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IMO-1975-3	https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_3	On the sides of an arbitrary triangle $ABC$, triangles $ABR, BCP, CAQ$ are constructed externally with $\angle CBP = \angle CAQ = 45^\circ, \angle BCP = \angle ACQ = 30^\circ, \angle ABR = \angle BAR = 15^\circ$. Prove that $\angle QRP = 90^\circ$ and $QR = RP$.	[ "Consider \$X\$ and \$Y\$ so that \$\\triangle CQA\\sim \\triangle CPX\$ and \$\\triangle CPB\\sim \\triangle CQY\$. Furthermore, let \$AX\$ and \$BY\$ intersect at \$F\$. Now, this means that \$\\angle PBC = 45 = \\angle QAC = \\angle PXC\$ and \$\\angle PCB = 15 = \\angle QCA = \\angle PCX\$, so \$\\triangle BPC\\cong \\triangle XPC\$. Thus, \$XC = BC\$ and \$\\angle BCX = 30 + 30 = 60\$, so \$\\triangle BCX\$ is equilateral. Similarly is \$\\triangle ACY\$. Yet, it is well-known that the intersection of \$BY\$ and \$AX\$, which is \$F\$, must be the Fermat Point of \$\\triangle ABC\$ if \$\\triangle AYC\$ and \$\\triangle BXC\$ are equilateral. Now, \$\\angle PBX = 60 - 45 = 15\$. Similarly, \$\\angle PXB = 15\$, so \$\\triangle BRA\\sim \\triangle BPX\$, so a spiral similarity maps \$\\triangle BPR\\sim \\triangle BXA\$. This implies that \$\\angle BRP = \\angle FAB\$. Similarly, \$\\angle ARQ = \\angle FBA\$, so \$\\angle BRP + \\angle ARQ = \\angle FBA + \\angle FAB = 180 - \\angle AFB = 60\$. Then, \$\\angle PRQ = (180 - 15 - 15) - 60 = 90\$. We also realize that \$\\frac {AX}{RP} = \\frac {AB}{RB} = \\frac {AB}{AR} = \\frac {BY}{RQ}\$. Now, \$Y\$ is a rotation of \$A\$ \$60\$ degrees around \$C\$ and \$B\$ is the same rotation of \$X\$ around \$C\$, so \$AX\$ maps to \$YB\$ from this rotation, so \$AX = YB\$. It follows that \$RP = RQ\$.\n\nThe above solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [1]", "Let \$\\triangle ABT\$ be the equilateral triangle constructed such that \$T\$ and \$R\$ are on the same side. \$\\triangle RTB \\sim \\triangle PCB \\sim \\triangle QCA\$. We have \$\\frac {AT}{AC} = \\frac {AR}{AQ}\$ from similarity. Also we have \$\\angle TAC = \\angle RAQ\$ . So \$\\triangle ACT \\sim \\triangle ARQ\$. Then \$\\angle ATC = \\angle ARQ = m\$ and \$\\frac {AR}{AT} = \\frac {RQ}{TC}\$. Similar calculations for \$B\$. We will have \$\\angle BTC = \\angle BRP = 60-m\$ and \$\\frac {BR}{AT} = \\frac {RP}{TC}\$. Also from the question we have \$AR = BR\$. So \$\\angle PRQ = 180 - (60-x) - (60-x) -m -(60-m) = 2x\$ and \$PR = RQ\$.\n\nThe above solution was posted and copyrighted by xeroxia. The original thread for this problem can be found here: [2]", "Define \$X\$ to be a point such that \$\\triangle AQC\$ is directly similar to \$\\triangle AXB\$.\n\nThen, it is trivial to show that \$\\angle AXR=60^{\\circ}\$ and that \$\\angle RXB=45^{\\circ}\$; that is, \$AX=AR=XR=RB\$ and \$\\triangle RBX\$ is a right isosceles triangle. If we prove that \$\\triangle RPQ\$ is similar to \$\\triangle RBX\$, then we will be done.\n\nAccording to the property of spiral similarity, it suffices to prove that \$\\triangle RBP\\sim\\triangle RXQ.\$\n\nSince \$\\triangle AQC\\sim \\triangle AXB\$, we have \$\\triangle AQX\\sim\\triangle ACB\$, and this gives \$\\angle RBP=60^{\\circ}+\\angle CBA=60^{\\circ}+\\angle QXA=\\angle RXQ\$. It remains to prove that \$\\frac{BP}{RB}=\\frac{XQ}{RX}\$.\n\nAs \$RB=RX\$, we must prove that \$BP=XQ\$. From Law of Sines on \$\\triangle BPC\$, we have \$BP=\\frac{a}{2\\cos 15^{\\circ}}\$. Since \$\\frac{XQ}{AX}=\\frac{a}{c}\$, it remains to prove that \$AX=AR=\\frac{c}{2\\cos 15^{\\circ}}\$, which is easily verified. We are done. \$\\blacksquare\$\n\nThe above solution was posted and copyrighted by fantasylover. The original thread for this problem can be found here: [3]" ]
IMO-1975-4	https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_4	When $4444^{4444}$ is written in decimal notation, the sum of its digits is $A$. Let $B$ be the sum of the digits of $A$. Find the sum of the digits of $B$. ($A$ and $B$ are written in decimal notation.)	[ "Note that\n\n\\[\n4444^{4444}<10000^{4444}=\\left(10^4\\right)^{4444}=10^{17776}\n\\]\n\nTherefore \$4444^{4444}\$ has fewer than 17776 digits. This shows that \$A<9\\cdot 17776=159984\$. The sum of the digits of \$A\$ is then maximized when \$A=99999\$, so \$B\\leq 45\$. Note that out of all of the positive integers less than or equal to 45, the maximal sum of the digits is 12.\n\nIt's not hard to prove that any base-10 number is congruent to the sum of its digits modulo 9. Therefore \$4444^{4444}\\equiv A\\equiv B(\\bmod{9})\$. This motivates us to compute \$X\$, where \$1\\leq X \\leq 12\$, such that \$4444^{4444}\\equiv X(\\bmod{9})\$. The easiest way to do this is by searching for a pattern. Note that\n\n\\[\n4444^1\\equiv 7(\\bmod 9)\\\\4444^2\\equiv 4(\\bmod 9)\\\\4444^3\\equiv 1(\\bmod 9)\n\\]\n\nand since \$4444=3\\times 1481+1\$,\n\n\\[\n4444^{4444}\\equiv 4444^{3\\times1481+1}\\equiv \\left(4444^3\\right)^{1481}\\times 4444\\equiv 1\\times 4444\\equiv 7(\\bmod{9})\n\\]\n\nThus, \$X=7\$, which means that the sum of the digits of \$B\$ is \$\\boxed{7}\$.\n\n~minor edits by KevinChen_Yay" ]
IMO-1975-5	https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_5	Determine, with proof, whether or not one can find $1975$ points on the circumference of a circle with unit radius such that the distance between any two of them is a rational number.	[ "Since there are infinitely many primitive Pythagorean triples, there are infinitely many angles \$\\theta\$ s.t. \$\\sin\\theta, \\cos\\theta\$ are both rational. Call such angles good. By angle-sum formulas, if \$a,b\$ are good, then \$a+b,a-b\$ are also good.\n\nFor points \$A,B\$ on the circle \$\\omega\$, let \$\\angle AB\$ be the angle subtended by \$AB\$. Now inductively construct points on \$\\omega\$ s.t. all angles formed by them are good; for 1,2 take any good angle. If there are \$n\$ points chosen, pick a good angle \$\\theta\$ and a marked point \$A\$ s.t. the point \$B\$ on \$\\omega\$ with \$\\angle AB=\\theta\$ is distinct from the \$n\$ points. Since there are infinitely many good angles but finitely many marked points, such \$\\theta\$ exists. For a previously marked point \$P\$ we have \$\\angle BP=\\pm \\angle AP\\pm \\angle AB\$ for suitable choices for the two \$\\pm\$. Since \$\\angle AP ,\\angle AB\$ are both good, it follows that \$\\angle BP\$ is good, which finishes induction by adding \$B\$.\n\nObserve that these points for \$n=1975\$ work: since \$AB=27sin\\angle AB\$ for \$A,B\$ on the circle, it follows that \$AB\$ is rational, and so we're done.\n\nThe above solution was posted and copyrighted by tobash_co. The original thread for this problem can be found here: [1]" ]
IMO-1975-6	https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_6	Find all polynomials $P$, in two variables, with the following properties: (i) for a positive integer $n$ and all real $t, x, y$ \[ P(tx, ty) = t^nP(x, y) \] (that is, $P$ is homogeneous of degree $n$), (ii) for all real $a, b, c$, \[ P(b + c, a) + P(c + a, b) + P(a + b, c) = 0, \] (iii) \[ P(1, 0) = 1. \]	[ "(i) If \$n = 0\$ : Clearly no solution (ii) If \$n = 1\$ : \$P(x, y) = ax+by \\implies\$ the identification yields directly \$P(x,y) = x-2y\$ (iii) If \$n > 1\$, \$a = 1, \\ b = 1, \\ c =-2 \\implies P(2,-2)+P(-1, 1)+P(-1, 1) = 0\$ \$\\implies ((-2)^{n}+2) P(-1, 1) = 0 \\implies P(-1, 1) = 0 \\implies P(-y, y) = 0 \\implies\$ \$P(x,y)\$ is divisible by \$(x+y)\$ It is then easy to see that \$\\frac{P(x, y)}{(x+y)}\$, of degree \$n-1\$ verifies all the equations.\n\nThe only solutions are thus \$P(x, y) = (x-2y)(x+y)^{n-1}\$\n\nThe above solution was posted and copyrighted by mathmanman. The original thread for this problem can be found here: [1]", "\$F(a,a,a) \\implies P(2a,a)=0 \\implies (x-2y)\$ is a factor of \$P(x,y)\$.\n\nWe may write \$P(x,y)=(x-2y)Q(x,y)\$\n\n\$F(a,b,b) \\implies 2P(a+b,b)+P(2b,a)=0 \\implies 2(a+b-2b)Q(a+b,b)+2(b-a)Q(2b,a) \\implies (a-b)(Q(a+b,b)-Q(2b,a))=0\$ Thus \$Q(a+b,b)=Q(2b,a) \\forall a \\neq b\$\n\nWe may rewrite it as \$Q(x,y)=Q(2y,x-y)=Q(2x-2y,3y-x=\\cdots\$ \$Q(x+d,y-d)-Q(x,y)\$ is a polynomial in \$d\$ of degree \$n-1\$ for any two fixed \$x,y\$,which has infinitely many zeroes,i.e,\$0,2y-x,x-2y,6y-3x,\\cdots\$.Thus \$Q(x+d,y-d)=Q(x,y)\$ holds for all \$d\$.In particular it holds for \$d=y\$,i.e, \$Q(x+y,0)=Q(x,y)\$.Now consider the polynomial \$R(x,y)=Q(x+y,0)-Q(x,y)\$.Suppose that its not the zero polynomial.Then its degree \$d\$ is defined.With \$t=\\frac{x}{y}\$ it can be wriiten as \$y^dS(t)=y^d(A(t)-B(t))\$.But \$S(t)\$ has infinitely many zeroes and this forces \$A(t)=B(t)\$,forcing \$R(x,y)\$ to be a zero polynomial.Contradiction!.Thus \$Q(x+y,0)\$ and \$Q(x,y)\$ are identical polynomials.This forces \$Q(x,y)=c(x+y)^{n-1}\$.With \$Q(1,0)=1\$ we get \$c=1\$.Thus\n\n\\[\nP(x,y)=(x-2y)(x+y)^{n-1}\n\\]\n\nThe above solution was posted and copyrighted by JackXD. The original thread for this problem can be found here: [2]" ]
IMO-1976-1	https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_1	In a convex quadrilateral (in the plane) with the area of $32 \text{ cm}^{2}$ the sum of two opposite sides and a diagonal is $16 \text{ cm}$. Determine all the possible values that the other diagonal can have.	[ "Label the vertices \$A\$, \$B\$, \$C\$, and \$D\$ in such a way that \$AB + BD + DC = 16\$, and \$\\overline{BD}\$ is a diagonal.\n\nThe area of the quadrilateral can be expressed as \$BD \\cdot ( d_1 + d_2 ) / 2\$, where \$d_1\$ and \$d_2\$ are altitudes from points \$A\$ and \$C\$ onto \$\\overline{BD}\$. Clearly, \$d_1 \\leq AB\$ and \$d_2 \\leq DC\$. Hence the area is at most \$BD \\cdot ( AB + DC ) / 2 = BD(16-BD) / 2\$.\n\nThe quadratic function \$f(x)=x(16-x)/2\$ has its maximum for \$x=8\$, and its value is \$f(8)=32\$.\n\nThe area of our quadrilateral is \$32\$. This means that we must have \$BD=8\$. Also, equality must hold in both \$d_1 \\leq AB\$ and \$d_2 \\leq DC\$. Hence both \$\\overline{AB}\$ and \$\\overline{DC}\$ must be perpendicular to \$\\overline{BD}\$. And in any such case it is clear from the Pythagorean theorem that \$AC = 8\\sqrt 2\$.\n\nTherefore the other diagonal has only one possible length: \$8\\sqrt 2\$." ]
IMO-1976-2	https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_2	Let $P_{1}(x) = x^{2} - 2$ and $P_{j}(x) = P_{1}(P_{j - 1}(x))$ for $j= 2,\ldots$ Prove that for any positive integer n the roots of the equation $P_{n}(x) = x$ are all real and distinct.	[ "I shall prove by induction that \$P_n(x)\$ has \$2^n\$ distinct real solutions, where \$2^{n-1}\$ are positive and \$2^{n-1}\$ are negative. Also, for ever root \$r\$, \$\|r\|<2\$.\n\nClearly, \$P_1(x)\$ has 2 real solutions, where 1 is positive and 1 is negative. The absolute values of these two solutions are also both less than 2. This proves the base case.\n\nNow assume that for some positive integer \$k\$, \$P_k(x)\$ has \$2^k\$ distinct real solutions with absolute values less than 2, where \$2^{k-1}\$ are positive and \$2^{k-1}\$ are negative.\n\nChoose a root \$r\$ of \$P_{k+1}(x)\$. Let \$P_1(r)=s\$, where \$s\$ is a real root of \$P_k(x)\$. We have that \$-2<s<2\$, so \$0<r^2<4\$, so \$r\$ is real and \$\|r\|<2\$. Therefore all of the roots of \$P_{k+1}\$ are real and have absolute values less than 2.\n\nNote that the function \$P_{k+1}(x)\$ is an even function, since \$P_1(x)\$ is an even function. Therefore half of the roots of \$P_{k+1}\$ are positive, and half are negative.\n\nNow assume for the sake of contradiction that \$P_{k+1}(x)\$ has a double root \$r\$. Let \$P_1(r)=s\$. Then there exists exactly one real number \$r\$ such that \$r^2-2=s\$. The only way that this could happen is when \$s+2=0\$, or \$s=-2\$. However, \$\|s\|<2\$ from our inductive hypothesis, so this is a contradiction. Therefore \$P_{k+1}(x)\$ has no double roots. This proves that that the roots of \$P_{k+1}(x)\$ are distinct.\n\nThis completes the inductive step, which completes the inductive proof.", "Let \$x=2\\cos\\theta\$. We then have that \$P_1\\left(x\\right)=2\\cos2\\theta\$, and we can prove using induction that \$P_n\\left(x\\right)=2\\cos2^n\\theta\$. Thus, we just need to solve \$2\\cos2^n\\theta=2\\cos\\theta\$, which happens when \$\\cos2^n\\theta=\\theta+2\\pi k\$ or \$\\cos2^n\\theta=-\\theta+2\\pi l\$. These give that \$\\theta=\\frac{2k}{2^n-1}\\cdot\\pi\$ or \$\\theta=\\frac{2l}{2^n+1}\\cdot\\pi\$. As we can choose the range \$0\\leq\\theta\\leq\\pi\$ to ensure no duplications, we get that, upon rearranging, \$0\\leq k\\leq2^{n-1}-\\frac{1}{2}\$ and \$0\\leq l\\leq2^{n-1}+\\frac{1}{2}\$. There are \$2^{n-1}\$ integers in the first range and \$2^{n-1}+1\$ in the second. However, exactly one time, they product the same \$\\theta\$: \$\\theta=0\$. Thus, there are \$2^{n-1}+2^{n-1}+1-1=2^n\$ distinct roots for \$\\cos\\theta=x\$. We can also prove using induction that there are \$2^n\$ roots of \$P_n\\left(x\\right)\$. Thus, all roots of \$P_n\\left(x\\right)\$ are distinct and real.\n\nQ.E.D." ]
IMO-1976-3	https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_3	A box whose shape is a parallelepiped can be completely filled with cubes of side $1.$ If we put in it the maximum possible number of cubes, each of volume $2$, with the sides parallel to those of the box, then exactly $40$ percent from the volume of the box is occupied. Determine the possible dimensions of the box.	[ "We name a,b,c the sides of the parallelepiped, which are positive integers. We also put\n\n\\[\n\\begin{align} x &= \\left\\lfloor\\frac{a}{\\sqrt[3]{2}}\\right\\rfloor \\\\ y &= \\left\\lfloor\\frac{b}{\\sqrt[3]{2}}\\right\\rfloor \\\\ z &= \\left\\lfloor\\frac{c}{\\sqrt[3]{2}}\\right\\rfloor \\\\ \\end{align}\n\\]\n\nIt is clear that \$xyz\$ is the maximal number of cubes with sides of length \$\\sqrt[3]{2}\$ that can be put into the parallelepiped with sides parallels to the sides of the box. Hence the corresponding volume is \$V_2=2\\cdot xyz\$. We need \$V_2=0.4\\cdot V_1=0.4\\cdot abc\$, hence\n\n\\[\n\\frac ax\\cdot \\frac by\\cdot \\frac cz=5\\ \\ \\ \\ \\ \\ \\ \\ (1)\n\\]\n\nWe give the values of \$x\$ and \$a/x\$ for \$a=1,\\dots ,8\$. The same table is valid for \$b,y\$ and \$c,z\$.\n\n\\[\n\\begin{tabular}{\|c\|c\|c\|} \\hline a & x & a/x \\\\ \\hline 1 & 0 & - \\\\ \\hline 2 & 1 & 2 \\\\ \\hline 3 & 2 & 3/2 \\\\ \\hline 4 & 3 & 4/3 \\\\ \\hline 5 & 3 & 5/3 \\\\ \\hline 6 & 4 & 3/2 \\\\ \\hline 7 & 5 & 7/5 \\\\ \\hline 8 & 6 & 4/3 \\\\ \\hline \\end{tabular}\n\\]\n\nBy simple inspection we obtain two solutions of \$(1)\$: \$\\{a,b,c\\}=\\{2,5,3\\}\$ and \$\\{a,b,c\\}=\\{2,5,6\\}\$. We now show that they are the only solutions.\n\nWe can assume \$\\frac ax\\ge \\frac by \\ge \\frac cz\$. So necessarily \$\\frac ax\\ge \\sqrt[3]{5}\$. Note that the definition of \$x\$ implies\n\n\\[\nx< a/\\sqrt[3]2 < x+1,\n\\]\n\nhence\n\n\\[\n\\sqrt[3]2< a/x < \\sqrt[3]2(1+\\frac 1x)\n\\]\n\nIf \$a\\ge 4\$ then \$x\\ge 3\$ and \$\\frac ax<\\sqrt[3]2(1+\\frac 1x)\\le \\sqrt[3]2(\\frac 43)<\\sqrt[3]5\$ since \$2\\cdot \\frac {4^3}{3^3}<5\$. So we have only left the cases \$a=2\$ and \$a=3\$. But for \$a=3\$ we have \$a/x=3/2<\\sqrt[3]5\$ and so necessarily \$a=2\$ and \$a/x=2\$. It follows\n\n\\[\n\\frac by \\cdot \\frac cz =\\frac 52 \\ \\ \\ \\ \\ \\ (2)\n\\]\n\nNote that the definitions of \$y,z\$ imply\n\n\\[\ny< b/\\sqrt[3]2 < y+1,\\ \\ \\textrm{and} \\ \\ z< c/\\sqrt[3]2 < z+1.\\ \\ \\ \\ (3)\n\\]\n\nMoreover we have from (2) and from \$b/y\\ge c/z\$ that\n\n\\[\n\\frac by \\ge \\sqrt{5/2}\\ \\ \\ \\ \\ (4)\n\\]\n\nIf \$b=2\$ then \$b/y=2\$ and we would have \$c/z=5/4<\\sqrt[3]2\$, which contradicts \$(3)\$.\n\nOn the other hand, if \$b>5\$ then \$y>4\$ and \$\\frac by<\\sqrt[3]2(1+\\frac 1y)\\le \\sqrt[3]2(\\frac 54)<\\sqrt{5/2}\$ since \$2^2\\cdot \\frac {5^6}{4^6}<\\frac{5^3}{2^3}\$ as \$5^3<2^7\$. So we have only left the cases \$b=3,4,5\$. But for \$b=3\$ we have \$b/y=3/2<\\sqrt{5/2}\$ and for \$b=4\$ we have \$b/y=4/3<\\sqrt{5/2}\$ and so necessarily \$b=5\$ and \$b/y=5/3\$ (\$>\\sqrt{5/2}\$)\n\nSo we arrive finally at \$a=2,b=5\$ and \$c/z=3/2\$. If \$c\\ge 8\$ then \$z\\ge 6\$ and \$\\frac cz<\\sqrt[3]2(1+\\frac 1z)\\le \\sqrt[3]2(\\frac 76)<\\frac 32\$ since \$2\\cdot \\frac {7^3}{6^3}<\\frac{3^3}{2^3}\$. On the other hand, for \$c\\le 7\$ there are the only two possible values \$c=3\$ and \$c=6\$ which yield the known solutions." ]
IMO-1976-4	https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_4	Determine the greatest number, who is the product of some positive integers, and the sum of these numbers is $1976.$	[ "Since \$33=222+1\$, 3's are more efficient than 2's. We try to prove that 3's are more efficient than anything:\n\nLet there be a positive integer \$x\$. If \$3\$ is more efficient than \$x\$, then \$x^3<3^x\$. We try to prove that all integers greater than 3 are less efficient than 3:\n\nWhen \$x\$ increases by 1, then the RHS is multiplied by 3. The other side is multiplied by \$\\dfrac{(x+1)^3}{x^3}\$, and we must prove that this is less than 3 for all \$x\$ greater than 3.\n\n\\[\n\\dfrac{(x+1)^3}{x^3}<3\\Rightarrow \\dfrac{x+1}{x}<\\sqrt[3]{3}\\Rightarrow 1<(\\sqrt[3]{3}-1)x\n\\]\n\n\\[\n\\dfrac{1}{\\sqrt[3]{3}-1}<x\n\\]\n\nThus we need to prove that \$\\dfrac{1}{\\sqrt[3]{3}-1}<4\$. Simplifying, we get \$5<4\\sqrt[3]{3}\\Rightarrow 125<643=192\$, which is true. Working backwards, we see that all \$x\$ greater than 3 are less efficient than 3, so we never use anything greater than 3. Therefore we use as many 3s in groups of 2 as possible, and since the remainder when 1976 is divided by 6 is 2, we can use 658 3s and one 2.\n\n\$\\dfrac{1976}{3}=658.6666\$, so the greatest product is \$\\boxed{3^{658}\\cdot 2}\$.", "(Non-rigorous) We demonstrate heuristically that 3's are the most efficient. As we know that the chosen numbers must be equal, by the AM-GM inequality, we wish to maximize\n\n\\[\nf(x) = \\left(\\frac{1976}{x}\\right)^{x}\n\\]\n\nSimple logarithmic differentiation shows that \$\\frac{S}{x} = e\$ maximizes the given function. As \$e\$ is approximately 2.71828, we use 2's and 3's only. 3's are optimal, but we must use one 2.", "Note: This solution uses the same strategy as Solution 1 (that having the largest possible number of three's is good), but approaches the proof in a different manner.\n\nLet \$f(S) \\triangleq \\prod_{x \\in S} x\$, and \$g(S) = \\sum_{x\\in S} x\$, where \$S\$ is a multiset. We fix \$g(S) = 1976\$, and aim to maximize \$f(S)\$. Since \$3(x-3) > x\$ for \$x \\geq 5\$, we notice that \$S\$ must only contain the integers \$1,2,3\$ and \$4\$. We can replace any occurrences of \$4\$ in \$S\$ by replacing it with a couple of \$2\$'s, without changing \$f(S)\$ or \$g(S)\$, so we may assume that \$S\$ only contains the integers \$1,2\$ and \$3\$. We may further assume that \$S\$ contains at most one \$1\$, since any two \$1\$'s can be replaced by a \$2\$ without changing \$g(S)\$, but with an increase in \$f(S)\$. If \$S\$ contains exactly one \$1\$, then it must also contain at least one \$2\$ (since \$1976 \\equiv 2\\;(\\mathrm{mod }\\;3)\$). We can then replace this pair of a \$1\$ and a \$2\$ with a \$3\$, thus keeping \$g(S)\$ constant, and increasing \$f(S)\$. Now we may assume that \$S\$ contains only \$2\$'s and \$3\$'s.\n\nNow, as observed in the last solution, any triplet of \$2\$'s can be replaced by a couple of \$3\$'s, with \$g(S)\$ constant, and an increase in \$f(S)\$. Thus, after repeating this operation, we will be left with at most two \$2\$'s. Since \$g(S) = 1976\$, and \$1976 \\equiv 2\\;(\\mathrm{mod }\\;3)\$, we therefore get that \$S\$ must have exactly one \$2\$ (since we already showed it consists only of \$2\$'s and \$3\$'s). Thus, we get \$\\boxed{f(S) = 2\\cdot 3^{658}}\$.", "We determine the greatest number, who is the product of some positive integers, and the sum of these numbers is \$1976.\$.\n\nFor this problem, we used induction where if we assumed that \$i, j\$ are greater than \$1\$. Then\n\n\\[\n2^i \\cdot 3^j > k\n\\]\n\nwhere \$k = i \\cdot 2 + j \\cdot 3\$. Knowing this implies that the optimal solution comprises of twos and threes as every integer greater than 1 can be expressed as a linear combination of twos and threes, we proceeded to show that given a configuration of\n\n\\[\n2^i \\cdot 3^k,\n\\]\n\n\\[\n2^{i+n} \\cdot 3^{\\,j-\\tfrac{2}{3}n}\n\\]\n\nis less than the original expression, implying that maximal multiplication by \$3\$ is maximizing under the given relationship with powers of two.\n\nMore rigorously, we can show that maximizing with \$3\$ is optimal by comparing the expression\n\n\\[\n2^i \\cdot 3^j\n\\]\n\nwith\n\n\\[\n2^{i+n} \\cdot 3^{\\,j-\\tfrac{2}{3}n}.\n\\]\n\nTaking the ratio gives\n\n\\[\n\\frac{2^{i+n} \\cdot 3^{\\,j-\\tfrac{2}{3}n}}{2^i \\cdot 3^j} = \\frac{2^n}{3^{\\tfrac{2}{3}n}}.\n\\]\n\nNow consider the \$n\$-th root of this factor:\n\n\\[\n\\left(\\frac{2^n}{3^{\\tfrac{2}{3}n}}\\right)^{\\!\\!1/n} = \\frac{2}{3^{2/3}} < 1.\n\\]\n\nThus multiplying by more \$3\$’s (rather than shifting weight into \$2\$’s) always gives a larger product, confirming that \$3\$ is maximizing under the given relationship.\n\nHowever, when we reach an edge, \$2 \\cdot 2\$ over \$3 \\cdot 1\$ is a superior solution. Given \$1976\$ we can have\n\n\\[\n3^{658} \\cdot 2.\n\\]" ]
IMO-1976-5	https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_5	We consider the following system with $q = 2p$: \[ \begin{matrix} a_{11}x_{1} + \ldots + a_{1q}x_{q} = 0, \\ a_{21}x_{1} + \ldots + a_{2q}x_{q} = 0, \\ \ldots , \\ a_{p1}x_{1} + \ldots + a_{pq}x_{q} = 0, \\ \end{matrix} \] in which every coefficient is an element from the set $\{ - 1,0,1\}$$.$ Prove that there exists a solution $x_{1}, \ldots,x_{q}$ for the system with the properties: a.) all $x_{j}, j = 1,\ldots,q$ are integers$;$ b.) there exists at least one j for which $x_{j} \neq 0;$ c.) $\|x_{j}\| \leq q$ for any $j = 1, \ldots ,q.$	[ "First of all note that we have \$(q + 1)^q - 1\$ possible nonzero vectors \$(x_1,\\cdots,x_q)\$ such that \$0\\leq x_i\\leq q\$ are integers.\n\nBut \$a_{j1}x_1 + \\cdots + a_{jq}x_q\$ can only assume \$q^2 + 1\$ different values, because if it is maximized/minimized by \$(M_1,M_2,\\cdots,M_q)/(m_1,m_2,\\cdots,m_q)\$, we have that \$\\sum_{i = 1}^{q}a_{ji}(M_i - m_i)\\leq q\\times q = q^2\$ (if \$a_{ji} = 0\$, it doesn't affect the sum, if it is \$1\$, \$M_i = q,m_i = 0\$, and if it is \$- 1\$, \$M_i = 0,m_i = q\$).\n\nFrom this we conclude that there are at most \$(q^2 + 1)^p = (q^2 + 1)^{q/2}\$ possible values for the vector \$(a_{11}x_{1} + \\ldots + a_{1q}x_{q},\\cdots,a_{p1}x_{1} + \\ldots + a_{pq}x_{q})\$. But we have that: \$(q + 1)^q - 1 = (q^2 + 1 + 2q)^{q/2} - 1 =\$ \$= (q^2 + 1)^{q/2} + \\left( - 1 + \\sum_{j = 1}^{q/2}{{q/2}\\choose j}(q^2 + 1)^{q/2 - j}(2q)^j\\right) > (q^2 + 1)^{q/2}\$\n\nWe conclude that by the pigeonhole principle there are two distinct vectors being mapped to the same vector. Taking their difference we have a vector with the desired properties.\n\nThe above solution was posted and copyrighted by Jorge Miranda. The original thread for this problem can be found here: [1]" ]
IMO-1976-6	https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_6	A sequence $(u_{n})$ is defined by \[ u_{0} = 2 \quad u_{1} = \frac {5}{2}, u_{n + 1} = u_{n}(u_{n - 1}^{2} - 2) - u_{1} \ \text{ for } n = 1,\cdots \] Prove that for any positive integer $n$ we have \[ \lfloor u_{n} \rfloor = 2^{\frac {(2^{n} - ( - 1)^{n})}{3}} \] (where $\lfloor x\rfloor$ denotes the smallest integer $\leq$ $x$)$.$	[ "Let the sequence \$(x_n)_{n \\geq 0}\$ be defined as\n\n\\[\nx_{0}=0,x_{1}=1, x_{n}=x_{n-1}+2x_{n-2}\n\\]\n\nWe notice\n\n\\[\nx_n=\\frac{2^n-(-1)^n}{3}\n\\]\n\nBecause the roots of the characteristic polynomial \$x_{n}=x_{n-1}+2x_{n-2}\$ are \$-1\$ and \$2\$.\n\nWe also see \$\\frac{2^1-(-1)^1}{3}=1=x_1\$, \$\\frac{2^2-(-1)^2}{3}=1=x_2\$ We want to prove\n\n\\[\n2^{x_{n}-2x_{n-1}}+2^{-x_{n}+2x_{n-1}}=2^{x_1}+2^{-x_1}\n\\]\n\nThis is done by induction\n\nBase Case: For \$n=1\$ ses det \$2^{1-0}+2^{0-1}=2^{1}+2^{-1}\$\n\nInductive step: Assume \$2^{x_{n-1}-2x_{n-2}}+2^{-x_{n-1}+2x_{n-2}}=2^{x_1}+2^{-x_1}\$ We notice\n\n\\[\n\\begin{align} 2^{x_{n}-2x_{n-1}}+2^{-x_{n}+2x_{n-1}} &=2^{x_{n-1}+2x_{n-2}-2x_{n-1}}+2^{-(x_{n-1}+2x_{n-2})+2x_{n-1}}\\\\ &=2^{-x_{n-1}+2x_{n-2}}+2^{x_{n-1}-2x_{n-2}}\\\\ &=2^{x_1}+2^{-x_1} \\end{align}\n\\]\n\nWe then want to show\n\n\\[\na_n=2^{x_n}+2^{-x_n}\n\\]\n\nThis can be done using induction\n\nBase Case\n\nFor \$n=1\$, it is clear that\n\n\\[\na_1=\\frac{5}{2}\n\\]\n\nand\n\n\\[\n2^{x_1}+2^{-x_1}=2^1+2^{-1}=2+\\frac{1}{2}=\\frac{5}{2}\n\\]\n\nTherefore, the base case is proved.\n\nInductive Step\n\nAssume for all natural \$k<n\$ at \$a_k=2^{x_k}+2^{-x_k}\$\\newline Then we have that:\n\n\\[\n\\begin{align} a_n &= a_{n}(a_{n-1}^{2}-2)-a_{1} \\\\ &= (2^{x_{n-1}}+2^{-x_{n-1}})((2^{x_{n-2}}+2^{-x_{n-2}})^2-2)-(2^{x_{1}}+2^{-x_{0}}) \\\\ &= (2^{x_{n-1}}+2^{-x_{n-1}})((2^{x_{n-2}})^2+(2^{-x_{n-2}})^2+22^{x_{n-2}}2^{-x_{n-2}}-2)-(2^{x_{1}}+2^{-x_{0}}) \\\\ &=(2^{x_{n-1}}+2^{-x_{n-1}})((2^{2x_{n-2}})+(2^{-2x_{n-2}})+22^{x_{n-2}-x_{n-2}}-2)-(2^{x_{1}}+2^{-x_{0}}) \\\\ &=(2^{x_{n-1}}+2^{-x_{n-1}})((2^{2x_{n-2}})+(2^{-2x_{n-2}})+2-2)-(2^{x_{1}}+2^{-x_{0}}) \\\\ &=2^{x_{n-1}}2^{2x_{n-2}}+2^{-x_{n-1}}2^{-2x_{n-2}}+2^{x_{n-1}}2^{-2x_{n-2}}+2^{-x_{n-1}}2^{2x_{n-2}}-2^{x_{1}}-2^{-x_{0}}\\\\ &=2^{x_{n-1}+2x_{n-2}}+2^{-(x_{n-1}+2x_{n-2}})+2^{x_{n-1}-2x_{n-2}}+2^{-x_{n-1}+2x_{n-2}}-2^{x_{1}}-2^{-x_{0}}\\\\ &=2^{x_{n}}+2^{-x_{n}}+2^{x_{n-1}-2x_{n-2}}+2^{-x_{n-1}+2x_{n-2}}-2^{x_{1}}-2^{-x_{0}} \\end{align}\n\\]\n\nFrom our first induction proof we have that:\n\n\\[\n2^{x_{n-1}-2x_{n-2}}+2^{-x_{n-1}+2x_{n-2}}=2^{x_1}+2^{-x_1}\n\\]\n\nThen:\n\n\\[\na_n=2^{x_n}+2^{-x_n}+2^{x_1}+2^{-x_1}-(2^{x_1}+2^{-x_1})=2^{x_n}+2^{-x_n}\n\\]\n\nWe notice \$\\left[a_n\\right]=\\left[2^{x_n}+2^{-x_n}\\right]=2^{x_n}\$, Because \$2^{x_n} \\in \\mathbb{N}\$ and \$2^{-x_n}<1\$, for all \$n>0\$ Finally we conclude\n\n\\[\n\\left[a_n\\right]=2^{\\frac{2^n-(-1)^n}{3}}\n\\]", "We note that there is a slightly easier claim that can be proved also by induction. By testing out smaller terms, one sees that\n\n\\[\nu_n= 2^{\\frac{2^n-(-1)^n}{3}} + 2^{- \\frac{2^n-(-1)^n}{3}}\n\\]\n\nfor \$n=1,2,3\$ (note that since we are using strong induction to prove this result this suffices as the base case).\n\nSuppose that the equation above holds for all \$k \\le n\$. Then for \$n+1\$, one have\n\n\\[\n\\begin{align} u_{n+1} &= u_n (u_{n-1} ^2 -2) -u_1 \\\\ &= (2^{\\frac{2^n-(-1)^n}{3}}+2^{- \\frac{2^n-(-1)^n}{3}}) \\cdot ( (2^{\\frac{2^{n-1}-(-1)^{n-1}}{3}}+2^{- \\frac{2^{n-1}-(-1)^{n-1}}{3}} )^2 -2) - \\frac{5}{2} \\\\ &= (2^{\\frac{2^n-(-1)^n}{3}}+2^{- \\frac{2^n-(-1)^n}{3}}) \\cdot ( (2^{\\frac{2^{n-1}-(-1)^{n-1}}{3}})^2 + (2^{- \\frac{2^{n-1}-(-1)^{n-1}}{3}})^2 )- \\frac{5}{2} \\\\ &= (2^{\\frac{2^n-(-1)^n}{3}}+2^{- \\frac{2^n-(-1)^n}{3}}) \\cdot (2^{\\frac{2^{n}-2 \\cdot (-1)^{n-1}}{3}} + 2^{- \\frac{2^{n}-2 \\cdot (-1)^{n-1}}{3}})- \\frac{5}{2} \\\\ &= 2^{\\frac{2^n-(-1)^n}{3} + \\frac{2^{n}-2 \\cdot (-1)^{n-1}}{3}} + 2^{- \\frac{2^n-(-1)^n}{3} - \\frac{2^{n}-2 \\cdot (-1)^{n-1}}{3}} + 2^{(-1)^n} + 2^{-(-1)^n} -\\frac{5}{2} \\\\ &= 2^{\\frac{2^n-(-1)^n}{3} + \\frac{2^{n}-2 \\cdot (-1)^{n-1}}{3}} + 2^{- \\frac{2^n-(-1)^n}{3} - \\frac{2^{n}-2 \\cdot (-1)^{n-1}}{3}} \\\\ &= 2^{\\frac{2^{n+1}-(-1)^{n+1}}{3}}+2^{- \\frac{2^{n+1}-(-1)^{n+1}}{3}} \\end{align}\n\\]\n\nas desired. The strong induction proves our claim. Now, it is rather obvious that \$2^{- \\frac{2^n-(-1)^n}{3}}\$ is between 0 and 1, so indeed our claim proves the result. \$\\square\$\n\n~Ddk001" ]
IMO-1977-1	https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_1	In the interior of a square $ABCD$ we construct the equilateral triangles $ABK, BCL, CDM, DAN.$ Prove that the midpoints of the four segments $KL, LM, MN, NK$ and the midpoints of the eight segments $AK, BK, BL, CL, CM, DM, DN, AN$ are the 12 vertices of a regular dodecagon.	[ "Just use complex numbers, with \$a = 1\$, \$b = i\$, \$c = - 1\$ and \$d = - i\$. With some calculations, we have \$k = \\frac {\\sqrt {3} - 1}{2}( - 1 - i)\$, \$l = \\frac {\\sqrt {3} - 1}{2}(1 - i)\$, \$m = \\frac {\\sqrt {3} - 1}{2}(1 + i)\$ and \$n = \\frac {\\sqrt {3} - 1}{2}( - 1 + i)\$. Now it's an easy job to calculate the twelve midpoints and to find out they are all of the form \$\\frac {\\sqrt {3} - 1}{2}e^{\\frac {k\\pi}{6}i}\$, with \$k\\in\\mathbb{N}: 0\\le k\\le 11\$, and the result follows.\n\nThe above solution was posted and copyrighted by Joao Pedro Santos. The original thread for this problem can be found here: [1]" ]
IMO-1977-2	https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_2	In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.	[ "Let \$x_1,x_2,\\ldots\$ be the given sequence and let \$s_n=x_1+x_2+\\ldots+x_n\$. The conditions from the hypothesis can be now written as \$s_{n+7}<s_n\$ and \$s_{n+11}>s_n\$ for all \$n\\ge 1\$. We then have: \$0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6<s_{17}<s_{10}<s_3<s_{14}<s_7<0,\$ a contradiction. Therefore, the sequence cannot have \$17\$ terms. In order to show that \$16\$ is the answer, just take 16 real numbers satisfying \$s_{10}<s_3<s_{14}<s_7<0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6\$. We have \$x_1=s_1\$ and \$x_n=s_n-s_{n-1}\$ for \$n\\ge 2\$. Thus we found all sequences with the given properties.\n\nThe above solution was posted and copyrighted by enescu. The original thread for this problem can be found here: [1]" ]
IMO-1977-3	https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_3	Let $n$ be a given number greater than 2. We consider the set $V_n$ of all the integers of the form $1 + kn$ with $k = 1, 2, \ldots$ A number $m$ from $V_n$ is called indecomposable in $V_n$ if there are not two numbers $p$ and $q$ from $V_n$ so that $m = pq.$ Prove that there exist a number $r \in V_n$ that can be expressed as the product of elements indecomposable in $V_n$ in more than one way. (Expressions which differ only in order of the elements of $V_n$ will be considered the same.)	[ "Lemma: there are \$\\infty\$ many prime numbers \$p \\not\\equiv 1 \\mod n\$.\n\nProof:\n\nAssume that there are only finitely many of them and let their product be \$k\$. Then all prime factors of \$nk - 1\$ must be \$\\equiv 1 \\mod n\$ since this number is coprime with \$k\$. But that would mean that \$- 1 \\equiv nk - 1 \\equiv 1 \\mod n\$ which is impossible for \$n > 2\$.\n\nNow we can tackle the problem: There are only finetely many residue classes \$\\mod n\$, thus we can find two primes \$p,q\$ with \$p \\equiv q \\not\\equiv 1 \\mod n\$ and coprime with \$n\$. Let \$s\$ be the smallest positive integer with \$p^s \\equiv 1 \\mod n\$, then the same property holds for \$q\$ too. Now we have that \$p^s, q^s, pq^{s - 1} , p^{s - 1}q \\equiv 1 \\mod n\$ are all indecomposable since all their nontrivial divisors are \$\\not\\equiv 1 \\mod n\$. But the product \$p^sq^s = (pq^{s - 1})(p^{s - 1}q)\$ gives a number that is represented in two ways.\n\nThe above solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]" ]
IMO-1977-4	https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_4	Let $a,b,A,B$ be given reals. We consider the function defined by \[ f(x) = 1 - a \cdot \cos(x) - b \cdot \sin(x) - A \cdot \cos(2x) - B \cdot \sin(2x). \] Prove that if for any real number $x$ we have $f(x) \geq 0$ then $a^2 + b^2 \leq 2$ and $A^2 + B^2 \leq 1.$	[ "\$f(x) = 1-\\sqrt{a^2+b^2}\\sin (x+\\arctan\\frac{a}{b}) - \\sqrt{A^2+B^2}\\sin (2x+\\arctan\\frac{A}{B}) \\geq 0\$. \$f(x+\\pi) = 1+\\sqrt{a^2+b^2}\\sin (x+\\arctan\\frac{a}{b}) - \\sqrt{A^2+B^2}\\sin (2x+\\arctan\\frac{A}{B}) \\geq 0\$\n\nTherefore, \$\\sqrt{A^2+B^2}\\sin (2x+\\arctan\\frac{A}{B}) \\leq 1\$. Since this identity is true for any real \$x\$, let the sine term be one, \$\\longrightarrow A^2+B^2 \\leq 1\$.\n\nTo get cancellation on the rightmost terms, note \$\\sin (x+\\pi/2) = \\cos x, \\sin (x-\\pi/2) = -\\cos x\$.\n\n\$f(x+\\pi/4) = 1-\\sqrt{a^2+b^2}\\sin (x+\\pi/4+\\arctan\\frac{a}{b}) - \\sqrt{A^2+B^2}\\cos2x+\\arctan\\frac{A}{B}) \\geq 0\$. \$f(x-\\pi/4) = 1-\\sqrt{a^2+b^2}\\sin (x-\\pi/4+\\arctan\\frac{a}{b}) + \\sqrt{A^2+B^2}\\cos2x+\\arctan\\frac{A}{B}) \\geq 0\$.\n\nLet \$x+\\arctan\\frac{a}{b} = y\$. Then \$\\sqrt{a^2+b^2}(\\sin (y+\\pi/4) + \\sin (y-\\pi/4)) \\leq 2\$ \$\\sqrt{a^2+b^2} \\leq \\dfrac{2}{\\sqrt{2}(\\sin y)}\$. Since it's valid for all real \$x\$ let \$\\sin y = 1\$, and we are done.\n\nThe above solution was posted and copyrighted by aznlord1337. The original thread for this problem can be found here: [1]" ]
IMO-1977-5	https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_5	Let $a,b$ be two natural numbers. When we divide $a^2+b^2$ by $a+b$, we the the remainder $r$ and the quotient $q.$ Determine all pairs $(a, b)$ for which $q^2 + r = 1977.$	[ "Using \$r=1977-q^2\$, we have \$a^2+b^2=(a+b)q+1977-q^2\$, or \$q^2-(a+b)q+a^2+b^2-1977=0\$, which implies \$\\Delta=7908+2ab-2(a^2+b^2)\\ge 0\$. Using AM-GM inequality and the fact a,b>0, we have 3(a+b)^2<=7908+8ab<=7098+8(a+b)^2/4, it follows \$a+b\\le 88\$. If \$q\\le 43\$, then \$r=1977-q^2\\ge 128\$, contradicting \$r<a+b\\le 88\$. But \$q\\le 44\$ from \$q^2+r=1977\$, thus \$q=44\$. It follows \$r=41\$, and we get \$a^2+b^2=44(a+b)+41\\Leftrightarrow (a-22)^2+(b-22)^2=1009\\in \\mathbb{P}\$. By Jacobi's two squares theorem, we infer that \$15^2+28^2=1009\$ is the only representation of \$1009\$ as a sum of squares. This forces \$\\boxed{(a,b)=(37,50) , (7, 50)}\$, and permutations. \$\\blacksquare\$\n\nThe above solution was posted and copyrighted by cobbler. The original thread for this problem can be found here: [1]" ]
IMO-1977-6	https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_6	Let $f(n)$ be a function $f: \mathbb{N}^{+}\to\mathbb{N}^{+}$. Prove that if \[ f(n+1) > f(f(n)) \] for each positive integer $n$, then $f(n)=n$.	[ "We will prove this via induction. First we will prove there is a \$t\$ such that \$f(t)=1\$ and then that \$t=1\$ is the only such solution.\n\nDefine the sequence \$a_n\$ with \$a_0>1\$ for \$a_0\\in \\mathbb{N}\$ and \$a_k=f(a_{k-1}-1)\$. By the given inequality we have that \$f(a_n)>f(a_{n+1})\$, this can be used to form a inequality chain of decreasing positive integers:\n\n\\[\nf(a_0)>f(a_1)>f(a_2)>...\n\\]\n\nBy Infinite Descent, this sequence must terminate, and the only way it can terminate is if we input something into \$f\$ that is outside of its range. This can only happen if \$a_n=1\$ since the range and domain of \$f\$ are the positive integers. Since \$a_0\\neq 1\$, there is a integer \$t\$ (\$a_{n-1}-1\$) such that \$f(t)=1\$.\n\nNow if \$t\\neq 1\$, then \$f(t)=1>f(f(t-1))\$, which is impossible since \$f(f(t-1))\\ge 1\$ by the range of \$f\$, so we have \$t=1\$ is the only time when \$f(t)=1\$.\n\nNow for the inductive step.\n\nAssume that \$f(n)=n\$ for all \$n<k\$ and these are the only times these values occur. We will prove that \$f(k)=k\$ and that this is the only time this value occurs. Define the sequence \$a_n\$ similarly, except that \$a_0>k\$, by the reasoning above, there is a \$a_m\$ such that \$f(a_m)=1\$, by the inductive assumption, this means that \$a_m=f(a_{m-1}-1)=1\$, we can repeat the inductive assumption to get that \$a_{m-k+1}=k\$. This implies that \$f(a_{m-k}-1)=k\$. Thus, there is a \$t\$ such that \$f(t)=k\$.\n\nNow for that \$t\$, we have \$k>f(f(t-1))\$, which means that \$k+1>t\$ by the inductive assumption which implies \$t=k\$ since we must have \$t>k-1\$, otherwise \$f(t)<k\$. So \$t=k\$ is the only time when \$f(t)=k\$\n\nSo the inductive step is complete. Therefore, by induction \$f(n)=n\$ for all positive integers \$n\$." ]
IMO-1978-1	https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_1	Let $m$ and $n$ be positive integers such that $1 \le m < n$. In their decimal representations, the last three digits of $1978^m$ are equal, respectively, to the last three digits of $1978^n$. Find $m$ and $n$ such that $m + n$ has its least value.	[ "We have \$1978^m\\equiv 1978^n\\pmod {1000}\$, or \$978^m-978^n=1000k\$ for some positive integer \$k\$ (if it is not positive just do \$978^n-978^m=-1000k\$). Hence \$978^n\\mid 1000k\$. So dividing through by \$978^n\$ we get \$978^{m-n}-1=\\frac{1000k}{978^n}\$. Observe that \$2\\nmid LHS\$, so \$2\\nmid RHS\$. So since \$2\|\| 978^n\$, clearly the minimum possible value of \$n\$ is \$3\$ (and then \$489^n\\mid k\$). We will show later that if \$n\$ is minimal then \$m\$ is minimal. We have \$978^{m-3}-1\\equiv 0\\pmod {125}\\Leftrightarrow 103^{m-3}\\equiv 1\\pmod {125}\$. Hence, \$m-3\\mid \\varphi(125)\\Rightarrow m-3\\mid 100\$. Checking by hand we find that only \$m-3=100\$ works (this also shows that minimality of \$m\$ depends on \$n\$, as claimed above). So \$m=103\$. Consequently, \$m+n=106\$ with \$\\boxed{(m,n)=(103,3)}\$.\n\nThe above solution was posted and copyrighted by cobbler and Andreas. The original thread for this problem can be found here: [1] and [2]" ]
