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+{"year": 1997, "problem_number": 1, "problem": "In the plane there is an infinite chessboard.\nFor any pair of positive integers $m$ and $n$,\nconsider a right-angled triangle with vertices at lattice points\nand whose legs, of lengths $m$ and $n$, lie along edges of the squares.\nLet $S_1$ be the total area of the black part of the triangle\nand $S_2$ be the total area of the white part.\nLet $f(m,n) = | S_1 - S_2 |$.\n\n[(a)]\nCalculate $f(m,n)$ for all positive integers $m$ and $n$\nwhich are either both even or both odd.\nProve that $f(m,n) \\leq \\frac 12 \\max \\{m,n\\}$ for all $m$ and $n$.\nShow that there is no constant $C$\nsuch that $f(m,n) < C$ for all $m$ and $ n$.", "solution": "In general, we say the discrepancy of a region in the plane\nequals its black area minus its white area.\nWe allow negative discrepancies,\nso discrepancy is additive and $f(m,n)$ equals the absolute value\nof the discrepancy of a right triangle with legs $m$ and $n$.\n\nFor (a), the answers are $0$ and $1/2$ respectively.\nTo see this, consider the figure shown below.\n[Figure omitted]\nNotice that triangles $APM$ and $BQM$ are congruent,\nand when $m \\equiv n \\pmod 2$, their colorings actually coincide.\nSo, the discrepancy of the triangle\nis exactly equal to the discrepancy of $CPQB$, which is an $m \\times n/2$\nrectangle and hence equal to $0$ or $1/2$ according to parity.\n\nFor (b), note that a triangle with legs $m$ and $n$, with $m$ even and $n$ odd,\ncan be dissected into one right triangle with legs $m$ and $n-1$\nplus a thin triangle of area $1/2$ which has height $m$ and base $1$.\nThe former region has discrepancy $0$ by (a),\nand the latter region obviously has discrepancy at most its area of $m/2$,\nhence $f(m,n) \\le m/2$ as needed.\n(An alternative slower approach, which requires a few cases,\nis to prove that two adjacent columns have at most discrepancy $1/2$.)\n\nFor (c), we prove:\n\nFor each $k \\ge 1$, we have\n$ f(2k, 2k+1) = \\frac{2k-1}{6}. $\n\nAn illustration for $k=2$ is shown below,\nwhere we use $(0,0)$, $(0,2k)$, $(2k+1,0)$ as the three vertices.\n[Figure omitted]\nWLOG, the upper-left square is black, as above.\nThe $2k$ small white triangles just below the diagonal have area sum\n$ \\frac12 \\cdot \\frac{1}{2k+1} \\cdot \\frac{1}{2k}\n[ 1^2 + 2^2 + \\dots + (2k)^2 ] = \\frac{4k+1}{12} $\nThe area of the $2k$ black polygons sums just below the diagonal to\n$ \\sum_{i=1}^{2k} ( 1\n- \\frac12 \\cdot \\frac{1}{2k+1} \\cdot \\frac{1}{2k} \\cdot i^2 )\n= 2k - \\frac{4k+1}{12} = \\frac{20k-1}{12}. $\nFinally, in the remaining $1+2+\\dots+2k$ squares,\nthere are $k$ more white squares than black squares.\nSo, it follows\n$ f(2k, 2k+1)\n= \\left| -k + \\frac{20k-1}{12} - \\frac{4k+1}{12} \\right|\n= \\frac{2k-1}{6}. $"}
+{"year": 1997, "problem_number": 2, "problem": "Let $ABC$ be a triangle with $\\angle A < \\min(\\angle B, \\angle C)$.\nThe points $B$ and $C$ divide the circumcircle of the triangle into two arcs.\nLet $U$ be an interior point of the arc between $B$ and $C$ which does not contain $A$.\nThe perpendicular bisectors of $ AB$ and $ AC$ meet the line $AU$ at $V$ and $W$, respectively.\nThe lines $BV$ and $CW$ meet at $T$.\n\nShow that $AU = TB + TC$.", "solution": "Let $BTV$ meet the circle again at $U_1$,\nso that $AU_1 UB$ is an isosceles trapezoid.\nDefine $U_2$ similarly.\n\n[Figure omitted]\nNow from the isosceles trapezoids we get\n$ AU = BU_1 = BT + TU_1 = BT + TC $\nas desired."}
+{"year": 1997, "problem_number": 3, "problem": "Let $x_1$, $x_2$, \\dots, $x_n$ be real numbers satisfying the conditions:\n\n|x_1 + x_2 + \\dots + x_n| &= 1 \\\\\n|x_i| &\\le \\frac{n+1}{2} \\qquad \\text{for } i= 1,2, \\dots, n\n\nShow that there exists a permutation $y_1$, $y_2$, \\dots, $y_n$\nof $x_1$, $x_2$, \\dots, $x_n$ such that\n$ | y_1 + 2 y_2 + \\dotsb + n y_n | \\leq \\frac {n + 1}{2}. $", "solution": "WLOG $\\sum x_i = 1$ (by negating $x_i$) and $x_1 \\le x_2 \\le \\dots \\le x_n$.\nNotice that\n\nThe largest possible value of the sum in question is\n$ A = x_1 + 2x_2 + 3x_3 + \\dots + nx_n. $\nwhile the smallest value is\n$ B = nx_1 + (n-1)x_2 + \\dots + x_n. $\nMeanwhile, the average value across all permutations is\n$ 1 \\cdot \\frac1n + 2 \\cdot \\frac1n + \\dots + n \\cdot \\frac1n = \\frac{n+1}{2}. $\n\nNow imagine we transform the sum $A$ to the sum $B$,\none step at a time, by swapping adjacent elements.\nEvery time we do a swap of two neighboring $u \\le v$, the sum decreases by\n$ (iu + (i+1)v) - (iv + (i+1)u) = v-u \\le n+1. $\n\nWe want to prove we land in the interval\n$ I = [ -\\frac{n+1}{2}, \\frac{n+1}{2} ] $\nat some point during this transformation.\nBut since $B \\le \\frac{n+1}{2} \\le A$ (since $\\frac{n+1}{2}$ was the average)\nand our step sizes were at most the length of the interval $I$,\nthis is clear."}
+{"year": 1997, "problem_number": 4, "problem": "An $n \\times n$ matrix whose entries come\nfrom the set $S = \\{1, 2, \\dots , 2n - 1\\}$\nis called a silver matrix if,\nfor each $i = 1, 2, \\dots , n$,\nthe $i$-th row and the $i$-th column together\ncontain all elements of $S$. Show that:\n[(a)]\nthere is no silver matrix for $n = 1997$;\nsilver matrices exist for infinitely many values of $n$.", "solution": "Solution to (a).: \nDefine a cross to be the union of the $i$th row and $i$th column.\nEvery cell of the matrix not on the diagonal is contained in exactly two crosses,\nwhile each cell on the diagonal is contained in one cross.\n\nOn the other hand, if a silver matrix existed for $n=1997$,\nthen each element of $S$ appears in all $1997$ crosses.\nSince $1997$ is odd, each number $s \\in S$ must appear on the diagonal an odd number of times.\n(For example, $s$ could appear on the diagonal once and off-diagonal $998$ times,\nor on the diagonal three times and off-diagonal $997$ times, etc.)\nIn particular, each number $s$ appears at least once on the diagonal.\n\nHowever, $|S| = 3993$ while there are only $1997$ diagonal cells.\nThis is a contradiction.\n\nSolution to (b).: \nWe construct a silver matrix $M_e$ for $n = 2^e$ for each $e \\ge 1$.\nWe write the first three explicitly for concreteness:\n\nM_1 &= \n1 & 2 \\\\ 3 & 1\n\\\\\nM_2 &= \n{1} & {2} & 4 & 5 \\\\\n{3} & {1} & 6 & 7 \\\\\n7 & 5 & {1} & {2} \\\\\n6 & 4 & {3} & {1}\n\\\\\nM_3 &= \n{1} & {2} & {4} & {5} & 8 & 9 & 11 & 12\\\\\n{3} & {1} & {6} & {7} & 10 & 15 & 13 & 14 \\\\\n{7} & {5} & {1} & {2} & 14 & 12 & 8 & 9 \\\\\n{6} & {4} & {3} & {1} & 13 & 11 & 10 & 15 \\\\\n15 & 9 & 11 & 12 & {1} & {2} & {4}\n& {5} \\\\\n10 & 8 & 13 & 14 & {3} & {1} & {6}\n& {7} \\\\\n14 & 12 & 15 & 9 & {7} & {5} & {1}\n& {2} \\\\\n13 & 11 & 10 & 8 & {6} & {4} & {3}\n& {1} \\\\\n\nThe construction is described recursively as follows.\nLet\n$\nM_e' = [\n{c|c}\n{M_{e-1}} & M_{e-1} + (2^e-1) \\\\ \\hline\nM_{e-1} + (2^e-1) & {M_{e-1}} \\\\\n\n].\n$\nThen to get from $M_e'$ to $M_e$,\nreplace half of the $2^e$'s with $2^{e+1}-1$:\nin the northeast quadrant, the even-indexed ones,\nand in the southwest quadrant, the odd-indexed ones.\n\nIn fact, it turns out silver matrices exist for all even dimensions.\nA claimed proof is outlined at ."}
+{"year": 1997, "problem_number": 5, "problem": "Find all pairs $(a,b)$ of positive integers satisfying\n$ a^{b^2} = b^a. $", "solution": "The answer is $(1,1)$, $(16,2)$ and $(27,3)$.\n\nWe assume $a,b > 1$ for convenience.\nLet $T$ denote the set of non perfect powers other than $1$.\n\nEvery integer greater than $1$\nis uniquely of the form $t^n$ for some $t \\in T$, $n \\in \\NN$.\n\nClear.\n\nLet $a = s^m$, $b = t^n$.\n$ s^{m \\cdot (t^n)^2} = t^{n \\cdot s^m}. $\nHence $s = t$ and we have\n$ m \\cdot t^{2n} = n \\cdot t^m\n\\implies t^{2n-m} = \\frac nm. $\nLet $n = t^e m$ and $2 \\cdot t^e m - m = e$, or\n$ e + m = 2t^e \\cdot m. $\nWe resolve this equation by casework\n\nIf $e > 0$, then $2t^e \\cdot m > 2e \\cdot m > e+m$.\nIf $e=0$ we have $m=n$ and $m = 2m$, contradiction.\nIf $e = -1$ we apparently have\n$ \\frac{2}{t} \\cdot m = m-1 \\implies\nm = \\frac{t}{t-2} $\nso $(t,m) = (3,3)$ or $(t,m) = (4,2)$.\nIf $e = -2$ we apparently have\n$ \\frac{2}{t^2} \\cdot m = m - 2\n\\implies m = \\frac{2}{1 - 2/t^2} = \\frac{2t^2}{t^2-2}. $\nThis gives $(t,m) = (2,2)$.\nIf $e \\le -3$ then let $k = -e \\ge 3$, so the equation is\n$ m-k = \\frac{2m}{t^k}\n\\iff m = \\frac{k \\cdot t^k}{t^k-2}\n= k + \\frac{2k}{t^k-2}. $\nHowever, for $k \\ge 3$ and $t \\ge 2$,\nwe always have $2k \\le t^k - 2$,\nwith equality only when $(t,k) = (2,3)$;\nthis means $m=4$, which is not a new solution."}
+{"year": 1997, "problem_number": 6, "problem": "For each positive integer $n$,\nlet $f(n)$ denote the number of ways of representing $n$\nas a sum of powers of 2 with nonnegative integer exponents.\nRepresentations which differ only in the ordering\nof their summands are considered to be the same.\nFor instance, $f(4) = 4$,\nbecause the number $4$ can be represented in the following four ways:\n$4$; $2+2$; $2+1+1$; $1+1+1+1$.\n\nProve that for any integer $n \\geq 3$\nwe have $2^{\\frac{n^2}{4}} < f(2^n) < 2^{\\frac{n^2}2}$.", "solution": "It's clear that $f$ is non-decreasing.\nBy sorting by the number of $1$'s we used,\nwe have the equation\n$ f(N) =\nf( \\left\\lfloor \\frac N2 \\right\\rfloor )\n+ f( \\left\\lfloor \\frac N2 \\right\\rfloor -1 )\n+ f( \\left\\lfloor \\frac N2 \\right\\rfloor -2 )\n+ \\dots\n+ f(1) + f(0). \\quad (\\bigstar)\n$\n\nUpper bound.: \nWe now prove the upper bound by induction.\nIndeed, the base case is trivial and for the inductive step\nwe simply use $(\\bigstar)$:\n$ f(2^n) = f(2^{n-1}) + f(2^{n-1}-1) + \\dots\n< 2^{n-1} f(2^{n-1})\n< 2^{n-1} \\cdot 2^{\\frac{(n-1)^2}{2}}\n= 2^{\\frac{n^2}{2} - 1/2}.\n$\n\nLower bound.: \nFirst, we contend that $f$ is convex.\nWe'll first prove this in the even case\nto save ourselves some annoyance:\n\n[$f$ is basically convex]\nIf $2 \\mid a+b$ then\nwe have $f(2a) + f(2b) \\ge 2 f( a+b )$.\n\nSince $f(2k+1) = f(2k)$, we will only prove the first equation.\nAssume WLOG $a \\ge b$ and use\n$(\\bigstar)$ on all three $f$ expressions here;\nafter subtracting repeated terms, the inequality then rewrites as\n$ \\sum_{(a+b)/2 \\le x \\le a} f(x)\n\\ge \\sum_{b \\le x \\le (a+b)/2} f(x). $\nThis is true since there are an equal number of terms on each side\nand $f$ is nondecreasing.\n\nFor each $1 \\le k < 2^{n-1}$, we have\n$ f(2^{n-1} - k) + f(k+1) \\ge 2f(2^{n-2}) $\n\nUse the fact that $f(2t+1)=f(2t)$ for all $t$\nand then apply convexity as above.\n\nNow we can carry out the induction:\n$ f(2^n) = f(2^{n-1}) + f(2^{n-1}-1) + \\dots\n> 2^{n-1} f(2^{n-2}) + f(0)\n> 2^{n-1} 2^{\\frac{(n-2)^2}{4}} = 2^{\\frac{n^2}{4}}.\n$"}
+{"year": 1998, "problem_number": 1, "problem": "A convex quadrilateral $ABCD$ has perpendicular diagonals.\nThe perpendicular bisectors of the sides $AB$ and $CD$ meet\nat a unique point $P$ inside $ABCD$.\nProve that the quadrilateral $ABCD$ is cyclic\nif and only if triangles $ABP$ and $CDP$ have equal areas.", "solution": "If $ABCD$ is cyclic, then $P$ is the circumcenter,\nand $\\angle APB + \\angle PCD = 180\\dg$.\nThe hard part is the converse.\n\n[Figure omitted]\n\nLet $M$ and $N$ be the midpoints of $AB$ and $CD$.\n\nUnconditionally, we have $\u2220 NEM = \u2220 MPN$.\n\nNote that $EN$ is the median of right triangle $\\triangle ECD$, and similarly for $EM$.\nHence $\u2220 NED = \u2220 EDN = \u2220 BDC$, while $\u2220 AEM = \u2220 ACB$.\nSince $\u2220 DEA = 90\\dg$, by looking at quadrilateral $XDEA$ where $X = CD \\cap AB$,\nwe derive that $\u2220 NED + \u2220 AEM + \u2220 DXA = 90\\dg$, so\n$ \u2220 NEM = \u2220 NED + \u2220 AEM + 90\\dg = -\u2220 DXA = -\u2220 NXM = -\u2220 NPM $\nas needed.\n\nHowever, the area condition in the problem tells us\n$ \\frac{EN}{EM} = \\frac{CN}{CM} = \\frac{PM}{PN}. $\nFinally, we have $\\angle MEN > 90\\dg$ from the configuration.\nThese properties uniquely determine the point $E$:\nit is the reflection of $P$ across the midpoint of $MN$.\n\nSo $EMPN$ is a parallelogram, and thus $ME \\perp CD$.\nThis implies $\u2220 BAE = \u2220 CEM = \u2220 EDC$ giving $ABCD$ cyclic."}
+{"year": 1998, "problem_number": 2, "problem": "In a competition, there are $a$ contestants\nand $b$ judges, where $b \\ge 3$ is an odd integer.\nEach judge rates each contestant as either ``pass'' or ``fail''.\nSuppose $k$ is a number such that for any two judges,\ntheir ratings coincide for at most $k$ contestants.\nProve that\n$ \\frac ka \\ge \\frac{b-1}{2b}. $", "solution": "This is a ``routine'' problem with global ideas.\nWe count pairs of coinciding ratings,\ni.e.\\ the number $N$ of tuples\n$(\\{J_1, J_2\\}, C) $\nof two distinct judges and a contestant\nfor which the judges gave the same rating.\n\nOn the one hand, if we count by the judges,\nwe have $ N \\le \\binom b2 k $\nby he problem condition.\n\nHowever, if we write $b=2m+1$ (so $m \\coloneq \\frac{b-1}{2}$), then each contestant $C$\ncontributes at least $\\binom{m}{2} + \\binom{m+1}{2} = m^2$ to $N$, and so\n$ N \\ge a \\cdot ( \\frac{b-1}{2} )^2 $\nPutting together the two estimates for $N$ yields the conclusion."}
+{"year": 1998, "problem_number": 3, "problem": "For any positive integer $n$,\nlet $\\tau(n)$ denote the number of its positive divisors (including $1$ and itself).\nDetermine all positive integers $m$ for which\nthere exists a positive integer $n$ such that\n$ \\frac{\\tau(n^{2})}{\\tau(n)}=m. $", "solution": "The answer is odd integers $m$ only.\nIf we write $n = p_1^{e_1} \\dots p_k^{e_k}$ we get\n$ \\prod \\frac{2e_i+1}{e_i+1} = m. $\nIt's clear now that $m$ must be odd,\nsince every fraction has odd numerator.\n\nWe now endeavor to construct odd numbers.\nThe proof is by induction, in which we are curating sets of\nfractions of the form $\\frac{2e+1}{e+1}$ that multiply\nto a given target.\n\nThe base cases are easy to verify by hand.\nGenerally, assume $p = 2^t k - 1$ is odd, where $k$ is odd.\nThen we can write\n$\n\\frac{2^{2t}k-2^t(k+1)+1}{2^{2t-1}k-2^{t-1}(k+1)+1}\n\\cdot\n\\frac{2^{2t-1}k-2^{t-1}(k+1)+1}{2^{2t-2}k-2^{t-2}(k+1)+1}\n\\cdot \\dots \\cdot\n\\frac{2^{t+1}k-2(k+1)+1}{2^tk-2^0(k+1)+1}.\n$\nNote that $2^{2t}k-2^t(k+1)+1 = (2^t k - 1)(2^t - 1)$,\nand $2^t k - k = k(2^t-1)$, so the above fraction simplifies to\n$ \\frac{2^t k - 1}{k} $\nmeaning we just need to multiply by $k$,\nwhich we can do using induction hypothesis."}
+{"year": 1998, "problem_number": 4, "problem": "Determine all pairs $(x,y)$ of positive integers\nsuch that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$.", "solution": "The answer is $(7k^2,7k)$ for all $k \\ge 1$,\nas well as $(11,1)$ and $(49,1)$.\n\nWe are given $xy^2+y+7 \\mid x^2y+x+y$.\nMultiplying the right-hand side by $y$ gives\n$ xy^2+y+7 \\mid x^2y^2+xy+y^2 $\nThen subtracting $x$ times the left-hand side gives\n$ xy^2+y+7 \\mid y^2-7x. $\nWe consider cases based on the sign of $y^2=7x$.\n\nIf $y^2 > 7x$, then $0 < y^2-7x < xy^2+y+7$,\ncontradiction.\nIf $y^2=7x$, let $y = 7k$, so $x = 7k^2$.\nPlugging this back in to the original equation reads\n$ 343k^4 + 7k + 7 \\mid 343k^5 + 7k^2 + 7k $\nwhich is always valid, hence these are all solutions.\nIf $y^2 < 7x$, then $|y^2-7x| \\le 7x$,\nso $y \\in \\{1,2\\}$.\n\nWhen $y=1$ we get\n$ x+8 \\mid x^2+x+1 \\iff x+8 \\mid 64-8+1=57.$\nThis has solutions $x=11$ and $x=49$.\n\nWhen $y=2$\n\n4x+9 \\mid 2x^2+x+2 \\\\\n\\implies 4x+9 &\\mid 16x^2+8x+16 \\\\\n\\implies 4x+9 &\\mid 81-18+16 = 79\n\nwhich never occurs."}
+{"year": 1998, "problem_number": 5, "problem": "Let $I$ be the incenter of triangle $ABC$.\nLet the incircle of $ABC$ touch\nthe sides $BC$, $CA$, and $AB$ at $K$, $L$, and $M$, respectively.\nThe line through $B$ parallel to $MK$ meets the lines\n$LM$ and $LK$ at $R$ and $S$, respectively.\nProve that angle $RIS$ is acute.", "solution": "Observe that $\\triangle MKL$ is acute with circumcenter $I$.\nWe now present two proofs.\n\nFirst simple proof (grobber).: \nThe problem is equivalent to showing $BI^2 > BR \\cdot BS$.\nBut from\n$ \\triangle BRK \\sim \\triangle MKL \\sim \\triangle BLS $\nwe conclude\n$ BR = t \\cdot \\frac{MK}{ML},\n\\qquad BS = t \\cdot \\frac{ML}{MK} $\nwhere $t = BK = BL$ is the length\nof the tangent from $B$.\nHence $BR \\cdot BS = t^2$.\nSince $BI > t$ is clear, we are done.\n\nSecond projective proof.: \nLet $N$ be the midpoint of $KL$,\nand let ray $MN$ meet the incircle again at $P$.\n\nNote that line $RBS$ is the polar of $N$.\nBy Brocard's theorem, lines $MK$ and $PL$ should thus\nmeet the polar of $N$, so we conclude $R = MK \\cap PL$.\nAnalogously, $S = ML \\cap PK$.\n\nAgain by Brocard's theorem, $\\triangle NRS$ is self-polar,\nso $N$ is the orthocenter of $\\triangle RIS$.\nSince $N$ lies between $I$ and $B$ we are done."}
+{"year": 1998, "problem_number": 6, "problem": "Classify all functions $f \\colon \\NN \\to \\NN$\nsatisfying the identity\n$ f(n^2 f(m)) = m f(n)^2. $", "solution": "Let $\\mathcal P$ be the set of primes,\nand let $g \\colon \\mathcal P \\to \\mathcal P$ be any involution on them.\nExtend $g$ to a completely multiplicative function on $\\NN$.\nThen $f(n) = d g(n)$ is a solution for any $d \\in \\NN$\nwhich is fixed by $g$.\n\nIt's straightforward to check these all work,\nsince $g \\colon \\NN \\to \\NN$ is an involution on them.\nSo we prove these are the only functions.\n\nLet $d = f(1)$.\n\nWe have $df(n) = f(dn)$ and $d \\cdot f(ab) = f(a) f(b)$.\n\nLet $P(m,n)$ denote the assertion in the problem statement.\nOff the bat,\n\n$P(1,1)$ implies $f(d) = d^2$.\n$P(n,1)$ implies $f(f(n)) = d^2n$.\nIn particular, $f$ is injective.\n$P(1,n)$ implies $f(dn^2) = f(n)^2$.\n\nThen\n\nf(a)^2 f(b)^2 &= f(da^2) f(b)^2 & \\text{by third bullet}\\\\\n&= f(b^2 f(f(da^2))) & \\text{by problem statement} \\\\\n&= f(b^2 \\cdot d^2 \\cdot da^2) & \\text{by second bullet} \\\\\n&= f(dab)^2 & \\text{by third bullet} \\\\\n\\implies f(a) f(b) &= f(dab).\n\nThis implies the first claim by taking $(a,b) = (1,n)$.\nThen $df(a) = f(da)$, and so we actually have\n$f(a) f(b) = d f(ab)$.\n\nAll values of $f$ are divisible by $d$.\n\nWe have\n\nf(n^2) &= \\frac 1d f(n)^2 \\\\\nf(n^3) &= \\frac{f(n^2) f(n)}{d} = \\frac{f(n)^3}{d^2} \\\\\nf(n^4) &= \\frac{f(n^3) f(n)}{d} = \\frac{f(n)^4}{d^3}\n\nand so on,\nwhich implies the result.\n\nThen, define $g(n) = f(n) / d$.\nWe conclude that $g$ is completely multiplicative, with $g(1) = 1$.\nHowever, $f(f(n)) = d^2n$ also implies $g(g(n)) = n$,\ni.e.\\ $g$ is an involution.\nMoreover, since $f(d) = d^2$, $g(d) = d$.\n\nAll that remains is to check that $g$ must map primes to primes\nto finish the description in the problem.\nThis is immediate; since $g$ is multiplicative and $g(1) = 1$,\nif $g(g(p)) = p$ then $g(p)$ can have at most one prime factor,\nhence $g(p)$ is itself prime.\n\nThe IMO problem actually asked for the least value of $f(1998)$.\nBut for instruction purposes,\nit is probably better to just find all $f$.\nSince $1998 = 2 \\cdot 3^3 \\cdot 37$,\nthis answer is $2^3 \\cdot 3 \\cdot 5 = 120$, anyways."}
+{"year": 1999, "problem_number": 1, "problem": "A set $S$ of points from the space will be called\ncompletely symmetric if it has at least three elements\nand fulfills the condition that for every two distinct points $A$ and $B$ from $S$,\nthe perpendicular bisector plane of the segment $AB$ is a plane of symmetry for $S$.\nProve that if a completely symmetric set is finite,\nthen it consists of the vertices of either a regular polygon,\nor a regular tetrahedron or a regular octahedron.", "solution": "Let $G$ be the centroid of $S$.\n\nAll points of $S$ lie on a sphere $\\Gamma$ centered at $G$.\n\nEach perpendicular bisector plane passes through $G$.\nSo if $A,B \\in S$ it follows $GA = GB$.\n\nConsider any plane passing through three or more points of $S$.\nThe points of $S$ in the plane form a regular polygon.\n\nThe cross section is a circle because we are intersecting a plane with sphere $\\Gamma$.\nNow if $A$, $B$, $C$ are three adjacent points on this circle,\nby taking the perpendicular bisector we have $AB=BC$.\n\nIf the points of $S$ all lie in a plane, we are done.\nOtherwise, the points of $S$ determine a polyhedron $\\Pi$ inscribed in $\\Gamma$.\nAll of the faces of $\\Pi$ are evidently regular polygons,\nof the same side length $s$.\n\nEvery face of $\\Pi$ is an equilateral triangle.\n\nSuppose on the contrary some face $A_1 A_2 \\dots A_n$ has $n > 3$.\nLet $B$ be any vertex adjacent to $A_1$ in $\\Pi$ other than $A_2$ or $A_n$.\nConsider the plane determined by $\\triangle A_1 A_3 B$.\nThis is supposed to be a regular polygon, but arc $A_1 A_3$ is longer than arc $A_1 B$,\nand by construction there are no points inside these arcs.\nThis is a contradiction.\n\nHence, $\\Pi$ has faces all congruent equilateral triangles.\nThis implies it is a regular polyhedron --- either\na regular tetrahedron, regular octahedron, or regular icosahedron.\nWe can check the regular icosahedron fails by\ntaking two antipodal points as our counterexample.\nThis finishes the problem."}
+{"year": 1999, "problem_number": 2, "problem": "Find the least constant $C$ such that for any integer $n > 1$ the inequality\n$\\sum_{1 \\le i < j \\le n} x_i x_j (x_i^2 + x_j^2)\n\\le C ( \\sum_{1 \\le i \\le n} x_i )^4$\nholds for all real numbers $x_1, \\dots, x_n \\ge 0$.\nDetermine the cases of equality.", "solution": "Answer: $C = \\frac 18$, with equality when two $x_i$ are equal\nand the remaining $x_i$ are equal to zero.\n\nWe present two proofs of the bound.\n\nFirst solution by smoothing.: \nFix $\\sum x_i = 1$.\nThe sum on the left-hand side can be interpreted as\n$\\sum_{i=1}^n x_i^3 \\sum_{j \\neq i} x_j = \\sum_{i=1}^n x_i^3(1-x_i)$,\nso we may rewrite the inequality as:\nThen it becomes $ \\sum_i (x_i^3 - x_i^4) \\le C. $\n\n[Smoothing]\nLet $f(x) = x^3 - x^4$.\nIf $u + v \\le \\frac 34$, then $f(u) + f(v) \\le f(0) + f(u+v)$.\n\nNote that\n\n(u^3-u^4)+(v^3-v^4) &\\le (u+v)^3-(u+v)^4 \\\\\n\\iff uv(4u^2+4v^2+6uv) &\\le 3uv(u+v)\n\nIf $u+v\\le \\frac 34$ this is obvious as $4u^2+4v^2+6uv \\le 4(u+v)^2$.\n\nObserve that if three nonnegative reals have pairwise sums\nexceeding $\\frac34$ then they have sum at least $\\frac 98$.\nHence we can smooth until $n-2$ of the terms are zero.\nHence it follows\n$ C = \\max_{a+b=1} (a^3+b^3-a^4-b^4) $\nwhich is routine computation giving $C = \\frac18$.\n\nSecond solution by AM-GM (Nairit Sarkar).: \nWrite\n\n\\text{LHS}\n&\\le ( \\sum_{1 \\le k \\le n} x_k^2 )\n( \\sum_{1 \\le i < j \\le n} x_i x_j )\n= 1/2 ( \\sum_{1 \\le k \\le n} x_k^2 )\n( \\sum_{1 \\le i < j \\le n} 2 x_i x_j ) \\\\\n&\\le 1/2 ( \\frac{\\sum_k x_k^2 + 2 \\sum_{i<j} x_i x_j}{2} )^2\n= \\frac 18 ( \\sum_{1 \\le i < n} x_i )^4\n\nas desired."}
+{"year": 1999, "problem_number": 3, "problem": "Let $n$ be an even positive integer.\nFind the minimal number of cells on the $n \\times n$ board\nthat must be marked so that any cell\n(marked or not marked) has a marked neighboring cell.", "solution": "For every marked cell, consider the marked cell adjacent\nto it; in this way we have a domino of two cells.\nFor each domino, its aura consists of all the cells\nwhich are adjacent to a cell of the domino.\nThere are up to eight squares in each aura,\nbut some auras could be cut off by the boundary of the board,\nwhich means that there could be as few as five squares.\n\nWe will prove that $1/2 n (n+2)$\nis the minimum number of auras\nneeded to cover the board\n(the auras need not be disjoint).\n\nA construction is shown on the left below,\nshowing that $1/2 n (n+2)$.\nColor the board as shown to the right into ``rings''.\nEvery aura takes covers exactly (!) four blue cells.\nSince there are $2n(n+2)$ blue cells, this implies the lower bound.\n\n[Figure omitted]\n\nNote that this proves that a partition into disjoint auras\nactually always has exactly $1/2 n (n+2)$ auras,\nthus also implying EGMO 2019/2."}
+{"year": 1999, "problem_number": 4, "problem": "Find all pairs of positive integers $(x,p)$\nsuch that $p$ is a prime and $x^{p-1}$ is a divisor of $(p-1)^{x}+1$.", "solution": "If $p = 2$ then $x \\in \\{1,2\\}$,\nand if $p = 3$ then $x \\in \\{1,3\\}$, since this is IMO 1990/3.\nAlso, $x = 1$ gives a solution for any prime $p$.\nWe show that there are no other solutions.\n\nAssume $x > 1$ and let $q$ be smallest prime divisor of $x$.\nWe have $q > 2$ since $(p-1)^x+1$ is odd.\nThen\n$ (p-1)^x \\equiv -1 \\pmod q \\implies (p-1)^{2x} \\equiv 1 \\pmod q $\nso the order of $p-1 \\bmod q$ is even and divides $\\gcd(q-1,2x) \\le 2$.\nThis means that\n$ p-1 \\equiv -1 \\pmod q \\implies p = q. $\n\nIn other words $p \\mid x$ and we get $x^{p-1} \\mid (p-1)^{x}+1$.\nBy exponent lifting lemma, we now have\n$ 0 < (p-1) \\nu_{p}(x) \\le 1 + \\nu_p(x). $\nThis forces $p=3$,\nwhich we already addressed."}
