = 0$, producing another solution, which, when summed with the original gives you 4.. So each 2 solutions represents a sum of 4 for, and being 4 solutions, the sum of all the roots must be 8.", "Solution_6": "\\boxed{E}\r\n\r\nif $r_1 = 2+a$ is a root, so is $r_2 = 2-a$ because they are symmetrical about $x=2$\r\n\r\n$(2-a)+(2+a) = 4$\r\neach pair has sum $4$.\r\n\r\nSince there are $4$ distinct roots, there are $2$ pairs\r\n\r\nthe sum of all roots is \\boxed{8}" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "[color=darkblue]Are well-known the following inequalities in a triangle $ABC\\ :$\n\n$\\{\\begin{array}{ccc}(-a+b+c)(a-b+c)(a+b-c) & \\le & abc\\\\\\ (b+c)(c+a)(a+b) & \\ge & 8\\cdot abc\\end{array}\\ .$ Prove that :\n\n$\\boxed{\\ (b+c)(c+a)(a+b)+(-a+b+c)(a-b+c)(a+b-c)\\ \\ge\\ 9\\cdot abc\\ }\\ .$[/color]", "Solution_1": "Expand -> It is well-known inequality\r\n$2(a+b+c)(a^{2}+b^{2}+c^{2})\\ge 3(a^{3}+b^{3}+c^{3}+3abc)$\r\nwhich is true by schur.\r\n$\\iff \\sum_{cyc}(b+c-a)(a-b)(a-c)\\ge0$\r\n\r\nor $a=y+z, b=x+z, c=x+y$ ->\r\n$LHS-RHS=2(x^{3}+y^{3}+z^{3}+3xyz-\\sum_{sym}x^{2}y)\\ge 0$.", "Solution_2": "[quote=\"Virgil Nicula\"][color=darkred]Are well-known the following inequalities in a triangle $ABC\\ :$\n\n$\\{\\begin{array}{c}(-a+b+c)(a-b+c)(a+b-c)\\ \\le\\ abc\\\\\\ (b+c)(c+a)(a+b)\\ \\ge\\ 8\\cdot abc\\end{array}\\ .$ Prove that :\n\n$\\boxed{\\ (b+c)(c+a)(a+b)+(-a+b+c)(a-b+c)(a+b-c)\\ \\ge\\ 9\\cdot abc\\ }\\ .$[/color][/quote]\r\n[color=darkblue][b]Proof.[/b] The well-known relation $9\\cdot IG^{2}=p^{2}+5r^{2}-16Rr$ $\\implies$ $p^{2}+5r^{2}\\ge 16Rr\\ .$\n\n$\\{\\begin{array}{c}abc=4Rrp\\\\\\\\ ab+bc+ca=p^{2}+r^{2}+4Rr\\\\\\\\ \\prod (-a+b+c)=8pr^{2}\\\\\\\\ \\prod (b+c)=\\sum a\\cdot\\sum bc-abc\\end{array}\\|$ $\\implies$\n\n$\\prod (b+c)+\\prod (-a+b+c)-9abc=$\n\n$2p\\cdot (p^{2}+r^{2}+4Rr)-4Rpr+8pr^{2}-36Rpr=$ $2p\\cdot (p^{2}+5r^{2}-16Rr)\\ge 0\\ .$[/color]", "Solution_3": "more from them :Prove that\r\n$(b+c)(c+a)(a+b)\\ge 8abc+\\frac{1}{3}r^{2}(R-2r)$\r\n:D" } { "Tag": [], "Problem": "In triangle ABC, m_a is the lenth of median from A,\r\n m_b is the lenth of median from B, \r\n m_c is the lenth of median from C.\r\n\r\nprove that:\r\n a/m_a + b/m_b + c/m_c >= 2*sqrt(3).", "Solution_1": "[hide=\"hint\"]\n$m_a^2=\\frac{1}{4}(2b^2+2c^2-a^2)$\n$m_a^2+m_b^2+m_c^2=\\frac{3}{4}(a^2+b^2+c^2)$[/hide]", "Solution_2": "HI amirhtlusa,\r\n \r\n From your hint, one may change the origin problem into a new problem relating a,b,c only(express m_a with a,b,c...), solving the new problem means solving the origin problem. However, I still could not find the direction or\r\nanything obvious that helps the proof, what need I watch out more??", "Solution_3": "doesn't $a$ in $\\frac{a}{m_a}$ stands for the lenght of the side, opposite the angle $A$?", "Solution_4": "Yes,it does, so we change a/m_a into 2a/sqrt(2b^2+2c^2-a^2) ??", "Solution_5": "I can prove a similar one:\r\n \r\n a/r_a + b/r_b +c/r_c>=2sqrt(3)" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Hey all\r\n\r\nI've been stuck on this problem involving commutators:\r\nSuppose A, B and C are all normal subgroups of some group G. Show [AB,C] = [A,B][B,C]\r\nI tried showing directly that any element in one group is an element of the other by judicious use of the normality properties, but haven't been able to get anywhere.\r\n\r\nAny advice would be greatly appreciated.", "Solution_1": "Use the following facts : let $ a,b,c\\in G$ where $ G$ is a group\r\n0. for subsets $ A,B$ of $ G$ then $ [A,B]\\equal{}[B,A]$\r\n1. $ [a,bc]\\equal{}[a,c](c^{\\minus{}1}[a,b]c)$\r\n2.$ [ab,c]\\equal{}(b^{\\minus{}1}[a,c]b)[b,c]$" } { "Tag": [ "algebra", "function", "domain", "limit", "calculus", "calculus computations" ], "Problem": "I woulk like to see the definitions in your school or Uni textbooks , of the symbol \r\n $\\sqrt[n] a$\r\n\r\n for the real number $a$. \r\n In Greece we introduce this symbol only for non negative numbers. In my school years $a$ could be any number if $n$ is odd.\r\n a) How do you use $\\sqrt[n] a$ ?\r\n b) Which is the domain of $f(x) = \\sqrt[3] x$ ? For us it is $A= [0,+\\infty)$ , although we know that $\\mathbb{R}$\r\n could work very nice.\r\n \r\n[color=red]Thanks in advance.[/color]\r\n\r\n Babis", "Solution_1": "What I have always used:\r\n\r\nFor $n$ odd:\r\n$a \\in \\mathbb{R}$\r\n\r\nfor $n$ even:\r\n$a \\geq 0$", "Solution_2": "I think conventionally that notation is used only when $n$ is an integer larger than 2. (The 2 is omitted for the square root.) However, one could certainly give a meaningful interpretation to the notation for any $n \\neq 0$ by setting $\\sqrt[n]a = a^{1/n}$.", "Solution_3": "Dear JBL , the question is for what kind of real numbers $a$ we define the n-nth root of $a$.For exaple , some define this root only for non negative numbers, even if n is odd. So , I'm trying to see in which other countries they have this convention. For example in USA I read that for odd n the n-nth root of $a$ has meaning for all real numbers $a$.\r\n But in Greece and in some other books they define it only for non negatives.\r\n Babis", "Solution_4": "What I can tell you is that in my books the domain is also $[0+\\infty)$.", "Solution_5": "But I read in your new school book by M. Ganga(Editura GIL) , that for odd $n$ ,\r\n\r\n he defines the $\\sqrt[n]a$ for all $a \\in \\mathbb{R}$.\r\n\r\n Which book are you refering to ?\r\n \r\n [u]Babis[/u]", "Solution_6": "For $n\\in N^{*}$ : if $n$- odd, then $\\sqrt [n]\\ : \\ \\mathcal R\\rightarrow \\mathcal R$ ; if $n$- even, then $\\sqrt [n]\\ : \\ [0,\\infty )\\rightarrow \\mathcal R$. [b]Convention.[/b] $\\sqrt [1] x=x$, for any $x\\in \\mathcal R$. Therefore,\r\n$1\\blacktriangleright$ If $n$- odd, then $\\sqrt [n]x=y\\Longleftrightarrow$ $\\{\\begin{array}{c}y\\in\\mathcal R\\\\\\ y^{n}=x\\in\\mathcal R\\end{array}$.\r\n$2\\blacktriangleright$ If $n$- even, then $\\sqrt [n]x=y\\Longleftrightarrow$ $\\{\\begin{array}{c}y\\ge 0\\\\\\ y^{n}=x\\ge 0\\end{array}$.\r\n[b]Examples.[/b]\r\n\r\n$\\sqrt[3]{xy}=\\sqrt [3]x\\cdot\\sqrt [3]y$ ;\r\n\r\n$xy\\ge 0\\implies\\sqrt [6]{xy}=\\sqrt [6]{|x|}\\cdot\\sqrt [6]{|y|}$ ;\r\n\r\n$\\sqrt [6]{x^{2}}=\\sqrt [3]{|x|}=|\\sqrt [3]x|$ ;\r\n\r\n$\\sqrt [5]{x^{15}y}=x^{3}\\sqrt [5]y$\r\n\r\n$\\sqrt [4]{x^{4}y}=|x|\\sqrt [4]y$.\r\n\r\n$x\\in \\mathcal R\\implies$ $x\\sqrt [3]y=\\sqrt [3]{x^{3}y}$ ;\r\n\r\n$x\\sqrt y=\\mathrm{sgn}(x)\\cdot\\sqrt{x^{2}y}$, where $x=|x|\\cdot\\mathrm{sgn}(x)$ a.s.o.\r\n\r\n[b]Exercises.[/b]\r\n$\\lim_{x\\to-\\infty}\\frac{x}{\\sqrt{x^{2}+1}}=-\\lim_{x\\to-\\infty}\\sqrt\\frac{x^{2}}{x^{2}+1}=-1$.\r\n\r\n$\\{\\begin{array}{c}x\\ge 0\\ ,\\ y\\in\\mathcal R\\\\\\\\ \\sqrt x+\\sqrt [3]y=6\\\\\\\\ \\sqrt [6]{x^{3}y^{2}}=8\\end{array}\\|$ $\\Longleftrightarrow$ $\\{\\begin{array}{c}\\sqrt x+\\sqrt [3]y=6\\\\\\\\ \\sqrt x\\cdot|\\sqrt [3]{y}|=8\\end{array}$ $\\Longleftrightarrow$ $\\{\\begin{array}{c}u\\ge 0\\ ,\\ v\\in\\mathcal R\\\\\\\\ u+v=6\\\\\\\\ u\\cdot |v|=8\\end{array}$, where $u=\\sqrt x$ and $v=\\sqrt [3]y$ a.s.o.", "Solution_7": "[quote=\"Virgil Nicula\"]For $n\\in N^{*}$, if $n$- even, then $\\sqrt [n]\\ : \\ [0,\\infty )\\rightarrow \\mathcal R$. [/quote]\r\nShould it be: if $n$ is even $\\sqrt [n]\\ : \\ [0,\\infty )\\rightarrow \\ [0,\\infty )$", "Solution_8": "Emiraga, why surjective ?!", "Solution_9": "[quote=\"Virgil Nicula\"]Emiraga, why surjective ?![/quote]It's difficult to say why for a [u]definition[/u]. Usually elementary [i]real[/i] functions are defined as single-valued. You said it yourself:\n[quote=\"Virgil Nicula\"]If $n$- even, then $\\sqrt [n]x=y\\Longleftrightarrow$ $\\{\\begin{array}{c}y\\ge 0\\\\\\ y^{n}=x\\ge 0\\end{array}$.[/quote]\r\n$y\\ge 0$", "Solution_10": "Is there any difference between $\\sqrt[n] x$ and $x^{\\frac{1}{n}}$? (because rational powers are defined only for positive numbers, I had this question in a contest )", "Solution_11": "[b]YES![/b]\r\n\r\n It is a different notation. For possitive numbers it is the same. For negatives we have problem. See for example this :\r\n\r\n $-1 = \\sqrt[3]{-1}= (-1)^{\\frac{1}{3}}= (-1)^{\\frac{2}{6}}= \\sqrt[6]{(-1)^{2}}= \\sqrt[6]1 = 1$ !\r\n\r\n For this reason we define the rational (fraction) powers only on possitive or nonnegative numbers. \r\n But I want to see the many opinions on this subject.\r\n\r\n[u] Babis [/u]" } { "Tag": [ "function", "ratio", "trigonometry", "complex numbers", "complex analysis", "complex analysis unsolved" ], "Problem": "I believe you have to use the Weiestrass [b]M-Test[/b] for these.\r\n\r\nShow that each of the following is an entire funtion of z:\r\n\r\nA)\r\ninf z^(n)\r\nSum ------------\r\nn=1 (n!)^(1/2)\r\n\r\nB) \r\ninf z^(n)\r\nSum --------------\r\nn=1 2^(n)^(2)\r\n\r\nC)\r\ninf 1 1\r\nSum -------- ---------\r\nn=1 2^(n) n^z)\r\n\r\nD) \r\ninf sin(nz)\r\nSum ------------\r\nn=1 n!\r\n\r\n\r\nThanks in advanced. Maybe the ratio test needs to be used too.", "Solution_1": "Well if you read what the weirstrass M-test is,\r\n\r\nhttp://en.wikipedia.org/wiki/Weierstrass_M-test\r\n\r\nThe first one you just need to prove that\r\n\r\n$ \\sum_{n\\equal{}1}^\\infty \\frac{|z|^n}{\\sqrt{n!}}$ converges. This is obvious by the ratio test.\r\n\r\nThen the original converges uniformly for any $ z$.\r\n\r\nThen presumably, since this function has a power series that converges everywhere, it is analytic. \r\n\r\nB) is basically the same.\r\n\r\nFor part C), you need to know what $ |n^z|$ is... If $ z\\equal{}re^i\\theta$, then $ |n^z|\\equal{}n^r$\r\n\r\nThen the rest is obvious.\r\n\r\nD), Consider the series that is\r\n\r\n$ \\frac{\\cos nz\\plus{}i\\sin nz}{n!}$ which is equal to $ \\frac{(e^z)^n}{n!}$ which is just $ e^{e^z}$ so that series converges so of course the real part converges. Then done as before." } { "Tag": [ "geometry", "calculus", "integration", "vector", "number theory", "Ring Theory", "superior algebra" ], "Problem": "Let $ k\\subseteq k'$ be fields and $ k'$ normal as a field extension. Let $ G\\equal{}Aut_k(k')$ and $ R$ an affine ring over $ k$ and let $ R'\\equal{}R\\otimes_k k'$. Show that the map $ Spec(R')\\rightarrow Spec(R)$ induces a bijection between $ Spec(R)$ and the $ G$-orbits of $ Spec R'$", "Solution_1": "The proof uses the intermediate fixed field $ P \\equal{} (k')^G$ for a two-step approach: $ k'|P$ is a Galois extension and $ P|k$ is purely inseparable (French: extension radicielle). The latter implies that $ Spec(P)\\to Spec(k)$ is universally injective, in particular $ Spec(R\\otimes_k P)\\hookrightarrow Spec (R)$ is injective (*).\r\n\r\nWhat I want to show is that $ Spec(R\\otimes_kk')/G\\overset{\\varphi_1}{\\to} Spec((R\\otimes_kk')^G)\\overset{\\varphi_2}{\\to}Spec(R\\otimes_kP) \\overset{\\varphi_3}{\\to} Spec(R)$ is a composition of bijections.\r\n\r\n\r\nA collection of facts proves the claim:\r\n\r\n1. Since $ k'|k$ is normal, in particular finite, hence integral, thus $ Spec(k')\\to Spec(k)$ is (universally) surjective. In particular $ Spec(R\\otimes_kk')\\to Spec(R)$ is surjective. This implies that all maps involved are surjective. Together with (*) we get that $ \\varphi_3$ is bijective.\r\n\r\n2. $ P$ is defined by the exact sequence of $ k$-vector spaces\r\n\\[ 0\\to (k')^G\\to k'\\rightrightarrows \\bigoplus_{\\sigma\\in G}k'\\]\r\nwhere the upper arrow is defined by $ a\\mapsto (a)_{\\sigma\\in G}$, the lower by $ a\\mapsto (\\sigma(a))_{\\sigma\\in G}$. Tensoring with the $ k$-vector space $ R$ yields the exact sequence defining $ (R\\otimes_kk')^G$\r\n\\[ 0\\to R\\otimes_k(k')^G\\to R\\otimes_kk'\\rightrightarrows \\bigoplus_{\\sigma\\in G}R\\otimes_kk'\\,.\\]\r\nThus we can read off $ (R\\otimes_kk')^G \\equal{} R\\otimes_k(k')^G \\equal{} R\\otimes_kP$, hence $ \\varphi_2$ is a bijection.\r\n\r\n3. To see that $ \\varphi_1$ is injective simply mimic the well known proof in number theory that prime ideals in the Galois extension lying above the same prime ideal are conjugated by a Galois homomorphism. The proof makes heavy use of the Chinese remainder theorem.\r\n\r\n\r\nNote that \r\n- it is true that all maps are homeomorphims which is a bit more than claimed. But is follows easily from that the morphisms are integral.\r\n- $ P \\equal{} k$ in characteristic 0." } { "Tag": [ "algebra", "polynomial", "system of equations" ], "Problem": "Solve the system of equations: \\[ \\left\\{\\begin{array}{lll} 2y(x^2\\minus{}y^2)&\\equal{}&3x\\\\ x(x^2\\plus{}y^2)&\\equal{}&10y\\end{array}\\right.\\]", "Solution_1": "if we multiply equations we find $ x^4 \\minus{} y^4 \\equal{} 15$ one of x and y are x= 2 and y=1.", "Solution_2": "Missing some solutions. Is there division by zero? :wink:", "Solution_3": "Furthermore, the problem doesn't specify a restriction to integers.", "Solution_4": "Well, if there isn't a restriction for only rational numbers, then shouldn't there be infinite amount of solutions?", "Solution_5": "[url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem]Bezout's theorem[/url] tells us that there are at most $ 9$ solutions. Note that this is equivalent to solving for one variable in terms of the other and plugging in to produce a polynomial of degree $ 9$ (although this construction is not always explicitly possible).\r\n\r\nEdit: [hide=\"Solution\"] Let $ a \\equal{} \\frac{y}{x}, b \\equal{} x^2$. This gives\n\n$ 2ab(1 \\minus{} a^2) \\equal{} 3$\n$ b(1 \\plus{} a^2) \\equal{} 10a \\implies b \\equal{} \\frac{10a}{1 \\plus{} a^2}$.\n\nSubstituting into the other equation, $ 20a^2(1 \\minus{} a^2) \\equal{} 3(1 \\plus{} a^2)$. Letting $ c \\equal{} a^2$, we have\n\n$ 20c \\minus{} 20c^2 \\equal{} 3 \\plus{} 3c \\implies$\n$ 20c^2 \\minus{} 17c \\plus{} 3 \\equal{} 0 \\implies$\n$ (5c \\minus{} 3)(4c \\minus{} 1) \\equal{} 0 \\implies$\n$ c \\equal{} \\frac{3}{5}, \\frac{1}{4}$\n\nfrom which the other solutions can be calculated. [/hide]", "Solution_6": "For $ y\\neq 0$, dividing both sides gives $ \\frac {2y(x^2 \\minus{} y^2)}{x(x^2 \\plus{} y^2)} \\equal{} \\frac {3}{10}\\cdot \\frac {x}{y}$\r\n\r\n$ \\Longleftrightarrow 2\\cdot \\frac {y}{x}\\cdot \\frac {\\left(\\frac {x}{y}\\right)^2 \\minus{} 1}{\\left(\\frac {x}{y}\\right)^2 \\plus{} 1} \\equal{} \\frac {3}{10}\\cdot \\frac {x}{y}$\r\n\r\nLet $ z \\equal{} \\frac {x}{y}$, we have $ 3z^4 \\minus{} 17z^2 \\plus{} 20 \\equal{} 0$, yielding $ z \\equal{} \\pm 2,\\ \\pm \\sqrt {\\frac {5}{3}}\\Longleftrightarrow \\frac {y}{x} \\equal{} \\pm 2,\\ \\pm \\sqrt {\\frac {5}{3}}$.\r\n\r\nNeedless to say, $ (x,\\ y) \\equal{} (0,\\ 0)$ is one of the solutions.", "Solution_7": "Are you guys sure? I multiply both to get $ x^4 \\minus{} y^4 \\equal{} 15$, so $ x \\equal{} \\pm \\sqrt [4]{y^4 \\plus{} 15}$. Also by the first equation we have $ 2yx^2 \\minus{} 3x \\minus{} 2y^3 \\equal{} 0$, so by the quad. formula, we have $ x \\equal{} (3 \\pm \\sqrt {16y^4 \\plus{} 9})/4y$. Setting both equal and simplifying gives $ 256y^4(y^4 \\plus{} 15) \\equal{} (3 \\pm \\sqrt {16y^4 \\plus{} 9})^4$. I just saw that $ y^4 \\equal{} 1$ is a soultion. Then, x would equal 2.", "Solution_8": "[quote=\"dgreenb801\"]Are you guys sure? I multiply both to get $ x^4 \\minus{} y^4 \\equal{} 15$, so $ x \\equal{} (y^4 \\plus{} 15)^(1/4)$.[/quote]\r\n\r\nIs limited $ x>0$?", "Solution_9": "Oh, nevermind, what I wrote matches what kunny said but doesn't include everything.", "Solution_10": "O.K. :)", "Solution_11": "$ 20y^2(x^2\\minus{}y^2)\\minus{}3x^2(x^2\\plus{}y^2)\\equal{}(3x^2\\minus{}5y^2)(4y^2\\minus{}x^2)\\equal{}0$", "Solution_12": "[quote=\"Altheman\"]$ 20y^2(x^2 \\minus{} y^2) \\minus{} 3x^2(x^2 \\plus{} y^2) \\equal{} (3x^2 \\minus{} 5y^2)(4y^2 \\minus{} x^2) \\equal{} 0$[/quote]\r\n\r\nOh! Best soution. :lol:" } { "Tag": [ "calculus", "integration", "quadratics", "function", "trigonometry", "calculus computations" ], "Problem": "For any quadratic functions $ f(x)$ such that $ f'(2)\\equal{}1$, evaluate $ \\int_{2\\minus{}\\pi}^{2\\plus{}\\pi}f(x)\\sin\\left(\\frac{x}{2}\\minus{}1\\right) dx$.", "Solution_1": "[hide=\"Straightforward.\"]This looks better with a change of variables. Let $ u\\equal{}x\\minus{}2$ and $ g(u)\\equal{}f(u\\plus{}2)$ so that $ g(x\\minus{}2)\\equal{}f(x)$ and $ g'(0)\\equal{}1.$\n\n$ \\int_{2\\minus{}\\pi}^{2\\plus{}\\pi}f(x)\\sin\\left(\\frac{x}{2}\\minus{}1\\right)\\,dx\\equal{}\\int_{\\minus{}\\pi}^{\\pi}g(u)\\sin\\left(\\frac{u}{2}\\right)\\,du$\n\n$ \\equal{}\\int_{\\minus{}\\pi}^{\\pi}(a\\plus{}u\\plus{}cu^{2})\\sin\\left(\\frac{u}{2}\\right)\\,du$\n\nBut $ \\int_{\\minus{}\\pi}^{\\pi}(a\\plus{}cu^{2})\\sin\\left(\\frac{u}{2}\\right)\\,du\\equal{}0$ because the integrand is odd.\n\nThat leaves $ \\int_{\\minus{}\\pi}^{\\pi}u\\sin\\left(\\frac{u}{2}\\right)\\,du\\equal{} 2\\int_{0}^{\\pi}u\\sin\\left(\\frac{u}{2}\\right)\\,du$\n\n$ \\equal{}\\left.\\minus{}4u\\cos\\left(\\frac{u}{2}\\right)\\right|_{0}^{\\pi}\\plus{}4\\int_{0}^{\\pi}\\cos\\left(\\frac{u}{2}\\right)\\,du$\n\n$ \\equal{}\\left.8\\sin\\left(\\frac{u}{2}\\right)\\right|_{0}^{\\pi}\\equal{}8.$[/hide]", "Solution_2": "That's right! :)" } { "Tag": [ "function" ], "Problem": "Consider the function $ f: R\\minus{}>R$ $ f(x)\\equal{}\\frac{1\\minus{}2x}{x\\plus{}2}$.Solve the equation $ (f\\circ f\\circ...\\circ f)(x)\\equal{}\\minus{}1$($ n$ times $ f$), where $ (f\\circ f)(x)\\equal{}f(f(x))$", "Solution_1": "First list the complete cycle of $ f(x)$.\r\n\r\n$ f(x) \\equal{} \\frac {1 \\minus{} 2x}{x \\plus{} 2}$\r\n\r\n$ f^2 (x) \\equal{} x$\r\n\r\nSo we find that \r\n\r\nIf $ n \\equal{} 2k \\plus{} 1$, $ x \\equal{} 3$.\r\n\r\nIf $ n \\equal{} 2k$, $ x \\equal{} \\minus{}1$." } { "Tag": [ "abstract algebra", "geometry" ], "Problem": "I was wondering if anyone has used any of the Add-On Curriculum modules for GSP? They seem to be very interesting but I would like to hear what others thought of them.", "Solution_1": "[quote=\"joml88\"]Add-On Curriculum modules for GSP[/quote]\r\nCan you please tell me what it is?", "Solution_2": "It's a geometry program (makes geometry drawings on the computer).", "Solution_3": "I have the Geometer's Sketchpad, but what is Add-On Curriculum modules?", "Solution_4": "http://www.keypress.com/sketchpad/getting_started/curriculum_modules.php" } { "Tag": [ "number theory", "relatively prime" ], "Problem": "Prove that the ten's digit of any power of 3 is even.", "Solution_1": "Ten's = tenth ?\r\n\r\n$3^{20} = *3*486784401$ :?: :?:", "Solution_2": "No, the tens digit is the last but one digit, the one coming just before the units digit.", "Solution_3": "We are going to look at 3^k mod 20. It repeats 3,9,7,1. Hence result.", "Solution_4": "[hide=\"more completely...\"]\n\nwe know that if $3^m \\mod{20} \\equiv n$ where $0>n>10$, then it has an even tens digit because the remainder will not effect the tens digit if it is less than 10, and if it is in mod 20, then any number times 20 has an even tens digit\n\nnote that 3 and 20 are relatively prime, so using euler's extention of FLT, (note that $\\phi(20)=8$) \nso $3^{8k+m} \\mod{20} \\equiv 3^m \\mod{20}$\n\nso we only need to go through m=0,1,2,3,4,5,6, and 7, and show that the remainders are all less than 10, then all the other powers are also proven\n\n$3^{0} \\mod{20} \\equiv 1$\n$3^{1} \\mod{20} \\equiv 3$\n$3^{2} \\mod{20} \\equiv 9$\n$3^{3} \\mod{20} \\equiv 7$\n$3^{4} \\mod{20} \\equiv 1$\n$3^{5} \\mod{20} \\equiv 3$\n$3^{6} \\mod{20} \\equiv 9$\n$3^{7} \\mod{20} \\equiv 7$\n\nthus every remainder $\\mod{20}$ for a power of $3^n$ is 1,3,7, or 9\nso the tens digit must be even because any number mod 20 where the remainder is less than 10 will have an even tens digit\n\n[/hide]", "Solution_5": "[quote=\"Altheman\"]$0>n>10$[/quote]\r\nSmall correction: it should be $0n>10$, then it has an even tens digit because the remainder will not effect the tens digit if it is less than 10, and if it is in mod 20, then any number times 20 has an even tens digit\n\nnote that 3 and 20 are relatively prime, so using euler's extention of FLT, (note that $\\phi(20)=8$) \nso $3^{8k+m} \\mod{20} \\equiv 3^m \\mod{20}$\n\nso we only need to go through m=0,1,2,3,4,5,6, and 7, and show that the remainders are all less than 10, then all the other powers are also proven\n\n$3^{0} \\mod{20} \\equiv 1$\n$3^{1} \\mod{20} \\equiv 3$\n$3^{2} \\mod{20} \\equiv 9$\n$3^{3} \\mod{20} \\equiv 7$\n$3^{4} \\mod{20} \\equiv 1$\n$3^{5} \\mod{20} \\equiv 3$\n$3^{6} \\mod{20} \\equiv 9$\n$3^{7} \\mod{20} \\equiv 7$\n\nthus every remainder $\\mod{20}$ for a power of $3^n$ is 1,3,7, or 9\nso the tens digit must be even because any number mod 20 where the remainder is less than 10 will have an even tens digit\n\n[/hide][/quote]\n\nwhat is euler's extension of FLT", "Solution_13": "[quote=RMO 1993 P2]Prove that the ten's digit of any power of 3 is even.[/quote]\n[b][color=#000]Solution:[/color][/b] Suppose that the power of $3$ is $3^k$. We'll do a quick induction on $k$. First check that $3^3=27$, $3^4=81$, $3^5=243$, $3^6=729$ and $3^7=2187$\n\nSuppose, for some $m \\in \\mathbb{N}$, $3^m$ has an even ten's digit, say $x$. So:\n1) If the unit's digit of $3^m$ is 1, then ten's digit of $3^{m+1}$ $\\implies$ $3x \\equiv \\ell \\pmod{10}$\n2) If the unit's digit of $3^m$ is 3, then ten's digit of $3^{m+1}$ $\\implies$ $3x \\equiv \\ell \\pmod{10}$\n3) If the unit's digit of $3^m$ is 7, then ten's digit of $3^{m+1}$ $\\implies$ $3x+2 \\equiv \\ell \\pmod{10}$\n4) If the unit's digit of $3^m$ is 9, then ten's digit of $3^{m+1}$ $\\implies$ $3x +2\\equiv \\ell \\pmod{10}$\n(Where, $\\ell \\in \\{0,2,4,6,8\\}$). Hence, our Induction is complete and all powers of $3$ have an even ten's digit! $\\qquad \\blacksquare$", "Solution_14": "Really sorry to bump this trivial problem but I thought this approach was nice.$\\pmod{20}\\Rightarrow$ there exists $t$ such that $20t<3^n<20t+10$ so $2t<\\frac{3^n}{10}<2t+1$ hence $\n [\\frac{3^n}{10}]=2t$ ($[.]$ is the floor function).Note that $[\\frac{3^n}{10}]$ is the number formed by removing the last digit of $3^n$ and moreover this is even.The result follows", "Solution_15": "Whoops, I fell asleep while solving this. Here's the solution:\n\nNotice that the last digit of every number in the form $3^k$ (where $k$ is a non-negative integer) goes in the pattern: $3, 9, 7, 1$. \n\nThe last digit will be $3$ when $k \\equiv 1\\pmod{4}$, the last digit will be $9$ when $k \\equiv 2\\pmod{4}$, the last digit will be $7$ when $k \\equiv 3\\pmod{4}$ and the last digit will be $1$ when $k \\equiv 0\\pmod{4}$. We can separate this problem into 4 different cases.\n\nCase 1: The last digit of $3^k$ is $1$ - This means that $k \\equiv 0\\pmod{4}$. $k$ can be written as $4n$ where $n$ is a non-negative integer. So, $3^k = 3^{4n} = 81^n$. We want the ten's digit to be even, so:\n$$\\frac{81^n - 1}{10} \\equiv 0\\pmod{2}$$\n$$\\implies 81^n \\equiv 1\\pmod{20}$$\nThis is clearly true as $81 \\equiv 1\\pmod{20}$. So we have proved our statement for the first case.\n\nCase 2: The last digit of $3^k$ is $3$ - This means that $k \\equiv 1\\pmod{4}$. $k$ can be written as $4n + 1$ and we can proceed to do the same thing as Case 1 to get that $81^n\\cdot3^1 \\equiv 3\\pmod{20}$ which is true.\n\nCase 3: The last digit of $3^k$ is $9$ - This means that $k \\equiv 2 \\pmod{4}$. $k$ can be written as $4n + 2$ and we can proceed to do the same thing as Case 1 to get that $81^n\\cdot3^2 \\equiv 9\\pmod{20}$ which is also true.\n\nCase 4: The last digit of $3^k$ is $7$ - This means that $k \\equiv 3\\pmod{4}$. $k$ can be written as $4n + 3$ and we can proceed to do the same thing as Case 1 to get that $81^n\\cdot3^3 \\equiv 7\\pmod{20}$ which is true as $81^n \\equiv 1\\pmod{20}$ and $3^3 \\equiv 7\\pmod{20}$.\n\nWe are done $\\blacksquare$" } { "Tag": [ "combinatorics solved", "combinatorics" ], "Problem": "HERE GOES THE PROBLEM\r\nGiven are 3 nonnegative integers: k, l and m. In one transformation one considers two of the three numbers, say WLOG k and l, and replaces them by k+l and |k\u2212l|. Find out if there it is possible to obtain the triplet (0, 0, 0)by making such transformations.", "Solution_1": "Do I miss something or the problem is trivial?\r\n\r\nIt is clear that if you start with (k,l,n) where k,l,n are nonnegative integers, the only way to obtain (0,0,0) is to start with (0,0,0) since :\r\nIf at some step we choose k and l with k \\geq l, then we replace them with k' = k+l and l' = k-l, which are nonnegative too and such that k'+l'+n' = k+l + k-l + n = 2k +n \\geq k + l + n.\r\nThen the operation does not decrease the sum of the numbers. So if we start from k+l+n > 0 we will never reach (0,0,0).\r\n\r\nPierre.", "Solution_2": "This reminds me of following problem:\r\n\r\nGiven are 3 non-negative integers: (a,b,c)\r\nOne transformation is to reduce the value of one integer x and double the size of integer y by replacing x with x-y and y with 2y. This is only allowed, when x >= y.\r\nIs it possible to make one field zero (reaching (x,y,0), the order of the integers doesn't matter)?\r\n\r\nThis shouldn't be so easy :D", "Solution_3": "i actually had solved the problem the same way as pierre but didnt think it would be that simple :D .that's why i made the post", "Solution_4": "Let's solve the Daniel's problem by the affirmative :\r\nLet a,b,c be the initial number. Wlog, we may assume that a \\leq b \\leq c, and that a > 0 if not the problem is trivial.\r\n\r\nWe will prove that after a finite number of steps we may reach a configuration (a',b',c') with min(a',b',c') \\leq a-1. Thus, repeating this process we will eventually reach (x,y,0) for some nonegative integers x,y after a finite number of steps.\r\n\r\nLet k be such that b = ka + p with 0 \\leq p < a, and k = k_0 + 2*k_1 + ... + 2 n *k_n be the binary expansion of k.\r\nThen b = p + sum a*k_i*2 i and c \\geq p + sum a*k_i*2 i.\r\n\r\nLet r be the least subscript such that k_r = 1. We use r times the operation (c,a) - > (c-a, 2a) to reach (2 r *a, b, c - 2 r *a + a).\r\nThen we use (b, 2 r *a) - > (b - 2 r *a, 2 r+1 *a), and reach (2 r+1 *a, p + sum{i > r} 2 i *a*k_i , c - 2 r *a + a).\r\nIf r = n, we have p + sum{i > r} 2 i *a*k_i = p < a, and we are done.\r\nIf r < n, then let r_2 be the least subscript > r such that k_(r_2) = 1, and we do it again as above....and so on, until we reach (a*2 n , a*2 n + p, c') and then we obtain (a*2 n+1 , p, c'), and we are done.\r\n\r\nPierre.", "Solution_5": "This problem from poland TST 2002 was a bit diffrent: to obtain the triplet (a, 0, 0) by making such transformations, a is nonnegative integer.", "Solution_6": "The immediate answer is that [b]not always [/b]can we do that. If we start with $0=l$.After the move the numbers look like : $k-l,k+l,m$.Their sum is:\n$k-l+k+l+m=2k+l>=k+l+m$.This way we see that sum can only raise after every move so we can never reach $(0,0,0)$.", "Solution_10": "it is not written here that 3 must come to zero.\nSome were asked if the two could come to zero.\nThis is a very wrong idea.", "Solution_11": "The condition of the issue is also different with Polish MO finals 2002 problem 3 it is said whether there can ever be two 0's", "Solution_12": "If this is the case, then the exact sum always increases and never reaches 3 zeros\n" } { "Tag": [ "ARML" ], "Problem": "I recently received the invitation to seminar, and from what i've read in the invitation letter, it sounds really fun and nice :) .\r\nHowever, there is a little problem:\r\n[QUOTE]RULES:5.Appropriate dress(eg jackets and ties for young men( :rotfl: )) is required for the banquets.[end quote]\r\nSo, this means there's some kind of dress code for banquets. I want you ppl who have gone to seminar before please write what i am supposed to wear???, cause i don't want to look like a total idiot during the banquet. I just ask this question that may seem stupid to some of you because I haven't attended any such banquets in my life.\r\nAnd btw, if others want ot ask the questions about the seminar(such as transportation arrangements, etc) or write any comments, feel free to post in this trend.", "Solution_1": "You already got an invitation? Wow, that's fast.\r\nYou can take a look at Yufei's photo to find out what to wear.\r\n[url]http://ca.pg.photos.yahoo.com/ph/z_yufei/my_photos[/url]\r\n___________________________________________________________________________\r\nEDITED: A grammatic error fixed. :blush: I looked at some photos. I should bring a blindfold this time.", "Solution_2": "Thanks a lot. now i know. :)\r\nEDIT: Now it not only sounds like fun, it looks like fun, too. :lol:", "Solution_3": "Seminar is THE BEST. End of discussion. You'll love it. I might even show up on a few of the nights (considering I'll be at school all day anyway, a walk over to renison after classes might not be a bad idea)\r\n\r\nBTW. If you make the IMO team, you have to start a post on here somewhere with 3 words that begin with the letters I, M, O, respectively (since you're not allowed to reveal the team yet). For example: \"It may only be a matter of time before seminar starts, but I can't wait!\"\r\n\r\nGet it?\r\n\r\nBWAHAHAHAHAH!\r\n\r\nOh yeah, and for the appropriate dress thing... be sure to wear sneakers like peng did.", "Solution_4": "[quote=\"Elyot\"]\nBTW. If you make the IMO team, you have to start a post on here somewhere with 3 words that begin with the letters I, M, O, respectively (since you're not allowed to reveal the team yet). For example: \"It may only be a matter of time before seminar starts, but I can't wait!\"\n\nGet it?\n\nBWAHAHAHAHAH!\n[/quote]\r\n[b]I[/b]n [b]m[/b]y [b]o[/b]pinion, rem will make the team without problem :D", "Solution_5": "[quote=\"lightrhee\"][quote=\"Elyot\"]\nBTW. If you make the IMO team, you have to start a post on here somewhere with 3 words that begin with the letters I, M, O, respectively (since you're not allowed to reveal the team yet). For example: \"It may only be a matter of time before seminar starts, but I can't wait!\"\n\nGet it?\n\nBWAHAHAHAHAH!\n[/quote]\n[b]I[/b]n [b]m[/b]y [b]o[/b]pinion, rem will make the team without problem :D[/quote]\r\n\r\ngeez... you couldn't be a little more subtle than that? I mean... you had to quote me AND use bold letters? \r\n\r\nMeh, we all knew you'd make the team anyway.", "Solution_6": "[quote=\"Elyot\"]\nOh yeah, and for the appropriate dress thing... be sure to wear sneakers like peng did.[/quote]\r\n\r\n\r\nHahahahaha. I laughed so hard at this :D", "Solution_7": "Can you get invited to seminar with 144 on Fermat?", "Solution_8": "I think me and Lee are the only ones from wci to get invited. That's what Burns told me. But if you get in or don't get in totally depends on how others do compared to you. There are really a lot of people who got 144 this year.", "Solution_9": "I got in with less that 144 last year, but then again, I'm a girl... they make it easier for us :P", "Solution_10": "They tend to invite provincial champions of each of the contests, provided you get above a certain cut-off, so there's a good geographical representation there. They seem to bias it toward girls for sure. So your best chance of getting there would be to like... be a female in PEI or something. Sadly enough, there were people there with 120-something on Fermat. Really though, I don't know many people who deserved an invite and didn't get one.", "Solution_11": "Ah, wow, its nearing seminar time again! Bleh, too bad I'll only get in if I did decently on Euclid. \r\n\r\nHehee, yes, its jacket and tie. Or no jacket if you don't want to bring one.", "Solution_12": "[quote=\"Sunny\"]Ah, wow, its nearing seminar time again! Bleh, too bad I'll only get in if I did decently on Euclid. [/quote]\r\n\r\nI'm hoping you'll get lucky enough to be invited though. They're kinder to girls when they select who gets to come to seminar.", "Solution_13": "The seminar news that more of you will REALLY care about!\r\n\r\n- Sportsworld is GONE! Kaput! Vamoosed! Essentially, they went bankrupt and sold off everything at a huge auction. So I think the annual seminar trip to Sportsworld will have to be replaced with something else!\r\n\r\n- Waterloo's DDR Machine is GONE! Yep, the upper cove management decided to move it to Toronto this summer, citing reasons of \"it's more profitable there\".\r\n\r\nAs a solution to both of the above problems, I propose that the annual trip to Sportsworld be replaced with a party at Elyot's (yes, I have DDR!)\r\n\r\nHahahah... just kidding, of course. I don't want to get too many people in trouble this year! Maybe instead I can make a visit and teach you all how to play The Settlers of Catan.\r\n\r\nIt might not be a bad idea to start a seminar roll call to see how many AOPS/Mathlink people make it this year!", "Solution_14": "___________________________________\r\nEDITED: Oops! I was going to quote myself, and I edited this thing!! So, the whole thing got deleted. :(", "Solution_15": "[quote=\"Elyot\"]The seminar news that more of you will REALLY care about!\n\n- Sportsworld is GONE! Kaput! Vamoosed! Essentially, they went bankrupt and sold off everything at a huge auction. So I think the annual seminar trip to Sportsworld will have to be replaced with something else!\n\n- Waterloo's DDR Machine is GONE! Yep, the upper cove management decided to move it to Toronto this summer, citing reasons of \"it's more profitable there\".\n\nAs a solution to both of the above problems, I propose that the annual trip to Sportsworld be replaced with a party at Elyot's (yes, I have DDR!)\n\nHahahah... just kidding, of course. I don't want to get too many people in trouble this year! Maybe instead I can make a visit and teach you all how to play The Settlers of Catan.\n\nIt might not be a bad idea to start a seminar roll call to see how many AOPS/Mathlink people make it this year![/quote]\r\n :o \r\nCan this get any worse?\r\nSo we will have to do Math all the time during the seminar. :)", "Solution_16": "[quote=\"Elyot\"]The seminar news that more of you will REALLY care about!\n\n- Sportsworld is GONE! Kaput! Vamoosed! Essentially, they went bankrupt and sold off everything at a huge auction. So I think the annual seminar trip to Sportsworld will have to be replaced with something else!\n\n- Waterloo's DDR Machine is GONE! Yep, the upper cove management decided to move it to Toronto this summer, citing reasons of \"it's more profitable there\".\n\nAs a solution to both of the above problems, I propose that the annual trip to Sportsworld be replaced with a party at Elyot's (yes, I have DDR!)\n\nHahahah... just kidding, of course. I don't want to get too many people in trouble this year! Maybe instead I can make a visit and teach you all how to play The Settlers of Catan.\n\nIt might not be a bad idea to start a seminar roll call to see how many AOPS/Mathlink people make it this year![/quote]\r\n\r\nWHAT?! NO!!! But but but.....DDR....Sportsworld....SEMINAR TRADITION!!!!!!!!!!!!!!!! *sniff*\r\n\r\nJust won't be the same.....", "Solution_17": "[quote=\"lightrhee\"]\nElyot(or anyone else who knows), do you know when the seminar starts and ends?\n[/quote]\r\nI need this information as fast as I can(especially the latter). Please answer if you know when it ends. :omighty:", "Solution_18": "June 11 - June 17.", "Solution_19": "Ahhhh well **** that then. SATs on the 17th, and Summer Program starting sometime the week of the Seminar. That's what I get for compromising with ARML ;) .\r\n\r\nHave fun guys. Eat cheese. :P", "Solution_20": "[quote=\"rem\"]June 11 - June 17.[/quote]\r\nThanks :thumbup:" } { "Tag": [ "induction", "pigeonhole principle", "strong induction" ], "Problem": "Hello,\r\n\r\nI struggled on the following problem for dozens of hours and still can't find the answer!! Please help me before I go crazy!!\r\n\r\nHere's the problem:\r\nGiven a set of n + 1 numbers out of the first 2n natural numbers 1,2...,2n, prove by induction that there are two numbers in the set, one of which divides the other.\r\n\r\nIt might be a strong induction proof!", "Solution_1": "It's not induction; it's pigeonhole.\r\n\r\nA hint: the quotient of some pair will be a power of 2.", "Solution_2": "But it can be proved with induction also.\r\n\r\nSuppose that the thesis works for $2n$. We will prove that it works for $2n+2$. So, suppose that there exist a set $A$ of $n+2$ numbers (from the $\\{1, 2,..., 2n+2\\}$) such that our claim is not true. If there were at least $n+1$ numbers from the set $\\{1, 2, ..., 2n\\}$ in $A$ then we would be done, because of the induction hypothesis. So, $2n+1, 2n+2 \\in A$. Now, it's clear that $n+1 \\not \\in A$ since $n+1|2n+2$. But, because of the induction hypothesis, we know that if we would add $n+1$ to the set $A$ the claim would be true, so there exist number $d$ such that $d|n+1$ or $n+1|d$. The latter is not possible since $d \\not = n+1$ and $d \\leq 2n < 2(n+1)$. So $d|n+1$. But then $d|n+1|2n+2$, a contradiction.", "Solution_3": "[hide=\"Pigeonhole Solution\"]\n\nLet $f(n)$ be the greatest odd factor of $n$. Observing that $f(2n)=f(n)$, there are exactly $n$ different values of $f(n)$ over the integers from $1$ to $2n$. Then, if we choose $n+1$ of these numbers, by the Pigeonhole Principle, we must have two distinct numbers $a= n sau produs din ceva >= n...sau ceva trigonometrie <= sau >= cu n\r\n\r\n:) :blush:", "Solution_3": "Si de accea postez rezolvari numai la astfel de inegalitati :)", "Solution_4": "Bine mai...asteapta ca o sa propun si d-alea de care zici.Intr-adevar, asta e mult prea usoara si cam lunga, poate d-aia nu vrei sa postezi. ;)", "Solution_5": "nu ma...dar nu-mi place :D poate sa fie ea usoara sau grea...problemele de genu asta le fac doar cand e nevoie neaparat :D acu la 2:37 a.m neee :)" } { "Tag": [], "Problem": "[b]Problem.[/b] Find the number of ordered triples of sets $(A,B,C)$ such that $A\\cup B\\cup C=\\{ 1,2,\\hdots, 2003\\}$ and $A\\cap B\\cap C=\\emptyset .$", "Solution_1": "[quote=\"boxedexe\"][b]Problem.[/b] Find the number of ordered triples of sets $(A,B,C)$ such that $A\\cup B\\cup C=\\{ 1,2,\\hdots, 2003\\}$ and $A\\cap B\\cap C=\\emptyset .$[/quote]\r\n[hide=\"...Er...\"]\nBy the second condition they all share an element, and there are 2003 possibilities for that... then every other element in $\\{1,2,\\ldots , 2003\\}$ have to be put in any of $A,B,C$. For each element there are 7 possible ways: $A,B,C,AB,BC,AC,ABC$ and there are 2002 of those...$2003\\cdot 7^{2002}$... :huh: [/hide]", "Solution_2": "[quote=\"ch1n353ch3s54a1l\"][quote=\"boxedexe\"][b]Problem.[/b] Find the number of ordered triples of sets $(A,B,C)$ such that $A\\cup B\\cup C=\\{ 1,2,\\hdots, 2003\\}$ and $A\\cap B\\cap C=\\emptyset .$[/quote]\n[hide=\"...Er...\"]\nBy the second condition they all share an element, and there are 2003 possibilities for that... then every other element in $\\{1,2,\\ldots , 2003\\}$ have to be put in any of $A,B,C$. For each element there are 7 possible ways: $A,B,C,AB,BC,AC,ABC$ and there are 2002 of those...$2003\\cdot 7^{2002}$... :huh: [/hide][/quote]\r\n\r\nAm I missing something? If they all share an element then the intersection would not be an empty set... :maybe:", "Solution_3": "oh oops, I thought it said $A\\cap B\\cap C\\neq \\emptyset$ :huh: wonder why...\r\n[hide=\"Tried getting somewhere...\"]\nSo then they can't share any elements... there are $2003!$ ways to arrange the numbers in the set and the number of ways to choose how many elements in A,B,C are the number of non-negative solutions to $A+B+C=2003$ which is $\\binom{2005}{2}$...Since each permutation of the 2003 elements is counted and you can count forward from 1 in the set to the value of A in $A+B+C=2003$ , then to the value of B, then the rest is C...but then we've over counted because the elements might be permuted within A,B,C....argh[/hide]", "Solution_4": "How about just this? It's basically the same thing you said earlier...\r\n\r\n[hide]\n\nEach number must be in atleast one set, but not in all 3. So there are 6 combinations of what it can be in, A, B, C, AB, AC, or BC. So then the answer is $6^{2003}$\n\n[/hide]", "Solution_5": "[quote=\"tjhance\"]How about just this? It's basically the same thing you said earlier...\n\n[hide]\n\nEach number must be in atleast one set, but not in all 3. So there are 6 combinations of what it can be in, A, B, C, AB, AC, or BC. So then the answer is $2003^{6}$\n\n[/hide][/quote]\r\n\r\nOh I see...thanks\r\n\r\nNot $6^{2003}$? :huh:", "Solution_6": "oops just a typo :blush:" } { "Tag": [ "trigonometry", "geometry proposed", "geometry", "U sviti matematyky" ], "Problem": "Points $ A, B, C$ and $ D$ are chosen on a circle in such a way that the point $ S$ of the intersection of the lines $ AB$ and $ CD$ lies outside the circle and the point $ T$ of intersection of the lines $ AC$ and $ BD$ lies inside the circle. Let points $ M$ and $ N$ lie on chords $ BD$ and $ AC$ respectively and $ K$ denotes point of intersection of the lines $ ST$ and $ MN$. Prove that $ \\frac{MK}{KN}\\equal{}\\frac{TM}{TN}.\\frac{BD}{AC}$.", "Solution_1": "I will assume the points $ M,N,T$ to be distinct (otherwise: guess what will happen :P)\r\n\r\nIt is clear that $ \\angle SBT = \\angle SCT$. Also, the triangles $ SBD$ and $ SCA$ are similar, so\r\n\\[ \\frac {SC}{SB} = \\frac {AC}{BD}.\r\n\\]\r\nAlso, using the law of sine on the triangles $ TMK$ and $ TNK$ respectively, we obtain\r\n\\[ \\frac {TM}{TN} \\cdot\\frac {KN}{KM} = \\frac {\\sin\\angle NTK}{\\sin\\angle MTK}.\r\n\\]\r\nThus,\r\n\\begin{align*} \\frac {TM}{TN}\\cdot\\frac {KN}{KM} & = \\frac {\\sin\\angle NTK}{\\sin\\angle MTK} \\\\\r\n& = \\frac {\\sin\\angle CTS}{\\sin\\angle BTS} \\\\\r\n& = \\frac {\\frac {SC}{ST}\\sin\\angle SCT}{\\frac {SB}{ST}\\sin\\angle SBT} = \\frac {SC}{SB} = \\frac {AC}{BD} \\end{align*}\r\n$ \\Box$" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ABCD$ is an inscribed quadrilateral.\r\n$K\\in{AC\\cap{BD}}$,in the triangle $AKD$ there is a point $P$ such that $\\angle{APC}=\\angle{ADC}+90^0$ and $\\angle{BPD}=\\angle{BAD}+90^0$.\r\nProve that the diogonals of the quadrilateral formed by projections of $P$ on the side of $ABCD$ are prpendicular.", "Solution_1": "Posted before as [url=http://www.mathlinks.ro/Forum/topic-80597.html]Nice problem[/url]." } { "Tag": [], "Problem": "(i) Consider two positive integers $a$ and $b$ which are such that $a^a b^b$ is divisible by $2000$. What is the least possible value of $ab$?\r\n(ii) Consider two positive integers $a$ and $b$ which are such that $a^b b^a$ is divisible by $2000$. What is the least possible value of $ab$?", "Solution_1": "$2000 = 2^4 \\cdot 5^3$\r\n\r\nIntroducing other prime factors would only increasing product uselessly.\r\n\r\n$10$ is an obvious minimum in the first case which is reached with $1^1 \\cdot 10^10$\r\n\r\n\r\nIn the second case, wlog, $2 | a$ and $5 | b$. Minimum is for $4^5 \\cdot 5^4$", "Solution_2": "Isn't the answer for problem 1\n$a=10, b=1$?", "Solution_3": "Yeah, the answers are 10 and 20 respectively.. \n" } { "Tag": [ "linear algebra", "matrix", "abstract algebra", "Ring Theory", "linear algebra unsolved" ], "Problem": "Find the number of homomorphisms of the rings $\\mathrm{Mat}_{2\\times2}(\\mathbb{C})\\to\\mathrm{Mat}_{3\\times3}(\\mathbb{C})$ such that the image of $2\\times2$ identity matrix is $3\\times3$ identity matrix.\r\n(V. Mazorchuk)", "Solution_1": "The general result is this:\r\n\r\n$(\\#)$ Let $k$ be a field, and let $m,n$ be positive integers. There is a ring homomorphism $\\varphi: \\mathcal M_{n}(k)\\to\\mathcal M_{m}(k)$ iff $n|m$. \r\n\r\n(here, all rings and algebras have a unit, and all ring or algebra homomorphisms are assumed to preserve units)\r\n\r\n\r\nThe version of $(\\#)$ with \"ring homomorphism\" replaced by \"$k$-algebra homomorphism\" has been discussed before on the forum (or at least the more difficult implication has :)). See, for example, the last message in [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=algebra&t=4923]this[/url] topic, and also [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=module+simple&t=90862]this[/url]. The stronger statement $(\\#)$ follows if we prove that whenever there is a ring homomorphism $\\varphi: \\mathcal M_{n}(k)\\to\\mathcal M_{m}(k)$, we can also define a $k$-algebra homomorphism $\\tilde\\varphi: \\mathcal M_{n}(k)\\to\\mathcal M_{m}(k)$:\r\n\r\nFor $i,j=\\overline{1,n}$, let $e_{ij}$ be the $n\\times n$ matrix having $1$ in position $(i,j)$ and $0$ everywhere else. The elements $e_{ij}$ of $\\mathcal M_{n}(k)$ form a $k$-basis, and satisfy the relations $e_{ij}e_{kl}=\\delta_{jk}e_{il}\\ (*)$ ($\\delta_{jk}$ being the Kronecker delta). If $f_{ij}=\\varphi(e_{ij})$, then the elements $f_{ij}\\in\\mathcal M_{m}(k)$ also satisfy $(*)$. Now simply define $\\tilde\\varphi\\left(\\sum\\alpha_{ij}e_{ij}\\right)$ to be $\\sum\\alpha_{ij}f_{ij}$ for all choices of scalars $\\alpha_{ij}\\in k,\\ i,j=\\overline{1,n}$. It's easy to check that this makes $\\tilde\\varphi$ a $k$-algebra homomorphism from $\\mathcal M_{n}(k)$ into $\\mathcal M_{m}(k)$." } { "Tag": [ "algorithm", "Divisibility Theory" ], "Problem": "Find all $n \\in \\mathbb{N}$ such that $3^{n}-n$ is divisible by $17$.", "Solution_1": "We have that $ 3^8\\equiv(3^3)^2*3^2\\equiv10^2*3^2\\equiv( \\minus{} 2)*3^2\\equiv \\minus{} 1\\bmod{17}$so 3 is primitive root mod17.\r\nClearly 17 does not divide n.Let $ n\\equiv{a}\\bmod{17}$where $ a\\in\\left\\{1,2,...16\\right\\}$.Because 3 is primitive root mod17 there is unique $ b\\in\\left\\{1,2,...16\\right\\}$ such that $ 3^b\\equiv{a}\\bmod{17}$.\r\nSo $ 3^n\\equiv{n}\\equiv{a}\\equiv3^b\\bmod{17}\\Rightarrow n\\equiv{b}\\bmod{16}$ We conclude that $ n\\equiv x\\bmod(16*17)$such that$ x\\equiv a\\bmod{17}$and$ x\\equiv{b}\\bmod{16}$.\r\nThis is the algorithm for finding the solutions mod272\r\nWe have $ 3\\equiv3\\mod{17},3^2\\equiv9\\bmod{17},3^3\\equiv10\\bmod{17},3^4\\equiv13\\bmod{17},3^5\\equiv5\\bmod{17},3^6\\equiv15\\bmod{17},3^7\\equiv11\\bmod{17},3^8\\equiv16\\bmod{17},3^9\\equiv14\\bmod{17},3^{10}\\equiv8\\bmod{17},3^{11}\\equiv7\\bmod{17},3^{12}\\equiv4\\bmod{17},3^{13}\\equiv12\\bmod{17},3^{14}\\equiv2\\bmod{17},3^{15}\\equiv6\\bmod{17},3^{16}\\equiv1\\bmod{17}$\r\nIf $ a \\equal{} 1\\Rightarrow b \\equal{} 16\\Rightarrow n\\equiv256\\bmod{272}$\r\nIf $ a \\equal{} 2\\Rightarrow b \\equal{} 14\\Rightarrow n\\equiv205\\bmod{272}$\r\nIf $ a \\equal{} 3\\Rightarrow b \\equal{} 1\\Rightarrow n\\equiv241\\bmod{272}$\r\nIf $ a \\equal{} 4\\Rightarrow b \\equal{} 12\\Rightarrow n\\equiv140\\bmod{272}$\r\nIf $ a \\equal{} 5\\Rightarrow b \\equal{} 5\\Rightarrow n\\equiv5\\bmod{272}$\r\nIf $ a \\equal{} 6\\Rightarrow b \\equal{} 15\\Rightarrow n\\equiv159\\bmod{272}$\r\nIf $ a \\equal{} 7\\Rightarrow b \\equal{} 11\\Rightarrow n\\equiv75\\bmod{272}$\r\nIf $ a \\equal{} 8\\Rightarrow b \\equal{} 10\\Rightarrow n\\equiv42\\bmod{272}$\r\nIf $ a \\equal{} 9\\Rightarrow b \\equal{} 2\\Rightarrow n\\equiv162\\bmod{272}$\r\nIf $ a \\equal{} 10\\Rightarrow b \\equal{} 3\\Rightarrow n\\equiv163\\bmod{272}$\r\nIf $ a \\equal{} 11\\Rightarrow b \\equal{} 7\\Rightarrow n\\equiv215\\bmod{272}$\r\nIf $ a \\equal{} 12\\Rightarrow b \\equal{} 13\\Rightarrow n\\equiv29\\bmod{272}$\r\nIf $ a \\equal{} 13\\Rightarrow b \\equal{} 4\\Rightarrow n\\equiv132\\bmod{272}$\r\nIf $ a \\equal{} 14\\Rightarrow b \\equal{} 9\\Rightarrow n\\equiv201\\bmod{272}$\r\nIf $ a \\equal{} 15\\Rightarrow b \\equal{} 6\\Rightarrow n\\equiv134\\bmod{272}$\r\nIf $ a \\equal{} 16\\Rightarrow b \\equal{} 8\\Rightarrow n\\equiv152\\bmod{272}$\r\n(I correct my wrongs in calculations)", "Solution_2": "So what is your final answer? Not all those n's work, for example $ n \\equal{} 3$ doesn't work?", "Solution_3": "all such $ n$ are in form $ n\\equal{}16(17t\\plus{}r\\minus{}3^r)\\plus{}r$ where $ r,t\\in\\mathbb Z$ which $ 0\\leq r\\leq 16$ and $ t\\geq \\frac{3^r\\minus{}r}{17}$. A boring problem huh? :maybe:" } { "Tag": [], "Problem": "Prove that [tex]F_{n+1}^3-4F_n^3-F_{n-1}^3=3(-1)^nF_n[/tex], where [tex]\\{F_n\\}^{\\infty}_{n=1}[/tex] is the well known Fibonacci sequence.", "Solution_1": "If we write it all in terms of [tex]F_n[/tex] and [tex]F_{n-1}[/tex], we get\r\n\r\n[tex]\\\\ (F_n+F_{n-1})^3-4F_n^3-F_{n-1}^3=3(-1)^nF_n\\\\\r\n=3F_n (F_{n-1}^2+F_n F_{n-1}-F_n^2)[/tex]\r\n\r\nTo deal with the parenthesized part we eliminate [tex]F_n[/tex]:\r\n\r\n[tex]\\\\ F_{n-1}^2+(F_{n+1}-F_{n-1}) F_{n-1}-(F_{n+1}-F_{n-1})^2\\\\\r\n= F_{n+1}F_{n-1}-(F_{n+1}-F_{n-1})^2\\\\\r\n= F_{n+1}F_{n-1}-F_n^2\\\\\r\n=(-1)^n[/tex]\r\n\r\nThis last identity is well known." } { "Tag": [ "email", "LaTeX" ], "Problem": "I'm having trouble getting the solutions template to work for me, so I'm going to send them each in one file. However, I then have 5 + 5 = 10 files and Yahoo! only let's me attach 5 at a time, so can I just send 5 in each of two emails?\r\n\r\nAlso, I'm going to mail in my solutions as well, since the email is just a back up, and do I have to include the latex form of each solution as well as the printed pdf version or what?", "Solution_1": "[quote=\"vinny919\"]I'm having trouble getting the solutions template to work for me, so I'm going to send them each in one file. However, I then have 5 + 5 = 10 files and Yahoo! only let's me attach 5 at a time, so can I just send 5 in each of two emails?[/quote]\nThat's fine. Just make sure that the subject lines of your emails say \"1 of 2\" and \"2 of 2\" or something like that, so we know to expect 2 messages.\n\n[quote]Also, I'm going to mail in my solutions as well, since the email is just a back up, and do I have to include the latex form of each solution as well as the printed pdf version or what?[/quote]\r\nIf you're sending stuff through the regular mail, you just need to mail us your printed output (i.e. don't send us printed pages of LaTeX code, just send us the printouts of the PDF files)." } { "Tag": [], "Problem": "Em mu\u1ed1n xin 1 s\u1ed1 t\u00e0i li\u1ec7u h\u00ecnh h\u1ecdc ph\u1eb3ng , vecto c\u0169ng \u0111c \u1ea1", "Solution_1": "http://forum.mathscope.org/forumdisplay.php?f=28", "Solution_2": "T\u00e0i li\u1ec7u gi\u00e1o khoa chuy\u00ean to\u00e1n l\u1edbp 10 ph\u1ea7n Hinh h\u1ecdc, tr\u00ecnh b\u00e0y c\u00f3 h\u1ec7 th\u1ed1ng, d\u1ec5 hi\u1ec3u, b\u00e0i t\u1eadp hay", "Solution_3": "Co mot it o day http://trungtuan.wordpress.com/2010/04/16/topic-42/" } { "Tag": [ "group theory", "abstract algebra", "inequalities", "induction", "Galois Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "$ \\text{1.}$Prove that a minimal normal subgroup $ N$ of a finite solvable group $ G$ is an elementary abelian $ p$-group for some prime $ p$.\r\n\r\n$ \\text{2.}$Let $ G$ be a group in which each subgroup is subnormal. If $ N$ is a normal subgroup of $ G$, $ Z(N)\\neq1$, and $ G/N$ is cyclic then prove that $ Z(G)\\neq1$\r\n\r\n$ \\text{3.}$If $ H$ is a normal subgroup of a group $ G$, and $ H$ and $ G/H^{\\prime}$ are nilpotent groups of classes $ k$ and $ m$, respectively, then $ G$ is nilpotent of class $ \\leq\\left(2^{k}\\minus{}1\\right)m\\minus{}2^{k\\minus{}1}\\plus{}1$.", "Solution_1": "1) Since $ G$ is solvable, we have $ N'\\equal{}[N,N] < N$ is normal in $ G$ as a characteristic subgroup of $ N$. Therefore $ N$ must be abelian by the minimality of $ N$. Moreover, it must be a $ p$-group, since any $ p$-Sylowgroup of a finite abelian group is characteristic. It remains to show that $ N$ has exponent $ p$. This follows from the fact that the Frattini subgroup is a proper characteristic subgroup in a $ p$-group, ie. it must be equal to $ 1$, otherwise it would form a smaller minimal normal subgroup of $ G$.\r\n\r\n2) Let $ G\\equal{}\\cdot N$ and set $ U\\equal{}$. Since $ Z(N)$ is normal in $ G$ as a characteristic subgroup of $ G$, we can consider the subgroup $ H\\equal{}U\\cdot Z(N)$. The cyclic subgroup $ U$ is subnormal in $ G$ and hence, it is sub-normal in $ H$. In particular, $ U \\subset N_H(U)\\leq H$. We claim that $ N_H(U)\\equal{}U\\cdot N_{Z(N)}(U)$ holds with $ N_{Z(N)}(U)$ a normal subgroup of $ N_H(U)$. Clearly, $ N_H(U)\\cap Z(N)\\equal{}N_{Z(N)}(U)$ is a normal subgroup of $ N_H(U)$, therefore, it remains to show that $ N_H(U)\\subseteq U\\cdot N_{Z(N)}(U)$ holds. Let $ h\\in N_H(U)\\leq H$ be arbitrary. Then there exists some $ z\\in Z(N)$ with $ h\\in U\\cdot z$. Since $ U\\subseteq N_H(U)$, the element $ z$ must be contained in $ N_H(U)$ as well, i.e. $ z\\in N_H(U)\\cap Z(N)$. One consequence of the now verified claim is, that $ 1\\not\\equal{} N_{Z(N)}(U)\\leq Z(N)$ holds. Note, that $ N_{Z(N)}(U)$ already commutes with $ N$. Since $ U$ and $ N_{Z(N)}(U)$ are both normal in $ N_H(U)$, we compute for their commutator $ C\\equal{}[U,N_{Z(N)}(U)]\\subseteq\\, U\\cap\\, N_{Z(N)}(U)$. If $ C\\equal{}1$ is valid, we have directly $ N_{Z(N)}(U)\\subseteq Z(G)$. If $ C\\not\\equal{}1$, then $ C$ commutes with $ U$ and with $ N$, ie. $ C\\subseteq Z(G)$.\r\n\r\n3) It is trivial to show that $ \\forall k\\in \\mathbb{N}$ the inequalities $ 2^{k\\minus{}1}\\geq k$ and $ 2^k > \\frac{k(k\\plus{}1)}{2}$ hold. E.g. for the second inequality we verify the claim easily for $ k\\equal{}1,2,3$ and for $ k\\plus{}1\\geq 4$ we use induction, i.e. $ \\frac{(k\\plus{}1)(k\\plus{}2)}{2}\\equal{}\\frac{k(k\\plus{}1)}{2}\\cdot \\frac{k\\plus{}2}{k}<2^k\\cdot \\frac{5}{3}<2^{k\\plus{}1}$. Therefore, we have for all $ k,m\\in\\mathbb{N}$:\r\n\r\n$ \\left(2^k\\minus{}1\\minus{}\\frac{k(k\\plus{}1)}{2}\\right)(m\\minus{}1)\\plus{}2^{k\\minus{}1}\\minus{}k\\geq 0$ which is equivalent to $ (m\\minus{}1)\\frac{k(k\\plus{}1)}{2}\\plus{}k\\leq m(2^k\\minus{}1)\\minus{}2^{k\\minus{}1}\\plus{}1$. By a famous result of P. Hall based on a repeated application of the Three Subgroup Lemma we know under the assumptions of the problem that $ G$ is nilpotent of class $ \\leq (m\\minus{}1)\\frac{k(k\\plus{}1)}{2}\\plus{}k$. This completes the proof." } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "Euler", "cyclic quadrilateral", "geometry solved" ], "Problem": "ABC is a triangle with orthocenter H. The median from A meets the circumcircle again at A1, and A2 is the reflection of A1 in the midpoint of BC. The points B2 and C2 are defined similarly. Show that H, A2, B2 and C2 lie on a circle. \r\n\r\nNow I think G, the centroid, must also lie on the circle trough H, A2, B2, C2, but I don't know how to prove this.", "Solution_1": "In fact the center of the circle trough H, A2, B2, C2 should be the midpoint of [HG] (where G is the centroid). So it suffices to prove that A2G is perpendicular to A2H. I'm trying to prove this now.", "Solution_2": "Here?s a soln involving inversion (I just love inversion). Let?s make some notations: H1 is the foot of the altitude from A, H2 for B and H3 for C, M is the midpoint of BC.\r\n\r\nConsider the inversion of pole A and power AH3*AB (=AH*AH1=AH2*AC, you can prove this easily). The circle H3H1MH2 (Euler?s circle) must be transformed in another circle which passes through B, C and H because these are the images of the pts H3, H2 and H1 respectively. The image of point M must also be on this circle (let?s call this image X). Because A, M and X are collinear we get that X is on the median AM (*). Because BHXC must be a cyclic quadrilateral angle BXC=angle BHC=pi-angle A=angle BA1C (**). From (*) and (**) we find that X is the reflection of A1 in the midpoint of BC, so X=A2=the image of pt M in the inversion we considered. This means that AH*AH1=AA2*AM, which in turn means that HH1MA2 is a cyclic quadrilateral, and because angle HH1M=90 deg we find angle HA2M=90 deg, so HA2 is perpendicular to the median from A=line A2G, which means that A2, B2, C2 are on a circle of diameter HG, exactly as you thought.", "Solution_3": "Does anyone have a solution without inversion ?\r\n(Nice job Grobber, but I don't like inversion and I never see how to apply it the right way)\r\nThe problem simplified a bit. It suffices to prove the following.\r\n\r\nLet ABC be a triangle with orthocenter H. Let F be the foot of the altitude which passes trough A. Let M be the midpoint of [BC], A(1) the intersection of AM with the circumcircle of ABC. Finally, let A(2) be the reflection of A(1) in M. Prove that A(2)H is perpendicular to AM.\r\n\r\nOne can try to prove that A(2)HFM is cyclic, or that triangles A(2)HA and AMF are similar. But up till now I didn't manage to do this ...", "Solution_4": "So, a non - inversion proof, anyone ?", "Solution_5": "It is easy to see, that A_2, B_2, C_2 --- projections of H on the medians of ABC. So A_2, B_2,C_2,H, G lies on the circle, having GH as a diameter", "Solution_6": "....without any inversion :D", "Solution_7": "[quote=\"Arne\"] It suffices to prove the following.\n\nLet ABC be a triangle with orthocenter H. Let F be the foot of the altitude which passes trough A. Let M be the midpoint of [BC], A(1) the intersection of AM with the circumcircle of ABC. Finally, let A(2) be the reflection of A(1) in M. Prove that A(2)H is perpendicular to AM.\n[/quote]\r\n\r\nAlso simple solution \r\n prove that AHA2 ~ AMF (triangles) <=> \r\nAM*AA2=AF*AH but AM*AA2=AM(AM-MA2)=AM(AM-MA1)=AM^2-P(M) where P(M) is the power of M. The rest is very simple computation.\r\n\r\nand AF=2S/a and AH=2RcosA\r\n\r\nfedor's is much more simple, but this is \"valid\" too" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Indeed problem seems to be easy but \u0131 coludn't solve it :(\r\n\r\n\r\nThink a number such that all of its digits are different from each other and also it is divisible by 11111 we call such numbers \"interesting numbers\". How many interesting number that consists of 10 different digits?", "Solution_1": "[quote=\"epsilon07\"]Indeed problem seems to be easy but \u0131 coludn't solve it :(\n\n\nThink a number such that all of its digits are different from each other and also it is divisible by 11111 we call such numbers \"interesting numbers\". How many interesting number that consists of 10 different digits?[/quote]\r\n\r\nThe sum of the $ 10$ digits of such numbers is $ 0\\plus{}1\\plus{}2\\plus{}3\\plus{}4\\plus{}5\\plus{}6\\plus{}7\\plus{}8\\plus{}9\\equal{}45$ and so they are divisible by $ 9$, so by $ 99999$ since $ 11111\\neq 0\\pmod 3$\r\n\r\nSo, the required numbers may be written $ x(10^5\\minus{}1)$ and $ x$ has exactly $ 5$ digits (else, $ x\\equal{}100000$ or $ 100001$ and $ x(10^5\\minus{}1)$ would be $ 9999900000$ or $ 9999999999$ and could not have all its digits different).\r\n\r\nIt's then immediate to see that $ x(10^5\\minus{}1)\\equal{}\\overline{abcde(9\\minus{}a)(9\\minus{}b)(9\\minus{}c)(9\\minus{}d)(9\\minus{}e)}$ for some $ a,b,c,d,e$ and so :\r\n\r\n$ a$ may take $ 9$ different values (I suppose that when the problem says \"$ 10$ digits numbers\", the leftmost one is non-zero)\r\n$ b$ may then take any of the $ 8$ remaining values (we exclude $ a$ and $ 9\\minus{}a$)\r\n$ c$ may then take any of the $ 6$ remaining values\r\n$ d$ may then take any of the $ 4$ remaining values\r\n$ e$ may then take any of the $ 2$ remaining values\r\n\r\nAnd the result is $ 9.8.6.4.2\\equal{}3456$ such numbers.", "Solution_2": "very well thanks a lot nice solution....", "Solution_3": "Let $n=\\overline{d_1d_2...d_{10}}$ be a 10-digit interesting number. Note that $\\{d_i\\}_{i=1}^{10}$ are $\\{i\\}_{i=0}^9$ in some order, and obviously $d_1\\not=0$. This means $\\sum_{i=1}^{10}d_i\\equiv 45\\equiv 0\\pmod 9$ so $9\\mid n$. Since and $\\gcd(9,11111)=1$, and we also know that any interesting number must be divisible by $11111$, it follows that $9(11111)=99999\\mid n$. Now let $x=\\overline{d_!d_2d_3d_4d_5}$ and let $y=\\overline{d_6d_7d_8d_9d_{10}}$, so that $n=10^5x+y\\equiv x+y\\equiv 0\\pmod {99999}$. But $x+y<99999+99999=2(99999)$ so in order for the congruence to be true we must have $x+y=(0,99999)$. The first one is clearly false so $x+y=99999$. This means that $d_i+d_{i+5}=9$ for $i=1,3,5,7,9$. There are $5!=120$ ways to distribute $(0,9),(1,8),...,(4,5)$ among $(d_1,d_6),(d_3,d_8),(d_4,d_9),(d_5,d_{10})$, and for each pair the numbers are interchangeable (e.g. $(d_2,d_7)$ can be $(0,9)$ [i]as well as[/i] $(9,0)$) in $2^5=32$ ways for a total of $32(120)=3840$ ways. However, we have over-counted a $\\tfrac{1}{10}$th of these numbers which have $a=0$. This is clearly not allowed, so there are actually $3840-\\tfrac{1}{10}(3840)=\\tfrac{9}{10}(3840)=9(384)=\\boxed{3456}$ interesting numbers.", "Solution_4": "I cant remember the year but thats from Turkish first round.", "Solution_5": "[quote=\"Mathematicalx\"]I cant remember the year but thats from Turkish first round.[/quote]\nWell actually the original source is an old Russia MO." } { "Tag": [ "induction" ], "Problem": "Prove that $ 21 | 4^{n\\plus{}1} \\plus{} 5^{2n\\minus{}1}$ for every natural number $ n$. \r\n\r\nI think I have it with mod 3 and mod 7, but I don't understand how you'd use induction to prove it...o_O", "Solution_1": "[hide=\"Inductive step\"] Let $ S_n \\equal{} 4^{n\\plus{}1} \\plus{} 5^{2n\\minus{}1}$. If $ 21 | S_k \\equal{} 4^{k\\plus{}1} \\plus{} 5^{2k\\minus{}1}$, then\n\n$ S_{k\\plus{}1} \\equal{} 4^{k\\plus{}2} \\plus{} 5^{2k\\plus{}1} \\equal{} 4 \\cdot 4^{k\\plus{}1} \\plus{} 25 \\cdot 5^{2k\\minus{}1} \\equal{} 4S_k \\plus{} 21 \\cdot 5^{2k\\minus{}1}$\n\nso $ 21 | S_{k\\plus{}1}$. [/hide]" } { "Tag": [ "conics", "parabola" ], "Problem": "[u][b]Problem 1:[/b][/u] Find all reals $ (x,y)$ satisfying\r\n\\begin{align} x^2 + 4y & = 21 \\\\\r\ny^2 + 4x & = 21 \\end{align}\r\n\r\n[hide=\"Answer\"]$ (3,3),( - 7, - 7),( - 1,5),(5 - 1)$[/hide]\n\n[hide=\"Hint\"]Try to build squares.[/hide]\n\n[hide=\"Solution\"]\n(1)-(2):$ x^2 - 4x - (y^2 - 4y) = 0\\Rightarrow x^2 - 4x + 4 - (y^2 - 4y - 4) = 0\\Rightarrow \\\\\n(x + 2)^2 - (y + 2)^2 = 0\\Rightarrow (x - y)(x + y - 4) = 0$\n$ x = y$ yields $ (3,3),( - 7, - 7)$, $ x + y = 4$ yields $ ( - 1,5),(5, - 1)$[/hide]\n\n[u][b]Problem 2:[/b][/u]The excircle $ c$ of an isosceles triangle $ \\Delta ABC$ $ \\left(|\\overline{AC}| = |\\overline{BC}|\\right)$ touches $ \\overline{BC}$ internally. Prove that its radius is the same as the height from $ C$ to $ \\overline{AB}$.\n\n[hide=\"Hint\"]Use $ A = \\frac {(b + c - a)r_a}2.$[/hide]\n\n[hide=\"Answer\"]$ b = a\\Rightarrow \\frac {ch_c}2 = A = \\frac {b + c - a)r_a}2 = \\frac {cr_a}2\\;\\;(a = b).$[/hide]\n\n[u][b]Problem 3:[/b][/u] Is there a magic square that starts with the numbers in the attachment?\n\n[hide=\"Answer\"]No.[/hide]\n\n[hide=\"Hint\"]Prove that there has to be a 13 in the top left corner, which is impossible.[/hide]\n\n[hide=\"Solution\"]We can find that on top of the 13 is a 1, on top of the 15 is a 5 and in the top right corner there's a 16. Now look at the different numbers we can put into the bottom left corner (2 and 3). No matter what we put there, we'll have to write a 13 in the top left corner.[/hide]\n\n[u][b]Problem 4:[/b][/u]We call a positive integer $ n$ [i]stout[/i] if there are exactly 2 pairs of positive integers $ (k,l)$ with $ \\frac 1n = \\frac 1k + \\frac 1l$ and $ k\\le l$. Prove that $ n$ is stout iff $ n$ is a prime.\n\n[hide=\"Hint\"]$ \\frac 1n = \\frac 1{2n} + \\frac 1{2n} = \\frac 1{n + 1} + \\frac 1{n(n + 1)}$ and $ \\frac 1n = \\frac 1k + \\frac 1l \\Rightarrow \\frac 1{mn} = \\frac 1{mk} + \\frac 1{ml} = \\frac 1{k'} + \\frac 1{l'}.$[/hide]\n\n[hide=\"Solution\"]\nFirst, we observe $ n < k\\le2n\\le l$, and a direct condequence is that 1 is not stout as $ k$ has to be 2. Now consider a non-prime number $ n$. Per definitionem $ n$ has at least one divisor $ 1 < x < n$. Then $ 1 < \\frac nx < n$ is a natural number, too. There are at least 2 ways to write $ x$ as $ \\frac 1k + \\frac 1l$ (look it up in the hint). Then for each such pair we have $ \\frac 1n = \\frac 1{xy} = \\frac 1{ky} + \\frac 1{ly}$, which are at least two compositions. But for each $ (k',l')$ We found this way $ \\gcd(n,k') = \\gcd(xy,ky) \\neq 1$ and $ \\gcd(n,l')\\neq 1$. However, $ \\gcd(n,n + 1) = 1$ wo we have to have another pair $ (n + 1,n(n + 1))$ which makes more than two pairs in total so $ n$ isn't stout.\n\nNow let $ p$ be a prime number and $ \\frac 1p = \\frac 1k + \\frac 1l \\Leftrightarrow p = \\frac {kl}{k + l}\\Leftrightarrow p|kl\\Leftrightarrow p|k \\or p|l.$\n[u]Case 1: [/u]$ p|k$\n$ p < k\\le 2p\\Rightarrow k = 2p \\Rightarrow (k,l) = (2p,2p)$\n[u]Case 2: [/u]$ p|l$\nWrite $ l = qp$; then $ \\frac 1p = \\frac 1k + \\frac 1{qp} \\Rightarrow k = \\frac {pq}{q - 1}\\in\\mathbb{N}.$ Now $ gcd(q,q - 1) = 1$ so we have to have $ (q - 1)|p\\Rightarrow q = 2 \\or q = p + 1.$ $ q = 2$ yields $ (2p,2p)$ and $ q = p + 1$ yields $ (p + 1,p(p + 1))$ which are all solutions. So we conclude that if $ n$ is prime it is stout and the problem is solved.\n[/hide]", "Solution_1": "No. 4\r\n[hide]\n$ \\frac {1}{n} \\equal{} \\frac {1}{k} \\plus{} \\frac {1}{l}$\n$ \\Rightarrow n \\equal{} \\frac {kl}{k \\plus{} l}$\n$ \\Rightarrow kl \\minus{} n(k \\plus{} l) \\equal{} 0$\n$ \\Rightarrow (k \\minus{} n)(l \\minus{} n) \\equal{} n^2$\nBecause $ k, l, n$ are all positive integers, there is,\n$ k \\minus{} n \\equal{} p$\n$ l \\minus{} n \\equal{} q$\nwhere $ p,q$ are integers and $ pq \\equal{} n^2$; because $ k \\leq l, p\\leq q$. Then,\n$ (k, l) \\equal{} (n \\plus{} p, n \\plus{} q)$\nBecause $ k > 0, l > 0 \\Rightarrow n > \\minus{} p, n > \\minus{} q.$\nIf $ p < 0 \\Rightarrow q < 0 \\Rightarrow n > |p|, n > |q| \\Rightarrow n^2 > |pq| \\equal{} pq$. Contradiction. Therefore, $ p > 0, q > 0.$\n\n[b]Case 1[/b]: $ n \\equal{} 1$. Then the only positive integer factoring of $ n^2 \\equal{} 1$ is $ (p, q) \\equal{} (1, 1)$. Only one solution.\n\n[b]Case 2[/b]: $ n$ is a prime. Then all the $ (p, q)$ are $ (1, n^2)$ and $ (n, n)$. Exactly two solutions.\n\n[b]Case 3[/b]: $ n$ is a composite. Let $ n \\equal{} a\\cdot m$, where $ a$ is a prime and $ m$ is not 1. Then besides the factorings mentioned in Case 2, there's at least one more factoring $ (a, a\\cdot m^2)$. More than two solutions.\n\nHence, iff $ n$ is a prime, the statement holds.[/hide]", "Solution_2": "[hide=\"1\"]We add the equations and add two fours:\n\n$ x^2\\plus{}4x\\plus{}4\\plus{}y^2\\plus{}4y\\plus{}4\\equal{}50$\n\n$ (x\\plus{}2)^2\\plus{}(y\\plus{}2)^2\\equal{}50$\n\nWe can easily graph that circle.[/hide]\r\n\r\n@bertram: The problem said reals, not integers.", "Solution_3": "[quote=\"1=2\"][hide=\"1\"]We add the equations and add two fours:\n\n$ x^2 \\plus{} 4x \\plus{} 4 \\plus{} y^2 \\plus{} 4y \\plus{} 4 \\equal{} 50$\n\n$ (x \\plus{} 2)^2 \\plus{} (y \\plus{} 2)^2 \\equal{} 50$\n\nWe can easily graph that circle.[/hide]\n\n@bertram: The problem said reals, not integers.[/quote]Check your work before posting. Not all real (in fact almost none) points in that circle [u]satisfy the original system[/u]", "Solution_4": "[hide=\"1\"] Adding the equations gives $ (x \\plus{} 2)^2 \\plus{} (y \\plus{} 2)^2 \\equal{} 50$.\n\nSubtracting, we get \n\n$ x^2 \\minus{} y^2 \\plus{} 4y \\minus{} 4x \\equal{} 0 \\Leftrightarrow$\n$ (x \\minus{} y)(x \\plus{} y \\minus{} 4) \\equal{} 0$\n\nWhen $ x \\equal{} y$ we have $ (x \\plus{} 2)^2 \\equal{} 25$, so $ (x, y) \\equal{} \\boxed{ (3, 3), (\\minus{}7, \\minus{}7) }$\n\nWhen $ 4 \\minus{} x \\equal{} y$ we have \n\n$ x^2 \\plus{} 4x \\plus{} 4 \\plus{} 36 \\minus{} 12x \\plus{} x^2 \\equal{} 2x^2 \\minus{} 8x \\plus{} 40 \\equal{} 50 \\Leftrightarrow$\n$ x^2 \\minus{} 4x \\minus{} 5 \\equal{} 0 \\Leftrightarrow$\n$ (x \\minus{} 5)(x \\plus{} 1) \\equal{} 0$\n\nSo $ (x, y) \\equal{} \\boxed{ (5, \\minus{}1), (\\minus{}1, 5) }$.\n\nGenerally, two parabolas intersect in four points. [/hide]", "Solution_5": "[quote=\"t0rajir0u\"][hide=\"1\"] Adding the equations gives $ (x \\plus{} 2)^2 \\plus{} (y \\plus{} 2)^2 \\equal{} 50$.\n\nSubtracting, we get \n\n$ x^2 \\minus{} y^2 \\plus{} 4y \\minus{} 4x \\equal{} 0 \\Leftrightarrow$\n$ (x \\minus{} y)(x \\plus{} y \\minus{} 4) \\equal{} 0$\n\nWhen $ x \\equal{} y$ we have $ (x \\plus{} 2)^2 \\equal{} 25$, so $ (x, y) \\equal{} \\boxed{ (3, 3), ( \\minus{} 7, \\minus{} 7) }$\n\nWhen $ 4 \\minus{} x \\equal{} y$ we have \n\n$ x^2 \\plus{} 4x \\plus{} 4 \\plus{} 36 \\minus{} 12x \\plus{} x^2 \\equal{} 2x^2 \\minus{} 8x \\plus{} 40 \\equal{} 50 \\Leftrightarrow$\n$ x^2 \\minus{} 4x \\minus{} 5 \\equal{} 0 \\Leftrightarrow$\n$ (x \\minus{} 5)(x \\plus{} 1) \\equal{} 0$\n\nSo $ (x, y) \\equal{} \\boxed{ (5, \\minus{} 1), ( \\minus{} 1, 5) }$.\n\nGenerally, two parabolas intersect in four points. [/hide][/quote]\r\nIf this problem was on an olympiad, could we find those four solutions and assert that since two distinct parabolas have at most four intersections points, we're done?", "Solution_6": "The solution I gave is complete without that last statement. Alternately, you could cite Bezout's Theorem, which in real affine space states that the [b]maximum[/b] number of intersections of a curve of degree $ n$ and a curve of degree $ m$ is $ mn$.", "Solution_7": "[quote=\"t0rajir0u\"]The solution I gave is complete without that last statement. Alternately, you could cite Bezout's Theorem, which in real affine space states that the [b]maximum[/b] number of intersections of a curve of degree $ n$ and a curve of degree $ m$ is $ mn$.[/quote]\r\n\r\nYou'd have to prove this theorem though, and that's definitely not worth it. :)\r\n\r\nYou could prove it for this problem by substituting $ y \\equal{} \\frac{21\\minus{}x^2}4$ and obtain an equation of degree 4, of which you can guess the roots -7, -1, 3, 5 and then compute the corresponding y's.", "Solution_8": "[quote=\"bertram\"][quote=\"t0rajir0u\"]The solution I gave is complete without that last statement. Alternately, you could cite Bezout's Theorem, which in real affine space states that the [b]maximum[/b] number of intersections of a curve of degree $ n$ and a curve of degree $ m$ is $ mn$.[/quote]\n\nYou'd have to prove this theorem though[/quote]\r\n\r\nBezout's Theorem is an extremely well-known result. As long as it's cited, it doesn't need to be proven, and the proof of the general theorem is much more difficult.", "Solution_9": "[quote=\"t0rajir0u\"][quote=\"bertram\"][quote=\"t0rajir0u\"]The solution I gave is complete without that last statement. Alternately, you could cite Bezout's Theorem, which in real affine space states that the [b]maximum[/b] number of intersections of a curve of degree $ n$ and a curve of degree $ m$ is $ mn$.[/quote]\n\nYou'd have to prove this theorem though[/quote]\n\nBezout's Theorem is an extremely well-known result. As long as it's cited, it doesn't need to be proven, and the proof of the general theorem is much more difficult.[/quote]\r\n\r\nActually, there's a disclaimer on top of the page with the problems that sais you're only allowed to use theorems if you know them from class. Another example where the problem would have been made trivial if you applied a well known theorem is finding all natural numbers x and y with $ (x\\minus{}1)^{2y}\\plus{}x^{2y} \\equal{} (x\\plus{}1)^{2y}$ which is extremely easy once you use fermat's last theorem... it's from the german imo qualification exams 1998, so you's guess you can't just prove it that easily ;)" } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "suppose$ D$ be a division ring.$ D^* \\equal{}D\\minus{}${0 } as the definition is a group with multiplication action and necesserily is abelian.show if $ charD\\equal{}p>0$ then every finite and abelian subgroup of $ D^*$ is cyclic.", "Solution_1": "Er, do you mean *not* necessarily abelian?\r\n\r\nEDIT: nevermind, maybe it's true, but I don't see it.", "Solution_2": "If $ D$ is a division ring and $ G\\le D^\\times$ finite, then $ G$ must be cyclic if either $ G$ abelian [b]or[/b] $ \\text{char} D\\equal{}p>0$.\r\nLook here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=201470[/url].", "Solution_3": "yes.it is not necessrilly abelian.the correct question was wrote by orilin." } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "K pasa con la comunida espa\u00f1ola!!! Nadie a propuesto problemas!!\r\nEspero k posteen algunos!!\r\n\r\nAki les pongo uno divertido:\r\n\r\nSea A el conjunto de enteros positivos representados de la forma a^2+2(b^2) para cualesquiera \"a\"y \"b\" enteros con \"b\" NO CERO.\r\nMuestra que si p^2 pertenece a A para p primo, entonces p tambien pertenece a A.", "Solution_1": "El problema que propones ya fue resuelto en el foro:\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=75320[/url]\r\n(Es uno de los \u00faltimos post)\r\n\r\nPero aqu\u00ed propongo otro problema:\r\n\r\nSuponga que $\\mathcal{S}$ es un subconjunto finito de los reales, con m\u00e1s de un elemento, con la siguiente propiedad: si tomamos dos elementos distintos de $\\mathcal{S}$, estos forman una progresi\u00f3n aritm\u00e9tica con otro elemento de $\\mathcal{S}$. Halle todos los valores que puede tomar el cardinal de $\\mathcal{S}$.\r\n\r\n(es una ligera variaci\u00f3n de un problema de la Olimpiada Nacional de Ir\u00e1n)\r\n\r\n$Tipe$", "Solution_2": "Gracias por el link, vi la solucion y no me gusto el problema.... krei k era mas bonito!!! \r\nAlguna sugerencia para tu problema?? Ah y a k te refieres con cardinalidad de S????", "Solution_3": "[quote=\"conejita\"]Ah y a k te refieres con cardinalidad de S????[/quote]\r\n\r\nEl cardinal de un conjunto es el n\u00famero de elementos que tiene.\r\n\r\n$Tipe$", "Solution_4": "Por fin alguien con ganas, aqui hay problemas que propuse y nadie resolvio\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77856[/url]\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81765[/url]\r\n\r\nVeo que has posteado muchos problemas, que te parece si por cada problema mio que resulevas yo posteo la solucion de uno tuyo(ya tengo dos soluciones a dos problemas) y ademas te ense\u00f1o a resolver problemas con inversion :roll: que tal?\r\n\r\nAceptas?\r\n\r\n$\\bigstar$ Capit\u00e1n Mandarina\r\n\r\nPD. $Tipe$ es extra\u00f1o, no entiendo como le puede gustar la mate discreta :|", "Solution_5": "Me parece muy wena tu idea, apenas aga alguno tuyo lo posteo, y de paso me vas a pasando los problemas de inversion!!\r\nY ojala y la comunidad espa\u00f1ola se anime a poner problemas, soluciones...\r\npa k nos envidien las demas... jaaja\r\n\r\n :D", "Solution_6": "Yo tambi\u00e9n ando con ganas de resolver problemas!!! pero me parecen dificiles y no los puedo resolver", "Solution_7": "bueno yo tengo unos problemas q no me salen :blush: quiza a ustedes si ( la experiencia :lol: ) aca van dos , ma\u00f1ana quiza postee m\u00e1s", "Solution_8": "En la primera ayudar\u00eda demostrar:\r\n\r\n$x^{2}+y^{2}+z^{2}\\geq 9+2(x+y+z)$, si: $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1$", "Solution_9": "Weno, ps los ultimos k pusieron, ya los ice, ai pongo los soluciones:\r\n\r\nEl primero:\r\nLa desigualdad es equivalente a:\r\na/c^2+b/c^2+a/b^2+c/b^2+c/a^2+^b/a^2 >= 2(1/a+1/b+1/c)\r\nLo anterior es facil de ver multiplicando todo por (abc)^2 y usando muirhead. \r\nAhora (a+b+c)(1/a+1/b+1/c) >= 9 de aqui que 1/a+1/b+1/c >= 9/(a+b+c).\r\nPor ultimo 2(1/a+1/b+1/c) >= 1/a+1/b+1/c + 9/(a+b+c)\r\nlo cual queriamos demostrar.\r\n\r\n\r\nEl segundo problema:\r\nCada una de los factores es equivalente a\r\n((a+b)+(b+c)/2(a+c)). Llamemos x=a+b, y=a+c, z=b+c, y tenemos que probar que\r\n((x+y)/2z)((y+z)/2x)((x+z)/2y) >= 1\r\nPor MG-MA el lado izquierdo es mayor o igual a\r\n (((xy)^1/2)/z)(((yz)^1/2)/x)(((xz)^1/2)/y) \r\ny Volviendo ausar MG-MA con todo ese termino obtenemos el resultado deseado.\r\n\r\nYa resolvi algunos, xfavor, k pongan los problemas de inversion!!!!! :D", "Solution_10": "[quote=\"M4RI0\"]En la primera ayudar\u00eda demostrar:\n\n$x^{2}+y^{2}+z^{2}\\geq 9+2(x+y+z)$, si: $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1$[/quote]\r\n\r\n$\\sum x^{2}\\geq 9+2(x+y+z)$, claramente, agrupando terminos tenemos que $\\sum (x-1)^{2}\\geq 12$, haciendo un cambio de variable $a=x-1$, $b=y-1$, $c=z-1$ se sigue que hay que demostrar que $\\sum a^{2}\\geq 12$. De la condicion tenemos que $\\sum \\frac{1}{a+1}=1$ y desarrllando se sigue que $a+b+c+2=abc$, de ahi que, por la desigualdad MA y MG se tiene que $(a+b+c)\\geq 3\\sqrt[3]{abc}$ con lo que, reemplazando $(a+b+c)^{3}\\geq 27abc$, esto es $(a+b+c)^{3}\\geq 27(a+b+c+2)$. Sea $k=a+b+c$, la ecuacion anterior se reduce a $k^{3}\\geq 27(k+2)$ con lo que $k^{3}-27(k+2)\\geq 0$ y de ahi que $k\\geq 6$. Por la desigualdad de Cauchy se sigue que $(a^{2}+b^{2}+c^{2})(1^{2}+1^{2}+1^{2}) \\geq (a+b+c)^{2}$ con lo que $\\sum a^{2}\\geq (a+b+c)^{2}/3\\geq 12$. $\\blacksquare$", "Solution_11": "K onda Hucht!!\r\n\r\nOye ya resolvi alguno de tus problemas, kreo k como 3, pero los otros son algo dificiles!!! Ya vas a poner la solucion a alguno de los mios??\r\n\r\n :lol:", "Solution_12": "[quote=\"conejita\"]K onda Hucht!!\n\nOye ya resolvi alguno de tus problemas, kreo k como 3, pero los otros son algo dificiles!!! Ya vas a poner la solucion a alguno de los mios??\n\n :lol:[/quote]\r\n\r\nPrimero postea las soluciones", "Solution_13": "no se si les parecio faciles los anteriores problemas , aver q les parece este :blush: :\r\n\r\nse tiene un tablero de nxn pintado como tablero de ajedrez . esta permitido efectuar la siguiente operacion : \r\nescoger un rectangulo en la cuadricula tal que las longitudes de sus lados sean ambas pares o ambas impares , pero que no sean las dos iguales a 1 al mismo tiempo , e invertir los colores de los cuadritos de ese rectangulo ( es decir , los cuadraditos del rectangulo que eran negros se convierten en blancos y los que eran blancos , se convierten en negros) \r\nencontrar para que valores de n es posible lograr que todos los cuadritos queden de un mismo color despues de haber efectuado la operacion un numero finito de veces ( Nota : las dimensiones de los rectanglos que se escogen pueden ir cambiando), espero sus soluciones :D", "Solution_14": "q paso con las soluciones al problema :D", "Solution_15": "A ver, ya avance algo en tu problema, ya probe que se puede para todos los valores de n, excepto cuando son una potencia de 2. Me imagino k sera probar algo de invarianza. \r\nLa forma de cambiar las fichas en los demas arreglos, es muy facil d encontrar. Si kieres k los ponga, dime. (aunk no te daria trabajo buscarla)", "Solution_16": "[quote=\"conejita\"]A ver, ya avance algo en tu problema, ya probe que se puede para todos los valores de n, excepto cuando son una potencia de 2. Me imagino k sera probar algo de invarianza. \nLa forma de cambiar las fichas en los demas arreglos, es muy facil d encontrar. Si kieres k los ponga, dime. (aunk no te daria trabajo buscarla)[/quote]\r\n\r\nPostealo! es mejor, al menos para que los dem\u00e1s tratemos de ersolverlo a partir de tu m\u00e9todo", "Solution_17": "Weno, ps te explico, para todos los impares (excepto 1), lo k acemos es, cambiar todas las colunmas impares, y luego cambiar toas las filas pares. Para los numeros pares (no potencias de dos) lo unico k acemos, es dividimos el cuadro en un subcuadro que sea impar, y cada uno de esos subcuadros realizamos el arreglo anterior!!! \r\nAhora solo falta para 2^n!!!!\r\n\r\nSi alguien tiene alguna idea...", "Solution_18": "Al fin he acabado tu problema. \r\nEscribire la solucion completa:\r\n\r\nPara n=1 la forma de acerlo es inmediata.\r\nPara n=2 es facil ver k no es posible de ningna forma.\r\nAhora, para todos los n impares, primero movemos todas las columnas impares completas, y luego todas las filas pares completas. Siempre obtendremos la cuadricula del mismo color.\r\nPara todos los numeros pares, exceptuando las potencias de 2, la forma es dividir el cuadro en subcuadros impares, y realizar un procedimiento similar al anterior en cada subcuadro.\r\nAhora, para los n potencias de 2, la formas es:\r\nFijemonos en la cuadricula de 4 x 4, a esta cuadricula dividamosla en 4 subcuadros de 2 x 2 cada uno. Al cuadro de 2x2 de arriba a la izquierda le acemos el cambio, igual le acemos el cambio al cuadro de 2x2 de abajo a la derecha. Luego acemos el cambio al cuadro central de 2x4 y para acabar se lo acemos al cuadro central de 4x2. Ya visto que se puede para el de 4x4, todos los demas k son potencias de 2, los dividimos un subcuadros de 4x4, y realizamos el procedimiento anterior las veces necesarias.\r\nDe esat manera demostramos que para todos los n (excepto 2) se puede acer lo pedido.\r\n :P", "Solution_19": "bien, asi me salio a mi tambien , mas bien ,conejita ,sabes como sale el problema 5 y 6 de la OCCM , si te salieron postealas porfavor :lol: , ma\u00f1ana quiza postee otro problema", "Solution_20": "Ola, aki te dejo este link, en el se encuentran las soluciones oficiales de la prueba.\r\n\r\nhttp://viiiomcc.opm.org.pa/Soluciondia2.pdf", "Solution_21": "ok gracias por el link mas bien ahora dos problemas de geo :D \r\n\r\n1) sean AD , BE, y CF las bisectrices interiores del triangulo ABC , demostrar que si uno de los angulos 0,$ $0$ for $x<0.$\r\n\r\nFrom this we get that on the interval $[-1,0]$ we have\r\n\r\n$f(x)=\\frac16[(x+2)^{3}-4(x+1)^{3}]$ and in particular,\r\n\r\n$f(0)=\\frac16(8-4)=\\frac23.$\r\n\r\nSo I now claim that \r\n\r\n$\\int_{-\\infty}^{\\infty}\\text{sinc}^{4}{\\xi}\\,d\\xi=\\frac23,$ or\r\n\r\n$\\int_{0}^{\\infty}\\left(\\frac{\\sin(\\pi x)}{\\pi x}\\right)^{4}\\,dx=\\frac13.$\r\n\r\n(One should be able to change variables from here to answer the original problem.)", "Solution_3": "This would have been easier with the Plancherel theorem, which states that $\\|f\\|_{2}^{2}=\\|\\widehat{f}\\|_{2}^{2}.$ In particular, in this case we get that \r\n\r\n$\\int_{\\mathbb{R}}\\text{sinc}^{4}(x)\\,dx=\\|\\text{sinc}^{2}\\|_{2}^{2}= \\|\\text{box}*\\text{box}\\|_{2}^{2}.$\r\n\r\nNow, it's not hard to compute that $\\text{box}*\\text{box}(x)=\\max(1-|x|,0).$ The square of the 2-norm of that is\r\n\r\n$\\int_{-1}^{1}(1-|x|)^{2}\\,dx=2\\int_{0}^{1}(1-x)^{2}\\,dx=\\frac23.$\r\n\r\nSo we arrive at the same result as in the previous post:\r\n\r\n$\\int_{-\\infty}^{\\infty}\\text{sinc}^{4}(x)\\,dx=\\frac23$ or $\\int_{0}^{\\infty}\\left(\\frac{\\sin \\pi x}{\\pi x}\\right)^{4}\\,dx=\\frac13.$\r\n\r\n---\r\n\r\nNow to finish the problem. Assume for now that $a>0.$ Make the change of variables $\\pi x=au$ or $x=\\frac{au}{\\pi}.$ This gives us\r\n\r\n$\\frac{a}{\\pi}\\int_{0}^{\\infty}\\left(\\frac{\\sin au}{au}\\right)^{4}\\,du=\\frac13$\r\n\r\n$\\int_{0}^{\\infty}\\left(\\frac{\\sin au}{u}\\right)^{4}\\,du=\\frac{\\pi a^{3}}3.$\r\n\r\nNoting that the integral is an even function of $a,$ we arrive at the final result of $\\frac{\\pi|a|^{3}}3,$ as marshell claimed." } { "Tag": [ "induction", "floor function" ], "Problem": "could someone find a solution not as the obvious one? thanks.", "Solution_1": "Induction is quite easy here.", "Solution_2": "[hide]\nThe given expression is equal to $\\frac{(2n)!}{n!}$, so dividing it by $2^{n}$ gives $\\frac{(2n)!}{2^{n}(n!)}$, which equals\n\\[\\frac{1\\cdot 2\\cdot 3\\cdot \\cdots \\cdot 2n}{2\\cdot 4\\cdot 6\\cdot \\cdots 2n}=1\\cdot 3\\cdot 5\\cdot \\cdots \\cdot (2n-1)\\]\nwhich is clearly an integer (and an odd integer as well, so the expression is not divisible by $2^{n+1}$).\n[/hide]", "Solution_3": "[quote=\"Farenhajt\"]Induction is quite easy here.[/quote]\r\n[hide=\"k here we go\"] Base case: $n=1$, $2$ clearly divides $2$. Yay. Let's now assume that $(n+1)(n+2)...(2n)$ divides $2^{n}$. Increasing $n$ by $1$ eliminates the $(n+1)$ term, and adds the terms $(2n+1)(2n+2)$, simplified to $2(2n+1)(n+1)$ Thus, increasing $n$ by 1 justs multiplies the whole product by $2(2n+1)$. Clearly, that is divisible by $2$, so we have one more factor of $2$. Thus, the new expression is divisible by $2^{n}\\cdot2=2^{n+1}$, and the induction is complete.[/hide]", "Solution_4": "[quote=\"13375P34K43V312\"][quote=\"Farenhajt\"]Induction is quite easy here.[/quote]\n[hide=\"k here we go\"] Base case: $n=1$, $2$ clearly divides $2$. Yay. Let's now assume that $(n+1)(n+2)...(2n)$ divides $2^{n}$. Increasing $n$ by $1$ eliminates the $(n+1)$ term, and adds the terms $(2n+1)(2n+2)$, simplified to $2(2n+1)(n+1)$ Thus, increasing $n$ by 1 justs multiplies the whole product by $2(2n+1)$. Clearly, that is divisible by $2$, so we have one more factor of $2$. Thus, the new expression is divisible by $2^{n}\\cdot2=2^{n+1}$, and the induction is complete.[/hide][/quote]\r\nthat is what Farenhajt means. thanks all. my solution:\r\n$\\left\\lfloor \\frac{2n}{2}\\right\\rfloor+\\left\\lfloor \\frac{2n}{4}\\right\\rfloor+...-(\\left\\lfloor \\frac{n}{2}\\right\\rfloor+\\left\\lfloor \\frac{n}{4}\\right\\rfloor+...)=n$" } { "Tag": [ "calculus", "derivative", "function", "logarithms", "calculus computations" ], "Problem": "Solve the equation $2^{x}=\\frac{2}{3}x^{2}+\\frac{x}{3}+1$", "Solution_1": "$x=0, \\,1,\\, 3$.. looking at the derivative, we see that there are no others..", "Solution_2": "How actually did you solve the question?\r\nI don't understand how the derivatives could tell these are all there is.", "Solution_3": "Derivatives just show that there are only 3 roots. I found the roots by inspection..", "Solution_4": "I wonder how does derivative [i]shows[/i] 3 roots?\r\n$\\ln(2) 2^{x}= 4x/3+1/3$", "Solution_5": "rolle's theorem can help us now: if the original equation had more than three roots, its derivative would have more than two roots, and its second derivative would have more than one... but the second derivative of the function is $2^{x}\\log^{2}2-\\frac43$, which is increasing, so it has at most one..." } { "Tag": [ "counting", "distinguishability" ], "Problem": "How many quintuples of positive integers $ (v,w,x,y,z)$ satisfy both equations: $ v\\plus{}x\\plus{}y\\plus{}z\\equal{}35$ and $ w\\plus{}x\\plus{}y\\plus{}z\\equal{}30$?", "Solution_1": "$ x\\plus{}y\\plus{}z \\equal{} 35\\minus{}v \\equal{} 30\\minus{}w \\le 29$. Since the value of $ v$ determines the value of the quantity $ x\\plus{}y\\plus{}z$, which in turn determines the value of $ w$, we only need to know how many solutions there are in the positive integers to $ x\\plus{}y\\plus{}z\\plus{}v \\equal{}30$.\r\n\r\nNow lets use a ball and divider argument. imagine 30 balls in a row. We are going to place dividers between them to section the balls into groups. The 1st group will have $ x$ balls, the second, $ y$, etc. We will need $ 4$ groups, 1 for each variable, so we must place $ 3$ dividers. By specifying the order that the groups are assigned variables, we insure that the variable are treated as distinguishable from eachother. If we place at most 1 divider between two adjacent balls, we are guaranteeing that each group has at least $ 1$ balls, i.e. every variable is an integer greater than $ 1$. There are $ 29$ places in which we can place either $ 0$ or $ 1$ dividers, so there are $ \\binom{29}{3} \\equal{} \\boxed{3654}$ to place the dividers, and thus the same number of solutions.\r\n\r\n\r\n\r\nSide note:\r\nNote that if we wanted to treat the variables as indistinguishable (i.e. count the number of partitions of an integer into a specified number of parts), we could not just divide our result by the number of groups factorialed (the standard way of changing from distinguishable to indistinguishable permutations), since this does not consider that there are fewer actual permutations to undistinguish when two or more of the variables are equal. for example, there are $ 7$ solutions to $ x\\plus{}y\\equal{}8$ with x,y positive integers: $ (1,7)$ $ (2,6)$, etc. But if we make x and y indistinguishable, we have the solutions $ 1,7$ $ 2,6$, $ 3,5$, $ 4,4$ - $ 4$ solutions. Our old method would ignore the fact that $ 4,4$ only appears once in the solutions when $ x$ and $ y$ are considered distinguishable. The stirling numbers of the second kind should be used if we want to count solutions with variables indistinguishable from one another.", "Solution_2": "but shouldn't it be 30C3 instead of 29C3?", "Solution_3": "no definitely 29. We have 30 balls (imagined in a line, not a circle), so there are $ 29$ spaces between the balls, and choose $ 3$ of these spaces to place dividers.\r\n\r\nBtw, I meant to use $ x \\plus{} y \\plus{} z \\plus{} w \\equal{} 30$ in my solution not $ v$. If we wanted to instead solve the problem in terms of $ x\\plus{}y\\plus{}z\\plus{}v \\equal{} 35$, we are restricted to $ v \\ge 6$ and so have to adjust the method to compensate. Define $ v'\\equal{}v\\minus{}5$ then $ x\\plus{}y\\plus{}z\\plus{}v' \\equal{} 30$ and we end up finding the number of solutions to this in positive integers, the exact same thing we did before except with $ w$ renamed as $ v'$", "Solution_4": "[quote=\"facis\"]no definitely 29. We have 30 balls (imagined in a line, not a circle), so there are $ 29$ spaces between the balls, and choose $ 3$ of these spaces to place dividers.\n\nBtw, I meant to use $ x \\plus{} y \\plus{} z \\plus{} w \\equal{} 30$ in my solution not $ v$. If we wanted to instead solve the problem in terms of $ x \\plus{} y \\plus{} z \\plus{} v \\equal{} 35$, we are restricted to $ v \\ge 6$ and so have to adjust the method to compensate. Define $ v' \\equal{} v \\minus{} 5$ then $ x \\plus{} y \\plus{} z \\plus{} v' \\equal{} 30$ and we end up finding the number of solutions to this in positive integers, the exact same thing we did before except with $ w$ renamed as $ v'$[/quote]\r\n\r\ni see, if w,x,y,z can be non-negative number, is the answer going to be 33 choose 3?", "Solution_5": "Yes, it would be.", "Solution_6": "cdymdcool, yes.\r\n\r\nHere's a related question: if $ a$, $ b$, $ c$ and $ d$ are positive integers, how many ordered quadruples $ (a,b,c,d)$ are solutions to $ a\\plus{}b\\plus{}c\\plus{}d \\le 12$?", "Solution_7": "[quote=\"facis\"]cdymdcool, yes.\n\nHere's a related question: if $ a$, $ b$, $ c$ and $ d$ are positive integers, how many ordered quadruples $ (a,b,c,d)$ are solutions to $ a \\plus{} b \\plus{} c \\plus{} d \\le 12$?[/quote]\r\n\r\n11C3+10C3+9C3......+3C3\r\nWhat's the fast way to do it?", "Solution_8": "right idea. One approach you could take is\r\n\r\n$ \\binom{11}{3} \\plus{} \\binom{10}{3} \\plus{} \\ldots \\plus{} \\binom{3}{3}$\r\n\r\nAnd if we don't want to compute each of these and add them, we need to use the hockeystick identity: $ \\sum_{x\\equal{}r}^n \\binom{x}{r} \\equal{} \\binom{n\\plus{}1}{r\\plus{}1}$\r\n\r\nSo\r\n$ \\binom{11}{3} \\plus{} \\binom{10}{3} \\plus{} \\ldots \\plus{} \\binom{3}{3} \\equal{} \\binom{12}{4} \\equal{} 495$ \r\n\r\nAlternatively, let $ e \\equal{} 13 \\minus{} (a\\plus{}b\\plus{}c\\plus{}d)$ so $ e$ must be a positive integer, and you are looking for the number of solutions to $ a\\plus{}b\\plus{}c\\plus{}d\\plus{}e\\equal{}13$, which is $ \\binom{12}{4} \\equal{} 495$." } { "Tag": [ "Functional Analysis", "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "let $E$, $F$ be two Banach spaces. Suppose that $T_n$ is a sequence of operators from $E$ to $F$ such that $dim(T_n(E))< \\infty$ for any $n \\geq 1$ and that $T_n$ converges towards $T$. Then show that the closure of $T(B)$ is compact where $B$ is the open unit ball of $E$.", "Solution_1": "We only need the fact that $T_n$ are compact (i.e. they take the unit ball onto a relatively compact set). In other words, we can prove that the set of compact operators from a Banach space $E$ to a Banach space $F$ is closed in the norm topology.\r\n\r\nIf $U$ is the closed unit ball of $E$, then the closure of $T(U)$ is a complete subset of $F$, being a closed subset of a complete metric space, so all we need in order to prove that it's compact is to show that it's totally bounded, i.e. for every $\\varepsilon>0$ it can be covered by finitely many balls of radius $\\varepsilon$ in $F$.\r\n\r\nFix some $\\varepsilon>0$, choose $n$ such that $\\|T_n-T\\|<\\frac\\varepsilon 3$, and let $(x_k)_{k=1}^t$ be points in $E$ (we may assume they're in $U$, to keep things simple) such that the $\\frac\\varepsilon 3$-balls centered at $T_nx_k$ cover $T_n(U)$. Now take some $x\\in U$. There is a $k\\in\\overline{1,t}$ such that $\\|T_nx-T_nx_k\\|<\\frac\\varepsilon 3$. We then have $\\|Tx-Tx_k\\|\\le\\|Tx-T_nx\\|+\\|T_nx-T_nx_k\\|+\\|T_nx_k-Tx_k\\|<\\frac\\varepsilon 3+\\frac\\varepsilon 3+\\frac\\varepsilon 3=\\varepsilon$, and we're done: for arbitrary $\\varepsilon>0$, we were able to cover $T(U)$ with finitely many $\\varepsilon$-balls in $Y$." } { "Tag": [ "geometry", "trapezoid", "ratio" ], "Problem": "Let $ABCD$ be a trapezoid with the measure of $AB$ twice that of base $DC$, and let $E$ be the point of intersection of the diagonals. If the measure of diagonal $AC$ is $11$, then find that of segment $EC$.", "Solution_1": "[hide=\"hint\"]If AB is on the $x$ axis, what is the ratio of the height of $E$ to the height of $CD$?[/hide]\n[hide=\"answer\"]The ratio is 1:3 because however long it takes the two diagonals to get together it takes twice as long to get apart. That means $EC=AC/3=11/3$.[/hide]" } { "Tag": [ "calculus", "integration", "algebra", "polynomial", "derivative", "function", "calculus computations" ], "Problem": "[b](1) Integration of log(sinx) from 0 to pi/2.