IMO-1978-2	https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_2	We consider a fixed point $P$ in the interior of a fixed sphere$.$ We construct three segments $PA, PB,PC$, perpendicular two by two$,$ with the vertexes $A, B, C$ on the sphere$.$ We consider the vertex $Q$ which is opposite to $P$ in the parallelepiped (with right angles) with $PA, PB, PC$ as edges$.$ Find the locus of the point $Q$ when $A, B, C$ take all the positions compatible with our problem.	[ "\\[\nIMO 1978 P2a.png\n\\]\n\nLet \$R\$ be the radius of the given fixed sphere.\n\nLet point \$O\$ be the center of the sphere.\n\nLet point \$D\$ be the 4th vertex of the face of the parallelepiped that contains points \$P\$, \$A\$, and \$B\$.\n\nLet point \$E\$ be the point where the line that passes through \$OP\$ intersects the circle on the side nearest to point \$A\$\n\nLet \$\\alpha=\\angle AOP,\\;\\beta=\\angle BPD,\\;\\theta=\\angle APE\$\n\nWe start the calculations as follows:\n\n\\[\n\\left\| AB \\right\|= \\left\| PD \\right\|\n\\]\n\n\\[\n\\left\| AB \\right\|^{2}=\\left\| PA \\right\|^{2}+\\left\| PB \\right\|^{2}\n\\]\n\nTherefore, \$\\left\| PD \\right\|^{2}=\\left\| PA \\right\|^{2}+\\left\| PB \\right\|^{2}\$ [Equation 1]\n\nUsing law of cosines:\n\n\\[\nR^{2}=\\left\| OP \\right\|^{2} + \\left\| PB \\right\|^{2} - 2 \\left\| OP \\right\| \\left\| PB \\right\| cos (\\angle OPB)\n\\]\n\n\\[\nR^{2}=\\left\| OP \\right\|^{2} + \\left\| PB \\right\|^{2} - 2 \\left\| OP \\right\| \\left\| PB \\right\| cos \\left( \\frac{\\pi}{2}-\\theta \\right)\n\\]\n\n\\[\nR^{2}=\\left\| OP \\right\|^{2} + \\left\| PB \\right\|^{2} - 2 \\left\| OP \\right\| \\left\| PB \\right\| sin (\\theta)\n\\]\n\n\$\\left\| PB \\right\|^{2} =R^{2}-\\left\| OP \\right\|^{2} + 2 \\left\| OP \\right\| \\left\| PB \\right\| sin (\\theta)\$ [Equation 2]\n\nUsing law of cosines again we also get:\n\n\\[\n\\left\| PA \\right\|^{2} =R^{2}+\\left\| OP \\right\|^{2} - 2 \\left\| OP \\right\| R cos(\\alpha)\n\\]\n\nSince \$R cos(\\alpha) = \\left\| PA \\right\| cos(\\theta) + \\left\| OP\\right\|\$, then\n\n\\[\n\\left\| PA \\right\|^{2} =R^{2}+\\left\| OP \\right\|^{2} - 2 \\left\| OP \\right\| \\left[ \\left\| PA \\right\| cos(\\theta) + \\left\| OP\\right\| \\right]\n\\]\n\n\$\\left\| PA \\right\|^{2} =R^{2}-\\left\| OP \\right\|^{2} - 2 \\left\| OP \\right\| \\left\| PA \\right\| cos(\\theta)\$ [Equation 3]\n\nSubstituting [Equation 2] and [Equation 3] into [Equation 1] we get:\n\n\$\\left\| PD \\right\|^{2}=2R^{2}-2\\left\| OP \\right\|^{2}+2\\left\| OP \\right\| \\left[ \\left\| PB \\right\| sin(\\theta) - \\left\| PA \\right\| cos(\\theta) \\right]\$ [Equation 4]\n\nNow we apply the law of cosines again:\n\n\\[\n\\left\| OD \\right\|^{2}=\\left\| OP \\right\|^{2}+\\left\| PD \\right\|^{2}-2\\left\| OP \\right\| \\left\| PD \\right\|cos(\\angle OPD)\n\\]\n\n\\[\n\\left\| OD \\right\|^{2}=\\left\| OP \\right\|^{2}+\\left\| PD \\right\|^{2}-2\\left\| OP \\right\| \\left\| PD \\right\| cos(\\angle OPB+\\angle BPD)\n\\]\n\n\\[\n\\left\| OD \\right\|^{2}=\\left\| OP \\right\|^{2}+\\left\| PD \\right\|^{2}-2\\left\| OP \\right\| \\left\| PD \\right\| cos \\left(\\frac{\\pi}{2}-\\theta+\\beta \\right)\n\\]\n\n\\[\n\\left\| OD \\right\|^{2}=\\left\| OP \\right\|^{2}+\\left\| PD \\right\|^{2}-2\\left\| OP \\right\| \\left\| PD \\right\| sin(\\theta-\\beta)\n\\]\n\n\\[\n\\left\| OD \\right\|^{2}=\\left\| OP \\right\|^{2}+\\left\| PD \\right\|^{2}-2\\left\| OP \\right\| \\left\| PD \\right\| \\left[sin(\\theta)cos(\\beta)-sin(\\beta)cos(\\theta) \\right]\n\\]\n\nSince, \$sin(\\beta)=\\frac{\\left\| PA \\right\|}{\\left\| PD \\right\|}\$ and \$cos(\\beta)=\\frac{\\left\| PB \\right\|}{\\left\| PD \\right\|}\$ then,\n\n\\[\n\\left\| OD \\right\|^{2}=\\left\| OP \\right\|^{2}+\\left\| PD \\right\|^{2}-2\\left\| OP \\right\| \\left\| PD \\right\| \\left[\\frac{\\left\| PB \\right\|}{\\left\| PD \\right\|}sin(\\theta)-\\frac{\\left\| PA \\right\|}{\\left\| PD \\right\|}cos(\\theta) \\right]\n\\]\n\n\$\\left\| OD \\right\|^{2}=\\left\| OP \\right\|^{2}+\\left\| PD \\right\|^{2}-2\\left\| OP \\right\| \\left[\\left\| PB \\right\|sin(\\theta)-\\left\| PA \\right\|cos(\\theta) \\right]\$ [Equation 5]\n\nSubstituting [Equation 4] into [Equation 5] we get:\n\n\\[\n\\left\| OD \\right\|^{2}=\\left\| OP \\right\|^{2}+2R^{2}-2\\left\| OP \\right\|^{2}+2\\left\| OP \\right\| \\left[ \\left\| PB \\right\| sin(\\theta) - \\left\| PA \\right\| cos(\\theta) \\right]-2\\left\| OP \\right\| \\left[\\left\| PB \\right\|sin(\\theta)-\\left\| PA \\right\|cos(\\theta) \\right]\n\\]\n\nNotice that all of the terms with \$\\theta\$ cancel and thus we're left with:\n\n\$\\left\| OD \\right\|^{2}=2R^{2}-\\left\| OP \\right\|^{2}\$ regardless of \$\\theta\$. [Equation 6]\n\nNow we need to find \$\\left\| PC \\right\|\$\n\nSince points \$O\$, \$P\$, and \$C\$ are on the plane perpendicular to the plane with points \$O\$, \$P\$, and \$A\$, then these points lie on the big circle of the sphere. Therefore the distance \$\\left\| PC \\right\|\$ can be found using the formula:\n\n\\[\nR^{2}=\\left\| OP \\right\|^{2}+\\left\| PC \\right\|^2\n\\]\n\nSolving for \$\\left\| PC \\right\|^2\$ we get:\n\n\$\\left\| PC \\right\|^2=R^{2}-\\left\| OP \\right\|^{2}\$ [Equation 7]\n\nNow we need to get \$\\left\| OQ \\right\|^{2}\$ which will be using the formula:\n\n\$\\left\| OQ \\right\|^{2}=\\left\| OD \\right\|^{2}+\\left\| PC \\right\|^2\$ [Equation 8]\n\nSubstituting [Equation 6] and [Equation 7] into [Equation 8] we get:\n\n\\[\n\\left\| OQ \\right\|^{2}=2R^{2}-\\left\| OP \\right\|^{2}+R^{2}-\\left\| OP \\right\|^{2}\n\\]\n\nThis results in:\n\n\\[\n\\left\| OQ \\right\|^{2}=3R^{2}-2\\left\| OP \\right\|^{2}\n\\]\n\nwhich is constant regardless of \$\\theta\$ and constant regardless of where points \$A\$, \$B\$, and \$C\$ are located as long as they're still perpendicular to each other.\n\nIn space, this is a sphere with radius \$\\left\| OQ \\right\|\$ which is equal to \$\\sqrt{3R^{2}-2\\left\| OP \\right\|^{2}}\$\n\nTherefore, the locus of vertex \$Q\$ is a sphere of radius \$\\sqrt{3R^{2}-2\\left\| OP \\right\|^{2}}\$ with center at \$O\$, where \$R\$ is the radius of the given sphere and \$\\left\| OP \\right\|\$ the distance from the center of the given sphere to point \$P\$\n\n~ Tomas Diaz. orders@tomasdiaz.com" ]
IMO-1978-3	https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_3	Let $0<f(1)<f(2)<f(3)<\ldots$ a sequence with all its terms positive$.$ The $n-th$ positive integer which doesn't belong to the sequence is $f(f(n))+1.$ Find $f(240).$	[ "Since the \$n\$-th number missing is \$f(f(n))+1\$ and \$f(f(n))\$ is a member of the sequence, it results that there are exactly \$n-1\$ \"gaps\" less than \$f(f(n))\$, which leads us to:\n\n\$f(f(n))=f(n)+n-1\$ \$(\\star)\$\n\nnow with a simple induction we can prove that \$f(F_{n}+1) = F_{n+1}+f(1)\$ , where \$F_{k}\$ is the Fibonacci sequence.\n\nour next step now is to prove that \$f(F_{n}+x) =F_{n+1}+f(x)\$, for all \$x\$ with \$1\\leq x\\leq F_{n-1}\$....and agaiiiiin induction(on \$n\$) : :P\n\nfor \$n=0\$ and \$n=1\$ it`s trivial and now we supose that it`s true for \$n-1\$ and to prove that holds for \$n\$ , in other words \$P(n-1) \\Rightarrow P(n)\$\n\n\$(i)\$ If \$x=f(y)\$ for some \$y\$, then by the inductive assumption and \$(\\star)\$ we have: \$f(F_{n}+x)=f(F_{n}+f(y))=f(f(F_{n-1}+y))=F_{n}+f(y)+F_{n-1}+y-1=F_{n+1}+f(x)\$\n\n\$(ii)\$ If \$x=f(f(y))+1\$ is a gap, then \$f(F_{n}+x-1)+1=F_{n+1}+f(x-1)+1\$ is a gap also: \$F_{n+1}+f(x)+1=F_{n+1}+f(f(f(y)))+1=f(F_{n}+f(f(y)))+1=f(f(F_{n-1}+f(y)))+1\$\n\nIt follows that : \$f(F_{n}+x)=F_{n+1}+f(x-1)+2=F{n+1}+f(x)\$\n\nnow since we know that each positive integer \$x\$ is expressible as: \$x=F_{k_{1}}+F_{k_{2}}+...+F_{k_{r}}\$ , where \$0<k_{r}\\neq 2\$, \$k_{i}\\geq k_{i+1}+2\$\n\nwe obtain that \$f(x) = F_{k_{1}+1}+...F_{k_{r}+1}\$ and now that we have the general form, pe particulary calculate for \$f(240)\$:\n\nwe can write \$240=233+5+2\$, therefore \$f(240)= 377+8+3=388\$\n\nRemark: it can be shown now that \$f(x)=[ax]\$ , where \$a=\\frac{1+\\sqrt{5}}{2}\$\n\nThe above solution was posted and copyrighted by pohoatza. The original thread for this problem can be found here: [1]" ]
IMO-1978-4	https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_4	In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$	[ "Denote \$a = BC, b = AB = AC\$ the triangle sides, \$r, R\$ the triangle inradius and circumradius and \$s\$, \$\\triangle\$ the triangle semiperimeter and area. Let \$S, T\$ be the tangency points of the incircle (I) with the sides \$AB, AC\$. Let \$K, L\$ be the midpoints of of \$BC, PQ\$ and \$M\$ the midpoint of the arc \$BC\$ of the circumcircle (O) opposite to the vertex \$A\$. The isosceles triangles \$\\triangle AST \\sim \\triangle APQ \\sim \\triangle ABC\$ are all centrally similar with the homothety center \$a\$. The homothety coefficient for the triangles \$\\triangle AST \\sim \\triangle ABC\$ is \$h_{13} = \\frac{AS}{AB} = \\frac{s - a}{b}\$. The homothety coefficient for the triangles \$\\triangle AST \\sim \\triangle APQ\$ is \$h_{12} = \\frac{AS}{AP}\$. The point K is the tangency point of the incircle (I) with the side BC, while the point M is the tangency point of the given circle with the triangle circumcircle (O). It follows that the triangles \$\\triangle ASK \\sim \\triangle APM\$ are also centrally similar with the homothety center A and the same homothety coefficient as the triangles \$\\triangle AST \\sim \\triangle APQ\$: \$h_{12} = \\frac{AS}{AP} = \\frac{AK}{AM} = \\frac{h}{2R}\$, where h = AK is the A-altitude of the triangle \$\\triangle ABC\$. The homothety coefficient of the triangles \$\\triangle APQ \\sim \\triangle ABC\$ is then \$h_{23} = \\frac{h_{12}}{h_{13}} = \\frac{s - a}{b} \\cdot \\frac{2R}{h}\$ Denote \$h' = AL\$ the \$A\$-altitude of the triangle \$\\triangle APQ\$. Then\n\n\\[\n\\frac{h'}{h} = h_{23} = \\frac{s - a}{b} \\cdot \\frac{2R}{h}\n\\]\n\n\\[\nKL = h - h' = h - 2R\\ \\frac{s - a}{b}\n\\]\n\nSubstituting \$h = \\frac{2 \\triangle}{a}\$, \$2R = \\frac{ab^2}{2 \\triangle}\$ and \$s - a = \\frac{\\triangle^2}{s(s - b)^2}\$, we get\n\n\\[\nKL = \\frac{2 \\triangle}{a} - \\frac{ab^2}{2 \\triangle} \\cdot \\frac{\\triangle^2}{bs(s - b)^2} =\n\\]\n\n(substituting 2s = 2b + a and \$s - b = \\frac a 2\$ for an isosceles triangle)\n\n\\[\n= \\frac{\\triangle}{s}\\left(\\frac{2s}{a} - \\frac{ab}{2(s - b)^2}\\right) = \\frac{\\triangle}{s} \\left(\\frac{2b}{a} + 1 - \\frac{2b}{a}\\right) = \\frac{\\triangle}{s} = r\n\\]\n\nwhich means that the point \$L \\equiv I\$ is identical with the incenter of the triangle \$\\triangle ABC\$.\n\nThe above solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [1]" ]
IMO-1978-5	https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_5	Let $f$ be an injective function from ${1,2,3,\ldots}$ in itself. Prove that for any $n$ we have: $\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.$	[ "We know that all the unknowns are integers, so the smallest one must greater or equal to 1.\n\nLet me denote the permutations of \$(k_1,k_2,...,k_n)\$ with \$(y_1,y_2,...,y_n)=y_i (*)\$.\n\nFrom the rearrangement's inequality we know that \$\\text{Random Sum} \\geq \\text{Reversed Sum}\$.\n\nWe will denote we permutations of \$y_i\$ in this form \$y_n \\geq ...\\geq y_1\$.\n\nSo we have \$\\frac{k_1}{1^2}+\\frac{k_2}{2^2}+...+\\frac{k_n}{n^2} \\geq \\frac{y_1}{1^2}+ \\frac{y_2}{2^2}+...+ \\frac{y_n}{n^2} \\geq 1+\\frac{1}{2}+...+\\frac{1}{n}\$.\n\nLet's denote \$\\frac{y_1}{1^2}+ \\frac{y_2}{2^2}+...+ \\frac{y_n}{n^2}=T\$ and \$1+\\frac{1}{2}+...+\\frac{1}{n}=S\$.\n\nWe have \$T \\geq S\$. Which comes from \$y_1 \\geq1, y_2 \\geq2, ...,y_n \\geq n\$.\n\nSo we are done.\n\nThe above solution was posted and copyrighted by Davron. The original thread for this problem can be found here: [1]" ]
IMO-1978-6	https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_6	An international society has its members from six different countries. The list of members has 1978 names, numbered $1, 2, \ldots, 1978$. Prove that there is at least one member whose number is the sum of the numbers of two (not necessarily distinct) members from his own country.	[ "Suppose the contrary. Then there exists a partition of \$\\{1,2,3,\\ldots, 1978\\}\$ into 6 difference-free subsets \$A,B,C,D,E,F\$. A set S is difference-free if there are not \$x,y,z \\in S\$ such that x = y - z.\n\nBy the Pigeonhole principle, one of these subsets, say \$A\$, has a minimum of \$\\left \\lfloor \\frac{1978}{6}\\right \\rfloor +1 = 330\$ members, \$a_1 < a_2 < \\ldots < a_{330}\$. Since \$A\$ is difference-free, each of the \$329\$ differences \$a_{330} - a_1, a_{330} - a_2, \\ldots, a_{330} - a_{329}\$ cannot belong to \$A\$, and so must belong to one of \$B, C, D, E, F\$.\n\nOne of these subsets, say \$B\$, has a minimum of \$\\left \\lfloor \\frac{329}{5}\\right \\rfloor +1 = 66\$ members, \$b_1 < b_2 < \\ldots < b_{66}\$. Since \$A\$ and \$B\$ are difference-free, each of the \$65\$ differences \$b_{66} - b_1, b_{66} - b_2, \\ldots, b_{66} - b_{65}\$ cannot belong to one of \$A, B\$, and so must belong to one of \$C, D, E, F\$.\n\nOne of these subsets, say \$C\$, has a minimum of \$\\left \\lfloor \\frac{65}{4}\\right \\rfloor +1 = 17\$ members, \$c_1 < c_2 < \\ldots < c_{17}\$. Since \$A, B, C\$ are difference-free, each of the \$16\$ differences \$c_{17} - c_1, c_{17} - c_2, \\ldots, c_{17} - c_{16}\$ cannot belong to one of \$A, B, C\$, and so must belong to one of \$D, E, F\$.\n\nOne of these subsets, say \$D\$, has a minimum of \$\\left \\lfloor \\frac{16}{3}\\right \\rfloor +1 = 6\$ members, \$d_1 < d_2 < \\ldots < d_6\$. Since \$A, B, C, D\$ are difference-free, each of the \$5\$ differences \$d_6 - d_1, d_6 - d_2, \\ldots, d_6 - d_5\$ cannot belong to one of \$A, B, C, D\$, and so must belong to one of \$E, F\$.\n\nOne of these subsets, say \$E\$, has a minimum of \$\\left \\lfloor \\frac{5}{2}\\right \\rfloor +1 = 3\$ members, \$e_1 < e_2 < e_3\$. Since \$A, B, C, D, E\$ are difference-free, each of the \$2\$ differences \$e_3 - e_1, e_3 - e_2\$ cannot belong to one of \$A, B, C, D, E\$, and so must belong to \$F\$.\n\nThen \$F\$ has at least two members, \$f_1 < f_2\$. The difference \$g = f_2 - f_1\$ cannot belong to one of \$A, B, C, D, E, F\$, a contradiction! \$\\blacksquare\$" ]
IMO-1979-1	https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_1	If $p$ and $q$ are natural numbers so that \[ \frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319}, \] prove that $p$ is divisible with $1979$.	[ "We first write\n\n\\[\n\\begin{align} \\frac{p}{q} &=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots-\\frac{1}{1318}+\\frac{1}{1319}\\\\ &=1+\\frac{1}{2}+\\cdots+\\frac{1}{1319}-2\\cdot\\left(\\frac{1}{2}+\\frac{1}{4}+\\cdots+\\frac{1}{1318}\\right)\\\\ &=1+\\frac{1}{2}+\\cdots+\\frac{1}{1319}-\\left(1+\\frac{1}{2}+\\cdots+\\frac{1}{659}\\right)\\\\ &=\\frac{1}{660}+\\frac{1}{661}+\\cdots+\\frac{1}{1319} \\end{align}\n\\]\n\nNow, observe that\n\n\\[\n\\begin{align} \\frac{1}{660}+\\frac{1}{1319}=\\frac{660+1319}{660\\cdot 1319}=\\frac{1979}{660\\cdot 1319} \\end{align}\n\\]\n\nand similarly \$\\frac{1}{661}+\\frac{1}{1318}=\\frac{1979}{661\\cdot 1318}\$ and \$\\frac{1}{662}+\\frac{1}{1317}=\\frac{1979}{662\\cdot 1317}\$, and so on. We see that the original equation becomes\n\n\\[\n\\begin{align} \\frac{p}{q} =\\frac{1979}{660\\cdot 1319}+\\frac{1979}{661\\cdot 1318}+\\cdots+\\frac{1979}{989\\cdot 990}=1979\\cdot\\frac{r}{s} \\end{align}\n\\]\n\nwhere \$s=660\\cdot 661\\cdots 1319\$ and \$r=\\frac{s}{660\\cdot 1319}+\\frac{s}{661\\cdot 1318}+\\cdots+\\frac{s}{989\\cdot 990}\$ are two integers. Finally consider \$p=1979\\cdot\\frac{qr}{s}\$, and observe that \$\\gcd(s,1979)=1\$ since s consists of products of positive integers less than \$1979\$. It follows that \$\\frac{qr}{s}\\in\\mathbb{Z}\$. Hence we deduce that \$p\$ is divisible with \$1979\$.\n\nThe above solution was posted and copyrighted by Solumilkyu. The original thread for this problem can be found here: [1]" ]
IMO-1979-2	https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_2	We consider a prism which has the upper and inferior basis the pentagons: $A_{1}A_{2}A_{3}A_{4}A_{5}$ and $B_{1}B_{2}B_{3}B_{4}B_{5}$. Each of the sides of the two pentagons and the segments $A_{i}B_{j}$ with $i,j=1,\ldots$,5 is colored in red or blue. In every triangle which has all sides colored there exists one red side and one blue side. Prove that all the 10 sides of the two basis are colored in the same color.	[ "Let us prove first that the edges \$A_1A_2,A_2A_3,\\cdots ,A_5A_1\$ are of the same color. Assume the contrary, and let w.l.o.g. \$A_1A_2\$ be red and \$A_2A_3\$ be green. Three of the segments \$A_2B_l, (l = 1, 2, 3, 4, 5)\$, say \$A_2B_i,A_2B_j,A_2B_k\$,have to be of the same color, let it w.l.o.g. be red. Then \$A_1B_i,A_1B_j,A_1B_k\$ must be green. At least one of the sides of triangle \$B_iB_jB_k\$, say \$B_iB_j\$ ,must be an edge of the prism. Then looking at the triangles \$A_1B_iB_j\$ and \$A_2B_iB_j\$ we deduce that \$B_iB_j\$ can be neither green nor red, which is acontradiction. Hence all five edges of the pentagon \$A_1A_2A_3A_4A_5\$ have thesame color. Similarly, all five edges of \$B_1B_2B_3B_4B_5\$ have the same color.We now show that the two colors are the same. Assume otherwise, i.e.,that w.l.o.g. the \$A\$ edges are painted red and the \$B\$ edges green. Let us call segments of the form \$A_iB_j\$ diagonal (\$i\$ and \$j\$ may be equal). We now count the diagonal segments by grouping the red segments based on their \$A\$ point, and the green segments based on their \$B\$ point. As above, the assumption that three of \$A_iB_j\$ for fixed \$i\$ are red leads to a contradiction. Hence at most two diagonal segments out of each \$A_i\$ may be red, which counts up to at most \$10\$ red segments. Similarly, at most \$10\$ diagonal segments can be green. But then we can paint at most \$20\$ diagonal segments out of \$25\$, which is a contradiction. Hence all edges in the pentagons \$A_1A_2A_3A_4A_5\$ and \$B_1B_2B_3B_4B_5\$ have the same color.\n\nThis solution was posted and copyrighted by Learner94. The original thread for this problem can be found here: [1]" ]
IMO-1979-3	https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_3	Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$	[ "Let \$O_1, O_2\$ be the centers of the 2 circles and let the points B, C move on the circles \$(O_1), (O_2)\$, recpectively. Let A' be the other intersection of the 2 circles different from the point A. Since the angles \$\\angle BO_1A = \\angle CO_2A\$ are equal, the isosceles triangles \$\\triangle AO_1B \\sim \\triangle AO_2C\$ are similar. Consequently, the angles \$\\angle BAO_1 = \\angle CAO_2\$ are equal and the angles\n\n\\[\n\\angle BAC = BAO_1 + \\angle O_1AO_2 + \\angle O_2AC =\n\\]\n\n\\[\n= BAO_1 + \\angle O_1AO_2 - \\angle CAO_2 = \\angle O_1AO_2\n\\]\n\nare also equal. The angles \$\\angle A'BA = \\frac{\\angle A'O_1A}{2} = \\angle O_2O_1A\$ are equal and the angles \$\\angle A'CA = \\frac{\\angle A'O_2A}{2} = \\angle O_1O_2A\$ are also equal. As a result, the point A' is on the side BC of the triangle \$\\triangle ABC\$, which is always similar to the fixed triangle \$\\triangle O_1AO_2\$.\n\nLemma (Yaglom's Geometric Transformations II, pp 68-69). Assume that a figure F' (the triangle \$\\triangle ABC\$) moves in such a way that it remains at all times similar to a fixed figure F (the triangle \$\\triangle O_1AO_2\$) and in addition, some point of the figure F' does not move at all (the point A). If a point B of the figure F' describes a curve \$\\Gamma\$ (the circle \$(O_1)\$), then any other point M of the figure F' describes a curve \$\\Gamma'\$ similar to the curve \$\\Gamma\$.\n\nConsequently, the midpoint M of the segment BC decribes a circle (Q). When the points B, C, become identical with the point A, the midpoint M of BC also becomes identical with A. Hence, the circle (Q) passes through the point A. The line BC always passes through A'. Just after the start of motion counter-clockwise, the points B, C, A' follow on the line BC in this order. The points B, C both pass through the point A' one after another and just before the end of one period of motion, the points A', B, C follow on the line BC In this order. From the continuity principle, the midpoint M of BC also passes through the point A' and consequently, the circle (Q), the locus of the midpoints M, also passes through the point A'. Thus the circle (Q) belongs to the same pencil with the circles \$(O_1), (O_2)\$ and it is centered on their center line \$O_1O_2\$. Let p be the perpendicular bisector of the segment BC intersectong the circle (Q) at a point P different from the point M. The point A' and the circle (Q) passing through this point are fixed and the angle \$\\angle A'MP = 90^\\circ\$ is right. Therefore, A'P is a fixed diameter of the circle (Q), which means that the point P is also fixed. Q.E.D.\n\nWhen the points B, C are both diametrally opposite to the point A on their respective circles, the segment \$O_1O_2\$ is the midline of the triangle \$\\triangle ABC\$, i.e., \$BC \\parallel O_1O_2\$, \$BC \\perp AA'\$. The cyclic quadrilateral AA'MP inscribed in the circle (Q) is then a rectangle. Its circumcenter Q is the intersection of its diagonals AM, A'P intersecting at their midpoint and the midpoint of AM is identical with the midpoint of \$O_1O_2\$. Thus the fixed point P is the reflection of the intersection A' of the circles \$(O_1), (O_2)\$ in the midpoint Q of the segment \$O_1O_2\$.\n\nThis solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [1]", "Let \$B\$ and \$C\$ be the antipodes of \$A\$ on the two circles and let \$D\$ be the foot of the altitude from \$A\$, which is the other intersection point of the two circles. Also, let \$M\$ be the midpoint of \$\\overline{BC}\$, and construct rectangle \$ADMZ\$. Our claim is that \$Z\$ is the fixed point. We let \$X\$ and \$Y\$ be the two points; by the condition the angles are the same. So we have a spiral similarity\n\n\\[\n\\triangle AXY \\sim \\triangle ABC.\n\\]\n\n\\[\n[asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A, B, C); pair M = midpoint(B--C); pair Z = A+M-D; pair X = midpoint(A--B)+dir(200)abs(A-B)/2; pair Y = -D+2foot(midpoint(A--C), X, D); draw(circumcircle(A, B, D), blue); draw(circumcircle(A, C, D), blue); draw(circumcircle(A, D, M), dashed+deepgreen); pair N = midpoint(X--Y); filldraw(A--X--Y--cycle, invisible, purple); filldraw(A--B--C--cycle, invisible, blue); draw(A--D, blue); draw(A--Z--M, deepgreen); dot(\"$A$\", A, dir(A)); dot(\"$B$\", B, dir(B)); dot(\"$C$\", C, dir(C)); dot(\"$D$\", D, dir(D)); dot(\"$M$\", M, dir(M)); dot(\"$Z$\", Z, dir(Z)); dot(\"$X$\", X, dir(X)); dot(\"$Y$\", Y, dir(Y)); dot(\"$N$\", N, dir(N)); /* TSQ Source: A = dir 110 B = dir 210 C = dir 330 D = foot A B C M = midpoint B--C Z = A+M-D X = midpoint(A--B)+dir(200)abs(A-B)/2 Y = -D+2foot midpoint A--C X D circumcircle A B D blue circumcircle A C D blue circumcircle A D M dashed deepgreen N = midpoint X--Y A--X--Y--cycle 0.1 red / purple A--B--C--cycle 0.1 yellow / blue A--D blue A--Z--M deepgreen */ [/asy]\n\\]\n\nNow let \$N\$ be the midpoint of \$\\overline{XY}\$. By spiral similarity, since \$N\$ maps to \$M\$, it follows \$M\$, \$N\$, \$A\$, and \$D\$ are cyclic too. So actually \$N\$ lies on the circumcircle of rectangle \$ADMZ\$, meaning \$\\overline{ZN} \\perp \\overline{XY}\$, hence \$ZX = ZY\$ as needed.\n\nRemark: The special point \$Z\$ can be identified by selecting the special case \$X \\to A\$, \$Y \\to A\$ and \$X=B\$, \$Y=C\$.\n\nThis solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]", "Let the points on the circles be \$E\$ and \$F\$. Then, let \$B\$ the the other intersection of the two circles, and \$C\$ and \$D\$ be the centers of the circles containing \$E\$ and \$F\$, respectively. WLOG, let us name \$E\$ and \$F\$ such that \$CA>DA\$. Let \$O\$ be the midpoint of \$CD\$, and the antipode of \$B\$ with respect to the circle centered at \$O\$ with radius \$OA\$ be \$P\$. Let \$M\$ be where this such circle hits \$EF\$. We first note that\n\n\\[\n\\angle ADF = 2\\angle ABF = 2 \\angle APM = \\angle AOM\n\\]\n\nand also\n\n\\[\n2 \\angle APM = 360^{\\circ} - 2\\angle MBA = 360^{\\circ} - 2\\angle EBA = \\angle ACE\n\\]\n\nmeaning that since \$AD=DF\$, \$AO=OM\$, and \$AC=CE\$, then SAS similarity gets us that \$\\triangle ADF \\sim \\triangle AOM \\sim \\triangle ACE\$. This means we have \$\\frac{AO}{AM} = \\frac{AC}{AE}\$ so \$\\frac{AO}{AC} = \\frac{AM}{AE}\$ and because \$\\angle MAE = \\angle OAC\$, then \$\\triangle AOC \\sim \\triangle AME\$ and similarly, we also have \$\\triangle AOD \\sim \\triangle AMF\$. This means that\n\n\\[\n\\frac{EM}{MA} = \\frac{OC}{OA} = \\frac{DO}{OA} = \\frac{FM}{MA} \\to EM=FM\n\\]\n\nmeaning that \$M\$ is the midpoint of \$EF\$. Now, since \$P\$ is the antipode of \$B\$, then \$\\angle BMP = 90^{\\circ} = \\angle EMP = \\angle FMP\$ so \$EP=FP\$ for all \$E\$ and \$F\$ for some fixed \$P\$ and we are done.\n\nThis solution was posted and copyrighted by kevinmathz. The original thread for this problem can be found here: [3]" ]
IMO-1979-4	https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_4	We consider a point $P$ in a plane $p$ and a point $Q \not\in p$. Determine all the points $R$ from $p$ for which \[ \frac{QP+PR}{QR} \] is maximum.	[ "Let \$T\$ be the orthogonal projection of the point \$Q\$ on the plane \$p\$. Then, the line \$PT\$ is the orthogonal projection of the line \$PQ\$ on the plane \$p\$, and thus forms the least angle with the line \$PQ\$ among all lines through the point \$P\$ which lie in the plane \$p\$; hence, \$\\measuredangle RPQ\\geq\\measuredangle TPQ\$, and equality holds if and only if the point \$R\$ lies on the ray \$PT\$ (the only exception is when \$PQ\\perp p\$; in this case, \$P = T\$, so the ray \$PT\$ is undefined, and equality holds for all points \$R\$ in the plane \$p\$, since we always have \$\\angle RPQ = \\angle TPQ = 90^{\\circ}\$).\n\nFrom \$\\measuredangle RPQ\\geq\\measuredangle TPQ\$, it follows that \$\\frac{\\measuredangle RPQ}{2}\\geq\\frac{\\measuredangle TPQ}{2}\$; also, since the angles \$\\angle RPQ\$ and \$\\angle TPQ\$ are \$180^{\\circ}\$, their half-angles \$\\frac{\\measuredangle RPQ}{2}\$ and \$\\frac{\\measuredangle TPQ}{2}\$ are < 90°, so that from \$\\frac{\\measuredangle RPQ}{2}\\geq\\frac{\\measuredangle TPQ}{2}\$ we can conclude that \$\\sin\\frac{\\measuredangle RPQ}{2}\\geq\\sin\\frac{\\measuredangle TPQ}{2}\$. Equality holds, as in the above, if and only if the point \$R\$ lies on the ray \$PT\$ (and, respectively, for all points \$R\$ in the plane \$p\$ if \$PQ\\perp p\$).\n\nOn the other hand, we obviously have \$\\cos\\frac{\\measuredangle QRP-\\measuredangle PQR}{2}\\leq 1\$ with equality if and only if \$\\frac{\\measuredangle QRP-\\measuredangle PQR}{2}=0^{\\circ}\$, i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base \$QR\$, i. e. if and only if \$PR = PQ\$, i. e. if and only if the point \$R\$ lies on the sphere with center \$P\$ and radius \$PQ\$.\n\nNow, applying the Mollweide theorem in triangle \$QPR\$, we get\n\n\$\\frac{QP+PR}{QR}=\\frac{\\cos\\frac{\\measuredangle QRP-\\measuredangle PQR}{2}}{\\sin\\frac{\\measuredangle RPQ}{2}}\$ \$\\leq\\frac{1}{\\sin\\frac{\\measuredangle RPQ}{2}}\$ (since \$\\cos\\frac{\\measuredangle QRP-\\measuredangle PQR}{2}\\leq 1\$) \$\\leq\\frac{1}{\\sin\\frac{\\measuredangle TPQ}{2}}\$ (since \$\\sin\\frac{\\measuredangle RPQ}{2}\\geq\\sin\\frac{\\measuredangle TPQ}{2}\$),\n\nand equality holds here if and only if equality holds in both of the inequalities \$\\sin\\frac{\\measuredangle RPQ}{2}\\geq\\sin\\frac{\\measuredangle TPQ}{2}\$ and \$\\cos\\frac{\\measuredangle QRP-\\measuredangle PQR}{2}\\leq 1\$ that we have used, i. e. if and only if the point \$R\$ lies both on the ray \$PT\$ (this condition should be ignored if \$PQ\\perp p\$) and on the sphere with center \$P\$ and radius \$PQ\$.\n\nHence, the point \$R\$ for which the ratio \$\\frac{QP+PR}{QR}\$ is maximum is the point of intersection of the ray \$PT\$ with the sphere with center \$P\$ and radius \$PQ\$ (or, respectively, it can be any arbitrary point on the intersection of the plane \$p\$ with the sphere with center \$P\$ and radius \$PQ\$ if \$PQ\\perp p\$).\n\nThis solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [1]" ]
IMO-1979-5	https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_5	Determine all real numbers a for which there exists non-negative reals $x_{1}, \ldots, x_{5}$ which satisfy the relations $\sum_{k=1}^{5} kx_{k}=a,$ $\sum_{k=1}^{5} k^{3}x_{k}=a^{2},$ $\sum_{k=1}^{5} k^{5}x_{k}=a^{3}.$	[ "Let \$\\Sigma_1= \\sum_{k=1}^{5} kx_{k}\$, \$\\Sigma_2=\\sum_{k=1}^{5} k^{3}x_{k}\$ and \$\\Sigma_3=\\sum_{k=1}^{5} k^{5}x_{k}\$. For all pairs \$i,j\\in \\mathbb{Z}\$, let\n\n\\[\n\\Sigma(i,j)=i^2j^2\\Sigma_1-(i^2+j^2)\\Sigma_2+\\Sigma_3\n\\]\n\nThen we have on one hand\n\n\\[\n\\Sigma(i,j)=i^2j^2\\Sigma_1-(i^2+j^2)\\Sigma_2+\\Sigma_3=\\sum_{k=1}^5(i^2j^2k-(i^2+j^2)k^3+k^5)x_k =\\sum_{k=1}^5k(i^2j^2-(i^2+j^2)k^2+k^4)x_k\n\\]\n\nTherefore \\\$1)\n\n\\[\n\\Sigma(i,j)=\\sum_{k=1}^5k(k^2-i^2)(k^2-j^2)x_k\n\\]\n\nand on the other hand \\\\ (2)\n\n\\[\n\\Sigma(i,j)=i^2j^2a-(i^2+j^2)a^2+a^3=a(a-i^2)(a-j^2)\n\\]\n\nThen from (1) we have\n\n\\[\n\\Sigma(0,5)=\\sum_{k=1}^5k^3(k^2-5^2)x_k\\leq 0\n\\]\n\nand from (2)\n\n\\[\n\\Sigma(0,5)=a^2(a-25)\n\\]\n\nso \\(a\\in [0,25]\$ Besides we also have from (1)\n\n\\[\n\\Sigma(0,1)=\\sum_{k=1}^5k^3(k^2-1)x_k\\geq 0\n\\]\n\nand from (2)\n\n\\[\n\\Sigma(0,1)=a^2(a-1)\\geq 0 \\implies a\\notin (0,1)\n\\]\n\nand for \$n=1,2,3,4\$\n\n\\[\n\\Sigma(n,n+1)=\\sum_{k=1}^5k(k^2-n^2)(k^2-(n+1)^2)x_k\n\\]\n\nwhere in the right hand we have that\n\n\\[\nk<n \\implies (k^2-n^2)<0, (k^2-(n+1)^2)<0\n\\]\n\n, so\n\n\\[\n(k^2-n^2)(k^2-(n+1)^2)>0\n\\]\n\n,\n\n\\[\nk=n,n+1 , \\implies (k^2-n^2)(k^2-(n+1)^2)=0\n\\]\n\nand\n\n\\[\nk>n \\implies (k^2-n^2)(k^2-(n+1)^2)>0\n\\]\n\n, so\n\n\\[\n\\Sigma(n,n+1)\\geq 0\n\\]\n\nfor \$n=1,2,3,4\$ From the latter and (2) we also have\n\n\\[\n\\Sigma(n,n+1)=a(a-n^2)(a-(n+1)^2))\\geq 0\\implies a\\notin (n^2,(n+1)^2)\n\\]\n\nSo we have that\n\n\\[\na\\in [0,25]-\\bigcup_{n=0}^4(n^2,(n+1^2))=\\{0,1,4,9,16,25\\}\n\\]\n\nIf \$a=k^2\$, \$k=0,1,2,3,4,5\$ take \$x_k=k\$, \$x_j=0\$ for \$j\\neq k\$. Then \$\\Sigma_1=k^2=a\$, \$\\Sigma_2=k^3k=k^4=a^2\$, and \$\\Sigma_3=k^5k=k^6=a^3\$", "Applying Cauchy-Schwarz on the three equations yields\n\n\\[\na\\cdot a^3 = \\left(\\sum_{k=1}^5 kx_k\\right)\\left(\\sum_{k=1}^5 k^5x_k\\right)\\geq \\left(\\sum_{k=1}^5 k^3x_k\\right)^2=(a^2)^2\n\\]\n\nSo this implies the following vectors are multiples of each other:\n\n\\[\n(x_1,2x_2,3x_3,4x_4,5x_5) =\\lambda (x_1,2^5x_2,3^5x_3,4^5x_4,5^5x_5)\\quad \\quad \\lambda\\in \\mathbb{R}^+\n\\]\n\nThis is only possible when at most one of the \$x_i\$'s are nonzero. Hence cycling through the cases where \$x_i\\neq 0\$, we get the choices of \$a=1,4,9,16,25\$ and the case of \$a=0\$." ]
IMO-1979-6	https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_6	Let $A$ and $E$ be opposite vertices of an octagon. A frog starts at vertex $A.$ From any vertex except $E$ it jumps to one of the two adjacent vertices. When it reaches $E$ it stops. Let $a_n$ be the number of distinct paths of exactly $n$ jumps ending at $E$. Prove that: \[ a_{2n-1}=0, \quad a_{2n}={(2+\sqrt2)^{n-1} - (2-\sqrt2)^{n-1} \over\sqrt2}. \]	[ "First the part \$a_{2n-1}=0\$ It's pretty obvious. Let's make a sequence \$b\$ of 1 and -1, setting 1 when the frog jumps left and -1 when it jumps right. If the frog would want to come to vertex E from vertex A, then \$\\sum b_{i}\$ from \$i =1\$ to \$i =2n-1\$ should be equal to either 4 or -4. But this sum is odd, so we have \$a_{2n-1}= 0\$\n\nNow the less obvious part. Let's name \$f(X,y)\$ the number of ways, in which we can come to vertex X in y moves.\n\nThen \$f(E,2n) = f(F,2n-1)+f(D, 2n-1) = f(C, 2n-2)+f(G, 2n-2) =\$ \$f(D, 2n-3)+f(B, 2n-3)+f(F, 2n-3)+f(H, 2n-3) =\$ \$2f(D, 2n-5)+2f(F, 2n-5)+4f(H,2n-5)+4f(B, 2n-5) =\$ \$4[ f(B,2n-5)+f(D, 2n-5)+f(F,2n-5)+f(H, 2n-5) ]-2 [ f(F,2n-5)+f(D,2n-5)] =\$ \$4f(E,2n-2)-2f(E,2n-4)\$\n\nNow we can find a solution to this recurrence or simply prove by induction the given answer.\n\nThis solution was posted and copyrighted by Myszon11. The original thread for this problem can be found here: [1]", "The given regular octagon is shown in Figure Since the number of edges in any path joining A and E is even, it is impossible to reach E from A in an odd number of jumps. Thus a2n – 1 = 0, for all n ≥ 1. It is obvious that a2 = 0. Also, as ABCDE and AHGFE are the only 2 different path of 4 jumps from A to E, we have a4 = 2. To find a recurrence relation for a2n\n\n, we introduce a new\n\nsupplementary sequence (bn\n\n) as follows. For each n ≥ 1, let bn be the number of paths of exactly n jumps from C (or G) to E. Starting at A there are 4 ways for the frog to move in the first 2 jumps, namely, A → B → A, A → H → A A → B → C, A → H → G. Thus, by definitions of (an\n\n) and (bn )\n\na2n = 2a2n – 2 + 2b2n – 2 ...(1) On the other hand, starting at C, there are 3 ways for the frog to move in the next 2 jumps if it does not stop at E, namely, C → B → C, C → D → C, C → B → A. Thus, b2n = 2b2n – 2 + a2n – 2.....(2) We shall now solve the system of linear recurrence relation (1) and (2) From (1), b2(n – 1) = a2n – a2(n – 1) .....(3) Substituting (3) into (2) gives a2(n + 1) – a2n = a2n – 2a2(n – 1) + a2(n – 1) or a2(n + 1) – 4a2n\n\n+ 2a2(n – 1) = 0....(4)\n\nLet dn = a2n\n\n. Then (4) may be written as\n\ndn + 1 – 4dn\n\n+ 2dn – 1 = 0 ...(5)\n\nThe characteristic equation of (5) is x 2 – 4x + 2 = 0 and its roots are 2 ± .root2 regards kislay kai" ]
IMO-1981-1	https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_1	$P$ is a point inside a given triangle $ABC$. $D, E, F$ are the feet of the perpendiculars from $P$ to the lines $BC, CA, AB$, respectively. Find all $P$ for which \[ \frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} \] is least.	[ "We note that \$BC \\cdot PD + CA \\cdot PE + AB \\cdot PF\$ is twice the triangle's area, i.e., constant. By the Cauchy-Schwarz Inequality,\n\n\\[\n{(BC \\cdot PD + CA \\cdot PE + AB \\cdot PF) \\left(\\frac{BC}{PD} + \\frac{CA}{PE} + \\frac{AB}{PF} \\right) \\ge ( BC + CA + AB )^2}\n\\]\n\n,\n\nwith equality exactly when \$PD = PE = PF\$, which occurs when \$P\$ is the triangle's incenter or one of the three excenters. But since we know \$P\$ is inside \$\\triangle ABC\$, we can say \$P\$ is the incenter. \$\\square\$" ]
IMO-1981-2	https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_2	Let $1 \le r \le n$ and consider all subsets of $r$ elements of the set $\{ 1, 2, \ldots , n \}$. Each of these subsets has a smallest member. Let $F(n,r)$ denote the arithmetic mean of these smallest numbers; prove that \[ F(n,r) = \frac{n+1}{r+1}. \]	[ "Clearly, the sum of the desired least elements is \$\\sum_{k=1}^{n} k {n-k \\choose r-1} = \\sum_{k=1}^{n} {k \\choose 1}{n-k \\choose r-1}\$, which we claim to be equal to \${n+1 \\choose r+1}\$ by virtue of the following argument.\n\nConsider a binary string of length \$n+1\$ which contains \$r+1\$ 1s. For some value of \$k\$ between 1 and \$n\$, inclusive, we say that the second 1 will occur in the \$(k+1)\$th place. Clearly, there are \${k \\choose 1}\$ ways to arrange the bits coming before the second 1, and \${n-k \\choose r-1}\$ ways to arrange the bits after the second 1. Our identity follows.\n\nSince the sum of the least elements of the sets is \${n+1 \\choose r+1}\$, the mean of the least elements is \$\\frac{{n+1 \\choose r+1}}{{n \\choose r}} = \\frac{n+1}{r+1}\$, Q.E.D.", "We proceed as in the previous solution, but we prove our identity using the following manipulations:\n\n\\[\n\\begin{matrix}\\sum_{k=1}^{n}k{n-k \\choose r-1 } &=&\\sum_{k=1}^{n-(r-1)}{k}{n-k \\choose r-1}\\\\ &=& \\sum_{i=1}^{n-(r-1)}\\sum_{k=i}^{n-(r-1)}{n-k \\choose r-1}\\\\ &=&\\sum_{i=1}^{n-(r-1)}{n-i+1 \\choose r}\\\\ &=&{n+1 \\choose r+1} \\end{matrix}\n\\]\n\nQ.E.D.", "We proceed by strong induction.\n\nWe define \$F(k, k-1)\$ to be zero (the empty sum).\n\nWe consider \$r\$ to be fixed. The assertion obviously holds for \$r = n\$. We now assume the problem to hold for values of \$n\$ less than or equal to \$k\$. By considering subsets containing \$k+1\$ and not containing \$k+1\$, respectively, we conclude that\n\n\\[\nF(k+1, r) = \\frac{{k \\choose r-1}F(k,r-1) + {k \\choose r}F(k,r)}{{k+1 \\choose r}} = 1 + \\frac{k-r+1}{r+1} = \\frac{k+2}{r+1}\n\\]\n\n.\n\nThis completes our induction, Q.E.D.", "Consider a bipartite graph \$G\$ with bipartition \$\\{A,B\\}\$. The vertices in \$A\$ are the \$(r+1)\$-element subsets of \$\\{0, \\dots , n\\}\$, and the vertices in \$B\$ are the \$r\$-element subsets of \$\\{1, \\dots , n\\}\$, and we draw an edge \$\\overline{ab}\$ iff the subset \$b \\in B\$ may be obtained from \$a \\in A\$ by deleting the smallest element in \$a\$.\n\nNote that\n\n\\[\n\|A\|=\\binom{n+1}{r+1}, \|B\|=\\binom{n}{r}, \|E(G)\|=\\binom{n+1}{r+1}=\\frac{n+1}{r+1} \\binom{n}{r}.\n\\]\n\nThe degree of a vertex in \$B\$ is the value of the least element of its corresponding subset. Hence\n\n\\[\nF(n,r)=\\frac{1}{\\binom{n}{r}} \\sum_{v \\in B} \\deg (v)= \\frac{n+1}{r+1}.\n\\]", "We will count how many times each element of the set \$\\{1, 2, \\ldots, n\\}\$ appear as the smallest element in a set, which will lead us to the result.\n\nTo count how many times \$1\$ appears as the smallest element of a subset, we need to choose an \$r-1\$ element subset from the remaining \$n - 1\$ numbers, as all of them are greater than \$1.\$ There are simply \$\\dbinom{n - 1}{r - 1}\$ ways to do this.\n\nSimilarly, to count how many times \$2\$ appears as the smallest element, we need to choose an \$r - 1\$ element subset from the remaining \$n - 2\$ numbers greater than \$2.\$ There are \$\\dbinom{n - 2}{r - 1}\$ ways to do this.\n\nAs there are \$\\dbinom{n}{r}\$ subsets, and thus smallest elements, in total, it now becomes clear that we need to evaluate the sum\n\n\\[\n\\frac{\\binom{n - 1}{r - 1} + 2\\binom{n - 2}{r - 1} + \\cdots + (n - r + 1)\\binom{r - 1}{r - 1}}{\\binom{n}{r}}.\n\\]\n\nWe can rearrange this sum as\n\n\\[\n\\frac{\\left(\\binom{n - 1}{r - 1} + \\binom{n - 2}{r - 1} + \\cdots + \\binom{r - 1}{r - 1}\\right)+\\left(\\binom{n - 2}{r - 1} + \\binom{n - 3}{r - 1} + \\cdots + \\binom{r - 1}{r - 1}\\right)+\\cdots+ \\binom{r-1}{r-1}}{\\binom{n}{r}}.\n\\]\n\nUsing the Hockey Stick Identity on each of the smaller sums on the numerator gives us\n\n\\[\n\\frac{\\binom{n}{r} + \\binom{n - 1}{r} + \\binom{n-2}{r} + \\cdots + \\binom{r}{r}}{\\binom{n}{r}}.\n\\]\n\nUsing Hockey stick again gives us\n\n\\[\n\\frac{\\binom{n + 1}{r + 1}}{\\binom{n}{r}},\n\\]\n\nwhich simplifies to\n\n\\[\n\\frac{\\frac{(n + 1)!}{(r + 1)!(n-r)!}}{\\frac{n!}{r!(n-r)!}} = \\left(\\frac{(n + 1)!}{(r + 1)!(n - r)!}\\right) \\left(\\frac{r!(n-r)!}{n!}\\right) = \\frac{n + 1}{r + 1},\n\\]\n\nas desired.\n\nSolution by Ilikeapos" ]