+{"year": 1999, "problem_number": 5, "problem": "Two circles $\\Omega_{1}$ and $\\Omega_{2}$ touch internally the circle\n$\\Omega$ in $M$ and $N$ and the center of $\\Omega_{2}$ is on $\\Omega_{1}$.\nThe common chord of the circles $\\Omega_{1}$ and $\\Omega_{2}$\nintersects $\\Omega$ in $A$ and $B$\nLines $MA$ and $MB$ intersects $\\Omega_{1}$ in $C$ and $D$.\nProve that $\\Omega_{2}$ is tangent to $CD$.", "solution": "Let $P$ and $Q$ be the centers of $\\Omega_1$ and $\\Omega_2$.\n\nLet line $MQ$ meet $\\Omega_1$ again at $W$,\nthe homothetic image of $Q$ under $\\Omega_1 \\to \\Omega$.\n\nMeanwhile, let $T$ be the intersection of segment $PQ$\nwith $\\Omega_2$, and let $L$ be its homothetic image on $\\Omega$.\nSince $PTQ \\perp AB$, it follows $LW$ is a diameter of $\\Omega$.\nLet $O$ be its center.\n\n[Figure omitted]\n\n$MNTQ$ is cyclic.\n\nBy Reim: $\u2220 TQM = \u2220 LWM = \u2220 LNM = \u2220 TNM$.\n\nLet $E$ be the midpoint of $AB$.\n\n$OEMN$ is cyclic.\n\nBy radical axis, the lines $MM$, $NN$, $AEB$ meet at a point $R$.\nThen $OEMN$ is on the circle with diameter $OR$.\n\n$MTE$ are collinear.\n\n$\u2220 NMT = \u2220 TQN = \u2220 LON = \u2220 NOE = \u2220 NME$.\n\nNow consider the homothety mapping $\\triangle WAB$ to $\\triangle QCD$.\nIt should map $E$ to a point on line $ME$\nwhich is also on the line through $Q$ perpendicular to $AB$; that is, to point $T$.\nHence $TCD$ are collinear, and it's immediate that $T$ is the desired tangency point."}
+{"year": 1999, "problem_number": 6, "problem": "Find all the functions $f \\colon \\RR \\to \\RR$ such that\n$f(x-f(y))=f(f(y))+xf(y)+f(x)-1$\nfor all $x,y \\in \\RR$.", "solution": "The answer is $f(x) = -1/2 x^2+1$\nwhich obviously works.\n\nFor the other direction, first note that\n$ P(f(y),y) \\implies 2f(f(y)) + f(y)^2 - 1 = f(0). $\nWe introduce the notation $c = \\frac{f(0)-1}{2}$,\nand $S = \\opname{img} f$.\nThen the above assertion says\n$ f(s) = -1/2 s^2 + (c + 1). $\nThus, the given functional equation can be rewritten as\n$ Q(x,s) : f(x-s)=-1/2 s^2 + sx + f(x) - c. $\n\n[Main claim]\nWe can find a function $g \\colon \\RR \\to \\RR$ such that\n$ f(x-z) = zx + f(x) + g(z). \\qquad (\\spadesuit). $\n\nIf $z \\neq 0$,\nthe idea is to fix a nonzero value $s_0 \\in S$ (it exists)\nand then choose $x_0$ such that $- 1/2 s_0^2 + s_0 x_0 - c = z$.\nThen, $Q(x_0, s)$ gives an pair $(u,v)$ with $u-v = z$.\n\nBut now for any $x$, using $Q(x+v,u)$ and $Q(x,-v)$ gives\n\nf(x-z)-f(x) &= f(x-u+v)-f(x)\n= f(x+v)-f(x) + u(x+v) - 1/2 u^2 + c \\\\\n&= -vx-1/2 s^2-c + u(x+v) - 1/2 u^2 + c \\\\\n&= -vx-1/2 v^2 + u(x+v) - 1/2 u^2 = zx + g(z)\n\nwhere $g(z) = -1/2(u^2+v^2)$ depends only on $z$.\n\nNow, let\n$ h(x) \\coloneq 1/2 x^2 + f(x) - (2c+1), $\nso $h(0) = 0$.\n\nThe function $h$ is additive.\n\nWe just need to rewrite $(\\spadesuit)$.\nLetting $x=z$ in $(\\spadesuit)$,\nwe find that actually $g(x)=f(0)-x^2-f(x)$.\nUsing the definition of $h$ now gives\n$ h(x-z) = h(x) + h(z). \\qedhere $\n\nTo finish, we need to remember that $f$, hence $h$, is known\non the image\n$ S = \\left\\{ f(x) \\mid x \\in \\RR \\right\\}\n= \\left\\{ h(x) - 1/2 x^2 + (2c+1) \\mid x \\in \\RR \\right\\}. $\nThus, we derive\n$ h( h(x)-1/2 x^2+(2c+1) ) = -c\n\\qquad \\forall x \\in \\RR. \\qquad(\\heartsuit) $\nWe can take the following two instances of $\\heartsuit$:\n\nh( h(2x)-2x^2+(2c+1) ) &= -c \\\\\nh( 2h(x)-x^2+2(2c+1) ) &= -2c.\n\nNow subtracting these and using $2h(x)=h(2x)$ gives\n$ c = h( -x^2 - (2c+1) ). $\nTogether with $h$ additive, this implies readily $h$ is constant.\nThat means $c=0$ and the problem is solved."}
+{"year": 2000, "problem_number": 1, "problem": "Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$.\nLet $AB$ be the line tangent to these circles at $A$ and $B$,\nrespectively, so that $M$ lies closer to $AB$ than $N$.\nLet $CD$ be the line parallel to $AB$\nand passing through the point $M$,\nwith $C$ on $G_1$ and $D$ on $G_2$.\nLines $AC$ and $BD$ meet at $E$; lines $AN$ and $CD$ meet at $P$;\nlines $BN$ and $CD$ meet at $Q$.\nShow that $EP = EQ$.", "solution": "First, we have $\u2220 EAB = \u2220 ACM = \u2220 BAM$\nand similarly $\u2220 EBA = \u2220 BDM = \u2220 ABM$.\nConsequently, $AB$ bisects $\\angle EAM$ and $\\angle EBM$,\nand hence $\\triangle EAB \\cong \\triangle MAB$.\n\n[Figure omitted]\n\nNow it is well-known that $MN$ bisects $AB$\nand since $AB \\parallel PQ$\nwe deduce that $M$ is the midpoint of $PQ$.\nAs $AB$ is the perpendicular bisector of $EM$,\nit follows that $EP = EQ$ as well."}
+{"year": 2000, "problem_number": 2, "problem": "Let $a$, $b$, $c$ be positive real numbers with $abc = 1$.\nShow that\n$\n( a - 1 + \\frac 1b )\n( b - 1 + \\frac 1c )\n( c - 1 + \\frac 1a )\n\\le 1.\n$", "solution": "Let $a = x/y$, $b = y/z$, $c = z/x$ for $x,y,z > 0$.\nThen the inequality rewrites as\n$ (-x+y+z)(x-y+z)(x+y-z) \\le xyz $\nwhich when expanded is equivalent to Schur's inequality.\nAlternatively, if one wants to avoid appealing to Schur,\nthen the following argument works:\n\nAt most one term on the left-hand side is negative;\nif that occurs we are done from $xyz > 0 > (-x+y+z)(x-y+z)(x+y-z)$.\nIf all terms in the left-hand side are nonnegative,\nlet us denote $m = -x+y+z \\ge 0$, $n = x-y+z \\ge 0$, $p = x+y-z \\ge 0$.\nThen it becomes\n$ mnp \\le \\frac{(m+n)(n+p)(p+m)}{8} $\nwhich follows by AM-GM."}
+{"year": 2000, "problem_number": 3, "problem": "Let $n \\ge 2$ be a positive integer\nand $\\lambda$ a positive real number.\nInitially there are $n$ fleas on a horizontal line,\nnot all at the same point.\nWe define a move as choosing two fleas at some points $A$ and $B$,\nwith $A$ to the left of $B$,\nand letting the flea from $A$ jump over the flea from $B$ to the point $C$\nso that $\\frac{BC}{AB} = \\lambda$.\n\nDetermine all values of $ \\lambda$ such that,\nfor any point $M$ on the line\nand for any initial position of the $n$ fleas,\nthere exists a sequence of moves that will take\nthem all to the position right of $M$.", "solution": "The answer is $\\lambda \\ge \\frac{1}{n-1}$.\n\nWe change the problem by replacing the fleas\nwith \\textbf{bowling balls} $B_1$, $B_2$, \\dots, $B_n$ in that order.\nBowling balls aren't exactly great at jumping,\nso each move can now be described as follows:\n\nSelect two indices $i < j$.\nThen ball $B_i$ moves to $B_{i+1}$'s location,\n$B_{i+1}$ moves to $B_{i+2}$'s location, and so on;\nuntil $B_{j-1}$ moves to $B_j$'s location,\nFinally, $B_j$ moves some distance forward;\nthe distance is at most $\\lambda \\cdot |B_j B_i|$\nand $B_j$ may not pass $B_{j+1}$.\n\nIf $\\lambda < \\frac{1}{n-1}$\nthe bowling balls have bounded movement.\n\nLet $a_i \\ge 0$ denote the initial distance\nbetween $B_i$ and $B_{i+1}$,\nand let $\\Delta_i$ denote the distance travelled by ball $i$.\nOf course we have\n$\\Delta_1 \\le a_1 + \\Delta_2$,\n$\\Delta_2 \\le a_2 + \\Delta_3$,\n\\dots,\n$\\Delta_{n-1} \\le a_{n-1} + \\Delta_n$\nby the relative ordering of the bowling balls.\nFinally, distance covered by $B_n$ is always\n$\\lambda$ times distance travelled by other bowling balls, so\n\n\\Delta_n &\\le \\lambda \\sum_{i=1}^{n-1} \\Delta_i\n\\le \\lambda \\sum_{i=1}^{n-1}\n( ( a_i + a_{i+1} + \\dots + a_{n-1} )\n+ \\Delta_n ) \\\\\n&= (n-1)\\lambda \\cdot \\Delta_n + \\sum_{i=1}^{n-1} i a_i\n\nand since $(n-1)\\lambda > 1$, this gives an upper bound.\n\nEquivalently, you can phrase the proof without\nbowling balls as follows:\nif $x_1 < \\dots < x_n$ are the positions of the fleas,\nthe quantity\n$ L = x_n - \\lambda(x_1 + \\dots + x_{n-1}) $\nis a monovariant which never increases;\ni.e.\\ $L$ is bounded above.\nSince $L > (1-(n-1)\\lambda) x_n$, it follows\n$\\lambda < \\frac{1}{n-1}$ is enough to stop the fleas.\n\nWhen $\\lambda \\ge \\frac{1}{n-1}$,\nit suffices to always jump the leftmost flea\nover the rightmost flea.\n\nIf we let $x_i$ denote the distance travelled by $B_1$\nin the $i$th step,\nthen $x_i = a_i$ for $1 \\le i \\le n-1$\nand $x_i = \\lambda(x_{i-1} + x_{i-2} + \\dots + x_{i-(n-1)})$.\n\nIn particular, if $\\lambda \\ge \\frac{1}{n-1}$\nthen each $x_i$ is at least the average of the previous $n-1$ terms.\nSo if the $a_i$ are not all zero,\nthen $\\{x_{n}, \\dots, x_{2n-2}\\}$ are all positive\nand thereafter $x_i \\ge \\min \\left\\{ x_n, \\dots, x_{2n-2} \\right\\} > 0$\nfor every $i \\ge 2n-1$.\nSo the partial sums of $x_i$ are unbounded, as desired.\n\nOther inductive constructions are possible.\nHere is the idea of one of them,\nalthough the details are more complicated.\n\nWe claim in general that given $n-1$ fleas at $0$\nand one flea at $1$,\nwe can get all the fleas arbitrarily close to\n$\\frac{1}{1-(n-1)\\lambda}$\n(or as far as we want if $\\lambda > \\frac{1}{n-1}$.).\nThe proof is induction by $n \\ge 2$;\nfor $n=2$ we get a geometric series.\nFor $n \\ge 3$, we leave one flea at zero\nand move the remainder close to $\\frac{1}{1-(n-2)\\lambda}$,\nthen jump the last flea to\n$\\frac{1+\\lambda}{1-(n-2)\\lambda}$.\n\nNow we're in the same situation,\nexcept we shifted $\\frac{1}{1-(n-2)\\lambda}$ right\nand have then scaled everything by\n$r = \\frac{\\lambda}{1-(n-2)\\lambda}$.\nIf we repeat this process again and check the geometric series,\nwe see the fleas converge to\n$ \\frac{1}{1-(n-2)\\lambda}\n( 1 + r + r^2 + r^3 + \\dots )\n= \\frac{1}{1-(n-2)\\lambda} \\cdot \\frac{1}{1-r}\n= \\frac{1}{1-(n-1)\\lambda}. $"}
+{"year": 2000, "problem_number": 4, "problem": "A magician has one hundred cards numbered $1$ to $100$.\nHe puts them into three boxes, a red one, a white one and a blue one,\nso that each box contains at least one card.\nA member of the audience draws two cards from two different boxes\nand announces the sum of numbers on those cards.\nGiven this information,\nthe magician locates the box from which no card has been drawn.\n\nHow many ways are there to put the cards\nin the three boxes so that the trick works?", "solution": "There are $2 \\cdot 3! = 12$ ways, which amount to:\n\nPartitioning the cards modulo $3$, or\nPlacing $1$ alone in a box,\n$100$ alone in a second box,\nand all remaining cards in the third box.\n\nThese are easily checked to work so we prove they are the only ones.\n\nFirst solution.: \nWe proceed by induction on $n \\ge 3$ with the base case being immediate.\n\nFor the inductive step,\nconsider a working partition of $\\{1, 2, \\dots, n\\}$.\nThen either $n$ is in its own box; or\ndeleting $n$ gives a working partition of $\\{1, 2, \\dots, n-1\\}$.\nSimilarly, either $1$ is in its own box; or\ndeleting $1$ gives a working partition of $\\{2, 3, \\dots, n\\}$,\nand we can reduce all numbers by $1$ to get\na working partition of $\\{1, 2, \\dots, n-1\\}$.\n\nTherefore, we only need to consider there cases.\n\nIf $1$ and $n$ are both in their own box,\nthis yields one type of solution we already found.\n\nIf $n$ is not in a box by itself,\nthen by induction hypothesis the cards $1$ through $n-1$\nare either arranged mod $3$,\nor as $\\{1\\} \\cup \\{2,3,\\dots,n-2\\} \\cup \\{n-1\\}$.\n\nIn the former mod $3$ situation,\nsince $n + (n-3) = (n-1) + (n-2)$,\nso $n$ must be in the same box as $n-3$.\nIn the latter case and for $n > 4$,\nsince $n + 1 = 2 + (n-1)$,\n$n$ must be in the same box as $1$.\nBut now $n + 2 = (n-1) + 3$ for $n > 4$, contradiction.\n\nThe case where $1$ is in a box by itself is analogous.\n\nThis exhausts all cases, completing the proof.\n\nSecond solution.: \nLet $A$, $B$, $C$ be the sets of cards in the three boxes.\nThen $A+B$, $B+C$, $C+A$ should be disjoint,\nand contained in $\\{3, 4, \\dots, 199\\}$.\nOn the other hand, we have the following famous fact.\n\nLet $X$ and $Y$ be finite nonempty sets of real numbers.\nWe have $|X+Y| \\ge |X|+|Y|-1$,\nwith equality if and only if $X$ and $Y$ are arithmetic\nprogressions with the same common difference,\nor one of $X$ and $Y$ is a singleton set.\n\nPutting these two together gives the estimates\n$ 197 \\ge |A+B| + |B+C| + |C+A|\n\\ge 2( |A|+|B|+|C| )-3 = 197. $\nSo all the inequalities must be sharp.\nConsequently we conclude that:\n\nEither the sets $A$, $B$, $C$ are disjoint arithmetic progressions\nwith the same common difference\n$d = \\min_{x \\neq y \\text{ in same set}} |x-y|$,\nor two of the sets are two singleton.\nMoreover, $\\{3, 4, \\dots, 199\\}\n= (A+B) \\sqcup (B+C) \\sqcup (C+A)$.\n\nFrom here it is not hard to deduce the layouts above are the only ones,\nbut there are some details.\nFirst, we make the preliminary observation that\n$3=1+2$, $4=1+3$, $198=98+100$, $199=99+100$\nand these numbers can't be decomposed in other ways;\nthus from the remark about the disjoint union:\n\n[Convenient corollary]\nThe pairs $(1,2)$, $(1,3)$, $(98,100)$, $(99,100)$\nare all in different sets.\n\nWe now consider the four cases.\n\nIf two of the boxes are singletons,\nit follows from the corollary that we should have $A = \\{1\\}$,\n$B = \\{100\\}$ and $C = \\{2, \\dots, 99\\}$, up to permutation.\nOtherwise $A$, $B$, $C$ are disjoint arithmetic\nprogressions with the same common difference $d$.\nAs two of $\\{1,2,3,4\\}$ are in the same box\n(by pigeonhole), we must have $d \\le 3$.\n\nIf $d=3$, then no two elements of different residues\nmodulo $3$ can be in the same box,\nso we must be in the first construction claimed earlier.\n\nIf $d=2$, then the convenient corollary\ntells us we may assume WLOG that $1 \\in A$ and $2 \\in B$,\nhence $3 \\in C$\n(since $3 \\notin A$ by convenient corollary,\nand $3 \\notin B$ because common difference $2$).\nThus we must have $A = \\{1\\}$, $B = \\{2, 4, \\dots, 100\\}$\nand $C = \\{3, 5, \\dots 99\\}$\nwhich does not work since $1+4 = 2+3$.\nTherefore there are no solutions in this case.\n\nIf $d=1$, then by convenient corollary\nthe numbers $1$ and $2$ are in different sets,\nas are $99$ and $100$.\nSo we must have $A = \\{1\\}$, $B = \\{2, \\dots, 99\\}$, $C = \\{100\\}$\nwhich we have already seen is valid."}
+{"year": 2000, "problem_number": 5, "problem": "Does there exist a positive integer $n$\nsuch that $n$ has exactly 2000 distinct prime divisors\nand $n$ divides $2^n + 1$?", "solution": "Answer: Yes.\n\nWe say that $n$ is Korean if $n \\mid 2^n+1$.\nFirst, observe that $n=9$ is Korean.\nNow, the problem is solved upon the following claim:\n\nIf $n > 3$ is Korean,\nthere exists a prime $p$ not dividing $n$\nsuch that $np$ is Korean too.\n\nI claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$,\nwhich exists by Zsigmondy theorem.\nObviously $p \\neq 2$.\nThen:\n\nSince $p \\nmid 2^{\\varphi(n)}-1$ it follows then that $p \\nmid n$.\nMoreover, $p \\mid 2^n+1$ since $p \\nmid 2^n-1$.\n\nHence $np \\mid 2^n+1 \\mid 2^{np} + 1$ by Chinese Theorem,\nsince $\\gcd(n,p) = 1$."}
+{"year": 2000, "problem_number": 6, "problem": "Let $AH_1$, $BH_2$, and $CH_3$ be the\naltitudes of an acute triangle $ABC$.\nThe incircle $\\omega$ of triangle $ABC$ touches the sides\n$BC$, $CA$ and $AB$ at $T_1$, $T_2$ and $T_3$, respectively.\nConsider the reflections of the lines $H_1H_2$, $H_2H_3$, and\n$H_3H_1$ with respect to the lines $T_1T_2$, $T_2T_3$, and $T_3T_1$.\nProve that these images form a triangle whose vertices lie on $\\omega$.", "solution": "We use complex numbers with $\\omega$ the unit circle.\nLet $T_1 = a$, $T_2 = b$, $T_3 = c$.\nThe main content of the problem is to show that\nthe triangle in question has vertices\n$ab/c$, $bc/a$, $ca/b$\n(which is evident from a good diagram).\n\nSince $A = \\frac{2bc}{b+c}$, we have\n$ H_1 = 1/2 ( \\frac{2bc}{b+c} + a + a - a^2 \\cdot\n\\frac{2}{b+c} )\n= \\frac{ab+bc+ca-a^2}{b+c}. $\nThe reflection of $H_1$ over $T_1 T_2$ is\n\nH_1^C &= a + b - ab H_1\n= a + b - b \\cdot \\frac{ac+ab+a^2-bc}{a(b+c)} \\\\\n&= \\frac{a(a+b)(b+c) - b(a^2+ab+ac-bc)}{a(b+c)}\n= \\frac{c(a^2+b^2)}{a(b+c)}.\n\nNow, we claim that $H_1^C$ lies on the chord joining\n$\\frac{ca}{b}$ and $\\frac{cb}{a}$;\nby symmetry so will $H_2^C$\nand this will imply the problem\n(it means that the desired triangle has vertices\n$ab/c$, $bc/a$, $ca/b$).\nA cartoon of this is shown below.\n[Figure omitted]\nTo see this, it suffices to compute\n\nH_1^C + ( \\frac{ca}{b} )( \\frac{cb}{a} ) H_1^C\n&= \\frac{c(a^2+b^2)}{a(b+c)}\n+ c^2 \\frac{\\frac 1c \\cdot \\frac{a^2+b^2}{a^2b^2}}\n{\\frac1a( \\frac{b+c}{bc} )} \\\\\n&= \\frac{c(a^2+b^2)}{a(b+c)}\n+ \\frac{c(a^2+b^2)}{abc\\inv(b+c)} \\\\\n&= \\frac{c(a^2+b^2)}{a(b+c)} ( \\frac{b+c}{b} ) \\\\\n&= \\frac{c(a^2+b^2)}{ab}= \\frac{ca}{b} + \\frac{cb}{a}\n\nas desired."}
+{"year": 2001, "problem_number": 1, "problem": "Let $ABC$ be an acute-angled triangle with $O$ as its circumcenter.\nLet $P$ on line $BC$ be the foot of the altitude from $A$.\nAssume that $\\angle BCA \\ge \\angle ABC + 30\\dg$.\nProve that $\\angle CAB + \\angle COP < 90\\dg$.", "solution": "The conclusion rewrites as\n\n\\angle COP &< 90\\dg - \\angle A = \\angle OCP \\\\\n\\iff PC &< PO \\\\\n\\iff PC^2 &< PO^2 \\\\\n\\iff PC^2 &< R^2 - PB \\cdot PC \\\\\n\\iff PC \\cdot BC &< R^2 \\\\\n\\iff ab \\cos C &< R^2 \\\\\n\\iff \\sin A \\sin B \\cos C & < \\frac14.\n\nNow\n$ \\cos C \\sin B\n= 1/2 ( \\sin(C+B)-\\sin(C-B) )\n\\le 1/2 ( 1 - 1/2 ) = \\frac 14 $\nwhich finishes when combined with $\\sin A < 1$.\n\nIf we allow $ABC$ to be right then\nequality holds when $\\angle A = 90\\dg$,\n$\\angle C = 60\\dg$, $\\angle B = 30\\dg$.\nThis motivates the choice of estimates\nafter reducing to a trig inequality."}
+{"year": 2001, "problem_number": 2, "problem": "Let $a$, $b$, $c$ be positive reals. Prove that\n$ \\frac{a}{\\sqrt{a^2+8bc}} + \\frac{b}{\\sqrt{b^2+8ca}} + \\frac{c}{\\sqrt{c^2+8ab}} \\ge 1. $", "solution": "By Holder, we have\n$\n( \\sum_{\\text{cyc}} \\frac{a}{\\sqrt{a^2+8bc}} )^2\n( \\sum_{\\text{cyc}} a(a^2+8bc) )\n\\ge (a+b+c)^3.\n$\nSo it suffices to show $(a+b+c)^3 \\ge a^3+b^3+c^3+24abc$ which is clear by expanding."}
+{"year": 2001, "problem_number": 3, "problem": "Twenty-one girls and twenty-one boys took part in a mathematical competition.\nIt turned out that each contestant solved at most six problems,\nand for each pair of a girl and a boy,\nthere was at least one problem that was solved by both the girl and the boy.\nShow that there is a problem that was solved by at least three girls and at least three boys.", "solution": "We will show the contrapositive.\nThat is, assume that\n\nFor each pair of a girl and a boy,\nthere was at least one problem that was\nsolved by both the girl and the boy.\nAssume every problem is either solved\nmostly by girls (at most two boys)\nor mostly by boys (at most two girls).\n\nThen we will prove that then some contestant\nsolved more than six problems.\n\nCreate a $21 \\times 21$ grid with boys as columns\nand girls as rows, and in each cell\nwrite the name of a problem solved by the pair.\nColor the cell \\textbf{green} if at most two girls solved that problem,\nand color it \\textbf{blue} if at most two boys solved that problem.\n(G for girl, B for boy.\nIt's possible both colors are used for some cell.)\n\nWLOG, there are more green cells than blue,\nso (by pigeonhole) take a column (boy) with that property.\nThat means the boy's column has at least $11$ green squares.\nBy hypothesis, those corresponds to at least $6$ different problems\nsolved. Now there are two cases:\n\nIf there are any blue-only squares,\nthen that square means a seventh distinct problems.\nIf the entire column is green,\nthen among the $21$ green squares\nthere are at least $11$ distinct problems solved\nin that column.\n\nThe number $21$ cannot be decreased.\nHere is an example of $20$ boys and $20$ girls\nwho solve problems named $A$-$J$\nand $0$-$9$, which motivates the solution to begin with.\n[Figure omitted]\n\nThis took me embarrassingly long,\nbut part of the work of the problem seemed to be\nfinding the right ``data structure'' to get a foothold.\nI think the grid is good because we want to fill each intersection,\nthen we consider for each cell a problem to put.\n\nI initially wanted to capture the full data by writing\nin each green cell the row index of the other girl who solved it,\nand similarly for the blue cells.\n(That led naturally to the colors, there were two types of cells.)\nThis was actually helpful for finding the equality case above,\nbut once I realized the equality case\nI also realized that I could discard the extra information\nand only remember the colors."}
+{"year": 2001, "problem_number": 4, "problem": "Let $n > 1$ be an odd integer and let $c_1$, $c_2$, \\dots, $c_n$ be integers.\nFor each permutation $a = (a_1, a_2, \\dots, a_n)$\nof $\\{1,2,\\dots,n\\}$, define $S(a) = \\sum_{i=1}^n c_i a_i$.\nProve that there exist two permutations $a \\neq b$\nof $\\{1,2,\\dots,n\\}$ such that $n!$ is a divisor of $S(a)-S(b)$.", "solution": "Assume for contradiction that all the $S(a)$ are distinct modulo $n!$.\nThen summing across all permutations gives\n\n1 + 2 + \\dots + n!\n&\\equiv \\sum_a S(a) \\\\\n&= \\sum_a \\sum_{i=1}^n c_i a_i \\\\\n&= \\sum_{i=1}^n c_i \\sum_a a_i \\\\\n&= \\sum_{i=1}^n c_i \\cdot ( (n-1)! \\cdot (1+\\dots+n) ) \\\\\n&= (n-1)! \\cdot \\frac{n(n+1)}{2} \\sum_{i=1}^n c_i \\\\\n&= n! \\cdot \\frac{n+1}{2} \\sum_{i=1}^n c_i \\\\\n&\\equiv 0\n\nsince $1/2(n+1)$ is an integer.\nBut on the other hand\n$1 + 2 + \\dots + n! = \\frac{n!(n!+1)}{2}$\nwhich is not divisible by $n!$ if $n > 1$,\nas the quotient is the non-integer $\\frac{n!+1}{2}$.\nThis is a contradiction."}
+{"year": 2001, "problem_number": 5, "problem": "Let $ABC$ be a triangle.\nLet $AP$ bisect $\\angle BAC$ and let $BQ$ bisect $\\angle ABC$,\nwith $P$ on $BC$ and $Q$ on $AC$.\nIf $AB + BP = AQ + QB$ and $\\angle BAC = 60\\dg$,\nwhat are the angles of the triangle?", "solution": "The answer is $\\angle B = 80\\dg$ and $\\angle C = 40\\dg$.\nSet $x = \\angle ABQ = \\angle QBC$, so that $\\angle QCB = 120\\dg - 2x$.\nWe observe $\\angle AQB = 120\\dg - x$ and $\\angle APB = 150\\dg - 2x$.\n\n[Figure omitted]\n\nNow by the law of sines, we may compute\n\nBP &= AB \\cdot \\frac{\\sin 30\\dg}{\\sin (150\\dg- 2x)} \\\\\nAQ &= AB \\cdot \\frac{\\sin x}{\\sin (120\\dg- x)} \\\\\nQB &= AB \\cdot \\frac{\\sin 60\\dg}{\\sin (120\\dg- x)}.\n\nSo, the relation $AB + BP = AQ + QB$ is exactly\n$ 1 + \\frac{\\sin 30\\dg}{\\sin (150\\dg - 2x)}\n= \\frac{\\sin x + \\sin 60\\dg}{\\sin (120\\dg - x)}. $\nThis is now a trig problem, and we simply solve for $x$.\nThere are many possible approaches\nand we just present one.\n\nFirst of all, we can write\n$ \\sin x + \\sin 60\\dg\n= 2\\sin ( 1/2 (x+60\\dg) )\n\\cos ( 1/2 (x-60\\dg) ). $\nOn the other hand, $\\sin (120\\dg - x) = \\sin(x+60\\dg)$ and\n$ \\sin (x+60\\dg)\n= 2 \\sin ( 1/2 (x+60\\dg) )\n\\cos ( 1/2(x+60\\dg) ) $\nso\n$ \\frac{\\sin x + \\sin 60\\dg}{\\sin (120\\dg - x)}\n= \\frac{\\cos ( 1/2 x - 30\\dg )}\n{\\cos ( 1/2 x + 30\\dg )}. $\nLet $y = 1/2 x$ for brevity now. Then\n\n\\frac{\\cos(y-30\\dg)}{\\cos(y+30\\dg)} - 1\n&= \\frac{\\cos(y-30\\dg)-\\cos(y+30\\dg)}{\\cos(y+30\\dg)} \\\\\n&= \\frac{2 \\sin (30\\dg) \\sin y}{\\cos(y+30\\dg)} \\\\\n&= \\frac{\\sin y}{\\cos (y+30\\dg)}.\n\nHence the problem is just\n\n\\frac{\\sin 30\\dg}{\\sin(150\\dg - 4y)} &= \\frac{\\sin y}{\\cos(y+30\\dg)}. \\\\\n\\intertext{Equivalently,}\n\\cos(y+30\\dg) &= 2\\sin y \\sin (150\\dg -4y) \\\\\n&= \\cos(5y-150\\dg) - \\cos(150\\dg-3y) \\\\\n&= -\\cos(5y+30\\dg) + \\cos(3y+30\\dg).\n\nNow we are home free, because $3y+30\\dg$\nis the average of $y+30\\dg$ and $5y+30\\dg$.\nThat means we can write\n$ \\frac{\\cos(y+30\\dg)+\\cos(5y+30\\dg)}{2} = \\cos(3y+30\\dg) \\cos(2y). $\nHence\n$ \\cos(3y+30\\dg) ( 2\\cos(2y)-1 ) = 0. $\nRecall that\n$ y = 1/2 x = \\frac 14 \\angle B\n< \\frac 14 ( 180\\dg - \\angle A ) = 30\\dg. $\nHence it is not possible that $\\cos(2y) = 1/2$,\nsince the smallest positive value of $y$\nthat satisfies this is $y=30\\dg$.\nSo $\\cos (3y+30\\dg) = 0$.\n\nThe only permissible value of $y$ is then $y = 20\\dg$,\ngiving $\\angle B = 80\\dg$ and $\\angle C = 40\\dg$."}