\nhere base = e[/b]\r\n\r\n[b](2) If f(x) is polynomial and satisfy the following properties\n (a) f(0) =2, f '(0) = 3\n (b) f '' (x) = f(x). \nThen find f(x) = \nhere f '(x) means first Derivative \nand f ''(x) means second Derivative.[/b]", "Solution_1": "#2: You contradict yourself. The only polynomial $ f$ satisfying $ f''\\equal{}f$ is identically zero.", "Solution_2": "[b] for 2 question \n sorry It is not a Polynomial function.[/b]\r\n \r\n\r\n(2) If f(x) is function satisfy the following properties \r\n(a) f(0) =2, f '(0) = 3 \r\n(b) f '' (x) = f(x). \r\nThen find f(x) = \r\nhere f '(x) means first Derivative \r\nand f ''(x) means second Derivative.", "Solution_3": "hello, for 2) we get $ f(x)\\equal{}\\minus{}1/2e^{\\minus{}x}\\plus{}5/2e^{x}$ as the solution.\r\nSonnhard.", "Solution_4": "[quote=\"jagdish\"][b](1) Integration of log(sinx) from 0 to pi/2.\nhere base = e[/b]\n\n[b](2) If f(x) is polynomial and satisfy the following properties\n (a) f(0) =2, f '(0) = 3\n (b) f '' (x) = f(x). \nThen find f(x) = \nhere f '(x) means first Derivative \nand f ''(x) means second Derivative.[/b][/quote]\r\nfor (1)\r\n$ \\int _{0}^{\\frac {\\pi}{2}} ln(sin x) dx \\equal{} \\minus{}\\frac {\\pi}{2} ln 2$", "Solution_5": "2) If f(x) is polynomial and satisfy the following properties\r\n(a) f(0) =2, f '(0) = 3\r\n(b) f '' (x) = f(x).\r\nThen find f(x) =\r\nhere f '(x) means first Derivative\r\nand f ''(x) means second Derivative.\r\n[color=blue]\nFor (2) My solution is as follows:\nLet $ f(x)=y$\nNow $ y''=y\\implies 2y'y''=2y'y\\implies ((y')^{2})'=(y^{2})'\\implies (y')^{2}=y^{2}+c$\nNow by using $ f(0) =2, f '(0) = 3$, we get $ c = 5$\n$ \\implies (y')^{2}=y^{2}+5\\implies y'=\\pm\\sqrt {y^{2}+5}\\implies y'=\\sqrt {y^{2}+5}$ as $ y'(0)$ is positive.\n$ \\implies\\int {\\frac {1}{\\sqrt {y^{2}+5}}dy =dx}$\n$ \\implies ln|y+\\sqrt {y^{2}+5}|=x+c_{1}$\nBy using $ f(0) =2, c_{1}=ln5$\n$ \\implies ln\\left|\\frac {y+\\sqrt {y^{2}+5}}{5}\\right|=x$\n$ \\implies \\frac {y+\\sqrt {y^{2}+5}}{5}= e^{x}$\n$ \\implies \\sqrt {y^{2}+5}+y = 5e^{x}$\n$ \\implies \\sqrt {y^{2}+5}-y = e^{-x}$\n$ \\implies \\boxed {f(x) = \\left(\\frac {5}{2}\\right)e^{x}-\\left(\\frac {1}{2}\\right)e^{-x}}$[/color]" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "email" ], "Problem": "Greeting,\r\n\r\nI'm thinking of making PDF Document with a lot of problems in many areas. I haven't decided yet but I might want to show it on the public? But my question is:\r\n\r\na) If I change the wording of the problem, like Kalva does, and post on the web or do something like that, is it allowed?\r\n\r\nb) If a) doesn't work, can I still use them by changing the values?\r\n\r\nBTW, how did you get the problems on the aops book? I always wondered how you got those problems. :?", "Solution_1": "This is a very tabu discussion, and it is impossible to give a perfect answer. If you completely change the wording and the numbers of the problems then you should be ok, but ... :? :?", "Solution_2": "[quote=\"Silverfalcon\"]Greeting,\n\nI'm thinking of making PDF Document with a lot of problems in many areas. I haven't decided yet but I might want to show it on the public? But my question is:\n\na) If I change the wording of the problem, like Kalva does, and post on the web or do something like that, is it allowed?\n\nb) If a) doesn't work, can I still use them by changing the values?\n\nBTW, how did you get the problems on the aops book? I always wondered how you got those problems. :?[/quote]Have you tried asking the owners of the problems? Tell them what you're going to do with the problems, and ask them if it is fine with them to do it.", "Solution_3": "[quote=\"Myself\"]Have you tried asking the owners of the problems? Tell them what you're going to do with the problems, and ask them if it is fine with them to do it.[/quote]\r\n\r\nI tried that when I was trying to post my Recopilation of Math Conquest Around the World, but people from the USAMO, CMO and APMO told me explicitely that I should prepare the book that I was planning to, and afterwards let the publisher to negociate the copyright. When I pointed out that it was a material that I wanted to put freely available on the web, no one else than CMO answer, and the guy from CMO understood that it was a webpage and asked me to just put a link to their page. When I explained that I wanted to do a PDF, again, they didn't ask (that was a couple of years ago).\r\n\r\nNow, I tried to investigate about the international laws of copyright, but it seems like it is very vague, but at least two years ago, even asking to a lawyer that happened to be on campus, it seems that if you find something that is posted online by the authors, and it is freely available in the site, you can make a recopilation without asking anyone, and you are able to place it in a P2P, because that was in some sence your own creation (like if you sing \"Billy Jean\" by yourself and put in on a P2P), but not in the web, by some weird law, because it is copyrighted material and you don't have the redistribution rights. You can also freely say that you have it, and that if anyone wants it they just need to email you, and send it by email to anyone (that is what I do with my recopilation), because it is not public, is a private comunications.\r\n\r\nYou can do your own research, and try if you have better luck with the authors of the problems.\r\n\r\nBest,", "Solution_4": "So, how about more SURE one:\r\n\r\nI change the wording. I change the value. I change everything. But!!! I get the idea from the other competition's problems.\r\n\r\nSo, the problem may look similar but not same.\r\n\r\nIs this more okay for sure?\r\n\r\nLike it's totally different. But I\"m getting \"ideas\" from them. Is \"ideas\" okay?", "Solution_5": "I don't see how it can't be okay. It's probably impossible to regulate it anyway since you can't really tell if the author's getting 'inspiration' if everything else is different. Besides, it doesn't seem ethically wrong to get ideas from others. What would this world be like if communication and trading of ideas didn't happen?" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Hi all,\r\nI have a question that I have diffuculty in solving:\r\n\r\nLet G be a finite group and H the subgroup generated by the elements of odd order. Show that H is normal and that the order of G/h is a power of 2.\r\n\r\nThanks", "Solution_1": "Let $E$ be the set of all elements of $G$ with odd order. Take $g \\in G$, and consider $\\Delta=gHg^{-1}$. $\\Delta$ is the subgroup of $G$ generated by the set $gEg^{-1}$. But $gEg^{-1}\\subseteq E$ since $\\forall e \\in E, geg^{-1}$ has same odd order than $e$. Whence $\\Delta \\subseteq H$, which proves the normality of $H$ in $G$.\r\n\r\nNow if $p|(G: H)$ where $m$ is odd and prime, then $G/H$ contains a subgroup $D$ of order $m$, and the preimage of $D$ by the canonical projection $p: G \\rightarrow G/H$ is a subgroup of $G$ with order $m|H|$. The $m$-Sylow of $G$ has odd order but is not included in $H$, which is absurd." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let n be a positive integer and M be a set containing n^2 + 1 positive integers with the property that in each n+1 numbers from the set M there are two numbers for which one is divisible by the other. Prove that the set M contains different numbers a_1, ..., a_(n+1) for which a_(i+1) divides a_i for i = 1,...,n", "Solution_1": "For every element $a$ in M, give it a value k which is defined as the maximum integer such that there exists $a_1, a_2, ...,a_k$ such that\r\n$\r\na_1BC.Q and P are points on line segments CD and AD respectively, such that CQ=AP.The intersection point of AQ and CP is S.Prove that S lies on the angle bisector of angle ABC.", "Solution_1": "See...\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=415147#p415147\r\nhttp://www.mathlinks.ro/Forum/topic-66410.html\r\nhttp://www.mathlinks.ro/Forum/topic-5873.html \r\nhttp://www.mathlinks.ro/Forum/topic-19182.html", "Solution_2": "I think I have come up with another solution which has not been posted before. :) \r\nLet the intersection of $BA$ and $CP$ be $X$, that of $BC$ and $AQ$ be $Y$ and that of $BS$ and $CD$ be $R$. It is easy to see that $\\angle ABS=\\angle CBS\\Leftrightarrow RC=AD$.\r\nConsider the transversal $SQY$ in $\\triangle RBC$ and $PSC$ in $\\triangle ADQ$ respectively. We get, by Menelaus' Theorem, that\r\n$\\frac{RS}{SB}\\cdot\\frac{BY}{YC}\\cdot\\frac{CQ}{QR}=1$ --- (1)\r\n$\\frac{AP}{PD}\\cdot\\frac{DC}{CQ}\\cdot\\frac{QS}{SA}=1$ --- (2)\r\nSince $\\frac{RS}{SB}=\\frac{QS}{SA}$ and $\\frac{BY}{YC}=\\frac{AY}{YQ}=\\frac{DC}{CQ}$, by comparing (1) and (2), we have $\\frac{CQ}{QR}=\\frac{AP}{PD}$. As $AP=CQ$, we now have $PD=QR$ and so $AD=RC$." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Can anyone give an algebraic proof of why F= ma or can it only be proven using laboratory experiments.", "Solution_1": "[quote=\"arbelos08\"]Can anyone give an algebraic proof of why F= ma or can it only be proven using laboratory experiments.[/quote]\r\nYou cant really prove laws from physics mathematically. You can sometimes provide explanation to why you expect it to be true mathematically, and in many cases by experiments. But overall there isnt a full rigorous mathematical proof to any of the laws of physics. (That is probably the reason why Hilbert asked, \"can we axiomize physics\").", "Solution_2": "There is no proof for the 3 Laws of Newton: they are just principles derived (and confirmed) from experimental observation, which are then used as axioms or first principles in the theory of Physics from which we can then derive logical conclusions. This is how some branches of science work. Another very important example is Thermodynamics, which is founded on four laws derived again from experimental observation (and which have been always confirmed until today): the theory of Thermodynamics on its applications to various systems are then logical conclusions that must follow from those four principles.", "Solution_3": "i won't tell u and pacify that there are proof s of the newton's law of motion \r\nthe Least action principle which agian is a axiom can be used to 'prove' newton's Laws\r\nAgain let me stress they are just a def,\r\nhave a look at [url=http://en.wikipedia.org/wiki/Least_Action_Principle]here[/url]", "Solution_4": "Isn't force defined as the change of momentum?", "Solution_5": "[quote=\"modeler\"]Isn't force defined as the change of momentum?[/quote]\r\n\r\nThats impulse, IIRC.\r\n\r\nAnd yes, F=ma is just an idea or concept which 'works' when we look at physics. I mean, we use maths to create physical models, not truths. Out of interest, isn't F=ma for certain situations (like slow speeds and reasonable gravity)", "Solution_6": "no any force is defined so as to follow this equation except third law which fails under certain situations" } { "Tag": [], "Problem": "This is in a book. I dont know how to organize the information, and also i dont know how to use the info about a loss in the share price, please if you can help me.\r\n\r\nAn investor has 3 shares, Delta Airlines, Hilton Hotels and Mcdonalds. He says that 2 days ago the shares went down 350 dollars and yesterday they went up 600 dollars. His stock broker remembers that 2 days ago Delta lost 1 dollar per share, Hilton lost 1.5 dollars, and Mc gained 0.5 dollars. He also remembered that yesterday Delta gained 1.5 dollars, Hilton lost again 0.5 dollars and Mc gained 1 dollar. Is it possible to know with this info how many shares of each company has the investor?\r\nIf the investor says he has 200 shares of Mc, how many does he have of the other 2?\r\n\r\nThanks to anyone who can give me a hand!", "Solution_1": "Basically what you have here is a system of linear equations. For the first day, you have -d - 1.5h + .5m = -350, and for the second day you have 1.5d - .5h + m = 600. Unfortunately, we have three variables and only two equations, so we cannot solve this system. But if we know he owns 200 shares of McDonald's, we can substitute this in and get -d - 1.5h + 100 = -350, or -d - 1.5h = -450, and 1.5d - .5h +200 = 600, or 1.5d - .5h = 400. Solving the first equation for d, we get d=450 - 1.5h. If we substitute this into the second equation, we have 675 - 2.25h - .5h = 400, which simplifies to -2.75h = -275, or h=100. If we substitute this back into the first equation, we get -d - 150 = -450, from which we get d = 300. So he owns [b]100[/b]shares of Hilton and [b]300[/b] shares of Delta.", "Solution_2": "Thank you very much!" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Let $n\\geq 3$ be an integer. Let also $P_1,P_2,...,P_n$ be different two-element-subsets of $M=\\{1,2,...,n\\}$, such that when for $i,j \\in M , i\\neq j$ the sets $P_i,P_j$ are not totally disjoint, then there is a $k \\in M$ with $P_k = \\{ i,j\\}$.\r\nProve that every element of $M$ occurse in exactly $2$ of these subsets.", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56611[/url]", "Solution_2": "Just for completness, here goes the 'official' solution:\r\n\r\nLet $a_i$ denote the number of sets $i$ is contained in.\r\nThen $a_1+a_2+...+a_n=2n$, since every subset contains two different numbers.\r\nWe also have $\\binom{a_1}{2} + \\binom{a_2}{2} + ... + \\binom{a_n}{2} \\leq n$, since there are at most $n$ (=all) sets that are 'implied' by the given condition.\r\nNow $(a_1^2-a_1) + (a_2^2-a_2) + ... + (a_n^2-a_n) \\leq 2n$ and subtracting $3$ times the equality we get\r\n$(a_1^2-4a_1) + (a_2^2-4a_2) + ... + (a_n^2-4a_n) \\leq -4n \\iff (a_1-2)^2 + (a_2-2)^2 + ... + (a_n-2)^2 \\leq 0$\r\nwhich gives immediatly $a_1=a_2=...=a_n=2$." } { "Tag": [ "inequalities" ], "Problem": "I read darij's post about majorization theory ( http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14975 ). I think there must be a flaw, because by his definition $(1,1,0) \\succ (\\frac{2}{3}, \\frac{2}{3}, \\frac{2}{3})$ and $(\\frac{2}{3}, \\frac{2}{3}, \\frac{2}{3}) \\succ (0,1,1)$. So are permutations of an array considered when majorizing?", "Solution_1": "Edit: \r\n\r\n[quote=\"darij grinberg\"][b]1.[/b] Let $x_{1}$, $x_{2}$, ..., $x_{n}$, $y_{1}$, $y_{2}$, ..., $y_{n}$ be arbitrary real numbers [b]satisfying $x_{1}\\geq x_{2}\\geq ... \\geq x_{n}$ and $y_{1}\\geq y_{2}\\geq ... \\geq y_{n}$. [/b][/quote]" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "$E$ is euclidean space such that $E=E_{1}\\oplus E_{2}$ just direct sum \r\n$p$ orthogonal projection on $E_{1}$\r\n$q$ orthogonal projection on $E_{2}$\r\n\r\n\r\n$r=p+q$ \r\n\r\nprove that $01$ pas age $ S$ bekhad aval bashe paranteze dovom bayad barabare $ 1$ bashe yeni khahim dasht:\r\n\r\n$ S=a+b+c+d \\Rightarrow a^{2}+b^{2}+c^{2}+d^{2}=a+b+c+d \\Rightarrow \\boxed{a=b=c=d=1}$\r\n\r\nke inam ba sharte $ ad=b^{2}+c^{2}+bc$ dar tanaghoze...\r\n\r\npas $ S$ nemitoone aval bashe...\r\n[hide=\"ellate inke natije shod a=b=c=d=1 ?\"]\nbe das avordim $ a^{2}+b^{2}+c^{2}+d^{2}=a+b+c+d$\namma tebghe miangine morabai hesabi darim:\n$ \\sqrt{\\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}}\\geq \\frac{a+b+c+d}{4}$\npas darim:\n$ \\sqrt{\\frac{S}{4}}\\geq \\frac{S}{4}$\n$ \\Rightarrow 4\\geq S$\n$ \\Rightarrow a+b+c+d\\leq 4$\namma $ a,b,c,d\\geq 1$ pas darim:\n$ 4\\leq a+b+c+d\\leq 4 \\Rightarrow a+b+c+d=4 \\Rightarrow a=b=c=d=1$[/hide]" } { "Tag": [ "algebra", "polynomial" ], "Problem": "For what b and c, $x=sqrt19+sqrt98$ is a root of $x^4+bx^2+c=0$", "Solution_1": "If $\\sqrt{19}+\\sqrt{98}$ is a root, then $\\pm\\sqrt{19}\\pm\\sqrt{98}$ are roots.\r\nExpanding $(x-\\sqrt{19}-\\sqrt{98})(x-\\sqrt{19}+\\sqrt{98})(x+\\sqrt{19}-\\sqrt{98})(x+\\sqrt{19}+\\sqrt{98})$, we get the polynomial $x^4-234x^2+6241$.\r\n\r\nEDIT: I have many multiplication issues.", "Solution_2": "Squaring both sides of $x-\\sqrt{19}=\\sqrt{98}$, we have $x^2-79=2\\sqrt{19}x$.Again squaring, we get $x^4-234x^2+6241=0$,yielding $b=-234,c=6241$" } { "Tag": [ "inequalities", "calculus", "integration", "geometry", "perimeter", "triangle inequality" ], "Problem": "How many scalene triangles have all sides of integral lengths and perimeter less than $13$?", "Solution_1": "[hide]Assume $ac$, and $a+b+c<13$\n\nBrute force:\n9: 2, 3, 4 = 1\n11: 2, 4, 5 = 1\n12: 3, 4, 5 = 1\n\n$1+1+1=\\boxed{3}$[/hide]", "Solution_4": "How do you brute-force this in a systematic way?", "Solution_5": "[quote=\"rnwang2\"]How do you brute-force this in a systematic way?[/quote]\r\n\r\n[hide]\nWell first, you realize if $a=1$ the triangle has to be isoceles or equilateral so that doesn't work. When $a=2$ the smallest is 2, 3, 4 which adds up to 9. The difference between 3 and 4 can't exceed 1 or else triangle inequality wouldn't be satisfied. So you add 2 to get 2, 4, 5, which adds up to 11 and can't add anymore. Then you try $a=3$ or 3, 4, 5. That is already at 12 so you can't add anymore and you have 3.[/hide]" } { "Tag": [ "email" ], "Problem": "i dunno if this is the appropriate forum for this, but...\r\n\r\nmy math team coach is freaking out because she didnt receive the mandelbrot contests (for the week of 10/17) in the mail yet. she knows they have a mailing address, but is wondering if they have an email address or some other way they can be contacted.\r\n\r\nthanks in advance.", "Solution_1": "http://www.mandelbrot.org/\r\n\r\nClick FAQ'S on lower left\r\n\r\nthen click Dialogue on upper right" } { "Tag": [ "trigonometry", "geometry" ], "Problem": "In triangle $ABC$, $\\tan \\angle CAB = 22/7$, and the altitude from $A$ divides $BC$ into segments of length 3 and 17. What is the area of triangle $ABC$?", "Solution_1": "let the lentgh of altitute be x; notice that the altitute divides the angle CAB at two different angle, and tangent of these angle are both 17/x and 3/x. using the tangent sum formula we got: 22/7 = (17/x+3/x)/(1-(17/x)(3/x)) multiplying we got:\r\n 140/x=22-1122/x^2 multiplying x^2 at both side, and then pass everything to left side we get:\r\n22x^2-140x-1122=0\r\nresolve that equation we get the altitute, and as we already got the base which is 17+3=20, we can easily get the area.\r\n\r\n(sorry for poor english)", "Solution_2": "[hide=\"my solution\"]ok i used b instead of x and got 22/7 = (20/b) / [(b^2-51)/b^2]\nof course then i got 22/7 = 20b^2 / (b^3 - 51b)\nthen 11b^3 - 70 b^2 - 561b = 0 \ni got b = 11\nso area is \n20 * 11 / 2 = 110[/hide]", "Solution_3": "[hide=\"Solution\"]Let the altitude from $A$ meet $BC$ at $H$ such that $BH=17,$ $CH=3,$ and $AH=x.$ Since we're given two right triangles and the tangent of an angle, it seems like a good idea to use trig. By the tangent addition formula, \\begin{eqnarray*}\\tan \\angle BAC &=& \\tan (\\angle BAH+\\angle CAH)\\\\ &=& \\cfrac{\\frac{17}{x}+\\frac{3}{x}}{1-\\left(\\frac{17}{x}\\right)\\left(\\frac{3}{x}\\right)}\\\\ \\\\ &=& \\frac{20x}{x^{2}-51}= \\frac{22}{7}. \\end{eqnarray*} Solving, $x=11$ and the area of triangle $ABC$ is $\\frac{1}{2}(BC)(AH)=110.$[/hide]", "Solution_4": "A slightly different way to look at this problem is to note the identity:\r\n\r\n$ \\tan A\\plus{}\\tan B\\plus{}\\tan C\\equal{}\\tan A\\tan B\\tan C$. From this, we get\r\n\r\n$ \\frac{22}{7}\\plus{}\\frac{h}{3}\\plus{}\\frac{h}{17}\\equal{}\\frac{22h^{2}}{7*3*17}\\iff 22*3*17\\plus{}h*20*7\\equal{}22h^{2}$. The answer is $ \\frac{1}{2}*h*20$. Note that $ 11|h$, checking this works. So the answer is $ 110$." } { "Tag": [ "inequalities", "geometry", "3D geometry", "inequalities open" ], "Problem": "Let x = (cube root of 1000) - (cube root of 999). what integer is closest to 1/x?", "Solution_1": "1/x=(cube root of 1000)^2 + (cube root of 999)(cube root of 1000)+(cube root of 999)^2\r\n\r\n100+(cube root of 999)(10+(cube root of 999))\r\n\r\nI claim it is closest to 300\r\n\r\nso (cube root of 999)(10+(cube root of 999)) is $ \\ge 9.99*19.99\\ge199.7>199.5$\r\n\r\nso $ 300\\le 1/x>299.5$\r\n\r\nAll that has to be shown is that (cube root of 999) is > 9.99 (not too hard)\r\nQ.E.D.", "Solution_2": "thank you \r\ngreat solution" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "prove there isn't exist $x,y,z \\in Z^+$ and$p\\in P;p=3(mod 4)$\r\n $x^4+py^4=z^4$", "Solution_1": "Find all$x,y,z \\in Z^+$ suchthat\r\n $x^4+py^4=z^4$ $(p is a prime number ,p=3(mod 4))$" } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "If $ z \\equal{} f(x,y)$ and $ x \\plus{} y \\equal{} (u \\plus{} v)^{3}$ and $ x \\minus{} y \\equal{} (u \\minus{} v)^{3}$ ,\r\nthen show that \r\n$ 9(x^{2} \\minus{} y^{2}) \\cdot (\\frac {\\partial^{2} z}{\\partial x^{2}} \\minus{} \\frac {\\partial^{2} z}{\\partial y^{2}}) \\equal{} (u^{2} \\minus{} v^{2}) \\cdot (\\frac {\\partial^{2} z}{\\partial u^{2}} \\minus{} \\frac {\\partial^{2} z}{\\partial v^{2}})$\r\n\r\n :coolspeak:", "Solution_1": "Plz Help someone.\r\nThank you.", "Solution_2": "OK\r\n After keep on trying about an hour, finally I got the answer.\r\n\r\nSolution : \r\nsolving , $ x + y = (u + v)^{3} , x - y = (u - v)^{3}$\r\n\r\n$ x = \\frac {(u + v)^{3} + (u - v)^{3}}{2}$ and\r\n \r\n$ y = \\frac {(u + v)^{3} - (u - v)^{3}}{2}$\r\n\r\n$ \\frac {\\partial z}{\\partial u} = \\frac {\\partial z}{\\partial x} \\cdot \\frac {\\partial x}{\\partial u} + \\frac {\\partial z}{\\partial y} \\cdot \\frac {\\partial y}{\\partial u}$\r\n \r\n $ = 3\\cdot \\frac {(u + v)^{2} + (u - v)^{2}}{2}\\cdot \\frac {\\partial z}{\\partial x} + 3\\cdot \\frac {(u + v)^{2} - (u - v)^{2}}{2}\\cdot \\frac {\\partial z}{\\partial y}$\r\n\r\n\r\nLet us say that\r\n $ 3\\cdot \\frac {(u + v)^{2} + (u - v)^{2}}{2} = k_1$ and\r\n $ 3\\cdot \\frac {(u + v)^{2} - (u - v)^{2}}{2} = k_2$\r\n\r\n$ \\frac {\\partial z}{\\partial u} = k_1 \\cdot \\frac {\\partial z}{\\partial x} + k_2 \\cdot \\frac {\\partial z}{\\partial y} ______________________ (1)$\r\n\r\nor $ \\frac {\\partial }{\\partial u} = k_1 \\cdot \\frac {\\partial }{\\partial x} + k_2 \\cdot \\frac {\\partial }{\\partial y}$\r\n\r\nSimilarly we get,\r\n\r\n$ \\frac {\\partial z}{\\partial v} = k_2 \\cdot \\frac {\\partial z}{\\partial x} + k_1 \\cdot \\frac {\\partial z}{\\partial y} ______________________ (2)$\r\n\r\nor $ \\frac {\\partial }{\\partial v} = k_2 \\cdot \\frac {\\partial }{\\partial x} + k_1 \\cdot \\frac {\\partial }{\\partial y}$\r\n\r\nsolving $ (1)$ and $ (2)$, we get,\r\n\r\n$ \\frac {\\partial z}{\\partial x} = \\frac { k_1 \\cdot \\frac {\\partial z }{\\partial u} - k_2 \\cdot \\frac {\\partial z }{\\partial v}}{k^{2}_{1} - k^{2}_{2} }$\r\n\r\n$ \\frac {\\partial z}{\\partial y} = \\frac { k_1 \\cdot \\frac {\\partial z}{\\partial v} - k_2 \\cdot \\frac {\\partial z}{\\partial u}}{k^{2}_{1} - k^{2}_{2}}$\r\n\r\n$ \\frac {\\partial }{\\partial x}(\\frac {\\partial z}{\\partial x}) = ( \\frac {k_{1}\\cdot \\frac {\\partial }{\\partial u} - k_{2}\\cdot \\frac {\\partial}{\\partial v}}{k^{2}_{1} - k^{2}_{2}}) \\cdot ( \\frac {k_{1}\\cdot \\frac {\\partial z}{\\partial u} - k_{2}\\cdot \\frac {\\partial z}{\\partial v}}{k^{2}_{1} - k^{2}_{2}})$\r\n\r\ntherefore,\r\n\r\n$ = \\frac {k^{2}_{1}\\frac {\\partial^{2}}{\\partial u^{2}} - 2k_{1}k_{2}\\frac {\\partial^{2}z}{\\partial x \\partial y} + k^{2}_{2}\\frac {\\partial^{2}z}{\\partial v^{2}}}{(k^{2}_{1} - k^{2}_{2})^2} ________________________________________(3)$\r\n\r\nsimilarly,\r\n\r\n$ \\frac {\\partial }{\\partial y}(\\frac {\\partial z}{\\partial y}) = \\frac {k^{2}_{2}\\frac {\\partial^{2}}{\\partial u^{2}} - 2k_{1}k_{2}\\frac {\\partial^{2}z}{\\partial x \\partial y} + k^{2}_{1}\\frac {\\partial^{2}z}{\\partial v^{2}}}{(k^{2}_{1} - k^{2}_{2})^2} ________________________________________(4)$\r\n \r\n$ (3) - (4)$ gives ,\r\n\r\n${ \\frac {\\partial^{2}z}{\\partial x^{2}} - \\frac {\\partial^{2}z}{\\partial y^{2}} = \\frac {1}{(k^{2}_{1} - k^{2}_{2})^2} \\cdot ((k^{2}_{1} - k^{2}_{2})\\cdot \\frac {\\partial^{2}z}{\\partial u^{2}} - (k^{2}_{1} - k^{2}_{2})\\cdot \\frac {\\partial^{2}z}{\\partial v^{2} }})$\r\n\r\nbut, $ k^{2}_{1} - k^{2}_{2} = 9 \\cdot (u^{2} - v^{2})^{2}$\r\nand $ x^{2} - y^{2} = (u^{2} - v^{2})^3$.\r\n\r\nhence,\r\n \r\n$ 9(x^{2} - y^{2})\\cdot (\\frac {\\partial^{2}z}{\\partial x^{2}} - \\frac {\\partial^{2}z}{\\partial y^{2}}) = (u^{2} - v^{2})\\cdot (\\frac {\\partial^{2}z}{\\partial u^{2}} - \\frac {\\partial^{2}z}{\\partial v^{2}})$\r\n\r\nIt's lengthy if any body has shorter solution plz post it.", "Solution_3": "[quote=Anandiscool]OK\n After keep on trying about an hour, finally I got the answer.\n\nSolution : \nsolving , $ x + y = (u + v)^{3} , x - y = (u - v)^{3}$\n\n$ x = \\frac {(u + v)^{3} + (u - v)^{3}}{2}$ and\n \n$ y = \\frac {(u + v)^{3} - (u - v)^{3}}{2}$\n\n$ \\frac {\\partial z}{\\partial u} = \\frac {\\partial z}{\\partial x} \\cdot \\frac {\\partial x}{\\partial u} + \\frac {\\partial z}{\\partial y} \\cdot \\frac {\\partial y}{\\partial u}$\n \n $ = 3\\cdot \\frac {(u + v)^{2} + (u - v)^{2}}{2}\\cdot \\frac {\\partial z}{\\partial x} + 3\\cdot \\frac {(u + v)^{2} - (u - v)^{2}}{2}\\cdot \\frac {\\partial z}{\\partial y}$\n\n\nLet us say that\n $ 3\\cdot \\frac {(u + v)^{2} + (u - v)^{2}}{2} = k_1$ and\n $ 3\\cdot \\frac {(u + v)^{2} - (u - v)^{2}}{2} = k_2$\n\n$ \\frac {\\partial z}{\\partial u} = k_1 \\cdot \\frac {\\partial z}{\\partial x} + k_2 \\cdot \\frac {\\partial z}{\\partial y}(1)$\n\nor $ \\frac {\\partial }{\\partial u} = k_1 \\cdot \\frac {\\partial }{\\partial x} + k_2 \\cdot \\frac {\\partial }{\\partial y}$\n\nSimilarly we get,\n\n$ \\frac {\\partial z}{\\partial v} = k_2 \\cdot \\frac {\\partial z}{\\partial x} + k_1 \\cdot \\frac {\\partial z}{\\partial y} (2)$\n\nor $ \\frac {\\partial }{\\partial v} = k_2 \\cdot \\frac {\\partial }{\\partial x} + k_1 \\cdot \\frac {\\partial }{\\partial y}$\n\nsolving $ (1)$ and $ (2)$, we get,\n\n$ \\frac {\\partial z}{\\partial x} = \\frac { k_1 \\cdot \\frac {\\partial z }{\\partial u} - k_2 \\cdot \\frac {\\partial z }{\\partial v}}{k^{2}_{1} - k^{2}_{2} }$\n\n$ \\frac {\\partial z}{\\partial y} = \\frac { k_1 \\cdot \\frac {\\partial z}{\\partial v} - k_2 \\cdot \\frac {\\partial z}{\\partial u}}{k^{2}_{1} - k^{2}_{2}}$\n\n$ \\frac {\\partial }{\\partial x}(\\frac {\\partial z}{\\partial x}) = ( \\frac {k_{1}\\cdot \\frac {\\partial }{\\partial u} - k_{2}\\cdot \\frac {\\partial}{\\partial v}}{k^{2}_{1} - k^{2}_{2}}) \\cdot ( \\frac {k_{1}\\cdot \\frac {\\partial z}{\\partial u} - k_{2}\\cdot \\frac {\\partial z}{\\partial v}}{k^{2}_{1} - k^{2}_{2}})$\n\ntherefore,\n\n$ = \\frac {k^{2}_{1}\\frac {\\partial^{2}}{\\partial u^{2}} - 2k_{1}k_{2}\\frac {\\partial^{2}z}{\\partial x \\partial y} + k^{2}_{2}\\frac {\\partial^{2}z}{\\partial v^{2}}}{(k^{2}_{1} - k^{2}_{2})^2}(3)$\n\nsimilarly,\n\n$ \\frac {\\partial }{\\partial y}(\\frac {\\partial z}{\\partial y}) = \\frac {k^{2}_{2}\\frac {\\partial^{2}}{\\partial u^{2}} - 2k_{1}k_{2}\\frac {\\partial^{2}z}{\\partial x \\partial y} + k^{2}_{1}\\frac {\\partial^{2}z}{\\partial v^{2}}}{(k^{2}_{1} - k^{2}_{2})^2} (4)$\n \n$ (3) - (4)$ gives ,\n\n${ \\frac {\\partial^{2}z}{\\partial x^{2}} - \\frac {\\partial^{2}z}{\\partial y^{2}} = \\frac {1}{(k^{2}_{1} - k^{2}_{2})^2} \\cdot ((k^{2}_{1} - k^{2}_{2})\\cdot \\frac {\\partial^{2}z}{\\partial u^{2}} - (k^{2}_{1} - k^{2}_{2})\\cdot \\frac {\\partial^{2}z}{\\partial v^{2} }})$\n\nbut, $ k^{2}_{1} - k^{2}_{2} = 9 \\cdot (u^{2} - v^{2})^{2}$\nand $ x^{2} - y^{2} = (u^{2} - v^{2})^3$.\n\nhence,\n \n$ 9(x^{2} - y^{2})\\cdot (\\frac {\\partial^{2}z}{\\partial x^{2}} - \\frac {\\partial^{2}z}{\\partial y^{2}}) = (u^{2} - v^{2})\\cdot (\\frac {\\partial^{2}z}{\\partial u^{2}} - \\frac {\\partial^{2}z}{\\partial v^{2}})$\n\n.[/quote]" } { "Tag": [ "AMC", "AIME", "AIME II" ], "Problem": "My school is taking the AIME II instead of the I due to spring break,\r\nbut I still want to take the I test (just for practise purposes :P) unofficially.\r\n\r\nIs this possible? or should I just wait until somebody posts the questions after taking the test?\r\n\r\nthx.", "Solution_1": "The questions will definitely be up within a few days afterwards, you'll have plenty of time to take it for practice.\r\n\r\nJust avoid looking at any threads right after this forum is unlocked that might spoil any of the questions before you get a chance to take it.", "Solution_2": "I will be on break, so I'll have the questions up *very* soon after the forum reopens.", "Solution_3": "[quote=\"worthawholebean\"]I will be on break, so I'll have the questions up *very* soon after the forum reopens.[/quote]\r\n\r\nHow come you're the only mod who does this?", "Solution_4": "I've taken it upon myself to do so. Other people either don't have time or aren't sufficiently obsessive." } { "Tag": [], "Problem": "The front and back covers of a book are each 3 mm thick, and a\nstack of 25 sheets of paper is 4 mm thick. Al's book is 38 mm thick,\nand all of the pages are numbered front and back starting with page 1. What is the number of the last page facing the back cover?", "Solution_1": "$ 38$ mm minus $ 3$ mm for the front and back covers gives $ 35$. $ 35$/$ 4$ gives 8.75. $ 25$ sheets of paper is $ 50$ pages. $ 8.75*50\\equal{}\\fbox{438}$.", "Solution_2": "No, $ 3$ mm for each of the front and back. So there's $ 32$ mm for the pages, dividing by $ 4$ yields $ 8 \\times 25\\equal{}200$ sheets, for $ \\boxed{400}$ pages." } { "Tag": [ "function" ], "Problem": "The equation $ (x^{2}\\minus{}4)(x^{2}\\minus{}1)\\equal{}k$ has four non-zero real roots, equally spaced on the number line. If $ x_{4}$ is the largest of the roots, find the value of $ kx_{4}$.", "Solution_1": "[hide]\n$ (x^{2}\\minus{}4)(x^{2}\\minus{}1)\\minus{}k \\equal{} 0$ has 4 equidistant roots\n\nLet $ 2a$ be the distance between two of these roots\n\n\nBecause that function is an even function, we can see that the roots must be:\n$ a,\\minus{}a, 3a,\\minus{}3a$\nnow, we plug in our info to solve for a\n\n$ (a^{2}\\minus{}4)(a^{2}\\minus{}1)\\minus{}k \\equal{} (9a^{2}\\minus{}4)(9a^{2}\\minus{}1)\\minus{}k \\equal{} 0$\n\n$ a^{4}\\minus{}5a^{2}\\plus{}4 \\equal{} 81a^{4}\\minus{}45a^{2}\\plus{}4$\n\n$ 77a^{4}\\minus{}40a^{2}\\equal{} 0$\n\na isn't 0 (because that wouldn't make sense)\n\n$ 77a^{2}\\minus{}40 \\equal{} 0$\n\n$ a \\equal{}\\sqrt{\\frac{40}{77}}\\equal{} 2\\sqrt{\\frac{10}{77}}$\n\nThe largest root is $ 3a \\equal{}\\boxed{6\\sqrt{\\frac{10}{77}}}$\n\nnow\n$ k \\equal{} (x^{2}\\minus{}4)(x^{2}\\minus{}1) \\equal{} (\\frac{360}{77}\\minus{}4)(\\frac{360}{77}\\minus{}1)$ \n\nmultiply those and you got it\n[/hide]" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "$G$ a group and $f$ an outer automorphism of $G$.\r\nProve that $G$ can be embedded in another group $H$ such that $f$ is the restriction of an inner automorphism of $H$.", "Solution_1": "Look at http://www.mathlinks.ro/Forum/viewtopic.php?highlight=semi-direct&t=34802" } { "Tag": [ "ARML", "geometry" ], "Problem": "Hello all,\r\n\r\nA little introduction: I'm a junior going into my senior year. I live in the Greater Hartford region and I'm currently in the process of starting a math team at the math and science school I go to. The director of the school is on board with the proposal I had and, if everything works out for next year, we'll be participating in the HMMT. I'm hoping we can extend the list of competitions we participate in, so if any of you have any suggestions, that would be great. After my finals, I'm going to start getting more information on the ARML that I know a lot of you guys participate in.\r\n\r\nAlso, if you guys are involved in any math teams at your schools that you feel are run and managed particularly well, or if any of you are captains of your math teams and have any tips to share, please do so!", "Solution_1": "Are you participating in the Hartford area math league?", "Solution_2": "If that's another name for the Capital Area Math League (CAML), then yes, but not with the math team I mentioned above (I go to two schools). If you're not referring to this, could you give a bit more information about it?\r\n\r\nAlso, another thing I forgot to mention in my last post: how many students do you guys have in your math teams? And of these, how many are very passionate about the team? Specifically, about how much time do your team members contribute to team meetings as well as self-study? Any information like this would be useful, since I'm conducting a survey about student interest in the math team, and I'd like some data to compare with. Thanks again!", "Solution_3": "We have probably 6ish people who are really involved in math team activities. But my school isn't really comparable to others, as it's private. There are maybe two or three people who do significant amounts of studying for the competitions we do; the others just do the assigned rounds.\r\n\r\nYeah, I meant the CAML, I just didn't know the name.", "Solution_4": "[quote=\"worthawholebean\"]We have probably 6ish people who are really involved in math team activities. But my school isn't really comparable to others, as it's private. There are maybe two or three people who do significant amounts of studying for the competitions we do; the others just do the assigned rounds.\n\nYeah, I meant the CAML, I just didn't know the name.[/quote]\r\n\r\nWho are the 2 or 3? :maybe:", "Solution_5": "[quote=\"Ignite168\"]Who are the 2 or 3? :maybe:[/quote]Dunno, that was just a number I pulled out of nowhere. Do YOU study a lot?", "Solution_6": "What is the CAML? Like, what kind of problems do you do and how does it work?\r\n\r\nAlso, is there one near the NW corner?", "Solution_7": "CAML is the Capitol Area Math League. I'm not sure how far in CT it ranges, but I'm pretty sure it's just schools in the Greater Hartford area. Each participating town sends a team of $ \\leq 5$ people to some host school every month (my town this month!), where each member does three of five rounds, with three questions (worth $ 1,2,3$ points respectively) per $ 10$-minute round.\r\n\r\nThe problems are supposed to increase in difficulty in any given round, but quite frankly they are all quite easy, maybe like AoPS1, some AoPS2, level. For example, the contest writers I think are in love with $ 60^o$ angles and $ 30\\minus{}60\\minus{}90$ triangles." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "[size=150]$ x,y,z\\geq 0, xy\\plus{}yz\\plus{}zx\\equal{}1,$ \n proof that: $ \\frac{1}{x\\plus{}y}\\plus{}\\frac{1}{y\\plus{}z}\\plus{}\\frac{1}{z\\plus{}x} \\geq\\frac{5}{2}$\n\n\n\n\n\n :) \nwhen $ x\\equal{}y\\equal{}o$ , $ z\\equal{}1$ the inequality becomes equal\n[/size]", "Solution_1": "[quote=\"admire9898\"][size=150]$ x,y,z\\geq 0, xy \\plus{} yz \\plus{} zx \\equal{} 1,$ \n proof that: $ \\frac {1}{x \\plus{} y} \\plus{} \\frac {1}{y \\plus{} z} \\plus{} \\frac {1}{z \\plus{} x} \\geq\\frac {5}{2}$\n\n\n\n\n\n :) \nwhen $ x \\equal{} y \\equal{} o$ , $ z \\equal{} 1$ the inequality becomes equal\n[/size][/quote]\r\n\r\nIt's old problem :) and very easy :P", "Solution_2": "[quote=\"admire9898\"][size=150]$ x,y,z\\geq 0, xy \\plus{} yz \\plus{} zx \\equal{} 1,$ \n proof that: $ \\frac {1}{x \\plus{} y} \\plus{} \\frac {1}{y \\plus{} z} \\plus{} \\frac {1}{z \\plus{} x} \\geq\\frac {5}{2}$\n :) \nwhen $ x \\equal{} y \\equal{} o$ , $ z \\equal{} 1$ the inequality becomes equal\n[/size][/quote]\r\n\r\nYou can find it on this uvw paper [url]http://www.mathlinks.ro/viewtopic.php?t=278791[/url], page 8.", "Solution_3": "It is MOSP 2000", "Solution_4": "[quote=\"Gibbenergy\"]\n[size=150]You can find it on this uvw paper [url]http://www.mathlinks.ro/viewtopic.php?t=278791[/url], page 8.[/quote]\n\nusing the uvw method,this problem can be transformed to proof\n$ 18u^{2} \\minus{} 15u \\plus{} 5w^{3} \\plus{} 2\\geq o$\n$ 3u \\equal{} x \\plus{} y \\plus{} z , w^{3} \\equal{} xyz$\n :P[/size]" } { "Tag": [], "Problem": "$ N$ and $ M$ are consecutive integers such that $ N < \\sqrt{942} < M$. Find $ N \\plus{} M$.", "Solution_1": "Educated guestimation. You [should] know that $ \\sqrt{900} \\equal{} 30$. Another nice thing to know is that $ (n\\plus{}1)^2 \\equal{} n^2 \\plus{} 2n \\plus{} 1$ (easy-to-prove yet helpful algebra). This yields $ 31^2 \\equal{} 900 \\plus{} 2(30) \\plus{} 1 \\implies 961$. Since $ \\sqrt{900} < \\sqrt{942} < \\sqrt{961}$, our answer is $ 30 \\plus{} 31$ or $ \\boxed{61}$." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "In this PDF http://www.math.uiuc.edu/~mjunge/540-material.pdf\r\n on page 26 he claims that \r\n\r\nf_n(x) + 2*epsilon/3 <= f_n(y) + epsilon near the end of the proof.\r\n\r\nI don't see how he gets this because the delta(x)'s he chose only work for that particular function in the sequence, and aren't guaranteed for functions after that function in the sequence.\r\n\r\nIs the fact that f_n(x) is increasing necessary? Is there a counterexample when this is not so?\r\n\r\nThanks,\r\nBeoulve\r\n\r\nEdit: forgot to include link", "Solution_1": "What is name of your book?", "Solution_2": "The book I'm using is Elementary Classical Analysis by Mardsen. The particular excercise this relates to is Excercise 8 at the end of Chapter 5.", "Solution_3": "Please post that proof here, because I haven't that book! :oops:", "Solution_4": "Are you having trouble accessing the pdf?", "Solution_5": "I don't see any PDF files here ... thankfully. In general, a question stated in your own words is more likely to get a response than an attached file. \r\n\r\nIt appears that we are to discuss the following theorem: if $f_{n}\\to f$ on $[a,b]$, $f$ is continuous, and $f_{n}(x)\\le f_{n+1}(x)$ for all $n$ and all $x\\in [a,b]$, then the convergence is uniform. If so, then your second question should answer the first one: $\\delta(x)$ work for subsequent terms because the sequence is increasing. \r\n\r\nA counterexample with $\\{f_{n}\\}$ not being an increasing sequence: $f_{n}(x)=\\max(0,1-n|x-1/n|)$ converge to $f\\equiv 0$ but not uniformly.", "Solution_6": "Ack! It seems I never attached the link.\r\n\r\nhttp://www.math.uiuc.edu/~mjunge/540-material.pdf\r\n\r\nThanks for answering the question mlok!" } { "Tag": [ "probability", "inequalities", "geometry", "function", "3D geometry", "tetrahedron", "expected value" ], "Problem": "Imagine that I keep a running total of random numbers from $[0,1]$ until the sum exceeds 1. So I pick a number, write it down, pick another number, add it to the number on my paper... stoping when the sum exceeds one. What is the expected number of numbers I add up until I exceed 1? The numbers are selected with uniform probability. I think it's either $1/2$ or $e$", "Solution_1": "You mean you think it's either $2$ or $e$.\r\n\r\n[hide=\"Possibility\"] Consider the $n$-dimensional graph of the inequality $x_{1}+x_{2}+...+x_{n}\\ge 1$ as a partition of the unit $n$-cube. \n\nIf you can characterize the area of this inequality as a function of $n$ then the expected number can be found. Unfortunately, while it's not so bad for $n = 2$ I don't see an obvious way to generalize... but it's late. [/hide]", "Solution_2": "It can't be 2 because with 2 numbers, you have a $\\frac{1}{2}$ chance of getting a sum of larger than 1, and with fewer than 2 numbers you have a 0 chance of breaking 1. So the expected value is at least $2\\cdot \\frac{1}{2}+3\\cdot \\frac{1}{2}= \\frac{5}{2}$.\r\n\r\n\r\nIn general, the probability that $n$ numbers randomly selected from $[0, 1]$ have sum greater than 1 appears to be $1-\\frac{1}{n!}$. Geometrically, the $\\frac{1}{n!}$ is the volume of a rectangular $n$-simplex, the analog of a tetrahedron where all angles between edges at a given vertex are right, each of whose edges at that vertex has length 1. Alternatively, it's what you get when you cut off the lower corner of the rectangular box $[0, 1]^{n}$, since the points $(0, 0, \\ldots, 1), (0, 0, \\ldots, 1, 0), \\ldots, (0, 1, \\ldots, 0), (1, 0, \\ldots, 0)$ define the hyper-plane $x_{1}+x_{2}+\\ldots+x_{n}= 1$.\r\n\r\nThen the expected value is $\\sum_{n = 2}^{\\infty}n((1-\\frac{1}{n!})-(1-\\frac{1}{(n-1)!})) = \\sum_{n = 2}^{\\infty}\\frac{1}{(n-2)!}= e$, as you conjectured.", "Solution_3": "[quote=\"JBL\"]\nIn general, the probability that $n$ numbers randomly selected from $[0, 1]$ have sum greater than 1 appears to be $1-\\frac{1}{n!}$. \n\n[/quote]\r\n\r\nwhy is this?", "Solution_4": "That was what the next paragraph attempted to explain, although actually it explained the complementary result, that the probability that $x_{1}+\\ldots+x_{n}\\leq 1$ is $\\frac{1}{n!}$. For the geometric reasoning, it helps to examine first the cases $n = 2$ and $n = 3$ (the square and cube).", "Solution_5": "[quote=\"JBL\"]That was what the next paragraph attempted to explain, although actually it explained the complementary result, that the probability that $x_{1}+\\ldots+x_{n}\\leq 1$ is $\\frac{1}{n!}$. For the geometric reasoning, it helps to examine first the cases $n = 2$ and $n = 3$ (the square and cube).[/quote]\r\n\r\nbut the whole: the volume of the n-simplex with blah blah right angles, etc\r\n\r\ni don't think thats clear...", "Solution_6": "Altheman - what JBL was saying about the volume of an n-simplex is just some basic definitional stuff. If you want a more in-depth definition of an n-simplex and its properties, you should be able to find one on Wikipedia or somesuch. \r\n\r\nTo anyone - I understand the geometric solution, but can anyone offer an algebraic one? (I'm not that comfortable with geometry in 2-space, much less n-space. :P)\r\n\r\nOh, and by the way, Watson - I told you it was e! I told you! Ha! :) (Okay, done gloating now.) Cya on friday!\r\n-Katie C.", "Solution_7": "May be this one is not geometric...\r\n\r\nLet's consider a similar problem. We will add natural numbers from the set $\\{1,2,...,n\\}$ and than take a limit.\r\nSuppose we already collected $n$ we have to wait 0 turns, $e_{n}=0$.\r\nSuppose we collected $n-1$, so $e_{n-1}=1$ (1 turn)\r\n$e_{n-2}=\\frac{1}{n}e_{n-1}+\\frac{n-1}{n}e_{n}+1=(1+\\frac{1}{n})$\r\n$e_{n-3}=\\frac{1}{n}e_{n-2}+\\frac{1}{n}e_{n-1}+\\frac{n-2}{n}e_{n}+1=(1+\\frac{1}{n})^{2}$\r\n...\r\n$e_{0}=(1+\\frac{1}{n})^{n-1}$\r\nThe limit of $e_{0}$ is $e$", "Solution_8": "er, nevermind what this post was before I edited it - I get it now. **sheepish** Thanks guys!" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "If $ r$ is a rational number given by $ r\\equal{}\\frac{m}{n}$, where $ m$ and $ n$ are integers and $ n>0$, let $ S_r(a)\\equal{}\\left\\{x\\in{R}: 0\\leq{x^n}\\leq{a^m}\\right\\}$, and define $ a^r\\equal{}\\sup{S_r(a)}$. ($ a>1$)\r\n Prove that $ z\\equal{}a^r$ is the unique positive root of the equation $ z^n\\equal{}a^m$.\r\n(Hint: there is a constant $ K$ such that if $ 0<\\varepsilon<1$, then $ (1\\plus{}\\varepsilon)^n<1\\plus{}K\\varepsilon$. Hence if $ x^n0$ such that $ x^n(1\\plus{}\\varepsilon)^n 0$, let $ S_r(a) = \\left\\{x\\in{R}: 0\\leq{x^n}\\leq{a^m}\\right\\}$, and define $ a^r = \\sup{S_r(a)}$. ($ a > 1$)\n Prove that $ z = a^r$ is the unique positive root of the equation $ z^n = a^m$.\n(Hint: there is a constant $ K$ such that if $ 0 < \\varepsilon < 1$, then $ (1 + \\varepsilon)^n < 1 + K\\varepsilon$. Hence if $ x^n < a^m < y^n$, there exists an $ \\varepsilon > 0$ such that $ x^n(1 + \\varepsilon)^n < a^m < \\frac {y^n}{(1 + \\varepsilon)^n}$.)\n\n I don't know how to use the hint. Please, help me. Thanks.[/quote]\r\nWarning : you can use the normal exponent operations only with integer exponents. \r\nWe have $ a^m > 0, n \\geq 1$ . It's easy to see that $ a^r > 0$\r\n$ 1 \\leq a^m \\Longrightarrow 1\\in S_r(a) \\Longrightarrow a^r > 0$\r\n$ 0 0$\r\n\r\nIf $ 0< \\varepsilon < 1$ then $ (1 + \\varepsilon)^n < 1 + 2^n\\varepsilon$. Let's suppose that $ (a^r)^n\\neq a^m$, so $ (a^r)^n< a^m$ or $ (a^r)^n> a^m$ . \r\nIf $ (a^r)^n< a^m$ , let's denote $ \\varepsilon=max(\\frac 12, \\frac{\\frac {a^m}{(a^r)^n}-1}{2^n})$ so $ (a^r)^n<(a^r)^n(1+\\varepsilon)^n<(a^r)^n(1+2^n\\varepsilon)\\leq a^m\\ \\Longrightarrow\\ a^r(1+\\varepsilon)\\in S_r(a)$ \r\nso $ a^r$ is not the superior bound of ${ S_r(a)}$ (contradiction)\r\n\r\nIf $ (a^r)^n> a^m$ , let's denote $ \\varepsilon=max(\\frac 12, \\frac{\\frac {(a^r)^n}{a^m}-1}{2^n})$ so $ a^m\\leq \\frac{(a^r)^n}{1+2^n\\varepsilon}< \\frac{(a^r)^n}{(1+\\varepsilon)^n} < (a^r)^n$, so ${ 0 \\leq x^n \\leq a^m \\Longrightarrow x <\\frac {a^r}{1+\\varepsilon}} a^m$ , let's denote $ \\varepsilon = min(\\frac 12, \\frac {\\frac {(a^r)^n}{a^m} - 1}{2^n})$ so $ \\ a^m\\leq \\frac {(a^r)^n}{1 + 2^n\\varepsilon} < \\frac {(a^r)^n}{(1 + \\varepsilon)^n} < (a^r)^n$, so ${ 0 \\leq x^n \\leq a^m \\Longrightarrow x < \\frac {a^r}{1 + \\varepsilon}} < a^r$\r\n...\r\n :cool:" } { "Tag": [ "geometry" ], "Problem": "I have come across this observation from one of my students. It states that \"If two n-sided polygons intersect each other, then the total number of line segments(diagonals) that can be drawn from the vertices of the ploygons to all the intesecting points is n(6n-11)\"\r\nHe wanted this to be checked for its validity and earlier existence in the History.", "Solution_1": "I think it's not correct. Try $ n\\equal{}3$.", "Solution_2": "Another case :\r\n\r\nAssume that $ n$-gons intersect only in one point(let that point be a common vertex of both polygons).", "Solution_3": "two $ n$-gons can intersect in almost any number of points, so there is not a formula depending only on $ n$.\r\n\r\nSuppose two $ n$-gons intersect in $ k$ points. For each point of intersection (which we assume is not a vertex of either polygon), we draw in $ n \\minus{} 2$ segments from the vertices of one of the polygons (disregard the endpoints of the side of the polygon that the point of intersection lies on). There are $ 2n$ vertices between the two polygons, but we only use $ n\\minus{}2$ from each polygon, giving $ 2(n \\minus{} 2)^2$ segments to each point of intersection, for a total of $ 2k(n \\minus{} 2)^2$ segments." } { "Tag": [ "LaTeX", "email" ], "Problem": "If we are using Latex, Are we allowed to winzip all of the files we're gonna use, and send them by email?", "Solution_1": "Yep. Don't even have to zip 'em if you don't want to." } { "Tag": [ "function", "floor function", "algebra unsolved", "algebra" ], "Problem": "For any real number $ x$, let $ [x]$ denote the greatest integer which is less than or equal to $ x$. Define\r\n\r\n$ q(n) \\equal{} [\\frac{n}{[\\sqrt{n}]}]$ for $ n\\equal{}1, 2, 3, ... .$\r\n\r\nDetermine all positive integers $ n$ for which $ q(n) > q(n\\plus{}1)$.", "Solution_1": "between successive squares, $ \\lfloor \\sqrt{n} \\rfloor$ is constant, and thus $ q(n)$ is non-decreasing. thus the only places where $ q$ can possibly decrease are in going from $ k^2\\minus{}1$ to $ k^2$, for some integer $ k$. in fact, $ q(k^2\\minus{}1)\\equal{}q(k^2)\\plus{}1$.\u00a0\r\n\r\nin conclusion, $ n$ satisfies the condition in question if and only if it is one less than a square.\u00a0" } { "Tag": [], "Problem": "The distance from the earth to the sun is 93,000,000 miles, and light travels at 186,000 miles per second. How many seconds does light from the sun take to reach the earth?", "Solution_1": "$ \\frac {93000000}{186000} \\equal{} \\frac {93000}{186} \\equal{} \\boxed{500 \\ \\text{seconds}}$." } { "Tag": [ "limit", "logarithms", "inequalities", "factorial", "calculus", "calculus computations" ], "Problem": "I don't get it \u00bfsame idea?\r\n\r\n$L=\\lim_{n\\rightarrow\\infty}\\frac{(n^{2}-1)(n^{2}-2)(n^{2}-3)...(n^{2}-n)}{(n^{2}+1)(n^{2}+3)(n^{2}+5)...(n^{2}+2n-1)}$\r\n\r\nTKS\r\n\r\n[hide]$L=e^{-3/2}$[/hide]", "Solution_1": "Yes, that's the right answer.\r\n\r\nWe can use this: $x-\\frac{x^{2}}2\\le\\ln(1+x)\\le x$ for $x>-1.$\r\n\r\n$L_{n}=\\prod_{k=1}^{n}\\left(1-\\frac{k}{n^{2}}\\right) \\prod_{k=1}^{\\infty}\\left(1+\\frac{2k-1}{n^{2}}\\right)^{-1}$\r\n\r\n$\\ln L_{n}=\\sum_{k=1}^{n}\\ln\\left(1-\\frac{k}{n^{2}}\\right)-\\sum_{k=1}^{n}\\ln\\left(1+\\frac{2k-1}{n^{2}}\\right)= A_{n}-B_{n}$\r\n\r\nFrom the above inequalities,\r\n\r\n$\\sum_{k=1}^{n}\\left(-\\frac{k}{n^{2}}-\\frac{k^{2}}{2n^{4}}\\right) \\le A_{n}\\le \\sum_{k=1}^{n}-\\frac{k}{n^{2}}$\r\n\r\n$-\\frac{n(n+1)}{2n^{2}}-\\frac{O(n^{3})}{n^{4}}\\le A_{n}\\le-\\frac{n(n+1)}{2n^{2}}$\r\n\r\nHence, $\\lim_{n\\to\\infty}A_{n}=-\\frac12.$\r\n\r\nBy a very similar argument, and the fact that $\\sum_{k=1}^{n}(2k-1)=n^{2},$ we find that $\\lim_{n\\to\\infty}B_{n}=1.$\r\n\r\nThus $\\ln L=\\lim\\ln L_{n}=-\\frac12-1=-\\frac32$ and $L=e^{-3/2}.$", "Solution_2": "Thanks, Kent :lol: . I was working for a long time with factorails, double factorials and Stirling's formula, but I did not get any conclusive thing" } { "Tag": [ "inequalities", "trigonometry", "inequalities proposed" ], "Problem": "Prove that in any acute tringle $ABC$ the following inequality holds:\r\n\r\n \\[ m_{a}\\leq\\frac{1}{2}\\cdot\\sqrt{(b+c)^{2} cos A+a^{2}}. \\]", "Solution_1": "using the fact that \r\n$4m_a^2=2(b^2+c^2)-a^2$, we see that the inequality to be proven is equivalent to\r\n$2(b^2+c^2)-a^2\\leq (b+c)^2\\cos A+a^2 \\\\\\Leftrightarrow 2(b^2+c^2-a^2)\\leq (b+c)^2 \\cos A \\\\\\Leftrightarrow 4bc\\cos A\\leq (b+c)^2 \\cos A \\\\\\Leftrightarrow 0\\leq (b-c)^2$\r\nand we're done!", "Solution_2": "Nice, campos. My solution is the same. ;)" } { "Tag": [ "quadratics", "algebra", "quadratic formula" ], "Problem": "Find $\\displaystyle n \\in \\mathbb N$, $\\displaystyle n \\geq 2$, and the digits $\\displaystyle a_1,a_2,\\ldots,a_n$ such that\r\n\\[ \\displaystyle \\sqrt{\\overline{a_1 a_2 \\ldots a_n}} - \\sqrt{\\overline{a_1 a_2 \\ldots a_{n-1}}} = a_n . \\]", "Solution_1": "[quote=\"perfect_radio\"]Find $\\displaystyle n \\in \\mathbb N$, $\\displaystyle n \\geq 2$, and the digits $\\displaystyle a_1,a_2,\\ldots,a_n$ such that\n\\[ \\displaystyle \\sqrt{\\overline{a_1 a_2 \\ldots a_n}} - \\sqrt{\\overline{a_1 a_2 \\ldots a_{n-1}}} = a_n . \\][/quote]\r\n\r\n[hide]\nWe see that $\\overline{a_1 a_2 \\ldots a_{n-1}}$ has to be a square, so let $\\overline{a_1 a_2 \\ldots a_{n-1}} = x^2$ and $a_n = y$.\n\nWe want to solve\n\n$\\sqrt{10x^2+y}-x = y$\n\n$9x^2-2xy+(y-y^2) = 0$.\n\nQuadratic formula says\n\n$x = \\frac{y}{9} \\pm \\frac{\\sqrt{10y^2-9y}}{9}$.\n\nThen obviously $y = 9$ so $x = 4, -2$. Checking, we see that $x = 4$ is right, so\n\n$\\sqrt{169}-\\sqrt{16} = 9$.[/hide]" } { "Tag": [ "modular arithmetic", "combinatorics unsolved", "combinatorics" ], "Problem": "A regular polygon of $ n$ sides ($ n\\geq3$) has its vertex numbered from 1 to $ n$. One draws all the diagonals of the polygon. Show that if $ n$ is odd, it is possible to assign to each side and to each diagonal an integer number between 1 and $ n$, such that the next two conditions are simultaneously satisfied:\r\n\r\n(a) The number assigned to each side or diagonal is different to the number assigned to any of the vertices that is endpoint of it. \r\n\r\n(b) For each vertex, all the sides and diagonals that have it as an endpoint, have different number assigned.", "Solution_1": "Label each edge with endpoints $i,j\\in\\overline{1,n-1}$ with the number in $\\overline{1,n}$ congruent to $i+j\\pmod n$. Now for each $i\\in\\overline{1,n-1}$ all numbers in $\\overline{1,n}$ appear on edges adjacent to $i$ except for one, which is different from $n$ (otherwise we'd have $2i\\equiv 0\\pmod n\\Rightarrow i\\equiv 0\\pmod n$, contradiction; the implication is a consequence of the fact that $n$ is odd). Label the edge $in$ with this number. This is the labeling we're looking for.", "Solution_2": "This is a little hard to explain, so it might be a little long. (For the sake of explanation I present the case when $n=11$ while detailing the solution).\n\nWLOG suppose all numbers are positioned clockwise ( if not then just relabel the numbers). Let us assign to every edge the number corresponding to the vertex opposite to it ( we can guarantee there is one since $n$ is odd) (see figure 1).\nNow pick a vertex, say 1, draw diagonals to the endpoints of the opposite edge, we will call these diagonals \"main diagonals\". As you can see, we end up with two symmetric shapes , each of which has $n$ numbers on it, specifically; $1, 2, 3, 4,..., n$ and there will be a \"middle\" or \"pivot\" number on each side, in our example 9 and 4 are pivot numbers(Remark: The pivot number could be either on an edge or a vertex), and that will be the assigned number to the main diagonal nearest to it. Now if you draw all diagonals from vertex 1, there will always be a pivot number for each diagonal (this because of the parity of $n$) and that's gonna be the one you'll assign to it (See figure 2). Finally, to see that every diagonal connected to 1 is different, notice by symmetry on our pick of the numbers of the edges that the endpoint (different from 1) of any diagonal is counted as an edge on the other shape formed by the other main diagonal i.e. the other symmetrical half, thus there can't be two diagonals with the same number." } { "Tag": [ "induction", "ceiling function", "function", "inequalities", "calculus", "integration", "5th edition" ], "Problem": "Ok, let's gather our solnutions. I will post my solution from home. Anyhow, can any one give solution of 3? It seemed easiest problem at first but oh! no! it was the hardest one! :o\r\n\r\nBy the way, valentine why haven't u posted round 4 problem? I am a bit puzzled!\r\n\r\nand one thing more, when will u give the result of edition 4? And when will u give result of edition 5? After IMO? :D", "Solution_1": "[quote=\"mahbub\"]By the way, valentine why haven't u posted round 4 problem? I am a bit puzzled!\n\nand one thing more, when will u give the result of edition 4? And when will u give result of edition 5? After IMO? :D[/quote]The problems have just been posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=239943#p239943]here[/url]. I had some troubles with my internet connection, thus the delay. \r\n\r\nI hope to publish soon the results of the 4th edition. The results of the 5th edition should also be available before the IMO :)", "Solution_2": "Oh! that's a nice news :lol:", "Solution_3": "So, it seemed trivial at first... To me too, when I found it, but then I realized the opposite. :P :P", "Solution_4": "I know this is a stupid question :blush: But I am not familiar with the notation $\\underset{1\\leq i1$.\r\n\r\n{\\bf Solution:} We will show that $x_n =1 $ for infinitely many $n's$, which is clearly sufficient. Define an integer sequence $\\{ y_n \\}_n$ such that $y_n = nx_n$, then $y_n=2y_{n-1}$ if $y_{n-1}\\leq n$ and $y_n=y_{n-1}-1$ if $y_{n-1}> n$. \r\n\r\n\tFirst, we show by induction that $y_n \\geq n-1$ for all $n$. This is obviously true for $n=1$ and 2. Suppose that it is true for $n=k$. If $y_k \\geq k+1$, then $y_{k+1} \\geq k =(k+1) -1$. Suppose that $k \\geq y_k \\geq k-1$, then $y_{k+1} \\geq 2(k-1) \\geq (k+1)-1$, since $k\\geq 2$. Now, everytime $y_n$ take the value of $n$ or $n-1$, we call it a \"pass\", and we call the two kinds of passes \"class n passes\" and \"class n-1 passes\" resepetively. For $n \\geq 4$, it is easy to see that as $n$ increases, $y_n$ will decrease by one each time until it makes a pass, after which it doubles and decrease by one each time again (since for $n \\geq 4$, $2(n-1) > n+1$). If the sequence makes a pass at index $n=k$, then let $t=y_k$. The sequence will make a pass again at $y_{k+1+ \\left\\lceil {2t-k-1 \\over 2} \\right\\rceil}$, since $y_n-n$ decreases by 2 each time after $y_n$ doubles. It follows that the sequence will make infinitely many passes.\r\n\r\n\tWe now show that the $y_n$ will make infinitely many class n passes. Suppose on the contrary that it dosn't, then there exists $M$ such that for all $n>M$, the $y_n$ never makes a class n pass. Since we've shown that the sequence will make infinitely many passes, all the passes after $n>M$ must be class n-1 passes. Let the sequence makes a pass at index $g$ where $g>M$. Let $y_g = g-1 = 2^mp$ where p is odd. Using the formula above, we see that the sequence will make a pass again at index $g'= g+1+\\left\\lceil {2(g-1)-g-1 \\over 2} \\right\\rceil=g+1+ \\left\\lceil {g-3 \\over 2} \\right\\rceil$, which has a value of $t_{g'}=2(g-1)- \\left\\lceil {g-3 \\over 2} \\right\\rceil$. Since this must be a class n-1 pass, we have $g+1+ \\left\\lceil {g-3 \\over 2}\\right\\rceil -1= 2(g-1)- \\left\\lceil {g-3 \\over 2}\\right\\rceil$, which implies that $g$ must be even. Now, $g'=g+1+ \\left\\lceil {g-3 \\over 2}\\right\\rceil=2^mp+2^{m-1}p= 2^{m-1}(3p)$, which divides a smaller power of 2 than $g$. Thus, we can repeat this argument $m$ times and get a $g''$ which is odd, which would be a contradiction since all the $g$'s must be even. Therefore, $y_n =n$ for infinitely many $n$'s and as a result, $x_n =1 $ for infinitely many $n's$. \r\n{\\bf Q.E.D}\r\n\r\n{\\bf Problem 2:} Let $0 145$, and 122 can be replaced by 147.\r\n\r\nBy also using consecutive triplets, this can be further improved. I am not sure what is the best bound achievable by this method, and I would be surprised if it is anywhere near the true bound.", "Solution_17": "Defining $y_n=nx_n$, we show that $y_n=n$ infinitely often. We have $y_{n+1} = y_n-1$ if $y_n>n$ or $2y_n$ otherwise. Consider $y_n$ just before such doublings. Either $y_n=n$, or $y_n=n-1$. Suppose that from some time on only the latter occurs.\r\n\r\nIf a doubling occurs at an even time $n$, then the next doubling occurs at time $3n/2$. It follows that eventually a doubling must occur at an odd time $n$, but then $y_m=m$ for $m=3n-1$.", "Solution_18": "i am a bit surprised to see 147 as a possible bound. really surprised! :huh: \r\n\r\nAnyhow, one thing not related to math. i think i saw once a spider emotion. but i can't see that in left list. why? and i think i saw champion emotion, can't see that also. were these elemenated?", "Solution_19": "[quote=\"mahbub\"]i am a bit surprised to see 147 as a possible bound. really surprised! :huh: [/quote]I was kinda expecting this :P :)", "Solution_20": "following the method.\r\nif we consider $b_i=b_i+b_{i+1}+b_{i+2}$\r\nwe can get $a_{16}-a_{1} \\geq 151$\r\nthen we can replaced by 152. :D", "Solution_21": "well zhaobin have u solved p3?\r\n\r\nNo one???\r\n\r\nValentin, can u give us some hint on that?", "Solution_22": "no :( ,sorry\r\nI try to use largrange multping,but I faided.\r\nthe another problem is 2nd-p2 ;)", "Solution_23": "I apologise for not posting the solution of nb 3 problem, but I could not enter mathlinks this week. It seems it wasn't solved by many people. Ok, here it is my idea:\r\n Consider the convex function $ f(x)=\\frac{1}{x+\\frac{1}{n-1}}$. Use generalized Popoviciu inequality $\\sum {f(x_i)}+n(n-2)\\cdot f(\\frac{x_1+...+x_n}{n})\\geq (n-1)\\sum {f(\\frac{x_2+...+x_n}{n-1})}$. After that it reduces to a very simple inequality. Do you manage to conclude with this hint?", "Solution_24": "[quote=\"harazi\"]\nConsider the convex function $ f(x)=\\frac{1}{x+\\frac{1}{n-1}}$. Use generalized Popoviciu inequality $\\sum {f(x_i)}+n(n-2)\\cdot f(\\frac{x_1+...+x_n}{n})\\geq (n-1)\\sum {f(\\frac{x_2+...+x_n}{n-1})}$. After that it reduces to a very simple inequality. Do you manage to conclude with this hint?[/quote]\r\n\r\ndid anyone manage to finish it off? i just get another inequality which looks as ugly as the first one :(", "Solution_25": "[quote=\"perfect_radio\"][quote=\"harazi\"]\nConsider the convex function $ f(x)=\\frac{1}{x+\\frac{1}{n-1}}$. Use generalized Popoviciu inequality $\\sum {f(x_i)}+n(n-2)\\cdot f(\\frac{x_1+...+x_n}{n})\\geq (n-1)\\sum {f(\\frac{x_2+...+x_n}{n-1})}$. After that it reduces to a very simple inequality. Do you manage to conclude with this hint?[/quote]\n\ndid anyone manage to finish it off? i just get another inequality which looks as ugly as the first one :([/quote]\r\nUsing the hint,\r\n$\\sum f(x_i) - \\sum f(\\frac{S-x_i}{n-1}) \\ge (n-2) (\\sum f(\\frac{S-x_i}{n-1}) - n f(\\frac{S}{n})) \\ge 0$. The last one is true by jensen.\r\nbtw, harazi, how can you think of popoviciu?! I only tried induction, mixing variables and your integration method... :(", "Solution_26": "[quote=\"siuhochung\"]btw, harazi, how can you think of popoviciu?! I only tried induction, mixing variables and your integration method... :([/quote]He invented the problem, so that's how he thought of popoviciu :D", "Solution_27": "[quote=\"siuhochung\"][quote=\"perfect_radio\"][quote=\"harazi\"]\nConsider the convex function $ f(x)=\\frac{1}{x+\\frac{1}{n-1}}$. Use generalized Popoviciu inequality $\\sum {f(x_i)}+n(n-2)\\cdot f(\\frac{x_1+...+x_n}{n})\\geq (n-1)\\sum {f(\\frac{x_2+...+x_n}{n-1})}$. After that it reduces to a very simple inequality. Do you manage to conclude with this hint?[/quote]\n\ndid anyone manage to finish it off? i just get another inequality which looks as ugly as the first one :([/quote]\nUsing the hint,\n$\\sum f(x_i) - \\sum f(\\frac{S-x_i}{n-1}) \\ge (n-2) (\\sum f(\\frac{S-x_i}{n-1}) - n f(\\frac{S}{n})) \\ge 0$. The last one is true by jensen.\nbtw, harazi, how can you think of popoviciu?! I only tried induction, mixing variables and your integration method... :([/quote]\r\n\r\n :huh: \r\nthe second sum is $(n-1)\\cdot$ RHS\r\nthe first sum is $(n-1)\\cdot \\frac{1}{(n-1)x_{i}+1}$\r\n\r\nhow does it imply the given ineq ?\r\n\r\nsorry if i'm missing smth very obvious", "Solution_28": "Can someone please show me what to do after the Popoviciu? Since the inequality you get afterwards is not simple to me at all. Thanks.", "Solution_29": "Actually, I figured it out. After applying Generalized Popoviciu, we need to prove:\r\n\\[\\sum{(n-1)^2 \\over n-1+x_i} \\geq \\sum{1 \\over x_i + {1 \\over n-1}} + {n(n-2) \\over {S \\over n} + {1 \\over n-1}}\\]\r\nSubtracting the first term on the right from the left, rearranging, the inequality becomes equivalent to:\r\n\\[\\sum{1 \\over x_i + {1 \\over x_i} + n-1 + {1 \\over n-1}} \\geq {1 \\over {S \\over n} + {1 \\over n-1}}\\]\r\nWe now apply jensen to $f(y_i)={1 \\over y_i + n-1 + {1 \\over n-1}}$ where $y_i = x_i + {1 \\over x_i}$, and the above follows from the fact that $S \\geq n$, which can be found using cauchy. Nice problem, Harazi.", "Solution_30": "Thanks! ;) . This is why I thought of generalized Popoviciu, because I created it.", "Solution_31": "[quote=\"harazi\"]Thanks! ;) . This is why I thought of generalized Popoviciu, because I created it.[/quote]\r\nOh, valentin and harazi, that explains the source of this ingenious solution. :P", "Solution_32": "I give a nice solution once a year on the forum and then Valentin must say that in fact I give the solution to a problem created by me. It's not fair! This demoralizes me... :( :( :P :P :P :P", "Solution_33": "OK next time I won't tell anyone that it's your problem :D", "Solution_34": ":rotfl: :rotfl: :rotfl:", "Solution_35": "well i would be glad if some one help me!\r\n\r\nIf we let $f(x)=\\frac{1}{x+\\frac{1}{n-1}}$ (according to harazi)\r\n\r\nthen according to popoviciu's generalized inequality we will get:\r\n\r\n$\\sum {f(x_i)} + n(n-2)f(\\frac{s}{n}) \\geq (n-1)\\sum{ f(\\frac{s-x_i}{n-1})} = (n-1)^2*RHS$\r\n\r\nSo how to finish the problem now? :( \r\n\r\nAnother thing, $f(x_1)=\\frac{n-1}{nx_1-x_1+1}$ not the one given in the problem. Again i have not used the fact: $\\sum{x_i} = \\sum{\\frac{1}{x_i}}$ fact yet! :(", "Solution_36": "Is the solution I posted above not correct?\r\nI think that it answers all your questions.", "Solution_37": "Well, after subtracting i got:\r\n$x_i+\\frac{x_i^2}{n-1}+\\frac{1}{n-1}+\\frac{x_i}{(n-1)^2}$ at denominator\r\n\r\nhow r u bringing that?", "Solution_38": "[quote=\"mahbub\"]how r u bringing that?[/quote]mahbub, please no chatspeak, ok? :) (write \"you\" instead of \"u\", \"are\" instead of \"r\", etc).", "Solution_39": "Ok. One thing more. Why do you say not to use only capital letter? (actually my exam is running. so i shortened my writing as possible.)", "Solution_40": "[quote=\"mahbub\"]Ok. One thing more. Why do you say not to use only capital letter? (actually my exam is running. so i shortened my writing as possible.)[/quote]\r\n\r\nwell, in general using capital letters for all alphabets is quite rude.", "Solution_41": "hey no one wanna cler me on that inequality?", "Solution_42": "[quote=\"mahbub\"]Well, after subtracting i got:\n$x_i+\\frac{x_i^2}{n-1}+\\frac{1}{n-1}+\\frac{x_i}{(n-1)^2}$ at denominator\n\nhow r u bringing that?[/quote]\r\n\r\nMultiply the top and bottom by $(n-1) / x_i$, then you get $n-1+{1\\over n-1}+ (x_i+{1\\over x_i})$ at the bottom. Now, you treat $x_i+{1\\over x_i}$ as a whole and apply jensen, following my previous post.\r\n\r\nI hope that this is clear now." } { "Tag": [], "Problem": "solve:\r\nall triples $ (a,b,c)$ such that ,\r\n$ a^2 \\minus{} 2b^2 \\equal{} 1 ,\r\n 2b^2 \\minus{} 3c^2 \\equal{} 1,\r\n ab \\plus{} bc \\plus{} ca \\equal{} 1$", "Solution_1": "Hello!!\r\n\r\nMaybe $ (\\sqrt{2},\\frac{1}{\\sqrt{2}},0)$ and $ (\\minus{}\\sqrt{2},\\frac{\\minus{}1}{\\sqrt{2}},0)$ are the only solution .", "Solution_2": "yep your final answer is correct .\r\nplz post ur solution", "Solution_3": "[hide=\"solution\"]\n$ ab\\plus{}bc\\plus{}ca\\equal{}1 \\Longrightarrow (b\\plus{}c)(b\\plus{}a)\\equal{}1\\plus{}b^2 \\qquad(1)$\n\n$ a^2\\minus{}2b^2\\equal{}1 \\Longrightarrow (a\\minus{}b)(a\\plus{}b)\\equal{}1\\plus{}b^2 \\qquad(2)$\n\n$ a\\plus{}b\\not \\equal{} 0$ otherwise $ c(a\\plus{}b) \\plus{} ab \\equal{} 1 \\Rightarrow ab\\equal{}1$ which is impossible\n\nSo from (1) and (2), $ a\\equal{}2b\\plus{}c (*)$\n\nSimilarly\n$ ab\\plus{}bc\\plus{}ca\\equal{}1 \\Longrightarrow (c\\plus{}b)(c\\plus{}a)\\equal{}1\\plus{}c^2 \\qquad(3)$\n\n$ 2b^2\\minus{}3c^2\\equal{}1 \\Longrightarrow 2(b\\minus{}c)(b\\plus{}c)\\equal{}1\\plus{}c^2 \\qquad(4)$\n\nTherefore $ a\\plus{}c\\equal{}2(b\\minus{}c) \\Rightarrow a\\equal{}2b\\minus{}3c (**)$\n\nFrom $ (*)$ and $ (**)$ clearly $ c\\equal{}0$\n\nHence $ a\\equal{}2b \\Longrightarrow (2b)b \\plus{} 0(b\\plus{}c) \\equal{} 1 \\Longrightarrow b\\equal{}\\pm \\frac{1}{\\sqrt{2}}$\n\nAnd $ a\\equal{}\\pm\\sqrt{2}$\n\nThis gives $ (a,b,c) \\equal{} (\\sqrt{2},\\frac{1}{\\sqrt{2}},0)$ or $ (\\minus{}\\sqrt{2},\\minus{}\\frac{1}{\\sqrt{2}},0)$ as the only solutions :) [/hide]", "Solution_4": "jinx! i got the same solution." } { "Tag": [ "geometry", "parallelogram", "geometry proposed" ], "Problem": "Given a ball $K.$ Find the locus of the vertices $A$ of all parallelograms $ABCD$ such that $AC\\leq BD,$ and the diagonal $BD$ lies completely inside the ball $K.$", "Solution_1": "Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url] :)" } { "Tag": [ "number theory", "prime numbers", "algebra proposed", "algebra" ], "Problem": "Solve in prime numbers: \r\n\r\n$ \\{\\begin{array}{c}\\ \\ 2a - b + 7c = 1826 \\\r\n3a + 5b + 7c = 2007 \\end{array}$", "Solution_1": "Multiply the first equation by $ 5$ and add to the second. We get $ 13a\\plus{}42c\\equal{}11137$. Since $ 42c$ and $ 11137$ are both divisible by $ 7$ and $ a$ is prime, must be $ a\\equal{}7$. Then easily follows $ c\\equal{}263$ and $ b\\equal{}29$.", "Solution_2": "It is funny I am author of this problem)", "Solution_3": "5x first eq + second gives 13a+42c=11137. RHS is 0 mod 7 so LHS mod 7 means 13a is divisible by 7, or a=7. Plug it into our equation and we get c=263 and b=29, which indeed satisfies the equation." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "I must prove that if $p$ is an odd prime and $p \\neq 5$, then $p^{2}+1$ or $p^{2}-1$ is divisible by $10$.\r\n\r\nThe way I am proving this is that $p$ must be a number such that the digit is the ones place is $1,3,7$ or $9$. Squaring any number with a $1$ or a $9$ in the ones place results in a number with a $1$ in the ones place and then $p^{2}-1$ will be divisible by $10$. Squaring any number with a $3$ or a $7$ in the ones place will result in a number with a $9$ in the ones place and then $p^{2}+1$ will be divisible by $10$.", "Solution_1": "[quote=\"illcrowflu\"]I must prove that if $p$ is an odd prime and $p \\neq 5$, then $p^{2}+1$ or $p^{2}-1$ is divisible by $10$.\n\nThe way I am proving this is that $p$ must be a number such that the digit is the ones place is $1,3,7$ or $9$. Squaring any number with a $1$ or a $9$ in the ones place results in a number with a $1$ in the ones place and then $p^{2}-1$ will be divisible by $10$. Squaring any number with a $3$ or a $7$ in the ones place will result in a number with a $9$ in the ones place and then $p^{2}+1$ will be divisible by $10$.[/quote]\r\nYes, It is true!" } { "Tag": [ "inequalities", "trigonometry", "inequalities solved" ], "Problem": "Here is the problem posted at\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=761\r\n\r\nBut I couldn't reply since the topic is locked. \r\n\r\nWhat is \"topic locked\" mean? This is the first time I encounter the problem.\r\n\r\n[quote]Let $x,y,z>0 $, then\n\\[ \\frac{x}{x+ \\sqrt {(x+y)(x+z)}}+ \\frac{y}{y+ \\sqrt{(y+z)(y+x)}} + \\frac{z}{z+ \\sqrt {(z+x)(z+y)}} \\leq 1. \\]\n[/quote]\r\n\r\nIt looks impossible, actually it's very simple. Here we go:\r\n\r\nBy AM-GM, we have ${\\displaystyle \\sqrt {(x+y)(x+z)}} \\leq \\frac{(x+y)+(x+z)}{2} = x + \\frac{y+z}{2}$.\r\n\r\nHence:\r\n\r\n(1) $\\displaystyle \\frac{x}{x+ \\sqrt {(x+y)(x+z)}} = \\frac{x( \\sqrt {(x+y)(x+z)} - x) }{xy+yz+zx} \\leq \\frac {x(y+z)}{2(xy+yz+zx)}$\r\n\r\nSimilarly,\r\n\r\n\r\n(2) $\\displaystyle \\frac{y}{y+ \\sqrt{(y+z)(y+x)}} \\leq \\frac {y(z+x)}{2(xy+yz+zx)}$\r\n\r\n(3) $\\displaystyle \\frac{z}{z+ \\sqrt {(z+x)(z+y)}} \\leq \\frac {z(x+y)}{2(xy+yz+zx)}$\r\n\r\nAdding (1), (2) and (3) together, we get the desired result.", "Solution_1": "If we let $a=\\sqrt{y+z}$, $b=\\sqrt{z+x}$ and $c=\\sqrt{x+y}$, then $a,b,c$ are sides of an acute triangle. And, we have:\r\n\r\n$x= (b^2+c^2-a^2)/2$, $y= (c^2+a^2-b^2)/2$ and $z= (a^2+b^2-c^2)/2$.\r\n\r\nThe inequality is equivalent to:\r\n\r\n$\\displaystyle \\frac{\\cos{A}}{1+\\cos{A}} + \\frac{\\cos{B}}{1+\\cos{B}} + \\frac{\\cos{C}}{1+\\cos{C}} \\leq 1$.\r\n\r\nOr equivalent to:\r\n\r\n$\\displaystyle \\frac{1}{1+\\cos{A}} + \\frac{1}{1+\\cos{B}} + \\frac{1}{1+\\cos{C}} \\geq 2$,\r\n\r\nwhich is equivalent to:\r\n\r\n$\\displaystyle \\sec^2 \\frac{A}{2} + \\sec^2 \\frac{B}{2} + \\sec^2 \\frac{C}{2} \\geq 4$.\r\n\r\nThe last one is one of the trig inequalities posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2873", "Solution_2": "we have (x+y)(x+z)=x^2+x(y+z)+yz>=2xsqrt(yz)+x(y+z)=x(sqrt(y)+sqrt(z))^2\r\nso x+sqrt[(x+y)(x+z)]>=sqrtx(sqrtx+sqty+sqrtz)\r\nand x|(x+sqrt[(x+y)(x+z)])<=sqrtx|sqrtx+sqrty+sqrtz\r\nwe have similar inequalities and adding them we get the result" } { "Tag": [ "geometry", "parallelogram", "vector", "cyclic quadrilateral", "geometric transformation" ], "Problem": "Let $ ABCD$ is a rectangle. Let $ N$ is midpoint of $ CD$, let $ H$ is project orthogonal of $ B$ on $ AC$, let $ M$ is midpoint of $ AH$. Prove that $ \\angle BMN \\equal{} 90^0$.", "Solution_1": "Project?\r\nWhat does that mean?", "Solution_2": "He meant [url=http://mathworld.wolfram.com/Projection.html]projection[/url].\r\n\r\nI see that $ \\triangle ADB\\sim \\triangle MNB$, but I don't see how I could prove this...", "Solution_3": "[hide=\"Solution\"]\nSince $ \\angle BAH \\equal{} \\angle BDC$, we have $ \\triangle BAH \\sim \\triangle BDC$. Since $ M$ is the midpoint of $ \\overline{AH}$ and $ N$ is the midpoint of $ \\overline{DC}$, it follows that $ \\triangle BMH \\sim \\triangle BNC$, and so $ \\angle BMC \\equal{} \\angle BNC$. Thus $ BMNC$ is a [[cyclic quadrilateral]], from which it follows that $ \\angle BMN \\equal{} 180 \\minus{} \\angle BCN \\equal{} 90^{\\circ}$. \n[/hide]\r\n\r\nEdit: Typo fix, thanks. \r\n\r\nMight it be possible to solve this through a spiral similarity (centered at $ B$, which sends $ \\triangle BHC$ to $ \\triangle BAD$)?", "Solution_4": "[quote=\"azjps\"]\nSince $ \\angle BAH \\equal{} \\angle BDA$, we have $ \\triangle BAH \\sim \\triangle BDC$. [...][/quote]\n\nI think you mean $ \\angle BAH \\equal{} \\angle BDC$, but otherwise nice solution.\n\n[quote=\"azjps\"]\nMight it be possible to solve this through a spiral similarity (centered at $ B$, which sends $ \\triangle BHC$ to $ \\triangle BAD$)?[/quote]\r\n\r\nThat's what I had in mind, except with $ \\triangle BMN$.", "Solution_5": "This is picture of problem.", "Solution_6": "Thank you very much. I will post my solution.", "Solution_7": "This is my solution:\r\n\r\n$ \\bullet$ Let $ P$ is midpoint of $ BH$ $ \\Longrightarrow MP // AB$ and $ MP \\equal{} \\frac {1}{2} AB$ $ \\Longrightarrow MP // NC$ and $ MP \\equal{} NC$ $ \\Longrightarrow MPCN$ is parallelogram $ \\Longrightarrow MN // CP$, (1).\r\n\r\n$ \\bullet$ $ MP // AB$ $ \\Longrightarrow MP \\perp BC \\Longrightarrow P$ is orthocentre point of $ \\triangle BCM$ $ \\Longrightarrow CP \\perp BM$, (2).\r\n\r\n$ \\bullet$ Since (1) & (2) $ \\Longrightarrow MN \\perp BM \\Longrightarrow \\angle BMN \\equal{} 90^0$.", "Solution_8": "This is a nice problem. Can you have other solution?", "Solution_9": "Solution 1: A spiral similarity maps $ AHB$ to $ DCB$ center $ B$. Since $ M$, $ N$ are midpoints, that spiral similarity maps $ MHB$ to $ NCB$, but that means that $ MNB$ is similar to $ HCB$, so we're done.\r\n\r\nSolution 2: A spiral similarity center $ B$ maps $ CHB$ to $ DAB$, so the average of these triangles, $ NMB$ is similar to the original triangles, so we're done.", "Solution_10": "[quote=\"Altheman\"]A spiral similarity center $ B$ maps $ CHB$ to $ DAB$, so the average of these triangles, $ NMB$ is similar to the original triangles, so we're done.[/quote]\r\n\r\nOk, so that argument does work. :)\r\n\r\nEdit: Yeah, I've seen mean geometry before, but not for a while, so I wasn't sure.", "Solution_11": "Realize that it is not an obvious result. I've heard of it called \"MGT\" or \"Mean Geometry Theorem.\" I believe that on an olympiad, you would want to cite outline a proof...it is not particularly long or difficult, but it is not obvious.", "Solution_12": "Using the dot product of vectors, we will get the result easily! I'll let you guys try solving it by this method!(I;m also busy right now) :wink:", "Solution_13": "Yes, we can done it easy by vector.", "Solution_14": "$ \\Vec {BM}.\\Vec {MN} \\equal{} 0 \\Longrightarrow \\angle BMN \\equal{}90^0$!! :rotfl:" } { "Tag": [ "linear algebra", "matrix", "induction", "combinatorics proposed", "combinatorics" ], "Problem": "Let $ n\\geq 2$ an integer and consider a $ n*n$ matrix with all its elements 1 or -1. Each element is the product of its neighbours (with a common side). Prove that we can find an infinity of n for wich all the numbers must be 1.", "Solution_1": "NObody? :( C'mon :P at least give me some hint! I will appreciate a solution using linear algebra.", "Solution_2": "well I didnt use linear algebra... :maybe: \r\n[hide=\"hint\"]note that if a $ n*n$ table had the propertyy,then a $ (2n-1)*(2n-1)$ table has the property too,also proved by induction I think...[/hide]", "Solution_3": "Look at [b]iandrei[/b]'s solution: \r\nhttp://www.mathlinks.ro/viewtopic.php?t=3819\r\nAs a resume, [b]iandrei[/b] proves that for $ n\\geq2$, for a $ 2^{n}-1\\times2^{n}-1$ board there is a single arrangement - when all numbers are 1." } { "Tag": [ "linear algebra", "matrix", "algorithm", "graph theory" ], "Problem": "Can we arrange the numbers $ 1$, $ 2$, ..., $ 16$ on a circle such that the sum of each two adjacent numbers is a perfect square? Same question for a line.", "Solution_1": "The only number that can be adjacent to $ 8$ is the number $ 3^2 \\minus{} 8 \\equal{} 1$, because \r\n$ 2^2 \\minus{} 8 \\equal{} \\minus{} 4 < 0$, \r\n$ 4^2 \\minus{} 8 \\equal{} 8$ and \r\n$ 5^2 \\minus{} 8 \\equal{} 17 > 16$\r\nSo we cannot put the numbers in a circle, because then we would need two adjacent numbers to $ 8$\r\nThe same thing happens with $ 16$, which can have only $ 9$ as adjcacent\r\n\r\nSo, if we put the numbers in a line, we should put $ 8$ and $ 16$ at the edges\r\n\r\n\r\nWe make a table\r\n[img]14363[/img]\r\nIn the row - [b]i[/b] we can see all the numbers that can be adjacent to the number [b]i[/b]. From the table we can see that the numbers $ 8,16$ have only one adjacent number each\r\n\r\nThe corresponding graph is the following: \r\n[img]14369[/img]\r\n\r\nFrom this graph we can see that there is only one solution. The order is $ 8 \\minus{} 1 \\minus{} 15 \\minus{} 10 \\minus{} 6 \\minus{} 3 \\minus{} 13 \\minus{} \\cdots \\minus{} 16$ (or the inverse order)", "Solution_2": "[quote=\"Altheman\"]Can we arrange the numbers $ 1$, $ 2$, ..., $ 16$ on a circle such that the sum of each two adjacent numbers is a perfect square? Same question for a line.[/quote]\r\nThis is a prime example of a question where reading carefully is helpful, since the way the question is phrased it's definite that the first is not possible (otherwise the second would be pointless)", "Solution_3": "Nice use of graph theory. I just use the same argument on 8 and 16 in words, and showed an example to prove the line part, basing it on the fact that 8 and 16 must be at the end.", "Solution_4": "Well done. My method was similar:\r\n\r\nFor each number $ 1$,$ 2$,...,$ 16$, associate one point with each number in a graph.\r\n\r\nLet $ M$ be the matrix where $ m_{ij}\\equal{}i\\plus{}j$. Then let $ f: \\mathbb{N}\\rightarrow \\{0,1\\}$ such that if $ x$ is a perfect square, $ f(x)\\equal{}1$, else $ f(x)\\equal{}0$. Then $ f(M)\\equal{}A$ where $ A$ is the adjacency matrix of the graph (of course $ A$ is symmetric since $ i\\plus{}j\\equal{}j\\plus{}i$). \r\n\r\nIn graph theory terms, the first question asks us to determine if there are any Hamiltonian cycles. However $ 8$ and $ 16$ each have degree $ 1$ (in terms of undirected edges) so there are no cycles.\r\n\r\nThe next question asks us to find a Hamiltonian path. There isn't any good algorithm for this, but given the graph, it is pretty easy. We know $ 8$ and $ 16$ must be on the edges, then we follow the path starting from $ 16$." } { "Tag": [], "Problem": "For $ a,b > 1,$ prove that there exists infinite number of pairs $ (a,b)$ such that $ \\displaystyle \\frac {a}{b} \\plus{} \\frac {a \\minus{} b}{\\frac {b}{a}}$ is an integer.", "Solution_1": "[hide=\"The bestest solution ever\"]Set $ a \\equal{} b$. :wink: [/hide]\n[hide=\"Or...\"]$ \\frac {a}{b} \\plus{} \\frac {a \\minus{} b}{\\frac {b}{a}} \\equal{} \\frac {a}{b} \\plus{} \\frac {a(a \\minus{} b)}{b} \\equal{} \\frac {a \\plus{} a(a \\minus{} b)}{b} \\equal{} \\frac {a(a \\minus{} b \\plus{} 1)}{b}$ So, even limiting yourself to $ a,b \\in \\mathbb{N}$, you can have $ b|a$ or $ b|a \\minus{} b \\plus{} 1 \\Rightarrow b|a \\plus{} 1$.[/hide]", "Solution_2": "Yes, that is well-done. I should've put restriction of $ a \\neq b$ but I forgot.", "Solution_3": "It is equivalent to $ b|(a^2\\plus{}a\\minus{}ab)$ or $ b|(a^2\\plus{}a)$ which is still trivial.", "Solution_4": "$ \\frac {a}{b} \\plus{} \\frac {a \\minus{} b}{\\frac {a}{b}} \\equal{} \\frac {a}{b} \\plus{} \\frac {a(a \\minus{} b)}{b}$???\r\n\r\nisnt it $ \\frac {a}{b} \\plus{} \\frac {a \\minus{} b}{\\frac {a}{b}} \\equal{} \\frac {a}{b} \\plus{} \\frac {b(a \\minus{} b)}{a}$????", "Solution_5": "Brut3Forc3 wrote the denominator of the second fraction incorrectly. Reread the original problem.", "Solution_6": "Oops. Edited. :blush:" } { "Tag": [ "AMC", "AIME", "geometry", "3D geometry", "pyramid", "USA(J)MO", "USAMO" ], "Problem": "How hard was the AMC? I'm not really in a position to judge. I thought it was a little tougher than the ones I had seen, but I may be wrong. I ended up with a 120, when I took the 2001 and got a 132 the night before. Maybe it was just testing conditions that hurt me.", "Solution_1": "It was easier for people aiming to make the AIME, but harder for people aiming for a perfect. This was because most questions were easy, but a few problems (particularly 18, 24, and 25) were very easy to mess up (I missed 24 and left 25 blank).", "Solution_2": "I left a couple blank that I shouldn't have, like 16 and 17. I wasn't quite sure if 1/3bh applied to non-right pyramids, so I didn't answer that. I think I was way too conservative, because I made 4 dumb mistakes last year, getting a 98.\r\n\r\nI got 132.5, and I got 125.5 on practice 2001 a few days earlier. If I had done the 2001 like I had done the 2005, I would probably have gotten around 118.5 on 2001.\r\n\r\nAnyone wanna guess what index will be needed to qualify for USAMO? Of course, AIME hasn't occurred, but maybe considering an AIME of average difficulty?", "Solution_3": "It was easier than the Mock AMC's (on average) for sure. :D", "Solution_4": "I can attest to the much easier than the MOCK ones, but for me, it was really easy. But again, easiness doesn't guarantee that I do well. I screwed this really easy one up alot.", "Solution_5": "I guessed well on 10.24. \r\n\r\n\r\n\r\n[hide=\"my solution\"]I figured that n is a square of a prime, since GPF= :sqrt: n, and thus we have :sqrt: prime^2+48 as the GPF of n+48. Now we see that if n!=4(plug it in), n+48 is not a square, since from n^2=>(n+1)^2 is n^2+n+(n+1) and 48 does not fit that formula, so :sqrt: n+48 is not integral, so [/hide]", "Solution_6": "I think 1-20 are okay(like last year except there's no tricky question about the dollars and cents ), \r\n\r\n21-23??? What's wrong? Why make them so easy? It really really upset me that I didn't attempt #21 and 23(i got both in 5 minutes after the test). \r\n\r\n#24 and 25 are still pretty hard.", "Solution_7": "It's significantly easier than the 12B was last year. I did Q19-24 in about 15 minutes combined (zabelman gave me the questions). I did a lot of casework on 18 though; can anyone write out the question for me again?\r\n\r\n..\r\n\r\nHahahaha, I just realized the shortcut to 18. Ignore me, people!", "Solution_8": "You did a lot of casework? Perhaps I did it incorrectly then.\r\n\r\n18) Call a number \"prime-looking\" if it is composite but not divisible by 2, 3, or 5. The three smallest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000?\r\n\r\na)100\r\nb)102\r\nc)104\r\nd)106\r\ne)108\r\n\r\n[hide]I just found multiples of 2, 3, and 5 of 1000, then add back the double counted ones (unless you never subtracted them to begin with), then subtract the number of primes. The answer is (a), I believe.[/hide]", "Solution_9": "Like I said, ignore me. It's straightforward inclusion-exclusion.", "Solution_10": "Yeah, but my friend actually wrote a program on his calculator to do that problem. It took 36 minutes.", "Solution_11": "On that problem I didn't realize that 2, 3, and 5 were both prime and multiplies of (yes, this sounds idiotic now) 2, 3, and 5. I also forgot about 1 being in its own group so I came up with 98. I knew my answer could not have been too far off so I said 100...thank god they didn't put 98 as an answer..phew.", "Solution_12": "At first I got 101, then I realized 1 is neither prime nor composite. If they put 101 as a choice, I might've gotten it wrong", "Solution_13": "[quote=\"blahblahblah\"]Like I said, ignore me. It's straightforward inclusion-exclusion.[/quote]\r\n\r\nah, my pardons. I tend to leave pages open in my browser and not refresh them.", "Solution_14": "[quote=\"frankthetank\"]Yeah, but my friend actually wrote a program on his calculator to do that problem. It took 36 minutes.[/quote]\r\n\r\n :) \r\n\r\nThat makes me smile. Yesterday before I went to bed, I programmed [i]my[/i] calculator ([url=http://education.ti.com/us/product/tech/v200/features/features.html]TI Voyage 200[/url]) and it took a couple of minutes. I'm not too familiar with the programming though, so I ended up making the program output a string with every number that was prime-like. Then I took the length of the string and after subtracting six (the three two digit prime-like number) and dividing by three, I got that in total there are 100 prime-like numbers\r\n\r\nOn the test I subtracted out the primes before I did the exclusion thing so I got some huge number like 400. Later I was informed that exclusion only works if you have an arithmatic sequence of numbers and that I was supposed to subtract out the primes first... Whoops...\r\n\r\nI know this is a strange question, but what calculator + code did your friend use?", "Solution_15": "Writing a program for this problem is very fun to do! So in the last 10 minutes I decided to write one. I came up with the following within a few mins (TI-83 Plus, Basic syntax):\r\n\r\n[code]0->K\nFor(A,2,1000)\nDisp A\nIf fPart(A/2) $\\neq$ 0 and fPart(A/3) $\\neq$ 0 and fPart(A/5) $\\neq$ 0\n1+k->k\nEnd\nDisp K[/code]\r\n\r\n(I put in the line \"Disp A\" so I could watch it count up to 1000: I *hate* waiting for a program to finish running without knowing if it is going at all.) Anyway, it took about 2 minutes to run, and then it told me 265. So after subtracting off the 165 leftover primes, thats 100. Yay, it worked!", "Solution_16": "[quote=\"blahblahblah\"]It's significantly easier than the 12B was last year. I did Q19-24 in about 15 minutes combined (zabelman gave me the questions). I did a lot of casework on 18 though; can anyone write out the question for me again?\n\n..\n\nHahahaha, I just realized the shortcut to 18. Ignore me, people![/quote]\r\n\r\nWhen I looked #18, I stored noticed that it gave the number of primes, but then thought to myself \"how willl this help\" and forgot it was even there. Then, when I realized I DID need the number of primes, I started listing them, and then stopped. Didn't want to count them. Boy, oh boy was I angry at myself!\r\n\r\nJB", "Solution_17": "[quote=\"blahblahblah\"]It's significantly easier than the 12B was last year. I did Q19-24 in about 15 minutes combined (zabelman gave me the questions). I did a lot of casework on 18 though; can anyone write out the question for me again?\n\n..\n\nHahahaha, I just realized the shortcut to 18. Ignore me, people![/quote]\r\n\r\nWhen I looked #18, I stored noticed that it gave the number of primes, but then thought to myself \"how willl this help\" and forgot it was even there. Then, when I realized I DID need the number of primes, I started listing them, and then stopped. Didn't want to count them. Boy, oh boy was I angry at myself!\r\n\r\nJB", "Solution_18": "When I first read that problem I accidently skipped over the part where they told you 168 primes so i was like...\"wtf, they expect us to know how many primes are less than 1000?\" So then I reread the problem and calmed down once I saw 168 :D .", "Solution_19": "[quote=\"avolfson\"][quote=\"frankthetank\"]Yeah, but my friend actually wrote a program on his calculator to do that problem. It took 36 minutes.[/quote]\n\nI know this is a strange question, but what calculator + code did your friend use?[/quote]\r\n\r\nI don't have his code, but he had a TI-83 plus, which is probably why it took 36 minutes. Fortunately, I lent him an additional calculator before the test started.", "Solution_20": "[quote=\"zabelman\"]Writing a program for this problem is very fun to do! So in the last 10 minutes I decided to write one. I came up with the following within a few mins (TI-83 Plus, Basic syntax):\n\n[code]0->K\nFor(A,2,1000)\nDisp A\nIf fPart(A/2) $\\neq$ 0 and fPart(A/3) $\\neq$ 0 and fPart(A/5) $\\neq$ 0\n1+k->k\nEnd\nDisp K[/code]\n\n(I put in the line \"Disp A\" so I could watch it count up to 1000: I *hate* waiting for a program to finish running without knowing if it is going at all.) Anyway, it took about 2 minutes to run, and then it told me 265. So after subtracting off the 165 leftover primes, thats 100. Yay, it worked![/quote]\r\n\r\nHah, that's quite good. In fact, it would have been faster to do this then inclusion-exlcusion by hand (work on another problem while the calc does a search). I wouldn't have thought of using fpart; mod(a,x) came to mind when people said code.", "Solution_21": "[quote=\"frankthetank\"]Yeah, but my friend actually wrote a program on his calculator to do that problem. It took 36 minutes.[/quote]\r\n\r\nOh, I definately should have thought to do that. I ended up with 96 because I forgot 1,2,3, and 5 :blush:. I imagine I could make that program in under 5 minutes on an 89, though it would have taken a little while to run.", "Solution_22": "When would be the scores for AMC A and AMC B available?\r\n\r\nI haven't heard from my teacher about my score.", "Solution_23": "[quote=\"Ultimatemathgeek\"]When would be the scores for AMC A and AMC B available?\n\nI haven't heard from my teacher about my score.[/quote]\r\n\r\nThe answer key is already posted on the UNL website\r\n\r\n[url]http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05AMC12AB-answers.html[/url] so you can go check your answers\r\n\r\nYou probably won't get your official score until another two weeks", "Solution_24": "[quote=\"zabelman\"]Writing a program for this problem is very fun to do! So in the last 10 minutes I decided to write one. I came up with the following within a few mins (TI-83 Plus, Basic syntax):\n\n[code]0->K\nFor(A,2,1000)\nDisp A\nIf fPart(A/2) $\\neq$ 0 and fPart(A/3) $\\neq$ 0 and fPart(A/5) $\\neq$ 0\n1+k->k\nEnd\nDisp K[/code]\n\n(I put in the line \"Disp A\" so I could watch it count up to 1000: I *hate* waiting for a program to finish running without knowing if it is going at all.) Anyway, it took about 2 minutes to run, and then it told me 265. So after subtracting off the 165 leftover primes, thats 100. Yay, it worked![/quote]\r\nRock on, Zack. I never thought to use such easy syntax. Wish I could program better--but on to the AIME. And my 89 from the night before the test (no, I didn't bother using it anyway) won't be useful there.\r\n\r\nI still don't see why the last problem was such a stress. It's not even maths--maybe that's the problem--hardly, just some counting applied to a picture tha makes some easy sums of 8s... All in all, this test was easy, but I made simpler mistakes than I thought possible and crashed 20 points from a near-perfect, best-in-State score. All well, I'll go try that program." } { "Tag": [ "function" ], "Problem": "let f(x) = 6x^4 - 12x^3 + 3x^2 + 3x + 3\r\n\r\nFind the value of x that maximizes the function", "Solution_1": "If I'm not mistaken, the function monotonically increases for $ \\approx x\\geq 3$ so the answer must be $ \\boxed{\\infty}$.", "Solution_2": "Perhaps you want us to find a minimum.", "Solution_3": "I forgot to include the interval is from 0 to 1." } { "Tag": [ "calculus", "integration", "probability", "function", "real analysis", "real analysis solved" ], "Problem": "Hey guys, \r\nI stumbled upon a problem, and I can't find a way to relate it to the FTC Part I:\r\n\r\nFor what value of x does: integral (from 0 to x) e^(-t^2) dt = 0.6? \r\n\r\nAny help would be appreciated. \r\n\r\n :cool:", "Solution_1": "This is going to have to be approximated one way or another. One approach is to use a table of the cumulative probability function of the standard normal distribution. Such tables are widely available. To do that, you would need to divide by $\\sqrt{2\\pi}.$ You would be looking on the table for the (approximate) value of $z$ such that $P(0TR$.\n\nYou can also use the Hinge Theorem some here[/hide]", "Solution_2": "I get part of it until moving the side over. were you assuming temporarily that QT perpendicular PR? and if you use the hinge theorem, then wouldn't it make the triangle PQT congruent to triangle RQT?" } { "Tag": [ "calculus", "integration", "function", "derivative", "calculus computations" ], "Problem": "Hello \r\n\r\nI am trying to solve for the constant $C$ in the following equation:\r\n\r\n$\\frac{dW}{dR}= \\frac{-0.02W+0.00002RW}{0.08R-0.001RW}$\r\n\r\nI want to show that $\\frac{R^{0.02}W^{0.08}}{e^{0.00002R}e^{0.001W}}= C$\r\n\r\nSo far I have gotten to: $(0.08-0.001W)\\frac{dW}{W}= (-0.02+0.00002R)\\frac{dR}{R}$\r\n\r\nWhat would I do next?\r\n\r\nAny help is appreciated.\r\n\r\nThanks", "Solution_1": "What's $C$? You haven't defined it.", "Solution_2": "$C$ is a constant. sorry about that.", "Solution_3": "Do you mean that $C$ is the constant of integration obtained when the given differential equation is solved?", "Solution_4": "If $C$ is a constant of integration, there are two points to make:\r\n1. $C$ depends strongly on the method of solution. There are many different ways we can do this, and many different constants we can write down.\r\n2. You need an initial condition to say anything.\r\n\r\nSo, here are two questions:\r\nWhat's the initial condition?\r\nWhat's the form of the solution?", "Solution_5": "There was no initial condition given. This is a model for rabbit and wolf populations. Also, it is impossible to solve for $W$ as an explicit function of $R$. Thats why I need to show that the constant obtained by solving the diff eq. is equaled to the above. But I cant show this without an initial condition?", "Solution_6": "Ah, that's it. You're not supposed to \"solve for $C$\" at all, and there's no need to solve the equation. You want to show that a certain quantity depending on $R$ and $W$ is constant in time. To do this, ignore everything you have so far toward solving the equation, and differentiate that quantity. If you can show its derivative is zero, you have what you want." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "$\\frac{a^2x^2}{(by+cz)(bz+cy)}+\\frac{b^2y^2}{(ax+cz)(az+cx)}+\\frac{c^2z^2}{(ax+by)(ay+bx)} \\geq \\frac{3}{4}$", "Solution_1": "You have forgotten an important condition, namely that the numbers a, b, c and x, y, z are such that the number arrays (a; b; c) and (x; y; z) are equally sorted. With this condition, the inequality is proved on http://www.mathlinks.ro/Forum/viewtopic.php?t=5517 .\r\n\r\n darij", "Solution_2": "it's a problem from Romania in the 90's." } { "Tag": [ "function", "integration", "calculus", "calculus computations" ], "Problem": "Let $ (X, \\mathcal{F})$ and $ (Y, \\mathcal{G})$ be measurable spaces. Suppose that $ K: X\\times \\mathcal{G}\\to [0,\\infty]$ is such that for each $ x\\in X$, the function $ B\\to K(x,B)$ is a measure on $ (Y, \\mathcal{G})$ and for each $ B\\in \\mathcal{G}$, the function $ x\\to K(x,B)$ is $ \\mathcal{F}$- measurable.\r\n\r\nGiven such a $ K$, a measure $ \\mu$ on $ (X, \\mathcal{F})$ and an $ \\mathcal{G}$- measurable nonnegative function $ f$ on $ Y$, I [b]have shown[/b] that \r\n1) $ \\nu(B)\\equal{}\\int K(x,B) \\mu(dx)$ is a measure on $ (Y, \\mathcal{G})$ and\r\n2) $ g(x)\\equal{}\\int f(y) K(x,dy)$ is an $ \\mathcal{F}$- measurable function on $ X$.\r\n\r\nCan anyone show that $ \\int f d\\nu \\equal{}\\int \\left [\\int f(y) K(x,dy)\\right ] \\mu(dx)$?\r\n\r\nThanks in advance.", "Solution_1": "This should immediately follow from what you've done for characteristic functions of sets. Now just use the linearity of the integral to extend it to all simple functions and the monotone convergence theorem to get the result for all non-negative functions." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let there's a function $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ \r\nFind all functions $ f$ that satisfies:\r\na) $ f(2u)\\equal{}f(u\\plus{}v)f(v\\minus{}u)\\plus{}f(u\\minus{}v)f(\\minus{}u\\minus{}v)$\r\nb) $ f(u)\\geq0$", "Solution_1": "[quote=\"wangsacl\"]Let there's a function $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ \nFind all functions $ f$ that satisfies:\na) $ f(2u) \\equal{} f(u \\plus{} v)f(v \\minus{} u) \\plus{} f(u \\minus{} v)f( \\minus{} u \\minus{} v)$\nb) $ f(u)\\geq0$[/quote]\r\nTake $ u\\to \\minus{} u ,v\\to \\minus{} v$ then we have \r\n$ f( \\minus{} 2u) \\equal{} f( \\minus{} u \\minus{} v)f(u \\minus{} v) \\plus{} f(v \\minus{} u)f(u \\plus{} v) \\equal{} f(2u)$\r\nTherefore $ f(x) \\equal{} f( \\minus{} x),\\forall x\\in R$\r\nSo $ f(u \\plus{} v) \\equal{} f( \\minus{} u \\minus{} v),f(u \\minus{} v) \\equal{} f(v \\minus{} u)$ :\r\n$ f(2u) \\equal{} 2f(u \\plus{} v)f(u \\minus{} v)$\r\nLet $ x \\equal{} u \\plus{} v,y \\equal{} u \\minus{} v$ then $ f(x \\plus{} y) \\equal{} 2f(x)f(y)$\r\nTake $ x\\to \\minus{} x$ then $ f(y \\minus{} x) \\equal{} 2f( \\minus{} x)f(y) \\equal{} 2f(x)f(y)$\r\nIt gives :$ f(y \\minus{} x) \\equal{} f(x \\plus{} y) \\Rightarrow f(x) \\equal{} 0$\r\nEasy to check $ f(x)\\equiv 0$ is solution of this equation .\r\nI think you have a typo :wink:", "Solution_2": "Ha?\r\nNo, I'm not...\r\n\r\nHm... $ f(x)\\equiv0$?\r\nWhat does it mean?\r\n\r\nI have another solution, $ f(x)\\equal{}\\frac{1}{2}$", "Solution_3": "I'm sorry . \r\n$ f(x)\\equiv c$ \r\nThen $ c\\equal{}2c^2\\Leftrightarrow c\\equal{}\\in\\{0,\\frac{1}{2}\\}$\r\nBut I think ,we don't need condition ii.", "Solution_4": "wangsacl, why don't you post the exact wording please?", "Solution_5": "@Johan : Because I'm not holding the copy of the problem. I just recall the problem from my memory. Sorry for my bad grammar.", "Solution_6": "Take $ u \\equal{} v \\equal{} 0$ and we have $ f(0)^2 \\minus{} f(0) \\equal{} 0$ so $ f(0)\\in{\\{0,\\frac {1}{2}\\}}$. If $ f(0) \\equal{} 0$ we have $ f(x) \\equal{} 0$. If $ f(0) \\equal{} \\frac {1}{2}$ and $ u \\equal{} v$ we get\r\n\\[ f(2u) \\equal{} \\frac {f(2u)}{2} \\plus{} \\frac {f( \\minus{} 2u)}{2}\\Rightarrow f(u) \\equal{} f( \\minus{} u)\r\n\\]\r\nThen we replace this in\r\n\\[ f(2u) \\equal{} f(u \\plus{} v)(u \\minus{} v) \\plus{} f(v \\minus{} u)f( \\minus{} u \\minus{} v)\\Rightarrow f(2u) \\equal{} 2f(u \\plus{} v)f(u \\minus{} v)\r\n\\]\r\nIf $ v \\equal{} 0$ we get $ f(2u) \\equal{} 2f(u)f(u)\\Rightarrow f(2u) \\equal{} 2f(u)^2$ and i'm trying to look for something more :D", "Solution_7": "Find all functions $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ so that \na) $ f(2u)=f(u+v)f(v-u)+f(u-v)f(-u-v)$ for all $u,v \\in \\mathbb{R}$ and\nb) $ f(u)\\ge 0$ for all $u \\in \\mathbb{R}$ \n\n[this is the exact wording]" } { "Tag": [ "quadratics", "number theory proposed", "number theory" ], "Problem": "Let $a_1,\\ldots,a_{2004}$ be integers with the property that for all $n\\geq 1$ the number $(a_1)^n+(a_2)^n+\\cdots+(a_{2004})^n$ is the square of an integer. Prove that at least $21$ numbers are $0$.", "Solution_1": "Noone? And I thought that this problem is so beautiful. It seems I was wrong. So be it. I will let you another 5 days and after that I will post a solution.", "Solution_2": "could you post solution please", "Solution_3": "Umm... why is it 21? Let k be the number of non-zero terms. For any prime p larger than all the a_i, k is a quadratic residue mod p (just consider n = p-1). I think it is a well-known result of QR that this implies k is a square.", "Solution_4": "probability1.01 not exacly :). We must take such $p$ that numbers $2004, 2004-1,...2004-21$ are all non quadratic residue $mod p$. Such p we pruduct using Chinese theory and Dirichlet theory.\r\n\r\nEDIT : sorry I wrongly read.\r\n\r\nI tell you why $k<2004-44^{2}$. In numbers $2004,2003,....$ its probably the situation such that we have 3 numbers form $a^{2}sl$, $b^{2}lm$, $c^{2}ms$. Where $s,l,m$ are prime numbers. This 3 numbers can be all at once non quadric residue.", "Solution_5": "Don't see what's incorrect with probability1.01's reasoning...", "Solution_6": "we're waiting for a slution !!!!!!!!:sleeping: can any one post a solution ?", "Solution_7": "Your problem is easy. Let $p>N=max(|a_{i}|,2004)$ is prime. Let $n=p-1$, then $\\sum_{i}a_{i}^{n}\\equiv k(mod \\ p)$, were k is number of terms, suth that $a_{i}\\not =0$. From your condition we get $(\\frac{k}{p})=1 \\ \\forall p>N$. Therefore k is perfect square.\r\nIt mean $k=l^{2}<2004\\Longrightarrow l\\le 44$ therefore at least $2004-1936=68$ terms are 0.\r\nFor example $a_{i}=m^{2},i=1,2,...,1936, a_{i}=0,i=1937,...,2004$, then $\\sum a_{i}^{n}=(44m^{n})^{2}.$" } { "Tag": [ "induction", "function", "algebra proposed", "algebra" ], "Problem": "$ 0 < x_{1}< 1$\r\n $ x_{n+1}= x_{n}+{\\frac{x_{n}}{n^{2}}}$\r\nProve that $ x_{n}$converge\r\n 2)$ x_{1}= 29$\r\n $ x_{n+1}=\\sqrt{3}+{\\frac{x_{n}}{\\sqrt{x_{n}^{2}-1}}}$\r\n Find a such that \r\n $ x_{2k}< a < x_{2k+1}$\r\n @to Rust:It is easy to know that a is limit of this sequence so can you post a full solution?", "Solution_1": "[quote=\"TTsphn\"]$ 0 < x_{1}< 1$\n $ x_{n+1}= x_{n}+{\\frac{x_{n}}{n^{2}}}$\nProve that $ x_{n}$converge\n 2)$ x_{1}= 29$\n $ x_{n+1}=\\sqrt{3}+{\\frac{x_{n}}{\\sqrt{x_{n}^{2}-1}}}$\n Find a such that \n $ x_{2k}< a < x_{2k+1}$\n @to Rust:It is easy to know that a is limit of this sequence so can you post a full solution?[/quote]\r\n1. $ x_{n+1}=x_{1}\\prod_{k=1}^{n}(1+\\frac{1}{k^{2}})1$.\r\nLet $ a$ is root $ a=f(a)$, Obviosly $ a>\\sqrt 3+1$, therefore $ -1 a$ we get $ y_{n}\\equal{} (x\\minus{}a_{n})(\\minus{}1)^{n\\minus{}1}>0$ and decrease.", "Solution_4": "2)$ x_{n}$is bound prove by induction\r\n3)You must concide function \r\n$ g(x)\\equal{}f(f(x))\\minus{}x$ and prove it monotone or increasing to prove the system has uniqge solution.\r\nMore than you must prove that my question ." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "FInd the prime factors of $ 9^8\\plus{}7^6\\plus{}5^4\\plus{}3^2\\plus{}1$. I need the detailed solution, not the answer. thanks", "Solution_1": "$ \\equal{}3*5*13*41*5399$.", "Solution_2": "how to find this without any aids (calculators, math software) and only use simple formula and or logic?" } { "Tag": [ "function", "algebra", "domain" ], "Problem": "Come on guys, comment on ur performance in FT 11 (the last test before the REAL TEST!!)", "Solution_1": "Start with yourself no? Dont worry i wont die of a stroke soon :D ive had quite a few number of shocks today so dont hesitate to even say if you got 380 odd :D", "Solution_2": "oops sorry!\r\ni forgot to do that!\r\nand don't worry, in such a test if i had got 380 odd marks, then it would have been me who would have a died of a stroke.\r\n\r\nanyway, my score is 221\r\n\r\nPHYSICS- 46 + 57 = 103\r\nCHEMISTRY- 13 + 23 = 36 (?!?!?!)\r\nMATHS - 23 + 49 = 72\r\n\r\nmy analysis would be- i could have scored more in physics, considering that i gave 1 1/2 hours in each paper to it. chemistry is always a problem, and maths is an improvement! most certainly my best performance in a FIITJEE Full Test.\r\n\r\nbtw in ur centre, did they take the answer sheet, or did they refuse saying that they won't evaluate it?", "Solution_3": "They refused :rotfl:\r\n\r\nWell anyway let the bigheads post their marks, then the puny fella (thats me :D ) shalt reveal mine :)", "Solution_4": "Got 233.......I think\r\n\r\nPhysics - 78 \r\nChemistry - 80\r\nMaths - 75\r\n\r\nNot much to boast about......... :| But yeah plenty of mistakes in solns to point out!", "Solution_5": "well probable errors and corrections for FIITJEE solns........!\r\n\r\npaper 1:\r\n\r\ngiven answer---------->correct answer\r\nPhysics:\r\nbicycle ques: B ------------->C\r\n(wonder how they will provide torque for front wheel!) :rotfl: \r\ncomprehension:\r\nq3(areal velocity): B---->C (confirmed by achu and chrono)\r\n\r\nChemistry:\r\n(rubber)\r\nmulti correct B,C---------->B,C,D\r\n(if C is true so is D........exothermic rxn. delta H is -ve....diff gives positive! [b]simple math[/b])\r\n\r\nMath:\r\ncompr. Q3\r\n(explain how inverse will exist for many one function) :maybe: \r\n\r\nHypothesis: Question paper setters are not the same people who provide solns...as far as FIITJEE FULL Tests are concerned this is always true! :rotfl:", "Solution_6": "Well chronoz got the right answer for the physics question as far as i know :)\r\n\r\nAnd seeing that he is god of logic ( :P ), ( i think its imperative that u know that by now :D ), he ob did it by physics.", "Solution_7": "got it right ? according to FIITJEE or according logic and physics?", "Solution_8": "the latter i believe :)", "Solution_9": "did you get the multi correct about NaF in paper 2 chem???????? :?:", "Solution_10": "can someone post the chem question plz. the bicycle thingie also if possible. \r\nforget it if its too long", "Solution_11": "Paper 2:\r\n\r\nPhy: \r\nAssert -reasoning (Doppler)....C---->D\r\nthere is no relative motion b/w source and observer along the line joining the two!\r\n\r\nChem:\r\nmatch the following: (thermo)\r\nfirst are three fine but the last.......getting A->r,B->r,C->p, D->p.......... :maybe:(help needed) \r\n\r\nNaF ques(multi correct).....B,C---->A,C :maybe: (help needed)\r\n\r\nequilibrium ques(multi)..A,C,D----->C,D\r\npressure will not affect solid-liquid equilibrium!\r\n\r\nMath:\r\nsin^-1[sec x] ques \r\nquery: A is also an answer!\r\n\r\ncomprehension:\r\nprob for periodic ques(Q2).....D--->B\r\nexpect many to falter here \r\nf1,f2,f4.taken two at a time + one func f3+f5=[x]-x=-{x} :ninja:\r\nhence 4 func(sample space), 3 with period two.......therefore, ans. (B)3/4\r\n\r\n(multi correct)\r\nGiven natural no. n........A,B,C---->A,C\r\nthey mised out '-' sign in soln hence B is infact wrong.......try it for a value say .5\r\n\r\nIn case any mistakes are found in the above mentioned errors...feel free to point out (as if that was necessary to mention.......people here at Aops are would gladly do that)!", "Solution_12": "bicycle is perfect ans pramod.... if yku want an expl call me at 100. or i'll ask rohit to post proper expl.. but the third ans in gravi comp is wrong crap.\r\n\r\nme 227\r\n91 + 67 + 69\r\n\r\nhorrible sols. worse than usual pathetic... im annoyed.\r\n\r\ncan you guys achu pramod phoenix post the other commendable performances inur city. chaappli got 253 here.", "Solution_13": "i am not aware of stunning performances, coz many didn't give it. maybe pramod can tell u of someone (he knows a lotta guys!!)\r\n\r\n\r\nand can u plz tell why there will be no friction in the front wheel??", "Solution_14": "i got 252\r\nthe twins from my place r getting 270 odd\r\nanit is getting 340\r\nhow did he get that i m myself surprised\r\nhe doesnt deserve it :mad:", "Solution_15": "@ashwath\r\n\r\nI rarely come to aops in academic pursuit...\r\nits a place for me to chill out... and spam the place with my gibberish...\r\n\r\nSo I am perfectly comfortable with the private fights fought publicly unless of course it becomes way too personal...\r\n\r\n[quote]I think that basically shows India in a poor light to all outsiders who read this forum. [/quote]\r\n\r\npoor light...hmmmmmmm...\r\nThey have to check if their bulbs glow dimmer when they come to the Indian forum...\r\nIts a problem with their [b]bulbs[/b] and not with us...(no pun intended)", "Solution_16": "poor light...hmmmmmmm...\r\nThey have to check if their bulbs glow dimmer when they come to the Indian forum...\r\nIts a problem with their [b]bulbs[/b] and not with us...(no pun intended)[/quote]\r\n\r\ni think they've switched off their bulbs for earth hour..\r\n\r\nachu imo a score of 230 if ft11 was jee would have got us something in the 500s.\r\n\r\npramod ill ask madness to post the answer. i'm confident its right.", "Solution_17": "chronoz thala U back hmmmmm......... so madness lied about that typing problem ah??? :P", "Solution_18": "No he uses an on screen keyboard :D\r\n\r\nAnd btw chronoz, i refuse to type it out unless i get a favour :D", "Solution_19": "[quote=\"madness\"]No he uses an on screen keyboard :D\n\nAnd btw chronoz, i refuse to type it out unless i get a favour :D[/quote]\r\n\r\nrohit havent u read that lesson i had written in our english book :P \r\n\r\n[i][u][b][color=red][size=150]\"NEVER HELP IN THE CUSTOMARY SENSE DEVELOP YOURSELF INTO A HARMONIUS INDIVIDUALS WHO WORKS TOWARDS THE BETTERMENT OF THE SOCIETY!!!![/size] [/color]:[/b][/u][/i] :police:", "Solution_20": "Hmmm, well you are going wayyyyyyyy off topic mr. einstien :huuh: So if you want to continue further, kindly do so in the metathread.\r\n\r\nAlso i might kindly want to point out that it should be develop yourselves into harmonius individuals or into a harmonius individual and not what you wrote :D", "Solution_21": "[quote=\"madness\"]Hmmm, well you are going wayyyyyyyy off topic mr. einstien :huuh: So if you want to continue further, kindly do so in the metathread.\n\nAlso i might kindly want to point out that it should be develop yourselves into harmonius individuals or into a harmonius individual and not what you wrote :D[/quote]\r\nno i am sure i am rite as you follow what i wrote but my actual edicts have been copied to the golden guide!!!! and this general discussion da not IITJEE thread for one not to enjoy or gossip as i come to mathlinks just like what soumya comes for the same reason refreshment!!!! :D :D :spider:", "Solution_22": "Enough of jalras and NV's. And by the way, why did u not come mr.einstein? Scared/playing safe. :evilgrin: \r\n\r\nI'm getting 266:\r\nPhysics-88\r\nMaths-82\r\nchem-96", "Solution_23": "[quote=\"chemrock\"]Enough of jalras and NV's. And by the way, why did u not come mr.einstein? Scared/playing safe. :evilgrin: \n\nI'm getting 266:\nPhysics-88\nMaths-82\nchem-96[/quote]\r\n\r\nspidey :spidy: is never scared of anything my friend i had fever thought could take some rest thats why!!!! :play_ball: :spider:", "Solution_24": "well in paper 2 math que one of my claims is wrong\r\nthe multi :( -correct ques is perfectly alright!...my mistake :mad: \r\n\r\nforgot to use chain rule :rotfl:", "Solution_25": "@madness,chronoz: wishing speedy recovery of whoever's faulty keyboard......\r\n\r\nhowever, kindly enlighten me with your so-called 'correct' soln. of the bicycle question..... :starwars: .....achu and i are very keen to be proved wrong by u guys!!!!!!\r\n\r\n\r\n[quote]speak up where do you see urself[/quote]\r\n\r\n@achu: at the beginning of the journey i saw myself at AIR 1 .....thought i was born to be AIR 1........until i was acquainted with people like u and our fellow mates @ aops!!!......Now, I'm seriously considering my chances of making it big if not exactly vanquishing IIT-JEE 2008.......let me just say..lets wait ad watch!!!!!!! :weightlift:", "Solution_26": "Honestly, i'd say that even if i get a rank, ill get only from 2000 - 5000 odd :(", "Solution_27": "[quote=\"pramod.best\"]@madness,chronoz: wishing speedy recovery of whoever's faulty keyboard......\n\nhowever, kindly enlighten me with your so-called 'correct' soln. of the bicycle question..... :starwars: .....achu and i are very keen to be proved wrong by u guys!!!!!!\n\n\n[quote]speak up where do you see urself[/quote]\n\n@achu: at the beginning of the journey i saw myself at AIR 1 .....thought i was born to be AIR 1........until i was acquainted with people like u and our fellow mates @ aops!!!......Now, I'm seriously considering my chances of making it big if not exactly vanquishing IIT-JEE 2008.......let me just say..lets wait ad watch!!!!!!! :weightlift:[/quote]\r\n\r\nHow can you pramod?? how dare you?\r\nYou better not even dream abt such stuff when I am around... :rotfl: :P \r\n :P", "Solution_28": "@Soumya, thats why he stopped dreaming. He woke up, saw you and realized the futility of his dream and the complications of reality.\r\n\r\nMadness, why being such a hypocrite especially in the sea of hypocrites. \r\n\r\n@pramod: Just because you got it wrong, doesn't mean answer is wrong?? Even though you maybe the lord of the physics of the 31 dimensions.\r\n\r\nOkay, its given that the cycle moves down with pure rolling. So at any instant, the mg force pulls you down and threatens to accelarate you, making you lose your Pure Rolling. So what do you do? You press the brakes, which clamps the back wheel's rolling thereby v>rw, so friction acts on the back wheel and is sufficient enough to counter mg sin. What about the front? Front is already in pure rolling and nobody has disturbed it. Moreover, friction on rear balances sin component of gravity. And so front wheel is not disturbed.\r\n\r\nThis was how I thought and wrote the answer.. and fiitjee agrees with me. So great minds think alike.\r\n\r\nOr\r\n\r\n[hide]Fools seldom differ[/hide]", "Solution_29": "@ chronoz:\r\n\r\ni think u are right. actually, when i agreed with pramod, i totally forgot that the question involved applying brakes, and thought that it was discussing a simple case of a bicycle rolling down the plane. (however, i got the answer wrong in the exam, because i rather foolishly thought that the front and rear wheels were connected through the chain!! :oops: )\r\nlong time since i rode a bicycle!!!" } { "Tag": [ "calculus", "integration" ], "Problem": "If $n$ and $a$ are positive integers, prove that $n^{4}+a$ will be composite for an infinite number of positive integers $a$.", "Solution_1": "[hide=\"hint\"]Sophie Germaine's Factorization: $x^{4}+4y^{4}= (x^{2}+2xy+2y^{2})(x^{2}-2xy+2y^{2})$. [/hide]", "Solution_2": "Define $a$ as $4m^{4}$, where $m>n, m\\in Z$ (of which there are an infinite number). Using the factorization Phelpedo posted, we just need to show that $n^{2}-2nm+2m^{2}> 1$ (since $n$ and $m$ are positive integers, other part of the factorization is obviously greater than 1).\r\n\r\nWe know $m>n$, therefore $2m^{2}>2nm$ and since all numbers are integral, $2m^{2}-2nm\\geq 1$. You then add on $n^{2}$, which is at least 1, and you get $2m^{2}-2nm+2n^{2}\\geq 2$, meaning $n^{4}+a$ factors in a way other than 1 and itself, and is therefore composite.", "Solution_3": "Um, as stated it is trivial: if $n$ is odd, $n^{4}+a$ is composite for all odd $a$ and if $n$ is even, $n^{4}+a$ is composite for all even $a$. I think you mean [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=363654#p363654]this problem[/url].", "Solution_4": "Yes, sorry, that came out as a VERY trivial proof. :P I was going by memory, and I didn't remember the problem..." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "$t_n$ defined by its first term $t_0$ positive integer $\\geq 2$ \r\n$t_{n+1}=2t_n^2-1$ \r\n\r\nCompute $\\sum_{n\\geq 0}\\frac{1}{2t_02t_1...2t_n}$", "Solution_1": "nice problem.\r\nlet $t_0=\\frac{x+1/x}{2}$ ;)" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "prove that if 3 lines are concurrent then their poles are collinear.", "Solution_1": "it must have been posted before but anyway,\r\nuse the fact that if a point $ P$ is on the polar of $ A$ then the polar of $ P$ passes through $ A$ .\r\nwe also have : if 3 points are collinear then their polars are concurrent", "Solution_2": "[quote]if 3 points are collinear then[/quote]\r\nthen what?", "Solution_3": "[quote]it must have been posted before but anyway, [/quote]\r\ncan anybody post the link to this problem", "Solution_4": "come on people reply", "Solution_5": "is it too easy or what?\r\njust post the link if so" } { "Tag": [ "calculus", "derivative", "analytic geometry", "number theory", "algebra", "linear equation", "number theory unsolved" ], "Problem": "Hello,\r\n\r\nI found this problem in a book and I haven't been able to solve it. It is Putnam's style (I think). It requires from calculus just the idea of the implicit derivative (to find a tangent) and it is on the edge between Number Theory and Algebra for the rest (at least that is what I think) Therefore, I thought this was the best place to post it. [b]Moderators[/b], if you consider that it should be moved, go ahead!\r\n\r\nThis is the problem. Suppose that in the curve $y^2=x^3-n^2x$ one has a point $(x_0,y_0)$ where both coordinates are rational. Prove that the tangent line to this curve at $(x_0,y_0)$ intersect the curve in a point $(x_1,y_1)$ such that $x_1,\\ x_1-n,\\ x_1+n$ are squares of rational numbers.\r\n\r\nNow, after almost a page of algebra, I get to the point where $x_1=(x_0^2+n^2)^2$. In this case it is evident that $x_1$ is the square of a rational number, but what about the other two. Now, I think my calculations are accurate, but may be I got something wrong, you may check if you want.\r\n\r\nBest,", "Solution_1": "If you have a linear equation for $x_1$ and $y_1$, you must have that $y_1$ is rational. Then we also know that $(x_1+n)(x_1-n)$ is a perfect square, which seems very close.", "Solution_2": "[quote=\"Fiachra\"]If you have a linear equation for $x_1$ and $y_1$, you must have that $y_1$ is rational. Then we also know that $(x_1+n)(x_1-n)$ is a perfect square, which seems very close.[/quote]\r\n\r\nIt is celar that $y_1$ is a rational (basically after you compute the tangent line), and therefore there is no doubt that $(x_1+n)(x_1-n)$ is a perfect square, but how is that very close to the fact that each of the factor is a perfect square. \r\n\r\nFor example, take $n=5$ and $x_1=\\frac{25}{4}$, then $y_1=\\frac{75}{8}$, but neither $x_1-n$ nor $x_1+n$ is a square (but $(x_1+n)(x_1-n)$ is). Of course, according to the problem, there can be no rational point $(x_0,y_0)$ in the curve such that the tangent through it passes through $(x_1,y_1)$.", "Solution_3": "I suppose you avoid $x_0=-n,\\, 0,\\, n$ (with $\\sqrt{2n}\\not \\in \\mathbb{Z}_+$). In other case, if $(x_0,y_0)$ is a point of the curve $y^2=x^3-n^2x$ which are rationals, then \r\n\\[\r\ny'_0=\\frac{x_0^2}{y_0}+\\frac{y_0}{x_0} \\ \\iff \\ y-y_0=\\left[\\frac{x_0^2}{y_0}+\\frac{y_0}{x_0}\\right](x-x_0).\r\n\\]\r\nThen, apply this equation in the original curve we get that\r\n\\[\r\ny_0^2+(3x_0^2-n^2)(x-x_0)+\\left(\\left[\\frac{x_0^2}{y_0}+\\frac{y_0}{x_0}\\right](x-x_0)\\right)^2=x^3-n^2x\r\n\\]\r\nthen applying algebraic properties of the solution $(x_0,y_0)$ we get that \r\n\\[\r\nx=\\left[\\frac{x_0^2}{y_0}-\\frac{y_0}{x_0}\\right]=\\frac{n^4}{y_0^2}\r\n\\]\r\nwhich is a square of a rational, and $y_1^2=x_1^3-n^2x_1$ then \r\n\\[\r\nx_1^2-n^2=\\frac{y_1^2}{x_1}\r\n\\]\r\nthen $(x_1-n)(x_1+n)$ is a square of a rational. And as djimenez told before we can find examples where $x_1-n$ is not a square of a rational.\r\n\r\nBest regards.\r\n\r\nRobertuX" } { "Tag": [ "inequalities", "AMC", "USA(J)MO", "USAMO", "inequalities solved" ], "Problem": "a=( m^(m+1) + n^(n+1) )/(m^m + n^n). If m,n,a are positive integers prove the inequality:\r\na^m + a^n >= m^m + n^n.", "Solution_1": "This one was already discussed! check the proposed problems topic!\r\n\r\ncheers! :D:D", "Solution_2": "Unfortunately I couldn't find this problem in Proposed Problems Section. Are u sure it is from that Section?", "Solution_3": "Yes! :D \r\n\r\nhttp://www.mathlinks.ro/phpBB/viewtopic.php?t=223\r\n\r\ncheers! :D :D", "Solution_4": "I saw that you discassed it but there is no solution. Taking n>=m IMHO is a dead-end. :(", "Solution_5": "I think this works.\r\n\r\nDefine $x = \\frac{a}{m} - 1, y = \\frac{a}{n} - 1$. We have $x = n^n \\cdot \\frac{\\frac{n}{m} - 1}{m^m + n^n}$, $y = m^m \\cdot \\frac{\\frac{m}{n} - 1}{m^m + n^n}$. Now, $(1+x)^m \\ge 1 + mx \\Rightarrow \\frac{a^m}{m^m} \\ge 1 + n^n \\cdot \\frac{n-m}{m^m+n^n} \\Rightarrow a^m \\ge m^m + m^mn^n\\frac{n-m}{m^m+n^n}$. Adding the similar inequality given by $(1+y)^n \\ge 1+ny$ gives the result.", "Solution_6": "It's from USAMO 1991: [url=http://www.kalva.demon.co.uk/usa/usoln/usol914.html]www.kalva.demon.co.uk/usa/usoln/usol914.html[/url]" } { "Tag": [ "geometry", "\\/closed" ], "Problem": "what are moderators and how do you become one?", "Solution_1": "Moderators are people who warn, edit, and delete useless, bad, and/or stupid posts.\r\nYou can become one if you're really really good in the forums :)", "Solution_2": "Moderators are people responsible for certain sections of the message board. The duty of the moderators is primarily to encourage useful discussion. Moderators of the math sections of the board will throw out problems or ideas for discussion a few times a week. Moderators will also have the ability to split threads, delete offensive posts, and move posts to other boards, and they can do each of these at their discretion.\r\n\r\nWe picked out the initial set of moderators based *very loosely* on the activity of folks on the site to date. If you haven't been asked yet, please don't be offended. As you can see, we haven't finished naming moderators. We expect to have 1-3 per board (depending on the activity of the board). If you are interested in being a moderator, pm me and let me know what areas you are interested in. We may not have a place for you right away, but as time goes by, we'll likely have attrition among our current moderators and have other opportunities." } { "Tag": [ "algebra", "polynomial", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $ A$ and $ B$ mtrices s.t $ A^HB\\equal{}0$\r\nShow that $ P_A\\equal{}I\\minus{}P_B$\r\n\r\n[hide=\"Idea\"]It seems that ranges of A and B are orthogonal, but I couldn't go further[/hide]", "Solution_1": "I'm not sure what your notation means. Does $ A^H$ denote transpose and $ P_A$ denote characteristic polynomial?", "Solution_2": "$ A^H$ is hermitian of $ A$, $ P_A$ denotes the projection matrix.", "Solution_3": "Any idea ? :(" } { "Tag": [ "\\/closed", "super old unlocked thread" ], "Problem": "hi folks, \r\n\r\nbecause the other wish list topic is getting crowded with different sorts of stuff, I am creating this other wish list, which I will more closely manage.\r\n\r\nany post without relevance will be erased without notice (ie wish for things that can already be set up in the options menu, etc.) \r\n\r\nfor now, on the To Do list there are the following item \r\n\r\n1. Each user chooses which category should be displayed on the main index [in progress]\r\n\r\n2. WAPgateway [released, still adding features]\r\n\r\n3. Integrated nav bar [in progress]\r\n\r\n4. User chooses style of unread / read posts. Normal mode, or keep unread posts. [on the waiting list]", "Solution_1": "Ah, great idea Valentin! :)\r\n\r\nI still got one suggestions unanswered: [list][*] About the structure: since the newer people complain they don't want subforums in pre-olympiad and such, perhaps it may be a good idea to allow people to pick a subject when posting? Like a topic status to pick from either \"Algebra\", \"Combinatorics\", etc.\nThat doesn't require any hassle on different forums, but it does have the advantage that you can immediately have what you're looking for.[/list]\r\n\r\nJust my 2 cents.\r\n\r\nGreetz, Peter", "Solution_2": "In the mathlinks style, is there a \"mark all posts read\" button on every forum? If not, can you add one?", "Solution_3": "yes, there is one in ML style. :)\r\n\r\nI'm not sure it works though with the flag keeping...", "Solution_4": "Do you think it would be possible to make it so that is hidden when you put it in spoiler? Right now, it is not. For example, the following is in spoiler:\n\n[hide]\n\n\n\n[/hide]\n\nFor now, whenever I want to post a solution without ruining it for other people, I am stuck using fractions like 1/2 instead of .", "Solution_5": "Some of us already had this idea. I think it is technically very hard, if not even impossible. But who knows...\r\n\r\n Darij", "Solution_6": "It is technically almost impossible to make this happen. Not in the way LaTeX is rendered on this site anyway.", "Solution_7": "It might be hard to make it come out in spoiler per se, but what happens if you include the color package so that people can post in white? Then again, if you tell us the colors of the backgrounds, we can make the $\\text{\\TeX}$ in exactly those colors.", "Solution_8": "It doesn't matter. It can't work because you need to change the image as people select it. Also there are 4 different backgrounds, and there are also users that don't use spoilers (selected the option not to appear). \r\n\r\nThis would mean something like rendering 5 images instead of 1, which would induce decrease of site speed and increase by 5 times the number of files in the /pictures/ directory, which is as is already huge!.", "Solution_9": "... perhaps you can help this people giving an option to hide all replies to a topic, and reveal the rest after a button click :maybe:\r\n\r\nMay sound stupid but not more stupid than the spoiler system as is... heck now they even got a good reason not to learn latex! :P", "Solution_10": "It's a good idea Peter. I think that this will indeed solve many of some people's problems. I will try to find if there is already a similar mod (I belive I saw such a thing on the VBulletin 3.0.3 board, where you could choose the style of viewing a thread - you can either use this style, or the old replies system like this one here http://lists.suse.com/archive/suse-linux-e/2001-Jun/0472.html) ;)", "Solution_11": "Perhaps even better: make [b]some parts[/b] of a reply revealable upon mouseclick (while the rest of the is automatically shown; in fact, not everything must be hidden). For instance:\r\n\r\n[quote=\"darij grinberg\"]Nice problem, thanks. There is a little typo: $a_i$ should be replaced by $a_j$.\n\n[u]Click here to see the solution.[/u][/quote]\r\n\r\n Darij", "Solution_12": "Darij, that would mean using dhtml into posts, which can be extremelly messy. I don't work with dhtml, neither does Richard, so I guess we can't pull this off (properly). In plus it would not work on some browsers. \r\n\r\nHopefully something like this will be available in phpbb 2.2 :) \r\n\r\nMeanwhile the collapsing forums mod is up and running. You can select what categories are open, and the computer remembers them. You can also save a specific state in the config, and when you want to use it, just push the restore button.", "Solution_13": "It would be nice if, in the my classes section, the listing of transcripts had names, and not just the date. Clearly, a lot (well, the Algebra class) already knows what they're discussing, b/c they say in the forum. However, it would be nice to see \"Synthetic Divison (9/30)\" in the transcript list instead of \"Class Transcript from 9/30\" so that you could easliy use the the transcripts for references.", "Solution_14": "[quote=\"Valentin Vornicu\"]Darij, that would mean using dhtml into posts, which can be extremelly messy. I don't work with dhtml, neither does Richard, so I guess we can't pull this off (properly). In plus it would not work on some browsers. \n\nHopefully something like this will be available in phpbb 2.2 :) \n\nMeanwhile the collapsing forums mod is up and running. You can select what categories are open, and the computer remembers them. You can also save a specific state in the config, and when you want to use it, just push the restore button.[/quote]\r\n\r\nCouldn't you just have two versions of the page, one where spoilers are displayed, and one where they aren't, and you would click a link to move between them. To avoid showing latex, you just wouldn't add the link to the image (ie, no tag in the source). (I haven't actually used the forum much, and certainly haven't looked at the source much, but that doesn't sound too hard)", "Solution_15": "I think I did this a long time ago :D", "Solution_16": "[quote=\"Valentin Vornicu\"]I think I did this a long time ago :D[/quote]\r\nOh, i see...i did not pay attention to the [b]go to page[/b] thing...... :blush: \r\nThanks for all of your help! :D", "Solution_17": "how do we sort members by number of posts? i was going to make a post about that because i can't figure it out. :oops:", "Solution_18": "Go to the Memberlist tab at the top. It should take you [url=http://www.artofproblemsolving.com/Forum/memberlist.php]here[/url]. It should already be sorted by post (descending).", "Solution_19": "[quote=\"jhollenbeck\"]how do we sort members by number of posts? i was going to make a post about that because i can't figure it out. :oops:[/quote]\r\n\r\nTo add to what joml88 said...there is a dropbox on the memberlist page that allows you to sort members in several ways, including post count.", "Solution_20": "What about an option for the author of a Thread to delete his thread :?: \r\n\r\nAlso sometimes a thread exists but a user does not notice it and starts a new one with the same topic.\r\n\r\nSo if the author can delete his own thread then things will be simplified.", "Solution_21": "If we let users delete their own threads, we would be handing the ability to censor directly to users. There are other problems with such an option, so it's unlikely we would ever institute it.", "Solution_22": "How do you become an administrator? :D ;)", "Solution_23": "i don't think they would let anyone be an admin becuase you didn't help build the site.", "Solution_24": "[quote=\"dogseatcheese\"]How do you become an administrator? :D ;)[/quote]\r\n\r\nAdministrators are AoPS employees.", "Solution_25": "[quote=\"MCrawford\"][quote=\"dogseatcheese\"]How do you become an administrator? :D ;)[/quote]\n\nAdministrators are AoPS employees.[/quote]\r\ncan i work for AoPS?", "Solution_26": "In light of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87291]these[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87476]two[/url] posts, I would like to officially request the creation of a Self Pity forum. This forum will serve a vital purpose in the community, giving complainers of all stripes an ideal location to commiserate. The AMC forum is likely to especially benefit from this in the aftermath of the administration of all AMC contests.\r\n\r\nAnyhow, think about it.", "Solution_27": "One qualification for an admn. is to be really old.....\r\n\r\nI wish the GFF posts are deleted soon and there is no country bias thing", "Solution_28": "[quote=\"shadysaysurspammed\"]One qualification for an admn. is to be really old.....[/quote]There is no such criterion.", "Solution_29": "[quote=\"Valentin Vornicu\"][quote=\"shadysaysurspammed\"]One qualification for an admn. is to be really old.....[/quote]There is no such criterion.[/quote]\r\n\r\nNo,by really old i meant about 20+ :D" } { "Tag": [ "geometry", "rhombus", "rectangle" ], "Problem": "ABCD is a rectangle with AB = 16 cm and BC = 12 cm. Points E and F lie on sides AB and CD so that AECF is a rhombus.\r\nWhat is the length of EF?", "Solution_1": "hello, the length of $ \\overline{EF}\\equal{}15cm$.\r\nSonnhard.", "Solution_2": "$ AE\\equal{}EC\\equal{}CF\\equal{}AF\\equal{}x$\r\nConsider $ \\triangle BEC$\r\nWe have $ EC^2\\equal{}BE^2\\plus{}BC^2$\r\n$ x^2\\equal{}(16\\minus{}x)^2\\plus{}144$\r\n$ 32x\\equal{}400$\r\n$ x\\equal{}12.5$\r\n\r\nAnd $ AC\\equal{}20$\r\n\r\nSo, $ EF\\equal{}2\\sqrt{12.5^2\\minus{}10^2}\\equal{}15$", "Solution_3": "Thanks IW@IT." } { "Tag": [], "Problem": "\\[ \\left\\{ \\begin{array}{l}\r\n 4x \\plus{} 3y \\plus{} 1 \\equal{} 3\\sqrt {\\left( {x \\plus{} y} \\right)\\left( {2x \\plus{} y \\plus{} 1} \\right)} \\\\ \r\n x^2 \\plus{} y^2 \\equal{} 2 \\\\ \r\n \\end{array} \\right.\r\n\\]", "Solution_1": "hello, squaring the first equation and simplifying we get $ y\\equal{}\\minus{}\\frac{2}{3}x\\plus{}\\frac{1}{3}$, substituting this \r\nin the second equation we have $ x^2\\plus{}(\\minus{}\\frac{2}{3}x\\plus{}\\frac{1}{3})^2\\equal{}2$. Solving the last equation by $ x$ we have $ x_1\\equal{}\\frac{17}{13}$ and $ x_2\\equal{}\\minus{}1$, from here we get $ y_1\\equal{}\\minus{}\\frac{7}{13}$ and $ y_2\\equal{}1$, these are all solutions of our system.\r\nSonnhard.", "Solution_2": "Squaring and factoring the first equation gives (2x+3y-1)(x+1)=0, so x = -1 and y = 1 (as y=-1 doesn't work) and then substitution in the second equation gives the other solution: x = 17/13 and y = -7/13.", "Solution_3": "[quote=\"toancapba\"]\n\\[ \\left\\{\\begin{array}{l} 4x \\plus{} 3y \\plus{} 1 \\equal{} 3\\sqrt {\\left( {x \\plus{} y} \\right)\\left( {2x \\plus{} y \\plus{} 1} \\right)} \\\\\nx^2 \\plus{} y^2 \\equal{} 2 \\\\\n\\end{array} \\right.\n\\]\n[/quote]\r\nPut\r\n\\[ \\begin{array}{l} \\left\\{\\begin{array}{l} u \\equal{} x \\plus{} y \\\\\r\nv \\equal{} 2x \\plus{} y \\plus{} 1 \\\\\r\n\\end{array} \\right. \\\\\r\n\\Rightarrow 2u \\plus{} v \\equal{} 3\\sqrt {uv} \\Leftrightarrow 2\\sqrt {\\frac {u}{v}} \\plus{} \\sqrt {\\frac {v}{u}} \\equal{}3 \\\\\r\n\\end{array}\r\n\\]" } { "Tag": [ "analytic geometry" ], "Problem": "We have a $ m\\times n$ grid situated in the cartesian plane, with a corner in $ (0,0)$ and the opposite corner in $ (m,n)$. We kill a point $ (i,j)$ with $ 0\\le i\\le m$ and $ 0\\le j\\le n$, and you can't pass over it. In how many ways you can travel from $ (0,0)$ to $ (m,n)$ if you can only move one point in the positive direction (to the right or up)? \r\n\r\n :(", "Solution_1": "[hide=\"Hint\"]Count the number of paths that [i]do[/i] pass through the point, and subtract this from the total number of paths.[/hide]", "Solution_2": "i have tried it\r\nbut i dont know how to count the paths :oops: !!!\r\nplease could you show your solution\r\nwith details please\r\nim really bad at combinatorics\r\ni would be very happy if you tell me\r\nwhere can i learn combinatorics\r\nto do this kind of problems\r\nwell, good bye :)", "Solution_3": "Well to get from $ (0, 0)$ to $ (i, j)$ we need to make $ i\\plus{}j$ moves, $ i$ of which are rights and $ j$ of which are ups. Similarly to get from $ (i, j)$ to $ (m, n)$ we need to make $ m\\plus{}n\\minus{}i\\minus{}j$ moves, $ m\\minus{}i$ of which are rights and $ n\\minus{}j$ of which are ups. From here it should be straightforward how to calculate the number of paths from $ (0, 0)$ to $ (m, n)$ going through $ (i, j)$.", "Solution_4": "you say that to get (2,2) we have 2+2=4 options, but\r\nthere are 6 :o \r\ni'm trying to see something interesting, but\r\ni feel i'm far to get it :oops:", "Solution_5": "The number of [i]moves[/i] (not the number of [i]ways[/i]) to get to $ (m,n)$ from $ (i,j)$ is $ m\\plus{}n\\minus{}i\\minus{}j$.\r\n\r\nEDIT: I made a careless mistake.", "Solution_6": "[quote=\"pelao_malo\"]We have a $ m\\times n$ grid situated in the cartesian plane, with a corner in $ (0,0)$ and the opposite corner in $ (m,n)$. We kill a point $ (i,j)$ with $ 0\\le i\\le m$ and $ 0\\le j\\le n$, and you can't pass over it. In how many ways you can travel from $ (0,0)$ to $ (m,n)$ if you can only move one point in the positive direction (to the right or up)? \n\n :([/quote]\r\n\r\n$ \\binom{m \\plus{} n}{m} \\minus{} \\binom{i \\plus{} j}{i} \\cdot \\binom{m \\plus{} n \\minus{} i \\minus{} j}{m \\minus{} i}$ is what you're lookin for\r\n\r\n\r\n\r\n$ \\binom{m \\plus{} n}{m}$ represents the number of ways to get from $ (0,0)$ to $ (m,n)$\r\n\r\n$ \\binom{i \\plus{} j}{i}$ represents the number of ways to get from $ (0,0)$ to $ (i,j)$\r\n\r\n$ \\binom{m \\plus{} n \\minus{} i \\minus{} j}{m \\minus{} i}$ represents the number of ways to get from $ (i,j)$ to $ (m,n)$\r\n\r\n\r\ni probably made a careless mistake or two in there, by the way" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "[color=blue]\n Let $a, b, c>0$ such that $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=1.$ \n Prove that\n \\[ a^{2}+b^{2}+c^{2}+1 < 2abc(a+b+c). \\]\n\n [/color]", "Solution_1": "[quote=\"cezar lupu\"][color=blue]\n Let $a, b, c>0$ such that $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=1.$ \n Prove that\n \\[ a^{2}+b^{2}+c^{2}+1 < 2abc(a+b+c). \\]\n\n [/color][/quote]\r\n\r\nLet $x=\\frac1{a}$, $y=\\frac1{b}$ and $z=\\frac1{c}$.\r\n\r\nThe inequality becomes $x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2<2(xy+yz+zx)$ with $x+y+z=1$\r\n\r\nFrom $x+y+z=1$ we have that $2(xy+yz+zx)=1-(x^2+y^2+z^2)$, so it remains to prove that:\r\n\r\n$x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2+x^2+y^2+z^2<1$\r\n\r\nCould anyone finish the prove?", "Solution_2": "$x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2<2(xy+yz+zx)$\r\n$\\Leftrightarrow 2(xy+yz+zx)(x+y+z)^4>(x^2y^2+y^2z^2+z^2x^2)(x+y+z)^2+x^2y^2z^2$\r\n$\\Leftrightarrow 2\\sum_{sym}x^5y+7\\sum_{sym}x^4y^2+10\\sum_{sym}x^3y^3+9\\sum_{sym}x^4yz+42\\sum_{sym}x^3y^2z\\frac{34}{3}\\sum_{sym}x^2y^2z^2>0$", "Solution_3": "$a,b,c>0, \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=1\\Rightarrow a,b,c>1.$\r\nIt follows $2abc(a+b+c)=2a^{2}bc+2ab^{2}c+2abc^{2}>2a^{2}+2b^{2}+2c^{2}>a^{2}+b^{2}+c^{2}+1.$", "Solution_4": "[quote=\"Incumbuss\"]$2a^{2}+2b^{2}+2c^{2}>a^{2}+b^{2}+c^{2}+1.$[/quote]\r\n\r\nThis is not true ....", "Solution_5": "A more harder one is $\\displaystyle \\sum a^2+1<\\frac{1}{2} abc(\\sum a)$", "Solution_6": "[quote=\"shyong\"][quote=\"Incumbuss\"]$2a^{2}+2b^{2}+2c^{2}>a^{2}+b^{2}+c^{2}+1.$[/quote]\n\nThis is not true ....[/quote]\r\n\r\nWhy?", "Solution_7": "[quote=\"shyong\"][quote=\"Incumbuss\"]$2a^{2}+2b^{2}+2c^{2}>a^{2}+b^{2}+c^{2}+1.$[/quote]\n\nThis is not true ....[/quote]\r\nNo no,that is true cause $a^2,b^2,c^2>1$" } { "Tag": [], "Problem": "This is my first time to chapter. I need your help to understand some rules. I know how to get the overall scores for the written contest(sprint & target), but I am not sure how to calculate the scores for the countdown. Suppose I got 40 points and placed 10th from the written. If I beat the 9th and 8th in the countdown, is my overall placement for the chapter jump to 8th? If this is the case, the countdown weights a lot, right?\r\nThanks a lot.", "Solution_1": "Yes, to all of your questions.\r\n\r\nIf your countdown is official:\r\n\r\n10th against 9th. Loser is 10th.\r\n\r\nWinner of previous match against 8th. Loser is 9th.\r\n\r\nEtc.\r\n\r\nSo the most you can drop is 1 spot but you can go all the way up to first place. Of course, you need to make top 10 first :D", "Solution_2": "That is only if the countdown is official. If it isn't, it doesn't matter at all in your final placing.", "Solution_3": "Yes, they must hold your chapter countdown this way to make it official. If it is any other way, then it's unofficial and just for fun. For example, my final rankings are just based upon the written competition, and the countdown is a single elimination bracket with 16 people.", "Solution_4": "for a countdown to be official, it has to be in the format mentioned (but i like a bracket with 12 people much better :) )", "Solution_5": "Thanks a lot. But how do I know if my chapter is official or not?", "Solution_6": "Your chapter? Um, if you can qualify for state, then I guess it's official... Are you talking about the entire competition, or just countdown? I recommend you first try very hard to do well, of course. Then, if you make it, try your best at countdown to do well. At chapter, ask your coordinator if countdown is official at state. Make sure you know if the state countdown is official. or not. Chapter should be okay, because a lot more people make the next level." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that there do not exist two consecutive perfect numbers (number is called perfect if it is equal to the sum of its proper divisors).", "Solution_1": "Doesn't that follow by the formula for generating perfect numbers with mersenne primes?", "Solution_2": "Basically, Euler's characterization is used in a proof but then you have to prove that neither $2^{p-1}(2^p-1) +1$ nor $2^{p-1}(2^p-1) -1$ can't be perfect.", "Solution_3": "Yes.Suppose n is perfect number ,we will prove that:n-1 and n+1 is not.Indeed:\r\n$n=2^{p-1}(2^p-1)=4^{\\frac{p-1}{2}}(2.4^{\\frac{p-1}{2}}-1)=4.(2.4-1)=4(\\mod 12)$\r\n->n-1\u2261-1(mod 4)\r\n\u03c3(n-1)=n-1/d+d\u22610(mod 4) but 2(n-1)\u2261-2(mod 4) .Contradiction.\r\nn+1\u2261-1(mod 3).Hence:\r\n\u03c3(n+1)=n+1/d+d\u22610(mod 3) but 2(n+1)\u2261-2(mod 3).It's also false." } { "Tag": [ "ratio", "trigonometry", "geometry", "trig identities", "Law of Sines" ], "Problem": "How does one prove that equal angles in two triangles implies ratios of corresponding sides equal? I think it has something to do with the law of sines, but I'm not sure exactly how to proceed.", "Solution_1": "Say you have two angles t1 and t2 in two triangles. For one triangle, call [b]a[/b] the side opposite to the angle t1 and [b]b[/b] the side opposite to t1. For the other triangle, call [b]A[/b] the side opposite to the angle t1 and [b]B[/b] the side opposite to t2. By the sine law:\r\n\r\na/sin(t1) = b/sin(t2) and\r\nA/sin(t1) = B/sin(t2)\r\n\r\nSo a/b = sin(t1)/sin(t2) = A/B\r\n\r\ni.e. the ratios are the same.", "Solution_2": "There's a bit of a problem with using the law of sines here - using sine at all essentially assumes AA Similarity. Try proving it without the law of sines. (1337 - if you're still stuck in 3 days, ask Coach Frost to borrow his Intro Geometry book and work through the similarity chapter. The proof is there.)", "Solution_3": "Ok...here is a pretty random proof (?):\r\n\r\nGiven $\\triangle{ABC}$ and $\\triangle{DEF}$, where $\\angle{A}\\cong\\angle{D}$, $\\angle{B}\\cong\\angle{E}$, and $\\angle{C}\\cong\\angle{F}$.\r\n\r\nLet E' and F' be points on sides AB and AC, respectively, such that AE'=DE and AF'=DF. \r\n\r\nBy SAS, $\\triangle{AE'F'}\\cong\\triangle{DEF}\\implies\\angle{AE'F'}\\cong\\angle{E}$. \r\n\r\nSince $\\angle{B}\\cong\\angle{E}$, $\\angle{B}\\cong\\angle{AE'F'}$.\r\n\r\nIf E'=B, then triangle ABC and AE'F' are the same triangle, and AB/DE = AC/DF because each fraction is equal to 1.\r\n\r\nIf E' and B are not equal, then lines E'F' and BC are parallel. It is not too hard to prove that the triangles must now be similar, by considering ratios of areas, same altitude, etc.", "Solution_4": "i have to say that the proof is pretty tricky...but cool...basically place the two triangles ontop of each other sides/vertex coincide, the other sides are parallel...the hard part is where to go from here...here is a hint: (as in try and figure it out now...) use area ratios and the parallel line shows two areas are equal...\r\n\r\nEDIT: this is basically what you want to do based on the last two sentences of the previous post" } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "homothety", "trigonometry", "Euler", "reflection" ], "Problem": "I have already posted this in [url=http://groups.yahoo.com/group/Hyacinthos/message/8510?expand=1]Hyacinthos message #8150[/url], but there was no reaction. Hope that this will be different here. The problem is really beautiful, and I have a rather nice synthetic solution. Here goes the problem:\r\n\r\nLet D, E, F be the feet of the three altitudes of a triangle ABC, and let D', E', F' be the midpoints of the segments AD, BE, CF. Let the line E'F' meet the lines AB and CA at the points $B_a$ and $C_a$, let the line F'D' meet the lines BC and AB at the points $C_b$ and $A_b$, and let the line D'E' meet the lines CA and BC at the points $A_c$ and $B_c$.\r\n\r\n[b](a) (original problem by A. Angelescu, Gazeta Matematica 35):[/b] If X', Y', Z' are the circumcenters of triangles $AB_aC_a$, $BC_bA_b$, $CA_cB_c$, respectively, then the points D', E', F', X', Y', Z' lie on one circle.\r\n\r\n[b](b) (my observations, which lead to a proof of (a)):[/b] Let L be the de Longchamps point of triangle ABC (this is the image of the orthocenter of triangle ABC in the homothety with center at the centroid of triangle ABC and factor -2). Let the lines AL, BL, CL meet the lines BC, CA, AB at the points X, Y, Z, respectively. Then, the points $B_a$, $C_a$ and D lie on the circle with diameter AX. Similarly for the circles with diameters BY and CZ.\r\n\r\n Darij", "Solution_1": "Denote $a = BC, b = CA, c = AB$ the triangle sides and $\\alpha = \\angle A, \\beta = \\angle B, \\gamma = \\angle C$ the triangle angles. WLOG, assume that the angle $\\gamma > \\alpha, \\beta$ is the largest one, the relatively trivial case of an isosceles triangle is disregarded. We will also assume that $CH < HF$, i.e., the points $C, H, F', F$ follow on the altitude $CF$ in this order. By considering an isosceles triangle with $CH = HF$,\r\n\r\n$CH = \\frac{a \\sin \\alpha}{2} = \\frac{a \\cos \\alpha}{\\tan \\alpha} = HF$\r\n\r\n$\\sin^2{\\alpha} = 2 \\cos^2{\\alpha},\\ \\ \\cos^2{\\alpha = \\frac 1 3,\\ \\ \\cos \\gamma = 1 - 2 \\cos^2{\\alpha} = \\frac 1 3,\\ \\ \\gamma \\doteq 70.53^o}$\r\n\r\nwe see that if $CH < HF$, $\\cos \\gamma < \\frac 1 3$ and $\\gamma > 70.53^o$ is the largest triangle angle. The case when $CH > HF$, while $\\gamma$ remains the largest triangle angle, is treated similarly.\r\n\r\n[color=red][b]Step 1:[/b][/color] Let $(O_B), (O_E)$ be the circumcircles of the triangles $\\triangle BE'F, \\triangle EE'F$ and let $A', B'$ be their intersections with the sides $CA, BC$ other than the points $E, B$, respectively. Using properties of the orthic triangle $\\triangle DEF$, the triangles $\\triangle EFA' \\sim \\triangle BFB'$ are similar, because the angles $\\angle A'EF = \\angle B'BF = \\beta$ are equal and the angles\r\n\r\n$\\angle BB'F = \\angle BE'F = 180^o - \\angle FE'E = \\angle EA'F$\r\n\r\nare also equal. The similarity coeficient of these 2 triangles is\r\n\r\n$\\frac{EF}{BF} = \\frac{a \\cos \\alpha}{a \\cos \\beta} = \\frac{\\cos \\alpha}{\\cos \\beta}$\r\n\r\nThe angles $\\angle EFA' = \\angle EE'A'$ spanning the same arc $FA'$ of the circle $(O_E)$ are equal and the angles $\\angle BFB' = \\angle BE'B'$ spanning the same arc $BB'$ of the circle $(O_B)$ are also equal. Due to similarity of the triangles $\\triangle EFA' \\sim \\triangle BFB'$, these angles are equal to each other, hence, the angles $\\angle EE'A' = \\angle BE'B'$ are vertical, which means that the points $A'E'B'$ are collinear. The triangles $\\triangle AFA' \\sim \\triangle DFB'$ are also similar, because the angles $\\angle FAA' = \\angle FDB' = \\alpha$ are equal and the angles $\\angle AA'F = 180^o - \\angle EA'F = 180^o - \\angle BB'F = \\angle DB'F$ are also equal. The similarity coefficient of these 2 triangles is also\r\n\r\n$\\frac{AF}{DF} = \\frac{b \\cos \\alpha}{b \\cos \\beta} = \\frac{\\cos \\alpha}{\\cos \\beta}$\r\n\r\nDenote $\\omega = \\angle BE'F$. Using the sine theorem for the triangle $\\triangle BFE'$,\r\n\r\n$\\frac{BF}{FE'} = \\frac{\\sin \\omega}{\\sin{(90^o - \\alpha)}} = \\frac{\\sin \\omega}{\\cos \\alpha}$\r\n\r\nThe segment $FE'$ is the median of the triangle $\\triangle BEF$ from the vertex $F$. Hence\r\n\r\n$FE'^2 = \\frac 1 4(2EF^2 + 2BF^2 - BE^2) =$\r\n\r\n$= \\frac{a^2}{4}(2\\cos^2{\\alpha} + 2 \\cos^2{\\beta} - \\sin^2{\\gamma}) = \\frac{a^2}{4}(\\cos^2{\\alpha} + \\cos^2{\\beta} - 2 \\cos \\alpha \\cos \\beta \\cos \\gamma)$\r\n\r\n$\\sin^2{\\omega} = \\frac{BF^2}{FE'^2}\\ \\cos^2{\\alpha} = \\frac{4 \\cos^2{\\alpha} \\cos^2{\\beta}}{\\cos^2{\\alpha} + \\cos^2{\\beta} - 2 \\cos \\alpha \\cos \\beta \\cos \\gamma}$\r\n\r\n$\\cot^2{\\omega} = \\frac{1 - \\sin^2{\\omega}}{\\sin^2{\\omega}} = \\frac{\\cos^2{\\alpha} + \\cos^2{\\beta} - 4 \\cos^2{\\alpha} \\cos^2{\\beta} + 2 \\cos^2 {\\alpha} \\cos^2{\\beta} - 2 \\sin \\alpha \\cos \\alpha \\sin \\beta \\cos \\beta}{4 \\cos^2{\\alpha} \\cos^2{\\beta}} =$\r\n\r\n$= \\frac{\\cos^2{\\alpha} (1 - \\cos^2{\\beta}) + \\cos^2{\\beta} (1 - \\cos^2{\\alpha}) - 2 \\sin \\alpha \\cos \\alpha \\sin \\beta \\cos \\beta}{4 \\cos^2{\\alpha} \\cos^2{\\beta}} =$\r\n\r\n$= \\frac 1 4 (\\tan^2{\\alpha} + \\tan^2{\\beta} - 2 \\tan \\alpha \\tan \\beta) = \\frac 1 4 (\\tan \\alpha - \\tan \\beta)^2$\r\n\r\nLet $C'$ be the intersection of the altitude $CF$ with the circle $(O_B)$ other than the point $F$. Since the angle $\\angle BFC' = 90^o$ is right, $BC' = 2O_BB$ is a diameter of the circle $(O_B)$ and\r\n\r\n$FC' = \\sqrt{BC'^2 - BF^2} = \\sqrt{4O_BB^2 - BF^2}$\r\n\r\nRadius of the circle $(O_B)$ is equal to $O_BB = \\frac{BF}{2 \\sin \\omega}$ and the power of the vertex $C$ to this circle is\r\n\r\n$CB \\cdot CB' = CF \\cdot CC',\\ \\ \\ CB (CB - BB') = CF (CF + FC')$\r\n\r\n$BB' = CB - \\frac{CF (CF + \\sqrt{4O_BB^2 - BF^2})}{CB} =$\r\n\r\n${= a - a (\\sin^2 \\beta + \\sin \\beta \\cos \\beta \\cot \\omega}) = a (\\cos^2{\\beta} - \\sin \\beta \\cos \\beta \\cot \\omega)$\r\n\r\nLet the line $A'B'$ cut the altitude $AD$ at a point $D''$. Using Menelaus' theorem for the triangle $\\triangle ACD$ cut by the line $A'B'$,\r\n \r\n$\\frac{A'A}{A'C} \\cdot \\frac{B'C}{B'D} \\cdot \\frac{D''D}{D''A} = 1$\r\n\r\n$\\frac{D''D}{D''A} = \\frac{A'C}{A'A} \\cdot \\frac{B'D}{B'C} = \\frac{\\cos \\beta}{\\cos \\alpha} \\cdot \\frac{A'C}{BD'} \\cdot \\frac{B'D}{B'C} = \\frac{\\cos \\beta}{\\cos \\alpha} \\cdot \\frac{A'E + EC}{BC - BB'} =$\r\n\r\n$= \\frac{\\cos \\beta}{\\cos \\alpha} \\cdot \\frac{BB' \\frac{\\cos \\alpha}{\\cos \\beta} + a \\cos \\gamma}{a - BB'} = \\frac{BB' \\cos \\alpha + a \\cos \\beta \\cos \\gamma}{(a - BB') \\cos \\alpha} =$\r\n\r\n$= \\frac{a (\\cos^2{\\beta} - \\sin \\beta \\cos \\beta \\cot \\omega) \\cos \\alpha + a \\cos \\beta \\cos \\gamma}{[a - a (\\cos^2{\\beta} - \\sin \\beta \\cos \\beta \\cot \\omega)] \\cos \\alpha} =$\r\n\r\n$= \\frac{(\\cos^2{\\beta} - \\sin \\beta \\cos \\beta \\cot \\omega) \\cos \\alpha + \\cos \\beta \\cos \\gamma}{(\\sin^2{\\beta} + \\sin \\beta \\cos \\beta \\cot \\omega) \\cos \\alpha} = \\frac{\\cot \\beta - \\cot \\omega - \\cot \\beta + \\tan \\alpha}{\\tan \\beta + \\cot \\omega} =$\r\n\r\n$= \\frac{\\tan \\alpha - \\cot \\omega}{\\tan \\beta + \\cot \\omega} = \\frac{\\tan \\alpha - \\frac 1 2 (\\tan \\alpha - \\tan \\beta)}{\\tan \\beta + \\frac 1 2 (\\tan \\alpha - \\tan \\beta)} = \\frac{\\frac 1 2 (\\tan \\alpha + \\tan \\beta)}{\\frac 1 2 (\\tan \\alpha + \\tan \\beta)} = 1$\r\n\r\nThus we proved that $D''A = D''D$, i.e., $D'' \\equiv D'$ is the midpoint of the altitude $AD$ lying on the line $A'B'$. Since the midpoint $E'$ of the altitude $BE$ also lies on the line $A'B'$, the points $A' \\equiv A_c$ and $B' \\equiv B_c$ are identical. As a result, the quadrilateral $CA_cFB_c$ is cyclic, because\r\n\r\n$\\angle A_cFB_c = \\angle A_cFE + \\angle EFD + \\angle DFB_c =$\r\n\r\n$= (180^o - 2 \\gamma) + \\angle A_cFE + \\angle AFA_c = 180^o - 2 \\gamma + \\gamma = 180^o - \\gamma$\r\n\r\nThis means that the circumcircle $(Z')$ of the triangle $\\triangle CA_cB_c$ passes through the foot of the altitude $CF$. Consequently, this circle is centered on the triangle midline $A_1B_1 \\parallel AB$, where $A_1, B_1$ are the midpoints of the sides $BC, CA$. Similarly, quadrilaterals $AB_aDC_a, BA_bEC_b$ are also cyclic, the circles $(X'), (Y')$ pass through the altitude feet $D, E$ and they are centered on the triangle midlines $B_1C_1 \\parallel BC, C_1A_1 \\parallel CA$, where $C_1$ is the midpoint of the remaining triangle side $AB$. The power of the orthocenter $H$ to the circlec $(X'), (Y'), (Z')$ is $HA \\cdot HD = HB \\cdot HE = HC \\cdot HF$, i.e., the orthocenter is the radical center of these 3 circles.\r\n\r\nTo be continued.\r\n\r\nYetti", "Solution_2": "[color=red][b]Step 2:[/b][/color] Earlier, we found that\n\n$BB_c = a (\\cos^2{\\beta} - \\sin \\beta \\cos \\beta \\cot \\omega) = a \\left(\\cos^2{\\beta} - \\sin \\beta \\cos \\beta\\ \\frac{\\tan \\alpha - \\tan \\beta}{2}\\right)$\n\nBy cyclic exchange, we obtain\n\n$CC_a = b \\left(\\cos^2 \\gamma - \\sin \\gamma \\cos \\gamma\\ \\frac{\\tan \\beta - \\tan \\gamma}{2}\\right)$\n\nComparing the segments $CC_a, AA_c$, we have\n\n$DB_c = BD - BB_c = c \\cos \\beta - a (\\cos^2{\\beta} - \\sin \\beta \\cos \\beta \\cot \\omega) =$\n\n$= a \\left(\\frac{\\sin \\gamma}{\\sin \\alpha} \\cos \\beta - \\cos^2{\\beta} + \\sin \\beta \\cos \\beta \\cot \\omega \\right) =$\n\n$= a (\\cos^2{\\beta} + \\cot \\alpha \\sin \\beta \\cos \\beta - \\cos^2{\\beta} + \\sin \\beta \\cos \\beta \\cot \\omega) =$\n\n$= a \\sin \\beta \\cos \\beta \\left(\\cot \\alpha + \\frac{\\tan \\alpha - \\tan \\beta}{2}\\right)$\n\nRemoving the angle $\\alpha$ from the expression for $AA_c$,\n\n$AA_c = \\frac{\\cos \\alpha}{\\cos \\beta}\\ DB_c = b \\sin \\alpha \\cos \\alpha \\left(\\cot \\alpha + \\frac{\\tan \\alpha - \\tan \\beta}{2}\\right) = $\n\n$= b \\left(\\cos^2{\\alpha} + \\frac{\\sin^2{\\alpha}}{2} - \\sin \\alpha \\cos \\alpha\\ \\frac{\\tan \\beta}{2}\\right) = $\n\n\n$= \\frac b 2 [1 + (\\cos \\beta \\cos \\gamma - \\sin \\beta \\sin \\gamma)^2\\ +$ \n\n${+ (\\sin \\beta \\cos \\gamma + \\cos \\beta \\sin \\gamma)(\\cos \\beta \\cos \\gamma - \\sin \\beta \\sin \\gamma) \\tan \\beta}] =$\n\n\n$= \\frac b 2 [1 + \\cos^2{\\beta} \\cos^2{\\gamma} - 2 \\sin \\beta \\sin \\gamma \\cos \\beta \\cos \\gamma + \\sin^2{\\beta}\\sin^2{\\gamma}\\ +$ \n\n$+ (\\sin \\beta \\cos \\beta \\cos^2{\\gamma} - \\sin^2{\\beta} \\sin \\gamma \\cos \\gamma + \\cos^2{\\beta} \\sin \\gamma \\cos \\gamma - \\sin \\beta \\cos \\beta \\sin^2{\\gamma}) \\tan \\beta] =$\n\n\n$= \\frac b 2 (1 + \\cos^2{\\beta} \\cos^2{\\gamma} - 2 \\sin \\beta \\sin \\gamma \\cos \\beta \\cos \\gamma + \\sin^2{\\beta}\\sin^2{\\gamma}\\ +$ \n\n$+ \\sin^2{\\beta} \\cos^2{\\gamma} - \\sin^2{\\beta} \\tan \\beta \\sin \\gamma \\cos \\gamma + \\sin \\beta \\cos \\beta \\sin \\gamma \\cos \\gamma - \\sin^2{\\beta} \\sin^2{\\gamma}) =$\n\n\n$= \\frac b 2 (1 + \\cos^2{\\gamma} - \\sin \\beta \\sin \\gamma \\cos \\beta \\cos \\gamma - \\sin^2{\\beta} \\tan \\beta \\sin \\gamma \\cos \\gamma) =$\n\n$= \\frac b 2 [2 \\cos^2{\\gamma} + \\sin \\gamma \\cos \\gamma (\\tan \\gamma - \\sin \\beta \\cos \\beta - \\sin^2{\\beta} \\tan \\beta)] =$\n\n$= b \\left(\\cos^2{\\gamma} - \\sin \\gamma \\cos \\gamma \\frac{\\tan \\beta - \\tan \\gamma}{2}\\right) = CC_a$\n\nThus we proved that the segments $AA_c = CC_a$ are not only equal, but either both inside or both outside of the triangle side $CA$ (this follows from the signs of the tangents). Similarly, we could show that the segments $BB_c = CC_b, AA_b = BB_a$ are equal and either both inside or both outside of the triangle sides $BC, AB$. As a result, the powers of the midpoint $C_1$ of the side $AB$ to the circles $(X'), (Y')$ are equal. For example, if the points $A_b, B_a$ are both outside of the side $AB$,\n\n$C_1A \\cdot C_1B_a = C_1A (C_1B + BB_a) = C_1B (C_1A + AA_b) = C_1B \\cdot C_1A_b$\n\nand consequently, the radical axis of the circles $(X'), (Y')$ passes through the midpoint $C_1$. Similarly, the radical axes of the circle pairs $(Y'), (Z')$ and $(Z'), (X')$ pass through the midpoints $A_1, B_1$ of the sides $BC, CA$.\n\nTo be continued.\n\nYetti", "Solution_3": "[color=red][b]Step 3:[/b][/color] The angles $\\angle B_cA_cF \\equiv \\angle E'A_cF = \\angle E'EF \\equiv \\angle BEF$ and the angles $\\angle A_cB_cF \\equiv \\angle E'B_cF = \\angle E'BF \\equiv \\angle EBF$ spanning the same arcs $E'F$ of the circles $(O_B), (O_E)$ are equal. Hence, the triangles $\\triangle ABH \\sim \\triangle EFB \\sim \\triangle AFD \\sim \\triangle A_cFB_c$ are all similar, because\n\n$\\angle FAD \\equiv \\angle BAH = 90^o - \\beta,\\ \\ \\angle B_cA_cF = \\angle BEF = 90^o - \\angle FEA = 90^o - \\beta$\n\n$\\angle FBE \\equiv \\angle ABH = 90^o - \\alpha,\\ \\ \\angle A_cB_cF = \\angle EBF \\equiv \\angle EBA = 90^o - \\alpha$\n\nThe segments $FE', HC_1$ are the corresponding medians of the similar triangles $\\triangle EFB \\sim \\triangle ABH$, hence, the angles $\\angle FC_1H \\equiv BC_1H = \\angle FE'E = 180^o - \\angle FE'B = \\omega$ are equal. The angles $\\angle FE'B = \\angle FB_cB = 180^o - \\omega$ spanning the the same arc $BF$ of the circle $(O_B)$ are also equal. Let the circle $(Z')$ intersect the line $AB$ at a point $Z$ different from $F$. Since the angle $\\angle CFZ \\equiv CFB = 90^o$ is right, $CZ$ is a diameter of the circle $(Z')$. The angles $\\angle FZC = \\angle FB_cC = 180^o - \\angle FB_cB = \\omega$ are also equal. Thus the right angle triangles $\\triangle CFZ \\sim \\triangle HFC_1$ are centrally similar with the similarity center $F$ and consequently, the lines $HC_1 \\parallel CZ$ are parallel. As a result, the circumcenter $Z'$ of the triangle $\\triangle CA_cB_c$ is identical with the intersection of the midline $A_1B_1$ with the line $CZ \\parallel HC_1$ parallel to the radical axis of the circles $(X'), (Y')$. Since the cevians $A_1H, B_1H, C_1H$ of the medial triangle $\\triangle A_1B_1C_1$, the pairwise radical axes of the circles $(X'), (Y'), (Z')$, meet at the orthocenter $H$ of the original triangle $\\triangle ABC$, by central similarity of these 2 triangles with the similarity center at the triangle centroid $G$ and similarity coefficient -2, the cevians $AX, BY, CZ$ of the triangle $\\triangle ABC$ parallel to $A_1H, B_1H, C_1H$ meet at a point $L$ on the line $GH$ (Euler line) opposite to the orthocenter $H$, such that $\\frac{GH}{LH} = 2$ (Longchamps point, never heard of it) or $\\frac{OH}{OL} = 1$, i.e., the reflection of the orthocenter $H$ in the circumcenter $O$. Since the angles $\\angle ADX = \\angle BEY = \\angle CFZ = 90^o$ spanned by the appropriate arcs $AX, BY, CZ$ of the circumcircles $(X'), (Y'), (Z')$ are right, the circumcenters $X', Y', Z'$ are the midpoints of the cevians $AX, BY, CZ$. \n \nThe right angle triangle $\\triangle CFZ$ is half the size of the right angle triangle $\\triangle CF'Z'$, hence, the triangles $\\triangle CFZ \\sim \\triangle HFC_1$ are also similar with the similarity coefficient\n\n$\\frac{CF'}{HF} = \\frac{CF}{2HF} = \\frac{a \\sin \\beta}{2 a \\cos \\beta \\cot \\alpha} = \\frac{\\tan \\alpha \\tan \\beta}{2}$\n\n$Z'F' = \\frac{\\tan \\alpha \\tan \\beta}{2}\\ C_1F = \\frac{\\tan \\alpha \\tan \\beta}{2} \\left(a \\cos \\beta - \\frac c 2\\right) = \\frac{a \\tan \\alpha \\tan \\beta}{4} \\left(2 \\cos \\beta - \\frac{\\sin \\gamma}{\\sin \\alpha}\\right) =$\n\n$= \\frac{a \\tan \\alpha \\tan \\beta}{4} \\left(2 \\cos \\beta - \\cos \\beta - \\frac{\\sin \\beta}{\\tan \\alpha}\\right) = \\frac{a \\sin \\beta}{4}(\\tan \\alpha - \\tan \\beta)$\n\n$A_1Z' = A_1F' - Z'F' = \\frac{a \\cos \\beta}{2} - \\frac{a \\sin \\beta}{4}(\\tan \\alpha - \\tan \\beta) =$\n\n$= \\frac{a}{4}[\\cos \\beta (1 - \\tan \\alpha \\tan \\beta) + \\cos \\beta (1 + \\tan^2{\\beta}) ] = \\frac{a}{4 \\cos \\beta} [\\cos^2{\\beta} (1 - \\tan \\alpha \\tan \\beta) + 1]$ \n \n$A_1F' \\cdot A_1Z' = \\frac{a \\cos \\beta}{2} \\cdot \\frac{a}{4 \\cos \\beta} [\\cos^2{\\beta} (1 - \\tan \\alpha \\tan \\beta) + 1] =$\n\n$= \\frac{a^2}{8} [\\cos^2{\\beta} (1 - \\tan \\alpha \\tan \\beta) + 1] = \\frac{a^2}{8} \\left(1 - \\frac{\\cos \\beta \\cos \\gamma}{\\cos \\alpha}\\right)$\n\nLet $Y$ be the intersection of the circle $(Y')$ with the line $CA$ other that the points $E$. For similar reasons as above, the right angle triangles $\\triangle BE'Y' \\sim \\triangle BEY \\sim \\triangle HEB_1$ are similar, the similarity coefficient of $\\triangle BE'Y' \\sim \\triangle HEB_1$ equal to\n\n$\\frac{BE'}{HE} = \\frac{BE}{2HF} = \\frac{a \\sin \\gamma}{2 a \\cos \\gamma \\cot \\alpha} = \\frac{\\tan \\gamma \\tan \\alpha}{2}$\n\n$Y'E' = \\frac{\\tan \\gamma \\tan \\alpha}{2}\\ B_1E = \\frac{\\tan \\gamma \\tan \\alpha}{2} \\left(\\frac b 2 - a \\cos \\gamma\\right) = \\frac{a \\tan \\gamma \\tan \\alpha}{4} \\left(\\frac{\\sin \\beta}{\\sin \\alpha} - 2 \\cos \\gamma\\right) =$\n\n$= \\frac{a \\tan \\gamma \\tan \\alpha}{4} \\left(\\frac{\\sin \\gamma}{\\tan \\alpha} + \\cos \\gamma - 2\\cos \\gamma\\right) = \\frac{a \\sin \\gamma}{4} \\left(\\tan \\gamma - \\tan \\alpha\\right)$\n\n$A_1Y' = AE' + Y'E' = \\frac{a \\cos \\gamma}{2} + \\frac{a \\sin \\gamma}{4}(\\tan \\gamma - \\tan \\alpha) =$\n\n$= \\frac{a}{4}[\\cos \\gamma (1 + \\tan^2{\\gamma) + \\cos \\beta (1 - \\tan \\gamma \\tan \\alpha) ] = \\frac{a}{4 \\cos \\gamma} [\\cos^2{\\gamma} (1 - \\tan \\gamma \\tan \\alpha) + 1]}$ \n \n$A_1E' \\cdot A_1Y' = \\frac{a \\cos \\gamma}{2} \\cdot \\frac{a}{4 \\cos \\gamma} [\\cos^2{\\gamma} (1 - \\tan \\gamma \\tan \\alpha) + 1] =$\n\n$= \\frac{a^2}{8} [\\cos^2{\\gamma} (1 - \\tan \\gamma \\tan \\alpha) + 1] = \\frac{a^2}{8} \\left(1 - \\frac{\\cos \\beta \\cos \\gamma}{\\cos \\alpha}\\right)$\n\nThus we proved that $A_1E' \\cdot A_1Y' = A_1F' \\cdot A_1Z'$, i.e., that the quadrilateral $E'F'Y'Z'$ is cyclic. In an entirely similar way, we can show that $B_1F' \\cdot B_1Z' = B_1D' \\cdot B_1X'$ and $C_1D' \\cdot C_1X' = C_1E' \\cdot C_1Y'$, i.e., that the quadrilaterals $F'D'Z'X', D'E'X'Y'$ are also cyclic. This still leaves the possibility that the circumcircles of these 3 cyclic quadrilaterals are different form each other and from the circumcircle of the triangle $\\triangle D'E'F'$. Assume that the circumcircles $(P_A), (P_B)$ of the cyclic quadrilaterals $E'F'Y'Z', F'D'Z'X'$ intersecting at the points $F', Z'$ are different from each other. Since $C_1D' \\cdot C_1X' = C_1E' \\cdot C_1Y'$, the point $C_1$ has the same power to the circles $(P_A), (P_B)$, which means that it lies on their radical axis $F'Z' \\equiv A_1B_1$, i.e., that the midpoints $A_1, B_1, C_1$ of the triangles sides $BC, CA, AB$ are collinear, which is clearly impossible. As a result, the circumcircles of the cyclic quadrilaterals $E'F'Y'Z', F'D'Z'X', D'E'X'Y'$ are identical. This concludes the proof of the problem proposition.\n\nPerhaps we should mention the case, when the angle $\\gamma = 90^0$ is right. The triangle $\\triangle D'E'F'$ then degenerates into a line segment, its circumcircle into the midline $A_1B_1$ of the right angle triangle $\\triangle ABC$, the circles $(X'), (Y')$ degenerate into the lines $CA, BC$ and the circle $(Z')$ becomes identical with the triangle 9-point circle. Hence, the circumcenters $X', Y'$ move to infinity and the circumcenter $Z'$, the midpoint of the segment $CC_1 \\equiv OH$ lies on the midline $A_1B_1$, the degenerate circumcircle of the degenerate triangle $\\triangle D'E'F'$. \n\nTo be continued.\n\nYetti", "Solution_4": "[color=red][b]Bonus:[/b][/color] Earlier, we proved that the circumcenters $X', Y', Z'$ lie of the (extended) triangle midlines $B_1C_1, C_1A_1, A_1B_1$. The midpoints $D', E', F'$ of the altitudes $AD, BE, CF$ also lie on these midlines. The cevians $A_1X', B_1Y', C_1Z'$ of the medial triangle $\\triangle A_1B_1C_1$ cut its (extended) sides at the points $X', Y', Z'$ in the ratios $\\frac{X'B_1}{X'C_1},\\ \\frac{Y'C_1}{Y'A_1},\\ \\frac{Z'A_1}{Z'B_1}$.\r\n\r\n$Z'B_1 = A_1B_1 - Z'A_1 = \\frac c 2 - \\frac{a}{4 \\cos \\beta} \\left[\\cos^2{\\beta} (1 - \\tan \\alpha \\tan \\beta) + 1\\right] = \\frac a 4 \\left(\\frac{2 \\sin \\gamma}{\\sin \\alpha} - \\cos \\beta (1 - \\tan \\alpha \\tan \\beta) - \\frac{1}{\\cos \\beta}\\right) =$\r\n\r\n$= \\frac a 4 \\left(2 \\cos \\beta + 2 \\sin \\beta \\cot \\alpha - \\cos \\beta + \\tan \\alpha \\sin \\beta - \\frac{1}{\\cos \\beta}\\right) =$\r\n\r\n$= \\frac{a \\sin \\beta}{4} (2 \\cot \\alpha + \\tan \\alpha - \\tan \\beta) = \\frac{a \\sin \\beta}{4 \\sin \\alpha \\cos \\alpha} \\left[\\cos^2{\\alpha}(1 - \\tan \\alpha \\tan \\beta) + 1\\right]$\r\n\r\n$\\frac{Z'A_1}{Z'B_1} = \\frac{\\sin \\alpha \\cos \\alpha}{\\sin \\beta \\cos \\beta } \\cdot \\frac{\\cos^2{\\beta} (1 - \\tan \\alpha \\tan \\beta) + 1}{\\cos^2{\\alpha} (1 - \\tan \\alpha \\tan \\beta) + 1}$\r\n\r\nBy cyclic exchange (and paying attention to the signs of the tangents),\r\n\r\n$\\frac{X'B_1}{X'C_1} \\cdot \\frac{Y'C_1}{Y'A_1} \\cdot \\frac{Z'A_1}{Z'B_1} =$\r\n\r\n$= \\frac{\\cos^2{\\gamma} (1 - \\tan \\beta \\tan \\gamma) + 1}{\\cos^2{\\beta} (1 - \\tan \\beta \\tan \\gamma) + 1} \\cdot \\frac{\\cos^2{\\alpha} (1 - \\tan \\gamma \\tan \\alpha) + 1}{\\cos^2{\\gamma} (1 - \\tan \\gamma \\tan \\alpha) + 1} \\cdot \\frac{\\cos^2{\\beta} (1 - \\tan \\alpha \\tan \\beta) + 1}{\\cos^2{\\alpha} (1 - \\tan \\alpha \\tan \\beta) + 1} =$ [color=white].[/color]$1$\r\n\r\nbecause all terms cancel out, for example,\r\n\r\n$(?)\\ \\ \\cos^2{\\gamma} (1 - \\tan \\beta \\tan \\gamma) + 1 = \\cos^2{\\alpha} (1 - \\tan \\alpha \\tan \\beta) + 1$\r\n\r\n$(?)\\ \\ \\cos^2{\\gamma} - \\cos^2{\\alpha} = (\\sin \\gamma \\cos \\gamma - \\sin \\alpha \\cos \\alpha) \\tan \\beta = \\frac{(\\sin \\gamma \\cos \\gamma - \\sin \\alpha \\cos \\alpha) (\\sin \\alpha \\cos \\gamma + \\cos \\alpha \\sin \\gamma)}{\\cos \\beta}$\r\n\r\n$(?)\\ \\ \\cos^2{\\gamma} - \\cos^2{\\alpha} = \\frac{(\\sin^2{\\gamma} - \\sin^2{\\alpha}) \\cos \\alpha \\cos \\gamma + (\\cos^2{\\gamma} - \\cos^2{\\alpha}) \\sin \\alpha \\sin \\gamma}{\\cos \\beta}$\r\n\r\n$(?)\\ \\ \\cos^2{\\gamma} - \\cos^2{\\alpha} = \\frac{(\\cos^2{\\gamma} - \\cos^2{\\alpha}) (\\sin \\alpha \\sin \\gamma - \\cos \\alpha \\cos \\gamma)}{\\cos \\beta}$\r\n\r\nand the last equation is an obvious identity. By Ceva's theorem, the cevians $A_1X', B_1Y', C_1Z'$ of the medial triangle $\\triangle A_1B_1C_1$ concur at a point $Q$, so-called Yetti's point of the triangle $\\triangle ABC$ (hey, everybody's got a triangle point and there's plenty more available :rotfl:).\r\n\r\n\r\n[color=blue][b]Proposition:[/b][/color] Prove that the common circumcenter $P$ of the triangles $\\triangle D'E'F', \\triangle X'Y'Z'$ lies on the line $QH$ connecting Yetti's point $Q$ with the triangle orthocenter $H$.\r\n\r\nYetti\r\n\r\n\r\n[b]Note:[/b] All trigonometric formulas in the entire proof check with the Sketchpad.", "Solution_5": "http://forumgeom.fau.edu/FG2004volume4/FG200405index.html\r\n\r\n The above article by JPE may bring some clearity into the subject roughed-up by previous\r\n replies by our own Yetti. A complete quad where the 4th side is a line thru midpoints of\r\n altitudes in a triangle made by the other 3 lines should yield some interesting properties.\r\n\r\n\r\n\r\n Maj. Pestich", "Solution_6": "I still hope that in some geo problem the Fox-Talbot theorem will be used. \r\n Is it possible this is the one?\r\n\r\n\r\n\r\n Maj. Pestich", "Solution_7": "[quote=\"pestich\"] The above article by JPE may bring some clearity into the subject roughed-up by previous\n replies by our own Yetti. A complete quad where the 4th side is a line thru midpoints of\n altitudes in a triangle made by the other 3 lines should yield some interesting properties.\n\n\n\n Maj. Pestich[/quote]\r\n\r\nLet's make one thing clear, comrade Pestich. You do not own me. \r\n\r\nAs for the problem, I solved it, not roughed up. If the solution was quick and easy, those guys at Hyacintos would take it apart in no time, trust me. If you do not understand the concept of solution (something you never will), then shut up. It seems that you just cannot let go and are looking for another trashing. Please seek help.\r\n\r\nYetti", "Solution_8": "My dear Yetti,\r\n\r\n You do not have to rough me up. I was the first one to agree that it was a wonderful solution indeed.\r\n Put your customary doubts about your own (I had to use this word again, sorry) solutions aside: this\r\n time it is a bona fide one. I'd say it is even more than a solution to a tough problem (those people at Hya-group\r\n wouldda never imagined the circumcenters on midlines), it is a novel to read, with a hero, villans, chapters, \r\n paragraphs and such. But you are a bit lazy, and if I do not tell you, nobody will. Where are the table of contents, \r\n index, and referrences? \r\n \r\n\r\n Sincerely,\r\n\r\n Maj. Pestich", "Solution_9": "[quote=\"pestich\"]My dear Yetti,\n\n You do not have to rough me up. I was the first one to agree that it was a wonderful solution indeed.\n Put your customary doubts about your own (I had to use this word again, sorry) solutions aside: this\n time it is a bona fide one. I'd say it is even more than a solution to a tough problem (those people at Hya-group\n wouldda never imagined the circumcenters on midlines), it is a novel to read, with a hero, villans, chapters, \n paragraphs and such. But you are a bit lazy, and if I do not tell you, nobody will. Where are the table of contents, \n index, and referrences? \n \n\n Sincerely,\n\n Maj. Pestich[/quote]\r\n\r\nNow you are highly embarrassing :blush: :blush: :blush: ... Please, don't try to be funny, try to be serious and prove something, because that makes you hilarious. I fondly remember your exquisite proof that [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=387&forum=DCForumID6#11]all quadrilaterals with perpendicular diagonals are concyclic[/url] :rotfl:. Trust me, I am not the only one on this forum, who knows that Golland = Pestich. Or when you [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=391&forum=DCForumID6#17]projected two parallel lines to infinity[/url] :rotfl:. Or when you repeatedly made a claim about [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=390&forum=DCForumID6#1]poles and polars[/url] equivalent to $a+b = b+a\\ \\Rightarrow\\ a = b$ :rotfl:. But let's make a table, if that's what you miss. It is the table of [b]proofs without words[/b] by no other than comrade Pestich.\r\n\r\n \r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16824&start=11]projections which lie on a circle[/url], reply 11\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18084&start=6]Perpendicular projections and acute-angled triangle[/url], replies 6, 8, 10, 12\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19133]Interesting geometry[/url], replies 4, 6\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19178]The multisolution problem![/url], reply 6\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16213]Poland 3[/url], replies 3, 9\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19770]IMO ShortList 1999, geometry problem 2[/url], reply 5\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3028]Pizza everybody!!![/url], reply 11\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14741]posted before? [a circumcircle tangent to the incircle][/url], reply 4\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19500]GEOMETRY[/url], replies 4, 6\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19718]Cyclic[/url], reply 5\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22633]Thailand TST[/url], replies 7, 11, 14\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23066]concurrent [PD, QE, RF concur at the incenter of ABC][/url], reply 4\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19468]Inradii problem[/url], reply 3\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5799]Concurrence of AA', BN and CM in the Gergonne configuration[/url], reply 17 [moderator edit: i've merged this topic and deleted pestich's post. darij]\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28877]Special triangle[/url], replies 6, 9\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29333]HELP!! [related to the nine-point circle of triangle ABC][/url], reply 3\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29782]Triangle center[/url], replies 4, 10\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30002]MN||OO'[/url], reply 3\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30036]Triangle + Circle problem[/url], replies 4, 9, 12\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30195]cyclic[/url], reply 3\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31420]Concurent[/url], reply 3\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31077]Locus problem [2][/url], reply 2\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18426]Please prove elementarily[/url], reply 2\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31921]convex quadrilateral and centroids[/url], reply 8\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32251]an easy problem!!!![/url], reply 3\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32412]Perpendicular Distances[/url], reply 3\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33050]actually a physics problem[not hard][/url], reply 2\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33174]Orthopole on incircle[/url], replies 3, 5\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33149]Please solve my problem!!![/url], reply 7\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33278]Circumcenter of ABC is same as incenter of CMN[/url], reply 2\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32582]acute triangle[/url], reply 4\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33205]Please help to prove this[/url], replies 3, 5, 7, 9\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33514]cyclic quadrilateral[/url], replies 8, 13 \r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33948]Angle bisector[/url], reply 4, 7\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34000]Circumcenter on radical axis[/url], replies 3, 6\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34316]classical triangle geo - points on circle[/url], reply 4\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30611]Simson line property[/url], reply 5\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36132]ugly geometry[/url], reply 6\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=37964]Unique[/url], reply 3\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4450]constant angle[/url], reply 3\r\n\r\n\r\nThat's about 22% - not bad for proofs without words, although these are a bit sketchy and I don't even think that you are correct in a single one. No claims are made as to the completeness of this table, it may and will grow at any time.\r\n\r\nLooking forward to more stuff from you, but please, include some math content (just to make it funny).\r\n\r\n:rotfl: :rotfl: :rotfl: \r\nYetti", "Solution_10": "jesus christ , jesus christ yetti\r\ni havent seen a stonger one in calculating than you in my life even never heard of\r\nand this skill of you is so valuable but please dont take these in a bad manner but \r\ndo you think that all of these are necessary :?: :?: :?: :?: \r\nin some cases some uses of geometrical ideas would really make the calculations shorter, but in all the ways no one can disagree your abilities :omighty: \r\n:omighty: :omighty: :omighty: :omighty: :omighty:", "Solution_11": "No, I do not think that all these calculations are necessary. The fact that everything came out in closed forms (i.e., I did not have to resort anywhere to numerical calculations) indicates that the problem has underlying symmetries, which can be used in the solution (reflections in lines?). After I identified the basic steps of the proof, I was not able to see those symmetries, hence, I used trigonometry. But trigonometry itself is a (powerful) [b]geometric[/b] tool. If the sine and cosine functions are defined as the ratios of the opposite and adjacent legs of a right angle triangle to the hypotenuse, geometric proofs of any trigonometric formulas do exist. Trigonometry works even if nothing else does, i.e., if the problem does not have underlying symmetries and you are willing to use numerical methods to get the result. Moreover, after the solution is found by any means, for example by trigonometry, it might be easier for someone else to see the underlying symmetries. In this particular case, it is a mute proposition, because the problem was not posted in the Unsolved Problems section and Darij said that he already has a synthetic solution.\r\n\r\nI will give you an example from my line of work. Suppose I have to transport a particle beam in an accelerator over huge distances in a beamline of a rather small diameter without losing too much by hitting the beamline walls. The beamline has to be evacuated and any increase in the beamline diameter calls for larger magnets, the principal item in the cost of building an accelerator. So, the beamline diameter really has to be rather small. The phase space volume of the beam is the \"volume\" in the space with one coordinate being the beam displacement from the beamline axis and the other coordinate the particle radial velocity toward or away from the beamline axis, proportional to the particle divergence, the angle between the particle path and the beamline axis. This phase space volume is called the beam emittance and it looks like an ellipse tilted to these coordinate axes. From Liouville's theorem of analytical mechanics, it follows that this space volume remains constant, i.e., it cannot be decreased by any focusing element in the beamline. The ellipse can be only rotated or \"squashed\" without decreasing its area. Consequently, to get a good beam, which can be successfully transported over those huge distances, the beam emittance has to be small to begin with, as the beam emerges from the ion source. Also, the beam emittance is an important beam parameter to measure. The measurement methods depend on whether the beam can be \"consumed\" for the measurement (for a while), or whether it has to be done on the fly and usually, they are very ingenious. The beam focusing elements are almost exclusively magnetic quadrupole lenses, because at relatively low particle energies, they overcome the focusing capabilities of other elements, such as electrostatic lenses, etc. The action of a magnetic quadrupole lens on the beam is described by a matrix operator with elements that are either normal or hyperbolic trigonometry functions. You can have tens to hundreds of these lenses, which rotate and squash the emittance ellipse in various ways. One of the non-destructive methods of measuring the beam emittance consists of slowly moving a thin wire across the beam and measuring the (very small) beam current. In principle, this yields the position of the vertical tangent to the emittance ellipse at this spot. Several thin wires at several different places in between the quadrupole lenses can be moved simultaneously, thus obtaining several vertical tangents to several different ellipses, which are all images of some original ellipse transformed by the quadrupole lenses. Using the inverse operators for the quadrupole lenses, these tangents can be transformed back to one spot and in this transformation, they become rotated and no longer vertical. To get the beam emittance, an ellipse has to be inscribed into these tangents and its area calculated. For this, at least 3 different tangents are necessary (because the ellipse center is known), but since all measurements carry some errors, more than 3 tangents are measured and the best possible ellipse is inscribed into these lines by the least squares method. This is an interesting application of geometry and I doubt that it could be done without both the normal and hyperbolic trigonometry.\r\n\r\nSo, [color=blue][b]three cheers for trigonometry !!![/b][/color] :thumbup: :thumbup: :thumbup:\r\n\r\nRegards, Yetti", "Solution_12": "Smeenk show (in Crux Mathematicorum, the problem nr. 2379/98) that the lines MD', NE', PF' are conccurent, where M, N, P are the middlepoints of the sides BC, CA, AB respectively. This fact may be used ?", "Solution_13": "After a few hours, I finally got a purely synthetic solution to Darij 's observation. It seems that I am getting very close to proving the 3 circumcenters lying on the circumcirlce of $D'E'F'$. I will post the solution in a next few days. I am very busy right now.", "Solution_14": "Finally, everything is clear. And here is the solution (guess this is the longest I have ever posted). I won't write just only the solutions of this problem but also others nice theorems relating to this wonderful sketch.\r\n So, let me first redenote some points so that the problem accords with my figure. (I am sure you don't mind :) ).\r\n Let $ABC$ be a triangle whose circumcenter and orthocenter are $O$ and $H$. Let L be the de Longchamp point, i.e the point on $HO$ such that $\\frac{OL}{OH} = -1$ \r\n Let $A'B'C'$ be the orthic triangle and $MNP$ be the medial triangle. Call $D, E, F$ the midpoints of $AA', BB', CC'$. Suppose the line $EF$ intersects the line $AB$ and $AC$ at $A_b$ and $A_c$ respectively. Analogously, we define the point $B_a$, $B_c$, $C_a$, $C_b$.\r\n Let $O_a, O_b, O_c$ be the circumcenters of $AA{}_bA{}_c, BB{}_aB{}_c, CC{}_aC{}_b$ respectively.\r\n All right ! Let 's do some small observation of the de Longchamp point for a moment. Well, by calling $T$ the antipode of $A$ in the circle $(O)$., we have $AHTL$ is a parallelogram since the diagonals bisect each other. On the other hand, we have $BHCT$ is a parallelogram as its opposite sides are parallel. Hence $H, M, T$ are collinear and therefore $HM // AL$. Suppose we call $O_a'$ and $A_1$ the intersections of $AL$ with $NP$ and $BC$, then surely we have $O_a'$ is the midpoint of $AA_1$. Thus if we call $U$ the midpoint of $HM$, since $HM // AA_1$,we obtain that $A', U, O_a'$ are collinear. (Later, we shall see that this point $O_a'$ is the point $O_a$). So let 's official state it as a theorem :\r\n [b] Theorem 1[/b] $A, U, O_a'$ are collinear.\r\n Let 's next consider the point $U$. Since $\\measuredangle{MEH} = \\measuredangle{MFH} = \\measuredangle{MA'H} = 90^0$, we obtain five points $H, E, A', M, E$ lie on the circle $(U)$ with diameter $HM$. Therefore $\\measuredangle{A'EF} = \\measuredangle{A'HF} = \\measuredangle{A'BA_b}$. This implies that $BA'EA_b$ is cylic. By similar argument, we obtain that $CA'FA_c$ is cyclic and we have a theorem\r\n [b]Theorem 2[/b] $BA'EA_b$ is inscribed in $(b)$ and $CA'FA_c$ is inscribed in $(c)$\r\n Let 's derive some more properties from this cool theorem. From theorem 2, we have $\\measuredangle{A'A{}_bA{}_c} = \\measuredangle{A'A_bE} = \\measuredangle{A'BE} = \\measuredangle{A'AA_c}$. Hence $AA_bA'A_c$ is cyclic, i.e the circumcircle $(O_a)$ of $AA{}_bA{}_c$ passes through $A'$. This means that $O_a$ lies on $NP$. Now let 's prove that this point lies on $AL$ by showing that $AO_a // HM$. Let $A_1$ be the second intersection different from $A'$ of the circle $(O_a)$ and $BC$. Then $AA_1$ is the diameter of $(O_a)$. Hence $\\measuredangle{A{}_1A{}_cC} = 90^0$. Now according to theorem 1 and 2, we have $\\measuredangle{A'AA_1} = \\measuredangle{A'A{}_cA{}_1} = \\measuredangle{A'A{}_c{}_C} - 90^0 = \\measuredangle{A'FC} - 90^0 = \\measuredangle{A'FM} = \\measuredangle{A'HM}$. This yields that $HM // AA_1$, and therefore $AO_a$ passes through the de Longchamp point $L$. And Darij 's observation is already proved. Let 's state it\r\n [b] Theorem 3[/b] The circumcircle $(O_a)$ of $AA{}_bA{}_c$ passes through $A'$ and $A, O_a, L$ are collinear.\r\n In my next posts, we will give some more observations on the relations of this circle with the circle $(U)$, $(b)$ and $(c)$.", "Solution_15": "OK. Let 's continue with our work.\r\n We have $\\measuredangle{A_bA'F} = \\measuredangle{A_bA'E} + \\measuredangle{EA'F} = \\measuredangle{A_bBE} + \\measuredangle{EHF} = \\measuredangle{ABB'} + \\measuredangle{B'AB} = 90^0$. Hence $A_bA'FC'$ is cylic. Similarly, we have $A'EB'A_c$ is cylic.\r\n [b]Theorem 4[/b] $A_bA'FC'$ and $A'EB'A_c$ are cylic. \r\n By applying the Menelaus 's theorem to triangle $ABB'$ and three collinear points $A_b, E, A_c$, we have $\\frac{AA_b}{AA_c} = \\frac{A_bB}{B'A_c}$. On the other hand, we have $\\frac{A_bB}{A'A_b} = \\frac{sin{ \\measuredangle{BA'A_b}}}{sinB} = \\frac{sin{ \\measuredangle{BEA_b}}}{sinB} = \\frac{sin{ \\measuredangle{B'A'A_c}}}{sin{ \\measuredangle{A'B'A_c}}} = \\frac{B'A_c}{A'A_c}$ \r\n [hide=\"Reason\"] From theorem 4, we have $ \\measuredangle{B'A'A_c} = \\measuredangle{B'EA_c} = \\measuredangle{HEA_c} = \\measuredangle{BEA_b}$ [/hide]\r\n Hence $\\frac{A'A_b}{A'A_c} = \\frac{A_bB}{B'A_c}$. Thus $\\frac{AA_b}{AA_c} = \\frac{A'A_b}{A'A_c}$. This implies that $AA_bA'A_c$ is a harmonic quadrilateral. We will see its application later to prove the original problem.\r\n [b]Theorem 5 [/b] $AA_bA'A_c$ is a harmonic quadrilateral.\r\n We will finish our proof in the next post.", "Solution_16": "Let 's consider the relationship between the two circles $(b)$ and $(c)$. We have $\\measuredangle{A'O_aA_c} = 2.\\measuredangle{A'AA_c} = 2.\\measuredangle{A'AC} = \\measuredangle{A'NA_c}$. Thus $A'O_aNA_c$ is cylic. Hence $\\measuredangle{O_aA'A_c} = \\measuredangle{O_aNA} = \\measuredangle{PNA} = \\measuredangle{BCA} = \\measuredangle{A'CA_c}$. This yields that $A'O_a$ touches the circle ${(c}$ at $A'$. Similarly we have $A'O_a$ touches the circle ${(b}$ at $A'$. And theorem 6 follows\r\n [b]Theorem 6[/b] $(b)$ and $(c)$ touches each other and the line $A'O_a$ at $A'$.\r\n All right, we are 90% done. \r\n Let $l$ be the line joining the centers of $(b)$ and $(c)$. Then $l$ passes through $A'$ and is perpendicular to $O_aA'$. Hence $l$ touches the circle $(O_a)$ at $A'$\r\n Recall theorem 1 that $A', U, O_a$ are collinear. Thus, the circle $(U)$ touches the circle $(O_a)$ at $A'$. Therefore theorem 7 follows\r\n [b]Theorem 7[/b] The circle $(O_a)$ and $(U)$ touches each other and the line $l$ at $A'$.\r\n Again, recall from theorem 5 that $AA_bA'A_c$ is harmonic, we obtain that the tangents at $A$ and at $A'$ (i.e the line $l$) are concurrent with the line $A_bA_c$ at $R$. Since $DO_a$ is the perpendicular bisector of the segment $AA'$, we obtain that $DO_a$ passes through $R$. Therefore $RA'^2 = RE.RF = RA_b . RA_c = RD. RO_a$, we obtain that $D, E, F, O_a$ lie on a circle. By similar argument, it follows that\r\n [b] Theorem 8 [/b] $D, E, F, O_a, O_b, O_c$ lie on a circle.\r\n Note that from the equality above, we also obtain that $A_bDO_aA_c$ is cyclic. Since $O_aA_b = O_aA_c$, we have \r\n [b]Theorem 9[/b] $DA'$ is the bisector of $\\measuredangle{A_bDA_c}$.\r\n All right ! Let me finish this post by introducing some more nice properties relating to this sketch in theorem 10 \r\n [b]Theorem 10[/b] \r\n[i](i)[/i] $R$ is the excenter of similitude of the two circles $(b)$ and $(c)$.\r\n[i] (ii)[/i]$(b)$ and $(U)$ is orthogonal, $(c)$ and $(U)$ is orthogonal. \r\n [i](iii)[/i] The circumcircle of $ADA_b$ touches $AC$ at $A$, The circumcircle of $ADA_c$ touches $AB$ at $A$\r\n [hide=\"Hint\"] The (i) and (ii) questions, we use angle chasing. The (iii) question, we will use a nice property of harmonic quadrilateral. Notice that $D$ is the midpoint of $AA'$ [/hide]\r\n Khoa Lu Nguyen.", "Solution_17": "Dear Mathlinkers,\r\nA new proof of \"the Mineur's circle\" has been put on my website\r\n\r\nhttp://perso.orange.fr/jl.ayme\r\n\r\nsee contenu, vol. 3\r\nSincerely\r\nJean-Louis" } { "Tag": [], "Problem": "A long thin strip of paper is 1024 units in length, 1 unit in width, and is divided into 1024 unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a 512 by 1 strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a 256 by 1 strip of quadruple thickness. This process is repeated 8 more times. After the last fold, the strip has become a stack of 1024 unit squares. How many of these squares lie below the square that was originally the 942nd square counting from the left?", "Solution_1": "[hide=\"Solution\"]\nThe general solution is messy and I haven't worked it out very well at all, so I will present a non-rigorous process that I used to arrive at a specific answer. (It is right, though, I checked the AMC website ;) )\n\nNumber the squares $0, 1, 2, 3, ... 2^{k} - 1$. In this case $k = 10$, but we will consider more generally to find an inductive solution. Call $s_{n, k}$ the number of squares below the $n$ square after the final fold in a strip of length $2^{k}$.