IMO-1981-3	https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_3	Determine the maximum value of $m^2 + n^2$, where $m$ and $n$ are integers satisfying $m, n \in \{ 1,2, \ldots , 1981 \}$ and $( n^2 - mn - m^2 )^2 = 1$.	[ "We first observe that since \$\\gcd(m,n)=1\$, \$m\$ and \$n\$ are relatively prime. In addition, we note that \$n \\ge m\$, since if we had \$n < m\$, then \$n^2 -nm -m^2 = n(n-m) - m^2\$ would be the sum of two negative integers and therefore less than \$-1\$. We now observe\n\n\\[\n(m+k)^2 -(m+k)m - m^2 = -(m^2 - km - k^2)\n\\]\n\n,\n\ni.e., \$(m,n) = (a,b)\$ is a solution iff. \$(b, a+b)\$ is also a solution. Therefore, for a solution \$(m, n)\$, we can perform the Euclidean algorithm to reduce it eventually to a solution \$(1,n)\$. It is easy to verify that if \$n\$ is a positive integer, it must be either 2 or 1. Thus by trivial induction, all the positive integer solutions are of the form \$(F_{n}, F_{n+1})\$, where the \$F_i\$ are the Fibonacci numbers. Simple calculation reveals \$987\$ and \$1597\$ to be the greatest Fibonacci numbers less than \$1981\$, giving \$987^2 + 1597^2=3524578\$ as the maximal value." ]
IMO-1981-4	https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_4	(a) For which values of $n>2$ is there a set of $n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $n-1$ numbers? (b) For which values of $n>2$ is there exactly one set having the stated property?	[ "Let \$k = \\prod p_i^{e_i}\$ be the greatest element of the set, written in its prime factorization. Then \$k\$ divides the least common multiple of the other elements of the set if and only if the set has cardinality at least \$\\max \\{ p_i^{e_i} \\}\$, since for any of the \$p_i^{e_i}\$, we must go down at least to \$k-p_i^{e_i}\$ to obtain another multiple of \$p_i^{e_i}\$. In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal to 3 must be divisible by a number that is greater than two and is a power of a prime.\n\nFor \$n > 3\$, we may let \$k = \\mbox{lcm} (n-1, n-2) = (n-1)(n-2)\$, since all the \$p_i^{e_i}\$ must clearly be less than \$n\$ and this product must also be greater than \$n\$ if \$n\$ is at least 4. For \$n > 4\$, we may also let \$k = \\mbox{lcm} (n-2, n-3) = (n-2)(n-3)\$, for the same reasons. However, for \$n = 4\$, this does not work, and indeed no set works other than \$\\{ 3,4,5,6 \\}\$. To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.\n\nQ.E.D.", "Let, for some \$n\$ and \$m\$ with \$m>n\$, \$m\\ \|\\ \\mbox{lcm}(m-1,m-2,\\cdots,m-n+1)\\ \|\\ (m-1)(m-2)\\cdots(m-n+1)\$.\n\nWe can trivially check that, there is no such \$m\$ for \$n=3\$, only \$m=3\$ works for \$n=4\$ and \$m=3,8\$ works for \$n=5\$.\n\nNow, consider, \$n>5\$. By Bertrand's postulate there is a prime \$p\$ such that \$2 \\le \\left\\lfloor\\frac{n}{2}\\right\\rfloor < p < 2\\left\\lfloor\\frac{n}{2}\\right\\rfloor\$.\n\nWhich implies, \$p< n \\le 2p\$.\n\nAs, \$n-1 \\ge p,3,2\$, there must be a multiple of \$p\$, a multiple of \$3\$ and a multiple of \$2\$ in the set, \$\\{m-1,m-2,\\cdots,m-n+1\\}\$.\n\nSo, \$2p\\ \|\\ \\mbox{lcm}(m-1,m-2,\\cdots,m-n+1)\$ and \$3p\\ \|\\ \\mbox{lcm}(m-1,m-2,\\cdots,m-n+1)\$.\n\nSo, \$m=2p\$ and \$m=3p\$ both work for \$n>5\$.\n\nSo,\n\nThere exists solution for all \$n>3\$,\n\nOnly one Solution for \$n=4\$.\n\nQ.E.D." ]
IMO-1981-5	https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_5	Three congruent circles have a common point $O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle and the point $O$ are collinear.	[ "Let the triangle have vertices \$A,B,C\$, and sides \$a,b,c\$, respectively, and let the centers of the circles inscribed in the angles \$A,B,C\$ be denoted \$O_A, O_B, O_C\$, respectively.\n\nThe triangles \$O_A O_B O_C\$ and \$ABC\$ are homothetic, as their corresponding sides are parallel. Furthermore, since \$O_A\$ lies on the bisector of angle \$A\$ and similar relations hold for the triangles' other corresponding points, the center of homothety is the incenter of both the triangles. Since \$O\$ is clearly the circumcenter of \$O_A O_B O_C\$, \$O\$ is collinear with the incenter and circumcenter of \$ABC\$, as desired.", "Suppose 3 congruent circles with centres P,Q,R lie inside ABC and are such that the circle with centre P touches AB & AC and the circle with centre Q touches CA & BC.an R with remaining 2.\n\nSince O lies in all 3 circles, PO=QO=RO. Therefore, O is circumcentre of PQR. let O' be circumcentre of ABC.\n\nSince BC is tangent to the circles with centers Q & R, the lengths of perpendiculars from Q & R, the lengths are equal. therefore, QR//BC,RP//CA,PQ//AB.\n\nAgain, since AB and AC both touch circle with centre P. Therefore P is equidistant from AB & AC. Therefore P lies on the internal bisector of angle A. Similarly Q & R lie internal bisectors of angle B and angle C respectively. Therefore, AP,BQ,CR produced meet at incenter I. Since, QR//BC,RP//CA,PQ//AB, it follows that I is also incentre of PQR, I being the centre of homothety. By the property of enlargements, O and O' must be co-linear with I , the centre of enlargement." ]
IMO-1981-6	https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_6	The function $f(x,y)$ satisfies (1) $f(0,y)=y+1,$ (2) $f(x+1,0)=f(x,1),$ (3) $f(x+1,y+1)=f(x,f(x+1,y)),$ for all non-negative integers $x,y$. Determine $f(4,1981)$.	[ "We observe that \$f(1,0) = f(0,1) = 2\$ and that \$f(1, y+1) = f(0, f(1,y)) = f(1,y) + 1\$, so by induction, \$f(1,y) = y+2\$. Similarly, \$f(2,0) = f(1,1) = 3\$ and \$f(2, y+1) = f(2,y) + 2\$, yielding \$f(2,y) = 2y + 3\$.\n\nWe continue with \$f(3,0) + 3 = 8\$; \$f(3, y+1) + 3 = 2(f(3,y) + 3)\$; \$f(3,y) + 3 = 2^{y+3}\$; and \$f(4,0) + 3 = 2^{2^2}\$; \$f(4,y) + 3 = 2^{f(4,y-1) + 3}\$.\n\nIt follows that \$f(4,1981) = 2^{2\\cdot ^{ . \\cdot 2}} - 3\$ when there are 1984 2s, Q.E.D.", "We can start by creating a list consisting of certain x an y values and their outputs.\n\n\\[\nf(0,0)=1, f(0,1)=2, f(0,2)=3, f(0,3)=4, f(0,4)=5\n\\]\n\nThis pattern can be proved using induction. After proving, we continue to setting a list when \$x=2\$.\n\n\\[\nf(1,0)=2, f(1,1)=3, f(1,2)=4, f(1,3)=5, f(1,4)=6\n\\]\n\nThis pattern can also be proved using induction. The pattern seems d up of a common difference of 1. Moving on to \$x=3\$\n\n\\[\nf(3,0)=5, f(3,1)=13, f(3,2)=29, f(3,3)=61, f(3,4)=125\n\\]\n\nAll of the numbers are being expressed in the form of \$3^a -3\$ where \$a=y+3\$. Lastly where x=4 we have\n\n\\[\nf(4,0)=13, f(4,1)=65533, f(4,2)=^5 2, f(3,3)=^6 2, f(4,4)=^7 2\n\\]\n\nwhere each term can be represented as \$^a 2\$ when \$a=y+2\$. In \$^a 2\$ represents tetration or 2 to the power 2 to the power 2 to the power 2 ... where \$a\$ amount of 2s. So therefore the answer is \$f(4,1981) = 2^{2\\cdot ^{ . \\cdot 2}} - 3\$ with 1984 2s, 2 tetration 1983. -Multpi12\n\nRemark. This function is well-studied, known widely as the Ackermann function." ]
IMO-1982-1	https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_1	The function $f(n)$ is defined on the positive integers and takes non-negative integer values. $f(2)=0,f(3)>0,f(9999)=3333$ and for all $m,n:$ \[ f(m+n)-f(m)-f(n)=0 \text{ or } 1. \] Determine $f(1982)$.	[ "Clearly \$f(1) \\ge 1 \\Rightarrow f(m+1) \\ge f(m)+f(1) \\ge f(m)+1\$ so \$f(9999) \\ge 9999\$.Contradiction!So \$f(1)=0\$.This forces \$f(3)=1\$.Hence \$f(3k+3) \\ge f(3k)+f(3)>f(3k)\$ so the inequality \$f(3)<f(6)<\\cdots<f(9999)=3333\$ forces \$f(3k)=k \\forall k \\le 3333\$.Now \$f(3k+2) \\ge k+1 \\Rightarrow f(6k+4) \\ge 2k+2 \\Rightarrow f(12k+8) \\ge 4k+4 \\le f(12k+9)=4k+3\$(Note:This is valid for \$12k+9 \\le 9999\$ or \$3k+2 \\le 2499\$).Contradiction!Hence the non-decreasing nature of \$f\$ gives \$f(3k+1)=k\$.Hence \$f(n)=\\lfloor\\frac{n}{3}\\rfloor \\forall 1\\le n \\le 2499\$.\n\nSo \$f(1982)=\\lfloor\\frac{1982}{3}\\rfloor=660\$.\n\nThis solution was posted and copyrighted by sayantanchakraborty. The original thread for this problem can be found here: [1]", "First observe that\n\n\\[\n3333=f(9999)\\geq f(9996)+f(3)\\geq f(9993)+2f(3)\\geq \\cdots\\geq 3333f(3)\n\\]\n\nSince \$f(3)\$ is a positive integer, we need \$f(3)=1\$. Next, observe that \\begin{align} 3333=f(9999)\\geq 5f(1980)+33f(3)=5f(1980)+33\\quad\\Longrightarrow\\quad f(1980)\\leq 660 \\end{align}On the other hand, \$f(1980)\\geq 660f(3)=660\$, so combine the two inequalities we obtain \$f(1980)=660\$. Finally, write\n\n\\[\nf(1982)=f(1980)+f(2)+0\\vee 1=660\\vee 661\n\\]\n\nSuppose that \$f(1982)=661\$, then \\begin{align} 3333=f(9999)\\geq 5f(1982)+29f(3)=3305+29=3334 \\end{align}a contradiction. Hence we conclude that \$f(1982)=660\$.\n\nThis solution was posted and copyrighted by Solumilkyu. The original thread for this problem can be found here: [2]", "We show that \$f(n) = [n/3]\$ for \$n <= 9999\$, where [ ] denotes the integral part. We show first that \$f(3) = 1\$. \$f(1)\$ must be \$0\$, otherwise \$f(2) - f(1) - f(1)\$ would be negative. Hence \$f(3) = f(2) + f(1) + 0\$ or \$1\$ = \$0\$ or \$1\$. But we are told \$f(3) > 0\$, so \$f(3) = 1\$. It follows by induction that \$f(3n) >= n\$. For \$f(3n+3) = f(3) + f(3n)\$ + \$0\$ or \$1 = f(3n) + 1\$ or \$2\$. Moreover if we ever get \$f(3n) > n\$, then the same argument shows that \$f(3m) > m\$ for all \$m > n\$. But \$f(3.3333) = 3333\$, so \$f(3n) = n\$ for all \$n <= 3333\$. Now \$f(3n+1) = f(3n) + f(1) + 0\$ or \$1\$ = \$n\$ or \$n + 1\$. But \$3n+1 = f(9n+3) >= f(6n+2) + f(3n+1) >= 3f(3n+1)\$, so \$f(3n+1) < n+1\$. Hence \$f(3n+1) = n.\$ Similarly, \$4f(3n+2) = n\$. In particular \$f(1982) = 660\$.\n\nThis solution was posted and copyrighted by Tega. The original thread for this problem can be found here: [3]", "Similar to solution 3.\n\nProof: Lemma 1: \$f(mn)\\geq nf(m)\$ Let, \$P(m,m)\$ be assertion.\n\n\\[\nf(2m) \\geq 2f(m)\n\\]\n\nSimilarly,we can induct to get \$f(nm)\\geq nf(m)\$.\n\n\\[\nf(3m) \\geq f(2m)+f(m) \\geq 3f(m)\\implies f(nm)\\geq \\underbrace{f(m(n-1))+f(m)\\geq .....}_{\\text{n times}}\\geq nf(m)\n\\]\n\nLemma proved.\n\nThen we see that,\n\n\\[\nf(9999)\\geq 3333f(3) \\implies f(3)=1:f(3)>0\n\\]\n\nThen,\n\n\\[\nf(9999) \\geq 5f(1980)+33f(3) \\implies 3300 \\geq 5f(1980) \\implies f(1980) \\leq 660\n\\]\n\n\\[\nf(1980) \\geq 660f(3) \\implies f(1980)=660\n\\]\n\nThen we can easily get,by assertion \$P(1980,2)\$\n\n\\[\nf(1982)=660 \\space or \\space f(1980)=661\n\\]\n\nHence, \$\\boxed{f(1980)=660,661}\$.And, we are done. \$\\blacksquare\$\n\nThis solution was posted and copyrighted by IMO2019. The original thread for this problem can be found here: [4]" ]
IMO-1982-2	https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_2	A non-isosceles triangle $A_{1}A_{2}A_{3}$ has sides $a_{1}$, $a_{2}$, $a_{3}$ with the side $a_{i}$ lying opposite to the vertex $A_{i}$. Let $M_{i}$ be the midpoint of the side $a_{i}$, and let $T_{i}$ be the point where the inscribed circle of triangle $A_{1}A_{2}A_{3}$ touches the side $a_{i}$. Denote by $S_{i}$ the reflection of the point $T_{i}$ in the interior angle bisector of the angle $A_{i}$. Prove that the lines $M_{1}S_{1}$, $M_{2}S_{2}$ and $M_{3}S_{3}$ are concurrent.	[ "Rename the vertices so that \$A_1=A,A_2=B\$ and \$A_3=C\$. Let \$I\$ be the incenter of \$\\triangle ABC\$.\n\nClaim 1: \$S_1, S_2, S_3\$ all lie on the incircle. Proof: By construction, \$IS_1\$ is the reflection of \$IT_1\$ with respect to the \$A\$ angle bisector since \$I\$ lies on the angle bisector of \$A\$. So, \$IS_1=IT_1\$ so \$S_1\$ lies on the incircle. Similarly, \$S_2\$ and \$S_3\$ lie on the incircle. \$\\Box\$\n\nNote that \$I\$ is the circumcenter of \$\\triangle S_1S_2S_3\$ because of equal radii due to Claim 1.\n\nClaim 2: \$\\angle S_1IS_2=2\\angle C.\$ Proof: In cyclic quadrilateral \$T_1IT_2C\$, \$\\angle T_1IT_2=180^{\\circ}-\\angle C\$. Let \$D\$ be the point where the \$A\$ angle bisector intersects \$BC\$. Due to the construction of \$S_1\$, \$\\angle S_1IT_1=2\\angle DIT_1.\$ But, \$\\angle ADC=180^{\\circ}-\\angle C-\\frac{1}{2}\\angle A\$ so \$\\angle DIT_1=\\frac{1}{2}\\angle A+\\angle C-90^{\\circ}.\$ Thus,\n\n\\[\n\\angle S_1IT_1=\\angle A+2\\angle C-180^{\\circ}.\n\\]\n\nSimilarly, \$\\angle S_2IT_2=\\angle B+2\\angle C-180^{\\circ}.\$ Therefore, \$\\angle S_1IS_2=\\angle S_1IT_1+\\angle T_1IT_2+\\angle T_2IS_2=2\\angle C.\$\$\\Box\$\n\nHence, we use Claim 2 to get \$\\angle S_1S_3S_2=\\frac{1}{2}\\angle S_1IS_2=\\angle C\$ and we similarly get \$\\angle S_3S_1S_2=\\angle A\$ and \$\\angle S_1S_2S_3=\\angle B.\$\n\nClaim 3: Let \$X_1\$ be the midpoint of \$S_2S_3.\$ Then, \$T_1IX_1\$ is collinear. Proof: We already derived that \$\\angle T_1IT_2=180^{\\circ}-\\angle C\$ and \$\\angle S_2IT_2=\\angle B+2\\angle C-180^{\\circ}.\$ Recall that a radius is perpendicular to the midpoint of a chord so \$S_2S_3\\perp IX_1.\$ Then, \$\\angle X_1IS_2=90^{\\circ}-\\angle X_1S_2I=90^{\\circ}-(90^{\\circ}-\\angle A)=\\angle A.\$ Thus,\n\n\\[\n\\angle T_1IX_1=\\angle X_1IS_2+\\angle S_2IT_2+\\angle T_2IT_1=\\angle A+\\angle B+2\\angle C-180^{\\circ}+180^{\\circ}-\\angle C=180^{\\circ},\n\\]\n\nas desired.\$\\Box\$\n\nThen, \$T_1IX_1\\perp S_2S_3\$ because a diameter is perpendicular to a chord at its midpoint and because \$T_1IX_1\$ is part of a diameter, by Claim 3. Hence, \$S_2S_3\\parallel BC.\$ But, \$M_2M_3\\parallel BC\$ so \$S_2S_3\\parallel M_2M_3\$. Similarly, we get \$S_1S_2\\parallel M_1M_2\$ and \$S_1S_3\\parallel M_1M_3\$ so \$\\triangle M_1M_2M_3\\sim \\triangle S_1S_2S_3\$ with corresponding parallel sides. Thus, there exists a homothety that maps \$M_1M_2M_3\$ to \$S_1S_2S_3\$ and \$M_1S_1, M_2S_2,\$ and \$M_3S_3\$ intersect at the center of the homothety. \$\\blacksquare\$\n\nThis solution was posted and copyrighted by Jzhang21. The original thread for this problem can be found here: [1]" ]
IMO-1982-3	https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_3	Consider infinite sequences $\{x_n\}$ of positive reals such that $x_0=1$ and $x_0\ge x_1\ge x_2\ge\ldots$. a) Prove that for every such sequence there is an $n\ge1$ such that: \[ {x_0^2\over x_1}+{x_1^2\over x_2}+\ldots+{x_{n-1}^2\over x_n}\ge3.999. \] b) Find such a sequence such that for all $n$: \[ {x_0^2\over x_1}+{x_1^2\over x_2}+\ldots+{x_{n-1}^2\over x_n}<4. \]	[ "By Cauchy, the LHS is at least: \$\\frac{(x_{0}+...+x_{n-1})^{2}}{x_{1}+...+x_{n}}.\$\n\nIt is clear that \$x_{0}=1\$ and \$x_{n}\\le \\frac{x_{1}+...+x_{n-1}}{n-1}\$\n\nWe thus have that the LHS is at least: \$(\\frac{1+2(x_{1}+...+x_{n-1})+(x_{1}+...+x_{n-1})^{2}}{x_{1}+...+x_{n-1}})(\\frac{n-1}{n})\$\n\nApplying 2-variable AM-GM to \$1\$ and \$(x_{1}+...+x_{n-1})^{2}\$ the LHS is at least \$4(\\frac{n-1}{n})\$\n\nWe then simply choose \$n\$ such that \$4(\\frac{n-1}{n}) \\ge 3.999\$.\n\nPart b) is likely to occur only if both the AM-GM and Cauchy equality conditions are close to met, or if \$1=x_{1}+x_{2}+...\$ and \$\\frac{x_{0}}{x_{1}}=\\frac{x_{1}}{x_{2}}=...\$, respectively. This points to the geometric series with common ratio \$\\frac{1}{2}\$ as the equality case.\n\nThis solution was posted and copyrighted by n^4 4. The original thread for this problem can be found here: [1]", "b) is immediate by taking \$x_i = 2^{-i}\$.\n\nFor a), we prove \$LHS \\ge 2^{2^0 + 2^{-1}+...+ 2^{2-n}}\$, and taking \$n\\to \\infty\$ will yield the result. (Here \$n=1\$ gives the empty sum, so \$LHS \\ge 2^0=1\$, which is trivial).\n\nThe proof is not hard. Let \$s_i\$ be the sum of the first \$i\$ terms, so \$s_n = s_{n-2} + \\dfrac{x_{n-2}^2}{x_{n-1}} + \\dfrac{x_{n-1}^2}{x_n} \\ge s_{n-2} + 2^{2^0} x_{n-2} \\sqrt{\\dfrac{x_{n-1}}{x_n}} \\ge s_{n-2} + 2^{2^0} x_{n-2}\$. This in turn is \$s_{n-3} + \\dfrac{x_{n-3}^2}{x_{n-2}} + 2^{2^0} x_{n-2}\\ge s_{n-3} + 2^{2^0 + 2^{-1}} x_{n-3}\$. By repeatedly applying AM-GM to the last two terms of the sum, collecting powers of two, and cancelling variables, we eventually obtain\n\n\$s_0 + \\dfrac{x_0^2}{x_1} + 2^{2^0 + 2^1 + ... + 2^{3-n}}x_1 \\ge 0 + 2^{2^0 + 2^1 + ... + 2^{2-n}}x_0\$, as desired.\n\nThis solution was posted and copyrighted by tastymath75025. The original thread for this problem can be found here: [2]" ]
IMO-1982-4	https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_4	Prove that if $n$ is a positive integer such that the equation $x^3-3xy^2+y^3=n$ has a solution in integers $x,y$, then it has at least three such solutions. Show that the equation has no solutions in integers for $n=2891$.	[ "Suppose the equation \$x^3-3xy^2+y^3=n\$ has solution in integers \$(x,y)\$ with \$y=x+k\$. Then, completing the cube yields \$(y-x)^3-3x^2y+2x^3\$. Using the substitution \$y=x+k\$ yields \$k^3-3kx^2-x^3=n\$. Notice that equality directly implies that \$(k,-x)\$ is also a solution to the original equation. Applying the transformation again yields that \$(-x-k,-k)\$ is also a solution. We show that these three solutions are indeed distinct: If \$(x,y)=(k,-x)\$ then \$x=k,x+k=-x\$ which only has solution \$x=k=0\$ which implies that \$n\$ is not a positive integer, a contradiction. Similarly, since the transformation from \$(k,-x)\$ to \$(-x-k,-k)\$ and \$(-x-k,-k)\$ to \$(x,y)\$ is the same as the transformation from \$(x,y)\$ to \$(k,-x)\$, we have that the three solutions are pairwise distinct.\n\nFor the case \$n=2891\$, notice that \$7\|n\$. Considering all solutions modulo \$7\$ of the equation \$x^3-3xy^2+y^3\\equiv0\\pmod{7}\$ yields only \$x\\equiv y\\equiv 0\\pmod{7}\$. But, this implies that \$7^3\$ divides \$2891\$ which is clearly not true." ]
IMO-1982-5	https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_5	The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that \[ {AM\over AC}={CN\over CE}=r. \] Determine $r$ if $B,M$ and $N$ are collinear.	[ "O is the center of the regular hexagon. Then we clearly have \$ABC\\cong COA\\cong EOC\$. And therefore we have also obviously \$ABM\\cong AOM\\cong CON\$, as \$\\frac{AM}{AC} =\\frac{CN}{CE}\$. So we have \$\\angle{BMA} =\\angle{AMO} =\\angle{CNO}\$ and \$\\angle{NOC} =\\angle{ABM}\$. Because of \$\\angle{AMO} =\\angle{CNO}\$ the quadrilateral \$ONCM\$ is cyclic. \$\\Rightarrow \\angle{NOC} =\\angle{NMC} =\\angle{BMA}\$. And as we also have \$\\angle{NOC} =\\angle{ABM}\$ we get \$\\angle{ABM} =\\angle{BMA}\$. \$\\Rightarrow AB=AM\$. And as \$AC=\\sqrt{3} \\cdot AB\$ we get \$r=\\frac{AM}{AC} =\\frac{AB}{\\sqrt{3} \\cdot AB} =\\frac{1}{\\sqrt{3}}\$.\n\nThis solution was posted and copyrighted by Kathi. The original thread for this problem can be found here: [1]", "Let \$X\$ be the intersection of \$AC\$ and \$BE\$. \$X\$ is the mid-point of \$AC\$. Since \$B\$, \$M\$, and \$N\$ are collinear, then by Menelaus Theorem, \$\\frac{CN}{NE}\\cdot\\frac{EB}{BX}\\cdot\\frac{XM}{MC}=1\$. Let the sidelength of the hexagon be \$1\$. Then \$AC=CE=\\sqrt{3}\$. \$\\frac{CN}{NE}=\\frac{CN}{CE-CN}=\\frac{\\frac{CN}{CE}}{1-\\frac{CN}{CE}}=\\frac{r}{1-r}\$ \$\\frac{EB}{BX}=\\frac{2}{\\frac{1}{2}}=4\$ \$\\frac{XM}{MC}=\\frac{AM-AX}{AC-AM}=\\frac{\\frac{AM}{AC}-\\frac{AX}{AC}}{1-\\frac{AM}{AC}}=\\frac{r-\\frac{1}{2}}{1-r}\$ Substituting them into the first equation yields \$\\frac{r}{1-r}\\cdot\\frac{4}{1}\\cdot\\frac{r-\\frac{1}{2}}{1-r}=1\$ \$3r^2=1\$ \$\\therefore r=\\frac{\\sqrt{3}}{3}\$\n\nThis solution was posted and copyrighted by leepakhin. The original thread for this problem can be found here: [2]", "Note \$x=m(\\widehat {EBM})\$. From the relation \$r=\\frac{AM}{AC}=\\frac{CN}{CE}\$ results \$\\frac{MA}{MC}=\\frac{NC}{NE}\$, i.e.\n\n\$\\frac{BA}{BC}\\cdot \\frac{\\sin (60+x)}{\\sin (60-x)}=\\frac{BC}{BE}\\cdot \\frac{\\sin (60-x)}{\\sin x}\$. Thus, \$2\\sin x\\sin (60+x)=\\sin^2(60-x)\\Longrightarrow\$ \$2[\\cos 60-\\cos (60+2x)]=1-\\cos (120-2x)\\Longrightarrow \\cos (60+2x)=0\\Longrightarrow x=15^{\\circ}.\$\n\nTherefore, \$\\frac{MA}{MC}=\\frac{\\sin 75}{\\sin 45}=\\frac{1+\\sqrt 3}{2}\$, i.e. \$r=\\frac{MA}{AC}=\\frac{1+\\sqrt 3}{3+\\sqrt 3}=\\frac{1}{\\sqrt 3}\\Longrightarrow r=\\frac{\\sqrt 3}{3}\$\n\nThis solution was posted and copyrighted by Virgil. The original thread for this problem can be found hercommunity/p398343]", "Let \$AM = CN = a\$. By the cosine rule,\n\n\\[\nAC = \\sqrt{AB^{2} + BC^{2} - 2 \\cdot AB \\cdot BC \\cdot \\cos \\angle BAC}\n\\]\n\n\\[\n= \\sqrt{1 + 1 - 2 \\cdot \\cos 120^{\\circ}}\n\\]\n\n\$= \\sqrt{3}\$.\n\n\\[\nBM = \\sqrt{a^{2} + 1 - 2a \\cdot \\cos 30^{\\circ}}\n\\]\n\n\\[\n= \\sqrt{a^{2} - \\sqrt{3} \\cdot a + 1}\n\\]\n\n\\[\nMN = \\sqrt{(\\sqrt{3} - a)^{2} + a^{2} - 2 \\cdot (\\sqrt{3} - a) \\cdot a \\cdot \\cos \\angle MCN}\n\\]\n\n\\[\n= \\sqrt{3 + a^{2} - 2\\cdot \\sqrt{3} \\cdot a + a^{2} - \\sqrt{3} \\cdot a + a^{2}}\n\\]\n\n\\[\n= \\sqrt{3a^{2} - 3\\sqrt{3}\\cdot a + 3}\n\\]\n\n\$= BM \\cdot \\sqrt{3}\$.\n\nNow if B, M, and N are collinear, then \$\\angle AMB = \\angle CMN\$\n\n\$\\implies \\sin \\angle AMB = \\sin \\angle CMN\$.\n\nBy the law of Sines,\n\n\\[\n\\frac{1}{\\sin \\angle AMB} = \\frac{BM}{sin 30^{\\circ}} = 2BM\n\\]\n\n\$\\implies \\sin \\angle AMB = \\frac{1}{2BM}\$.\n\nAlso,\n\n\\[\n\\frac{a}{\\sin \\angle CMN} = \\frac{\\sqrt{3} \\cdot BM}{\\sin 60^{\\circ}} = 2BM\n\\]\n\n\$\\implies \\sin \\angle CMN = \\frac{a}{2BM}\$.\n\nBut \$\\sin \\angle AMB = \\sin \\angle CMN\$\n\n\$\\implies \\frac{1}{2BM} = \\frac{a}{2BM}\$, which means \$a = 1\$. So, r = \$\\frac{1}{\\sqrt{3}}\$.", "We can assign coordinates to solve this question. First, WLOG, let the side length of the hexagon be 1. We also know that each angle of the hexagon is \$\\frac{(6-2) \\cdot 180}{6} = 120\$. From Law of Cosines, we find \$AC = \\sqrt{1 + 1 - 2cos(120)} = \\sqrt3\$.\n\nNow, let point \$E\$ be located at \$(0, 0)\$. Since \$AE = AC,\$ \$A\$ is located at \$(0, \\sqrt3)\$ meaning \$B\$ is located at \$(1, \\sqrt3)\$. Using the ratios given to us, \$CN = r\\sqrt3\$ so \$EN = (1-r)\\sqrt3\$. Let \$O\$ be defined on line \$ED\$ such that \$EO\$ is perpendicular to \$NO\$. Since \$\\bigtriangleup ENO\$ is a \$30-60-90\$ triangle, we can find that point N is located at \$(\\frac{3(1-r)}{2}, \\frac{(1-r)\\sqrt3}{2})\$. Similarly, \$M\$ would be located at \$(\\frac{3r}{2}, \\sqrt3(1 - \\frac{r}{2}))\$.\n\nSince B, M, N, are collinear, they have the same slope. So,\n\n\\[\n\\frac{\\sqrt3 - \\frac{(1-r)\\sqrt3}{2}}{1 - \\frac{3(1-r)}{2}} = \\frac{\\sqrt3 - \\sqrt3(1 - \\frac{r}{2})}{1 - \\frac{3r}{2}}\n\\]\n\n\\[\n\\frac{\\sqrt3 + r\\sqrt3}{3r - 1} = \\frac{r\\sqrt3}{2 - 3r}\n\\]\n\n\\[\n2\\sqrt3 + 2r\\sqrt3 - 3r\\sqrt3 - 3r^2\\sqrt3 = 3r^2\\sqrt3 - r\\sqrt3\n\\]\n\n\\[\n6r^2\\sqrt3 = 2\\sqrt3\n\\]\n\n\\[\n\\boxed{r = \\frac{\\sqrt3}{3}}\n\\]\n\n~SID-DARTH-VATER" ]
IMO-1982-6	https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_6	Let $S$ be a square with sides length $100$. Let $L$ be a path within $S$ which does not meet itself and which is composed of line segments $A_0A_1,A_1A_2,A_2A_3,\ldots,A_{n-1}A_n$ with $A_0=A_n$. Suppose that for every point $P$ on the boundary of $S$ there is a point of $L$ at a distance from $P$ no greater than $\frac {1} {2}$. Prove that there are two points $X$ and $Y$ of $L$ such that the distance between $X$ and $Y$ is not greater than $1$ and the length of the part of $L$ which lies between $X$ and $Y$ is not smaller than $198$.	[ "Let the square be A'B'C'D'. The idea is to find points of L close to a particular point of A'D' but either side of an excursion to B'.\n\nWe say L approaches a point P' on the boundary of the square if there is a point P on L with PP' ≤ 1/2. We say L approaches P' before Q' if there is a point P on L which is nearer to A0 (the starting point of L) than any point Q with QQ' ≤ 1/2.\n\nLet A' be the first vertex of the square approached by L. L must subsequently approach both B' and D'. Suppose it approaches B' first. Let B be the first point on L with BB' ≤ 1/2. We can now divide L into two parts L1, the path from A0 to B, and L2, the path from B to An.\n\nTake X' to be the point on A'D' closest to D' which is approached by L1. Let X be the corresponding point on L1. Now every point on X'D' must be approached by L2 (and X'D' is non-empty, because we know that D' is approached by L but not by L1). So by compactness X' itself must be approached by L2. Take Y to be the corresponding point on L2. XY ≤ XX' + X'Y ≤ 1/2 + 1/2 = 1. Also BB' ≤ 1/2, so XB ≥ X'B' - XX' - BB' ≥ X'B' - 1 ≥ A'B' - 1 = 99. Similarly YB ≥ 99, so the path XY ≥ 198." ]
IMO-1983-1	https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_1	Find all functions $f$ defined on the set of positive reals which take positive real values and satisfy the conditions: (i) $f(xf(y))=yf(x)$ for all $x,y$; (ii) $f(x)\to0$ as $x\to \infty$.	[ "Let \$x=y=1\$ and we have \$f(f(1))=f(1)\$. Now, let \$x=1,y=f(1)\$ and we have \$f(f(f(1)))=f(1)f(1)\\Rightarrow f(1)=[f(1)]^2\$ since \$f(1)>0\$ we have \$f(1)=1\$.\n\nPlug in \$y=x\$ and we have \$f(xf(x))=xf(x)\$. If \$a=1\$ is the only solution to \$f(a)=a\$ then we have \$xf(x)=1\\Rightarrow f(x)=\\frac{1}{x}\$. We prove that this is the only function by showing that there does not exist any other \$a\$:\n\nSuppose there did exist such an \$a\\ne1\$. Then, letting \$y=a\$ in the functional equation yields \$f(xa)=af(x)\$. Then, letting \$x=\\frac{1}{a}\$ yields \$f(\\frac{1}{a})=\\frac{1}{a}\$. Notice that since \$a\\ne1\$, one of \$a,\\frac{1}{a}\$ is greater than \$1\$. Let \$b\$ equal the one that is greater than \$1\$. Then, we find similarly (since \$f(b)=b\$) that \$f(xb)=bf(x)\$. Putting \$x=b\$ into the equation, yields \$f(b^2)=b^2\$. Repeating this process we find that \$f(b^{2^k})=b^{2^k}\$ for all natural \$k\$. But, since \$b>1\$, as \$k\\to \\infty\$, we have that \$b^{2^k}\\to\\infty\$ which contradicts the fact that \$f(x)\\to0\$ as \$x\\to \\infty\$.", "Let \$x=1\$ so\n\n\\[\nf(f(y))=yf(1).\n\\]\n\nIf \$f(a)=f(b),\$ then \$af(1)=f(f(a))=f(f(b))=bf(1)\\implies a=b\$ because \$f(1)\$ goes to the real positive integers, not \$0.\$ Hence, \$f\$ is injective. Let \$x=y\$ so\n\n\\[\nf(xf(x))=xf(x)\n\\]\n\nso \$xf(x)\$ is a fixed point of \$f.\$ Then, let \$y=1\$ so \$f(xf(1))=f(x)\\implies f(1)=1\$ as \$x\$ can't be \$0\$ so \$1\$ is a fixed point of \$f.\$ We claim \$1\$ is the only fixed point of \$f.\$ Suppose for the sake of contradiction that \$a,b\$ be fixed points of \$f\$ so \$f(a)=a\$ and \$f(b)=b.\$ Then, setting \$x=a,y=b\$ in (i) gives\n\n\\[\nf(ab)=f(af(b))=bf(a)=ab\n\\]\n\nso \$ab\$ is also a fixed point of \$f.\$ Also, let \$x=\\frac{1}{a},y=a\$ so\n\n\\[\n1=f(1)=f(\\frac{1}{a}\\cdot a)=f(\\frac{1}{a}\\cdot f(a))=af(\\frac{1}{a})\\implies f(\\frac{1}{a})=\\frac{1}{a}\n\\]\n\nso \$\\frac{1}{a}\$ is a fixed point of \$f.\$ If \$f(a)=a\$ with \$a>1,\$ then \$f(a^n)\$ is a fixed point of \$f\$, contradicting (ii). If \$f(a)=a\$ with \$0<a<1,\$ then \$f(\\frac{1}{a^n})=\\frac{1}{a^n}\$ so \$\\frac{1}{a^n}\$ is a fixed point, contradicting (ii). Hence, the only fixed point is \$1\$ so \$xf(x)=1\$ so \$f(x)=\\boxed{\\frac{1}{x}}\$ and we can easily check that this solution works.", "Let \$x=1\$. So \$f(f(y)) = y f(1)\$. This clearly implies \$f\$ is injective. Let \$y = 1\$. Then \$f(f(y)) = f(1)\$. Applying \$f^{-1}\$ yields \$f(1) = 1\$, so \$f(f(y)) = y\$. So \$f\$ is surjective. Applying \$f^{-1}\$ yields \$f(y) = f^{-1}(y)\$.\n\nLet \$u=x\$ and \$y = f^{-1}(v)\$. (i) becomes\n\n\\[\nf(uv) = f^{-1}(v)f(u) = f(u)f(v)\n\\]\n\nsince \$f = f^{-1}\$. So \$f\$ is a group isomorphism on the group \$(0,\\infty)\$ with binary operation multiplication. One gets \$f(u)^n = f(u^n)\$. Letting \$v = u^n\$ implies \$f(v^{1/n}) = f(v)^{1/n}\$. Combining these identities,\n\n\\[\nf(u)^{\\frac{m}{n}} = f(u^{\\frac{m}{n}}).\n\\]\n\nSuppose \$u>1\$. Taking \$\\frac{m}{n} \\rightarrow \\infty\$, the RHS goes to \$0\$ by (ii). Considering the LHS, necessarily \$0 < f(u) < 1\$. So \$f(u) < 1\$ for \$u>1\$. We claim \$f\$ is strictly monotonically decreasing. Indeed, if \$0<x<y\$ then\n\n\\[\nf(y) = f\\Big(x \\frac{y}{x}\\Big) = f(x)f\\Big(\\frac{y}{x}\\Big) < f(x)\n\\]\n\nsince \$y/x > 1\$ thus \$f(y/x) < 1\$. Now suppose \$r > 0\$. We claim \$f(x^r) = f(x)^r\$. Indeed let \$m_1/n_1 < r < m_2/n_2\$. If \$x > 1\$ then\n\n\\[\nf(x)^{\\frac{m_1}{n_1}} = f(x^{\\frac{m_1}{n_1}}) > f(x^r) > f(x^{\\frac{m_2}{n_2}}) = f(x)^{\\frac{m_2}{n_2}}\n\\]\n\nand letting \$\\frac{m_1}{n_1}, \\frac{m_2}{n_2} \\rightarrow r\$ implies \$f(x^r) = f(x)^r\$ by the squeeze theorem. The case \$x < 1\$ is similar. The case \$x=1\$ is trivial since \$f(1) = 1\$. Suppose now \$f(e) = c\$. Then\n\n\\[\nf(x) = f(e^{\\ln x}) = f(e)^{\\ln x} = c^{\\ln x}.\n\\]\n\nSolving \$y = c^{\\ln x}\$ for \$x\$ yields \$x = \\exp\\big((\\ln y)/(\\ln c)\\big)\$. Since \$f = f^{-1}\$ it follows\n\n\\[\nc^{\\ln x} = \\exp\\bigg(\\frac{\\ln x}{\\ln c}\\bigg),\n\\]\n\nwhich implies \$(\\ln c)^2 = 1\$ when \$x \\neq 1\$. So either \$c = e\$ or \$c = 1/e\$. Since \$x>1\$ implies \$f(x) < 1\$ we have \$c = f(e) < 1\$, so \$c = 1/e\$. In which case \$f(x) = c^{\\ln x} = 1/x\$.\n\n~not_detriti" ]
IMO-1983-2	https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_2	Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. One of the common tangents to the circles touches $C_1$ at $P_1$ and $C_2$ at $P_2$, while the other touches $C_1$ at $Q_1$ and $C_2$ at $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2=\angle M_1AM_2$.	[ "Let \$A\$ be one of the two distinct points of intersection of two unequal coplanar circles \$C_1\$ and \$C_2\$ with centers \$O_1\$ and \$O_2\$ respectively. Let \$S\$ be such point on line \$O_1O_2\$ so that tangents on \$C_1\$ touches it at \$P_1\$ and \$Q_1\$ and tangents on \$C_2\$ touches it at \$P_2\$ and \$Q_2\$. Let \$M_1\$ be the midpoint of \$P_1Q_1\$ and \$M_2\$ the midpoint of \$P_2Q_2\$. Prove that \$\\angle O_1AO_2 = \\angle M_1AM_2\$.\n\nProof: Since \$S\$ is image of \$M_1\$ under inversion wrt circle \$C_1\$ we have:\n\n\\[\n\\angle O_1AM_1 = \\angle O_1M_1'A'= \\angle O_1SA\n\\]\n\nSince \$S\$ is image of \$M_2\$ under inversion wrt circle \$C_2\$ we have:\n\n\\[\n\\angle O_2SA= \\angle O_2A'S'= \\angle O_2AM_2\n\\]\n\nImage of \$A\$ is in both cases \$A\$ itself, since it lies on both circles. Since \$\\angle O_1SA = \\angle O_2SA\$ we have:\n\n\\[\n\\angle M_1AO_1=\\angle M_2AO_2\n\\]\n\nNow:\n\n\\[\n\\angle O_1AO_2 = \\angle M_1AM_2-\\angle M_1AO_1+\\angle M_2AO_2 = \\angle M_1AM_2\n\\]\n\nThis solution was posted and copyrighted by Number1. The original thread for this problem can be found here: [1]", "Let \$P_1P_2\$ and \$Q_1Q_2\$ meet at \$R\$. Let \$RA\$ meet \$C_2\$ at \$B\$. Now, it is well-known that \$O_1O_2\$, \$P_1P_2\$, and \$Q_1Q_2\$ are concurrent at \$R\$, the center of homothety between \$C_1\$ and \$C_2\$. Now, it is well-known that \$O_1O_2\$ bisects \$\\angle P_1RQ_1\$. Since \$P_1R = Q_1R\$, we have that \$O_1O_2R\$ meets \$P_1Q_1\$ at its midpoint, \$M_1\$, and \$P_1Q_1\$ is perpendicular to \$O_1O_2R\$. Similarly, \$O_1O_2R\$ passes through \$M_2\$ and is perpendicular to \$P_2Q_2\$. Since \$O_2P_2\\perp P_2R\$, we have that \$RA\\cdot RB = RP_2^2 = RM_2\\cdot RO_2\$, which implies that \$ABM_2O_2\$ is cyclic. Yet, since \$A\$ and \$B\$ lie on \$C_1\$ and \$C_2\$ respectively and are collinear with \$R\$, we see that the homothety that maps \$C_1\$ to \$C_2\$ about \$R\$ maps \$A\$ to \$B\$. Also, \$O_1\$ is mapped to \$O_2\$ by this homothety, and since \$M_1\$ and \$M_2\$ are corresponding parts in these circles, \$M_1\$ is mapped to \$M_2\$ by this homothety, so \$\\angle O_1AM_1 = \\angle O_2BM_2 = \\angle O_2AM_2\$, from which we conclude that \$\\angle O_2AO_1 = \\angle M_2AOM_1\$.\n\nThis solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [2]", "Let \$B\$ be the other intersection point of these two circles. Let \$P_1P_2\$, \$Q_1Q_2\$, \$O_1O_2\$ meet at \$Q\$. Let \$AB\$ meet \$P_1P_2\$ at \$P\$. Clearly, \$P_1Q_1\$ and \$O_1O_2\$ are perpendicular at \$M_1\$; \$P_2Q_2\$ and \$O_1O_2\$ are perpendicular at \$M_2\$. Since \$\\triangle O_1P_1M_1 \\sim \\triangle O_2P_2M_2\$,\n\n\\[\n\\dfrac{O_1P_1}{O_1M_1} = \\dfrac{O_2P_2}{O_2M_2} \\Rightarrow \\dfrac{O_1A}{O_1M_1} = \\dfrac{O_2A}{O_2M_2} \\qquad{()}\n\\]\n\nSince \$P\$ is on the radical axis, \$PP_1 = PP_2\$, so in the right trapezoid \$P_1P_2M_2M_1\$, \$AB\$ is the midsegment. So we have \$M_1A = AM_2\$ and \$\\angle AM_1O_1 = \\angle AM_2O_1\$. Let \$M\$ be a point on \$O_1O_2\$ such that \$\\triangle AM_1O_1 \\cong \\triangle AM_2M\$ (which means \$AM=AO_1\$ ve \$MM_2 = M_1O_1\$). So, from \$()\$, we get\n\n\\[\n\\dfrac{AM}{MM_2} = \\dfrac{O_2A}{O_2M_2}\n\\]\n\nThis means, \$AM_2\$ is the angle bisector of \$\\angle MAO_2\$. So,\n\n\\[\n\\angle O_2AM_2 = \\angle M_2AM= \\angle M_1AO_1 \\Longrightarrow \\angle M_1AM_2 = \\angle O_1AO_2.\n\\]\n\nThis solution was posted and copyrighted by matematikolimpiyati. The original thread for this problem can be found here: [3]", "Let, \$P_1P_2,Q_1Q_2,O_1O_2\$ concur at \$Q\$.Then a homothety with centre \$Q\$ that sends \$C_1\$ to \$C_2\$.Let \$QA \\cap C_1 =B\$.Under the homothety \$A\$ is the image of \$B\$.So, \$\\angle M_1BO_1 =\\angle M_2AO_2\$ and \$\\triangle QP_1O_1 \\sim \\triangle QM_1P_1 \\implies \\frac{QO_{1}}{QP_{1}} = \\frac{QP_{1}}{QM_{1}}\$.so \$QO_1.QM_1 = Q.P^2_1 = QA .QB\$ Hence \$A,M_1,B,O_1\$ are concyclic.so \$\\angle O_1BM_1=\\angle O_1AM_1 \\implies \\angle O_1AM_1= \\angle O_2AM_2 \\implies \\angle O_1AO_2=\\angle M_1AM_2\$\n\nThis solution was posted and copyrighted by spikerboy. The original thread for this problem can be found here: [4]", "It can be easily solved with this lemma: Let \$A\$-symmedian of \$\\triangle ABC\$ intersect its circumscribed circle \$u\$ at \$D, P\$ on \$AD\$ satisfying \$AP = PD\$. Let us define center of \$u\$ as \$O\$. Then \$(B, C, P, O)\$ are concyclic. Now let \$S\$ be intersection of tangents to circles from the problem, \$AS\$ intersects \$C_1\$ at \$B\$. One can prove that \$\\square AP_1BQ_1\$ is harmonic qudrilateral, so \$P_1Q_1\$ is symmedian of \$\\triangle AP_1B\$. By lemma we get that \$(B, A, O_1, M_1)\$ are concyclic, thus \$\\angle O_1AM_1\$ is equivalent to \$\\angle O_1BM_1\$. By homothety we get \$\\angle O_1BM_1 = \\angle O_2AM_2\$. \$Q. E. D.\$\n\nThis solution was posted and copyrighted by LeoLeon. The original thread for this problem can be found here: [5]" ]