+{"year": 2001, "problem_number": 6, "problem": "Let $a > b > c > d > 0$ be integers satisfying\n$ ac + bd = (b+d+a-c)(b+d-a+c). $\nProve that $ab + cd$ is not prime.", "solution": "The problem condition is equivalent to\n$ ac + bd = (b+d)^2 - (a-c)^2 $\nor\n$ a^2-ac+c^2 = b^2+bd+d^2. $\n\nLet us construct a quadrilateral $WXYZ$ such that\n$WX = a$, $XY = c$, $YZ = b$, $ZW = d$,\nand $ WY = \\sqrt{a^2-ac+c^2} = \\sqrt{b^2+bd+d^2}.$\nThen by the law of cosines, we obtain $\\angle WXY = 60\\dg$\nand $\\angle WZY = 120\\dg$.\nHence this quadrilateral is cyclic.\n\n[Figure omitted]\n\nBy the more precise version of Ptolemy's theorem,\nwe find that\n$ WY^2 = \\frac{(ab+cd)(ad+bc)}{ac+bd}. $\n\nNow assume for contradiction that that $ab+cd$ is a prime $p$.\nRecall that we assumed $a > b > c > d$.\nIt follows, for example by rearrangement inequality, that\n$ p = ab+cd > ac+bd > ad+bc. $\nLet $y = ac+bd$ and $x = ad+bc$ now.\nThe point is that $ p \\cdot \\frac xy $\ncan never be an integer if $p$ is prime and $x < y < p$.\nBut $WY^2 = a^2-ac+c^2$ is clearly an integer, and this is a contradiction.\n\nHence $ab+cd$ cannot be prime.\n\nIt may be tempting to try to apply the more typical form of\nPtolemy to get $ab+cd = WY \\cdot XZ$;\nthe issue with this approach is that $WY$ and $XZ$ are usually not integers."}
+{"year": 2002, "problem_number": 1, "problem": "Let $n$ be a positive integer.\nLet $T$ be the set of points $(x,y)$ in the plane\nwhere $x$ and $y$ are non-negative integers with $x+y<n$.\nEach point of $T$ is coloured red or blue, subject to the following condition:\nif a point $(x,y)$ is red,\nthen so are all points $(x',y')$ of $T$ with $x'\\leq x$ and $y'\\leq y$.\nLet $A$ be the number of ways to choose $n$ blue points with distinct $x$-coordinates,\nand let $B$ be the number of ways to choose $n$ blue points with distinct $y$-coordinates.\nProve that $A=B$.", "solution": "Let $a_x$ denote the number of blue points\nwith a given $x$-coordinate.\nDefine $b_y$ to be the number of blue points\nwith a given $y$-coordinate.\n\nWe actually claim that\n\nThe multisets $\\mathcal A \\coloneq \\{ a_x \\mid x \\}$\nand $\\mathcal B \\coloneq \\{ b_y \\mid y \\}$ are equal.\n\nBy induction on the number of red points.\nIf there are no red points at all,\nthen $\\mathcal A = \\mathcal B = \\{1, \\dots, n\\}$.\n[Figure omitted]\nThe proof consists of two main steps.\nFirst, suppose we color a single point $P = (x,y)$\nfrom blue to red (while preserving the condition).\nBefore the coloring, we have $a_x = b_y = n-(x+y)$;\nafterwards $a_x = b_y = n-(x+y)-1$\nand no other numbers change, as desired.\n\nWe also must show that this operation\n(repeatedly adding a single point $P$) reaches all\npossible shapes of red points.\nThis is well-known as the red points form a Young tableaux;\nfor example, one way is to add all the points with $x=0$\nfirst one by one, then all the points with $x=1$, and so on.\nSo the induction implies the result.\n\nFinally, $ A = \\prod_{x=0}^{n-1} a_x = \\prod_{y=0}^{n-1} b_y = B. $"}
+{"year": 2002, "problem_number": 2, "problem": "Let $BC$ be a diameter of circle $\\omega$ with center $O$. Let $A$ be a point\nof circle $\\omega$ such that $0^\\circ < \\angle AOB < 120^\\circ$. Let $D$ be the midpoint\nof arc $AB$ not containing $C$. Line $\\ell$ passes through $O$ and is\nparallel to line $AD$. Line $\\ell$ intersects line $AC$ at $J$.\nThe perpendicular bisector of segment $OA$ intersects circle $\\omega$ at $E$\nand $F$. Prove that $J$ is the incenter of triangle $CEF$.", "solution": "By construction, $AEOF$ is a rhombus with $60\\dg$-$120\\dg$ angles.\nConsequently, we may set $s = AO = AE = AF = EO = EF$.\n[Figure omitted]\n\nWe have $AJ = s$ too.\n\nIt suffices to show $AJ = AO$ which is angle chasing.\nLet $\\theta = \\angle BOD = \\angle DOA$,\nso $\\angle BOA = 2\\theta$.\nThus $\\angle CAO = 1/2 \\angle BOA = \\theta$.\nHowever $\\angle AOJ = \\angle OAD = 90\\dg - 1/2 \\theta$,\nas desired.\n\nThen, since $AE = AJ = AF$,\nwe are done by the infamous Fact 5."}
+{"year": 2002, "problem_number": 3, "problem": "Find all pairs of positive integers $m, n \\ge 3$ for which there exist infinitely many\npositive integers $a$ such that\n$ \\frac{a^m+a-1}{a^n+a^2-1} $\nis itself an integer.", "solution": "The condition is equivalent to $a^n+a^2-1$\ndividing $a^m+a-1$ as polynomials.\nThe big step is the following analytic one.\n\nWe must have $m \\le 2n$.\n\nAssume on contrary $m > 2n$\nand let $0 < r < 1$ be the unique real number\nwith $r^n+r^2 = 1$, hence $r^m+r = 1$.\nBut now\n\n0 &= r^m + r - 1 \\le r^{2n+1} + r - 1 \\\\\n&= r( (1-r^2)^2+1 ) - 1 \\\\\n&= -(1-r)( r^4+r^3-r^2-r+1 ) \\\\\n&= -(1-r)( r^4 + (1-r)(1-r^2) ).\n\nHewer, $1-r > 0$ and $r^4 + (1-r)(1-r^2) > 0$ are both obvious for $0 < r < 1$.\nSo this is a contradiction.\n\nNow for the algebraic part.\nObviously $m > n$.\n\na^n+a^2-1 &\\mid a^m+a-1 \\\\\n\\iff a^n+a^2-1 &\\mid (a^m+a-1)(a+1) = a^m(a+1) + (a^2-1) \\\\\n\\iff a^n+a^2-1 &\\mid a^m(a+1) - a^n \\\\\n\\iff a^n+a^2-1 &\\mid a^{m-n}(a+1) - 1.\n\nThe right-hand side has degree $m-n+1 \\le n+1$,\nand the leading coefficients are both $+1$.\nSo the only possible situations are\n\na^{m-n}(a+1) - 1 &= (a+1)( a^n+a^2-1 ) \\\\\na^{m-n}(a+1) + 1 &= a^n+a^2-1.\n\nThe former fails by just taking $a=-1$;\nthe latter implies $(m,n) = (5,3)$.\nAs our work was reversible, this also implies $(m,n) = (5,3)$ works, done."}
+{"year": 2002, "problem_number": 4, "problem": "Let $n \\ge 2$ be a positive integer\nwith divisors $1 = d_1 < d_2 < \\dots < d_k = n$.\nProve that $d_1d_2 + d_2d_3 + \\dots + d_{k-1} d_k$ is always less than $n^2$,\nand determine when it is a divisor of $n^2$.", "solution": "We always have\n\nd_k d_{k-1} + d_{k-1} d_{k-2} + \\dots + d_2 d_1\n&< n \\cdot \\frac n2 + \\frac n2 \\cdot \\frac n3 + \\dots \\\\\n&= ( \\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\dots ) n^2 = n^2.\n\nThis proves the first part.\n\nFor the second, we claim that this only happens\nwhen $n$ is prime (in which case we get $d_1 d_2 = n$).\nAssume $n$ is not prime (equivalently $k \\ge 2$)\nand let $p$ be the smallest prime dividing $n$.\nThen\n$ d_k d_{k-1} + d_{k-1} d_{k-2} + \\dots + d_2 d_1\n> d_k d_{k-1} = \\frac{n^2}{p} $\nexceeds the largest proper divisor of $n^2$,\nbut is less than $n^2$, so does not divide $n^2$."}
+{"year": 2002, "problem_number": 5, "problem": "Find all functions $f \\colon \\RR \\to \\RR$ such that\n$ (f(x)+f(z))(f(y)+f(t))\n= f(xy-zt)+f(xt+yz) $\nfor all real numbers $x$, $y$, $z$, $t$.", "solution": "The answer is $f(x) \\equiv 0$, $f(x) \\equiv 1/2$\nand $f(x) \\equiv x^2$ which are easily seen to work.\nLet's prove they are the only ones;\nwe show two solutions.\n\nFirst solution (multiplicativity).: \nLet $P(x,y,z,t)$ denote the given statement.\n\nBy comparing $P(x,1,0,0)$ and $P(0,0,1,x)$\nwe get $\\boxed{f \\text{ even}}$.\nBy $P(0,y,0,t)$ we get for nonconstant $f$\nthat $f(0) = 0$.\nIf $f$ is constant we get the solutions earlier,\nso in the sequel assume $\\boxed{f(0) = 0}$.\nBy $P(x,y,0,0)$ we get $\\boxed{f(xy) = f(x) f(y)}$.\nNote in particular that for any real number $x$ we now have\n$ f(x) = f(|x|) = f( \\sqrt{|x|} )^2 \\ge 0 $\nthat is, $f \\ge 0$.\n\nFrom $P(x,y,y,x)$ we now have\n$ f(x^2 + y^2) = ( f(x) + f(y) )^2\n= f(x^2) + 2f(x)f(y) + f(y^2) \\ge f(x^2) $\nso $f$ is weakly increasing.\nCombined with $f$ multiplicative and nonconstant,\nthis implies $f(x) = |x|^r$ for some real number $r$.\n\nFinally, $P(1,1,1,1)$ gives $f(2) = 4f(1)$,\nso $f(x) \\equiv x^2$.\n\nSecond solution (ELMO).: \nLet $P(x,y,z,t)$ denote the statement.\nAssume $f$ is nonconstant,\nas before we derive that $f$ is even, $f(0) = 0$,\nand $f(x) \\ge 0$ for all $x$.\n\nNow comparing $P(x,y,z,t)$ and $P(z,y,x,t)$ we obtain\n$ f(xy-zt) + f(xt+yz) =\n( f(x)+f(z) )\n( f(y)+f(t) )\n= f(xy+zt) + f(xt-yz) $\nwhich in particular implies that\n$ f(a-d) + f(b+c) = f(a+d) + f(b-c)\n\\qquad \\text{ if } ad=bc \\text{ and } a,b,c,d > 0. $\nThus the restriction of $f$ to $(0,\\infty)$ satisfies\n\\textbf{ELMO 2011, problem 4}\nwhich implies that $f(x) = kx^2+\\ell$ for constants $k$ and $\\ell$.\nFrom here we recover the original.\n\n(Minor note: technically ELMO 2011/4 is $f \\colon (0,\\infty) \\to (0,\\infty)$\nbut we only have $f \\ge 0$,\nhowever the proof for the ELMO problem\nworks as long as $f$ is bounded below;\nwe could also just apply the ELMO problem to $f+0.01$ instead.)"}
+{"year": 2002, "problem_number": 6, "problem": "Let $n \\ge 3$ be a positive integer.\nLet $C_1$, $C_2$, \\dots, $C_n$ be unit circles in the plane,\nwith centers $O_1$, $O_2$, \\dots, $O_n$ respectively.\nIf no line meets more than two of the circles, prove that\n$ \\sum_{1 \\le i < j \\le n } \\frac{1}{O_i O_j}\n\\le \\frac{(n-1)\\pi}{4}. $", "solution": "For brevity, let $d_{ij}$ be the length of $O_{ij}$\nand let $\\angle(ijk)$ be shorthand for $\\angle O_i O_j O_k$\n(or its measure in radians).\n\nFirst, we eliminate the circles completely\nand reduce the problem to angles using the following fact\n(which is in part motivated by the mysterious\npresence of $\\pi$ on right-hand side,\nand also brings $d_{ij}\\inv$ into the picture).\n\nFor any indices $i$, $j$, $m$ we have the inequalities\n$ \\angle (imj) \\ge\n\\max ( \\frac{2}{d_{mi}}, \\frac{2}{d_{mj}} )\n\\quad\\text{and}\\quad\n\\pi - \\angle (imj) \\ge\n\\max ( \\frac{2}{d_{mi}}, \\frac{2}{d_{mj}} ). $\n\nWe first prove the former line.\nConsider the altitude from $O_i$ to $O_m O_j$.\nThe altitude must have length at least $2$,\notherwise its perpendicular bisector passes\nintersects all of $C_i$ , $C_m$, $C_j$.\nThus\n$ 2 \\le d_{mi} \\sin \\angle(imj) \\le \\angle(imj) $\nproving the first line.\nThe second line follows by considering the external\nangle formed by lines $O_m O_i$ and $O_m O_j$\ninstead of the internal one.\n\nOur idea now is for any index $m$\nwe will make an estimate on\n$\\sum_{\\substack{1 \\le i \\le n \\\\ i \\neq b}} \\frac{1}{d_{bi}}$\nfor each index $b$.\nIf the centers formed a convex polygon,\nthis would be much simpler,\nbut because we do not have this assumption some more care is needed.\n\nSuppose $O_a$, $O_b$, $O_c$ are consecutive vertices\nof the convex hull.\nThen\n$ \\frac{n-1}{n-2} \u2220(abc) \\ge \\frac{2}{d_{1b}} + \\frac{2}{d_{2b}}\n+ \\dots + \\frac{2}{d_{nb}} $\nwhere the term $\\frac{2}{d_{bb}}$ does not appear (obviously).\n\nWLOG let's suppose $(a,b,c) = (2,1,n)$ and\nthat rotating ray $O_2 O_1$ hits $O_3$, $O_4$, \\dots, $O_n$\nin that order.\nWe have\n\n\\frac{2}{d_{12}} &\\le \\angle(213) \\\\\n\\frac{2}{d_{13}} &\\le \\min \\left\\{ \\angle(213), \\angle(314) \\right\\} \\\\\n\\frac{2}{d_{14}} &\\le \\min \\left\\{ \\angle(314), \\angle(415) \\right\\} \\\\\n&\\vdotswithin= \\\\\n\\frac{2}{d_{1(n-1)}} &\\le\n\\min \\left\\{ \\angle((n-2)1(n-1)), \\angle((n-1)1n) \\right\\} \\\\\n\\frac{2}{d_{1n}} &\\le \\angle( (n-1) 1 n ).\n\nOf the $n-1$ distinct angles appearing on the right-hand side,\nwe let $\\kappa$ denote the smallest of them.\nWe have $\\kappa \\le \\frac{1}{n-2} \\angle(21n)$\nby pigeonhole principle.\nThen we pick the minimums on the right-hand side in\nthe unique way such that summing gives\n\n\\sum_{i=2}^n \\frac{2}{d_{1i}}\n&\\ge ( \\angle(213)+\\angle(314)+\\dots+\\angle( (n-1)1n ) )\n+ \\kappa \\\\\n&\\ge \\angle(21n) + \\frac{1}{n-2} \\angle(21n) = \\frac{n-1}{n-2} \\angle(21n)\n\nas desired.\n\nNext we show a bound that works for any center,\neven if it does not lie on the convex hull $\\mathcal H$.\n\nFor any index $b$ we have\n$ \\frac{n-1}{n-2} \\pi \\ge \\frac{2}{d_{1b}} + \\frac{2}{d_{2b}}\n+ \\dots + \\frac{2}{d_{nb}} $\nwhere the term $\\frac{2}{d_{bb}}$ does not appear (obviously).\n\nThis is the same argument as in the previous proof,\nwith the modification that\nbecause $O_b$ could lie inside the convex hull now,\nour rotation argument should use lines instead of rays\n(in order for the angle to be $\\pi$ rather than $2\\pi$).\nThis is why the first lemma is stated with two cases;\nwe need it now.\n\nAgain WLOG $b=1$.\nConsider line $O_{1} O_2$ (rather than just the ray)\nand imagine rotating it counterclockwise through $O_2$;\nsuppose that this line passes through $O_3$, $O_4$, \\dots, $O_{n}$\nin that order before returning to $O_{2}$ again.\nWe let $\u2220 (i1j) \\in \\{ \\angle (i1j), \\pi-\\angle(i1j) \\} \\in [0, \\pi)$ \nbe the counterclockwise rotations obtained in this way,\nso that\n$ \u2220(21n) = \u2220(213) + \u2220(314) +\n+ \\dots + \u2220((n-1)1n). $\n(This is not ``directed angles'', but related.)\n\nThen we get bounds\n\n\\frac{2}{d_{12}} &\\le \u2220(213) \\\\\n\\frac{2}{d_{13}} &\\le \\min \\left\\{ \u2220(213), \u2220(314) \\right\\} \\\\\n&\\vdotswithin= \\\\\n\\frac{2}{d_{1(n-1)}} &\\le\n\\min \\left\\{ \u2220((n-2)1(n-1)), \u2220((n-1)1n) \\right\\} \\\\\n\\frac{2}{d_{1n}} &\\le \u2220\\left\\{ (n-1) 1 n \\right\\}\n\nas in the last proof, and so as before we get\n$ \\sum_{i=1}^n \\frac{2}{d_{1i}} \\le \\frac{n-1}{n-2} \u2220(21n) $\nwhich is certainly less than $\\frac{n-1}{n-2} \\pi$.\n\nNow suppose there were $r$ vertices in the convex hull.\nIf we sum the first claim across all $b$ on the hull,\nand the second across all $b$ not on the hull (inside it),\nwe get\n\n\\sum_{1 \\le i<j \\le n} \\frac{2}{d_{ij}}\n&= 1/2 \\sum_b \\sum_{i \\neq b} \\frac{2}{d_{bi}} \\\\\n&\\le 1/2 \\cdot \\frac{n-1}{n-2} ( (r-2)\\pi + (n-r)\\pi ) \\\\\n&= \\frac{(n-1)\\pi}{4}\n\nas needed (with $(r-2)\\pi$ being the sum of angles in the hull)."}
+{"year": 2003, "problem_number": 1, "problem": "Let $A$ be a $101$-element subset of $S=\\{1,2,\\dots,10^6\\}$.\nProve that there exist numbers $t_1$, $t_2, \\dots, t_{100}$ in $S$ such that the sets\n$ A_j=\\{x+t_j\\mid x\\in A\\},\\qquad j=1,2,\\dots,100 $\nare pairwise disjoint.", "solution": "A greedy algorithm works: suppose we have picked\n$ T = \\left\\{ t_1, \\dots, t_n \\right\\} $\nas large as possible, meaning it's impossible to add any more elements to $T$.\nThat means, for each $t \\in \\left\\{ 1, \\dots, 10^6 \\right\\}$ either $t \\in T$ already\nor there exists two distinct elements $a, b \\in A$ and $t_i \\in T$ such that\n$ t = t_i + b - a \\qquad (\\star). $\n\nThere are at most\n$|T| \\cdot |A| \\cdot ( |A|-1 ) = n \\cdot 101 \\cdot 100$\npossible values for the right-hand side of $(\\star)$.\nSo we therefore must have\n$ 101 \\cdot 100 \\cdot n + n \\ge 10^6 $\nwhich implies $n > 99$, as desired.\n\nIt is possible to improve the bound significantly with a small optimization;\nrather than adding any $t$, we require that $t_1 < \\dots < t_n$\nand that at each step we add the least $t \\in S$ which is permitted.\nIn that case, one finds we only need to consider $b > a$ in $(\\star)$,\nand so this will save us a factor of $2+o(1)$\nas the main term $101 \\cdot 100$ becomes $\\binom{101}{2}$ instead.\nThis proves it's possible to choose $198$ elements.\n\nSee, e.g., for such a write-up."}
+{"year": 2003, "problem_number": 2, "problem": "Determine all pairs of positive integers $(a,b)$ such that\n$ \\frac{a^2}{2ab^2-b^3+1} $\nis a positive integer.", "solution": "The answer is $(a,b) = (2\\ell, 1)$, $(a,b) = (\\ell, 2\\ell)$\nand $(a,b) = (8\\ell^4-\\ell, 2\\ell)$, for any $\\ell$.\nCheck these work.\n\nIn the sequel, assume $b > 1$,\nand integers $a$, $b$, $k$ obey $k = \\frac{a^2}{2ab^2-b^3+1}$.\nExpanding, we have the polynomial\n$ X^2 - 2kb^2 \\cdot X + k(b^3-1) = 0 $\nhas two integer roots, one of which is $X = a$.\nThis means solutions to the original problem come in pairs\n(even with $k$ fixed):\n$ (a,b) \\longleftrightarrow\n( 2kb^2 - a, b)\n= ( \\frac{k(b^3-1)}{a}, b). $\n(Here, the first representation ensures\n$2kb^2-a \\in \\ZZ$,\nwhile the latter representation and the hypothesis $b > 1$ ensures\nthat $\\frac{k(b^3-1)}{a} > 0$.)\n\nOn the other hand, we claim that:\n\nFor any solution $(a,b)$,\neither $2a = b$ or $a > b$.\n\nSince the denominator is positive, $a \\ge b/2$.\nNow,\n$ a^2 \\ge 2ab^2 - b^3 + 1 \\iff a^2 \\ge b^2(2a-b) + 1 $\nand so if $2a - b > 0$ then $a^2 > b^2 \\implies a > b$.\n\nNow assume we have pair $(a_1, b)$ and $(a_2, b)$\nof solutions with $b \\neq 2a_1, 2a_2$.\nThen assume $a_1 > a_2 > b$ and\n\na_1 + a_2 &= 2k \\cdot b^2 \\\\\na_1a_2 &= k(b^3-1)\n\nThat's impossible, since then $a_1 > \\frac{a_1+a_2}{2} = k b^2$\nand hence $a_1a_2 > kb^2 \\cdot b = kb^3$.\nThus the only solutions are the ones we claimed at the beginning.\n\nImportant to notice that the problem is positive divides,\nnot just divides.\nThere is an implicit inequality built in to the problem\nstatement and it is essentially impossible to solve without.\nI would be interested in a pair $(a,b)$\nfor which $k < 0$, $k \\in \\ZZ$ yet $a, b > 0$."}
+{"year": 2003, "problem_number": 3, "problem": "Each pair of opposite sides of convex hexagon has the property that\nthe distance between their midpoints is $\\frac{\\sqrt3}{2}$\ntimes the sum of their lengths.\nProve that the hexagon is equiangular.", "solution": "Unsurprisingly, this is a geometric inequality.\nDenote the hexagon by $ABCDEF$.\nThen we have that\n$\n\\left|\n\\frac{\\vec D + \\vec E}{2} - \\frac{\\vec A + \\vec B}{2}\n\\right|\n= \\sqrt3 \\cdot \\frac{\\left| \\vec B - \\vec A \\right|\n+ \\left| \\vec E - \\vec D \\right|}{2}\n\\ge \\sqrt 3 \\cdot\n\\left| \\frac{(\\vec B - \\vec A) - (\\vec E - \\vec D)}{2} \\right|\n$\nand cyclic variations.\nSuppose we define the right-hand sides as variables\n\n\\vec x = (\\vec B - \\vec A) - (\\vec E - \\vec D) \\\\\n\\vec y = (\\vec D - \\vec C) - (\\vec A - \\vec F) \\\\\n\\vec z = (\\vec F - \\vec E) - (\\vec C - \\vec B).\n\nThen we now have\n\n\\left| \\vec y - \\vec z \\right|\n&\\ge \\sqrt 3 \\left| \\vec x \\right| \\\\\n\\left| \\vec z - \\vec x \\right|\n&\\ge \\sqrt 3 \\left| \\vec y \\right| \\\\\n\\left| \\vec x - \\vec y \\right|\n&\\ge \\sqrt 3 \\left| \\vec z \\right|.\n\nWe square all sides (using\n$\\left| \\vec v \\right|^2 = \\vec v \\cdot \\vec v$)\nand then sum to get\n$ \\sum_{\\text{cyc}} (\\vec y - \\vec z) \\cdot (\\vec y - \\vec z)\n\\ge 3 \\sum_{\\text{cyc}} \\vec x \\cdot \\vec x $\nwhich rearranges to\n$- \\left| \\vec x + \\vec y + \\vec z \\right|^2 \\ge 0. $\nThis can only happen if $\\vec x + \\vec y + \\vec z =0$,\nand moreover all the inequalities above were actually equalities.\nThat means that our triangle inequalities above were actually sharp\n(and already we have $AB \\parallel DE$ and so on).\n\nWorking with just $x$ and $y$ now we have\n\n3 (\\vec x \\cdot \\vec x) &= (2 \\vec y - \\vec x) \\cdot (2 \\vec y - \\vec x) \\\\\n&= \\vec x \\cdot \\vec x - 4 \\vec y \\cdot \\vec x + 4 \\vec y \\cdot \\vec y \\\\\n\\implies\n-\\vec x \\cdot \\vec x + 2 (\\vec y \\cdot \\vec y) &= 2 \\vec x \\cdot \\vec y \\\\\n2 (\\vec x \\cdot \\vec x) - \\vec y \\cdot \\vec y &= 2 \\vec x \\cdot \\vec y.\n\nwhich implies $\\vec x \\cdot \\vec x = \\vec y \\cdot \\vec y$,\nthat is, $\\vec x$ and $\\vec y$ have the same magnitude.\nIn this way we find $\\vec x$, $\\vec y$, $\\vec z$ all\nhave the same magnitude,\nand since $\\vec x + \\vec y + \\vec z = 0$\nthey are related by $120\\dg$ rotations, as desired.\n\nIn fact one can show further that the equiangular hexagons\nwhich work are exactly those formed by taking an equilateral triangle\nand cutting off equally sized corners.\nThis equality case helps motivate the solution.\n\nOne can note this ``must'' be an inequality\nbecause the space of such hexagons is $2$-dimensional,\neven though a priori the space of hexagons satisfying\nthree given conditions should have dimension $9-3=6$."}
+{"year": 2003, "problem_number": 4, "problem": "Let $ABCD$ be a cyclic quadrilateral.\nLet $P, Q$ and $R$ be the feet of perpendiculars\nfrom $D$ to lines $BC$, $CA$ and $AB$, respectively.\nShow that $PQ = QR$ if and only if the\nbisectors of angles $ABC$ and $ADC$ meet on segment $AC$.", "solution": "Let $\\gamma$ denote the circumcircle of $ABCD$.\nThe condition on bisectors is equivalent to $(AC;BD)_\\gamma = -1$.\nMeanwhile if $\\infty$ denotes the point at infinity along Simson line $PQR$\nthen $PQ = QR$ if and only if $(PR;Q\\infty) = -1$.\n\nLet rays $BQ$ and $DQ$ meet the circumcircle again at $F$ and $E$.\n\n[Figure omitted]\n\n[EGMO Proposition 4.1]\nThen $BE \\parallel PQR$.\n\nSince $\u2220 DPR = \u2220 DAR = \u2220 DAB = \u2220 DEB$.\n\nNow we have\n$ (PR;Q\\infty) \\overset{B}{=} (CA;FE)_\\gamma\n\\overset{Q}{=} (AC;BD)_\\gamma $\nas desired."}
+{"year": 2003, "problem_number": 5, "problem": "Let $n$ be a positive integer and\nlet $x_1 \\le x_2 \\le \\dots \\le x_n$ be real numbers.\nProve that\n$ (\\sum_{i=1}^{n}\\sum_{j=1}^{n} |x_i - x_j|)^2\n\\le \\frac{2(n^2-1)}{3}\\sum_{i=1}^{n}\\sum_{j=1}^{n} (x_i - x_j)^2 $\nwith equality if and only if $x_1$, $x_2$, \\dots, $x_n$\nform an arithmetic sequence.", "solution": "Let $d_1 = x_2 - x_1$, \\dots, $d_{n-1} = x_n - x_{n-1}$.\nThe inequality in question becomes:\n$\n( \\sum_i i(n-i) d_i )^2\n\\le\n\\frac{n^2-1}{3} \\cdot\n( \\sum_i i(n-i) d_i^2 + 2\\sum_{i<j} i(n-j) d_i d_j ).\n$\nClearing the square on the right-hand side we want to show\n$\n\\sum_{i<j} ( 3ij(n-i)(n-j) - (n^2-1)i(n-j) ) \\cdot 2d_i d_j\n\\le \\sum_i (n^2-1-3i(n-i)) \\cdot i(n-i) d_i^2.\n$\nWe use AM-GM directly on $2d_i d_j \\le d_i^2 + d_j^2$:\nthis actually solves the problem.\nThe annoying part is to check that the coefficients actually match:\n\n[Big bash]\nFor an index $1 \\le k \\le n-1$, we have\n\n&\\sum_{i<k} ( 3ik(n-i)(n-k) - (n^2-1)i(n-k) ) \\\\\n+& \\sum_{j>k} ( 3kj(n-k)(n-j) - (n^2-1)k(n-j) ) \\\\\n= & (n^2-1-3k(n-k)) \\cdot k(n-k).\n\nRewrite as:\n\n3k(n-k) ( -k(n-k) + \\sum_i i(n-i) )\n&= (n^2-1)( (n-k)\\sum_{i<k} i + k\\sum_{j>k} (n-j) ) \\\\\n&+ (n^2-1-3k(n-k)) \\cdot k(n-k) \\\\\n\\iff 3k(n-k) \\sum_i i(n-i)\n&= (n^2-1) ( (n-k)\\sum_{i<k} i + k\\sum_{j>k}(n-j) ) \\\\\n&+ (n^2-1)k(n-k) - 3k^2(n-k)^2 \\\\\n\\iff 3k(n-k) ( \\sum_{i} i(n-i) )\n&= (n^2-1) ( (n-k)\\sum_{i \\le k} i + k \\sum_{i < n-k} i ) \\\\\n\\iff 3k(n-k) \\frac{(n-1)n(n+1)}{6}\n&= (n^2-1) ( (n-k)\\frac{k(k+1)}{2} ) \\\\\n& + (n^2-1) ( k \\frac{(n-k)(n-k-1)}{2} ) \\\\\n\\iff 3k(n-k) \\frac{(n-1)n(n+1)}{6}\n&= (n^2-1)k(n-k) \\cdot \\frac n2\n\nwhich is visibly true.\n\nEquality occurs only if all $d_i$ are equal\nbecause the coefficient of $d_i d_j$ is nonzero\nfor any $i \\le n/2$ and $j \\ge n/2$."}
+{"year": 2003, "problem_number": 6, "problem": "Let $p$ be a prime number.\nProve that there exists a prime number $q$ such that for every integer $n$,\nthe number $n^p-p$ is not divisible by $q$.", "solution": "By orders, we must have $q=pk+1$ for this to be possible\n(since if $q \\not \\equiv 1 \\pmod p$, then $n^p$ can be any residue modulo $q$).\nSince $p \\equiv n^p \\pmod q \\implies p^k \\equiv 1 \\pmod q$,\nit suffices to prevent the latter situation from happening.\n\nSo we need a prime $q \\equiv 1 \\pmod p$ such that $p^k \\not\\equiv 1 \\pmod q$.\nTo do this, we first recall the following lemma.\n\nLet $\\Phi_p(X) = 1 + X + X^2 + \\dots + X^{p-1}$.\nFor any integer $a$, if $q$ is a prime divisor of $\\Phi_p(a)$ other than $p$,\nthen $a \\pmod q$ has order $p$. (In particular, $q \\equiv 1 \\pmod p$.)\n\nWe have $a^p-1 \\equiv 0 \\pmod q$, so either the order is $1$ or $p$.\nIf it is $1$, then $a \\equiv 1 \\pmod q$, so $q \\mid \\Phi_p(1) = p$, hence $q = p$.\n\nNow the idea is to extract a prime factor $q$\nfrom the cyclotomic polynomial\n$ \\Phi_p(p) = \\frac{p^p-1}{p-1} \\equiv 1+p \\pmod{p^2} $\nsuch that $q \\not\\equiv 1 \\pmod{p^2}$;\nhence $k \\not\\equiv 0 \\pmod p$,\nand as $p \\pmod q$ has order $p$ we have $p^k \\not\\equiv 1 \\pmod q$."}
+{"year": 2004, "problem_number": 1, "problem": "Let $ABC$ be an acute-angled triangle with $AB\\neq AC$.\nThe circle with diameter $BC$ intersects the sides $AB$ and $AC$\nat $M$ and $N$ respectively.\nDenote by $O$ the midpoint of the side $BC$.\nThe bisectors of the angles $\\angle BAC$ and $\\angle MON$ intersect at $R$.\nProve that the circumcircles of the triangles $BMR$ and $CNR$\nhave a common point lying on the side $BC$.", "solution": "By Miquel's theorem it's enough to show $AMRN$ is cyclic.\n[Figure omitted]\nIn fact, since the bisector of $\\angle MON$\nis just the perpendicular bisector of $MN$,\nthe point $R$ is actually just the arc midpoint\nof $\\widehat{MN}$ of $(AMN)$ as desired."}
+{"year": 2004, "problem_number": 2, "problem": "Find all polynomials $P$ with real coefficients such that\nfor all reals $a$, $b$, $c$ such that $ab+bc+ca = 0$, we have\n$ P(a-b) + P(b-c) + P(c-a) = 2P(a+b+c). $", "solution": "The answer is $ P(x) = \\alpha x^4 + \\beta x^2 $\nwhich can be checked to work, for any real numbers $\\alpha$ and $\\beta$.\n\nIt is easy to obtain that $P$ is even and $P(0) = 0$.\nThe trick is now to choose $(a,b,c) = (6x,3x,-2x)$\nand then compare the leading coefficients to get\n$ 3^n + 5^n + 8^n = 2 \\cdot 7^n $\nfor $n = \\deg f$ (which is even).\nAs $n \\ge 7 \\implies (8/7)^n > 2$, this means that we must have $n \\le 6$.\nMoreover, taking modulo $7$ we have $3^n + 5^n \\equiv 6 \\pmod 7$\nwhich gives $n \\equiv 2, 4 \\pmod 6$.\n\nThus $\\deg P \\le 4$, which (combined with $P$ even) resolves the problem."}