\n\nNow, consider the strip of length $1024$. The problem asks for $s_{941, 10}$. We can derive some useful recurrences for $s_{n, k}$ as follows: Consider the first fold. Each square $s$ is now paired with the square $2^{k} - s - 1$. Now, imagine that we relabel these pairs with the indices $0, 1, 2, 3... 2^{k-1} - 1$ - then the $s_{n, k}$ value of the pairs correspond with the $s_{n, k-1}$ values - specifically, double, and maybe $+1$ (if the member of the pair that you're looking for is the top one at the final step).\n\nSo, after the first fold on the strip of length $1024$, the $941$ square is on top of the $82$ square. We can then write\n\n$s_{941, 10} = 2s_{82, 9} + 1$\n\n(We add one because $941$ is the odd member of the pair, and it will be on top. This is more easily visually demonstrated than proven.) We can repeat this recurrence, adding one every time we pair an odd to an even (but ignoring the pairing if our current square is the smaller of the two):\n\n$s_{82, 9} = 2s_{82, 8} = 4s_{82, 7} = 8s_{127-82, 6} = 8s_{45, 6}$\n$s_{45, 6} = 2s_{63-45, 5} + 1 = 2s_{18, 5} + 1 = 4s_{31-18, 4} + 1 = 4s_{13, 4} + 1$\n$s_{13, 4} = 2s_{15-13, 3} + 1 = 2s_{2, 3}$\n\nWe can easily calculate $s_{2, 3} = 4$ from a diagram. Plugging back in,\n\n$s_{13, 4} = 9$ (this can be verified from a diagram as well)\n$s_{45, 6} = 37$\n$s_{82, 9} = 296$\n$s_{941, 10} = \\boxed{593}$\n\nWhew.\n[/hide]\r\n\r\n$1024^{th}$ post! :P I couldn't post again until I solved this problem, it was too good of an opportunity.", "Solution_2": "Yeah, that's what I got too by reconstructing the sequence from smaller cases and figuring out the position square 941 is in.\r\n\r\nPS. In 7th period right now posting this :P.", "Solution_3": "[quote=\"t0rajir0u\"](We add one because $941$ is the odd member of the pair, and it will be on top. This is more easily visually demonstrated than proven.) We can repeat this recurrence, adding one every time we pair an odd to an even (but ignoring the pairing if our current square is the smaller of the two)[/quote]\nJust curious how would you prove this?" } { "Tag": [ "function", "integration", "calculus", "derivative", "algebra", "domain", "algebra proposed" ], "Problem": "does there exist a function $f$ such that $f$ is defined on an interval. so that $f'$=$f^{-1}$\r\n(where $f^{-1}$=inverse of $f$.for example if$f(x)=2x$ then the inverse is $\\frac{x}{2}$)", "Solution_1": "from $f'(x) = f^{-1}(x)$ we derive two equations \\[ f(x) +C=\\int f'(x)\\,dx = \\int f^{-1}(x)\\,dx \\] \\[ f(f'(x))=f(f^{-1}(x))=x \\]\r\ntherefore we get\\[ x+C = \\int f^{-1}(f('x))\\,dx \\]\r\nand hence\\[ f^{-1}(f('x))= 1 \\]\r\nsince $f^{-1}$ needs to be injectiv (otherwise $f$ wouldn't be a function) we find (let $f: N\\to M$ ) \\[ f'(N) = {a} \\qquad\\text{,where \\ensuremath{a} is a constant} \\]\r\ntherefore it is only possible, that\\[ f(x) = ax+b \\]\r\nbut then $f'(x) = a$ and $f^{-1}(x) = \\frac{x-b}{a}$, so \\[ \\forall x \\in N: a = \\frac{x-b}{a} \\iff x= a^2+b \\]\r\nhence there is no such interval if u dont consider $[x_0,x_0]$ as an interval\r\n\r\nif you do, then one'd have to find an $f$ so that\\[ \\exists x \\in \\mathbb R : f'(x)= f^{-1}(x) \\]\r\n(but i dont want to do that right now ;) )", "Solution_2": "Huh.\r\n\r\nI see an extremely messy solution by writing $f(f'(x)) = x$ in terms of Taylor Series $f(x) = a_0 + a_1 x + \\frac{a_2}{2!} x^2 + ...$ and $f'(x) = a_1 + a_2 x + \\frac{a_3}{2!} x^2 + ...$. :(", "Solution_3": "@dashmiz: is my solution right anyway?\r\n@anybody: is my proof valid? (i did alot of little (or big) mistakes on functions in the last days ;) )", "Solution_4": "i do not think it is completly right.\r\ni would check it man :)", "Solution_5": "In general if we have that $(f^{-1})'(f(x)) \\ne 0$ it is true that $f'(x) =\\frac{1}{(f^{-1})'(f(x))}$. Given that $f^{-1} = f'$ we get the differential equation $f'(x) =\\frac{1}{f''(f(x))}$\r\n\r\nI don't know how to work with differential equations, but I'm sure someone who has done a lot of calculus could tell us why there are no solutions to this thing.\r\n\r\nedit: Oh, we have that $f''(f(x))f'(x) = 1$. This is a seperable equation, so we now get $f'(f(x)) + C = x$. However, we know that $f' = f^{-1}$ so that equation says that $C = 0$, so all the information we have gotten is that $f'(f(x)) = x$, which we already knew.\r\n\r\npeeta, I don't understand how you get $x + C = \\int f^{-1}(f'(x))dx$. Please clarify.", "Solution_6": "[quote=\"Kalle\"]In general if we have that $(f^{-1})'(f(x)) \\ne 0$ it is true that $f'(x) =\\frac{1}{(f^{-1})'(f(x))}$. Given that $f^{-1} = f'$ we get the differential equation $f'(x) =\\frac{1}{f''(f(x))}$\n\nI don't know how to work with differential equations, but I'm sure someone who has done a lot of calculus could tell us why there are no solutions to this thing.\n\nedit: Oh, we have that $f''(f(x))f'(x) = 1$. This is a seperable equation, so we now get $f'(f(x)) + C = x$. However, we know that $f' = f^{-1}$ so that equation says that $C = 0$, so all the information we have gotten is that $f'(f(x)) = x$, which we already knew.\n\npeeta, I don't understand how you get $x + C = \\int f^{-1}(f'(x))dx$. Please clarify.[/quote]\r\n\r\noh sorry ... i skipped a step or two there:\r\n\r\nwe already got\\[ f(x) + C = \\int f^{-1}(x)\\,dx \\]\r\nplugging in $f'(x)$ for $x$ (note: $f' \\in D_f$ because $f'=f^{-1}\\in D_f$) we get\\[ f(f'(x)) + C = \\int f^{-1}(f'(x))\\,dx \\]\r\nsince we already know that $f(f'(x)) = x$ we get\\[ x + C = \\int f^{-1}(f'(x))\\,dx \\]", "Solution_7": "Hey, what about $f(x)=\\phi^{-\\frac{1}{\\phi}}x^{\\phi}$? $f^{-1}(x)=f'(x)=\\phi^{-\\frac{1}{\\phi^2}}x^{-\\frac{1}{\\phi}}$, where $\\phi$ is the golden ratio.", "Solution_8": "[quote=\"peeta\"]\nplugging in $f'(x)$ for $x$ (note: $f' \\in D_f$ because $f'=f^{-1}\\in D_f$) we get\\[ f(f'(x)) + C = \\int f^{-1}(f'(x))\\,dx \\][/quote]\r\nThis is where your error is. You cannot make substitutions like that. Try for example $x^2 = \\int_0^x 2xdx$. Now make essentially any substitution like you just did and it won't work. Of course what you might actually mean is that we find the primitive and then plug in $f'(x)$ instead of $x$, but then your differentiation later in the text won't work (you must use chain rule).", "Solution_9": "and about you peeta\r\nkalle said right.you can not just substitute evrything.(remembering $f(g(x))'$) :D", "Solution_10": "isn;t there any complete solution for that?", "Solution_11": "[quote=\"dashmiz\"]but Agrippina .i think you'v made a mistake.\ncheck it once more.\nand about you peeta\nkalle said right.you can not just substitute evrything.(remembering $f(g(x))'$)[/quote]\r\nyeah thats true i understood that argument, i was just too much into \"functional equations\" (not diff equ), where one can normally do that ;) im thinking about a new (diffrent) solution, but not today ;)\r\n(but Agrippina seems to have a method to find the solutions, although i dont think either, that his function is a solution, it is pretty close to one as it seems ;) )", "Solution_12": "[quote=\"Agrippina\"]Hey, what about $f(x)=\\phi^{-\\frac{1}{\\phi}}x^{\\phi}$? $f^{-1}(x)=f'(x)=\\phi^{-\\frac{1}{\\phi^2}}x^{-\\frac{1}{\\phi}}$, where $\\phi$ is the golden ratio.[/quote]\r\n\r\nYou're onto something here. Let's try something more general. $f(x) = ax^b \\implies f'(x) = abx^{b-1}, f^{-1} (x) = a^{ \\frac{-1}{b} } x^{-b}$ so we need to solve\r\n\r\n$ab = a^{ \\frac{-1}{b} }$\r\n$b-1 = -b$\r\n\r\nThis gives $2b = 1 \\implies b = \\frac{1}{2}$ and $a^3 = 2 \\implies a = \\sqrt[3]{2}$, so our function is $\\boxed{ f(x) = \\sqrt[3]{2} \\sqrt{x}}$. I think. :)", "Solution_13": "t0rajir0u, rather than $f^{-1} (x) = a^{ \\frac{-1}{b} } x^{-b}$, you should have $f^{-1} (x) = a^{ \\frac{-1}{b} } x^{\\frac{1}{b}}$. (For example, if $f(x)=x^2$ then $f^{-1}(x)=x^{\\frac{1}{2}}$, not $x^{-2}$.) This produces the function I posted before.\r\n\r\nTo check, if $f(x)=\\phi^{-\\frac{1}{\\phi}}x^{\\phi}$, then $f'(x)=\\phi^{1-\\frac{1}{\\phi}}x^{\\phi}$, while $f^{-1}(x)=[\\phi^{\\frac{1}{\\phi}}x]^{\\frac{1}{\\phi}}$. Since $1-\\frac{1}{\\phi}=\\frac{1}{\\phi^2}$ and $\\phi-1=\\frac{1}{\\phi}$, we have $f'(x)=f^{-1}(x)$.\r\n\r\nBut as for [i]all[/i] solutions, I haven't a clue. :lol:", "Solution_14": "Oh. Sorry. I was doing that during lunch and someone else was pelting me with questions about metaphors :D", "Solution_15": "but as you mentioned i think the answer to all solutions need higher mathematics.\r\ni would ask it if i find some one.", "Solution_16": "[quote=\"Agrippina\"]\nTo check, if $f(x)=\\phi^{-\\frac{1}{\\phi}}x^{\\phi}$, then $f'(x)=\\phi^{1-\\frac{1}{\\phi}}x^{\\phi}$:lol:[/quote]\r\n\r\n Please , see the derivative again. The exponent of $x$ is $\\phi - 1$ and not $\\phi$ . I think that many years ago I had proved- if the problem was really the same - that f can not have as domain the set of real numbers, but I have forgotten the proof. \r\n\r\n[u] Babis[/u]", "Solution_17": "but stergiu what if the domain(as was mentioned)is a domain.\r\ni think it is then true. :?:", "Solution_18": "isn't there a solution?", "Solution_19": "Let me try to fix mine.\r\n\r\nWe will assume the function is of the form $f(x) = ax^b$ since such functions have the property that both their inverses and derivatives are power functions.\r\n\r\n$f'(x) = ab x^{b-1}$\r\n$f^{-1}(x) = \\left( \\frac{x}{a} \\right)^{ \\frac{1}{b} } = a^{ \\frac{-1}{b} } x^{ \\frac{1}{b} }$\r\n\r\n$b-1 = \\frac{1}{b} \\implies b^2 - b = 1 \\implies b = \\phi$\r\n$a \\phi = a^{ \\frac{-1}{\\phi} }$\r\n\r\n$\\phi^2 = \\phi + 1$\r\n$\\phi = 1 + \\frac{1}{\\phi}$\r\n$\\frac{-1}{\\phi} = 1 - \\phi$\r\n\r\n$a \\phi = a^{1 - \\phi}$\r\n$a^{\\phi} \\phi = 1$\r\n$a = \\left( \\frac{1}{\\phi} \\right)^{ \\frac{1}{\\phi} } = (\\phi - 1)^{\\phi - 1}$\r\n\r\n$\\boxed{ f(x) = (\\phi-1)^{\\phi - 1} x^{ \\phi } }$\r\n\r\n :D\r\n\r\nEdit: So, er, yes, this is equivalent to the previous solution posted, but even so!", "Solution_20": "As shown , if the domain is an interval , then there exists such a function.For domain IR , I had seen the problem in an old Romanian book(unsolved) and I spent much time to prove the non existence.\r\n\r\n Babis", "Solution_21": "It was posted before :) : http://www.mathlinks.ro/Forum/viewtopic.php?t=37916 ." } { "Tag": [ "function" ], "Problem": "Let $\\displaystyle y = f(x) = \\frac{a^{x-1}+5}{a^{x-3}+3a}$ be a function, where a is a positive real parameter.\r\nDetermine the value of a, such that the range of the function doesn\u2019t contain any odd integer numbers?", "Solution_1": "does $x$ have to be integer ? Or all real number ?\r\n\r\nwhat about $a=1$ :)", "Solution_2": "$x$ belongs to the set of real numbers." } { "Tag": [ "LaTeX", "Support" ], "Problem": "Several times when I have been making documents in Texnic Center, the computer will say there are three errors even though there aren't any. Also, if I add errors and click Ctrl-F7, the number of errors doesn't change. \r\nI have been able to make the 3-error message go away by adding errors intentionally and deleting them, but this only works sometimes. Has anyone else gotten this false error message? What can be done to prevent or get rid of it?\r\nCan some1 help? (My challenge set is due Thurs :( )!!!!", "Solution_1": "Why don't you tell us what these errors are so we can advise?\r\n[quote=\"baldcypress\"]the computer will say there are three errors even though there aren't any. Also, if I add errors and click Ctrl-F7, the number of errors doesn't change. \nI have been able to make the 3-error message go away by adding errors intentionally and deleting them, but this only works sometimes. Has anyone else gotten this false error message?[/quote]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example please![/url] since we are not mind-readers.\r\n\r\nThere are two types of errors that TexnicCenter gives:\r\n\r\n1. The important ones are on the last line which says:\r\nLaTeX-Result: [i]k[/i] Error(s), [i]l[/i] Warning(s), [i]m[/i] Bad Boxe(s), [i]n[/i] Page(s)\r\nand you need to get [i]k[/i] and [i]l[/i] to be 0.\r\n\r\n2. Higher up in the output window TexnicCenter tells you if it has found citation, bibdata and bibstyle commands in the aux file. If it hasn't then it says there are errors. Obviously if you have no bibliographic data then these errors aren't real errors at all.", "Solution_2": "This is only from my own experience, but I highly suspect the problem is that you are not closing the PDF (if you're using something else to view it, I don't know) of the previously compiled version before compiling the next version (after you edit some more) of the document.\r\n\r\nSorry I don't know the correct terminology.", "Solution_3": "Huh? I can always view the PDF, then re-compile without errors.", "Solution_4": "What leoxnlin says happens to me too. When you keep your PDF viewer with the compiled LaTeX document open and then you build your LaTeX document, it gives you an error saying that it can't write on the compiled document.", "Solution_5": "Wait, what compiler are you using? I'm using TeXShop (for OSX) so it might be different for me.", "Solution_6": "That explains it. I use TeXnicCenter lol.", "Solution_7": "Hi sunehra,\r\n\r\nit depends also on the pdf viewer you are using. For instance the [url=http://blog.kowalczyk.info/software/sumatrapdf/]Sumatra PDF viewer[/url] would work like expected. If you use the Adobe Reader you could adjust the output profile in a way that the reader will be closed before compilation by a DDE command and will be opened again.\r\n\r\nStefan", "Solution_8": "Ok, thanks! I didn't know that. So with other PDF viewers, they don't lock the file or something so that they can't be accessed by other programs?", "Solution_9": "Adobe 8.0 had this lockng mechanism which was a bug. Versions before and after that are fine. Even if you have version 8.0 there is a workaround. In Build, Define Output Profiles, Viewer add [DocOpen(\"%bm.pdf\")] to the 2 instances of [DocOpen(\"%bm.pdf\")][FileOpen(\"%bm.pdf\")] so it reads [DocOpen(\"%bm.pdf\")][DocOpen(\"%bm.pdf\")][FileOpen(\"%bm.pdf\")].\r\n\r\nOne advantage of the Sumatra pdf viewer is that it supports [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=pdfsync]pdfsync[/url] which allows you to go to specific places in the pdf and back to the relevant source code. I haven't tried it so I don't know how well it works or if TexnicCenter supports it." } { "Tag": [ "inequalities", "national olympiad" ], "Problem": "Can anyone translate the Polish MO 2009 Round 2?", "Solution_1": "First day.\r\n\r\n1. $ a_1 \\geq a_2 \\geq \\ldots \\geq a_n > 0$. Prove the inequality\r\n$ a_1a_2\\ldots a_{n\\minus{}1} \\plus{} (2a_2\\minus{}a_1)(2a_3\\minus{}a_2) \\ldots (2a_n\\minus{}a_{n\\minus{}1}) \\geq 2a_2a_3 \\ldots a_n.$\r\n\r\n2. There are given integers numbers $ a, b$ such that $ a>b>1$, $ ab\\plus{}1$ is divisible by $ a\\plus{}b$ and $ ab\\minus{}1$ is divisible by $ a\\minus{}b$. Prove that $ a < b \\sqrt{3}$.\r\n\r\n3. Disjoint circles $ o_1, o_2$, with centers $ I_1, I_2$ respectively, are tangent to the line $ k$ at $ A_1, A_2$ respectively and they lie on the same side of this line. Point $ C$ lies on segment $ I_1I_2$ and $ \\angle A_1CA_2 \\equal{} 90^{\\circ}$. Let $ B_1$ be the second intersection of $ A_1C$ with $ o_1$, and let $ B_2$ be the second intersection of $ A_2C$ with $ o_2$. Prove that $ B_1B_2$ is tangent to the circles $ o_1, o_2$.\r\n\r\nSecond day.\r\n\r\n4. Segment $ AB$ is diameter of the circle $ o$, outscribed on the convex quadrilateral $ ABCD$. Let $ E$ be the intersection of diagonals of this quadrilateral and let $ P$ be the intersection of tangents to the circle $ o$ at points $ C, D$. Prove that $ PC\\equal{}PE$.\r\n\r\n5. Find all integer numbers $ n \\geq 4$ which satisfy the following condition: From every $ n$ different $ 3$-element subsets of $ n$-element set it is possible to choose $ 2$ subsets, which have exactly one element in common.\r\n\r\n6. For every integer $ n \\geq 3$ find all sequences of real numbers $ (x_1, x_2, \\ldots, x_n)$ such that\r\n$ \\sum_{i\\equal{}1}^{n}x_i \\equal{} n$ and $ \\sum_{i\\equal{}1}^{n} (x_{i\\minus{}1}\\minus{}x_i\\plus{}x_{i\\plus{}1})^2 \\equal{} n$,\r\nwhere $ x_0\\equal{}x_n$ and $ x_{n\\plus{}1}\\equal{}x_n$." } { "Tag": [], "Problem": "A draftsperson makes a scale drawing of a 100 meter $ \\times$ 30 meter building, where 1 centimeter represents 2.5 meters. How many centimeters are in the smaller dimension of the drawing of the building?", "Solution_1": "Complying with the scale, $ \\frac{30}{2.5}\\equal{}\\fbox{12}$ centimeters." } { "Tag": [ "inequalities", "LaTeX", "inequalities unsolved" ], "Problem": "please help me with the following problems.\r\n\r\n1. for any convex quadrilateral ABCD, AC^2+BD^2<=AB^2+BC^2+CD^2+DA^2\r\n(i have a proof of this in \"Inequalities\" by Radmila Bulajich Manfrino and two more writers but i want another proof.)\r\n\r\n2. for positive reals a,b,c such that abc=1, ab+bc+ca>=a+b+c\r\n(i proved this when i was little(!) but i cant find it, so...)", "Solution_1": "cant someone solve these easy problems? i've waited for so long and i cant wait any more :mad:", "Solution_2": "ok here's my solution:\r\n(ab+bc+ca)/(a+b+c)>=[(a^b)(b^c)(c^a)]^[1/(a+b+c)] by weighted AM-GM. now if i assume that a=[(a^a)(b^a)(c^a)]^[1/(a+b+c)]=1.\r\nmy question is, is it safe to assume a 1 $, such\r\nthat $ ab=n $ and $ gcd(a,b) > 1 $. Then consider a group $ G =\r\n\\mathbb{Z}_{a} \\oplus \\mathbb{Z}_{b} $, which is not cyclic and\r\naction of $ G $ on itself.\r\n\r\nI hope it is correct.", "Solution_4": "I really think it's the same thing, man :).", "Solution_5": "Yes, you are right :)" } { "Tag": [ "algebra", "polynomial", "Vieta", "algebra proposed" ], "Problem": "The product of two of the four zeros of the quadric equation \r\n\r\n$ x^4\\minus{}18x^3\\plus{}kx^2\\plus{}200x\\minus{}1984\\equal{}0$\r\n\r\nis $ \\minus{}32$.Find $ k$", "Solution_1": "Just use Vieta's theorem and we get $ k\\equal{}86$:\r\nsuppose $ x_1,..,x_4$ ae the roots and we have $ x_1x_2\\equal{}\\minus{}32$ then;\r\n$ p\\plus{}q\\equal{}18$\r\n$ 62\\plus{}pq\\minus{}32\\equal{}k$\r\n$ \\minus{}62p\\plus{}32q\\equal{}200$\r\nwhere $ q\\equal{}x_3\\plus{}x_4$ and $ p\\equal{}x_1\\plus{}x_2$." } { "Tag": [ "Mafia", "videos", "FTW", "blogs", "big brother" ], "Problem": "Hi!! \r\n\r\nI'm creating Big Brother!!!\r\n\r\n[i](scattered applause)[/i]\r\n\r\n[size=9]I'm sorry for bringing down Amazing Race...[/size]\r\n\r\n1.\r\n2.\r\n3.\r\n4.\r\n5.\r\nand so on and so forth... (maybe 10-15)\r\n\r\nThere are challenges... THIS WAS DEFINITELY NOT COPIED\r\n\r\nHead of Household (HoH): This position is the most powerful in the house. It takes place at the start of every week, and everyone competes except for the previous week's HoH. The HoH has the responsibility to nominate two Houseguests for eviction, the worst place to be in the house. If you are still nominated (see Power of Veto) by the Eviction vote, the rest of the house will vote, and potentially send you home. \r\n\r\nPower of Veto (PoV): This competition takes place after the HoH has made nominations. There are (at the beginning of the game) 6 competitors: The HoH, the two nominees, and three HGs selected by random draw. The winner of this competition has the ability to Veto one of the HoH's nominees. They do not have to, however. If a nominee is Vetoed, then the HoH is required to put up a replacement nominee. \r\nThere are also several ground rules, which must be followed at all times by everyone:\r\n\r\n1. Communication, both inside the thread and out, is allowed at all times, except the following:\r\na. HoH Competitions\r\nb. PoV Competitions, unless otherwise specified\r\nc. Eviction Votes\r\nd. Any other time I tell you so\r\n2. I expect that most of the game should take place outside this thread: I also require that any discussion groups be made known to me.\r\n3. You are NOT allowed to discuss the game with anyone not in the game, unless you are told otherwise.\r\n4. Once you are evicted, you will be sent special instructions about your future role in the game. As a rule, though, don't talk to anyone still in the game once evicted.\r\n5. Use your good judgment: don't do anything you think I would disallow.\r\n6. This is a game: have fun, and make it fun for others. No flaming, etc.\r\n\r\nIf you ever have any questions about what you are and aren't allowed to do, send me a pm, or ask in-thread if you don't care that others know. I will never reveal anything someone pms me that isn't a competition.\r\n\r\nAND I'm bringing back the Coup d'Etat with a little twist... :lol: SO JOIN!!\r\n\r\n[size=200]IF YOU ARE GOING TO JOIN, MAKE SURE YOU ARE ACTIVE FOR THE NEXT MONTH OR TWO OR THREE OR...[/size]", "Solution_1": "Maybach joins.", "Solution_2": "okay i will join! i dropped outof amzing race cuz i didnt liek the idea very much :blush:", "Solution_3": "/in..........", "Solution_4": "I join. :jump:", "Solution_5": "I will join.", "Solution_6": "1. Maybach\r\n2. Nachosaurus\r\n3. EggyLv.999\r\n4. FlyAgaric\r\n5. spaceguy524\r\n6.\r\n7.\r\nand so on and so forth...", "Solution_7": "HAHA this is so funny most of these peeps just exited the old big brother", "Solution_8": "/in\r\n\r\nI don't watch the show, but this looks interesting.", "Solution_9": "1. Maybach\r\n2. Nachosaurus\r\n3. EggyLv.999\r\n4. FlyAgaric\r\n5. spaceguy524\r\n6. westiepaw\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12??\r\n13??\r\n14??\r\n15??", "Solution_10": "Hehee..\r\n/in", "Solution_11": "I'll join.\r\nMWAHAHHAHAHAheheheheheuheuheuheuahemahemaheamHUUURRAUUGH.\r\nGagging just after laughing is not fun. :(", "Solution_12": "1. Maybach\r\n2. Nachosaurus\r\n3. EggyLv.999\r\n4. FlyAgaric\r\n5. spaceguy524\r\n6. westiepaw\r\n7.[hide=\" \"]If you are reading this, you are granted an extra clue to the Coup d'Etat. The Coup d'Etat will give you a special power in this game. Keep looking for clues in messages such as this one. Good luck! [/hide]meewhee009\r\n8. 1=2\r\n9.\r\n10.\r\n11.\r\n12??\r\n13??\r\n14??\r\n15??", "Solution_13": "Each of you must PM me a paragraph (6+ sentences)[hide=\" \"]Good! You're starting to notice things! [/hide]about how you will go about winning the game and your personality.\r\n[b]\nWRITE IN COMPLETE SENTENCES.\n\nPM this ASAP [/b]", "Solution_14": "For people who have joined early, there's a helpful thing!\r\n\r\n[b]Coup d\u2019Etat [/b]\r\n\r\n \r\n\r\nYep, the Coup d\u2019Etat is back!!\r\n\r\n\r\nThe Coup d\u2019\u00e9tat is a power only available to one person in the entire game. That is, this is the only time it will be offered, and only one[hide=\" \"]or not...[/hide]of you can win it. If won, you will acquire a great power in the house, but you won\u2019t know what it is until you win. Also, no one else will know who won, or its power, until it is used. I will tell you that it could completely turn your game around...The way to win it is to submit to me the correct answer to a puzzle. Occasionally, I will post a new clue. You get only one guess the entire game, so answer wisely. You may guess it at ANY time, including during other competitions, and during voting periods. I will continue posting clues even after it is won. If you answer, I will tell you either you won or you didn\u2019t: not if someone else already won, or the validity of your answer. PM me your answers. Good luck!", "Solution_15": "And our winner is[hide=\"...\"]\n\nI don't know yet. FlyAgaric has not PMed me yet. Whatever...\n\nThere was a 2-2 tie. \n\nSo, now, the winner is[hide=\"...\"]\n\n[b]westiepaw[/b] and [b]meewhee009[/b]\n\nThank you all for another successful season and a successful first game I have modded.[/hide][/hide]", "Solution_16": "A [b]TIE?![/b] This is completely unacceptable! I declare a mis-vote! Another vote or another test shall be scheduled!", "Solution_17": "Nope. Wont do. Prod FlyAgaric. We cant leave big brother with a tie", "Solution_18": "Well then, prod he/she/it.", "Solution_19": "[hide=\"WE HAVE A WINNER\"]\n[hide=\"westiepaw\"]\n[hide=\"you\"]\n\nnot won. Congratulations [b]meewhee009[/b]!!!!!!!!!!!\n\nNice season.[/hide][/hide][/hide]", "Solution_20": ":( \r\nOh.\r\nMy.\r\nGosh.\r\nI would like to thank my family,\r\nall of AoPS,\r\nall of the contestants on Big Brother,\r\nAndrewy,\r\nmy imaginary dog,\r\nand ice cream. If it wasn't for ice cream I wouldn't have been here right now.", "Solution_21": "lol big brotha aops season three ftw", "Solution_22": "[b]1=2[/b] was in fact in Season 2, but was evicted first.\r\n\r\nI thank everyone again for a great season.", "Solution_23": "Wait, I was???\r\nI don't remember that.\r\nI must have been really unpopular back then.", "Solution_24": "[quote=\"1=2\"]Wait, I was???\nI don't remember that.\nI must have been really unpopular back then.[/quote]\r\n\r\nWow...\r\n\r\nYou forgot...\r\n\r\n :rotfl:", "Solution_25": "Also... [size=200]VISIT MY BLOG!!!\n[/size]\r\nSiggy...", "Solution_26": "is this game over?\r\n[size=200]in 14 pages?![/size]", "Solution_27": "[quote=\"piegraph\"]is this game over?\n[size=200]in 14 pages?![/size][/quote]\r\n\r\nyeah...\r\n\r\npiegraph, that post made me think that you just wanted to raise your post count...\r\n\r\nand this isn't one of your RPG games.", "Solution_28": "i know.\r\nand that's what abcak said.", "Solution_29": "[b]SSHHHH.[/b]" } { "Tag": [], "Problem": "This forum is great and I want to know if there are other math forums similar to this one.\r\n\r\nAlso, I'm looking for any Biology, Physics, and Chemistry related forums.\r\n\r\nIf you know any, can you give me some suggestions to which one to go to?", "Solution_1": "Yeah I've been wondering if there were any sites like this one also. I have found this one [url]http://www.scienceforums.net/forums/[/url] It has science and math mostly science though. It looks really good. The only problem is that a lot of the stuff is a bit on the advanced side for me (at least what Ive looked at). I'll keep looking at it because I'm sure I can learn a lot from it just like I have here.", "Solution_2": "yea I just joined that forum...I'm lost on it. very lost." } { "Tag": [ "HMMT", "Harvard", "college", "MIT", "algebra", "polynomial", "AMC" ], "Problem": "Source: Harvard-MIT Math Tournament - Algebra 2003 #10\r\n\r\nSuppose that P(x) is a polynomial such that P(1) = 1 and \r\n\r\nP(2x)/P(x+1) = 8 - 56/(x+7)\r\n\r\nfor all real x for which both sides are defined (i.e. P(x+1) \u2260 0 and(x+7) \u2260 0). Find P(-1)", "Solution_1": "Um. [hide]1[/hide]?", "Solution_2": "I would be willing to wager that in the initial problem, the given condition was supposed to be P(0) = 1, not P(1) = 1. Try the problem that way.", "Solution_3": "But then, the problem doesn't work, cuz it becomes 1/P(1) = 0, so it's not a polynomial.", "Solution_4": "Whoops, good call. Hmm.\r\n\r\nOkay, well, lets play around with it anyways. \r\nP(2x)/P(x+1) = 8 - 56/(x+7) = 8x/(x+7)\r\n8x*P(x+1) = (x+7)*P(2x). Now, let P(x) = a*x^n + . . . + c. We can see by checking the coefficient of x^(n+1) in that equation that P is a cubic polynomial. So P(x) = ax^3 + bx^2 + cx + d.\r\n\r\n8x*P(x) = 8ax^4 + 8bx^3 + 8cx^2 + 8dx\r\n(x+7)*P(2x) = (x+7)*(8ax^3 + 4bx^2 + 2cx + d) = \r\n = 8ax^4 + (4b+56a)x^3 + (2c + 28b)x^2 + (d + 14c)x + 7d.\r\n\r\nThese two expressions are equal for all x (except possibly -7 and the roots of P(x+1), but those won't matter). Therefore they have equal coefficients. Thus:\r\n7d = 0\r\n8d = 14c + d\r\n8c = 2c + 28b\r\n8b = 56a + 4b\r\n8a = 8a\r\n\r\nThis seems problematic to me. Someone else catch an error on my part? Hmm.", "Solution_5": "According to the document I have, the problem states P(1) = 1.", "Solution_6": "Heh, maybe its a trick question then?", "Solution_7": "But I still don't think any such polynomial exists. Can you verify my work, Stephen?", "Solution_8": "I think you changed a P(x+1) into a P(x) JBL.\r\nBut doing what you did would give you the solution as ax^3 + 8ax^2 + 12ax, or ax(x+2)(x+6). Then you'd substitute in the given value and work out a, and then find P(1).\r\n\r\nI'd like to know the actual question though.. sometimes there is an easier way to attack a problem.", "Solution_9": "I found the question, and it was a bit different indeed.\r\n\r\nSuppose P(x) is a polynomial with P(1)=1 and \r\n\r\nP(2x)/P(x+1)=8-56/(x+7)\r\n\r\nfor all real x for which both sides are defined. Find P(-1)." } { "Tag": [ "conics", "ellipse", "trigonometry", "ratio", "geometry", "geometric transformation", "homothety" ], "Problem": "Let $Z$ be a circle with centre $O$. Let $A$ and $B$ be two points on this circle $Z$ such that $AB$ is a diameter of $Z$. Select a point $C$ on the segment $AB$. We construct another circle, say $Y$, with diameter $AC$. Let $D$ be a point on the circumference of $Z$ such that $DCB$ is an isosceles triangle with $CD=BD$. Draw a circle $X$ (inside $Z$) which is tangent to $Z$, to $Y$ and to the segment $CD$. Let $E$ be the centre of this circle $X$. Show that $EC$ is perpendicular to $AB$.", "Solution_1": "We can look at it backwards: let's show that a circle tangent to the two circles passing through $A$ and with its center $E$ satisfying $EC\\perp AB$ must also be tangent to $DC$.\r\n\r\nLet $F$ be the projection of $D$ on $AB$. If $r$ is the length of the radius of the circle $x$, then we need to show $\\frac{EC^2}{r^2}=\\frac{AD^2}{DF^2}$.\r\n\r\n$E$ lies on the ellipse having $O$ and $O'$ (the center of $y$) as foci and having $AF$ as its large axis. Call $a,c$ its half of the large axis and distance from the center to a focus, respectively (usual notations). The computation is now fairly simple. If it's unclear, I'll do it here.", "Solution_2": "I have quite long solution.\r\n\r\nConsider inversion with center at p. $C$. circle $Z$ becomes new circle $Z$, circle $Y$ becomes line $Y$ and $Y\\perp AB$. Since $X$ tangents $Z$, $Y$ and $CD$ before inversion we conclude new circle $X$ tangents lines $EA$ and $CD$ and circle $Z$. It implies that $F$ lies on bisector of $AED$.\r\nLet $h$ be perpendicular to $AB$ at point $C$. We need to show that $F$ lies on $h$. \r\nWe will prove [b]equivalent[/b] statement: Let $F$ be point of intersection of bisector $AED$ and $h$, then distances from $F$ to $AE$, $ED$ and $Z$ are equal, i.e. $FG=FK$ and $O'F=R(Z)+FG$ [b](0)[/b], where $R(Z)$ is a radius of new $Z$. (It means that circle this center $F$ and radius $FG$ tangents lines $AE$, $EC$ and circle $Z$)\r\n\r\nFirst of all, we state that $CB=DB$ [b](1)[/b]. \r\nIndeed, before inversion we had $DC=DB$, so after inversion new length $BC$ is $\\frac{R^2}{BC}$ and new length $DB$ is $\\frac{R^2\\cdot DB}{CD\\cdot CB}=\\frac{R^2}{BC}$, where $R$ is the radius of inversion.\r\n\r\nThus we have $\\angle CBD=\\angle DCB$ $\\Rightarrow$ $\\angle ACE=\\angle ACH=\\angle HAC$ and triangle $HAC$ is isosceles. Since $AEC$ is right triangle we obtain that $HC=EC/2$ [b](2)[/b].\r\n\r\nSince $EF$ is bisector of $AEC$ and $AE||CF$ we conclude $\\angle CEF=\\angle AEF=\\angle EFC$ $\\Rightarrow$ triangle $EFC$ is isosceles and $CF=EC$ [b](3)[/b].\r\n\r\nSince $EF$ is bisector and $AE||CF$ we obtain $GF=GK=AC$ [b](4)[/b].\r\n\r\nSuppose WLOG radius $Z$ is 1. \r\n\r\nLet $\\angle CBD=x$. Then from right triangle $ABD$ we see that $BD=2\\cos x$. From [b](1)[/b] we conclude $BC=2\\cos x$. It follows that $DC=4\\cos x\\sin\\frac{x}{2}$. Since triangles $HAC$ and $CBD$ are similar we have $HC=BC\\cdot \\frac{AC}{DC}=2\\cos x\\cdot\\frac{2-2\\cos x}{4\\cos x\\sin\\frac{x}{2}}$. So from [b](2)[/b] and [b](3)[/b] we obtain $FC=4\\sin\\frac{x}{2}$ (we use $1-\\cos x=2\\sin^2\\frac{x}{2}$).\r\n\r\nLast step. Using Piphagorean theorem for $O'CF$ we obtain $O'F^2=O'C^2+FC^2$. Accordingly to [b](0)[/b] (and using [b](4)[/b]) we need to show that\r\n\\[O'F^2=(1-2\\cos x)^2+16\\sin^2\\frac{x}{2}\\mbox{ and }(1+FK)^2=(1+(2-2\\cos x))^2\\mbox{ are equal.}\\]\r\nIt is very simple trigonometry exercise.", "Solution_3": "grobber, sorry if I am super slow, but could you explain more of your solution? (i.e. Why does that ratio do the job?) And I don't really see the computations easy...\r\n\r\nThanks in advance. :P", "Solution_4": "Don't worry. I expected this, and it's not because you would be \"super slow\" or anything, but because my answer realy was cryptic :). \r\n\r\nIf we let $E'$ be the projection of $E$ on $DC$, we need to show that $E'$ lies on $x$, or, in other words, that $EE'=r$. This means that we must show that a right triangle having $EC$ as a hypothenuse and $r$ as one of its legs is similar with the right triangle $ADF$, with $r$ corresponding to $DF$ (thus the ratios; I put squares because it's easier to compute $r^2,EC^2$, etc.).\r\n\r\nThe computations aren't easy either, I exaggerated when I said that :). $FD,AD$ are, however, easy to compute in terms of $a,c$: $DF^2=FA\\cdot FB=FA\\cdot FC=2a\\cdot 2c$ and $AD^2=AF^2+DF^2=(2a)^2+2a\\cdot 2c$, so $\\frac{AD^2}{DF^2}=\\frac{a+c}c\\ (*)$.\r\n\r\n$EC^2$ is simply computed from the equation of the ellipse: $\\frac{EC^2}{a^2-c^2}+\\frac{O''C^2}{a^2}=1$, where $O''$ is the center of the ellipse, i.e. the midpoint of $OO'$. We use the fact that $O''C=|2c-a|$. \r\n\r\nConsider now the inversion of pole $A$ which sends $C$ to $B$. The circles $y,z$ are turned into lines through $B,C$ respectively and perpendicular to $AB$, and $x$ is turned into a circle tangent to these two parallel lines, which can also be obtained from $x$ through a homothety of center $A$. If $r'$ is the radius of the image of $x$, then $r'=\\frac{BC}2=2a-2c$ and $\\frac{r'}r=\\frac{AF}{AC}=\\frac{2a}{2c}$. We thus get $r=c(1-\\frac ca)\\ (**)$.\r\n\r\nWe can now use $(*),(**)$ to show the equality of those two ratios mentioned in my first post.", "Solution_5": "Now I think my solution is not so long... :)", "Solution_6": "Thanks grobber, your solution is indeed valid. (Sorry for the delay) :)", "Solution_7": "Dear Mathlinkers,\r\nfinally, you can see a completely synthetical proof of this nice san Gaku on my site :\r\nhttp://perso.orange.fr/jl.ayme vol. 4 La fameuse san Gaku de lma prefecture de Gumma (1803)\r\nSincerely\r\nJean-Louis" } { "Tag": [ "function", "quadratics", "algebra", "domain", "calculus", "calculus computations" ], "Problem": "Hello,\r\nI am working on a project for functional iterations. I study the function $ y\\equal{}\\frac{x\\plus{}10}{x\\plus{}1}$. I know from the cob-web plot that the value $ x\\equal{}sqrt(10)$ is a fixed point. My question is, how do i prove mathematically that this function has a UNIQUE fixed point and how do i know if this point is attracting or repeling?\r\nMaybe all this sound very elementary, but i haven't done something simar until now. Thanks for any help.", "Solution_1": "The equation $ x \\equal{} \\frac {x \\plus{} 10}{x \\plus{} 1}$ can be rearranged to a quadratic equation, and that quadratic equation has [i]two[/i] roots: $ \\pm\\sqrt {10}.$\r\n\r\nA fixed point of $ x \\equal{} f(x)$ is (at least locally) attracting if $ |f'(x)| < 1$ at that point.\r\n\r\nIn this case $ f'(x) \\equal{} \\frac { \\minus{} 9}{(x \\plus{} 1)^2}.$ Evaluating that at both $ \\sqrt {10}$ and $ \\minus{} \\sqrt {10}$ shows that the fixed point at $ \\sqrt {10}$ is attracting and the fixed point at $ \\minus{} \\sqrt {10}$ is repelling.\r\n\r\nThe only way this would have a unique fixed point would be if you started by restricting the domain.\r\n\r\nNow: suppose we restrict ourselves to $ [0,\\infty).$ We can show that $ f$ maps this domain into itself; we can show by the intermediate value theorem that there is at least one fixed point (or explicitly solve for that point), and the very fact that $ f$ is a decreasing function on $ [0,\\infty)$ means there cannot be more than one fixed point.\r\n\r\nThere is a sequence of points $ \\{\\minus{}1,\\minus{}\\tfrac{11}2,\\minus{}\\tfrac{31}9,\\dots\\}$ such that if you start the iteration at any of those points, some member of the sequence will become undefined. Otherwise, the sequence should converge to $ \\sqrt{10}.$", "Solution_2": "I assume then that also the condition that you state there is also the other way ? (for the point being attractive)", "Solution_3": "If $ |f'(x)|>1,$ the point is repelling. If $ |f'(x)|\\equal{}1,$ we need a more delicate analysis.", "Solution_4": "I don't understant why $ \\left[ 0,\\infty \\right)$ is mapped to itself..I think that when $ x \\geq 0$ then $ f(x) \\in \\left( 0,10 \\right)$. Where is my mistake?\r\nThanks", "Solution_5": "Into, not onto. We only require that $ f([0,\\infty))\\subseteq[0,\\infty)$, not that it equal $ [0,\\infty).$", "Solution_6": "Thanks for all your answers. There is something final i would like to ask. The definition of a repelling point is that it is a fixed point p, for which the sequence of iterations diverge when starting from any value sufficiently close, but not equal to p.\r\nWhen i evaluate this seqence in Matlab for different starting values, even if i start from a value near $ \\minus{}\\sqrt{10}$, i still see that the series converge to $ \\sqrt{10}$. Shouldn't the members of the sequence diverge, by the definition,since $ \\minus{}\\sqrt{10}$ is a repelling point?\r\nThanks", "Solution_7": "Saying that $ \\minus{}\\sqrt{10}$ is repelling simply means that sequences won't converge to $ \\minus{}\\sqrt{10}.$ Any sequence that starts out close to that point will, sooner or later, cease to be close to $ \\minus{}\\sqrt{10}.$ That doesn't tell you that such a sequence will diverge - it just tells you that the repelling point will eventually become irrelevant to the convergence or divergence of that sequence. The sequence will just go somewhere else, and perhaps be captured by the influence of a different fixed point." } { "Tag": [], "Problem": "There's a marathon in all of the other forums, so I thought I'd bring it here:\r\n\r\n\r\n[size=150]1. Answer a question\n2. Post a question about music or art\n 3.:play_ball: :lol: [/size]\r\n\r\nI'll start:\r\n\r\nWho's your favorite composer?", "Solution_1": "Yay, a music and art marathon!! :lol: \r\n\r\nMy favorite composer's Franz Liszt. \r\n\r\nWhat instrument do you play and how long have you been playing it?", "Solution_2": "Piano, 4 years\r\n\r\nWhat is your favorite piece (what a typical question)?", "Solution_3": "[quote=\"236factorial\"]Piano, 4 years\n\nWhat is your favorite piece (what a typical question)?[/quote]\r\n\r\nBeethoven's Moonlight. First movement and Third movement are completely different, and they both RULE :) :lol: \r\n\r\nwhat's your favorite music period?", "Solution_4": "Romantic era.\r\n\r\nWhat's your least favorite musical time period?", "Solution_5": "Baroque period\r\n\r\nDo you like quicker songs or slower songs?", "Solution_6": "[quote=\"236factorial\"]Baroque period[/quote]\r\nHaha, me too. :D \r\n\r\nI like quicker songs - they're more fun.\r\n\r\nWhat's the hardest piece you've ever played?", "Solution_7": "moonlight sonata-not only was it hard to get, it was hard for my ears to listen to it. :roll: \r\n\r\nfavorite artist?", "Solution_8": "Da Vinci!!!\r\n\r\nSorry for double posting, but this topic's gotta keep going.", "Solution_9": "$Hendrix$\r\n\r\ncompare and constrast soloing versus written instrumentals", "Solution_10": "I'll assume that by \"soloing,\" you mean improvising.\r\n\r\nImprovisation can be really expressive if done well (I'm listening to Miles Davis right now :D), but so are most well-written pieces. I would say that Improvisation demonstrates a greater knowledge and intuition involving harmonic structure and relative sound, as well as a spontaneous creativity that verbatim rehearsal can't emulate, but I think the ideas of written pieces are sometimes clearer and more mature than those embodied by improv.\r\n\r\n\r\nYour favorite impressionist? (in music or art, or both, if you please)[/b]", "Solution_11": "Debussy (I love Golliwog's cakewalk).\r\n\r\nDo you think that piano is harder or easier than other instruments?", "Solution_12": "I guess it is easier.\r\n\r\nMy question : where exactly on mathlinks is it appropriate to discuss movies?\r\n\r\nMovies you just saw is a topic at games and fun\r\n\r\nbut I'd say : movies are a very popular form of art?", "Solution_13": "Movies are a form of art. By previous statement of the board owners, discussions along the lines of \"What is your favorite band?\" belong on fun and games, and discussions of art as art forms (and film is an art form) belong here. \r\n\r\nCan everyone sing?", "Solution_14": "Thank you.\r\nHowever experience leads to me thinking that here most music people reside here : piano players, classical music fans\r\n\r\nI must say I know nearly nothing about that.\r\n\r\nMost people discuss films in fun and games, like harry potter, star wars etc...\r\n\r\n\r\n\r\nSo when I would want to ask a specific question about movies, I would have to do it here, where most movie experts won't notice my question?\r\n\r\nOr am i seeing things incorrectly?" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $x,y,z \\in R$ so that $x+y+z=3$. Prove that $ x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2} \\geq 3xyz$. \r\n\r\nmade by Ghenghiu R.\r\n\r\ncheers! :D :D :cool:", "Solution_1": "We want to prove $x^2y^2+y^2z^2+z^2x^2 \\geq xyz(x+y+z)$. This follows directly from Muirhead's theorem, as $\\sum_\\text{sym} x^2y^2 \\geq \\sum_\\text{sym} |x^2yz| \\geq \\sum_\\text{sym} x^2yz$.", "Solution_2": "Somehow I wonder why this is posted into the \"Unsolved Problems\" section while the proof is more or less banal. If we \"de-homogenize\" the inequality, we get $x^2 y^2 + y^2 z^2 + z^2 x^2 \\geq xyz \\left(x+y+z\\right)$ for three arbitrary real numbers x, y, z (the condition x + y + z = 3 is not necessary anymore). Now, this inequality is trivial if we use Muirhead; if we don't use Muirhead, it is not difficult either:\r\n\r\n$x^2 y^2 + y^2 z^2 + z^2 x^2$\r\n$= \\frac12 \\left(x^2 y^2 + x^2 z^2 + y^2 z^2 + y^2 x^2 + z^2 x^2 + z^2 y^2\\right)$\r\n$= \\frac12 \\left(x^2 \\left(y^2 + z^2\\right) + y^2 \\left(z^2 + x^2\\right) + z^2 \\left(x^2 + y^2\\right)\\right)$\r\n$\\geq \\frac12 \\left(x^2 \\cdot 2yz + y^2 \\cdot 2zx + z^2 \\cdot 2xy\\right)$\r\n$= x^2 yz + y^2 zx + z^2 xy = xyz \\left(x+y+z\\right)$.\r\n\r\nDarij", "Solution_3": "It's $a^2+b^2+c^2 \\geq ab+bc+ca$ again!!! where $a=xy, b=yz, c=zx$.\r\n\r\n :D :D :D", "Solution_4": "Well,in fact we can take the condition $ a^2+b^2+c^2=3 $ to make it more stronger,however it is still too trivial. :)" } { "Tag": [ "number theory", "greatest common divisor", "least common multiple", "combinatorics proposed", "combinatorics", "games", "Combinatorial games" ], "Problem": "Two players play a game as follows: there $n > 1$ rounds and $d \\geq 1$ is fixed. In the first round A picks a positive integer $m_1$, then B picks a positive integer $n_1 \\not = m_1$. In round $k$ (for $k = 2, \\ldots , n$), A picks an integer $m_k$ such that $m_{k-1} < m_k \\leq m_{k-1} + d$. Then B picks an integer $n_k$ such that $n_{k-1} < n_k \\leq n_{k-1} + d$. A gets $\\gcd(m_k,n_{k-1})$ points and B gets $\\gcd(m_k,n_k)$ points. After $n$ rounds, A wins if he has at least as many points as B, otherwise he loses.\r\n\r\nFor each $(n, d)$ which player has a winning strategy?", "Solution_1": "B always has a winning strategy.\nLet $L$ be the least common multiple of $m_1+1 , m_1+2 , ... , m_1 + (n-1)d$, which are preciesly all numbers such that A can possibly choose during the game. Firstly, let B choose $n_1 = L+m_1$.\nEach turn, if A chooses $m_{k-1}+c$, for some $1\\leq c\\leq d$, then let B choose $n_{k-1}+c$.\nThen, we always have $n_k = m_k + L$. But since $m_k|L$, we have $\\gcd (m_k , n_k )=m_k$. Therefore, B earns $m_2 + ... + m_n$ points at the end.\nIf A wants to win this game, we must have $\\gcd (m_2 , n_1)+\\ldots+\\gcd (m_n, n_{n-1})\\geq m_2 + ...+m_n$,\nso we must have $\\gcd (m_k , n_{k-1}) = m_k$, or equivalently, $m_k | n_{k-1}$. In particular, we must have $m_2 | n_1$, which is impossible, since $m_2 | L$, but $m_2 \\not| m_1$" } { "Tag": [ "function", "trigonometry", "calculus", "derivative", "complex analysis" ], "Problem": "(1) Find a power series of $ z \\plus{} 2i$ whose sum is the function $ \\frac {1}{1 \\minus{} z}$, and calculate its radius of convergence.\r\n\r\n\r\n(2) (a) Find the power series about 0 for the function $ f(z) \\equal{} e^{z^3}$, and give its radius of convergence.\r\n\r\n(b) using the series obtained in part (a), compute $ f^{(20)}(0)$ and $ f^{(21)}(0)$.", "Solution_1": "(3) Find the Maclaurin series for each of the following functions, together with the respective radius of convergence:\r\n\r\n(a) $ \\cos^3z$\r\n\r\n(b) $ (z^2\\plus{}1) \\sin z$\r\n\r\n(c) $ e^z \\cos z$\r\n\r\n(d) $ \\sin(2z)$\r\n\r\n(e) $ \\cos(z^3)$\r\n\r\n(f) $ z \\sinh(z^2)$\r\n\r\n(g) $ \\frac{1}{(2\\minus{}z)^3}$", "Solution_2": "1. Let $ w \\equal{} z \\plus{} 2i$ so the function may be expressed as $ \\frac{1}{1 \\plus{} 2i \\minus{} w} \\equal{} \\frac{1 \\minus{} 2i}{5} \\frac{1}{1 \\minus{} \\frac{1 \\minus{} 2i}{5}w} \\equal{} \\frac{1 \\minus{} 2i}{5} \\Big(1 \\plus{} \\frac{1 \\minus{} 2i}{5}w \\plus{} \\Big(\\frac{1 \\minus{} 2i}{5}\\Big)^2w^2 \\plus{} \\cdots \\Big)$ $ \\equal{}$ $ \\sum_{n \\equal{} 0}^{\\infty} \\Big(\\frac{1 \\minus{} 2i}{5}\\Big)^{n\\plus{}1} w^n$. The radius of convergence is the distance from the center, $ \\minus{}2i$, to the nearest singularity or branch point, so $ \\sqrt{5}$.\r\n\r\n2. Substitute $ z^3$ into the power series for $ \\exp z$; it will converge everywhere. The $ n$th derivatives of $ f$ evaluated at $ 0$ will vanish for $ n$ not a multiple of $ 3$ and for $ n \\equal{} 3k$ will be something like $ \\frac{(3k)!}{k!}$." } { "Tag": [ "\\/closed" ], "Problem": "i recall there being 63 users on for the max", "Solution_1": "Yes, that was after 21 November. Unfortunately we lost much of the data from 21 November after in the attack, so some posts and data was lost. See the global annoucements." } { "Tag": [ "calculus", "real analysis", "real analysis theorems" ], "Problem": "I just got Spivak's book a couple days ago, I've flipped through it and was looking through the Index. Does Spivak cover Related Rates or Optimization? I did not find it at all, so I am curious and wondering if it's under some other name. Thanks!", "Solution_1": "See P.204", "Solution_2": "[quote=\"kunny\"]See P.204[/quote]Aw thank you, now that mystery has been solved." } { "Tag": [ "AMC" ], "Problem": "I saw a solution posted in another thread, but I wanted to ask why this solution did not work for me:\r\n\r\n[hide]You can set the sides equal to $a^{2}-b^{2}$, $2ab$, and $a^{2}+b^{2}$.\n\nTherefore, $(a-b)(a+b)(ab) = 3[2a(a+b)]$\n\nThis cancels to $(a-b)(b)=6$, giving you [size=125][b]4[/b] [/size]answers. [/hide]", "Solution_1": "The Pythagorean triple generating formulae do not generate all triples. They only generate all [i]primitive[/i] triples. For example, $(9, 12, 15)$ cannot be generated.\r\n\r\nThe [b]full[/b] Pythagorean triple generating formulae are $c(a^{2}-b^{2}), 2abc, c(a^{2}+b^{2})$ where $c$ cannot be represented as a sum of two squares." } { "Tag": [ "inequalities" ], "Problem": "For any positive integer $ n$, let $ N_n$ be the set of integers from 1 to $ n$, i.e. $ N_n \\equal{} \\{1, 2, 3, \\cdots, n\\}$. Now assume that $ n \\ge 10$. Determine the maximum value of $ n$ such that the following inequality\r\n\\[ \\text{min}_{\\substack{\r\na, b \\in A \\\\\r\na \\neq b\r\n}} |a \\minus{} b| \\leq 10\r\n\\] holds for each $ A \\subset N_n$ with $ |A| \\ge 10$.", "Solution_1": "i think the answer is n = 99 :)", "Solution_2": "What is your solution?", "Solution_3": "[hide=\"Solution\"]\nAssume $ n >\\equal{} 100$ Clearly we have a contradiction since there will always be the following set which will not hold\n\n\\[ \\{ 1, 12, 23, 34, 45, 56, 67, 78, 89, 100 \\} \\]\n\nWhen $ n \\equal{} 99$ \n\nConsider the $ 9$ pigeon holes\n\n$ [1,11], [12, 22], [23, 33], \\ldots, [89, 99]$\n\nIf $ |A| >\\equal{} 10$ clearly one hole must contain more than one number, and their difference will be less than 10\n\nTherefore $ \\boxed{99}$ suffices[/hide]", "Solution_4": "But mustn't the condition hold for all subsets of $ N_n$? In that case the answer would be 11 right :P Or am I misunderstanding the problem", "Solution_5": "The problem says that $ |A| \\geq 10$.", "Solution_6": "This means that the number of elements in $ A$ is greater than or equal to 10 right?", "Solution_7": "Yes, so only those sets with at least 10 elements have to satisfy the given inequality.", "Solution_8": "Shucks I misunderstood... I get it now thanks :)" } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Hello,\r\nHere is a question, which is probably not so hard, but I'm not sure how answer:\r\n\r\nLet $ H_1, H_2$ be Hilbert spaces and $ A_1 \\subset H_1, A_2 \\subset H_2$ their subsets.\r\nSuppose there is a transformation $ T: A_1 \\to A_2$ such that for all $ x_1, x_2 \\in A_1$,\r\n$ < Tx_1, Tx_2 > \\equal{} < x_1, x_2 >$. (The corresponding inner/scalar product in H_2 and H_1.)\r\nThen there exists a unitary extension of T (to all H_1). (It is proved.)\r\nThe question is:\r\nIn case $ \\bar{span{A_1}} \\equal{} H_1$ (that is [b]closure[/b] of span of A_1 equals to H_1),\r\nprove that this extension is unique.\r\n\r\nHow do we know it's unique?\r\n\r\nThanks!", "Solution_1": "[hide=\"Hint\"]If there are two candidates $ T_1$ and $ T_2$, then define $ A'\\equal{}\\{x\\in H_1|T_1(x)\\equal{}T_2(x)\\}$. Now show that the closed set $ A'$ is dense in $ H_1$.[/hide]" } { "Tag": [ "integration", "Ring Theory" ], "Problem": "\u0395\u03bd\u03b1 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03bf\u03bd \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03c9\u03bc\u03b1: :wink: \r\n\r\n[img]http://glyk.kwikphp.com/forum/img/Maths/mimetex1.gif[/img]\r\n\r\n\u03a4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03b8\u03b1 \u03c4\u03b7\u03bd \u03b3\u03c1\u03ac\u03c8\u03c9 \u03c3\u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03bc\u03bf\u03c5(\u03ad\u03bd\u03b1 \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03b3\u03b9\u03b1 \u03bc\u03b1\u03b8\u03b7\u03c4\u03ad\u03c2 \u03c4\u03c1\u03af\u03c4\u03b7\u03c2 \u03bb\u03c5\u03ba\u03b5\u03af\u03bf\u03c5) [url]http://glyk.kwikphp.com/forum/viewtopic.php?id=19[/url]", "Solution_1": "Mailo..na se rotiso kati..ti rolo pezi pu to iporizo tu deyteru olokliromatos ine grameno me aytai tin morfi?afu grafete os..$Inx$ :huh: .....Meta..arkei na 8esis....\r\n$x=e^{u^{2}}$ \r\n+telioses.... :wink: (an dn exo kani kana la8os...)\r\n\r\n :D", "Solution_2": "Mallon mperdeuei tous mathites etsi kai isws gi'auto na egine i metatropi auti. Ki egw etsi tin antimetwpisa RAKAR kai esteila ti lisi sto parapanw forum tou kainouriou filou mas. Meta eriksa mia matia oti to en logw oloklirwma (meta pou tha to ftiakseis) einai mia askisi tou diagwnismou tou ASEP tou 2002 (akrivws idia). Nomizw arketa einai arketa vati gia kapoion pou gnwrizei oti apo mono tou to oloklirwma $\\int e^{x^{2}}dx$ den iparxei (se analitiki morfi), opote anagastika tha prepei na xrisimopoiisei me kapoio tropo kai to 2o oloklirwma mplekontas ta metaksi tous.\r\n\r\nAlexandros", "Solution_3": "Cretanman,ontos,dn ixa di tn lisi sto alo forum....(etsi akrivos tin elisa+ego)\r\nOntos to olokliroma ayto....(anaferome sto$\\int e^{x^{2}}dx$ )ine arketa ... +tis perisoteres fores i emfanisi tu+mono (logo tu oti dn exi analitiki morfi) ine ena kinitro gia to yaksimo mias idiomorfis lisis.... :wink: \r\n\r\n :D", "Solution_4": "[quote=\"cretanman\"]Mallon mperdeuei tous mathites etsi kai isws gi'auto na egine i metatropi auti. Ki egw etsi tin antimetwpisa RAKAR kai esteila ti lisi sto parapanw forum tou kainouriou filou mas.\nAlexandros[/quote]\r\n\r\n\u039d\u03b1\u03b9 \u03c4\u03b7\u03bd \u03bc\u03b5\u03c4\u03b1\u03c4\u03c1\u03bf\u03c0\u03ae \u03b1\u03c5\u03c4\u03ae $\\sqrt{lnx^{2}+ln\\frac{1}{x}}$ \u03c4\u03b7\u03bd \u03ad\u03ba\u03b1\u03bd\u03b1 \u03b5\u03b3\u03ce \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03c0\u03b5\u03c1\u03b4\u03b5\u03cd\u03b5\u03b9 \u03c4\u03bf\u03c5\u03c2 \u03bc\u03b1\u03b8\u03b7\u03c4\u03ad\u03c2 :roll: \r\n\r\n\u03a0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03b1\u03c2, \u03b1\u03bb\u03bb\u03ac \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03ac\u03bb\u03bb\u03bf\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2, \u03c3\u03c7\u03b5\u03b4\u03cc\u03bd \u03af\u03b4\u03b9\u03bf\u03c2, \u03b1\u03c1\u03ba\u03b5\u03af \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03ae\u03c3\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03bf\u03b9 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 $e^{x^{2}}$ \u03ba\u03b1\u03b9 $\\sqrt{lnx}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03c1\u03bf\u03c6\u03b5\u03c2.\r\n\r\n\u03a5\u0393. \u0391\u03bb\u03b5\u03be\u03b1\u03b4\u03c1\u03b5 \u03c3\u03b5 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03c5 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b5\u03c2 \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03c3\u03c4\u03bf \u03c6\u03bf\u03c1\u03bf\u03c5\u03bc \u03bc\u03bf\u03c5 :P", "Solution_5": "\u0391\u03bb\u03b5\u03be\u03b1\u03bd\u03b4\u03c1\u03b5, \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5\u03b9\u03c2 \u03c0\u03c9\u03c2 \u03b2\u03b3\u03b7\u03ba\u03b5 \u03b1\u03c5\u03c4\u03bf:\r\n\r\n$\\left[xe^{x^{2}}\\right]_{0}^{1}=1$", "Solution_6": "[quote=\"mailo\"]\u0391\u03bb\u03b5\u03be\u03b1\u03bd\u03b4\u03c1\u03b5, \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5\u03b9\u03c2 \u03c0\u03c9\u03c2 \u03b2\u03b3\u03b7\u03ba\u03b5 \u03b1\u03c5\u03c4\u03bf:\n\n$\\left[xe^{x^{2}}\\right]_{0}^{1}=1$[/quote]\r\n\r\nKata lathos. To diorthwsa idi. Fisika einai iso me $e$.\r\n\r\nAlexandros", "Solution_7": "\u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03b5, \u03ad\u03ba\u03b1\u03bd\u03b1 \u03b1\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7 \u03c3\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b9\u03bf \u03ba\u03b1\u03c4\u03b1\u03bd\u03bf\u03b7\u03c4\u03ae. \u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 \u03ba\u03b1\u03b9 \u03ac\u03bb\u03bb\u03b7 \u03bc\u03b9\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03b4\u03b9\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b4\u03b9\u03ba\u03b9\u03ac \u03c3\u03bf\u03c5 \u03ac\u03bb\u03bb\u03b1 \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03cc \u03c4\u03c1\u03cc\u03c0\u03bf:\r\n\r\n[url]http://glyk.kwikphp.com/forum/viewtopic.php?pid=39#p39[/url]", "Solution_8": "\u039c\u03b9\u03b1 \u03b5\u03c1\u03c9\u03c4\u03b7\u03c3\u03b7. \u039c\u03c0\u03bf\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03c0\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 :?: \r\n$\\int_{0}^{1}e^{x^{2}}=I-\\int_{1}^{e}\\sqrt{lnx}=I-\\left[\\frac{2}{3}x\\sqrt{(lnx)^{3}}\\right]_{1}^{e}=e-\\frac{2}{3}e=\\frac{e}{3}$" } { "Tag": [ "function", "number theory proposed", "number theory" ], "Problem": "If $ 3^k | n$ then prove that $ 3^{2k}| (1\\plus{}C_2^1\\plus{}C_4^2\\plus{}....\\plus{}C_{2n\\minus{}2}^{n\\minus{}1})$", "Solution_1": "I think that the solution of this require an explicit formula for the sumation. Please some body knows a formula? \r\nIs know that C(0;0) + C(2;1)x + C(4;2)x^2 + ... = 1/(1-4x)^1/2 (generating functions)\r\nBut a formula for finitely many terms is unknow for me." } { "Tag": [ "inequalities", "algebra unsolved", "algebra" ], "Problem": "For every positive integer $ n,p$, prove that\r\n\r\n$ (1\\plus{}\\frac{2006p^2}{1\\plus{}2007p})(1\\plus{}\\frac{2006p^2}{2(2\\plus{}2007p)})...(1\\plus{}\\frac{2006p^2}{n(n\\plus{}2007p)}<\\frac{(2007p)!}{p!(2006p)!}$.\r\n\r\nIs it possible to replace $ \\frac{(2007p)!}{p!(2006p)!}$ with some smaller number?", "Solution_1": "\\[ (1 \\plus{} \\frac {2006p^2}{1 \\plus{} 2007p})(1 \\plus{} \\frac {2006p^2}{2(2 \\plus{} 2007p)})...(1 \\plus{} \\frac {2006p^2}{n(n \\plus{} 2007p)})\\equal{}\\prod_{k\\equal{}1}^n\\frac{(n\\plus{}p)(n\\plus{}2006p)}{n(n\\plus{}2007p)}\\equal{}\\frac{(2007p)!}{p!(2006p)!}\\prod_{k\\equal{}1}^p\\frac{1}{1\\plus{}\\frac{2006p}{n\\plus{}k}}<\\frac{(2007p!)}{p!(2006p)!}.\\]\r\nTherefore it is best evaluation." } { "Tag": [ "trigonometry", "rotation", "geometry", "geometric transformation", "reflection" ], "Problem": "I've got a couple problems from homework that I'm not sure how to solve. I've tried manipulating them some and doing substitutions, but math isn't one of my big strengths :( \r\n\r\nIf anyone could help solve these, it'd be greatly appreciated.\r\n\r\n#1: Verify the identity (trig proof): cos(pi+x)=-cosx\r\n\r\n\r\n\r\n#2: Write the product as a sum or difference: 4sinxcos3x\r\n\r\n\r\nThanks in advance! :roll: \r\n\r\n[url=http://www.math123xyz.com]Interactive Multimedia Math Resources[/url]", "Solution_1": "For the first one use $ \\cos(\\alpha \\plus{} \\beta)\\equal{}\\cos \\alpha \\cos \\beta \\minus{}\\sin \\alpha \\sin \\beta$\r\n\r\nFor second one use $ \\sin \\alpha \\cos \\beta\\equal{}\\frac {1}{2} (\\sin (\\alpha \\minus{} \\beta)\\plus{}\\sin(\\alpha \\plus{}\\beta))$", "Solution_2": "For #1, you can also note that since:\r\n\r\n$ \\cos \\equal{} \\minus{} \\cos(\\pi \\minus{} x)$\r\n\r\nSo,\r\n\r\n$ \\cos(\\pi \\plus{} x)$\r\n$ \\equal{} \\minus{} \\cos(\\pi \\minus{} (\\pi \\plus{} x))$\r\n$ \\equal{} \\minus{} \\cos( \\minus{} x)$\r\n$ \\equal{} \\minus{} \\cos(x)$", "Solution_3": "Carbon57, the identity you cite is equivalent to the identity desired. \r\n\r\n#1 has a very intuitive interpretation in terms of the unit circle: the point $ (x, y) \\equal{} (r \\cos \\theta, r \\sin \\theta)$, when rotated $ 180^{\\circ}$, becomes the point $ ( \\minus{} x, \\minus{} y) \\equal{} (r \\cos (\\theta \\plus{} \\pi), r \\sin (\\theta \\plus{} \\pi))$. Note that a rotation by $ 180^{\\circ}$ is just a reflection through the origin." } { "Tag": [ "calculus", "integration", "derivative", "vector" ], "Problem": "A body of mass m was slowly hauled up the hill.(Fig. 1.29) by a force F which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base l, and the coefficient of friction k.\r\nI have trouble understanding the solution of this problem since it involves quite a bit of calculus. So, I would like to see a more elementary solution, if possible, for this problem. The reason I don't understand much about the solution from the book is because I don't have any background of calculus in Physics. I meant I don't know if integral sth turns out to be sth else except for rectilinear motion.\r\nHere's a solution as well as the figure( all included in the following file):", "Solution_1": "I agree the solution is a bit confusing but I'll see if I can explain it to you:\r\nIt seems to me that the book is using $ A$ as its notation for work. Work is defined as the dot product of force and distance: $ A\\equal{}\\vec{F}\\cdot\\vec{r}$, which is a scalar. The book is taking the amount of work done at each infinitesimal point and then adding them together, which is what integration is doing. So $ A\\equal{}\\int\\vec{F}\\cdot d\\vec{r}$, which basically means that the total work is the sum of the work done at each point on the path.\r\n\r\nYou may want to learn some basic differentiation and integration before you continue with physics-you'll need it.", "Solution_2": "[quote=\"JRav\"]I agree the solution is a bit confusing but I'll see if I can explain it to you:\nIt seems to me that the book is using $ A$ as its notation for work. Work is defined as the dot product of force and distance: $ A \\equal{} \\vec{F}\\cdot\\vec{r}$, which is a scalar. The book is taking the amount of work done at each infinitesimal point and then adding them together, which is what integration is doing. So $ A \\equal{} \\int\\vec{F}\\cdot d\\vec{r}$, which basically means that the total work is the sum of the work done at each point on the path.\n\nYou may want to learn some basic differentiation and integration before you continue with physics-you'll need it.[/quote]\r\nI understand the last part in the book's solution. However, I can't quite get why $ dA\\equal{}\\vec{F}\\cdot d\\vec{r}$. Shouldn't it be the integral of the whole thing:$ d vec{F}.vec{r}$??\r\n@JRav: I learned differentiation and integration already though, but I just learn them at the most basic level. However, I have no clue what does it mean in Physics and Irodov's book shows a lot of solutions like this one. Ca you show me some meanings of these integral and differentiation in Physics? Like in rectilinear motion, differentiation of $ s(t)\\equal{}v(t)$ and integral of $ v(t)\\equal{}s(t)$. I just learned the meaning of integral and differentiation for rectilinear motion and haven't learned any other meanings for other things(like trajectory, circular motion,...) yet :( .", "Solution_3": "It's still integration and differentiation like you learned. That is just notation. $ \\int\\vec{F}\\cdot\\,d\\vec{r}$ simply means the integral of the product of force, and some infinitesimal distance. So let's say we wanted the work done in moving an object in some gravitational field, of mass $ m$, gravitational constant $ g$, and distance $ h$. You should already suspect that the answer is just $ mgh$ but here's how:\r\n$ \\vec{F}$ is the force we need to exert, which is just the weight, $ mg$. The infinitesimal distance is $ dy$ in the vertical direction. We are integrating this vertical distance from 0 to $ h$ so we have\r\n$ W\\equal{}\\int\\vec{F}\\cdot\\,d\\vec{r}\\equal{}\\int_{0}^{h}mg\\,dy\\equal{}mg\\bigg|_{0}^{h}\\equal{}mgh\\minus{}0\\equal{}mgh$\r\nDoes that make any sense?", "Solution_4": "[quote=\"JRav\"]It's still integration and differentiation like you learned. That is just notation. $ \\int\\vec{F}\\cdot\\,d\\vec{r}$ simply means the integral of the product of force, and some infinitesimal distance. So let's say we wanted the work done in moving an object in some gravitational field, of mass $ m$, gravitational constant $ g$, and distance $ h$. You should already suspect that the answer is just $ mgh$ but here's how:\n$ \\vec{F}$ is the force we need to exert, which is just the weight, $ mg$. The infinitesimal distance is $ dy$ in the vertical direction. We are integrating this vertical distance from 0 to $ h$ so we have\n$ W \\equal{} \\int\\vec{F}\\cdot\\,d\\vec{r} \\equal{} \\int_{0}^{h}mg\\,dy \\equal{} mg\\bigg|_{0}^{h} \\equal{} mgh \\minus{} 0 \\equal{} mgh$\nDoes that make any sense?[/quote]\r\nThank you very much JRav! Your explantation helps me a lot. :) Can you show me more integral's application for other things such as the net accleration in circular motion? Irodov shows that it is equal to $ \\sqrt{angular acceleration^2\\plus{}linear acceleration^2}$, which I kind of assuming that he got it from vector component in the xy-plane when an object moves around the circle without friction.", "Solution_5": "Well yes, that's the magnitude of total acceleration. But I wouldn't use an integral to determine \"net acceleration\". When you get into E&M, there are a lot of applications of integrals." } { "Tag": [ "geometry", "3D geometry", "induction", "algebra", "polynomial", "limit", "system of equations" ], "Problem": "what is general formula for \r\n\r\n1^4+2^4+3^4.............\r\n\r\nif anything is there \r\n\r\ncan anybody please post the derivation", "Solution_1": "$\\binom{k+1}{5}-\\binom{k}{5}=\\binom{k}{4}$ for all $k\\in\\mathbb{N}$.\r\nSum up this equations to get $\\binom{n+4}{5}=\\binom{n+4}{5}-\\binom{4}{5}=\\sum_{k=4}^{n+3}\\binom{k}{4}=\\sum_{k=1}^{n}\\binom{k+3}{4}$ for all $n\\in\\mathbb{N}$.\r\nNow multiply this with $4!$ to get:\r\n${n(n+1)(n+2)(n+3)(n+4)\\over 5}=\\sum_{k=1}^{n}k(k+1)(k+2)(k+3)\\\\ \\hspace*{2in}=\\sum_{k=1}^{n}(k^{4}+6k^{3}+11k^{2}+6k)$\r\nNow use formulas for $\\sum k$,$\\sum k^{2}$ and $\\sum k^{3}$ to get $\\sum k^{4}$.\r\n\r\nThere is an explicit general formula for $\\sum k^{m}$ using Bernoulli numbers, but can't remember it right now.", "Solution_2": "can we do without using any combinatorics\r\n\r\nusing algebra \r\n\r\nhowever what is the simple formula to calculate 1^4+2^4+3^4.......\r\n\r\nplease do the derivation using algebra\r\n\r\ni am not familiar with combinatorics", "Solution_3": "What do you mean by using algebra, do you know of any way of deriving it for cubes using only algebra?\r\n\r\n\r\nDo you mean method of solving system of equations for coefficients in the induction step?", "Solution_4": "[quote=\"olorin\"] \nThere is an explicit general formula for $\\sum k^{m}$ using Bernoulli numbers, but can't remember it right now.[/quote]\r\n\r\n$\\sum_{j=1}^{n}j^{k}= \\sum_{i=1}^{k}S(k,i)\\binom{n+1}{i+1}i!$ :)", "Solution_5": "That's the formula using Stiling Numbers of the second kind, not Bernoulli numbers.", "Solution_6": "But there is one with Bernoulli numbers. I should have just wikied it. :blush: \r\n[url]http://en.wikipedia.org/wiki/Bernoulli_number[/url].", "Solution_7": "You can also try to use Abel method: let $f(x)=\\sum_{k=0}^{n}x^{k}=\\frac{x^{n+1}-1}{x-1}$. Now step by step $f^{'}(x)=\\sum_{k=1}^{n}kx^{k-1}$, $f^{'}(1)=\\sum_{k=1}^{n}k$ $f^{j}(x)=\\sum_{k=j}^{n}k(k-1)\\dots(k-j+1)x^{k-j}$.", "Solution_8": "[quote=\"boxedexe\"]That's the formula using Stiling Numbers of the second kind, not Bernoulli numbers.[/quote]\r\n\r\ndid i say that was the formula using Bernoulli numbers? :wink: :wink: \r\n\r\nwow...", "Solution_9": "Logically, when you quote someone referring to Bernoulli numbers, and you immediately answer them without any explanation, then people think that you are answering to the exact question (or something related to it). :wink: Sorry if you felt insulted because of that.", "Solution_10": "[quote=\"Yustas\"]You can also try to use Abel method: let $f(x)=\\sum_{k=0}^{n}x^{k}=\\frac{x^{n+1}-1}{x-1}$. Now step by step $f^{'}(x)=\\sum_{k=1}^{n}kx^{k-1}$, $f^{'}(1)=\\sum_{k=1}^{n}k$ $f^{j}(x)=\\sum_{k=j}^{n}k(k-1)\\dots(k-j+1)x^{k-j}$.[/quote]\r\n\r\nNice Idea!\r\nBut $f^{(j)}(x)$ has the form ${G_{j}(x)\\over (x-1)^{j+1}}$ for some polynomial $G_{j}$, \r\nso you need to apply l'H\u00f4pital's rule $(j+1)$-times to get $f^{(j)}(1)$\r\n\r\n\r\nDefine a sequence of polynomials $\\{P_{m}\\}_{m=1}^{\\infty}$, with \r\n$P_{1}(x)=\\sum_{k=0}^{n}x^{k}=\\frac{x^{n+1}-1}{x-1}$ and $P_{m+1}(x)=xP_{m}'(x)$ for all $m\\geq 1$.\r\nThen $P_{m+1}(x)=\\sum_{k=1}^{n}k^{m}x^{k}$ and $P_{m+1}(1)=\\sum_{k=1}^{n}k^{m}$.\r\n\r\nNow define $Q_{m}(x)=(x-1)^{m}P_{m}(x)$.\r\nThen $Q_{1}(x)=x^{n+1}-1$ and $Q_{m+1}(x)=x(x-1)Q_{m}'(x)-mxQ_{m}(x)$ for all $m\\geq 1$, with\r\n$\\sum_{k=1}^{n}k^{m}=P_{m+1}(1)=\\lim_{x\\rightarrow 1}{Q_{m+1}(x)\\over (x-1)^{m+1}}={Q_{m+1}^{(m+1)}(1)\\over (m+1)!}$." } { "Tag": [], "Problem": "Ed takes five 100-point tests in his algebra class. He scores 87, 85 and 87 points on the first three tests. If the scores of his last two tests differ by three points and he has a $ 90\\%$ average on his five tests, what was his highest test score?", "Solution_1": "87 is 3 under 90\r\n85 is 5 under 90\r\n87 is 3 under 90\r\n\r\nIn order to maintain an average of 90, the last two test scores must have been 3+5+3=11 over 90. Since one of them is higher by three, the higher one is (11+3)/2=7 over 90, whichi is 97", "Solution_2": "We could also do:\r\n\r\nIf the average is 90, then the sum of his 5 test scores is $ 90 \\cdot 5 \\equal{} 450$. We subtract the sum of his other 3 test scores to get $ 191$. We then have:\r\n\r\n$ x \\plus{} x \\plus{} 3 \\equal{} 191$\r\n\r\n$ 2x \\plus{} 3 \\equal{} 191$\r\n\r\n$ 2x \\equal{} 188$\r\n\r\n$ x \\equal{} 94$\r\n\r\nSo the higher test score is $ 94 \\plus{} 3 \\equal{} 97$." } { "Tag": [ "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Show that for any positive sequence $ \\{a_n\\}_{n\\equal{}1}^{\\infty}$,we have $ \\limsup_{n\\rightarrow\\infty} (\\frac{1\\plus{}a_{n\\plus{}1}}{a_n})^n>e$.", "Solution_1": "$ >e$ or $ \\geqq e$?", "Solution_2": "$ >e$ :)", "Solution_3": "already posted .. :lol:", "Solution_4": "To prove $ \\limsup_{n\\rightarrow\\infty} (\\frac {1 \\plus{} a_{n \\plus{} 1}}{a_n})^n\\geq e$ is trivial,but I strongly guess the equlity can never occur. :)", "Solution_5": "still no one? :( Now I have prove it if $ \\{a_{n\\plus{}1}\\}$ does not tend to $ \\infty$", "Solution_6": "When $ a_n \\equal{} n \\ln n$, I believe the lim sup is $ e$.", "Solution_7": "[quote=\"Ravi B\"]When $ a_n \\equal{} n \\ln n$, I believe the lim sup is $ e$.[/quote]\r\nIf $ a_n \\equal{} n \\left(\\ln n\\right)^k$ with $ k > 0$, then the [b]limit[/b] is $ e$.", "Solution_8": "thank Ravi B and hpe! :)" } { "Tag": [], "Problem": "Are there any schools near Chapel Hill, NC that would allow me to take the B test there? Just curious.", "Solution_1": "You could always move down into the lower part of South Carolina and join my school :D . Then we would win EVERY (local) competition." } { "Tag": [ "graph theory", "combinatorics unsolved", "combinatorics" ], "Problem": "There are $100$ cities in a country, some of them being joined by roads. Any four cities are connected to each other by at least two roads. Assume that there is no path passing through every city exactly once. Prove that there are two cities such that every other city is connected to at least one of them.", "Solution_1": "I dont remember the exact solution,but as far as I remember the fbegining was as wollow...\r\n[hide]first of all name the graph $G$...\nin the case that $G$ isnt connected,the problem is easy to solve...\nso assume that $G$ is connected,now name the vertices of $G$ from $1$ to $100$\nnow if any $2$ vertex were connected the we would have a hamiltonian path,so there exist $2$ vertices that are not connecte, WLOG name them $1$ and $(($,now if every other vertex was connected to at least one of $1,99$ then the problem is solved,so there exists a vertex that is not connected to $1$ and not to $99$,again WLOG name it $100$...\nso any other vertex $i$,must be connected to at least $2$ of the $1,99,100$ beacause if not,then vetices $1,i,99,100$ are $4$ vertices with less than $2$ edges...\nso every $i$ is connected to at least tow of $1,99,100$ so it must be connected to at least one of the vertices $1,99$...\nso every other vertices (other than $100,99,1$) must be connected to $1$ or $99$ or both of them...\nnow take the longest path between $1,99$ (such path exist since we assumed that the graph is connected...)\nso this path must cover every vertices between $1,...,99$...\nnow WLOG name these vertices in increasing order $1,2,...,99$\nnow we prove that if $i$ is connected to $99$ then so is $i+1$,becuaes otherwise it must be connected to $1$ so the cycle $1,...,i,99,98,...,i+1,1$ is a hamiltonian cycle between $99$ vertices of $G$ so somhow $G$ would have a hamiiltonian path...\nthis contradiction shows that if $i$ is connected to $99$ then so is $i+1$[/hide]", "Solution_2": "I dont understand this step:\r\n\r\nso this path must cover every vertices between $1,...,99$...\r\n\r\nWhy this path cover every vertices and does vertex $100$ belong to this path(I think it should)? :maybe:", "Solution_3": "I think this will work:\r\ntake the longest path in $G$,and name its ends $1,99$ now we may assume that $1,99$ are not connected,we will prove this:\r\nproof:assume that $1,99$ are connected,so the longest path is actually a cycle,now in this cycle,if there were two un-connected vertices like $i,j$ we would take these vertices as the ends of the path...so every 2 vertices of this cycle are connected,so we have a complete graph,now for any other vertex not in this cycle (there exist such vertex) like $k$,we see that $k$ could not be connected to any vertex from this cycle...(otherwise we would have a longer path,which is a contradiction)\r\nthis contradiction shows that the ends of the longest path in this graph are not connected...\r\nname these ends $1,99$ now if every other vertex was connected to at least one of these 2 vertex,then the problem is solved,so there exists a vertex $100$ such that it isnt connected to $1,99$ we want to gain a contradiction:\r\nso there isnt any edge between vertices $1,99,100$,now for any other vertex like $i$ we get that $i$ must be connected to at least one of the vertices $1,99$...\r\nand also because $1,99$ are the ends of the longest path,so their neighbours must be in this path,and from the last thing that I said,we get that every vertex except $100$ is in the path...\r\nnow WLOG name the vertices of this path in increasing order $1,...,99$ now note that if $i$ was connected to $99$ as shown in the last post,$i+1$ must be connected to $99$ to... otherwise we would have a cycle along vertices $1,...,99$ and we kknew that $G$ was connected so vertex number $100$ has a neighbour along vertices $1,...,99$ let it be $t$ so the path $100,t,t+1,...,99,1$ will be a longer path and also a hamiltonian path,which is a contradiction...\r\nso if $i$ was connected to $99$ then is also $i+1$...\r\nnow we just have to prove that this path is a cycle,because if we show this then as showed above,we can show that the graphb has a hamiltonian path which is a contradiction..." } { "Tag": [ "vector" ], "Problem": "If $ a\\equal{}\\minus{}2$, the largest number in the set $ \\left \\{ \\minus{}3a,4a,\\frac{24}{a},a^2,1 \\right \\}$ is\r\n\r\n\\[ \\textbf{(A)}\\ \\minus{}3a \\qquad\r\n\\textbf{(B)}\\ 4a \\qquad\r\n\\textbf{(C)}\\ \\frac{24}{a} \\qquad\r\n\\textbf{(D)}\\ a^2 \\qquad\r\n\\textbf{(E)}\\ 1\r\n\\]", "Solution_1": "Since $ a$ is negative, you'd need the scalar in front to be negative as well to maximze direction. So it is $ \\minus{}3a$. $ a^{2}$ is also positive, but that is only 4, whereas $ \\minus{}3a\\equal{}6$.", "Solution_2": "[hide=\"I dont really understand what scalar and direction are, so...\"]\n\\begin{align*}\n-3a&=6 \\\\ 4a&=-8 \\\\ \\frac{24}{a}&=-12 \\\\ a^2&=4 \\\\ 1&=1\n\\end{align*}\n\nClearly the largest is $ -3a$, or $ \\boxed{\\textbf{(A)}}$.\n[/hide]", "Solution_3": "A scalar is a coefficient, and direction is either positive or negative.", "Solution_4": "Scalar doesn't have a direction; vector is the one which has directions.", "Solution_5": "Every number has a magnitude and a direction. Magnitude is size. So if I ask whether 6 or 10 has a bigger magnitude, it's 10. Also, if I ask whether -10 or 3 has a bigger magnitude, it's 10, despite the fact that it's negative. Magnitude doesn't care whether it's negative or positive, just how big. Direction cares about negative and positive. So when you ask which number is the largest, you need to take both magnitude and direction into account. So $ \\left|\\minus{}10\\right|>\\left|2\\right|$, but because 2 has positive direction, it is larger.", "Solution_6": "I dont really understand what scalar and direction are, so...", "Solution_7": "[quote=KrrishPradhani]I dont really understand what scalar and direction are, so...[/quote]\n\nSeems you copied the name of the hide tag in @Post 3. Try rereading the thread." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Note: I'm looking only for solutions by AM-GM. I'm sure that these inequalities are easily proven by stronger inequalities than AM-GM.\r\n\r\n1. Given $ x,y,z>0$, prove that $ \\frac{1}{3x\\plus{}y}\\plus{}\\frac{1}{3y\\plus{}z}\\plus{}\\frac{1}{3z\\plus{}x}\\ge\\frac{1}{2x\\plus{}y\\plus{}z}\\plus{}\\frac{1}{2y\\plus{}z\\plus{}x}\\plus{}\\frac{1}{2z\\plus{}x\\plus{}y}$.\r\n\r\n2. Given $ a,b,c>0$ and $ abc\\equal{}1$, prove that $ \\frac{ab}{ab\\plus{}a^5\\plus{}b^5}\\plus{}\\frac{bc}{bc\\plus{}b^5\\plus{}c^5}\\plus{}\\frac{ca}{ca\\plus{}c^5\\plus{}a^5}\\le 1$.", "Solution_1": "Well, here are the answers for you expectation.\r\nThe first problem, I can only give you a hint\r\n$ \\frac{1}{3x\\plus{}y}\\plus{}\\frac{1}{2z\\plus{}x\\plus{}y} \\ge \\frac{4}{4x\\plus{}2y\\plus{}2z}\\equal{}\\frac{2}{2x\\plus{}y\\plus{}z}$\r\nand after that, you sum up all the inequalities.\r\n\r\nFor the second problem, \r\nwe have $ a^5\\plus{}b^5 \\ge ab(a^3\\plus{}b^3)$ and $ a^3\\plus{}b^3 \\ge ab(a\\plus{}b)\\equal{}\\frac{a\\plus{}b}{c}$( because $ abc\\equal{}1$)\r\nIt follows\r\n$ \\frac{ab}{a^5\\plus{}b^5\\plus{}ab} \\le \\frac{ab}{ab\\plus{}ab(a^3\\plus{}b^3)}\\equal{}\\frac{1}{a^3\\plus{}b^3\\plus{}1}\\le \\frac{1}{\\frac{a\\plus{}b}{c}\\plus{}1}\\equal{}\\frac{c}{a\\plus{}b\\plus{}c}$\r\nFinally, you keep summing up all the similar inequalities as shown.\r\nHave fun!", "Solution_2": "See number two here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=247194" } { "Tag": [ "probability", "calculus" ], "Problem": "Three real numbers are chosen between 0 and 10. What is the probability that their sum is less than 20??\r\n\r\nI've seen this kind of problem many times before, but never knew how to set it up....thanks in advance!", "Solution_1": "That problem was discussed [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=20506]here[/url] a little bit. blahblahblah's solution uses calculus but its basically the same as finding volume." } { "Tag": [ "induction", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $ A,B \\in M_n(R)$ such that $ Rank(AB\\minus{}BA)\\equal{}1,(A\\plus{}B)^n\\equal{}0,A^n\\equal{}0$ Prove that $ B^n\\equal{}0$", "Solution_1": "$ rank(AB\\minus{}BA)\\leq{1}$ implies that $ A,B$ are simultaneously triangularizable......", "Solution_2": "[quote=\"loup blanc\"]$ rank(AB \\minus{} BA)\\leq{1}$ implies that $ A,B$ are simultaneously triangularizable......[/quote]\r\n I don't known prove this simultaneously triangularizable . Please Prove .... Thanks !", "Solution_3": "QuyBac,\r\n\r\nI thought the result that loup blanc mentioned was on this forum, but I can't find it.\r\n\r\nI know the result it true in $ \\mathcal{M}_n(\\mathbb{C})$, not so sure in $ \\mathcal{M}_n(\\mathbb{R})$, but it doesn't matter for the inital question, you show that the matrices are nilpotent in $ \\mathcal{M}_n(\\mathbb{C})$, then they are nilpotent in $ \\mathcal{M}_n(\\mathbb{R})$.\r\n\r\nTo prove that $ \\mathrm{rank}(AB \\minus{} BA)\\leqslant 1$ implies that $ A,B$ are simultaneously triangularizable, you start to show that that $ B(\\ker A) \\subset \\ker A$ or $ B(\\mathrm{Im} A) \\subset \\mathrm{Im} A$, then you proceed by induction over $ n$.\r\n\r\nloup blanc, \r\n\r\nAre you sure that if $ A,B \\in \\mathcal{M}_n(\\mathbb{R})$, $ \\mathrm{rank}(AB \\minus{} BA)\\leqslant 1$ implies that $ A,B$ are simultaneously triangularizable in $ \\mathcal{M}_n(\\mathbb{R})$?", "Solution_4": "i think you don't need such a theorem to solve that problem the proof is more elementary....well i think i have such a proof i will check my proof and i will post it", "Solution_5": "Hi JC_math,\r\nof course the field must be algebraically closed. Here if $ A,B\\in\\mathcal{M}_n(K)$ then we prove the result in $ \\mathcal{M}_n(\\bar{K})$ where $ \\bar{K}$ is the algebraical closure of $ K$.\r\nReferences:\r\n1) Laffey / Simultaneous triangularization of matrices low rank cases and non derogatory case. Linear and multilinear algebra 6 (1978) p. 269-305.\r\nAbout the simult. Triang., the bible is:\r\n2) Radjavi, Rosenthal / Simultaneous triangularization, Springer, 2000", "Solution_6": "Hi loup blanc \r\n\r\nRadjavi and Rosenthal's book is very nice, and a good source for problems (that I used a few times :wink: )", "Solution_7": "[quote=\"othman\"]i think you don't need such a theorem to solve that problem the proof is more elementary....well i think i have such a proof i will check my proof and i will post it[/quote]\r\n I will wait a proof of you !\r\nThanks! :roll:" } { "Tag": [ "function" ], "Problem": "The repeating reciprocal digits function ($RRD(n)$ for short) is a discrete function that returns the number of digits it takes for $\\frac{1}{n}$ to repeat and returns $0$ if it doesn't repeat. For example,\r\n$\\frac{1}{2}= .5$, so $RRD(2) = 0$.\r\n$\\frac{1}{3}= .3333 \\hdots$, so $RRD(3) = 1$.\r\n$\\frac{1}{37}= .027027027 \\hdots$, so $RRD(37) = 3$.\r\n\r\n$a)$ Compute or describe all integers $n$ for which $RRD(n) = 3$.\r\n$b)$ Compute or describe all integers $n$ for which $RRD(n) = 4$.\r\n$c)$ Compute $RRD(2007)$.\r\n$d)$ Prove that $RRD(n) = 2007$ has no solutions.\r\n$e)$ Compute the smallest number $k > 2007$ for which $RRD(n) = k$ has one or more solutions.\r\n$f)$ Describe the set of integers $k$ for which $RRD(n) = k$ has one or more solutions.", "Solution_1": "[hide=\"a\"]\nSuppose $RRD(n) = 3$, then we can write $\\frac{1}{n}= 0.abcabcabc...$ (where a, b and c denote digits).\nNow $\\frac{1}{999}= 0.001001001$, so $\\frac{1}{n}= 0.abcabcabc... = \\frac{abc}{999}$.\nSince we have to be able to write $\\frac{abc}{999}$ as $\\frac{1}{n}$, abc has to divide 999 (and $n = \\frac{999}{abc}$.\nSo all divisors of 999 will be ok for abc, and thus for n, except for those with $a=b=c$ (because then $RRD(n) = 1$).\n\nNow we can easily find all divisors of 999 by its prime factorisation ($999 = 3^{3}*37$).\n\nRemark: as originally stated, $RRD(37) = 3$.\n[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$ a,b,c\\in \\mathbb{R}_\\plus{}$ such that $ a\\plus{}b\\plus{}c\\equal{}1$, prove that \r\n\r\n$ \\frac{1}{a^2\\plus{}b}\\plus{}\\frac{1}{b^2\\plus{}c}\\plus{}\\frac{1}{c^2\\plus{}a}\\geq\\frac{27}{4}$", "Solution_1": "[quote=\"tobeno_1\"]$ a,b,c\\in \\mathbb{R}_ \\plus{}$ such that $ a \\plus{} b \\plus{} c \\equal{} 1$, prove that \n\n$ \\frac {1}{a^2 \\plus{} b} \\plus{} \\frac {1}{b^2 \\plus{} c} \\plus{} \\frac {1}{c^2 \\plus{} a}\\geq\\frac {27}{4}$[/quote]\r\nIt's wrong. See here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=88792" } { "Tag": [ "geometry", "search" ], "Problem": "Does anyone know of a good site online that gos through the basics of geometry (that is, the stuff normally taught in a high school geometry course)? I need to brush up on some of the very basics (ie when triangles are congruent, when they are similar, what happens in cases like AAS where the triangle is not congruent, etc...). :mrgreen: :mrgreen: :mrgreen:", "Solution_1": "cant you just read through the parts in aops again?", "Solution_2": "I only have AoPS 2, not 1 (which I really don't need since I know most of the stuff in it).", "Solution_3": "[quote=\"Fierytycoon\"]Does anyone know of a good site online that gos through the basics of geometry (that is, the stuff normally taught in a high school geometry course)?[/quote]\r\n\r\nHi, Fiery, prompted by your question, and by the fact that my son has a geometry final coming up in about a month, I just Googled \"geometry review\" and found quite a few interesting sites, with varying levels of depth. Next I'll try Googling \"geometry end of course examination\" or \"geometry lecture notes\" or other such search terms. \r\n\r\nHope this helps!", "Solution_4": "tokenadult:\r\n\r\nWell, what a surprise! I had thought that googling up such a general topic like that would yield a boatload of useless and poorly-made websites (having had many such previous experiences), but I actually found a few decent resources in the first 20 matches or so. Thanks." } { "Tag": [ "geometry", "Euler", "Gauss", "articles" ], "Problem": "Hello everybody,\r\n\r\n\r\nI am sad to inform you that Murray Klamkin (1921-2004) passed away on August 6th 2004 because of an intestinal tumour and heart failure.\r\n\r\nHe was arguably the most well-known mathematics problem-solver in the whole world. He left his mark in every major journal which had a problem section (American Mathematical Monthly, Crux Mathematicorum, SIAM Review,....etc).\r\n\r\nWe will all miss him.", "Solution_1": "Unfortunately, I did not read this message before. Yeah, we all will miss him in the problem-solving community. How comes that you get to know from it ? Is there any obituary for him ? I am really interested in it. He made so many valuable contributions !!! :)\r\n\r\nBTW what do you think are the mos well-known people in the problem-solving world ? I am thinking about Pl Erds and Arthur Engel. Any comments requested. :)", "Solution_2": "Well, Pal Erdos was not really a problem-solver but a professional mathematician. There is no doubt that he was one of the greatest mathematicians in the 20th century.\r\n\r\nNow, concerning problem-solving, there are -fortunately- several very talented people. I will just give an example with names of problem-solvers specialized in geometry:\r\n\r\nChris Fisher (Canada), Michael Lambrou (Greece), Da", "Solution_3": "Oops! sorry for this error: here is the list of names:\r\n\r\nChris Fisher (Canada), Michael Lambrou (Greece), Dan Branzei (Romania), Francisco Bellot-Rosado (Spain), Sergey Ivanov (Russia), Michel Bataille (France), Waldemar Pompe (Poland), Istvan Reiman (Hungary), Toshio Seimiya (Japan),...etc\r\n\r\nAnd, in few years, I will be happy to include the name of Darij Grinberg", "Solution_4": "I feel flattered... but why??!! Just because I'm posting day and night on ML? I don't know what I will do in a few years, but there are many people about whom I know well that I will always remain under their level. Why haven't you included the name of Christian Reiher in the list (he is actually doing well with solving [i]open[/i] problems)? And what about Andrey Badzyan, Wolfgang Burmeister, Titu Andreescu, Valentin Vornicu, Reid Barton? I think they have and will have solved much more and much harder problems than me, maybe with the difference that they don't post every of their solutions into a maths forum...\r\n\r\n Darij", "Solution_5": "Darij: all the people you cited are very talented, but I am talking about people who spend 90% of their time thinking about geometry. Furthermore, you are the one who has already published very important results (sometimes with the collaboration of professional mathematician) in a specialized geometric journal. Finally, do not forget that you are \"still\" 16 years old. CONGRATULATIONS!", "Solution_6": "I still assert that Pl Erds is a great problem solver and even greater mathematician. I read some books about famous mathematicians and other scientists including P. Erds, S. Ramanujan, K. Gdel, A. Turing, L. Euler, C.F. Gauss, A. Wiles, A. Einstein, W. Heisenberg, E. Noether, J. Nash, B. Gates etc. Maybe it is just me or do I love reading biographies... :)", "Solution_7": "Well I see people discuss other things in this thread now. But does anyone have any reports or articles about the life of Murray Klamkin ? Does anybody know a nice web site ? :)", "Solution_8": "I read a biography of Erdos, but I also meet him personaly two times (in September 1995 and May 1996). I can tell you that he is most known as a problem-proposer rather than a problem-solver!!", "Solution_9": "[quote=\"Kantor\"]\nChris Fisher (Canada), Michael Lambrou (Greece), Dan Branzei (Romania), Francisco Bellot-Rosado (Spain), Sergey Ivanov (Russia), Michel Bataille (France), Waldemar Pompe (Poland), Istvan Reiman (Hungary), Toshio Seimiya (Japan),...etc [/quote]\r\n\r\nI see you often read Crux Mathematicorum and papers by the Forum Geometricorum. :D", "Solution_10": "Does anyone know of a biography or obituary for Murray Klamkin?", "Solution_11": "[quote=\"orl\"]Well I see people discuss other things in this thread now. But does anyone have any reports or articles about the life of Murray Klamkin ? Does anybody know a nice web site ? :)[/quote]\n\nHi EveryOne,\n\nWhen I read the first post in this thread I wanted to confirm that Klamkin did indeed pass away before I wrote anything. I had trouble finding any death notices, etc., but I finally came across one this evening on the Mathematical Association of Americas website [What follows if from < [url]http://www.maa.org/news/inmemoriam.html[/url] >]:\n\n[quote][b]Murray Klamkin[/b], a long time member of the MAA and one of the original organizers, leaders, and coaches for the American Mathematics Competitions program, died on August 6, 2004 at the age of 83. Klamkin received a Certificate of Merit from the Association in 1978. He was a member of MAA since 1948. [/quote]\r\n\r\n :( \r\n\r\nI'm a novice problem solver, but I have heard about many of the contributions made by Klamkin (for example, his \"Klamkin Quickies\" in Crux, etc.). I was not fortunate enough to meet this man. Even though I only new Klamkin as a name mentioned in periodicals, I think we should honor his memory by continuing what he seem to have liked best --- solving math problems!\r\n\r\nBy the way, I would bet that in the issues of some of next month's mathematical related periodicals, there will be eulogies, etc., about Klamkin.\r\n\r\nYours truly,\r\n\r\nmathfreak", "Solution_12": "Oh...I wasn't aware of Murray's dead...\r\nYou are right : We will miss him.\r\n\r\nPierre.", "Solution_13": "I am very sorry for his death and I think his beautiful problems published \r\n\r\non AMM,CRUX,Mathematics Magazine,.... will miss every people keen on \r\n\r\nproblem solving.", "Solution_14": "[b]Curiosity:[/b]\r\nIn a personal letter, I appreciate from period 1982-1986 (??) , Murray Klamkin has informed me that he had the possibility to study papers written in Romanian.The reason was that his wife has the ancestors in Romania - [Bucharest]." } { "Tag": [ "calculus", "integration", "function", "Putnam", "algebra", "domain", "logarithms" ], "Problem": "evaluate(find a formula -- not value in terms of functions and known constatnts)\r\n\r\n$ \\int_{0}^{1} \\frac{ln(1\\plus{}x^2)}{(1\\plus{}x)} .dx$\r\n\r\n........", "Solution_1": "Well, it's not [i]exactly[/i] the same as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64436]Putnam 2005 A5[/url].\r\n\r\nA similar trick might work: $ \\ln(x^2 \\plus{} 1) \\equal{} \\int_0^{x^2}\\frac1{1 \\plus{} t}\\,dt$, so the integral becomes $ \\int_0^1\\int_0^{x^2}\\frac1{1 \\plus{} x}\\cdot\\frac1{1 \\plus{} t}\\,dt$.\r\nSubstituting $ t \\equal{} x^2y,dt \\equal{} x^2\\,dy$ to make the domain a square, we have\r\n$ \\int_0^1\\int_0^1 \\frac1{1 \\plus{} x}\\cdot\\frac {x^2}{1 \\plus{} x^2y}\\,dy\\,dx \\equal{} \\int_0^1\\int_0^1 \\frac1{1\\plus{}y}\\left(\\frac1{1\\plus{}x}\\plus{}\\frac{yx}{1\\plus{}yx^2}\\minus{}\\frac{y}{1\\plus{}yx^2}\\right) \\,dx\\,dy$\r\n$ \\equal{}\\int_0^1 \\frac1{1\\plus{}y}\\cdot\\left[\\ln(1\\plus{}x)\\plus{} \\frac12\\ln(1\\plus{}yx^2)\\minus{}\\sqrt{y}\\arctan(\\sqrt{y}x)\\right]_{x\\equal{}0}^{x\\equal{}1}\\,dy$\r\n$ \\equal{} \\int_0^1 \\frac{\\ln 2}{1\\plus{}y}\\plus{}\\frac{\\ln(1\\plus{}y)}{2(1\\plus{}y)}\\minus{} \\frac{\\sqrt{y}\\arctan(\\sqrt{y})}{1\\plus{}y}\\,dy$\r\n\r\nAll three terms have elementary antiderivatives.\r\nFor the second: $ \\int_0^1 \\frac{\\ln (1\\plus{}y)}{1\\plus{}y}\\,dy\\equal{}\\int_0^{\\ln 2}u\\,du\\equal{}\\frac12(\\ln 2)^2$ by the substitution $ u\\equal{}\\ln(1\\plus{}y)$.\r\nFor the third: $ \\int_0^1 \\arctan{\\sqrt{y}} \\cdot\\frac{\\sqrt{y}}{1\\plus{}y^2}\\,dy\\equal{}\\int_0^{\\frac{\\pi}{4}}2t\\tan^2 t\\,dt$ by the substitution $ y\\equal{}\\tan^2 t$. We can find that $ \\frac{d}{dt}\\left(t^2\\plus{}2t\\tan t\\plus{}2\\ln(\\cos t)\\right)\\equal{}2t\\tan^2 t$ by integration by parts; the integral of this piece is $ 1\\plus{}\\frac{\\pi}{2}\\plus{}2\\ln\\frac1{\\sqrt{2}}$\r\n\r\nPutting the parts together, the integral is $ \\frac54(\\ln 2)^2\\plus{}1\\plus{}\\frac{\\pi}{2}\\minus{}\\ln 2$.\r\n\r\nI like the Putnam problem better.", "Solution_2": "[quote=\"jmerry\"]\n$ \\int_0^1\\int_0^1 \\frac1{1 \\plus{} x}\\cdot\\frac {x^2}{1 \\plus{} x^2y}\\,dy\\,dx \\equal{} \\int_0^1\\int_0^1 \\frac1{1 \\plus{} y}\\left(\\frac1{1 \\plus{} x} \\plus{} \\frac {yx}{1 \\plus{} yx^2} \\minus{} \\frac {y}{1 \\plus{} yx^2}\\right) \\,dx\\,dy$\n...\nPutting the parts together, the integral is $ \\frac54(\\ln 2)^2 \\plus{} 1 \\plus{} \\frac {\\pi}{2} \\minus{} \\ln 2$.\n.[/quote]\r\nThis is not correct??? the $ y$ on RHS should not be there with $ x$ i.e it should be:-....\r\n$ \\int_0^1\\int_0^1 \\frac1{1 \\plus{} x}\\cdot\\frac {x^2}{1 \\plus{} x^2y}\\,dy\\,dx \\equal{} \\int_0^1\\int_0^1 \\frac1{1 \\plus{} y}\\left(\\frac1{1 \\plus{} x} \\plus{} \\frac {x}{1 \\plus{} yx^2} \\minus{} \\frac {1}{1 \\plus{} yx^2}\\right) \\,dx\\,dy$\r\nand the furthur antiderivatives are not trivial....", "Solution_3": "I was hesitating the same but I didn't post 'cause I thought may be I was mistaken.\r\n\r\n :ninja:", "Solution_4": "OK, I see it- I thought the zeros of the denominator were at $ \\pm\\sqrt{\\minus{}y}$, not $ \\pm\\sqrt{\\minus{}1/y}$.\r\n\r\nNew version: the second and third terms are $ \\int_0^1 \\frac{\\ln(1\\plus{}y)}{2y(1\\plus{}y)}\\,dy$ and $ \\minus{}\\int_0^1 \\frac{\\arctan(\\sqrt{y})}{\\sqrt{y}(1\\plus{}y)}\\,dy$. That last term is easier after the change- we don't need to integrate by parts, and the value is $ \\left(\\frac{\\pi}{4}\\right)^2$.