IMO-1983-3	https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_3	Let $a$, $b$ and $c$ be positive integers, no two of which have a common divisor greater than $1$. Show that $2abc - ab - bc- ca$ is the largest integer which cannot be expressed in the form $xbc + yca + zab$, where $x$, $y$ and $z$ are non-negative integers.	[ "First off, I prove \$2abc-ab-bc-ca\$ is un-achievable. Also, assume WLOG \$a\\ge b\\ge c\$.\n\nAssume \$2abc-bc-ca=xbc+yca+zab\$, then take the equation \$\\pmod{ab}\$ to give us \$(x+1)bc+(y+1)ac\\equiv 0\\pmod{ab}\$. By CRT and \$\\gcd(a,b)=1\$, we take this equation mod a and mod b to give us \$y+1\\equiv 0\\pmod{b}\$, and \$x+1\\equiv 0\\pmod{a}\$ (and using all numbers are relatively prime pairwise). Substituting this back into the original equation gives us \$z\\le -1\$, contradiction (this is where the \$2abc\$ part comes in).\n\nNow, let \$2abc-bc-ca-ab+k=xbc+yca+zab\$, and if a solution exists where \$bc-1\\ge k\\ge 1\$ then we are done because we can add \$1\$ to \$x\$ always.\n\nConsider the set \$x=a-m\$, where \$m\\in \\{1,2,\\cdots, a-1, a\\}\$ and \$y=b-n\$ where \$n\\in \\{1,2,\\cdots, b-1, b\\}\$.\n\nTherefore, we get \$2abc-bc-ca-ab+k=(a-m)bc+(b-n)ca+zab\$. This is the same as \$bc(m-1)+ca(n-1)+k=(z+1)ab\$ after doing some simplifying. By CRT, this must hold mod a and mod b and because \$\\gcd(a,b)=1\$. Mod \$a\$ gives us \$bc(m-1)+k\\equiv 0\\pmod{a}\$, for which a value of \$m-1\$ obviously exists mod \$a\$, which can be chosen from the set of values we have assigned for \$m\$. Similar method shows a value of \$n-1\$ exists mod \$b\$ from the set we have given to \$n\$.\n\nNow, we already know that \$x,y\\ge 0\$. Also, the LHS of the equation \$bc(m-1)+ca(n-1)+k\$ is at least \$1\$, therefore \$z\\ge 0\$ and we are done.\n\nThis solution was posted and copyrighted by binomial-theorem. The original thread for this problem can be found here: [1]", "First we will prove \$2abc-ab-bc-ca\$ is unattainable, as such: Suppose \$xbc+yca+zab=2abc-ab-bc-ca\$. Then, taking this mod \$a\$, we have that \$xbc\\equiv 0\\pmod{a}\$, so \$x\\equiv 0\\pmod{a}\$, and \$a\|x\$. Similarly, \$b\|y\$, and \$c\|z\$, so \$x\\ge a\$, \$y\\ge b\$, and \$z\\ge c\$. Thus, \$xbc+yca+zac\\ge 3abc\$, so \$2abc\\ge 3abc\$, which is a contradiction.\n\nNow we will prove all \$n>2abc-ab-bc-ca\$ is attainable, as such: consider the integer \$r\$ such that \$r\\equiv \\frac{n}{bc} \\pmod{a}\$ and \$0\\le r\\le a-1\$. Rearranging the equation \$xbc+yca+zab=n\$ yields \$\\frac{n-xbc}{a}=yc+zb\$, so set \$x=r\$. We see that \$xbc-n=rbc-n\\equiv 0 \\pmod{a}\$, so \$\\frac{n-xbc}{a}\$ is a positive integer (obviously \$n-xbc>0)\$. Now, note that since \$x\\le a-1\$, we have that \$2abc-ab-ac-bc=(a-1)bc+abc-ab-ac\\ge xbc+abc-ab-ac\$, so \$n>xbc+abc-ab-ac\$. Thus, \$n-xbc>abc-ab-ac\$ so \$\\frac{n-xbc}{a}>bc-b-c\$, so by Chicken Mc-Nugget theorem, there exist \$b, c\$ that satisfy the equation and are now done. \$\\blacksquare\$\n\nThis solution was posted and copyrighted by Stormersyle. The original thread for this problem can be found here: [2]" ]
IMO-1983-4	https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_4	Let $ABC$ be an equilateral triangle and $\mathcal{E}$ the set of all points contained in the three segments $AB$, $BC$ and $CA$ (including $A$, $B$ and $C$). Determine whether, for every partition of $\mathcal{E}$ into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle. Justify your answer.	[ "The answer is positive; there always a class of the partition will contain the three vertices of a right-angled triangle.\n\nFirst notice that if there are at least two points of the same color on a side, then all the points in \$\\mathcal{E}\$ which project orthogonally onto points of that color on the respective side must have the opposite color if we are to have no monochromatic right-angled triangles.\n\nNow assume we can find a side on which there is at most one point bearing one of the colors (blue, say; we take the two colors to be red and blue). Then, by the observation above, it's obvious that the other two sides contain three blue vertices of a right-angled triangle, and we're done (all the points of those two sides are blue, except maybe for the endpoints which they have in common with the other side and the point which projects orthogonally onto the unique blue point on the third side). We can now assume all three sides contain at least two points of each color. Take \$U\\in BC\$ s.t. \$BC=3BU\$. \$U\$ projects orthogonally onto \$V\\in CA\$, and \$V\$ projects orthogonally onto \$T\\in AB\$. We can easily see that \$T\$ projects orthogonally onto \$U\$. Again, by the observation in the previous paragraph, the points \$U,V,T\$ must have different colors if we are to have no monochromatic right-angled triangle, and this is impossible (we only have two colors).\n\nThis solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]" ]
IMO-1983-5	https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_5	Is it possible to choose $1983$ distinct positive integers, all less than or equal to $10^5$, no three of which are consecutive terms of an arithmetic progression? Justify your answer.	[ "The answer is yes. We will construct a sequence of integers that satisfies the requirements.\n\nTake the following definition of \$a_n\$: \$a_n\$ in base 3 has the same digits as \$n\$ in base 2. For example, since \$6=110_2\$, \$a_6=110_3=12\$.\n\nFirst, we will prove that no three \$a_n\$'s are in an arithmetic progression. Assume for sake of contradiction, that \$i < j < k\$ are numbers such that \$2a_j=a_i + a_k\$. Consider the base 3 representations of both sides of the equation. Since \$a_j\$ consists of just 0's and 1's in base three, \$2a_j\$ consists of just 0's and 2's base 3. No whenever \$2a_j\$ has a 0, both \$a_i\$ and \$a_k\$ must have a 0, since both could only have a 0 or 1 in that place. Similarly, whenever \$2a_j\$ has a 2, both \$a_i\$ and \$a_k\$ must have a 1 in that place. This means that \$a_i=a_k\$, which is a contradiction.\n\nNow since \$1983=11110111111_2\$, \$a_{1983} = 11110111111_3 < \\frac{3^{11}}{2} < 10^5\$. Hence, the set \$\\{a_1,a_2,...a_{1983}\\}\$ is a set of 1983 integers less than \$10^5\$ with no three in arithmetic progression." ]
IMO-1983-6	https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_6	Let $a$, $b$ and $c$ be the lengths of the sides of a triangle. Prove that \[ a^{2}b(a - b) + b^{2}c(b - c) + c^{2}a(c - a)\ge 0. \] Determine when equality occurs.	[ "By Ravi substitution, let \$a = y+z\$, \$b = z+x\$, \$c = x+y\$. Then, the triangle condition becomes \$x, y, z > 0\$. After some manipulation, the inequality becomes:\n\n\$xy^3 + yz^3 + zx^3 \\geq xyz(x+y+z)\$.\n\nBy Cauchy, we have:\n\n\$(xy^3 + yz^3 + zx^3)(z+x+y) \\geq xyz(y+z+x)^2\$ with equality if and only if \$\\frac{xy^3}{z} = \\frac{yz^3}{x} =\\frac{zx^3}{y}\$. So the inequality holds with equality if and only if x = y = z. Thus the original inequality has equality if and only if the triangle is equilateral.", "Without loss of generality, let \$a \\geq b \\geq c > 0\$. By Muirhead or by AM-GM, we see that \$a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \\geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)\$.\n\nIf we can show that \$a^3 b + b^3 c+ c^3 a \\geq a^3 c + b^3 a + c^3 b\$, we are done, since then \$2(a^3 b + b^3 c+ c^3 a ) \\geq a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \\geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)\$, and we can divide by \$2\$.\n\nWe first see that, \$(a^2 + ac + c^2) \\geq (b^2 + bc + c^2)\$, so \$(a-c)(b-c)(a^2 + ac + c^2) \\geq (a-c)(b-c)(b^2 + bc + c^2)\$.\n\nFactoring, this becomes \$(a^3 - c^3)(b-c) \\geq (a-c)(b^3 - c^3)\$. This is the same as:\n\n\$(a^3 - c^3)(b-c) + (b^3 - c^3)(c-a) \\geq 0\$.\n\nExpanding and refactoring, this is equal to \$a^3 (b-c) + b^3(c-a) + c^3 (a-b) \\geq 0\$. (This step makes more sense going backwards.)\n\nExpanding this out, we have\n\n\$a^3b + b^3 c + c^3 a \\geq a^3 c + b^3 a + c^3 b\$,\n\nwhich is the desired result.", "Let \$s\$ be the semiperimeter, \$\\frac{a+b+c}{2}\$, of the triangle. Then, \$a=s-\\frac{-a+b+c}{2}\$, \$b=s-\\frac{a-b+c}{2}\$, and \$c=s-\\frac{a+b-c}{2}\$. We let \$x=\\frac{-a+b+c}{2},\$ \$y=\\frac{a-b+c}{2}\$, and \$z=\\frac{a+b-c}{2}.\$ (Note that \$x,y,z\$ are all positive, since all sides must be shorter than the semiperimeter.) Then, we have \$a=s-x\$, \$b=s-y\$, and \$c=s-z\$. Note that \$x+y+z=s\$, so\n\n\\[\na=y+z,b=x+z,c=x+y.\n\\]\n\nPlugging this into\n\n\\[\na^2b(a-b)+b^2c(b-c)+c^2a(c-a)\\geq0\n\\]\n\nand doing some expanding and cancellation, we get\n\n\\[\n2x^3z+2xy^3+2yz^3-2x^2yz-2xy^2z-2xyz^2\\geq0.\n\\]\n\nThe fact that each term on the left hand side has at least two variables multiplied motivates us to divide the inequality by \$2xyz\$, which we know is positive from earlier so we can maintain the sign of the inequality. This gives\n\n\\[\n\\frac{x^2}{y}-x+\\frac{y^2}{z}-y-z+\\frac{z^2}{x}\\geq0.\n\\]\n\nWe move the negative terms to the right, giving\n\n\\[\n\\frac{x^2}{y}+\\frac{y^2}{z}+\\frac{z^2}{x}\\geq x+y+z.\n\\]\n\nWe rewrite this as\n\n\\[\n\\sum_{cyc}\\frac{x^2}{y}\\geq\\sum_{cyc}rx+(1-r)y.\n\\]\n\nwhere \$r\$ is any real number. (This works because if we evaulate the cyclic sum, then as long as the coefficients of \$x\$ and \$y\$ on the right sum to 1 the right side will be \$x+y+z\$.", "\"Solution from 111 problems in Algebra and Number Theory\"" ]
IMO-1984-1	https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_1	Let $x$, $y$, $z$ be nonnegative real numbers with $x + y + z = 1$. Show that $0 \leq xy+yz+zx-2xyz \leq \frac{7}{27}$	[ "Note that this inequality is symmetric with x,y and z.\n\nTo prove\n\n\\[\nxy+yz+zx-2xyz\\geq 0\n\\]\n\nnote that \$x+y+z=1\$ implies that at most one of \$x\$, \$y\$, or \$z\$ is greater than \$\\frac{1}{2}\$. Suppose \$x \\leq \\frac{1}{2}\$, WLOG. Then, \$xy+yz+zx-2xyz=yz(1-2x)+xy+zx\\geq 0\$ since \$(1-2x)\\geq 0\$, implying all terms are positive.\n\nTo prove \$xy+yz+zx-2xyz \\leq \\frac{7}{27}\$, suppose \$x \\leq y \\leq z\$. Note that \$x \\leq y \\leq \\frac{1}{2}\$ since at most one of x,y,z is \$\\frac{1}{2}\$. Suppose not all of them equals \$\\frac{1}{3}\$-otherwise, we would be done. This implies \$x \\leq \\frac{1}{3}\$ and \$z \\geq \\frac{1}{3}\$. Thus, define\n\n\\[\nx' =\\frac{1}{3}\n\\]\n\n,\n\n\\[\ny'=y\n\\]\n\n\\[\nz'=x+y-\\frac{1}{3}\n\\]\n\n\\[\n\\epsilon = \\frac{1}{3}-x\n\\]\n\nThen, \$x'=x+\\epsilon\$, \$z'=z-\\epsilon\$, and \$x'+y'+z'=1\$. After some simplification,\n\n\\[\nx'y'+y'z'+z'x'-2x'y'z'=xy+yz+zx-2xyz+(1-2y)(z-x-\\epsilon)>xy+yz+zx-2xyz\n\\]\n\nsince \$1-2y>0\$ and \$z-x-\\epsilon=z-\\frac{1}{3}>0\$. If we repeat the process, defining\n\n\\[\nx'' =x'=\\frac{1}{3}\n\\]\n\n\\[\ny''=\\frac{1}{3}\n\\]\n\n\\[\nz''=z'+y'-\\frac{1}{3}=\\frac{1}{3}\n\\]\n\nafter similar reasoning, we see that\n\n\\[\nxy+yz+zx-2xyz\\leq x'y'+y'z'+z'x'-2x'y'z' \\leq x''y''+y''z''+z''x''-2x''y''z''=\\frac{7}{27}\n\\]\n\n.", "By the method of Lagrangian multipliers. Let \$f(x,y,z) = xy + yz + zx - 2xyz\$ and \$\\phi(x,y,z) = x+y+z-1\$. We will find the local maxima/minima of \$f\$ over \$S = \\{(x,y,z) \\in \\mathbb{R}^2 : 0 \\leq x,y,z \\leq 1\\}\$ subject to \$\\phi = 0\$.\n\nLet \$S' = S \\cap \\{(x,y,z) \\in \\mathbb{R}^2 : \\phi(x,y,z) = 0\\}\$. \$S'\$ is bounded and the intersection of two closed sets, hence compact. So every sequence in \$S'\$ has a convergent subsequence. Hence if \$(a_k)_k \\subset S'\$ is such that \$f(a_k)\$ converges to the infimum or supremum of \$f\$ in \$S'\$, there will be a convergent subsequence \$a_{k_j} \\rightarrow a_\\infty \\in S'\$ and \$f(a_\\infty) = \\lim_{j \\rightarrow \\infty}f(a_{k_j})\$ by the continuity of \$f\$. Hence the infimum/supremum of \$f\$ over \$S'\$ will also be a local minima/maxima of \$f\$ over \$S'\$.\n\nWe must solve \$\\nabla f - \\lambda \\nabla \\phi = 0\$. This is equivalent to \\begin{align} x+y - 2xy &= -2uv + 1/2 = \\lambda \\\\ y+z - 2yz &= -2vw + 1/2 = \\lambda \\\\ z+x - 2zx &= -2wu + 1/2 = \\lambda \\end{align} where \$u = 1/2 - x, v = 1/2 - y, w = 1/2 - z\$. If \$\\lambda = 1/2\$ then \$uv = vw = wu = 0\$. WLOG \$u= 0\$ and we have \$vw = 0\$. Then WLOG \$v = 0\$. These imply \$x = y = 1/2\$. Then \$z = 0\$ since \$x+y+z = 1\$. We get \$f(1/2,1/2,0) = 1/4 < 7/27\$.\n\nIf \$\\lambda \\neq 1/2\$ then letting \$\\lambda' = 1/4 - \\lambda/2 \\neq 0\$ one gets \$uv = vw = wu = \\lambda'\$ which imply \$u=v=w\$ since \$u,v,w \\neq 0\$. This implies \$x=y=z=1/3\$ since \$x+y+z=1\$. And \$f(1/3,1/3,1/3) = 7/27\$.\n\nNow to check the boundary of \$0 \\leq x,y,z \\leq 1\$. WLOG we must consider the cases \$z=1\$ and \$z=0\$. If \$z=1\$ then \$x=y=0\$ by \$x+y+z=1\$ so \$f(x,y,z) = 0\$. If \$z = 0\$ substituting \$y=1-x\$ in \$f(x,y,0)\$ yields\n\n\\[\nf(x,1-x,0) = x(1-x) = -(1/2-x)^2+1/4.\n\\]\n\nwhich is between \$0\$ and \$1/4<7/27\$ since \$0 \\leq x \\leq 1\$. Considering all the values found we find \$0 \\leq f(x,y,z) \\leq 7/27\$.\n\n~not_detriti" ]
IMO-1984-2	https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_2	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 $7^7$.	[ "So we want \$7 \\nmid ab(a+b)\$ and \$7^7 \| (a+b)^7-a^7-b^7 = 7ab(a+b)(a^2+ab+b^2)^2\$, so we want \$7^3 \| a^2+ab+b^2\$. Now take e.g. \$a=2,b=1\$ and get \$7\|a^2+ab+b^2\$. Now by some standard methods like Hensels Lemma (used to the polynomial \$x^2+x+1\$, so \$b\$ seen as constant from now) we get also some \$\\overline{a}\$ with \$7^3 \| \\overline{a}^2+\\overline{a}b+b^2\$ and \$\\overline{a} \\equiv a \\equiv 2 \\mod 7\$, so \$7\\nmid \\overline{a}b(\\overline{a}+b)\$ and we are done. (in this case it gives \$\\overline{a}=325\$)\n\nThis solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]" ]
IMO-1984-3	https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_3	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 $C(X)$.	[ "Let \$a_{n},\\ n\\ge 1\$ be a sequence of positive reals such that \$\\left(\\sum_{n\\ge 1}a_{n}\\right)^{2}<2\\pi\\ ()\$. For each \$n\\ge 1\$, let \$\\mathcal C_{n}\$ be the circle centered at \$O\$ with radius \$\\sum_{i=1}^{n}a_{i}\$.\n\nBecause of \$()\$, we can find points \$x_{i,j}\\in\\mathcal C_{i},\\ i,j\\ge 1\$ such that for all \$i,j\\ge 1\$ we have \$\\mathcal C(x_{i,j})=\\mathcal C_{i+j}\$. We now forget about all the other points, and work only with the matrix \$M=(x_{i,j})\$.\n\nSuppose we use \$n\$ colors. There must be one, \$c_{1}\$, which appears infinitely many times on the first row of \$M\$, in, say, points \$x_{1,j_{1}},x_{1,j_{2}},\\ldots\$. Then \$c_{1}\$ cannot appear on the lines \$j_{k}+1\$, \$k\\ge 1\$. Next, there is a color \$c_{2}\$ which appears infinitely often among the points \$x_{j_{1}+1,j_{k}-j_{1}}\$, \$k\\ge 2\$. But then \$c_{2}\$ cannot appear on the lines \$j_{k}+1\$ for such \$k\$. Repeating this procedure, we reach a stage where we have a row of \$M\$ (infinitely many actually) on which none of our \$n\$ colors \$c_{1},\\ldots,c_{n}\$ can appear. This is a contradiction.\n\nThis solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]" ]
IMO-1984-4	https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_4	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.	[ "First, we prove that if \$BC\$ and \$AD\$ are parallel then the claim is true: Let \$AB\$ and \$CD\$ intersect at \$E\$ (assume \$E\$ is closer to \$AD\$, the other case being analogous). Let \$M,N\$ be the midpoints of \$AB,CD\$ respectively. Let the length of the perpendicular from \$N\$ to \$AB\$ be \$r\$. It is known that the length of the perpendicular from \$M\$ to \$CD\$ is \$\\frac{1}{2}AB\$. Let the foot of the perpendicular from \$C\$ to \$AB\$ be \$H\$, and similarly define \$G\$ for side \$CD\$. Then, since triangles \$MNE\$ and \$BCE\$ are similar, we have \$\\frac{CH}{r}=\\frac{BG}{\\frac{1}{2}AB}\$. This gives an expression for \$r\$:\n\n\\[\nr=\\frac{1}{2}AB\\cdot\\frac{CH}{BG}\n\\]\n\nNoticing that \$CH=BC\\sin EBC,BG=BC\\sin ECB\$ simplifies the expression to\n\n\\[\nr=\\frac{1}{2}AB\\cdot\\frac{\\sin EBC}{\\sin ECB}\n\\]\n\nBy the Law of Sines, \$\\frac{\\sin EBC}{\\sin ECB}=\\frac{EC}{EB}\$. Since triangles \$EDA,ECB\$ are similar, we have \$\\frac{EC}{EB}=\\frac{CD}{AB}\$ and thus we have\n\n\\[\nr=\\frac{1}{2}AB\\cdot\\frac{CD}{AB}=\\frac{1}{2}CD\n\\]\n\nand we are done.\n\nNow to prove the converse. Suppose we have the quadrilateral with \$BC\$ parallel to \$AD\$, and with all conditions satisfied. We shall prove that there exists no point \$T\$ on \$CD\$ such that \$T\$ is a midpoint of a side \$CD^\\prime\$ of a quadrilateral \$ABCD^\\prime\$ which also satisfies the condition. Suppose there was such a \$T\$. Like before, define the points \$E,M,N\$ for quadrilateral \$ABCD\$. Let \$t\$ be the length of the perpendicular from \$T\$ to \$AB\$. Then, using similar triangles, \$\\frac{ET}{t}=\\frac{EN}{\\frac{1}{2}CD}\$. This gives\n\n\\[\nt=\\frac{\\frac{1}{2}CD\\cdot ET}{EN}\n\\]\n\nBut, we must have \$t=DT\$. Thus, we have\n\n\\[\nDT=\\frac{\\frac{1}{2}CD\\cdot ET}{EN}\n\\]\n\n\\[\n\\Rightarrow \\frac{1}{2}CD\\cdot EN+NT\\cdot EN=\\frac{1}{2}CD\\cdot (EN+NT)\n\\]\n\n\\[\n\\Rightarrow NT\\cdot EN=\\frac{1}{2}CD\\cdot NT\n\\]\n\nSince \$\\frac{1}{2}CD\\ne EN\$, we have \$NT=0\$ as desired.", "Let \$M\$ and \$N\$ be midpoints of \$AB\$ and \$CD\$, respectively. Let \$[X]\$ denote the area of figure \$X\$.\n\nSuppose that \$AD // BC\$ and the circle centered at \$M\$ with diameter \$AB\$ touches \$CD\$ at \$T\$. We know that midsegment \$MN // AD\$. It follows that \$[AMN] = [DMN]\$, so \$\\frac{1}{2} AM \\cdot d = \\frac{1}{2} MT \\cdot DN\$, where \$d\$ is the distance from \$N\$ to \$AB\$. But we have \$MT = MA\$, so \$DN = d\$. It follows that the circle centered at \$N\$ with diameter \$CD\$ touches \$AB\$.\n\nIf on the other hand we have the circle with diameter \$CD\$ touching \$AB\$ at \$U\$(as well as the circle with diameter \$AB\$ touching \$CD\$ at \$T\$), we must have \$\\frac{1}{2} DN \\cdot MT = \\frac{1}{2} AM \\cdot UN\$ because of the equivalence of radii in circles. Hence \$[ANM] = [DMN]\$, so \$A\$ and \$D\$ are equidistant from \$MN\$ (as \$MN\$ is the common base). Hence \$AB // MN\$. Similarly \$CD // MN\$, and so \$AB // CD\$, as desired." ]
IMO-1984-5	https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_5	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: \[ n-3<{2d\over p}<\Bigl[{n\over2}\Bigr]\cdot\Bigl[{n+1\over 2}\Bigr]-2, \] where $[x]$ denotes the greatest integer not exceeding $x$.	[ "Consider all the pairs of non-adjacent sides in the polygon. Each pair uniquely determines two diagonals which intersect in an interior point. The sum of these two diagonals is greater than the sum of the two sides (triangle inequality). Each side will appear in a pair \$n-3\$ times. And each time the side appears it also includes two diagonals. Thus, \$(n-3)p < 2d\$.\n\nNote that for any diagonal, its length is less than the sum of the sides on either side of it (triangle inequality). Consider all the diagonals from a fixed vertex and using the least number of sides in each application of the triangle inequality. We can sum everything up over all the diagonals considering the two cases (even or odd) to get the result.\n\nThis solution was posted and copyrighted by aznlord1337. The original thread for this problem can be found here: [1]" ]
IMO-1984-6	https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_6	Let $a,b,c,d$ be odd integers such that $0<a<b<c<d$ and $ad=bc$. Prove that if $a+d=2^k$ and $b+c=2^m$ for some integers $k$ and $m$, then $a=1$.	[ "Let \$f:[1,b]\\rightarrow \\mathbb{R},\\ f(x)=x+\\dfrac{bc}{x}\$. As \$f^\\prime (x)=1-\\dfrac{bc}{x^2}\\le 0\$, we infer that \$f(x)\\ge f(b)=b+c,\\ \\forall x\\in [1,b]\$; in particular, \$a+d=f(a)\\ge b+c\\Leftrightarrow k\\ge m\$.\n\nNow, \$ad=bc\\Leftrightarrow a(2^k-a)=b(2^m-b)\\Leftrightarrow (b-a)(a+b)=2^m(b-2^{k-m}a)\$, hence \$2^m\|(b-a)(a+b)\$. It is easy to see that for \$x,y\\in \\mathbb{Z}\$, if \$v_2(x\\pm y)\\ge 2\$, then \$v_2(x\\mp y)=1\$. If \$v_2(b-a)\\ge 2\$, \$v_2(a+b)=1\$, so \$v_2(b-a)\\ge m-1\\Rightarrow b>2^{m-1}\$, which is in contradiction with the fact that \$b<\\dfrac{b+c}{2}=2^{m-1}\$. Thereby, \$v_2(a+b)\\ge m-1,\\ v_2(b-a)=1\$.\n\nWrite \$a+b=2^{m-1}\\alpha\$. If \$\\alpha \\ge 2\\Rightarrow2^m\\le a+b<b+c=2^m\$, contradiction; so \$a+b=2^{m-1}\$ and \$b-a=2\\beta\$, or equivalently \$a=2^{m-2}-\\beta,\\ b=2^{m-2}+\\beta\$ ( \$m>2\$ otherwise \$b+c=2\\Leftrightarrow b=c=1\$ or \$b+c=4\\Leftrightarrow c=3,b=1\\Rightarrow a=0\$ , contradiction )\n\nSubstituting back, \$(b-a)(a+b)=2^m(b-2^{k-m}a)\\Leftrightarrow 2^m\\beta=2^m(2^{m-2}+\\beta-2^{k-m}a)\\Leftrightarrow 2^{k-m}a=2^{m-2}\$\n\nAs \$m>2\$ and \$a\$ is odd, we get that \$a=1\$. Furthermore, \$k=2m-2\$, hence \$d=2^{2m-2}-1\$. Now \$b(2^m-b)=2^{2m-2}-1\\Leftrightarrow \\left ( b-(2^{m-1}-1) \\right ) \\left ( b-(2^{m-1}+1)\\right )=0\$ which together with the fact that \$b<c\$, we get that \$b=2^{m-1}-1,\\ c=2^{m-1}+1\$, so the family of the solutions \$(a,b,c,d)\$ is described by the set\n\n\\[\nM=\\{ \\left (1,2^{m-1}-1,2^{m-1}+1,2^{2m-2}-1 \\right )\|\\ m\\in \\mathbb{N},m\\ge 3 \\}\n\\]\n\nThis solution was posted and copyrighted by TheFunkyRabbit. The original thread for this problem can be found here: [1]", "Note that, if \$k\\leqslant m\$, then using \$a+d\\leqslant b+c\$, and \$ad=bc\$, we get \$(d-a)^2\\leqslant (c-b)^2\$, which is clearly impossible. Thus, \$k>m\$ must hold.\n\nNext, observe that, \$b+c =2^m\$ and \$b<c\$ implies \$b<2^{m-1}\$. Keeping this in mind, we now observe that, \$bc=b(2^m-b)=ad=a(2^k-a)\$, implies \$b\\cdot 2^m - a\\cdot 2^k = (b-a)(b+a)\$. Thus, \$2^m \\mid (b-a)(b+a)\$. Now, note also that, if \$d={\\rm gcd}(b-a,b+a)\$, then \$d\\mid 2b\$, and thus, the largest power of \$2\$ dividing either \$b-a\$ or \$b+a\$ is exactly \$2\$. Clearly, \$b-a<2^{m-1}\$, and thus, \$2^{m-1}\\mid b+a\$. Moreover, if \$b+a\\geqslant 2\\cdot 2^{m-1}\$, then we again have a contradiction, as \$b,a<2^{m-1}\$. Thus, \$b+a=2^{m-1}\$. This yields, \$b-a = 2(b-a\\cdot 2^{k-m})\$, which brings us, \$b=(2^{k-m+1}-1)a\$. Now, using \$b+a=2^{k-m+1}a = 2^{m-1}\$, we immediately obtain \$a=2^{2m-k-2}\$. This immediately establishes (since \$a\$ is odd), that \$a=1\$.\n\nThis solution was posted and copyrighted by grupyorum." ]
IMO-1985-1	https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_1	A circle has center on the side $AB$ of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD + BC = AB$.	[ "Let \$O\$ be the center of the circle mentioned in the problem. Let \$T\$ be the second intersection of the circumcircle of \$CDO\$ with \$AB\$. By measures of arcs, \$\\angle DTA = \\angle DCO = \\frac{\\angle DCB}{2} = \\frac{\\pi}{2} - \\frac{\\angle DAB}{2}\$. It follows that \$AT = AD\$. Likewise, \$TB = BC\$, so \$AD + BC = AB\$, as desired.", "Let \$O\$ be the center of the circle mentioned in the problem, and let \$T\$ be the point on \$AB\$ such that \$AT = AD\$. Then \$\\angle DTA = \\frac{ \\pi - \\angle DAB}{2} = \\angle DCO\$, so \$DCOT\$ is a cyclic quadrilateral and \$T\$ is in fact the \$T\$ of the previous solution. The conclusion follows.", "Let the circle have center \$O\$ and radius \$r\$, and let its points of tangency with \$BC, CD, DA\$ be \$E, F, G\$, respectively. Since \$OEFC\$ is clearly a cyclic quadrilateral, the angle \$COE\$ is equal to half the angle \$GAO\$. Then\n\n\\[\n\\begin{matrix} {CE} & = & r \\tan(COE) \\\\ & = &r \\left( \\frac{1 - \\cos (GAO)}{\\sin(GAO)} \\right) \\\\ & = & AO - AG \\\\ \\end{matrix}\n\\]\n\nLikewise, \$DG = OB - EB\$. It follows that\n\n\\[\n{EB} + CE + DG + GA = AO + OB\n\\]\n\n,\n\nQ.E.D.", "We use the notation of the previous solution. Let \$X\$ be the point on the ray \$AD\$ such that \$AX = AO\$. We note that \$OF = OG = r\$; \$\\angle OFC = \\angle OGX = \\frac{\\pi}{2}\$; and \$\\angle FCO = \\angle GXO = \\frac{\\pi - \\angle BAD}{2}\$; hence the triangles \$OFC, OGX\$ are congruent; hence \$GX = FC = CE\$ and \$AO = AG + GX = AG + CE\$. Similarly, \$OB = EB + GD\$. Therefore \$AO + OB = AG + GD + CE + EB\$, Q.E.D.", "This solution is incorrect. The fact that \$BC\$ is tangent to the circle does not necessitate that \$B\$ is its point of tangency. -Nitinjan06\n\nFrom the fact that AD and BC are tangents to the circle mentioned in the problem, we have \$\\angle{CBA}=90\\deg\$ and \$\\angle{DAB}=90\\deg\$.\n\nNow, from the fact that ABCD is cyclic, we obtain that \$\\angle{BCD}=90\\deg\$ and \$\\angle{CDA}=90\\deg\$, such that ABCD is a rectangle.\n\nNow, let E be the point of tangency between the circle and CD. It follows, if O is the center of the circle, that \$\\angle{OEC}=\\angle{OED}=90\\deg\$\n\nSince \$AO=EO=BO\$, we obtain two squares, \$AOED\$ and \$BOEC\$. From the properties of squares we now have\n\n\\[\nAD+BC=AO+BO=AB\n\\]\n\nas desired.", "Lemma. Let \$I\$ be the in-center of \$ABC\$ and points \$P\$ and \$Q\$ be on the lines \$AB\$ and \$BC\$ respectively. Then \$BP + CQ = BC\$ if and only if \$APIQ\$ is a cyclic quadrilateral.\n\nSolution. Assume that rays \$AD\$ and \$BC\$ intersect at point \$P\$. Let \$S\$ be the center od circle touching \$AD\$, \$DC\$ and \$CB\$. Obviosuly \$S\$ is a \$P\$-ex-center of \$PDB\$, hence \$\\angle DSI=\\angle DSP = \\frac{1}{2} \\angle DCP=\\frac{1}{2} \\angle A=\\angle DAI\$ so DASI is concyclic." ]
IMO-1985-2	https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_2	Let $n$ and $k$ be given relatively prime natural numbers, $k < n$. Each number in the set $M = \{ 1,2, \ldots , n-1 \}$ is colored either blue or white. It is given that (i) for each $i \in M$, both $i$ and $n-i$ have the same color; (ii) for each $i \in M, i \neq k$, both $i$ and $\|i-k\|$ have the same color. Prove that all the numbers in $M$ have the same color.	[ "We may consider the elements of \$M\$ as residues mod \$n\$. To these we may add the residue 0, since (i) may only imply that 0 has the same color as itself, and (ii) may only imply that 0 has the same color as \$k\$, which put no restrictions on the colors of the other residues.\n\nWe note that (i) is equivalent to saying that \$i\$ has the same color as \$-i\$, and given this, (ii) implies that \$i\$ and \$(-i + k)\$ have the same color. But this means that \$i, -i\$, and \$i+k\$ have the same color, which is to say that all residues of the form \$i + mk \\; (m \\in \\mathbb{N}_0)\$ have the same color. But these are all the residues mod \$n\$, since \$k\$ and \$n\$ are relatively prime. Q.E.D." ]
IMO-1985-3	https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_3	For any polynomial $P(x) = a_0 + a_1 x + \cdots + a_k x^k$ with integer coefficients, the number of coefficients which are odd is denoted by $w(P)$. For $i = 0, 1, \ldots$, let $Q_i (x) = (1+x)^i$. Prove that if $i_1, i_2, \ldots , i_n$ are integers such that $0 \leq i_1 < i_2 < \cdots < i_n$, then \[ w(Q_{i_1} + Q_{i_2} + \cdots + Q_{i_n}) \ge w(Q_{i_1}) \] .	[ "We first observe that \$(1+x)^{2^m} \\equiv 1 + x^{2^m} \\pmod{2}\$, so for any polynomial \$P\$ of degree less than \$2^m\$, \$w(P\\cdot Q_{2^m}) = 2w (P)\$.\n\nLet \$k\$ be the largest power of 2 that is less than or equal to \$i_n\$. We proceed by induction on \$k\$.\n\nFor \$k = 0\$, the problem is trivial.\n\nNow assume that the problem holds for \$k-1\$. We now have two cases: \$i_1 \\ge k\$, and \$i_1 < k\$.\n\nIn the first case, we note that \$w \\left( \\sum_{j=1}^{n}Q_{i_j} \\right) = w \\left( (1+x)^k \\sum_{j=1}^{n}Q_{i_j - k} \\right) = 2 w \\left( \\sum_{j=1}^{n}Q_{i_j - k} \\right)\$, which is greater than or equal to \$w( Q_{i_1} )\$ by assumption.\n\nIn the second case, we use the division algorithm to note that\n\n\\[\n\\sum_{j=1}^{n}Q_{i_j} = \\sum_{j=0}^{k-1}a_j x^j + (1+x)^k \\sum_{j=0}^{k-1} b_j x^j \\equiv \\sum_{j=0}^{k-1}\\left[ (a_j + b_j)x^j + b_j x^{j+k} \\right] \\pmod{2}\n\\]\n\n.\n\nBut by assumption, \$w \\left( \\sum_{j=0}^{k-1}a_j x^j \\right) \\ge w( Q_{i_1})\$, and for each odd \$a_j\$, at least one of \$(a_j + b_j )\$ and \$b_j\$ must be odd, Q.E.D." ]
IMO-1985-4	https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_4	Given a set $M$ of $1985$ distinct positive integers, none of which has a prime divisor greater than $23$, prove that $M$ contains a subset of $4$ elements whose product is the $4$th power of an integer.	[ "We have that \$x\\in M\\Rightarrow x=2^{e_1}3^{e_2}\\cdots 19^{e_8}23^{e_9}\$. We need only consider the exponents. First, we consider the number of subsets of two elements, such that their product is a perfect square. There are \$2^9=512\$ different parity cases for the exponents \$e_1,e_2,...,e_9\$. Thus, we have at least one pair of elements out of \$1985\$ elements. Removing these two elements yields \$1983\$ elements. By applying the Pigeon Hole Principle again, we find that there exists another such subset. Continuing on like this yields at least \$734\$ pairs of elements of \$M\$ whose product is a perfect square. Let \$S\$ be the set of the square roots of the products of each pair. Then, by the Pigeon Hole Principle again, there exist at least two elements whose product is a perfect square. Let the elements be \$x,y\$ and let \$x=\\sqrt{ab},y=\\sqrt{cd}\$ where \$a,b,c,d\\in M\$. Then, we have \$xy=z^2\$ for some \$z\$ which implies \$abcd=z^4\$ and the claim is proved." ]
IMO-1985-5	https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_5	A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$). Prove that $\angle OMB = 90^{\circ}$.	[ "\$M\$ is the Miquel Point of quadrilateral \$ACNK\$, so there is a spiral similarity centered at \$M\$ that takes \$KN\$ to \$AC\$. Let \$M_1\$ be the midpoint of \$KA\$ and \$M_2\$ be the midpoint of \$NC\$. Thus the spiral similarity must also send \$M_1\$ to \$M_2\$ and so \$BMM_1 M_2\$ is cyclic. \$OM_1 B M_2\$ is also cyclic with diameter \$BO\$ and thus \$M\$ must lie on the same circumcircle as \$B\$, \$M_1\$, and \$M_2\$ so \$\\angle OMB = 90^{\\circ}\$.", "Let \$\\Omega, \\Omega', \\omega\$ and \$O,O',O''\$ be the circumcircles and circumcenters of \$AKNC, ABC, BNKM,\$ respectively.\n\nLet \$\\angle ACB = \\gamma, AKNC\$ is cyclic \$\\implies \\angle BKN = \\gamma.\$\n\nThe radius of \$\\omega\$ is \$MO'' = BO'' = \\frac {BN}{2 \\sin \\gamma}.\$\n\nLet \$D\$ and \$E\$ be midpoints of \$BC\$ and \$NC\$ respectively.\n\n\$OE \\perp BC, OD \\perp BC, OO' \\perp AC, DE = \\frac {BC}{2} - \\frac {NC}{2} = \\frac {BN}{2}\$ \$\\implies OO' = \\frac {DE}{\\sin \\gamma} = \\frac {BN}{2 \\sin \\gamma} = MO''.\$\n\n\$M\$ is the Miquel Point of quadrilateral \$ACNK,\$ so \$MO''O'O\$ is cyclic. \$MO''O'O\$ is trapezium \$\\implies O''O' \|\| MO.\$ \$O''O' \\perp BM \\implies MO\\perp BM\$ as desired.\n\nvladimir.shelomovskii@gmail.com, vvsss", "Consider \$\\triangle MKA\$ and \$\\triangle MNC\$, they are similar because \$\\angle MAK\$ = \$\\angle MCN\$, and also \$\\angle MKA = \\angle MNC\$.\n\nNow draw \$OP \\perp AB\$, and intersecting \$AB\$ at \$P\$; \$OQ \\perp BC\$, at \$Q\$. Naturally \$OP\$ bisects \$AK\$, and \$OQ\$ bisects \$CN\$. We claim \$\\triangle MAP \\sim \\triangle MCQ\$, because \$\\frac {AP}{CQ} = \\frac {AK}{CN} = \\frac {AM}{CB}.\$\n\nThus \$\\angle AMP = \\angle CMQ\$, this implies \$\\angle PMQ = \\angle AMC = \\angle ABC = \\angle PBQ\$. Obviously BMPQ is cyclic, and so is BPOQ. Finally, we have \$OM \\perp MB\$. (by gougutheorem)" ]
IMO-1985-6	https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_6	For every real number $x_1$, construct the sequence $x_1,x_2,\ldots$ by setting $x_{n+1}=x_n \left(x_n + \frac{1}{n}\right)$ for each $n \geq 1$. Prove that there exists exactly one value of $x_<x_n<x_{n+1}<1$ for every $n$.	[ "By recursive substitution, one can write \$x_n=P_n(x_1)\$ , where \$P_n\$ is a polynomial with non-negative coefficients and zero constant term. Thus, \$P_n(0)=0\$, \$P_n\$ is strictly increasing in \$[0,+\\infty)\$ , and \$\\displaystyle \\lim_{x_1 \\rightarrow + \\infty} P_n(x_1)=+\\infty\$. We can therefore define the inverse \$P_n^{-1}\$ of \$P_n\$ on \$[0,+\\infty)\$. It follows that \$x_1=P_n^{-1}(x_n)\$, \$P_n^{-1}(0)=0\$, \$P_n^{-1}\$ is strictly increasing in \$[0,+\\infty)\$, and \$\\displaystyle \\lim_{x_1 \\rightarrow + \\infty} P_n^{-1}(x_1) =+\\infty\$.\n\nDenote by \$\\displaystyle a_n=P_n^{-1}(1-\\frac{1}{n})\$ and \$b_n=P_n^{-1}(1)\$. By the monotonicity of \$P_n^{-1}\$ we have \$a_n<b_n\$ for each \$n\$. Note that:\n\n(a) \$\\displaystyle x_n<x_{n+1} \\Leftrightarrow x_n>1-\\frac{1}{n} \\Leftrightarrow P_n^{-1}(x_n)>P_n^{-1}(1-\\frac{1}{n}) \\Leftrightarrow x_1>a_n\$; (b) \$\\displaystyle x_n<1 \\Leftrightarrow P_n^{-1}(x_n)<P_n^{-1}(1) \\Leftrightarrow x_1<b_n\$.\n\nThus, \$0<x_n<x_{n+1}<1,\\forall n\$ holds if and only if \$a_n<x_1<b_n,\\forall n\$, or \$\\displaystyle x_1 \\in \\bigcap_{n=1}^{+\\infty}(a_n,b_n)\$. We need to show that \$\\displaystyle \\bigcap_{n=1}^{+\\infty}(a_n,b_n)\$ is a singleton. We have:\n\n(c) if \$x_1=a_n\$, then \$x_n=1-\\frac{1}{n}\$, which implies that \$x_{n+1}=1-\\frac{1}{n}<1-\\frac{1}{n+1}=P_{n+1}(a_{n+1})\$, and \$x_1<a_{n+1}\$. It follows that \$a_n<a_{n+1},\\forall n\$; and (d) if \$x_1=b_n\$, then \$x_n=1\$, which implies that \$x_{n+1}=1+\\frac{1}{n}>1=P_{n+1}(b_{n+1})\$, and \$x_1>b_{n+1}\$. It follows that \$b_n>b_{n+1},\\forall n\$; and\n\nThus, \$a_n<a_{n+1}<b_{n+1}<b_n, \\forall n\$. Therefore, the two sequences \$\\{a_n\\}_{n=1}^{+\\infty}\$ and \$\\{b_n\\}_{n=1}^{+\\infty}\$ converge, and their limits \$a\$ and \$b\$ satisfy \$a \\leq b\$. Hence, \$\\displaystyle \\bigcap_{n=1}^{+\\infty}(a_n,b_n)=[a,b]\$ is non-empty, which demonstrates the existence of \$x_1\$.\n\nNow, suppose that \$a \\leq x_1 \\leq x_1' \\leq b\$. We have \$x_{n+1}'-x_{n+1} = (x_n'-x_n)(x_n'+x_n+\\frac{1}{n}) \\geq (x_n'-x_n)(2-\\frac{1}{n}) \\geq (x_n'-x_n)\$ for each \$n\$, so that \$x_n'-x_n \\geq x_1'-x_1\$ for each \$n\$. However, \$1-\\frac{1}{n}<x_n \\leq x_n'<1\$, so that \$0 \\leq x_n'-x_n<\\frac{1}{n}\$, which implies that \$\\displaystyle \\lim_{n \\rightarrow +\\infty}(x_n'-x_n)=0\$. Therefore, \$x_1' \\leq x_1\$, proving unicity.\n\nThis solution was posted and copyrighted by DAFR. The original thread for this problem can be found here: [1]", "For each \$n \\ge 1\$ let \$I_n\$ be the interval of real numbers \$0 \\textless x \\textless 1\$ such that if \$x=x_1\$, we will have \$x_n \\textless x_{n+1} \\textless 1\$. (That is, the values of \$x_1\$ which make \$x_{n+1}\$ \"work\"). We must prove these intervals intersect at one point.\n\nFirst, I prove they intersect. Notice that as \$x_1\$ increases, \$x_n\$ and \$x_{n+1}\$ do too. So if \$I_n=(a_n,b_n)\$ it is easy to see \$a_n=x_1\$ will give \$x_n=x_{n+1}=1-1/n\$ and \$b_n=x_1\$ will give \$x_{n+1}=1\$ (the \"extremal\" values \$x_{n+1}\$ can take are given by the extremal values \$x_1\$ can take).\n\nBut if \$a_n=x_1\$ we see that \$x_{n+1} \\textless 1-1/(n+1)\$ and so \$x_{n+1} \\textgreater x_{n+2}\$ and similarly \$x_1=b_n\$ gives \$x_{n+1}=1\$ which gives a \$x_{n+2}\$ too big. So basically when \$x_1\$ only barely works for \$x_{n+1}\$, it won't work for \$x_{n+2}\$. So \$I_{n+1}\$ is a subset of \$I_n\$.\n\nSo we have an infinite sequence of open intervals, each contained inside the previous one. Therefore their intersection must contain at least one number. This guarantees the existence of \$x_1\$.\n\nBut, we must prove that \$x_1\$ can't take two different values. Suppose it could. Suppose \$x_1=a,b\$ both work. Let \$x_i\$ be the sequence generated by \$a\$ and \$x_i'\$ the sequence generated by \$b\$, and let \$l_i=x_i'-x_i\$ (wlog \$a \\textless b\$). Then we see \$l_{n+1}=l_n(2x_n+l_n+1/n) \\textgreater l_n\$ for \$n \\ge 2\$ since \$2x_n \\textgreater 1\$ since \$x_n \\textgreater 1-1/n \\ge 1/2\$. So there exists a constant \$C\$ such that \$x_k' \\ge x_k+C\$ for \$k \\ge 2\$. But since \$x_k, x_k' \\in (1-1/k, 1)\$, this is impossible. So we're done.\n\nThis solution was posted and copyrighted by JuanOrtiz. The original thread for this problem can be found here: [2]", "Consider, now, the following observations:\n\n(a) If, at any point, \$x_{n+1}\\le x_n\$, then from here we know that\n\n\\[\nx_n\\le 1-\\frac{1}{n},\\quad x_{n+1}\\le 1-\\frac{1}{n} < 1-\\frac{1}{n+1}\\to x_{n+2}<x_{n+1}, \\cdots x_m\\le 1-\\frac{1}{n} < 1-\\frac{1}{m}\\to x_{m+1}<x_m, \\forall m>n\n\\]\n\nand therefore the sequence \$\\{x_n\\}\$ will be monotonically decreasing after one point.\n\n(b) If some \$x\$ does satisfy \$0<x_n<x_{n+1}<1\$ for all \$n\$, then \$1-\\frac{1}{n}<x_n<1\$ for all \$n\$, and therefore by Squeeze's theorem \$\\lim_{n\\to\\infty}x_n = 1\$.\n\n(c) If \$x_n\\ge 1\$ for some \$n\$, then \$x_{n+1}=x_n(x_n+\\frac 1n)>x_n^2\\ge 1\$ and so \$x_{n+m}>(x_{m+1})^{2^{m-1}}\$ which then gives \$\\lim_{n\\to\\infty}x_n=+\\infty\$.\n\n(d) If \$x_1=0\$, then for each \$n\$, \$x_n=0\$. If \$x_1\\to\\infty\$ then for each \$n\$ (fixed), \$x_n\\to\\infty\$. Thus, we can denote a mapping \$f_n:\\mathbb{R}^{\\ge 0}\\to \\mathbb{R}^{\\ge 0}\$ that maps \$x_1\$ to \$x_n\$, which is continuous and monotonically increasing, with \$\\lim_{x\\to +\\infty} f_n(x)=+\\infty\$ so \$f_n\$ is bijective.\n\nLet's first show uniqueness. Suppose that \$\\{x_n\\}\$ and \$\\{y_n\\}\$ are both such sequences. We have \$\\lim_{n\\to\\infty}x_n=\\lim_{n\\to\\infty}y_n=1\$ and suppose that \$y_1>x_1\$. Then for all \$n\$,\n\n\\[\n\\frac{y_{n+1}}{x_{n+1}} = \\frac{y_n(y_n+\\frac 1n)}{x_n(x_n+\\frac 1n)}>\\frac{y_n}{x_n}\n\\]\n\nso inductively, \$\\frac{y_{n+1}}{x_{n+1}}>(\\frac{y_{1}}{x_{1}})^n\$ with \$\\lim_{n\\to\\infty}\\frac{y_{n+1}}{x_{n+1}}=+\\infty\$. However, \$\\lim_{n\\to\\infty}x_n=\\lim_{n\\to\\infty}y_n=1\$ gives \$\\lim_{n\\to\\infty}\\frac{y_{n+1}}{x_{n+1}}=\\frac{1}{1}=1\$, contradiction. Hence, the \$x_1\$ that satisfies this must be unique.\n\nNext, let's show existence. We have seen from above that, \$y_1>x_1\$ implies \$y_n>x_n\$, and that if \$x_n>1\$ for some \$n\$ then \$\\{x_n\\}\$ is monotonically increasing after some point. Suppose no such \$x_1\$ exists. Let\n\n\\[\nA=\\{x_1: \\exists n: x_{n+1}<x_n\\}\\qquad B=\\{x_1: \\exists n: x_n>1\\}\n\\]\n\nthen if \$x\\in A\$, \$y\\in A\$ for all \$y<x\$ and similarly \$x\\in B\$, \$y\\in B\$ for all \$y>B\$. Notice also an wasy fact that \$1\\in B\$, so \$A\$ is bounded. Define, now, \$c=glb(A)\$. As we assumed \$A\\cup B=\\mathbb{R}^+\$, this \$c\$ implies that \$x_1<c\\to x_1\\in A\$ and \$x_1>c\\to x_1\\in B\$. It remains to ask whether \$c\\in A\$ or \$c\\in B\$.\n\nIf \$c\\in A\$, then for this \$x_1:=c\$, \$x_n\\le 1-\\frac{1}{n}\$ and so \$x_{n+1}<1-\\frac{1}{n+1}\$. Let \$y_{n+1}\$ be such that \$x_{n+1}<y_{n+1}<1\$. By above, there's exactly one \$y_1\$ with \$f_{n+1}(y_1)<y_{n+1}\$, and notice that \$y_1>x_1=c\$ by the monotonicity of \$f_{n+1}\$. This means that there exists \$y_1>c\\in A\$, contradicting the definition of glb. A similar contradiction (but opposite direction) can also be established for the case \$c\\in B\$.\n\nHence \$c\$ is neither in \$A\$ or \$B\$, means that \$x_1=c\$ should satisfy the problem condition. Q.E.D.\n\nThis solution was posted and copyrighted by Anzoteh. The original thread for this problem can be found here: [3]" ]