+{"year": 2004, "problem_number": 3, "problem": "Define a ``hook'' to be a figure made up of six unit squares\nas shown below in the picture,\nor any of the figures obtained by applying rotations\nand reflections to this figure.\n[Figure omitted]\nWhich $m \\times n$ rectangles can be tiled by hooks?", "solution": "The answer is that one requires:\n\n$\\{1,2,5\\} \\notin \\{m,n\\}$,\n$3 \\mid m$ or $3 \\mid n$,\n$4 \\mid m$ or $4 \\mid n$.\n\nFirst, we check all of these work, in fact we claim:\n\nAny rectangle satisfying these conditions\ncan be tiled by $3 \\times 4$ rectangles (and hence by hooks).\n\nIf $3 \\mid m$ and $4 \\mid n$, this is clear.\nElse suppose $12 \\mid m$ but $3 \\nmid n$, $4 \\nmid n$.\nThen $n \\ge 7$, so it can be written in the form\n$3a+4b$ for nonnegative integers $a$ and $b$, which permits a tiling.\n\nWe now prove these conditions are necessary.\nIt is not hard to see that $m,n \\notin \\{1,2,5\\}$ is necessary.\n\nWe thus turn our attention to divisibility conditions.\nEach hook has a hole, and if we associate each hook with\nthe one that fills its hole, we get a bijective pairing of hooks.\nThus the number of cells is divisible by $12$,\nand the cells come grouped into two possible shapes,\nwhich we will call \\textbf{tiles} shown below,\n(rotations and reflections permitted).\n[Figure omitted]\n\nIn particular, the total number of cells is divisible by $12$.\nThus the problem is reduced to proving that:\n\nif a $6a \\times 2b$ rectangle is tiled by tiles,\nthen at least one of $a$ and $b$ is even.\n\nNote that the tiles can be further classified into two types:\n\n\\textbf{First type}: These tiles have exactly four columns,\neach with exactly three cells (an odd number).\nMoreover, all rows have an even number of cells (either $2$ or $4$).\n\\textbf{Second type}: vice-versa.\nThese tiles have exactly four rows,\neach with exactly three cells (an odd number).\nMoreover, all rows have an odd number of cells.\n\nWe claim that any tiling uses an even number of each type, which is enough.\n\nBy symmetry we prove an even number of first-type tiles.\nColor red every fourth column of the rectangle.\nThe number of cells colored red is even.\nThe tiles of the second type cover an even number of red cells,\nand the tiles of the first type cover an odd number of red cells.\nHence the number of tiles of the first type must be even.\n\nThis shows that a rectangle can be tiled by hooks\nif and only if it can be tiled by $3 \\times 4$ rectangles.\nBut there exist tilings which do not decompose into $3 \\times 4$;\nsee e.g.\\ ."}
+{"year": 2004, "problem_number": 4, "problem": "Let $n \\ge 3$ be an integer\nand $t_1$, $t_2$, \\dots, $t_n$ positive real numbers such that\n$ n^2+1 > (t_1 + t_2 + \\dots + t_n)\n( \\frac{1}{t_1} + \\frac{1}{t_2} + \\dots + \\frac{1}{t_n} ). $\nShow that $t_i$, $t_j$, $t_k$ are the sides of a triangle\nfor all $i$, $j$, $k$ with $1 \\le i < j < k \\le n$.", "solution": "Let $a = t_1$, $b = t_2$, $c = t_3$.\nExpand:\n\nn^2+1 &> (t_1 + t_2 + \\dots + t_n)\n( \\frac{1}{t_1} + \\dots + \\frac{1}{t_n} ) \\\\\n&= n + \\sum_{1 \\le i < j \\le n}\n( \\frac{t_i}{t_j} + \\frac{t_j}{t_i} ) \\\\\n&= n + \\sum_{1 \\le i < j \\le n}\n( \\frac{t_i}{t_j} + \\frac{t_j}{t_i} ) \\\\\n&\\ge n + \\sum_{1 \\le i < j \\le 3}\n( \\frac{t_i}{t_j} + \\frac{t_j}{t_i} )\n+ \\sum_{\\substack{1 \\le i < j \\le n \\\\ j > 3 }} 2 \\\\\n&= n + 2( \\binom n2-3 )\n+ ( \\frac ab + \\frac ba )\n+ \\frac{a+b}{c} + \\frac{c}{b} + \\frac{c}{a} \\\\\n&\\ge n + 2( \\binom n2-3 ) + 2\n+ \\frac{a+b}{c} + c \\cdot \\frac{4}{a+b}\n\nSo, we conclude that\n$ \\frac{a+b}{c} + \\frac{4c}{a+b} < 5 $\nwhich rearranges to\n$ ( 4c-(a+b) )( c-(a+b) ) < 0. $\nThis is enough to imply $c \\le a+b$.\n\nA variant of the same argument allows one to improve\nthe left-hand side to $(n+\\sqrt{10}-3)^2$.\nOne does so by writing\n$ \\text{RHS} \\ge ( \\sqrt{( a+b+c )\n( \\frac1a+\\frac1b+\\frac1c )} + (n-3) )^2 $\nand estimating the square root as in the previous approach.\n\nIn addition, $(n+\\sqrt{10}-3)^2$ is best possible,\nas seen by taking $(a,b,c) = (2,1,1)$ and $t_4 = t_5 = \\dots = \\frac25 \\sqrt{10}$."}
+{"year": 2004, "problem_number": 5, "problem": "In a convex quadrilateral $ABCD$,\nthe diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$.\nThe point $P$ lies inside $ABCD$ and satisfies\n$\\angle PBC=\\angle DBA \\quad\\text{and}\\quad \\angle PDC=\\angle BDA. $\nProve that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$.", "solution": "We show two solutions.\nWe note that the first hypothesis cannot be dropped, because $ABCD$ being a kite\nwith $BA=BC$ and $DA=DC$ with center $P$ is a counterexample\n(where $AP=CP$ but $ABCD$ is not cyclic).\nHowever, the condition that $BD$ bisects neither angle $\\angle B$ nor $\\angle D$\nis equivalent to requiring that the point $P$ does not lie on line $BD$\nand we assume this henceforth.\n\nBarycentric solution (from teenager Evan).: \nApply barycentric coordinates to $\\triangle PBD$\n(nondegenerate by the non-bisect condition)\nwith $P = (1,0,0)$, $B = (0,1,0)$ and $D = (0,0,1)$.\nDefine $a = BD$, $b = DP$ and $c = PB$.\n\nSince $A$ and $C$ are isogonal conjugates with respect to $\\triangle PBD$,\nwe set $ A = (au : bv : cw) \\qquad\\text{and}\\qquad\nC = ( \\frac au : \\frac bv : \\frac cw ). $\nFor brevity define $M = au + bv + cw$ and $N = avw + bwu + cuv$.\n\nWe now compute each condition.\n\nQuadrilateral $ABCD$ is cyclic if and only if $N^2 = u^2 M^2$.\n\nW know a circle through $B$ and $D$\nis a locus of points\nwith $ \\frac{a^2yz+b^2zx+c^2xy}{x(x+y+z)} $\nis equal to some constant.\nTherefore quadrilateral $ABCD$ is cyclic if and only if\n$\\frac{abc \\cdot N}{au \\cdot M}$\nis equal to $\\frac{abc \\cdot uvw \\cdot M}{avw \\cdot N}$\nwhich rearranges to $N^2 = u^2M^2$.\n\nWe have $PA = PC$ if and only if $N^2 = u^2 M^2$.\n\nWe have the displacement vector\n$\\overrightarrow{PA}\n= \\tfrac{1}{M} ( bv+cw , -bv ,-cw )$.\nTherefore,\n\nM^2 \\cdot \\left| PA \\right|^2\n&= -a^2(bv)(cw) + b^2(cw)(bv+cw) + c^2(bv)(bv+cw) \\\\\n&= bc(-a^2vw + (bw+cv)(bv+cw)).\n\nIn a similar way\n(by replacing $u$, $v$, $w$ with their inverses)\nwe have\n\n( \\frac{N}{uvw} )^2 \\cdot \\left| PC \\right|^2\n&= (vw)^{-2} \\cdot bc(-a^2vw + (bv+cw)(bw+cv)) \\\\\n\\iff N^2 \\cdot \\left| PC \\right|^2\n& = u^2 bc(-a^2vw + (bw+cv)(bv+cw))\n\nThese are equal if and only if $N^2 = u^2M^2$, as desired.\n\nAngle chasing solution (Evan Chen and Petko Lazarov).: \nThe solution consists of two parts.\nThe first part is that by angle chasing, we will prove that\n$ \u2220 PAC = \u2220 ACP \\iff \u2220 BAC + \u2220 CBD + \u2220 DCA + \u2220 ADB = 0.\n\\qquad (\\spadesuit). $\nA careless reader would be forgiven for thinking that $(\\spadesuit)$\nimplies the problem or at least one direction,\nbut it turns out the situation is more subtle.\nThe second part analyzes the angle conditions more carefully\nand provides a complete proof.\n\n\\subparagraph{Proof of the equivalence $(\\spadesuit)$ by angle chasing.}\nWe start with the following unconditional claim, valid for any quadrilateral.\n\n[Isogonal conjugation]\nLet $ABCD$ and $P$ be as in the problem statement.\nThen $\u2220 APB + \u2220 CPD = 180\\dg$.\n\nThe angles in the statement imply that $A$ and $C$\nare isogonal conjugates with respect to $\\triangle PBC$.\nThus, lines $PA$ and $PC$ are isogonal with respect to $\\angle BPC$, as needed.\n\n[Figure omitted]\n\nNext we rewrite the two angles $\u2220 APB$ and $\u2220 CPD$ in the claim\n(colored green with three rings)\nso that their only dependence on $P$\nis through the angles $\u2220 PAC$ and $\u2220 CAP$ (colored red with two rings),\nas follows:\n\n-\u2220 APB &= \u2220 PBA + \u2220 BAP = \u2220 PBA + (\u2220 BAC - \u2220 PAC) \\\\\n&= \u2220 CBD + \u2220 BAC - \u2220 PAC \\\\\n-\u2220 CPD &= \u2220 PDC + \u2220 DCP = \u2220 PDC + (\u2220 DCA + \u2220 ACP) \\\\\n&= \u2220 ADB + \u2220 DCA + \u2220 ACP.\n\nSince the claim says $\u2220 APB + \u2220 CPD = 0$, summing\nlets us finally rewrite $\u2220 PAC - \u2220 APC$ in terms of only $A$, $B$, $C$, $D$:\n\n0 &= (\u2220 ACP - \u2220 PAC) + \u2220 ADB + \u2220 DCA + \u2220 CBD + \u2220 BAC \\\\\n\\implies \u2220 PAC - \u2220 ACP &= \u2220 ADB + \u2220 DCA + \u2220 CBD + \u2220 BAC.\n\nThese four latter angles are colored blue with one ring in the figure.\nThis proves $(\\spadesuit)$.\n\n\\subparagraph{Quasi-harmonic quadrilaterals.}\nTo interpret the condition $(\\spadesuit)$, we define a new term:\na quadrilateral $ABCD$ is quasi-harmonic if $AB \\cdot CD = BC \\cdot DA$.\n(See IMO 2018/6 for another problem involving quasi-harmonic quadrilaterals.)\nThe following two lemmas show why this condition is relevant:\n\nThe condition\n$ \u2220 BAC + \u2220 CBD + \u2220 DCA + \u2220 ADB = 0 $\nis equivalent to $ABCD$ being either cyclic or quasi-harmonic or both.\n\nA clean approach with complex numbers\n(posted at ) goes as follows:\ndefine complex numbers $\\mu = (a-b)(c-d) \\neq 0$ and $\\nu = (b-c)(d-a) \\neq 0$.\nThen $ \\mu-\\nu = (a-c)(b-d) \\neq 0. $\n\nNow the angle condition is equivalent to\n\n\\RR &\\ni \\frac{a-b}{d-b} \\cdot \\frac{b-c}{a-c} \\cdot \\frac{c-d}{b-d} \\cdot \\frac{d-a}{c-a} \\\\\n&= \\frac{\\mu \\nu}{(\\mu - \\nu)^2} \\\\\n&= \\frac{1}{\\frac{\\mu}{\\nu} + \\frac{\\nu}{\\mu} - 2} \\\\\n\\iff \\RR &\\ni \\frac{\\mu}{\\nu} + \\frac{\\nu}{\\mu}.\n\nIn general, though $z+1/z$ is real exactly when either $|z|=1$ or $z \\in \\RR$.\nNote $\\mu/\\nu \\in \\RR$ is equivalent to $ABCD$ cyclic\nand $|\\mu| = |\\nu|$ is equivalent to $ABCD$ being quasiharmonic; the proof follows.\n\nThis can be also proved by inversion at $A$, though we don't write the details here.\n(See also ).\n\nQuadrilateral $ABCD$ is quasi-harmonic if and only if $P$ lies on line $AC$.\n\nLet $X = BD \\cap AC$.\nIf the isogonal of line $BD$ with respect to $\\angle B$ meets line $AC$ at $Y$, then\n$\\frac{AX}{CX} \\frac{AY}{CY} = ( \\frac{BA}{BC} )^2$.\nSimilarly if the isogonal to $\\angle D$ meets line $AC$ at $Y'$, then\n$\\frac{AX}{CX} \\frac{AY'}{CY'} = ( \\frac{DA}{DC} )^2$.\nHence $ABCD$ is quasi-harmonic if and only if $Y = Y'$ (that is, $Y=Y'=P$).\n\n\\subparagraph{Wrap-up.}\nWe now show that $PA = PC$ if and only if $ABCD$ is cyclic\nby cases on whether $P$ lies on $AC$.\n\nIf $ABCD$ is not quasi-harmonic,\nthen $(\\spadesuit)$ implies the problem statement immediately.\nIndeed, $\u2220 PAC = \u2220 ACP$ if and only if $PA = PC$\n(as $\\triangle PAC$ is not degenerate)\nand the second lemma turns our angle condition into $ABCD$ cyclic.\n\nNow assume $ABCD$ is quasi-harmonic and $P$ lies on line $AC$.\nWe ignore $(\\spadesuit)$.\nInstead, note that if $ABCD$ is also cyclic then\n$BD$ is a symmedian of $\\triangle ABC$ and hence $P$ is the midpoint.\nConversely, suppose we know $BD$ is a symmedian of $\\triangle ABC$.\nLet $D' \\neq B$ be the point for which $ABCD'$ is cyclic and harmonic;\nthen $B$, $D$, $D'$ are collinear and $\\frac{BD}{CD} = \\frac{BD'}{CD'} = \\frac{BA}{CA}$.\nSo $D' = D$ (the corresponding Apollonian circle only meets line $BD$ twice), as needed."}
+{"year": 2004, "problem_number": 6, "problem": "We call a positive integer alternating if every two consecutive digits\nin its decimal representation are of different parity.\nFind all positive integers $n$ which have an alternating multiple.", "solution": "If $20 \\mid n$, then clearly $n$ has no alternating\nmultiple since the last two digits are both even.\nWe will show the other values of $n$ all work.\n\nThe construction is just rush-down do-it.\nThe meat of the solution is the two following steps.\n\n[Tail construction]\nFor every even integer $w \\ge 2$,\n\nthere exists an even alternating multiple $g(w)$ of $2^{w+1}$\nwith exactly $w$ digits, and\nthere exists an even alternating multiple $h(w)$ of $5^{w}$\nwith exactly $w$ digits.\n\n(One might note this claim is implied by the problem, too.)\n\nWe prove the first point by induction on $w$.\nIf $w = 2$, take $g(2) = 32$.\nIn general, we can construct $g(w+2)$ from $g(w)$\nby adding some element in\n$ 10^w \\cdot \\{10, 12, 14, 16, 18, 30, \\dots, 98\\} $\nto $g(w)$, corresponding to the digits\nwe want to append to the start.\nThis multiple is automatically divisible by $2^{w+1}$,\nand also can take any of the four possible values modulo $2^{w+3}$.\n\nThe second point is a similar induction,\nwith base case $h(2) = 50$.\nThe same set above consists of numbers divisible by $5^w$,\nand covers all residues modulo $5^{w+2}$.\nCareful readers might recognize the second part\nas essentially USAMO 2003/1.\n\n[Head construction]\nIf $\\gcd(n,10) = 1$, then for any $b$,\nthere exists an even alternating number $f(b \\bmod n)$ which is\n$b \\pmod n$.\n\nA standard argument shows that\n$ 10 \\cdot \\frac{100^m-1}{99}\n= \\underbrace{1010\\dots10}_{m\\ 10\\text{'s}}\n\\equiv 0 \\pmod n $\nfor any $m$ divisible by $\\varphi(99n)$.\nTake a very large such $m$,\nand then add on $b$ distinct numbers of the form $10^{\\varphi(n)r}$\nfor various even values of $r$; these all are $1 \\pmod n$\nand change some of the $1$'s to $3$'s.\n\nNow, we can solve the problem.\nConsider three cases:\n\nIf $n = 2^k m$ where $\\gcd(m,10) = 1$ and $k \\ge 2$ is even,\nthen the concatenated number\n$ 10^k f( -\\frac{g(k)}{10^k} \\bmod m ) + g(k) $\nworks fine.\n\nIf $n = 5^k m$ where $\\gcd(m,10) = 1$ and $k \\ge 2$ is even,\nthen the concatenated number\n$ 10^k f( -\\frac{h(k)}{10^k} \\bmod m ) + h(k) $\nworks fine.\n\nIf $n = 50m$ where $\\gcd(m,10) = 1$,\nthen the concatenated number\n$ 100 f( -\\frac{1}{2} \\bmod m ) + 50 $\nworks fine.\n\nSince every non-multiple of $20$ divides such a number, we are done."}
+{"year": 2005, "problem_number": 1, "problem": "Six points are chosen on the sides of an equilateral triangle $ABC$:\n$A_1$, $A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$,\nsuch that they are the vertices of a\nconvex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths.\nProve that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent.", "solution": "The six sides of the hexagon, when oriented, comprise\nsix vectors with vanishing sum.\nHowever note that $ \\overrightarrow{A_1A_2}\n+ \\overrightarrow{B_1B_2}\n+ \\overrightarrow{C_1C_2} = 0. $\nThus\n$ \\overrightarrow{A_2B_1} + \\overrightarrow{B_2C_1} +\n\\overrightarrow{C_2A_1} = 0 $\nand since three unit vectors with vanishing sum\nmust be rotations of each other by $120\\dg$,\nit follows they must also form an equilateral triangle.\n\n[Figure omitted]\n\nConsequently, triangles $A_1A_2B_1$, $B_1B_2C_1$, $C_1C_2A_1$\nare congruent, as $\\angle A_2 = \\angle B_2 = \\angle C_2$.\nSo triangle $A_1 B_1 C_1$ is equilateral and the\ndiagonals are concurrent at the center."}
+{"year": 2005, "problem_number": 2, "problem": "Let $a_1$, $a_2$, \\dots\\ be a sequence of integers\nwith infinitely many positive and negative terms.\nSuppose that for every positive integer $n$\nthe numbers $a_1$, $a_2$, \\dots, $a_n$\nleave $n$ different remainders upon division by $n$.\nProve that every integer occurs exactly once in the sequence.", "solution": "Obviously every integer appears at most once\n(otherwise take $n$ much larger).\nSo we will prove every integer appears at least once.\n\nFor any $i < j$ we have $\\left| a_i-a_j \\right| < j$.\n\nOtherwise, let $n = \\left| a_i-a_j \\right| \\neq 0$.\nThen $i,j \\in [1,n]$ and $a_i \\equiv a_j \\pmod n$,\ncontradiction.\n\nFor any $n$, the set $\\{a_1, \\dots, a_n\\}$\nis of the form $\\{k+1, \\dots, k+n\\}$ for some integer $k$.\n\nBy induction, with the base case $n=1$ being vacuous.\nFor the inductive step,\nsuppose $\\{a_1, \\dots, a_n\\} = \\{k+1, \\dots, k+n\\}$ are determined.\nThen\n$ a_{n+1} \\equiv k \\pmod{n+1}. $\nMoreover by the earlier claim we have\n$ \\left| a_{n+1}-a_1 \\right| < n+1. $\nFrom this we deduce $a_{n+1} \\in \\{k, k+n+1\\}$ as desired.\n\nThis gives us actually a complete description\nof all possible sequences satisfying the hypothesis:\nchoose any value of $a_1$ to start.\nThen, for the $n$th term,\nthe set $S = \\{a_1, \\dots, a_{n-1}\\}$\nis (in some order) a set of $n-1$ consecutive integers.\nWe then let $a_n = \\max S + 1$ or $a_n = \\min S - 1$.\nA picture of six possible starting terms is shown below.\n\n[Figure omitted]\n\nFinally, we observe that the condition that\nthe sequence has infinitely many positive and negative terms\n(which we have not used until now)\nimplies it is unbounded above and below.\nThus it must contain every integer."}
+{"year": 2005, "problem_number": 3, "problem": "Let $x,y,z > 0$ satisfy $xyz\\geq 1$. Prove that\n$ \\frac { x^5-x^2 }{x^5+y^2+z^2}\n+ \\frac {y^5-y^2}{x^2+y^5+z^2}\n+ \\frac {z^5-z^2}{x^2+y^2+z^5} \\geq 0. $", "solution": "Negating both sides and adding $3$ eliminates the minus signs:\n$ \\sum_{\\text{cyc}} \\frac{1}{x^5+y^2+z^2}\n\\le \\frac{3}{x^2+y^2+z^2}. $\nThus we only need to consider the case $xyz = 1$.\n\nDirect expansion and Muirhead works now!\nAs advertised, once we show it suffices to analyze if $xyz=1$\nthe inequality becomes more economically written as\n$ S = \\sum_{\\text{cyc}} x^2(x^2-yz)(y^4+x^3z+xz^3)(z^4+x^3y+xy^3)\n\\overset{?}{\\ge} 0. $\nSo, clearing all the denominators gives\n\nS &= \\sum_{\\text{cyc}} x^2(x^2-yz)\n[ y^4z^4 + x^3y^5 + xy^7 + x^3z^5 + x^6yz + x^4y^3z\n+ xz^7 + x^4yz^3 + x^2y^3z^3 ] \\\\\n&= \\sum_{\\text{cyc}}\n[ x^4y^4z^4 + x^7y^5 + x^5y^7 + x^7z^5 + x^{10}yz + x^8y^3z\n+ x^5z^7 + x^8yz^3 + x^6y^3z^3 ] \\\\\n&- \\sum_{\\text{cyc}}\n[ x^2y^5z^5 + x^5y^6z + x^3y^8z + x^5yz^6 + x^8y^2z^2 + x^6y^4z^2\n+ x^3yz^8 + x^6y^2z^4 + x^4y^4z^4 ] \\\\\n&= \\sum_{\\text{cyc}}\n[ x^7y^5 + x^5y^7 + x^7z^5 + x^{10}yz\n+ x^5z^7 + x^6y^3z^3 ] \\\\\n&- \\sum_{\\text{cyc}}\n[ x^2y^5z^5 + x^5y^6z + x^5yz^6 + x^8y^2z^2 + x^6y^4z^2\n+ x^6y^2z^4 ]\n\nIn other words we need to show\n$\n\\sum_{\\text{sym}} ( 2x^7y^5\n+1/2 x^{10}yz + 1/2 x^6y^3z^3 )\n\\ge\n\\sum_{\\text{sym}} ( 1/2 x^8 y^2 z^2\n+ 1/2 x^5 y^5 z^2 + x^6 y^4 z^2 + x^6 y^5 z ).\n$\nwhich follows by summing\n\n\\sum_{\\text{sym}} \\frac{x^{10} yz + x^6 y^3 z^3}{2}\n&\\ge \\sum_{\\text{sym}} x^8 y^2 z^2 \\\\\n1/2 \\sum_{\\text{sym}} x^8 y^2 z^2\n&\\ge 1/2 \\sum_{\\text{sym}} x^6 y^4 z^2 \\\\\n1/2 \\sum_{\\text{sym}} x^7y^5\n&\\ge 1/2 \\sum_{\\text{sym}} x^5 y^5 z^2 \\\\\n1/2 \\sum_{\\text{sym}} x^7y^5\n&\\ge 1/2 \\sum_{\\text{sym}} x^6 y^4 z^2 \\\\\n\\sum_{\\text{sym}} x^7y^5\n&\\ge \\sum_{\\text{sym}} x^6 y^5 z.\n\nThe first line here comes from AM-GM,\nthe rest come from Muirhead.\n\nMore elegant approach is to use Cauchy in the form\n$ \\frac{1}{x^5+y^2+z^2} \\le \\frac{x\\inv+y^2+z^2}{(x^2+y^2+z^2)^2}. $"}
+{"year": 2005, "problem_number": 4, "problem": "Determine all positive integers relatively\nprime to all the terms of the infinite sequence\n$ a_n = 2^n+3^n+6^n-1, \\quad n \\ge 1. $", "solution": "The answer is $1$ only (which works).\n\nIt suffices to show there are no primes.\nFor the primes $p=2$ and $p=3$, take $a_2=48$.\nFor any prime $p \\ge 5$ notice that\n\na_{p-2} &= 2^{p-2} + 3^{p-2} + 6^{p-2} - 1 \\\\\n&\\equiv \\frac 12 + \\frac 13 + \\frac 16 - 1 \\pmod p \\\\\n&\\equiv 0 \\pmod p\n\nso no other larger prime works."}
+{"year": 2005, "problem_number": 5, "problem": "Let $ABCD$ be a fixed convex quadrilateral\nwith $BC=DA$ and $BC \\nparallel DA$.\nLet two variable points $E$ and $F$ lie on the\nsides $BC$ and $DA$, respectively, and satisfy $BE=DF$.\nThe lines $AC$ and $BD$ meet at $P$,\nthe lines $BD$ and $EF$ meet at $Q$,\nthe lines $EF$ and $AC$ meet at $R$.\nProve that the circumcircles of the triangles $PQR$,\nas $E$ and $F$ vary, have a common point other than $P$.", "solution": "Let $M$ be the Miquel point of complete quadrilateral $ADBC$;\nin other words, let $M$ be the second intersection point\nof the circumcircles of $\\triangle APD$ and $\\triangle BPC$.\n(A good diagram should betray this secret;\nall the points are given in the picture.)\nThis makes lots of sense since we know $E$ and $F$\nwill be sent to each other under the spiral similarity too.\n\n[Figure omitted]\n\nThus $M$ is the Miquel point of complete quadrilateral $FACE$.\nAs $R = FE \\cap AC$ we deduce $FARM$ is a cyclic quadrilateral\n(among many others, but we'll only need one).\n\nNow look at complete quadrilateral $AFQP$.\nSince $M$ lies on $(DFQ)$ and $(RAF)$,\nit follows that $M$ is in fact the Miquel point of $AFQP$ as well.\nSo $M$ lies on $(PQR)$.\n\nThus $M$ is the fixed point that we wanted.\n\nNaturally, the congruent length\ncondition can be relaxed to $DF/DA = BE/BC$."}
+{"year": 2005, "problem_number": 6, "problem": "In a mathematical competition $6$ problems were posed to the contestants.\nEach pair of problems was solved by more than $\\frac{2}{5}$ of the contestants.\nNobody solved all 6 problems.\nShow that there were at least $2$ contestants\nwho each solved exactly $5$ problems.", "solution": "Assume not and at most one contestant solved five problems.\nBy adding in solves,\nwe can assume WLOG that one contestant solved problems one through five,\nand every other contestant solved four of the six problems.\n\nWe split the remaining contestants based on whether they solved P6.\nLet $a_i$ denote the number of contestants who solved\n$\\{1,2,\\dots,5\\} \\setminus \\{i\\}$ (and missed P6).\nLet $b_{ij}$ denote the number of contestants who solved\n$\\{1,2,\\dots,5,6\\} \\setminus \\{i,j\\}$, for $1 \\le i < j \\le 5$\n(thus in particular they solved P6).\nThus\n$ n = 1 + \\sum_{1 \\le i \\le 5} a_i + \\sum_{1 \\le i < j \\le 5} b_{ij} $\ndenotes the total number of contestants.\n\nConsidering contestants who solved P1/P6 we have\n$ t_1 \\coloneq b_{23} + b_{24} + b_{25} + b_{34} + b_{35} + b_{45} \\ge \\frac25n + \\frac15 $\nand we similarly define $t_2$, $t_3$, $t_4$, $t_5$.\n(We have written $\\frac25n+\\frac15$ since we know\nthe left-hand side is an integer strictly larger than $\\frac25n$.)\nAlso, by considering contestants who solved P1/P2 we have\n$ t_{12} = 1 + a_{3} + a_{4} + a_{5} + b_{34} + b_{35} + b_{45}\n\\ge \\frac25n + \\frac15 $\nand we similarly define $t_{ij}$ for $1 \\le i < j \\le 5$.\n\nThe number $\\frac{2n+1}{5}$ is equal to some integer $k$,\nfourteen of the $t$'s are equal to $k$,\nand the last one is equal to $k+1$.\n\nFirst, summing all fifteen equations gives\n\n6n+4 = 10 + 6(n-1) &= 10\n+ \\sum_{1 \\le i \\le 5} 6a_i + \\sum_{1 \\le i < j \\le 5} 6b_{ij} \\\\\n&= \\sum_{1 \\le i \\le 5} t_i + \\sum_{1 \\le i < j \\le 5} t_{ij}.\n\nThus the sum of the $15$ $t$'s is $6n+4$.\nBut since all the $t$'s are integers at least\n$\\frac{2n+1}{5} = \\frac{6n+3}{15}$, the conclusion follows.\n\nHowever, we will also manipulate the equations to get the following.\n\nWe have\n$ t_{45} \\equiv 1 + t_1 + t_2 + t_3 + t_{12} + t_{23} + t_{31} \\pmod 3. $\n\nThis follows directly by computing the coefficient of the $a$'s and $b$'s.\nWe will nonetheless write out a derivation of this equation, to\nmotivate it, but the proof stands without it.\n\nLet $B = \\sum_{1 \\le i < j \\le 5} b_{ij}$ be the sum of all $b$'s.\nFirst, note that\n\nt_1 + t_2 &= B + b_{34} + b_{45} + b_{35} - b_{12} \\\\\n&= B + ( t_{12}-1-a_3-a_4-a_5 ) - b_{12} \\\\\n\\implies b_{12} &= B - (t_1 + t_2) + t_{12} - 1 - (a_3+a_4+a_5).\n\nThis means we have more or less solved for each $b_{ij}$\nin terms of only $t$ and $a$ variables.\nNow\n\nt_{45} &= 1 + a_1 + a_2 + a_3 + b_{12} + b_{23} + b_{31} \\\\\n&= 1 + a_1 + a_2 + a_3 \\\\\n&+ [B - (t_1 + t_2) + t_{12} - 1 - (a_3+a_4+a_5)] \\\\\n&+ [B - (t_2 + t_3) + t_{23} - 1 - (a_1+a_4+a_5)] \\\\\n&+ [B - (t_3 + t_1) + t_{13} - 1 - (a_2+a_4+a_5)] \\\\\n&\\equiv 1 + t_1 + t_2 + t_3 + t_{12} + t_{23} + t_{31} \\pmod 3\n\nas desired.\n\nHowever, we now show the two claims are incompatible\n(and this is easy, many ways to do this).\nThere are two cases.\n\nSay $t_5 = k+1$ and the others are $k$.\nThen the equation for $t_{45}$ gives that $k \\equiv 6k+1 \\pmod 3$.\nBut now the equation for $t_{12}$ give $k \\equiv 6k \\pmod 3$.\nSay $t_{45} = k+1$ and the others are $k$.\nThen the equation for $t_{45}$ gives that $k+1 \\equiv 6k \\pmod 3$.\nBut now the equation for $t_{12}$ give $k \\equiv 6k+1 \\pmod 3$.\n\nIt is significantly easier to prove that there\nis at least one contestant who solved five problems.\nOne can see it by dropping the $+10$ in the proof of the claim,\nand arrives at a contradiction.\nIn this situation it is not even necessary to set up\nthe many $a$ and $b$ variables;\njust note that the expected number of contestants\nsolving any particular pair of problems is\n$\\frac{\\binom42n}{\\binom62} = \\frac25n$.\n\nThe fact that $\\frac{2n+1}{5}$\nshould be an integer also follows quickly,\nsince if not one can improve the bound to $\\frac{2n+2}{5}$\nand quickly run into a contradiction.\nAgain one can get here without setting up $a$ and $b$.\n\nThe main difficulty seems to be the precision required\nin order to nail down the second $5$-problem solve.\n\nThe second claim may look miraculous,\nbut the proof shows that it is not too unnatural\nto consider $t_1 + t_2 - t_{12}$ to isolate $b_{12}$\nin terms of $a$'s and $t$'s.\nThe main trick is: why mod $3$?\n\nThe reason is that if one looks closely, for a\nfixed $k$ we have a system of $15$ equations in $15$ variables.\nUnless the determinant $D$ of that system happens to be zero,\nthis means there will be a rational solution in $a$ and $b$,\nwhose denominators are bounded by $D$.\nHowever if $p \\mid D$ then we may conceivably run into mod $p$\nissues.\n\nThis motivates the choice $p = 3$,\nsince it is easy to see the determinant is divisible by $3$,\nsince constant shifts of $\\vec a$ and $\\vec b$ are also solutions mod $3$.\n(The choice $p = 2$ is a possible guess as well for this reason,\nbut the problem seems to have better $3$-symmetry.)"}