\r\n\r\nThe other term can be broken up by partial fractions; there's $ \\minus{}\\frac12\\int_0^1\\frac{\\ln(1\\plus{}y)}{1\\plus{}y}\\,dy$, which we know how to do, and $ \\frac12\\int_0^1\\frac{\\ln(1\\plus{}y)}{y}\\,dy$, which is a well-known function without an elementary antiderivative. For this term, we can use the power series of the logarithm; we get $ \\frac12\\int_0^1 1\\minus{}\\frac{y}{2}\\plus{}\\frac{y^2}{3}\\minus{}\\frac{y^3}{4}\\plus{}\\cdots\\,dy\\equal{} \\frac12\\left(1\\minus{}\\frac14\\plus{}\\frac19\\minus{}\\frac1{16}\\plus{}\\cdots\\right)\\equal{}\\frac{\\pi^2}{24}$\r\n\r\nThe original integral is $ \\frac34(\\ln 2)^2\\minus{}\\frac{\\pi^2}{48}.$" } { "Tag": [ "inequalities" ], "Problem": "Prove that if $ 0 0$, which is trivial.", "Solution_2": "Maybe not trivial for beginners.", "Solution_3": "i think he just means the trivial inequality:\r\n\r\nx^2 >0 is true for all real x.", "Solution_4": "Well, that's really trivial. But i suppose that solution is not complete. Am i right?", "Solution_5": "Completing anderssonsw's solution\r\n\r\n$ \\frac {a \\plus{} b}{2} \\minus{} \\sqrt {ab} < \\frac {(b \\minus{} a)^2}{8a} \\iff \\\\\r\n\\frac {b \\minus{} 2\\sqrt {ab} \\plus{} a}{2} < \\frac {(b \\minus{} a)^2}{8a} \\iff \\\\\r\n(\\sqrt {b} \\minus{} \\sqrt {a})^2 < \\frac {(b \\minus{} a)^2}{4a} \\iff \\\\\r\n\\frac {(b \\minus{} a)^2 \\minus{} 4a(\\sqrt {b} \\minus{} \\sqrt {a})^2}{4a} > 0 \\iff \\\\\r\n\\frac {(\\sqrt {b} \\minus{} \\sqrt {a})^2[(\\sqrt {b} \\plus{} \\sqrt {a})^2 \\minus{} 4a]}{4a} > 0 \\iff \\\\\r\n\\frac {(\\sqrt {b} \\minus{} \\sqrt {a})^2(b \\plus{} 2\\sqrt {ab} \\minus{} 3a)}{4a}$\r\n\r\nThus it suffices to prove\r\n\r\n$ 2\\sqrt {ab} \\ge 3a \\minus{} b$.\r\n\r\n$ 2\\sqrt {ab} > 2a > 3a \\minus{} b \\iff \\sqrt {ab} > a$, which is trivial. Did I make any algebraic mistakes?\r\n\r\nEDIT: Of course that's easier... You just need to remark that $ a \\neq b$ (so the equality cannot be achieved).", "Solution_6": "Wouldn't it be easier to go\r\n\r\n$ (\\sqrt{b}\\minus{}\\sqrt{a})^2 \\le \\dfrac{(b\\minus{}a)^2}{4a}$\r\n\r\n$ \\Rightarrow 2\\sqrt{a}(\\sqrt{b}\\minus{}\\sqrt{a}) \\le b\\minus{}a$ which we can do becasue $ b>a$\r\n\r\n$ \\sqrt{ab} \\le \\frac{a\\plus{}b}{2}$\r\n\r\nWhich is true by AM-GM" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "It is well known that a convex quadrilateral ABCD is tangential iff AB+CD=BC+AD.\r\nWhat about concave quadrilateral? is it true that for a concave quadrilateral ABCD, there exists a circle touching AB, BC, CD & DA or their extensions iff AB+CD=BC+AD?\r\nAnd what about self-intersecting quadrilaterals?", "Solution_1": "[quote=\"zvi\"]It is well known that a convex quadrilateral ABCD is tangential iff AB+CD=BC+AD.\nWhat about concave quadrilateral? is it true that for a concave quadrilateral ABCD, there exists a circle touching AB, BC, CD & DA or their extensions iff AB+CD=BC+AD?\nAnd what about self-intersecting quadrilaterals?[/quote]\r\n\r\nit is also true for concave quadrilaterals.", "Solution_2": "See http://www.mathlinks.ro/Forum/viewtopic.php?p=359769#359769 , the proof given by nttu", "Solution_3": "Indeed, a beautiful proof, which applies also to a concave quadrilateral." } { "Tag": [ "vector", "geometry", "parallelogram", "Euler", "geometry unsolved" ], "Problem": "$A,B,C,D$ are points on a circle in this order clockwise. If $AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}$, find $\\vec{AC}\\cdot\\vec{BD}$ in term of $AB$ and $BC$.", "Solution_1": "[quote=\"OHO\"]$AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}$, [/quote]\r\n\r\nThis holds if and only if quadrilateral $ABCD$ is a parallelogram (that follows from Eulers theorem), but because $ABCD$ is cyclic, we conclude that $ABCD$ is a rectangle.\r\n\r\nNow, it's not hard to see that wanted value is $AB^{2}+BC^{2}$\r\n\r\nBye", "Solution_2": "What's Euler's theorem?", "Solution_3": "If $ABCD$ is a quadrilateral with sidelengths $a,b,c,d$ and diagonal length $e$ and $f$ then $a^{2}+b^{2}+c^{2}+d^{2}= e^{2}+f^{2}+4m^{2}$ where $m$ is length of the segment which connects two midpoints of diagonals. So $m=0$ iff $ABCD$ is a parallelogram." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$x,y,z>0$.Prove that:\r\n$\\sum_{cyclic}\\frac{yz}{(y+z)^2}+2\\sum_{cyclic}\\frac{yz}{(x+y)(x+z)} \\le \\frac{9}{4}$", "Solution_1": "$\\frac{9}{4}-\\sum_{cyc}\\frac{yz}{(y+z)^2}-2\\sum_{cyc}\\frac{yz}{(x+y)(x+z)}=\\frac{(x-y)^2(x-z)^2(y-z)^2}{4(x+y)^2(x+z)^2(y+z)^2}.$ :)", "Solution_2": "[quote=\"arqady\"]$\\frac{9}{4}-\\sum_{cyc}\\frac{yz}{(y+z)^2}-2\\sum_{cyc}\\frac{yz}{(x+y)(x+z)}=\\frac{(x-y)^2(x-z)^2(y-z)^2}{4(x+y)^2(x+z)^2(y+z)^2}.$ :)[/quote]\r\nNice solution!\r\nHere's my solution:\r\nLet $a=\\frac{y}{x+y},b=\\frac{z}{y+z},c=\\frac{x}{z+x}$\r\nRewrite the original ineqiality as\r\n$\\sum b(1-b)+2\\sum a(1-c)\\leq \\frac{9}{4}$\r\n$\\Leftrightarrow \\sum a-\\sum a^2 +2\\sum a-2\\sum ab\\leq \\frac{9}{4}$\r\n$\\Leftrightarrow (\\sum a-\\frac{3}{2})^2\\geq 0$" } { "Tag": [ "vector", "geometry", "area of a triangle", "geometry proposed" ], "Problem": "Let ABCDE be a convex pentagon, and let M, N, P, Q, R be the midpoints of its sides AB, BC, CD, DE, EA respectively.\r\nIf the lines AP, BQ, CR, DM all pass through the same point O, prove that the line EN also passes through this point O.", "Solution_1": "Something like vectors or affine transformations might work.", "Solution_2": "Just a generalization of Ceva 's theorem : Poncelet 's theorem. :D", "Solution_3": "Here is my solution of the problem. It uses a generalization of the Ceva theorem indeed (by the way, I have never seen it being called Poncelet theorem, but who knows):\r\n\r\n[color=blue][b]Theorem 1.[/b] Let n be a positive integer, and let $A_1A_2...A_{2n+1}$ be any (2n + 1)-gon. We work with indices cyclic modulo 2n + 1, so that $A_{2n+1+i}=A_i$ for any i. Let P be a point in the plane; for every index i, let the line $A_iP$ meet the line $A_{n+i}A_{n+1+i}$ at a point $X_i$. Then,\n\n$\\prod_{i=1}^{2n+1}\\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=1$.\n\nHereby, we work with directed segments.[/color]\r\n\r\n[i]Proof of Theorem 1.[/i] Let's use not only directed segments, but also directed areas, and let's denote the directed area of a triangle $P_1P_2P_3$ by $\\left[P_1P_2P_3\\right]$.\r\n\r\nLet S and T be the orthogonal projections of the points $A_{n+i}$ and $A_{n+1+i}$ on the line $A_iP$. Then, the two lines $A_{n+i}S$ and $A_{n+1+i}T$ are parallel to each other (since they are both perpendicular to $A_iP$); if we direct these two lines in the same way, then Thales yields $\\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=\\frac{A_{n+i}S}{A_{n+1+i}T}$. On the other hand, the segments $A_{n+i}S$ and $A_{n+1+i}T$ are the altitudes of the triangles $A_iPA_{n+i}$ and $A_iPA_{n+1+i}$ to the common side $A_iP$; since the area of a triangle equals $\\pm\\frac12\\cdot\\text{ side }\\cdot\\text{ corresponding altitude}$, where the $\\pm$ sign is + or - depending on the orientation of the triangle, it follows that $\\left[A_iPA_{n+i}\\right]=\\pm\\frac12\\cdot A_iP\\cdot A_{n+i}S$ and $\\left[A_iPA_{n+1+i}\\right]=\\pm\\frac12\\cdot A_iP\\cdot A_{n+1+i}T$; note that the two $\\pm$ signs are either both + or both - (since the lines $A_{n+i}S$ and $A_{n+1+i}T$ are directed in the same way). Hence, $\\frac{\\left[A_iPA_{n+i}\\right]}{\\left[A_iPA_{n+1+i}\\right]}=\\frac{A_{n+i}S}{A_{n+1+i}T}$. Comparing this with $\\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=\\frac{A_{n+i}S}{A_{n+1+i}T}$, we obtain $\\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=\\frac{\\left[A_iPA_{n+i}\\right]}{\\left[A_iPA_{n+1+i}\\right]}$. Thus,\r\n\r\n$\\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=-\\frac{A_{n+i}X_i}{A_{n+1+i}X_i}=-\\frac{\\left[A_iPA_{n+i}\\right]}{\\left[A_iPA_{n+1+i}\\right]}=\\frac{\\left[A_iPA_{n+i}\\right]}{\\left[A_{n+1+i}PA_i\\right]}=\\frac{\\left[A_iPA_{n+i}\\right]}{\\left[A_{n+1+i}PA_{n+\\left(n+1+i\\right)}\\right]}$,\r\n\r\nwhere in the last step, we have used the fact that $A_{n+\\left(n+1+i\\right)}=A_{2n+1+i}=A_{i}$ (since indices are cyclic modulo 2n + 1). This identity holds for every index i; multiplying all such identities, we get\r\n\r\n$\\prod_{i=1}^{2n+1}\\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=\\prod_{i=1}^{2n+1}\\frac{\\left[A_iPA_{n+i}\\right]}{\\left[A_{n+1+i}PA_{n+\\left(n+1+i\\right)}\\right]}=\\prod_{i=1}^{2n+1}\\left[A_iPA_{n+i}\\right]\\left/\\prod_{i=1}^{2n+1}\\left[A_{n+1+i}PA_{n+\\left(n+1+i\\right)}\\right]\\right.$.\r\n\r\nBut actually, $\\prod_{i=1}^{2n+1}\\left[A_iPA_{n+i}\\right]$ and $\\prod_{i=1}^{2n+1}\\left[A_{n+1+i}PA_{n+\\left(n+1+i\\right)}\\right]$ are the same product, just with a different order of factors. Hence,\r\n\r\n$\\prod_{i=1}^{2n+1}\\frac{A_{n+i}X_i}{X_iA_{n+1+i}}=\\prod_{i=1}^{2n+1}\\left[A_iPA_{n+i}\\right]\\left/\\prod_{i=1}^{2n+1}\\left[A_{n+1+i}PA_{n+\\left(n+1+i\\right)}\\right]\\right.=1$,\r\n\r\nand Theorem 1 is proven.\r\n\r\nNow, let's reformulate the problem (note that the pentagon does not need to be convex):\r\n\r\n[color=blue][b]Theorem 2.[/b] Let $A_1A_2A_3A_4A_5$ be an arbitrary pentagon, and let $M_1$, $M_2$, $M_3$, $M_4$, $M_5$ be the midpoints of its sides $A_3A_4$, $A_4A_5$, $A_5A_1$, $A_1A_2$, $A_2A_3$, respectively. If the lines $A_1M_1$, $A_2M_2$, $A_3M_3$ and $A_4M_4$ have a common point O, then the line $A_5M_5$ also passes through this point O.[/color]\r\n\r\n[i]Proof of Theorem 2.[/i] Since the lines $A_1M_1$, $A_2M_2$, $A_3M_3$ and $A_4M_4$ pass through the point O, the points $M_1$, $M_2$, $M_3$, $M_4$ are the points where the lines $A_1O$, $A_2O$, $A_3O$, $A_4O$ meet the lines $A_3A_4$, $A_4A_5$, $A_5A_1$, $A_1A_2$, respectively. Let the line $A_5O$ meet the line $A_2A_3$ at a point $N_5$. Then, by Theorem 1, we have\r\n\r\n$\\frac{A_3M_1}{M_1A_4}\\cdot\\frac{A_4M_2}{M_2A_5}\\cdot\\frac{A_5M_3}{M_3A_1}\\cdot\\frac{A_1M_4}{M_4A_2}\\cdot\\frac{A_2N_5}{N_5A_3}=1$.\r\n\r\nBut since $M_1$, $M_2$, $M_3$, $M_4$ are the midpoints of the sides $A_3A_4$, $A_4A_5$, $A_5A_1$, $A_1A_2$, we have $\\frac{A_3M_1}{M_1A_4}=1$, $\\frac{A_4M_2}{M_2A_5}=1$, $\\frac{A_5M_3}{M_3A_1}=1$, $\\frac{A_1M_4}{M_4A_2}=1$. Thus, we also have $\\frac{A_2N_5}{N_5A_3}=1$, so the point $N_5$ must be the midpoint of the side $A_2A_3$. Thus, $N_5=M_5$. Now, from the definition of the point $N_5$, it is clear that the line $A_5N_5$ passes through the point O; hence, the line $A_5M_5$ passes through the point O, and Theorem 2 is proven.\r\n\r\n Darij", "Solution_4": "using affine transformations is not a good idea since we have $R^2$ and $4$ vectors to \"control\"", "Solution_5": "We can use area method to prove it just like the Ceva theorem " } { "Tag": [ "quadratics", "geometry", "similar triangles" ], "Problem": "Given: Triangle ABD, Angle BAD = 90* altitude AC (C lying on segment BD) \r\nAD = 6, BC = 5.. \r\n\r\nTo Prove: AB and CD \r\nProof: NOW HERES WHERE I GET STUCK... plz som1 help i have no idea what to do..", "Solution_1": "Are you trying to find AB and CD, or prove AB = CD?", "Solution_2": "err im going to assume he means find $AB$ and $CD$, because $AB \\not = CD$\r\n\r\n[hide]\nwe know that in this triangle\n\n$AB^2 = BC(BD)$\n\nsetting $CD = x$, we have $AB^2 = 5(5+x)$\n\nthen we also know $AD^2 = CD(BD) \\Rightarrow 36=x(5+x) = x^2 + 5x$\n\nso $x^2 + 5x - 36 = 0$, $x = 4 or -9$ but we can toss out the negative value of $x$, so $x=CD=4$\n\nplugging that back into the equation for $AB$, we have $AB^2 = 5(5+4) = 45$\n\nso $AB = 3\\sqrt5$\n\nthe formulas i used (i.e. $AB^2 = BC(BD)$ ) can be derived using the similar triangles that you make when you draw the altitude, so if u arent familiar with them jacquz you can work them out that way \n\n[/hide]", "Solution_3": "hah $\\$*%# som1 hah :lol: havent used this internet talk for a long time since joining MathLinks :D\r\n\r\nthis problem has been posted twice in same forum so jaquz please avoid from doing that in the future\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=45195]http://www.mathlinks.ro/Forum/viewtopic.php?t=45195[/url]", "Solution_4": "Yeah, i set $AC=x$ and then used similar triangles and quadratic formula to solve..." } { "Tag": [], "Problem": "For what value of $ x$ does $ \\frac{0.\\overline{09}}{x} \\equal{} 11^{\\minus{}1}$?", "Solution_1": "We know that $ {0.\\overline{09}=\\frac{1}{11},}$ so we have that $ \\frac{\\frac{1}{11}}{x}=\\frac{1}{11}\\implies\\boxed{x=1}.$" } { "Tag": [ "inequalities", "algebra", "polynomial", "calculus", "derivative", "inequalities proposed" ], "Problem": "If $ a,b,c>0$ and $ abc=1$ prove that:\r\n \\[ (a-b)^{2}+(b-c)^{2}+(c-a)^{2}+\\frac{5}{2}(\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca})+\\frac{1}{2}(\\sqrt{a}+\\sqrt{b}+\\sqrt{c})\\geq 2(a+b+c)+3\\]", "Solution_1": "[hide]\n$ \\frac{5}{2}\\sum_{cyc}\\sqrt{ab}+\\frac{1}{2}\\sum_{cyc}\\sqrt{a}\\geq 9$ by $ AM-GM$\nso we need to prove\n$ a^{2}+b^{2}+c^{2}+3\\geq a+b+c+ab+bc+ca$ \nset $ a=x^{3},b=y^{3},c=z^{3}$ and manipulate the inequality into\n$ \\sum_{cyc}x^{6}+3x^{2}y^{2}z^{2}\\geq \\sum_{cyc}x^{4}yz+\\sum_{cyc}x^{3}y^{3}$\nwhich is true because $ LHS\\geq \\sum_{sym}x^{4}y^{2}$ by schurr and the rest is easy...\n[/hide]", "Solution_2": "For proving $ x^{2}+y^{2}+z^{2}+3\\geq x+y+z+xy+yz+zx$ you can mix variables, or simply use the Turkevici inequality:\r\n\r\n\\[ \\sum a^{4}+2abcd\\geq \\sum_{sym}a^{2}b^{2}\\] for \\[ d=1, abc=1, x=\\sqrt{a}, y=\\sqrt{b}, z=\\sqrt{c}\\]", "Solution_3": "Here's a stronger one:\r\n\r\n[i]If $ a,b,c>0$ prove that:[/i]\r\n\\[ (a-b)^{2}+(b-c)^{2}+(c-a)^{2}+\\frac{5ab+1}{a+b}+\\frac{5bc+1}{b+c}+\\frac{5ca+1}{c+a}\\geq 2(a+b+c)+3\\]", "Solution_4": "ok, i think i found a solution, but i need a check up.\r\n[hide]\nwe want to prove that \n$ (a-b)^{2}+\\frac{5ab+1}{a+b}\\geq a+b+1$ which is equivalent to proving\n$ (a-b)^{2}(a+b-1)+(a-1)(b-1)\\geq 0$. now if both $ a,b$ are $ \\geq 1$ it's obvious.\nif both are $ \\leq 1$ it's safe to assume that $ a+b-1\\leq 0$ otherwise we are done. in this case we have\n$ (1-a)(1-b)=1-a-b+ab\\geq 1-a-b\\geq (1-a-b)(a-b)^{2}$ because $ 1-a-b$ is positive and $ (a-b)^{2}\\leq 1$ because $ 0\\leq a,b\\leq 1$\nnow the third case is $ a\\geq 1\\geq b$. assume at first that $ a+b-1\\geq \\frac{1}{4}$\nnow we have $ (a-b)^{2}(a+b-1)\\geq \\frac{1}{4}((a-1)+(1-b))^{2}\\geq (a-1)(1-b)$\nassume now that $ a+b-1<\\frac{1}{4}$. look at the polynomial of variable $ b$\n$ f(b)=b^{3}-b^{2}(a+1)+b(3a-a^{2}-1)+(a-1)^{2}(a+1)$.we can assume $ a\\not=1$ because then it's trivial. we want to prove that $ f(b)\\geq 0$ for all $ b\\geq 0$. lets assume different. then $ f$ has at least one positive real root. since $ f(0)>0$ it has at least one negative root and we conclude that it has three real roots, one negative and two positive. we can also assume that the positive roots are different otherwise we are done. now we have\n$ f(b)=(b+p)(b-q)(b-r)$ with $ p,q,r\\geq 0$ comparing coefficients with $ b^{2}$ we have $ p-q-r=-a-1$ or $ q+r=1+a+p$. WLOG assume $ q>r$ so we have $ 2q>q+r=1+a+p>1$ so $ q>\\frac{1}{2}$.\nwe assumed there exists $ b_{0}$ such that $ f(b_{0})<0$. if $ b_{0}>q$ then $ f$ would have another root greater than $ q$ which cannot be so $ q>b_{0}$.\nif there exists $ \\epsilon>0$ such that for all $ x$ from $ (q-\\epsilon,q)$ we have $ f(x)>0$ then $ q$ would be a local minimum of $ f$ and it's derivative would be $ 0$ and $ q$ would be a double root of $ f$ and that isn't true(i think you can do this part without derivatives but it's just shorter for me).\nnow pick $ \\epsilon$ such that $ \\epsilon<\\frac{1}{4}$ now we have that $ f(b_{0})<0$ where $ b_{0}>\\frac{1}{4}$ but that's impossible because then we would have $ b_{0}+a-1>\\frac{1}{4}$\n[/hide]" } { "Tag": [ "geometry", "trapezoid" ], "Problem": "How many square units are in the area of polygon $ ABCD$ with vertices $ A(\\minus{}4,4), B(1,\\minus{}4), C(1,\\minus{}1)$ and $ D(\\minus{}4,\\minus{}1)$?", "Solution_1": "The key to this problem is to recognize the trapezoid with bases AD and BC and perpendicular leg CD. Since AD=5, BC=3, and CD=5, the answer is (1/2)(5+3)(5)=[b]20[/b].", "Solution_2": "Hmm...the review says its $25$ but I got $20$ too.", "Solution_3": "I agree with 20\n\nyou can use the shoelace method (see [url=http://mathworld.wolfram.com/PolygonArea.html]a reference site[/url])with the points in clockwise order (ACBD) as follows\n\n-4 4\n1 -1\n1 -4\n-4 -1\n-4 4\n\nLeft products in order: 4,-1,16,4\nRight products in order: 4,-4,-1,-16\n\nleft sum: 23\nright sum: -17\n23--17=40\n40/2=[b][i][u]20[/u][/i][/b]", "Solution_4": "I took a different approach but still got the same answer as previous posters:\n\nDraw a very simple graph, just enough to get and idea of where the points. As you can see the graph is divided into 2 triangles which has a very clear base and a very clear height. Triangle 1: (-4,4)(-4,-1)(1,-1) triangle 2: (1,-1)(1,-4)(-4,-1)\n\nSimple mind calc: triangle 1: 12.5units^2 triangle 2: 7.5units^2\n\nAdd these up and you get 20, however the computer said the answer is 25, going to report this problem now." } { "Tag": [ "function", "induction", "algebra", "functional equation", "algebra unsolved" ], "Problem": "Find all continuous functions $f: \\mathbb R \\to \\mathbb R$ so that $ f(x+f(x))=f(x)$ for all $ x \\in \\mathbb R$.", "Solution_1": "If we assume the uniform continuity on $ \\mathbb{R}$ (not just the continuity), then I can prove that $ f$ is constant... :wink: \r\n\r\nSo, can we deduce the uniform continuity from the functional equation (which is just to prove that $ f$ has limit in $ \\plus{}\\infty$ and $ \\minus{}\\infty$)?\r\n\r\nPierre.", "Solution_2": "[quote=\"duytungct\"]Find all continuous function $ f: R \\to R$ so that $ f(x \\plus{} f(x)) \\equal{} f(x)$ for all $ x \\in R$[/quote]\r\n\r\n$ 1)$ First, it is easy to show with induction that $ f(x\\plus{}nf(x))\\equal{}f(x)$ $ \\forall x\\in \\mathbb{R}$ and $ \\forall n\\in\\mathbb{N}$\r\n\r\n$ 2)$ Constant functions $ f(x)\\equal{}c$ are obviously solutions.\r\n\r\n$ 3)$ Assume it exists $ f(x)$ non-constant continuous solution. Then \r\nIt exists $ x_0x_0$ such that $ f(x)\\equal{}f(x_1)\\}$\r\n$ A\\neq_\\emptyset$ (since $ x_1\\in A$) and $ x>x_0$ $ \\forall x\\in A$. Let then $ x_2\\equal{}\\inf(A)$.\r\n\r\nSince $ f(x)$ is continuous, $ f(x_2)\\equal{}f(x_1)$, $ x_2>x_0$ and $ \\boxed{f(x)\\neq f(x_1)\\forall x\\in[x_0,x_2)}$\r\n\r\nSince $ f(x_2)\\neq f(x_0)$, $ \\exists n\\neq p\\in\\mathbb{N}$ such that either $ x_0\\plus{}nf(x_0)x_2\\plus{}pf(x_2)>x_2\\plus{}pf(x_2)$\r\nSince $ f(x)$ is continuous, so is $ x\\plus{}nf(x)$, so $ \\exists x_3\\in(x_0,x_2)$ such that $ x_3\\plus{}nf(x_3)\\equal{}x_2\\plus{}pf(x_2)$, so $ f(x_3\\plus{}nf(x_3))\\equal{}f(x_2\\plus{}pf(x_2))$, and so $ f(x_3)\\equal{}f(x_2)\\equal{}f(x_1)$\r\n\r\nBut this is a contradiction since we know (see boxed statement above) that $ f(x)\\neq f(x_1)$ $ \\forall x\\in[x_0,x_2)$\r\n\r\nSo the only solutions are constant functions.", "Solution_3": "Ok, I got it!\r\n\r\nLet $ f$ be a solution.\r\nClearly, $ f$ cannot be injective.\r\n\r\nLet $ g(x)\\equal{}f(x)\\plus{}x$.\r\nLet $ g^n$ denotes the $ n^{th}$ iterate of $ g$.\r\n\r\nIt is easy to verify that $ g$ is continuous and $ g^2(x)\\equal{}2g(x)\\minus{}x$ for all $ x$.\r\nThus, $ g$ is injective.\r\nSince $ g$ is continuous and injective, we deduce that $ g$ is strictly monotone. But, if $ g$ is decreasing then $ f(x)\\equal{}g(x)\\minus{}x$ is (strictly) decreasing too, so that $ f$ would be injective, a contradiction.\r\nThus, $ g$ is increasing.\r\n\r\nA straightforward induction leads to $ g^n(x) \\equal{} ng(x)\\minus{}(n\\minus{}1)x \\equal{} nf(x)\\plus{}x$ for all $ x$ and positive integer $ n$.\r\n\r\nLet $ a,b$ such that $ a0$.\r\nTaking limit on $ n$, leads to $ f(a)\\minus{}f(b) \\leq 0$, so that $ f$ is non-decreasing.\r\n\r\nA straightforward induction leads to $ f(x\\plus{}nf(x)) \\equal{} f(x)$ for all $ x$ and non-negative integer $ n$\r\nSince $ f$ is non-decreasing, it follows that $ f$ is constant on each intervall with endpoints $ x$ and $ x\\plus{}nf(x)$ for each $ x$ $ n >0$.\r\n\r\nAssume that there exists $ a$ such that $ f(a) > 0$ and $ c$ such that $ f(c)\\equal{}0$.\r\nLet $ M$ be the sup of the set $ S$ of the $ x$ such that $ f(x) \\equal{}0$ (for all such $ x$, we have $ x < a$ since $ f$ is non-decreasing).\r\nUsing a sequence of element of $ S$ which has limit $ M$ and continuity of $ f$, we clearly deduce that $ f(M)\\equal{}0$.\r\nFor each $ e>0$, we have $ f(M\\plus{}e)>0$ so that $ f$ is constant on $ [M\\plus{}e,\\plus{}\\infty[$. Clearly, when $ e$ varies, the constant value is always the same, say $ y$. Thus, for each $ e>0$, we have $ 00$. A contradiction.\r\n\r\nAssume there exist $ b$ such that $ f(b) < 0$ and $ c$ such that $ f(c)\\equal{}0.$\r\nThus, from above $ f$ is constant on $ ]\\minus{}\\infty , b]$, and we get the same contradiction as above by considering $ m \\equal{} \\inf S$.\r\n\r\nTherefore $ f$ is the zero function, or $ f$ has always the same sign on $ \\mathbb{R}$.\r\nIn the latter case, if $ f(a) > 0$ for all $ a$, then $ f$ is constant on each $ [a, \\plus{} \\infty[$ for all $ a$, so that $ f$ is constant.\r\nIf $ f(b) < 0$ for all $ b$ then $ f$ is constant on each $ ]\\minus{} \\infty , b]$ for all $ b$, so that $ f$ is constant.\r\n\r\nThus, in each case $ f$ is constant.\r\n\r\nConversly, if $ f$ is constant then $ f$ is clearly a solution.\r\n\r\nPierre.", "Solution_4": "Aaargh... a bit too slow in writing... :oops: \r\n\r\nPierre.", "Solution_5": "That's a bit too complicated, Pierre. :D Once you know that $ g(g(x))\\equal{}2g(x)\\minus{}x$ you easily deduce that $ g$ is bijective (by continuity and monotony), so that relation $ g^n(x)\\equal{}ng(x)\\minus{}(n\\minus{}1)x$ actually holds for all integers $ n$. Next write $ g^n(x)0$\r\n\r\nprove\r\n\r\n$ \\frac{3\\minus{}a^2}{b\\plus{}c}\\plus{}\\frac{3\\minus{}b^2}{c\\plus{}a}\\plus{}\\frac{3\\minus{}c^2}{a\\plus{}b} \\geq a\\plus{}b\\plus{}c$", "Solution_1": "[hide=\"Solution\"]From the given condition, we have $ \\frac{3\\minus{}a^{2}}{b\\plus{}c}\\plus{}\\frac{3\\minus{}b^{2}}{c\\plus{}a}\\plus{}\\frac{3\\minus{}c^{2}}{a\\plus{}b}\\equal{}\\frac{a^2\\plus{}b^2}{a\\plus{}b}\\plus{}\\frac{b^2\\plus{}c^2}{b\\plus{}c}\\plus{}\\frac{c^2\\plus{}a^2}{c\\plus{}a}$.\n\nHence, it suffices to prove that $ \\sum_{\\text{cyc}}\\frac{a^2\\plus{}b^2}{a\\plus{}b}\\ge \\sum_{\\text{cyc}}\\frac{a\\plus{}b}{2}$, which follows from AM-GM Inequality.\n\nWe are done.[/hide]", "Solution_2": "[quote]\n\\[ \\sum\\limits_{cyc} {\\frac {{3 \\minus{} a^2 }} {{b \\plus{} c}}} \\equal{} \\sum\\limits_{cyc} {\\frac {{b^2 \\plus{} c^2 }} {{b \\plus{} c}} \\equal{} } \\sum\\limits_{cyc} {\\frac {{\\left( {b \\plus{} c} \\right)^2 \\minus{} 2bc}} {{b \\plus{} c}} \\equal{} \\sum\\limits_{cyc} {b \\plus{} c \\minus{} \\frac {{2bc}} {{b \\plus{} c}}} } \\geqslant a \\plus{} b \\plus{} c\\]\n[/quote]\n\n[quote]\n\\[ b \\plus{} c \\minus{} \\frac {{2bc}} {{b \\plus{} c}} \\plus{} a \\plus{} b \\minus{} \\frac {{2ab}} {{a \\plus{} b}} \\plus{} c \\plus{} a \\minus{} \\frac {{2ca}} {{c \\plus{} a}} \\geqslant a \\plus{} b \\plus{} c\\]\n[/quote]\n\n[quote]\n\\[ a \\plus{} b \\plus{} c \\equal{} \\frac {{\\left( {a \\plus{} b} \\right) \\plus{} \\left( {b \\plus{} c} \\right) \\plus{} \\left( {c \\plus{} a} \\right)}} {2} \\geqslant \\frac {{2bc}} {{b \\plus{} c}} \\plus{} \\frac {{2ab}} {{a \\plus{} b}} \\plus{} \\frac {{2ca}} {{c \\plus{} a}}\\]\n[/quote]\nes directo ya que\n[quote]\\[ \\sum\\limits_{cyc} {\\frac{{a \\plus{} b}}\n{2}} \\geqslant \\sum\\limits_{cyc} {\\frac{2}\n{{\\frac{1}\n{a} \\plus{} \\frac{1}\n{b}}}} \\equal{} \\sum\\limits_{cyc} {\\frac{{2ab}}\n{{a \\plus{} b}}}\\][/quote]", "Solution_3": "hello, your inequality is equivalent to\r\n$ \\frac{b^2\\plus{}c^2}{b\\plus{}c}\\plus{}\\frac{c^2\\plus{}a^2}{c\\plus{}a}\\plus{}\\frac{a^2\\plus{}b^2}{a\\plus{}b}\\geq a\\plus{}b\\plus{}c$\r\nand this is equivalent with\r\n$ a(b^2\\minus{}c^2)(b\\minus{}c)\\plus{}b(a^2\\minus{}c^2)(a\\minus{}c)\\plus{}c(a^2\\minus{}b^2)(a\\minus{}b)\\geq0$\r\nand this is true.\r\nSonnhard.", "Solution_4": "$ \\frac{b^{2}\\plus{}c^{2}}{b\\plus{}c}\\plus{}\\frac{c^{2}\\plus{}a^{2}}{c\\plus{}a}\\plus{}\\frac{a^{2}\\plus{}b^{2}}{a\\plus{}b}\\geq a\\plus{}b\\plus{}c$\r\n\r\nchevychev's inequality works , we have $ \\sum \\frac{a^{2}\\plus{}b^{2}}{a\\plus{}b} \\ge \\sum \\frac{a\\plus{}b}{2}\\equal{}a\\plus{}b\\plus{}c$" } { "Tag": [ "parameterization", "quadratics", "LaTeX", "algebra", "quadratic formula" ], "Problem": "Solve the equation with the parameter $ a\\in\\mathbb{R}$ :\r\n\r\n$ x \\plus{} \\sqrt {a \\plus{} \\sqrt {x}} \\equal{} a$\r\n\r\nThis is a really annoying problem... I've done the multiple squaring and the existence conditions but I can't get anywhere, could anyone post a [i]solution[/i] if you can find the time?", "Solution_1": "[hide]Here's a \"slick\" solution:\nLet $ t \\equal{} sqrt(a \\plus{} sqrt{x})$. We have: $ x \\plus{} t \\equal{} t^2 \\minus{} \\sqrt {x}$. \nThis is equivalent to: $ (t \\minus{} 1/2)^2 \\equal{} (\\sqrt {x} \\plus{} 1/2)^2$. \nSo either $ t \\minus{} 1/2$ = $ \\sqrt {x} \\plus{} 1/2$ or $ 1/2 \\minus{} t \\equal{} \\sqrt {x} \\plus{} 1/2$.\nSince $ t\\geq 0$, we see that the 2nd cases occurs at only $ t \\equal{} 0$ and $ x \\equal{} 0$.\nNow,we have left with the 1st case, which is: $ t\\equal{}\\sqrt{x}\\plus{}1$.Replace $ t \\equal{} \\sqrt{a \\plus{} \\sqrt{x}}$ and squaring both sides with $ t\\geq 1$, we get the quadratic equation:\n$ x \\plus{} \\sqrt {x} \\plus{} 1 \\minus{} a \\equal{} 0$\nUsing quadratic formula we get: $ x \\equal{} ( \\minus{} 1 \\plus{} ( \\minus{} )\\sqrt {4a \\minus{} 3})/2$. Since x must be greater than or equal to 0, $ x$ = $ ( \\minus{} 1 \\plus{} \\ sqrt(4a \\minus{} 3))/2$($ a\\geq 3/4$). Now plugging back in this value of x into the equation to find any possibly essential restrictions of a.[/hide]", "Solution_2": "ghjk, thank you for your solution. I understand what you did and I don't see any mistakes, but could you recheck what you did? I have as answers:\r\n\r\n$ \\begin{cases}a<0 \\text{ or }00)$.\r\n\r\n1972 Tokyo Medical and Dental University entrance exam" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "1)Prove that for any $a,b,c>0$ and $k\\ge 2$ we always have\r\n\\[\\frac{a^{2}}{ka^{2}+bc}+\\frac{b^{2}}{kb^{2}+ca}+\\frac{c^{2}}{kc^{2}+ab}\\le \\frac{3}{k+1}\\]\r\nA related inequality:\r\n2)Prove that for any $a,b,c>0$ we always have\r\n\\[\\frac{a^{2}}{2b^{2}+ca}+\\frac{b^{2}}{2c^{2}+ab}+\\frac{c^{2}}{2a^{2}+bc}\\ge 1 \\]\r\nAn open question: Is the general of (2) true?", "Solution_1": "2)\r\n\r\nBy Cauchy, we have $LHS \\geq \\frac{(a^{2}+b^{2}+c^{2})^{2}}{\\sum_{cyc}2a^{2}b^{2}+a^{3}c}$.\r\n\r\nTo prove that $a^{4}+b^{4}+c^{4}\\geq a^{3}c+b^{3}a+c^{3}b$, only use rearrangement!", "Solution_2": "[hide=\"1\"]\nThis is equivalent to\n\\[\\sum_\\text{cyc}\\frac{bc}{ka^{2}+bc}\\geq \\frac{3}{k+1}\\]\nafter multiplying by $k$ and subtracting each term from 1. We can prove this with Cauchy:\n\\[\\left( \\sum_\\text{cyc}\\frac{bc}{ka^{2}+bc}\\right) \\left( \\sum_\\text{cyc}bc(ka^{2}+bc)\\right) \\geq (ab+bc+ca)^{2}\\]\n\\[\\Rightarrow \\left( \\sum_\\text{cyc}\\frac{bc}{ka^{2}+bc}\\right) \\geq \\frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc(a+b+c)}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+kabc(a+b+c)}\\]\nShowing that the right hand side is at least $\\frac{3}{k+1}$ is equivalent to $(k-2)(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}-abc(a+b+c))\\geq 0$, which is a simple AM-GM.\n[/hide]\n\n[hide=\"2\"]\nThe inequality\n\\[\\sum_\\text{cyc}\\frac{a^{2}}{kb^{2}+ca}\\geq \\frac{3}{k+1}\\]\nholds for any nonnegative $k$, since by Cauchy,\n\\[\\left( \\sum_\\text{cyc}\\frac{a^{2}}{kb^{2}+ca}\\right) \\left( \\sum_\\text{cyc}a^{2}(kb^{2}+ca)\\right) \\geq (a^{2}+b^{2}+c^{2})^{2}\\]\n\\[\\Rightarrow \\sum_\\text{cyc}\\frac{a^{2}}{kb^{2}+ca}\\geq \\frac{(a^{2}+b^{2}+c^{2})^{2}}{k(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})+a^{3}c+b^{3}a+c^{3}b}\\]\nNow we need to show the right hand side at least $\\frac{3}{k+1}$; this follows from\n\\[k[a^{4}+b^{4}+c^{4}-a^{2}b^{2}-b^{2}c^{2}-c^{2}a^{2}]+[(a^{2}+b^{2}+c^{2})^{2}-3(a^{3}c+b^{3}a+c^{3}b)]\\geq 0\\]\nBut maybe there is a way to prove it without resorting to such a strong result as Vasc's inequality.\n[/hide]", "Solution_3": "[quote=\"The soul of rock\"]2)Prove that for any $a,b,c>0$ we always have\n\\[\\frac{a^{2}}{2b^{2}+ca}+\\frac{b^{2}}{2c^{2}+ab}+\\frac{c^{2}}{2a^{2}+bc}\\ge 1 \\]\nAn open question: Is the general of (2) true?[/quote]\r\nThe sharper for (2):\r\n\\[\\sum\\frac{a^{2}}{2b^{2}+ca}\\ge \\frac{3(a^{3}+b^{3}+c^{3})}{(a+b+c)(a^{2}+b^{2}+c^{2})}\\]\r\nwith equality with $a=b=c$ and $a=b,c=0$. :) My solution for it is too urgly. :(", "Solution_4": "Sorry, my inequality is not true (I have some mistakes in calculations). :( But the orginal problem is true, and we can prove it easily by Cauchy Schwarz\r\n\\[\\sum\\frac{a^{2}}{2b^{2}+ca}\\ge \\frac{(a^{2}+b^{2}+c^{2})^{2}}{2\\sum a^{2}b^{2}+\\sum ab^{3}}\\ge \\frac{(a^{2}+b^{2}+c^{2})^{2}}{2\\sum a^{2}b^{2}+\\sum a^{4}}=1 \\]\r\n:)" } { "Tag": [ "limit", "number theory solved", "number theory" ], "Problem": "For each positive integer n, let $a_n$ denote the number of integer solutions (x,y) to the equation $n^2+x^2=y^2$ such that $x\\ge n,y\\ge n$.\r\n\r\na) Prove that for every M, $a_n > M$ is satisfied by at least one positive integer n.\r\nb) Is it true that $\\lim_{n->\\infty}a_n = \\infty$?", "Solution_1": "a) We can take the sequence $n_k=p_1p_2\\ldots p_k$, where $\\{p_k\\}$ is the sequence of primes. For any $M$, we will eventually have $a_{n_k}>M$ for some $k$.\r\n\r\nb) No, it doesn't tend to $\\infty$. We can take $n_k=p_k^2$. There is at most one decomposition of $n_k$ as the difference of two squares satisfying those conditions." } { "Tag": [], "Problem": "I found this game site a while ago, and I come to realy love it!\r\n[url]http://www.kongregate.com/?referrer=jackdillon[/url]\r\ntell me what you guys think!\r\nalso, any flash game creators can publish there games on the site.", "Solution_1": "i know what that site is as well, and i have an account too\r\n\r\nplease dont post it as a referral link\r\n\r\n[url]http://www.kongregate.com[/url]\r\n\r\nthere are cards and badges for people who get certain goals in games" } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Let $ (O)$ be a circle with diameter $ AB$.$ M$ is a point on $ OB$.$ CD$ is a chord through $ M$($ CD \\neq AB$).Suppose that $ O_1$ and $ O_2$ be circumcenter of triangle of $ AMC$ and $ BDM$.Suppose that $ E$ is midpoint of $ O_1O_2$\r\nProve that $ E$ lies on $ AB$", "Solution_1": "$ \\angle{O_2MB}\\equal{}90^o \\minus{} \\angle{BDM}\\equal{}90^o\\minus{}\\angle{BDC}\\equal{}90^o\\minus{}\\angle{BAM}$. Hence $ O_2M \\perp AC$.\r\n\r\nSince $ OO_1 \\perp AC$, so $ O_2M \\parallel OO_1$. Similarly, we have $ OO_2 \\parallel O_1M$. Therefore $ O_1OMO_2$ is a parallelogram. Hence the midpoint of $ O_1O_2$ is always on $ AB$." } { "Tag": [], "Problem": "I took a practice test SAT at Kaplan yesterday and I don't know how to solve this problem,it's a grid in.\r\n\r\nAmbrose,working alone, can paint a certain room in 8 hours. Either one of Benedict or Charles,working alone,can paint a room in 4 hours.If all 3 people, working at these rates, work together to paint the room, what fraction of the room is painted by Ambrose?", "Solution_1": "charles and bob work $ 2$ times as fast as ambrose. Thus together they work 4 times a fast and will have done 4 times as much work as ambrose when they finish painting the room $ \\implies$ ambrose does $ \\frac{1}{5}$ of the work." } { "Tag": [], "Problem": "Two students set their digital watches to 10:00. One watch runs one minute per hour too slow, and the other watch runs 2 minutes per hour too fast. What time will the slow watch show when it is exactly one hour behind the fast watch?", "Solution_1": "The slow one falls behind a total of 3 minutes every real hour. So it takes 20 hours for the slow one to be behind the fast one by 1 hour. Now, since the slow one loses one minute per hour, it will be 19 hours and 40 minutes past 10:00 on the slow one when the fast one is 1 hour ahead. This is the same as 19:40 mod 12:00 = 7:40 later. After adding that to 10:00 we get 10:00 + 7:40 = 17:40 = $ \\boxed{5: 40}$ mod 12:00.", "Solution_2": "Wait, if we were to use variables, how would we set the equation up?" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "$ P,Q \\in R[x]$ of polynomials with real coefficients such that $ P(\\frac{1}{2010}\\plus{}x\\plus{}Q^{4}(x))\\equal{}Q(\\frac{1}{2010}\\plus{}x\\plus{}P^{4}(x))$ $ x \\in R$ \r\nProve that $ P\\equal{}Q$\r\n\r\n\r\n\r\n_________________________________________\r\nAzerbaijan Land of Fire :ninja:", "Solution_1": "Where is the solution ???\r\nWho can manage this problem ???" } { "Tag": [ "conics", "ellipse", "parabola", "geometry", "rectangle", "parallelogram", "cyclic quadrilateral" ], "Problem": "I was thinking about some problem and wondered: Can any given quadrilateral be inscribed in an ellipse? For certain quadrilaterals, does there exist only one such ellipse? Can someone give a proof/explanation as to what the case is, and why it is true or false? Thanks :)", "Solution_1": "I think the answer is yes. The non-rigorous way of showing it would be to divide the quadrilateral into two triangles. For example in ABCD consider ABC and DBC. The problem is equivalent to stretching the altitudes of the triangles to BC to make a cyclic quadrilateral A'BCD'. As you stretch A and D perpendicularly to BC, the angles A and D are monotonically increasing, and their sum is 360 and 0 at the two extremes. Thus at some point the sum is 180.", "Solution_2": "you don't know whether that sum of opposite angles = 180 property holds for quads inscribed in an ellipse...maybe some sort of projection would help in this prob...", "Solution_3": "hrm. the problem with projections is that it may not be an ellipse; it may be a parabola or hyperbola. The general property in this case is that 5 points (in general configuration) uniquely define a conic (so, in general, a quadrilateral would not uniquely define an ellipse).", "Solution_4": "[quote=\"s0mp\"]you don't know whether that sum of opposite angles = 180 property holds for quads inscribed in an ellipse...maybe some sort of projection would help in this prob...[/quote]\r\n\r\n\r\n:? read my post again. If what MysticTerminator said is true, then it seems that a quadrilateral must be inscribable in an ellipse, if not uniquely.", "Solution_5": "You guys are thinking too deeply about this. [hide]This problem asks if there is a unique ellipse determined by four coplanar points.[/hide]", "Solution_6": "Equation for ellipse:\r\n\r\n$\\frac{(x-h)^2}{a^2}+\\frac{(y-k)^2}{b^2} = 1$.\r\n\r\nSince we have four variables, there should be just one ellipse (then again, they are squared, so that might introduce more solutions)?", "Solution_7": "umm...Magnara, yea. How does that simplify anything I've said?", "Solution_8": "[quote=\"MysticTerminator\"]umm...Magnara, yea. How does that simplify anything I've said?[/quote]\r\n\r\nHeh. I didn't read that post.... :)\r\n\r\nPaladin, don't forget that that's an ellipse with major and minor axes parallel to the x and y axis.", "Solution_9": "Yeah, for a rectangle, there are an infinite number of ellipses. However, for, say, a more general parallelogram, i think it may be unique...dunno.\r\nBut yeah, the general equation for a conic is $ax^2+by^2+cx+dy+e=0$; five coefficients, five points, five equations, one solution, makes sense...", "Solution_10": "Hm, for the ellipse, doesn't a length of the minor and major axis, ie. 2 variables, define a ellipse.\r\nI was thinking about this problem as a circle. 3 pts define a circle because it can define a radius, and that's all the circle needs.\r\n\r\nOr maybe I'm wrong, but I'm just saying that I don't think the center/position of the ellipse doesn't really matter.", "Solution_11": "K8: you forget the xy term; you didn't read my post ... it's not unique in general", "Solution_12": "[quote=\"MysticTerminator\"]K8: you forget the xy term; you didn't read my post ... it's not unique in general[/quote]\r\no ya\r\nBut it does appear that for any quadrilateral, there is at least one ellipse." } { "Tag": [ "geometry", "pigeonhole principle", "algebra unsolved", "algebra" ], "Problem": "Show that it is impossible to cover a unit square with five equal squares with side $s<\\frac{1}{2}$.", "Solution_1": "maybe you mean $s<\\frac{1}{2}$.", "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?t=67555", "Solution_3": "[quote=\"jin\"]maybe you mean $s<\\frac{1}{2}$.[/quote]\r\n\r\nYes you're right !!! \r\n\r\n\r\nThanks", "Solution_4": "Assume we can cover it with 5 squares. However look at a side of the unit square , since $s<\\frac{1}{2}$ it must be covered by at least 3 of the 5 squares. Analogously the opposite side must be covered by at least 3 squares. Therefore at least one of these squares must cover both sides but this is impossible since its diagonal is less than $\\frac{\\sqrt{2}}{2}$ which is in turn less than 1. Therefore we have reached a contradiction and we are done.\nCorrect me if I'm wrong please.", "Solution_5": "Yes, unfortunately you're wrong, since for $s<1/2$, but large enough, it is easy to use two squares to cover a side. A start towards a solution is to look at the nine points given by the vertices of the unit square, the midpoints of its sides, and its centre. No square of side $s<1/2$ may cover three of them, since if collinear they stretch on a length of at least $1 > s\\sqrt{2}$, while otherwise they determine a triangle of area at least $1/8 > s^2/2$. Therefore by pigeonhole principle, we need at least five small squares.", "Solution_6": "Ok now I see that my mistake was assuming we purposefully had to place the squares inside the unit square.\nHowever your solution is not clear to me, you mention the fact that we can't cover those nine points simultaneously with the squares and then you prove that we can't do so with [b]4[/b] squares? Since if you mean it's not possible with 5 I have an easy counterexample for that.\nThe interesting thing about this problem is that it is so simple it should be possible to generalize.\nI'm sorry if I misunderstood your proof." } { "Tag": [ "number theory", "number theory unsolved" ], "Problem": "I'm don't know who moved my post, but if I did I would have private messaged that person, so I apologize for making this public.\r\n\r\nI posted a question only a few hours ago, and it was apparently moved to the high school intermediate level forum.\r\n\r\nWith all due respect, this is a question that was assigned to a 400 level college class. I'm sorry if you think it's too simplistic for you to bother having it in your forum, or if you think it's below you to have it here.\r\n\r\nI respect that you want to keep your forums clean, but please keep in mind that just because you (and your genius level intellect) think it may be a simple problem, does not mean that it is easy for everyone.\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=187826&sid=b457e0747df626c726c7bb7c2237dbea[/url]", "Solution_1": "It was me.\r\n\r\nI've no clue what a \"400 level college class\" is, but this problem can be done with intermediate methods. And it doesn't require much work, too. Thus it fits there. Possibly, Pre-Olympiad fits also, but I considered it better there.\r\nCompared to the other problems in the Number Theory forum, it's really easy. It's simple diversion of difficulty. If the criterium for a problem to be kept here is \"it's not easy for some person\", then we must keep problems like \"solve x^2=4\", too. So this is no argument.\r\n\r\nAnd yes, I want my forums to be clean :wink:", "Solution_2": "Just to fill you in, I'm in my final semester as a mathematics major, and 400 level classes are \"upper level\" or \"senior level\" classes (typically the more difficult).\r\n\r\nI realize now that the answer is not that difficult, but every now and again, I come across problems, that at first glance, cause a bit of \"writer's block\", and I just can't seem to get the ball rolling.\r\n\r\nThank you for your respectful tone. :wink:", "Solution_3": "The main question is:\r\nWhat else should I do\u00bf\r\nAs it looks, the only alternative is doing nothing. And it's not like your post is the only one ever moved, I move about 3 threads a day, I think (maybe I should start counting). Additional duties of a moderator are: deleting multiposts, searching/merging/linking reposts, stopping trolls and flames, helping on general questions, renaming threads, et cetera (and I'm far away from doing everything that should be done). Most of this isn't that much fun, and I've real world stuff (like university) to do, too.\r\nWriting a personal message to every user in these cases would even make the time needed 5 times that much. Thus this is only the last resort if some user acts to extreme.\r\n\r\nAbout 90% of the work I do is caused by users that in some sense act not \"good\". So in fact, most of this work could be avoided, and if that would happen, I would truly have more time to answer each remaining thing personally.\r\n\r\nIn this special case, it sounds like you feel discriminated or considered stupid because I moved it. Or that I think that the question is stupid. That's simply not true. But please understand that it does not fit here as this subforum is for problems around olympiad level or beyond, and these are normally really more advanced." } { "Tag": [ "projective geometry", "Olimpiada de matematicas" ], "Problem": "Aca les dejo los examenes de la OMCC 2006 realizada en Panama del 30 de julio al 5 de agosto.", "Solution_1": "\u00bfAlguien tiene los resultados por pa\u00edses?", "Solution_2": "Los resultados se encuentran en la pagina oficial de la VIII centro, los resultados de la delegacion mexicana fueron:\r\n\r\nOro:\r\nPaul y Jose Daniel\r\n\r\nPlata: \r\nAndres Emilsson\r\n\r\nVamos Mexico!!!!", "Solution_3": "Mis soluciones para los tres primeros.", "Solution_4": "Los problemas de la OMCC 2006 ya me salieron, pero con excepcion del 6, la vdd ya he hecho varias cosas, y todas me resultan contradictorias. Me gustaria saber si alguien lo ha hecho, para k me de cuenta de n k ando fallando. \r\nSi kieren k escriba lo k he hecho asta aora, k me digan y con gusto lo pongo!!\r\n\r\nY no mas como comentario, creo k esta OMCC no fue dificil!!\r\n\r\nEspero sus soluciones :oops:", "Solution_5": "EXELENTE PRIMER PERFECTO EN OMCC FELICIDADES DANIEL!!!!", "Solution_6": "[quote=\"conejita\"]\nSi kieren k escriba lo k he hecho asta aora, k me digan y con gusto lo pongo!![/quote]\r\n\r\nLos que podr\u00edas hacer es abrir 6 topics, cada uno con un problema, y ahi compartir la soluci\u00f3n", "Solution_7": "seria bueno q alguien ponga la soluciones , especialmente para el 5 y la 6 en mi caso :blush: ,", "Solution_8": "[quote]Sean $ \\Gamma$ y $ \\Gamma'$ dos circunferencias de igual radio con centros $ O$ y $ O',$ respectivamente. $ \\Gamma$ y $ \\Gamma'$ se cortan en dos puntos y $ A$ es uno de ellos. Se escoge un punto $ B$ cualquiera en $ \\Gamma.$ Sea $ C$ el otro punto de corte de la recta $ AB$ con $ \\Gamma'$ y $ D$ un punto en $ \\Gamma'$ tal que $ OBDO'$ es un paralelogramo. Demuestre que la longitud de $ CD$ es constante, es decir, no depende de la elecci\u00f3n de $ B.$[/quote]\nSea $ E \\equiv BD \\cap \\Gamma,$ diferente de $ B.$ Basta observar que como $ \\Gamma \\cong \\Gamma',$ entonces por evidente simetr\u00eda si $ BD\\parallel OO',$ los arcos $ AE$ y $ AD$ en $ \\Gamma$ y $ \\Gamma'$ son congruentes $\\Longrightarrow$ $ \\angle ABE \\equal{} \\angle ACD.$ As\u00ed el tri\u00e1ngulo $ \\triangle BDC$ es is\u00f3sceles con \u00e1pice $ D$ $ \\Longrightarrow$ $ CD \\equal{} BD \\equal{} OO' \\equal{} \\text{const}.$", "Solution_9": "[quote]Sea $ ABCD$ un cuadril\u00e1tero convexo. Sea $ I$ el punto de intersecci\u00f3n de las diagonales $ AC$ y $ BD.$ Sean $ E, H, F$ y $ G$ puntos sobre los segmentos $ AB, BC, CD$ y $ DA,$ respectivamente, tales que $ EF$ y $ GH$ se cortan en $ I.$ Sea $ M$ el punto de intersecci\u00f3n de $ EG$ y $ AC$ y $ N$ el punto de intersecci\u00f3n de $ HF$ y $ AC.$ Demuestre que $ \\frac {_{AM}}{^{IM}} \\cdot \\frac {_{IN}}{^{CN}} \\equal{} \\frac {_{IA}}{^{IC}}.$[/quote]\nNote que los tri\u00e1ngulos $ \\triangle AEG$ y $ \\triangle CFH$ son perspectivos a trav\u00e9s del perspector $ I,$ entonces por teorema de Desargues las intersecciones $ X \\equiv AD \\cap BC,$ $ Y \\equiv AB \\cap DC$ y $ P \\equiv EG \\cap HF$ son colineales. Consid\u00e9rese $ Q \\equiv AC \\cap XY.$ Luego por teorema de Menelao en los tri\u00e1ngulos $ \\triangle MQP$ y $ \\triangle NQP$ cortados por las rectas $ AB$ y $ CD,$ tenemos\n\n$ \\frac {YQ}{YP} \\cdot \\frac {PE}{EM} \\cdot \\frac {MA}{QA} \\equal{} 1 \\ , \\ \\frac {YQ}{YP} \\cdot \\frac {PF}{NF} \\cdot \\frac {NC}{QC} \\equal{} 1$ \n\n$ \\Longrightarrow \\frac {MA}{NC} \\cdot \\frac {PE}{EM} \\cdot \\frac {NF}{PF} \\equal{} \\frac {QA}{QC} \\ (*)$\n\nPero por teorema de Menelao en $ \\triangle MNP$ cortado por $ \\overline{FIE},$ se tiene \n\n$ \\frac {PE}{EM} \\cdot \\frac {NF}{PF} \\equal{} \\frac {IN}{IM}.$ Combinando con $ (*)$ resulta\n\n$ \\frac {MA}{NC} \\cdot \\frac {IN}{IM} \\equal{} \\frac {QA}{QC}.$ Luego como $ (A,C,I,Q) \\equal{} \\minus{} 1,$ se sigue que \n\n$\\frac {AM}{IM} \\cdot \\frac {IN}{CN} \\equal{} \\frac {QA}{QC} \\equal{} \\frac {IA}{IC}.$", "Solution_10": "El problema anterior puede ser generalizado de la siguiente forma\n\n[b]Proposici\u00f3n:[/b] $ABCD$ es un cuadril\u00e1tero convexo y $J$ es un punto en la diagonal $AC.$ Una recta por $J$ corta a $AB,CD$ en $P,Q$ y otra recta por $J$ corta a $BC,DA$ en $R,S.$ $AC$ corta a $PS,RQ$ en $M,N.$ Entonces $\\frac{_{IA}}{^{IC}}=\\frac{_{AM}}{^{JM}} \\cdot \\frac{_{JN}}{^{CN}}.$\n\nProy\u00e9ctese la recta a trav\u00e9s de $AD \\cap BC$ y $AB \\cap DC$ al infinito. Denotando los puntos projectados con sub\u00edndice cero, $ABCD$ con intercepto $I$ de diagonales se transforma en un paralelogramo $A_0B_0C_0D_0$ con centro $I_0$ y claramente $\\triangle A_0P_0S_0$ y $\\triangle C_0Q_0R_0$ son homot\u00e9ticos a trav\u00e9s de $J_0.$ Como $I_0A_0=I_0C_0$ y $\\frac{_{J_0M_0}}{^{J_0N_0}}=\\frac{_{A_0M_0}}{^{C_0N_0}},$ se sigue que\n\n$ \\left (\\frac{I_0A_0}{I_0C_0} \\cdot \\frac{M_0C_0}{M_0A_0} \\right) \\cdot \\left (\\frac{J_0M_0}{J_0N_0} \\cdot \\frac{C_0N_0}{C_0M_0} \\right)=(A_0,C_0,I_0,M_0) \\cdot (M_0,N_0,J_0,C_0) =1$\n\nAs\u00ed, $(A,C,I,M) \\cdot (M,N,J,C) =1 \\Longrightarrow \\frac{IA}{IC} \\cdot \\frac{CN}{JN} \\cdot \\frac{JM}{MA}=1.$" } { "Tag": [ "geometry", "geometric transformation", "reflection", "function" ], "Problem": "question: give the equation of the reflection of y = 2x-5 into y=3x\r\ni am not sure if i got this right or not, and looking from the graph, it does not seem to be right, i would like a second opinion, my work is as follows, \r\n\r\n[hide]consider point line distance formula for $ax+by+c=0$ and a point (x,y)\nif we have 2 functions, and they reflect over a line, the distance from one line to to line it is being reflected into is always the same as the distance from the reflection and the line it is being reflected into for it to be a reflection, \nso for the first function, it can be an ordered pair, (x,2x-5), and that distance from that point to $3x-y+0=0$ is the same as another linear function if it is to be an inverse, let our other function be $y=mx+b$ or (x,mx+b)\nso setting them equal, we get\n$\\frac{|3x-y|}{\\sqrt{3^2+1^2}} = \\frac{|3x-y|}{\\sqrt{3^2+1^2}}$ simplifying, and putting in our 2 functions,\n$|3x-(2x-5)| = |3x -(mx+b)|$\n$|x+5|=|(3-m)x-b|$, note that the previous function where m=2, and b=-5 satisfies this, but we need 2 other values of m and b for this, by inspection, we see that m=4, and b=5 also works, so the answer is\n$\\boxed{y=4x+5}$, but as i said, looking at the graph on my calc does not seem to show this? my answer seems logical?, is the graph on the calc just unreliable, or what?[/hide]", "Solution_1": "[quote=\"Altheman\"]question: give the equation of the reflection of y = 2x-5 into y=3x\ni am not sure if i got this right or not, and looking from the graph, it does not seem to be right, i would like a second opinion, my work is as follows, \n\n[hide]consider point line distance formula for $ax+by+c=0$ and a point (x,y)\nif we have 2 functions, and they reflect over a line, the distance from one line to to line it is being reflected into is always the same as the distance from the reflection and the line it is being reflected into for it to be a reflection, \nso for the first function, it can be an ordered pair, (x,2x-5), and that distance from that point to $3x-y+0=0$ is the same as another linear function if it is to be an inverse, let our other function be $y=mx+b$ or (x,mx+b)\nso setting them equal, we get\n$\\frac{|3x-y|}{\\sqrt{3^2+1^2}} = \\frac{|3x-y|}{\\sqrt{3^2+1^2}}$ simplifying, and putting in our 2 functions,\n$|3x-(2x-5)| = |3x -(mx+b)|$\n$|x+5|=|(3-m)x-b|$, note that the previous function where m=2, and b=-5 satisfies this, but we need 2 other values of m and b for this, by inspection, we see that m=4, and b=5 also works, so the answer is\n$\\boxed{y=4x+5}$, but as i said, looking at the graph on my calc does not seem to show this? my answer seems logical?, is the graph on the calc just unreliable, or what?[/hide][/quote]\r\n\r\nIf a line $L_1$ is reflection of the line $L_3$ in the line $L_2$, then image of every point on $L_1$ w.r.t $L_2$ lies on $L_3$ and vice versa. And $L_2$ is one of the bisectors of angle between $L_1$ and $L_3$.\r\nThe lines $L_1 \\equiv {\\rm 2x-y - 5=0}$ and ${\\rm L}_{\\rm 2} \\equiv {\\rm 3x-y=0}$meet at (-5, -15)\r\npick some point, say (5, 5) on the line $L_1 \\equiv {\\rm 2x-y - 5=0}$\r\nFind the image of this point in the line ${\\rm L}_{\\rm 2} \\equiv {\\rm 3x-y=0}$\r\nit gives us (-1, 7). \r\n\r\nThe required line is nothing but the line joining the points (-5, -15) and (-1, 7) \r\nand the answer is $11x-2y+25=0$" } { "Tag": [], "Problem": "I guess it is 2,2-dimethyl bromocyclopentane...", "Solution_1": "No it isnt it. There s another methyl shift to produce a more stable tert carbocation.\r\nSo it is 1-bromo-1,2-dimethylcyclopentane.\r\nAshwath plz no your question.", "Solution_2": "2-bromo 1,1-dimethyl cyclopentane is the answer....", "Solution_3": "Yeah there is one more rearrangement. Raghavendra is right. So much for my paltry question..." } { "Tag": [ "geometry" ], "Problem": "is it possible to cut a square into several obtuse triangles?\r\n\r\nif you can, can you give a proof for it?", "Solution_1": "i meant is it possible to dissect a square into only obtuse triangles?", "Solution_2": "Since any square can be dissected into four right triangles and that any right triangle can be dissected into three obtuse triangles, it is possible to dissect a square into obtuse triangles. A right triangle can be dissected into obtuse triangles by connecting the orthocenter to the vertices.", "Solution_3": "Just curious,\r\nif you knew the solution why did you post it as \"can't do it\"? :(", "Solution_4": "[quote=\"maximocapcom\"] A right triangle can be dissected into obtuse triangles by connecting the orthocenter to the vertices.[/quote]\r\nBut, the orthocenter, here does not lie inside the (right) triangle :(\r\n\r\n[hide](Of course, you can dissect into 2 (non right triangles) and one which is not obtuse can then be dissected into obtuse triangles :) )\n[/hide]\n(BTW, A related problem is to dissect a square into all acute triangles) The answer is yes (Not too hard) and of course, a acute triangle can be cut into obtuse triangle by the above method)\n\n(Hint: The question has been asked and answered in AoPS before )\n\n(Peek only if you want to see the answer) \n\n[hide] [url=http://www.ics.uci.edu/~eppstein/junkyard/acute-square/]click here[/url][/hide]", "Solution_5": "Posted and discussed before: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=obtuse&t=105119\r\n\r\n[quote=\"thealchemist\"]Just curious,\nif you knew the solution why did you post it as \"can't do it\"? :([/quote]\r\n\r\nHis original post is from January 2006. He solved it a few days ago. There's nothing wrong with that." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Simplify the following polynomial:\r\n\r\n(X \u2013 A)(X \u2013 B)(X \u2013 C)(X \u2013 D) . . . (X \u2013 Z)\r\n\r\nis therea easy way to do this?", "Solution_1": "$=ABCDE......X^{26}$", "Solution_2": "[quote=\"moogra\"]Simplify the following polynomial:\n\n(X \u2013 A)(X \u2013 B)(X \u2013 C)(X \u2013 D) . . . (X \u2013 Z)\n\nis therea easy way to do this?[/quote]\r\n\r\nLook at all of the seperate factors. Is there one that will make this problem much easier? :D [hide](This is really a trick question)[/hide]", "Solution_3": "[quote=\"moogra\"]Simplify the following polynomial:\n\n(X \u2013 A)(X \u2013 B)(X \u2013 C)(X \u2013 D) . . . (X \u2013 Z)\n\nis therea easy way to do this?[/quote]\r\n[hide=\"Easy\"]The answer is $0$ because one of the parts is $(X-X).$[/hide]", "Solution_4": "[hide=\"My Solution\"]\n\nWe know that $(X-X)$ is also in there, and x-x=0, so the whole thing is 0.[/hide]", "Solution_5": "[hide=\"hint\"]Look at the last 5 terms or so[/hide]\n[hide=\"answer\"]$x-x$ is one of the factors...[/hide]\r\nCan some mod please move this to puzzles and brainteasers (although I don't think moogra knew that this was a trick question)?", "Solution_6": "Solution. [hide]0 because of (x-x).[/hide]\r\n\r\nSeems every high school does this question!", "Solution_7": "Actually, there was just a problem exactly like this in Puzzles and Brainteasers :D", "Solution_8": "right thanks. so its just when x-x=0 then 0*the rest is 0. ok and yeah i got this from a whitehouse.gov site..." } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "Prove that in any acute angle triangle ABC we have:\r\n\r\nsum(1/(sinA/2))>=2(sum(1/cos((A-B)/4))).", "Solution_1": "In the left member is it \\sum (1/(sin(A/2)) ?", "Solution_2": "Yaap .. soory about that! the ineq is like so:\r\n\r\nProve that in any acute angle triangle ABC we have: \r\n\r\nsum(1/(sin(A/2)))>=2(sum(1/cos((A-B)/4)))." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Show that the equation $z^{n}+z+1=0$ has a solution with $|z|=1$ if and only if $n-2$ is divisble by $3$.", "Solution_1": "If $|z| = 1$ then $|z^{n}| = |z+1| = 1$, hence $z$ is a primitive third root of unity, and of course $\\omega^{n}+\\omega+1 = 0 \\Leftrightarrow 3 | n-2$.", "Solution_2": "This must have been the most trivial problem ever to be proposed in the NMO" } { "Tag": [ "algebra", "function", "domain", "complex analysis", "complex analysis unsolved" ], "Problem": "Let $ D$ be a bounded domain, and let $ f(z)$ and $ h(z)$ be meromorphic functions on $ D$ that extend to be analytic on $ \\partial D$. Suppose that $ |h(z)|< |f(z)|$ on $ \\partial D$. Show by example that $ f(z)$ and $ f(z)\\plus{}h(z)$ can have different numbers of zeros on $ D$. What can be said about $ f(z)$ and $ f(z)\\plus{}h(z)$? Prove your assertion.\r\n\r\nI do not see what example I can use here. I think what we can say is that they have the same number of zeros minus poles. I just don't see an example here. Thanks.", "Solution_1": "You are almost done. Since the number (zeros-poles) is the same for both functions, take two functions that have different number of poles, and they'll have different number of zeroes." } { "Tag": [ "geometry", "calculus", "calculus computations" ], "Problem": "The side of an Equilateral Triangle decreases at the rate of 2m/s\r\nAt what rate is the Area Decreasing if the area is 100 m^2?", "Solution_1": "Show us what you've done so far.\r\n\r\nIn particular, can you give an equation relating the side and area of the triangle?", "Solution_2": "[img]http://www.mathwords.com/a/a_assets/area%20equilateral%20triangle%20formula.gif[/img]\r\n\r\nI used that formula and found out that each side is ~15.1967...m\r\nNow I am stuck on how to find the rate", "Solution_3": "That specific value of $s$ is \"snapshot\" information: it's only true at one instant, and does not remain true as things change. So it's not something you can differentiate. However, the equation $A=\\frac{s^{2}\\sqrt{3}}4$ does remain true even as things change. Both $A$ and $s$ depend on $t,$ time. Take $\\frac{d}{dt}$ of both sides of that equation. Once you've done that, what pieces of information do you already know and what are you trying to find?", "Solution_4": "[quote=\"V4Vendetta\"]The side of an Equilateral Triangle decreases at the rate of 2m/s\nAt what rate is the Area Decreasing if the area is 100 m^2?[/quote]\r\nWe know: $\\frac{ds}{dt}=-2 \\frac{m}{sec}$ and that $A=\\frac{s^{2}\\sqrt{3}}{4}$.\r\n\r\n[hide]$A=\\frac{s^{2}\\sqrt{3}}{4}$\n$\\iff 100=\\frac{s^{2}\\sqrt{3}}{4}$\n$\\iff \\frac{400}{ \\sqrt{3}}=s^{2}\\implies s=\\frac{20}{3^{\\frac{1}{4}}}$\n\nNow:\n\n$A'=\\frac{s \\sqrt{3}}{2}\\cdot \\frac{ds}{dt}$\n$\\iff A'=\\frac{\\left( \\frac{20}{3^{\\frac{1}{4}}}\\right) \\sqrt{3}}{2}\\cdot-2$\n${\\iff \\frac{dA}{dt}=-20 \\cdot 3^{\\frac{1}{4}}}$\n$\\implies$ that $A$ is decreasing at a rate of $\\boxed{20 \\cdot 3^{\\frac{1}{4}}\\frac{m^{2}}{sec}}$[/hide]", "Solution_5": "I think Skimnc hit the spot. Though i think he made a mistake on the last step.\r\nShouldn't it be\r\n20/ 3^(1/2)", "Solution_6": "skinmc's calculation is correct (although I wish he'd held off and made V4Vendetta work a little more). Note that $\\frac{20}{\\sqrt[4]3}$ is the approximately $15.197$ that V reported earlier.", "Solution_7": "[quote=\"V4Vendetta\"]The side of an Equilateral Triangle decreases at the rate of 2m/s\nAt what rate is the Area Decreasing if the area is 100 m^2?[/quote]\r\n\r\n[hide=\"slightly different\"][size=200]$\\frac{dA}{dt}=\\frac{\\frac{ds}{dt}}{\\frac{ds}{dA}}$[/size]\n\n[color=blue]$\\frac{ds}{dt}=-2$[/color]\n\n$A=\\frac{s^{2}\\sqrt 3}{4}$ it follows that $s=\\frac 2{\\sqrt[4]3}\\sqrt{A}$\n\n$\\frac{ds}{dA}=\\frac{1}{\\sqrt[4]3\\sqrt A}$ $A=100$ so, $\\frac{ds}{dA}=\\frac 1{10\\sqrt[4]3}$\n\nso the answer is $\\frac{dA}{dt}=\\frac{2}{\\frac 1{10\\sqrt[4]3}}=20\\sqrt[4]3$ $m^{2}/s$[/hide]" } { "Tag": [], "Problem": "An ice cream shop offers three toppings. If a customer's order may include none, one, two or all of the toppings, how many combinations of the toppings can a customer choose?", "Solution_1": "0 toppings- 1combo\r\n1 topping- 3combo\r\n2 toppings- 3combo\r\n3 toppings- 1combo\r\n\r\n\r\n1+3+3+1= 8\r\n\r\n\r\n :surf:", "Solution_2": "or you could notice that it fits the description of a set and just apply $ 2^3\\equal{}\\boxed{8}$" } { "Tag": [ "modular arithmetic" ], "Problem": "These types questions have been bothering me:\r\nWhat is the remainder when 13^(13)+5 is divided by 6.\r\n\r\nThanks", "Solution_1": "[hide=\"Solution\"]\n$ 13^{13} \\plus{} 5\\equiv 1^{13} \\plus{} 5\\equiv 0\\mod 6$\n\nAnswer: 0[/hide]", "Solution_2": "[url=http://en.wikipedia.org/wiki/Modular_arithmetic]Modular[/url] [url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction]arithmetic[/url] is what is used to solve these problems.\r\n\r\nIn this particular one, $ 13^{13}\\plus{}5\\equiv1^{13}\\plus{}(\\minus{}1)\\equiv1\\minus{}1\\equiv\\boxed0\\pmod6$." } { "Tag": [ "number theory", "least common multiple" ], "Problem": "Here's an LCM (Least common multiple) question.\r\n\r\nWhat is lcm(1,2,3,4,5,6,7,8,9,10)?", "Solution_1": "[hide]\n2520\n[/hide]", "Solution_2": "sort by prime factors; the biggest exponent on the factors of 2 is 3, on 3 it's 2, on 5, 1, on 7, 1. This gives 8*9*5*7" } { "Tag": [], "Problem": "What is your favorite team and who do you think will win the BCS Championship?\r\n\r\nI like Tennessee the most (I know they're not that great right now).\r\nOther teams that I like include UTEP, Colorado State, and Ohio State.\r\n\r\nI think that Ohio State has a chance at winning the BCS Championship? What do you think?", "Solution_1": "I don't like football, but I'll say this\r\nOSU WILL PWN EVERYONE", "Solution_2": "In my opinion football is great.\r\nWhy do you like OSU?", "Solution_3": "My daddy and mommy went there :P\r\nAND they had the BEST chinese food at the cafeteria", "Solution_4": "Go OSU!!!\r\n\r\nThey re beating Texas!!!\r\n\r\nYAY!\r\n\r\nI used to live in Ohio." } { "Tag": [], "Problem": "proof :\r\n$ 1.3^2.5^2.7^2. . . . .(2n\\plus{}1)^2. . . \\equal{} (2.4)(4.2^2).(4.3^2). . . (4.n^2). . .$. \r\nHelp me. Thanhks", "Solution_1": "People need to stop using periods as multiplication signs; the code is either \\cdot or \\times.\r\n\r\nThis doesn't even look like it's true. You can't multiply together odd numbers and expect to get back an even one, as you have. If this is not what you meant, it's even more reason not to use periods since no one can understand what you're trying to say.", "Solution_2": "proof: \r\n$ 1\\times 3^2\\times 5^2\\times 7^2\\times . . . \\times (2n\\plus{}1)^2\\times . . . \\equal{} (2\\times 4)(4\\times 2^2)(4\\times 3^2)\\times . . . \\times (4\\times n^2)\\times . ..$.\r\nhelp me. Thanks", "Solution_3": "As JRav said, this is clearly impossible. Let $ n\\equal{}1$, for example. You get that $ 3 \\equal{} 4$, which is false. In general, the LHS is odd, but the RHS is even. \r\n\r\nI suspect the LHS numbers are all supposed to be even, in which case the proof is trivial." } { "Tag": [ "USAMTS" ], "Problem": "By the new rules, if we only send the PDF and not the TEX, how will solutions get published then? (i.e. on the solutions page [url=http://www.usamts.org/Problems/U_ProblemsPast.php]here[/url])", "Solution_1": "Somewhere on the changes page, it says that they will no longer be publishing students' solutions, and will just publish their own official solutions. This, along with no longer commending solutions, will supposedly speed up the process of grading. (Yes!!!)\r\n\r\nedit: Here is the link. [url]http://usamts.org/year20changes.php[/url]", "Solution_2": "Indeed, we will not be publishing students' solutions anymore, for 2.5 reasons:\r\n\r\n1) It was rather time-consuming for the USAMTS staff to select solutions and edit them for publication. Plus, we had to keep track of all the TeX source files. \r\n\r\n2) Instead, we're going to post \"official\" solutions a few days after each round is due. That way, you'll get to see more quickly what the solutions to the problems are, instead of having to wait a month or more.\r\n\r\n2.5) If you think you have a nice solution, you can always post it on this message board." } { "Tag": [ "algebra", "polynomial", "Vieta" ], "Problem": "Solve the equation for any integer x\r\nx\u00b3 + x\u00b2 + 5 is congruent to 0 (mod 7\u00b3)", "Solution_1": "Interesting problem! All I can think of is using the cubic formula but it requires TONS of time as is extremely UGLY.", "Solution_2": "Short solution without ugly computation:\r\nFind all of them $\\mod 7$ and use Hensel.", "Solution_3": "Well, you could also use Tschirnhaus and Vieta substitutions; it doesn't require much ugly computations. :)" } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "I imported the last six years (2003-2008) list into a database and ran some queries. \r\n\r\n count | school \r\n-------+-------------------------------------------------\r\n 76 | Phillips Exeter Academy\r\n 75 | Thomas Jefferson HS Sci/Tech\r\n 47 | Montgomery Blair HS\r\n 43 | Academy Advancement Sci/Tech\r\n 40 | Stuyvesant HS\r\n 31 | Lynbrook HS\r\n 26 | Lexington HS\r\n 26 | Phillips Academy\r\n 24 | Henry M. Gunn HS\r\n 24 | Saratoga HS\r\n 22 | Illinois Math & Sci Academy\r\n 17 | Clements HS\r\n 17 | Mission San Jose HS\r\n 17 | The Harker School\r\n 16 | Palo Alto HS\r\n 16 | Detroit Country Day School\r\n 14 | Naperville North HS\r\n 14 | Vestavia Hills HS\r\n 12 | East Chapel Hill HS\r\n 11 | James Rickards HS\r\n 11 | Lakeside School\r\n 10 | Vincent Massey Secondary School\r\n 10 | Choate Rosemary Hall\r\n\r\nNote, the same person qualifying for 3 years in a row count as 3.", "Solution_1": "This has basically already been posted, in the form of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199100]this thread[/url]. Those tables also incorporate some details that the automatic system would have missed." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "i think this is a Lemoine P point with Soddy circle but i don't know how :(", "Solution_1": "What is the Soddy circle?", "Solution_2": "Apollonius-Soddy circles" } { "Tag": [ "geometry", "angle bisector", "similar triangles", "power of a point", "radical axis", "geometry proposed" ], "Problem": "Let $\\triangle ABC$ and two mobile points $M\\in (BA,\\ N\\in (CA,\\ BM=CN.$\r\nAscertain the geometrical locus of the intersection $L\\in CM\\cap BN$.\r\n\r\n[u]Remark.[/u] $X\\in (UV\\Longleftrightarrow X\\in (UV]\\ \\vee \\ V\\in (UX)$, i.e. $(UV$ is a [b]ray[/b].", "Solution_1": "Let $AD$ be the interior angle bisector of $\\triangle ABC$.\r\n\r\nThe perallel lines from $M,N,L$ to $AD$ intersect the line $BC$ at the points $M',N',L'$ respectively.\r\nWe will show that point $L'$ is fixed point\r\n\r\n$\\triangle ABD \\sim \\triangle MBM' \\Rightarrow \\frac{AB}{BM} = \\frac{BD}{BM'} = \\frac{AD}{MM'}$\r\n\r\n$\\triangle ACD \\sim \\triangle NCN' \\Rightarrow \\frac{AC}{CN} = \\frac{CD}{CN'} = \\frac{AD}{NN'}$ \r\n\r\nFrom these equations we get\r\n\r\n$BM' = CN'$ and\r\n\r\n$\\frac{MM'}{NN'} = \\frac{AC}{BC}$\r\n\r\n$BM' = CN'$ means that the points $M',N'$ are symmetric with respect to the midpoint of $BC$ \r\n[we are sure for this because $M' \\in (BC$ and $N' \\in (CB$]\r\n\r\nSo we have also $BN' = CM'$ \r\n\r\n\r\n\r\n\r\nFrom the other similar triangles :\r\n\r\n$\\triangle BNN' \\sim \\triangle BLL' \\Rightarrow \\frac{LL'}{NN'} = \\frac{BL'}{BN'}$\r\n\r\n\r\n$\\triangle CMM' \\sim \\triangle CLL' \\Rightarrow \\frac{LL'}{MM'} = \\frac{CL'}{CM'}$\r\n\r\ndividing these two equations we get\r\n\r\n$\\frac{MM'}{NN'} = \\frac{BL'}{CL'} \\cdot \\frac{CM'}{BN'} = \\frac{BL'}{CL'}$\r\n\r\n\r\nfinally $\\frac{BL'}{CL'} = \\frac{AC}{BC} = \\frac{CD}{BD}$\r\n\r\n\r\nHence $L'$ is symmetric of $D$ through the midpoint of $BC$", "Solution_2": "Pontios, I erred the address. Here is the shortest solution of the proposed problem from the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=72366\r\n$ML\\parallel PN\\Longrightarrow \\widehat {LMP}\\equiv \\widehat {MPN}\\equiv \\widehat {RPN}\\equiv \\widehat {RAN}\\equiv \\widehat {LAN}\\Longrightarrow$\r\n$\\widehat {LMP}\\equiv \\widehat {LAN}\\Longrightarrow \\triangle LMR\\sim \\triangle LAM\\Longrightarrow \\frac{LM}{LA}=\\frac{MR}{AM}=\\frac{RL}{ML}\\Longrightarrow$\r\n$LM^2=LR\\cdot LA\\ .$ Therefore, the point $L$ has same power w.r.t. the crcles\r\n$w_1$ and $w_2$, i.e. the point $L$ belongs to the radical axis between these circles." } { "Tag": [], "Problem": "There are positive integers that have these properties:\\\\\nI. The sum of the squares of their digits is 50, and\\\\\nII. Each digit is larger than the one on its left.\\\\\nWhat is the product of the digits of the largest integer with both\nproperties?", "Solution_1": "$ 1^2 \\plus{} 2^2 \\plus{} 3^2;4^2 \\plus{} 5^2 \\equal{} 55 \\neq 50$\r\n$ 2^2 \\plus{} 3^2 \\plus{} 4^2 \\plus{} 5^2 \\equal{} 54 \\neq 50$\r\n$ 1^2 \\plus{} 3^2 \\plus{} 4^2 \\plus{} 5^2 \\equal{} 51 \\neq 50$\r\n$ 1^2 \\plus{} 2^2 \\plus{} 4^2 \\plus{} 5^2 \\equal{} 46 \\neq 50$\r\n$ 3^2 \\plus{} 4^2 \\plus{} 5^2 \\equal{} 50$\r\n\r\nThus $ \\boxed{345}$.", "Solution_2": "The official solution found a larger one...", "Solution_3": "[quote=\"AIME15\"]The official solution found a larger one...[/quote]\r\nI have found the smallest number in a big digit...\r\nDo we found the largest number in the largest digit?\r\n\r\nIf so, The answer is $ \\boxed{543}$.", "Solution_4": "What about 6321? 6^2+3^2+2^2+1^2=36+9+4+1=45+5=50\r\n\r\nAlso, we're looking for the product of the digits so it's 36.\r\n\r\nHow can we prove it's the largest, though?", "Solution_5": "We know it can't be 10 digits (only 9 numbers! we cant use 0 here)\r\n\r\n9-Sum of squares is too large. (1^2+2^2....+9^2>50)\r\n\r\nLet's keep with this process.\r\n\r\n1^2+2^2+3^2+4^2+5^2>50. Thus it must be 4 digits.\r\n\r\nThe first digit MUST be 1. Can you see why?\r\n\r\nThus we have 3 numbers whose squares add up to 49.\r\n\r\nIn order to maximize we want the smallest of the three to be as large as possible.\r\n\r\nThe largest can at most be 7, meaning the smallest can at most be 5.\r\n\r\n5-MInimum possible is 5,6,7 too big\r\n\r\n4-4^2+5^2+6^2 is too big\r\n\r\n3-3^2+4^2+5^2 is 50! So close!\r\n\r\n2-2^2+3^2+4^2 fits in our range!\r\n\r\nThus if its 4 digits our number is 12xx\r\n\r\nNow the sum must be 45. Let's use a quick trick! 45 is 0mod3. All squares are 0 or 1 mod 3. The only way a sum of two squares is 0 mod 3 is if both squares are 0mod3. Thus our only possiblities contain 3,6,9. But 9^2 is already greater than 45. Trying 3^2+6^2 we see it work so our number is 1236, whose digits product is 36.", "Solution_6": "I'm sorry.\r\nI jumped to a misleading conclusion and seem to have done big misunderstanding.\r\n\r\nI correct the answer as follows.\r\n\r\n$ x^2 \\le 50$ i.e. $ x\\equal{}1, 2, 3, 4, 5, 6, 7$.\r\n$ 50\\minus{}7^2\\equal{}1$.\r\n$ 50\\minus{}6^2\\equal{}14\\equal{}3^2\\plus{}2^2\\plus{}1$. (Because, $ x^2<14$ and $ x \\neq 6$ i.e. $ x\\equal{}1, 2, 3$. similar after this...)\r\n$ 50\\minus{}5^2\\equal{}25\\equal{}4^2\\plus{}3^2$. (Because, $ x^2<25$ and $ x \\neq 5$ i.e. $ x\\equal{}1, 2, 3, 4$. similar after this...)\r\n\r\nTherefore, $ 17$, $ 1236$, $ 345$. (Because property II.)\r\nThus, $ \\boxed{1236}$.", "Solution_7": "Then we read the problem, and see that it asks for the product of the digits, so we do [hide]1*2*3*6=36[/hide].", "Solution_8": "[hide=solution]Since we want the largest number possible, we also want the number with the most digits. Thus, we want to use the smallest digits possible.\nLet's start with $1$, leaving $49$ \"points\" left over. We could place a $7$, but it is possible that we could get more digits.\nThen, place $2$, leaving $45$ points.\nThen, place $3$, leaving $36$ points.\nWe know we can't place a $4$ and $5$, because this would cost more than $36$ points. Thus, we place a $6$.\nOur number is $1236$, but let's make sure that there aren't any larger numbers. As we proved, there aren't enough points to have a 5 digit number. Additionally, we can't do $1245$, or any $4$ digit number containing $4,5$, as there aren't any other numbers to switch such that it will sum to $50$ points. Finally, we know we can't do $13xy$, since we wouldn't reach exactly $50$ points.[/hide]" } { "Tag": [ "modular arithmetic", "ceiling function", "number theory unsolved", "number theory" ], "Problem": "Find all pairs if positive integers $ (m,n)$ so that $ m|a^n\\minus{}1$ $ \\forall$ $ a \\in \\{1,2,...,n\\}$.", "Solution_1": "Solutions are $ (m,1)$ for any $ m$ and $ (p,p \\minus{} 1)$ for any prime $ p$.\r\n\r\nSuppose that $ (m,n)$ is such that $ m|a^n \\minus{} 1$ $ \\forall a \\in \\{1,2,\\dots,n\\}$ and $ n > 1$. Then:\r\ni) $ n < p$ where $ p$ is the smallest prime divisor of $ m$ (since $ m$ does not divide $ p^n \\minus{} 1$);\r\nii) for every prime divisor $ q|m$, $ n \\equal{} q \\minus{} 1$.\r\n\r\nTo prove ii) one needs to notice that according to i), $ n < q$, and the number of solutions to the congruence $ x^n\\equiv 1\\pmod{q}$ equals $ \\gcd(n,q \\minus{} 1)$. Since $ 1,2,\\dots,n$ represent distinct solutions to this congruence, $ n\\leq \\gcd(n,q \\minus{} 1)$, implying that $ n$ divides $ q \\minus{} 1$. But if $ n < q \\minus{} 1$ (i.e., $ n\\leq \\frac {q \\minus{} 1}{2}$) then $ 2\\cdot \\left\\lceil \\frac {n \\plus{} 1}{2} \\right\\rceil > n$ represents yet another solution to the aforementioned congruence. This contradiction proves that $ n \\equal{} q \\minus{} 1$.\r\n\r\nThe statement ii) implies that $ m$ cannot have more then one distinct prime divisor. Therefore, $ m \\equal{} p^k$ for some prime $ p$ and integer $ k > 0$. According to ii), we have $ n \\equal{} p \\minus{} 1$. Since the number of solutions to $ x^{p \\minus{} 1}\\equiv 1\\pmod{p^k}$ equals $ p \\minus{} 1$, the numbers $ 1,2,\\dots,p \\minus{} 1$ represent all such solutions. But for $ k > 1$, there is yet another solution $ 2\\cdot \\left\\lceil \\frac {p}{2} \\right\\rceil > p \\minus{} 1$. The contradiction proves that $ k \\equal{} 1$." } { "Tag": [ "factorial", "probability", "number theory open", "number theory" ], "Problem": "Hi everybody,\r\n\r\nLet a integer >2\r\n\r\ns=int(sqrt(a))\r\ns! is factorial s \r\n\r\nP= a + s!\r\n\r\nIf P is prime then a is prime.\r\n\r\nHow to prove it?\r\nIf a is composite we know for sure that P is composite but is there another way to prove it?", "Solution_1": "That's by far the most straightforward way. Are you asking for a way to \"reverse\" the proof so that it doesn't rely on proving the contrapositive or are you asking for a genuinely different proof?", "Solution_2": "I'm just asking if there is another way to prove it.", "Solution_3": "I'm thinking to Wilson theorem.\r\nMaybe we could use it...", "Solution_4": "Here is the sequence (a,P) where a and P are both prime.\r\n\r\n2,3\r\n5,7\r\n11,17\r\n13,19\r\n17,41\r\n19,43\r\n23,47\r\n29,149\r\n31,151\r\n37,757\r\n41,761\r\n59,5099\r\n61,5101\r\n67,40387\r\n89,362969\r\n97,362977\r\n131,39916931\r\n139,39916939\r\n167,479001767\r\n\r\nI did not find the sequence in Sloane\r\n\r\n2,5,11,13,17,19,23,29,31,37,41,59,61,67,89,97,131,139,167\r\n\r\nI remind you that \r\n\r\nP=a+s!\r\n\r\nwhere s=int(sqrt(a))", "Solution_5": "Let us build a sequence as follows \r\n\r\np is prime \r\n\r\nU(0)=p\r\nU(1)=U(0)+int(sqrt(U(0)))!\r\n......\r\nU(n)=U(n-1) + int(sqrt(U(n-1))!\r\n\r\nExample \r\n\r\nU(0)=41\r\nU(1)=41+int(sqrt(41))!=41+6! =41+720=761 is prime\r\nU(2)=761+27!=10888869450418352160768000761 is not prime = 229 * 47549648255101974501170309 so k=2\r\nand so on \r\n\r\nCan we prove that at the step n=k \r\nU(n) is composite with certainty?", "Solution_6": "[quote=\"Matharith\"]\nLet a integer >2\n\ns=int(sqrt(a))\ns! is factorial s \n\nP= a + s!\n\nIf P is prime then a is prime.\n\nHow to prove it?\n[/quote]\r\nOK, I'm rewriting:\r\nIf $ a \\plus{} ([\\sqrt {a}])!$ is prime, then $ a$ is prime.\r\nSolution...\r\n[hide]\nSuppose that it isn't true, so $ a \\equal{} d_{1}d_{2}$, where $ 1 < d_1 \\leq \\sqrt {a}$.\nIt's easy to see that $ d_1 \\leq [\\sqrt {a}]$.\nWe have that $ ([\\sqrt {a}])!$ is divisible by $ d_1$.\nSo, $ a \\plus{} ([\\sqrt {a}])! \\equal{} d_{1} d_{2} \\plus{} 1*2*...*d_1*...*[\\sqrt {a}]$ is divisible by $ d_1 \\neq 1$, but not equal to $ d_1$.\nContradiction.[/hide]", "Solution_7": "P=a+s! \r\n\r\nwhere s=int(sqrt(a))\r\n\r\nIf we express \r\na= s^2 + r\r\n\r\nP= s(s+(s-1)!) + r \r\n\r\nCan we find forms of s and r such as P will have a high probability to be prime?" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $ S^{n}=\\{(x_{0},\\ \\cdots ,\\ x_{n})\\in{{\\mathbb{R}}^{n+1}| x_{0}^{2}+\\cdots x_{n}^{2}=1\\}}$ be a n-dimensional sphere.\r\n\r\n(1) Let $ f: S^{n}\\rightarrow\\mathbb{R}$ be a continuous mapping for $ n\\geq 1$. Prove that there exists $ x\\in{S^{n}}$ satisfying $ f(x)=f(-x)$.\r\n\r\n(2) Let $ f: S^{n}\\rightarrow S^{1}$ be a continuous mapping for $ n\\geq 2$. Prove that there exists $ x\\in{S^{n}}$ satisfying $ f(x)=f(-x)$.", "Solution_1": "For (1), let $ g(x) \\equal{} f(x)\\minus{}f(\\minus{}x)$. Then $ g$ is a continuous mapping $ S^{n}$ into $ \\mathbb{R}$.\r\n\r\nChoose $ a\\in S^{n}$. If $ g(a) \\equal{} 0$, then it is done. If not, then $ g(\\minus{}a) \\equal{}\\minus{}g(a)$ and one of $ g(a)$ and $ \\minus{}g(a)$ is positive, and the other is negative.\r\n\r\nSince continuous mapping preserves connectedness and $ S^{n}$ is connected, $ g(S^{n})$ is connected and hence $ g(S^{n})$ is an interval containing both $ g(a)$ and $ \\minus{}g(a)$. So $ [\\minus{}g(a), g(a)]\\subseteq g(S^{n})$ and $ 0\\in g(S^{n})$, which proves (1)." } { "Tag": [ "geometry", "geometric transformation", "email" ], "Problem": "I'm sorry this was asked before but can anyone tell me where I can get 2005-2006 COMC? I don't participate in COMC but I'll be adding the problems to the resources page so if anyone can give them to me, please PM or post here.", "Solution_1": "http://cemc.uwaterloo.ca/english/contests/past_contest/2006/comc.pdf", "Solution_2": "If you want some translation of Korean Mathematics Olympiad, I could help you out. I already translated some of them. :)", "Solution_3": "[quote=\"Singular\"]http://cemc.uwaterloo.ca/english/contests/past_contest/2006/comc.pdf[/quote]\r\n\r\nSo, 2006 COMC hasn't taken place yet (thanks for 2005 test)? I know the great deals about the contests around the world but time is something I always don't remember.\r\n\r\nAlso, I'm pretty sure that COMC goes before 1996. If so, can you see if you can find any extra tests from your school or teachers? I remember you knowing a lot of CMO and COMC stuffs so that's why I'm asking. Thanks.\r\n\r\nTo lightrhee, yes, I'll very appreciate them. Shoot me a private message and I'll give you my \"math\" email address. I can't guarantee on when the problems will be added but I'll do them ASAP so send them if you translated them.", "Solution_4": "COMC takes place in November. And the 1996 COMC is the first COMC there ever was.", "Solution_5": "I thought there was a contest before 1996 that determined qualification to the CMO. Wasn't it like the Canadian 11th Grade...I ono..." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let n blue lines, no two of which are parallel and no three concurrent, be drawn on a plane. An intersection of two blue lines is colored blue. Through any two blue points that have not alreadz been joined by a blue line, a red line is drawn. An intersection of two red points is colored red, and an intersection of a red line and a blue line is colored purple. What is the maximum possible number of purple points?", "Solution_1": "This could be wrong as I am in a hurry, but I have calculated it to be $ \\frac {n{(n \\minus{} 2)^{2}}(n \\minus{} 1)(n \\plus{} 1)}{8}\\minus{}n(n\\minus{}2)$", "Solution_2": "Now that I have some more time I looked at the problem again and I'll give the sketch of a solution:\r\n\r\nStep 1: The blue points are $ \\frac {n(n \\minus{} 1)}2$\r\n\r\nStep 2: Every red point \"creates\" $ \\frac {n(n \\minus{} 1)}2 \\minus{} (n \\minus{} 1)$\r\n\r\nStep 3: The number of red lines is $ \\frac {n{(n \\minus{} 1)^{2}}(n \\minus{} 2)}{8}$\r\n\r\nStep 4: Red-blue lines multiplication principle\r\n\r\nStep 5: Final answer $ \\frac {{n^2}{(n \\minus{} 1)^{2}}(n \\minus{} 2)}{8}$ and not $ \\frac {n{(n \\minus{} 2)^{2}}(n \\minus{} 1)(n \\plus{} 1)}{8} \\minus{} n(n \\minus{} 2)$" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Thursday night, same time? Tinytim please comment!", "Solution_1": "ok, I will be there...", "Solution_2": "Where are u guys going lol? Top-secret Mathcounts meeting???", "Solution_3": "*007math gets mew's address and plans an mission to make them fail along with his friends who failed at states*" } { "Tag": [ "category theory", "advanced fields", "advanced fields theorems" ], "Problem": "let $ S$ be a scheme; a $ \\mathcal{O}_S$-algebra $ A$ is called quasi-coherent if it is a sheaf and for every open affine $ U \\subseteq S$ there is a $ \\mathcal{O}_S(U)$-algebra $ B(U)$, whose associated $ \\mathcal{O}_U$-algebra is isomorphic to $ A|_U$, that is $ A|_U \\cong \\widetilde{B(U)}$. obviously, we may assume $ B(U) = A(U)$.\r\n\r\nI try to understand a construction of $ Spec(A)$ in a book, which glues the ordinary spectra $ Spec(A(U))$ together. when I try to fill in the details, many problems arise, which the author does not mention at all:\r\n\r\nlet $ S = \\cup_{i \\in I} S_i$ be an open affine covering of $ S$. put $ X_i = \\text{Spec~}A(S_i)$. now $ \\mathcal{O}_S(S_i) \\to A(S_i)$ induces $ p_i : X_i \\to \\text{Spec~}\\mathcal{O}_S(S_i) \\cong S_i$. put $ X_{ij} = p_i^{-1}(S_i \\cap S_j)$. the first claim is that there is a canonical ([i]what does canonical mean _here_?[/i]) isomorphism $ X_{ij} \\to X_{ji}$. first choose an open affine covering $ S_i \\cap S_j = \\cup_{\\lambda \\in \\Lambda_{ij}} S_{ij \\lambda}$ such that $ S_{ij \\lambda} \\cap S_{ij \\mu}$ is affine for all $ \\lambda, \\mu \\in \\Lambda_{ij}$ (for example, we may take basic-open subsets $ D_{S_i}(f)$). for $ f \\in \\mathcal{O}_S(S_i)$, we have\r\n\r\n(1) $ ~~~p_i^{-1}(D_{S_i}(f)) = D_{\\text{Spec~}A(S_i)}(f) \\cong \\text{Spec~}A(S_i)_f$\r\n$ ~~~ \\cong \\text{Spec~}A(S_i) \\otimes_{\\mathcal{O}_S(S_i)} \\mathcal{O}_S(S_i)_f = \\text{Spec~}\\widetilde{A(S_i)}(D_{S_i}(f)) \\cong \\text{Spec~}A(D_{S_i}(f))$ \r\n\r\n[i]honestly, what justifies the author to write always equalities instead of isomorphisms here?[/i]\r\n\r\nin particular, $ p_i^{-1}(S_{ij \\lambda}) \\cong \\text{Spec~}A(S_{ij \\lambda})$. [i]why do we have the same result with $ p_j$[/i]? I think we may assume $ \\Lambda_{ij} = \\Lambda_{ji}$, $ S_{ij \\lambda} = S_{ji \\lambda}$, and that the $ S_{ij \\lambda}$ are basic-open simultanuously for $ S_i$ and $ S_j$ (I omit the proof for that). but that's not what the author had in mind. \r\n\r\nthere is an isomorphism $ \\phi_{ij \\lambda} : p_i^{-1}(S_{ij \\lambda}) \\to p_j^{-1}(S_{ij \\lambda})$. now the author claims that the same holds for the intersections $ S_{ij \\lambda} \\cap S_{ij \\lambda'}$. [i]how do you see this directly? probably (1) can be extended to arbitrary open affines in $ S_i$?[/i] I wrote $ S_{ij \\lambda} = D_{S_i}(f_\\lambda)$, and checked by the explicit definitions that the isomorphism $ p_i^{-1}(D_{S_i}(f_\\lambda) \\to \\text{Spec~}A(D_{S_i}(f_{\\lambda'}))$ restricts to the isomorphism for $ f_\\lambda * f_\\lambda'$, which comes down to the fact that the $ \\phi_{ij \\lambda}$ restrict to isomorphisms $ p_i^{-1}(S_{ij \\lambda} \\cap S_{ij \\lambda'}) \\to p_j^{-1}(S_{ij \\lambda} \\cap S_{ij \\lambda'})$. thus the $ \\phi_{ij \\lambda}$ glue together to an isomorphism $ \\phi_{ij} : X_{ij} \\to X_{ji}$. clearly, we have $ \\phi_{ii} = id$ and $ \\phi_{ij} = \\phi_{ji}^{-1}$. in order to apply the gluing lemma and to optain $ Spec(A)$, [i]why does the cocycle-condition hold?[/i] that is, why does $ \\phi_{ij}$ restrict to an isomorphism $ \\phi_{ij}^{(k)} : X_{ij} \\cap X_{ik} \\to X_{ji} \\cap X_{jk}$ such that $ \\phi_{jk}^{(i)} \\circ \\phi_{ij}^{(k)} = \\phi_{ik}^{(j)}$? the author gives the following reason: over open affine parts, the isomorphisms are induced by identity maps on sections of $ A$. well, I don't see any identity maps, have no idea why this justifies the cocycle condition and my calculation will be nasty. \r\n\r\nfinally, the author writes that is easy to see that $ Spec(A)$ - up to [i]canonical[/i] (?) isomorphy - does not depend on the covering of $ S$. I have not really thought over this, but probably we become the same $ Spec(A)$ if we refine $ (S_i)$. then choose a common refinement of two open coverings of $ S$. but at the moment, I'm very unsure if this works and I would have to do redundant, tedious calculations. instead, I ask if someone knows some sort of [i]universal property[/i] of $ Spec(A)$ and if it is possible to avoid these calculations. there are many other gluing constructions which work similar, which start with a functor {affine schemes} -> {schemes}. can we trace a general condition which yields by gluing a functor {schemes} -> {schemes}?\r\n\r\nthe experienced reader will probably lough that I have difficulties with such natural constructions, but I have not seen this naturality yet ;)", "Solution_1": "also cleared :)" } { "Tag": [ "function", "conics", "ellipse", "rotation", "ratio", "geometry", "geometric transformation" ], "Problem": "Hello all,\r\n\r\nI'm baffled over this problem...\r\n\r\nFor a webproject I'm trying to move objects in a trajectory like a planet: an elliptic track that slowly rotates. The ellipse is defined by the following functions:\\[x(t) = sin(t) \\]\\[y(t) = .2cos(t) \\]These equations deliver a flat ellipse, now I would like this ellipse to rotate while keeping it's height-width ratio the same. A rotation should have a period of 20pi.\r\nI've tried all I can, I know it should be possible, but I can't figure it out. \r\nDoes anyone here know the trick?\r\n\r\nThanks for any help,\r\nSmakkie.", "Solution_1": "Rotating how? Around one of its axes, making a 3D solid? Or within its plane - if that, whether around the center or around a focus or around some other point?", "Solution_2": "I would like a 'flat' rotation around the center. Somewhat like a flower's leaves.", "Solution_3": "Then just put $x=\\sin(t-\\phi), y=0.2\\cos(t-\\phi)$ where $\\phi$ is the angle of rotation. Creating a loop with increasing values of $\\phi$ would make a \"flower\".", "Solution_4": "That's not quite what I mean or I misunderstand the use of $\\phi$.\r\nWith those equations, all I get is points ranging between -.2 and .2 on the $y$-axis, creating a 'flattened' flower.\r\nI would like it to look like a flower from above, so the tips of the 'leaves' are in a circle.", "Solution_5": "Oh, sorry, my bad.\r\n\r\nThe correct equations are\r\n\r\n$x=\\sin t\\cos \\phi-0.2\\cos t\\sin\\phi$\r\n$y=\\sin t\\sin\\phi+0.2\\cos t\\cos\\phi$", "Solution_6": "That does the trick perfectly!\r\nThank you very much." } { "Tag": [], "Problem": "[quote]The New 'Da Vinci Code' Code \n\nUpdated 1:30 AM ET April 28, 2006 \n\nA British High Court judge who rejected claims that \"The Da Vinci Code\" author Dan Brown had plagiarized his novel managed to inject a bit more mystery into the international bestseller. It emerged on Wednesday that the judge at the center of the copyright trial has hidden his own code in his judgment -- and he seems to be daring anyone to try to crack it. Dan Tench, an observant lawyer at the London-based law firm Olswang, noticed 38 random italicized letters in the 71-page court document. He said at first he didn't believe it was anything more than a problem with the judge's computer. \"One doesn't really think that a High Court judge is going to encipher a message in the text of a judgement, so I thought it was probably just a word processing error,\" Tench told ABC News. Justice Peter Smith sent Tench an e-mail confirming his suspicions and told him to go back and look at the first paragraph of the judgement. When Tench did, he realized that the first italicized letters spell out \"Smith Code.\" For three weeks in March, Smith had presided over the case brought by authors Michael Baigent and Richard Leigh, who claimed that Brown plagiarized their book, \"The Holy Blood and Holy Grail.\" \"It's clear from the judgement itself, the text of the judgement, that the judge got very much into the substance of the two books in question and it was very much in the spirit of those two books that he's obviously encoded this message,\" Tench said. If Tench's reading of the judge's ruling is correct, it would suggest that the code the jusge worked into his decision is related to Brown's theory that the descendents of Jesus Christ and Mary Magdalene are still alive. Although the judge would not reveal his code and its meaning, he did say he would probably confirm it if someone cracked it, which was \"not a difficult thing to do.\" The first nine letters have been deciphered, but beyond that is anyone's guess. These are the letters: s, m, i, t, h, c, o, d, e, J, a, e, i, e, x, t, o, s, t, p, s, a, c, g, r, e, a, m, q, w, f, k, a, d, p, m, q, z. It can only come as welcome public relations not only for the book, which has now sold more than 40 million copies worldwide, but also for the much-anticipated movie version of the novel, starring Tom Hanks as historian Robert Langdon, which is due to be released next month. [/quote]", "Solution_1": "this could well be a fraud." } { "Tag": [ "geometry" ], "Problem": "In triangle ABC, AC=12, AB=8, and BC=10. Point T is on BC, AT bisects angle BAC, and AM is the median from A to side BC. What is the area of triangle AMT?", "Solution_1": "since the point T bisects the angle BAC, BT:TC = 8:12= 4:6\r\nand M is the median of BC so BM:MC=5:5\r\n\r\nconsequently, point T and M is on the line BC in order\r\nand BT=4, TM=1, MC=5\r\n\u2192area of the triangle AMT is one tenth of triangle ABC\r\nand the area of triangle ABC can be found by the Heron's formula." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Hi guys,I remember an interesting and wery easy exersise :D \r\nHer it.\r\n[color=blue]Prove that $f(x)=\\sqrt[3]{x}$ is elementary function[/color] :rotfl:.\r\nI am don't joking :blush:", "Solution_1": "Do you mean $\\sqrt [3] x = e^{\\frac{1}{3}ln(x)}$ \u00bf", "Solution_2": "[quote=\"ZetaX\"]Do you mean $\\sqrt [3] x = e^{\\frac{1}{3}ln(x)}$ \u00bf[/quote]\r\nThis equaliti is true only if $x>0$,isn't it?", "Solution_3": "It's also true for $x<0$ (or can easily be adapted to include this).\r\nBut by complex continuation thats all neglegtable :P \r\n\r\nBut why even prove it, the definition from MathWorld says:\r\n[quote=\"MathWorld\"]A function built up of a finite combination of constant functions, field operations (addition, multiplication, division, and root extractions--the elementary operations)--and algebraic, exponential, and logarithmic functions and their inverses under repeated compositions (Shanks 1993, p. 145; Chow 1999). Among the simplest elementary functions are the logarithm, exponential function (including the hyperbolic functions), power function, and trigonometric functions.[/quote]\r\nSo the inverse of $x^3$ is allowed.", "Solution_4": "By the way, is $f(x)=\\sqrt{x^2}, f: \\mathbb{R}\\to\\mathbb{R}$ elementary ?:)", "Solution_5": "How I know $\\sqrt[n]{x}={x}^{\\frac{1}{n}}$ in real plane dose holde if and only if$x\\geq 0$.\r\nSo we can say that $\\sqrt{x^2}=(x^2)^{\\frac{1}{2}}$\r\nBut $x^2=x*x$ is elementary =>$\\sqrt{x^2}={x^2}^{\\frac{1}{2}}$ is elementary too.=>$\\sqrt{x^2}=|x|$is elementary=>$\\frac{x+|x|}{2}$ and $\\frac{x-|x|}{2}$\r\nis elementary functions,also we can easy see that$\\frac{x+|x|}{2}\\geq 0$ and $\\frac{x-|x|}{2}\\geq 0$for all $x\\in R$=>$\\sqrt[3]{x}=\\sqrt[3]{\\frac{x+|x|}{2}}+\\sqrt[3]{\\frac{x-|x|}{2}}=(\\frac{x+|x|}{2})^{\\frac{1}{3}}+(\\frac{x-|x|}{2})^{\\frac{1}{3}}$\r\nis elementary function :)", "Solution_6": "You also use $x^{\\frac{1}{3}}$ here...\r\nHow do you define $\\sqrt[n] x$ and $x^\\frac{1}{n}$\u00bf", "Solution_7": "[quote=\"ZetaX\"]You also use $x^{\\frac{1}{3}}$ here...\nHow do you define $\\sqrt[n] x$ and $x^\\frac{1}{n}$\u00bf[/quote]\r\nWhat do you mean? :?" } { "Tag": [], "Problem": "I'm going on holiday to italy in a few weeks (going to rome then some other places then staying in Chianciano in Tuscany for 9 days)... is maths quite a big thing in chianciano... not that im gonna do any maths on holiday :P", "Solution_1": "NO you won't find anything about maths in chianciano. Enjoy Italy :)" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "A quite beautiful and challenging problem:\r\nThree magicians show a trick. They give you a deck of cards labelled from $ 1$ to $ 2n \\plus{} 1$ ($ n > 6$). You pick one of them and give the remaining ones [i]equally[/i] to the first $ 2$ magicians. Each of these $ 2$ magicians independently of each other and independently of the third one picks some $ 2$ of his $ n$ cards, and gives them as an ordered pair to the third one. This magician, after looking at the received $ 4$ cards, correctly guesses the number of the card you have taken. Explain how this trick may be shown.\r\n\r\n[hide] According to the statistics from the book, nobody solved the problem at the contest.\nAlso, you might want to try the following good problem [unsolved at the moment this problem is posted]:\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=162659[/hide]", "Solution_1": "I know two similar problems,both of them can be solved by [b]hall theory[/b] in graphs...\r\nso this one might also have the same idea..." } { "Tag": [ "ratio", "geometry", "perimeter", "geometry unsolved" ], "Problem": "In a triangle $ ABC$ the excircle of $ [BC]$ is tangent to the lines $ BC$ , $ CA$ , $ AB$ at $ A_1$ , $ B_1$ , $ C_1$. The excircle of $ [CA]$ is tangent to the lines $ BC$ , $ CA$ , $ AB$ at $ A_2$ , $ B_2$ , $ C_2$. the excircle of $ [AB]$ is tangent to the lines $ BC$ , $ CA$ , $ AB$ at $ A_3$ , $ B_3$ , $ C_3$. Find the greatest ratio between the sum of perimeters of triangles $ A_1B_1C_1 , A_2B_2C_2 , A_3B_3C_3$ with the circumradii of triangle $ ABC$.\r\n\r\nSorry for my bad english.", "Solution_1": "[quote=\"mustafa-yafes\"]In a triangle $ ABC$ the excircle of $ [BC]$ is tangent to the lines $ BC$ , $ CA$ , $ AB$ at $ A_1$ , $ B_1$ , $ C_1$. The excircle of $ [CA]$ is tangent to the lines $ BC$ , $ CA$ , $ AB$ at $ A_2$ , $ B_2$ , $ C_2$. the excircle of $ [AB]$ is tangent to the lines $ BC$ , $ CA$ , $ AB$ at $ A_3$ , $ B_3$ , $ C_3$. Find the greatest ratio between the sum of perimeters of triangles $ A_1B_1C_1 , A_2B_2C_2 , A_3B_3C_3$ with the circumradii of triangle $ ABC$.