IMO-1986-1	https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_1	Let $d$ be any positive integer not equal to $2, 5$ or $13$. Show that one can find distinct $a,b$ in the set $\{2,5,13,d\}$ such that $ab-1$ is not a perfect square.	[ "We do casework with modular arithmetic.\n\n\$d\\equiv 0,3 \\pmod{4}: 13d-1\$ is not a perfect square.\n\n\$d\\equiv 2\\pmod{4}: 2d-1\$ is not a perfect square.\n\nTherefore, \$d\\equiv 1, \\pmod{4}.\$ Now consider \$d\\pmod{16}.\$\n\n\$d\\equiv 1,13 \\pmod{16}: 13d-1\$ is not a perfect square.\n\n\$d\\equiv 5,9\\pmod{16}: 5d-1\$ is not a perfect square.\n\nAs we have covered all possible cases, we are done. ~Shen kislay kai", "Proof by contradiction:\n\nSuppose \$p^2=2d-1\$, \$q^2=5d-1\$ and \$r^2=13d-1\$. From the first equation, \$p\$ is an odd integer. Let \$p=2k-1\$. We have \$d=2k^2-2k+1\$, which is an odd integer. Then \$q^2\$ and \$r^2\$ must be even integers, denoted by \$4n^2\$ and \$4m^2\$ respectively, and thus \$r^2-q^2=4m^2-4n^2=8d\$, from which\n\n\\[\n2d=m^2-n^2=(m+n)(m-n)\n\\]\n\ncan be deduced. Since \$m^2-n^2\$ is even, \$m\$ and \$n\$ have the same parity, so \$(m+n)(m-n)\$ is divisible by \$4\$. It follows that the odd integer \$d\$ must be divisible by \$2\$, leading to a contradiction. We are done." ]
IMO-1986-2	https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_2	Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define $A_s=A_{s-3}$ for all $s\ge4$. Construct a set of points $P_1,P_2,P_3,...$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^\circ$ clockwise for $k=0,1,2,...$. Prove that if $P_{1986}=P_0$, then the triangle $A_1A_2A_3$ is equilateral.	[ "Consider the triangle and the points on the complex plane. Without loss of generality, let \$A_1=0\$, \$A_2=1\$, and \$A_3=a\$ for some complex number \$a\$. Then, a rotation about \$A_1\$ of \$120^\\circ\$ sends point \$P\$ to point \$e^{\\frac{4i\\pi}{3}}P\$. For \$A_2\$, the rotation sends \$P\$ to \$e^{\\frac{4i\\pi}{3}}(P-1)+1\$ and for \$A_3\$ the rotation sends \$P\$ to \$e^{\\frac{4i\\pi}{3}}(P-a)+a\$. Thus the result of all three rotations sends \$P\$ to\n\n\\[\ne^{\\frac{4i\\pi}{3}}(e^{\\frac{4i\\pi}{3}}(e^{\\frac{4i\\pi}{3}}P-1)+1-a)+a\n\\]\n\n\\[\n=P-e^{\\frac{2i\\pi}{3}}+e^{\\frac{4i\\pi}{3}}-ae^{\\frac{4i\\pi}{3}}+a\n\\]\n\n\\[\n=P-i\\sqrt{3}-ae^{\\frac{4i\\pi}{3}}+a\n\\]\n\nSince the transformation \$P\\to P-i\\sqrt{3}-ae^{\\frac{4i\\pi}{3}}+a\$ occurs \$1986/3=662\$ times, to obtain \$P_{1986}\$. But, we have \$P_{1986}=P_0\$ and so we have\n\n\\[\nP_0=P_0+662(-i\\sqrt{3}-ae^{\\frac{4i\\pi}{3}}+a)\n\\]\n\n\\[\n\\Rightarrow -i\\sqrt{3}-ae^{\\frac{4i\\pi}{3}}+a=0\n\\]\n\n\\[\n\\Rightarrow a(-e^{\\frac{4i\\pi}{3}}+1)=i\\sqrt{3}\n\\]\n\n\\[\n\\Rightarrow a=\\frac{i\\sqrt{3}}{-e^{\\frac{4i\\pi}{3}}+1}\n\\]\n\n\\[\n\\Rightarrow a=\\frac{1}{2}+i\\frac{\\sqrt{3}}{2}\n\\]\n\nNow it is clear that the triangle \$A_1A_2A_3\$ is equilateral. Shen kislay kai" ]
IMO-1986-3	https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_3	To each vertex of a regular pentagon an integer is assigned, so that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers $x,y,z$ respectively, and $y<0$, then the following operation is allowed: $x,y,z$ are replaced by $x+y,-y,z+y$ respectively. Such an operation is performed repeatedly as long as at least one of the five numbers is negative. Determine whether this procedure necessarily comes to an end after a finite number of steps.	[ "The algorithm always stops. Indeed, consider the five numbers on the vertices of the pentagon as forming a \$5\$-tuple \$\\mathbf{x} = (x_1, x_2, x_3, x_4, x_5) \\in \\mathbb{Z}^5\$. The sum of the five entries of this \$5\$-tuple is positive (by assumption), and stays unchanged through the entire process (since our operation does not change the sum); thus, it is always positive. Now, define the function \$f : \\mathbb{Z}^5 \\to \\mathbb{Z}\$ by\n\n\\[\nf(x_1, x_2, x_3, x_4, x_5)=\\sum_{i=1}^5(x_i-x_{i+2})^2, \\text{ where } x_6=x_1 \\text{ and } x_7=x_2.\n\\]\n\nClearly, \$f \\geq 0\$ always. Now, let \$\\mathbf{x}_{\\text{old}} = (x_1, x_2, x_3, x_4, x_5)\$ and \$\\mathbf{x}_{\\text{new}}\$ be two consecutive \$5\$-tuples obtained in this process. Thus, \$\\mathbf{x}_{\\text{new}}\$ is obtained from \$\\mathbf{x}_{\\text{old}}\$ by applying our operation once. Suppose, WLOG, that \$y=x_4<0\$, and that the last three entries \$x_3, x_4, x_5\$ of \$\\mathbf{x}_{\\text{old}}\$ are replaced by \$x_3 + x_4, -x_4, x_5 + x_4\$ in \$\\mathbf{x}_{\\text{new}}\$. Let \$S = x_1 + x_2 + x_3 + x_4 + x_5\$ be the sum of all entries of \$\\mathbf{x}_{\\text{old}}\$ (or \$\\mathbf{x}_{\\text{new}}\$). Then, a straightforward computation shows that \$f(\\mathbf{x}_{\\text{new}}) - f(\\mathbf{x}_{\\text{old}}) = 2Sx_4 < 0\$ (since \$S>0\$ and \$x_4 < 0\$). Thus, if the algorithm does not stop, we can find an infinite decreasing sequence of nonnegative integers \$f_0>f_1>f_2>\\cdots\$ (by applying \$f\$ to each of the \$5\$-tuples). This is impossible, so the algorithm must stop. \$\\Box\$\n\nThis solution was posted and copyrighted by tc1729. The original thread for this problem can be found here: [1]", "The key idea of this proof is to create an positive integer valued semi-invariant that decreases (strictly) as we preform the operation. Then the existence of the semi-invariant would guarantee the finiteness (note that the integer valued condition is needed).\n\nWe see that the sum of all five integers is a invariant, so we see if the sum of the absolute values work. Unfortunately, as the operation is performed, the sum of the absolute values decrease by \$\|x\|+\|z\|-\|x+y\|-\|y+z\|\$ after each operation (which is not always positive). This suggest we use pairwise sums as well. Upon testing that, we note that sums of triples and quadriples should be included as well. We see the semi-invariant\n\n\\[\nS=\|v\|+\|w\|+\|x\|+\|y\|+\|z\|+\|v+w\|+\|w+x\|+\|x+y\|+\|y+z\|+\|v+w+x\|+\|w+x+y\|+\|x+y+z\|+\|y+z+w\|+\|z+v+w\|+\|v+w+x+y\|+\|w+x+y+z\|+\|x+y+z+v\|+\|y+z+v+w\|+\|z+v+w+x\|\n\\]\n\nis reduced by \$\|(v+w+x+y+z)-y\|-\|(v+w+x+y+z)+y\|\$ after the operation, and that is positive. (Note that not all sums appeared in the semi-invariant) The result follows.\n\nTyped by Ddk001.\n\nNote that J. Keane, a former member of the US team, was awarded a special prize for this solution." ]
IMO-1986-4	https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_4	Let $A,B$ be adjacent vertices of a regular $n$-gon ($n\ge5$) with center $O$. A triangle $XYZ$, which is congruent to and initially coincides with $OAB$, moves in the plane in such a way that $Y$ and $Z$ each trace out the whole boundary of the polygon, with $X$ remaining inside the polygon. Find the locus of $X$.	[ "Let \$C \\not = A\$ the vertex which is adjacent to \$B\$. While \$XYZ\$ moves from \$OAB\$ to \$OBC\$, it is easy to see \$XYBZ\$ is cyclic. Thus \$X\$ lies on the bisector of \$\\angle YBZ = \\angle ABC\$. Moreover, \$X\$ is the intersection of a circle passing through \$B\$ (the circumcircle of \$XYBZ\$) and with a fixed radius (the radius is a function of \$\\triangle XYZ\$). Therefore \$X\$ varies in a line segment ended in \$O\$. When \$Y\$ and \$Z\$ pass through the other sides, we get as locus \$n\$ distinct line segments, each passing throught \$O\$ and contained in \$OV\$ (but not in \$\\vec{OV}\$) for some vertex \$V\$ of the polygon. Each two of these lines are obtained one from another by a rotation with center \$O\$.\n\nThis solution was posted and copyrighted by feliz. The original thread for this problem can be found here: [1]" ]
IMO-1986-5	https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_5	Find all (if any) functions $f$ taking the non-negative reals onto the non-negative reals, such that (a) $f(xf(y))f(y) = f(x+y)$ for all non-negative $x$, $y$; (b) $f(2) = 0$; (c) $f(x) \neq 0$ for every $0 \leq x < 2$.	[ "For all \$x+y\\ge2\$, there exists a nonnegative real \$t\$ such that \$t+2=x+y\$ and \$f(x+y)=f(t+2)=f(tf(2))f(2)=0\$. Hence, \$f(x)=0\$ for all \$x\\ge 2\$.\n\nFor \$x<2\$, take \$0=f(2)=f((2-x)+(x))=f((2-x)f(x))f(x)\$. Since \$f(x)\\neq0\$ for \$x<2\$, then \$f((2-x)f(x))=0\$. Thus, \$(2-x)f(x)\\ge2\\Leftrightarrow f(x)\\ge\\frac{2}{2-x}\$ for all \$x<2\$.\n\nSuppose \$f(a)>\\frac{2}{2-a}\$ so that \$f(a)=\\frac{2}{2-a-\\epsilon}\$ for some \$\\epsilon>0\$ and \$0\\le a<2\$. Then \$f\\left((2-a-\\epsilon)f(a)\\right)f(a)=f(2-\\epsilon)\$. The left hand side is equivalent to \$f\\left((2-a-\\epsilon)\\left(\\frac{2}{2-a-\\epsilon}\\right)\\right)f(a)=f(2)=0\$, but the right hand side is nonzero, since \$2-\\epsilon<2\$ and \$f(x)\\neq0\$ for all \$x<2\$.\n\nHence, \$f(x)=\\frac{2}{2-x}\$ for \$0\\le a<2\$. For \$x+y<2\$,\\begin{eqnarray}f(xf(y))f(y) &=& f\\left((x)\\left(\\frac{2}{2-y}\\right)\\right)\\left(\\frac{2}{2-y}\\right)\\\\ &=&\\left(\\frac{2}{2-\\frac{2x}{2-y}}\\right)\\left(\\frac{2}{2-y}\\right)\\\\ &=&\\frac{4}{4-2y-2x}\\\\ &=&\\frac{2}{2-y-x}\\\\ f(x+y)&=&\\frac{2}{2-y-x}.\\end{eqnarray} Indeed, \$f(xf(y))f(y)=f(x+y)\$, so the desired function is\n\n\\[\nf(x)=\\begin{cases}\\frac{2}{2-x},& 0\\le x<2\\\\0,& x\\ge2.\\end{cases}\n\\]\n\nThis solution was posted and copyrighted by Shen Kislay Kai. The original thread for this problem can be found here: [1]", "Taking \$y=2\$ gives \$f(x+2)=0 \\forall x\\ge 0\$, or equivalently \$f(x)=0 \\forall x\\ge 2\$. Taking \$y=2-x\$ for \$0\\le x<2\$ yields\n\n\\[\nf(xf(2-x))f(2-x)=f(2)=0\n\\]\n\nSince \$0<2-x\\le 2\$, we must have \$f(xf(2-x))=0\$, which forces\n\n\\[\nxf(2-x)\\ge 2\\implies f(x)\\ge \\frac{2}{2-x}\\forall 0\\le x<2.\n\\]\n\nNow taking \$x=2/f(y)\$, for \$0\\le y<2\$ gives,\n\n\\[\nf(y+2/f(y))=f(2)f(y)=0\\implies y+2/f(y)\\ge 2.\n\\]\n\nThus,\n\n\\[\nf(y)\\le \\frac{2}{2-y}\\forall 0\\le y<2\n\\]\n\nHence, this forces \$f(x)=2/(2-x)\$ for \$0\\le x<2\$, giving a final answer of\n\n\\[\nf(x)=\\begin{cases} 0 & \\text{ if }x \\ge 2 \\\\ \\frac{2}{2-x} & \\text{ if } 0 \\le x < 2 . \\end{cases}\n\\]\n\nThis solution was posted and copyrighted by winnertakeover. The original thread for this problem can be found here: [2]", "Let \$P(x,y)\$ be the assertion. Note that \$P(x,2)\\implies f(x+2)=0.\$\n\nSince \$x\$ is defined for non-negative reals, we have \$x+2\\geq 2.\$ So, \$\\boxed{f(x)=0\\qquad \\forall x\\geq2}.\$\n\nNow, \$P(2-y,y)\\implies f((2-y)f(y))f(y)=0.\$\n\nHere, we are working with \$0\\leq y<2.\$ Since we may not have \$f(y)=0\$ due to condition (iii), we must have \$f((2-y)f(y))=0.\$\n\nFrom our first solution, it is imperative that \$(2-y)f(y)\\geq 2.\$ So, \$f(y)\\geq \\dfrac{2}{2-y} \\qquad (1).\$\n\nNow consider \$P(\\dfrac{2}{f(y),y}\$. We get,\n\n\\[\n0=f\\left(\\dfrac{2+yf(y)}{f(y)}\\right).\n\\]\n\nSimilar to the previous step, we have \$\\dfrac{2+yf(y)}{f(y)}\\geq 2.\$ Re arranging this, we get \$f(y)\\leq \\dfrac{2}{2-y}\\qquad (2).\$\n\nSince \$f(y)\$ satisfies both \$(1)\$ and \$(2),\$ we must have \$\\boxed{f(x)=\\dfrac{2}{2-x}\\qquad \\forall x\\in [0,2)}.\$\n\nOf course, we plug both solution back to see if they satisfy the three conditions. Sure enough, they do!\n\nThis solution was posted and copyrighted by proshi. The original thread for this problem can be found here: [3]" ]
IMO-1986-6	https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_6	Given a finite set of points in the plane, each with integer coordinates, is it always possible to color the points red or white so that for any straight line $L$ parallel to one of the coordinate axes the difference (in absolute value) between the numbers of white and red points on $L$ is not greater than $1$?	[ "I'll use a well-known result: if a connected graph has \$2k>0\$ vertices of odd degree, then its edge set can be partitioned into \$k\$ paths, and if all its vertices have even degree, then it has an eulerian circuit.\n\nWe have a bipartite graph with bipartition \$A,B\$ here, constructed as follows: the horizontal lines which contain points from our set represent the vertices in \$A\$, the vertical lines represent the vertices in \$B\$, and the point represent the edges between the vertices corresponding to the rows and columns which contain them. What we must prove is that we can color the edge set of a bipartite graph in two colors s.t. the difference between the number of red edges and white edges adjacent to each vertex is \$\\le 1\$ in absolute value. It suffices to prove this for connected bipartite graphs, since then we can apply the result to each connected component of the graph.\n\nIf all the vertices have even degree, then we can find an eulerian circuit. The graph is bipartite, so this circuit has an even number of edges, and we can thus color the edges in the circuit alternatively red and white so that two edges which are consecutive in the circuit have different colors. It's clear that this coloration satisfies the requirements.\n\nIf, on the other hand, the graph has \$2k>0\$ vertices with odd degre, then we partition the edge set into \$k\$ paths, and in each path we color the edges alternatively red and white. Again, it's easy to verify the required properties of the coloration.\n\nThis solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]", "We induct on number of points. The small cases are easily checked. Let there exist such a function for \$n\$ points. We will show there is a function for \$n+1\$ points.\n\nIf there exists a line \$\\ell\$, parallel to any of the coordinate axes (from the next time, any line will be parallel to either of the coordinate axes, unless otherwise mentioned ), containing odd number of points, then choose a point \$P_x \\in \\ell\$, and consider \$S \\setminus P_x\$. By inductive hypothesis there exists such a function \$f : P \\to \\left \\{ -1, 1 \\right \\}\$ for \$S \\setminus P_x\$. Since \$P \\in \\ell, P \\ne P_x\$ contain even number of points, we have \$\\sum_{\\substack{P \\in \\ell \\\\ P \\ne P_x}} f(P) = 0\$. Let \$\\ell '\$ be the line \$\\perp\$ to \$\\ell\$, passing through \$P_x\$. Let \$\\sum _{\\substack{P \\in \\ell ' \\\\ P \\ne P_x}} f(P) = t\$, where \$t \\in \\left \\{ -1,0 , 1 \\right \\}\$. Now for \$S\$ (with \$n+1\$ points) define \$g\$ as \$g(P) = f(P)\$ for \$P \\in S \\setminus P_x\$, and \$g(P_x) = -t\$, if \$t \\ne 0\$, or \$1\$ or \$-1\$ if \$t=0\$. It indeed works as such a function for \$n+1\$ points.\n\nIf the above is not the case, i.e. if all the lines contain even number (greater than zero) of points, pick up an arbitrary point \$P_y \\in S\$. Let \$\\ell\$ and \$\\ell '\$ be the two lines containing \$P_y\$. Also \$\\ell \\perp \\ell '\$. Consider \$S \\setminus P_y\$, for this set , by the inductive hypothesis, there exists such a function \$f\$. In \$S \\setminus P_y\$ , \$\\ell\$ and \$\\ell '\$ contains odd number of points and if \$L\$ is a line different from \$\\ell\$ and \$\\ell '\$, then it contains even number of points. So \$\\sum_{\\substack{P \\in L \\\\ L \\ne \\ell, \\ell '}} f(P) = 0\$. Therefore it is seen that\n\n\$\\sum_{\\substack{P \\in \\ell \\\\ P \\ne P_y}} f(P)\$ \$= - \\sum _{\\substack{P \\in S \\\\ P \\notin \\ell , \\ell ' }} f(P)\$ \$= \\sum_{\\substack{P \\in \\ell ' \\\\ P \\ne P_y}} f(P) = t\$.\n\nSince \$\\ell \\setminus P_y\$ contain odd number of points, \$t\$ is either \$-1\$ or \$1\$. Wlog \$t = 1\$. Now for \$S\$, (containing \$n+1\$ points) define \$g\$ as : \$g(P) = f(P)\$, for \$P \\in S \\setminus P_y\$ and \$g(P_y) = -t\$.\n\nSo the induction is complete, and the statement is established.\n\nThis solution was posted and copyrighted by Learner94. The original thread for this problem can be found here: [2]" ]
IMO-1987-1	https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_1	Let $p_n (k)$ be the number of permutations of the set $\{ 1, \ldots , n \} , \; n \ge 1$, which have exactly $k$ fixed points. Prove that \[ \sum_{k=0}^{n} k \cdot p_n (k) = n!. \] (Remark: A permutation $f$ of a set $S$ is a one-to-one mapping of $S$ onto itself. An element $i$ in $S$ is called a fixed point of the permutation $f$ if $f(i) = i$.)	[ "The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element \$i\$ of the set, there are \$(n-1)!\$ permutations which have \$i\$ as a fixed point. Therefore\n\n\\[\n\\sum_{k=0}^{n} k \\cdot p_n (k) = n!\n\\]\n\n,\n\nas desired.\n\nSlightly Clearer Solution\n\nFor any \$k\$, if there are \$p_n(k)\$ permutations that have \$k\$ fixed points, then we know that each fixed point is counted once in the product \$k \\cdot p_n{k}\$. Therefore the given sum is simply the number of fixed points among all permutations of \$\\{ 1, \\ldots , n \\}\$. However, if we take any \$x\$ such that \$1 \\le x \\le n\$ and \$x\$ is a fixed point, there are \$(n-1)!\$ ways to arrange the other numbers in the set. Therefore our desired sum becomes \$n \\cdot (n-1)! = n!\$, so we are done.", "The probability of any number \$i\$ where \$1\\le i\\le n\$ being a fixed point is \$\\frac{1}{n}\$. Thus, the expected value of the number of fixed points is \$n\\times \\frac{1}{n}=1\$.\n\nThe expected value is also \$\\sum_{k=0}^{n} \\frac{k \\cdot p_n (k)}{n!}\$.\n\nThus,\n\n\\[\n\\sum_{k=0}^{n} \\frac{k \\cdot p_n (k)}{n!}=1\n\\]\n\nor\n\n\\[\n\\sum_{k=0}^{n} k \\cdot p_n (k) = n!.\n\\]", "Call a permutation of \$k\$ objects a derangement if the permutation has \$0\$ fixed points. Consider \$d_k\$ the number of derangements of \$k\$ objects. Notice that\n\n\\[\n\\sum_{k=0}^n {n \\choose k}d_{n-k} = n!.\n\\]\n\nSpecifically, for \$n\$ objects being permuted, there are \${n \\choose k}\$ ways of holding \$k\$ points fixed and \$d_{n-k}\$ ways of deranging the other \$n-k\$ points. And summing these possibilities from \$0\$ points held fixed to \$n\$ points held fixed yields all \$n!\$ permutations of \$n\$ objects. On the other hand clearly \$p_n(k) = {n \\choose k} d_{n-k}\$. Since\n\n\\[\n{n \\choose k} = \\frac{n}{k}{n-1 \\choose k-1}\n\\]\n\nfor \$n \\geq k \\geq 1\$ (taking \${0 \\choose 0} = 1\$) we find\n\n\\[\n\\begin{align} \\sum_{k=0}^n k \\cdot p_n(k) &= \\sum_{k=1}^n k {n \\choose k} d_{n-k} \\\\ &= n\\sum_{k=1}^n {n-1 \\choose k-1} d_{n-k} \\\\ &= n \\sum_{k=0}^{n-1} {n-1 \\choose k} d_{n-1-k} \\\\ &= n \\cdot (n-1)! \\end{align}\n\\]\n\nby our first equation as desired.\n\n~not_detriti\n\n## Note\n\nMaybe try and find a formula for \$p_n(k)\$. It is quite elementary if you know basic properties of binomial coefficients and stuff. For instance, how many ways can we choose \$k\$ fixed points out of the \$n\$ total digits? Well, it can be done in \$\\binom{n}{k}\$ ways. Now since we want exactly \$k\$ fixed points, what do we do with the remaining \$(n-k)\$ digits? Well we don't want any of those fixed. Clearly, of the \$(n-k)\$ spots left to put these \$(n-k)\$ points, we can put where it started off. So we have then \$(n-k-1)\$ spots to put one of the remaining \$(n-k)\$ points. Continuing on, we actually obtain a formula for \$p_n(k)\$, namely, \$\\binom{n}{k}(n-k-1)(n-k-2)\\dots1\$. Now we have to be careful, because now, what about for \$k=n-1\$? We see that no matter how we choose \$n-1\$ fixed points, we always have to put the remain point into the last possible spot, which was the spot it started on. Therefore, we must eliminate the case \$k=n-1\$.\n\n~th1nq3r\n\n## Note 2\n\n\$p_n(k)\$ is actually \$\\binom{n}{k}d(n-k)\$, where \$d\$ is the derangement counter function (See https://oeis.org/A000166)" ]
IMO-1987-2	https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_2	In an acute-angled triangle $ABC$ the interior bisector of the angle $A$ intersects $BC$ at $L$ and intersects the circumcircle of $ABC$ again at $N$. From point $L$ perpendiculars are drawn to $AB$ and $AC$, the feet of these perpendiculars being $K$ and $M$ respectively. Prove that the quadrilateral $AKNM$ and the triangle $ABC$ have equal areas.	[ "We are to prove that \$[AKNM]=[ABC]\$ or equivilently, \$[ABC]+[BNC]-[KNC]-[BMN]=[ABC]\$. Thus, we are to prove that \$[BNC]=[KNC]+[BMN]\$. It is clear that since \$\\angle BAN=\\angle NAC\$, the segments \$BN\$ and \$NC\$ are equal. Thus, we have \$[BNC]=\\frac{1}{2}BN^2\\sin BNC=\\frac{1}{2}BN^2\\sin A\$ since cyclic quadrilateral \$ABNC\$ gives \$\\angle BNC=180-\\angle A\$. Thus, we are to prove that\n\n\\[\n\\frac{1}{2}BN^2\\sin A=[KNC]+[BMN]\n\\]\n\n\\[\n\\Leftrightarrow \\frac{1}{2}BN^2\\sin A=\\frac{1}{2}CN\\cdot CK\\sin NCA+\\frac{1}{2}BN\\cdot BM\\sin NBA\n\\]\n\n\\[\n\\Leftrightarrow BN\\sin A=CK\\sin NCA+BM\\sin NBA\n\\]\n\nFrom the fact that \$\\angle BNC=180-\\angle A\$ and that \$BNC\$ is iscoceles, we find that \$\\angle NBC=\\angle NCB=\\frac{1}{2}A\$. So, we have \$BN\\cos\\frac{1}{2}A=\\frac{1}{2}BC\\Rightarrow BN=\\frac{BC}{2\\cos \\frac{1}{2}A}\$. So we are to prove that\n\n\\[\n\\frac{BC\\sin A}{2\\cos \\frac{1}{2}A}=CK\\sin NCA+BM\\sin NBA\n\\]\n\n\\[\n\\Leftrightarrow BC\\sin \\frac{1}{2}A=CK\\sin (C+ \\frac{1}{2}A)+BM\\sin (C+ \\frac{1}{2}A)\n\\]\n\n\\[\n\\Leftrightarrow BC=CK(\\sin C\\cot\\frac{1}{2}A+\\cos C)+BM(\\sin B\\cot\\frac{1}{2}A+\\cos B)\n\\]\n\nWe have \$\\sin C=\\frac{KL}{CL}\$,\$\\cos C=\\frac{CK}{CL}\$, \$\\cot\\frac{1}{2}A=\\frac{AK}{KL}=\\frac{AM}{LM}\$, \$\\sin B=\\frac{LM}{BL}\$,\$\\cos B=\\frac{BM}{ML}\$, and so we are to prove that\n\n\\[\nBC=CK(\\frac{KL}{CL}\\frac{AK}{KL})+\\frac{CK}{CL})+BM(\\frac{LM}{BL}\\frac{AM}{LM}+\\frac{BM}{ML})\n\\]\n\n\\[\n\\Leftrightarrow BC=CK(\\frac{AK}{CL}+\\frac{CK}{CL})+BM(\\frac{AM}{BL}+\\frac{BM}{ML})\n\\]\n\n\\[\n\\Leftrightarrow BC=\\frac{CK\\cdot AC}{CL}+\\frac{BM\\cdot AB}{BL}\n\\]\n\n\\[\n\\Leftrightarrow BC=AC\\cos C+AB\\cos B\n\\]\n\nWe shall show that this is true: Let the altitude from \$A\$ touch \$BC\$ at \$A^\\prime\$. Then it is obvious that \$AC\\cos C=CA^\\prime\$ and \$AB\\cos B=A^\\prime B\$ and thus \$AC\\cos C+AB\\cos B=BC\$.\n\nThus we have proven that \$[AKNM]=[ABC]\$.", "Clearly, \$AKLM\$ is a kite, so its diagonals are perpendicular. Furthermore, we have triangles \$ABN\$ and \$ALC\$ similar because two corresponding angles are equal.\n\nHence, we have \$[AKNM] = \\frac{1}{2} AN \\cdot KM = \\frac{1}{2} \\frac{AB \\cdot AC}{AL} \\cdot KM.\$ Notice that we used the fact that a quadrilateral's area is equal to half the product of its perpendicular diagonals (if they are, in fact, perpendicular).\n\nBut in (right) triangle \$AKL\$, we have \$<LAB = <A/2\$. Furthermore, if \$Q\$ is the intersection of diagonals \$AL\$ and \$KM\$ we have \$Q\$ the midpoint of \$KM\$ and \$KQ\$ an altitude of \$AKL\$, so\n\n\\[\n\\frac{KM}{2} = \\frac{AK \\cdot KL}{AL} = \\frac{AL \\sin <A/2 \\cdot AL \\cos <A/2}{AL} = \\frac{AL \\sin <A}{2},\n\\]\n\nso \$\\frac{KM}{AL} = \\sin <A\$. Hence\n\n\\[\n[AKNM] = \\frac{1}{2} \\frac{AB \\cdot AC}{AL} \\cdot KM = \\frac{1}{2} AB \\cdot AC \\sin <A = [ABC],\n\\]\n\nas desired.", "Proceed as in Solution 2. To prove that \$\\frac{KM}{AL} = \\sin A\$, consider that\n\n\\[\nAL = \\frac{AK}{\\sin <KLA} = \\frac{AK}{\\sin <AKM} = \\frac{KM}{\\sin A}\n\\]\n\nvia usage of definition of sine, equal angles in a right triangle if its altitude is drawn, and the Law of Sines." ]
IMO-1987-3	https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_3	Let $x_1 , x_2 , \ldots , x_n$ be real numbers satisfying $x_1^2 + x_2^2 + \cdots + x_n^2 = 1$. Prove that for every integer $k \ge 2$ there are integers $a_1, a_2, \ldots a_n$, not all 0, such that $\| a_i \| \le k-1$ for all $i$ and \[ \|a_1x_1 + a_2x_2 + \cdots + a_nx_n\| \le \frac{ (k-1) \sqrt{n} }{ k^n - 1 } \] .	[ "We first note that by the Power Mean Inequality, \$\\sum_{i=1}^{n} x_i \\le \\sqrt{n}\$. Therefore all sums of the form \$\\sum_{i=1}^{n} b_i x_i\$, where the \$b_i\$ is a non-negative integer less than \$k\$, fall in the interval \$[ 0 , (k-1)\\sqrt{n} ]\$. We may partition this interval into \$k^n - 1\$ subintervals of length \$\\frac{ (k-1)\\sqrt{n} }{k^n - 1}\$. But since there are \$k^n\$ such sums, by the pigeonhole principle, two must fall into the same subinterval. It is easy to see that their difference will form a sum with the desired properties.", "This solution is very similar to Solution 1 but uses a slightly different approach for the first part. It suffices to find \${a_i}\$ where \$a_i\$ is positive. Let \$f(a_1, a_2, ..., a_n) = a_1\| x_1\|+a_2 \|x_2\|+\\cdots+a_n \|x_n\|\\geq0\$. By the Cauchy-Schwarz Inequality,\n\n\\[\n\\begin{split} \\left(a_1 x_1+a_2 x_2+...+a_n x_n\\right)^2 &\\leq \\left(a_1^2+a_2^2+...+a_n^2\\right)\\cdot\\left(x_1^2+x_2^2+...+x_n^2\\right) \\\\ &=a_1^2+a_2^2+...+a_n^2 \\\\ &\\leq n(k-1)^2 \\end{split}\n\\]\n\nThis implies that \$\\left(a_1 x_1+a_2 x_2+...+a_n x_n\\right)^2\\leq \\sqrt{n} (k-1)\$, and hence the codomain of \$f(a_1, a_2, ..., a_n)\$ is \$\\left[0, \\sqrt{n} (k-1)\\right]\$. The rest of the proof is similar to Solution 1.\n\n~Iraevid13" ]
IMO-1987-4	https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_4	Prove that there is no function $f$ from the set of non-negative integers into itself such that $f(f(n)) = n + 1987$ for every $n$.	[ "We prove that if \$f(f(n)) = n + k\$ for all \$n\$, where \$k\$ is a fixed positive integer, then \$k\$ must be even. If \$k = 2h\$, then we may take \$f(n) = n + h\$.\n\nSuppose \$f(m) = n\$ with \$m \\equiv n \\mod k\$. Then by an easy induction on \$r\$ we find \$f(m + kr) = n + kr\$, \$f(n + kr) = m + k(r+1)\$. We show this leads to a contradiction. Suppose \$m < n\$, so \$n = m + ks\$ for some \$s > 0\$. Then \$f(n) = f(m + ks) = n + ks\$. But \$f(n) = m + k\$, so \$m = n + k(s - 1) \\ge n\$. Contradiction. So we must have \$m \\ge n\$, so \$m = n + ks\$ for some \$s \\ge 0\$. But now \$f(m + k) = f(n + k(s+1)) = m + k(s + 2)\$. But \$f(m + k) = n + k\$, so \$n = m + k(s + 1) > n\$. Contradiction.\n\nSo if \$f(m) = n\$, then \$m\$ and \$n\$ have different residues \$\\pmod k\$. Suppose they have \$r_1\$ and \$r_2\$ respectively. Then the same induction shows that all sufficiently large \$s \\equiv r_1 \\pmod k\$ have \$f(s) \\equiv r_2 \\pmod k\$, and that all sufficiently large \$s \\equiv r_2 \\pmod k\$ have \$f(s) \\equiv r_1 \\pmod k\$. Hence if \$m\$ has a different residue \$r \\mod k\$, then \$f(m)\$ cannot have residue \$r_1\$ or \$r_2\$. For if \$f(m)\$ had residue \$r_1\$, then the same argument would show that all sufficiently large numbers with residue \$r_1\$ had \$f(m) \\equiv r \\pmod k\$. Thus the residues form pairs, so that if a number is congruent to a particular residue, then \$f\$ of the number is congruent to the pair of the residue. But this is impossible for \$k\$ odd.", "Solution by Sawa Pavlov:\n\nLet \$\\mathbb{N}\$ be the set of non-negative integers. Put \$A = \\mathbb{N} - f(\\mathbb{N})\$ (the set of all \$n\$ such that we cannot find \$m\$ with \$f(m) = n\$). Put \$B = f(A)\$.\n\nNote that \$f\$ is injective because if \$f(n) = f(m)\$, then \$f(f(n)) = f(f(m))\$ so \$m = n\$. We claim that \$B = f(\\mathbb{N}) - f(f(\\mathbb{N}))\$. Obviously \$B\$ is a subset of \$f(\\mathbb{N})\$ and if \$k\$ belongs to \$B\$, then it does not belong to \$f(f(\\mathbb{N}))\$ since \$f\$ is injective. Similarly, a member of \$f(f(\\mathbb{N}))\$ cannot belong to \$B\$.\n\nClearly \$A\$ and \$B\$ are disjoint. They have union \$\\mathbb{N} - f(f(\\mathbb{N}))\$ which is \$\\{0, 1, 2, \\ldots , 1986\\}\$. But since \$f\$ is injective they have the same number of elements, which is impossible since \$\\{0, 1, \\ldots , 1986\\}\$ has an odd number of elements.", "Consider the function \$g: \\mathbb{Z}_{1987} \\rightarrow \\mathbb{Z}_{1987}\$ defined by \$g(x) = f(x\\; {\\rm mod }\\; 1987) \\;{\\rm mod }\\; 1987\$. Notice that we have \$f(k) + 1987 = f(f(f(k))) = f(k + 1987)\$, so that \$f(x) = f(y) \\;{\\rm mod }\\; 1987\$ whenever \$x = y \\;{\\rm mod }\\; 1987\$, and hence \$g\$ is well defined.\n\nNow, we observe that \$g\$ satisfies the identity \$g(g(x)) = x\$, for \$x \\in Z_{1987}\$. Thus, \$g\$ is an invertible function on a finite set of odd size, and hence must have a fixed point, say \$a \\in \\mathbb{Z}_{1987}\$. Identifying \$a\$ with its canonical representative in \$\\mathbb{Z}\$, we therefore get \$f(a) = a + 1987k\$ for some non-negative integer \$k\$.\n\nHowever, we then have \$f(f(a)) = a + 1987\$, while \$f(f(a)) = f(a + 1987k) = 1987k + f(a) = 2\\cdot1987k + a\$ (where we use the identity \$f(x + 1987) = f(x) + 1987\$ derived above, along with \$f(a) = f(a) + 1987\$. However, these two equations imply that \$k = \\dfrac{1}{2}\$, which is a contradiction since \$k\$ is an integer. Thus, such an \$f\$ cannot exist.\n\nNote: The main step in the proof above is that the function \$g\$ can be shown to have a fixed point. This step works even if 1987 is replaced with any other odd number larger than 1. However, for any even number \$c\$, \$f(x) = x + \\dfrac{c}{2}\$ satisfies the condition \$f(f(x)) = x + c\$.\n\n--Mahamaya 21:15, 21 May 2012 (EDT)" ]
IMO-1987-5	https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_5	Let $n$ be an integer greater than or equal to 3. Prove that there is a set of $n$ points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.	[ "Consider the set of points \$S = \\{ (x,x^2) \\mid 1 \\le x \\le n , x \\in \\mathbb{N} \\}\$ in the \$xy\$-plane.\n\nThe distance between any two distinct points \$(x_1,x^2_1)\$ and \$(x_2,x^2_2)\$ in \$S\$ (with \$x_1 \\neq x_2\$) is:\n\n\$d = \\sqrt{(x_1-x_2)^2+\\left(x^2_1-x^2_2\\right)^2} =\$ \$\\sqrt{(x_1-x_2)^2+(x_1-x_2)^2(x_1+x_2)^2} = \|x_1-x_2\|\\sqrt{1+(x_1+x_2)^2}\$.\n\nSince \$1+(x_1+x_2)^2\$ is an integer and not a perfect square, \$\\sqrt{1+(x_1+x_2)^2}\$ is irrational. Since \$\|x_1-x_2\|\$ is a non-zero integer, \$d\$ is irrational as desired.\n\nAll the points in \$S\$ lie on the parabola \$y = x^2\$. Thus, it is impossible of any set of three points to be collinear, since no line can intersect a parabola at more than two points. Therefore, any triangle with all vertices in \$S\$ must be non-degenerate as desired.\n\nSince all the points in \$S\$ are lattice points, by Pick's Theorem, the area of any triangle with all vertices in \$S\$ must be in the form \$A = I + \\dfrac{B}{2} - 1\$ where \$I\$ and \$B\$ are integers. Thus, the area of the triangle must be rational as desired.\n\nThis completes the proof." ]
IMO-1987-6	https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_6	Let $n$ be an integer greater than or equal to 2. Prove that if $k^2 + k + n$ is prime for all integers $k$ such that $0 \leq k \leq \sqrt{n/3}$, then $k^2 + k + n$ is prime for all integers $k$ such that $0 \leq k \leq n - 2$.	[ "First observe that if \$m\$ is relatively prime to \$b+1\$, \$b+2\$, \$\\cdots\$, \$2b\$, then \$m\$ is relatively prime to any number less than \$2b\$. Since if \$c\\leq b\$, then we can choose some \$i\$ to make \$2^ic\$ lies in range \$b+1,b+2,\\cdots,2b\$, so \$2^ic\$ is relatively prime to \$m\$. Hence \$c\$ is also. If we also have \$(2b+1)^2>m\$, then we can conclude that \$m\$ is a prime. Since there must be a factor of \$m\$ less than \$\\sqrt{m}\$.\n\nLet \$n=3r^2+h\$ where \$0\\leq h<6r+3\$, so \$r\$ is the greatest integer less than or equal to \$\\sqrt{n/3}\$.(to see this, just let \$r=\\lfloor\\sqrt{n/3}\\rfloor\$, then we can write \$n=3(r+\\epsilon)^2(0\\leq\\epsilon< 1)\$, so \$h=6r\\epsilon+3\\epsilon^2\\leq 6r+3\$).\n\nAssume that \$n+k(k+1)\$ is prime for \$k=1,2,3\\cdots,r\$. We show that \$N=n+(r+s)(r+s+1)\$ is prime for \$s=0,1,2,\\cdots,n-r-2\$. By our observation above, it is sufficient to show that \$(2s+2r+t)^2>N\$ and \$N\$ is relatively prime to all of \$r+s+1,r+s+2,\\cdots,2r+2s\$. We have \$(2r+2s+1)^2=4r^2+4s^2+8rs+4r+4s+1\$. Since \$s,t\\ge1\$, we have \$4s+1>s+2\$, \$4s^2>s^2\$ and \$6rs>3r\$. Hence\n\n\\[\n(2s+2r+1)^2>4r^2+2rs+s^2+7r+s+2=3r^2+6r+2+(r+s)(r+s+1)\\ge N\n\\]\n\nNow if \$N\$ has a factor which divides \$2r-i\$ in the range \$-2s\$ to \$r-s-1\$, then so does \$N-(i+2s+1)(2r-i)=n+(r-i-s-1)(r-i-s)\$ which have the form \$n+s'(s'+1)\$ with \$s'\$ in range \$0\$ to \$r\$." ]
IMO-1988-1	https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_1	Consider 2 concentric circles with radii $R$ and $r$ ($R>r$) with center $O$. Fix $P$ on the small circle and consider the variable chord $AP$ of the small circle. Points $B$ and $C$ lie on the large circle; $B,P,C$ are collinear and $BC$ is perpendicular to $AP$. 1. For which values of $\angle OPA$ is the sum $BC^2+CA^2+AB^2$ extremal? 2. What are the possible positions of the midpoints $U$ of $AB$ and $V$ of $AC$ as $A$ varies?	[ "1. We claim that the value \$BC^2+CA^2+AB^2\$ stays constant as \$\\angle OPA\$ varies, and thus achieves its maximum at all value of \$\\angle OPA\$. We have from the Pythagorean Theorem that \$CA^2=AP^2+PC^2\$ and \$AB^2=AP^2+PB^2\$ and so our expression becomes \\[ BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\\cdot PC \\] Since \$PB\\cdot PC\$ is the power of the point \$P\$, it stays constant as \$A\$ varies. Thus, we are left to prove that the value \$BC^2+AP^2\$ stays constant as \$\\angle OPA\$ varies. Let \$G\$ be the midpoint of \$AP\$ and let \$H\$ be the midpoint of \$BC\$. Since \$OG\$ is perpendicular to \$AP\$, we find that \$PG=r\\cos OPA\$. Similarly, we find that \$OH=r\\sin OPC=r\\cos OPA\$. Thus, by the Pythagorean Theorem, we have \\[ \\begin{align} BC^2& =4(R^2-r^2\\cos^2 OPA) \\\\ AP^2&=4r^2\\cos^2OPA \\end{align} \\] Now it is obvious that \$BC^2+AP^2=4R^2\$ is constant for all values of \$\\angle OPA\$.\n2. We claim that all points \$U,V\$ lie on a circle centered at the midpoint of \$OP\$, \$M\$ with radius \$\\frac{R}{2}\$. Let \$T\$ be the midpoint of \$UV\$. Since \$H\$ is the midpoint of \$BC\$, it is clear that the projection of \$T\$ onto \$BC\$ is the midpoint of \$H\$ and \$P\$ (the projection of \$A\$ onto \$BC\$). Thus, we have that \$MT\$ is perpendicular to \$UV\$ and thus the triangle \$MUV\$ is isosceles. We have \\[ \\begin{align} UT &=\\frac{1}{2}UV=\\frac{1}{4}BC \\mbox{ and } \\\\ MT &=\\frac{1}{2}PG=\\frac{1}{4}AP. \\end{align} \\] Thus, from the Pythagorean Theorem we have \\[ MV^2=MU^2=\\frac{1}{16}\\left(BC^2+AP^2\\right). \\] Since we have shown already that \$BC^2+AP^2=4R^2\$ is constant, we have that \$MV=MU=\\frac{R}{2}\$ and the locus of points \$U,V\$ is indeed a circle of radius \$\\frac{R}{2}\$ with center \$M\$." ]
IMO-1988-2	https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_2	Let $n$ be a positive integer and let $A_1, A_2, \cdots, A_{2n+1}$ be subsets of a set $B$. Suppose that (a) Each $A_i$ has exactly $2n$ elements, (b) Each $A_i\cap A_j$ $(1\le i<j\le 2n+1)$ contains exactly one element, and (c) Every element of $B$ belongs to at least two of the $A_i$. For which values of $n$ can one assign to every element of $B$ one of the numbers $0$ and $1$ in such a way that $A_i$ has $0$ assigned to exactly $n$ of its elements?	[ "Answer: All \$n\$ such that \$4\|n\$\n\nWe first make the following \$2\$ claims:\n\nClaim \$1\$: Each element of union belongs to exactly \$2\$ subsets.\n\nProof:\n\nConsider a subset \$A_i\$. Assume that some element \$x \\in\\ A_i\$ also \$\\in A_k, A_l\$. There are \$n-1\$ elements remaining in \$A_i\$ and there are \$n-2\$ subsets to choose from. By pigeon hole principle, at least \$1\$ of the remaining elements in \$A_i\$ must \$\\in A_k\$ or \$\\in A_l\$. This contradicts the assumption that any \$2\$ subsets have only \$1\$ element in common.\n\nClaim \$2\$: \$4\|n\$\n\nProof:\n\nNow, since each element in \$A_i \\in\$ exactly \$1\$ other subset, total number of elements present in the union = \$n(n+1)/2\$. If each subset must have \$n/2\$ elements assigned a value of \$1\$, the total number of elements assigned value of \$1\$ = \$n/2(n+1)/2 = n(n+1)/4\$. Thus \$4\$ must divide \$n\$.\n\nNow we make our final claim:\n\nClaim \$3\$: \$4\|n\$ is a sufficient condition to assign every element of the union one of the numbers 0 and 1 in such a manner that each of the sets has exactly \$\\frac {n}{2}\$ zeros.\n\nProof:\n\nConsider a regular polygon consisting of \$n+1\$ vertices where each line joining two vertices \$A_i, A_j\$ represents the element which \$\\in A_i, A_j\$. Clearly there are a total of \$n(n+1)/2\$ such lines representing the total number of elements of the union where each vertex is connected to \$n\$ vertices, meaning each of the \$n\$ elements of \$A_i\$ is part of \$1\$ other subset.\n\nStarting with \$i = 1\$, let us start coloring all lines joining vertices \$A_i\$, \$A_{i+1}\$ with color Red, all lines joining \$A_i\$, \$A_{i+2}\$ with color White, \$A_i\$, \$A_{i+3}\$ with color Red, \$A_i\$, \$A_{i+4}\$ with color White and so on ... \$A_i\$, \$A_{i+n/2}\$ with color White.\n\nClearly each line from vertex \$A_i\$ alternates Red, White for first \$n/2\$ lines and then alternates White, Red for remaining \$n/2\$ lines implying that we could have exactly \$n/2\$ red lines emanating from each vertex \$A_i\$. But these \$n/2\$ lines represent \$n/2\$ elements of each subset \$A_i\$ which could each be assigned a value of \$0\$. This completes the proof.\n\n- Kris17" ]