+{"year": 2006, "problem_number": 1, "problem": "Let $ABC$ be a triangle with incenter $I$.\nA point $P$ in the interior of the triangle satisfies\n$ \\angle PBA + \\angle PCA = \\angle PBC + \\angle PCB. $\nShow that $AP \\ge AI$ and that equality holds if and only if $P=I$.", "solution": "The condition rewrites as\n$\n\\angle PBC + \\angle PCB\n= (\\angle B - \\angle PBC)\n+ (\\angle C - \\angle PCB)\n\\implies\n\\angle PBC + \\angle PCB = \\frac{\\angle B + \\angle C}{2}\n$\nwhich means that\n$ \\angle BPC = 180\\dg - \\frac{\\angle B + \\angle C}{2}\n= 90\\dg + \\frac{\\angle A}{2}\n= \\angle BIC.\n$\nSince $P$ and $I$ are both inside $\\triangle ABC$\nthat implies $P$ lies on the circumcircle of $\\triangle BIC$.\n\nIt's well-known (by ``Fact 5'') that the circumcenter\nof $\\triangle BIC$ is the arc midpoint $M$ of $\\widehat{BC}$.\nTherefore\n$ AI + IM = AM \\le AP + PM \\implies AI \\le AP $\nwith equality holding iff $A$, $P$, $M$ are collinear, or $P=I$."}
+{"year": 2006, "problem_number": 2, "problem": "Let $P$ be a regular $2006$-gon.\nA diagonal is called good if its endpoints\ndivide the boundary of $P$ into two parts,\neach composed of an odd number of sides of $P$.\nThe sides of $P$ are also called good.\nSuppose $P$ has been dissected into triangles by $2003$ diagonals,\nno two of which have a common point in the interior of $P$.\nFind the maximum number of isosceles triangles having two good\nsides that could appear in such a configuration.", "solution": "Call a triangle with the desired property special.\nWe prove the maximum number of special triangles is $1003$,\nachieved by paring up the sides of the polygon.\n\nWe present two solutions for the upper bound.\nBoth of them rely first on two geometric notes:\n\nIn a special triangle, the good sides are congruent\n(and not congruent to the third side).\nNo two isosceles triangles share a good side.\n\nSolution using bijections.: \nCall a good diagonal \\textbf{special} if it's part of a special triangle;\nspecial diagonals come in pairs.\nConsider the minor arc cut out by a special diagonal $d$,\nwhich has an odd number of sides.\nSince special diagonals come in pairs,\none can associate to $d$ a side of the polygon\nnot covered by any special diagonals from $d$.\nHence there are at most $2006$ special diagonals,\nso at most $1003$ special triangles.\n\nSolution using graph theory.: \nConsider the tree $T$ formed by the $2004$\ntriangles in the dissection, with obvious adjacency.\nLet $F$ be the forest obtained by deleting\nany edge corresponding to a good diagonal.\nThen the resulting graph $F$ has only degrees $1$ and $3$,\nwith special triangles only occurring at degree $1$ vertices.\n\nIf there are $k$ good diagonals drawn,\nthen this forest consists of $k+1$ trees.\nA tree with $n_i$ vertices ($0 \\le i \\le k$)\nconsequently has $\\frac{n_i+2}{2}$ leaves.\nHowever by the earlier remark at least $k$ leaves\ndon't give special triangles\n(one on each side of a special diagonal);\nso the number of leaves that do give good triangles is at most\n$ -k + \\sum_i \\frac{n_i+2}{2}\n= -k + \\frac{2004 + 2(k+1)}{2} = 1003. $"}
+{"year": 2006, "problem_number": 3, "problem": "Determine the least real number $M$ such that the inequality\n$ \\left| ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2) \\right|\n\\leq M( a^2+b^2+c^2 )^2 $\nholds for all real numbers $a$, $b$ and $c$.", "solution": "It's the same as\n$ \\left| (a-b)(b-c)(c-a)(a+b+c) \\right|\n\\le M ( a^2+b^2+c^2 )^2. $\nLet $x=a-b$, $y=b-c$, $z=c-a$, $s=a+b+c$.\nThen we want to have\n$\n\\left| xyzs \\right|\n\\le \\frac{M}{9} (x^2+y^2+z^2+s^2)^2.\n$\nHere $x+y+z=0$.\n\nNow if $x$ and $y$ have the same sign,\nwe can replace them with the average\n(this increases the LHS and decreases RHS).\nSo we can have $x=y$, $z=-2x$.\nNow WLOG $x > 0$ to get\n$ 2x^3 \\cdot s \\le \\frac{M}{9} ( 6x^2+s^2 )^2. $\nAfter this routine calculation gives\n$M = \\frac{9}{32}\\sqrt2$ works and is optimal\n(by $6x^2+s^2 = 2x^2 + 2x^2 + 2x^2 + s^2$ and AM-GM)."}
+{"year": 2006, "problem_number": 4, "problem": "Determine all pairs $(x,y)$ of integers such that\n$ 1 + 2^x + 2^{2x+1} = y^2. $", "solution": "Answers: $(0, \\pm 2)$, $(4, \\pm 23)$, which work.\n\nAssume $x \\ge 4$.\n$ 2^x ( 1 + 2^{x+1} )\n= 2^x + 2^{2x+1} = y^2 - 1 = (y-1)(y+1). $\nSo either:\n\n$y = 2^{x-1} m + 1$ for some odd $m$, so\n$ 1 + 2^{x+1} = m( 2^{x-2}m+1 )\n\\implies 2^x = \\frac{4(1-m)}{m^2-8}. $\n$y = 2^{x-1} m - 1$ for some odd $m$, so\n$ 1 + 2^{x+1} = m( 2^{x-2}m-1 )\n\\implies 2^x = \\frac{4(1+m)}{m^2-8}. $\n\nIn particular we need $4|1 \\pm m| \\ge 2^4 |m^2-8|$,\nwhich is enough to imply $m < 5$.\nFrom here easily recover $x = 4$, $m = 3$ as the last solution\n(in the second case)."}
+{"year": 2006, "problem_number": 5, "problem": "Let $P(x)$ be a polynomial of degree $n > 1$\nwith integer coefficients and let $k$ be a positive integer.\nConsider the polynomial\n$ Q(x) = P(P(\\dots P(P(x)) \\dots )) $ where $P$ occurs $k$ times.\nProve that there are at most $n$ integers $t$ such that $Q(t) = t$.", "solution": "First, we prove that:\n\n[Putnam 2000 et al]\nIf a number is periodic under $P$\nthen in fact it's fixed by $P \\circ P$.\n\nLet $x_1$, $x_2$, \\dots, $x_n$ be a minimal orbit.\nThen\n$ x_i - x_{i+1} \\mid P(x_i) - P(x_{i+1})\n= x_{i+1} - x_{i+2} $\nand so on cyclically.\n\nIf any of the quantities are zero we are done.\nElse, we must eventually have $x_i - x_{i+1} = -(x_{i+1} - x_{i+2})$,\nso $x_i = x_{i+2}$ and we get $2$-periodicity.\n\nThe tricky part is to study the $2$-orbits.\nSuppose there exists a fixed pair $u \\neq v$\nwith $P(u) = v$, $P(v) = u$.\n(If no such pair exists, we are already done.)\nLet $(a,b)$ be any other pair with $P(a) = b$, $P(b) = a$,\npossibly even $a = b$, but $\\{a,b\\} \\cap \\{u,v\\} = \\varnothing$.\nThen we should have\n$ u-a \\mid P(u)-P(a) = v-b\n\\mid P(v) - P(b) = u-a $\nand so $u-a$ and $v-b$ divide each other (and are nonzero).\nSimilarly, $u-b$ and $v-a$ divide each other.\n\nHence $u-a = \\pm (v-b)$ and $u-b = \\pm (v-a)$.\nWe consider all four cases:\n\nIf $u-a = v-b$ and $u-b = v-a$\nthen $u-v = b-a = a-b$, contradiction.\nIf $u-a = -(v-b)$ and $u-b = -(v-a)$\nthen $u+v = u-v = a+b$.\nIf $u-a = -(v-b)$ and $u-b = v-a$,\nwe get $a+b = u+v$ from the first one\n(discarding the second).\nIf $u-a = v-b$ and $u-b = -(v-a)$,\nwe get $a+b = u+v$ from the second one\n(discarding the first one).\n\nThus in all possible situations we have\n$ a+b = c \\coloneq u+v $\na fixed constant.\n\nTherefore, any pair $(a,b)$ with $P(a) = b$\nand $P(b) = a$ actually satisfies $P(a) = c-a$.\nAnd since $\\deg P > 1$,\nthis means there are at most $n$ roots to $a+P(a)=c$, as needed."}
+{"year": 2006, "problem_number": 6, "problem": "Assign to each side $b$ of a convex polygon $P$\nthe maximum area of a triangle that has $b$ as a side and is contained in $P$.\nShow that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.", "solution": "We say a polygon in almost convex\nif all its angles are at most $180\\dg$.\n\nNote that given any convex or almost convex polygon,\nwe can take any side $b$ and add another vertex on it, and the sum of the labels doesn't change\n(since the label of a side is the length of the side times the distance of the farthest point).\n\nLet $N$ be an even integer.\nThen any almost convex $N$-gon with area $S$\nshould have an inscribed triangle with area at least $2S/N$.\n\nThe main work is the proof of the lemma.\n\nLabel the polygon $P_0 P_1 \\dots P_{N-1}$.\nConsider the $N/2$ major diagonals of the almost convex $N$-gon,\n$P_0 P_{N/2}$, $P_1 P_{N/2+1}$, et cetera.\nA butterfly refers to a self-intersecting quadrilateral\n$P_i P_{i+1} P_{i+1+N/2} P_{i+N/2}$.\nAn example of a butterfly is shown below for $N=8$.\n[Figure omitted]\n\nEvery point $X$ in the polygon is contained in the wingspan of some butterfly.\n\nConsider a windmill-like process which\n\nstarts from some oriented red line $P_0 P_{N/2}$, oriented to face $P_0 P_{N/2}$\nrotates through $P_0 P_{N/2} \\cap P_1 P_{N/2+1}$ to get line $P_1 P_{N/2+1}$,\nrotates through $P_1 P_{N/2+1} \\cap P_2 P_{N/2+2}$ to get line $P_2 P_{N/2+2}$,\n\\dots et cetera, until returning to line $P_{N/2} P_0$,\nbut in the reverse orientation.\n\nAt the end of the process, every point in the plane has switched sides with our moving line.\nThe moment that $X$ crosses the moving red line, we get it contained in a butterfly, as needed.\n\nIf $ABDC = P_i P_{i+1} P_{i+1+N/2} P_{i+N/2}$ is a butterfly,\none of the triangles $ABC$, $BCD$, $CDA$, $DAB$\nhas area at least that of the butterfly.\n\nLet the diagonals of the butterfly meet at $O$,\nand let $a = AO$, $b = BO$, $c = CO$, $d = DO$.\nIf we assume WLOG $d = \\min(a,b,c,d)$\nthen it follows $[ABC] = [AOB] + [BOC] \\ge [AOB] + [COD]$, as needed.\n\nNow, since the $N/2$ butterflies cover an area of $S$,\nit follows that one of the butterflies\nhas area at least $S / (N/2) = 2S/N$,\nand so that butterfly gives a triangle with area at least $2S/N$,\ncompleting the proof of the lemma.\n\nMain proof.: \nLet $a_1$, \\dots, $a_n$ be the numbers assigned to the sides.\nAssume for contradiction $a_1 + \\dots + a_n < 2S$.\nWe pick even integers $m_1$, $m_2$, \\dots, $m_n$ such that\n\n\\frac{a_1}{S} &< \\frac{2m_1}{m_1 + \\dots + m_n} \\\\\n\\frac{a_2}{S} &< \\frac{2m_2}{m_1 + \\dots + m_n} \\\\\n&\\vdotswithin\\le \\\\\n\\frac{a_n}{S} &< \\frac{2m_n}{m_1 + \\dots + m_n}.\n\nwhich is possible by rational approximation,\nsince the right-hand sides sum to $2$ and the left-hand sides sum to strictly less than $2$.\n\nNow we break every side of $P$ into $m_i$ equal parts\nto get an almost convex $N$-gon, where $N = m_1 + \\dots + m_n$.\n\nThe main lemma then gives us a triangle $\\Delta$ of the almost convex $N$-gon\nwhich has area at least $\\frac{2S}{N}$.\nIf $\\Delta$ used the $i$th side then it then follows the label $a_i$ on that side should be\nat least $m_i \\cdot \\frac{2S}{N}$, contradiction."}
+{"year": 2007, "problem_number": 1, "problem": "Real numbers $a_1$, $a_2$, \\dots, $a_n$ are fixed.\nFor each $1 \\le i \\le n$ we let\n$d_i = \\max\\{a_j : 1 \\le j \\le i\\} - \\min\\{a_j : i \\le j \\le n\\}$\nand let $d = \\max \\{d_i : 1 \\le i \\le n\\}$.\n\n[(a)]\nProve that for any real numbers $x_1 \\le \\dots \\le x_n$ we have\n$\n\\max \\left\\{ \\left| x_i - a_i \\right| :\n1 \\le i \\le n \\right\\}\n\\ge 1/2 d.\n$\nMoreover, show that there exists some\nchoice of $x_1 \\le \\dots \\le x_n$ which achieves equality.", "solution": "Note that we can dispense of $d_i$ immediately\nby realizing that the definition of $d$ just says\n$ d = \\max_{1 \\le i \\le j \\le n} ( a_i - a_j ). $\n\nIf $a_1 \\le \\dots \\le a_n$ are already nondecreasing\nthen $d = 0$ and there is nothing to prove\n(for the equality case, just let $x_i = a_i$),\nso we will no longer consider this case.\n\nOtherwise, consider any indices $i < j$ with $a_i > a_j$.\nWe first prove (a) by applying the following claim\nwith $p = a_i$ and $q = a_j$:\n\nFor any $p \\le q$, we have\neither $|p - a_i| \\ge 1/2(a_i-a_j)$\nor $|q - a_j| \\ge 1/2(a_i-a_j)$.\n\nAssume for contradiction both are false.\nThen $p > a_i - 1/2(a_i-a_j)\n= a_j + 1/2(a_i-a_j) > q$, contradiction.\n\nAs for (b), we let $i < j$ be any indices for which\n$a_i - a_j = d > 0$ achieves the maximal difference.\nWe then define $x_\\bullet$ in three steps:\n\nWe set $x_k = \\frac{a_i + a_j}{2}$ for $k = i, \\dots, j$.\nWe recursively set $x_{k} = \\max(x_{k-1}, a_k)$\nfor $k = j+1, j+2, \\dots$.\nWe recursively set $x_{k} = \\min(x_{k+1}, a_k)$\nfor $k = i-1, i-2, \\dots$.\n\nBy definition, these $x_\\bullet$ are weakly increasing.\nTo prove this satisfies (b) we only need to check that\n$ \\left| x_k - a_k \\right| \\le \\frac{a_i-a_j}{2} \\qquad\n(\\star) $\nfor any index $k$ (as equality holds for $k = i$ or $k = j$).\n\nWe note $(\\star)$ holds for $i < k < j$ by construction.\nFor $k > j$, note that $x_k \\in \\{a_j, a_{j+1}, \\dots, a_k\\}$\nby construction, so $(\\star)$ follows from our choice of $i$ and $j$\ngiving the largest possible difference; the case $k < i$ is similar."}
+{"year": 2007, "problem_number": 2, "problem": "Consider five points $A$, $B$, $C$, $D$ and $E$\nsuch that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral.\nLet $\\ell$ be a line passing through $A$.\nSuppose that $\\ell$ intersects the interior of the segment $DC$ at $F$\nand intersects line $BC$ at $G$.\nSuppose also that $EF = EG = EC$.\nProve that $\\ell$ is the bisector of angle $ DAB$.", "solution": "Let $M$, $N$, $P$ denote the midpoints of $CF$, $CG$, $AC$\n(noting $P$ is also the midpoint of $BD$).\n\nBy a homothety at $C$ with ratio $1/2$,\nwe find $MNP$ is the image of line $\\ell \\equiv AGF$.\n\n[Figure omitted]\n\nHowever, since we also have $EM \\perp CF$\nand $EN \\perp CG$ (from $EF=EG=EC$)\nwe conclude $PMN$ is the Simson line of $E$ with respect to $\\triangle BCD$,\nwhich implies $EP \\perp BD$.\nIn other words, $EP$ is the perpendicular bisector of $BD$,\nso $E$ is the midpoint of arc $\\widehat{BCD}$.\n\nFinally,\n\n\u2220(AB, \\ell) &= \u2220(CD, MNP) = \u2220 CMN = \u2220 CEN \\\\\n&= 90\\dg - \u2220 NCE = 90\\dg + \u2220 ECB\n\nwhich means that $\\ell$ is parallel to a bisector of $\\angle BCD$,\nand hence to one of $\\angle BAD$.\n(Moreover since $F$ lies on the interior of $CD$,\nit is actually the internal bisector.)"}
+{"year": 2007, "problem_number": 3, "problem": "In a mathematical competition some competitors are (mutual) friends.\nCall a group of competitors a clique if each two of them are friends.\nGiven that the largest size of a clique is even,\nprove that the competitors can be arranged into two rooms\nsuch that the largest size of a clique contained in one room\nis the same as the largest size of a clique contained in the other room.", "solution": "Take the obvious graph interpretation $G$.\nWe paint red any vertices in one of the maximal cliques $K$,\nwhich we assume has $2r$ vertices, and paint the remaining vertices green.\nWe let $\\alpha(\\bullet)$ denote the clique number.\n\nInitially, let the two rooms $A = K$, $B = G-K$.\n\nWe can move at most $r$ vertices of $A$ into $B$\nto arrive at $\\alpha(A) \\le \\alpha(B) \\le \\alpha(A)+1$.\n\nThis is actually obvious by discrete continuity.\nWe move one vertex at a time, noting $\\alpha(A)$ decreases by one at each step,\nwhile $\\alpha(B)$ increases by either zero or one at each step.\n\nWe stop once $\\alpha(B) \\ge \\alpha(A)$, which happens before we have moved $r$ vertices\n(since then we have $\\alpha(B) \\ge r = \\alpha(A)$).\nThe conclusion follows.\n\nSo let's consider the situation\n$ \\alpha(A) = k \\ge r \\qquad\\text{and}\\qquad \\alpha(B) = k+1. $\n\nAt this point $A$ is a set of $k$ red vertices,\nwhile $B$ has the remaining $2r-k$ red vertices (and all the green ones).\nAn example is shown below with $k=4$ and $2r = 6$.\n\n[Figure omitted]\n\nNow, if we can move any red vertex from $B$ back to $A$\nwithout changing the clique number of $B$, we do so, and win.\n\nOtherwise, it must be the case that every\n$(k+1)$-clique in $B$ uses every red vertex in $B$.\nFor each $(k+1)$-clique in $B$ (in arbitrary order), we do the following procedure.\n\nIf all $k+1$ vertices are still green, pick one and re-color it blue.\nThis is possible since $k+1 > 2r-k$.\nOtherwise, do nothing.\n\nThen we move all the blue vertices from $B$ to $A$,\none at a time, in the same order we re-colored them.\nThis forcibly decreases the clique number of $B$ to $k$,\nsince the clique number is $k+1$ just before the last blue vertex is moved,\nand strictly less than $k+1$ (hence equal to $k$) immediately after that.\n\nAfter this, $\\alpha(A) = k$ still holds.\n\nAssume not, and we have a $(k+1)$-clique\nwhich uses $b$ blue vertices and $(k+1)-b$ red vertices in $A$.\nTogether with the $2r-k$ red vertices already in $B$ we then get a clique of size\n$ b + ( (k+1-b) ) + ( 2r-k ) = 2r + 1 $\nwhich is a contradiction.\n\nDragomir Grozev posted the following motivation on \\href{https://dgrozev.wordpress.com/2019/12/05/splitting-the-cliques-of-a-graph-imo-2007-p3/}{his blog}:\n\nI think, it's a natural idea to place all students in one room and begin\nmoving them one by one into the other one.\nThen the max size of the cliques in the first and second room\nincrease (resp.\\ decrease) at most with one.\nSo, there would be a moment both sizes are almost the same.\nAt that moment we may adjust something.\n\nTrying the idea, I had some difficulties\nkeeping track of the maximal cliques in the both rooms.\nIt seemed easier all the students in one of the rooms to comprise a clique.\nIt could be achieved by moving only the members of the maximal clique.\nFollowing this path the remaining obstacles can be overcome naturally."}
+{"year": 2007, "problem_number": 4, "problem": "In triangle $ABC$ the bisector of $\\angle BCA$\nmeets the circumcircle again at $R$,\nthe perpendicular bisector of $BC$ at $P$,\nand the perpendicular bisector of $AC$ at $Q$.\nThe midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$.\nProve that the triangles $RPK$ and $RQL$ have the same area.", "solution": "We first begin by proving the following claim.\n\nWe have $CQ = PR$ (equivalently, $CP = QR$).\n\nLet $O = LQ \\cap KP$ be the circumcenter.\nThen\n$ \u2220 OPQ = \u2220 KPC = 90\\dg - \u2220 PCK\n= 90\\dg - \u2220 LCQ = \u2220 \u2220 CQL = \u2220 PQO. $\nThus $OP = OQ$.\nSince $OC = OR$ as well, we get the conclusion.\n\nDenote by $X$ and $Y$ the feet from $R$ to $CA$\nand $CB$, so $\\triangle CXR \\cong \\triangle CYR$.\nThen, let $t = \\frac{CQ}{CR} = 1 - \\frac{CP}{CR}$.\n\n[Figure omitted]\n\nThen it follows that\n$ [RQL] = [XQL] = t(1-t) \\cdot [XRC]\n= t(1-t) \\cdot [YCR] = [YKP] = [RKP] $\nas needed.\n\nTrigonometric approaches are very possible\n(and easier to find) as well:\nboth areas work out to be $\\frac 18 ab \\tan 1/2 C$."}
+{"year": 2007, "problem_number": 5, "problem": "Let $a$ and $b$ be positive integers.\nShow that if $4ab - 1$ divides $(4a^{2} - 1)^{2}$, then $a = b$.", "solution": "As usual,\n$ 4ab-1 \\mid (4a^2-1)^2 \\iff 4ab-1 \\mid (4ab \\cdot a-b)^2\n\\iff 4ab-1 \\mid (a-b)^2. $\nThen we use a typical Vieta jumping argument.\nDefine $ k = \\frac{(a-b)^2}{4ab-1}. $\nNote that $k = 0 \\iff a = b$.\nSo we will prove that $k > 0$ leads to a contradiction.\n\nIndeed, suppose $(a, b)$ is a minimal solution with $a > b$\n(we have $a \\neq b$ since $k \\neq 0$).\nBy Vieta jumping, $(b, \\frac{b^2+k}{a})$ is also such a solution.\nBut now\n\n\\frac{b^2+k}{a} \\ge a &\\implies k \\ge a^2 - b^2 \\\\\n&\\implies \\frac{(a-b)^2}{4ab-1} \\ge a^2-b^2 \\\\\n&\\implies a-b \\ge (4ab-1)(a+b)\n\nwhich is absurd for $a,b \\in \\ZZ_{>0}$.\n(In the last step we divided by $a-b > 0$.)"}
+{"year": 2007, "problem_number": 6, "problem": "Let $n$ be a positive integer.\nConsider\n$ S = \\left\\{ (x,y,z) \\mid\nx,y,z \\in \\{ 0, 1, \\dots, n\\}, \\;\nx+y+z > 0 \\right\\} $\nas a set of $(n+1)^3-1$ points in the three-dimensional space.\nDetermine the smallest possible number of planes,\nthe union of which contains $S$ but does not include $(0,0,0)$.", "solution": "The answer is $3n$.\nHere are two examples of constructions with $3n$ planes:\n\n$x+y+z=i$ for $i=1,\\dots,3n$.\n$x=i$, $y=i$, $z=i$ for $i=1,\\dots,n$.\n\nSuppose for contradiction we have $N < 3n$ planes.\nLet them be $a_i x + b_i y + c_i z + 1 = 0$, for $i = 1, \\dots, N$.\nDefine the polynomials\n\nA(x,y,z) &= \\prod_{i=1}^n (x-i) \\prod_{i=1}^n (y-i) \\prod_{i=1}^n (z-i) \\\\\nB(x,y,z) &= \\prod_{i=1}^N ( a_i x + b_i y + c_i z + 1 ).\n\nNote that $A(0,0,0) = (-1)^n (n!)^3 \\neq 0$\nand $B(0,0,0) = 1 \\neq 0$,\nbut $A(x,y,z) = B(x,y,z) = 0$ for any $(x,y,z) \\in S$.\nAlso, the coefficient of $x^n y^n z^n$ in $A$ is $1$,\nwhile the coefficient of $x^n y^n z^n$ in $B$ is $0$.\n\nNow, define\n$ P(x,y,z) \\coloneq A(x,y,z) - \\lambda B(x,y,z). $\nwhere $\\lambda = \\frac{A(0,0,0)}{B(0,0,0)} = (-1)^{n} (n!)^3$.\nWe now have that\n\n$P(x,y,z) = 0$ for any $x,y,z \\in \\left\\{ 0,1,\\dots,n \\right\\}^3$.\nBut the coefficient of $x^n y^n z^n$ is $1$.\n\nThis is a contradiction to Alon's combinatorial nullstellensatz."}
+{"year": 2008, "problem_number": 1, "problem": "Let $H$ be the orthocenter of an acute-angled triangle $ABC$.\nThe circle $\\Gamma_{A}$ centered at the midpoint of $BC$ and passing\nthrough $H$ intersects the sideline $BC$ at points $A_1$ and $A_2$.\nSimilarly, define the points $B_1$, $B_2$, $C_1$, and $C_2$.\nProve that six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.", "solution": "We show two solutions.\n\nFirst solution using power of a point.: \nLet $D$, $E$, $F$ be the centers of $\\Gamma_A$, $\\Gamma_B$, $\\Gamma_C$\n(in other words, the midpoints of the sides).\n\nWe first show that $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.\nIt suffices to prove that $A$\nlies on the radical axis of the circles $\\Gamma_B$ and $\\Gamma_C$.\n\n[Figure omitted]\n\nLet $X$ be the second intersection of $\\Gamma_B$ and $\\Gamma_C$.\nClearly $XH$ is perpendicular to the line\njoining the centers of the circles, namely $EF$.\nBut $EF \\parallel BC$, so $XH \\perp BC$.\nSince $AH \\perp BC$ as well,\nwe find that $A$, $X$, $H$ are collinear, as needed.\n\nThus, $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.\nSimilarly, $C_1$, $C_2$, $A_1$, $A_2$ are concyclic,\nas are $A_1$, $A_2$, $B_1$, $B_2$.\nNow if any two of these three circles coincide, we are done;\nelse the pairwise radical axii are not concurrent, contradiction.\n(Alternatively, one can argue directly that $O$ is the center of all\nthree circles, by taking the perpendicular bisectors.)\n\nSecond solution using length chase (Ritwin Narra).: \nWe claim the circumcenter $O$ of $\\triangle ABC$\nis in fact the center of $(A_1A_2B_1B_2C_1C_2)$.\n\nDefine $D$, $E$, $F$ as before.\nThen since $OD \\perp A_1A_2$ and $DA_1 = DA_2$,\nwhich means $OA_1 = OA_2$. Similarly, we have $OB_1 = OB_2$ and $OC_1 = OC_2$.\n\nNow since $DA_1 = DA_2 = DH$, we have $OA_1^2 = OD^2 + HD^2$.\nWe seek to show\n$ OD^2 + HD^2 = OE^2 + HE^2 = OF^2 + HF^2. $\nThis is clear by Appollonius's Theorem\nsince $D$, $E$, and $F$ lie on the nine-point circle,\nwhich is centered at the midpoint of $OH$."}
+{"year": 2008, "problem_number": 2, "problem": "Let $x$, $y$, $z$ be real numbers with $xyz = 1$, all different from $1$.\nProve that\n$ \\frac{x^2}{(x-1)^2} + \\frac{y^2}{(y-1)^2} + \\frac{z^2}{(z-1)^2} \\ge 1 $\nand show that equality holds for infinitely many choices\nof rational numbers $x$, $y$, $z$.", "solution": "Let $x=a/b$, $y=b/c$, $z=c/a$, so we want to show\n$ (\\frac{a}{a-b})^2+(\\frac{b}{b-c})^2\n+(\\frac{c}{c-a})^2\\ge 1.$\nA boring computation shows this is equivalent to\n$ \\frac{(a^2b+b^2c+c^2a-3abc)^2}{(a-b)^2(b-c)^2(c-a)^2} \\ge 0 $\nwhich proves the inequality\n(and it is unsurprising we are in such a situation,\ngiven that there is an infinite curve of rationals).\n\nFor equality, it suffices to show there are infinitely many integer solutions to\n$ a^2b+b^2c+c^2a=3abc \\iff \\frac ac + \\frac ba + \\frac ca = 3 $\nor equivalently that there are infinitely many rational solutions to\n$ u + v + \\frac{1}{uv} = 3. $\nFor any $0 \\neq u \\in \\QQ$ the real solution for $u$ is\n$ v = \\frac{-u + (u-1)\\sqrt{1-4/u} + 3}{2} $\nand there are certainly infinitely many rational numbers $u$\nfor which $1-4/u$ is a rational square\n(say, $u = \\frac{-4}{q^2-1}$ for $q \\neq \\pm 1$ a rational number)."}