\n\nSorry for my bad english.[/quote]\r\nIt's easy\r\n$ p_A\\equal{}2R_A(sin(\\frac{C}{2})\\plus{}sin(\\frac{b}{2})\\plus{}cos(\\frac{A}{2})$\r\nThe expression that we need to find the greatest value is\r\n$ 2(sinA\\plus{}sinB\\plus{}sinC)(sin(\\frac{C}{2})\\plus{}sin(\\frac{b}{2})\\plus{}sin(\\frac{A}{2}))\\plus{}2(sinA\\plus{}sinB\\plus{}sinC)\\{tan(\\frac{A}{2})[sin(\\frac{C}{2})\\plus{}sin(\\frac{B}{2})]\\plus{}tan(\\frac{B}{2})[sin(\\frac{A}{2})\\plus{}sin(\\frac{C}{2})]\\plus{}tan(\\frac{C}{2})[sin(\\frac{A}{2})\\plus{}sin(\\frac{B}{2})]\\}$ $ \\leq 2.\\frac{{3\\sqrt 3 }}{2}.\\frac{3}{2} \\plus{} 2.\\frac{{3\\sqrt 3 }}{2}.\\sqrt 3 \\equal{} 9 \\plus{} \\frac{{9\\sqrt 3 }}{2}$", "Solution_2": "thanks this solution... :)" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Solve the equation for the reals (x,y):\r\n $ {x}^{2}\\plus{}20x\\plus{}351\\equal{}\\frac{2008}{{y}^{2}\\plus{}30y\\plus{}233}$", "Solution_1": "L.H.S=>251 for all x, R.H.S<= 251 for all y. So the only possible solution is L.H.S=R.H.S=251, which is the case for x=-10, y=-15.\r\n(x,y)=(-10,-15) is the solution.\r\nGood and easy question", "Solution_2": "Thanks for your post & comments" } { "Tag": [], "Problem": "For what value of $ b$ is $ \\frac{b}{13} \\equal{} \\frac{16}{52}$?", "Solution_1": "$ \\frac{b}{13} = \\frac{16}{52}$\r\n\\begin{eqnarray*}\r\nb\r\n& = & \\frac{16}{52} \\times 13 \\\\\r\n& = & \\frac{4 \\times 4}{4 \\times 13} \\times 13 \\\\\r\n& = & \\frac{\\color{red}\\not\\color{black}4 \\times 4}{\\color{red}\\not\\color{black}4 \\times \\color{blue}\\not\\color{black}{13}} \\times \\color{blue}\\not\\color{black}{13} \\\\\r\n& = & \\boxed{4} \\\\\r\n\\end{eqnarray*}", "Solution_2": "Or, we can simply notice that $ 13 \\times 4\\equal{}52$. Thus, we divide $ 16$ by $ 4$ to get $ 16 \\div 4\\equal{}\\boxed{4}$." } { "Tag": [ "geometry", "trapezoid", "ratio", "FTW" ], "Problem": "A trapezoid has base angles of $ 45^\\circ$ and bases that measure 8 inches and 18 inches. The trapezoid is folded as shown. What is the ratio of the area of the black section to the area of the white section? Express your answer as a common fraction.", "Solution_1": "We can't solve this, due to the lack of a picture.", "Solution_2": "On FTW! there is a picture. \n[asy]\ndraw((1,0)--(4,0)--(3,1)--(1,1)--cycle);\ndraw((2,0)--(1,1));\nfill((1,0)--(2,0)--(1,1)--cycle);\ndraw((0,0)--(1,0)--(1,1)--cycle,dashed);\n[/asy]\nThis is the picture. Since we have the top base $8$ and bottom base $18$, we can see that the ratio of the shaded area to the white area is \\[\\boxed{\\frac{5}{16}}\\]" } { "Tag": [ "LaTeX" ], "Problem": "\u03b1 \u03b2 \u03b3 \u03b4 \u03b5 \u03b6 \u03b7 \u03b8 \u03b9 \u03ba \u03bb \u03bc \u03bd \u03be \u03bf \u03c0 \u03c1 \u03c3 \u03c4 \u03c5 \u03c6 \u03c7 \u03c8 \u03c9 \r\n\u0391 \u0392 \u0393 \u0394 \u0395 \u0396 \u0397 \u0398 \u0399 \u039a \u2227 \u039c \u039d \u039e \u039f \u220f \u03a1 \u2211 \u03a4 \u03a5 \u03a6 \u03a7 \u03a8 \u03a9 \r\n\u2022 \u00a8 \u2026 \u2236 \u2237 \u2234 \u2235 \r\n\u222b \u222e \u00f7 \u00b1 \u2208 \u223d \u2248 \u224c \u221d \u2260 \u221e \u2211 \u220f \u222a \u2229 \r\n\u3013\uff01\u2248 \u2260 \u2261 \u2264 \u2265 \u2264 \u2265 \u226e \u226f\u300e\u300f\u3016\u3017\u3010\u3011\u3014\u3015\u3014\u3015\uff5b\uff5d \r\n\u2016 \u22a5 \u22bf \u221f \u2220 \u2016 \u2227 \u2228 \u2229 \u222a \u2223 \r\n\u221a \uff0a \uff20 \uff20 \u221e \u2295 \u2312 \u2299 \u2030 \uff06 \uff1f\uff01 \uff05 \u2105 \u2109 \r\n\u2103 \uff04 \uffe0 \uffe1 \u2030 \u00a7 \u2116 \u2606 \u2605 \u203b \uffe5 \uff05 \uff06", "Solution_1": "Lol, we use Latex......", "Solution_2": "\u03b1 \u03b2 \u03b3 \u03b4 \u03b5 \u03b6 \u03b7 \u03b8 \u03b9 \u03ba \u03bb \u03bc \u03bd \u03be \u03bf \u03c0 \u03c1 \u03c3 \u03c4 \u03c5 \u03c6 \u03c7 \u03c8 \u03c9 \r\n\u0391 \u0392 \u0393 \u0394 \u0395 \u0396 \u0397 \u0398 \u0399 \u039a \u2227 \u039c \u039d \u039e \u039f \u220f \u03a1 \u2211 \u03a4 \u03a5 \u03a6 \u03a7 \u03a8 \u03a9 \r\nHow to print them in Latex?", "Solution_3": "$\\alpha,\\beta$\u7b49\u7b49", "Solution_4": "[quote=\"dingdongdog\"]\u03b1 \u03b2 \u03b3 \u03b4 \u03b5 \u03b6 \u03b7 \u03b8 \u03b9 \u03ba \u03bb \u03bc \u03bd \u03be \u03bf \u03c0 \u03c1 \u03c3 \u03c4 \u03c5 \u03c6 \u03c7 \u03c8 \u03c9 \n\u0391 \u0392 \u0393 \u0394 \u0395 \u0396 \u0397 \u0398 \u0399 \u039a \u2227 \u039c \u039d \u039e \u039f \u220f \u03a1 \u2211 \u03a4 \u03a5 \u03a6 \u03a7 \u03a8 \u03a9 \nHow to print them in Latex?[/quote]\r\n\u039e: $\\equiv$\r\n\u220f: $\\prod$\r\n\u2211: $\\sum$\r\n\u03be: $\\zeta$\r\n\u0393: $\\Gamma$\r\n\u03bb: $\\lambda$\r\n$\\cdots \\cdots$", "Solution_5": "[quote=\"shobber\"][quote=\"dingdongdog\"]\u03b1 \u03b2 \u03b3 \u03b4 \u03b5 \u03b6 \u03b7 \u03b8 \u03b9 \u03ba \u03bb \u03bc \u03bd \u03be \u03bf \u03c0 \u03c1 \u03c3 \u03c4 \u03c5 \u03c6 \u03c7 \u03c8 \u03c9 \n\u0391 \u0392 \u0393 \u0394 \u0395 \u0396 \u0397 \u0398 \u0399 \u039a \u2227 \u039c \u039d \u039e \u039f \u220f \u03a1 \u2211 \u03a4 \u03a5 \u03a6 \u03a7 \u03a8 \u03a9 \nHow to print them in Latex?[/quote]\n\u039e: $\\equiv$\n[/quote]\r\nI think you are mistaken...\r\n\u039e is the capital of \u03be..." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "$ a$ and $ n$ are natural numbers such that $ a>1$ and $ n|a^n\\minus{}1$. Prove that $ (a\\minus{}1)n|a^n\\minus{}1$.", "Solution_1": "this follows very easily from the lemma \r\n\r\n$ a\\equiv b\\pmod{p^k}\\Rightarrow a^p\\equiv b^p\\pmod{p^{k\\plus{}1}}$.", "Solution_2": "Explain it please :wink:", "Solution_3": "well, pick a prime and let $ p^k$ be the highest power of $ p$ dividing $ a\\minus{}1$ and $ p^l$ the highest power of $ p$ dividing $ n$. then $ a\\equiv 1\\pmod{p^k}$. now applying the lemma $ l$ times, we get $ a^{p^l}\\equiv 1\\pmod{p^{k\\plus{}l}}$. since $ p^l|n$, it then follows that $ a^n\\equiv 1\\pmod{p^{k\\plus{}l}}$. since $ p^{k\\plus{}l}$ is the highest power of $ p$ dividing $ n(a\\minus{}1)$, the desired result follows.", "Solution_4": "Oh.. :blush: Thanks :)" } { "Tag": [ "algebra", "polynomial", "quadratics", "function" ], "Problem": "Let $f(x)=ax^{2}+bx+c$ be a non constant polynomial, and assume that $a$, $b$, and $c$ are integers.\r\n\r\nProve that there are infinitely many integers $n$ such that $f(n)$ is composite.\r\n\r\n\r\n(I'm shameless -- this is a homework problem I'm stuck on >_<. Any and all help is appreciated!)", "Solution_1": "[hide=\"hint 1\"] Quadratic Equation [/hide]\n\n[hide=\"hint 2\"] Factoring [/hide]", "Solution_2": "[quote=\"G-UNIT\"][hide=\"hint 1\"] Quadratic Equation [/hide]\n\n[hide=\"hint 2\"] Factoring [/hide][/quote]\r\nI don't understand how your hints help: consider $x^2+x+1$. This cannot be factored and has integer coefficients and it is composite for infinitely many values of $n\\in\\mathbb{Z}$", "Solution_3": "I may have found a way to prove it, but I don't really know much about setting up proofs, so it'll be a bit sloppy.\r\n\r\nLet $n=kc$, where $c$ is the constant term in the polynomial, and k is any integer $\\ge 1$.\r\n\r\nSo, $f(kc) = ak^{2}c^{2} + bkc + c\\\\f(kc) = c(ak^{2}c+bk+1)$\r\n\r\nIt follows that $c|c(ak^{2}c+bk+1)$, so $f(kc)$ must be composite.\r\n\r\nBecause $k$ can be any positive integer, $f(kc)$ or $f(n)$ has an infinite amount of $n$ values that make it composite.\r\n\r\n\r\n\r\nIt's really sloppy, but I think it works. Did I make any wrong/illegal moves? Is this sufficient form for a proof?", "Solution_4": "[quote] Is this sufficient form for a proof? [/quote]\r\nGood work but you may have some more work to do.. what if c=1? Hope that helps.\r\n[hide](Goes without saying that if c divdes a number , the number need not be composit if c=1)[/hide]", "Solution_5": "well since it goes through all numbers, consider the numbers that are a multiple of c, so $x=nc$, that gives us\r\na(nc)^2 +b(nc)+c\r\nc(acn+bn+1), and clearly that has 2 factors unless the second part is 1 or 0, but that cannot happen for every value of n, therefore we have infinitely many numbers that are composite because there are infinite values of n that can be input, and there may be 2 y values for a given composite, but there cannot be more repeates than that due to the nature of the function", "Solution_6": "But you still have the problem for c=1 and $an^2+bn+1$ a prime.", "Solution_7": "[quote]c(acn+bn+1), and clearly that has 2 factors unless the second part is 1 or 0, but that cannot happen for every value of n,[/quote]\r\nAgain question is why? To start with let $c=1$ , now you have only one factor :) You have to show/prove \" but that cannot happen\" (remember c is given and it could be 1 for all values of n) ..... Hope this points in the right direction.. Yes, still there is some work needed before the proof can be counted as a good proof." } { "Tag": [ "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "I need some help with this one.\r\n\r\nLet V and W be isomorphic vector spaces. Prove that if V' is a subspace of V, then V' is isomorphic to a subspace W' of W.", "Solution_1": "since V and W are isomorphic we can find an isomorphism T from V into V. now choose basis vectors for V'. look at the images of these basis vectors. the space spanned by these vectors is the W' u r lookin for", "Solution_2": "What if V hasn't a finite dimension ?\r\nDoes this result remain true ?", "Solution_3": "Yes. The restriction of an isomorphism is an isomorphism, so $T(V')$ is isomorphic to $V'$. $T(V')$ is a subspace of $W'$, and we are done." } { "Tag": [ "quadratics", "algebra unsolved", "algebra" ], "Problem": "Solve system in positive reals:\r\n\r\n$2\\sqrt{a}+3b=4$;\r\n$4\\sqrt{2(2-b)}=5+3a$.", "Solution_1": "From the first equation, $a=\\frac{9b^{2}-24b+16}{4}$. Inserting into the second gives $2(2-b)=\\frac{1}{256}\\left(4624-9792b+8856b^{2}-3888b^{3}+729 b^{4}\\right)$, and tidying this up we get\r\n$\\frac{1}{256}(9b-10)^{2}\\left(9b^{2}-28b+36\\right)=0$. The discriminant of the quadratic term is negative, so the only solution in positive reals is $b=\\frac{10}{9}$, which gives $a=\\frac{1}{9}$.", "Solution_2": "[quote=\"Diarmuid\"]..., and tidying this up we get ...[/quote]\r\nWhich trick did you use to factor it?", "Solution_3": "I think he found it searching for rational roots :wink:" } { "Tag": [ "inequalities" ], "Problem": "prove that:\r\n$ \\frac{1\\minus{}sin(\\frac{\\pi}{14})}{2*sin(\\frac{\\pi}{14})}>\\sqrt{3*cos(\\frac{\\pi}{7})}$", "Solution_1": "after 2 hours.I can solve this ineq. :D \r\nuse this lemma;\r\nsin(pi/14) is the root of this equation:\r\n$ 8x^3\\minus{}4x^2\\minus{}4x\\plus{}1\\equal{}0$" } { "Tag": [ "induction", "number theory", "number theory proposed" ], "Problem": "$ (y_n) , n \\geq 2 $ is a sequence defined : \r\n $ y_2=y_3 =1 $ ; $ (n+1)(n-2)y_{n+1} = n(n^2-n-1)y_n - (n-1)^3y_{n-1} $ . \r\n Find all $ n $ such that : $ y_n \\in Z $ ;)", "Solution_1": "This recurrence reminds me Aperi's proof of the irrationality of $\\zeta(3)$ :)", "Solution_2": "Oh well.\r\n\r\nIt's easy to prove by induction that $ny_n = (n - 1)! + 1$.\r\n\r\nBy Wilson's theorem, this implies that $n$ must be prime.", "Solution_3": "[quote=\"Arne\"]Oh well.\n\nIt's easy to prove by induction that $ny_n = (n - 1)! + 1$.\n\nBy Wilson's theorem, this implies that $n$ must be prime.[/quote]\r\n Great ! ;) My solution is the same with yours :)" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "This picture is a part of Hungerford Book . In above lemma needs the $ |S|\\equal{}p^n,p \\ is \\ prime$, but in the proof of Cauchy , the order of $ |S|\\equal{}n^{p\\minus{}1}$, and $ n$ needs not be a prime number . Could u clarify this point for me please?\r\n\r\nThank you in advanced :)", "Solution_1": "We have $ |S| \\equal{} n^{p \\minus{} 1}$ as we have no matter how do we choose the first $ p \\minus{} 1$ elements. The last is fixed by the uniqueness of the inverse. I don't see how Lemma 5.1 is used in Hungerford's book. I have learned the proof which uses orbit-stabilizer theorem.", "Solution_2": "Okay \r\n\r\nIn Hungerford's book , The lemma in the picture depends on finite $ p\\minus{}group$ , for some prime $ p$ , in the proof of Cauchy , Hungerford depended on that lemma , and he considered $ S\\equal{}\\Omega$ as u wrote it . Now, $ |\\Omega|$ which is $ |S|$ in hungerford is $ n^{p\\minus{}1}$ elements ( No problem here for me ), but he used the lemma during the proof on the set $ |\\Omega|$.\r\n\r\nHow he used that lemma since $ |\\Omega|$ is not necessary to be of order prime to some power , i.e. $ n$ is not necessary to be prime ?\r\n\r\n :)", "Solution_3": "Does he uses the theorem at all? I don't see where it is required.", "Solution_4": "Oh, sorry :blush: , I forgot to put the rest of a proof :D \r\n\r\nHere is the full proof :", "Solution_5": "I can't see where he defines the set $ S_0$. However, $ |S|$ is divisible by $ p$ as the order of $ G$ is divisible by $ p$. The sentence\r\n[quote]$ (a_1,\\ldots,a_p)\\in S_0$ if and only if $ a_1\\equal{}a_2\\equal{}\\ldots \\equal{}a_p$[/quote] implies that the number of elements of $ S_0$ is a multiple of $ p$, again as the order of $ G$ is divisible by $ p$.", "Solution_6": "Guys , I understood what are u said :whistling: \r\n\r\nTo see the definition of $ S_0$ , take a look at first picture .\r\n\r\n[b]MY QUESTION :[/b] How he used the lemma , when $ |S|$ [b]is not necessary[/b] to be some power of prime , i.e. $ n$ is not necessary to be a prime , so we couldn't say $ |S|$ is $ p$-subgroup .\r\n\r\nThe first necessary condition to use lemma is the group or subgroup is a $ p \\minus{} group$.\r\n\r\n :)", "Solution_7": "$ S$ is not even a group, much less a p-group! The $ H$ of the lemma is $ Z_p$, $ S$ is merely the set on which it acts.", "Solution_8": "Oh, I should pay attention more in next time :cleaning: \r\n\r\nThank you LydianRain :)" } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Given a connected, weighted graph G with n nodes and m edges, and the cost value for each edge is also given. Then we can construct the Minimum Spanning Tree T for this graph G. \r\n\r\nNow the question is:\r\n\r\n1) If a new node added into graph G, with some new edges connected with some nodes in the old graph. Then how to simply modify the constructed MST T to get a MST for the new graph.\r\n\r\n2) If one node of graph G is deleted, the remaining n-1 nodes still connected. Then how to simply construct the MST for the new graph.\r\n\r\n\r\nThe similar problem for the directed graph.\r\n\r\n\r\nPlease give me any hints, ideas or reference papers if you have.\r\nThanks.", "Solution_1": "1) I would add the edges to the new vertex one by one in order of weight, and always remove the edge of largest weight in the unique emerging cycle.\r\n\r\n2) The original ST is still good, but maybe not connected, so continue adding the cheapest edges until it does.", "Solution_2": "Exactly, the solution you give is practical and will produce a good ST.\r\nBut the solution is not necessarily optimal, right? i.e. it is not Minimum Spanning Tree any more.\r\n\r\nSo I wonder what is the simple way to generate the Minimum Spanning Tree for the new graph by modifying the old MST to adapt the topology change due to vertex inserting or vertex removing. \r\n\r\nThe method should be much simple (from the point of computation complexity) than generating the MST for the new graph without using the old MST information.", "Solution_3": "My spanning trees are optimal, at least if the original ones have been derived with the greedy algorithm.", "Solution_4": "Yes, I aggre with you that your method is optimal. Thanks for your answering my questions.\r\n\r\nWould you please consider the following question further:\r\n\r\nWhat I exactly want to do is move one of the vertex in the graph, thus weights of the edges corresponding to this vertex will be changed, especially we consider the Euclidean graph (the weight of each edge is the squared distance between the two end vertex). \r\n\r\nOne of the way to solve the problem is to remove the vertex which will be moved, and construct the MST, and then add a vertex at the new position where the vertex will move to, and construct the MST. This is why I asked the two questions before.\r\n\r\nSo now my question is whether there is a simple method to directly modify the original MST to take into account position and weights changeing of one vertex.\r\n\r\nFurthermore, if we can move the given vertex to anywhere, where is the optimal position to minimize the total weights of the MST tree?" } { "Tag": [ "trigonometry", "Taylor series", "calculus", "Inequality", "Trigonometric inequality", "IMO Shortlist", "IMO Longlist" ], "Problem": "Prove the trigonometric inequality $\\cos x < 1 - \\frac{x^2}{2} + \\frac{x^4}{16},$ when $x \\in \\left(0, \\frac{\\pi}{2} \\right).$", "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": "Little Lemma:\r\n$\\frac{x^{2n}}{(2n)!}>\\frac{x^{2n+2}}{(2n+2)!}$ where $x \\in (0,\\frac{\\pi}{2})$ and $n \\in N$\r\n\r\nBy multiplying we get:\r\n\r\n$(2n+1)(2n+2)x^{2n}>x^{2n+2}$\r\n$(2n+1)(2n+2)>x^2$\r\nwhich is obvious because $x \\in (0,\\frac{\\pi}{2})$ and $n \\in N$\r\n\r\nComing back to our problem:\r\n\r\n$cos(x) = 1-\\frac{x^2}{2}+\\frac{x^4}{24}-\\frac{x^6}{6!}+...<1-\\frac{x^2}{2}+\\frac{x^4}{16}$\r\n$-\\frac{x^4}{48}-\\frac{x^6}{6!}+...<0$\r\nwhich is true using our lemma", "Solution_3": "Since $\\sin x\\text{trace}A\\ge\\minus{}n.$\r\n\r\nThe specific thing I want to comment on is that if $ A\\ne0,$ then $ A$ cannot be nilpotent. We get this from the iterating the equation above: $ A^3\\equal{}A,A^4\\equal{}\\minus{}A,$ and in general, $ A^k\\equal{}(\\minus{}1)^kA,$ which is assumed not to be zero." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "If the equation $ ax^2+(c-b)x+(e-d)=0$ has real roots greater than $1$, prove that the equation $ax^4+bx^3+cx^2+dx+e=0$ has at least one real root.", "Solution_1": "Let $ r$ be the root of the first equation and $ f(x)$ be the second polynomial.\r\nIt is easy to verify $ f(\\sqrt r)\\equal{}(br\\plus{}d)(\\sqrt r\\plus{}1)$ and $ f(\\minus{}\\sqrt r)\\equal{}(br\\plus{}d)(\\minus{}\\sqrt r\\plus{}1)$.\r\nHence, $ f(\\sqrt r)f(\\minus{}\\sqrt r)\\le0$.\r\nThen there is a root on $ [\\minus{}\\sqrt r,\\sqrt r]$ to $ f(x)\\equal{}0$." } { "Tag": [ "geometry", "trigonometry", "geometry unsolved" ], "Problem": "Given a circumference D, a circular arc G joins two distincts points on D and is inside the circle C contained in D. Prove that, if the two regions in which G divides C have equal area, then G's lenght is greater then D's diameter.", "Solution_1": "No one? :(", "Solution_2": "Disclaimer: the usual user of this account is not the author of the following (which basically means it will be not very well done).\r\n\r\nWe fix two points on the circle, seperated by angle $ \\theta$. We must determine the curve for which area is split in half. WLOG, raduis is 1 hence we want a bounded area of $ \\frac{\\pi}{2}$. We divide the cut off (inner) region into two sectorish parts of a circle - each, when viewed with correct circle, is a sector minus a triangle. Plugging through the algebra (and considering a new circle with arc on the circle acting as critical boundary) we find the constraint to be:\r\n\\[ \\frac{\\pi}{2}=\\left(\\frac{\\sin \\frac{\\theta}{2}}{\\sin \\frac{\\phi}{2}}\\right)^{2}\\cdot\\left(\\frac{\\phi-\\sin \\phi}{2}\\right)+\\frac{\\theta-\\sin \\theta}{2}\\]\r\nIt remains to prove, under this constraint, that $ \\phi\\sin \\frac{\\theta}{2}\\geq 2\\sin \\frac{\\phi}{2}$\r\nThe last two statements are equivalent to the problem." } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "let $G$ be a finite semigroup such that:\r\nif $xy=yz$ then $x=z$\r\nprove that $G$ is a abelian group.", "Solution_1": "We have to assume $G$ to be non-empty. Pick an element $a$. There are $i>j>0$ so that $a^{i}= a^{j}$, since $G$ is finite. Consider $e : = a^{i-j}$. We have $ae a^{j}= a a^{i}= a a^{j}= a^{j}a$, hence $ae = a$ by assumption. Now let $x \\in G$ be arbitrary. Then $ae = a \\Rightarrow axa = aexa \\Rightarrow xa = aex \\Rightarrow x = ex$ by applying twice the assumption. Similary, we optain $ea = a$ and then $x = xe$. Thus, $e$ is an identity element.\r\n\r\nTo find an inverse element for $x$, choose again $i>j>0$ so that $x^{i}= x^{j}$. Then $x^{i-j}x^{j}= x^{i}= x^{j}= x^{j}e$, hence $x^{i-j}= e$ by assumption. Since $i-j-1 \\geq 0$, we can consider $y : = x^{i-j-1}$. It follows $x y x^{j+1}= x x^{i}= x x^{j}= x^{j+1}= x^{j+1}e$, thus $x y = e$ by assumption.\r\n\r\nWe conclude, that $G$ is a group. Finally, for all $x,y \\in G$, we have $xy = y(y^{-1}xy) \\Rightarrow x = y^{-1}xy \\Rightarrow yx = xy$, so that $G$ is abelian." } { "Tag": [ "function" ], "Problem": "How would the model: v(t) = 100(1.04)^x change if your money will never be worth 100 or less.?", "Solution_1": "What is the function $ v(t)$? What does it describe?", "Solution_2": "Assuming $ t$ is time and $ x$ is money, $ x>100\\implies 1.04^x>1.04^100\\approx 50.5\\implies v(t)>5050.5$.\r\nAnd $ v(t)$ diverges as $ x\\to\\infty$ (albeit slowly) so I guess your answer is it starts at about 5050 and then increases.", "Solution_3": "When you were born, your Grandmother gives you 100 dollars to put in the bank. Fortunately, at that time, Nick Bank is offering a 4% annual interest rate. \r\n1.\tIdentify the multiplier. _____________ Why?\r\n2.\tWrite an Exponential Model, V(t), for the given information.\r\nModel: ________________________ Explain:\r\n3.\tIdentify the first 6 terms of the function to the nearest cent if t=0 in 1992.\r\nt\t\tV(t)\r\n t=0\r\n t=1\r\n\r\n\r\n4.\tProve these values belong to the function V(t).\r\n\r\n\r\n5.\tAlgebraically determine the value of your $ 100 today.\n6.\tDetermine with the aid of your calculator when your money will be worth hundred times its initial value. \n7.\tHow would the model change if your Grandmother told you that your money will never be worth$100 or less? \r\n\r\n......................Does number seven make sense?", "Solution_4": "These are clearly homework problems.\r\nStop trying to cheat using AoPS and try asking about the concepts rather than just getting answers to problems.", "Solution_5": "I do not want the answers to the homework problems, i am not trying to cheat, i am giving you the other questions in order to help me on #7. Those other questions serve as a reference in order to answer number #7. All i want is information on #7, the others were just clarify the steps in which i had to solve to get to the last one.", "Solution_6": "Which one is #7?", "Solution_7": "Number seven is the last question. Please note the homework questions i put are there just there to show you the work i had to do in order to get to #7. Essentially, showing where the function was dervived.", "Solution_8": "[quote=\"ydog\"]How would the model: v(t) = 100(1.04)^x change if your money will never be worth 100 or less.?[/quote]\r\n\r\nIt's unclear what your variables, or your constants for that matter, represent in this equation. For instance, $ v$ is a function of $ t$, but there is no $ t$ on the right hand side of the equals symbol.\r\n\r\nDespite its defects, the equation bears a vague resemblance to the formula for the future value of a single deposit, held at 4% interest compounded annually. If that is so then the equation should be $ v(x) \\equal{} 100(1.04)^x$, where $ x$ is the number of years the deposit is held.\r\n\r\nAlso this means the answer to your question is that the model would not have to change in any way for your money to never be worth less than $ 100.$ It will be worth exactly $ 100$ at $ x \\equal{} 0$, i.e. when you first put it in the bank. Further, as far as this model is concerned $ x$ can only take on whole number values." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $p_i$ denote the $i$-th prime. Show that $p_1p_2...p_n + 1$ is never a square!", "Solution_1": "Hmm... it's just $\\mod{4}$?", "Solution_2": "[quote=\"TomciO\"]Hmm... it's just $\\mod{4}$?[/quote]\r\nYes, and that shows that it's even not the sum of two squares ;)" } { "Tag": [ "quadratics", "algebra", "polynomial", "sum of roots", "product of roots" ], "Problem": "[color=red][size=200]In the name of GOD[/size][/color]\r\n\r\n[color=blue][size=150]Dear Brothers and sisters :\n\nIf A is the root of the equation X^2 + 5X + 7 = 0. Find an equation whose new root is A+1 .\n\n\n(please attach the explanation )[/size][/color]", "Solution_1": "Have $ P(x) \\equal{} x^2 \\plus{} 5x \\plus{} 7 \\equal{} (x \\minus{} A)(x \\minus{} B) \\equal{} ((x\\plus{}1) \\minus{} (A\\plus{}1))((x\\plus{}1) \\minus{} (B\\plus{}1))$. Then $ A\\plus{}1$ is a root of $ Q(x) \\equal{} P(x\\plus{}1) \\equal{} (x\\plus{}1)^2 \\plus{} 5(x\\plus{}1) \\plus{} 7 \\equal{} x^2 \\plus{} 7x \\plus{} 13 \\equal{} 0$.", "Solution_2": "Stop typing in large colored text...", "Solution_3": "yes\r\nthank you mavropnevma\r\nthis is true answer\r\n\r\nYou are a respectable", "Solution_4": "But...the equation X^2+5X+7=0 doesn't have any roots....", "Solution_5": "[quote=\"Layla Fu\"]But...the equation X^2+5X+7=0 doesn't have any roots....[/quote]\r\n\r\n$ x^2 \\plus{} 5x \\plus{} 7 \\equal{} 0$\r\n$ x \\equal{} \\frac {\\minus{}5\\plus{}\\sqrt {3}i}{2}$ or$ \\frac{\\minus{}5\\minus{}\\sqrt{3}i}{2}$", "Solution_6": "[quote=\"mavropnevma\"]Have $ P(x) \\equal{} x^2 \\plus{} 5x \\plus{} 7 \\equal{} (x \\minus{} A)(x \\minus{} B) \\equal{} ((x \\plus{} 1) \\minus{} (A \\plus{} 1))((x \\plus{} 1) \\minus{} (B \\plus{} 1))$. Then $ A \\plus{} 1$ is a root of $ Q(x) \\equal{} P(x \\plus{} 1) \\equal{} (x \\plus{} 1)^2 \\plus{} 5(x \\plus{} 1) \\plus{} 7 \\equal{} x^2 \\plus{} 7x \\plus{} 13 \\equal{} 0$.[/quote]\r\n\r\n$ P(A) \\equal{} 0$\r\n$ A \\plus{} 1$ is a root of $ Q(x) \\equal{} P(x \\minus{} 1)$ instead of $ P(x \\plus{} 1)$\r\n$ P(x \\minus{} 1) \\equal{} (x \\minus{} 1)^2 \\plus{} 5(x \\minus{} 1) \\plus{} 7 \\equal{} x^2 \\plus{} 3x \\plus{} 3$", "Solution_7": "Thought X belonged to R...\r\nThx though :oops:", "Solution_8": "let's think about this situation with an equation in the general form: ax^2 +bx+c=0 ???\r\nMaybe or maybe not?\r\nplease instruct me :D", "Solution_9": "If the general quadratic equation $ ax^{2} \\plus{} bx\\plus{}c\\equal{}0$ has roots $ x\\equal{}\\alpha$ and $ x\\equal{}\\beta$, then the equation with roots $ y\\equal{}\\alpha \\plus{} 1$ and $ y\\equal{}\\beta \\plus{}1$ is obtained by writing $ y\\equal{}x\\plus{}1$, that is, $ x\\equal{}y\\minus{}1$, giving $ a(y\\minus{}1)^{2} \\plus{} b(y\\minus{}1) \\plus{} c \\equal{} 0$, that is $ ay^{2} \\plus{}(b\\minus{}2a)y \\plus{} a \\minus{} b\\plus{}c\\plus{}0$ or simply $ ax^{2} \\plus{} (b\\minus{}2a)x \\plus{} a \\minus{}b \\plus{}c\\equal{}0$.\r\n\r\nIn terms of the graph of $ ax^{2} \\plus{} bx \\plus{} c$, we want to move it $ 1$ unit to the right; this is equivalent ot replacing $ x$ by $ x\\minus{}1$.\r\n\r\nAlternatively, if the roots of the original equation are $ \\alpha$ and $ \\beta$, the the roots of the new equation are $ \\alpha\\plus{}1$ and $ \\beta \\plus{}1$. The sum of the new roots is $ \\alpha \\plus{} \\beta \\plus{}2$ and the product of tghe new roots is $ (\\alpha\\plus{}1)(\\beta\\plus{}1)$ and we can substitute these into $ x^{2} \\minus{} Sx \\plus{} P \\equal{} 0$, where $ S$ is the sum of the roots and $ P$ is the product of the roots.", "Solution_10": "x^2 + 5x + 7 = 0\r\n\r\nLet the other root be B\r\n\r\nSum of roots= A+B= -5\r\nProduct of roots= AB= 7\r\n\r\nLet the new equation have roots A+1 and B+1\r\n\r\nNew sum of roots= A+1+B+1= -3\r\nNew product of roots= (A+1)(B+1)= AB+A+B+1=3\r\n\r\nThen the new equation is x^2+3x+3=o\r\n\r\nIs my answer acceptable?" } { "Tag": [ "geometry", "perimeter", "pigeonhole principle", "geometry unsolved" ], "Problem": "I am having trouble with this problem.\r\n\r\nn points are chosen from the perimeter or the interior of a regular hexagoin with sides of unit length, so that the distance between any two of them is not less than $\\sqrt{2}$. What is the largest natural number n for which this is possible?\r\n\r\nCould you please help?\r\n\r\nThanks.\r\n\r\nPS, no need for pigeonhole right?", "Solution_1": "what do you mean by not less?\r\n\r\nit's greater than it? i suposse so, if it the opposite we will have a strange problem\r\n\r\nwhat's the perimeter of this polygon?\r\n\r\nsplit the poligon to $\\sqrt{2}$ and take the integer part. you will get n", "Solution_2": "[quote=\"yagaron\"]what do you mean by not less?\n\nit's greater than it? i suposse so, if it the opposite we will have a strange problem\n\nwhat's the perimeter of this polygon?\n\nsplit the poligon to $\\sqrt{2}$ and take the integer part. you will get n[/quote]\r\n\r\nThat means that the distance between any two points has to be greater than or equal to $\\sqrt{2}$.\r\n\r\nYeah, so you were right.\r\n\r\nThe perimeter of this polygon is just the circumference of the polygon, as in the points lie on the sides of the polygon.\r\n\r\nThat should be good enough." } { "Tag": [], "Problem": "a) Howard and Samantha are sitting in a boat in a small lake. There is a large rock in the boat. Howard throws the rock overboard and it sinks to the bottom of the lake. Does the water level in the lake rise, fall or remain the same? \r\nb) Samantha throws Howard overboard, but he\u2019s wearing a lifejacket and thus he floats. Does the water level in the lake rise, fall or remain the same?\r\n\r\na) seems straightforward but theres probably a trick to this... is therE?", "Solution_1": "It is worth mentioning that question a) was once proposed to proeminent physicists, including Robert Oppenheimer, George Gamow, and Felix Bloch, and all them gave wrong answers. However, the answer can be easily reached by comparing the volume of water displaced by the rock when it is 1) in the boat, and 2) in the water.\r\n\r\nFor question b), a similar approach can be used." } { "Tag": [], "Problem": "Can any one help me about this topic? My project is about this topic. I got stuck about quantization. What is it? How it work?. And about arithmetic coding, can any one give some useful link about this? \r\nTHANK YOU :)", "Solution_1": "Don't look at me! :whistling:" } { "Tag": [ "geometry", "parallelogram", "geometry proposed" ], "Problem": "Let $ ABC$ be a triangle.$ M$ is a point in this triangle.Let $ A_o;B_o;C_o$ be midpoints of $ BC;CA: AB$,respectively.$ AA_o;BB_o;CC_o$ concurrency at $ M$. Take the points $ A';B';C'$ such that $ A_o;B_o;C_o$ be midpoints of $ MA';MB';MC'$,res.Prove that $ AA';BB';CC'$ concurrency at $ M'$ and $ GM'\\equal{}\\frac{GM}{2}$", "Solution_1": "[quote=\"chien than\"]Let $ ABC$ be a triangle.$ M$ is a point in this triangle.Let $ A_o;B_o;C_o$ be midpoints of $ BC;CA: AB$,respectively.[u]$ AA_o;BB_o;CC_o$ concurrency at $ M$[/u]. Take the points $ A';B';C'$ such that $ A_o;B_o;C_o$ be midpoints of $ MA';MB';MC'$,res.Prove that $ AA';BB';CC'$ concurrency at $ M'$ and $ GM' \\equal{} \\frac {GM}{2}$[/quote]\r\n\r\nCould you explain if $ M$ is a random point or is it the centroid of the triangle?(then what do you mean by the phrase I underlined?). Also is G the centroid or another point?\r\nPlease make more clear these points...", "Solution_2": "Let $ ABC$ be a triangle.$ M$ is a point in this triangle.Let $ A_o;B_o;C_o$ be midpoints of $ BC;CA: AB$,respectively.[u]$ AA_o;BB_o;CC_o$ concurrency at $ G$[/u]. Take the points $ A';B';C'$ such that $ A_o;B_o;C_o$ be midpoints of $ MA';MB';MC'$,res.Prove that $ AA';BB';CC'$ concurrency at $ M'$ and $ GM' \\equal{} \\frac {GM}{2}$", "Solution_3": "It 's not so difficult.\r\n\r\nA', B', C' are images of M after the symetry transformation about $ A_0, B_0, C_0$\r\n\r\nWe have:\r\n\r\n$ AC' \\equal{} BM$ ( $ AMBC'$ is a parallelogram)\r\n$ BM \\equal{} CA'$ ( $ BMCA'$ is a parallelogram)\r\n\r\nThen $ AC' \\equal{} CA'$ or $ AC'A'C$ is a parallelogram. Therefore, $ CC'$ passes through the midpoint of $ AA'$ (1)\r\n\r\nSimilarly, $ BB'$ passes through the midpoint of $ AA'$. (2)\r\n\r\n(1) and (2) means the three segments $ AA', BB', CC'$ intersects each other at the midpoint or they are concurent at the common midpoint.\r\n\r\n$ M'$ is the midpoint of AA' then $ MM'$ and $ AA_0$ are 2 medians of $ \\Delta MAA'$. So they intersects at the centroid $ G$ of $ \\Delta MAA'$ which means $ GM' \\equal{} \\frac{GM}{2}$", "Solution_4": "[quote=\"chien than\"]Let $ ABC$ be a triangle.$ M$ is a point in this triangle.Let $ A_o;B_o;C_o$ be midpoints of $ BC;CA: AB$,respectively.[u]$ AA_o;BB_o;CC_o$ concurrency at $ G$[/u]. Take the points $ A';B';C'$ such that $ A_o;B_o;C_o$ be midpoints of $ MA';MB';MC'$,res.Prove that $ AA';BB';CC'$ concurrency at $ M'$ and $ GM' \\equal{} \\frac {GM}{2}$[/quote]\r\n1.We have $ \\vec{MA} \\plus{} \\vec{MA'} \\equal{} \\vec{MB} \\plus{} \\vec{MB'} \\equal{} \\vec{MC} \\plus{} \\vec{MC'}( \\equal{} \\vec{MA} \\plus{} \\vec{MB} \\plus{} \\vec{MC})$\r\nSo $ AA';BB';CC'$ concurrency at $ M'$ in there $ M'$ be midpoints of $ AA';BB';CC'$\r\n2.We have $ 3\\vec{MG} \\equal{} \\vec{MA} \\plus{} \\vec{MB} \\plus{} \\vec{MC} \\equal{} 2\\vec{MM'}$\r\nSo $ GM' \\equal{} \\frac {GM}{2}$\r\n :blush: :D" } { "Tag": [ "search", "Harvard", "college", "percent", "geometry", "quadratics", "algebra" ], "Problem": "I think that homeworks are often useless(not just to top but also to other students).\r\n\r\nThey don't teach you anything new and they only 'review' of school stuff.\r\n\r\nSo do you think that homeworks are good?Do you think that some things about homeworks are good and some are not?", "Solution_1": "I voted for the third option, because in my experience, I've had homework that was extremely helpful in continuing my understanding of a particular subject, or to introduce a new concept for another day. \r\n\r\nOf course, busiwork is never a good thing, because the sole purpose is to keep someone busy, and therefore it's unlikely you'll learn something.", "Solution_2": "I agree if homework means re-thinking and doing similar problems to make sure you really understand something ... but sometimes you need to have that sort of review and sometimes not. And in the most textbooks tasks are boring . Often the tasks are bacisly the same exept they put in different numbers . I think that it's easy to make someone hate maths if you make him doing homework.", "Solution_3": "I also voted for the third option, but I have never heard of this \"busiwork\" you speak of...Heres the rest of my post from the other topic. :wink: \r\n\r\n[quote]Homework can help boost your status of the current unit, as example, fractions. At a very young age fractions can be difficult to master and certainly make you confused. My math teacher thinks it as a review, to make sure a few hours later that you still obtain the knowledge. I think that they shouldn't have as much homework unless it's reading a chapter, taking notes, finishing up classwork because it would only waste time. You could cover the worksheet or the book work in class and ask your teachers or classmates about any questions or concerns you might have. Also, my math/social studies teacher believes that homework would only clutter on the stress of after school activities, going to the park or playing a sport with friends. Maybe that's why there is obesity in the U.S, too much homework and not enough time to run around.\nI don't care much about homework unless it's a whole lot to be done in one day. I'd be happy to do it as long as I'm sure I still got the information in my head.[/quote]", "Solution_4": "WELL.. homework is good if it's useful homework..\r\nUSEFUL is like: helps you understand a concept.\r\n\r\nHowever, whenever homework is gone over during class, it usually takes up half the class to explain the homework and basically go over what was supposed to be understood the day before.\r\nThat's one of the reasons I don't want \"stupid\" people in my class.\r\ni don't mind if homework is gone over in 5 minutes, and the rest is used for learning.\r\n\r\nand homework can be extremely stupid such as word scrambles and word searches.\r\nHow are they supposed to help?\r\n\r\nWith word searches I understand if you give the definition and you search for the word, but for word scrambles, you are given a bunch of letters and you know the list of words it comes from.\r\n\r\nand for reading passages, most people, like me, look only for the answers. never for the story itself.", "Solution_5": "All the work I do at school can be done at home in 10 minutes with a textbook.\r\n:wink:", "Solution_6": "Meh. What everyone else said.\r\n\r\n...Sometimes, it makes me wish I could just be homeschooled. It would go a lot faster, even if I did every problem.", "Solution_7": "I believe that homework is a good thing. It has been assigned to students for quite a while, and it seems to be one of those things that children with their ARGH RAGE AGAINST THE MACHINE IF IT IS CONVENIENT TO DO SO mindset enjoy viewing as worthless. It rarely takes as long as one claims it does, and it has absolutely no negative effects. (Procrastinate less if it cuts into your sleep?) It additionally has the benefit of getting kids who care somewhat about grades, but are not motivated enough to self-study - like myself - to review materials and learn something outside of the limited time that is given in class.\r\n\r\nAlso, the \"That's one of the reasons I don't want 'stupid' people in my class\" argument kinda bothers me. You are the unusual one for learning at an accelerated pace with respect to your classmates, yet you are placing blame upon the average students while expecting special treatment? Other kids do not learn at the same pace as the rare talented and hardworking child. Sometimes the learning basic action is difficult enough without simultaneously being force-fed the abstract concepts behind this action - just so the \"stupid\" people can learn at the same pace as the student multiple standard deviations above the norm.\r\n\r\n\r\nJust my two cents. Sorry kimmystar94 if it sounds a little ad hominem; don't take it that way.", "Solution_8": "[quote=\"Lazarus\"]Also, the \"That's one of the reasons I don't want 'stupid' people in my class\" argument kinda bothers me. You are the unusual one for learning at an accelerated pace with respect to your classmates, yet you are placing blame upon the average students while expecting special treatment? Other kids do not learn at the same pace as the rare talented and hardworking child. Sometimes the learning basic action is difficult enough without simultaneously being force-fed the abstract concepts behind this action - just so the \"stupid\" people can learn at the same pace as the student multiple standard deviations above the norm.[/quote]\r\n\r\nI think you're missing the point. There's a reason \"stupid\" is in quotes. It's not because the people are actually stupid. Some people who have little academic motivation think that it's the teacher's job to lay a mound of knowledge in front of them on a silver platter, with fine champagne and a side of caviar. They are so apathetic and unmotivated that they do not want to invest the thought to determine the value or the point of the homework, and they make the teacher tell them.\r\n\r\nSecondly, I don't think an education should focus on the average student at the expense of the advanced students. Perhaps I'm biased, but the notion that advanced students are there to help the teachers teach the average students is somewhat bogus. This is because, often, the average students are in that position because they really don't care to do better. They have other priorities.", "Solution_9": "[quote]I think you're missing the point. There's a reason \"stupid\" is in quotes. It's not because the people are actually stupid. Some people who have little academic motivation think that it's the teacher's job to lay a mound of knowledge in front of them on a silver platter, with fine champagne and a side of caviar. They are so apathetic and unmotivated that they do not want to invest the thought to determine the value or the point of the homework, and they make the teacher tell them. [/quote]\n\nThis remainded me abot another disatventage of homeworks.Many people say that they teach you the working habits. I don't agree. Because homework doesn't teach you that you need to do something to get what you want, but it teaches you that you'll have to do things if an only if someone tells you to do them.\n\n[quote]\nSecondly, I don't think an education should focus on the average student at the expense of the advanced students. Perhaps I'm biased, but the notion that advanced students are there to help the teachers teach the average students is somewhat bogus. This is because, often, the average students are in that position because they really don't care to do better. They have other priorities.[/quote]\r\n\r\n$ 100$% true.", "Solution_10": "[quote=\"The Zuton Force\"]I don't think an education should focus on the average student at the expense of the advanced students.[/quote]\r\n\r\nIt's not clear why an educational system should focus on the needs of any particular group at the expense of any other group, but if it must do so, I'd prefer it focus on \"average students\" (the largest group, by far) at the expense of people who are likely to succeed even with a less-than-ideal education. I also wonder whether you think you will feel this way when you are no longer one of the smartest people around. \r\n\r\nThis latter musing is phrased in the future tense because I think it extremely likely that you will find yourself in this situation at some point (regardless of how smart you actually are -- in particular, please don't think this a slight on your intelligence). I think this because, for example, a very large number of undergraduates at Harvard were essentially the smartest person they'd ever met until the first day of freshman year, and most of them found themselves in the middle of the pack once they arrived.\r\n\r\n\r\nAnd, to give this post a little on-topic veneer: I'm hard-pressed to imagine a successful educational system that does not include required work of some sort, and I imagine this is true of the author as well. Then the question is whether \"homework in its current form\" is good or bad, and the major objection here seems to be that it's not very interesting. But since homework doesn't prevent you from going to study interesting questions in your own free time, this doesn't really strike me as compelling. (That isn't to say, of course, that it wouldn't be nice if homework were more interesting, just that this does not seem to be a fault of homework in general but rather of the particular homework that you get assigned.)", "Solution_11": "[quote]Then the question is whether \"homework in its current form\" is good or bad,[/quote]\n\nthat was also meant (however I think that you can learn people some stuff without making them do it)\n\n[quote]But since homework doesn't prevent you from going to study interesting questions in your own free time,[/quote] \r\n\r\nhere i don't agree. The main problem is that it waists your time.That time could be used to do that interesting stuff....", "Solution_12": "[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the post immediately preceding yours.[/color]\r\n\r\nIn all honesty, how many people are going to spend the time that would have been spent doing homework on something both interesting and educational?", "Solution_13": "[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the post immediately preceding yours. (Although you get half-credit for cutting out the older quote in the post you were quoting. :P )[/color]\r\n\r\nAgreed, I mean if it was me, I'd spend the time on my Xbox on most days, and possibily doing AoPS on others.", "Solution_14": "How can you be so sure of new material that you don't need any reinforcing exercises?", "Solution_15": "Depends how new it is.\r\n\r\nFor example, we covered partial fractions a few weeks back at school, and since I'd already done some work on them I didn't do any re-enforcing exercises. (And still fare fine with them)", "Solution_16": "[quote=\"The Zuton Force\"]I think you're missing the point. There's a reason \"stupid\" is in quotes. It's not because the people are actually stupid. Some people who have little academic motivation think that it's the teacher's job to lay a mound of knowledge in front of them on a silver platter, with fine champagne and a side of caviar. They are so apathetic and unmotivated that they do not want to invest the thought to determine the value or the point of the homework, and they make the teacher tell them.[/quote]\r\n\r\nI thought of this immediately today when talking to such a student. He sleeps through class but complains about teachers not teaching. He was talking about how one of his teachers is \"the worst teacher ever\" (probably with a swear word in there somewhere) for having the students create a presentation on the subjects instead of him just teaching the information.\r\n\r\nIn fact, he was sleeping through our English class's discussion on the previous night's homework, which was actually had some pretty difficult analytical questions about the book we were reading. I would rate the worksheet on the \"better\" side of homework because it forced you to think about the book, instead of being just busywork, despite that it was annoying to do (but the question of how I feel about the usefulness of studying these books is another topic...)", "Solution_17": "I believe that if homework were to be removed completely, school performance would go up in general. Why? Because now, homework is a quite big part of one's grade (and projects). If homework were to be abolished, then the only thing left for the teacher to fill in grades with is tests. If one does need to reinforce what they have learned, then they will be forced to seek practice eventually, or fail due to believing more in \"other priorities\", like The Zuton Force said. I don't think anyone will be not motivated, because they will be forced to realize that they do need to know the material to do well. Anyone not motivated to earn a good test grade is not motivated to do homework either.\r\nThe only problem I see with that would be the \"unfairness\" of certain testing methods, or the likes, but that is not too hard to do in my opinion.", "Solution_18": "Your reasoning is solid, serialk11r, but in fact, I believe school performance would go [i]down[/i]. Not everyone would spend their time in education, in fact many are only at school because they are \"forced\" too. Taking tests would only result in their failure. Only the people that will take time in their education will be doing well.", "Solution_19": "If you took away homework, school performance would go [i]up[/i]? That would be true in some cases except that my teachers check if you actually [i]tried[/i], and not wrote down random numbers. That should be the case for every school.", "Solution_20": "Well that may also happen, but I think it depends on each person's view on education and the purposes of it.\r\nAt my school I see many many people who may not be very smart, and may not get good grades, but they still try to some extent. They DO study for tests, and they study very much for them. The people failing school are pretty much all the people who sit by the back of the school and smoke weed, and I don't see any hope in them already.\r\nIf you believe that the purpose of education is to educate society as a whole, then no homework would be bad. However I don't think that the people who aren't motivated are doing well in school anyways...\r\nIf you believe that the purpose of education is to provide for those who are motivated, then I see no reason why abolishing homework would be bad.\r\n\r\nPersonally, I would really love to have no homework. I pick up on info really fast, and homework is pretty useless for me. In math I already know everything, in history there really isn't that much info in the book to memorize, bio I already know everything, etc. so it would be really great if I had no homework, then I could focus on math more and piano more. I really hate this effort thing on homework, I really don't put a whole lot of effort into homework/projects because I think it's redundant, and it hurts my grade. If grades were completely based off test grades (with the exception of english) then my grades would be a lot higher. I think a lot of people here possibly feel the same way.", "Solution_21": "@serialk11r's second paragraph: I agree.\r\n\r\nBut I think they should spend more money on those less gifted than the more advanced students, because advanced students can progress on their own and independence is a factor of life, is it not?", "Solution_22": "[quote=\"serialk11r\"]\nIf you believe that the purpose of education is to educate society as a whole, then no homework would be bad. However I don't think that the people who aren't motivated are doing well in school anyways...\nIf you believe that the purpose of education is to provide for those who are motivated, then I see no reason why abolishing homework would be bad.\n\nPersonally, I would really love to have no homework.[/quote]\r\n\r\nAnd that's not a selfish attitude?\r\n\r\nThe reason education is state funded is mainly due to the economic positive externalities of educating the [i]whole[/i] of society. Therefore homework has some value\r\n\r\nWRT to the performance argument.\r\n\r\nPlease can we have some studies or leave it, hypothesis mean nothing.", "Solution_23": "I'm not talking about advanced students. Students who are advanced enough at math for example, will not benefit at all from going to math class. What I see is that anyone who is motivated, and can learn from class, will benefit. I really don't see any way to convince those people who look like they'll be dropping out of school to try harder and learn, so I think it is pointless. The average student seems to care to a good extent about doing well in school. Optional, light homework may be a solution to this, so those who feel like they do not have a good grasp on the material may do some problems to study. In bio for example, almost everyone in my class thinks it's a hard class, so when the teacher says \"do you guys want the review sheet to be graded?\" they all say \"YES!!!!\" and I'm the only one saying \"NOOOOOO!!!\". Well another reason is because they want the extra points to boost their grade, but its only a measly 2 points. Maybe they just don't realize that it won't increase their grade much, but still, they WANTED more homework.", "Solution_24": "[quote]In all honesty, how many people are going to spend the time that would have been spent doing homework on something both interesting and educational?[/quote]\n not many. :lol:\n\n [quote]How can you be so sure of new material that you don't need any reinforcing exercises?[/quote]\n\nbeacuse I often already know this stuff + whenever I learn something new i often do it by solving some problem \n\n[quote]Well that may also happen, but I think it depends on each person's view on education and the purposes of it.\nAt my school I see many many people who may not be very smart, and may not get good grades, but they still try to some extent. They DO study for tests, and they study very much for them. The people failing school are pretty much all the people who sit by the back of the school and smoke weed, and I don't see any hope in them already.\nIf you believe that the purpose of education is to educate society as a whole, then no homework would be bad. However I don't think that the people who aren't motivated are doing well in school anyways...\nIf you believe that the purpose of education is to provide for those who are motivated, then I see no reason why abolishing homework would be bad.\n\nPersonally, I would really love to have no homework. I pick up on info really fast, and homework is pretty useless for me. In math I already know everything, in history there really isn't that much info in the book to memorize, bio I already know everything, etc. so it would be really great if I had no homework, then I could focus on math more and piano more. I really hate this effort thing on homework, I really don't put a whole lot of effort into homework/projects because I think it's redundant, and it hurts my grade. If grades were completely based off test grades (with the exception of english) then my grades would be a lot higher. I think a lot of people here possibly feel the same way.[/quote]\n\n[quote]\t\nI'm not talking about advanced students. Students who are advanced enough at math for example, will not benefit at all from going to math class. What I see is that anyone who is motivated, and can learn from class, will benefit. I really don't see any way to convince those people who look like they'll be dropping out of school to try harder and learn, so I think it is pointless. The average student seems to care to a good extent about doing well in school. Optional, light homework may be a solution to this, so those who feel like they do not have a good grasp on the material may do some problems to study. In bio for example, almost everyone in my class thinks it's a hard class, so when the teacher says \"do you guys want the review sheet to be graded?\" they all say \"YES!!!!\" and I'm the only one saying \"NOOOOOO!!!\". Well another reason is because they want the extra points to boost their grade, but its only a measly 2 points. Maybe they just don't realize that it won't increase their grade much, but still, they WANTED more homework.[/quote]\r\n\r\n\r\n\r\nthis is is what i'm talking about!!!\r\n\r\nschool is meant to educate avrege student. Ok I can live with that. However I really don't like when school forces me to do stuff i don't need to do.\r\n\r\nI'm not form USA and I was completly shocked when I heard that a part of grade in USA come from these homeworks and projects.\r\n\r\nI really don't know why school is more happy if you're doing something , and it doesn't matter wether it is going to bring any result, instead if you do a little less but there are some results at the end. \r\n\r\nfor example my teacher is always insainly happy when she sees that one of my school frineds copys every note from lesson in another notebook. this student still gets C in that subject...\r\n\r\ni don't even write things said at lessons and I still get an A ...{but the teacher doesn't like me not doing anything} \r\n\r\nyes I know making a good example however I don't know form where came this assumption that people all learn in the same way ?", "Solution_25": "I don't know about elsewhere, but homework is only 10-20% of our grade. And this is taking in the standard \"gifted\" course. \r\n\r\nNow if you're in an AP class, there is absolutely no homework at all, and your grade is based on tests.\r\n\r\nIn my experience, having homework being a marginally small percent of your overall grade lets you \"choose\" which homework is needed. For example, in my geometry class, every homework is assigned a 5 point value, with a test being a 100 point value. The final grade is dividing the points you recieved over the total points possible. I have found that rarely doing your homework, but acing the tests, can still give you a good grade. We recieved a small report of our geometry grades today. I still have an A in that class, but my homework is ranked as a C.", "Solution_26": "so it isn't standard ... it can be between 10-20 %\r\n\r\nWhat would be your grade if you wouldn't do any homework?", "Solution_27": "Well assuming that homework is 15% of your grade, and assuming you didn't do any of it, your maximum grade would be an 85%. Now, if your teacher uses a point system like my geometry teacher, then it is different.", "Solution_28": "[quote=\"Lazarus\"]\nAlso, the \"That's one of the reasons I don't want 'stupid' people in my class\" argument kinda bothers me. You are the unusual one for learning at an accelerated pace with respect to your classmates, yet you are placing blame upon the average students while expecting special treatment? Other kids do not learn at the same pace as the rare talented and hardworking child. Sometimes the learning basic action is difficult enough without simultaneously being force-fed the abstract concepts behind this action - just so the \"stupid\" people can learn at the same pace as the student multiple standard deviations above the norm.[/quote]\n\nI understand about the hard-working people. I can't stand the people who go to school because their parents force them to.\nand I understand that the stupid thing was a bit mean of me. I really do. Sometimes I post things without even rechecking.\n\nand the thing is, I have to say, I am very impatient. I try to do things as fast as possible, and that, I think, is what drives me. It drives me to accelerate.\n\nBACK TO THE TOPIC!\n\n[quote]All the work I do at school can be done at home in 10 minutes with a textbook. [/quote]\n\nexactly my point. We need to have homework that's useful. Ten minutes with a textbook isn't useful. You search for the answer and that's about it.\n\n[quote]Personally, I would really love to have no homework. I pick up on info really fast, and homework is pretty useless for me.[/quote]\r\n\r\nYeah. My case.", "Solution_29": ":\\\r\n\r\nI am not quite sure if you guys are getting the fact that homework - as trivial as it may seem to you - is [i]actually difficult for average students[/i]. \r\n\r\nLet this sink in, and reevaluate your arguments.\r\n\r\nYou all are generally pushing for homework that is challenging. However, for most students, not all, not you, homework is actually challenging.\r\n\r\nMaking homework even more difficult for the other students is unrealistic.\r\n\r\nSimilarly, although it is sad that this is the case, giving gifted students specially-tailored homework assignments to fit their needs is often unrealistic as well. Homeschooled students and privately schooled students are more lucky in this respect. It is a shame that public schools cannot do this, for funding is more necessary in other areas, but I mean this is something you will have to live with.\r\n\r\nTake up a school sport if you want to see taxpayer money put to good use. Academics? Funding?\r\n\r\n\r\n\r\nand on a totally unrelated note, GOOD LUCK with exeter, kimmystar94. from what i hear it's a cluster**** of homework assignments, impossibly difficult classes, and actual mental stimulation. who would have known.", "Solution_30": "[quote]I am not quite sure if you guys are getting the fact that homework - as trivial as it may seem to you - is actually difficult for average students. [/quote]\r\n\r\nSchool is meant for avrege student. I can live with that.\r\nSchool can't provide more interesting taks for me. I can live with that either.\r\n\r\nBut i really don't like that school is making me do this trivial tasks. Ok they can't find anything interesting so why not just leave me alone?", "Solution_31": "Homework is 70% of our grade in history :O Which is why my grade sucks ***. Homework is >80% of our grade in english, but that's because everything pretty much has to be done at home.\r\n\r\nYes I realize that I'm one of the few people in the whole grade at my school who can easily pull As on bio tests, and that many people are struggling to get a B. Still, there should be some kind of option to not do homework, like if your grade is below a B or something, homework is mandatory, then if your grade is higher, you can skip it. A lot of people in my class would end up doing all the homework anyways, even people with As (hard earned ones...).\r\nI've heard from some seniors that they have teachers who don't believe in homework, or they have optional homework, and I really think this is good, because it allows each student to be able to choose the way they want to study.", "Solution_32": "[quote]Yes I realize that I'm one of the few people in the whole grade at my school who can easily pull As on bio tests, and that many people are struggling to get a B. Still, there should be some kind of option to not do homework, like if your grade is below a B or something, homework is mandatory, then if your grade is higher, you can skip it. A lot of people in my class would end up doing all the homework anyways, even people with As (hard earned ones...).\nI've heard from some seniors that they have teachers who don't believe in homework, or they have optional homework, and I really think this is good, because it allows each student to be able to choose the way they want to study.[/quote]\r\n\r\noptional homework is a very good idea(and I have it at some subjects!) :)", "Solution_33": "This is why external examination is a much better idea. In the UK it doesn't really matter what grades you get at school (to a certain extent) since the only results which universities will care about are examined externally so all that homework I missed 10 years ago wont have too much of an effect when I apply to universities.", "Solution_34": "well the main problem with the American education system is that good students are paired with bad students- and it makes the bad students lag behind while the good ones are bored.\r\n\r\nGiving more/less homework isn't the answer. Many times, it is helpful to lots of people, and helps the re-enforce topics. The problem is that students just skim through homework for the answers, while not really finding the meaning in it. Another problem is that while advanced students think it is easy, bad students find they can't even do it and many times don't even try.\r\n\r\nThe solution to this is to have more levels. If you have 8 periods including lunch, there would be about 5-6 academic classes. They should have a honors class, a level 2 class, a level 3 class,etc. in each subject so the best students get what they need to expand their horizons and the worst students can get extra help so they will actually understand the material.", "Solution_35": "Several thoughts:\r\n\r\nFirstly, I've had an English/History teacher that gave us the following option for making study materials for tests that is somewhat close to optional homework. Essentially, if you get at least 85% on the test, she'll assume that you studied for it and give you an A on the study materials. If you didn't get at least 85%, she looks at your study materials and talks with you about them.\r\n\r\nSecondly, what would really be nice is if teachers could give an independant study option for the class, and have the people using it only take the tests. Then, they couldy study/work on problems without distracting everyone around them while being bored, but the teacher could still confirm that they knew the basic curriculum.\r\n\r\nMy favorite math teacher of all time had a nice method: she would tell you the homework at the beginning of class, and then ask people what problems from the last homework they had trouble on, and go over them on the projector. Meanwhile, she didn't really care if we just did the next day's homework instead of actually listening to her, especially if we didn't have any issues with the homework she was talking about. (We didn't have any difference between homework and classwork; whatever you didn't get done in class became homework.) It was really great because she didn't care too much about what she did as long as we got the work done. We probably spent only about 7 or 8 days out of 180 taking notes, and most of the time actually working on math. :o \r\n\r\nOne of my pet peeves is teachers (especially math) that feel like they have to force-feed you every part of the curriculum, and will waste their class time going over the previous homework even if everybody understands it anyways. \r\n\r\nJust because homework is boring doesn't mean that it's necessarily useless (although it should be part of a teacher's job to make their subject interesting). Especially with basics in math (i.e. factoring quadratic equations, arithmetic), it is important to be able to run through it quickly so that you can focus on the more interesting parts of problems, instead of getting bogged down. It's even worse when they actually expect everyone to be paying attention. :P \r\n\r\n[quote]well the main problem with the American education system is that good students are paired with bad students- and it makes the bad students lag behind while the good ones are bored. [/quote] \r\n\r\nI wouldn't quite agree with that. Especially in my experience, teachers tend to teach down to the slowest student in the class. Even as one of the \"smarter\" students in my class, I'm not particularly bothered if other people ask questions about things that seem easy to me. What bothers me is when the teacher has to go over everything multiple times because people aren't paying attention.\r\n\r\nI have willingly spent a great deal of time working on homework and preparing for tests in some of my history classes in middle school, because I felt like the topics were interesting and I was actually getting out of it. However, I think one of the issues for us nowadays is that a majority of the homework we get [i]is[/i] busywork, so that even when we get non-busywork, nobody's really willing to do it.\r\n\r\nThat comes back to how to define busywork. What may be busywork for one person might not be busywork for somebody else. One possibility: as long as you get above a certain percentage on tests, you don't have to do the homework. Once you drop below that percentage, you have to do it.\r\n\r\nBTW, I've survived class by making origami, playing with embroidery floss, and doodling wallpaper designs (http://simathcircle.org/extras_wallpaper.aspx). I know friends that have actually worked on cross-stitching in class and the teachers don't care.\r\n\r\nThis post is getting ridiculously long now, so I'll stop.", "Solution_36": "personally, I think SCHOOL is pretty pointless. everything that we learn in a year at school, I could probably master in a month or so just from a textbook.", "Solution_37": "I wish instead of giving homework, they extended the school day." } { "Tag": [ "geometry", "3D geometry", "sphere" ], "Problem": "Jemima was given a spherical popsicle with radius 3. She decided to evenly remove popsicle from all sides of the popsicle until only 1/2 the popsicle was left (by volume) and give brother Jack the rest. What is the radius of the popsicle given to Jack? Express your answer in simplest radical form.", "Solution_1": "[hide]The volume of the original popsicle sphere is $3^3(\\frac{4}{3})\\pi=36\\pi$. THus half of the sphere's volume is $18\\pi$. The radius of this is $\\sqrt[3]{18(\\frac{3}{4})}=\\frac{3}{\\sqrt[3]{2}}=\\frac{3\\sqrt[3]{4}}{2}$. [/hide]", "Solution_2": "[hide]The area of the total popsicle is $3^3\\cdot\\frac43\\pi=36\\pi$. Half of that area is $\\frac{36\\pi}{2}=18\\pi$. The radius of the small popsicle is $x$. That means that $18\\pi=\\frac{4\\pi r^3}{3}$\n$54=4r^3$\n$4r^3=54$\n$r^3=\\frac{27}{2}$\n$r=\\sqrt[3]{\\frac{27}{2}}$\n$r=\\frac{3\\sqrt[3]{4}}{2}$[/hide]\r\n\r\n236factorial, I think you have a typo.", "Solution_3": "Pay no attention to this post. :whistling:", "Solution_4": "[quote=\"JesusFreak197\"]Yes, he has the correct answer, but one of his fractions is flipped partway through.[/quote]\r\n\r\nAre you talking about the $\\frac{3}{4}$?", "Solution_5": "Err... I don't know what you're talking about. :whistling:", "Solution_6": "[hide]\nThe area of the lollipop is $\\frac{4}{3}\\pi27=36\\pi$\nHalf of $36\\pi=18\\pi$\n$\\sqrt[3]{18\\cdot\\frac{3}{4}}=\\sqrt[3]{\\frac{27}{2}}=\\boxed{\\boxed{\\frac{3\\sqrt[3]{4}}{2}}}$[/hide]" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "let explain contradiction:\r\n solve equation : \r\nsolve :\r\n${ x^{x^{{x}^...^{x}}}} = a$ or $ x^a = a$ hence : $ x = \\sqrt [a]{a}$ \r\n if where $ a = 2$ or$ a = 4$then we accepted $ x = \\sqrt {2}.$\r\n but retry roots then appear 1 contracdiction: \r\n$ \\sqrt {2}^{{\\sqrt {2}}^{...^{\\sqrt {2}}}} = 2 = 4$ \r\n let explain why ?", "Solution_1": "Well $ \\sqrt{2}^{\\sqrt{2}^{\\sqrt{2}^...}}\\neq 2\\neq 4$. If you have an infinite number of powers, then the expression will be infinitely large.", "Solution_2": "[quote=\"JRav\"]Well $ \\sqrt {2}^{\\sqrt {2}^{\\sqrt {2}^...}}\\neq 2\\neq 4$. If you have an infinite number of powers, then the expression will be infinitely large.[/quote]\r\n maybe explain for contradiction problem's not right", "Solution_3": "I don't understand what you mean.", "Solution_4": "I'm pretty sure your solution is correct. However, for $ x\\geq1$, the exponent will blow up. If $ x\\neq1$, then the exponent will converge to 1. So I believe $ x\\equal{}a\\equal{}1$ is the only value that works.", "Solution_5": "Actually since $ (\\sqrt {2}) ^{\\sqrt {2}} < (\\sqrt {2})^2 \\equal{} 2$ , you can prove that the expression with powers of $ \\sqrt2$ taken finite times is less than 2. In the limit it becomes 2", "Solution_6": "[quote=\"JRav\"]Well $ \\sqrt {2}^{\\sqrt {2}^{\\sqrt {2}^...}}\\neq 2\\neq 4$. If you have an infinite number of powers, then the expression will be infinitely large.[/quote]\r\n\r\nThis is incorrect JRav. This expression will converge for $ 1/e \\leq x \\leq e$ and diverge otherwise. The proof of this can be found (somewhere) on this forum.\r\n\r\nThe mistake in your original solution Hong Quy is that $ x \\equal{} \\sqrt[a]{a}$ is a solution only if $ a$ exists in the first place (i.e. the series converges). It would be like saying $ 1\\minus{}1\\plus{}1\\minus{}1\\plus{}... \\equal{} a$ so $ a\\minus{}1 \\equal{} \\minus{}a$, and therefore $ a \\equal{} 1/2$ (which is not true).", "Solution_7": "Consider the sequence $ a_1 \\equal{} x$ and $ a_{n \\plus{} 1} \\equal{} x^{a_n}$\r\n\r\nFor $ 0 < x < 1$, the sequence is bounded above by $ 1$. Also, we can see that the sequences $ \\left\\langle a_{2n}\\right\\rangle$ and $ \\left\\langle a_{2n \\plus{} 1}\\right\\rangle$ are seperately monotonic, and I believe that they have the same limit.\r\n\r\nFor $ 2 < x$, we can write $ a_{n \\plus{} 1} \\equal{} (1 \\plus{} h)^{a_n}\\geq 1 \\plus{} a_nh$\r\ni.e. $ a_{n \\plus{} 1} \\minus{} a_n\\geq 1 \\plus{} (h \\minus{} 1)a_n\\geq 1$, ad hence, sequence is not convergent.\r\n\r\nIt seems that $ \\exists \\alpha$ such that the sequence is convergent if and only if $ 0\\frac1M$\r\n\r\nLet $c_0=0$. Let $c_1$ be an arbitrary integer not in $B$. For $1< k c_{k-1}+1$. Then, for any $k>i$, $c_k-c_i\\not\\in B$. The $M$ sets $A_k=A+c_k$ are therefore disjoint.\r\n\r\nBy the density condition, we can find some $t>0$ such that the set of $x$ for which $\\frac{N(x)}{x}\\ge \\frac1M+t$ is unbounded. Find such an integer $x$ with the additional property that $x>\\frac{c_{M-1}}{Mt}$.\r\nNow, we count the elements of $A_0\\cup A_1\\cup\\cdots\\cup A_{M-1}$ in $\\{1,2,\\dots,x+c_{M-1}\\}$. There are at least $x\\left(\\frac1M+t\\right)$ elements of each $A_k$ in this range, so in total there are at least $x+tMx$ elements in the set because the $A_k$ are disjoint. On the other hand, there are at most $x+c_{M-1}$ elements in this union, so $tMx\\le c_{M-1}$, contradicting our choice of $x$.\r\n\r\nThe only way out of this contradiction is to say that the sets were not disjoint, and $b_{j-1}-b_j$ was in fact bounded." } { "Tag": [ "calculus", "integration", "algebra", "polynomial", "calculus computations" ], "Problem": "If $ I_{n}$ denotes $ \\frac{1}{n!}\\int_{0}^{x} t^{n}e^{\\minus{}t}\\ dt$, find a relation connecting $ I_{n}$ and $ I_{n\\minus{}1}$. Deduce that the value of $ e^{x}$ exceeds its Maclaurin polynomial of degree $ n$ by $ \\frac{e^{x}}{n!}\\int_{0}^{x} t^{n}e^{\\minus{}t}\\ dt$.", "Solution_1": "$ I_n \\equal{} \\frac{1}{n!} \\int_0^xt^ne^{\\minus{}t}\\,dt \\equal{} \\frac{1}{n!} \\left[\\minus{}x^ne^{\\minus{}x} \\plus{} \\int_0^x nt^{n\\minus{}1}e^{\\minus{}t}\\,dt\\right]$\r\n\r\n$ \\equal{} \\minus{} \\frac{x^n}{n!}e^{\\minus{}x} \\plus{} \\frac{1}{(n\\minus{}1)!}\\int_0^x nt^{n\\minus{}1}e^{\\minus{}t}\\,dt$\r\n\r\n$ \\equal{}> I_n \\equal{} I_{n\\minus{}1} \\minus{} \\frac{x^n}{n!}e^{\\minus{}x}$.\r\n\r\nNow we have\r\n\r\n$ I_1 \\equal{} I_0 \\minus{} x e^{\\minus{}x} \\equal{} 1 \\minus{} e^{\\minus{}x} \\minus{} xe^{\\minus{}x}$\r\n\r\n$ I_2 \\equal{} I_1 \\minus{} \\frac{x^2}{2!} e^{\\minus{}x} \\equal{} 1 \\minus{} e^{\\minus{}x} \\minus{} xe^{\\minus{}x} \\minus{} \\frac{x^2}{2!} e^{\\minus{}x}$\r\n\r\n$ I_3 \\equal{} I_2 \\minus{} \\frac{x^3}{3!} e^{\\minus{}x} \\equal{} 1 \\minus{} e^{\\minus{}x} \\minus{} xe^{\\minus{}x} \\minus{} \\frac{x^2}{2!} e^{\\minus{}x} \\minus{} \\frac{x^3}{3!} e^{\\minus{}x}$\r\n\r\n$ ...$\r\n\r\n$ I_n \\equal{} 1 \\minus{} e^{\\minus{}x} \\minus{} xe^{\\minus{}x} \\minus{} \\frac{x^2}{2!} e^{\\minus{}x} \\minus{} ... \\minus{} \\frac{x^{n\\minus{}1}}{(n\\minus{}1)!} e^{\\minus{}x} \\minus{} \\frac{x^n}{n!} e^{\\minus{}x}$\r\n\r\nand denoting the Maclaurin polynomial of $ e^x$ of order n by $ P_n$ this is\r\n\r\n$ I_n \\equal{} 1\\minus{}P_ne^{\\minus{}x} <\\equal{}> e^x \\equal{} P_n \\plus{} e^x I_n \\equal{} P_n \\plus{} \\frac{e^x}{n!} \\int_0^xt^ne^{\\minus{}t}\\,dt$." } { "Tag": [ "topology", "vector", "function", "Functional Analysis", "absolute value", "advanced fields", "advanced fields open" ], "Problem": "Hi!\r\n\r\nFirst of all, a very simple problem.\r\n1. Suppose $\\{A_n\\}$ is a family of subsets of metric space $X$. It is known that for any increasing sequence $\\{n_k\\}$ there is a subsequence $\\{n_{k_l}\\}$ s.t. it is possible to select $a_{n_{k_l}}\\in A_{n_{k_l}}$ and $a_{n_{k_l}}\\to a$. \r\nProve that it is possible to select $a_n\\in A_n$ in such a way that $a_n\\to a$.\r\n\r\nAnd now the problem on which I got stuck.\r\nThe same condition as above, but we replace the metric space $X$ with reflexive separable banach space and the strong convergence to the weak convergense. Is the satement still true?\r\nNow formally.\r\n2. Suppose $\\{A_n\\}$ is a family of subsets of reflexive separable banach space $X$. It is known that for any increasing sequence $\\{n_k\\}$ there is a subsequence $\\{n_{k_l}\\}$ s.t. it is possible to select $a_{n_{k_l}}\\in A_{n_{k_l}}$ and $a_{n_{k_l}}\\rightharpoonup a$. \r\nIs it true that we can select $a_n\\in A_n$ in such a way that $a_n\\rightharpoonup a$?\r\n\r\nFortunately, I avoided to prove it, since all my sets $A_n$ consists of single points. But the general question seems to be very interesting. I have a feeling that I am missing something simple here. :?", "Solution_1": "Ops... :blush: \r\nIt was really easy. The weak topology of separable banach space is metrizable. So [b]2[/b] comes immediately from [b]1[/b].\r\n\r\nBTW, these are interesting problems themself to show that $X^*$ is separable and then to show how to metrize the weak topology of $X$.", "Solution_2": "[quote=\"Myth\"]\nIt was really easy. The weak topology of separable banach space is metrizable. So [b]2[/b] comes immediately from [b]1[/b].[/quote]\r\nTo be honest, I thought so for the first ten minutes after seeing your question too and almost posted it :blush: but then I realized that it is not that easy. You can metrize the unit ball (or a ball of any finite radius), but not the entire space. So, if all $A_n$ are all contained in some fixed ball, then the metrization argument is perfect. But if we do not assume such a uniform bound a priori, there are counterexamples. To describe one of them, let us assume that $X$ is just a separable Hilbert space. Let $a=0$. Let $\\{e_j\\}_{j\\geq 0}$ be an orthonormal basis in $X$ and let $p(x)=\\sum_{j\\geq 0}2^{-j}|(x,e_j)|$ be the seminorm that is used for metrization of balls. Then $x_n$ weakly converges to $0$ if and only if the sequence $\\|x_n\\|$ is bounded and the sequence $p(x_n)$ converges to $0$. For each pair of positive integers $k,\\ell$ choose some vector $v_{k\\ell}$ such that $\\|v_{k,\\ell}\\|=k$, $p(v_{k\\ell})=\\frac1\\ell$. Choose also some bijection $\\psi:\\mathbb N\\to \\mathbb Q$. The sets $A_n$ will consist only of vectors $v_{k\\ell}$ and are defined according to the following rule: $v_{k,\\ell}\\in A_n$ if and only if $\\psi(n)\\notin[k,\\ell)$. \r\n\r\nNote that the sequence $v_{k_n\\ell_n}$ weakly converges to $0$ if and only if $k_n$ are bounded and $\\ell_n\\to\\infty$. So, if we have a sequence $v_{k_n\\ell_n}\\in A_n$ weakly converging to $0$, we would have some $K$ such that all $A_n$ except finitely many contain a vector $v_{k,\\ell}$ with $k\\leq K$, $\\ell\\geq K+1$. But it cannot happen if $\\psi(n)\\in [K,K+1)$ and the set of such $n$ is infinite. On the other hand, if $M\\subset\\mathbb N$ is an infinite subset, then either $\\psi(M)\\cap [\\ell,+\\infty)$ is non-empty for all $\\ell$, and then we can choose vectors $v_{1,r}$ in some $A_{n_r}$, $n_r\\in M$, which will give us a subsequence weakly converging to $0$, or $\\psi(M)\\subset (-\\infty,K)$ for some $K$ and then $v_{K,n}\\in A_n$ for all $n\\in M$.", "Solution_3": "Yes, I forgot to say about this aspect of metrization :(\r\nTo be honest, by some speculations in my mind I came to conclusion that $A_n$ are necessarily bounded. :(\r\nIt happens very often to me, since I very rare write anything to a piece of paper and all my solutions live only in my head, and it leads to a number of errors.\r\n\r\nYour construction is very complicated. How did you imagine it?\r\n\r\nInitially, in my paper I defined $A_n$ as sets lying in $\\varepsilon$-neighbour of some point $u\\in W_q^2(\\Omega)$, so clearly their were bounded. Further, I slightly changed my approach and got that $A_n$ are just single points.\r\n\r\nAnyway, the problem I formulated is very nice (in my opinion, of course).", "Solution_4": "[quote=\"Myth\"]Your construction is very complicated. How did you imagine it? \n [/quote] I was lucky to start thinking in the right way. Since weak convergence is all about boundedness of norms and $p(x_n)$ tending to $0$, I defined the sets $N(R,\\delta)=\\{n\\in \\mathbb N: A_n\\cap B(R)\\cap\\{p<\\delta\\}=\\varnothing\\}$ and tried to express the given conditions in terms of these sets. After some time I managed to do it and the problem reduced to the following: does there exist a collection of sets $N(R,\\delta)$ that is decreasing in both $R$ and $\\delta$ and such that a) for every $R$ there exists $\\delta$ such that $N(R,\\delta)$ is infinite and b) For every function $\\delta(R)$ there exists $\\rho$ such that $\\bigcap_{R<\\rho}N(R,\\delta(R))$ is finite. This reminded me of the well-known problem about construction of a family of subsets of $\\mathbb N$ of cardinality continuum with the property that every 2 sets have finite intersection. The rest was fairly straightforward, but it took me almost the entire evening to finish the solution.\n\n[quote=\"Myth\"]Anyway, the problem I formulated is very nice [/quote]\nYes, it was beautiful. The same can be said about many other problems you proposed (in my humble opinion, of course).\n\n[quote=\"Myth\"] I very rare write anything to a piece of paper and all my solutions live only in my head[/quote] I envy you here. I often need to write the statement of the problem down before I can start solving it. As a result, my office is usually a complete mess with paper lying (and flying) everywhere :)", "Solution_5": "I will include both cases (bounded and unbounded) in my future functional analysis course.", "Solution_6": "I know reviving old threads isn't always welcome, but while leafing through Paul Halmos's wonderful \"A Hilbert Space Problem Book\" I came across the following problem: The weak topology of an infinite-dimensional Hilbert space is not metrizable.\r\n\r\nSo fedja's counterexample is the rule, not the exception :). I hope the following is a correct proof of the theorem.\r\n\r\nIf the weak topology of an infinite-dimensional hilbert space $H$ were metrizable, the space would be first-countable under this topology. It thus suffices to prove that $0$ does not have a countable basis of weakly open neighborhoods. This is equivalent to proving the following:\r\n\r\nWe cannot find a sequence $x_1,x_2,\\ldots$ of vectors in $H$ s.t. for every vector $x$ and every $\\varepsilon>0$ there is a $k$ and an $n$ s.t. $|(y,x_i)|<\\frac 1n,\\ \\forall i\\in\\overline{1,k}\\Rightarrow |(y,x)|<\\varepsilon$.\r\n\r\nSuppose the contrary. Extract a linearly independent subset from $(x_i)_i$ and perform the Gram-Schmidt orthonormalization on it to get an orthonormal system $(e_i)_i$ s.t. the span of $e_1,\\ldots,e_k$ equals the span of $x_1,\\ldots,x_{m_k}$ for some $m_k\\ge k$, for all $k$. Complete this orthonormal system to a maximal one in $H$. Choose $\\varepsilon>0,n$ and $x_1,\\ldots,x_k$ as described above. Now choose $x$ s.t. it has only non-zero Fourier coefficients when represented in the orthonormal basis constructed above, and consider a vector $y$ orthogonal to all the $e_i$ 's which appear in the decompositions of $x_1,\\ldots,x_k$ (there are finitely many such $e_i$'s), but not orthogonal to $x$. For a complex number $\\alpha$ with large enough absolute value, we will have $|(\\alpha y,x_i)|=0<\\frac 1n,\\ \\forall i\\in\\overline{1,k}$, but $|(\\alpha y,x)|>\\varepsilon$, contradicting the existence of the countable basis. \r\n\r\nIs it correct? :?" } { "Tag": [], "Problem": "If someone has 10 bills in their wallet and in all they equal 40 dollars, if they have one more 5 dollar bill than 10 dollar bill, and two more 1 dollar bills than 5 dollar bills...\r\n\r\nhow many of each do they have?", "Solution_1": "[hide] Let $ x$ be the number of $ 10$ dollar bills. Then, $ x\\plus{}(x\\plus{}1)\\plus{}(x\\plus{}3)\\equal{}10 \\implies x\\equal{}2$. Thus, he has $ 2$ $ \\$10$ dollar bills, $ 3$ $ \\$5$ dolllar bills, and $ 5$ $ \\$1$ dollar bills (this also checks with the value of the money that we were given).[/hide]", "Solution_2": "[quote=\"sheeprule\"]If someone has 10 bills in their wallet and in all they equal 40 dollars, if they have one more 5 dollar bill than 10 dollar bill, and two more 1 dollar bills than 5 dollar bills...\n\nhow many of each do they have?[/quote]\r\n\r\nThis question gives too much information. You don't need to be told the relationship \r\nof the number of 1-dollar bills to the number of 5-dollar bills.\r\n\r\n It could certainly could have been:\r\n\r\n\"Suppose someone has 10 bills in their wallet, and in all they equal 40 dollars.\r\nIf they have one more 5-dollar bill than 10-dollar bill and at least one 1-dollar bill, \r\nthen how many of each type of bill do they have?\"\r\n\r\nCandidates:\r\n\r\nA) 1 ten & 2 fives\r\nB) 2 tens & 3 fives\r\n\r\nBecause there is a total of 10 bills (given), then scenario A) would \r\nmean there are 7 ones, but that totals to only 27 dollars.\r\n\r\nBut in scenario B), there would have to be 5 ones using the same reasoning\r\nbecause there is a total of 5 bills between the tens and fives. That equals to \r\n(tens) 2 bills + (fives) 3 bills + (ones) 5 bills = 40 dollars." } { "Tag": [ "algebra", "binomial theorem" ], "Problem": "solve for x\r\n\r\n\\[ \\minus{}1\\plus{}x\\equal{}\\sum_{k\\equal{}1}^{2000} \\binom{2000}{k}\\]", "Solution_1": "hello, $ x\\equal{}\\sum_{k\\equal{}1}^{2000}\\binom{2000}{k}\\plus{}1$.\r\nSonnhard.", "Solution_2": "EDIT: Wow, epic failed.\r\n\r\n[hide=\"Solution\"]\nWe have $ x \\equal{} \\sum_{k \\equal{} 1}^{2000}\\binom{2000}{k} \\plus{} 1$.\n\nIt is known that $ \\sum_{k \\equal{} 0}^{n}\\binom{n}{k} \\equal{} 2^n$.\n\nThus, $ x \\equal{} 2^{2000} \\plus{} 1\\minus{}\\binom{2000}{0}\\equal{}2^{2000}$.\n\nAnswer: $ 2^{2000}$[/hide]", "Solution_3": "Not exactly.", "Solution_4": "[hide=\"solution\"]\n\\begin{eqnarray*}\n- 1 + x &=& \\sum_{k = 1}^{2000} \\binom{2000}{k}\\\\\nx &=& \\sum_{k = 1}^{2000} \\binom{2000}{k} + 1\\\\\nx &=& \\sum_{k = 0}^{2000} \\binom{2000}{k} - \\binom{2000}{0} + 1\\\\\nx &=& 2^{2000} - 1 + 1\\\\\nx &=& \\boxed{2^{2000}}\n\\end{eqnarray*}\n\nNotice that $ \\sum_{k = 0}^{n} \\binom{n}{k} = 2^n$ not $ \\sum_{k = 1}^{n} \\binom{n}{k}$\n[/hide]", "Solution_5": "so is it 2^2000 or 2^2000 +1?", "Solution_6": "Yep, it's 2^2000 as tonypr said because k starts from 0, not 1.", "Solution_7": "Time to be stupid here and ask...how does that work?? D:", "Solution_8": "Hint: how many subsets exist of a set with n elements? Count it two different ways.", "Solution_9": "[quote=\"ckck\"]Time to be stupid here and ask...how does that work?? D:[/quote]\r\n\r\nUsing binomial theorem, we have $ (x\\plus{}1)^n\\equal{}\\binom{n}{0}x^n\\plus{}\\binom{n}{1}x^{n\\minus{}1}\\plus{}\\cdots\\plus{}\\binom{n}{n\\minus{}1}x^{1}\\plus{}\\binom{n}{n}x^0$\r\n\r\nIf $ x\\equal{}1$, we will have $ 2^n\\equal{}\\binom{n}{0}\\plus{}\\binom{n}{1}\\plus{}\\cdots\\plus{}\\binom{n}{n\\minus{}1}\\plus{}\\binom{n}{n}.$", "Solution_10": "[quote=\"ckck\"]Time to be stupid here and ask...how does that work?? D:[/quote]\r\nAnother way would be to look at Pascal's triangle. Knowing that each number on the triangle is the number of ways you can get to it from the top moving either left or right, try summing the numbers in the rows of the triangle. If you're not seeing it,\r\n[hide=\"Hint\"]Count the number of ways you can get to any number on a certain row.\n[hide=\"Another\"]Now count it in a different way.[/hide][/hide]\r\nThe idea is really similar to the one behind ThinkFlow's method.\r\n\r\nOnce you know the identity, the problem is really straightforward. Just plug in and isolate x." } { "Tag": [ "geometry", "geometric transformation", "rotation", "reflection", "floor function", "search", "group theory" ], "Problem": "\"In how many ways may we color the edges of a (regular) (2n + 1)-gon\r\nfree to move in three-dimensional space so that n edges are colored\r\nred and n + 1 edges are colored blue? Your answer may depend on\r\nwhether n is even or odd.\"\r\n\r\nThis is a question from a textbook of mine and I was shocked that they would include such a hard problem. Do you guys think this is olympiad level or is it more routine than that?", "Solution_1": "Well, it depends what the content of the textbook is. This is a pretty standard problem from Polya theory and we can hit it directly with the lemma that isn't Burnside's. Our set $ X$ is the set of $ n$-subsets of $ [2n \\plus{} 1]$ (the red vertices), and our group $ G$ is the dihedral group acting in the obvious way. The identity element fixes all $ \\binom{2n \\plus{} 1}{n}$ elements of $ X$. Since $ (n, 2n \\plus{} 1) \\equal{} 1$, the non-identity rotations don't fix any element of $ X$, and we only have to consider the reflections. If $ n$ is even then each reflection fixes $ \\binom{n}{n/2}$ subsets, while if $ n$ is odd then each reflection fixes $ \\binom{n}{(n \\minus{} 1)/2}$ subsets. Thus, the number of orbits (which is what we're after) is given by\r\n\\[ \\frac {\\binom{2n \\plus{} 1}{n} \\plus{} (2n \\plus{} 1) \\cdot \\binom{n}{\\lfloor n/2 \\rfloor}}{2(2n \\plus{} 1)}.\\]\r\nThis is sequence [url=http://www.research.att.com/~njas/sequences/A007123]A007123[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].\r\n\r\nHere are some other threads that use a similar technique:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=319975\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=314174\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=288514\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=14125\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=259872\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=252879\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=228658\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=225043\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=194846\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=192829\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=179833\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=106466\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=106467\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=109090" } { "Tag": [ "analytic geometry" ], "Problem": "The student radio station has a broadcasting range of 24 miles. Describe the locus of points which represents the outer edge of the broadcasting range.", "Solution_1": "[quote=\"Interval\"]The student radio station has a broadcasting range of 24 miles. Describe the locus of points which represents the outer edge of the broadcasting range.[/quote]\r\n[hide]I think it is a circle with radius of 24 miles. [/hide]", "Solution_2": "hard extention,\r\n\r\nif the locus was graphed on a cartesion coordinate plane and the center at the origin, where each unit square was a house, how many houses would the locus pass through?" } { "Tag": [], "Problem": "The prime factorization of 2160 is $ 2^4 \\times 3^3 \\times 5$. How many of its positive integer factors are perfect squares?", "Solution_1": "Squares include even exponent of 2,3 and 5. \r\nFor 2: 0,2,4 can be the exponents\r\n3: 0,2\r\n5: 0\r\n\r\nThat's 3*2*1=6 possibilities" } { "Tag": [ "inequalities", "algebra", "function", "domain", "induction" ], "Problem": "How do you prove Bernoulli's Inequality, which says that for x>-1, x not equal to 0, and n a positive integer greater than 1, then:\r\n\r\n$(1+x)^{n}>1+nx$.", "Solution_1": "[hide=\"Proof\"]Let $f\\left(x\\right)=\\left(1+x\\right)^{n}-1-nx$. Then you have \\[f'\\left(x\\right)=n\\left(1+x\\right)^{n-1}-n.\\] Given the domain, $f'\\left(x\\right)>0$ for $x>-1$ and $x\\neq 0$. Also, we have $f\\left(0\\right)=0$ so the function is greater than $0$ since it is increasing. This proves the inequality.\n\nAlternatively, you could consider the binomial series which would give you the same result.[/hide]", "Solution_2": "You can also prove it by induction in $n$ (more simply than the other proof above):\r\n\r\nBase case: $n = 2$:\r\n$(1+x)^{2}> 1+2x$\r\n$1+2x+x^{2}> 1+2x$\r\n$x^{2}> 0$ which is true for all real $x \\not= 0$\r\n\r\nInductive Step: assume $(1+x)^{n}> 1+nx$\r\nNow because $x >-1$, we can multiply both sides of the inequality by $1+x$, so:\r\n$(1+x)^{n+1}> 1+(n+1)x+nx^{2}> 1+(n+1)x$, so \r\n$(1+x)^{n+1}> 1+(n+1)x$, which completes the inductive proof.", "Solution_3": "yes but the induction proof only proves that it is true for integers and not all real values.", "Solution_4": "[quote=\"maokid7\"]yes but the induction proof only proves that it is true for integers and not all real values.[/quote]\r\n\r\nn[i] is [/i]a positive integer", "Solution_5": "oops sorry.\r\nnormally when i see this inequality it stats that $n$ is a real number such that\r\n$n\\geq 1$", "Solution_6": "[quote=\"amcavoy\"]Let $f\\left(x\\right)=\\left(1+x\\right)^{n}-1-nx$. Then you have \\[f'\\left(x\\right)=n\\left(1+x\\right)^{n-1}-n.\\] Given the domain, $f'\\left(x\\right)>0$ for $x>-1$ and $x\\neq 0$. Also, we have $f\\left(0\\right)=0$ so the function is greater than $0$ since it is increasing. \n[/quote]\r\nI would rather think $f'(x)<0$ if $-10$ if $x>0$, hence it has a (nonnegative) minimum at x=0." } { "Tag": [], "Problem": "Exista inel cu 2006 elemente si exact 10 elemente inversabile?", "Solution_1": "[quote=\"Cezar Lupu\"]Exista inel cu 2006 elemente si exact 10 elemente inversabile?[/quote]\r\n\r\nFie cele zece elemente inversabile $x_{1},x_{2},...,x_{10}$. Desigur unul dintre elemente este $1$ deci din cele 9 (inpar) elemente ramase una trebuie sa fie egal cu inversa sa. Fie asta $x\\ne 1$ Deci $x=x^{-1}$ deci $x^{2}=x$ rezulta $x(x-1)=0\\Rightarrow x=1$ imposibil", "Solution_2": "hm...\r\nnu cred ca daca $x=x^{-1}$ rezulta ca $x^{2}=x$... :blush:", "Solution_3": "da :) si in plus nu te-ai folosit deloc de ipoteza ca inelul are 2006 elemente, sau o fi doar o capcana :P ma mai gandesc :roll:", "Solution_4": "[quote=\"maky\"]hm...\nnu cred ca daca $x=x^{-1}$ rezulta ca $x^{2}=x$... :blush:[/quote]\r\n\r\nDa am observat dupa ce am trimis:) Rusinica:((", "Solution_5": "cred ca ce a vrut cckek sa spuna este ca daca din 9 elemente exista\r\nsigur 2 elemente care au acelasi invers(deoarece 9 impar) astfel ar rezulta\r\nca daca au acelasi invers sunt identice deci multimea n-ar mai fi de 9\r\nelemente ci defapt doar de 8 si impreuna cu elementul unitate n-ar mai\r\nface 10 ci doar 9." } { "Tag": [ "inequalities", "LaTeX", "inequalities proposed" ], "Problem": "Hi ! \r\nBelow you can find an inequality invented by me and I know it's really easy, but the problem [imaginary anyway] is to solve it in the most original way you can. Interesting ? I hope so :) Enjoy :D \r\n\r\na,b,c are non-negative reals:\r\n\r\n$a^3 + b^3 + c^3 \\geq \\sqrt{a^2c^2(b+a)(b+c)+b^2c^2(a+b)(a+c)}$\r\n\r\n.", "Solution_1": "is it SO EASY that nobody wanna try ? :maybe:", "Solution_2": "[quote=\"Megus\"]a,b,c are non-negative reals:\n\n$a^3 + b^3 + c^3 \\geq \\sqrt{a^2c^2(b+a)(b+c)+b^2c^2(a+b)(a+c)}$[/quote]\r\n\r\nIf this is really easy, please post your solution. I don't find it easy; it took me a week until I found ANY solution. Well, here is my solution. No doubt it is very original. Enjoy.