IMO-1988-3	https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_3	A function $f$ defined on the positive integers (and taking positive integers values) is given by: \[ \begin{matrix} f(1) = 1, f(3) = 3 \\ f(2 \cdot n) = f(n) \\ f(4 \cdot n + 1) = 2 \cdot f(2 \cdot n + 1) - f(n) \\ f(4 \cdot n + 3) = 3 \cdot f(2 \cdot n + 1) - 2 \cdot f(n), \end{matrix} \] for all positive integers $n.$ Determine with proof the number of positive integers $\leq 1988$ for which $f(n) = n.$	[ "Considering that \$f(n)=f(2n)\$, the two last equations give : \$f(4n + 1)-f(4n) = 2(f(2n + 1) - f(2n))\$ \$f(4n + 3)-f(4n+2)= 2(f(2n + 1) - f(2n))\$\n\nAnd so, if \$n\$ is even and \$2^{p+1}>n\\geq 2^p>1\$, we have \$f(n+1)-f(n)=2^p\$\n\nSo, if we have an even \$n=\\sum_{i=1}^{k} 2^{a_i}\$, where \$\\{a_i\\}\$ is a strictly increasing sequence with \$a_1>0\$ (\$n\$ even) : \$f(n+1)=2^{a_k}+f(n)\$ Then \$f(n)=f(\\sum_{i=1}^{k} 2^{a_i})\$ \$=f(\\sum_{i=1}^{k} 2^{a_i-a_1})\$ \$=2^{a_k-a_1}+f(\\sum_{i=2}^{k} 2^{a_i-a_1})\$ And so \$f((\\sum_{i=1}^{k} 2^{a_i})+1)=2^{a_k}+2^{a_k-a_1}+f(\\sum_{i=2}^{k} 2^{a_i-a_1})\$\n\nAnd it is easy to conclude that \$f(\\sum_{i=1}^{k} 2^{a_i})=\\sum_{i=1}^{k} 2^{a_k-a_i}\$ and that applying \$f(n)\$ means reversing the order of binary representation of n (and this could be also easily shown with induction).\n\nSo \$f(n)=n\$ occurs if and only if the binary representation of n is symmetrical.\n\nIt remains to count these \"symetric\" numbers. We have exactly \$2^{\\lceil\\frac{m-1}{2}\\rceil}\$ such numbers in \$[2^m,2^{m+1})\$. So : We have exactly 1 such numbers in \$[1,2)\$ We have exactly 1 such numbers in \$[2,4)\$ We have exactly 2 such numbers in \$[4,8)\$ We have exactly 2 such numbers in \$[8,16)\$ We have exactly 4 such numbers in \$[16,32)\$ We have exactly 4 such numbers in \$[32,64)\$ We have exactly 8 such numbers in \$[64,128)\$ We have exactly 8 such numbers in \$[128,256)\$ We have exactly 16 such numbers in \$[256,512)\$ We have exactly 16 such numbers in \$[512,1024)\$\n\nSince \$1988=B11111000100\$, positions 2 to 6 may be any between \$00000\$ and \$11101\$, and so : We have exactly 30 such numbers in \$[1024,1988]\$\n\nAnd so the requested number is \$1+1+2+2+4+4+8+8+16+16+30=92\$\n\nThis solution was posted and copyrighted by pco. The original thread for this problem can be found here: [1]", "The main claim is following.\n\nClaim: \$f(n)\$ is equal to the result when \$n\$ is written in binary and its digits are reversed. Proof. Follows directly by induction. \$\\blacksquare\$\n\nSo the question asks for the number of binary palindromes which are at most \$1988 = 11111000100_2\$. For \$k = 1, 2, \\dots, 10\$ there are \$2^{\\left\\lceil k/2 \\right\\rceil-1}\$ binary palindromes with \$k\$ bits (note the first bit must be \$1\$). For \$k=11\$, the number of binary palindromes which are also less than \$1988\$ is \$2^5 - 2\$ (only \$11111011111\$ and \$11111111111\$ are missing). Hence the final count is\n\n\\[\n2^0+2^0 + 2^1+2^1 + 2^2+2^2 + 2^3+2^3 + 2^4+2^4 + (2^5-2) = 92.\n\\]\n\nThis solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]", "We claim that \$f(x)\$ is the reversal of the digits of \$x\$ in binary. We proceed by induction, because given \$f(1)=f(2)=1\$ and \$f(3)=3\$, we can determine all other values, and this property holds for \$f(1)\$, \$f(2)\$, and \$f(3)\$.\n\nWe prove that all operations conserve this property. For \$f(2n)=f(n)\$, we are adding a zero at the end, thus obviously preserving this property. For \$f(4n+1) = 2f(2n+1)-f(n)\$, we see that letting \$4n+1=\\overline{a_1a_2\\ldots a_k01}_2\$ gets that assuming this holds true for \$f(2n+1)\$ and \$f(n)\$, then\n\n\\[\nf(4n+1) = 2f(2n+1)-f(n)=2f(\\overline{a_1a_2\\ldots a_k1}_2)-f(\\overline{a_1a_2\\ldots a_k}_2)\n\\]\n\n\\[\n= 2 \\cdot\\overline{1a_ka_{k-1}\\ldots a_2a_1}_2 - \\overline{a_ka_{k-1}\\ldots a_2a_1}_2 = \\overline{10a_ka_{k-1}\\ldots a_2a_1}_2\n\\]\n\nwhich keeps the property since \$\\overline{10a_ka_{k-1}\\ldots a_2a_1}_2\$ is the reversal of \$\\overline{a_1a_2\\ldots a_k01}_2\$ in binary. Similarly, letting \$4n+1=\\overline{a_1a_2\\ldots a_k11}_2\$, then\n\n\\[\nf(4n+3) = 3f(2n+1)-2f(n)=3f(\\overline{a_1a_2\\ldots a_k1}_2)-2f(\\overline{a_1a_2\\ldots a_k}_2)\n\\]\n\n\\[\n= 3 \\cdot\\overline{1a_ka_{k-1}\\ldots a_2a_1}_2 - 2 \\cdot \\overline{a_ka_{k-1}\\ldots a_2a_1}_2 = \\overline{11a_ka_{k-1}\\ldots a_2a_1}_2\n\\]\n\nwhich keeps the property since \$\\overline{11a_ka_{k-1}\\ldots a_2a_1}_2\$ is the reversal of \$\\overline{a_1a_2\\ldots a_k11}_2\$. Due to this, we have proved that this property holds for all \$n\$ which means we need to find the number of binary palindromes below \$1988\$. This number is just going to be, for \$n<11\$-digit numbers, \$2^{\\lfloor \\frac {n-1}2 \\rfloor}\$ ways, so summing up for less than \$11\$ digits gets \$62\$ ways. For \$11\$-digit numbers, we have \$32\$ ways but need to subtract a few in which are greater than \$1988\$; there are two numbers we don't count, namely \$2015\$ and \$2047\$. As a result, our answer is \$62+30=\\boxed{92}\$.\n\nThis solution was posted and copyrighted by kevinmathz. The original thread for this problem can be found here: [3]" ]
IMO-1988-4	https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_4	Show that the solution set of the inequality \[ \sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4} \] is a union of disjoint intervals, the sum of whose length is $1988$.	[ "Consider the graph of \$f(x)=\\sum_{k=1}^{70}\\frac{k}{x-k}\\ge\\frac{5}{4}\$. On the values of \$x\$ between \$n\$ and \$n+1\$ for \$n\\in\\mathbb{N}\$ \$1\\le n\\le 69\$, the terms of the form \$\\frac{k}{x-k}\$ for \$k\\ne n,n+1\$ have a finite range. In contrast, the term \$\\frac{n}{x-n}\$ has an infinite range, from \$+\\infty\$ to \$n\$. Similarly, the term \$\\frac{n+1}{x-n-1}\$ has infinite range from \$-n-1\$ to \$-\\infty\$. Thus, since the two undefined values occur at the distinct endpoints, we can deduce that \$f(x)\$ takes on all values between \$+\\infty\$ and \$-\\infty\$ for \$x\\in(n,n+1)\$. Thus, by the Intermediate Value Theorem, we are garunteed a \$n<r_n<n+1\$ such that \$f(r_n)=\\frac{5}{4}\$. Additionally, we have that for \$x>70\$, the value of \$f(x)\$ goes from \$+\\infty\$ to \$0\$, since as \$x\$ increases, all the terms go to \$0\$. Thus, there exists some \$r_{70}>70\$ such that \$f(r_{70})=\\frac{5}{4}\$ and so \$f(x)\\ge\\frac{5}{4}\$ for \$x\\in(70,r_{70})\$.\n\nSo, we have \$70\$ \$r_i\$ such that \$f(r_i)=\\frac{5}{4}\$. There are obviously no other such \$r_i\$ since \$f(x)=\\frac{5}{4}\$ yields a polynomial of degree \$70\$ when combining fractions. Thus, we have that the solution set to the inequality \$f(x)\\ge\\frac{5}{4}\$ is the union of the intervals \$(n,r_n]\$ (since if \$f(x)<\\frac{5}{4}\$ for \$x\\in(n,r_n)\$ then there would exist another solution to the equation \$f(x)=\\frac{5}{4}\$.\n\nThus we have proven that the solution set is the union of disjoint intervals. Now we are to prove that the sum of their lengths is \$1988\$.\n\nThe sum of their lengths is \$r_1+r_2+\\cdots+r_{70}-(1+2+\\cdots+70)=r_1+r_2+\\cdots+r_{70}-35\\cdot71\$. We have that the equation \$f(x)=\\frac{5}{4}\$ yields a polynomial with roots \$r_i\$. Thus, opposite of the coeficient of \$x^{69}\$ divided by the leading coefficient is the sum of the \$r_i\$. It is easy to see that the coefficient of \$x^{69}\$ is \$-5(1+2+\\cdots+70)-4(1+2+\\cdots+70)=-9\\cdot35\\cdot 71\$. Thus, since the leading coefficient is \$5\$ we have \$r_1+r_2+\\cdots+r_{70}=9\\cdot7\\cdot71\$. Thus, the sum of the lengths of the intervals is \$63\\cdot71-35\\cdot71=28\\cdot71=1988\$ as desired." ]
IMO-1988-5	https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_5	In a right-angled triangle $ABC$ let $AD$ be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles $ABD, ACD$ intersect the sides $AB, AC$ at the points $K,L$ respectively. If $E$ and $E_1$ dnote the areas of triangles $ABC$ and $AKL$ respectively, show that \[ \frac {E}{E_1} \geq 2. \]	[ "Lemma: Through the incenter \$I\$ of \$\\triangle{ABC}\$ draw a line that meets the sides \$AB\$ and \$AC\$ at \$P\$ and \$Q\$, then:\n\n\\[\n\\frac{AB}{AP} \\cdot AC + \\frac{AC}{AQ} \\cdot AB = AB+BC+AC\n\\]\n\nProof of the lemma: Consider the general case: \$M\$ is any point on side \$BC\$ and \$PQ\$ is a line cutting AB, AM, AC at P, N, Q. Then:\n\n\\[\n\\frac{AM}{AN}=\\frac{S_{APMQ}}{\\triangle{APQ}}=\\frac{\\triangle{APM}+\\triangle{AQM}}{\\triangle{PQA}}=\\frac{\\frac{AP}{AB}\\triangle{ABM}+\\frac{AQ}{AC}\\triangle{ACM}}{\\frac{AP\\cdot AQ}{AB \\cdot AC}}=\n\\]\n\n\\[\n=\\frac{AC}{AQ}\\cdot \\frac{BM}{BC}+\\frac{AB}{AP}\\cdot \\frac{CM}{BC}\n\\]\n\nIf \$N\$ is the incentre then \$\\frac{AM}{AN}=\\frac{AB+BC+CA}{AB+AC}\$, \$\\frac{BM}{BC}=\\frac{AB}{AB+AC}\$ and \$\\frac{CM}{BC}=\\frac{AC}{AC+AB}\$. Plug them in we get:\n\n\\[\n\\frac{AB}{AP} \\cdot AC + \\frac{AC}{AQ} \\cdot AB = AB+BC+AC\n\\]\n\nBack to the problem Let \$I_1\$ and \$I_2\$ be the areas of \$\\triangle{ABD}\$ and \$\\triangle{ACD}\$ and \$E\$ be the intersection of \$KL\$ and \$AD\$. Thus apply our formula in the two triangles we get:\n\n\\[\n\\frac{AD}{AE} \\cdot AB + \\frac{AB}{AK} \\cdot AD = AB+BD+AD\n\\]\n\nand\n\n\\[\n\\frac{AD}{AE} \\cdot AC + \\frac{AC}{AL} \\cdot AD = AC+CD+AD\n\\]\n\nCancel out the term \$\\frac{AD}{AE}\$, we get:\n\n\\[\n\\frac{AB+BD+AD-\\frac{AB}{AK} \\cdot AD }{AC+CD+AD- \\frac{AC}{AL} \\cdot AD }=\\frac{AB}{AC}\n\\]\n\n\\[\nAB \\cdot CD + AB \\cdot AD - \\frac{AB \\cdot AC \\cdot AD}{AL}=AC \\cdot BD+ AC \\cdot AD -\\frac{AB \\cdot AC \\cdot AD}{AK}\n\\]\n\n\\[\nAB+AB \\cdot \\frac{CD}{AD}-\\frac{AB \\cdot AC}{AL}=AC+ AC \\cdot \\frac{BD}{AD} - \\frac{AB \\cdot AC}{AK}\n\\]\n\n\\[\nAB+AC - \\frac{AB \\cdot AC}{AL}=AB+AC - \\frac{AB \\cdot AC}{AK}\n\\]\n\n\\[\n\\frac{AB \\cdot AC}{AK} = \\frac{AB \\cdot AC}{AL}\n\\]\n\nSo we conclude \$AK=AL\$.\n\nHence \$\\angle{AKI_1}=45^o=\\angle{ADI_1}\$ and \$\\angle{ALI_2}=45^o=\\angle{ADI_2}\$, thus \$\\triangle{AK_1} \\cong \\triangle{ADI_1}\$ and \$\\triangle{ALI_2} \\cong \\triangle{ADI_2}\$. Thus \$AK=AD=AL\$. So the area ratio is:\n\n\\[\n\\frac{E}{E_1}=\\frac{AB \\cdot AC}{AD^2} = \\frac{BC}{AD} =\\frac{BD+CD}{\\sqrt{BD \\cdot CD}}\\geq 2\n\\]\n\nThis solution was posted and copyrighted by shobber. The original thread for this problem can be found here: [1]" ]
IMO-1988-6	https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_6	Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^{2} + b^{2}$. Show that $\frac {a^{2} + b^{2}}{ab + 1}$ is the square of an integer.	[ "Choose integers \$a,b,k\$ such that \$a^2+b^2=k(ab+1)\$ Now, for fixed \$k\$, out of all pairs \$(a,b)\$ choose the one with the lowest value of \$\\min(a,b)\$. Label \$b'=\\min(a,b), a'=\\max(a,b)\$. Thus, \$a'^2-kb'a'+b'^2-k=0\$ is a quadratic in \$a'\$. Should there be another root, \$c'\$, the root would satisfy: \$b'c'\\leq a'c'=b'^2-k<b'^2\\implies c'<b'\$ Thus, \$c'\$ isn't a positive integer (if it were, it would contradict the minimality condition). But \$c'=kb'-a'\$, so \$c'\$ is an integer; hence, \$c'\\leq 0\$. In addition, \$(a'+1)(c'+1)=a'c'+a'+c'+1=b'^2-k+b'k+1=b'^2+(b'-1)k+1\\geq 1\$ so that \$c'>-1\$. We conclude that \$c'=0\$ so that \$b'^2=k\$.\n\nThis construction works whenever there exists a solution \$(a,b)\$ for a fixed \$k\$, hence \$k\$ is always a perfect square.", "We proceed by way of contradiction.\n\nWLOG, let \$a\\geq{b}\$ and fix \$c\$ to be the nonsquare positive integer such that such that \$\\frac{a^2+b^2}{ab+1}=c,\$ or \$a^2+b^2=c(ab+1).\$ Choose a pair \$(a, b)\$ out of all valid pairs such that \$a+b\$ is minimized. Expanding and rearranging,\n\n\\[\nP(a)=a^2+a(-bc)+b^2-c=0.\n\\]\n\nThis quadratic has two roots, \$r_1\$ and \$r_2\$, such that\n\n\\[\n(a-r_1)(a-r_2)=P(a)=0.\n\\]\n\nWLOG, let \$r_1=a\$. By Vieta's, \$\\textbf{(1) } r_2=bc-a,\$ and \$\\textbf{(2) } r_2=\\frac{b^2-c}{a}.\$ From \$\\textbf{(1)}\$, \$r_2\$ is an integer, because both \$b\$ and \$c\$ are integers.\n\nFrom \$\\textbf{(2)},\$ \$r_2\$ is nonzero since \$c\$ is not square, from our assumption.\n\nWe can plug in \$r_2\$ for \$a\$ in the original expression, because \$P(r_2)=P(a)=0,\$ yielding \$c=\\frac{r^2_2+b^2}{r_2b+1}\$. If \$c>0,\$ then \$r_2b+1>0,\$ and \$r_2b+1\\neq{0},\$ and because \$b>0, r_2\$ is a positive integer.\n\nWe construct the following inequalities: \$r_2=\\frac{b^2-c}{a}<a,\$ since \$c\$ is positive. Adding \$b\$, \$r_2+b<a+b,\$ contradicting the minimality of \$a+b.\$\n\n-Benedict T (countmath1)", "Given that \$ab+1\$ divides \$a^2+b^2\$, we have \$a^2+b^2=k(ab+1)\$ for some integer \$k\$.\n\nExpanding the right side, we get \$a^2+b^2=kab+k\$. Rearranging terms, we have \$a^2-kab+b^2-k=0\$.\n\nConsider this as a quadratic equation in \$a\$. By the quadratic formula, we have\n\n\\[\na=\\frac{kb\\pm\\sqrt{k^2b^2-4(b^2-k)}}{2}.\n\\]\n\nFor \$a\$ to be an integer, the discriminant \$k^2b^2-4(b^2-k)\$ must be a perfect square. Let \$k^2b^2-4(b^2-k)=m^2\$ for some integer \$m\$.\n\nRearranging terms, we get \$m^2=k^2b^2-4b^2+4k\$. Factoring the right side, we have \$m^2=(kb-2)^2\$ (??? this factoring does not seem right!!! By J Xu).\n\nThus, \$m=kb-2\$ and \$a=\\frac{kb\\pm(kb-2)}{2}=b\$ or \$a=kb-b\$. In either case, we have \$a=kb-b\$.\n\nSubstitute \$a=kb-b\$ back into \$a^2+b^2=kab+k\$, we get \$b^2+k^2b^2-2kb^2+b^2=kb^2-kb+k\$.\n\nSimplifying, we have \$b^2=k\$. Therefore, \$\\frac{a^2+b^2}{ab+1}=\\frac{(kb-b)^2+b^2}{b(k)+1}=\\frac{b^2}{b+1}=b\$, which is the square of an integer. (it seems something minor issue here, too. By J Xu) By M. Nazaryan." ]
IMO-1989-1	https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_1	Prove that in the set $\{1,2, \ldots, 1989\}$ can be expressed as the disjoint union of subsets $A_i, \{i = 1,2, \ldots, 117\}$ such that i.) each $A_i$ contains 17 elements ii.) the sum of all the elements in each $A_i$ is the same.	[ "Let us start pairing numbers in the following fashion, where each pair sums to \$1990\$: \$(1, 1990-1), (2, 1990-2), \\ldots (936, 1054)\$\n\nThere are a total of \$1178\$ pairs above. Let us start putting these pairs into each of the \$117\$ subsets starting with the first pair \$(1, 1990-1)\$ going into \$A_1\$, \$(2, 1990-2)\$ into \$A_2\$ and so on \$\\ldots\$ with \$(k, 1990-k)\$ going into \$A_{k(mod 117)}\$.\n\nNow we have \$16\$ numbers present in each of the \$117\$ subsets all of which have the same total sum. We need \$1\$ more number to be filled in each subset from the remaining \$117\$ numbers \$(937, 938, ... 1053)\$.\n\nLet us arrange these \$117\$ numbers in the following manner: \$995-58, 995-57, \\ldots, 995-1, 995, 995+1, 995+2, \\ldots, 995+57, 995+58\$ \$==> (1)\$\n\nNow notice that for each number that's off by \$x\$ from \$995\$, there's a counter number off by \$x\$ from \$995\$ in the opposite direction.\n\nSo we need to create similar imbalances in the Subsets \$A_1\$ to \$A_{117}\$, so that we could add the above \$117\$ numbers to those imbalances to keep the total sum unchanged.\n\nNow start swapping the \$1st\$ number of each pair that we added to subsets \$A_1\$ to \$A_{117}\$ to create the above imbalances:\n\nSwap \$1\$ from \$A_1\$ with \$59\$ from \$A_{59}\$ - this creates an imbalance of \$+58\$ and \$-58\$\n\nSwap \$2\$ from \$A_2\$ with \$58\$ from \$A_{58}\$ - this creates an imbalance of \$+56\$ and \$-56\$\n\n\\[\n\\ldots\n\\]\n\nSwap \$29\$ from \$A_{29}\$ with \$31\$ from \$A_{31}\$ - creates an imbalance of \$+2\$ and \$-2\$\n\nLeave \$30\$ in \$A_{30}\$ as it is - \$0\$ imbalance.\n\nSimilarly,\n\nSwap \$60\$ from \$A_{60}\$ with \$117\$ from \$A_{117}\$ - this creates an imbalance of \$+57\$ and \$-57\$\n\nSwap \$61\$ from \$A_{61}\$ with \$116\$ from \$A_{116}\$ - this creates an imbalance of \$+55\$ and \$-55\$\n\n\\[\n\\ldots\n\\]\n\nSwap \$88\$ from \$A_{88}\$ with \$89\$ from \$A_{89}\$ - this creates an imbalance of \$+1\$ and \$-1\$\n\nNow start adding the \$117\$ numbers from \$(1)\$ to the subsets \$A_1\$ to \$A_{117}\$ to counter the imbalances \$-58\$ to \$+58\$ so that the sum remains unchanged. Notice that each subset now has \$17\$ elements with total sum = \$81990 + 995 = 16915\$.\n\n\\[\n- Kris17\n\\]" ]
IMO-1989-2	https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_2	$ABC$ is a triangle, the bisector of angle $A$ meets the circumcircle of triangle $ABC$ in $A_1$, points $B_1$ and $C_1$ are defined similarly. Let $AA_1$ meet the lines that bisect the two external angles at $B$ and $C$ in $A_0$. Define $B_0$ and $C_0$ similarly. Prove that the area of triangle $A_0B_0C_0 = 2 \cdot$ area of hexagon $AC_1BA_1CB_1 \geq 4 \cdot$ area of triangle $ABC$.	[ "Notice that since \$A_1C\$ and \$A_1B\$ substend the same angle in the circle, and so are equal. Thus \$A_1BC\$ is iscoceles, and similarly for triangles \$AB_1C\$ and \$ABC_1\$. Also, since \$AA_1BC\$ is a cyclic quadrilatera, \$\\angle A_1BC=\\angle A_1AC=\\frac{A}{2}\$, and similarly for the other triangles. Thus, the area of triangle \$A_1BC=\\frac{1}{4}a^2\\tan\\frac{A}{2}=\\frac{1}{4}a^2\\left(\\frac{2r}{b+c-a}\\right)\$ and similarly for the other triangles. Thus, the area of the hexagon is equal to \$[ABC]+\\sum\\frac{1}{4}a^2\\left(\\frac{2r}{b+c-a}\\right)\$.\n\nNow we shall find the area of triangle \$A_0B_0C_0\$. It is obvious that the points \$A_1,B,C_1\$ are collinear since the angle the make at \$B\$ is \$\\frac{1}{2}(180-B)+\\frac{1}{2}(180-B)+B=180\$, and similarly for the other points. Thus, it suffices to find the area of each of the triangles \$A_0BC\$. It is well known that \$A_0\$ is the center of the excircle \$O_a\$ with radius \$R_a=r\\left(\\frac{a+b+c}{b+c-a}\\right)\$. Thus the area of triangle \$A_0BC\$ is \$\\frac{1}{2}ar\\left(\\frac{a+b+c}{b+c-a}\\right)\$ and so the area of triangel \$A_0B_0C_0\$ is \$\\sum\\frac{1}{2}ar\\left(\\frac{a+b+c}{b+c-a}\\right)\$.\n\nNow we can prove that \$A_0B_0C_0 = 2 \\cdot\$ area of hexagon \$AC_1BA_1CB_1\$ by simplifying. We have\n\n\$A_0B_0C_0 -2 \\cdot\$ area of hexagon \$AC_1BA_1CB_1\$\n\n\\[\n=-[ABC]+\\sum\\left(\\frac{1}{2}ar\\left(\\frac{a+b+c}{b+c-a}\\right)-\\frac{1}{2}a^2\\left(\\frac{2r}{b+c-a}\\right)\\right)\n\\]\n\n\\[\n=-[ABC]+\\sum\\left(\\frac{1}{2}ar\\left(\\frac{a+b+c-2a}{b+c-a}\\right)\\right)\n\\]\n\n\\[\n=-[ABC]+\\sum\\left(\\frac{1}{2}ar\\right)\n\\]\n\n\\[\n=0\n\\]\n\nas desired.\n\nAs for the inequality, notice that it is equivalent to\n\n\\[\n[ABC]+\\sum\\frac{1}{2}ar\\left(\\frac{a+b+c}{b+c-a}\\right)\\ge 4[ABC]\n\\]\n\n\\[\n\\Leftrightarrow \\sum\\frac{1}{2}ar\\left(\\frac{a+b+c}{b+c-a}\\right)\\ge \\frac{3}{2}r(a+b+c)\n\\]\n\n\\[\n\\Leftrightarrow \\sum\\frac{a}{b+c-a}\\ge 3\n\\]\n\nLetting \$a=x+y,b=y+z,c=z+x\$ for \$x,y,z\$ positive reals, the inequality becomes\n\n\\[\n\\sum\\frac{x+y}{2z}\\ge 3\n\\]\n\n\\[\n\\Leftrightarrow \\sum\\frac{x+y+z}{z}\\ge 9\n\\]\n\nwhich is true by AM-HM." ]
IMO-1989-3	https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_3	Let $n$ and $k$ be positive integers and let $S$ be a set of $n$ points in the plane such that i.) no three points of $S$ are collinear, and ii.) for every point $P$ of $S$ there are at least $k$ points of $S$ equidistant from $P.$ Prove that: \[ k < \frac {1}{2} + \sqrt {2 \cdot n} \]	[ "Let \$\\{A_i\\}_{i=1}^n\$ be our set of points. We count the pairs \$(A_i,\\{A_j,A_\\ell\\}),\\ i\\ne j\\ne \\ell\\ne i\$ such that \$A_iA_j=A_iA_\\ell\$. There are \$n\$ choices for \$A_i\$, and for each \$A_i\$ there are \$\\ge\\frac{k(k-1)}2\$ choices for \$\\{A_j,A_\\ell\\}\$, so the number of such pairs is \$\\ge\\frac{nk(k-1)}2\$. On the other hand, there are \$\\frac{n(n-1)}2\$ choices for \$\\{A_j,A_\\ell\\}\$, and for each \$\\{A_j,A_\\ell\\}\$ there are at most two choices for \$A_i\$ (because we can't have \$\\ge 3\$ points on the perpendicular bisector of \$A_jA_\\ell\$). This means that the number of pairs is \$\\le n(n-1)\$. We have thus found \$n(n-1)\\ge\\frac{nk(k-1)}2\\Rightarrow 2(n-1)\\ge k(k-1)\\ ()\$. If \$k\\ge \\sqrt{2n}+\\frac 12\$, then \$k(k-1)\\ge 2n-\\frac 14>2n-2\$, contradicting the last inequality in \$()\$, so we're done.\n\nThis solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]", "So obvious simplification and the fact that \$n\$ is an integer gives that what we want is \$n\\ge\\dbinom{k}{2}+1\$ Now for each point, draw a circle around it that contains at least \$k\$ points in \$S\$. For any point \$P\$ in \$S\$, let \$d_P\$ be the number of circles it's on. For any point \$O\$ in \$S\$, let \$f_O\$ be the number of points on the defined circle with center \$O\$, so \$f_O\\ge k\$ for all \$O\$. Since the function \$\\dbinom{x}{2}\$ is increasing for \$x\\ge 1\$ an integer, \$\\dbinom{f_O}{2}\\ge \\dbinom{k}{2}\$ for all \$O\$. So \$E(\\dbinom{f}{2})\\ge \\dbinom{k}{2}\$. But notice that since any pair of two points shares at most 2 circles (otherwise we have 3 circles with centers on their perpendicular bisector, which is not allowed), \$\\sum_{O\\in S}\\dbinom{f_O}{2}\\le 2\\dbinom{n}{2}\$, since the left side counts the number of point-pairs that share a circle (with possible circle multiplicity) in \$P\$ while the right side upper-bounds the number of point-pairs that can share a circle (with possible circle multiplicity). So now \$\\dbinom{k}{2}\\le E(\\dbinom{f}{2})\\le (2/n)((n-1)(n)/2)=n-1\$, which is what we want.\n\nThis solution was posted and copyrighted by antimonyarsenide. The original thread for this problem can be found here: [2]" ]
IMO-1989-4	https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_4	Let $ABCD$ be a convex quadrilateral such that the sides $AB,AD,BC$ satisfy $AB=AD+BC$. There exists a point $P$ inside the quadrilateral at a distance $h$ from the line $CD$ such that $AP=h+AD$ and $BP=h+BC$. Show that: \[ \frac{1}{\sqrt{h}}\ge\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}} \]	[ "Without loss of generality, assume \$AD\\ge BC\$.\n\nDraw the circle with center \$A\$ and radius \$AD\$ as well as the circle with center \$B\$ and radius \$BC\$. Since \$AB=AD+BC\$, the intersection of circle \$A\$ with the line \$AB\$ is the same as the intersection of circle \$B\$ with the line \$AB\$. Thus, the two circles are externally tangent to each other.\n\nNow draw the circle with center \$P\$ and radius \$h\$. Since \$h\$ is the shortest distance from \$P\$ to \$CD\$, we have that the circle \$P\$ is tangent to \$CD\$. From the conditions \$AP=h+AD\$ and \$BP=h+BC\$, by the same reasoning that we found circles \$A,B\$ are tangent, we find that circles \$P,A\$ and \$P,B\$ are externally tangent also.\n\nFix the two externally tangent circles with centers \$A\$ and \$B\$ and let the points \$C\$ and \$D\$ vary about their respective circles (without loss of generality, assume that one can read the letters \$ABCD\$ in counter clockwise order). Notice that the value \$\\frac{1}{\\sqrt{AD}}+\\frac{1}{\\sqrt{BC}}\$ is fixed. Thus, in order to prove the claim presented in the problem, we must maximize the value of \$h\$. It is clear that all possible circles \$P\$ are contained within the region created by circles \$A,B\$ as well as their common tangent line. Now it is clear that \$h\$ is maximized when circle \$P\$ is tangent to the common tangent of circles \$A,B\$ as well as to the circles \$A,B\$. We shall prove that the value of \$h\$ at this point is \$\\frac{1}{\\sqrt{AD}}+\\frac{1}{\\sqrt{BC}}\$.\n\nFix \$C,D\$ so that \$CD\$ is the common tangent to circles \$A,B\$. Since \$\\angle ADC,\\angle BCD\$ are right, the length \$CD\$ is equal to the length of the segment from \$B\$ to the projection of \$B\$ onto \$AD\$ which by the Pythagorean Theorem is \$\\sqrt{(BC+AD)^2-(AD-BC)^2}=2\\sqrt{BC\\cdot AD}\$. Let the projection of \$P\$ onto \$CD\$ be \$x\$ units away from \$D\$. Then, as we found the length of \$CD\$, we find\n\n\\[\nx^2+(AD-h)^2=(AD+h)^2\\Rightarrow x^2=4AD\\cdot h\n\\]\n\n\\[\n(2\\sqrt{BC\\cdot AD}-x)^2+(BC-h)^2=(BC+h)^2\\Rightarrow (2\\sqrt{BC\\cdot AD}-x)^2=4BC\\cdot h\n\\]\n\nSubtracting the second equation from the first yields\n\n\\[\n4x\\sqrt{BC\\cdot AD}-4BC\\cdot AD=4AD\\cdot h-4BC\\cdot h\n\\]\n\n\\[\n\\Rightarrow x=h\\left(\\sqrt{\\frac{AD}{BC}}-\\sqrt{\\frac{BC}{AD}}\\right)+\\sqrt{BC\\cdot AD}\n\\]\n\nSince \$x^2=2AD\\cdot h\$, we have\n\n\\[\n0=h\\left(\\sqrt{\\frac{AD}{BC}}-\\sqrt{\\frac{BC}{AD}}\\right)-2\\sqrt{AD\\cdot h}+\\sqrt{BC\\cdot AD}\n\\]\n\nSolving for \$\\sqrt{h}\$ yields\n\n\\[\n\\sqrt{h}=\\frac{2\\sqrt{AD}\\pm\\sqrt{4AD-4(AD-BC)}}{2\\left(\\sqrt{\\frac{AD}{BC}}-\\sqrt{\\frac{BC}{AD}}\\right)}\n\\]\n\n\\[\n\\Rightarrow \\sqrt{h}=\\frac{\\sqrt{AD}\\pm\\sqrt{BC}}{\\sqrt{\\frac{AD}{BC}}-\\sqrt{\\frac{BC}{AD}}}\n\\]\n\nSince \$\\sqrt{h}\\ne \\frac{\\sqrt{AD}+\\sqrt{BC}}{\\sqrt{\\frac{AD}{BC}}-\\sqrt{\\frac{BC}{AD}}}\$, we have\n\n\\[\n\\sqrt{h}=\\frac{\\sqrt{AD}-\\sqrt{BC}}{\\sqrt{\\frac{AD}{BC}}-\\sqrt{\\frac{BC}{AD}}}\n\\]\n\n\\[\n=\\frac{\\sqrt{AD\\cdot BC}}{\\sqrt{AD}+\\sqrt{BC}}\n\\]\n\nAnd since this is the maximimal value of \$\\sqrt{h}\$, the minimal value of \$\\frac{1}{\\sqrt{h}}\$ is \$\\frac{1}{\\sqrt{AD}}+\\frac{1}{\\sqrt{BC}}\$ as desired." ]
IMO-1989-5	https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_5	Prove that for each positive integer $n$ there exist $n$ consecutive positive integers none of which is an integral power of a prime number.	[ "There are at most \$1+\\sqrt[2]{n}+\\sqrt[3]{n}+\\sqrt[4]{n}+...+\\sqrt[\\left\\lfloor \\log_2(n)\\right\\rfloor]{n} \\leq 1+ \\sqrt n log_2(n)\$ 'true' powers \$m^k , k\\geq 2\$ in the set \$\\{1,2,...,n\\}\$. So when \$p(n)\$ gives the amount of 'true' powers \$\\leq n\$ we get that \$\\lim_{n \\to \\infty} \\frac{p(n)}{n} = 0\$.\n\nSince also \$\\lim_{n \\to \\infty} \\frac{\\pi(n)}{n} = 0\$, we get that \$\\lim_{n \\to \\infty} \\frac{p(n)+\\pi(n)}{n} = 0\$. Now assume that there is no 'gap' of lenght at least \$k\$ into the set of 'true' powers and the primes. Then this would give that \$\\frac{p(n)+\\pi(n)}{n} \\geq \\frac{1}{k}\$ for all \$n\$ in contrary to the above (at east this proves a bit more).\n\nEdit: to elementarize the \$\\lim_{n \\to \\infty} \\frac{\\pi(n)}{n} = 0\$ part: Look \$\\mod (k+1)!\$. Then all numbers in the residue classes \$2,3,4,...,k+1\$ are not primes (except the smallest representants sometimes). So when there wouldn't exist a gap of length \$k\$, there has to be a 'true' power in each of these gaps of the prime numbers, so at least one power each \$(k+1)!\$ numbers, again contradicting \$\\lim_{n \\to \\infty} \\frac{p(n)}{n} = 0\$.\n\nThis solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]", "By the Chinese Remainder Theorem, there exists \$x\$ such that: \$x \\equiv -1\\;mod\\;p_1 q_1\\\\ x \\equiv -2\\;mod\\;p_2 q_2\\\\ x \\equiv -3\\;mod\\;p_3 q_3\\\\ ...\\newline x \\equiv -n\\;mod\\;p_n q_n\$ where \$p_1, p_2, ..., p_n, q_1, q_2, ..., q_n\$ are distinct primes. The \$n\$ consecutive numbers \$x+1, x+2, ..., x+n\$ each have at least two prime factors, so none of them can be expressed as an integral power of a prime.", "OG, If \$n=1\$, then select any composite number \$Claim:\$ If \$n>1\$, then the consectutive number \$(2n+3)!+2, (2n+3)!+3 \\cdots (2n+3)!+(n+1)\$, give the desired \$n\$ consecutive numbers as each of the numbers are divisible by at least 2 primes. Proof: Each of the numbers are \$(2n+3)!+k, 2 \\leq k \\leq n+1\$, Now since \$2k \\leq (2n+3)! \\implies k^2\|(2n+3)\$, Thus \$(2n+3)!+k= k(kq+1)\$, which clearly has atleast 2 prime divisors. Hence DONE, OG" ]
IMO-1989-6	https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_6	A permutation $\{x_1, \ldots, x_{2n}\}$ of the set$\{1,2, \ldots, 2n\}$ where $\mbox{n}$ is a positive integer, is said to have property $T$ if $\|x_i - x_{i + 1}\| = n$ for at least one $i \in \{1,2, \ldots, 2n - 1\}$. Show that, for each $\mbox{n}$, there are more permutations with property $T$ than without.	[ "So for the specific case when \$n = 2\$.\n\nWe have the set \$\\{1,2,3,4\\}\$\n\nTo satisfy the condition, the 2 numbers must be adjacent and we can have either \$(1,3), (3,1), (2,4), (4,2)\$ where \$(x,y)\$ represents an adjacent pair.\n\nTo find those that satisfy \$T\$ we need to find: \$(1,3) \\cup (3,1) \\cup (2,4) \\cup (4,2)\$\n\nUsing PIE we can find those that doesn't satisfy \$T\$ \$\\overline{(1,3) \\cup (3,1) \\cup (2,4) \\cup (4,2)} = \\overline{(1,3)} \\cap \\overline{(3,1)} \\cap \\overline{(2,4)} \\cap \\overline{(4,2)}\$\n\nLet \$Q = \|\\overline{(1,3)} \\cap \\overline{(3,1)} \\cap \\overline{(2,4)} \\cap \\overline{(4,2)}\|\$\n\nDefining an indicator function \$I_y(x)\$ where \$y \\in \\{(1,3), (3,1), (2,4), (4,2)\\}\$ with domain \$x \\in \\mathbb{U}\$ such that \$\\mathbb{U}\$ contains all sets \$(x,y)\$\n\n\\[\nI_y(x) = \\begin{cases} 0, & \\mbox{if} \\ \\mbox{x} \\ \\notin \\ \\mbox{y} \\\\ 1, & \\mbox{if} \\ \\mbox{x} \\ \\in \\ \\mbox{y} \\end{cases}\n\\]\n\n\\[\nQ = \\sum_{x \\in \\mathbb{U}} I_{\\overline{(1,3)}}(x) I_{\\overline{(3,1)}}(x)I_{\\overline{(2,4)}}(x)I_{\\overline{(4,2)}}(x)\n\\]\n\n\\[\nQ = \\sum_{x \\in \\mathbb{U}} (1-I_{(1,3)}(x))(1-I_{(3,1)}(x))(1-I_{(2,4)}(x))(1-I_{(4,2)}(x))\n\\]\n\n\\[\nQ = \\sum_{x \\in \\mathbb{U}} 1 - \\sum_{x \\in \\mathbb{U}}I_{(1,3)}(x)+I_{(3,1)}(x)+I_{(2,4)}(x)+I_{(4,2)}(x)+\\sum_{x \\in \\mathbb{U}} I_{(1,3)}(x)I_{(3,1)}(x) + I_{(1,3)}(x)I_{(2,4)}(x)+I_{(1,3)}(x)I_{(4,2)}(x) + I_{(3,1)}(x)I_{(2,4)}(x) +I_{(3,1)}(x)I_{(4,2)}(x)+I_{(2,4)}(x)I_{(4,2)}(x)-\\sum_{x \\in \\mathbb{U}}I_{(1,3)}(x)I_{(3,1)}(x)I_{(2,4)}(x)+I_{(1,3)}(x)I_{(2,4)}(x)I_{(4,2)}(x)+I_{(3,1)}(x)I_{(2,4)}(x)I_{(4,2)}(x)+I_{(1,3)}(x)I_{(3,1)}(x)I_{(4,2)}(x) + \\sum_{x \\in \\mathbb{U}}I_{(1,3)}(x)I_{(3,1)}(x)I_{(2,4)}(x)I_{(4,2)}(x)\n\\]\n\nNow to work out the cardinality of each consider the set \$\\{(1,3), (3,1)\\}\$ and \$\\{(2,4), (4,2)\\}\$\n\nThe first sum is obvious: \$\\sum_{x \\in \\mathbb{U}} 1 = 4!\$\n\nThe second sum is also pretty obvious: \$\\sum_{x \\in \\mathbb{U}}I_{(1,3)}(x)+I_{(3,1)}(x)+I_{(2,4)}(x)+I_{(4,2)}(x) = 4!\$\n\nThe third sum is not so obvious since we have terms that equal 0, eg, \$I_{(1,3)}(x)I_{(3,1)}(x) = 0\$.\n\nThus we need to pick any 2 pairs from the 1st set and any 2 pairs from the 2nd set.\n\nSo there are \$2 \\times 2 = 4\$ non-zero pairs. Each pair however has 2 ways to rearrange. So the third sum equals \$2 \\times 4 =8\$\n\nThe fourth sum is 0 since we need 3 sets but we only have 2 to pick from.\n\nThe fifth sum is also 0 by the same argument as above.\n\nNow we can generalise.\n\nConsider the set \$\\{1,2,3, \\cdots, 2n\\}\$\n\nLet \$(1,1+n), (1+n,1), (2,2+n), (2+n,2), \\cdots, (n,2n),(2n,n)\$ represent the adjacent pairs. There are a total of 2n pairs.\n\nTo find those that satisfy T we need to find \$(1,1+n) \\cup (1+n,1) \\cup (2,2+n) \\cup (2+n,2) \\cup \\cdots \\cup (n,2n) \\cup (2n,n)\$\n\nUsing PIE we can find the complement of T: \$\\overline{(1,1+n) \\cup (1+n,1) \\cup (2,2+n) \\cup (2+n,2) \\cup \\cdots \\cup (n,2n) \\cup (2n,n)} = \\overline{(1,1+n)} \\cap \\overline{(1+n,1)} \\cap \\overline{(2,2+n)} \\cap \\overline{(2+n,2)} \\cap \\cdots \\cap \\overline{(n,2n)} \\cap \\overline{(2n,n)}\$\n\nLet \$Q_n = \|\\overline{(1,1+n)} \\cap \\overline{(1+n,1)} \\cap \\overline{(2,2+n)} \\cap \\overline{(2+n,2)} \\cap \\cdots \\cap \\overline{(n,2n)} \\cap \\overline{(2n,n)}\|\$\n\nNow we define an indicator function \$I_y(x)\$ where \$y \\in \\{(1,1+n), (1+n,1), (2,2+n), (2+n,2), \\cdots, (n,2n),(2n,n)\\}\$\n\n\\[\nI_y(x) = \\begin{cases} 0, & \\mbox{if} \\ \\mbox{x} \\ \\notin \\ \\mbox{y} \\\\ 1, & \\mbox{if} \\ \\mbox{x} \\ \\in \\ \\mbox{y} \\end{cases}\n\\]\n\n\\[\nQ_n = \\sum_{x \\in \\mathbb{U}} (1-I_{(1,1+n)}(x))(1-I_{(1+n,1)}(x))(1-I_{(2,2+n)}(x))(1-I_{(2+n,2)}(x)) \\cdots (1-I_{(n,2n)}(x))(1-I_{(2n,n)}(x))\n\\]\n\n\\[\nQ_n = \\sum_{x \\in \\mathbb{U}} 1 - \\sum_{x \\in \\mathbb{U}} I_{(1,1+n)}(x)+I_{(1+n,1)}(x)+ \\cdots + I_{(n,2n)}(x) + I_{(2n,n)}(x) + \\sum_{x \\in \\mathbb{U}} I_{(1,1+n)}(x)I_{(1+n,1)}(x)+I_{(1,1+n)}(x)I_{(2,2+n)}(x)+ \\cdots + I_{(1+n,n)}(x)I_{(2n,n)}(x) \\cdots - \\sum_{x \\in \\mathbb{U}}I_{(1,1+n)}(x)I_{(1+n,1)}(x)I_{(2,2+n)}(x) \\cdots+\\sum_{x \\in \\mathbb{U}} (I_{(1,1+n)}(x))(I_{(1+n,1)}(x))(I_{(2,2+n)}(x))(I_{(2+n,2)}(x)) \\cdots (I_{(n,2n)}(x))(I_{(2n,n)}(x))\n\\]\n\nThe first sum equals \$(2n)!\$\n\nThe second sum equals \$\\binom{2n}{1}(2n-1)!\$\n\nThe third sum is a bit tricky since some pairs equal 0, thus consider all the different pairs placed into sets like this:\n\n\\[\n\\{(1,1+n), (1+n,1)\\}, \\{(2,2+n), (2+n,2)\\}, \\{(3,3+n), (3+n,n)\\}, \\cdots, \\{(n,2n), (2n,n)\\}\n\\]\n\nWe need 2 pairs, since there are \$\\mbox{n}\$ sets, we need to pick 2 sets first \$\\implies \\binom{n}{2}\$. But each set contains 2 terms, thus we can have \$2^2\$ different pairings for each 2 sets.\n\nTherefore this sum equals \$2^2\\binom{n}{2}(2n-2)!\$\n\nThe fourth sum is equal to \$2^3\\binom{n}{3}(2n-3)!\$\n\nThe fifth sum is equal to \$2^4\\binom{n}{4}(2n-4)!\$\n\n. . .\n\nThe last sum is equal to \$2^{2n}\\binom{n}{2n}(2n-2n)!\$\n\nIn total we have \$Q_n = (2n)!-(2n)! + 2^2\\binom{n}{2}(2n-2)! - 2^3\\binom{n}{3}(2n-3)! + 2^4\\binom{n}{4}(2n-4)! - \\cdots + 2^{2n}\\binom{n}{2n}(2n-2n)!\$\n\n\\[\nQ_n = \\sum_{i=2}^{n}(-1)^{i}2^i\\binom{n}{i}(2n-i)!\n\\]\n\nSo that means there are a total of \$Q_n\$ sets which does not satisfy \$T\$.\n\nNow we just have to prove that the number of sets that satisfy \$T\$ is larger than those that don't.\n\nThe number of sets that satisfies \$T\$ is equal to \$(2n)!-Q_n\$.\n\nSo we need to prove \$(2n)!-Q_n > Q_n\$\n\n\\[\n\\implies (2n)! > 2Q_n\n\\]\n\nFirst let \$t_i\$ represent \$\|(-1)^{i}2^i\\binom{n}{i}(2n-i)!\|\$\n\n\\[\n\\therefore Q_n = t_2-t_3+t_4-t_5+ \\cdots -t_{2n-1} + t_{2n}\n\\]\n\nWe see that \$2t_2 = (-1)^{2}2^3\\binom{n}{2}(2n-2)! = (2^2)\\left(\\frac{n!}{(n-2)!2!}\\right)(2n-2)! = (2^3)\\left(\\frac{n(n-1)}{2}\\right)(2n-2)! = 2n(2n-2)(2n-2)!\$\n\nBut \$(2n)! = (2n)(2n-1)(2n-2)!\$\n\n\\[\n(2n)(2n-1)(2n-2)! > 2n(2n-2)(2n-2)!\n\\]\n\n\\[\n\\implies (2n)! > 2t_2\n\\]\n\nNext take \$t_3\$ and \$t_4\$ for example.\n\n\$t_3\$ means at least 3 pairs satisfy T and \$t_4\$ means at least 4 pairs satisfy \$T\$.\n\nBut at least 4 pairs is a subset of at least 3 pairs which means \$t_4 \\subset t_3 \\implies \|t_3\| \\ge \|t_4\|\$\n\nGeneralising this leads to \$t_i \\ge t_{i+1}\$\n\nSo \$2Q_n = 2t_2 + 2(t_4-t_3) + \\cdots 2(t_{2n} - t_{2n-1})\$\n\n\\[\n\\{(t_4-t_3), \\cdots, (t_{2n}-t_{2n-1})\\} \\le 0\n\\]\n\n\\[\n\\implies 2t_2 + 2(t_4-t_3) + \\cdots 2(t_{2n} - t_{2n-1}) \\le 2t_2 < (2n)!\n\\]\n\n\\[\n\\therefore 2Q_n < (2n)!\n\\]\n\n\\[\n\\mbox{Q.E.D}\n\\]", "Let \$F\$ be the number of permutations with property \$T\$, and \$F_i\$ be the set of permutations \$\\{ x_1, x_2, \\ldots, x_{2n} \\}\$ such that \$\|x_i - x_{i+1}\| = n\$. By the inclusion-exclusion principle,\n\n\\[\n\|F\| = \\bigg\|\\bigcup_{i=1}^n F_i \\bigg\| = \\sum_{i=1}^{2n-1} \|F_i\| - \\sum_{1\\leq i<j \\leq 2n-1} \|F_i \\cap F_j\| + E \\tag{1}\n\\]\n\nfor some \$E \\geq 0\$. Let’s calculate the first two sums on the far right.\n\nFor any \$i\$, \$\|F_i\| = 2n \\cdot (2n - 2)!\$, since there are are \$2n\$ choices for \$x_i\$, which fixes \$x_{i+1}\$, and \$(2n-2)!\$ choices for the remaining elements. Thus\n\n\\[\n\\sum_{i=1}^{2n - 1}\|F_i\| = 2n \\cdot (2n - 2)! \\cdot (2n - 1) = (2n)!\n\\]\n\nNow let’s compute \$\|F_i \\cap F_j\|\$, where \$1 \\leq i < j \\leq 2n - 1\$. It’s not possible that \$\|x_i - x_{i+1}\| = \|x_{i+1}- x_{i+2}\| = n\$, since that would imply \$x_i = x_{i+2}\$. So \$\|F_i \\cap F_j\| = 0\$ if \$j = i + 1\$. If \$j > i + 1\$, then \$\|F_i \\cap F_j\| = 2n \\cdot (2n - 2) \\cdot (2n - 4)!\$ (there are \$2n\$ choices for \$x_i, x_{i+1}\$, \$2n - 2\$ for \$x_j, x_{j+1}\$, and \$(2n - 4)!\$ for everything else.)\n\nHow often is \$\|F_i \\cap F_j\| \\neq 0\$? We know \$i\$ is at least \$1\$ and at most \$2n - 3\$, and for any value of \$i\$, there are \$2n - 2 - i\$ possible values for \$j\$. So the number of non-empty intersections \$\|F_i \\cap F_j\|\$ is\n\n\\[\n\\sum_{i=1}^{2n-3} 2n - 2 - i = \\sum_{i=1}^{2n-3}i = \\frac{(2n-3)(2n-2)}{2}.\n\\]\n\nNow we can compute\n\n\\[\n\\begin{align} \\sum_{1\\leq i < j \\leq 2n-1} \|F_i \\cap F_j\| &= 2n \\cdot (2n - 2) \\cdot (2n - 4)! \\cdot \\frac{(2n - 3)(2n - 2)}{2} \\\\ &= \\frac{2n \\cdot (2n-2)^2 \\cdot (2n - 3)!}{2} = \\frac{(2n)!}{2} \\cdot \\frac{2n - 2}{2n - 1} \\\\ &< \\frac{(2n)!}{2}. \\end{align}\n\\]\n\nLet’s plug this into \$(1)\$:\n\n\\[\n\|F\| = \\sum_{i=1}^{2n - 1} \|F_i\| - \\sum_{1\\leq i < j \\leq 2n-1} \|F_i \\cap F_j\| + E > (2n)! - \\frac{(2n)!}{2} + E \\geq \\frac{(2n)!}{2}.\n\\]\n\nSo more than half of the \$(2n)!\$ permutations of \$\\{1, \\ldots, 2n\\}\$ have property \$T\$.", "Let \$A_n\$ be the set of permutations of \$1,2, \\ldots,2n\$ not having property \$T\$ and let \$B_n\$ be the set of permutations of \$1,2, \\ldots,2n\$ having exactly one value of \$k\\geq2\$ such that \$\|x_k-x_{k+1}\|=n\$. We will prove a 1-1 correspondence between \$A_n\$ and \$B_n\$.\n\nConsider any permutation in \$A_n\$. Now just take \$x_1\$ and for the unique value \$k\$ such that \$\|x_1-x_k\|=n\$, put \$x_1\$ in the spot right before \$x_k\$. This will give us a permutation that belongs in the set \$B_n\$. Now consider a permutation in \$B_n\$. If you take \$x_k\$ and put it as the first term, you'll get a permutation that belongs in set \$A_n\$.Therefore, we've proved a 1-1 correspondence between the two.\n\nBecause \$\|A_n\|\$ is clearly less than the number of permutations with property \$T\$, we finally have that the number of permutations with property \$T\$ is greater than the number of permutations without \$T\$.\n\n~BennettHuang" ]