+{"year": 2008, "problem_number": 3, "problem": "Prove that there are infinitely many positive integers $n$\nsuch that $n^2+1$ has a prime factor greater than $2n + \\sqrt{2n}$.", "solution": "The idea is to pick the prime $p$ first!\n\nSelect any large prime $p \\ge 2013$,\nand let $h = \\left\\lceil \\sqrt p \\right\\rceil$.\nWe will try to find an $n$ such that\n$ n \\le \\frac 12 (p-h) \\quad \\text{and} \\quad p \\mid n^2+1. $\nThis implies $p \\ge 2n+\\sqrt{p}$\nwhich is enough to ensure $p \\ge 2n + \\sqrt{2n}$.\n\nAssume $p \\equiv 1 \\pmod 8$ henceforth.\nThen there exists some $\\frac 12 p < x < p$\nsuch that $x^2 \\equiv -1 \\pmod p$,\nand we set $ x = \\frac{p+1}{2} + t. $\n\nWe have $t \\ge \\frac{h-1}{2}$ and hence may take $n = p-x$.\n\nAssume for contradiction this is false; then\n\n0 &\\equiv 4(x^2+1) \\pmod{p} \\\\\n&= ( p+1+2t )^2 + 4 \\\\\n&\\equiv (2t+1)^2 + 4 \\pmod{p} \\\\\n&< h^2+4\n\nSo we have that $(2t+1)^2+4$ is positive and divisible by $p$,\nyet at most $\\left\\lceil \\sqrt{p} \\right\\rceil^2 + 4 < 2p$.\nSo it must be the case that $(2t+1)^2+4 = p$,\nbut this has no solutions modulo $8$."}
+{"year": 2008, "problem_number": 4, "problem": "Find all functions $f$ from the positive reals to the positive reals such that\n$ \\frac{f(w)^2 + f(x)^2}{f(y^2)+f(z^2)} = \\frac{w^2+x^2}{y^2+z^2} $\nfor all positive real numbers $w$, $x$, $y$, $z$ satisfying $wx=yz$.", "solution": "The answers are $f(x) \\equiv x$ and $f(x) \\equiv 1/x$.\nThese work, so we show they are the only ones.\n\nFirst, setting $(t,t,t,t)$ gives $f(t^2) = f(t)^2$.\nIn particular, $f(1) = 1$.\nNext, setting $(t, 1, \\sqrt t, \\sqrt t)$ gives\n$ \\frac{f(t)^2 + 1}{2f(t)} = \\frac{t^2 + 1}{2t} $\nwhich as a quadratic implies $f(t) \\in \\{t, 1/t\\}$.\n\nNow assume $f(a) = a$ and $f(b) = 1/b$.\nSetting $(\\sqrt a, \\sqrt b, 1, \\sqrt{ab})$ gives\n$ \\frac{a + 1/b}{f(ab) + 1} = \\frac{a+b}{ab+1}. $\nOne can check the two cases on $f(ab)$ each imply\n$a=1$ and $b=1$ respectively.\nHence the only answers are those claimed."}
+{"year": 2008, "problem_number": 5, "problem": "Let $n$ and $k$ be positive integers with $k \\geq n$ and $k - n$ an even number.\nThere are $2n$ lamps labelled $1$, $2$, \\dots, $2n$ each of which can be either on or off.\nInitially all the lamps are off.\nWe consider sequences of steps: at each step one of the lamps is switched\n(from on to off or from off to on).\nLet $N$ be the number of such sequences consisting of $k$ steps\nand resulting in the state where lamps $1$ through $n$ are all on,\nand lamps $n + 1$ through $2n$ are all off.\nLet $M$ be number of such sequences consisting of $k$ steps,\nresulting in the state where lamps $1$ through $n$ are all on,\nand lamps $n + 1$ through $2n$ are all off,\nbut where none of the lamps $n + 1$ through $2n$ is ever switched on.\nDetermine $\\frac{N}{M}$.", "solution": "The answer is $2^{k-n}$.\n\nConsider the following map $\\Psi$ from $N$-sequences to $M$-sequences:\n\nchange every instance of $n+1$ to $1$;\nchange every instance of $n+2$ to $2$;\n[$\\vdots$]\nchange every instance of $2n$ to $n$.\n\n(For example, suppose $k=9$, $n=3$;\nthen $144225253 \\mapsto 111222223$.)\n\nClearly this is map is well-defined and surjective.\nSo all that remains is:\n\nEvery $M$-sequence has exactly $2^{k-n}$ pre-images under $\\Psi$.\n\nIndeed, suppose that there are $c_1$ instances of lamp $1$.\nThen we want to pick an odd subset of the $1$'s to change to $n+1$'s,\nso $2^{c_1 - 1}$ ways to do this.\nAnd so on.\nHence the number of pre-images is\n$ \\prod_i 2^{c_i - 1} = 2^{k-n}. \\qedhere $"}
+{"year": 2008, "problem_number": 6, "problem": "Let $ABCD$ be a convex quadrilateral with $BA \\neq BC$.\nDenote the incircles of triangles $ABC$ and $ADC$\nby $\\omega_1$ and $\\omega_2$ respectively.\nSuppose that there exists a circle $\\omega$ tangent\nto ray $BA$ beyond $A$ and to the ray $BC$ beyond $C$,\nwhich is also tangent to the lines $AD$ and $CD$.\nProve that the common external tangents to\n$\\omega_1$ and $\\omega_2$ intersect on $\\omega$.", "solution": "By the external version of Pitot theorem, the existence\nof $\\omega$ implies that\n$ BA + AD = CB + CD. $\nLet $PQ$ and $ST$ be diameters of $\\omega_1$ and $\\omega_2$\nwith $P, T \\in AC$.\nThen the length relation on $ABCD$ implies that $P$ and $T$\nare reflections about the midpoint of $AC$.\n\nNow orient $AC$ horizontally and let $K$ be the ``uppermost'' point of $\\omega$, as shown.\n\n[Figure omitted]\n\nConsequently, a homothety at $B$ maps $Q$, $T$, $K$ to each other\n(since $T$ is the uppermost of the excircle, $Q$ of the incircle).\nSimilarly, a homothety at $D$ maps $P$, $S$, $K$ to each other.\nAs $PQ$ and $ST$ are parallel diameters\nit then follows $K$ is the exsimilicenter of $\\omega_1$ and $\\omega_2$."}
+{"year": 2009, "problem_number": 1, "problem": "Let $n, k \\ge 2$ be positive integers and let $a_1$, $a_2$, $a_3$, \\dots, $a_k$\nbe distinct integers in the set $\\left\\{ 1,2,\\dots,n \\right\\}$\nsuch that $n$ divides $a_i(a_{i+1} - 1)$ for $i = 1,2,\\dots,k-1$.\nProve that $n$ does not divide $a_k(a_1 - 1)$.", "solution": "We proceed indirectly and assume that\n$ a_i (a_{i+1}-1) \\equiv 0 \\pmod n $\nfor $i = 1, \\dots, k$ (indices taken modulo $k$).\nWe claim that this implies all the $a_i$ are equal modulo $n$.\n\nLet $q = p^e$ be any prime power dividing $n$.\nThen, $a_1 (a_2 - 1) \\equiv 0 \\pmod q$, so $p$ divides either $a_1$ or $a_2-1$.\n\nIf $p \\mid a_1$, then $p \\nmid a_1 - 1$. Then\n$ a_k (a_1-1) \\equiv 0 \\pmod q \\implies a_k \\equiv 0 \\pmod q. $\nIn particular, $p \\mid a_k$.\nSo repeating this argument,\nwe get $a_{k-1} \\equiv 0 \\pmod q$, $a_{k-2} \\equiv 0 \\pmod q$, and so on.\n\nSimilarly, if $p \\mid a_2 - 1$ then $p \\nmid a_2$, and\n$ a_2 (a_3-1) \\equiv 0 \\pmod q \\implies a_3 \\equiv 1 \\pmod q. $\nIn particular, $p \\mid a_3 - 1$.\nSo repeating this argument,\nwe get $a_4 \\equiv 0 \\pmod q$, $a_5 \\equiv 0 \\pmod q$, and so on.\n\nEither way, we find $a_i \\pmod q$ is constant (and either $0$ or $1$).\n\nSince $q$ was an arbitrary prime power dividing $n$,\nby Chinese remainder theorem we conclude that $a_i \\pmod n$ is constant as well.\nBut this contradicts the assumption of distinctness."}
+{"year": 2009, "problem_number": 2, "problem": "Let $ABC$ be a triangle with circumcenter $O$.\nThe points $P$ and $Q$ are interior points of the sides $CA$ and $AB$ respectively.\nLet $K$, $L$, $M$ be the midpoints of $BP$, $CQ$, $PQ$.\nSuppose that $PQ$ is tangent to the circumcircle of $\\triangle KLM$.\nProve that $OP = OQ$.", "solution": "By power of a point, we have $-AQ \\cdot QB = OQ^2 - R^2$\nand $-AP \\cdot PC = OP^2 - R^2$.\nTherefore, it suffices to show $AQ \\cdot QB = AP \\cdot PC$.\n\n[Figure omitted]\n\nAs $ML \\parallel AC$ and $MK \\parallel AB$ we have that\n\n\u2220 APQ &= \u2220 LMP = \u2220 LKM \\\\\n\u2220 PQA &= \u2220 KMQ = \u2220 MLK\n\nand consequently we have the (opposite orientation) similarity\n$ \\triangle APQ \\overset{-}{\\sim} \\triangle MKL. $\nTherefore\n$ \\frac{AQ}{AP} = \\frac{ML}{MK} = \\frac{2ML}{2MK} = \\frac{PC}{QB} $\nid est $AQ \\cdot QB = AP \\cdot PC$, which is what we wanted to prove."}
+{"year": 2009, "problem_number": 3, "problem": "Suppose that $s_1,s_2,s_3, \\dotsc$ is a strictly increasing sequence of\npositive integers such that the sub-sequences\n$s_{s_1}$, $s_{s_2}$, $s_{s_3}$, \\dots\nand $s_{s_1 + 1}$, $s_{s_2 + 1}$, $s_{s_3 + 1}$, \\dots\nare both arithmetic progressions.\nProve that the sequence $s_1$, $s_2$, $s_3$, \\dots\\ is itself an arithmetic progression.", "solution": "We present two solutions.\n\nFirst solution (Alex Zhai).: \nLet $s(n) \\coloneq s_n$ and write\n\ns(s(n)) &= Dn + A \\\\\ns(s(n)+1) &= D'n + B.\n\nIn light of the bounds $s(s(n)) \\le s(s(n)+1) \\le s(s(n+1))$\nwe right away recover $D = D'$ and $A \\le B$.\n\nLet $d_n = s(n+1)-s(n)$.\nNote that $\\sup d_n < \\infty$ since $d_n$ is bounded above by $A$.\n\nThen we let\n$ m \\coloneq \\min d_n, \\qquad M \\coloneq \\max d_n. $\nNow suppose $a$ achieves the maximum, meaning $s(a+1)-s(a) = M$.\nThen\n\n\\underbrace{d_{s(s(a))} + \\dots + d_{s(s(a+1))-1}}_{D \\text{ terms}}\n&= \\boxed{s(s(s(a+1))) - s(s(s(a)))} \\\\\n&= (D \\cdot s(a+1) + A) - (D \\cdot s(a) + A) = DM.\n\nNow $M$ was maximal hence $M = d_{s(s(a))} = \\dots = d_{s(s(a+1))-1}$.\nBut $d_{s(s(a))} = B-A$ is a constant.\nHence $M = B-A$.\nIn the same way $m = B-A$ as desired.\n\nSecond solution.: \nWe retain the notation $D$, $A$, $B$ above,\nas well as $m = \\min_n s(n+1)-s(n) \\ge B-A$.\nWe do the involution trick first as:\n$ D = \\boxed{s(s(s(n)+1)) - s(s(s(n)))} = s(Dn+B) - s(Dn+A) $\nand hence we recover $D \\ge m(B-A)$.\n\nThe edge case $D = B-A$ is easy since then $m=1$\nand $D = s(Dn+B)-s(Dn+A)$ forces $s$ to be a constant shift.\nSo henceforth assume $D > B-A$.\n\nThe idea is that right now the $B$ terms are ``too big'',\nso we want to use the involution trick in a way that makes as many\n``$A$ minus $B$'' shape expressions as possible.\nThis motivates considering $s(s(s(n+1))) - s(s(s(n)+1)+1) > 0$,\nsince the first expression will have all $A$'s\nand the second expression will have all $B$'s.\nCalculation gives:\n\ns(D(n+1)+A) - s(Dn+B+1)\n&= \\boxed{s(s(s(n+1))) - s(s(s(n)+1)+1)} \\\\\n&= (Ds(n+1) + A) - (D(s(n)+1) + B) \\\\\n&= D( s(n+1)-s(n) ) + A-B-D.\n\nThen by picking $n$ achieving the minimum $m$,\n$ \\underbrace{m(D+A-B-1)}_{>0} \\le s(s(s(n+1))) - s(s(s(n)+1)+1) \\le Dm + A-B-D $\nwhich becomes\n$ ( D-m(B-A) ) + ( (B-A)-m ) \\le 0. $\nSince both of these quantities were supposed to be nonnegative,\nwe conclude $m = B-A$ and $D = m^2$.\nNow the estimate $D = s(Dn+B) - s(Dn+A) \\ge m(B-A)$ is actually sharp,\nso it follows that $s(n)$ is arithmetic."}
+{"year": 2009, "problem_number": 4, "problem": "Let $ABC$ be a triangle with $AB = AC$.\nThe angle bisectors of $\\angle CAB$ and $\\angle ABC$\nmeet the sides $BC$ and $CA$ at $D$ and $E$, respectively.\nLet $K$ be the incenter of triangle $ADC$.\nSuppose that $\\angle BEK = 45^\\circ$.\nFind all possible values of $\\angle CAB$.", "solution": "Here is the solution presented in my book EGMO.\n\nLet $I$ be the incenter of $ABC$,\nand set $\\angle DAC = 2x$ (so that $0\\dg < x < 45\\dg$).\nFrom $\\angle AIE = \\angle DIC$, it is easy to compute\n$\n\\angle KIE = 90\\dg - 2x, \\;\n\\angle ECI = 45\\dg -x, \\;\n\\angle IEK = 45\\dg, \\;\n\\angle KEC = 3x. $\nHaving chased all the angles we want, we need a relationship.\nWe can find it by considering the side ratio $\\frac{IK}{KC}$.\nUsing the angle bisector theorem,\nwe can express this in terms of triangle $IDC$;\nhowever we can also express it in terms of triangle $IEC$.\n\n[Figure omitted]\n\nBy the law of sines, we obtain\n$ \\frac{IK}{KC} = \\frac\n{\\sin 45\\dg \\cdot \\frac{EK}{\\sin ( 90\\dg - 2x )}}\n{\\sin ( 3x ) \\cdot \\frac{EK}{\\sin ( 45\\dg-x )}}\n= \\frac{\\sin 45\\dg \\sin ( 45\\dg - x )}\n{\\sin ( 3x ) \\sin ( 90\\dg - 2x )}. $\nAlso, by the angle bisector theorem on $\\triangle IDC$,\nwe have\n$ \\frac{IK}{KC} = \\frac{ID}{DC}\n= \\frac{\\sin ( 45\\dg-x )}{\\sin( 45\\dg+x )}. $\nEquating these and cancelling $\\sin ( 45\\dg-x ) \\neq 0$ gives\n$ \\sin 45\\dg \\sin ( 45\\dg + x )\n= \\sin 3x \\sin ( 90\\dg - 2x ). $\n\nApplying the product-sum formula\n(again, we are just trying to break down things as much as possible),\nthis just becomes\n$ \\cos( x ) - \\cos ( 90\\dg + x )\n= \\cos ( 5x-90\\dg ) - \\cos ( 90\\dg+x ) $\nor $\\cos x = \\cos ( 5x-90\\dg )$.\n\nAt this point we are basically done;\nthe rest is making sure we do not miss any solutions\nand write up the completion nicely.\nOne nice way to do this is by using product-sum in reverse as\n$ 0 = \\cos ( 5x-90\\dg ) - \\cos x\n= 2 \\sin ( 3x - 45\\dg ) \\sin ( 2x-45\\dg ). $\nThis way we merely consider the two cases\n$ \\sin ( 3x-45\\dg ) = 0 \\text{ and }\n\\sin ( 2x - 45\\dg ) = 0. $\nNotice that $\\sin\\theta = 0$ if and only $\\theta$\nis an integer multiple of $180\\dg$.\nUsing the bound $0\\dg < x < 45\\dg$,\nit is easy to see that that the permissible values of $x$\nare $x = 15\\dg$ and $x = \\frac{45}{2}\\dg$.\nAs $\\angle A = 4x$, this corresponds to $\\angle A = 60\\dg$\nand $\\angle A = 90\\dg$, which can be seen to work."}
+{"year": 2009, "problem_number": 5, "problem": "Find all functions $f \\colon \\ZZ_{>0} \\to \\ZZ_{>0}$\nsuch that for positive integers $a$ and $b$, the numbers\n$ a, \\qquad f(b), \\qquad f(b+f(a)-1) $\nare the sides of a non-degenerate triangle.", "solution": "The only function is the identity function (which works).\nWe prove it is the only one.\n\nLet $P(a,b)$ denote the given statement.\n\nWe have $f(1) = 1$, and $f(f(n)) = n$.\n(In particular $f$ is a bijection.)\n\nNote that $ P(1,b) \\implies f(b) = f(b+f(1)-1). $\nOtherwise, the function $f$ is periodic modulo $N = f(1)-1 \\ge 1$.\nThis is impossible since we can fix $b$ and let $a$ be arbitrarily\nlarge in some residue class modulo $N$.\n\nHence $f(1)=1$, so taking $P(n,1)$ gives $f(f(n)) = n$.\n\nLet $\\delta = f(2)-1 > 0$.\nThen for every $n$,\n$ f(n+1) = f(n) + \\delta\n\\quad\\text{ or }\\quad f(n-1) = f(n) + \\delta $\n\nUse\n$ P(2, f(n)) \\implies n-2 < f( f(n) + \\delta ) < n+2. $\nLet $y = f(f(n)+\\delta)$, hence $n-2 < y < n+2$\nand $f(y) = f(n)+\\delta$.\nBut, remark that if $y = n$, we get $\\delta = 0$, contradiction.\nSo $y \\in \\{n+1, n-1\\}$ and that is all.\n\nWe now show $f$ is an arithmetic progression\nwith common difference $+\\delta$.\nIndeed we already know $f(1) = 1$ and $f(2) = 1+\\delta$.\nNow suppose $f(1)=1$, \\dots, $f(n) = 1 + (n-1)\\delta$.\nThen by induction for any $n \\ge 2$,\nthe second case can't hold,\nso we have $f(n+1) = f(n)+\\delta$, as desired.\n\nCombined with $f(f(n)) = n$, we recover that $f$ is the identity."}
+{"year": 2009, "problem_number": 6, "problem": "Let $a_1$, $a_2$, \\dots, $a_n$ be distinct positive integers and\nlet $M$ be a set of $n-1$ positive integers not containing $s = a_1 + \\dots + a_n$.\nA grasshopper is to jump along the real axis, starting at the point $0$ and\nmaking $n$ jumps to the right with lengths $a_1$, $a_2$, \\dots, $a_n$ in some order.\nProve that the order can be chosen in such a way that\nthe grasshopper never lands on any point in $M$.", "solution": "The proof is by induction on $n$.\nAssume $a_1 < \\dots < a_n$ and call each element of $M$ a mine.\nLet $x = s - a_n$.\nWe consider four cases, based on whether $x$ has a mine\nand whether there is a mine past $x$.\n\nIf $x$ has no mine, and there is a mine past $x$,\nthen at most $n-2$ mines in $[0, x]$ and so we use induction to reach $x$,\nthen leap from $x$ to $s$ and win.\n\nIf $x$ has no mine but there is also no mine to the right of $x$,\nthen let $m$ be the maximal mine.\nBy induction hypothesis on $M \\setminus \\{m\\}$, there is a path to $x$\nusing $\\{a_1, \\dots, a_{n-1}\\}$ which avoids mines except possibly $m$.\nIf the path hits the mine $m$ on the hop of length $a_k$,\nwe then swap that hop with $a_n$, and finish.\n\nIf $x$ has a mine, but there are no mines to the right of $x$,\nwe can repeat the previous case with $m = x$.\n\nNow suppose $x$ has a mine, and there is a mine past $x$.\nThere should exist an integer $1 \\le i \\le n-1$\nsuch that $s-a_i$ and $y = s-a_i-a_n$ both have no mine.\nBy induction hypothesis, we can then reach $y$ in $n-2$ steps\n(as there are two mines to the right of $y$); then $y \\to s-a_i \\to s$ finishes.\n\nIt seems much of the difficulty of the problem is realizing induction will actually work.\nAttempts at induction are, indeed, a total minefield (ha!),\nand given the position P6 of the problem, it is expected that many contestants\nwill abandon induction after some cursory attempts fail."}
+{"year": 2010, "problem_number": 1, "problem": "Find all functions $f \\colon \\RR \\to \\RR$ such that for all $x,y \\in \\RR$,\n$ f(\\left\\lfloor x\\right\\rfloor y) = f(x)\\left\\lfloor f(y)\\right\\rfloor. $", "solution": "The only solutions are $f(x) \\equiv c$,\nwhere $c = 0$ or $1 \\le c < 2$.\nIt's easy to see these work.\n\nPlug in $x=0$ to get $f(0) = f(0) \\left\\lfloor f(y) \\right\\rfloor$,\nso either\n$ 1 \\le f(y) < 2 \\quad \\forall y\n\\qquad\\text{or}\\qquad f(0) = 0 $\n\nIn the first situation,\nplug in $y=0$ to get $f(x) \\left\\lfloor f(0) \\right\\rfloor = f(0)$,\nthus $f$ is constant.\nThus assume henceforth $f(0) = 0$.\n\nNow set $x=y=1$ to get\n$ f(1) = f(1) \\left\\lfloor f(1) \\right\\rfloor $\nso either $f(1) = 0$ or $1 \\le f(1) < 2$.\nWe split into cases:\n\nIf $f(1) = 0$, pick $x=1$ to get $f(y) \\equiv 0$.\nIf $1 \\le f(1) < 2$,\nthen $y=1$ gives\n$ f(\\left\\lfloor x \\right\\rfloor) = f(x) $\nfrom $y=1$, in particular $f(x) = 0$ for $0 \\le x < 1$.\nChoose $(x,y) = ( 2, 1/2 )$ to get\n$f(1) = f(2) \\left\\lfloor f( 1/2 ) \\right\\rfloor = 0$."}
+{"year": 2010, "problem_number": 2, "problem": "Let $I$ be the incenter of a triangle $ABC$ and let $\\Gamma$ be its circumcircle.\nLet line $AI$ intersect $\\Gamma$ again at $D$.\nLet $E$ be a point on arc $\\widehat{BDC}$ and $F$ a point on side $BC$ such that\n$ \\angle BAF = \\angle CAE < \\tfrac12 \\angle BAC. $\nFinally, let $G$ be the midpoint of $IF$.\nProve that $DG$ and $EI$ intersect on $\\Gamma$.", "solution": "Let $EI$ meet $\\Gamma$ again at $K$.\nThen it suffices to show that $KD$ bisects $IF$.\nLet $AF$ meet $\\Gamma$ again at $H$, so $HE \\parallel BC$.\nBy Pascal theorem on $ AHEKDD $\nwe then obtain that $P = AH \\cap KD$ lies on a line through $I$\nparallel to $BC$.\n\nLet $I_A$ be the $A$-excenter,\nand set $Q = I_AF \\cap IP$, and $T = AIDI_A \\cap BFC$.\nThen\n$ -1 = (AI;TI_A) \\overset{F} = (IQ;\\infty P) $\nwhere $\\infty$ is the point at infinity along $IPQ$.\nThus $P$ is the midpoint of $IQ$.\nSince $D$ is the midpoint of $II_A$ by ``Fact 5'',\nit follows that $DP$ bisects $IF$.\n\n[Figure omitted]"}
+{"year": 2010, "problem_number": 3, "problem": "Find all functions $g \\colon \\ZZ_{>0} \\to \\ZZ_{>0}$ such that\n$ ( g(m)+n )( g(n)+m ) $\nis always a perfect square.", "solution": "For $c \\ge 0$, the function $g(n) = n+c$ works; we prove this is the only possibility.\n\nFirst, the main point of the problem is that:\n\nWe have $g(n) \\equiv g(n') \\pmod p \\implies n \\equiv n' \\pmod p$.\n\nPick a large integer $M$ such that\n$ \\nu_p(M+g(n)), \\quad \\nu_p(M+g(n')) \\quad \\text{are both odd}. $\n(It's not hard to see this is always possible.)\nNow, since each of\n\n( M + g(n) )&( n + g(M) ) \\\\\n( M + g(n') )&( n' + g(M) )\n\nis a square, we get $n \\equiv n' \\equiv -g(M) \\pmod p$.\n\nThis claim implies that\n\nThe numbers $g(n)$ and $g(n+1)$ differ by $\\pm 1$ for any $n$, and\nThe function $g$ is injective.\n\nIt follows $g$ is a linear function with slope $\\pm 1$, hence done."}
+{"year": 2010, "problem_number": 4, "problem": "Let $P$ be a point interior to triangle $ABC$ (with $CA \\neq CB$).\nThe lines $AP$, $BP$ and $CP$ meet again its circumcircle $\\Gamma$\nat $K$, $L$, $M$, respectively.\nThe tangent line at $C$ to $\\Gamma$ meets the line $AB$ at $S$.\nShow that from $SC = SP$ follows $MK = ML$.", "solution": "We present two solutions using harmonic bundles.\n\nFirst solution (Evan Chen).: \nLet $N$ be the antipode of $M$, and let $NP$ meet $\\Gamma$ again at $D$.\nFocus only on $CDMN$ for now (ignoring the condition).\nThen $C$ and $D$ are feet of altitudes in $\\triangle MNP$;\nit is well-known that the circumcircle of $\\triangle CDP$\nis orthogonal to $\\Gamma$\n(passing through the orthocenter of $\\triangle MPN$).\n[Figure omitted]\nNow, we are given that point $S$ is such that $SC$\nis tangent to $\\Gamma$, and $SC = SP$.\nIt follows that $S$ is the circumcenter of $\\triangle CDP$,\nand hence $SC$ and $SD$ are tangents to $\\Gamma$.\n\nThen $-1 = (AB;CD) \\overset{P}{=} (KL;MN)$.\nSince $MN$ is a diameter, this implies $MK = ML$.\n\nI think it's more natural to come up with\nthis solution in reverse.\nNamely, suppose we define the points the other way:\nlet $SD$ be the other tangent, so $(AB;CD) = -1$.\nThen project through $P$ to get $(KL;MN) = -1$,\nwhere $N$ is the second intersection of $DP$.\nHowever, if $ML = MK$ then $KMLN$ must be a kite.\nThus one can recover the solution in reverse.\n\nSecond solution (Sebastian Jeon).: \nWe have $ SP^2 = SC^2 = SA \\cdot SB\n\\implies\n\u2220 SPA = \u2220 PBA = \u2220 LBA = \u2220 LKA = \u2220 LKP $\n(the latter half is Reim's theorem).\nTherefore $SP$ and $LK$ are parallel.\n\nNow, let $SP$ meet $\\Gamma$ again at $X$ and $Y$,\nand let $Q$ be the antipode of $P$ on $(S)$.\nThen\n$ SP^2 = SQ^2 = SX \\cdot SY\n\\implies (PQ;XY) = -1 \\implies \\angle QCP = 90\\dg $\nthat $CP$ bisects $\\angle XCY$.\nSince $XY \\parallel KL$,\nit follows $CP$ bisects to $\\angle LCK$ too."}
+{"year": 2010, "problem_number": 5, "problem": "Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$\ninitially contains one coin.\nThe following two types of operations are allowed:\n\nChoose a non-empty box $B_j$, $1\\leq j \\leq 5$,\nremove one coin from $B_j$ and add two coins to $B_{j+1}$;\nChoose a non-empty box $B_k$, $1\\leq k \\leq 4$,\nremove one coin from $B_k$ and swap the contents\n(possibly empty) of the boxes $B_{k+1}$ and $B_{k+2}$.", "solution": "First,\n\n(1,1,1,1,1,1) &\\to (0,3,1,0,3,1) \\to (0,0,7,0,0,7) \\\\\n&\\to (0,0,6,2,0,7) \\to (0,0,6,1,2,7) \\to (0,0,6,1,0,11) \\\\\n&\\to (0,0,6,0,11,0) \\to (0,0,5,11,0,0).\n\nand henceforth we ignore boxes $B_1$ and $B_2$,\nlooking at just the last four boxes;\nso we write the current position as $(5,11,0,0)$.\n\nWe prove a lemma:\n\nLet $k \\ge 0$ and $n > 0$.\nFrom $(k,n,0,0)$ we may reach $(k-1,2^n,0,0)$.\n\nWorking with only the last three boxes for now,\n\n(n,0,0) &\\to (n-1, 2, 0) \\to (n-1, 0, 4) \\\\\n&\\to (n-2, 4, 0) \\to (n-2, 0, 8) \\\\\n&\\to (n-3, 8, 0) \\to (n-3, 0, 16) \\\\\n&\\to \\dots \\to (1, 2^{n-1}, 0) \\to (1, 0, 2^n) \\to (0, 2^n, 0).\n\nFinally we have $(k,n,0,0) \\to (k,0,2^n,0) \\to (k-1,2^n, 0,0)$.\n\nNow from $(5,11,0,0)$ we go as follows:\n(5,11,0,0) &\\to (4, 2^{11}, 0, 0)\n\\to (3, 2^{2^{11}}, 0, 0)\n\\to (2, 2^{2^{2^{11}}}, 0, 0) \\\\\n&\\to ( 1, 2^{2^{2^{2^{11}}}}, 0, 0)\n\\to (0, 2^{2^{2^{2^{2^{11}}}}}, 0, 0).\n\nLet $A = 2^{2^{2^{2^{2^{11}}}}} > 2010^{2010^{2010}} = B$.\nThen by using move 2 repeatedly on the fourth box\n(i.e., throwing away several coins by swapping the empty $B_5$ and $B_6$),\nwe go from $(0,A,0,0)$ to $(0,B/4,0,0)$.\nFrom there we reach $(0,0,0,B)$."}
+{"year": 2010, "problem_number": 6, "problem": "Let $a_1, a_2, a_3, \\dots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that\n$\na_n =\n\\max \\{ a_k + a_{n-k} \\mid 1 \\leq k \\leq n-1 \\}\n\\text{ for all $n > s$}.\n$\nProve there exist positive integers $\\ell \\leq s$ and $N$, such that\n$\na_n =\na_{\\ell} + a_{n - \\ell} \\text{ for all $n \\ge N$}.\n$", "solution": "Let $ w_1 = \\frac{a_1}{1}, \\quad w_2 = \\frac{a_2}{2},\n\\quad \\dots, \\quad w_s = \\frac{a_s}{s}. $\n(The choice of the letter $w$ is for ``weight''.)\nWe claim the right choice of $\\ell$\nis the one maximizing $w_\\ell$.\n\nOur plan is to view each $a_n$ as a linear combination\nof the weights $w_1, \\dots, w_s$ and track their coefficients.\n\nTo this end, let's define an $n$-type\nto be a vector $T = \\left< t_1, \\dots, t_s\\right>$\nof nonnegative integers such that\n\n$n = t_1 + \\dots + t_s$; and\n$t_i$ is divisible by $i$ for every $i$.\n\nWe then define its valuation as $v(T) = \\sum_{i=1}^s w_i t_i$.\n\nNow we define a $n$-type to be valid\naccording to the following recursive rule.\nFor $1 \\le n \\le s$ the only valid $n$-types are\n\nT_1 &= \\left< 1, 0, 0, \\dots, 0 \\right> \\\\\nT_2 &= \\left< 0, 2, 0, \\dots, 0 \\right> \\\\\nT_3 &= \\left< 0, 0, 3, \\dots, 0 \\right> \\\\\n&\\vdotswithin= \\\\\nT_s &= \\left< 0, 0, 0, \\dots, s \\right>\n\nfor $n = 1, \\dots, s$, respectively.\nThen for any $n > s$, an $n$-type is valid\nif it can be written as the sum of a valid $k$-type\nand a valid $(n-k)$-type, componentwise.\nThese represent the linear combinations possible in the recursion;\nin other words the recursion in the problem is phrased as\n$ a_n = \\max_{T \\text{ is a valid $n$-type}} v(T). $\n\nIn fact, we have the following description of valid $n$-types:\n\nAssume $n > s$.\nThen an $n$-type $\\left< t_1, \\dots, t_s \\right>$ is valid\nif and only if either\n\nthere exist indices $i < j$ with $i+j > s$,\n$t_i \\ge i$ and $t_j \\ge j$; or\nthere exists an index $i > s/2$\nwith $t_i \\ge 2i$.\n\nImmediate by forwards induction on $n > s$\nthat all $n$-types have this property.\n\nThe reverse direction is by downwards induction on $n$.\nIndeed if $\\sum_i \\frac{t_i}{i} > 2$,\nthen we may subtract off on of $\\{T_1, \\dots, T_s\\}$\nwhile preserving the condition;\nand the case $\\sum_i \\frac{t_i}{i} = 2$\nis essentially by definition.\n\nThe claim is a bit confusingly stated in its two cases;\nreally the latter case should be thought of as the situation\n$i=j$ but requiring that $t_i/i$ is counted with multiplicity.\n\nNow, for each $n > s$ we pick a valid $n$-type $T_n$\nwith $a_n = v(T_n)$;\nif there are ties, we pick one for which the $\\ell$th\nentry is as large as possible.\n\nFor any $n > s$ and index $i \\neq \\ell$,\nthe $i$th entry of $T_n$\nis at most $2s + \\ell i$.\n\nIf not, we can go back $i\\ell$ steps to get\na valid $(n-i\\ell)$-type $T$\nachieved by decreasing the $i$th entry of $T_n$ by $i \\ell$.\nBut then we can add $\\ell$ to the $\\ell$th entry $i$\ntimes to get another $n$-type $T'$ which obviously\nhas valuation at least as large,\nbut with larger $\\ell$th entry.\n\nNow since all other entries in $T_n$ are bounded,\neventually the sequence $(T_n)_{n > s}$\njust consists of repeatedly\nadding $1$ to the $\\ell$th entry, as required.\n\nOne big step is to consider $w_k = a_k / k$.\nYou can get this using wishful thinking\nor by examining small cases.\n(In addition this normalization makes it easier\nto see why the largest $w$ plays an important role,\nsince then in the definition of type,\nthe $n$-types all have a sum of $n$.\nUnfortunately, it makes the characterization\nof valid $n$-types somewhat clumsier too.)"}