\r\n\r\nWe have to prove that\r\n\r\n$a^{3}+b^{3}+c^{3}\\geq \\sqrt{a^{2}c^{2}\\left( b+a\\right) \\left( b+c\\right) +b^{2}c^{2}\\left( a+b\\right) \\left( a+c\\right) }$.\r\n\r\nIn other words, we must prove\r\n\r\n$\\left( a^{3}+b^{3}+c^{3}\\right) ^{2}\\geq a^{2}c^{2}\\left( b+a\\right) \\left( b+c\\right) +b^{2}c^{2}\\left( a+b\\right) \\left( a+c\\right) $.\r\n\r\nWe will solve a more symmetric formulation of this inequality:\r\n\r\n[i]If we denote $x=b^{2}c^{2}\\left( a+b\\right) \\left( a+c\\right)$, $y=c^{2}a^{2}\\left( b+c\\right) \\left( b+a\\right)$, $z=a^{2}b^{2}\\left( a+b\\right) \\left( a+c\\right)$, then prove that all three inequalities $\\left( a^{3}+b^{3}+c^{3}\\right) ^{2}\\geq y+z$, $\\left( a^{3}+b^{3}+c^{3}\\right) ^{2}\\geq z+x$ and $\\left( a^{3}+b^{3}+c^{3}\\right) ^{2}\\geq x+y$ hold.[/i]\r\n\r\nThis problem is symmetric in a, b and c; hence, we can WLOG assume $a\\leq b\\leq c$. Then, we can denote u = b - a and v = c - b, and get $u\\geq 0$ and $v\\geq 0$. On the other hand, since u = b - a and v = c - b, we have b = a + u and c = b + v; hence we can express b, c, x, y and z through a, u and v, and thus, after a short calculation,\r\n\r\n$\\left( a^{3}+b^{3}+c^{3}\\right) ^{2}-\\left( x+y\\right) $\r\n$=10a^{4}v^{2}+68au^{3}v^{2}+62au^{2}v^{3}+60a^{2}uv^{3}+91a^{2}u^{2}v^{2}$\r\n$+40a^{3}u^{2}v+52a^{3}uv^{2}+6a^{4}uv+40au^{4}v+6uv^{5}+15u^{2}v^{4}$\r\n$+6av^{5}+v^{6}+30auv^{4}+a^{4}\\left( a-v\\right) ^{2}+15a^{2}v^{4}+21u^{3}v^{3}$\r\n$+18u^{4}v^{2}+9u^{5}v+16au^{5}+32a^{2}u^{4}+13a^{4}u^{2}+30a^{3}u^{3}$\r\n$+20a^{3}v^{3}+3u^{6}+2a^{5}u+63a^{2}u^{3}v$,\r\n\r\nwhat is clearly $\\geq 0$. Hence, $\\left( a^{3}+b^{3}+c^{3}\\right) ^{2}\\geq x+y$. Now, since $a\\leq b\\leq c$, we have $c\\geq a$, so that $c^{2}\\left( b+c\\right) \\geq a^{2}\\left( b+a\\right) $, thus also $c^{2}a^{2}\\left( b+c\\right) \\left( b+a\\right) \\geq a^{2}b^{2}\\left( a+b\\right) \\left( a+c\\right) $, or, in other words, $y\\geq z$. Similarly, $x\\geq z$. Hence, the inequality $\\left( a^{3}+b^{3}+c^{3}\\right) ^{2}\\geq x+y$ implies both $\\left( a^{3}+b^{3}+c^{3}\\right) ^{2}\\geq z+x$ and $\\left( a^{3}+b^{3}+c^{3}\\right) ^{2}\\geq y+z$. And the proof is complete.\r\n\r\nThis solution shows that the inequality is very crude. Equality is only achieved for a = b = c = 0. Probably there are some sharper versions of this inequality. Most likely these sharper versions are much easier to prove.\r\n\r\nI am running out of time. Please post your elegant solutions.\r\n\r\n Darij", "Solution_3": "First case a+b < 2c then\r\n\r\na^3+b^3+c^3 \u2265 c\u221a(a+b)\u221a(c(a^2+b^2)+ 2abc) \u2265 c\u221a(a+b)\u221a(c(a^2+b^2)+ab(a+b))\r\n\r\nso a^3 +b^3+c^3 \u2265 c(a+b)\u221a(c(a+b)^2) = c\u221ac *(a+b)\u221a(a+b)\r\n\r\nUsing a^3 +b^3 \u2265 [(a+b)^3]/4\r\n\r\na^3 +b^3+c^3 \u2265 [(a+b)^3]/4 +c^3 \u2265 c\u221ac *(a+b)\u221a(a+b)\r\n(equality when a = b = c = 0 ).\r\n\r\nSecond case a+b>2c then inequality is equalent to :\r\n\r\n(a+b)^3 +c^3 \u2265 3abc(a+b)+ c\u221a(a+b)*\u221a(c(a^2+b^2)+ab(a+b)) or\r\n\r\n(a+b)^3 +c^3 \u2265 3abc(a+b)+ c\u221a(a+b)*\u221a(c(a+b)^2+ab(a+b-2c))\r\n\r\nWe can consider mixing variables or Sturm metod - pair (a,b) change with \r\n((a+b)/2,(a+b)/2) \r\nRight part will remain constant, left part increase, because a*b increases.\r\nSo we can say a=b and using our second case a>c.\r\n\r\nSo 2a^3 +c^3 \u2265 2ac\u221aa(a+c). \r\n\r\nSquaring : 4a^6 +c^6 \u2265 4(a^4)*(c^2) remembering that a > c - obliviously.\r\n(equality when a = b = c = 0 ).", "Solution_4": "Well, this problem is not too difficult. We can use mixing variables.\r\nLet $ f(a,b,c)=a^3+b^3+c^3-\\sqrt{a^2c^2(a+b)(b+c)+b^2c^2(a+b)(a+c)}$\r\n $ =a^3+b^3+c^3-c\\sqrt{(a+b)[ab(a+b)+c(a^2+b^2)]} $\r\nAnd it is easy to see that:\r\n$ f(a,b,c) \\geq f(\\sqrt{\\frac{a^2+b^2}{2}},\\sqrt{\\frac{a^2+b^2}{2}},c) $.\r\nNow,we just need prove for $ a=b $.That means:\r\n$ 2a^3+c^3 \\geq 2\\sqrt{a^3c^2(a+c)} $\r\nBut this is clear by Cauchy: \r\n$ 2a^3+c^3 \\geq 2\\sqrt{a^3c^2 \\cdot 2c} \\geq2\\sqrt{a^3c^2(a+c)}$ (if $a \\leq c $)\r\nor\r\n$ a^3+(a+c)(a^2-ac+c^2) \\geq 2\\sqrt{a^3(a+c)(a^2-ac+c^2)} \\geq 2\\sqrt{a^3c^2(a+c)} $.(if $ a \\geq c $). :)", "Solution_5": "Ok, here is my solution :\r\n$2(a^3 + b^3 + c^3) \\geq a^2(b+c) + b^2(a+c) + c^2(a+b)$\r\nbecause {3,0,0} majorizes {2,1,0} this inequality is true due to the Muirhead's Theorem, so we have [using Cauchy]:\r\n$a^3 + b^3 + c^3 \\geq \\frac{a^2(b+c) + b^2(a+c) + c^2(a+b)}{2} \\geq \\sqrt{(a^2(b+c)+b^2(a+c))c^2(a+b)} = \\sqrt{a^2c^2(a+b)(b+c) + b^2c^2(a+c)(a+b)}$\r\n\r\nDarij: What do you think of my solution ? :D \r\nprowler: use LateX - it's not so hard....\r\nAnh Cuong: nice solution :)", "Solution_6": "Muuuh. I knew you would have something like this... And, when one reads the solution, it looks really simple... :D :D :D\r\n\r\nJust a remark: Here -\r\n\r\n[quote=\"Megus\"]$2(a^3 + b^3 + c^3) \\geq a^2(b+c) + b^2(a+c) + c^2(a+b)$\nbecause {3,0,0} majorizes {2,1,0} this inequality is true due to the Muirhead's Theorem,[/quote]\r\n\r\nyou needn't even to apply Muirhead. Just remember that $b^3+c^3\\geq b^2c+c^2b$ (this is clear by rearrangement, or by a simple application of $b^3+c^3=\\left(b+c\\right)\\left(b^2-bc+c^2\\right)$), and similarly $c^3+a^3\\geq c^2a+a^2c$ and $a^3+b^3\\geq a^2b+b^2a$; adding these three inequalities together yields\r\n\r\n$\\left(b^3+c^3\\right)+\\left(c^3+a^3\\right)+\\left(a^3+b^3\\right)\\geq \\left(b^2c+c^2b\\right)+\\left(c^2a+a^2c\\right)+\\left(a^2b+b^2a\\right)$,\r\n\r\nor, in other words,\r\n\r\n$2\\left(a^3+b^3+c^3\\right)\\geq a^2\\left(b+c\\right)+b^2\\left(c+a\\right)+c^2\\left(a+b\\right)$,\r\n\r\nqed..\r\n\r\n Darij", "Solution_7": "Yeah, I needn't but it's really making life simpler :) I love majorization :P - ok, I'll try to invent a new problem [inequality of course :D ], today or tomorrow.", "Solution_8": "If $x,y,z$ are non-negative reals, then\r\n\r\n$z^{2}x^{2}(y+z)(y+x)+x^{2}y^{2}(z+x)(z+y)\\leq\\frac{173+49\\sqrt{7}}{972}(x^{2}+y^{2}+z^{2})^{3}$,\r\n\r\nwhere $\\frac{173+49\\sqrt{7}}{972}=0.31135989119564294129071\\cdots$ is the best possible.", "Solution_9": "Ji Chen, do you have a nice proof for it?", "Solution_10": "Megus, the proof is ugly as follows:\r\n\r\n$27(x^{2}+y^{2}+z^{2})^{3}-2(173-49\\sqrt{7}) [z^{2}x^{2}(y+z)(y+x)+x^{2}y^{2}(z+x)(z+y)]$\r\n\r\n$=\\frac{[y+z-(\\sqrt{7}-1)x]^{2}}{8}[12(4+\\sqrt{7})x^{4}+4(11+5\\sqrt{7})x^{3}(y+z)$\r\n\r\n$+2(59+17\\sqrt{7})x^{2}(y+z)^{2}+54(\\sqrt{7}-1)x(y+z)^{3}+27(y+z)^{4}]$\r\n\r\n$+\\frac{(y-z)^{2}}{8}\\big\\{\\frac{4x^{2}}{81}[81x-(173-49\\sqrt{7})(y+z)]^{2}+\\frac{2x^{2}}{81}[(25970\\sqrt{7}-45763)(y^{2}+z^{2})$\r\n\r\n$+10(5194\\sqrt{7}-11777)yz]+27(7y^{4}+8y^{3}z+18y^{2}z^{2}+8yz^{3}+7z^{4})\\big\\}\\geq0$,\r\n\r\nwiht equality if and only if $y=z=\\frac{(\\sqrt{7}-1)x}{2}$.\r\n[hide=\"Remark.\"]\n$(x^{2}+y^{2}+z^{2})^{3}\\leq3(x^{3}+y^{3}+z^{3})^{2}$.\n[/hide]" } { "Tag": [ "email", "MATHCOUNTS" ], "Problem": "when can you take the next PMWC qualifying test?", "Solution_1": "Mostly likely Nov, 2008. you may email them to ask, see this:\r\n\r\nhttp://www.txstate.edu/mathworks/contact.html", "Solution_2": "Erm... what is PMWC?\r\n\r\nSorry if this is considered as spam.", "Solution_3": "I don't really know the full form (and i'm too lazy to google it) but its an international mathematics competition for middle schoolers a wee bit harder than mathcounts", "Solution_4": "It's the Primary World Mathematics Competition.", "Solution_5": "Ernie, look here:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=215511" } { "Tag": [ "floor function" ], "Problem": "Jasmine wishes to purchase some trading cards. She has $ \\$7.50$ and the cards each cost $ \\$0.85$, tax included. What is the most number of cards she can buy?", "Solution_1": "$ \\lfloor\\frac{7.50}{0.85}\\rfloor\\equal{}8$\r\n\r\nanswer : 8" } { "Tag": [], "Problem": "Ben performed the following incorrect operations on a number. First he added -5 instead of subtraction -5. Then he multiplied his result by $ \\frac{1}{4}$ instead of dividing by $ \\frac{1}{4}$. Finally he squared the last result instead of taking the square root. Ben's final result was $ \\frac{225}{16}$. If Ben had performed the correct operations, what would the result have been?", "Solution_1": "We can work backwards to solve this problem. To get $ \\frac{225}{16}$, the last step Ben performed was to square. Therefore, we take the square root to get $ \\frac{15}{4}$. Since he multiplied by $ \\frac{1}{4}$, we'll divide $ \\frac{15}{4}$ by $ \\frac{1}{4}$ to get $ 15$. He added $ \\minus{}5$, so we'll subtract $ \\minus{}5$ to get $ 15 \\minus{} (\\minus{}5) \\equal{} 20$. The original number was then $ 20$.\r\n\r\nHowever, we want to know what the correct result should have been. First we subtract $ 5$ to get $ 15$. Then, we divide by $ \\frac{1}{4}$ to get $ 60$. Finally, we take the square root to get $ 2\\sqrt{15}$.", "Solution_2": "[quote=\"isabella2296\"]\nHowever, we want to know what the correct result should have been. First we subtract $ 5$ to get $ 15$. Then, we divide by $ \\frac {1}{4}$ to get $ 60$. Finally, we take the square root to get $ 2\\sqrt {15}$.[/quote]\r\n\r\nActually, I think that we are supposed to subtract $ \\minus{}5$ to get $ 25$ instead. Then, we divide by $ \\frac {1}{4}$ to get $ 100$. Taking the square root of $ 100$ we get $ \\boxed {10}$" } { "Tag": [], "Problem": "Find 2005 positive integers, not necessarily distinct, whose sum is equal to their product\r\n\r\n :starwars: i just think the starwors thing is cool :gleam:", "Solution_1": "[hide]try smaller cases:\n$2+2 = 2*2$\n$2+3+1 = 2*3*1$\n$2+4+1+1 = 2*4*1*1$\n$\\vdots$\n$2+n+(n-2)*1 = 2*n*1^{n-2}$\n$2+2005+1+\\dots + 1 = 2*2005*1*\\dots*1$ (2003 ones)[/hide]", "Solution_2": "very nice now try my challange problem in the olympiad section.\r\nThe title is very hard problem. :D", "Solution_3": "[hide=\"A description of all such sets of integers\"]\nI worked on this problem before. Any such set can be generated by taking an arbitrary set of numbers, $a_n$, and adding to this set a number of $1$s equal to $\\prod_{k=1}^{n} a_k - \\sum_{k=1}^{n} a_k$. \n[/hide]", "Solution_4": "[hide]1 1 1 ... 1 2 2005[/hide]", "Solution_5": "[hide=\"Does it help that\"]They are the roots of $x^{2005}-kx^{2004}+k=0$? :?[/hide]", "Solution_6": "[quote=\"t0rajir0u\"]$\\prod_{k=1}^{\\infty} a_k - \\sum_{k=1}^{\\infty} a_k$[/quote]\r\nTypo Error in $\\infty$ :D" } { "Tag": [ "limit", "logarithms", "calculus", "calculus computations" ], "Problem": "$ \\lim_{x\\to\\infty}(1\\plus{}\\dfrac{1}{x})^{2x}$\r\n\r\ngood luck!", "Solution_1": "Do you regard $ \\lim_{x\\to\\infty}\\left(1\\plus{}\\frac1x\\right)^x$ as famous enough that you should just know it at a glance?", "Solution_2": "This $ \\lim_{x \\to \\infty} x^{\\frac {1}{x}} \\equal{} 1$ can easily be shown by taking the log and then by taking the exponentiation of the result. \r\n\r\nWe know that the log is differentiable at 1, which gives us:\r\n\r\n$ \\lim_{h \\to 0} \\log \\left((1 \\plus{} h)^{\\frac {1}{h}} \\right) \\equal{} \\lim_{h \\to 0} \\frac {\\log \\left(1 \\plus{} h\\right)}{h} \\equal{} \\lim_{h \\to 0} \\frac {\\log \\left(1 \\plus{} h\\right) \\minus{} \\log 1}{h} \\equal{} 1$\r\n\r\nTaking the exponential gives: $ \\lim_{h \\to 0} \\left(1 \\plus{} h \\right)^{\\frac {1}{h}} \\equal{} e$. If you set $ h \\equal{} \\frac {1}{x}$, you get $ \\lim_{x \\to \\infty} \\left(1 \\plus{} \\frac {1}{x} \\right)^{x} \\equal{} e$.\r\n\r\nNow you have $ \\left(1 \\plus{} \\frac {1}{x} \\right)^{2x} \\equal{} \\left(1 \\plus{} \\frac {1}{x} \\right)^{x} \\cdot \\left(1 \\plus{} \\frac {1}{x} \\right)^{x}$", "Solution_3": "is how to do the same way that other:\r\n\r\n$ \\lim_{x\\to\\infty}(1\\plus{}\\dfrac{1}{x})^{x\\plus{}2}$", "Solution_4": "$ 2^{m\\plus{}n}\\equal{}2^{m}2^{n}$.", "Solution_5": "tha answer is $ e^2$\r\n\r\n$ \\lim_{x\\to\\infty}(1+\\frac{1}{x})^{x+2}$ for this answer is e because it is $ \\lim_{x\\to\\infty}(1+\\frac{1}{x})^{x}$ $ \\lim_{x\\to\\infty}(1+\\frac{1}{x})^{2} \n= e . 1 = e$", "Solution_6": "Right. My hint was vague?", "Solution_7": "$ \\lim_{x\\to\\infty}(1 + \\frac {1}{x})^{x + 2}$\r\n$ = \\lim_{x\\to\\infty}(1 + \\frac {1}{x})^{\\frac {x(x + 2)}{x}}$\r\n$ = \\lim_{x\\to\\infty} e^{\\frac {x+2}{x}}$\r\n\r\n$ = e$\r\n\r\n(Just for another way of doing it)" } { "Tag": [ "geometry" ], "Problem": "I wish to disguise myself for a few days with the use of possibly makeup, wigs, clothing, etc. Instead of a wig I might get a haircut, but I wish to be completely indistinguishable. I know this is random, but if there are any tips, it would be much appreciated. Thanks!", "Solution_1": "Hmm well, I think they say that the triangular region with vertices at the eyes and nostrils are the \"defining\" characteristics of a face, so you might want to do something with those areas.", "Solution_2": "[quote=\"CheesyP18\"]I wish to be completely indistinguishable.[/quote]\r\n\r\nIndistinuishable from what? Perhaps \"unrecognizable,\" instead?", "Solution_3": "[quote=\"JBL\"][quote=\"CheesyP18\"]I wish to be completely indistinguishable.[/quote]\n\nIndistinuishable from what? Perhaps \"unrecognizable,\" instead?[/quote]\r\n\r\nI apologize. Unrecognizable was the word I should have used.", "Solution_4": "Try new insoles in your shoes. It will give you a totally different gait, if you get something with insanely high or low arches (opposite of what you're used to). Have fun!", "Solution_5": "i dont know whether you are a girl or a guy, but girls will pay close attention to facial details (i.e eyelashes, eyebrows, makeup, hairline, as in pulled back or not) try a different body spray. if you are known for wearing abercrombie, wear american eagle. people, esp girls pay attention to that kind of thing. \r\n\r\nare colored contacts an option?\r\n\r\nnails? painted, long, bitten\r\n\r\nglasses?\r\n\r\nears (earrings or not, how many piercings)\r\n\r\nbirthmarks? (cover/ make using makeup)\r\n\r\nthis sounds fun. good luck!", "Solution_6": "[quote=\"eryaman\"]Hmm well, I think they say that the triangular region with vertices at the eyes and nostrils are the \"defining\" characteristics of a face, so you might want to do something with those areas.[/quote]\r\n\r\nCongratulations, you turned makeup into a geometry problem. Weird. . .", "Solution_7": "To answer star 99:\r\n\r\nI'm a guy who usually wears khaki pants or shorts, polo shirts or just random t-shirts that I find at Goodwill.\r\n\r\nI wear glasses, but I have another set of glasses that I'm probably going to use.\r\n\r\nI bite my nails, so I guess I should stop that.\r\n\r\nNo piercings, and all of my birthmarks are covered by clothes.\r\n\r\nI like the insoles idea, and I think it'll increase my height just a little.", "Solution_8": "Put a paper bag over your head. Nobody will recognize you. Or try the cardboard box trick.\r\n\r\nIf you can get your eyebrows going a different way, that would help.\r\nOf course, a haircut is mandatory.\r\nThe clothes you wear sound somewhat like mine...you could try jeans with plad, and that would drmatically change your appearance :D", "Solution_9": "I heard somewhere that concentrated sulphuric acid, if allowed to work upon your face, makes it nicely unrecognizable. But do not try that, you won't enjoy the experience.\r\n\r\nTry wearing a burkha. It should be effective.", "Solution_10": "Be a cross dresser!\r\n\r\nThe army got fooled with that trick.\r\n\r\nMake yourself look older. Like, say, 63.\r\n\r\nMake sure to have a different voice. Like Mrs. Doubtfire!", "Solution_11": "Why did the moderator move this to the Games & Fun Factory forum? I want serious replies, and I had more serious replies over at the Round Table. Could someone move this thread back over there?", "Solution_12": "long hair <--> short hair\r\nflat hair <--> spiked hair\r\n\r\nyou could dress like a girl...that would be much easier...so many more things to do...", "Solution_13": "[quote=\"star99\"]long hair <--> short hair\nflat hair <--> spiked hair\n\nyou could dress like a girl...that would be much easier...so many more things to do...[/quote]\r\n\r\nNot necessarily...\r\nThere are some pretty weird outfits you could wear out there...", "Solution_14": "[quote=\"CheesyP18\"]Why did the moderator move this to the Games & Fun Factory forum? I want serious replies, and I had more serious replies over at the Round Table. Could someone move this thread back over there?[/quote]\r\n\r\nThe moderator that forum must have thought G & FF was a better fit for this topic.", "Solution_15": "Relax cheesy. If you want serious replies, give us a low-down on the situation. We must know the circumstances under which you want to go around incognito before offerign any real help. And if you hide facts from your friends, it means you don't consider them your friends. Two kinds of people can give helpful advice on such things - professional experts and friends. We are not the former, you know that very well. We are willing to be the latter (I am, at least) but only if you are open enough. Oterwise, you are free to seek out experts. We won't mind.", "Solution_16": "Come on! Buy makeup, plaster, call your Uncle Fred and Aunt Jack, and be Mrs. Doubtfire! That's as good as it can get!", "Solution_17": "Of course I'm trying to hide a fact from a friend, but it has no malicious intent. I only need to be unnoticed for a few days, after which I can safely reveal myself. I was thinking insoles and a full wig/beard/mustache setup. I know where to get the insoles, but I'm having a problem locating a wig/beard/mustache mask.", "Solution_18": "[quote=\"star99\"]long hair <--> short hair\nflat hair <--> spiked hair\n\nyou could dress like a girl...that would be much easier...so many more things to do...[/quote]\r\n\r\ncough girlslockerroom cough", "Solution_19": "ewwwwww.\r\n\r\nlets not go there. sorry for suggesting that.\r\n\r\nbut schools out in most places.\r\n\r\n\r\newwwwwwww.\r\n\r\nthere's still the bathroom problem.", "Solution_20": "wow - cheesy has 42 posts exactly! What a lucky boy!\r\nBut you can dress like a girl (unless you have huge beards or bushy moustaches - in which case you have to shave these off). As I suggested in the first post, burkha isgenerally the champion's choice in these cases." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "If $\\alpha$ and $\\beta$ are two irrational numbers and $A =${$[n \\alpha ] : n \\in N$}, $B =${$[n \\beta ] : n \\in N$}. Find the condition that $A$ and $B$ form a partition of $N$.\r\n$[x]$ represents the greatest integer less than or equal to $x$", "Solution_1": "It's $\\frac 1\\alpha+\\frac 1\\beta=1$. Try searching the forum (and the web, in general) for \"Beatty sequence\". It's been discussed here a number of times.\r\n\r\nP.S.\r\n\r\nClearly, this is not an \"open question\", and it's hard for me to believe you thought it was unsolved :). If you don't have a solution to such a problem, you should post it in the \"Unsolved Problems\" section, and if you do have a solution and you're posting it as a challenge, it should go into the \"Proposed and Own Problems\" section." } { "Tag": [ "Euler", "limit", "logarithms", "algebra", "polynomial" ], "Problem": "$1/2+1/3+1/5+1/7+1/11+1/13+1/17+1/19+1/23+1/29+\\cdots=$\r\n\r\nthe denominator of each term is prime number", "Solution_1": "Well-known. In fact, Euler showed that it diverges.\r\n\r\n[hide=\"Demonstration\"] We make use of the Euler product\n\n$\\lim_{n \\to \\infty}\\sum_{k=1}^{n}\\frac{1}{k}= \\lim_{n \\to \\infty}\\prod_{p \\in \\mathbb{P}, p \\le n}\\frac{1}{1-\\frac{1}{p}}\\propto \\ln n$\n\nWe know that the LHS diverges. Taking natural logs,\n\n$\\lim_{n \\to \\infty}\\sum_{p \\in \\mathbb{P}, p \\le n}-\\ln (1-\\frac{1}{p}) \\propto \\ln \\ln n$\n\nNote that the Taylor polynomial for $-\\ln (1-\\frac{1}{p})$ begins with a term $\\frac{1}{p}$, and that subsequent terms are negligible because their sum converges. Hence\n\n$\\lim_{n \\to \\infty}\\sum_{p \\in \\mathbb{P}, p \\le n}\\frac{1}{p}\\propto \\ln \\ln n = \\infty$ [/hide]" } { "Tag": [ "Gauss", "number theory unsolved", "number theory" ], "Problem": "Sequence $(a_n)$ is given such that \r\n $2^n=\\sum_{d|n}{a_d}$ for all $n\\geq 1$.\r\nProve that $a_n$ divises by $n$. :lol:", "Solution_1": "I think if $m$ divise $n$ then $a_m$ divise $a_n$ :P", "Solution_2": "By Mobius inversion formula we have \\[ a_n=\\sum_{d|n}2^d\\mu\\left(\\frac{n}{d}\\right) \\] and the result follows from Gauss' theorem.", "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?t=65816", "Solution_4": "it is posted :lol: ,thank you Zetax" } { "Tag": [], "Problem": "I know that if two proportions are equal the following equality holds:\r\n\r\n$ \\frac{a}{b}\\equal{}\\frac{c}{d}\\equal{}\\frac{a\\plus{}kc}{b\\plus{}kd}$\r\n\r\nCan you do the same if you are trying to compare two expressions?\r\n\r\nex) $ \\frac{a\\plus{}c}{b\\plus{}d}?\\frac{n\\plus{}c}{m\\plus{}d}?\\frac{a\\minus{}n}{b\\minus{}m}$\r\n\r\nThank you for your help.", "Solution_1": "im not really sure what u mean. could u give a numerical example?", "Solution_2": "Yes, in the following sense.\r\n\r\n[b]Lemma:[/b] For [b]positive[/b] $ a, b, c, d$, if $ \\frac{a}{b}>\\frac{c}{d}$ then $ \\frac{a}{b}>\\frac{a\\plus{}c}{b\\plus{}d}>\\frac{c}{d}$\r\n\r\nNote the strict inequalities." } { "Tag": [], "Problem": "I've heard Waterloo's Problems Problems Problems series only contain problems and solutions but no actual texts. Can you still learn some good techniques just from reading the solutions? Or would it be better to get some texts like the Art of Problem Solving?", "Solution_1": "I have a copy of one of the books and I think the solution is understandable. Although I have no copy of Art of Problem Solving but I know that it is one of the best books ever written when it comes to problem solving. I wish I could have a copy of it.", "Solution_2": "It's a good thing to use to study for FGH, PCF or Euclid." } { "Tag": [ "geometry", "3D geometry", "frustum" ], "Problem": "What are frustums, and can they vary in size and shape? \r\n\r\nI posted this to have a post to look for to learn about frustums and their formulas. If anyone knows any frustum formulas, hopefully you will list them. :)", "Solution_1": "Start from\r\n\r\n[url]http://en.wikipedia.org/wiki/Frustum[/url]\r\n[url]http://www.mathwords.com/f/frustum.htm[/url]\r\n\r\nand for advanced reading go to\r\n\r\n[url]http://mathworld.wolfram.com/PyramidalFrustum.html[/url]\r\n[url]http://mathworld.wolfram.com/ConicalFrustum.html[/url]", "Solution_2": "frustums are just cones with the top cut off. you can calculate the volume or surface area by just subtracting the big cone volume by the smaller cone volume and etc.", "Solution_3": "yep and u just subtract yea i looked on wiki, wiki rocks!\r\n\r\nlol frustums ./.... I just think of them as yoplait yogurt containers :roll:" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For $a,b,c$ sides of a triangle, prove $\\sum\\sqrt{ab+(a-c)(b-c)}\\ge \\sum\\sqrt{ab}\\ge\\sum \\sqrt{ab-(a-c)(b-c)}$ :D", "Solution_1": "Nobody? [hide=\"Hint\"] \nAfter the standard substitution $a=x+y$ etc. and renaming $x,y,z$ by $a,b,c$ the inequality chain writes with the same notation as in http://www.mathlinks.ro/Forum/viewtopic.php?t=67457 \nsimply as the $n=\\frac12$ case of $\\sum S^n\\ge\\sum\\left(\\frac{S+T}2\\right)^n\\ge\\sum T^n$. (See now how I came up with this Vornicu-Schur variation :) ? And the door is wide open for composing more variations :) :) )\nFrom here it should be clear how to proceed, at least for the right inequality which can in fact easily be shown to be true for all $n\\ge0$. Note that the left one is false for $00 then\r\n3(a/b+b/c+c/a) \\geq 9+ \\sum ab(a-b) 2 /(a+c)(b+c)\r\nsorry about my bad English..", "Solution_1": "i don't think that the inequality is good\r\ntake a=2b=2c and b large\r\nyou will have\r\n3(2+1+1/2)\\geq 9+2b^2/3 which is not true for b very large" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let $ x$ , $ y$ , $ z$ $ \\in ]0,\\plus{}\\infty[$ , prove that :\r\n\r\n$ \\frac{x\\plus{}y}{x^2\\plus{}xy\\plus{}y^2}$ $ \\plus{}$ $ \\frac{y\\plus{}z}{y^2\\plus{}yz\\plus{}z^2}$ $ \\plus{}$ $ \\frac{z\\plus{}x}{z^2\\plus{}zx\\plus{}x^2}$ $ \\geq \\frac{1}{x\\plus{}y\\plus{}z}$\r\n\r\ncreated by yassine", "Solution_1": "$ \\sum {\\frac{x+y}{x^2+xy+y^2}}\\geq\\sum {\\frac{1}{x+y}\\geq\\frac{9}{2(x+y+z)}>\\frac{1}{x+y+z}}$", "Solution_2": "thank's Marius Mainea\r\n\r\nnice prof even your result is the best" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Souce: USSR Olympiad\r\n165. Factor the following expression:\r\n$a^{10}+a^5+1$.", "Solution_1": "[hide]Let $w$ be a cubic root of unity. Then $w^3 = 1$ and $w^2 + w + 1 = 0$. So we have $w^3 \\cdot w^3 \\cdot w + w^3 \\cdot w^2 + 1 = w^2 + w + 1 = 0$. Hence $w$ and $w^2$ are roots of the polynomial. So $(a - w)(a - w^2) = a^2 + a + 1$ is a factor. Then use division of polynomials to get the others.[/hide]" } { "Tag": [ "ratio" ], "Problem": "When fitting \"ten-speed\" gears to a bicycle, you choose two front sprockets, which must have different numbers of teeth between 40 and 62 (inclusive) and five back sprockets, which must have different numbers of teeth between 11 and 35 (inclusive). The gear ratios available to you are then obtained by dividing the number of front sprocket teeth by the number of back sprocket teeth. For example. if you choose 44 and 60 tooth front sprockets and 14, 18, 22, 26, and 30 tooth back sprockets, your ratios are\r\n\r\n44/14\r\n44/18\r\n44/22\r\n44/26\r\n44/30\r\n60/14\r\n60/18\r\n60/22\r\n60/26\r\n60/30\r\n\r\nIf you choose badly, some of the ratiuos will be the same (for example, 44/22 = 60/30 above) and you will not obtain ten different ratios. Supposing you choose as badly as possible, what is the smallest number of ratios you wil end up with?", "Solution_1": "6? \r\nMy numbers are: 48,60 and 12,15,16,20,25" } { "Tag": [ "probability" ], "Problem": "out of 16 people who book a flight 1 do not turn up. What is the probability that in a 48-seat plane, if 50 people book for that flight, that there will not be enough seats?", "Solution_1": "[hide]\nLet $p$ be the probability that a person does not turn up. $p = \\frac{1}{16}$.\n\nSo if $49$ people show up, the probability of that is $\\binom{50}{49}(1-p)^{49}p$.\nAnd if $50$ people do, the probability is $(1-p)^{50}$.\n\nThen add them up, and I don't want to do that.[/hide]" } { "Tag": [ "AwesomeMath", "summer program" ], "Problem": "I'm doing test A, but I don't think I'm going to apply for early decision. Should I still include the 20 bucks grading fee when I send my solutions in? or do I include that later with the application?", "Solution_1": "I would also like to know the answer to this question.", "Solution_2": "Just send it with the rest of the application (to avoid having to write multiple checks.) We usually don't review your application until it's mostly complete, anyway.\r\n\r\nGood question!", "Solution_3": "I took test a but plan on also taking test b (deadline problem, see my post in \"Acceptance\" thread). Should I pay a grading fee for both tests, or just for one?\r\n\r\nIronically, this forum is a much better place to go for information pertaining to awesomemath than is its own forum (especially since that forum is now locked).", "Solution_4": "[quote=\"roadnottaken\"]I took test a but plan on also taking test b (deadline problem, see my post in \"Acceptance\" thread). Should I pay a grading fee for both tests, or just for one?\n\nIronically, this forum is a much better place to go for information pertaining to awesomemath than is its own forum (especially since that forum is now locked).[/quote]\r\nI think it's 20 bucks if you do multiple admin tests.", "Solution_5": "Do you mean in addition to the $\\$50$?", "Solution_6": "[quote=\"roadnottaken\"]Do you mean in addition to the $\\$50$?[/quote]\nI think so.\n[quote]There is an application fee of \\$50 (\\$30 for processing the application materials, and \\$20 for grading the admission test). If a student wants to take a second test (if time allows and spots are still available), an additional\\$20 will be charged.[/quote]\r\nIt's on the bottom of [url]http://www.awesomemath.org/application.shtml[/url]" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "1. how can we construct a sequence whose set of limit points is exactly the set of integers?\r\n\r\n2. can there exist a sequence whose set of limit points is exactly 1, 1/2, 1/3, ...?", "Solution_1": "The set of limit points must be closed (why?), so the answer to (2) is no.\r\n\r\nSuppose $(a_{n})$ is a sequence converging to $0$. Then the sequence $a_{1},a_{1}+1,a_{2},a_{2}+1,a_{3},a_{3}+1,\\dots$ has limit points at $0$ and $1$. This generalizes to any finite set of limit points, but you want an infinite one. For this, consider the sequence of sequences\r\n$a_{1},a_{2},a_{3},\\dots$\r\n$a_{1}+1,a_{2}+1,a_{3}+1,\\dots$\r\n$a_{1}+2,a_{2}+2,a_{3}+2,\\dots$\r\n$\\dots$\r\nand arrange all of these numbers into a single sequence." } { "Tag": [ "modular arithmetic", "quadratics", "number theory", "prime factorization" ], "Problem": "$ 2^{x} \\plus{} 3^{y} \\equal{} Z^{2}$\r\n\r\nSolve for positive integers X,Y,Z.", "Solution_1": "[hide=\"Hint\"]\nTreat the two cases x = 2k, x=2k+1 separately.\n[/hide]", "Solution_2": "Case $ x \\equal{} 2k \\plus{} 1$\r\n[hide]\nWe have $ x^{2k \\plus{} 1} \\plus{} 3^y \\equal{} z^2 \\implies 2 \\cdot 4^k \\plus{} 3^y \\equal{} z^2$. Examining this mod 6, we have $ 2 \\cdot 4^k \\equiv 2 \\pmod 6$ and $ 3^y \\equiv 3 \\pmod 6$, whic means that $ z^2 \\equiv 5 \\pmod 6$. However, 5 is not a quadratic residue mod 6, so there is no solution for $ x \\equal{} 2k \\plus{} 1$.\n[/hide]\n\nCase $ x \\equal{} 2k$:\n[hide=\"Progress\"]\nWe have $ 2^{2k} \\plus{} 3^y \\equal{} z^2 \\implies z^2 \\minus{} (2^k)^2 \\equal{} 3^y \\implies (z \\minus{} 2^k)(z \\plus{} 2^k) \\equal{} 3^y$\n\nThus since $ 3^y$ is always odd and positive if $ y$ is an integer, this implies that $ z$ must be odd and greater than $ 2^k$ (so that the $ (z \\minus{} 2^k)$ term does not become negative). Also, since the product of $ (z \\minus{} 2^k)$ and $ (z \\plus{} 2^k)$ yields a number with only 3's in its prime factorization, the terms also must have only 3's in its prime factorization.\n\nQuick brute-forcing yields $ 2^4 \\plus{} 3^2 \\equal{} 5^2$ is a solution. Working...\n[/hide]", "Solution_3": "[quote=\"Yongyi781\"]\nWe have $ x^{2k \\plus{} 1} \\plus{} 3^y \\equal{} z^2 \\implies 2 \\cdot 4^k \\plus{} 3^y \\equal{} z^2$. Examining this mod 6, we have $ 2 \\cdot 4^k \\equiv 2 \\pmod 6$ and $ 3^y \\equiv 3 \\pmod 6$, whic means that $ z^2 \\equiv 5 \\pmod 6$. However, 5 is not a quadratic residue mod 6, so there is no solution for $ x \\equal{} 2k \\plus{} 1$.\n[/quote]\r\n\r\nalternatively, for case $ x \\equal{} 2k \\plus{} 1$\r\n\r\n$ 2^x \\equiv 2\\mod3$\r\n\r\n$ 3^y \\equiv 0\\mod3$\r\n\r\n$ 2\\mod3$ isn't a square residue (:", "Solution_4": "Bah, I have no idea why I chose mod 6 :P", "Solution_5": "To continue from where Yongyi left off...\r\n\r\n[hide=\"Hint\"]Basically you have reduced the problem to $ 3^{a}\\plus{}2^{k\\plus{}1}\\equal{}3^{b}$. But then $ a\\equal{}0$, or else we have two terms divisible by 3 and one term that isn't. Now we have $ 3^{b}\\minus{}2^{k\\plus{}1}\\equal{}1$.[/hide]", "Solution_6": "[hide]\nFirstly, since $ z^2 \\not \\equiv 2 \\pmod{3}$, we have $ 2^x\\equiv 1\\pmod{3}$, so $ x\\equiv 0\\pmod{2}$.\nNow, since $ 2^x$ is at least 4, and $ z^2 \\equiv 1 \\pmod{4}$, we have $ 3^y\\equiv 1\\pmod{4}$, so $ y\\equiv0\\pmod{2}$ as well.\nNow, we can write $ x \\equal{} 2x_1$ and $ y \\equal{} 2y_1$ for some positive integers $ x_1$,$ y_1$.\nThus, the original equation becomes $ (2^{x_1})^2 \\plus{} (3^{y_1})^2 \\equal{} z^2$.\nIt is well known that the solutions to $ a^2 \\plus{} b^2 \\equal{} c^2$ are $ a \\equal{} 2rs$, $ b \\equal{} r^2 \\minus{} s^2$, $ c \\equal{} r^2 \\plus{} s^2$, for arbitrary integers $ r, s$.\nSince $ 3^{y_1}$ is always odd, $ a \\equal{} 2^{x_1}$ and $ b \\equal{} 3^{y_1}$, so we have:\n$ 2^{x_1} \\equal{} 2rs$, $ 3^{y_1} \\equal{} r^2 \\minus{} s^2$, $ z \\equal{} r^2 \\plus{} s^2$.\nFrom the first equation, $ r, s$ are both powers of 2. Then, since $ 3^{y_1}$ is odd, $ s \\equal{} 1$.\nSo $ 2^{x_1} \\equal{} 2r$ and $ 3^{y_1} \\equal{} r^2 \\minus{} 1 \\equal{} (r \\plus{} 1)(r \\minus{} 1)$, which means that both $ r \\plus{} 1, r \\minus{} 1$ are powers of 3, so $ r \\equal{} 2$, since r is also a power of 2.\nFrom this, we see that the only solution is $ 2^{x_1} \\equal{} 4$,$ 3^{y_1} \\equal{} 3$, $ z \\equal{} 2^2 \\plus{} 1 \\equal{} 5$, so $ (x,y,z) \\equal{} \\boxed{(4,2,5)}$.[/hide]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Determine all the integers $n >1$ such that $\\sum_{i=1}^{n}x_i^2\\ge x_n\\cdot \\sum_{i=1}^{n-1}x_i$ for all real numbers $x_1,x_2,\\cdot \\cdot \\cdot ,x_n$.", "Solution_1": "Terribly sorry! this inequality is so stupid. (I have just solve it :mad: )" } { "Tag": [ "function", "geometry", "complex analysis" ], "Problem": "When looking at boundary behaviour of analytic functions defined on $ \\mathbb{D} \\equal{} \\{z \\in \\mathbb{C} : |z| < 1\\}$, it turns out that it's very giving to look at so called non-tangential limits at the boundary. We say that a sequence $ (z_n)$ in $ \\mathbb{D}$ converges nontangentially to $ 1$ if there is an angle with vertex at $ 1$, symmetric with respect to $ [0,1)$ and with opening less than $ \\pi$ such that $ z_n$ belongs to this angle for all\u00a0$ n$ large. (Of course, we also require the sequence to converge in the normal fashion).\r\n\r\nAn elementary characterization is the following. $ z_n \\equal{} r_ne^{i\\theta_n}$ converges nontangentially to $ 1$ if and only if $ \\limsup_{n\\to\\infty} \\frac {|\\theta_n|}{1 \\minus{} r_n} < \\infty$. This notation simply means that the sequence is bounded.\r\n\r\nNow, I have shown this with an extremely messy brute-force calculation. However, it is easy to see that the condition is equivalent to the alternative condition that $ \\limsup_{n\\to\\infty} \\frac {|1 \\minus{} z_n|}{1 \\minus{} |z_n|} < \\infty$. During the lecture my professor commented that this condition is equivalent to the following statement, which should follow from elementary geometry:\r\n(*) In an angle of the above type, with a fixed opening, there is a $ C > 0$ such that $ |z \\minus{} 1| \\leq C(1 \\minus{} |z|)$ for all $ z$ in the angle.\r\n\r\nNow I'm certainly not familiar with elementary geometry, so I don't see why this follows from geometric considerations, but I know you guys are. So I ask if you can tell me why (*) holds? Basically, I can show that it holds by writing $ z \\equal{} re^{it}$ as $ z \\equal{} 1 \\plus{} \\rho e^{i\\theta}$ for suitable $ \\rho$ and $ \\theta$ so that $ z$ is in the angle, and then just brute force it to show the inequality. What I am looking for is a more succinct argument.\r\n\r\nTo mods: I thought this topic was fun to put here, since I really might get elementary solutions. If you feel that it's inappropriate, move it to complex analysis.", "Solution_1": "[quote=\"Kalle\"]analytic functions... converges nontangentially... During the lecture my professor... If you feel that it's inappropriate, move it to complex analysis.[/quote]Yea, you might get better responses if this is moved to Complex Analysis forum.", "Solution_2": "I guess I might have added too much background to the problem, making it look scary. It's really not an advanced problem, and I think you could solve it. Anyway, I realized earlier that basically the thing I ask for is more or less obvious if we look at an angle against a straight line instead of a circle. So if we just approximate the circle with its tangent it's quite clear." } { "Tag": [], "Problem": "Did anyone here take the physics B exam. How would you rate it? I think the free response was ok and the multiple choice was pretty hard.", "Solution_1": "I'm taking it next year. What review book or resource do you think was/would be the best for Physics B? Just want to know before I take the class.", "Solution_2": "[quote=\"hello\"]Did anyone here take the physics B exam. How would you rate it? I think the free response was ok and the multiple choice was pretty hard.[/quote]\r\nI took it last year ('03), after learning the curriculum from a review book over the course of a month. My algebra is strong, but I'd assume that if you're here, so is yours. I got a 5 without much difficulty, and I didn't know a few of the topics.", "Solution_3": "I felt the same about the multiple choice and the free response. BTW-I already had a thread on this..." } { "Tag": [ "logarithms", "floor function", "number theory proposed", "number theory" ], "Problem": "Find the first digit of $ 2^{81}$.", "Solution_1": "$ 2^0 \\equal{} 1$\r\n$ 2^1 \\equal{} 2$\r\n$ 2^2 \\equal{} 4$\r\n$ 2^3 \\equal{} 8$\r\n$ 2^4 \\equal{} 16$\r\n$ 2^5 \\equal{} 32$\r\n$ 2^6 \\equal{} 64$\r\n$ 2^7 \\equal{} 128$\r\n$ 2^8 \\equal{} 256$\r\n$ 2^9 \\equal{} 512$\r\n$ 2^{10} \\equal{} 1024$\r\n$ 2^{11} \\equal{} 2048$\r\n$ 2^{12} \\equal{} 4096$\r\n$ 2^{13} \\equal{} 8192$\r\n$ 2^{14} \\equal{} 16384$\r\n$ 2^{15} \\equal{} 32768$\r\n$ 2^{16} \\equal{} 65536$\r\n$ 2^{17} \\equal{} 131072$\r\n$ 2^{18} \\equal{} \\text{You guessed it, starts with a }2$\r\n\r\netc....first digit of $ 2^{81} \\equal{} 2$", "Solution_2": "To do this by hand...\r\n\r\nI remember, from all of those years ago when I used a base-10 log table to compute with, that $ \\log 2\\approx.30103.$ (It's an easy number to remember.)\r\n\r\nHence, $ \\log(2^{81})\\approx 81\\cdot.30103\\approx24.38$ (that much arithmetic is easy to do by hand). But if $ \\log2\\approx.3,$ then $ \\log2.5\\equal{}\\log10\\minus{}2\\log 2\\approx .4.$ Hence, $ 2\\times 10^{24}<2^{81}<2.5\\times 10^{24}.$", "Solution_3": "[quote=\"Euclid's Turtle\"]$ 2^0 \\equal{} 1$\n$ 2^1 \\equal{} 2$\n$ 2^2 \\equal{} 4$\n$ 2^3 \\equal{} 8$\n$ 2^4 \\equal{} 16$\n$ 2^5 \\equal{} 32$\n$ 2^6 \\equal{} 64$\n$ 2^7 \\equal{} 128$\n$ 2^8 \\equal{} 256$\n$ 2^9 \\equal{} 512$\n$ 2^{10} \\equal{} 1024$\n$ 2^{11} \\equal{} 2048$\n$ 2^{12} \\equal{} 4096$\n$ 2^{13} \\equal{} 8192$\n$ 2^{14} \\equal{} 16384$\n$ 2^{15} \\equal{} 32768$\n$ 2^{16} \\equal{} 65536$\n$ 2^{17} \\equal{} 131072$\n$ 2^{18} \\equal{} \\text{You guessed it, starts with a }2$\n\netc....first digit of $ 2^{81} \\equal{} 2$[/quote]Cool. I didn't know that the first digit of $ 2^n$ is periodic. However, you still need to prove it, right?", "Solution_4": "Bulgarian TST for IMO 2008.\r\nFor every natural number $ n$ we mark the first digit of the number $ 2^{n}$ with $ a_{n}$. Is the number $ 0,a_{1}a_{2}a_{3}\\cdots$ rational?", "Solution_5": "[quote=\"Johan Gunardi\"]Cool. I didn't know that the first digit of $ 2^n$ is periodic. However, you still need to prove it, right?[/quote]\r\nIt simply isn't.", "Solution_6": "[quote=\"Johan Gunardi\"]Find the first digit of $ 2^{81}$.[/quote]\r\n\r\nthis the simple way of solving this kind of question :\r\n\r\nwe have to find $ 0 < k < 10$ knowing that $ 2^{81} \\equal{} k[10]$.\r\n\r\n$ 2^3 \\equal{} 8 \\equal{} \\minus{} 2[10]$==> $ 2^9 \\equal{} 2[10]$ so $ 2^{81} \\equal{} 2[10]$ :mad:", "Solution_7": "$ 2^81\\equal{}2417851639229258349412352$", "Solution_8": "[quote=\"Ibn Rochde\"]\nthis the simple way of solving this kind of question :\n\nwe have to find $ 0 < k < 10$ knowing that $ 2^{81} \\equal{} k[10]$.\n\n$ 2^3 \\equal{} 8 \\equal{} \\minus{} 2[10]$==> $ 2^9 \\equal{} 2[10]$ so $ 2^{81} \\equal{} 2[10]$ :mad:[/quote]\r\n\r\nCan someone explain this solution?", "Solution_9": "That solution is wrong: it finds the _last_ digit instead of the _first_ digit.", "Solution_10": "Could you explain what he was doing? I don't understand the $ []$ at all, unless he was implying that meant modular? :S\r\n\r\nAlso, the left-most digits are not periodic for $ 2^n$ ? Mhmm, is there a proof for this? They look it up to whatever exponent I used. >_<", "Solution_11": "The $ [n]$ are some (rather unused and old) way to express $ \\mod n$.\r\n\r\n\r\nAssume the leading digit of $ 2^n$ is periodic with period $ m$. This means that all the elements of the sequence $ a_k \\equal{} \\left( (k \\cdot m \\cdot \\log_{10}(2)) \\mod 1 \\right)$ are in the interval $ I \\equal{} [0,\\log_{10}(2)[$ (here I used $ \\mod 1$ to express fractional parts).\r\nBut for any irrational $ \\alpha$, the numbers $ (\\alpha \\cdot k) \\mod 1$ are dense in $ [0,1]$, thus not restricted to $ I$. As $ m \\cdot log_{10}(2)$ is irrational, we reached a contradiction.", "Solution_12": "On a more intuitive level, you'll notice that the second digit is increasing between \"periods\" (compare $ 64$ and $ 65536$). You can verify directly by calculator that if you repeat this calculation the second digit is going to \"overflow.\"", "Solution_13": "[quote=\"ZetaX\"]\n\nAssume the leading digit of $ 2^n$ is periodic with period $ m$. This means that all the elements of the sequence $ a_k \\equal{} \\left( (k \\cdot m \\cdot \\log_{10}(2)) \\mod 1 \\right)$ are in the interval $ I \\equal{} [0,\\log_{10}(2)[$ (here I used $ \\mod 1$ to express fractional parts).\nBut for any irrational $ \\alpha$, the numbers $ (\\alpha \\cdot k) \\mod 1$ are dense in $ [0,1]$, thus not restricted to $ I$. As $ m \\cdot log_{10}(2)$ is irrational, we reached a contradiction.[/quote]\r\nCould someone explain this in more detail? I do not doubt the proof at all, I tried looking it up and finding various methods, all awfully similar but I don't understand why the whole $ \\log_{10}{x}$ and what does $ k$ mean? And why is $ m$ multiplying with the $ \\log$?\r\netc... THanks a plenty :)", "Solution_14": "The leading digit of a number $ n$ is determined by the fractional part of $ \\log_{10} n$. The integer part tells you how many digits there are; once you ignore that, you're left with information about the rest of the number. If the leading digit of $ 2^n$ has period $ m$, then the sequence of numbers\r\n\r\n$ A_k \\equal{} 2^{km}$\r\n\r\nall have the same leading digit (which, by plugging in $ k \\equal{} 0$, is $ 1$), which means that the sequence of fractional parts\r\n\r\n$ a_k \\equal{} \\left( \\log_{10} a_k \\right) \\bmod 1 \\equal{} (km \\log_{10} 2) \\bmod 1$\r\n\r\nall tell you the same leading digit. Specifically, the relevant calculation is that this sequence is in the interval ZetaX gave. ZetaX then cites a lemma that tells us this is impossible.", "Solution_15": "Mhmm sorry to bring this up again :P but could you explain this some more?\r\n\r\nI grabbed a calculator and tried $ \\log_{10}{49985275}$. This has 8 digits. The number on the calculator reads $ 7.698824086...$. I tried a few other examples and it seems the integer part gives me the number of digits $ \\minus{} 1$. To relate it to my example, I got a $ 7$ for the integer part but the number really has $ 8$ digits. Do they differ by one for all numbers? I tried 5 and they all seemed to work this way.\r\n\r\nNow the fractional part, $ 0.698824086...$ and the leading digit $ 4$... I can't see the relation. Even with the few other cases I tried. :(. \r\nThe rest of your explanation helped significantly though. Thanks very much, to ZetaX as well :)\r\n\r\nP.S. -> Does the lemma that he cited have a specific name? So I may study these sort of questions more.\r\n\r\nAgain, thanks :)", "Solution_16": "You are correct. It is the number of digits minus one. If a number $ a$ has $ n$ digits, then\r\n\r\n$ 10^{n \\minus{} 1}\\le a < 10^{n}$\r\n\r\n(Because $ 10^{n \\minus{} 1}$ is the first $ n$ digit number and $ 10^{n}$ is the first $ n \\plus{} 1$ digit number). Taking the log,\r\n\r\n$ \\log(10^{n \\minus{} 1})\\le \\log a < \\log(10^{n})$\r\n$ n \\minus{} 1\\le \\log a < n$\r\n\r\nFor your second question, you say the fractional part is $ .698824086$. Well $ \\log 4\\approx .6002$ and $ \\log 5\\approx .6989$, so\r\n\r\n$ \\log 4\\le 0.698824086 < \\log 5$.\r\n\r\nAs for the name of that lemma, I don't know.", "Solution_17": "The lemma goes by the name of the [url=http://en.wikipedia.org/wiki/Equidistribution_theorem]equidistribution theorem[/url].", "Solution_18": "Mhmm, my issue with that $ \\log$ stuff is unless one has them memorized, a calculator is needed >_<.\r\n\r\nBut aside ffrom thank you guys very much :) I am such an amateur but I shan't give up xD\r\n\r\nAlso, for the fractional part stuff. Is it safe to say... :\r\n\r\nlet $ n \\equal{} n_1n_2...n_n.0$ (0s just for simplicity)\r\nThen $ n_1 \\equal{} x$ where $ \\log_{10}{x} \\leq \\log_{10}{n } [\\mod 1]$. Here I am borrowing ZetaX's notation for the fractional part (ie. $ \\mod 1$) and $ x$ is the greatest such integer. That is, the greatest integer $ x$ such that $ \\log_{10}{x} \\leq \\log_{10}{n} [\\mod 1]$\r\n\r\nHopefully I didn't confuse myself and get screwed up. My apologies if I did. :P", "Solution_19": "[hide=\"my solution\"]\n\n$ 2^{80}\\equal{}1024^8>1000^8\\equal{}10^{24}$\n\n$ \\frac{2^{80}}{1000^8}\\equal{}(\\frac{1024}{1000})^8<(\\frac{41}{40})^8<\\frac{41}{40}\\cdot\\frac{40}{39}...\\frac{34}{33}\\equal{}\\frac{41}{33}<1,3$\n\n$ 2^{80}<1,3\\times10^{24}$\n\nso $ 10^{24}<2^{80}<1,3\\times10^{24}$, which implies that $ 2^{80}$ has 1 as the first digit and the second digit is less than 3. so $ 2^{81}\\equal{}2\\times2^{80}$ begins with $ 2$.[/hide]", "Solution_20": "This problem has got a large number of posts for such a specific, and non-generalised problem.\r\n\r\nInstead...anyone got a solution to the following generalisation?\r\n\r\nWhat is the first digit of $ a^b$ where $ a$ and $ b$ are natural numbers base $ n$? Giving your answer base $ n$.", "Solution_21": "This is an example of a question where a sufficiently detailed specific solution already contains all of the ingredients of the general solution; that is, ZetaX's discussion tells you immediately that the leading digit is the unique positive integer $ d$ such that $ \\log_n d < \\{ b \\log_n a \\} < \\log_n (d \\plus{} 1)$, where $ \\{ x \\}$ denotes the fractional part of $ x$. In other words,\r\n\r\n$ d \\equal{} \\lfloor n^{\\{ b \\log_n a \\} } \\rfloor$.\r\n\r\n(Are you expecting a \"simpler\" expression?) A similar expression is readily obtained for the first $ k$ digits, from which it follows that the first $ k$ digits of an exponential sequence such as $ 2^n$ are equidistributed; in other words, one can find a power of two beginning with any prescribed digit sequence." } { "Tag": [ "trigonometry", "geometric series" ], "Problem": "Show that $ 1-(cosn\\theta+isinn\\theta)$=$ -2isin\\frac{n\\theta}{2}(cos\\frac{n\\theta}{2}+isin\\frac{n\\theta}{2})$\r\n\r\nFind the sum of the series $ z+z^{2}+z^{3}+...+z^{n}$ for $ z\\neq1$\r\n\r\n(iii)If $ z=cos\\theta+isin\\theta$, show that for $ sin\\frac{\\theta}{2}\\neq0$,\r\n\r\n$ cos\\theta+cos2\\theta+cos3\\theta+...+cosn\\theta$=[url=http://img95.imageshack.us/my.php?image=39986391td8.png][img]http://img95.imageshack.us/img95/1276/39986391td8.th.png[/img][/url]", "Solution_1": "anyone willing to have a try?", "Solution_2": "The first question is a trivial use of a double angle formula for cosine.\r\n\r\nTreat your second problem as a geometric sum.\r\n\r\n$ \\sum_{k=1}^{n}\\cos k\\theta=\\Re\\sum_{k=1}^{n}e^{ik\\theta}=\\frac{1}{2}\\sum_{k=1}^{n}(e^{ik\\theta}+e^{-ik\\theta})$.", "Solution_3": "$ \\sin{n\\theta}= 2 \\sin{\\frac{n \\theta}{2}}\\cos{\\frac{n \\theta}{2}}$ and\r\n\r\n$ \\cos{n \\theta}= 1-2 \\sin^{2}{\\frac{n \\theta}{2}}$. Plug these identities in and the first part is easily solved.\r\n\r\n$ z+z^{2}+z^{3}+.........+z^{n}= \\frac{z^{n+1}-z}{z-1}$, is a geometric series summation.\r\n\r\nNow $ \\cos{\\theta}+\\cos{2 \\theta}+..........+\\cos{n \\theta}= Re(z+z^{2}+.......+z^{n}) = Re(\\frac{z^{n+1}-z}{z-1})$ and the relationship can be obtained in this way.\r\n\r\nEdit: Oops, boxedexe beat me to it." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "2n+1 segments are marked on a line .Each of the segments intersects at least n other segments.prove that one of the segments intersects all the other segments", "Solution_1": "assume the segment $ AB$ (from left to right) we call $ A,B$ the start and end point of segment $ AB$ respectively...\r\ntake the first segment that ended...\r\nname it $ a$ now there are $ n$ other segments $ x_{1},...,x_{n}$ which intersect $ a$,now between $ x_{i}$s choose the segment that ended last and name it $ x$,we claim that $ x$ is the desired segment...\r\nnow if we name the end point of $ a$ as $ A$ then $ A \\in x_{1},...,x_{n}$ so $ x_{i}$s must intersect each other,so $ x$ intersects $ a,x_{i}$s now note that other than $ a,x_{i}$s there are $ n$ segments left,name them $ y_{1},...,y_{n}$ now we know that every $ y_{i}$ must intersect at least on of $ a,x_{i}$s so it mst intersect $ x$...\r\nso $ x$ intersects $ a,x_{i}$s,$ y_{i}$s and we are done...\r\n[hide=\"another similar problem\"]\nin a graph with $ 2n+1$ vertices,every $ n$ vertices have a common neighbour,prove that there exists a vertex which is connedcted to every other vertices...[/hide]" } { "Tag": [ "modular arithmetic", "pigeonhole principle", "combinatorics proposed", "combinatorics" ], "Problem": "Let $ A$ be a set containing $ 2001$ positive integers. Prove that $ A$ contains a subset $ B$, such that \r\n\r\n1. $ B$ has at least $ 668$ elements.\r\n\r\n2. For any $ a,b \\in B$, $ a\\plus{}b$ doesn't belong to $ B$.($ a$ and $ b$ are not necessary distinct).", "Solution_1": "By Dirichlet's Theorem, there exists a prime $ p\\equiv2\\pmod3$ which is greater than any element of $ A$. Call a number [i]good[/i] if and only if its least positive residue modulo $ p$ lies in $ \\left(\\frac{p}3,\\frac{2p}3\\right)$, i.e., $ \\left[\\frac{p\\plus{}1}3,\\frac{2p\\minus{}1}3\\right]$.\r\n\r\nNote that for all $ a_i\\in A$, $ a_i,2a_i,\\ldots,(p\\minus{}1)a_i$ are different residues modulo $ p$. So there are $ \\frac{p\\plus{}1}3$ integers $ k\\in[1,p\\minus{}1]$ for which $ ka_i$ is good. Hence there are $ 2001\\cdot\\frac{p\\plus{}1}3\\equal{}667(p\\plus{}1)$ pairs of $ (k,a_i)$ for which $ ka_i$ is good. By Pigeonhole Principle, there exists $ k$ such that the set $ B\\equal{}\\{a_i|ka_i\\text{ is good}\\}$ contains 668 elements.\r\n\r\nWe shall prove that $ B$ is sum-free. It is enough to prove that $ k$ times the sum of any pair in $ B$ is not good. But $ k$ times the sum of any pair is in the interval $ \\left[\\frac{2p\\plus{}2}3,\\frac{4p\\minus{}2}3\\right]$, i.e., $ \\left[0,\\frac{p\\minus{}2}3\\right]\\cup\\left[\\frac{2p\\plus{}2}3,p\\minus{}1\\right]$ which is not good. Therefore the set $ B$ satisfies the desired conditions.", "Solution_2": "Thank you, but this is actually the official solution, replicated word for word, well, even if it is not - I've been already aware of it. Any other solutions?" } { "Tag": [ "AMC", "AMC 10", "AMC 10 B", "AIME", "AMC 10 A" ], "Problem": "DOes anyone know if 117 (the cutoff for amc10a) is going to be the cutoff for aime on amc10b?", "Solution_1": "im pretty sure the cutoff for amc10a doesnt carry over to amc10b...", "Solution_2": "Yes, the cutoff for the 10A does not carry over to the 10B. The cutoff is determined by the lowest score of the top 5% of the people who took the test, I believe.", "Solution_3": "i thought amc10 cutoff was determined by the top 1%..? not 5? o.O", "Solution_4": "Yep, the AMC 10 cutoff is the top 1%. Though I think that should be even further below 120 for the 10B than it was for the 10A.", "Solution_5": "really..why should it be lower...o.O ah well..i'm hoping its lower =X all i want to get into is aime...=( i don't care if i get a 0 on it..lol", "Solution_6": "Yeah, me too, i got 117 :ninja: \r\nfour stupid mistakes, two blank", "Solution_7": "I got 114. IMAGINE that.", "Solution_8": "Well, you might still be able to make it with 114 or 117. They might lower the cutoff again.", "Solution_9": "Why must all the B tests that have been administrated be harder than the A tests?!? They canceled the A test because it was cold outside.. because i didnt see any snow or ice.. i am sad", "Solution_10": "The difficulty of the tests don't usually matter, it's how well you did in comparison to other people in the US (or taiwan, ...). So if you think it was hard, a lot of other people will too.", "Solution_11": "is it sad if i scored 114 both times? \r\nfirst attempt: 4 unanswered, and 18 correct. \r\nsecond attempt: 19 correct, and 6 wrong...\r\n\r\nif the cutoff is lowered to 115.5 for the B test, i swear i'm going to just break down in tears... :(", "Solution_12": "[quote=\"shen02\"]is it sad if i scored 114 both times? \nfirst attempt: 4 unanswered, and 18 correct. \nsecond attempt: 19 correct, and 6 wrong...\n\nif the cutoff is lowered to 115.5 for the B test, i swear i'm going to just break down in tears... :([/quote]\r\n\r\n\r\nTHAT HAPPENED TO ME TOO! I GOT 114 ON A AND B!!!\r\n\r\n\r\n\r\nand same here! for attempts\r\n\r\n8th grader here.", "Solution_13": "ah that sucks. yea my score differential is pretty enormous. 109.5 on amc A and 132 on amc B. but i thought the B was harder for some reason...", "Solution_14": "I thought B was EXTREMELy difficult compared to the A.\r\n\r\n138 on the B anyways though [officially]. I took A officially and had 144", "Solution_15": "I didnt take the amc10a, so i dont know about difficulties. but from what ive heard so far, i think the cutoff may be ~114.", "Solution_16": "Well, theres a kid at my school who got a 115 on the amc 10a(missed by 1.5, fewest you can miss by) and a 99 on the amc 12b(so if the cutoff isnt lowered, thats super unlucky for him).. sad thing is i dont want him to qualify because hes annoying as hell", "Solution_17": "lol maybe because it was my first time taking AMC10 on A, but I thought B was much easier... and I actually studied for A. I got 112.5 on A and 133.5 on B... but then again, i did miss number 1 on A =]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ ABC$ is a triangle. Prove that: $ \\boxed{p^2\\le5R^2\\minus{}Rr\\plus{}9r^2}$", "Solution_1": "[quote=\"thanhnam2902\"]Let $ ABC$ is a triangle. Prove that: $ \\boxed{p^2\\le5R^2\\minus{}Rr\\plus{}9r^2}$[/quote]\nAfter Ravi's substitution we need to prove that\n$\\sum_{cyc}(5x^4y^2+5x^4z^2+10x^3y^3-6x^4yz-22x^3y^2z-22x^3z^2y+30x^2y^2z^2)\\geq0$,\nwhich is true by easy SOS." } { "Tag": [], "Problem": "Let $ n$ be a positive integer such that all its digits are the same (for example $ 222$, $ 555555$),and it is dividible by $ 1998$.\r\n\r\nDetermine the minimum $ n$.", "Solution_1": "[quote=\"Bovy11\"]Let $ n$ be a positive integer such that all its digits are the same (for example $ 222$, $ 555555$),and it is dividible by $ 1998$.\n\nDetermine the minimum $ n$.[/quote]\r\n\r\n[hide=\" This may be true\"]\n\nMin value of n $ \\equal{}\\underbrace{222\\cdots222}_{27}\\equal{}1998\\times111222333444555666777889$\nReference:\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=236702[/url][/hide]", "Solution_2": "uh wat lol\r\n\r\n1998 = 999 * 2\r\n\r\nobv the smallest value is a lot of 2's, cuz 2...2|{4...4, 6...6, 8...8} given each thing has the same amount of digits\r\n\r\nso now if 1998 | 2...2, it must be 9k digits long\r\n\r\n\r\n222,222,222/1998 = 111,111,111/999 = 1001001/9\r\n\r\nnow that doesn't work\r\n\r\nso let's try 18 digits\r\n\r\nand we get that turns out to be like 1001001001001001/9\r\n\r\nwhich doesn't work either\r\n\r\nso we try 27 digits\r\n\r\nwhich is like 1001001001001001001001001/9 which is an integer so we good, smallest thing is 2...2 with 27 digits", "Solution_3": "[quote]obv the smallest value is a lot of 2's, cuz 2...2|{4...4, 6...6, 8...8} given each thing has the same amount of digits [/quote]\r\n\r\nCompletely flawed. It is entirely possible that 1998 does not divide 222,222,222, but it does divide 3 times that.\r\n\r\nAnd it does.[b] 666,666,666[/b] is smallest.", "Solution_4": "my bad totally didn't consider that lol" } { "Tag": [ "Gauss" ], "Problem": "Is missouri bad for your health?", "Solution_1": "yea it is. Very bad!", "Solution_2": "This is like 6 months old.", "Solution_3": "Now it was made in 2006 so it's 2 years and 6 months old.", "Solution_4": "Seriously who keeps rating my posts spam when they aren't?", "Solution_5": "Lol, rhetorial question? Or sarcastical?", "Solution_6": "Maybe cuz ur occupation is spamming?", "Solution_7": "My occupation?", "Solution_8": "I thought math154's occupation was n00b,", "Solution_9": "That is true.", "Solution_10": "I thought mine was terrorist...", "Solution_11": "Seriously, yes. All the people here are nonsense.", "Solution_12": "Ha you posted here so you are nonsense too.", "Solution_13": "I just realized that I have posted about 500 posts in October. *gulp* :o\r\n\r\n\r\nEDIT: Checking, I actually DID post about 500 posts in October. My guesstimate was correct!!!\r\n\r\nI just looked at my posts and the October posts go back to the 26th page, and I've only posted 67 pages.\r\n\r\n$ \\frac{26}{67}\\cdot1333\\approx517$. Yay.", "Solution_14": "What's the 517 mean?", "Solution_15": "I have no idea.... :huh:", "Solution_16": "Ha gauss still hasnt replied.", "Solution_17": "Now I did.\r\nSo shut up.\r\n\r\n[hide=\"and lay off\"] :D :D :D [/hide]", "Solution_18": "Hey! Meanie!", "Solution_19": "Spam isn't cool. That's why people not from Missouri are invading your happy little forum. So it isn't happy anymore.", "Solution_20": "Except this isnt spam. This is the most intelligent people in Missouri posting the most intelligent things they know. Missouri is mostly rural you know.", "Solution_21": "Well, the rural people made up Framer's Rule, so they can't be that stupid.", "Solution_22": "Lol like Kramers rule with farmers. Thats pretty punny.", "Solution_23": "Can you post the expanded version of Cramer's Rule for 1337 variables again? I forgot it. :maybe:", "Solution_24": "Uh i thought 1337 \"didnt pwn\"", "Solution_25": "I know! It's the only one I can't remember!" } { "Tag": [ "geometry", "3D geometry", "geometric transformation", "reflection", "vector", "AMC", "AIME" ], "Problem": "Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12.$ A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P,$ which is 7 units from $\\overline{BG}$ and 5 units from $\\overline{BC}.$ The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\\sqrt{n},$ where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n.$", "Solution_1": "[hide=\"Answer\"]It moves $5$ to the side and $7$ up, and we must end up at another vertex. Every time it reflects, it will move $5$ units to the side and $7$ units either up or down. If it hits a face partway through a reflection, it will continue the remaining number of units in the opposite direction. We first have that both $5x$ and $7x$ are multiples of $12$ for $x=12$. The length $AP$ is $\\sqrt{5^2+7^2+12^2}=\\sqrt{218}$, so we multiply that by $12$, and we have $12\\sqrt{218}\\Rightarrow 12+218=\\boxed{230}$.[/hide]", "Solution_2": "Your answer is correct but I think you should explain more. Unless someone knows how to do the problem already, your solution doesn't shed much light on the essence of the problem.\r\n\r\n[hide=\"Hint\"]\nReflect the cube the face that the beam hits. Keep reflecting each time the beam hits another side.[/hide]\r\n\r\nI'll post a solution if no one else does and I have time later...", "Solution_3": "Well, just ignore this if you haven't done the problem yet.\r\n\r\n[hide]\n\nSplit vector AP up into its components. You have a vertical vector of 7 and a horizontal vector of 5, if we just look at what happens on face BCPG. Of course, since this is 3D, there is a depth vector, but it's not necessary since it is simply the side length of the cube. Each reflection of the vector AP travels 12 units deep. So it's negligible. Well, now we want after a certain number of reflections, which are always at the original angle (so the same vector is preserved throughout), to have the components to both be divisible by 12, as this will guarantee a corner. \n\nThe reason for this is because the vertical vector of magnitude 7 units moves up and down the BCPG face if we look at it in 2 dimensions. And after moving up and down and up and down, we want it to land on either side BG or side CP (from the 2D perspective). We want the horizontal vector of magnitude 5 units to move left and right, bouncing boppity bop on the walls, until it lands on PG or BC in the 2D perspective. Thus, for some minimum value of $x$, $5x$ and $7x$ are divisible by $12$. Well, clearly $x=12$.\n\nSo the actual vector reflects exactly 12 times, so the length is 12 times the length of the original vector, which by the Pythagorean theorem is: $\\sqrt{218}$. So $12\\sqrt{218}$ is the answer. And in AIME, it's 230 I guess, but that doesn't matter that much. :P \n\nHaha... I was high on sugar when I wrote this so it's pretty overkill.\n\n[/hide]", "Solution_4": "[url]http://www.artofproblemsolving.com/Wiki/index.php/2002_AIME_I_Problems/Problem_11[/url]\n\nFor the solution on this page, I don't understand how the photons are in exactly the same position relative to the cubes they are at any moment after the space if divided. I understand the rest, however, by similar triangles. Can somebody help me out?", "Solution_5": "bump please\ni am really struggling with this problem. is my question too confusing?", "Solution_6": "What it's saying is that since you've divided the entire space into cubes, at any moment the 2nd photon is in a cube, and you can take that cube and rotate/flip it or whatever and make it look like the cube the 1st photon is in.", "Solution_7": "Sorry for the bump, but I just wanted to share this with everyone because I find it super cool :-D\n\nI coded a 3D simulation of this in Python. It requires both Pygame (so you'll need to copy into your own editor to run it, you can't just run it on AoPS) and some patience as it takes a bit of time before the laser actually hits the corner again, but it's cool xD.\n\nIf you want to see the laser at a position just press the space bar to pause the simulation and press the space bar a second time to resume it.\n\nEnjoy haha\n[pywindow]\nimport pygame\nimport random\npygame.init()\n\n\ndef dot_product(vector1, vector2):\n value = vector1[0]*vector2[0]+vector1[1]*vector2[1]+vector1[2]*vector2[2]\n return value\n\n\ndef add_vector(shape, vector):\n # shape is a list of vectors of the same length as the given vector\n\n new_shape = []\n for vertex in shape:\n new_vertex = []\n for i in range(len(vertex)):\n new_vertex.append(vertex[i] + vector[i])\n new_shape.append(tuple(new_vertex))\n\n return new_shape\n\n\ndef project(vector):\n plane = (0, 0, 1)\n direction = (1, 1, -3)\n\n t = -(dot_product(plane, vector)/dot_product(plane, direction))\n\n projected_vector = (vector[0]+t*direction[0],\n vector[1]+t*direction[1])\n # vector[2]+t*direction[2])\n\n return projected_vector\n\n\nside_length = 500\ncube = [\n (0, 0, 0),\n (side_length, 0, 0),\n (side_length, side_length, 0),\n (0, side_length, 0),\n (0, 0, side_length),\n (side_length, 0, side_length),\n (side_length, side_length, side_length),\n (0, side_length, side_length)]\n\nstart_vector = (100, 75, 0)\ncube = add_vector(cube, start_vector)\n\nprojected_cube = []\nfor vertex in cube:\n projected_cube.append(project(vertex))\n\nwindow = pygame.display.set_mode((1000, 800))\npygame.display.set_caption('Cube')\n\nprint(projected_cube)\ncolor_of_cube = (255, 255, 255)\nline_width = 3\n\nfor i in range(7):\n if i != 3:\n pygame.draw.line(window, color_of_cube, projected_cube[i], projected_cube[i+1], line_width)\n\npygame.draw.line(window, color_of_cube, projected_cube[0], projected_cube[3], line_width)\npygame.draw.line(window, color_of_cube, projected_cube[4], projected_cube[7], line_width)\n\nfor i in range(4):\n pygame.draw.line(window, color_of_cube, projected_cube[i], projected_cube[i+4], line_width)\n\npygame.display.flip()\n\nspeed = 1/72\n\nlaser_dir = [speed*12, -speed*5, -speed*7]\nlaser_location = [0, 0, 0]\n\npaused = False\ncolor = [255, 0, 0]\n\nwhile True:\n if paused == False:\n laser_location_projected = project(tuple(laser_location))\n laser_location_projected_rounded = (int(laser_location_projected[0]+0.5), int(laser_location_projected[1]+0.5))\n laser_location_projected_shifted = list(add_vector([laser_location_projected_rounded], start_vector)[0])\n\n \"\"\"\n for i in range(3):\n if color[i] >= 253:\n color[i] = 0\n else:\n color[i] += (i+1)\n \"\"\"\n\n pygame.draw.circle(window, tuple(color), laser_location_projected_shifted, 1)\n\n triple_bounce = [False, False,False]\n for i in range(len(laser_location)):\n if laser_location[i] < 0 or laser_location[i] > side_length:\n triple_bounce[i] = True\n laser_dir[i] *= -1\n if triple_bounce == [True, True, True]:\n pass\n # color = (random.randint(0, 255), random.randint(0, 255), random.randint(0, 255))\n\n for i in range(len(laser_location)):\n laser_location[i] += laser_dir[i]\n\n pygame.display.flip()\n\n for event in pygame.event.get():\n if event.type == pygame.QUIT:\n pygame.quit()\n break\n elif event.type == pygame.KEYDOWN:\n if event.key == pygame.K_SPACE:\n paused = not paused\n[/pywindow]", "Solution_8": "@above the python code has an error in it\n[quote=Python]ImportError: No module named pygame on line 1[/quote]\n@below I'll try running it on my computer I guess", "Solution_9": "you probably have to run it in repl.it or another python runner", "Solution_10": "My computer still gave me that error...", "Solution_11": "[quote=HumanCalculator9]My computer still gave me that error...[/quote]\n\nSame here", "Solution_12": "[quote=peace09][quote=HumanCalculator9]My computer still gave me that error...[/quote]\n\nSame here[/quote]\n\nSame here", "Solution_13": "@above2 no need to say Same here thats what the upvote button is for :|", "Solution_14": "Yeah I know\n\nAoPS doesn\u2019t support the module Pygame, so you\u2019ll need to copy the code into your own editor that has Pygame and run it there.\n\nSorry, if I could make it run here I would but I can\u2019t :/", "Solution_15": "I think I know why my computer gave me an error...\nMaybe my python version isn't new enough", "Solution_16": "No, you probably just don\u2019t have Pygame installed onto your computer...\n\nIf you want to somehow download it just search up Pygame and it\u2019ll give you their website with instructions how to download it.\n\nI\u2019ll post a screenshot of the sim here later if you\u2019re too lazy to download it, I can\u2019t do it right now because I\u2019m on mobile." } { "Tag": [ "function", "modular arithmetic", "number theory", "greatest common divisor", "group theory", "number theory open" ], "Problem": "Maybe this is boring, but I asked myself this question a few days ago and couldn't find a very satisfying answer:\r\n\r\nLet $ m$ be a positive integer that[b] has no primitive roots[/b] and let $ d$ be a positive integer. How many integers $ x$ with $ 1\\leq x \\leq m$ and $ gcd(x,m)=1$ exist, so that $ ord_{m}(x)=d$?.\r\n\r\nPersonally, I didn't get much further than this (and these results are trivial :oops: ):\r\n[hide]\nIt is obvious and well known, that there is some residue $ x$ with order $ d$, if and only if $ d\\mid \\lambda(m)$ where $ \\lambda$ is the carmichael function; in particular, there is some $ x$ with $ ord_{m}(x)=\\lambda(m)$. The obvious lower boundary for the number of such residues is $ \\prod_{p^{k}\\parallel m}\\varphi(\\varphi(p^{k}))$ (number of primitive roots modulo $ p^{k}$, respectively number of residues $ x$ modulo $ 2^{v_{2}(m)}$ with $ ord_{2^{v_{2}(m)}}(x)=2^{v_{2}(m)-2}$ if $ v_{2}(m)\\geq 3$). However, this boundary is very rudimentary since such a residue must not neccessarily be congruent to a primitive root modulo every $ p^{k}$. Also, for every $ d\\mid \\lambda(m)$ there are at least $ \\varphi(d)$ residues with order $ d$, but this is not very sharp as well since altogether, we would only cover $ \\lambda(m)<\\varphi(m)$ residues.\n\nMaybe this can be killed by some theorem in group theory, but I must admit that I'm not very familiar with group theory (yes, I know, how embarrassing :blush: ).\n[/hide]", "Solution_1": "First let's find how many are the solutions to:\r\n$ x^{d}\\equiv 1 \\pmod m$\r\nso, the integers whose order is a divisor of d.\r\nLet's write $ m = p_{1}^{e_{1}}\\cdots p_{k}^{e_{k}}$\r\nIt is a solution (mod m) if and only if it is a solution mod every $ p_{i}^{e_{i}}$.\r\nAnd modulo $ p_{i}^{e_{i}}$, as I think you know, there are exactly $ \\mbox{gcd}(\\phi(p_{i}^{e_{i}}), d)$ solutions.\r\nSo, the solutions to $ x^{d}\\equiv 1 \\pmod m$ are:\r\n$ \\prod_{1 \\le i \\le k}\\mbox{gcd}(\\phi(p_{i}^{e_{i}}), d)$.\r\n\r\nIf we define $ \\psi(d)$ as the number of residues with order d, what we got is:\r\n$ \\sum_{a \\mid d}\\psi(a) =$ the formula described above.\r\nTo get a closed formula for $ \\psi$, we use Moebius' inversion formula:\r\n$ \\psi(d) = \\sum_{a \\mid d}\\mu(\\frac{d}{a}) \\prod_{1 \\le i \\le k}\\mbox{gcd}(\\phi(p_{i}^{e_{i}}), a)$\r\n\r\nThis is what I got... It's not a very good formula, I see, but I don't think you can find something very simple.", "Solution_2": "Ouch... :blush: \r\nThis was staring just right into my face, how could I fail to think of the M\u00f6bius' inversion formula? :wallbash: \r\n\r\nBut, however...\r\n[quote=\"edriv\"]And modulo $ p_{i}^{e_{i}}$, as I think you know, there are exactly $ \\frac{\\phi(p_{i}^{e_{i}})}{\\mbox{gcd}(\\phi(p_{i}^{e_{i}}), d)}$ solutions.[/quote]\r\nEuhm...ok, it's already a bit late, so maybe my brain has already gone asleep, but shouldn't it be ${ gcd(\\varphi(p_{i}^{e_{i}}), d)}$ rather than $ \\frac{\\phi(p_{i}^{e_{i}})}{gcd(\\phi(p_{i}^{e_{i}}), d)}$?\r\n\r\nAnyway...thanks... :oops:", "Solution_3": "Oh, you're absolutely right, now I'll edit my last post... :blush: \r\nYour brain is certainly more awake than mine!\r\n\r\nP.S. See you next week in Vietnam! :)" } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "can we calculate sin20 as a real number ?", "Solution_1": "Yes, $\\frac{\\sqrt{3}}{2}=\\sin{60^{0}}=3\\sin{20^{0}}-4\\sin^{3}{20^{0}}$.", "Solution_2": "i've already known this equation but my question is what sin20 is...\r\ndo you know its value?", "Solution_3": "$0.342020143325669...$ or what do you wan't to hear\u00bf\r\nYou can always express $\\sin\\left( r \\cdot \\pi \\right)$ in terms of roots (radicals) if $r$ is rational.", "Solution_4": "As Zetax said:\r\nLet $\\alpha$ a \"well chosen\" cubic root of $-4\\sqrt3+4i$.\r\nThen $sin(\\frac{\\pi}{9})=\\frac{\\alpha}{4}+\\frac{1}{\\alpha}$.", "Solution_5": "thank you loup blanc this is what i want to hear\u00bf" } { "Tag": [ "trigonometry" ], "Problem": "I don't really know anything at all about astronomy, but I'm going into 9th grade and they have a full year course, Advanced Astronomy. It said that \"there is a strong mathematical approach in this subject.\" I was wondering what kind of math is used in astronomy, and how useful it is. Thanks for any help.", "Solution_1": "if u look at most college course catalogs, astronomy is listed with physics and has a very heavy math requirement cuz it's a math-based science. in cosmology and more advanced astronomy it's almost entirely math-based, but if you're taking a 9th grade class that emphasizes that there will be little math, it probably just means that you'll be doing fairly straightforward stuff like calculating magnitude (that would be logs) and distance using parallax (trig), distance modulus (more logs), etc.\r\nprobably won't be a problem.\r\nbest of luck! :lol:" } { "Tag": [ "integration", "function", "inequalities unsolved", "inequalities" ], "Problem": "Let $a_1,a_2,...,a_n \\in R$. Prove that:\r\n$\\displaystyle \\sum_{i=1}^n \\sum_{j=1}^n \\frac {a_ia_j}{i+j} \\geq 0$.", "Solution_1": "let $f(x)=a_1.x+a_2.x^2+\\dots +a_nx^n$ then we have $\\int _{0}^{1} \\frac{f(x)^2}{x} dx= LHS\\geq 0$", "Solution_2": "-Another similarly problem:\r\nFor any $a_1,a_2,...,a_{n}$ we have:\r\n$\\sum_{i=1}^n\\sum_{j=1}^n\\frac{a_{i}.a_{j}}{1+|i-j|} \\geq 0$\r\nTry it! :D", "Solution_3": "[quote=\"hxtung\"]Let $a_1,a_2,...,a_n \\in R$. Prove that:\n$\\displaystyle \\sum_{i=1}^n \\sum_{j=1}^n \\frac {a_ia_j}{i+j} \\geq 0$.[/quote]\r\n\r\nThis is very similar to problem 4 of the Baltic Way 1990. That problem 4 demands to prove that\r\n\r\n$\\displaystyle \\sum_{i=1}^n \\sum_{j=1}^n \\frac {a_ia_j}{i+j-1} \\geq 0$.\r\n\r\nThe solution is almost the same as Sam-n gave to Hxtung's problem; the only thing different is that you have to integrate the function $\\displaystyle \\frac{f^2\\left(x\\right)}{x^2}$ instead of $\\displaystyle \\frac{f^2\\left(x\\right)}{x}$.\r\n\r\n Darij" } { "Tag": [], "Problem": "What is the twelfth number (the 2nd missing number) in the following pattern?\r\n1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ____, ____", "Solution_1": "[hide=\"Answer\"]144. It's the Fibonacci sequence.[/hide]", "Solution_2": "[hide] Fibanocci sequence: 144 [/hide]", "Solution_3": "[hide]fibonacci (every term is equal to the sum of the 2 preceding it, except the first two which are defined). 144. [/hide][/hide]", "Solution_4": "[hide]144, fibonnaci sequence. I just found the sum of the first 10 terms and subtracted by 1.[/hide]", "Solution_5": "its 144 from the famous Fibanocci sequence", "Solution_6": "[quote=\"4everwise\"][hide]144, fibonnaci sequence. I just found the sum of the first 10 terms and subtracted by 1.[/hide][/quote]\r\n\r\nI think you added one. \r\nby the way, nice observation, 4everwise. nth term in this sequence is sum of the first n-2 numbers plus one.", "Solution_7": "[quote=\"math92\"]What is the twelfth number (the 2nd missing number) in the following pattern?\n1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ____, ____[/quote]\r\n\r\nThe next number is the sum of the previous two numbers.\r\n\r\nSo the next two numbers are $34+55=89$\r\n\r\nAnd the next is $55+89=144$", "Solution_8": "[quote=\"math92\"]What is the twelfth number (the 2nd missing number) in the following pattern?\n1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ____, ____[/quote]\r\n\r\nThis is the fibianacci sequence.\r\n\r\n[hide]\n34+55=79\n55+79=134\n[/hide]", "Solution_9": "[hide]The Fibonacci Sequence\n\n$F_n=F_{(n-2)}+F_{(n-1)}$\n\nTherefore, the two Missing terms are $34+55=89$, and $55+79=144$[/hide]" } { "Tag": [], "Problem": "A database of famous mathematical proofs. If I knew any html I would make this myself. It could work just by having viewers submit proofs, and having viewers also be able to bring attention to false proofs. So one of you guys should make it, because right now all I have is google.", "Solution_1": "Google is a powerfull tool for the one that knows how to use it ;) \r\n\r\nThere is already such a site: http://planetmath.org ..." } { "Tag": [ "quadratics", "number theory", "Diophantine equation", "number theory unsolved" ], "Problem": "Find all integer solutions of the equation $ y^2 \\minus{} 5x^2 \\equal{} 44.$", "Solution_1": "Four solutions are $ (\\pm 1,\\pm 7)$, so there exists solutions. If $ y^2\\leq 5x^2$ or $ y^2< 44$, then there are no solutions. \r\nIf $ (x_1,y_1)$ are solutions, then $ y_1^2\\minus{}44\\equiv 0\\: (\\mod 5)$ i.e. $ y_1^2 \\plus{}1\\equiv 0 (\\mod 5)$.\r\nBy brute force, the quadratic residues are $ \\overline{2}\\equal{}5\\mathbb{Z}\\plus{}2$ and $ \\overline{3}\\equal{}5\\mathbb{Z}\\plus{}3$.\r\nSo the solution set of the diophantine equation is $ \\{(\\pm\\sqrt{\\frac{y^2\\minus{}44}{5}},\\pm y)\\: |\\: y\\equiv 2(\\mod 5)\\: \\: \\text{or}\\: \\: y\\equiv 3(\\mod 5), y^2\\geq 49\\}$.", "Solution_2": "$ y \\equal{} 12 \\equal{} 2\\mod 5\\to x \\equal{} \\sqrt {20}$ not $ \\in \\mathbb{Z}.$ In what cases we should take $ y \\equal{} 5k \\plus{} 2$ and in what $ y \\equal{} 5k \\plus{} 3$ ;) ? Equation has infinitely many solutions. I think there exist a recurence formula for them. Ok, I found a solution: [url]http://mathworld.wolfram.com/PellEquation.html[/url]. Use this technique to get all pairs of solutions.", "Solution_3": "I dont think this problem has infinitely many solutions, i think there are only 3 (+-5,+-13), (+-1,+-7) and (+-2.+-8). \r\nFirst we can easily see $ 5(x\\minus{}1)(x\\plus{}1) \\equal{} (y\\plus{}7)(y\\minus{}7)$ and from here on there is just som divisibility theory to get the result. Maybe i made a mistake.", "Solution_4": "If you clearly read my link you understand, that it has infinitely many solutions.", "Solution_5": "Interesting theory, i didnt read your link, because i tried to solve it alone. It really was to hard for me.", "Solution_6": "Pell's equations always have infinitely many solutions if they have one. For example, the next solution is $ (7, 17)$. The theory is worth learning.", "Solution_7": "I think it's not so worth :D \r\n\r\nFirst solve \"unit\" form of this equation $ r^2 \\minus{} 5s^2 \\equal{} 1$. First solution is $ (9, 4)$, then use recurence formula $ r_{i \\plus{} 1} \\equal{} r_1 r_i \\plus{} 5s_1 s_i$ and $ s_{i \\plus{} 1} \\equal{} r_1 s_i \\plus{} s _1 r_i$ to get all other solutions. \r\n\r\nThen build set of soutions of $ t^2 \\minus{} 5m^2 \\equal{} 44$ using it's set of fundamental solutions:$ A \\equal{} {(t, m)} \\equal{} {(7, 1), (8, 2), (13, 5), (17, 7), (32, 14), (43, 19)}$ and using the identity $ t^2 \\minus{} 5m^2 \\equal{} (t^2 \\minus{} 5m^2)(r^2 \\minus{} 5s^2) \\equal{} (tr \\plus{} 5ms)^2 \\minus{} 5(ts \\plus{} mr)^2 \\equal{} 44$. \r\n\r\nThis gives $ tr \\plus{} 5ms$ for the new value of $ t$ and $ ts \\plus{} mr$ for $ s$. \r\n\r\nSo for every start point from $ A$ we get: $ t_{i,\\;k \\plus{} 1} \\equal{} t_{i,\\;1} r_{k} \\plus{} 5m_{i,\\;1} s_{k}$ and $ m_{i,\\;k \\plus{} 1} \\equal{} t_{i,\\;1} s_{k} \\plus{} m_{i,\\;1} r_{k}$ for $ 1\\leq i\\leq 6.$ Then merge solutions for all $ i$." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "x is a real numbers , then show that\r\n\r\n\\[ x^{6}\\minus{}x^{5}\\plus{}x^{4}\\minus{}x^{3}\\plus{}x^{2}\\minus{}x\\plus{}1\\geq \\frac{1}{2}\\]", "Solution_1": "\\[ x^{6}\\minus{}x^{5}\\plus{}x^{4}\\minus{}x^{3}\\plus{}x^{2}\\minus{}x\\plus{}1\\equal{}(x^3\\minus{}\\frac{x^2}{2})^2\\plus{}\\frac{3}{4}(x^2\\minus{}\\frac{2x}{3})^2\\plus{}\\frac{2}{3}(x\\minus{}\\frac{3}{4})^2\\plus{}\\frac{5}{8}\\ge \\frac{5}{8}\\]" } { "Tag": [ "modular arithmetic", "combinatorics proposed", "combinatorics" ], "Problem": "An even number of politicians are sitting at a round table. After a break, they come back and sit down again in arbitrary places. Show that there must be two people with the same number of people sitting between them as before the break..\r\n\r\n[b]Additional problem:[/b]\r\nSolve the problem when the number of people is in a form $6k+3$.", "Solution_1": "For the first one, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34679]this[/url].", "Solution_2": "1. It comes from polish final round in 1989.", "Solution_3": "The second one can be solved by using the same idea (but it requires a bit more work). It all comes down to this:\r\n\r\nIf $n=6k+3$ and $\\sigma$ is a permutation of the residues modulo $n$, then we can't have both $\\{\\sigma(i)-i\\ |\\ i\\in\\overline{0,n-1}\\}=\\overline{0,n-1}$ and $\\{\\sigma(i)+i\\ |\\ i\\in\\overline{0,n-1}\\}=\\overline{0,n-1}$ (everything takes place in $\\mathbb Z_n$).\r\n\r\nAssume the contrary (and remember, all the equalities take place in $\\mathbb Z_n$; it's because I don't want to write $\\pmod n$ all the time). We then have $\\sum i^2=\\sum(\\sigma(i)+i)^2=\\sum(\\sigma(i)-i)^2\\Rightarrow 4\\sum\\sigma(i)i=0\\Rightarrow\\sum\\sigma(i)i=0$. This means that $\\sum i^2=\\sum\\sigma(i)^2+\\sum i^2=2\\sum i^2\\Rightarrow\\sum i^2=0$. This yields a contradiction, since $\\sum i^2=\\frac{n(n+1)(2n+1)}6$ is easily seen to be different from $0$ modulo $n$.\r\n\r\nI hope it's correct.", "Solution_4": "And it looks like for all $n=6k+1,6k+5$ we can find a counterexample. \r\n\r\nWith the notations above, for all $i\\in\\mathbb Z_n$, take $\\sigma(i)=2i$. We get $\\sigma(i)-i=i,\\ \\forall i$, and $\\sigma(i)+i=3i,\\ \\forall i$, and the requirement that both sets $\\{\\sigma(i)-i\\}$ and $\\{\\sigma(i)+i\\}$ are equal to $\\mathbb Z_n$ is met." } { "Tag": [ "trigonometry" ], "Problem": "john, who is 5 foot tall, sees the top of a mountain at 34 degrees.\r\nif he was 50 miles away, how high is the mountain?", "Solution_1": "Wrong forum. ;) \r\n\r\n[hide=\"wrong solution\"]$\\tan34=\\frac{x}{50}$\n$x=33.7$\nabout 178075 feet. [/hide]", "Solution_2": "it isnt thaat hard tho...\r\ni learned this in my school a year ago... :huh:", "Solution_3": "this belongs in high school, despite easiness ;)", "Solution_4": "( in a childish voice) okay, mommy :lol:", "Solution_5": "[quote=\"aznmonkey1992\"]( in a childish voice) okay, mommy :lol:[/quote]\r\nlol. :rotfl: \r\ni agree that this does not belong here", "Solution_6": "dont you need a calculator to solve problems like these?\r\n\r\nDoes anyone know how to figure out Tan, Sin, Cos without a calculator?", "Solution_7": "I thought I did but I don't remember clearly so I won't say anything in case i'm wrong.*zips lips*", "Solution_8": "[quote=\"B4k4ka\"]dont you need a calculator to solve problems like these?\n\nDoes anyone know how to figure out Tan, Sin, Cos without a calculator?[/quote]\r\n\r\nOnly certain values, otherwise either a table or calculator is necessary.", "Solution_9": "Trigonometry is incredibly difficult for this forum...unless it's easy.\r\n\r\nSine, Cosine and Tangent require a calculator, but how do you figure it out with nothing but addition? (calculators do that)", "Solution_10": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47804]do not click this[/url]\r\n\r\nThis is getting off topic. May a mod please move this to a higher forum?", "Solution_11": "[quote=\"236factorial\"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47804]do not click this[/url]\n\n[/quote]i didn't click it yet. Will something bad happen if i click it? :stink:", "Solution_12": "Haha no, it's just a 'nasty' demonstration about trigo :)\r\n(i.e. : calculate sin 3\u00b0 without a calculator..)", "Solution_13": "[quote=\"mathmanman\"]Haha no, it's just a 'nasty' demonstration about trigo :)\n(i.e. : calculate sin 3\u00b0 without a calculator..)[/quote]\r\noh, i'll click it now..." } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "Find: $ lim\\displaystyle \\left ( \\frac{cos^{2}1}{1.2}\\plus{} \\frac{cos^{2}2}{2.3}\\plus{}...\\plus{} \\frac{cos^{2}n}{n(n\\plus{}1)}\\right)$", "Solution_1": "[quote=\"thanhnam2902\"]Find: $ lim\\displaystyle \\left ( \\frac {cos^{2}1}{1.2} \\plus{} \\frac {cos^{2}2}{2.3} \\plus{} ... \\plus{} \\frac {cos^{2}n}{n(n \\plus{} 1)}\\right)$[/quote]\r\n\r\nHere it is! :lol: (specify your limit boundary / point)", "Solution_2": "Denote\r\n\\[ f(z)\\equal{}\\sum_{n\\equal{}1}^\\infty\\frac{z^n}{n(n\\plus{}1)}\\equal{}\\frac{\\ln(1\\minus{}z)\\plus{}z}{z}\\minus{}\\ln(1\\minus{}z),\\]\r\nthen the answer is\r\n\\[ \\frac12(f(1)\\plus{}\\mathop{\\mathrm{Re}}f(e^{2i}))\\equal{}1\\minus{}\\sin^21\\cdot\\ln(2\\sin1)\\plus{}\\left(\\frac12\\minus{}\\frac\\pi4\\right)\\sin2.\\]" } { "Tag": [ "geometry" ], "Problem": "does anyone know a website with old math team tests/practice math team tests. Not AMC or Mathcounts. it would be best if the tests were like the 25 question + 3 tiebrakers at the end test. \r\n(BTW, this is for ALABAMA geometry math team [i.e. Vestavia tourny, grissom tourny, hoover tourny, state tourny, etc.) thus, any help would be greatly appreciated :lol:", "Solution_1": "http://www.mistacademy.com/contests.htm#alcontests\r\n\r\nJust click around on the links.", "Solution_2": "are there anymore websites?", "Solution_3": "http://www.quia.com/pages/pizitztourn.html\r\n\r\nfor pizitz.\r\n\r\nhttp://www.hoover.k12.al.us/hhs/Math/PKustos/tournament/tests.html\r\n\r\nfor hoover.\r\n\r\nI think I have more somewhere but i kant find them. hope this helps.", "Solution_4": "OHHHH YEA, that helps (as in all 3 sites help)\r\n\r\nare there anymore sites?", "Solution_5": "are you being sarcastic??? (I can never tell on the internet...) You could, you know, google for tests yourself...", "Solution_6": "i tried googleing them, but i didnt get \"math team\" tests. i got tests with like 5 questions and tests that are NOT what i am looking for. \r\n\r\nthus, i came to ask the people of the AoPS forum if they knew any...\r\n\r\nso, i nEEEEEEDDD MOREEEE!!!!" } { "Tag": [ "Online Math Open", "LaTeX" ], "Problem": "\u039a\u03b1\u03bb\u03b7\u03c3\u03c0\u03b5\u03c1\u03af\u03b6\u03c9 .\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03b2\u03ac\u03b6\u03c9 \u03c4\u03b1 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1 \u03c0 \u03bc\u03c0\u03ae\u03ba\u03b1\u03bd \u03c3\u03ae\u03bc\u03b5\u03c1\u03b1 \u03c3\u03c4\u03bf\u03bd \u039c\u03b5\u03c3\u03bf\u03b3\u03b5\u03b9\u03b1\u03ba\u03cc \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03cc.\r\n\u0398\u03b1 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03bf\u03cd\u03c3\u03b1 \u03cc\u03bb\u03bf\u03c5\u03c2 \u03b1\u03c5\u03c4\u03bf\u03cd\u03c2 \u03c0\u03bf\u03c5 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03b5\u03af\u03c7\u03b1\u03bd \u03c3\u03c4\u03bf\u03bd \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03cc \u03bd\u03b1 \u03b2\u03ac\u03bb\u03bb\u03bf\u03c5\u03bd \u03c4\u03b9\u03c2 \u0394\u03b9\u03ba\u03b5\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03c3\u03c4\u03b1 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1 (\u03b3\u03b9\u03b1\u03c4\u03af \u03ba\u03b1\u03b8\u03b5\u03bc\u03b9\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03be\u03b5\u03c7\u03c9\u03c1\u03b9\u03c3\u03c4\u03ae) \u03c4\u03b1 \u03bf\u03c0\u03bf\u03af\u03b1 \u03bc\u03bf\u03c5 \u03c6\u03ac\u03bd\u03b7\u03ba\u03b1\u03bd \u03c0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03b1 \u03c9\u03c2 \u03c0\u03c1\u03ce\u03c4\u03b7 \u03bc\u03b1\u03c4\u03b9\u03ac .\r\n\r\n[color=brown]1[/color] \u039a\u03ac\u03b8\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u03c7\u03c1\u03c9\u03bc\u03b1\u03c4\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03b4\u03cd\u03bf \u03c7\u03c1\u03ce\u03bc\u03b1\u03c4\u03b1 \u03bc\u03c0\u03bb\u03b5 \u03ba\u03b1\u03b9 \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03bf. \u03a5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03ad\u03bd\u03b1 \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03bf \u03ba\u03b1\u03b9 \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03ad\u03bd\u03b1 \u03bc\u03c0\u03bb\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf .\r\n\u0395\u03af\u03bd\u03b1\u03b9 \u03b4\u03c5\u03bd\u03b1\u03c4\u03cc\u03bd \u03ba\u03ac\u03b8\u03b5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1\u03c2 1cm \u03bd\u03b1 \u03c0\u03b5\u03c1\u03b9\u03ad\u03c7\u03b5\u03b9 \u03b5\u03c0\u03af \u03c4\u03b7\u03c2 \u03c0\u03b5\u03c1\u03b9\u03c6\u03ad\u03c1\u03b5\u03b9\u03ac\u03c2 \u03c4\u03bf\u03c5 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2:\r\n\u03b9)\u03ad\u03bd\u03b1 \u03bc\u03c0\u03bb\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf\r\n\u03b9\u03b9)\u03b4\u03cd\u03bf \u03bc\u03c0\u03bb\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1\r\n\r\n[color=green]2[/color] \u0388\u03c3\u03c4\u03c9 $P$ \u03ad\u03bd\u03b1 \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b5\u03bd\u03cc\u03c2 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 \u0391\u0392C \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 \r\n$A_{1}B_{2}$, $B_{1}C_{2}$, $C_{1}A_{2}$ \u03bf\u03b9 \u03c0\u03b1\u03c1\u03ac\u03bb\u03bb\u03b7\u03bb\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03ac\u03b3\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf P \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b9\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 AB,BC,CA \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 \u03bc\u03b5 $A_{1},A_{2}\\in BC$ , $B_{1},B_{2}\\in AC$ \u03ba\u03b1\u03b9 $C_{1},C_{2}\\in AB$ .\u039d\u03b4\u03bf $(A_{1}A_{2}B_{1}B_{2}C_{1}C_{2})\\geq\\frac{2}{3}(ABC)$ \u03cc\u03c0\u03bf\u03c5 \u03bc\u03b5 (...) \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03b5\u03bc\u03b2\u03b1\u03b4\u03cc\u03bd\r\n\r\n [color=darkblue]3[/color] \u0398\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 ABC \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03c3\u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03c4\u03b1 \u03bc\u03ae\u03ba\u03b7 \u03c4\u03c9\u03bd \u03c0\u03bb\u03b5\u03c5\u03c1\u03ce\u03bd a,b,c \u03b5\u03af\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03af \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03b9 \u03ce\u03c3\u03c4\u03b5 \u039c\u039a\u0394(a,b,c)=1 .H \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ae \u03b4\u03b9\u03c7\u03bf\u03c4\u03cc\u03bc\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03b3\u03c9\u03bd\u03af\u03b1\u03c2 \u0391 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd \u0392C \u03c3\u03c4\u03bf D .\u039d\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 \r\nTo \u0391\u0392D \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bc\u03bf\u03b9\u03bf \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf \u0391\u0392C \u03b1\u03bd \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b1\u03bd \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 c \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03ad\u03bb\u03b5\u03b9\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf\r\n\r\n[color=orange] 4[/color] \u0388\u03c3\u03c4\u03c9 $m,n$ \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03b9 \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $x_{i,j}\\in\\left[ 0,1\\right]$ for i=1,2,....,m \u03ba\u03b1\u03b9\r\nj=1,2,....,n\r\n\r\n\u039d\u03b4\u03bf $\\prod_{j=1}^{n}(1-\\prod_{i=1}^{m}x_{i,j})+\\prod_{i=1}^{m}(1-\\prod_{j=1}^{n}(1-x_{i,j}))\\geq 1$\r\n\r\n\u0397 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03bc\u03bf\u03b9\u03ac\u03b6\u03b5\u03b9 \u03bb\u03af\u03b3\u03bf \u03bc\u03b5 Weirstrass II \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae \u03c4\u03b7\u03c2. \u0398\u03b1 \u03c4\u03bf \u03b4\u03c9.", "Solution_1": "\u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03b5 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03c9 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03b8\u03b5\u03c3\u03b7 \u03c4\u03c9\u03bd \u03b8\u03b5\u03bc\u03b1\u03c4\u03c9\u03bd !!\r\n\r\n\u039c\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b2\u03ac\u03bb\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03c9\u03bd \u03bc\u03b9\u03ba\u03c1\u03ce\u03bd??\r\n\r\n :)", "Solution_2": "Ta themata den itan asxima kai mallon arketa prosvasima. Elpizo na pigate kala osi simeteixate.\r\nSas grafo tis liseis mou :\r\n1)\r\n(i) Iparxei toulaxiston ena mple simeio, me kentro afto grapste kiklo aktinas 1, ean iparxei pano ston kiklo ena mple simeio tote apexei apo to kentro akrivos 1, ara iparxei (giati ?) kiklos aktinas 1 pou dierxete kai apo ta dio antifasi.\r\n(ii) As bapsoume mple omokentrous kiklous, ton opioon i aktines diaferoun kata 2 monades. To ipoloipo epipedo kokkino. Tote profanos kathe kiklos aktinas 1 , eite tha temnei enan apo autous se dio simeia, eite tha efaptetai dio apo aftous.\r\n\r\n2)\r\nAs apodeiksoume prota to parakato limma.\r\nEsto trigono ABC , metablito simeio D stin BC , kai fero paralliles DK, DL pros tis AB, AC antistoixa (me K stin AC kai L stin AB). Tote to embadon (LDK) einai megiso otan to D ginei to meso tis BC.\r\nApodeiksi :\r\n$\\frac{LD}{AC} \\cdot \\frac{DK}{AB} = \\frac{ BD}{BC} \\cdot \\frac{DC}{BC} \\leq \\frac{(BD+DC)^2}{4 \\cdot BC^2} =\\frac{1}{4}$ Me isotita otan D einai meso.\r\n\r\nAra (piso sto problima) an to P kinitai pano stin $C_2 B_1$ tote $(C_2 B A_1 P) +(B_1 C A_2 P)$ menei analoioto kai simfona me to parapano limma to $(A C_1 P B_2)$ tha ginei megisto otan to P brisketai sto meso tou tmimatos pou kinitai. Ara to P brisketai pano stin diameso AM (M meso tou BC) omoios gia tis alles pleures...\r\n\r\n3)\r\nTa dio trigona einai omoia an kai mono an A =2C (gonies) ara isxuei oti $a^2-c^2=bc$ (Gnosti tautotita , tin afino os askisi). Ara $a^2=c \\cdot (b+c)$. Tote prepei (b,c)=1, eidalos o koinos diaretis ton b,c tha diairouse to a , alla (a,b,c)=1. Tote omos (c,b+c)=1, ara to c tha prepei na einai teleio tetragono (giati ?)\r\nTo antistrofo den isxuei p.x parte to pithagorio trigono me pleures 3,4,5. (Silouane, mipos itheles na grapseis kati allo ?)\r\n\r\n4)\r\nExo dio liseis giauto, tha grapso tin pio amesi kai tha sas doso mia mikri ipodeiksi gia tin alli.