IMO-1990-1	https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_1	Chords $AB$ and $CD$ of a circle intersect at a point $E$ inside the circle. Let $M$ be an interior point of the segment $\overline{EB}$. The tangent line at $E$ to the circle through $D, E$, and $M$ intersects the lines $\overline{BC}$ and ${AC}$ at $F$ and $G$, respectively. If $\frac{AM}{AB} = t$, find $\frac{EG}{EF}$ in terms of $t$.	[ "With simple angle chasing, we find that triangles \$CEG\$ and \$BMD\$ are similar.\n\nso, \$\\frac{MB}{EC} = \\frac{MD}{EG}\$. .... \$(1)\$\n\nAgain with simple angle chasing, we find that triangles \$CEF\$ and \$AMD\$ are similar.\n\nso, \$\\frac{MA}{EC} = \\frac{MD}{EF}\$. .... \$(2)\$\n\nso, by \$(1)\$ and \$(2)\$, we have \$\\frac{EG}{EF} = \\frac{MA}{MB} = \\frac{t}{1-t}\$.\n\nThis solution was posted and copyrighted by e.lopes. The original thread for this problem can be found here: [1]", "This problem can be bashed with PoP and Ratio Lemma. Rewriting the given ratio gets \$\\frac{MA}{MB}=\\frac{t}{1-t}\$. By Ratio Lemma, \$\\frac{FB}{FC}=\\frac{BE}{CE} \\cdot \\frac{\\sin{\\angle{FEB}}}{\\sin{\\angle{FEC}}}=\\frac{DE}{AE} \\cdot \\frac{\\sin{\\angle{EDM}}}{\\sin{\\angle{DME}}}=\\frac{DE}{AE} \\cdot \\frac{EM}{DE}=\\frac{ME}{EA}\$. Similarly, \$\\frac{GA}{GC}=\\frac{ME}{EB}\$. We can rewrite these equalities to get \$\\frac{AM}{EM}=\\frac{BC}{FB}\$ and \$\\frac{BM}{EM}=\\frac{AC}{GC}\$. Using Ratio Lemma, \$\\frac{GE}{\\sin{\\angle{ACD}}}=\\frac{GC}{\\sin{\\angle{GED}}}\$ and \$\\frac{EF}{\\sin{\\angle{BCD}}}=\\frac{FC}{\\sin{\\angle{FEC}}}\$. Since \$\\angle{GED}=\\angle{FEC}\$, we have \$\\frac{FE}{GE}=\\frac{FC}{GC} \\cdot \\frac{\\sin{\\angle{BCD}}}{\\sin{\\angle{ACD}}}\$ (eq 1). Note that by Ratio Lemma, \$\\frac{\\sin{\\angle{BCD}}}{\\sin{\\angle{ACD}}}=\\frac{CA}{CB} \\cdot \\frac{EB}{EA}\$. Plugging this into (eq 1), we get \$\\frac{EF}{GE}=\\frac{FC}{GC} \\cdot \\frac{CA}{CB} \\cdot \\frac{EB}{EA}=\\frac{\\frac{EA}{EM} \\cdot FB}{\\frac{EB}{EM} \\cdot GA} \\cdot \\frac{CA}{CB} \\cdot \\frac{EB}{EA}=\\frac{FB}{GA} \\cdot \\frac{CA}{CB}=\\frac{EM}{AM} \\cdot \\frac{MB}{EM}=\\frac{MB}{MA}=\\frac{1-t}{t}\$. So \$\\frac{EG}{EF}=\\boxed{\\frac{t}{1-t}}\$.\n\nThis solution was posted and copyrighted by AIME12345. The original thread for this problem can be found here: [2]" ]
IMO-1990-2	https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_2	Let $n\ge3$ and consider a set $E$ of $2n-1$ distinct points on a circle. Suppose that exactly $k$ of these points are to be colored black. Such a coloring is “good” if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly $n$ points from $E$. Find the smallest value of $k$ so that every such coloring of $k$ points of $E$ is good.	[ "We form a graph on the \$2n-1\$ points by connecting two of these points iff one of the arcs they determine has exactly \$n\$ points in its interior. If \$3\\not\|2n-1\$, then this graph is a cycle, and the question becomes: what's the minimal \$k\$ s.t. whenever we color \$k\$ vertices of a cycle of length \$2n-1\$, there are two adjacent colored vertices? Obviously, the answer is \$k=n\$. On the other hand, when \$3\|2n-1\$, the graph is a disjoint union of three cycles of length \$\\frac{2n+1}3\$, and the answer will be, in this case, \$3\\cdot\\left\\lfloor\\frac{2n-1}6\\right\\rfloor+1\$.\n\nThis solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]" ]
IMO-1990-3	https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_3	Determine all integers $n > 1$ such that $\frac{2^n+1}{n^2}$ is an integer.	[ "Let \$N = \\{ n\\in\\mathbb{N} : 2^n\\equiv - 1\\pmod{n^2} \\}\$ be the set of all solutions and \$P = \\{ p\\text{ is prime} : \\exists n\\in N, p\|n \\}\$ be the set of all prime factors of the solutions.\n\nIt is clear that the smallest element of \$P\$ is 3. Assume that \$P\\ne\\{3\\}\$ and let's try to determine the second smallest element \$q = \\min (P\\setminus\\{3\\})\$.\n\nLet \$n\\in N\$ be a multiple of \$q\$. It is important to notice that \$9\\not\|n\$ (otherwise it is easy to get that any power of \$3\$ divides \$n\$, a non-sense). Therefore, \$n = 3^t n'\$ where \$t = 0\$ or \$1\$ and \$n'\$ does not have prime divisors smaller than \$q\$. Since \$2^{2n}\\equiv 1\\pmod{q}\$, the multiplicative order \$r = ord_q(2)\$ of 2 modulo \$q\$ divides \$2n\$. Moreover, \$r\$ must be even, since otherwise we would have \$2^n \\equiv 1\\pmod{q}\$, a contradiction to the required \$2^n \\equiv - 1\\pmod{q}\$. Since \$r < q\$ we must have \$r = 2\$ or \$2\\cdot 3 = 6\$. But the numbers \$2^2 - 1 = 3\$ and \$2^6 - 1 = 63\$ deliver only one new prime factor \$7\$, implying that \$q = 7\$. However, in this case \$r = ord_7(2) = 3\$, a contradiction. This contradiction proves that \$P = \\{3\\}\$ and thus \$N = \\{1,3\\}\$.\n\nThis solution was posted and copyrighted by maxal/orl. The original thread for this problem can be found here: [1]", "Let \$p\$ be the smallest prime divisor of \$n\$. It is easy to check that, \$n=3\$ is obviously solution. Let \$p^{a} \|\| n\$ . \$p \| 2^{2n}-1\$ and \$p \| 2^{p-1} - 1\$ (By the fermat's theorem), we obtain that \$p=3\$. It is also obvious that n is an odd number. Lemma: For all \$n\$ positive integers, \$2\$ is a primitive root modulo \$3^{n}\$. \$3^{2a} \| 2^{2n} - 1\$. Using the lemma, we get that \$\\phi(3^{2a}) \| 2n\$. Using the power of three, we obtain that \$3^{2a-1} \| 3^{a}\$. This is only possible when \$a \\geq 2a-1\$. So \$a=1\$. Now \$q\$ be the second smallest prime divisor of \$n\$. \$(2n,q-1) = 2 , 6\$. If this equals to 2, we get \$q=3\$ which is a contradiction.If \$(2n,q-1) = 6\$ then \$q\|63\$. We know that \$q\$ is different from \$3\$. Hence \$q\$ must be \$7\$. But this is impossible since \$2^{n} +1\\equiv 2\\mod 7\$ when \$n\$ is divisible by \$3\$. Hence the answer is \$n = 3\$.\n\nThis solution was posted and copyrighted by grumpyorum. The original thread for this problem can be found here: [2]", "Trivially \$n=1\$ is a solution. Now assume \$n\\neq 1\$ and define \$\\pi(n)\$ to be the smallest prime divisor of \$n\$. Let \$\\pi(n)=p\\neq 2\$. Then we have:\n\n\\[\n\\begin{eqnarray} p\\mid 2^n+1\\mid 2^{2n}-1 \\text{ and } p\\mid 2^{p-1}-1 \\\\ \\implies p\\mid 2^{\\gcd(2n, p-1)}-1 \\end{eqnarray}\n\\]\n\nNow if \$r\\neq 2\\mid n\$ then we can't have \$r\\mid p-1\$ because then \$r\\le p-1\$ contradiction. Therefore \$r=2\$ and since \$n\$ is odd \$\\gcd(2n, p-1)=2\$. Hence\n\n\\[\np\\mid 2^2-1 \\implies p=3\n\\]\n\nLet \$v_3(n)=k\$. By lifting the exponent we must have\n\n\\[\nv_3(2^n+1)=1+k\\ge v_3(n^2)=2k \\implies k=1\n\\]\n\nLet \$n=3n_1\$. \$n_1=1\$ is a solution (\$2^3+1=3^2=9\$). Assume \$n_1\\neq 1\$ and let \$\\pi(n_1)=q\\neq 3\$. By Chinese Remainder Theorem since \$q\\neq 3\$ we have:\n\n\\[\n\\begin{eqnarray} q\\mid 8^{n_1}+1\\mid 8^{2n_1}-1 \\text{ and } q\\mid 8^{q-1}-1 \\\\ \\implies q\\mid 8^{\\gcd(2n_1, q-1)}-1=63 \\text{ so } q=7 \\end{eqnarray}\n\\]\n\nHowever \$2^n+1\\equiv 8^{n_1}+1\\equiv 2\\pmod{7}\$ contradiction.\n\nThe solutions are henceforth \$n=1, 3\$. This solution was posted and copyrighted by binomial-theorem. The original thread for this problem can be found here: [3]", "et \$n>1\$ and \$p\$ be the smallest prime factor of \$n\$.Then \$p\|2^{2n}-1,p\|2^{p-1}-1 \\Rightarrow p\|2^{\\text{gcd}(2n,p-1)}-1=2^2-1=3\\Rightarrow \\boxed{p=3}\$. Thus \$n=3^xy\$ for some positive \$x\$ and \$y\$. Also \$n^2\|2^n+1 \\Rightarrow 2x=v_3(n^2) \\le v_3(2^n+1)=v_3(3)+v_3(n)=x+1 \\Rightarrow \\boxed{x=1}\$.\n\nThus \$n=3y\$ for some positive integer \$y\$. Now let \$y>1\$ and \$q\$ be the smallest prime divisor of \$y\$.Then we can deduce that \$q\|2^{\\text{gcd}(2n,q-1)}-1=2^{2\\text{gcd}(n,q-1)}-1=2^6-1=63=73^2 \\Rightarrow \\boxed{q=7}\$(Note that \$\\text{gcd}(n,q-1)\$ is \$1\$ or \$3\$).\n\nBut this gives \$7\|2^n+1\$ which is false as \$2^n+1\$ leaves remainders \$1,2,4\$ modulo \$7\$.\n\nThus \$y=1\$ and \$\\boxed{n=3}\$ if \$n>1\$.\n\nThus the solutions are \$\\boxed{n=1}\$ and \$\\boxed{n=3}\$\n\nThis solution was posted and copyrighted by sayantanchakraborty. The original thread for this problem can be found here: [4]\n\nMany more solutions can be found in the thread in Contest Collections." ]
IMO-1990-4	https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_4	Let $\mathbb{Q^+}$ be the set of positive rational numbers. Construct a function $f :\mathbb{Q^+}\rightarrow\mathbb{Q^+}$ such that $f(xf(y)) = \frac{f(x)}{y}$ for all $x, y\in{Q^+}$.	[ "If we let \$y = x = 1\$ this implies \$f(f(1)) = f(1)\$. Plugging \$f(1) = y\$ in with this new fact. We get \$f(f(f(1))) = \\frac{f(1)}{f(1)} = 1\$. Using \$f(f(1)) = f(1)\$ again, we see that \$f(f(f(1))) = f(f(1)) = f(1) = 1\$. Now plugging in \$x = 1\$ we get \$f(f(y)) = \\frac{1}{y}\$. Such a function can not be continuous as if \$f(x)\$ is increasing or decreasing on some observable interval, \$f(f(x))\$ will be increasing on that interval but \$\\frac{1}{x}\$ is decreasing on all positive intervals. This indicates the function is discrete which means we can assign a \"4-chain\" of \$f(a) = b\$, \$f(b) = \\frac{1}{a}\$, \$f(\\frac{1}{a}) = \\frac{1}{b}\$, and \$f(\\frac{1}{b}) = a\$. It is obvious to see that any function where \$f(f(x)) = \\frac{1}{x}\$. Must follow such a structure. The problem occurs that we do not know whether our current value of \$x\$ is a \$b\$ or an \$a\$. To make non-trivial progress on this we must split all rational numbers into two sets, \$S_1\$ and \$S_2\$, such that if \$E \\in S_1\$ then \$\\frac{1}{E} \\in S_1\$. As long as \$S_1\$ and \$S_2\$ have the same size, so the bijection \$f(\\textrm{an element from }S_1)\$ = \$(\\textrm{an element from } S_2)\$ can hold, there will exist a piece-wise function (possibly with a greater than countable infinity number of pieces) that satisfies f(x).\n\nTo find one of these functions we can limit our searching by proving that \$f(ab) = f(a)f(b)\$. To do this simply set \$x = \\frac{1}{f(y)}\$ yielding: \\newline \$f(1) = f(f(y) \\times \\frac{1}{f(y)}) = \\frac{f(\\frac{1}{f(y)})}{y} = 1\$. This means that \$f(x) = y\$. Since \$X,Y \\in Q+\$ there will always be an \$x\$ that suffices this and a corresponding \$y\$. Thus \$f(ab)\$ with \$b = f(c)\$ \$\\leftrightarrow\$ \$c = \\frac{1}{f(b)}\$ expands into \$f(af(c)) = \\frac{f(a)}{c} = f(a)f(b)\$.\n\nUtilizing \$f(ab) = f(a)f(b)\$ we can see that if we break-down every ratio into it's prime factorization of the numerator and denominator. Our \"4-chain\" is simply \$f(p) = q\$, \$f(q) = \\frac{1}{p}\$, \$f(\\frac{1}{p}) = \\frac{1}{q}\$, \$f(\\frac{1}{q}) = p\$ where \$p\$ and \$q\$ are unique primes. Thus our 2 sets \$S_1\$ and \$S_2\$ is simply 2 sets of equal size containing only primes and their inverse. A good way to do this assuming we aren't given the sequence of all infinite primes (as this would require an explicit formula that does not yet exist) is to split the primes into 2 and 1 mod 6 being in one set with 3 and 5 mod 6 in the other. These sets have equal sizes from a strong form of Dirichlet's theorem on arithmetic progressions.\n\nThus we have created our two sets and to show that this actually works we will let \$x = P_1^{E_{1x}} \\times P_2^{E_{2x}} \\times P_3^{E_{3x}} \\times \\dots\$ where \$P_i\$ is a prime and \$E_{ix}\$ are integers. Define \$y\$ similarly. For the sake of simplicity if \$i\$ is odd then \$P_i \\in S_1\$ and if \$i\$ is even \$P_i \\in S_2\$. Where our \"4-chain\" is \$f(p_n) = p_{n+1}\$, \$f(p_{n+1}) = \\frac{1}{p_n}\$, \$f(\\frac{1}{p_n}) = \\frac{1}{p_{n+1}}\$, \$f(\\frac{1}{p_{n+1}}) = p_n\$ with \$n\$ being odd. Then \$f(y) = P_2^{E_{1y}} \\times P_1^{-E_{2y}} \\times P_4^{E_{3y}} \\times P_3^{-E_{4y}}\\dots\$ (remember \$f(ab) = f(a)f(b)\$).\n\nThen \$xf(y) = P_1^{E_{1x} - E_{2y}} \\times P_2^{E_{2x} + E_{1y}} \\times P_3^{E_{3x} - E_{4y}} \\times \\dots\$\n\n\\[\nf(xf(y)) = P_2^{E_{1x} - E_{2y}} \\times P_1^{-E_{2x} - E_{1y}} \\times P_4^{E_{3x} - E_{4y}} \\times P_3^{-E_{4x} - E_{3y}} \\dots\n\\]\n\n\\[\nf(x) = P_2^{E_{1x}} \\times P_1^{-E_{2x}} \\times P_4^{E_{3x}} \\times P_3^{-E_{4x}}\\dots\n\\]\n\n\\[\n\\frac{f(x)}{y} = P_2^{E_{1x}- E_{2y}} \\times P_1^{-E_{2x} - E_{1y}} \\times P_4^{E_{3x} - E_{4y}} \\times P_3^{-E_{4x} - E_{3y}}\\dots\n\\]\n\nThus \$f(xf(y)) = \\frac{f(x)}{y}\$ QED", "We show first that f(1) = 1. Taking x = y = 1, we have f(f(1)) = f(1). Hence f(1) = f(f(1)) = f(1 f(f(1)) ) = f(1)/f(1) = 1.\n\nNext we show that f(xy) = f(x)f(y). For any y we have 1 = f(1) = f(1/f(y) f(y)) = f(1/f(y))/y, so if z = 1/f(y) then f(z) = y. Hence f(xy) = f(xf(z)) = f(x)/z = f(x) f(y).\n\nFinally, f(f(x)) = f(1 f(x)) = f(1)/x = 1/x.\n\nWe are not required to find all functions, just one. So divide the primes into two infinite sets S = {p1, p2, ... } and T= {q1, q2, ... }. Define f(pn) = qn, and f(qn) = 1/pn. We extend this definition to all rationals using f(xy) = f(x) f(y): f(pi1pi2...qj1qj2.../(pk1...qm1...)) = pm1...qi1.../(pj1...qk1...). It is now trivial to verify that f(x f(y)) = f(x)/y." ]
IMO-1990-5	https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_5	Given an initial integer $n_{0}\textgreater1$, two players, $\mathbb{A}$ and $\mathbb{B}$, choose integers $n_{1}, n_{2},n_{3}$, . . . alternately according to the following rules: Knowing $n_{2k}$, $\mathbb{A}$ chooses any integer $n_{2k+1}$ such that \[ n_{2k}\leq n_{2k+1}\leq n_{2k}^2 \] . Knowing $n_{2k+1}$, $\mathbb{B}$ chooses any integer $n_{2k+2}$ such that \[ \frac{n_{2k+1}}{n_{2k+2}} \] is a prime raised to a positive integer power. Player $\mathbb{A}$ wins the game by choosing the number 1990; player $\mathbb{B}$ wins by choosing the number 1. For which $n_{0}$ does: (a) $\mathbb{A}$ have a winning strategy? (b) $\mathbb{B}$ have a winning strategy? (c) Neither player have a winning strategy?	[ "If \$n_0<6\$ it is clear that B wins, because A can only choose numbers with at most 2 prime factors (30 is the smallest with three and \$30>5^2\$), which B can either make smaller (choose the lesser of the two) or the number contains one prime factor in which case B wins. Once \$n_{2k}=2\$, A can only choose prime powers, so B wins.\n\nIf \$n_0=6,7\$ it's a draw, because the only numbers that give A a chance of winning are those with three prime factors (for reasons discussed above). Less than 49, this gives 30 and 42 as the only choices which won't cause A to lose. However, if A chooses 30, B may choose 6, and we have a stalemate. If A chooses 42, B may choose 6, and we still have a stalemate.\n\nIf \$n_0 \\geq 8\$, if \$n_0\$ is less than 45 by choosing (perhaps starting later) 345, then 357, then 2357, and then 3457 A can force B to make \$n_{2k} \\geq 60\$ for some k.\n\nIf \$45 \\leq n_{2k} \\leq 1990\$ for some k, then A wins (he may pick 1990). If \$n_{2k}>1990\$ pick the least natural value of m such that \$2^m 47\\geq n_{2k}\$. The best B can do is half it, taking it to \$n_{2k+2} = 2^{m-1} 47 < n_{2k}\$. However, B cannot make \$n_{2k+2}\$ drop below 45 because \$47*32<1990 \\Rightarrow 2^m \\geq 64\$. Thus we will get a descending sequence of integers until we end up with \$45 \\leq n_{2k} \\leq 1990\$, at which point A wins.\n\nThis solution was posted and copyrighted by Ilthigore. The original thread for this problem can be found here: [1]" ]
IMO-1990-6	https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_6	Prove that there exists a convex $1990\text{-gon}$ with the following two properties: (a) All angles are equal. (b) The lengths of the $1990$ sides are the numbers $1^2, 2^2,3^2,\dots, 1990^2$ in some order.	[ "Let \$\\{a_1,a_2,\\cdots,a_{1990}\\}=\\{1,2,\\cdots,1990\\}\$ and \$\\theta=\\frac{\\pi}{995}\$. Then the problem is equivalent to that there exists a way to assign \$a_1,a_2,\\cdots,a_{1990}\$ such that \$\\sum_{i=1}^{1990}a_i^2e^{i\\theta}=0\$. Note that \$e^{i\\theta}+e^{i\\theta+\\pi}=0\$, then if we can find a sequence \$\\{b_n\\}\$ such that \$b_i=a_p^2-a_q^2, (p\\not=q)\$ and \$\\sum_{i=1}^{995}b_ie^{i\\theta}=0\$, the problem will be solved. Note that \$2^2-1^2=3,4^2-3^2=7,\\cdots,1990^2-1989^2=3979\$, then \$3,7,\\cdots,3979\$ can be written as a sum from two elements in sets \$A=\\{3,3+4\\times199,3+4\\times398,3+4\\times597,3+4\\times796\\}\$ and \$B=\\{0,4\\times1,4\\times2,\\cdots,4\\times 198\\}\$. If we assign the elements in \$A\$ in the way that \$l_{a1}=3,l_{a2}=3+4\\times199,l_{a3}=3+4\\times398,l_{a4}=3+4\\times597,l_{a5}=3+4\\times796\$, then clearly \$\\sum_{i=1}^{199}l_{aj}e^{i\\frac{2\\pi}{199}}=0, (j\\in\\{1,2,3,4,5\\})\$ Similarly, we could assign elements in \$B\$ in that way (\$l_{b1}=0,l_{b2}=4,\\cdots\$) to \$e^{i\\frac{2\\pi}{5}}\$. Then we make \$b_i\$ according to the previous steps. Let \$i\\equiv j\$ (mod 5), \$i\\equiv k\$ (mod 199), and \$b_i=l_{aj}+l_{bk}\$, then each \$b_i\$ will be some \$a_ p^2-a_q^2\$ and \$\\sum_{i=1}^{995}b_ie^{i\\theta}=\\sum_{j=1}^5\\sum_{i=1}^{199}l_{aj}e^{i\\frac{2\\pi}{199}}+\\sum_{k=1}^{199}\\sum_{i=1}^{5}l_{bk}e^{i\\frac{2\\pi}{5}}=0\$ And we are done.\n\nThis solution was posted and copyrighted by YCHU. The original thread for this problem can be found here: [1]", "Throughout this solution, \$\\omega\$ denotes a primitive \$995\$th root of unity.\n\nWe first commit to placing \$1^2\$ and \$2^2\$ on opposite sides, \$3^2\$ and \$4^2\$ on opposite sides, etc. Since \$2^2-1^2=3\$, \$4^2-3^2=7\$, \$6^2-5^2=11\$, etc., this means the desired conclusion is equivalent to\n\n\\[\n0 = \\sum_{n=0}^{994} c_n \\omega^n\n\\]\n\nbeing true for some permutation \$(c_0, \\dots, c_{994})\$ of \$(3, 7, 11, \\dots, 3981)\$.\n\nDefine \$z = 3 \\omega^0 + 7 \\omega^{199} + 11\\omega^{398} + 15 \\omega^{597} + 19 \\omega^{796}\$. Then notice that\n\n\\[\n\\begin{align} \tz &= 3 \\omega^0 + 7 \\omega^{199} + 11\\omega^{398} + 15 \\omega^{597} + 19 \\omega^{796} \\\\ \t\\omega^5 z &= 23 \\omega^5 + 27 \\omega^{204} + 31\\omega^{403} + 35 \\omega^{602} + 39 \\omega^{801} \\\\ \t\\omega^{10} z &= 43 \\omega^{10} + 47 \\omega^{209} + 51\\omega^{408} + 55 \\omega^{607} + 59 \\omega^{806} \\\\ \t&\\vdotswithin{=} \\end{align}\n\\]\n\nand so summing yields the desired conclusion, as the left-hand side becomes\n\n\\[\n(1+\\omega^5+\\omega^{10}+\\cdots+\\omega^{990})z = 0\n\\]\n\nand the right-hand side is the desired expression.\n\nThis solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]" ]
IMO-1991-1	https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_1	Given a triangle $ABC$ let $I$ be the center of its inscribed circle. The internal bisectors of the angles $A,B,C$ meet the opposite sides in $A^\prime,B^\prime,C^\prime$ respectively. Prove that \[ \frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27} \]	[ "We have \$\\prod\\frac{AI}{AA^\\prime}=\\prod\\frac{1}{1+\\frac{IA^\\prime}{IA}}\$. From Van Aubel's Theorem, we have \$\\frac{IA}{IA^\\prime}=\\frac{AB^\\prime}{B^\\prime C}+\\frac{AC^\\prime}{C^\\prime B}\$ which from the Angle Bisector Theorem reduces to \$\\frac{b+c}{a}\$. We find similar expressions for the other terms in the product so that the product simplifies to \$\\prod\\frac{1}{1+\\frac{a}{b+c}}=\\prod\\frac{b+c}{a+b+c}\$. Letting \$a=x+y,b=y+z,c=z+x\$ for positive reals \$x,y,z\$, the product becomes \$\\frac{1}{8}\\prod\\frac{x+2y+z}{x+y+z}=\\frac{1}{8}\\prod\\left(1+\\frac{y}{x+y+z}\\right)\$. To prove the right side of the inequality, we simply apply AM-GM to the product to get\n\n\\[\n\\frac{1}{8}\\prod\\left(1+\\frac{y}{x+y+z}\\right)\\le\\frac{1}{8}\\left(\\frac{\\sum1+\\frac{y}{x+y+z}}{3}\\right)^3=\\frac{8}{27}\n\\]\n\nTo prove the left side of the inequality, simply multiply out the product to get\n\n\\[\n\\frac{1}{8}\\left(1+\\sum\\frac{y}{x+y+z}+\\sum\\frac{yz}{(x+y+z)^2}+\\frac{xyz}{(x+y+z)^3}\\right)\n\\]\n\n\\[\n>\\frac{1}{8}\\left(1+\\sum\\frac{y}{x+y+z}\\right)=\\frac{1}{4}\n\\]\n\nas desired.\n\nRemark: To prove the right side of the inequality, a quicker way might be to use Gergonne's and AM-GM." ]
IMO-1991-2	https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_2	Let $\,n > 6\,$ be an integer and $\,a_{1},a_{2},\cdots ,a_{k}\,$ be all the natural numbers less than $n$ and relatively prime to $n$. If \[ a_{2} - a_{1} = a_{3} - a_{2} = \cdots = a_{k} - a_{k - 1} > 0, \] prove that $\,n\,$ must be either a prime number or a power of $\,2$.	[ "We use Bertrand's Postulate: for \$u\\ge 2\$ is a positive integer, there is a prime in the interval \$(u,2u)\$.\n\nClearly, \$a_2\$ must be equal to the smallest prime \$p\$ which does not divide \$n\$. If \$p=2\$, then \$n\$ is a prime since the common difference \$a_{i+1}-a_i\$ is equal to \$1\$, i.e. all positive integers less than \$n\$ are coprime to \$n\$. If, on the ther hand, \$p=3\$, we find \$n\$ to be a power of \$2\$: the positive integers less than \$n\$ and coprime to it are precisely the odd ones. We may thus assume that \$p\\ge 5\$. Furthermore, since \$n>6\$, the positive integers less than \$n\$ and coprime to it cannot be \$a_1,a_2\$ alone, i.e. \$k=\\varphi(n)\\ge 3\$.\n\nBy Bertrand's Postulate, the largest prime less than \$p\$ is strictly larger than \$\\frac{p-1}2\$, so it cannot divide \$p-1\$. We will denote this prime by \$q\$. We know that \$q\|n\$. It's easy to check that for \$p\\le 31\$ there is a prime \$r\$ strictly between \$a_2=p,\\ a_3=2p-1\$. \$rq\|n\$, so, in particular, \$pq<rq\\le n\$. On the other hand, if \$p>32\$, the two largest primes \$r_1<r_2\$ which are smaller than \$q\$ satisfy \$r_2\\ge\\frac p4,\\ r_1\\ge\\frac{r_2}2\\ge\\frac p8\$ (again, Bertrand's Postulate), so \$pq<\\frac{p^2}{32}q\\le r_1r_2q\\le n\$ so, once again, we have \$pq<n\$ (this is what I wanted to prove here).\n\nNow, \$q\\not\|p-1\$, and all the numbers \$1,1+p-1,\\ldots,1+(q-1)(p-1)\$ are smaller than \$n\$ (they are smaller than \$pq\$, which is smaller than \$n\$, according to the paragraph above). One of these numbers, however, must be divisible by \$q\|n\$. Since our hypothesis tells us that all of them must be coprime to \$n\$, we have a contradiction.\n\nThis solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]\n\n## Note\n\nBertrand actually allows for the bounds \$(a,2a-2)\$ for \$a>3\$; thus, since \$a_2\$ is the smallest prime not dividing \$n\$, \$a_3\$ must equal \$2a_2-1\$. Hence there cannot exist a prime in the bounds \$(a_2,2a_2-1)\$, clearly untrue for \$a_2>3\$ by Bertrand. Then \$a_2=2\$ is clearly not possible, and \$a_2=3\$ does not work (unless \$n\$ is a power of \$2\$); hence the conclusion holds.\n\n~eevee9406", "Let \$d=a_{i+1}-a_i\$. Then we have \$a_i=1_1+d(i-1)=1+d(i-1)\$. Since \$\\gcd(n-1,n)=1\$, so \$n-1\$ is the largest positive integer smaller than \$n\$ and co-prime to \$n\$. Thus, \$a_k=n-1\$ and \$1+d(k-1)=n-1\$ which shows \$d(k-1)=n-2\$ and so \$d\$ divides \$n-2\$. We make two cases. Case 1: \$n\$ is odd. Then \$\\gcd(n-2,n)=1\$. So, every divisor of \$n-2\$ is co-prime to \$n\$ too, so is \$d\$. Thus, there is a \$j\$ such that \$d=a_j=1+d(j-1)\$ which implies \$d\|1\$. Therefore, in this case, \$d=1\$ and we easily find that all numbers less than \$n\$ are co-prime to \$n\$, so \$n\$ must be a prime. Case 2: \$n\$ is even. But then \$\\gcd(n-2,n)=2\$ and also \$d=1\$ is ruled out. So \$d>1\$. Say, \$n=2l\$. Then, \$d\$divides \$2(l-1)\$. If \$d\$ is odd, \$d\|l\$ and \$d\|l-1\$ forcing \$d\|1\$, contradiction! So, \$d=2s\$ with \$s\|l-1\$. If \$l=2r+1\$ for some \$r\$, then \$\\gcd(r,n)=1\$. So, again \$r=1+d(j-1\$ for a \$j\$. If \$s\$ is odd, then \$s\|r\$ and thus, \$s\|1\$. Similarly, we find that actually \$d\|2\$ must occur. We finally have, \$d=2\$. But then all odd numbers less than \$n\$ are co-prime to \$n\$. So, \$n\$ does not have any odd factor i.e. \$n=2^k\$ for some \$k\\in\\mathbb N\$.\n\nThis solution was posted and copyrighted by ngv. The original thread for this problem can be found here: [2]", "We use Bonse's Inequality: \$p_{m+1}^2 < p_1p_2\\cdots p_m\$, for all \$m \\ge 4\$, \$p_1 = 2\$.\n\nLet \$p\$ denote the smallest prime which does not divide \$n\$.\n\nCase 1) \$p = 2\$. This implies \$n\$ is a prime. Case 2) \$p = 3\$. This implies \$n = 2^k\$ for some \$k \\in \\mathbb{N}\$. Case 3) \$p = 5\$. Since \$n > 6\$, we have that \$n > 9\$. Therefore \$a_3 = 9\$, but \$\\gcd (n, 9) > 1\$. Case 4) \$p \\ge 7\$. Write \$n = p_1^{\\alpha_1}p_2^{\\alpha_2}\\cdots p_k^{\\alpha_k}\$. We have \$(p - 2)^2 < p_1p_2\\cdots p_k \\le n = p_1^{\\alpha_1}p_2^{\\alpha_2}\\cdots p_k^{\\alpha_k}\$. Hence \$(p-2)^2 = a_i\$, a contradiction since \$\\gcd(p-2, n) > 1\$.\n\nThis solution was posted and copyrighted by SFScoreLow. The original thread for this problem can be found here: [3]", "Clearly, \$a_1 = 1\$. Now let \$p\$ be the smallest prime number which is not a divisor of \$n\$. Hence, \$a_2 = p\$, and the common difference \$d = p-1\$. Observe that since \$a_k = n-1\$, we have that\n\n\\[\nd \\mid a_k - a_1\n\\]\n\n, or\n\n\\[\np - 1 \\mid a_k - a_1 = n-2\n\\]\n\n. This means that all prime factors of \$p-1\$ are prime factors of \$n-2\$. However, due to the minimality of \$p\$, all primes factors of \$p-1\$ are prime factors of \$n\$. By the Euclidean algorithm, the only prime factors of \$p-1\$ are \$2\$, so \$p = 2^{2^k} + 1\$ or \$p = 2\$, by the theory of Fermat primes. We will now do casework on \$p\$.\n\nCase 1: \$p = 2\$ This case is relatively easy. \$a_2 = 2\$, so all numbers less than \$n\$ must be relatively prime to \$n\$. However, if \$n = pq\$, where \$p > 1\$ and \$q > 1\$, observe that \$p\$ is not relatively prime to \$n\$, so \$n\$ must be prime for this case to work.\n\nCase 2: \$k = 0\$ and \$p = 3\$. This case is somewhat tricky. Observe that \$n\$ is even. Since \$a_2 = 3\$, we have that all numbers less than \$n\$ which are odd are relatively prime to \$n\$. However, if \$n\$ has a odd divisor \$p > 1\$, clearly \$p\$ is not relatively prime to \$n\$, so \$n\$ must be a power of two.\n\nCase 3: \$k \\ge 1\$. Note that \$3\$ must be a divisor of \$n\$, so \$a_k\$ cannot be a multiple of \$3\$ for any \$k\$. I will contradict this statement. First, I claim that \$k = \\phi(n) > 2\$ for all integers \$n > 6\$. \$\\phi{n} = 1\$ is trivial, so let me show that \$\\phi(n) = 2\$ has only solutions less than or equal to \$6\$. Since \$\\phi\$ is a multiplicative function, observe that \$n\$ cannot be a multiple of a prime greater than \$3\$, because for such primes \$p\$ we have that \$\\phi(p^k) = (p-1)p^{k-1} \\ge p-1 > 2\$, which is contradiction. Hence, \$n\$ is just composed of factors of \$2\$ and \$3\$. If it divides both, and is of the form \$2^m 3^n\$ for \$m\$ and \$n\$ positive, we get that we need \$2^{m} 3^{n-1} = 2\$, so \$n = 1\$ and \$m = 1\$ in this case. If it is of the form \$n = 2^k\$, we get \$k = 2\$. And if it is of the form \$3^k\$, we get \$k = 1\$. \$3\$, \$4\$, and \$6\$ are all less than or equal to \$6\$, so \$\\phi(n) > 2\$. Now we're almost done. Since \$k > 2\$, the number \$a_3 = 2p-1\$ must be in this sequence. However, \$a_3 = 2p-1\$ is of the form \$2^{2^n+1} + 1\$, which is a multiple of \$3\$, and we have contradiction, so we are done.\n\nThis solution was posted and copyrighted by va2010. The original thread for this problem can be found here: [4]", "Clearly that \$a_1=1\$. Let \$a_{i+1}-a_i = d \\; (1 \\le i \\le \\varphi(n)-1)\$ then\n\n\\[\na_{i}=(i-1)d+1 \\; \\; (1 \\le i \\le \\varphi(n)). \\qquad (1)\n\\]\n\nWe can see that if \$n\$ is a prime then \$n\$ satisfies the condition. We consider the case when \$n\$ is a composite number. Let \$p\$ be the smallest prime divisor of \$n\$ then \$p \\le \\sqrt{n}\$. Note that \$\\varphi (n) > \\sqrt{n} \\ge p\$ for all prime \$n>6\$ since \$p_i^{k_i}-p_i^{k_i-1} \\ge p_i^{k_i/2}\$ for all \$p \\ge 3\$. Therefore, if \$p \\nmid d\$ then there will exists \$1 \\le i \\le \\varphi(n)\$ such that \$(i-1)d \\equiv -1 \\pmod{p}\$ or \$p \\mid a_i\$, a contradiction since \$\\gcd (a_i,n)=1\$. Thus, \$p \\mid d\$. We consider two cases:\n\nIf \$d \\ge 2p\$ then \$p+1 \\ne a_i \\; (1 \\le i \\le \\varphi(n))\$. Hence, \$\\gcd (p+1,n)>1\$. This means there exists a prime \$q>p \\ge 2\$ such that \$q \\mid p+1\$. Since \$q>p\$ so we get \$q=p+1\$ or \$p=2,q=3\$, a contradiction since \$n>6\$. Thus, \$d <2p\$ or \$d=p\$.\n\nIf \$p=d=2\$ then all odd numbers \$a_i \\le n\$ are relatively prime to \$n\$. This can only happen when \$n=2^x \\; (x \\in \\mathbb{N})\$.\n\nIf \$p=d >3\$ then \$p+3 \\ne a_i \\; \\forall 1 \\le i \\le \\varphi(n)\$ since \$a_i \\equiv 1 \\pmod{p} \\; \\forall 1 \\le i \\le \\varphi(n)\$. Thus, \$\\gcd (p+3,n)>1\$ or there exist a prime \$q>p>3\$ such that \$q \\mid p+3\$. Note that \$2 \\mid p+3\$ so \$q \\le \\tfrac{p+3}{2}<p\$, a contradiction.\n\nThus, \$n\$ must be either a prime number or a power of \$2\$. \$\\blacksquare\$\n\nThis solution was posted and copyrighted by shinichiman. The original thread for this problem can be found here: [5]\n\nSolution 6\n\nWe begin firstly by showing that if \$c\$ be the number of numbers smaller than and coprime to a number \$n\$, such that \$n > 2\$, then \$c\$ is even. This is very easily visible; if \$q\$ is coprime to \$n\$, then \$n-q\$ is also coprime to \$n\$. We see that all the coprime numbers are kind of symmetric around \$\\frac{n}{2}\$, and for \$n > 2\$, \$\\frac{n}{2}\$ is never coprime with \$n\$, and it is easy to see that \$c\$ is always even for \$n > 2\$.\n\nThis question essentially asks us that for \$n > 6\$, the numbers smaller than and coprime to n never form an arithmetic progression unless n is a power of two or n is prime.\n\nNow, we break this question into three scenarios:\n\nScenario 1: \$n\$ is odd. If a number is odd, then it is coprime with \$2\$. All numbers are coprime with \$1\$. So, the numbers smaller than and coprime to \$n\$ begin with 1, 2, .. If this is truly an arithemtic sequence up to \$n\$, then that means that \$n\$ is coprime to all number smaller than itself. But, if \$n\$ is not prime, then it has a prime factor smaller than itself, and the arithmetic sequence would break, as that prime factor won't be included in the coprime numbers. So, if \$n\$ is odd, and the numbers smaller than and coprime to \$n\$ form an arithemtic sequence, then \$n\$ is a prime.\n\nScenario 2: \$n\$ is divisible by \$4\$. It seems like we have nothing to argue here. So, we do something else here; we assume that the numbers smaller than \$n\$ form an arithemtic sequence like \$1\$, \$1+d\$, ... , \$1+(c-1)d\$ (where \$c\$ is the number of numbers smaller than and coprime to \$n\$), and sum them. The sum is \$c + \\frac{c(c-1)(d)}{2}\$. If you remember the part where we proved that all these numbers are symmetric around \$\\frac{n}{2}\$, we can see that we can sum these numbers another way, and find their sum to be \$\\frac{c(n)}{2}\$. So, \$c + \\frac{c(c-1)(d)}{2}\$ = \$\\frac{c(n)}{2}\$, which means \$2 = n - d(c-1)\$. \$n\$ here is even, \$c-1\$ is odd, so \$d\$ must be even too, otherwise the LHS will not be even. Also, gcd(\$n\$, \$d\$) = \$2\$, and as \$n\$ is divisible by \$4\$, it means that \$d\$ is only divisible by \$2\$, and not \$4\$, because then gcd(\$n\$, \$d\$) = \$4\$, and the LHS would be a multiple of \$4\$, which it is not. Before going on further, we'll see another very important thing, that \$d\$ is made up of prime factors that all divide \$n\$, the proof of which is very trivial. But, if there is another prime which divides \$n\$ and \$d\$, then that would mean that gcd(\$n\$, \$d\$) != 2, which is false, thus \$d\$ must be divisible by \$2\$, and not \$4\$ or any other prime number, which just means \$d = 2\$. So, Our arithmmetic progression is like \$1\$, \$3\$, \$5\$, .. , \$n-1\$, implying that \$n\$ is coprime to all the odd numbers smaller than it. But, if \$n\$ has some odd prime factors, then that would mean that that odd prime is not coprime to \$n\$, a contradiction, so, \$n\$ is only made up of powers of \$2\$, that is, \$n\$ is a power of two greater than or equal to \$4\$. The problem with \$4\$ is that it is only coprime to two numbers smaller than itself, which obviously form an arithemtic progression. But, as \$n > 6\$ here, we are on the safe side.\n\nScenario 3: \$n\$ is divisible by \$2\$, but not \$4\$. Let \$n\$ = \$2x\$, where \$x\$ is odd. Before making anymore arguments, we see that \$x-2\$ and \$x+2\$ are consecutive coprime numbers to \$2x\$. So, the arithemtic progression must have difference \$4\$. Our number \$n\$ must be coprime to some prefix of the sequence \$1\$, \$5\$, \$9\$, ..., and only that prefix. The moment \$n > 6\$, \$n\$ must be coprime to \$9\$ (it's easy to see why), and thus also coprime to \$3\$, which breaks this progression. Thus, for \$n > 6\$, we can see that no number of the form \$2x\$ satisfies the conditions of the problem. The proof is done Courtesy of EobardThawne" ]
IMO-1991-3	https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_3	Let $S = \{1,2,3,\cdots ,280\}$. Find the smallest integer $n$ such that each $n$-element subset of $S$ contains five numbers which are pairwise relatively prime.	[ "Let's look at this another way. We aim to find the least number of integers we can remove from S to leave a set which does not contain 5 pairwise coprime integers.\n\nLet \$S_p = \\{p^k \| p^k \\leq 280\\}\$ And more generally \$S_{p_1,p_2,...,p_l} = \\{n = p_1^{q_1}p_2^{q_2}....p_l^{q_l}\| n \\leq 280 \\}\$ And finally \$S_1 = \\{1\\}\$ It is clear that these sets taken over all collections of primes and 1 forms a partition of S.\n\nIf there are representatives from five sets \$S_p\$, there are five mutually coprime integers, so all but four of the sets \$S_p\$ must be completely removed. It is clear that (with the exception of 1, which can clearly be removed first) if \$p>q, \|S_p\| \\leq \|S_q\|\$ so it is worth removing \$S_p\$ before removing \$S_q\$ (if we are striving for a minimum). Furthermore, if we remove two sets \$S_p, S_q\$, we must also (to stop there from being 5 mutually coprime integers) remove \$S_{p,q}\$ and so on, and these have similar ordering relations. So once we have removed everything we need to keeping sizes of sets removed to a minimum, we are left with \$S_2,S_3,S_5,S_7\$ and the multi-index sets not coprime to all of these. In other words, we have the set of multiples of 2,3,5 and 7, which can be calculated (by lots of inclusion-exclusion) to have cardinality \$216\$. Therefore, a subset of \$S\$ of size \$217\$ must contain 5 coprime integers.\n\nThis solution was posted and copyrighted by Ilthigore. The original thread for this problem can be found here: [1] Firstly note that \$\\lfloor\\sqrt{280}\\rfloor = 16\$. Therefore, we only need to check primes \$p\\le 16\$ to check if a number from \$1\$ to \$280\$ is prime.", "I claim the answer is \$\\boxed{217}\$. First, let's look at a subset of \$S\$ that consists only of multiples of \$2,3,5,7\$. It is easy to calculate by PIE that the total number of elements in this set is\n\n\\[\n\\left\\lfloor\\tfrac{280}{2}\\right\\rfloor+\\left\\lfloor\\tfrac{280}{7}\\right\\rfloor+\\left\\lfloor\\tfrac{280}{3}\\right\\rfloor+\\left\\lfloor\\tfrac{280}{5}\\right\\rfloor-\\left\\lfloor\\tfrac{280}{14}\\right\\rfloor-\\left\\lfloor\\tfrac{280}{6}\\right\\rfloor-\\left\\lfloor\\tfrac{280}{10}\\right\\rfloor-\\left\\lfloor\\tfrac{280}{15}\\right\\rfloor-\\left\\lfloor\\tfrac{280}{21}\\right\\rfloor-\\left\\lfloor\\tfrac{280}{35}\\right\\rfloor+\\left\\lfloor\\tfrac{280}{70}\\right\\rfloor+\\left\\lfloor\\tfrac{280}{105}\\right\\rfloor+\\left\\lfloor\\tfrac{280}{42}\\right\\rfloor+\\left\\lfloor\\tfrac{280}{30}\\right\\rfloor-\\left\\lfloor\\tfrac{280}{210}\\right\\rfloor=216.\n\\]\n\nNow, note that no matter what \$5\$ element subset we consider of this subset we just made, consisting of multiples of \$2,3,5,7\$, we will always have at least \$2\$ of those elements that exist in the subset that share a factor greater than \$1\$ by the Pigeonhole Principle since \$\\left\\lfloor\\tfrac{5}{4}\\right\\rfloor+1=2\$. Thus, we know \$n>216\$ from this. We also know that there are \$212\$ composite numbers and \$4\$ prime numbers (obviously just \$2,3,5,7\$) in this subset of \$S\$. Now we do casework on the numbers \$11,13\$ (we don't have to check higher numbers because remember \$p\\le 16\$!) to find the other composite numbers. We find that all of numbers, \$121,143,169,187,209,221,247,253\$ fills in the rest of the composite numbers from \$1\$ to \$280\$. This gives a total of \$8\$ numbers when counted. So we can now count up the total as \$220\$ composite numbers and the remaining must be prime so \$59\$ prime numbers (\$1\$ not included as prime or composite). Now suppose \$X\$ is a subset of our set \$S\$ such that \$\|X\|=217\$ (\$\|X\|\$ represents its cardinality or the number of elements in the subset \$X\$). Also suppose that there is no \$5\$ element subset of \$X\$ such that all \$5\$ elements are relatively prime to each other. This means we would have to have at most \$4\$ prime numbers and at least \$213\$ composite numbers to make this subset \$X\$. This means the set \$S-X\$ (everything that is in \$S\$ but not in \$X\$) has at most \$220-213=7\$ composite numbers. Now consider the following \$8\$ sets and notice that at least one of these must be a subset of \$X\$!:\n\n\\[\n\\{2^2,3^2,5^2,7^2,13^2\\}\n\\]\n\n\\[\n\\{2\\times31,3\\times29,5\\times23,7\\times19,11\\times17\\}\n\\]\n\n\\[\n\\{2\\times47,3\\times43,5\\times41,7\\times37,13\\times19\\}\n\\]\n\n\\[\n\\{2\\times41,3\\times37,5\\times31,7\\times29,11\\times23\\}\n\\]\n\n\\[\n\\{2\\times23,3\\times19,5\\times17,7\\times13,11^2\\}\n\\]\n\n\\[\n\\{2\\times43,3\\times41,5\\times37,7\\times31,13\\times17\\}\n\\]\n\n\\[\n\\{2\\times37,3\\times31,5\\times29,7\\times23,11\\times19\\}\n\\]\n\n\\[\n\\{2\\times29,3\\times23,5\\times19,7\\times17,11\\times13\\}\n\\]\n\nAnd obviously each of these sets has \$5\$ elements that are all relatively prime numbers, as desired.\n\nThis solution was posted and copyrighted by Wave-Particle. The original thread for this problem can be found here: [2]" ]