+{"year": 2011, "problem_number": 1, "problem": "Given any set $A = \\{a_1, a_2, a_3, a_4\\}$ of four distinct\npositive integers, we denote the sum $a_1+a_2+a_3+a_4$ by $s_A$.\nLet $n_A$ denote the number of pairs $(i,j)$ with $1 \\le i < j \\le 4$\nfor which $a_i + a_j$ divides $s_A$.\nFind all sets $A$ of four distinct positive integers which achieve\nthe largest possible value of $n_A$.", "solution": "There are two curves of solutions,\nnamely $\\{x,5x,7x,11x\\}$ and $\\{x,11x,19x,29x\\}$, for any positive integer $x$,\nachieving $n_A = 4$ (easy to check).\nWe'll show that $n_A \\le 4$ and equality holds only in one of the curves.\n\nLet $A = \\{a < b < c < d\\}$.\n\nWe have $n_A \\le 4$ with equality iff\n$ a+b \\mid c+d, \\quad a+c \\mid b+d, \\quad a+d = b+c. $\n\nNote $a+b \\mid s_A \\iff a+b \\mid c+d$ etc.\nNow $c+d \\nmid a+b$ and $b+d \\nmid a+c$ for size reasons, so we already have $n_A \\le 4$;\nmoreover $a+d \\mid b+c$ and $b+c \\mid a+d$ if and only if $a+d = b+c$.\n\nWe now show the equality curve is the one above.\n$ a+c \\mid b+d \\iff a+c \\mid -a+2b+c \\iff a+c \\mid 2(b-a). $\nSince $a+c > |b-a|$, so we must have $a+c=2(b-a)$.\nSo we now have\n\nc &= 2b-3a \\\\\nd &= b+c-a = 3b+c-4a.\n\nThe last condition is\n$ a+b \\mid c+d = 5b-7a \\iff a+b \\mid 12a. $\nNow, let $x = \\gcd(a,b)$.\nThe expressions for $c$ and $d$ above imply that\n$x \\mid c,d$ so we may scale down so that $x = 1$.\nThen $\\gcd(a+b,a) = \\gcd(a,b) = 1$ and so $a+b \\mid 12$.\n\nWe have $c > b$, so $3a < b$.\nThe only pairs $(a,b)$ with $3a < 2b$, $\\gcd(a,b) = 1$ and $a+b \\mid 12$\nare $(a,b) \\in \\left\\{ (1,5), (1,11) \\right\\}$ which give the solutions earlier."}
+{"year": 2011, "problem_number": 2, "problem": "Let $\\mathcal{S}$ be a finite set of at least two points in the plane.\nAssume that no three points of $\\mathcal S$ are collinear.\nA windmill is a process that starts with a\nline $\\ell$ going through a single point $P \\in \\mathcal S$.\nThe line rotates clockwise about the pivot $P$ until the first time\nthat the line meets some other point belonging to $\\mathcal S$.\nThis point, $Q$, takes over as the new pivot,\nand the line now rotates clockwise about $Q$,\nuntil it next meets a point of $\\mathcal S$.\nThis process continues indefinitely.\n\nShow that we can choose a point $P$ in $\\mathcal S$ and\na line $\\ell$ going through $P$ such that the resulting windmill\nuses each point of $\\mathcal S$ as a pivot infinitely many times.", "solution": "Orient $\\ell$ in some direction,\nand color the plane such that its left half is red\nand right half is blue.\nThe critical observation is that:\n\nThe number of points on the red side of $\\ell$ does not change,\nnor does the number of points on the blue side\n(except at a moment when $\\ell$ contains two points).\n\nThus, if $|\\mathcal S| = n+1$,\nit suffices to pick the initial configuration\nso that there are $\\left\\lfloor n/2 \\right\\rfloor$\nred and $\\left\\lceil n/2 \\right\\rceil$ blue points.\nThen when the line $\\ell$ does a full $180\\dg$ rotation,\nthe red and blue sides ``switch'',\nso the windmill has passed through every point.\n\n(See official shortlist for verbose write-up;\nthis is deliberately short to make a point.)"}
+{"year": 2011, "problem_number": 3, "problem": "Let $f \\colon \\RR \\to \\RR$ be a real-valued function\ndefined on the set of real numbers that satisfies\n$ f(x+y) \\leq yf(x) + f(f(x))$\nfor all real numbers $x$ and $y$.\nProve that $f(x) = 0$ for all $x \\leq 0$.", "solution": "We begin by rewriting the given as\n$ f(z) \\le (z-x)f(x) + f(f(x)) \\quad\n\\forall x,z \\in \\RR \\qquad (\\heartsuit) $\n(which is better anyways since control over inputs to $f$ is more valuable).\nWe start by eliminating the double $f$:\nlet $z = f(w)$ to get\n$ f(f(w)) \\le (f(w)-x)f(x) + f(f(x)) $\nand then use the symmetry trick to write\n$ f(f(x)) \\le (f(x)-w)f(w) + f(f(w)) $\nso that when we sum we get\n$ wf(w) + xf(x) \\le 2f(x)f(w). $\n\nNext we use cancellation trick: set $w = 2f(x)$ in the above to get\n$ xf(x) \\le 0 \\quad \\forall x \\in \\RR. \\qquad (\\spadesuit) $\n\nFor every $p \\in \\RR$, we have $f(p) \\le 0$.\n\nAssume $f(p) > 0$ for some $p \\in \\RR$.\nThen for any negative number $z$,\n$ 0 \\overset{(\\spadesuit)}{\\le} f(z)\n\\overset{(\\heartsuit)}{\\le} (z-p)f(p) + f(f(p)). $\nwhich is false if we let $z \\to -\\infty$.\n\nTogether with $(\\spadesuit)$ we derive $f(x) = 0$ for $x < 0$.\nFinally, letting $x$ and $z$ be any negative numbers\nin $(\\heartsuit)$, we get $f(0) \\ge 0$, so $f(0) = 0$ too.\n\nAs another corollary of the claim, $f(f(x)) = 0$ for all $x$.\n\nA nontrivial example of a working $f$ is to take\n$ f(x) = \n-\\exp(\\exp(\\exp(x))) & x > 0 \\\\\n0 & x \\le 0.\n$\nor some other negative function growing rapidly in absolute value for $x > 0$."}
+{"year": 2011, "problem_number": 4, "problem": "Let $n > 0$ be an integer.\nWe are given a balance and $n$ weights of weight $2^0$, $2^1$, \\dots, $2^{n-1}$.\nWe are to place each of the $n$ weights on the balance, one after another,\nin such a way that the right pan is never heavier than the left pan.\nAt each step we choose one of the weights\nthat has not yet been placed on the balance,\nand place it on either the left pan or the right pan,\nuntil all of the weights have been placed.\nDetermine the number of ways in which this can be done.", "solution": "The answer is $a_n = (2n-1)!!$.\n\nWe refer to what we're counting as a valid $n$-sequence:\nan order of which weights to place,\nand whether to place them on the left or right pan.\n\nWe use induction, with $n=1$ being obvious.\nNow consider the weight $2^0 = 1$.\n\nIf we delete it from any valid $n$-sequence,\nwe get a valid $(n-1)$-sequence with all weights doubled.\nGiven a valid $(n-1)$-sequence with all weights doubled,\nwe may insert $2^0 = 1$ it into $2n-1$ ways.\nIndeed, we may insert it anywhere, and designate it either left or right,\nexcept we may not designate right if we choose to insert\n$2^0 = 1$ at the very beginning.\n\nConsequently, we have that\n$ a_n = (2n-1) \\cdot a_{n-1}. $\nSince $a_1 = 1$, the conclusion follows.\n\n[Gabriel Levin]\nAn alternate approach can be done by considering the heaviest weight $2^{n-1}$\ninstead of the lightest weight $2^0=1$.\nThis gives a more complicated recursion:\n$ a_n = \\sum_{k=0}^{n-1} \\binom{n-1}{k} 2^{n-k-1} (n-k-1)! a_k $\nby summing over the index $k$ corresponding to the number of weights\nplaced before the heaviest weight $2^{n-1}$ is placed.\n\nThis recursion can then be rewritten as\n\na_n &= \\sum_{k=0}^{n-1} \\binom{n-1}{k} 2^{n-k-1} (n-k-1)! a_k \\\\\n&= (n-1)! \\sum_{k=0}^{n-1} \\frac{2^{n-k-1}a_k}{k!} \\\\\n&= 2(n-1)! \\sum_{k=0}^{n-1} \\frac{2^{(n-1)-k-1}a_k}{k!} \\\\\n&= 2(n-1)(n-2)! ( \\sum_{k=0}^{(n-1)-1} \\frac{2^{(n-1)-k-1}a_k}{k!} ) + 2(n-1)! \\frac{\\frac12 a_{n-1}}{(n-1)!} \\\\\n&= 2(n-1) a_{n-1} + a_{n-1} \\\\\n&= (2n-1)a_{n-1}\n\nand since $a_1 = 1$, we have $a_n = (2n-1)!!$ by induction on $n$."}
+{"year": 2011, "problem_number": 5, "problem": "Let $f \\colon \\ZZ \\to \\ZZ_{>0}$ be a function such that\n$f(m-n) \\mid f(m) - f(n)$ for $m,n \\in \\ZZ$.\nProve that if $m,n \\in \\ZZ$ satisfy $f(m) \\le f(n)$\nthen $f(m) \\mid f(n)$.", "solution": "Let $P(m,n)$ denote the given assertion.\nFirst, we claim $f$ is even.\nThis is straight calculation:\n\n$P(x,0) \\implies f(x) \\mid f(x)-f(0)\n\\implies f(x) \\mid M \\coloneq f(0)$.\n$P(0,x) \\implies f(-x) \\mid M-f(x) \\implies\nf(-x) \\mid f(x)$.\nAnalogously, $f(x) \\mid f(-x)$.\nSo $f(x) = f(-x)$ and $f$ is even.\n\nLet $x$, $y$, $z$ be integers with $x+y+z=0$.\nThen among $f(x)$, $f(y)$, $f(z)$,\ntwo of them are equal and divide the third.\n\nLet $a = f(\\pm x)$, $b = f(\\pm y)$, $c = f(\\pm z)$\nbe positive integers.\nNote that\n\na &\\mid b-c \\\\\nb &\\mid c-a \\\\\nc &\\mid a-b\n\nfrom $P(y,-z)$ and similarly.\nWLOG $c = \\max(a,b,c)$; then $c > |a-b|$\nso $a=b$. Thus $a=b \\mid c$ from the first two.\n\nThis implies the problem,\nby taking $x$ and $y$ in the previous claim\nto be the integers $m$ and $n$.\n\nAt ,\nDavi Medeiros gives the following characterization\nof functions $f$ satisfying the hypothesis.\n\nPick $f(0)$, $k$ positive integers,\na chain $d_1 \\mid d_2 \\mid \\dots \\mid d_k$ of divisors of $f(0)$,\nand positive integers $a_1,a_2,\\dots,a_{k-1}$,\ngreater than $1$ (if $k=1$, $a_i$ doesn't exist, for every $i$).\nWe'll define $f$ as follows:\n\n$f(n)=d_1$, for every integer $n$ that is not divisible by $a_1$;\n$f(a_1n)=d_2$, for every integer $n$ that is not divisible by $a_2$;\n$f(a_1a_2n)=d_3$, for every integer $n$ that is not divisible by $a_3$;\n$f(a_1a_2a_3n)=d_4$, for every integer $n$ that is not divisible by $a_4$;\n\\dots\n$f(a_1a_2 \\dots a_{k-1}n)=d_k$, for every integer $n$;"}
+{"year": 2011, "problem_number": 6, "problem": "Let $ABC$ be an acute triangle with circumcircle $\\Gamma$.\nLet $\\ell$ be a tangent line to $\\Gamma$, and let $\\ell_a$, $\\ell_b$, $\\ell_c$ be the lines obtained\nby reflecting $\\ell$ in the lines $BC$, $CA$, and $AB$, respectively.\nShow that the circumcircle of the triangle determined by the lines $\\ell_a$, $\\ell_b$, and $\\ell_c$\nis tangent to the circle $\\Gamma$.", "solution": "This is a hard problem with many beautiful solutions.\nThe following solution is not very beautiful but not too hard to find during an olympiad,\nas the only major insight it requires is the construction of $A_2$, $B_2$, and $C_2$.\n\n[Figure omitted]\n\nWe apply complex numbers with $\\omega$ the unit circle and $p=1$. Let $A_1 = \\ell_B \\cap \\ell_C$, and let $a_2 = a^2$ (in other words, $A_2$ is the reflection of $P$ across the diameter of $\\omega$ through $A$). Define the points $B_1$, $C_1$, $B_2$, $C_2$ similarly.\n\nWe claim that $A_1A_2$, $B_1B_2$, $C_1C_2$ concur at a point on $\\Gamma$.\n\nWe begin by finding $A_1$. If we reflect the points $1+i$ and $1-i$ over $AB$, then we get two points $Z_1$, $Z_2$ with\n\nz_1 &= a+b-ab(1-i) = a+b-ab+abi \\\\\nz_2 &= a+b-ab(1+i) = a+b-ab-abi. \\\\\n\\intertext{Therefore,}\nz_1 - z_2 &= 2abi \\\\\nz_1z_2 - z_2{z_1}\n&= -2i ( a+b+\\frac1a+\\frac1b-2 ).\n\nNow $\\ell_C$ is the line $Z_1Z_2$,\nso with the analogous equation $\\ell_B$ we obtain:\n\na_1 &= \\frac{ -2i( a+b+\\frac1a+\\frac1b-2 )( 2ac i ) +\n2i( a+c+\\frac1a+\\frac1c-2 )(2abi) }\n{ ( -\\frac{2}{ab}i )\n( 2ac i ) - ( -\\frac{2}{ac}i ) ( 2abi )} \\\\\n&= \\frac{[ c-b ]a^2 + [ \\frac cb - \\frac bc - 2c + 2b ]a + (c-b) }{\\frac cb - \\frac bc} \\\\\n&= a + \\frac{(c-b)[ a^2-2a+1 ]}{(c-b)(c+b)/bc} \\\\\n&= a + \\frac{bc}{b+c} (a-1)^2.\n\nThen the second intersection of $A_1A_2$ with $\\omega$ is given by\n\n\\frac{a_1-a_2}{1-a_2a_1}\n&= \\frac{a+\\frac{bc}{b+c}(a-1)^2-a^2}{1-a-a^2 \\cdot \\frac{(1-1/a)^2}{b+c}} \\\\\n&= \\frac{a + \\frac{bc}{b+c}(1-a)}{1 - \\frac{1}{b+c}(1-a)} \\\\\n&= \\frac{ab+bc+ca - abc}{a+b+c-1}.\n\nThus, the claim is proved.\n\nFinally, it suffices to show $A_1B_1 \\parallel A_2B_2$.\nOne can also do this with complex numbers;\nit amounts to showing $a^2-b^2$, $a-b$, $i$\n(corresponding to $A_2 B_2$, $A_1 B_1$, $PP$)\nhave their arguments an arithmetic progression, equivalently\n$ \\frac{(a-b)^2}{i(a^2-b^2)} \\in \\RR\n\\iff\n\\frac{(a-b)^2}{i(a^2-b^2)}\n= \\frac{( \\frac 1a-\\frac1b )^2}\n{\\frac1i(\\frac{1}{a^2}-\\frac{1}{b^2})}\n$\nwhich is obvious.\n\nOne can use directed angle chasing for this last part too.\nLet $BC$ meet $\\ell$ at $K$ and $B_2C_2$ meet $\\ell$ at $L$.\nEvidently\n\n-\u2220 B_2LP &= \u2220 LPB_2 + \u2220 PB_2L \\\\\n&= 2 \u2220 KPB + \u2220 PB_2C_2 \\\\\n&= 2 \u2220 KPB + 2\u2220 PBC \\\\\n&= -2\u2220 PKB \\\\\n&= \u2220 PKB_1\n\nas required."}
+{"year": 2012, "problem_number": 1, "problem": "Let $ABC$ be a triangle and $J$ the center of the $A$-excircle.\nThis excircle is tangent to the side $BC$ at $M$,\nand to the lines $AB$ and $AC$ at $K$ and $L$, respectively.\nThe lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G$.\nLet $S$ be the point of intersection of the lines $AF$ and $BC$,\nand let $T$ be the point of intersection of the lines $AG$ and $BC$.\nProve that $M$ is the midpoint of $ST$.", "solution": "We employ barycentric coordinates with reference $\\triangle ABC$.\nAs usual $a = BC$, $b = CA$, $c = AB$, $s = 1/2(a+b+c)$.\n\nIt's obvious that $K = ( -(s-c): s : 0)$, $M = ( 0 : s-b : s-c)$.\nAlso, $J = (-a : b : c)$.\nWe then obtain\n$ G = ( -a: b : \\frac{-as + (s-c)b}{s-b} ). $\nIt follows that\n$ T = ( 0 : b : \\frac{-as + (s-c)}{s-b} ) = ( 0 : b(s-b) : b(s-c) - as). $\nNormalizing, we see that $T = ( 0, -\\frac{b}{a}, 1 + \\frac{b}{a} )$,\nfrom which we quickly obtain $MT = s$.\nSimilarly, $MS = s$, so we're done."}
+{"year": 2012, "problem_number": 2, "problem": "Let $a_2$, $a_3$, \\dots, $a_n$ be positive reals with product $1$,\nwhere $n \\ge 3$.\nShow that\n$ (1+a_2)^2 (1+a_3)^3 \\dots (1+a_n)^n > n^n. $", "solution": "Try the dumbest thing possible: by AM-GM,\n\n(1 + a_2)^2 &\\ge 2^2 a_2 \\\\\n(1 + a_3)^3 = ( 1/2 + 1/2 + a_3 )^3 &\\ge \\frac{3^3}{2^2} a_3 \\\\\n(1 + a_4)^4 = ( \\frac13 + \\frac13 + \\frac13 + a_4 )^4\n&\\ge \\frac{4^4}{3^3} a_4 \\\\\n&\\vdotswithin=\n\nand so on.\nMultiplying these all gives the result.\nThe inequality is strict since it's not possible\nthat $a_2 = 1$, $a_3 = 1/2$, et cetera."}
+{"year": 2012, "problem_number": 3, "problem": "The liar's guessing game is a game played between two players $A$ and $B$.\nThe rules of the game depend on two fixed positive integers $k$ and $n$\nwhich are known to both players.\n\nAt the start of the game $A$\nchooses integers $x$ and $N$ with $1 \\le x \\le N$.\nPlayer $A$ keeps $x$ secret, and truthfully tells $N$ to player $B$.\nPlayer $B$ now tries to obtain information about $x$\nby asking player $A$ questions as follows:\neach question consists of $B$ specifying an arbitrary set $S$\nof positive integers (possibly one specified in some previous question),\nand asking $A$ whether $x$ belongs to $S$.\nPlayer $B$ may ask as many questions as he wishes.\nAfter each question, player $A$ must immediately answer\nit with yes or no, but is allowed to lie as many times as she wants;\nthe only restriction is that, among any $k+1$ consecutive answers,\nat least one answer must be truthful.\n\nAfter $B$ has asked as many questions as he wants,\nhe must specify a set $X$ of at most $n$ positive integers.\nIf $x$ belongs to $X$, then $B$ wins;\notherwise, he loses.\nProve that:\n\n[(a)]\nIf $n \\ge 2^k$, then $B$ can guarantee a win.\nFor all sufficiently large $k$,\nthere exists an integer $n \\ge (1.99)^k$\nsuch that $B$ cannot guarantee a win.", "solution": "Call the players Alice and Bob.\n\n\\textbf{Part (a)}: We prove the following.\n\nIf $N \\ge 2^k+1$, then in $2k+1$ questions,\nBob can rule out some number in $\\{1, \\dots, 2^k+1\\}$\nform being equal to $x$.\n\nFirst, Bob asks the question $S_0 = \\{ 2^k+1 \\}$\nuntil Alice answers ``yes''\nor until Bob has asked $k+1$ questions.\nIf Alice answers ``no'' to all of these then Bob rules out $2^k+1$.\nSo let's assume Alice just said ``yes''.\n\nNow let $T = \\{1, \\dots, 2^k\\}$.\nThen, he asks $k$-follow up questions $S_1$, \\dots, $S_k$\ndefined as follows:\n\n$S_1 = \\{1, 3, 5, 7, \\dots, 2^k-1\\}$ consists of all numbers\nin $T$ whose least significant digit in binary is $1$.\n$S_2 = \\{ 2, 3, 6, 7, \\dots, 2^k-2, 2^k-1\\}$\nconsists of all numbers in $T$ whose second least\nsignificant digit in binary is $1$.\nMore generally $S_i$\nconsists of all numbers in $T$ whose $i$th least\nsignificant digit in binary is $1$.\n\nWLOG Alice answers these all as ``yes'' (the other cases are similar).\nAmong the last $k+1$ answers at least one must be truthful,\nand the number $2^k$ (having zeros in all relevant digits)\ndoes not appear in any of $S_0$, \\dots, $S_k$ and is ruled out.\n\nThus in this way Bob can repeatedly find non-possibilities for $x$\n(and then relabel the remaining candidates $1$, \\dots, $N-1$)\nuntil he arrives at a set of at most $2^k$ numbers.\n\n\\textbf{Part (b)}:\nIt suffices to consider $n = \\left\\lceil 1.99^k \\right\\rceil$\nand $N = n+1$ for large $k$.\nAt the $t$th step, Bob asks some question $S_t$;\nwe phrase each of Alice's answers in the form ``$x \\notin B_t$'',\nwhere $B_t$ is either $S_t$ or its complement.\n(You may think of these as ``bad sets'';\nthe idea is to show we can avoid having any number\nappear in $k+1$ consecutive bad sets,\npreventing Bob from ruling out any numbers.)\n\nMain idea: for every number $1 \\le x \\le N$,\nat time step $t$ we define its weight\nto be $ w(x) = 1.998^e $\nwhere $e$ is the largest number\nsuch that $x \\in B_{t-1} \\cap B_{t-2} \\cap \\dots \\cap B_{t-e}$.\n\nAlice can ensure the total weight never exceeds $1.998^{k+1}$\nfor large $k$.\n\nLet $W_{t}$ denote the sum of weights after the $t$th question.\nWe have $W_0 = N < 1000n$.\nWe will prove inductively that $W_t < 1000n$ always.\n\nAt time $t$, Bob specifies a question $S_t$.\nWe have Alice choose $B_t$ as whichever of $S_t$ or $S_t$\nhas lesser total weight, hence at most $W_t/2$.\nThe weights of for $B_t$ increase by a factor of $1.998$,\nwhile the weights for $B_t$ all reset to $1$.\nSo the new total weight after time $t$ is\n$ W_{t+1} \\le 1.998 \\cdot \\frac{W_t}{2}\n+ \\# B_t \\le 0.999 W_t + n. $\nThus if $W_t < 1000n$ then $W_{t+1} < 1000n$.\n\nTo finish, note that\n$1000n < 1000 ( 1.99^k + 1 ) < 1.998^{k+1}$\nfor $k$ large.\n\nIn particular, no individual number can have weight $1.998^{k+1}$.\nThus for every time step $t$ we have\n$ B_t \\cap B_{t+1} \\cap \\dots \\cap B_{t+k} = \\varnothing. $\nThen once Bob stops, if he declares a set of $n$ positive integers,\nand $x$ is an integer Bob did not choose,\nthen Alice's question history is consistent with $x$ being Alice's number,\nas among any $k+1$ consecutive answers\nshe claimed that $x \\in B_t$ for some $t$ in that range.\n\n[Motivation]\nIn our $B_t$ setup, let's think backwards.\nThe problem is equivalent to avoiding $e = k+1$ at any time step $t$,\nfor any number $x$.\nThat means\n\nhave at most two elements with $e = k$ at time $t-1$,\nthus have at most four elements with $e = k-1$ at time $t-2$,\nthus have at most eight elements with $e = k-2$ at time $t-3$,\nand so on.\n\nWe already exploited this in solving part (a).\nIn any case it's now natural to try letting $w(x) = 2^e$,\nso that all the cases above sum to ``equally bad'' situations:\nsince $8 \\cdot 2^{k-2} = 4 \\cdot 2^{k-1} = 2 \\cdot 2^k$, say.\n\nHowever, we then get $W_{t+1} \\le 1/2 (2W_t) + n$,\nwhich can increase without bound due to contributions\nfrom numbers resetting to zero.\nThe way to fix this is to change the weight to $w(x) = 1.998^e$,\ntaking advantage of the little extra space we have\ndue to having $n \\ge 1.99^k$ rather than $n \\ge 2^k$."}
+{"year": 2012, "problem_number": 4, "problem": "Find all functions $f \\colon \\ZZ \\to \\ZZ$ such that,\nfor all integers $a$, $b$, $c$ that satisfy $a+b+c=0$,\nthe following equality holds:\n$ f(a)^2+f(b)^2+f(c)^2 = 2f(a)f(b)+2f(b)f(c)+2f(c)f(a). $", "solution": "Answer: for arbitrary $k \\in \\ZZ$, we have\n[(i)]\n$f(x) = kx^2$,\n$f(x) = 0$ for even $x$, and $f(x) = k$ for odd $x$, and\n$f(x) = 0$ for $x \\equiv 0 \\pmod 4$,\n$f(x) = k$ for odd $x$, and $f(x) = 4k$ for $x \\equiv 2 \\pmod 4$.\n\nThese can be painfully seen to work.\n(It's more natural to think of these as\n$f(x) = x^2$, $f(x) = x^2 \\pmod 4$, $f(x) = x^2 \\pmod 8$,\nand multiples thereof.)\n\nSet $a=b=c=0$ to get $f(0)=0$.\nThen set $c=0$ to get $f(a) = f(-a)$, so $f$ is even.\nNow $ f(a)^2 + f(b)^2 + f(a+b)^2\n= 2f(a+b)( f(a)+f(b) ) + 2f(a)f(b) $\nor\n$ ( f(a+b) - ( f(a)+f(b) ) )^2\n= 4f(a)f(b). $\nHence $f(a)f(b)$ is a perfect square for all $a,b \\in \\ZZ$.\nSo there exists a $\\lambda$ such that $f(n) = \\lambda g(n)^2$, where $g(n) \\ge 0$.\nFrom here we recover\n$ \\boxed{g(a+b) = \\pm g(a) \\pm g(b)}. $\nAlso $g(0) = 0$.\n\nLet $k = g(1) \\neq 0$.\nWe now split into cases on $g(2)$:\n\n$g(2) = 0$. Put $b = 2$ in original to get\n$g(a+2) = \\pm g(a) = +g(a)$.\n$g(2) = 2k$. Cases on $g(4)$:\n\n$g(4) = 0$, then we get\n$(g(n))_{n\\ge0} = (0,1,2,1,0,1,2,1,\\dots)$. This works.\n$g(4) = 4k$. This only happens when\n$g(1) = k$, $g(2) = 2k$, $g(3) = 3k$, $g(4) = 4k$.\nThen\n\n$g(5) = \\pm 3k \\pm 2k = \\pm 4k \\pm k$.\n$g(6) = \\pm4k \\pm 2k = \\pm5k \\pm k$.\n\\dots\n\nand so by induction $g(n) = nk$."}
+{"year": 2012, "problem_number": 5, "problem": "Let $ABC$ be a triangle with $\\angle BCA = 90\\dg$,\nand let $D$ be the foot of the altitude from $C$.\nLet $X$ be a point in the interior of the segment $CD$.\nLet $K$ be the point on the segment $AX$ such that $BK = BC$.\nSimilarly, let $L$ be the point on the segment $BX$ such that $AL = AC$.\nLet $M = AL \\cap BK$.\nProve that $MK = ML$.", "solution": "Let $\\omega_A$ and $\\omega_B$ be the circles through $C$\ncentered at $A$ and $B$;\nextend rays $AK$ and $BL$ to hit $\\omega_B$ and $\\omega_A$ again at $K^\\ast$, $L^\\ast$.\nBy radical center $X$,\nwe have $KLK^{\\ast}L^{\\ast}$ is cyclic,\nsay with circumcircle $\\omega$.\n\n[Figure omitted]\n\nBy orthogonality of $(A)$ and $(B)$ we find that\n$AL$, $AL^\\ast$,\n$BK$, $BK^\\ast$ are tangents to $\\omega$\n(in particular, $KLK^{\\ast}L^{\\ast}$ is harmonic).\nIn particular $MK$ and $ML$ are tangents to $\\omega$,\nso $MK = ML$."}
+{"year": 2012, "problem_number": 6, "problem": "Find all positive integers $n$\nfor which there exist non-negative integers $a_1, a_2, \\dots, a_n$\nsuch that\n$ \\frac{1}{2^{a_1}} + \\frac{1}{2^{a_2}} + \\dots + \\frac{1}{2^{a_n}}\n= \\frac{1}{3^{a_1}} + \\frac{2}{3^{a_2}} + \\dots + \\frac{n}{3^{a_n}}\n= 1. $", "solution": "The answer is $n \\equiv 1, 2 \\pmod 4$.\nTo see these are necessary,\nnote that taking the latter equation modulo $2$ gives\n$ 1 = \\frac{1}{3^{a_1}} + \\frac{2}{3^{a_2}} + \\dots + \\frac{n}{3^{a_n}}\n\\equiv 1 + 2 + .. + n \\pmod 2. $\n\nNow we prove these are sufficient.\nThe following nice construction was posted on AOPS\nby the user \\texttt{cfheolpiixn}.\n\nIf $n = 2k-1$ works then so does $n = 2k$.\n\nReplace\n$ \\frac{k}{3^r} = \\frac{k}{3^{r+1}} + \\frac{2k}{3^{r+1}}.\n\\qquad(\\ast) \\qedhere $\n\nIf $n = 4k+2$ works then so does $n = 4k + 13$.\n\nFirst use the identity\n$\n\\frac{k+2}{3^r} = \\frac{k+2}{3^{r+2}}\n+ \\frac{4k+ 3}{3^{r+3}}\n+ \\frac{4k+ 5}{3^{r+3}}\n+ \\frac{4k+ 7}{3^{r+3}}\n+ \\frac{4k+ 9}{3^{r+3}}\n+ \\frac{4k+11}{3^{r+3}}\n+ \\frac{4k+13}{3^{r+3}}\n$\nto fill in the odd numbers.\nThe even numbers can then be instantiated with $(\\ast)$ too.\n\nThus it suffices to construct base cases\nfor $n = 1$, $n = 5$, $n = 9$.\nThey are\n\n1 &= \\frac{1}{3^0} \\\\\n&= \\frac{1}{3^2} + \\frac{2}{3^2} + \\frac{3}{3^2}\n+ \\frac{4}{3^3} + \\frac{5}{3^3} \\\\\n&= \\frac{1}{3^2} + \\frac{2}{3^3} + \\frac{3}{3^3}\n+ \\frac{4}{3^3} + \\frac{5}{3^3} + \\frac{6}{3^4}\n+ \\frac{7}{3^4} + \\frac{8}{3^4} + \\frac{9}{3^4}."}