\r\nI parasasti sta aristera einai grammiki se kathe tis metabliti ara pernei ta akrotata tis sta akra ton diastimaton pou orizete. (p.x an statheropoisete oles tis metablites ektos tis x11, tote einai gramiki sinartisi os pros aftin ..k.t.l )\r\nAra ola ta x_ij einai eite 0 eite 1 otan exoume elaxisto. An ola ta X_i1,X_i1,...X_i1 einai isa me 1 gia kapoio i, tote o protos oras tou athroismatos einai isos me 0 kai o deuteros isos me 1. An gia kathe i den isxuei to apopano tote o protos oros tou athroismatos einai panta toulaxtiston 1 kai o allos thetikos. Ara to elaxisto einai to 1 ;) (mporeite na breite to megisto ?)\r\n\r\nMia alli proseggisi tha itan na ergastoume me pithanotites .... Ti mporei na antiprosopeoun ta x_ij ?\r\n\r\n\r\nStergios", "Solution_3": "\u03a6\u0399\u039b\u039f\u0399 \u03a0\u03a1\u0395\u03a0\u0395\u0399 \u039d\u0391 \u0394\u0395\u0399\u03a4\u0395 \u03a4\u0399\u03a3 \u03a0\u03a1\u039f\u03a4\u0395\u0399\u039d\u039f\u039c\u0395\u039d\u0395\u03a3 \u039b\u03a5\u03a3\u0395\u0399\u03a3 \u0391\u03a0\u039f \u03a4\u0397\u039d \u0395\u03a4\u0391\u0399\u03a1\u0395\u0399\u0391 \u0393\u0399\u0391 \u03a4\u039f \u03a0\u03a1\u039f\u0392\u039b\u0397\u039c\u0391 3 \u03a0\u03a1\u0395\u03a0\u0395\u0399 \u039d\u0391 \u0399\u03a3\u03a7\u03a5\u0395\u0399 \u039a\u0399 \u03a4\u039f \u0391\u039d\u03a4\u0399\u03a3\u03a4\u03a1\u039f\u03a6\u039f \u039f\u039a? \r\n\u03a7\u0391\u0399\u03a1\u0395\u03a4\u0391\u0399", "Solution_4": "\u03a4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03b8\u03ad\u03bc\u03b1, \u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03c4\u03c1\u03af\u03c4\u03bf \u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03c3\u03ba\u03ad\u03bb\u03bf\u03c2 \u03c4\u03b1 \u03ad\u03bb\u03c5\u03c3\u03b1 \u03c4\u03b7\u03bd \u03ce\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd.\u0393\u03b9\u03b1 \u03c4\u03bf \u03c4\u03c1\u03af\u03c4\u03bf,\u03c4\u03bf \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03c1\u03bf\u03c6\u03bf \u03b4\u03b5\u03bd \u03ba\u03b1\u03c4\u03ac\u03c6\u03b5\u03c1\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c9 \u03ba\u03b1\u03b9 \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03bf Anto \u03ad\u03c7\u03b5\u03b9 \u03b4\u03af\u03ba\u03b9\u03bf \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03bc\u03b5 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 3,4,5 \u2026 :o (\u03ac\u03bb\u03bb\u03bf\u03c3\u03c4\u03b5 \u03bf\u03b9 \u03c0\u03c1\u03ce\u03c4\u03b5\u03c2 \u03b5\u03c0\u03af\u03c3\u03b7\u03bc\u03b5\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03b2\u03b3\u03ae\u03ba\u03b1\u03bd \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03c9\u03b8\u03b5\u03af \u03b1\u03ba\u03cc\u03bc\u03b1).\u0392\u03c1\u03ae\u03ba\u03b1 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03bc\u03b9\u03b1 \u03c9\u03c1\u03b1\u03af\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03c4\u03bf \u03c4\u03ad\u03c4\u03b1\u03c1\u03c4\u03bf \u03c0\u03bf\u03c5 (\u03cc\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03ad\u03c1\u03c5\u03c3\u03b9) \u03c4\u03bf \u03ad\u03bb\u03c5\u03c3\u03b1 \u03c3\u03c4\u03bf \u03c4\u03c1\u03ad\u03bd\u03bf \u03c4\u03bf\u03c5 \u03b3\u03c5\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd,\u03b1\u03bb\u03bb\u03ac \u03bb\u03cc\u03b3\u03c9 \u03c0\u03b5\u03c1\u03b9\u03bf\u03c1\u03b9\u03c3\u03bc\u03ad\u03bd\u03c9\u03bd \u03b3\u03bd\u03ce\u03c3\u03b5\u03c9\u03bd Latex \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c4\u03b7 \u03b3\u03c1\u03ac\u03c8\u03c9.\u0397 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03c0\u03b1\u03bd\u03c4\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03cc\u03bb\u03b1 \u03c4\u03b1 x_ij \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bd \u03c4\u03b9\u03c2 \u03b1\u03ba\u03c1\u03b1\u03af\u03b5\u03c2 \u03c4\u03b9\u03bc\u03ad\u03c2 0 \u03ae1 .\r\n\r\n1)ii)\u0395\u03ba\u03c4\u03cc\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03bf\u03c5 Anto \u03bc\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03bf\u03bc\u03cc\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf\u03c5\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03bc\u03c0\u03bb\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf\u03bd \u03b1\u03c5\u03c4\u03ac \u03b1\u03bd\u03ae\u03ba\u03bf\u03c5\u03bd \u03c3\u03b5 \u03c0\u03b1\u03c1\u03ac\u03bb\u03bb\u03b7\u03bb\u03b5\u03c2 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \u03bc\u03b5 \u03b1\u03c0\u03cc\u03c3\u03c4\u03b1\u03c3\u03b7 2cm.\r\n\r\n3)\u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c4\u03b9 $c=d^2$ ,\u03cc\u03c0\u03bf\u03c5 $d=(a,c)$ .\r\n\r\n\u039f\u03b9 \u03ac\u03bb\u03bb\u03b5\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03bc\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bc\u03bf\u03b9\u03b5\u03c2 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03ad\u03c2 \u03c4\u03bf\u03c5 Anto.\r\n\r\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2... :)\r\n________________________________", "Solution_5": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03b1\u03b3\u03b1\u03c0\u03b7\u03c4\u03bf\u03af \u03c6\u03af\u03bb\u03bf\u03b9...\r\n\u03a4\u03b1 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1\u03c4\u03bf\u03c5 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd \u03ae\u03c4\u03b1\u03bd \u03c0\u03ac\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd \u03cc\u03bc\u03bf\u03c1\u03c6\u03b1 \u03b1\u03bd \u03b5\u03be\u03b1\u03b9\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf \u03c0\u03bf\u03c5 \u03ba\u03b1\u03c4\u03b1 \u03c4\u03b7 \u03b3\u03bd\u03ce\u03bc\u03b7 \u03bc\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b5\u03bd\u03b4\u03b5\u03b9\u03ba\u03bd\u03c5\u03c4\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03c5 \u03b5\u03af\u03b4\u03bf\u03c5\u03c2 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03cc!\u0391\u03bb\u03bb\u03ac:\r\n1)\u03bf Silouan \u03bc\u03b5\u03c4\u03ad\u03c6\u03b5\u03c1\u03b5 \u03bc\u03b5 \u03b1\u03ba\u03c1\u03af\u03b2\u03b5\u03b9\u03b1 \u03c4\u03b9\u03c2 \u03b4\u03b9\u03b1\u03c4\u03c5\u03c0\u03ce\u03c3\u03b5\u03b9\u03c2 (\u0391\u03bd\u03c4\u03bf \u03b5\u03ba\u03c0\u03bb\u03ae\u03c3\u03c3\u03b5\u03c3\u03b1\u03b9???)\r\n2)\u03c3\u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03b7 \u03c3\u03c9\u03c3\u03c4\u03ae \u03b4\u03b9\u03b1\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b5\u03be\u03ae\u03c2:\r\n[b]\u03a5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c7\u03c1\u03c9\u03bc\u03b1\u03c4\u03b9\u03c3\u03bc\u03cc\u03c2 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03c2 \u03ce\u03c3\u03c4\u03b5...[/b] \u0393\u03b9\u03b1 \u03c4\u03bf \u03b2 \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b9 \u03ac\u03bb\u03bb\u03b7 \u03bc\u03af\u03b1 -\u03c0\u03b9\u03bf \u03b1\u03c6\u03b7\u03c1\u03b7\u03bc\u03ad\u03bd\u03b7 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ae..(\u03c0\u03ce\u03c2??)\r\n3)\u03a3\u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03b8\u03ad\u03bc\u03b1 \u03c0\u03bf\u03bb\u03cd \u03b1\u03c0\u03bb\u03ac \u03b8\u03ad\u03c4\u03bf\u03c5\u03bc\u03b5: \u0392\u03911=\u03c7, \u03911\u03912=\u03c8 \u03ba\u03b1\u03b9 \u03912\u0393=\u03b6 \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bc\u03b5 \u03c4\u03bf \u03b5\u03be\u03ae\u03c2:\r\n3(x^2+\u03c8^2+\u03b6^2)>=(\u03c7+\u03c8+\u03b6)^2\r\n4)[b][u]\u03a4\u03ce\u03b1 \u03b8\u03b1 \u03b3\u03af\u03bd\u03c9 \u03ba\u03b1\u03c5\u03c3\u03c4\u03b9\u03ba\u03cc\u03c2!\u0394\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03b1\u03bd \u03c6\u03c4\u03b1\u03af\u03bd\u03b5 \u03bf\u03b9 \u0399\u03c3\u03c0\u03b1\u03bd\u03bf\u03af \u03ae \u03b7 \u03bc\u03b5\u03c4\u03ac\u03c6\u03c1\u03b1\u03c3\u03b7 \u03c4\u03b7\u03c2 \u0395\u039c\u0395 \u03b1\u03bb\u03bb\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ad\u03c2 \u03cc\u03c4\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b5\u03be' \u03b1\u03c5\u03c4\u03ce\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u0391\u03a3\u03a7\u0395\u03a4\u039f\u03a3![/u][/b].\u0393\u03b9\u03b1 \u03c4\u03bf \u03b1 \u03c9\u03c1\u03b1\u03af\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03bf \u0391\u03bd\u03c4\u03bf \u03ba\u03b1\u03b9 \u03bf \u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2 (\u03b5\u03af\u03c7\u03b1 \u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03bb\u03cd\u03c3\u03b7)!\u0393\u03b9\u03b1 \u03c4\u03bf \u03b2 \u03b5\u03af\u03bd\u03b1\u03b9 [u][b]\u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03b5\u03c2 \u03cc\u03c4\u03b9 \u03bf \u0391\u03bd\u03c4\u03bf \u03ad\u03c7\u03b5\u03b9 \u03b4\u03af\u03ba\u03b9\u03bf!!!\u0397\u03c4\u03b1\u03bd \u03b1\u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03ba\u03c4\u03bf!![/b][/u]\r\n\u039a\u03c1\u03af\u03bc\u03b1 \u03b3\u03b9\u03ac\u03c4\u03b9 \u03bc\u03ad\u03c1\u03b9\u03ba\u03bf\u03af \u03b8\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03bf\u03cd\u03c3\u03b1\u03bd \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bd \u03ba\u03ac\u03c4\u03b9 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9!!\r\n5)\u03b3\u03b9\u03b1 \u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf \u03c4\u03b9 \u03bd\u03b1 \u03c0\u03c9??\u038c\u03c0\u03c9\u03c2 \u03b5\u03af\u03c0\u03b1 \u03bc\u03cc\u03bd\u03bf \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03bc\u03bf\u03c5 \u03ac\u03c1\u03b5\u03c3\u03b5 ... \u03b1\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03ac \u03c3\u03c5\u03bc\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bd \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b5\u03c2 \u03bf\u03b9\u03ba\u03bf\u03b3\u03ad\u03bd\u03b5\u03b9\u03b5\u03c2!!\r\n\r\n(\u0391\u03bd\u03c4\u03bf \u03c5\u03c0\u03bf\u03bd\u03bf\u03b5\u03af\u03c2 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 \u03b3\u03b9\u03b1 \u03ba\u03ad\u03c1\u03bc\u03b1\u03c4\u03b1) ???????? :roll: \r\n\r\n\u0395\u03b9\u03c2 \u03c4\u03bf \u03b5\u03c0\u03b1\u03bd\u03b9\u03b4\u03b5\u03af\u03bd!", "Solution_6": "\u0393\u03b9\u03b1 \u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf ,\u03c4\u03c1\u03af\u03c4\u03bf \u03ba\u03b1\u03b9 \u03c4\u03ad\u03c4\u03b1\u03c1\u03c4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03bf\u03b9 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03bc\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c7\u03b5\u03b4\u03cc\u03bd \u03cc\u03bc\u03bf\u03b9\u03b5\u03c2 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03ad\u03c2 \u03c4\u03bf\u03c5 \u0391nto \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03ad\u03c7\u03c9 \u03bc\u03b9\u03b1 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7 .\r\n\r\n\u0388\u03c3\u03c4\u03c9 $M_1=(C_2PA_1)$, $M_2=(B_2PC_1)$ , $M_3=(B_1A_2P)$ , $T_1=(A_1A_2P)$,\r\n$T_2=(B_1B_2P)$, $T_3=(C_1C_2P)$ ,$T=(ABC)$. \r\n\r\n\u0395\u03cd\u03ba\u03bf\u03bb\u03b1 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $(C_2A_1B)=M_1$ , $M_3=(A_2B_1C)$ kai $M_2=(AC_1B_2)$.\r\n\r\n\u03a0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03bf\u03cd\u03bc\u03b5 \u03cc\u03c4\u03b9 $2(M_1+M_2+M_3)+T_1+T_2+T_3=T$ (1)\r\n\r\n\u03ac\u03c1\u03b1 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b3\u03c1\u03ac\u03c6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1 (\u03bc\u03b5 \u03c4\u03b7 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03b1 \u03c4\u03b7\u03c2 (1)) \u03b1\u03c5\u03c4\u03ae \u03c0\u03bf\u03c5 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b8\u03bd\u03b4\u03bf $3(T_1+T_2+T_3)\\geq T$ \u03cc\u03bc\u03c9\u03c2 \u03b1\u03c5\u03c4\u03cc \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b1\u03bd \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03af\u03b7\u03c3\u03bf\u03c5\u03bc\u03b5 Cauchy \u03c3\u03c4\u03bf \u03b3\u03bd\u03c9\u03c3\u03c4\u03cc \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03bb\u03ad\u03b5\u03b9 \u03cc\u03c4\u03b9 \r\n$\\sqrt{T}=\\sqrt{T_1}+\\sqrt{T_2}+\\sqrt{T_3}$ :lol: ;) \r\n\r\nBTW ,\u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf (\u03bf\u03c5\u03c3\u03b9\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac ) \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 3 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bb\u03ac\u03b8\u03bf\u03c2 ?????", "Solution_7": "\u03a6\u0399\u039b\u039f\u0399 \u0391\u03a0\u039f \u039f\u03a4\u0399 \u0395\u03a7\u03a9 \u039a\u0399 \u03a4\u0399\u03a3 \u0395\u03a0\u0399\u03a3\u0397\u039c\u0395\u03a3 \u039b\u03a5\u03a3\u0395\u0399\u03a3 \u0395\u0399\u039d\u0391\u0399 \u0394\u03a5\u039d\u0391\u03a4\u039f\u039d \u03a4\u039f \u0391\u039d\u03a4\u0399\u03a3\u03a4\u03a1\u039f\u03a6\u039f \u0391\u039d\u0391\u039a\u039f\u0399\u039d\u03a9\u03a3\u0395 \u0397 \u0395\u039c\u0395 \u039f\u03a4\u0399 \u03a4\u039f \u0391\u039d\u03a4\u0399\u03a3\u03a4\u03a1\u039f\u03a6\u039f \u0394\u0395\u039d \u0393\u0399\u039d\u0395\u03a4\u0391\u0399? \u0393\u0399\u0391\u03a4\u0399 \u0391\u039d \u039f\u03a7\u0399 \u0398\u0391 \u0395\u0399\u039d\u0391\u0399 \u0395\u0393\u039a\u039b\u0397\u039c\u0391 \u03a3\u0395 \u0392\u0391\u03a1\u039f\u03a3 \u0391\u03a5\u03a4\u03a9\u039d \u03a0\u039f\u03a5 \u0394\u0395\u039d \u0398\u0391 \u03a0\u0395\u03a1\u0391\u03a3\u039f\u03a5\u039d \u039a\u0399 \u03a0\u03a1\u0395\u03a0\u0395\u0399 \u039d\u0391 \u03a0\u039f\u03a6\u0395\u03a5\u03a7\u0398\u0395\u0399 \u03a0\u0391\u03a1\u0391\u039a\u0391\u039b\u03a9 \u039d\u0391 \u039c\u0395 \u0395\u039d\u0397\u039c\u0395\u03a1\u03a9\u03a3\u0395\u03a4\u0395 \u03a3\u03a5\u039d\u03a4\u039f\u039c\u0391 \u0393\u0399\u0391 \u03a4\u039f \u0391\u039d \u0393\u0399\u039d\u0395\u03a4\u0391\u0399 \u0397 \u038c\u03a7\u0399 \u03a4\u039f \u0391\u039d\u03a4\u0399\u03a3\u03a4\u03a1\u039f\u03a6\u039f\r\n\u03a7\u0391\u0399\u03a1\u0395\u03a4\u0391\u0399" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "n points are given on a plane. For every integer k (0k-1. for p-prime.", "Solution_2": "Suppose n>5 and k>1.. Use Newton formula to find that n!=n^k+kn^(k-1)+...+k(k-1)/2n^2+kn. So, (n-1)!=n^(k-1)+...+k(k-1)/2n+k. Suppose n is not a prime. Then n|(n-1)!. So, n|k and so k>=n. So, n!+1>=(n+1)^n, false. So n is a prime. Since Mihai proved n+1 is a prime, n=2. So n<=4 or k=1 and now it's easy.", "Solution_3": "Remark: This was also discussed at http://www.mathlinks.ro/viewtopic.php?t=4591 .\r\n\r\n darij", "Solution_4": "So, we have\r\n\r\n\\[ n,k \\in \\mathbb{Z^+} \\qquad (n+1)^k=n!+1 \\]\r\n\r\n[b]Fact 1.[/b]If $l$ is composite and $l>4$ then we have $(l-1)!\\equiv0 \\pmod{l}$\r\n\r\n[b]Proof[/b] Let be $l=p_1^{a_1}p_2^{a_2}...p_j^{a_j}$. If $j>1$ then all the $p_1^{a_i}$ figure in the product $(l-1)!$ cause they're less than $l$ and so we're done. Let instead be $l=p^a\\mbox{ with }a>1$. It suffice that $p$ figures al least $a$ times in $(l-1)!$. That is\r\n\r\n\\[ ap2$ that is true cause we supposed $l>4$\n\n$a1$ the $RHS$ is odd. So $n$ must be even, $n=2m$. We have\r\n\r\n\\[ (n+1)^k=n!+1 \\Longrightarrow n^k+\\binom{k}{1}n^{k-1}...+\\binom{k}{k-1}n+1=n!+1 \\]\r\n\\[ (n-1)!=n^{k-1}+kn^{k-2}...+k \\Longrightarrow (2m-1)!=(2m)^{k-1}+k(2m)^{k-2}...+k \\]\r\n\r\nLet's consider the last equation. By [b]Fact 1[/b] we deduce $k\\equiv0 \\pmod{2m}$ and so $2mLHS$. The only solutions are $(1;1),(2;1),(4;2)$.", "Solution_5": "Oh nice, we just had this in our TST...", "Solution_6": "Let 1. $n+1$ - not prime.\\[n+1=ab\\] , $n>a,b>1$ well done. $a\\mid (n!)$ and $\\rightarrow a\\mid1$ impossible. $n+1=p$ must be prime. $n=p-1$ , \\[p^k-1=(p-1)!\\] $\\rightarrow$ $p=2, 3, 5$ well done. $p>5$ impossible. $k=1, 2$. \\[(k,n)=(1,1), (1,2), (2,4)\\].", "Solution_7": "it is USAMO", "Solution_8": "why don't use AM-GM for an upper bound for k and v2(n!) to get some cases", "Solution_9": "The answers are $(n,k)=(1,1),(2,1),(4,2)$.\nFor sufficiently large $n$, the number of ending zeroes in $n!$ is simply the number of factors of 5's, or $\\sum_{i=1}^\\infty \\left(\\lfloor \\frac{n}{5^i}\\rfloor\\right)$. The maximum number of ending zeroes on RS is simply the number of digits of $n-1$ or $\\lfloor (logn) \\rfloor$, due to the Binomial Theorem of $k$, and then subtracting $1$. Clearly, the number of zeroes on LS is bigger than this maximum on RS for $n\\ge5$, and testing for $n=1,2,3,4$ yields the only cases.", "Solution_10": "[quote=harazi]Suppose n>5 and k>1.. Use Newton formula to find that n!=n^k+kn^(k-1)+...+k(k-1)/2n^2+kn. So, (n-1)!=n^(k-1)+...+k(k-1)/2n+k. Suppose n is not a prime. Then n|(n-1)!. So, n|k and so k>=n. So, n!+1>=(n+1)^n, false. So n is a prime. Since Mihai proved n+1 is a prime, n=2. So n<=4 or k=1 and now it's easy.[/quote]\n\nI understand the logic, but why does it only work if $n>5$ and $k>1?$ Isn't $n!+1 < (n+1)^n$ true for all positive integer $n>1?$" } { "Tag": [], "Problem": "A suitcase lock has 3 dials with the digits $ 0, 1, 2,..., 9$ on each.\nHow many different settings are possible if all three digits have\nto be different?", "Solution_1": "Given the previous dials, there are $ 10 \\cdot 9 \\cdot 8 \\equal{} 720$ settings where all the digits are different." } { "Tag": [ "calculus", "integration", "logarithms", "trigonometry", "calculus computations" ], "Problem": "Evaluate\r\n\r\n$ \\int_{0}^{\\frac{\\pi}{4}}\\frac{\\ln(\\cot x)}{(\\sin^{2009}x\\plus{}\\cos^{2009}x)^{2}}\\cdot \\sin^{2008}(2x)dx$", "Solution_1": "\\[ \\begin{array}{l}\r\n\\displaystyle \\int_0^{\\frac{\\pi }{4}} {\\frac{{\\ln (\\cot x)}}{{{{({{\\sin }^{2009}}x \\plus{} {{\\cos }^{2009}}x)}^2}}}} \\cdot{\\sin ^{2008}}(2x)dx \\equal{} \\\\\\displaystyle \r\n \\equal{}\\displaystyle \\int_0^{\\frac{\\pi }{4}} {\\frac{{\\ln (\\cot x)}}{{{{(1 \\plus{} {{\\cot }^{2009}}x)}^2}}}} \\cdot\\frac{{{2^{2008}}{{\\sin }^{2008}}x \\cdot {{\\cos }^{2008}}x}}{{{{\\sin }^{2009}}x \\cdot {{\\sin }^{2009}}x}}dx \\equal{} \\\\ \r\n \\equal{}\\displaystyle \\minus{} \\frac{{{2^{2008}}}}{{{{2009}^2}}}\\int_0^{\\frac{\\pi }{4}} {\\frac{{\\ln {{(\\cot x)}^{2009}}}}{{{{(1 \\plus{} {{\\cot }^{2009}}x)}^2}}}} \\cdot\\frac{{2009{{\\cot }^{2008}}x}}{{\\left( { \\minus{} {{\\sin }^2}x} \\right)}}dx \\equal{} \\\\\\displaystyle \r\n \\mathop \\equal{} \\limits_{ \\minus{} \\frac{{2009{{\\cot }^{2008}}xdx}}{{{{\\sin }^2}x}} \\equal{} du}^{{{\\cot }^{2009}}x \\equal{} u} \\frac{{{2^{2008}}}}{{{{2009}^2}}}\\int_{ \\plus{} \\infty }^1 { \\minus{} \\frac{{\\ln (u)}}{{{{(1 \\plus{} u)}^2}}}} du \\equal{} \\\\ \r\n \\equal{} \\frac{{{2^{2008}}}}{{{{2009}^2}}}\\left[ {\\frac{{\\ln u}}{{1 \\plus{} u}}} \\right]_{ \\plus{} \\infty }^1 \\minus{} \\frac{{{2^{2008}}}}{{{{2009}^2}}}\\int_{ \\plus{} \\infty }^1 {\\frac{1}{{u\\left( {1 \\plus{} u} \\right)}}} du \\equal{} \\\\ \r\n \\equal{} \\displaystyle \\minus{} \\frac{{{2^{2008}}}}{{{{2009}^2}}}\\int_{ \\plus{} \\infty }^1 {\\left[ {\\frac{1}{u} \\minus{} \\frac{1}{{u \\plus{} 1}}} \\right]} du \\equal{} \\\\ \r\n \\displaystyle \\equal{}\\displaystyle \\minus{} \\frac{{{2^{2008}}}}{{{{2009}^2}}}\\left[ {\\ln \\frac{u}{{u \\plus{} 1}}} \\right]_{ \\plus{} \\infty }^1 \\equal{} \\frac{{{2^{2008}}\\ln 2}}{{{{2009}^2}}} \\\\ \r\n \\end{array}\\]" } { "Tag": [ "inequalities" ], "Problem": "Hey anyone got any kind of idea of how to become super genious in math by CMO? i got 65 in COMC if anyone is wondering. and CMOR was a peice of cake after you sleep on the questions for a few days. suggestion from pros would be nice :lol: by the way how many do u have to solve to make division 1.", "Solution_1": "The best advice is simply do lots and lots of problems. There're no magic tricks. At the very least, go through every problem on the past ten years of CMO, perhaps even more.", "Solution_2": "[quote=\"eddyliu\"]Hey anyone got any kind of idea of how to become super genious in math by CMO? i got 65 in COMC if anyone is wondering. and CMOR was a peice of cake after you sleep on the questions for a few days. suggestion from pros would be nice :lol: by the way how many do u have to solve to make division 1.[/quote]\r\n\r\nNostalgic. :) I suggest you read the thread below: \r\n[url]http://www.mathlinks.ro/viewtopic.php?t=80147[/url]", "Solution_3": "[quote=\"Mathgod\"]\nNostalgic. :) I suggest you read the thread below: \n[url]http://www.mathlinks.ro/viewtopic.php?t=80147[/url][/quote]\r\n\r\nya nice...if i was still in grade 10 when i asked this. grade 12 last shot :(", "Solution_4": "[quote=\"eddyliu\"][quote=\"Mathgod\"]\nNostalgic. :) I suggest you read the thread below: \n[url]http://www.mathlinks.ro/viewtopic.php?t=80147[/url][/quote]\n\nya nice...if i was still in grade 10 when i asked this. grade 12 last shot :([/quote]\r\n\r\nIt's ok. Don't give up.", "Solution_5": "hey u guys got anymore of those crash course lectures like the one on inequality that Yufei posted?" } { "Tag": [ "geometry", "circumcircle", "incenter", "geometry proposed" ], "Problem": "Let a triangle $ ABC$. A point $ M$ lies on the interior of the triangle. Denote $ d_a,d_b,d_c$ are the distances from $ M$ to the sides of the triangle. Prove that $ d_ad_bd_c\\ge r^3 \\iff OM\\le OI$ where $ O,I$ be the circumcenter and incenter, $ r$ is the inradius.", "Solution_1": "[quote=\"N.N.Trung\"]Let a triangle $ ABC$. A point $ M$ lies on the interior of the triangle. Denote $ d_a,d_b,d_c$ are the distances from $ M$ to the sides of the triangle. Prove that $ d_ad_bd_c\\ge r^3 \\iff OM\\le OI$ where $ O,I$ be the circumcenter and incenter, $ r$ is the inradius.[/quote]\r\n\r\n [geogebra]3219c3811ab86c14bb941da9d21d3ab9d846ec4d[/geogebra] \r\n\r\nI think this problem is wrong... See picture. $ d_ad_bd_c \\equal{} 0.5433 < r^3 \\equal{} 2.4061$ but $ OM < OI$" } { "Tag": [ "floor function", "pigeonhole principle", "number theory unsolved", "number theory" ], "Problem": "Prove there exists infinitely many positive integers n such that\r\n$\\lfloor x^{3/2}\\rfloor+\\lfloor y^{3/2}\\rfloor = n $\r\nhas at least 1980 solutions in positive integers.", "Solution_1": "Consider all pairs $(x,y)$ s.t. $x,y\\in \\{1,2,3,...,m\\}$, where m is a large integer.\r\nFor these $m^2$ pairs we have $2 \\leq [x^{3/2}]+[y^{3/2}]\\leq 2[m^{3/2}]$. Using pigeonhole principle we obtain that there is $n\\in \\{2,3,...,2[m^{3/2}]\\}$ s.t. equation $[x^{3/2}]+[y^{3/2}]=n$ has at least $\\sqrt{m}/2$ solutions. On the other hand, equation $[x^{3/2}]+[y^{3/2}]=n$ has at most $n$ solutions.\r\nIt implies result." } { "Tag": [ "inequalities" ], "Problem": "if $abc=1 ,a,b,c>0$prove it:\r\n$\\frac{1}{a^{2}(b+c)}+\\frac{1}{b^{2}(c+a)}+\\frac{1}{c^{2}(a+b)}\\ge\\frac{3}{2}$", "Solution_1": "Put $x=\\frac{1}{a}$ and the inequality is equivalent to \r\n$\\frac{x}{y+z}+\\frac{y}{z+x}+\\frac{z}{x+y}\\geq \\frac{3}{2}$ which is Nesbitt one and can be proved by Cauchy in Engel form.", "Solution_2": "we can prove Nesbitt inequality by Algebraic substitution too.$a=y+z ,b=x+z,c=x+y$", "Solution_3": "[quote=\"parinaz.sa\"]we can prove Nesbitt inequality by Algebraic substitution too.$a=y+z ,b=x+z,c=x+y$[/quote]\r\nOr by Jensen, or by $\\mathrm{AM-HM}$, etc., etc., etc., ...", "Solution_4": "also we can apply muirhead after we do some long paperwork :)", "Solution_5": "Or rearrangement!", "Solution_6": "[quote=\"pohoatza\"]also we can apply muirhead after we do some long paperwork :)[/quote]\r\nIt's like killing a fly with a tank. :wink:", "Solution_7": "[quote=\"Rzeszut\"][quote=\"pohoatza\"]also we can apply muirhead after we do some long paperwork :)[/quote]\nIt's like killing a fly with a tank. :wink:[/quote]\r\n\r\n :rotfl:", "Solution_8": "[color=darkblue]Often and I say in such cases a bit otherwise : [/color][color=darkred]It's like killing a [b][u]butterfly[/u][/b] with the atomic bomb ![/color]", "Solution_9": "[quote=\"Virgil Nicula\"][color=darkblue]Often and I say in such cases a bit otherwise : [/color][color=darkred]It's like killing a [b][u]butterfly[/u][/b] with the atomic bomb ![/color][/quote]\nGood point! :lol:\n[quote=\"pohoatza\"][quote=\"Rzeszut\"][quote=\"pohoatza\"]also we can apply muirhead after we do some long paperwork :)[/quote]\nIt's like killing a fly with a tank. :wink:[/quote]\n\n :rotfl:[/quote]\r\nMuirhead can be a good theorem as just a piece of art. It can be good to play with generalisations. But not to kill beautiful inequalities. Do you really think that solving by Muirhead an inequality which becomes the trivial Nesbitt after substitution $a=\\frac1x$ etc. is a good idea? Similar to proving $x^{2}+y^{2}\\geqslant \\frac12$ for $x+y=1$ by Lagrange multiplier. Some others' (mainly moderators') opinions about Muirhead:\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=373993&highlight=hate+muirhead#373993[/url]\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=521106&highlight=hate+muirhead#521106[/url]\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=299450&highlight=hate+muirhead#299450[/url]\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=18894&highlight=hate+muirhead#18894[/url]\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18937&highlight=hate+muirhead[/url]\r\nAnd so on." } { "Tag": [ "function", "induction", "calculus", "derivative", "Pascal\\u0027s Triangle", "AMC" ], "Problem": "To find out how many possible subsets a set with n elements has, is it 2^n? For example if the set is {1,2,3,4} does that mean there are 2^4 subsets? Can someone prove that? Thanks.", "Solution_1": "Just thought about it again... is this correct?\r\n\r\nEach element has two choices; it is either in or not in the subset. Thus there are 2^n subsets?", "Solution_2": "Yeah it's $2^n$. Here's a proof.\r\n\r\nThe number of subsets are the ways to take 0 elements, 1 element, 2 elements, ..., n elements. This can be represented\r\n$\\binom n 0 + \\binom n 1 + \\binom n 2 + \\cdots + \\binom n {n-1} + \\binom n n$\r\nThis is the sum of the elements of the nth row of pascal's triangle, which we know is $2^n$. You can prove that by doing binomial expansion on $(1+1)^n$\r\n\r\nedit: Hmm, your proof is intriguing. I've never seen that before, I like it.", "Solution_3": "[quote=\"deej21\"]\n\nedit: Hmm, your proof is intriguing. I've never seen that before, I like it.[/quote]\r\n\r\nIt's a very easy subsets proof of the \"sum-of-the-nth-row-of-Pascal's-triangle\" combinatorics identity :) (Speaking of which, I have two other proofs - one based on generating Pascal's triangle through addition, and another by generating functions)\r\n\r\nJust realize that $2^n$ includes the empty set, so if a problem asks for non-empty subsets you'd have to subtract $1$.", "Solution_4": "Do you think you could show me your two other proofs? Thanks.", "Solution_5": "[quote=\"Calculus\"]Just thought about it again... is this correct?\n\nEach element has two choices; it is either in or not in the subset. Thus there are 2^n subsets?[/quote]\r\nThat's the simplest and most straightforward proof of this identity. Any other proofs using Pascal's triangle seem to be to be a little more contrived. You can also use a binomial expansion and prove the identity by induction, but that's much less fun.", "Solution_6": "I just like to think of it as\r\n\r\n$(1+1)^n$\r\n\r\nFor the sum of the terms in Pascals triangle. \r\n\r\nPerhaps its a little \"cheaper\"", "Solution_7": "[quote=\"Pakman2012\"]I just like to think of it as\n\n$(1+1)^n$\n\nFor the sum of the terms in Pascals triangle. \n\nPerhaps its a little \"cheaper\"[/quote]\r\nFirst you have to prove that the binomial expansion is indeed the proper expansion of $(x + y)^n$, which usually proceeds by induction, though can also be done using a combinatoric proof.", "Solution_8": "Oh. Lame. \r\n\r\nThat makes sense though haha.", "Solution_9": "[quote=\"Calculus\"]Do you think you could show me your two other proofs? Thanks.[/quote]\r\n\r\nWell, the first proof requires showing that adding consecutive entries in Pascal's triangle generates the correct binomial expansion, which can be done inductively or by a subsets proof - then you reason that, because each entry in a row is used twice to construct the entries in the next row, the sum of the entries in the next row is twice the sum of the entries in the previous row. :)\r\n\r\n(There is also an induction-based subsets proof. Namely, if there are $2^n$ subsets for an $n$-size set, adding an element to the set doubles the number of possible subsets because for each subset of the $n$-size set you can either add the new element or keep it out. )\r\n\r\nThe generating functions proof is identical to the expansion of $(1 + 1)^n$, but the great thing about it is it can also be used to show ${n \\choose 0} - {n \\choose 1} + {n \\choose 2} - {n \\choose 3} +-... = 0$, the expansion of $(1 - 1)^n$.", "Solution_10": "[quote=\"Calculus\"]\nThe generating functions proof is identical to the expansion of $(1 + 1)^n$, but the great thing about it is it can also be used to show ${n \\choose 0} - {n \\choose 1} + {n \\choose 2} - {n \\choose 3} +-... = 0$, the expansion of $(1 - 1)^n$.[/quote]\r\nAnother great consequence of this is if you take the derivative of $(x + 1)^n$ to get\r\n\r\n$n(x + 1)^{n-1} = n \\binom{n}{0}x^{n-1} + (n - 1) \\binom{n}{1} x^{n-2} + ... + \\binom{n}{n-1}$\r\n\r\nwhich leads to the cool identity\r\n\r\n$n2^{n-1} = n \\binom{n}{0} + (n - 1) \\binom{n}{1} + ... + \\binom{n}{n-1}$, or, equivalently,\r\n\r\n$n2^{n-1} = \\binom{n}{1} + 2 \\binom{n}{2} + ... + (n - 1) \\binom{n}{n - 1} + n \\binom{n}{n} = \\sum_{k=1}^n k \\binom{n}{k}$.", "Solution_11": "This is basically the elevator problem:\r\nHow many ways can a person starting on the zeroth floor get to the nth floor\r\nHe can stop at any number of floors but must always go up.\r\nThen its just 2^n, because he can either stop at a each floor or not, which is the same as the set. \r\nEach set can either include each number or not." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all functions $f: \\mathbb R\\to \\mathbb R$ such that\n\\[\\left | \\sum_{k=1}^{n}2^{k}(f(x+ky)-f(x-ky)) \\right |\\leq 1 ,\\]\n for all integers $n \\geq 0$ and all $x,y \\in \\mathbb R$.", "Solution_1": "Hello spix !\r\n[quote=\"spix\"]Find all $f: R\\to R$ such that:\n $\\left | \\sum_{k=1}^{n}2^{k}(f(x+ky)-f(x-ky)) \\right |\\leq 1 , \\forall n \\in N^{\\star}, \\forall x,y \\in R$[/quote]\r\n\r\nLet $A= \\sum_{k=1}^{n}2^{k}(f(x+ky)-f(x-ky))$\r\n\r\nWe have $\\left | A \\right |\\leq 1$\r\n\r\nLet $B= \\sum_{k=1}^{n+1}2^{k}(f(x+ky)-f(x-ky)) = A+2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y))$\r\n\r\nWe have $\\left | B \\right |\\leq 1$, so $\\left | A+2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \\right |\\leq 1$\r\n\r\nBut : $\\left | A+2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \\right |$ $\\geq$ $\\left | 2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \\right |-\\left | A\\right |$\r\n\r\nSo : $1 \\geq$ $\\left | 2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \\right |-\\left | A\\right |$\r\n\r\nSo : $1+\\left | A\\right |\\geq$ $\\left | 2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \\right |$\r\n\r\nSo : $\\left | 2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \\right | \\leq 2$\r\n\r\nAnd : $\\left | f(x+(n+1)y)-f(x-(n+1)y) \\right | \\leq 2^{-n}, \\forall n \\in N^{\\star}, \\forall x,y \\in R$\r\n\r\nLet a and b two reals.\r\n\r\nLet $x=\\frac{a+b}{2}$ and $y=\\frac{b-a}{2(n+1)}$ \r\n\r\nThen $\\left | f(x+(n+1)y)-f(x-(n+1)y) \\right | \\leq 2^{-n}$ becomes $\\left | f(b)-f(a) \\right | \\leq 2^{-n}, \\forall n \\in N^{\\star}, \\forall a,b \\in R$\r\n\r\nAnd f is any constant function.\r\n\r\n-- \r\nPatrick" } { "Tag": [ "vector", "algebra", "polynomial", "linear algebra", "linear algebra unsolved" ], "Problem": "If you have the following basis vectors for a polynomial vector space\r\nc_1*[-1,1,0]\r\nc_2*[-1,0,1]\r\n\r\nthen how do you get to the form\r\n\r\n(x-1)\r\n(x^2 - 1)\r\n\r\nas representing the same basis?", "Solution_1": "Those \"vector\" forms only make sense with respect to a standard basis for the polynomials. What do you think this standard basis is?", "Solution_2": "The standard basis is \r\n{1,x,x^2}" } { "Tag": [ "number theory", "greatest common divisor", "induction" ], "Problem": "There is a row of real numbers, infinite in both directions. The square of each number is $ 1$ more than the product of the numberson either side. Prove that if four adjacent numbers are integers, then at least one number in the row is $ 0$.", "Solution_1": "[hide]\nWithout loss of generality, let our four numbers be $ a_1,a_2,a_3,a_4$ with $ |a_1| > |a_2|$. Then we have \\begin{align*}\na_2^2 &= a_1a_3 + 1 \\\\\na_2a_4 + 1 &= a_3^2\n\\end{align*} Summing yields $ a_2(a_2+a_4) = a_3(a_1+a_3)$. But $ \\text{gcd}\\,(a_2,a_3) = 1$, and as all of the quantities are integers, it follows that $ a_3 | a_2+a_4$ and $ a_2 | a_1 + a_3$. Hence, \\[ k = \\frac{a_1+a_3}{a_2} = \\frac{a_2+a_4}{a_3}\\] where $ k \\in \\mathbb{Z}$, and further for any $ n$, $ \\frac{a_{n-1}+a_{n+1}}{a_n} = k$. Assuming that $ a_{n-1},a_n$ are integers, then $ a_{n+1} = ka_n - a_{n-1}$ must also be an integer, so by induction all the elements are integers. Also, assuming that $ |a_{n-1}| > |a_n|$, then $ |a_{n+1}| = \\frac{a_{n}^2-1}{|a_{n-1}|} < \\frac{a_{n}^2-1}{|a_{n}|} < |a_{n}|$ (unless $ a_n = 0$, but then we're finished). It follows that the sequence $ |a_i|$ is integer and strictly decreasing, so $ 0$ must appear in the sequence.[/hide]", "Solution_2": "[quote=\"outback\"]There is a row of real numbers, infinite in both directions. The square of each number is $ 1$ more than the product of the numberson either side. Prove that if four adjacent numbers are integers, then at least one number in the row is $ 0$.[/quote]\r\nThis was on the 2008 ELMO at MOP. Here is my solution (quite similar to [b]azjps[/b]'s but a little bit different in the beginning.) \r\n[hide=\"Solution\"]\nAssume that $ 0$ does not appear. Let $ k$, $ a$, $ b$, and $ l$ be the integers that are consecutive. We then have that $ \\frac {a^2 \\minus{} 1}{k} \\equal{} b$, so\\[ b^2 \\minus{} 1 \\equal{} \\frac {a^4 \\minus{} 2a^2 \\plus{} 1}{k^2} \\minus{} 1 \\equal{} \\frac {a^4 \\minus{} 2a^2 \\minus{} (k^2 \\minus{} 1)}{k^2}\\]Yet, $ al \\equal{} b^2 \\minus{} 1$, so we have that $ alk^2 \\equal{} a^2 \\minus{} 2a^2 \\minus{} (k^2 \\minus{} 1)$. Since $ a|(a^4 \\minus{} 2a^2)$, we have that $ a|(k^2 \\minus{} 1)$. This means that the number to the left of $ k$, say $ r$ is $ \\frac {k^2 \\minus{} 1}{a}$, which is an integer. Similarly (inductively) we can show that all of the numbers in the sequence are integers. Now, we consider a number (integer) in the sequence, say $ m$. Let the numbers to the left and right of $ m$ be $ x$ and $ y$ respectively. Then, $ xy \\equal{} m^2 \\minus{} 1 < m^2$. Without loss of generality, let $ |x| < |y|$ (if $ |x| \\equal{} |y|$, then either $ \\minus{} x^2 \\equal{} m^2 \\minus{} 1$, so $ x^2 \\plus{} m^2 \\equal{} 1$, so either $ x$ or $ m$ is $ 0$, which is a contradiction or $ x^2\\equal{}m^2\\minus{}1$, so $ x\\equal{}0$, which is a contradiction as well.) Then, $ |x| < m$ since we otherwise have that $ |x\\parallel{}y| > m^2$, so $ xy$ is nonpositive, so $ m^2 \\minus{} 1\\le 0$, so $ m \\equal{} \\pm 1$ or $ m \\equal{} 0$. The second case is a contradiction, so we have that $ m \\equal{} \\pm 1$, so $ xy \\equal{} m^2 \\minus{} 1 \\equal{} 0$, which means that $ x$ or $ y$ is $ 0$, which is a contradiction. Hence, $ |x| < m$. Similarly, if the number to the left of $ x$ is $ w$, then $ |w| < |x|$, and we keep going. Eventually, we must reach $ 0$, so we have a contradiction. Thus, $ 0$ shows up somewhere. [/hide]" } { "Tag": [ "MATHCOUNTS", "\\/closed" ], "Problem": "I have posted a question on mathcounts, and I knew that some people answered it. But I can't find it now. May anyone please tell me the reason?", "Solution_1": "Step 1: Login.\r\nStep 2: click to ''View your posts''\r\nIf you not find it then It is deleted :P", "Solution_2": "Most likely, it was inappropriate (e.g. too hard) for the forum it was placed in and a moderator moved it.", "Solution_3": "I have viewed my posts and searched for it, but I cannot find it.", "Solution_4": "[quote=\"N.T.TUAN\"]Step 1: Login.\nStep 2: click to ''View your posts''\nIf you not find it then It is deleted :P[/quote]It think this explanation is very clear." } { "Tag": [], "Problem": "108/root(x^2 - 2916) = [378 - x - root(x^2 -2916)] / (x+4)", "Solution_1": "$ \\frac{108}{\\sqrt{x^2\\minus{}2916}}\\equal{}\\frac{378\\minus{}x\\minus{}\\sqrt{x^2\\minus{}2916}}{x\\plus{}4}$\r\n\r\nIs this what you mean? Because the \"/\" at 108 makes me think that it is also a fraction...", "Solution_2": "108/root(x^2 - 2916) = [378 - x - root(x^2 -2916)] / (x+4) :[b]note 108/root(x^2 - 2916) mean 108 division by root(x^2 - 2916)[/b]", "Solution_3": "Could the answers be \"58,5\" and \"62,5\" by any chance?", "Solution_4": "hello, i think he meant\r\n$ \\frac{108}{\\sqrt{x^2\\minus{}2916}}\\equal{}\\frac{378\\minus{}x\\minus{}\\sqrt{x^2\\minus{}2916}}{x\\plus{}4}$.\r\nSonnhard.", "Solution_5": "please show me your solution", "Solution_6": "hello, multiplying by $ \\sqrt{x^2-2916}$ and $ x+4$ and simplifying we have the equation\r\n$ x^2+108x-2484=(378-x)\\sqrt{x^2-2916}$. It must be $ x^2+108x-2484\\geq 0$ and $ 378-x\\geq 0$ and $ x^2-2916>0$ and $ x+4 \\ne0$. These conditions give $ x\\le 6(-9-5\\sqrt{6})$ or $ 5454 or x<-54, x!=(that's java code for \"does not equal)-4, [378-x-root(x^2-2916)]/(x+4)>0\r\n\r\nCheck if the solutions you got fit in all of these conditions\r\n\r\n58.616768>54, 58.616768!=-4, [378-58.616768-root(58.616768^2-2916)]/(58.616768+4) = 4.736453 > 0\r\n\r\n-55.417633<-54, -55.417633!=-4, [378-(-55.417633)-root((-55.417633)^2-2916)]/(-55.417633+4) = -8.187136 < 0 The last condition has failed to be satisfied - ignore this solution\r\n\r\n133.911976>54, 133.911976!=-4, [378-133.911976-root(133.911976^2-2916)]/(133.911976+4) = .881334 > 0\r\n\r\nSo the solutions are 58.616768 and 133.911976. :idea:" } { "Tag": [ "number theory", "AoPS Books" ], "Problem": "for the answer to one of my problems in the \"introduction to number theory\" book (on page 12 of the answer book) it simply says this;\r\n\r\n9409 = 1000 - 2(3)(100) + 3^2 = (100 - 3)^2 = 97^2\r\n\r\nhow did it get from the first step to the second step?!", "Solution_1": "This is simply about recognizing patterns; note that $ 10000$ and $ 9$ are both perfect squares, so it makes sense to try something with it: we end up with $ 10000\\plus{}9\\minus{}600$, or $ 10000\\minus{}2(3)(100)\\plus{}3^2\\equal{}(100\\minus{}3)^2\\equal{}97^2$.", "Solution_2": "It helps to know that...\r\n\r\n1. $ (a \\plus{} b)^2 \\equal{} a^2 \\plus{} 2ab \\plus{} b^2$\r\n2. $ (a \\minus{} b)^2 \\equal{} a^2 \\minus{} 2ab \\plus{} b^2$\r\n3. $ (a \\plus{} b)(a \\minus{} b) \\equal{} a^2 \\minus{} b^2$\r\n\r\n... and the other way around.\r\n\r\n(HINT: The 2nd formula-like thing is the one you're looking for!)\r\n\r\nBTW I think this is mentioned in one of the AoPS books, but I can't remember which", "Solution_3": "It's from the Intro to Number Theory book.\r\n\r\nEDIT: its probably in both as well as intro to algebra...", "Solution_4": "It is? I could have sworn it was from AoPS Vol 1 :huh:", "Solution_5": "Shouldn't it be $ 10000$ and not $ 1000$?", "Solution_6": "oh, I get it if it is 10,000! I like aops, but have they never heard of commas?!\r\n\r\n(thx for all your help guys)", "Solution_7": "I generally don't put commas - I find they waste time. :roll: \r\n\r\nJust my opinion.", "Solution_8": "[quote=\"Waffle\"]I generally don't put commas - I find they waste time. :roll: \n\nJust my opinion.[/quote]\r\n\r\n\r\n\r\nNot to sound rude, arrogant, \"goody two-shoes\" or anything, but that post was a waste of time for you to type and everyone to read. Please don't do that again; mods or admins may punish you.\r\n\r\nSincerely\r\n\r\n---EDIT---\r\nOops! I'm sorry; I meant to PM waffle, but I accidentally placed this post in the forum :blush: ! Sorry!" } { "Tag": [ "FTW" ], "Problem": "NO MORE ORCHESTRA PRACTICE RECORDS!!! (well, for a while, at least)", "Solution_1": "[quote=\"math154\"]NO MORE ORCHESTRA PRACTICE RECORDS!!! (well, for a while, at least)[/quote]\r\n\r\nIs there one due tomorrow?", "Solution_2": "Yeah, but I finished it while waiting for the extremely late bus. BTW, I got home at 5:00 and I was the FIRST stop. Waiting for the bus almost took the entire time you were in Anatomy. :wink: :roll:", "Solution_3": "LOL public transportation FTW!" } { "Tag": [ "topology", "real analysis", "real analysis unsolved" ], "Problem": "I was fiddling with this proof given (in more general form) by jmerry a long time ago but for some reason this application is leading to a false conclusion.\r\n\r\nThe complete metric space $ [1,\\infty [$ (in R, under the standard metric) has the representation $ [1,\\infty [ \\equal{} \\bigcup\\limits_{n \\equal{} 1} {[n,n \\plus{} 1]}$ and the Baire Category theorem (overkill here, but bare with me) shows that $ Y \\equal{} \\bigcup\\limits_{n \\equal{} 1} {]n,n \\plus{} 1[}$ is a nonempty proper open subset of $ [1,\\infty [$. $ A \\equal{} [1,\\infty [\\backslash Y$ is closed and nowhere dense in the space but $ A \\ne \\bigcup\\limits_{n \\equal{} 1} {([n,n \\plus{} 1] \\cap A)}$ because A is of second category in itself.\r\n\r\nBut this cannot be true.\r\n\r\nIt should be a simple error in reasoning somewhere but I cannot detect it. :blush:\r\n\r\nEDIT: I found it. I was thinking nowhere dense means in every open set there is an open subset which does not intersect the set but (I'm guessing) the empty set doesn't count as an 'open subset'. If any set were to be nowhere dense you'd think it'd be N...", "Solution_1": "An empty set is open. Therefore, the condition should read \"in every open set there is a [b]nonempty[/b] open subset...\" \r\n\r\nI prefer a different wording: a set is nowhere dense if its closure has empty interior." } { "Tag": [ "geometry" ], "Problem": "When you approach a math contest problem, what do you think in your head while you're reading? \r\nDo you reread the question to scan for small details and important words?\r\n\r\nDo you immediately try to classify what type of problem it is such as logic or algebra?\r\n\r\nWhat clues and signals do you see in the problem to use certain tools? For example, if I see a bunch of large equations, I might think to manipulate them all or some of them to get a prettier equation. If I see lots of shared or congruent angles, I might think of similar triangles. If I see logical statements in the forms if A, then B, then I think of the contrapositive and other stuff... \r\n\r\nWhat are some other common hints in problems use certain tools or pursue certain tactics?\r\n\r\nAfter you've found an answer to the problem, do you reread the problem to make sure that you've found what it's asking for? Do you tend to spend more time on understanding the problem than actually going through the steps to solve the problem?\r\n\r\nWhen you check over your work on a math competition that you're doing under a timed situation, how do you determine which questions to check? Do you just check them all, and when do you decide to go back and check your work? [/youtube]", "Solution_1": "I don't really know what I think...I just try many things that could be useful for that type of problem.\r\nAlso, I don't really double check :P", "Solution_2": "[quote=\"xpmath\"]I don't really know what I think...I just try many things that could be useful for that type of problem.\nAlso, I don't really double check :P[/quote]\r\n\r\nYeah, double checking is something I've never got in the habit of doing.\r\n\r\nI normally look for the easy marks first, so I'll read through as far as possible and then answer in the order of difficult which I consider for each question. I do categorise them, but my categories are \"Geometry and Combinatorics\" and \"Everything else\". If its in everything else, I'll take a swing at it, otherwise I will leave it until the very end", "Solution_3": "Well I'm not as good as a lot of people here on AoPS, but posting here helps me learn A LOT.\r\n\r\nHow do I solve problems...\r\n\r\nWhen reading I rewrite any givens and underline what seems to be the real 'problem' in each question. I've found that understanding what the problem is asking [i]is[/i] realizing the solution. So be sure to be able to identify what sort of math you're dealing with. And to that end, try to be well rounded in non-discrete and discrete mathematics. \r\n\r\nI do my best to improve my weaknesses, but I play to my strengths when the chips are down. On any test where all problems have equivalent point value I solve the easiest first. To me 'the easiest' doesn't mean the lowest level of math, it means the math I'm most comfortable with. I'm not saying that you shouldn't try the questions that seem ridiculous, I'm just saying work on them later.\r\n\r\nI know this sounds trivial, but be able to do things like square 37 without a calculator. Another good idea is learning divisibility rules. I believe calculators have been banned on the AMC....\r\n\r\nNow as for checking questions, I normally 'come back' if I realize there's a problem. Generally this happens when I look at the possible answers (I only work backwards if I have to).\r\n\r\nI visit AoPS every day. :lol:", "Solution_4": "igiul said:[quote]Now as for checking questions, I normally 'come back' if I realize there's a problem. Generally this happens when I look at the possible answers (I only work backwards if I have to).\n[/quote]\r\n\r\nWhat are some signals that there is a problem? How do you realize that there is a problem without going back to check answers?", "Solution_5": "[quote=\"aufha\"]igiul said:[quote]Now as for checking questions, I normally 'come back' if I realize there's a problem. Generally this happens when I look at the possible answers (I only work backwards if I have to).\n[/quote]\n\nWhat are some signals that there is a problem? How do you realize that there is a problem without going back to check answers?[/quote]\r\n\r\nSometimes I wont \"feel\" right about an answer. It may seem too . $ x$, where $ x \\in {}$obvious, small, low, ... generic quanitifier${ }$\r\n\r\nOther times, I will realize that I made a computational mistake later on for no apparent reason.", "Solution_6": "[quote=\"aufha\"]What are some signals that there is a problem? How do you realize that there is a problem without going back to check answers?[/quote]\r\n\r\nWell occasionally I too feel uneasy about a problem, but as I continue with the test I often relate and apply apparently irrelevant details back to those problems. Maybe the details come from the other problems themselves when you are looking at the same subject in a different manner. I always try to 'justify' my answer.\r\n\r\nMost often the this is due to contradiction, so I may not know which answer is right, but I do know that mine is wrong. If this is the case, its back to base one on that problem if I feel I can accommodate it. But always if this happens, change your answer on the answer sheet!\r\n\r\nHope that helps!" } { "Tag": [ "function", "geometry", "geometric transformation", "rotation", "integration", "calculus", "real analysis" ], "Problem": "Given $ k \\equal{} 1,\\ldots,n$ (we are working in $ \\mathbf R^n$), find Fourier transform of the function\r\n\\[ f(x) \\equal{} \\frac {x_k}{|x|}.\r\n\\]\r\nAdded: in the sense of tempered distributions.", "Solution_1": "Note that $ f$ is not integrable, and not in $ L^2.$ What you're asking for only makes sense if you're asking for it in the sense of tempered distributions.\r\n\r\nSince $ f$ has mean value zero over any ball centered at the origin, $ \\widehat{f}$ shouldn't contain any point masses. Since $ f$ is positively homogeneous of degree $ 0,$ we should have that $ \\widehat{f}$ is positively homogeneous of degree $ \\minus{} n.$ (The Dirac delta is positively homogeneous of degree $ \\minus{} n$ but we have already eliminated it.) Investigating how this behaves under rotations leads us to this:\r\n\\[ \\widehat{f}(\\xi) \\equal{} \\text{p.v. }\\frac {C\\xi_k}{|\\xi|^{n \\plus{} 1}}\r\n\\]\r\nfor some (pure imaginary) constant $ C$ which depends on $ n.$\r\n\r\nThe only thing left to do will be to compute $ C.$ (I leave that out in part because I don't know your Fourier transform conventions.) The notation \"p.v.\" means that this is to be interpreted as the distribution given by a principal value integral.\r\n\r\nFor more, look up the words \"Riesz transform.\" (That's one generalization of the $ n\\equal{}1$ object called the Hilbert transform.)", "Solution_2": "Here's a more constructive argument.\r\n\r\nTo establish the conventions, define\r\n\r\n$ \\widehat{f}(\\xi)\\equal{}\\int_{\\mathbb{R}^n}e^{\\minus{}2\\pi i\\xi\\cdot x}f(x)\\,dx$\r\n\r\nA specific computation is that the Fourier transform of $ e^{\\minus{}\\pi|x|^2}$ is $ e^{\\minus{}\\pi|\\xi|^2}.$ And if we put that in the definition above and take $ \\frac{\\partial}{\\partial \\xi_k},$ we get that \r\n\r\nif $ f(x)\\equal{}x_ke^{\\minus{}\\pi|x|^2},$ then $ \\widehat{f}(\\xi)\\equal{}\\minus{}i\\xi_ke^{\\minus{}\\pi|\\xi|^2}.$\r\n\r\nThen there is the scaling principle: if $ g(x)\\equal{}f(ax)$ for some $ a>0,$ then $ \\widehat{g}(\\xi)\\equal{}\\frac1{a^n}\\widehat{f}\\left(\\frac{\\xi}{a}\\right).$ In particular,\r\n\r\nthe transform of $ \\sqrt{t}x_ke^{\\minus{}\\pi t|x|^2}$ is $ \\minus{}i\\,\\frac{1}{t^{n/2}}\\cdot\\frac{\\xi_k}{\\sqrt{t}}e^{\\minus{}\\pi|\\xi|^2/t}.$\r\n\r\nWrite $ \\frac{x_k}{|x|}\\equal{}\\int_0^{\\infty}t^{\\minus{}1/2}x_ke^{\\minus{}\\pi t|x|^2}\\,dx\\equal{}\\int_0^{\\infty}\\frac1t\\left(\\sqrt{t}x_k\\right)e^{\\minus{}\\pi t|x|^2}\\,dt$\r\n\r\nTake the Fourier transform of this expression. (Consider it to converge weakly, in the distributional sense, and use that to justify what we're doing.) We get that the Fourier transform of $ \\frac{x_k}{|x|}$ is\r\n\r\n$ \\int_0^{\\infty}\\frac1t\\cdot\\frac{1}{t^{n/2}} \\cdot\\frac{\\xi_k}{\\sqrt{t}}e^{\\minus{}\\pi|\\xi|^2/t}\\,dt$\r\n\r\nCollect together the powers of $ t,$ make the appropriate change of variables in this integral, and if I didn't make any mistakes there, we get\r\n\r\n$ \\minus{}i\\,\\frac{\\Gamma\\left(\\frac{n\\plus{}1}{2}\\right)}{\\pi^{\\frac{n\\plus{}1}2}} \\cdot\\frac{\\xi_k}{|\\xi|^{n\\plus{}1}}$ as the transform we seek." } { "Tag": [ "geometry", "rectangle" ], "Problem": "how many squares are in 8 * 8 chess board?\r\n\r\n[hide=\"hint\"]not 64[/hide]", "Solution_1": "[hide][quote]how many squares are in 8 * 8 chess board? [/quote]\n\nI wasn't sure whether the question was only for the squares on the grid lines, or diagonal as well. I did only those on the grid lines.\n\nsize of squares :arrow: number of them\n\n1x1 :arrow: 64\n2x2 :arrow: 49\n3x3 :arrow: 36\n4x4 :arrow: 25\n5x5 :arrow: 16\n6x6 :arrow: 9\n7x7 :arrow: 4\n8x8 :arrow: 1\n\nIts important to count the overlapping ones.\n\nSince they are all squares, does any one know a formula for their sum? I added them all up. \n\nThe answer is [b]204 squares[/b], not counting diagonals.[/hide]", "Solution_2": "[hide=\"my answer\"]64+49+36+25+16+9+4+1=[b]204[/b][/hide]", "Solution_3": "Follow up question: how many squares in an $n$-by-$n$ chessboard?", "Solution_4": "[hide=\"hmmmmm...\"]maybe $n^2+(n-1)^2+(n-2)^2+(n-3)^3 \\cdots$\nyou would do this until you get n-x=1[/hide]", "Solution_5": "yeah, the sum comes out to be\r\n\r\nS = n*(n+1)*(2n+1)/6\r\n\r\ncomes out to 204 for n=8 as other people showed.", "Solution_6": "Follow up question:\r\n\r\nHow many rectangles are in a 8 x 8 chess board?", "Solution_7": "response to follow-up question of how many RECTANGLES:\r\n\r\n(9c2)^2 = 1296.\r\n\r\nthat's the slickest approach I know. now can someone tell me where I got that from? :D", "Solution_8": "Youre choosing the nine points and choosing to of them and then nine points on the side and choosing 2, making 4 points all with right angles", "Solution_9": "I think you got it. the way I thought of it was that in an 8x8 chessboard, there's really a grid of 9x9 lines. you have to choose any 4 of those (2 in horizontal, 2 in vertical) to define a rectangle.", "Solution_10": "64............................................................", "Solution_11": "64............................................................", "Solution_12": "What question you're answer, kishiman1??? Be more specific next time...", "Solution_13": "sorry about that. I didnt read that hint at the top.", "Solution_14": "[quote=\"kishiman1\"]sorry about that. I didnt read that hint at the top.[/quote]\r\n\r\nOk, for the next time....." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "[i] We know {1,3} = 0,3\n But { - 1,3} = ?\n \n\n {x} = x - [x] with all $ x$[/i]?", "Solution_1": "as you said $ \\{x\\} \\equal{} x \\minus{} [x]$,hence $ \\{\\minus{}1,3\\} \\equal{} \\minus{} 1,3 \\minus{} ( \\minus{} 2) \\equal{} 0,7$." } { "Tag": [ "geometry", "circumcircle" ], "Problem": "Let $ ABC$ be an isosceles triangle inscribed in a circle and $ M$ a point in the line $ BC$ ($ M$ out of the base $ BC$). Prove that $ \\overline{MA}^2 \\equal{} \\overline{AB}^2 \\plus{} \\overline{MB}\\cdot\\overline{MC}$.", "Solution_1": "[quote=\"Virgil Nicula\"]$ \\boxed {\\Longrightarrow}$ [color=darkred]Let $ ABC$ be a $ A$-isosceles triangle. Prove that for any point $ X$\n\nof the sideline $ BC$ exists the identity $ AX^2 \\equal{} AB^2 \\plus{} \\overline{BX}\\cdot\\overline{CX}$. \n\n$ \\boxed {\\Longleftarrow}$ Let $ ABC$ be a triangle for which exists a point $ X$ of the sideline $ BC$\n\nso that $ AX^2 \\equal{} bc \\plus{} \\overline{BX}\\cdot\\overline{CX}$. Prove that the triangle $ ABC$ is $ A$-isosceles\n\nor the ray $ [AX$ is the bisector of the angle $ \\widehat {BAC}\\ .$\n\nSee http://www.mathlinks.ro/viewtopic.php?t=177757 . [/color][/quote]\n[color=darkblue][b][u]Proof.[/u][/b]\n\n$ \\boxed {\\Longrightarrow}$ If $ AB \\equal{} AC$, then the [b]Stewart's relation[/b]\n\n$ AB^2\\cdot\\overline {CX} \\plus{} AC^2\\cdot \\overline {XB} \\plus{} AX^2\\cdot \\overline {BC} \\plus{} \\overline {CX}\\cdot\\overline {XB}\\cdot\\overline {BC} \\equal{} 0\\ \\ (*)$\n\nbecomes $ AB^2\\cdot \\overline {CB} \\minus{} AX^2\\cdot\\overline {CB} \\minus{} \\overline {CX}\\cdot\\overline {XB}\\cdot\\overline {CB} \\equal{} 0\\Longleftrightarrow$\n\n$ AB^2 \\minus{} AX^2 \\plus{} \\overline {BX}\\cdot\\overline {CX} \\equal{} 0\\Longleftrightarrow AX^2 \\equal{} AB^2 \\plus{} \\overline {BX}\\cdot\\overline {CX}\\ .$\n\n$ \\boxed {\\Longleftarrow}$ If exists a point $ X$ of the sideline $ BC$ so that $ AX^2 \\equal{} bc \\plus{} \\overline{BX}\\cdot\\overline{CX}$, \n\nthen from the [b]Stewart's relation[/b] $ (*)$ obtain :\n\n$ c^2\\cdot\\overline {CX} \\minus{} b^2\\cdot\\overline {BX} \\plus{} \\left(bc \\plus{} \\overline {BX}\\cdot\\overline {CX}\\right)\\cdot \\overline {BC} \\minus{} \\overline {CX}\\cdot\\overline {BX}\\cdot\\overline {BC} \\equal{} 0\\Longleftrightarrow$\n\n$ c^2\\cdot\\overline {CX} \\minus{} b^2\\cdot\\overline {BX} \\plus{} bc\\cdot\\overline {BC} \\equal{} 0\\Longleftrightarrow$\n\n$ c^2\\cdot\\overline {CX} \\minus{} b^2\\cdot\\overline {BX} \\plus{} bc\\cdot\\left(\\overline {BX} \\minus{} \\overline {CX}\\right) \\equal{} 0\\Longleftrightarrow$\n\n$ b(c \\minus{} b)\\cdot \\overline {BX} \\plus{} c(c \\minus{} b)\\cdot\\overline {CX} \\equal{} 0\\Longleftrightarrow$ $ b \\equal{} c$ or $ b\\cdot\\overline {BX} \\plus{} c\\cdot\\overline {CX} \\equal{} 0$, i.e.\n\n$ \\triangle\\ ABC$ is $ A$-isosceles or $ \\frac {\\overline {XB}}{\\overline {XC}} \\equal{} \\minus{} \\frac cb$ (the ray $ [AX$ is the bisector of the angle $ \\widehat {BAC}$).[/color]\n\n[quote=\"Virgil Nicula\"] [color=darkred][b][u]A similar problem.[/u][/b] Prove that for the triangle $ ABC$ exists a point $ X$ which belongs\n\nto the sideline $ BC$ so that $ AX^2 \\equal{} \\overline {BX}\\cdot\\overline {CX} \\minus{} bc$ if and only if $ b\\ne c$ and in this case\nthe point $ X$ is uniquelly (the ray $ [AX$ is the exterior bisector of the angle $ \\widehat {BAC}$). [/color][/quote]\n\n[quote=\"Virgil Nicula\"][color=darkred][b][u]An easy extension.[/u][/b] Let $ ABC$ be a triangle. Acertain for any $ m\\in\\mathcal R$ the set $ D(m)$\n\nof the points $ X$ which belongs to the sideline $ BC$ and $ AX^2 \\equal{} mbc \\plus{} \\overline{BX}\\cdot\\overline{CX}$.[/color][/quote] \r\n\r\n[color=darkblue][b][u]Notation.[/u][/b] The [b]Chasles' relation[/b] : for the points $ X(x)$ and $ Y(y)$\n\nthe algebraic measure of the orientated segment $ \\overline {XY}$ is $ y \\minus{} x$. [/color]", "Solution_2": "Thanks Virgil. I edited the sign of minus in my first post. When i finished the proof i saw your solution. Got one more mistake in the book where i found it.", "Solution_3": "Let AM intersect the circumcircle in D. $ \\angle BMD \\equal{} \\frac {\\widehat{AB}\\minus{}\\widehat{CD}}{2}\\equal{}\\frac{\\widehat{AC}\\minus{}\\widehat{CD}}{2}\\equal{}\\angle AMC$.\r\nOr $ \\Delta ACD \\sim \\Delta AMC \\Rightarrow \\frac{AC}{AM}\\equal{}\\frac {AD}{AC}$, so $ AC^2\\equal{}AM.AD\\equal{}AM(AM\\minus{}DM)\\equal{}AM^2\\minus{}AM.DM\\equal{}AM^2\\minus{}MC.MB$", "Solution_4": "Beautiful, delta :thumbup:" } { "Tag": [ "geometry", "rhombus" ], "Problem": "What is the area enclosed by the graph of $ |3x| \\plus{} |4y| \\equal{} 12$?\r\n\r\n$ \\textbf{(A)}\\ 6\\qquad\r\n\\textbf{(B)}\\ 12\\qquad\r\n\\textbf{(C)}\\ 16\\qquad\r\n\\textbf{(D)}\\ 24\\qquad\r\n\\textbf{(E)}\\ 25$", "Solution_1": "[hide=\"Answer\"]The equation goes through the axes at points $(0,\\pm 3)$ and $(\\pm 4,3)$. Therefore, we have four congruent right triangles with legs $3$ and $4$, so the answer is $4\\cdot \\frac{1}{2}\\cdot 3\\cdot 4=24\\Rightarrow \\boxed{D}$.[/hide]", "Solution_2": "actually, you made a typo :)\r\n\r\nI think you meant $ (\\pm 4, 0)$ instead of $ (\\pm4, 3)$", "Solution_3": "I know I am reviving a thread that is so old that I have not seen a thread that is older than this one, but I thought I have a valid justification: I have an alternative solution. \n\n[hide=\"My Solution\"]\nAs [b]JesusFreak197[/b] mentioned, the graph passes through the vertices $(0, \\pm3)$ and $(\\pm4, 0)$. This is the same as a rhombus with diagonals being the segment between $(0,3)$ and $(0,-3)$ and the segment between $(-4,0)$ and $(4,0)$. Thus, the diagonals have lengths $6$ and $8$. The area of a rhombus with diagonals $a$ and $b$ is $\\tfrac{1}{2}ab$. Thus, the desired area here is: \\[ \\dfrac {1}{2} \\cdot 6 \\cdot 8 = \\boxed {\\textbf {(D) } 24} \\] \n[/hide]", "Solution_4": "l3xl + l4yl = 12\n=> 3x + 4y =12;\n=> x/(12/3) + y/(12/4)=1;\nThus, it cuts x-axis at 4 units, and it cuts y-axis at 3 units.\nAgain evaluating the equation by removing modulus, we get 4 straight lines which intercept at x and y axes forming 4 congruent triangles. thus, the total area enclosed is:-\n 4*[(1/2)*4*3]\n=4*6\n=24 units. \nThus; answer is (D)" } { "Tag": [ "absolute value" ], "Problem": "A sequence of positive integers begins 2009, 2008, 1, 2007, etc. Beginning with the\r\nthird term, each term is the absolute value of the difference of the previous two\r\nterms. What is the 640th term of this sequence?", "Solution_1": "[hide=\"Hint\"]\n\nNotice the pattern: \n\n$ 2009,2008,1,2006,2005,1,2004,2003,\\cdots$\n\n[/hide]\r\n\r\nIs the answer $ 1582$?", "Solution_2": "I think its 1583, I kinda made the problem up so.... ya\r\n\r\nHere is another problem:\r\n\r\n1,2,4,7,8,10,11,13,14,....,17291729 excludes all integers that are either multiples of 3 or 5. What is the 1337th term in this sequence?\r\n\r\n[hide]\n2506?\n[/hide]" } { "Tag": [ "probability", "geometry", "trigonometry", "circumcircle", "calculus", "integration", "inequalities" ], "Problem": "Three points are selected inside square ABCD with side length 12. What is the probability that all three lie inside a circle of radius 3?\r\n\r\nAnyone have a solution?", "Solution_1": "[hide]The probability of one randomly chosen point being inside the circle is $ \\frac{\\text{area of circle}}{\\text{area of square}}\\equal{}\\frac {9\\pi}{144} \\equal{} \\frac {\\pi}{16}$.\n\nFor three randomly chosen points to be inside the circle, the probability is $ \\left(\\frac {\\pi}{16}\\right)^3 \\equal{} \\frac {\\pi^3}{4096}$.[/hide]", "Solution_2": "I think you misinterpreted -- the circle is not a particular, predefined circle. Here's a rephrase:\r\n\r\n[i]What is the probability of choosing 3 points for which there exists at least one circle of radius 3 that contains all three chosen points?[/i]\r\n\r\nAnd this is comparatively quite nontrivial, I'd think :wink:", "Solution_3": "[quote=\"blackbelt14253\"]Three points are selected inside square ABCD with side length 12. What is the probability that all three lie inside a circle of radius 3?\n\nAnyone have a solution?[/quote]\r\n\r\nWell this is probably (definitely) not right... whatever\r\n[hide=\"augh\"]Call the points complex numbers $ z_1, z_2, z_3$. The points lie inside a circle of radius 3 iff there exists a point $ z$ such that\n\\[ |z_1 \\minus{} z|\\le 3,\\]\n\\[ |z_2 \\minus{} z|\\le3,\\]\n\\[ |z_3 \\minus{} z|\\le 3.\\]\nIf they do lie inside a circle of radius 3, then $ z \\equal{} \\frac {z_1 \\plus{} z_2 \\plus{} z_3}{3}$ should work as a possible point (center of gravity? iono).\n\nBut yeah so plug $ z$ in and get\n\\[ |2z_1 \\minus{} z_2 \\minus{} z_3|\\le 9,\\]\n\\[ | \\minus{} z_1 \\plus{} 2z_2 \\minus{} z_3|\\le 9,\\]\n\\[ | \\minus{} z_1 \\minus{} z_2 \\plus{} 2z_3|\\le 9\\]\nbut then let $ |z_1 \\minus{} z_2| \\equal{} a$, $ |z_2 \\minus{} z_3| \\equal{} b$, $ |z_3 \\minus{} z_1| \\equal{} c$ be the side lengths of the triangle formed by the three points, then\n\\[ |a \\plus{} b|\\le 9,\\]\n\\[ |b \\plus{} c|\\le 9,\\]\n\\[ |c \\plus{} a|\\le 9.\\]\nAssume $ a\\ge b\\ge c$ then $ b \\plus{} c > a$ and then we just need $ |a \\plus{} b|\\le 9$ and $ |a \\plus{} c|\\le 9$. \n\nAnd then I finished off my amazing reasoning with a bogus argument, used some geometric probability, and got an answer of $ \\frac {2}{3}$ which seems absurdly large. >_>\" [/hide]\n\nedit: oh yeah I also tried other stuff but couldn't get anything out of it:\n\n[hide]Law of sines: $ \\frac {a}{\\sin A} \\equal{} \\frac {b}{\\sin B} \\equal{} \\frac {c}{\\sin C} \\equal{} 2R$ \nBut if there exists a point which is the center of a circle with radius 3 that meets the requirements, then the circumcenter of the triangle formed by the three points should work again: $ \\frac {a}{\\sin A} \\equal{} \\frac {b}{\\sin B} \\equal{} \\frac {c}{\\sin C}\\le 6$ \n\nAlso, using $ [ABC] \\equal{} \\frac {abc}{4R}$, get $ [ABC] \\equal{} \\frac {abc}{4R}\\ge \\frac {abc}{12}$ or $ 12[ABC]\\ge abc$ and then I stopped there[/hide]", "Solution_4": "I'd be shocked if there was a nice solution. If you take one particular point, then the circle of radius 3 containing that point can be cut off by the side of the square...this makes for a nasty integral.", "Solution_5": "oh right...the circle is arbitrary", "Solution_6": "and it nowhere says circle has to be inside square so that complicates things further", "Solution_7": "[quote=\"Flame\"]and it nowhere says circle has to be inside square so that complicates things further[/quote]\r\n\r\nThe center of the circle might not be in the square, but some part of the circle must be in the square (since all three points are in the square)", "Solution_8": "[quote]The center of the circle might not be in the square, but some part of the circle must be in the square (since all three points are in the square)[/quote]\r\n\r\nI don't think it matters whether the center is restricted to being in the square.", "Solution_9": "Yeah it does. Say your three points form a right triangle with sidelengths $ 3/\\sqrt{2}, \\; 3/\\sqrt{2}, \\; 3$, and the hypotenuse of the triangle is part of an edge :wink:", "Solution_10": "In any case, not more than a half of the circle has to be outside the square, so the probability should be $ P(x) \\geq \\frac {\\frac {9\\pi}{2}}{144} \\equal{} \\frac {9\\pi}{288} \\equal{} \\frac {\\pi}{32}$ (shouldn't it?).\r\n\r\nOr perhaps we should focus more on the distances of the points, since the points have to be within a distance of 6 from each other (unless someone can prove that you can choose two points inside a circle with a distance greater than the diameter of said circle)?", "Solution_11": "A rather lax upper bound comes about if we fix one point in the center and find the probability that two other points are both within 3 units of this point.\r\n\r\nThis gives a very lax upper bound of $ \\left( \\frac {9 \\pi}{144} \\right)^2 \\approx 0.04$. This, of course, is [b]much [/b]higher than the actual probability (though it knocks out the answer above :) ).\r\n\r\nHere's an approach that's probably nicer than using a six-way integral, but doesn't work when you get close to the edge.\r\n\r\nConsider two points that are 3 units or less apart. Draw the two circles of radius 3 that pass through these two points, and their union is the area of the plane in which the third can reside. We can now vary the first point, bound and vary the second point, and then bound, and then bound and vary the two-circle union area for the third point. Of course, it'll be mega-piecewise (at least 9 pieces) because of the square edges.\r\n\r\nWould the problem be neater if the three points were chosen from a circle? There would still be a boundary, but it would be more consistent. And we could assume that the first point lies on a particular radius.", "Solution_12": "Well, now that I think it a bit more, the probability should actually be $ P(x) \\geq \\left(\\frac {\\pi}{32}\\right)^{2} \\equal{} \\frac {\\pi^{2}}{1024} \\approx 0.01$, since we're supposed to choose [i]three[/i] points, so perhaps we could set the range of $ P(x)$ to be something similar to $ 0.01 \\leq P(x) \\leq 0.04$?", "Solution_13": "well if anyone can figure it out (WITHOUT CALCULUS, i havent learned that yet) can you generalize it to a square with side m and circle with size n? (and maybe even k dots)", "Solution_14": "Suggestion: Consider the triangle connecting the three points. If its circumcircle has radius greater than 3, then no circle of radius 3 contains them all. If its circumcircle has radius less than or equal to 3, then take the circle of radius 3 centered at the circumcenter, and you've got a circle of radius 3 containing them all.\r\n\r\nUsing this, you can (at the least) write an inequality that holds if and only if the coordinates of the three points have the required circle. But I'm guessing that this idea can lead to a better approach than that.\r\n\r\nMaybe this as a start: You have two points, $ x$ and $ y$ that are $ a$ units apart. Then for what points $ z$ will the circumradius of tri. $ XYZ$ be less than 6? Well, $ a/\\sin{\\alpha} \\leq 6$ iff $ \\sin{\\alpha} \\geq a/6$. So we're looking for points that will make certain angles with the first two we've chosen. I'm no geometry expert, so maybe someone else can pick up on this.", "Solution_15": "Actually... that won't work. Having circumradius greater than 3 does not imply they won't fit into some circle of radius 3. Oh well, back to 6-integral methods." } { "Tag": [], "Problem": "Find all numbers whose prime divisors are $2$ and/or $5$ such that $25$ added to that number gives a prefect square.", "Solution_1": "[hide=\"Solution\"] We'll work backwards. We want $n$ such that $n^{2}-25$ only has prime divisors of $2$ and $5$.\n\nWell, that factors to $(n+5)(n-5)$. So $n-5$ must have only prime divisors $2$ and $5$. Let $n-5 = 2^{a}5^{b}$.\n\n[b]Case:[/b] $b = 0$. Then $n+5 = 2^{a}+10 = 2(2^{a-1}+5)$. The second factor is odd, but it cannot be a power of $5$, so there are no solutions in this case.\n\n[b]Case:[/b] $b = 1$. Then $n+5 = 10(2^{a-1}+1)$. The second factor is odd, and is a power of $2$ when $a = 1$ and a power of $5$ when $a = 1, 3$, which gives $n+5 = 20, n^{2}-25 = \\boxed{ 200 }$ and $n+5 = 50, n^{2}-25 = \\boxed{ 2000 }$. For $a > 3$ the expression is odd so it must e some power of $5$; call it $5^{b}$. We require that $5^{b}-1 \\equiv 0 \\bmod 8 \\implies b \\equiv 0 \\bmod 2$. But then $3 | 5^{b}-1$, so there are no more solutions in this case.\n\n[b]Case:[/b] $b > 1$. Then $n+5 = 10(2^{a-1}5^{b-1}+1)$. If $a > 1$ the second factor isn't divisible by $2$ or by $5$, so $a = 1$. But $5^{b-1}+1 \\equiv 2 \\bmod 4$, and it is larger than $2$, so it cannot be a power of $2$ and there are no more solutions in this case. [/hide]" } { "Tag": [ "integration", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "I have little knowledge in differential,\r\nSo how to solved this differential equation:\r\n$ f'(x)\\equal{}1\\minus{}(f(x))^2$,such that $ f(0)\\equal{}0$.\r\nWhere $ f(x)$ is a differential in $ \\mathbb{R}$.\r\n\r\nI know the answer may should by $ f(x)\\equal{}\\frac{e^x\\minus{}e^{\\minus{}x}}{e^x\\plus{}e^{\\minus{}x}}$", "Solution_1": "You could use the notation:\r\n\r\n$ f'(x)\\equal{}\\frac{df}{dx}$\r\n\r\nThen you can write:\r\n\r\n$ \\frac{df}{dx}\\equal{}1\\minus{}f^2$\r\n\r\nor\r\n\r\n$ \\frac{df}{1\\minus{}f^2}\\equal{}dx$\r\n\r\nNow, integrate:\r\n\r\n$ \\int\\frac{df}{1\\minus{}f^2}\\equal{}x\\plus{}C$\r\n\r\nYou could use the fact that:\r\n\r\n$ \\frac{1}{1\\minus{}f^2}\\equal{}\\frac{1}{2}\\left(\\frac{1}{1\\plus{}f}\\plus{}\\frac{1}{1\\minus{}f}\\right)$\r\n\r\nand then fact that:\r\n\r\n$ \\int \\frac{df}{1\\plus{}f}\\equal{}\\ln(1\\plus{}f)$\r\n\r\n$ \\int \\frac{df}{1\\minus{}f}\\equal{}\\minus{}\\ln(1\\minus{}f)$\r\n\r\nSo the we have:\r\n\r\n$ \\int\\frac{df}{1\\minus{}f^2}\\equal{}\\frac{1}{2}\\ln(1\\plus{}f)\\minus{}\\frac{1}{2}\\ln(1\\minus{}f)\\equal{}\\frac{1}{2}\\ln{\\frac{1\\plus{}f}{1\\minus{}f}}\\equal{}x\\plus{}C$\r\n\r\n$ \\frac{1\\plus{}f}{1\\minus{}f}\\equal{}Ce^{2 x}$\r\n\r\nFor the ease of writing, denote $ Ce^{2 x}\\equal{}A$\r\n\r\n$ \\frac{1\\plus{}f}{1\\minus{}f}\\equal{}A$\r\n\r\n$ 1\\plus{}f\\equal{}A(1\\minus{}f)$\r\n\r\n$ 1\\plus{}f\\equal{}A\\minus{}A f$\r\n\r\n$ f(1\\plus{}A)\\equal{}A\\minus{}1$\r\n\r\n$ f(x)\\equal{}\\frac{A\\minus{}1}{A\\plus{}1}\\equal{}\\frac{A\\minus{}1}{A\\plus{}1}\\equal{}\\frac{Ce^{2 x}\\minus{}1}{Ce^{2 x}\\plus{}1}\\equal{}\\frac{Ce^{x}\\minus{}e^{\\minus{}x}}{Ce^{x}\\plus{}e^{\\minus{}x}}$\r\n\r\nSince:\r\n\r\n$ f(0)\\equal{}\\frac{C\\minus{}1}{C\\plus{}1}\\equal{}0$ and so $ C\\equal{}1$.\r\n\r\nTherefore:\r\n\r\n$ f(x)\\equal{}\\frac{e^{x}\\minus{}e^{\\minus{}x}}{e^{x}\\plus{}e^{\\minus{}x}}$", "Solution_2": "Hi,I believe you are right this time,Thank you very much.\r\nBut you have use the fact $ (f(x))^2<1$ in the solution,\r\nI think it also need a explain although it is not hard :)", "Solution_3": "I have? It's just a way of solving these types of differential equations I've learned in college. I do not see where have I used that fact.. Could you point it out for me, please!", "Solution_4": "[quote=\"milin\"]\n\n$ \\int \\frac {df}{1 \\plus{} f} \\equal{} \\ln(1 \\plus{} f)$\n\n$ \\int \\frac {df}{1 \\minus{} f} \\equal{} \\minus{} \\ln(1 \\minus{} f)$\n\n[/quote]\r\n\r\nYou have used here.", "Solution_5": "he actually didn't use that fact.\r\nhe just forgot to put absolute values ;)", "Solution_6": "But how can we do when there is some point x such that $ f(x)\\equal{}1$???", "Solution_7": "[quote=\"zhaobin\"]But how can we do when there is some point x such that $ f(x) \\equal{} 1$???[/quote]\r\n\r\nIn fact, $ f(x) \\ne 1$ for all $ x$.", "Solution_8": "For $ f'(x) \\equal{} 1 \\minus{} \\{f(x)\\}^2$, by $ f(0) \\equal{} 0$, we have that $ f(x)$ isn't always 0. Thus we have \r\n\r\n$ \\frac {f'(x)}{1 \\minus{} \\{f(x)\\}^2} \\equal{} 1\\Longleftrightarrow \\frac {f'(x)}{1 \\plus{} f(x)} \\plus{} \\frac {f'(x)}{1 \\minus{} f(x)} \\equal{} 2$\r\n\r\n$ \\therefore \\frac {d}{dx} \\ln \\left|\\frac {1 \\plus{} f(x)}{1 \\minus{} f(x)}\\right| \\equal{} 2$\r\n\r\n$ \\therefore \\ln \\left|\\frac {1 \\plus{} f(x)}{1 \\minus{} f(x)}\\right| \\equal{} 2x \\plus{} A$\r\n\r\n$ \\therefore \\frac {1 \\plus{} f(x)}{1 \\minus{} f(x)} \\equal{} Ce^{2x}$ where $ C \\equal{} \\pm e^A$.\r\n\r\n$ f(0) \\equal{} 0\\Longleftrightarrow C \\equal{} 1$, yielding $ f(x) \\equal{} \\frac {e^{2x} \\minus{} 1}{e^{2x} \\plus{} 1}$", "Solution_9": "[quote=\"bookworm_vn\"][quote=\"zhaobin\"]But how can we do when there is some point x such that $ f(x) \\equal{} 1$???[/quote]\n\nIn fact, $ f(x) \\ne 1$ for all $ x$.[/quote]\r\nBut when you haven't show that $ f(x) \\equal{} \\frac {e^x \\minus{} e^{ \\minus{} x}}{e^x \\plus{} e^{ \\minus{} x}}$,You can't assert that,although there is some way to show that,But we when you write solution you should give a reason.Ok,maybe there is something explain in the differential equation course,but I don't know,I am sorry.\r\nMy way to show $ f(x) \\ne 1$,is that,if there is some point $ f(x_1)\\equal{}1$,then $ f'(x_1)\\equal{}0$,Then try to show $ f(x)\\equal{}1$ for any $ x$.\r\n\r\n\r\nThank you kunny.But in your solution you also assume $ \\frac{1\\plus{}f(x)}{1\\minus{}f(x)}>0$ without any reason.", "Solution_10": "[quote=\"zhaobin\"]\nThank you kunny.But in your solution you also assume $ \\frac {1 \\plus{} f(x)}{1 \\minus{} f(x)} > 0$ without any reason.[/quote]\r\n\r\nIt isn't so, $ \\ln\\left|\\frac{1\\plus{}f(x)}{1\\minus{}f(x)}\\right|\\equal{}2x\\plus{}A\\Longleftrightarrow \\left|\\frac{1\\plus{}f(x)}{1\\minus{}f(x)}\\right|\\equal{}e^{2x\\plus{}A}$\r\n\r\n$ \\Longleftrightarrow \\frac{1\\plus{}f(x)}{1\\minus{}f(x)}\\equal{}\\pm e^{2x\\plus{}A}\\equal{}\\pm e^A\\cdot e^{2x}\\equal{}Ce^{2x}$ where $ C\\equal{}\\pm e^{A}$.", "Solution_11": "Ok,Thank you,I am wrong in reading this.", "Solution_12": ":)", "Solution_13": "Once we have a solution, we can invoke the existence-uniqueness theorem. $ \\tanh$ is the only solution satisfying the initial conditions.\r\n\r\nWhat do the general solutions (over $ \\mathbb{R}$) look like? There's $ y \\equal{} \\tanh(x \\minus{} a)$, $ y \\equal{} 1$, $ y \\equal{} \\minus{} 1$, and $ y \\equal{} \\coth(x \\minus{} a)$.\r\n\r\nThe latter doesn't cover all of $ \\mathbb{R}$; the formula is a positive solution for $ x > a$ and a negative solution for $ x < a$.", "Solution_14": "[quote=\"zhaobin\"][quote=\"bookworm_vn\"][quote=\"zhaobin\"]But how can we do when there is some point x such that $ f(x) \\equal{} 1$???[/quote]\n\nIn fact, $ f(x) \\ne 1$ for all $ x$.[/quote]\nBut when you haven't show that $ f(x) \\equal{} \\frac {e^x \\minus{} e^{ \\minus{} x}}{e^x \\plus{} e^{ \\minus{} x}}$,You can't assert that,although there is some way to show that,But we when you write solution you should give a reason.Ok,maybe there is something explain in the differential equation course,but I don't know,I am sorry.\nMy way to show $ f(x) \\ne 1$,is that,if there is some point $ f(x_1) \\equal{} 1$,then $ f'(x_1) \\equal{} 0$,Then try to show $ f(x) \\equal{} 1$ for any $ x$.[/quote]\r\n\r\nOK, I have a question for you. Under what Cauchy condition there exists $ x$ such that $ f(x) \\equal{} 1$. I mean $ f(0)$ equals to what?", "Solution_15": "[quote=\"zhaobin\"][quote=\"bookworm_vn\"][quote=\"zhaobin\"]But how can we do when there is some point x such that $ f(x) \\equal{} 1$???[/quote]\n\nIn fact, $ f(x) \\ne 1$ for all $ x$.[/quote]\nBut when you haven't show that $ f(x) \\equal{} \\frac {e^x \\minus{} e^{ \\minus{} x}}{e^x \\plus{} e^{ \\minus{} x}}$,You can't assert that,although there is some way to show that,But we when you write solution you should give a reason.Ok,maybe there is something explain in the differential equation course,but I don't know,I am sorry.\nMy way to show $ f(x) \\ne 1$,is that,if there is some point $ f(x_1) \\equal{} 1$,then $ f'(x_1) \\equal{} 0$,Then try to show $ f(x) \\equal{} 1$ for any $ x$.\n\n\nThank you kunny.But in your solution you also assume $ \\frac {1 \\plus{} f(x)}{1 \\minus{} f(x)} > 0$ without any reason.[/quote]\r\nif a solution starts at a non-critical value, it can not reach a critical point in a finite amount of time, as Jmerry said this would contradict the existence-uniqueness theorem. :)", "Solution_16": "Hi,guys,Thank you very much.But I have said that I am unfamiliar with the differential equaitons.And thanks to jmerry's comment.", "Solution_17": "[quote=\"zhaobin\"]Hi,guys,Thank you very much.But I have said that I am unfamiliar with the differential equaitons.And thanks to jmerry's comment.[/quote]\r\n\r\noh, that's why you should improve your ODEs right now :P", "Solution_18": "[quote=\"bookworm_vn\"]\noh, that's why you should improve your ODEs right now :P[/quote]\r\nThanks for your advice. :)" } { "Tag": [ "modular arithmetic" ], "Problem": "Find the smallest positive integer $ x$ that satisfies the congruence $ 723x \\equiv 117 \\pmod {720}.$", "Solution_1": "$ 723x\\equiv 720x \\plus{} 3x\\equiv 3x\\pmod{720}$, so we have $ 3x\\equiv 117\\pmod{720}$\r\n\r\nDividing by 3, we have $ x\\equiv 39\\pmod {240}$, so $ x \\equal{} 240k \\plus{} 39,\\;k\\in\\mathbb{Z}$.\r\n\r\nTo minimize $ x$ we set $ k \\equal{} 0$ and have $ \\boxed{x \\equal{} 39}$.\r\n\r\nChecking, $ 723\\cdot 39\\equiv 28197\\equiv 28080 \\plus{} 117\\equiv 117\\pmod{720}$.", "Solution_2": "you can't divide in a congruence.... well you can but with restrictions.", "Solution_3": "Um...the restriction is simple...\r\n\r\nIf we have $ a\\equiv b\\pmod m$, then $ \\frac ab\\equiv1\\pmod{\\frac m{\\gcd(b,m)}}$." } { "Tag": [], "Problem": "Hmm, i found this from Gabriel Carol.\r\nGod this creature is a real genuis,just look at these:\r\nhttp://www.people.fas.harvard.edu/~gcarroll/archives/puzzles/panalogies.htm\r\n\r\nOops.I think i should take this to Round table ,sorry", "Solution_1": "that site has nice questions", "Solution_2": "Those anologies are sure different from the one's on the SAT's. They have weird variations. Atomic number? Are you kidding me? :D", "Solution_3": "they are hard. that's what i mean by nice :)" } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "Let $a,b,x,y>0$. Prove that we have:\r\n\r\n$(a \\frac{x}{y}+b)^2+(a \\frac{y}{x}+b)^2 \\geq 2(a+b)^2$.\r\n\r\ncheers! :D :D :cool:", "Solution_1": "With a straight forward application of AM-QM, we get\r\n\r\n(ax/y+b) 2 +(ay/x+b) 2 \\geq 2 [ a(x/y+y/x)/2 + b ] 2 \\geq 2 (a+b) 2 \r\n\r\n(the last step follows from a well known fact that r + 1/r \\geq 2)\r\n\r\nEquality occurs iff x=y." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Can you prove it without SOS ?", "Solution_1": "[quote=\"duc 95\"]Can you prove it without SOS ?[/quote]\r\n\r\n\r\ntry $ a \\equal{} \\frac {1}{4},b \\equal{} \\frac {1}{2},c \\equal{} 1.$", "Solution_2": "Do you think this is wrong?", "Solution_3": "I think it is right", "Solution_4": "[quote=\"duc 95\"]I think it is right[/quote]\n\n\nthis is wrong.\n\n[quote=\"hedeng123\"][quote=\"duc 95\"]Can you prove it without SOS ?[/quote]\n\n\ntry $ a \\equal{} \\frac {1}{4},b \\equal{} \\frac {1}{2},c \\equal{} 1.$[/quote]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b,c\\geq0$.Prove that:\r\n$ a^3\\plus{}b^3\\plus{}c^3\\plus{}3abc \\geq bc\\sqrt{2(b^2\\plus{}c^2)}\\plus{}ab\\sqrt{2(a^2\\plus{}b^2)}\\plus{}ca\\sqrt{2(c^2\\plus{}a^2)}$", "Solution_1": "See here : http://www.mathlinks.ro/viewtopic.php?t=6317", "Solution_2": "If $ x,y,z>0,$ then\r\n\r\n$ x^3 \\plus{} y^3 \\plus{} z^3 \\plus{} 3xyz\\geq\\frac {2(yz \\plus{} zx \\plus{} xy)(x^2 \\plus{} y^2 \\plus{} z^2)}{x \\plus{} y \\plus{} z}$\r\n\r\n$ \\geq yz\\sqrt {2y^2 \\plus{} 2z^2} \\plus{} zx\\sqrt {2z^2 \\plus{} 2x^2} \\plus{} xy\\sqrt {2x^2 \\plus{} 2y^2}.$\r\n\r\nSee also : http://www.mathlinks.ro/viewtopic.php?t=216948", "Solution_3": "Sorry, dear Ji Chen, can you please explain why this inequality is true?\r\n\r\nThank you very much. :)\r\n\r\n\r\n$ \\frac {2(yz + zx + xy)(x^2 + y^2 + z^2)}{x + y + z}\n\\geq yz\\sqrt {2y^2 + 2z^2} + zx\\sqrt {2z^2 + 2x^2} + xy\\sqrt {2x^2 + 2y^2}.$", "Solution_4": "Dear manlio, \r\n\r\n$ \\sum{yz\\sqrt {2y^2 + 2z^2}}\\leq\\sum{\\frac{yz(3y^2+2yz+3z^2)}{2(y+z)}}$\r\n\r\n\r\n$ =\\frac {2(yz + zx + xy)(x^2 + y^2 + z^2)}{x + y + z} -\\sum{\\frac{yz(y-z)^2[x(y+z-x)^2+yz(x+y+z)]}{2(y+z)(z+x)(x+y)(x+y+z)}}.$\r\n\r\nBy the way,\r\n\r\n$ (x+y+z)(x^3+y^3+z^3+3xyz)-2(yz + zx + xy)(x^2 + y^2 + z^2)$\r\n\r\n$ =\\sum{x^2(x-y)(x-z)}=\\frac{1}{2}\\sum{(y-z)^2(y+z-x)^2\\geq0}$\r\n\r\nholds for all real numbers $ x,y,z.$", "Solution_5": "[quote=\"Ji Chen\"]Dear manlio, \n$ \\sum{yz\\sqrt {2y^2 \\plus{} 2z^2}}\\leq\\sum{\\frac {yz(3y^2 \\plus{} 2yz \\plus{} 3z^2)}{2(y \\plus{} z)}}$\n[/quote]\r\nBeautiful hint, Jichen :lol: \r\nI tried to use: $ \\sum{yz\\sqrt {2y^2 \\plus{} 2z^2}}\\leq \\sum\\ \\frac{2yz(y^2\\plus{}z^2)}{y\\plus{}z}$, and don't obtain a true result :oops:", "Solution_6": "[quote=\"Ji Chen\"]Dear manlio, \n\n$ \\sum{yz\\sqrt {2y^2 + 2z^2}}\\leq\\sum{\\frac {yz(3y^2 + 2yz + 3z^2)}{2(y + z)}}$\n\n\n$ = \\frac {2(yz + zx + xy)(x^2 + y^2 + z^2)}{x + y + z} - \\sum{\\frac {yz(y - z)^2[x(y + z - x)^2 + yz(x + y + z)]}{2(y + z)(z + x)(x + y)(x + y + z)}}.$\n\nBy the way,\n\n$ (x + y + z)(x^3 + y^3 + z^3 + 3xyz) - 2(yz + zx + xy)(x^2 + y^2 + z^2)$\n\n$ = \\sum{x^2(x - y)(x - z)} = \\frac {1}{2}\\sum{(y - z)^2(y + z - x)^2\\geq0}$\n\nholds for all real numbers $ x,y,z.$[/quote]\r\n\r\nWonderful, Ji Chen. :) \r\n\r\nThank you very much :)" } { "Tag": [ "LaTeX" ], "Problem": "This isn't exactly related to $\\LaTeX$, but how do you scan several files onto your computer and then merge them into one PDF file?", "Solution_1": "You can merge PDF's with Adobe's Acrobat \"Standard\" or \"Professional\" product. It's not particularly cheap though. Adobe gives away the \"Reader\" hoping folks like you will decide at some point that merging and \"editing\" PDF's will be worth a few hundred dollars (educational and other discounts can make the price reasonable though.) \r\n\r\nThe \"editing\" feature allows you to delete pages from a document. The software is capable of creating a new document using subsets of the pages from several documents. I've used this software and it works reliably.\r\n\r\nThere are other sources of \"free\" software that perform some of the same tasks but probably won't be as friendly to use. Try Google on \"Merge PDF\" or similar terms to find the \"freebies.\" I wouldn't put one on my computer unless someone I trusted had good experience with the product. \r\n\r\n- Greg" } { "Tag": [ "calculus", "integration", "logarithms", "function", "geometry", "limit", "trigonometry" ], "Problem": "find $\\int_{0}^{oo}\\frac{\\ln{x}dx}{x^{3}+1}$=$?$", "Solution_1": "\\begin{eqnarray*}\\int_{0}^{\\infty}\\frac{\\ln x}{1+x^{3}}\\, dx & = & \\int_{0}^{1}\\frac{\\ln x}{1+x^{3}}\\, dx+\\int_{1}^{\\infty}\\frac{\\ln x}{1+x^{3}}\\, dx\\\\ & = & \\int_{0}^{1}\\frac{\\ln x}{1+x^{3}}\\, dx+\\int_{0}^{1}\\frac{-t \\ln t}{1+t^{3}}\\, dt\\\\ & = & \\int_{0}^{1}\\frac{(1-x) \\ln x}{1+x^{3}}\\, dx\\\\ & = & \\int_{0}^{1}\\left( \\frac{1}{1+x}-\\frac{x^{2}}{1+x^{3}}\\right) \\ln x \\, dx\\\\ & = & \\int_{0}^{1}\\frac{1}{1+x}\\ln x \\, dx-\\int_{0}^{1}\\frac{x^{2}}{1+x^{3}}\\ln x \\, dx\\\\ & = & \\int_{0}^{1}\\frac{1}{1+x}\\ln x \\, dx-\\frac{1}{9}\\int_{0}^{1}\\frac{1}{1+u}\\ln u \\, du\\\\ & = & \\frac{8}{9}\\int_{0}^{1}\\frac{1}{1+x}\\ln x \\, dx\\\\ & = &-\\frac{8}{9}\\int_{0}^{\\infty}\\frac{v e^{-v}}{1+e^{-v}}\\, dv\\\\ & = &-\\frac{8}{9}\\int_{0}^{\\infty}v \\left( e^{-v}-e^{-2v}+e^{-3v}-e^{-4v}+\\cdots \\right) \\, dv\\\\ & = &-\\frac{8}{9}\\left( 1-\\frac{1}{2^{2}}+\\frac{1}{3^{2}}-\\frac{1}{4^{2}}+\\cdots \\right)\\\\ & = &-\\frac{8}{9}\\left( \\frac{\\pi^{2}}{12}\\right)\\\\ & = &-\\frac{2\\pi^{2}}{27}\\end{eqnarray*}\r\n\r\nwhere $t = 1/x$, $u = x^{3}$ and $x = e^{-v}$.", "Solution_2": "Thanks sos440. Later I will post sulution in more different way by complex variable", "Solution_3": "Umm...I think that ${\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{n^{2}}=\\frac{\\pi^{2}}{8}}$...or I miss something :?:", "Solution_4": "You missed something.", "Solution_5": "[quote=\"Extremal\"]find $\\int_{0}^{oo}\\frac{\\ln{x}dx}{x^{3}+1}$=$?$[/quote]\r\nin general: \r\n$\\int_{0}^{oo}\\frac{\\ln{x}dx}{x^{n}+1}$=$?$ :lol:", "Solution_6": "Generally, for $n>1$, we have\r\n\r\n$\\int_{0}^{\\infty}\\frac{\\ln{x}\\,dx}{x^{n}+1}=-\\frac{\\pi^{2}}{n^{2}}\\cot\\left(\\frac{\\pi}{n}\\right)\\csc\\left(\\frac{\\pi}{n}\\right)$\r\n\r\nExamples include\r\n\r\n$\\int_{0}^{\\infty}\\frac{\\ln{x}\\,dx}{x^{2}+1}= 0$\r\n$\\int_{0}^{\\infty}\\frac{\\ln{x}\\,dx}{x^{3}+1}=-\\frac{2\\pi^{2}}{27}$\r\n$\\int_{0}^{\\infty}\\frac{\\ln{x}\\,dx}{x^{4}+1}=-\\frac{\\pi^{2}}{8\\sqrt{2}}$", "Solution_7": "[quote=\"sos440\"]Generally, for $n>1$, we have\n\n$\\int_{0}^{\\infty}\\frac{\\ln{x}\\,dx}{x^{n}+1}=-\\frac{\\pi^{2}}{n^{2}}\\cot\\left(\\frac{\\pi}{n}\\right)\\csc\\left(\\frac{\\pi}{n}\\right)$\n\nExamples include\n\n$\\int_{0}^{\\infty}\\frac{\\ln{x}\\,dx}{x^{2}+1}= 0$\n$\\int_{0}^{\\infty}\\frac{\\ln{x}\\,dx}{x^{3}+1}=-\\frac{2\\pi^{2}}{27}$\n$\\int_{0}^{\\infty}\\frac{\\ln{x}\\,dx}{x^{4}+1}=-\\frac{\\pi^{2}}{8\\sqrt{2}}$[/quote]\r\ncan you post your sulution?", "Solution_8": "I integrated the function $f(z) = \\frac{\\ln z}{1+z^{n}}$ along the path $C$, the boundary of pizza-shaped area, travelled counter-clockwise:\r\n\r\n[img]http://myhome.shinbiro.com/~sosQED/nsol088.gif[/img]\r\n\r\nwhere $\\theta = \\frac{2\\pi}{n}$. For $n = 2, 3, ...$, this area contains exactly one root of $z^{n}+1 = 0$, namely $\\omega = e^{i\\theta/2}$. And then compare two sides of the equation\r\n\r\n$2\\pi i \\underset{z=\\omega}{\\text{Res}}\\, \\frac{\\ln z}{1+z^{n}}= \\int_{C}\\frac{\\ln z}{1+z^{n}}\\, dz$ $\\quad \\Longrightarrow \\quad$ $\\lim_{R\\to\\infty}\\int_{C}\\frac{\\ln z}{1+z^{n}}\\, dz = 2\\pi i \\frac{\\ln \\omega}{n \\omega^{n-1}}= \\frac{2\\pi^{2}\\omega}{n^{2}}$\r\n\r\nIf we use the fact that the integral of $f(z)$ over the arc part of the path $C$ becomes zero, then \r\n\r\n\\begin{eqnarray*}\\lim_{R\\to\\infty}\\int_{C}\\frac{\\ln z}{1+z^{n}}\\, dz & = & \\int_{0}^{\\infty}\\frac{\\ln z}{1+z^{n}}\\, dz+\\int_{\\omega^{2}\\infty}^{0}\\frac{\\ln z}{1+z^{n}}\\, dz\\\\ & = & \\int_{0}^{\\infty}\\frac{\\ln x}{1+x^{n}}\\, dx-\\int_{0}^{\\infty}\\frac{\\ln (\\omega^{2}x)}{1+(\\omega^{2}x)^{n}}\\, \\omega^{2}dx\\\\ & = & (1-\\omega^{2})\\int_{0}^{\\infty}\\frac{\\ln x}{1+x^{n}}\\, dx-\\frac{2\\pi i \\omega^{2}}{n}\\int_{0}^{\\infty}\\frac{1}{1+x^{n}}\\, dx \\end{eqnarray*}\r\n\r\nFinally, use the equation $\\int_{0}^{\\infty}\\frac{dx}{1+x^{n}}= \\frac{\\pi}{n}\\cot \\frac{\\pi}{n}$ then we have desired result.", "Solution_9": "Or: \r\n\r\nLet $I(\\alpha)=\\int_{0}^{\\infty}\\frac{x^{\\alpha}}{x^{n}+1}dx$.\r\n\r\nA calculation shows that:\r\n\r\n${I(\\alpha)=\\frac{\\pi}{n}\\frac{1}{\\sin \\frac{(\\alpha+1)\\pi}{n}}}$.\r\n\r\nThen,\r\n \r\n$I'(\\alpha)=\\int_{0}^{\\infty}\\frac{x^{\\alpha}\\ln x}{x^{n}+1}dx=-\\frac{\\pi^{2}}{n^{2}}\\frac{\\cos \\frac{(\\alpha+1)\\pi}{n}}{\\sin^{2}\\frac{(\\alpha+1)}{n}}$.\r\nNow, let $\\alpha \\rightarrow 0$, and you are done.", "Solution_10": "[quote=\"didilica\"]$I'(\\alpha)=\\int_{0}^{\\infty}\\frac{x^{\\alpha}\\ln x}{x^{n}+1}dx=-\\frac{\\pi^{2}}{n^{2}}\\frac{\\cos \\frac{(\\alpha+1)\\pi}{n}}{\\sin^{2}\\frac{(\\alpha+1)}{n}}$.[/quote]\r\nDo you mean:\r\n\r\n$I'(\\alpha)=\\int_{0}^{\\infty}\\frac{x^{\\alpha}\\ln x}{x^{n}+1}dx=-\\frac{\\pi^{2}}{n^{2}}\\frac{\\cos \\frac{(\\alpha+1)\\pi}{n}}{\\sin^{2}\\frac{(\\alpha+1)\\pi}{n}}$ ?", "Solution_11": "I proved next:\r\n$\\int_{0}^{oo}\\frac{\\ln{x}dx}{x^{n}+1}=-1+2\\sum_{k=1}^{oo}\\frac{1+k^{2}n^{2}}{(1-k^{2}n^{2})^{2}}(-1)^{k+1}$ \r\nand how can I find this sum?\r\n\r\n\r\nsos440: (I don't know some defination and theorem in complex analsisy, therefore I have some simpl question).\r\nHow ? \r\n1) $2\\pi i \\frac{\\ln \\omega}{n \\omega^{n-1}}= \\frac{2\\pi^{2}\\omega}{n^{2}}$\r\n2)$\\begin{eqnarray*}\\lim_{R\\to\\infty}\\int_{C}\\frac{\\ln z}{1+z^{n}}\\, dz & = & \\int_{0}^{\\infty}\\frac{\\ln z}{1+z^{n}}\\, dz+\\int_{\\omega^{2}\\infty}^{0}\\frac{\\ln z}{1+z^{n}}\\, dz\\\\ &$\r\nWhat is it: $\\omega^{2}\\infty$ ?\r\n3) How? $\\& \\int_{0}^{\\infty}\\frac{\\ln x}{1+x^{n}}\\, dx-\\int_{0}^{\\infty}\\frac{\\ln (\\omega^{2}x)}{1+(\\omega^{2}x)^{n}}\\, \\omega^{2}dx\\\\ = (1-\\omega^{2})\\int_{0}^{\\infty}\\frac{\\ln x}{1+x^{n}}\\, dx-\\frac{2\\pi i \\omega^{2}}{n}\\int_{0}^{\\infty}\\frac{1}{1+x^{n}}\\, dx \\end{eqnarray*}$\r\n4) Hoe do you find it? ${I(\\alpha)=\\frac{\\pi}{n}\\frac{1}{\\sin \\frac{(\\alpha+1)\\pi}{n}}}$.\r\n$\\alpha$ is real?\r\n5)how? $I'(\\alpha)=\\int_{0}^{\\infty}\\frac{x^{\\alpha}\\ln x}{x^{n}+1}dx$", "Solution_12": "1) $\\omega^{n}= (e^{i\\pi/n})^{n}= e^{i\\pi}=-1$, so $\\omega^{1-n}=-\\omega$. And also, $\\ln \\omega = \\ln (e^{i\\pi/n}) = \\frac{i\\pi}{n}$.\r\n\r\n2) $\\int_{\\omega^{2}\\infty}^{0}f(z) \\, dz$ is the conventional notation for the integral of $f(z)$ along the path $C = \\{\\omega^{2}x : x \\in [0, \\infty)\\}$, travelled centripetal direction." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "i hope i have not asked this question before... :| can you show :D\r\n\r\n$ \\sum_{k\\equal{}1}^{\\infty}\\;\\;\\frac{(\\minus{}1)^{k\\plus{}1}}{k^{3}\\cdot\\dbinom{2k}{k}}\\;\\;\\equal{}\\;\\;\\boxed{\\frac{2}{5}\\,\\zeta{(3)}}$", "Solution_1": "see : http://pagesperso-orange.fr/gery.huvent/articlespdf/Autour_primitive.pdf\r\npage 11 formula (7)\r\n\r\nApery formula : http://pagesperso-orange.fr/gery.huvent/articlespdf/article3.pdf" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Assume that a_i>0 for i=1,2,....n with a_n+1=a_n. Prove that\r\n\r\n\r\n \\sum (i=1,..,n) (a_i/(a_i+1))^n \\geq \\sum (i=1,...,n)(a_i+1/a_i)", "Solution_1": "Could you please write it more explicitely? The problem that I understand cannot be true. Thank you very much.", "Solution_2": "Sorry Harazi, I have forgotten brackets and also made a terrible mistake.\r\n\r\nThe correct inequality is: \r\n\r\nAssume that a_i>0 for i=1,2,....n with a_(n+1)=a_1. Prove that \r\n\r\n\r\n sum(i=1,..,n) (a_i/(a_(i+1))^n \\geq sum (i=1,...,n)(a_(i+1)/a_i)", "Solution_3": "Now everything is normal and -fortunately-I can solve it. It obviously reduces to proving that if a_1*a_2*...*a_n=1, then sum a^n_i>=sum 1/a_i. But this comes down to AM-GM and Cebasev, since 1/a_1=a_2*...*a_n<= 1/(n-1)(a^(n-1)_2+...+a^(n-1)_n). So sum 1/a_i<= sum a^(n-1)_i<=sum a^n_i. The last one follows from the fact that sum a^n_i>=(sum a^(n-1)_i)*(sum a_i)/n and of course sum a_i>=n." } { "Tag": [], "Problem": "Sam, John, and Sarah bought six books costing $ 5, 6, 7, 8, 10,$ and $ 15$ dollars. Sam bought two of these books, while John and Sarah bought the remaining books. If Sam paid four fifths of the amount paid by John, and Sarah did not spend more than $ 14$ dollars, then the amount paid by Sarah is:\r\n\r\n$ \\textbf{(A)}\\ 5 \\qquad \\textbf{(B)}\\ 6 \\qquad \\textbf{(C)}\\ 8 \\qquad \\textbf{(D)}\\ 11 \\qquad \\textbf{(E)}\\ 13$", "Solution_1": "[hide=\"Solution\"]\nLet $ S_m,J,S_w$ be the amounts that Sam, John and Sarah used, respectively.\n\n$ S_m\\plus{}J\\plus{}S_w\\equal{}51$\n\nWe know that $ S_m\\cdot \\dfrac{5}{4}\\equal{}J$ and that $ 12\\le S_m\\le 20\\implies 15\\le J\\le 25$.\n\nSince $ S_m$ and $ J$ are integers, $ S_m\\equal{}\\{12,16,20\\}$ and $ J\\equal{}\\{15,20,25\\}$, the only pair that makes $ S_w\\le 14$ is when $ S_m\\equal{}20$ and $ J\\equal{}25$ and $ S_w\\equal{}6$.\n\nSo the answer is $ \\mathbf{(B)}$\n\nNote: The books that Sam buys are the ones that cost \\$5 and \\$15. The books that John buys are the \\$7, \\$8 and \\$10 ones.\n[/hide]" } { "Tag": [ "analytic geometry", "combinatorial geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "There are $n$ distinct points on the plane. We color the middle of each line $A_iA_j, i \\neq j$ red. Find the minimum number od red points", "Solution_1": "I also remember something about $2n-3$, but I don't know if it can always be reached. To show that we have at least $2n-3$ midpoints, proceed by induction. Assume that the last time we added a point, we added it on the boundary of or outside the convex hull of the other $n-1$ points. It's easy to show that we add at least two more midpoints, between our new point and two of the vertices of the convex hull.\r\n\r\nEdit: this answer was meant to address [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26300]this[/url].", "Solution_2": "Hmm;.. I splitted the topic. To which problem are you answering? :D \r\n\r\nPierre.", "Solution_3": "I have edited my post to answer your question, Pierre :).", "Solution_4": "That's what I suspected while reading your answer. :P \r\n\r\nPierre.", "Solution_5": "[quote=\"grobber\"]I also remember something about $2n-3$, but I don't know if it can always be reached. [/quote]\r\n???\r\nTake $n$ points on line on the same distance from each other.", "Solution_6": "Oh, cool! :) I had a tendency to disregard degenerate cases :).", "Solution_7": "As for showing that there must be at least $2n-3$ distinct midpoints, we can set up coordinate axes such that the points have pairwise distinct $x$-coordinates. WLOG, we can number the points $A_1, A_2, \\ldots, A_n$ in order of increasing $x$-coordinate. Then it's easy to observe that the midpoints of the segments $A_1A_2, A_1A_3, \\ldots, A_1A_n, A_2A_n, A_3A_n, \\ldots, A_{n-1}A_n$ are all distinct.", "Solution_8": "or you can project the points on a line, not perpendicular to a line formed by two of them, and clearly the number of midpoints is less in the line.", "Solution_9": "thanks:)" }