IMO-1991-4	https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_4	Suppose $\,G\,$ is a connected graph with $\,k\,$ edges. Prove that it is possible to label the edges $1,2,\ldots ,k\,$ in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1.	[ "We provide an algorithmic approach for labelling the edges: We start off with the set \$V\$ of all vertices in \$G\$. For each step in the process we remove a number of vertices from the set \$V\$ if their exists an edge sorrounding it which is labelled. Additionally, we shall use up the numbers \$1,2,...,k\$ in the order from least to greatest.\n\nHere is the process: Let \$u\$ be the least number not used yet. Take the longest path that uses any vertex once formed solely by vertices from \$V\$ and call it path \$P=(v_1,v_2,...,v_n)\$. If \$v_1\$ has more than one edge, then since \$P\$ is maximal, it must be connected to some other edge \$v_0\$ which could be one of the vertices \$v_i\$ or one of the vertices not in \$V\$. Similarly, if \$v_n\$ has more than one edge, it must be connected to a vertex \$v_{n+1}\$ which is either one of the \$v_i\$ or one of the vertices not in \$V\$. Notice that since \$v_1,v_n\\in V\$ the edge connecting \$v_0\$ and \$v_1\$ and the edge connecting \$v_n\$ and \$v_{n+1}\$ are not labelled. Now, if \$v_0\$ exists, we shall label the edge joining \$v_i\$ and \$v_{i+1}\$ with the number \$u+i\$. If \$v_0\$ does not exist then we shall label the edge joining \$v_i\$ and \$v_{i+1}\$ with the number \$u+i-1\$.\n\nIf \$v_0\$ or \$v_{n+1}\$ does not exist, then the vertex they should have been connected to has degree 1, and so the labels of the edges sorrounding it do not matter. Otherwise and the case of any other vertex in the path, our process garuntees that the vertex will be sorrounded by two edges labelled by 2 consecutive numbers. Thus, that vertex will be sorrounded by edges whose greates common factor is 1, no matter how the other edges sorrounding it are labelled. Notice that from our definition of \$V\$, the edges \$v_1,v_2,...,v_n\$ are removed from \$V\$. Thus, it is clear that all vertices not in \$V\$, at any time, have edges sorrounding them that are relatively prime.\n\nWe repeat the process described until the only vertices in \$V\$ are those which are not connected to any other vertices in \$V\$. Let the least number we have not yet used by \$u\$. If any of them are of degree 1, we do not care. If there is a vertex \$v\$ of degree at least 2, then we simply label 2 of its edges \$u\$ and \$u+1\$ and now its edges are relatively prime, and we can remove it from \$V\$. Notice that this doesn't remove any other vertices from \$V\$ since \$v\$ is not connected to any vertices from \$V\$. We continue like this for the other vertices in \$V\$.\n\nNow, all the vertices in \$V\$ are those whose edges do not matter, while all vertices not in \$V\$ are those with edges relatively prime. Thus we have the desired configuration, and we simply distribute the remaining numbers in any way we wish." ]
IMO-1991-5	https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_5	Let $\,ABC\,$ be a triangle and $\,P\,$ an interior point of $\,ABC\,$. Show that at least one of the angles $\,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $30^{\circ }$.	[ "Let \$A_{1}\$ , \$A_{2}\$, and \$A_{3}\$ be \$\\angle CAB\$, \$\\angle ABC\$, \$\\angle BCA\$, respectively.\n\nLet \$\\alpha_{1}\$ , \$\\alpha_{2}\$, and \$\\alpha_{3}\$ be \$\\angle PAB\$, \$\\angle PBC\$, \$\\angle PCA\$, respcetively.\n\nUsing law of sines on \$\\Delta PAB\$ we get: \$\\frac{\\left\| PA \\right\|}{sin(A_{2}-\\alpha_{2})}=\\frac{\\left\| PB \\right\|}{sin(\\alpha_{1})}\$, therefore, \$\\frac{\\left\| PA \\right\|}{\\left\| PB \\right\|}=\\frac{sin(A_{2}-\\alpha_{2})}{sin(\\alpha_{1})}\$\n\nUsing law of sines on \$\\Delta PBC\$ we get: \$\\frac{\\left\| PB \\right\|}{sin(A_{3}-\\alpha_{3})}=\\frac{\\left\| PC \\right\|}{sin(\\alpha_{2})}\$, therefore, \$\\frac{\\left\| PB \\right\|}{\\left\| PC \\right\|}=\\frac{sin(A_{3}-\\alpha_{3})}{sin(\\alpha_{2})}\$\n\nUsing law of sines on \$\\Delta PCA\$ we get: \$\\frac{\\left\| PC \\right\|}{sin(A_{1}-\\alpha_{1})}=\\frac{\\left\| PA \\right\|}{sin(\\alpha_{3})}\$, therefore, \$\\frac{\\left\| PC \\right\|}{\\left\| PA \\right\|}=\\frac{sin(A_{1}-\\alpha_{1})}{sin(\\alpha_{3})}\$\n\nMultiply all three equations we get: \$\\frac{\\left\| PA \\right\|}{\\left\| PB \\right\|}\\frac{\\left\| PB \\right\|}{\\left\| PC \\right\|}\\frac{\\left\| PC \\right\|}{\\left\| PA \\right\|}=\\frac{sin(A_{2}-\\alpha_{2})}{sin(\\alpha_{1})}\\frac{sin(A_{3}-\\alpha_{3})}{sin(\\alpha_{2})}\\frac{sin(A_{1}-\\alpha_{1})}{sin(\\alpha_{3})}\$\n\n\\[\n1=\\frac{sin(A_{2}-\\alpha_{2})}{sin(\\alpha_{1})}\\frac{sin(A_{3}-\\alpha_{3})}{sin(\\alpha_{2})}\\frac{sin(A_{1}-\\alpha_{1})}{sin(\\alpha_{3})}\n\\]\n\n\\[\n\\prod_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}=1\n\\]\n\nUsing AM-GM we get:\n\n\\[\n\\frac{1}{3}\\sum_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}\\ge \\sqrt[3]{\\prod_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}}\n\\]\n\n\\[\n\\frac{1}{3}\\sum_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}\\ge 1\n\\]\n\n\$\\sum_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}\\ge 3\$. [Inequality 1]\n\n\\[\n\\sum_{i=1}^{3}\\frac{sin(A_{i})cos(\\alpha_{i})-cos(A_{i})sin(\\alpha_{i})}{sin(\\alpha_{i})}\\ge 3\n\\]\n\n\\[\n\\sum_{i=1}^{3}\\left[ sin(A_{i})cot(\\alpha_{i})-cos(A_{i})\\right]\\ge 3\n\\]\n\nNote that for \$0<\\alpha_{i}<180^{\\circ}\$, \$cot(\\alpha_{i})\$ decreases with increasing \$\\alpha_{i}\$ and fixed \$A_{i}\$\n\nTherefore, \$\\left[ sin(A_{i})cot(\\alpha_{i})-cos(A_{i})\\right]\$ decreases with increasing \$\\alpha_{i}\$ and fixed \$A_{i}\$\n\nFrom trigonometric identity:\n\n\$sin(x)+sin(y)=2sin\\left( \\frac{x+y}{2} \\right)cos\\left( \\frac{x-y}{2} \\right)\$,\n\nsince \$-1\\le cos\\left( \\frac{x-y}{2} \\right) \\le 1\$, then:\n\n\\[\nsin(x)+sin(y) \\le 2sin\\left( \\frac{x+y}{2} \\right)\n\\]\n\nTherefore,\n\n\\[\nsin(A_{1}-30^{\\circ})+sin(A_{2}-30^{\\circ}) \\le 2sin\\left( \\frac{A_{1}+A_{2}-60^{\\circ}}{2} \\right)\n\\]\n\nand also,\n\n\\[\nsin(A_{3}-30^{\\circ})+sin(30^{\\circ}) \\le 2sin\\left( \\frac{A_{3}}{2} \\right)\n\\]\n\nAdding these two inequalities we get:\n\n\\[\nsin(30^{\\circ})+\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le 2\\left[ sin\\left( \\frac{A_{1}+A_{2}-60^{\\circ}}{2} \\right)+sin\\left( \\frac{A_{3}}{2} \\right) \\right]\n\\]\n\n\\[\n\\frac{1}{2}+\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le 2\\left[ 2sin\\left( \\frac{A_{1}+A_{2}+A_{3}-60^{\\circ}}{4} \\right) \\right]\n\\]\n\n\\[\n\\frac{1}{2}+\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le 2\\left[ 2sin\\left( \\frac{180^{\\circ}-60^{\\circ}}{4} \\right) \\right]\n\\]\n\n\\[\n\\frac{1}{2}+\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le 4sin\\left( 30^{\\circ} \\right)\n\\]\n\n\\[\n\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le \\frac{3}{2}\n\\]\n\n\$2\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le 3\$.\n\n\$\\sum_{i=1}^{3}\\frac{sin(A_{i}-30^{\\circ})}{sin(30^{\\circ})}\\le 3\$. [Inequality 2]\n\nCombining [Inequality 1] and [Inequality 2] we see the following:\n\n\\[\n\\sum_{i=1}^{3}\\frac{sin(A_{i}-30^{\\circ})}{sin(30^{\\circ})}\\le \\sum_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}\n\\]\n\nThis implies that for at least one of the values of \$i=1\$,\$2\$,or \$3\$, the following is true:\n\n\\[\n\\frac{sin(A_{i}-30^{\\circ})}{sin(30^{\\circ})}\\le \\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}\n\\]\n\nor\n\n\\[\n\\frac{sin(\\alpha_{i})}{sin(A_{i}-\\alpha_{i})}\\le \\frac{sin(30^{\\circ})}{sin(A_{i}-30^{\\circ})}\n\\]\n\nWhich means that for at least one of the values of \$i=1\$,\$2\$,or \$3\$, the following is true:\n\n\\[\n\\alpha_{i} \\le 30^{\\circ}\n\\]\n\nTherefore, at least one of the angles \$\\,\\angle PAB,\\;\\angle PBC,\\;\\angle PCA\\,\$ is less than or equal to \$30^{\\circ }\$.\n\n~Tomas Diaz, orders@tomasdiaz.com", "At least one of \$\\angle ABC, \\angle BCA, \\angle CAB \\ge 60^\\circ\$. Without loss of generality, assume that \$\\angle BCA \\ge 60^\\circ\$\n\nIf \$\\angle PAB > 30^\\circ\$ and \$\\angle PBC > 30^\\circ\$\n\nDraw a circle \$R\$ centered at \$O\$ and passing through \$A, P, B\$. Since \$P\$ is an interior point of \$\\triangle ABC\$, thus \$C\$ is outside the circle \$R\$\n\nDraw two lines \$CD, CE\$ passing through \$C\$ and tangent to \$R\$. Line \$CD\$ intersect \$R\$ at \$D\$, and line \$CE\$ intersect \$R\$ at \$E\$. Choose \$D\$ near \$A\$, and choose \$E\$ near \$B\$\n\nExtends line \$BC\$, and intersect \$R\$ at \$F\$ other than \$B\$ when \$BC\$ is not tangent to \$R\$. If \$BC\$ is tangent to \$R\$, we have \$B = E\$ be the tangent point, and simply let \$F = B = E\$\n\nDraw the segment \$OE\$, and choose a point \$G\$ on \$R\$ such that \$\\angle GOE = 60^\\circ\$. There are two possible points, we choose \$G\$ near point \$P\$. Draw segments \$OG, GE\$, thus \$\\triangle GOE\$ is an equilateral triangle\n\nDraw segments \$OP, OC, OB, OF, PB, GC\$\n\n\$\\angle OCE = \\dfrac{1}{2} \\angle DCE \\ge \\dfrac{1}{2} \\angle BCA \\ge 30^\\circ\$. Then we have \$\\angle COE = 90^\\circ - \\angle OCE \\le 60^\\circ = \\angle GOE\$\n\n\$\\angle POB = 2 \\angle PAB > 60^\\circ, \\angle POF = 2 \\angle PBC > 60^\\circ\$, since we have either \$\\angle POE \\ge \\angle POB\$ or \$\\angle POE \\ge \\angle POF\$, thus \$\\angle POE > 60^\\circ = \\angle GOE\$\n\nThus we have \$\\angle COE \\le \\angle GOE < \\angle POE\$, then \$\\angle OCE \\le \\angle GCE < \\angle PCE\$\n\nBecause \$\\angle GCE \\ge \\angle OCE \\ge 30^\\circ = \\angle GEC\$, thus \$GC \\le GE = OG\$, and \$\\angle GCO \\ge \\angle GOC\$\n\nFinally, \$\\angle PCA = \\angle ACE - \\angle PCE < \\angle ACE - \\angle GCE = \\angle ACO - \\angle GCO\$\n\nSince \$\\angle ACO \\le \\angle DCO\$, and \$\\angle GCO \\ge \\angle GOC\$, thus we have \$\\angle PCA < \\angle ACO - \\angle GCO \\le \\angle DCO - \\angle GOC = 90^\\circ - \\angle COE - \\angle GOC = 90^\\circ - \\angle GOE = 30^\\circ\$\n\nWe have proved that when \$\\angle PAB > 30^\\circ\$ and \$\\angle PBC > 30^\\circ\$, the angle \$\\angle PCA\$ must be less than \$30^\\circ\$. Thus at least one of \$\\angle PAB, \\angle PBC, \\angle PCA\$ should less than or equal to \$30^\\circ\$\n\n~Joseph Tsai, mgtsai@gmail.com" ]
IMO-1991-6	https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_6	An infinite sequence $x_0, x_1, x_2,\ldots$ of real numbers is said to be bounded if there is a constant $C$ such that $\|x_i\| \leq C$ for every $i \geq 0$. Given any real number $a > 1$, construct a bounded infinite sequence $x_0,x_1,x_2,\ldots$ such that $\|x_i-x_j\|\cdot \|i-j\|^a\geq 1$ for every pair of distinct nonnegative integers $i,j$.	[ "Since \$a>1\$, the series \$\\sum_{k=1}^\\infty\\frac{1}{k^a}\$ is convergent; let \$L\$ be the sum of this convergent series. Let \$I\\subset \\mathbb{R}\$ be the interval \$[-L,L]\$ (or any bounded subset of measure \$\\geq 2L\$).\n\nSuppose that we have chosen points \$x_0,x_1,\\ldots,x_{m-1}\$ satisfying\n\n\\[\n\\qquad(\\ast)\\quad \|x_i-x_j\|\\cdot \|i-j\|^a\\geq 1\n\\]\n\nfor all distinct \$i,j<m\$. We show that we can choose \$x_m\\in I\$ such that \$(\\ast)\$ holds for all distinct \$i,j\\leq m\$. The only new cases are when one number (WLOG \$i\$) is equal to \$m\$, so we must guarantee that \$\|x_m-x_j\|\\cdot \|m-j\|^a\\geq 1\$ for all \$0\\leq j<m\$.\n\nLet \$U_j\$ be the interval \$(x_j-\\frac{1}{\|m-j\|^a},x_j+\\frac{1}{\|m-j\|^a})\$, of length \$\\frac{2}{\|m-j\|^a}\$. The points that are valid choices for \$x_m\$ are precisely the points of \$I\\setminus(U_0\\cup \\cdots \\cup U_{m-1})\$, so we must show that this set is nonempty. The total length \$\\mu(U_0\\cup \\cdot\\cup U_{m-1})\$ is at most the sum of the lengths \$\\mu(U_0)+\\cdots+\\mu(U_{m-1})=\\frac{2}{m^a}+\\frac{2}{(m-1)^a}+\\cdots+\\frac{2}{2^a}+\\frac{2}{1^a}\$. This is \$2\\sum_{k=1}^m\\frac{1}{k^a}<2\\sum_{k=1}^\\infty \\frac{1}{k^a}=2L\$.\n\nTherefore the total measure of \$U_0\\cup \\cdots \\cup U_{m-1}\$ is \$<2L=\\mu(I)\$, so \$I\\setminus(U_0\\cup \\cdots \\cup U_{m-1})\$ has positive measure and thus is nonempty. Choosing any \$x_m\\in I\\setminus(U_0\\cup \\cdots \\cup U_{m-1})\$ and continuing by induction constructs the desired sequence.", "The argument above would not work for \$a=1\$, since \${\\textstyle\\sum \\frac{1}{k^a}}\$ only converges for \$a>1\$. But Osmun Nal argues in this video that \$x_k=C\\cdot\\left(k\\sqrt{2}-\\lfloor k\\sqrt{2}\\rfloor\\right), C=2\\sqrt2+1\$ satisfies the stronger inequality \$\|x_i-x_j\|\\cdot \|i-j\|\\geq 1\$ for all distinct \$i,j\$; in other words, this sequence simultaneously solves the problem for all \$a\\geq 1\$ simultaneously.", "Let \$n = \\max \\left( \\left\\lceil \\dfrac{1}{a - 1} \\right\\rceil, 3 \\right)\$, then we have \$\\dfrac{\\ln \\left( n + 1 \\right)}{\\ln n} = \\dfrac{\\ln n + \\ln \\left( 1 + 1/n \\right)}{\\ln n} < \\dfrac{\\ln n + 1/n}{\\ln n} = 1 + \\dfrac{1}{n \\ln n} < 1 + \\dfrac{1}{n} \\le a\$\n\nThus \$n^a > n + 1\$\n\nFor any non-negative integer \$i\$, represent it as base-\$n\$ positional notation:\n\n\$i = i_m n^m + i_{m-1} n^{m-1} + \\cdots + i_2 n^2 + i_1 n + i_0\$, where \$i_m, i_{m-1}, \\cdots, i_2, i_1, i_0 \\in \\left\\{ 0, 1, 2, \\cdots, n - 1 \\right\\}\$\n\nWe directly construct \$x_i = n \\left( i_0 + i_1 \\left( n + 1 \\right)^{-1} + i_2 \\left( n + 1 \\right)^{-2} + \\cdots + i_m \\left( n + 1 \\right)^{-m} \\right)\$\n\nWe allow leading zeros for representing \$i\$ due to the leading zeros do not effect the value of \$x_i\$\n\nThen \$0 \\le x_i \\le n \\left( \\left( n - 1 \\right) + \\left( n - 1 \\right) \\left( n + 1 \\right)^{-1} + \\cdots + \\left( n - 1 \\right) \\left( n + 1 \\right)^{-m} \\right) < n \\left( n - 1 \\right) \\dfrac{1}{1 - \\left( n + 1 \\right)^{-1}} = n^2 - 1\$\n\nThus \$\\left\| x_i \\right\| < n^2 - 1\$, which is bounded\n\nFor any two distinct non-negative integers \$i \\neq j\$, represent them as base-\$n\$ positional notation:\n\n\\[\ni = i_m n^m + i_{m-1} n^{m-1} + \\cdots + i_2 n^2 + i_1 n + i_0\n\\]\n\n\\[\nj = j_m n^m + j_{m-1} n^{m-1} + \\cdots + j_2 n^2 + j_1 n + j_0\n\\]\n\nwhere \$i_m, i_{m-1}, \\cdots, i_2, i_1, i_0, j_m, j_{m-1}, \\cdots, j_2, j_1, j_0 \\in \\left\\{ 0, 1, 2, \\cdots, n - 1 \\right\\}\$\n\nLet \$\\ell\$ be the minimum number such that \$i_\\ell \\neq j_\\ell\$, that is, \$i_0 = j_0, i_1 = j_1, \\cdots, i_{\\ell - 1} = j_{\\ell - 1}, i_\\ell \\neq j_\\ell\$\n\nThen we have \$\\left\| i - j \\right\| \\ge n^\\ell\$\n\nWithout loss of generality, assume that \$i_\\ell > j_\\ell\$, then we have \$i_\\ell - j_\\ell \\ge 1, j_{\\ell + 1} - i_{\\ell + 1}, j_{\\ell + 2} - i_{\\ell + 2}, \\cdots, j_m - i_m \\le n - 1\$\n\n\\[\n\\left\| x_i - x_j \\right\| = n \\left( \\left( i_\\ell - j_\\ell \\right) \\left( n + 1 \\right)^{-\\ell} - \\left( j_{\\ell+1} - i_{\\ell+1} \\right) \\left( n + 1 \\right)^{-\\ell-1} - \\left( j_{\\ell+2} - i_{\\ell+2} \\right) \\left( n + 1 \\right)^{-\\ell-2} - \\cdots - \\left( j_m - i_m \\right) \\left( n + 1 \\right)^{-m} \\right) \\\\~\\\\ \\ge n \\left( n + 1 \\right)^{-\\ell} - n \\left( n - 1 \\right) \\left( \\left( n + 1 \\right)^{-\\ell-1} + \\left( n + 1 \\right)^{-\\ell-2} + \\cdots + \\left( n + 1 \\right)^{-m} \\right) \\\\~\\\\ > n \\left( n + 1 \\right)^{-\\ell} - n \\left( n - 1 \\right) \\left( n + 1 \\right)^{-\\ell-1} \\dfrac{1}{1 - \\left( n + 1 \\right)^{-1}} = \\left( n + 1 \\right)^{-\\ell}\n\\]\n\nThus we have \$\\left\| x_i - x_j \\right\| \\left\| i - j \\right\|^a > \\left( n + 1 \\right)^{-\\ell} n^{a \\ell} > \\left( n + 1 \\right)^{-\\ell} \\left( n + 1 \\right)^\\ell = 1\$\n\n~Joseph Tsai, mgtsai@gmail.com" ]
IMO-1992-1	https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_1	Find all integers $a$, $b$, $c$ satisfying $1 < a < b < c$ such that $(a - 1)(b -1)(c - 1)$ is a divisor of $abc - 1$.	[ "\\[\n1<\\frac{abc-1}{(a-1)(b-1)(c-1)}<\\frac{abc}{(a-1)(b-1)(c-1)}\n\\]\n\n\\[\n1<\\frac{abc-1}{(a-1)(b-1)(c-1)}<\\left(\\frac{a}{a-1}\\right) \\left(\\frac{b}{b-1}\\right) \\left(\\frac{c}{c-1}\\right)\n\\]\n\nWith \$1<a<b<c\$ it implies that \$a \\ge 2\$, \$b \\ge 3\$, \$c \\ge 4\$\n\nTherefore, \$\\frac{a}{a-1}=1+\\frac{1}{a-1}\$\n\nwhich for \$a\$ gives: \$\\frac{a}{a-1} \\le 1+\\frac{1}{2-1}\$, which gives :\$\\frac{a}{a-1} \\le 2\$\n\nfor \$b\$ gives: \$\\frac{b}{b-1} \\le 1+\\frac{1}{3-1}\$, which gives :\$\\frac{b}{b-1} \\le \\frac{3}{2}\$\n\nfor \$c\$ gives: \$\\frac{c}{c-1} \\le 1+\\frac{1}{4-1}\$, which gives :\$\\frac{c}{c-1} \\le \\frac{4}{3}\$\n\nSubstituting those inequalities into the original inequality gives:\n\n\\[\n1<\\frac{abc-1}{(a-1)(b-1)(c-1)}<\\left( 2 \\right) \\left(\\frac{3}{2}\\right) \\left(\\frac{4}{3}\\right)\n\\]\n\n\\[\n1<\\frac{abc-1}{(a-1)(b-1)(c-1)}<4\n\\]\n\nSince \$\\frac{abc-1}{(a-1)(b-1)(c-1)}\$ needs to be integer,\n\nthen \$\\frac{abc-1}{(a-1)(b-1)(c-1)}=2\$ or \$\\frac{abc-1}{(a-1)(b-1)(c-1)}=3\$\n\nCase 1: \$\\frac{abc-1}{(a-1)(b-1)(c-1)}=2\$\n\n\\[\nabc-1=2(a-1)(b-1)(c-1)=2abc-2(ab+bc+ac)+2(a+b+c)-2\n\\]\n\nCase 1, subcase \$a=2\$:\n\n\$2bc-1=2bc-2(b+c)+2\$ gives: \$2(b+c)=3\$ which has no solution because \$2(b+c)\$ is even.\n\nCase 1, subcase \$a=3\$:\n\n\\[\n3bc-1=4bc-4(b+c)+4\n\\]\n\n\\[\nbc-4b-4c+5=0\n\\]\n\n\\[\n(b-4)(c-4)=11\n\\]\n\n\$b-4=1\$ and \$c-4=11\$ provides solution \$(a,b,c)=(3,5,15)\$\n\nCase 2: \$\\frac{abc-1}{(a-1)(b-1)(c-1)}=3\$\n\n\\[\nabc-1=3(a-1)(b-1)(c-1)=3abc-3(ab+bc+ac)+3(a+b+c)-3\n\\]\n\nCase 2, subcase \$a=2\$:\n\n\\[\n2bc-1=3bc-3(b+c)+3\n\\]\n\n\\[\nbc-3b-3c+4=0\n\\]\n\n\\[\n(b-3)(c-3)=5\n\\]\n\n\$b-3=1\$ and \$c-3=5\$ provides solution \$(a,b,c)=(2,4,8)\$\n\nCase 2, subcase \$a=3\$:\n\n\\[\n3bc-1=6bc-6(b+c)+6\n\\]\n\nSince \$(3bc-1\$) mod \$3 = -1\$ and \$(6bc-6(b+c)+6)\$ mod \$3 = 0\$, then there is no solution for this subcase.\n\nNow we verify our two solutions:\n\nwhen \$(a,b,c)=(2,4,8)\$\n\n\$abc-1=(2)(4)(8)-1=63\$ and \$(a-1)(b-1)(c-1)=(1)(3)(7)=21\$\n\nSince \$21\$ is a factor of \$63\$, this solutions is correct.\n\nwhen \$(a,b,c)=(3,5,15)\$\n\n\$abc-1=(3)(5)(15)-1=224\$ and \$(a-1)(b-1)(c-1)=(2)(4)(14)=112\$\n\nSince \$112\$ is a factor of \$224\$, this solutions is also correct.\n\nThe solutions are: \$(a,b,c)=(2,4,8)\$ and \$(a,b,c)=(3,5,15)\$\n\n~ Tomas Diaz. orders@tomasdiaz.com", "Let \$x = a-1, y = b-1, z = c-1\$. So \$1 \\leq x < y < z\$. We're asked to solve\n\n\\[\n(k+1)xyz = (x+1)(y+z)(z+1)-1\n\\]\n\nwhere \$k\$ is a non-negative integer.. Dividing by \$xyz\$\n\n\\[\n\\begin{align} k+1 &= \\Big(1+\\frac{1}{x}\\Big)\\Big(1+\\frac{1}{y}\\Big)\\Big(1+\\frac{1}{z}\\Big) - \\frac{1}{xyz} \\\\ &< (1+1)\\Big(1+\\frac{1}{2}\\Big)\\Big(1+\\frac{1}{3}\\Big) = 4 \\end{align}\n\\]\n\nSo \$k=0,1\$ or \$2\$. Simplifying the first equation\n\n\\[\nkxyz = xy + yz + zy + x + y + z.\n\\]\n\n\$k=0\$ is impossible. Case \$k=2\$: if \$x \\geq 2\$, dividing the previous equation by \$xy\$\n\n\\[\n\\begin{align} 2z &= 1 + \\frac{1}{x} + \\frac{1}{y} + z\\Big(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{xy}\\Big) \\\\ &\\leq 1 + \\frac{1}{2}+\\frac{1}{3} + z\\Big(\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}\\Big) \\\\ &< 2 + z \\nonumber \\end{align}\n\\]\n\nSo \$z < 2\$ which is impossible. If \$x = 1\$ its easy to show \$y=2\$ leads to a contradiction. Solving \$(1)\$ for \$z\$\n\n\\[\nz = 2 + \\frac{5}{y-2}\n\\]\n\nwhich can only work if \$y = 3, z = 7\$. So \$x=1,y=3,z=7\$ is a solution. Case \$k=1\$: considering the version of \$(1)\$ with \$z\$ on the LHS instead of \$2z\$, if \$x = 1\$ then\n\n\\[\n\\begin{align} z &= 1 + \\frac{1}{x} + \\frac{1}{y} + z \\Big(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{xy}\\Big) \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (3) \\\\ &= 2 + \\frac{1}{y} + z\\Big(1+\\frac{1}{y}+\\frac{1}{xy}\\Big) \\end{align}\n\\]\n\nwhich is impossible. Similarly to \$(2)\$, if \$x \\geq 3\$ then\n\n\\[\nz \\leq 1 + \\frac{1}{3} + \\frac{1}{4} + z \\frac{8}{12}\n\\]\n\ni.e. \$z < 5\$ which is impossible since \$3 \\leq x < y < z\$. So \$x= 2\$. Its easy to show \$y=3\$ leads to a contradiction. Solving \$(3)\$ for \$z\$ with \$x=2\$ gives\n\n\\[\nz = 3 + \\frac{11}{y-3}\n\\]\n\nwhich can only work if \$y=4, z=14\$. So \$x=2,y=4,z=14\$ is a solution. Our two solutions give \$a = 2,b=4,c=8\$ and \$a=3,b=5,c=15\$.\n\n~not_detriti" ]
IMO-1992-2	https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_2	Let $\mathbb{R}$ denote the set of all real numbers. Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that \[ f\left( x^{2}+f(y) \right)= y+(f(x))^{2} \hspace{0.5cm} \forall x,y \in \mathbb{R} \]	[ "[quote=probability1.01]Set x = 0 to get \$f(f(y)) = y+f(0)^{2}\$. We'll let \$c = f(0)^{2}\$, so \$f(f(y)) = y+c\$. Then\n\n\$f(a^{2}+f(f(b))) = f(b)+f(a)^{2}\$ \$f(a^{2}+b+c) = f(b)+f(a)^{2}\$ \$f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}\$ \$a^{2}+b+2c = b+(a+c)^{2}\$ \$2c = c^{2}+2ac\$\n\nSince this holds for all \$a\$, it follows that \$c = 0\$. Now we have \$f(0) = 0 \\implies f(f(y)) = y\$. Note that f must be surjective since we may let y vary among all reals, and f must be injective since if \$f(a) = f(b)\$, then \$a+f(x)^{2}= f(x^{2}+f(a)) = f(x^{2}+f(b)) = b+f(x)^{2}\$. Finally, if \$u > v\$, then there is some \$t\$ s.t. \$u = t^{2}+v\$, and so \$f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)\$. Hence f is strictly increasing. It is now clear that since \$f(f(y)) = y\$, we must have \$f(x) = x\$ for all x.[/quote]" ]
IMO-1992-3	https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_3	Consider nine points in space, no four of which are coplanar. Each pair of points is joined by an edge (that is, a line segment) and each edge is either colored blue or red or left uncolored. Find the smallest value of $n$ such that whenever exactly n edges are colored, the set of colored edges necessarily contains a triangle all of whose edges have the same color.	[ "We show that for \$n = 32\$ we can find a coloring without a monochrome triangle. Take two squares \$R_1R_2R_3R_4\$ and \$B_1B_2B_3B_4\$. Leave the diagonals of each square uncolored, color the remaining edges of \$R\$ red and the remaining edges of \$B\$ blue. Color blue all the edges from the ninth point \$X\$ to the red square, and red all the edges from \$X\$ to the blue square. Color \$R_iB_j\$ red if \$i\$ and \$j\$ have the same parity and blue otherwise. Clearly \$X\$ is not the vertex of a monochrome square, because if \$XY\$ and \$XZ\$ are the same color then, \$YZ\$ is either uncolored or the opposite color. There is no triangle within the red square or the blue square, and hence no monochrome triangle. It remains to consider triangles of the form \$R_iR_jB_k\$ and \$B_iB_jR_k.\$ But if \$i\$ and \$j\$ have the same parity, then \$R_iR_j\$ is uncolored (and similarly \$B_iB_j\$), whereas if they have opposite parity, then \$R_iB_k\$ and \$R_jB_k\$ have opposite colors (and similarly \$B_iR_k\$ and \$B_jR_k\$). It remains to show that for \$n = 33\$ we can always find a monochrome triangle. There are three uncolored edges. Take a point on each of the uncolored edges. The edges between the remaining \$6\$ points must all be colored. Take one of these, \$X.\$ At least \$3\$ of the \$5\$ edges to \$X\$, say \$XA\$, \$XB\$, \$XC\$ must be the same color (say red). If \$AB\$ is also red, then \$XAB\$ is monochrome. Similarly, for \$BC\$ and \$CA.\$ But if \$AB\$, \$BC\$ and \$CA\$ are all blue, then \$ABC\$ is monochrome." ]
IMO-1992-4	https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_4	In the plane let $C$ be a circle, $l$ a line tangent to the circle $C$, and $M$ a point on $l$. Find the locus of all points $P$ with the following property: there exists two points $Q$, $R$ on $l$ such that $M$ is the midpoint of $QR$ and $C$ is the inscribed circle of triangle $PQR$.	[ "Note: This is an alternate method to what it is shown on the video. This alternate method is too long and too intensive in solving algebraic equations. A lot of steps have been shortened in this solution. The solution in the video provides a much faster solution,\n\nLet \$r\$ be the radius of the circle \$C\$.\n\nWe define a cartesian coordinate system in two dimensions with the circle center at \$(0,0)\$ and circle equation to be \$x^{2}+y{2}=r^{2}\$\n\nWe define the line \$l\$ by the equation \$y=-r\$, with point \$M\$ at a distance \$m\$ from the tangent and cartesian coordinates \$(m,-r)\$\n\nLet \$d\$ be the distance from point \$M\$ to point \$R\$ such that the coordinates for \$R\$ are \$(m+d,-r)\$ and thus the coordinates for \$Q\$ are \$(m-d,-r)\$\n\nLet points \$S\$, \$T\$, and \$U\$ be the points where lines \$PQ\$, \$PR\$, and \$l\$ are tangent to circle \$C\$ respectively.\n\nFirst we get the coordinates for points \$S\$ and \$T\$.\n\nSince the circle is the incenter we know the following properties:\n\n\\[\n\\left\| RU \\right\| = \\left\| RT \\right\|=(m+d)\n\\]\n\nand\n\n\\[\n\\left\| QU \\right\| = \\left\| QS \\right\|=(m-d)\n\\]\n\nTherefore, to get the coordinates of point \$T=(T_{x},T_{y})\$, we solve the following equations:\n\n\\[\nT_{x}^{2}+T_{y}^2=r^{2}\n\\]\n\n\\[\n\\left\| RT \\right\|^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2\n\\]\n\n\\[\n(m+d)^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2\n\\]\n\nAfter a lot of algebra, this solves to:\n\n\\[\nT_{x}=\\frac{2r^{2}(m+d)}{(m+d)^{2}+r^{2}}\n\\]\n\n\\[\nT_{y}=\\frac{r\\left[ (m+d)^{2}-r^{2} \\right]}{(m+d)^{2}+r^{2} }\n\\]\n\nNow we calculate the slope of the line that passes through \$PR\$ which is perpendicular to the line that passes from the center of the circle to point \$T\$ as follows:\n\n\\[\nSlope_{PR}=\\frac{-T_{x}}{T_{y}}=\\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\n\\]\n\nThen, the equation of the line that passes through \$PR\$ is as follows:\n\n\\[\nLine_{PR}\\colon \\; y+r=\\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\\left( x-(m+d) \\right)\n\\]\n\nNow we get the coordinates of point \$S=(S_{x},S_{y})\$, we solve the following equations:\n\n\\[\nS_{x}^{2}+S_{y}^2=r^{2}\n\\]\n\n\\[\n\\left\| QT \\right\|^{2}=(m-d-S_{x})^{2}+(r+S_{y})^2\n\\]\n\n\\[\n(m-d)^{2}=(m-d-S_{x})^{2}+(r+S_{y})^2\n\\]\n\nAfter a lot of algebra, this solves to:\n\n\\[\nS_{x}=\\frac{2r^{2}(m-d)}{(m-d)^{2}+r^{2}}\n\\]\n\n\\[\nS_{y}=\\frac{r\\left[ (m-d)^{2}-r^{2} \\right]}{(m-d)^{2}+r^{2} }\n\\]\n\nNow we calculate the slope of the line that passes through \$PQ\$ which is perpendicular to the line that passes from the center of the circle to point \$S\$ as follows:\n\n\\[\nSlope_{PQ}=\\frac{-S_{x}}{S_{y}}=\\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\n\\]\n\nThen, the equation of the line that passes through \$PQ\$ is as follows:\n\n\\[\nLine_{PQ}\\colon \\; y+r=\\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\\left( x-(m-d) \\right)\n\\]\n\nNow we solve for the coordinates for point \$P=(P_{x},P_{y})\$ by calculating the intersection of \$Line_{PR}\$ and \$Line_{PQ}\$ as follows:\n\n\\[\n\\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\\left( P_{x}-(m+d) \\right)=\\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\\left( P_{x}-(m-d) \\right)\n\\]\n\nSolving for \$P_{x}\$ we get:\n\n\\[\nP_{x}=\\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}\n\\]\n\nSolving for \$P_{y}\$ we get:\n\n\\[\nP_{y}=\\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\\left( \\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}-(m-d) \\right)-r\n\\]\n\n\\[\nP_{y}=\\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}\n\\]\n\nNow we need to find the limit of \$P_{x}\$ and \$P_{y}\$ as \$d\$ approaches infinity:\n\n\\[\nP_{x_{d \\to \\infty}}=\\lim_{d \\to \\infty} \\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}=0\n\\]\n\n\\[\nP_{y_{d \\to \\infty}}=\\lim_{d \\to \\infty} \\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}=r\n\\]\n\nThis means that the locus of \$P\$ starts at point \$(0,r)\$ on the circle \$C\$ but that point is not included in the locus as that is the limit.\n\nIf we assume that the locus is a ray that starts at \$(0,r)\$ let's calculate the slope of such ray:\n\n\\[\nSlope_{locus}=\\frac{P_{y}-r}{P_{x}}\n\\]\n\n\\[\nSlope_{locus}=\\frac{\\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}-r}{\\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}}\n\\]\n\n\\[\nSlope_{locus}=\\frac{-r^{2}}{2m}\n\\]\n\nSince the calculated slope of such locus at any point \$P\$ is not dependent on \$d\$ and solely dependent on fixed \$r\$ and \$m\$, then this proves the slope is fixed and thus the locus is a ray that starts at \$(0,r)\$ excluding that point and with a slope of \$\\frac{-r^{2}}{2m}\$ in the cartesian coordinate system moving upwards to infinity.\n\nWe can also write the equation of the locus as: \$y=\\frac{-r^{2}}{2m}x+r,\\;\\;\\forall y>r\\;\$and \$x<0\$\n\n~Tomas Diaz. orders@tomasdiaz.com", "We give a synthetic solution to this problem. Let \$D\$ be the tangency point of \$C\$ and \$l\$, and let \$I\$ be the center of \$C\$. Let \$X\$, \$D'\$ be the reflections of \$D\$ over \$M\$ and \$I\$ respectively. We claim that the locus is the line \$D'X\$.\n\nLet \$PQR\$ be a triangle with the given properties. Then \$X\$ is the tangency point of the \$P\$-excircle of \$PQR\$ with \$l\$, so by homothety at \$P\$, the points \$P\$, \$D'\$, \$X\$ are collinear.\n\nConversely, let \$Q\\ne R\$ be two points on \$l\$ such that \$PQ\$ and \$PR\$ are tangent to \$C\$. A homothety at \$P\$ sending \$D'\$ to \$X\$ sends \$C\$ to the \$P\$-excircle, so \$QD=RX\$ and \$M\$ is the midpoint of \$QR\$.\n\n~Alexander Chew. alexanderchew2021@gmail.com" ]
IMO-1992-5	https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_5	Let $S$ be a finite set of points in three-dimensional space. Let $S_{x}$,$S_{y}$,$S_{z}$, be the sets consisting of the orthogonal projections of the points of $S$ onto the $yz$-plane, $zx$-plane, $xy$-plane, respectively. Prove that \[ \|S\|^{2} \le \|S_{x}\| \cdot \|S_{y}\| \cdot \|S_{z}\|, \] where $\|A\|$ denotes the number of elements in the finite set $\|A\|$. (Note: The orthogonal projection of a point onto a plane is the foot of the perpendicular from that point to the plane)	[ "Let \$Z_{i}\$ be planes with index \$i\$ such that \$1 \\le i \\le n\$ that are parallel to the \$xy\$-plane that contain multiple points of \$S\$ on those planes such that all points of \$S\$ are distributed throughout all planes \$Z_{i}\$ according to their \$z\$-coordinates in common.\n\nLet \$a_{i}\$ be the number of unique projected points from each \$Z_{i}\$ to the \$yz\$-plane\n\nLet \$b_{i}\$ be the number of unique projected points from each \$Z_{i}\$ to the \$xz\$-plane\n\nThis provides the following: \$\|Z_{i}\| \\le a_{i}b_{i}\\;\$ [Equation 1]\n\nWe also know that \$\|S\|=\\sum_{i=1}^{n}\|Z_{i}\|\\;\$ [Equation 2]\n\nSince \$a_{i}\$ be the number of unique projected points from each \$Z_{i}\$ to the \$yz\$-plane,\n\nif we add them together it will give us the total points projected onto the \$yz\$-plane.\n\nTherefore, \$\|S_{x}\|=\\sum_{i=1}^{n}a_{i}\\;\$ [Equation 3]\n\nlikewise, \$\|S_{y}\|=\\sum_{i=1}^{n}b_{i}\\;\$ [Equation 4]\n\nWe also know that the total number of elements of each \$Z_{i}\$ is less or equal to the total number of elements in \$S_{z}\$\n\nThat is, \$\|Z_{i}\| \\le \|S_{z}\|\\;\$ [Equation 5]\n\nMultiplying [Equation 1] by [Equation 5] we get:\n\n\\[\n\|Z_{i}\|^{2} \\le a_{i}b_{i}\|S_{z}\|\n\\]\n\nTherefore, \$\|Z_{i}\| \\le \\sqrt{a_{i}b_{i}}\\sqrt{\|S_{z}\|}\$\n\nAdding all \$\|Z_{i}\|\$ we get:\n\n\$\\sum_{i=1}^{n}\|Z_{i}\| \\le \\sqrt{\|S_{z}\|}\\sum_{i=1}^{n}\\sqrt{a_{i}b_{i}}\\;\$[Equation 6]\n\nSubstituting [Equation 2] into [Equation 6] we get:\n\n\\[\n\|S\| \\le \\sqrt{\|S_{z}\|}\\sum_{i=1}^{n}\\sqrt{a_{i}b_{i}}\n\\]\n\n\\[\n\|S\|^{2} \\le \|S_{z}\| \\left( \\sum_{i=1}^{n}\\sqrt{a_{i}b_{i}} \\right)^{2}\n\\]\n\nSince, \$\\sum_{i=1}^{n}\\sqrt{a_{i}b_{i}} \\le \\sqrt{\\sum_{i=1}^{n}a_{i}}\\sqrt{\\sum_{i=1}^{n}b_{i}}\$,\n\nThen, \$\|S\|^{2} \\le \|S_{z}\| \\left( \\sqrt{\\sum_{i=1}^{n}a_{i}}\\sqrt{\\sum_{i=1}^{n}b_{i}} \\right)^{2}\$\n\n\$\|S\|^{2} \\le \|S_{z}\| \\left( \\sum_{i=1}^{n}a_{i}\\right)\\left( \\sum_{i=1}^{n}b_{i}\\right)\\;\$ [Equation 7]\n\nSubstituting [Equation 3] and [Equation 4] into [Equation 7] we get:\n\n\\[\n\|S\|^{2} \\le \|S_{x}\| \\cdot \|S_{y}\| \\cdot \|S_{z}\|\n\\]\n\n~Tomas Diaz. orders@tomasdiaz.com" ]
IMO-1992-6	https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_6	For each positive integer $n$, $S(n)$ is defined to be the greatest integer such that, for every positive integer $k \le S(n)$, $n^{2}$ can be written as the sum of $k$ positive squares. (a) Prove that $S(n) \le n^{2}-14$ for each $n \ge 4$. (b) Find an integer $n$ such that $S(n)=n^{2}-14$. (c) Prove that there are infinitely many integers $n$ such that $S(n)=n^{2}-14$.	[ "(a) Let \$n \\geq 4\$ be a positive integer. We will prove that \$S(n) \\leq n^2 - 14\$.\n\nAssume for the sake of contradiction that there exists a positive integer \$n \\geq 4\$ such that \$S(n) > n^2 - 14\$. Then, there exists a positive integer \$m\$ such that \$S(n) = n^2 - 14 + m\$.\n\nConsider the number \$n^2 - 14 + m\$. By definition of \$S(n)\$, for every positive integer \$k \\leq S(n)\$, \$n^2\$ can be written as the sum of \$k\$ positive squares. In particular, \$n^2\$ can be written as the sum of \$n^2 - 14 + m\$ positive squares.\n\nHowever, it is a well-known result that any positive integer can be expressed as the sum of at most \$4\$ positive squares. Therefore, \$n^2\$ cannot be expressed as the sum of \$n^2 - 14 + m\$ positive squares, which is a contradiction. Hence, \$S(n) \\leq n^2 - 14\$ for each \$n \\geq 4\$.\n\n(b) To find an integer \$n\$ such that \$S(n) = n^2 - 14\$, we need to show that \$n^2 - 14\$ can be expressed as the sum of \$n^2 - 14\$ positive squares.\n\nConsider the number \$n^2 - 14\$. We can express it as the sum of \$n^2 - 15\$ perfect squares of \$1\$ and \$1\$ perfect square of \$n-3\$. Therefore, \$S(n) = n^2 - 14\$.\n\n(c) To prove that there are infinitely many integers \$n\$ such that \$S(n) = n^2 - 14\$, note that for any integer \$n = 4 + 15k\$ where \$k\$ is a non-negative integer, we have \$S(n) = n^2 - 14\$. Since there are infinitely many non-negative integers \$k\$, there are infinitely many integers \$n\$ such that \$S(n) = n^2 - 14\$. By M. Nazaryan." ]

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