+{"year": 2013, "problem_number": 1, "problem": "Let $k$ and $n$ be positive integers.\nProve that there exist positive integers $m_1$, \\dots, $m_k$\nsuch that\n$ 1 + \\frac{2^k-1}{n} = ( 1 + \\frac{1}{m_1} ) ( 1 + \\frac{1}{m_2} )\n\\dots ( 1 + \\frac{1}{m_k} ). $", "solution": "By induction on $k \\ge 1$.\nWhen $k = 1$ there is nothing to prove.\n\nFor the inductive step, if $n$ is even, write\n$\n\\frac{n + (2^k-1)}{n}\n= ( 1 + \\frac{1}{n + (2^k-2)} ) \\cdot \\frac{\\frac n2 + (2^{k-1}-1)}{\\frac n2}\n$\nand use inductive hypothesis on the second term.\nOn the other hand if $n$ is odd then write\n$\n\\frac{n + (2^k-1)}{n}\n= ( 1 + \\frac{1}{n} ) \\cdot \\frac{\\frac{n+1}{2} + (2^{k-1}-1)}{\\frac{n+1}2}\n$\nand use inductive hypothesis on the second term."}
+{"year": 2013, "problem_number": 2, "problem": "A configuration of $4027$ points in the plane is called\nColombian if it consists of $2013$ red points and $2014$ blue points,\nand no three of the points of the configuration are collinear.\nBy drawing some lines, the plane is divided into several regions.\nAn arrangement of lines is good for a Colombian configuration\nif the following two conditions are satisfied:\n\n[(i)] No line passes through any point of the configuration.\n[(ii)] No region contains points of both colors.", "solution": "The answer is $k \\ge 2013$.\n\nTo see that $k = 2013$ is necessary,\nconsider a regular $4026$-gon and alternatively color the\npoints red and blue,\nthen place the last blue point anywhere\nin general position (it doesn't matter).\nEach side of the $4026$ is a red-blue line segment\nwhich needs to be cut by one of the $k$ lines,\nand each line can cut at most two of the segments.\n\nNow, we prove that $k = 2013$ lines is sufficient.\nConsider the convex hull of all the points.\n\nIf the convex hull has any red points,\ncut that red point off from everyone else by a single line.\nThen, for each of the remaining $2012$ red points,\nbreak them into $1006$ pairs arbitrarily,\nand for each pair $\\{A, B\\}$ draw two lines\nparallel to $AB$ and close to them.\n\nIf the convex hull has two consecutive blue points,\ncut those two blue points off from everyone else by a single line.\nThen repeat the above construction\nfor the remaining $2012$ blue points.\n\nThe end."}
+{"year": 2013, "problem_number": 3, "problem": "Let the excircle of triangle $ABC$ opposite\nthe vertex $A$ be tangent to the side $BC$ at the point $A_1$.\nDefine the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously,\nusing the excircles opposite $B$ and $C$, respectively.\nSuppose that the circumcenter of triangle $A_1B_1C_1$ lies\non the circumcircle of triangle $ABC$.\nProve that triangle $ABC$ is right-angled.", "solution": "We ignore for now the given condition\nand prove the following important lemma.\n\nLet $(AB_1C_1)$ meet $(ABC)$ again at $X$.\nFrom $BC_1 = B_1C$ follows $XC_1 = XB_1$,\nand $X$ is the midpoint of major arc $\\widehat{BC}$.\n\nThis follows from the fact that we have\na spiral similarity $\\triangle XBC_1 \\sim \\triangle XCB_1$\nwhich must actually be a spiral congruence\nsince $BC_1 = B_1C$.\n\nWe define the arc midpoints $Y$ and $Z$ similarly,\nwhich lie on the perpendicular bisectors of\n$A_1 C_1$, $A_1 B_1$.\n\n[Figure omitted]\n\nWe now turn to the problem condition\nwhich asserts the circumcenter $W$ of $\\triangle A_1B_1C_1$\nlies on $(ABC)$.\n\nWe may assume WLOG that $W = X$.\n\nThis is just configuration analysis,\nsince we already knew that the arc midpoints\nboth lie on $(ABC)$ and the relevant perpendicular bisectors.\n\nPoint $W$ lies on $(ABC)$ and hence outside $\\triangle ABC$,\nhence outside $\\triangle A_1 B_1 C_1$.\nThus we may assume WLOG that $\\angle B_1 A_1 C_1 > 90\\dg$.\nThen $A$ and $X$ lie on the same side of line $B_1 C_1$,\nand since $W$ is supposed to lie both on $(ABC)$\nand the perpendicular bisector of $B_1C_1$ it follows $W = X$.\n\nConsequently, $XY$ and $XZ$\nare exactly the perpendicular bisectors\nof $A_1 C_1$, $A_1 B_1$.\nThe rest is angle chase, the fastest one is\n\n\\angle A &= \\angle C_1 X B_1\n= \\angle C_1 X A_1 + \\angle A_1 X B_1\n= 2 \\angle YXA_1 + 2 \\angle A_1 X Z \\\\\n&= 2 \\angle YXZ = 180\\dg - \\angle A\n\nwhich solves the problem.\n\nAngle chasing is also possible even without\nthe points $Y$ and $Z$, though it takes much longer.\nIntroduce the Bevan point $V$ and use the fact\nthat $VA_1B_1C$ is cyclic (with diameter $VC$)\nand similarly $VA_1C_1B$ is cyclic;\na calculation then gives $\\angle CVB = 180\\dg - 1/2 \\angle A$.\nThus $V$ lies on the circle with diameter $I_b I_c$."}
+{"year": 2013, "problem_number": 4, "problem": "Let $ABC$ be an acute triangle with orthocenter $H$,\nand let $W$ be a point on the side $BC$, between $B$ and $C$.\nThe points $M$ and $N$ are the feet of the altitudes\ndrawn from $B$ and $C$, respectively.\nSuppose $\\omega_1$ is the circumcircle of triangle $BWN$\nand $X$ is a point such that $WX$ is a diameter of $\\omega_1$.\nSimilarly, $\\omega_2$ is the circumcircle of triangle $CWM$\nand $Y$ is a point such that $WY$ is a diameter of $\\omega_2$.\nShow that the points $X$, $Y$, and $H$ are collinear.", "solution": "We present two solutions, an elementary one\nand then an advanced one by moving points.\n\nFirst solution, classical.: \nLet $P$ be the second intersection of $\\omega_1$ and $\\omega_2$;\nthis is the Miquel point, so $P$ also lies on the circumcircle of $AMN$,\nwhich is the circle with diameter $AH$.\n\n[Figure omitted]\n\nWe now contend:\n\nPoints $P$, $H$, $X$ collinear.\n(Similarly, points $P$, $H$, $Y$ are collinear.)\n\n[Proof using power of a point]\nBy radical axis on $BNMC$, $\\omega_1$, $\\omega_2$,\nit follows that $A$, $P$, $W$ are collinear.\nWe know that $\\angle APH = 90\\dg$, and also\n$\\angle XPW = 90\\dg$ by construction.\nThus $P$, $H$, $X$ are collinear.\n\n[Proof using angle chasing]\nThis is essentially Reim's theorem:\n$ \u2220 NPH = \u2220 NAH = \u2220 BAH = \u2220 ABX = \u2220 NBX = \u2220 NPX $\nas desired.\nAlternatively, one may prove $A$, $P$, $W$ are collinear by\n$\u2220 NPA = \u2220 NMA = \u2220 NMC = \u2220 NBC = \u2220 NBW = \u2220 NPW$.\n\nSecond solution, by moving points.: \nFix $\\triangle ABC$ and vary $W$.\nLet $\\infty$ be the point at infinity perpendicular to $BC$\nfor brevity.\n\nBy spiral similarity, the point $X$ moves linearly on $B\\infty$\nas $W$ varies linearly on $BC$.\nSimilarly, so does $Y$.\nSo in other words, the map $ X \\mapsto W \\mapsto Y $\nis linear.\nHowever, the map $ X \\mapsto Y' \\coloneq XH \\cap C\\infty $\nis linear too.\n\nTo show that these maps are the same,\nit suffices to check it thus at two points.\n\nWhen $W = B$, the circle $(BNW)$\ndegenerates to the circle through $B$ tangent to $BC$,\nand $X = CN \\cap B\\infty$.\nWe have $Y = Y' = C$.\nWhen $W = C$, the result is analogous.\nAlthough we don't need to do so,\nit's also easy to check the result if $W$\nis the foot from $A$\nsince then $XHWB$ and $YHWC$ are rectangles."}
+{"year": 2013, "problem_number": 5, "problem": "Suppose a function $f \\colon \\QQ_{>0} \\to \\RR$ satisfies:\n\n[(i)] If $x,y \\in \\QQ_{>0}$, then $f(x)f(y) \\ge f(xy)$.\n[(ii)] If $x,y \\in \\QQ_{>0}$, then $f(x+y) \\ge f(x) + f(y)$.\n[(iii)] There exists a rational number $a > 1$ with $f(a) = a$.", "solution": "First, we dispense of negative situations by proving:\n\nFor any integer $n > 0$, we have $f(n) \\ge n$.\n\nNote by induction on (ii) we have $f(nx) \\ge n f(x)$.\nTaking $(x,y) = (a,1)$ in (i) gives $f(1) \\ge 1$,\nand hence $f(n) \\ge n$.\n\nThe $f$ takes only positive values,\nand hence by (ii) is strictly increasing.\n\n[Proof, suggested by Gopal Goel]\nLet $p,q > 0$ be integers.\nThen $f(q) f(p/q) \\ge f(p)$, and since both $\\min(f(p), f(q)) > 0$\nit follows $f(p/q) > 0$.\n\nFor any $x > 1$ we have $f(x) \\ge x$.\n\nNote that\n$ f(x)^N \\ge f(x^N) \\ge f( \\left\\lfloor x^N \\right\\rfloor )\n\\ge \\left\\lfloor x^N \\right\\rfloor > x^N-1 $\nfor any integer $N$.\nSince $N$ can be arbitrarily large, we conclude $f(x) \\ge x$ for $x > 1$.\n\nOn the other hand, $f$ has arbitrarily large fixed points\n(namely powers of $a$) so from (ii) we're essentially done.\nFirst, for $x > 1$ pick a large $m$ and note\n$ a^m = f(a^m) \\ge f(a^m-x) + f(x) \\ge (a^m-x)+x = a^m. $\nFinally, for $x \\le 1$ use\n$ nf(x) = f(n)f(x) \\ge f(nx) \\ge nf(x) $\nfor large $n$.\n\nNote that $a > 1$ is essential;\nif $b \\ge 1$ then $f(x) = bx^2$ works with unique fixed point $1/b \\le 1$."}
+{"year": 2013, "problem_number": 6, "problem": "Let $n \\ge 3$ be an integer, and consider a circle with $n + 1$ equally spaced points marked on it.\nConsider all labellings of these points with the numbers\n$0, 1, \\dots , n$ such that each label is used exactly once;\ntwo such labellings are considered to be the same if\none can be obtained from the other by a rotation of the circle.\nA labelling is called beautiful if, for any four labels $a < b < c < d$ with $a + d = b + c$,\nthe chord joining the points labelled $a$ and $d$\ndoes not intersect the chord joining the points labelled $b$ and $c$.\nLet $M$ be the number of beautiful labellings,\nand let $N$ be the number of ordered pairs $(x, y)$ of positive integers\nsuch that $x + y \\le n$ and $\\gcd(x, y) = 1$.\nProve that $M = N + 1$.", "solution": "First, here are half of the beautiful labellings up to reflection for $n = 6$,\njust for concreteness.\n\n[Figure omitted]\n\nAbbreviate ``beautiful labelling of points around a circle'' to ring.\nMoreover, throughout the solution we will allow degenerate\nchords that join a point to itself;\nthis has no effect on the problem statement.\n\nThe idea is to proceed by induction in the following way.\nA ring of $[0,n]$ is called linear\nif it is an arithmetic progression modulo $n+1$.\nFor example, the first two rings in the diagram\nand the last one are linear for $n = 6$,\nwhile the other three are not.\n\nOf course we can move from any ring on $[0,n]$\nto a ring on $[0,n-1]$ by deleting $n$.\nWe are going to prove that:\n\nEach linear ring on $[0,n-1]$ yields exactly two\nrings of $[0,n]$, and\nEach nonlinear ring on $[0,n-1]$ yields exactly one\nrings of $[0,n]$.\n\nIn light of the fact there are obviously $\\varphi(n)$ linear rings on $[0,n]$,\nthe conclusion will follow by induction.\n\nWe say a set of chords (possibly degenerate) is pseudo-parallel\nif for any three of them, one of them separates the two.\n(Pictorially, one can perturb the endpoints along the circle\nin order to make them parallel in Euclidean sense.)\nThe main structure lemma is going to be:\n\nIn any ring, the chords of sum $k$\n(even including degenerate ones) are pseudo-parallel.\n\nBy induction on $n$.\nBy shifting, we may assume that one of the chords is $\\{0,k\\}$\nand discard all numbers exceeding $k$; that is, assume $n = k$.\nSuppose the other two chords are $\\{a, n-a\\}$ and $\\{b, n-b\\}$.\n[Figure omitted]\nWe consider the chord $\\{u,v\\}$ directly above $\\{0,n\\}$, drawn in blue.\nThere are now three cases.\n\nIf $u+v = n$, then delete $0$ and $n$\nand decrease everything by $1$.\nThen the chords $\\{a-1, n-a-1\\}$, $\\{b-1, n-b-1\\}$, $\\{u-1, v-1\\}$\ncontradict the induction hypothesis.\n\nIf $u+v < n$, then search for the chord\n$\\{u+v, n-(u+v)\\}$.\nIt lies on the other side of $\\{0, n\\}$\nin light of chord $\\{0,u+v\\}$.\nNow again delete $0$ and $n$ and decrease everything by $1$.\nThen the chords $\\{a-1, n-a-1\\}$, $\\{b-1, n-b-1\\}$, $\\{u+v-1, n-(u+v)-1\\}$\ncontradict the induction hypothesis.\n\nIf $u+v > n$, apply the map $t \\mapsto n-t$ to the entire ring.\nThis gives the previous case as now $(n-u)+(n-v) < n$.\n\\qedhere\n\nNext, we give another characterization of linear rings.\n\nA ring on $[0,n-1]$ is linear if and only if the point $0$\ndoes not lie between two chords of sum $n$.\n\nIt's obviously true for linear rings.\nConversely, assume the property holds for some ring.\nNote that the chords with sum $n-1$ are pseudo-parallel and encompass every point,\nso they are actually parallel.\nSimilarly, the chords of sum $n$ are actually parallel\nand encompass every point other than $0$.\nSo the map\n$ t \\mapsto n-t \\mapsto (n-1)-(n-t) = t-1 \\pmod n $\nis rotation as desired.\n\nEvery nonlinear ring on $[0,n-1]$ induces exactly one ring on $[0,n]$.\n\nBecause the chords of sum $n$ are pseudo-parallel,\nthere is at most one possibility for the location $n$.\n\nConversely, we claim that this works.\nThe chords of sum $n$ (and less than $n$) are OK by construction, so\nassume for contradiction that there exists $a,b,c \\in \\{1,\\dots,n-1\\}$\nsuch that $a + b = n + c$.\nThen, we can ``reflect'' them using the (pseudo-parallel)\nchords of length $n$ to find that $(n-a) + (n-b) = 0 + (n-c)$,\nand the chords joining $0$ to $n-c$ and $n-a$ to $n-b$ intersect,\nby definition.\n[Figure omitted]\nThis is a contradiction that the original numbers on $[0,n-1]$ form a ring.\n\nEvery linear ring on $[0,n-1]$ induces\nexactly two rings on $[0,n]$.\n\nBecause the chords of sum $n$ are pseudo-parallel,\nthe point $n$ must lie either directly to the left or right of $0$.\nFor the same reason as in the previous proof, both of them work."}
+{"year": 2014, "problem_number": 1, "problem": "Let $a_0 < a_1 < a_2 < \\dotsb$ be an infinite sequence of positive integers.\nProve that there exists a unique integer $n\\geq 1$ such that\n$ a_n < \\frac{a_0+a_1+a_2+\\dotsb+a_n}{n} \\le a_{n+1}. $", "solution": "Fedor Petrov presents the following nice solution.\nLet us define the sequence\n$ b_n = ( a_n - a_{n-1} ) + \\dots + ( a_n - a_1 ). $\nSince $(a_n)_n$ is increasing, we get $(b_n)_n$ is strictly increasing,\nand moreover $b_1 = 0$.\nThe problem requires an $n$ such that\n$ b_n < a_0 \\le b_{n+1} $\nwhich obviously exists and is unique."}
+{"year": 2014, "problem_number": 2, "problem": "Let $n \\ge 2$ be an integer.\nConsider an $n \\times n$ chessboard consisting of $n^2$ unit squares.\nA configuration of $n$ rooks on this board is peaceful\nif every row and every column contains exactly one rook.\nFind the greatest positive integer $k$ such that,\nfor each peaceful configuration of $n$ rooks,\nthere is a $k \\times k$ square which does not\ncontain a rook on any of its $k^2$ unit squares.", "solution": "The answer is $k = \\left\\lfloor \\sqrt{n-1} \\right\\rfloor$.\n\nProof that the property holds when $n \\ge k^2+1$.: \nFirst, assume $n > k^2$ for some $k$.\nWe will prove we can find an empty $k \\times k$ square.\nIndeed, let $R$ be a rook in the uppermost column,\nand draw $k$ squares of size $k \\times k$ directly below it, aligned.\nThere are at most $k-1$ rooks among these squares, as desired.\n\n[Figure omitted]\n\nConstruction for all $n \\le k^2$.: \nWe first give an example where for $n = k^2$ showing there may be no empty $k \\times k$ square.\nWe draw the example for $k=3$ (with the generalization being obvious);\n\n[Figure omitted]\n\nTo show that this works,\nconsider for each rook drawing an $k \\times k$ square of $X$'s\nwhose bottom-right hand corner is the rook (these may go off the board).\nThese indicate positions where one cannot\nplace the upper-left hand corner of any square.\nIt's easy to see that these cover the entire board,\nexcept parts of the last $k-1$ columns,\nwhich don't matter anyways.\n\nIt remains to check that $n \\le k^2$ also all work\n(omitting this step is a common mistake).\nFor this, we can delete rows and column to get an $n \\times n$ board,\nand then fill in any gaps where we accidentally deleted a rook."}
+{"year": 2014, "problem_number": 3, "problem": "Convex quadrilateral $ABCD$ has $\\angle ABC = \\angle CDA = 90\\dg$.\nPoint $H$ is the foot of the perpendicular from $A$ to $BD$.\nPoints $S$ and $T$ lie on sides $AB$ and $AD$,\nrespectively, such that $H$ lies inside triangle $SCT$ and\n$ \\angle CHS - \\angle CSB = 90^{\\circ},\n\\quad \\angle THC - \\angle DTC = 90^{\\circ}. $\nProve that line $BD$ is tangent to the circumcircle of triangle $TSH$.", "solution": "First solution (mine).: \nFirst we rewrite the angle condition in a suitable way.\n\nWe have $\\angle ATH = \\angle TCH + 90\\dg$.\nThus the circumcenter of $\\triangle CTH$ lies on $AD$.\nSimilarly the circumcenter of $\\triangle CSH$ lies on $AB$.\n\n\u2220 ATH &= \u2220 DTH \\\\\n&= \u2220 DTC + \u2220 CTH \\\\\n&= \u2220 DTC - \u2220 THC - \u2220 HCT \\\\\n&= 90\\dg - \u2220 HCT = 90\\dg + \u2220 TCH.\n\nwhich implies conclusion.\n\n[Figure omitted]\n\nLet the perpendicular bisector of $TH$ meet $AH$ at $P$ now.\nIt suffices to show that $\\frac{AP}{PH}$ is symmetric in $b = AD$ and $d=AB$,\nbecause then $P$ will be the circumcenter of $\\triangle TSH$.\nTo do this, set $AH = \\frac{bd}{2R}$ and $AC=2R$.\n\nLet $O$ denote the circumcenter of $\\triangle CHT$.\nUse the Law of Cosines on $\\triangle ACO$ and $\\triangle AHO$,\nusing variables $x=AO$ and $r=HO$.\nWe get that\n$ r^2 = x^2 + AH^2 - 2x \\cdot AH \\cdot \\frac{d}{2R} = x^2 + (2R)^2 - 2bx. $\nBy the angle bisector theorem,\n$\\frac{AP}{PH} = \\frac{AO}{HO}$.\n\nThe rest is computation: notice that\n$ r^2 - x^2 = h^2 - 2xh \\cdot \\frac{d}{2R} = (2R)^2 - 2bx $\nwhere $h = AH = \\frac{bd}{2R}$, whence\n$ x = \\frac{(2R)^2-h^2}{2b - 2h \\cdot \\frac{d}{2R}}. $\nMoreover,\n$ \\frac{1}{2} ( \\frac{r^2}{x^2}-1 )\n= \\frac{1}{x} ( \\frac 2x R^2 - b ). $\nNow, if we plug in the $x$ in the\nright-hand side of the above, we obtain\n$ \\frac{2b-2h \\cdot \\frac{d}{2R}}{4R^2-h^2}\n( \\frac{2b-2h \\cdot \\frac{d}{2R}}{4R^2-h^2} \\cdot 2R^2 - b)\n= \\frac{2h}{(4R^2-h^2)^2} ( b- h \\cdot \\frac{d}{2R} )\n( -2hdR + bh^2 ). $\nPulling out a factor of $-2Rh$ from the rightmost term,\nwe get something that is symmetric in $b$ and $d$, as required.\n\nSecond solution (Victor Reis).: \nHere is the fabled solution using inversion at $H$.\nFirst, we rephrase the angle conditions in the following ways:\n\n$AD \\perp (THC)$,\nwhich is equivalent to the claim from the first solution.\n$AB \\perp (SHC)$, by symmetry.\n$AC \\perp (ABCD)$, by definition.\n\nNow for concreteness we will use a negative inversion at $H$\nwhich swaps $B$ and $D$ and overlay it on the original diagram.\nAs usual we denote inverses with stars.\n\nLet us describe the inverted problem.\nWe let $M$ and $N$ denote the midpoints of $A^\\ast B^\\ast$\nand $A^\\ast D^\\ast$, which are the centers of\n$(HA^\\ast B^\\ast)$ and $(HA^\\ast D^\\ast)$.\nFrom $T^\\ast C^\\ast \\perp (HA^\\ast D^\\ast)$,\nwe know have $C^\\ast$, $M$, $T^\\ast$ collinear.\nSimilarly, $C^\\ast$, $N$, $S^\\ast$ are collinear.\nWe have that $(A^\\ast H C^\\ast)$ is orthogonal to $(ABCD)$ which remains fixed.\nWe wish to show $T^\\ast S^\\ast$ and $MN$ are parallel.\n\n[Figure omitted]\n\nLot $\\omega$ denote the circumcircle of $\\triangle A^\\ast H C^\\ast$,\nwhich is orthogonal to the original circle $(ABCD)$.\nIt would suffices to show $(A^\\ast H C^\\ast)$\nis an $H$-Apollonius circle with respect to $MN$,\nfrom which we would get $C^\\ast M / H M = C^\\ast N / H N$.\n\nHowever, $\\omega$ through $H$ and $A$,\nhence it center lies on line $MN$.\nMoreover $\\omega$ is orthogonal to $(A^\\ast MN)$\n(since $(A^\\ast MN)$ and $(A^\\ast BD)$ are homothetic).\nThis is enough (for example, if we let $O$ denote the center of $\\omega$,\nwe now have $\\mathrm{r}(\\omega)^2 = OH^2 = OM \\cdot ON$).\n(Note in this proof that the fact that $C^\\ast$ lies on $(ABCD)$\nis not relevant.)"}
+{"year": 2014, "problem_number": 4, "problem": "Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$\nsuch that $\\angle PAB=\\angle BCA$ and $\\angle CAQ=\\angle ABC$.\nLet $M$ and $N$ be points on $AP$ and $AQ$,\nrespectively, such that $P$ is the midpoint of $AM$\nand $Q$ is the midpoint of $AN$.\nProve that $BM$ and $CN$ meet on the\ncircumcircle of $\\triangle ABC$.", "solution": "We give three solutions.\n\nFirst solution by harmonic bundles.: \nLet $BM$ intersect the circumcircle again at $X$.\n\n[Figure omitted]\n\nThe angle conditions imply that the tangent to $(ABC)$ at $B$\nis parallel to $AP$.\nLet $\\infty$ be the point at infinity along line $AP$.\nThen $ -1 = (AM;P\\infty) \\overset{B}{=} (AX;BC). $\nSimilarly, if $CN$ meets the circumcircle at $Y$\nthen $(AY;BC) = -1$ as well.\nHence $X=Y$, which implies the problem.\n\nSecond solution by similar triangles.: \nOnce one observes $\\triangle CAQ \\sim \\triangle CBA$,\none can construct $D$ the reflection of $B$ across $A$,\nso that $\\triangle CAN \\sim \\triangle CBD$.\nSimilarly, letting $E$ be the reflection of $C$ across $A$,\nwe get $\\triangle BAP \\sim \\triangle BCA\n\\implies \\triangle BAM \\sim \\triangle BCE$.\nNow to show $\\angle ABM + \\angle ACN = 180\\dg$\nit suffices to show $\\angle EBC + \\angle BCD = 180\\dg$,\nwhich follows since $BCDE$ is a parallelogram.\n\nThird solution by barycentric coordinates.: \nSince $PB = c^2/a$ we have\n$ P = (0 : a^2-c^2 : c^2) $\nso the reflection $\\vec M = 2\\vec P - \\vec A$ has coordinates\n$ M = (-a^2 : 2(a^2-c^2) : 2c^2). $\n\nSimilarly $N = (-a^2 : 2b^2 : 2(b^2-a^2))$. Thus\n$ BM \\cap CN = (-a^2 : 2b^2 : 2c^2) $\nwhich clearly lies on the circumcircle,\nand is in fact the point identified in the first solution."}
+{"year": 2014, "problem_number": 5, "problem": "For every positive integer $n$,\nthe Bank of Cape Town issues coins of denomination $\\frac 1n$.\nGiven a finite collection of such coins (of not necessarily different denominations)\nwith total value at most $99 + \\frac12$, prove that it is possible to split\nthis collection into $100$ or fewer groups, such that each group has total value at most $1$.", "solution": "We'll prove the result\nfor at most $k - \\frac{k}{2k+1}$ with $k$ groups.\nFirst, perform the following optimizations.\n\nIf any coin of size $\\frac{1}{2m}$ appears twice,\nthen replace it with a single coin of size $\\frac{1}{m}$.\nIf any coin of size $\\frac{1}{2m+1}$ appears $2m+1$ times,\ngroup it into a single group and induct downwards.\n\nApply this operation repeatedly until it cannot be done anymore.\n\nNow construct boxes $B_0$, $B_1$, \\dots, $B_{k-1}$.\nIn box $B_0$ put any coins of size $\\tfrac 12$ (clearly there is at most one).\nMore generally for $m \\ge 0$, $B_m$, put coins of size\n$\\frac{1}{2m+1}$ and $\\frac{1}{2m+2}$\n(there at most $2m$ of the former and at most one of the latter).\nNote that\n$ \\text{total weight in $B_m$} \\le 2m \\cdot \\frac{1}{2m+1} + \\frac{1}{2m+2} < 1. $\nFinally, place the remaining ``light'' coins of size at most $\\frac{1}{2k+1}$ in a pile.\n\nOne way to explain where this boxing comes from is to imagine\nwhat happens after applying the operation repeatedly until it can't be done any more.\nWe expect the most troublesome situation would be if the leftover coins are\nas large as possible, which looks like\n$ 1/2, \\qquad \\frac13 \\times 2, \\qquad \\frac14, \\qquad \\frac15 \\times 4,\n\\qquad \\frac16, \\qquad \\frac17 \\times 6, \\qquad \\frac 18, \\qquad \\dots $\nSeeing this the boxing $B_i$ is quite reasonable because it groups these\ninto groups of almost $1$ without too much effort:\n\nB_0 &\\longrightarrow 1/2 \\\\\nB_1 &\\longrightarrow \\frac13 \\cdot 2 + \\frac 14 = \\frac{11}{12} \\\\\nB_2 &\\longrightarrow \\frac15 \\cdot 4 + \\frac 16 = \\frac{29}{30} \\\\\nB_3 &\\longrightarrow \\frac17 \\cdot 6 + \\frac 18 = \\frac{55}{56} \\\\\n&\\vdotswithin\\longrightarrow\n\nThen just toss coins from the pile into the boxes arbitrarily,\nother than the proviso that no box should have its weight exceed $1$.\nWe claim this uses up all coins in the pile.\nAssume not, and that some coin remains in the pile\nwhen all the boxes are saturated.\nThen all the boxes must have at least $1 -\\frac{1}{2k+1}$,\nmeaning the total amount in the boxes is strictly greater than\n$ k ( 1 - \\frac{1}{2k+1} ) > k - \\tfrac 12 $\nwhich is a contradiction.\n\nThis gets a stronger bound $k - \\frac{k}{2k+1}$ than the requested $k-\\tfrac 12$."}
+{"year": 2014, "problem_number": 6, "problem": "A set of lines in the plane is in general position\nif no two are parallel and no three pass through the same point.\nA set of lines in general position cuts the plane into regions,\nsome of which have finite area; we call these its finite regions.\nProve that for all sufficiently large $n$,\nin any set of $n$ lines in general position\nit is possible to colour at least $\\sqrt{n}$ lines blue\nin such a way that none of its finite regions\nhas a completely blue boundary.", "solution": "Suppose we have colored $k$ of the lines blue, and that\nit is not possible to color any additional lines.\nThat means any of the $n-k$ non-blue lines\nis the side of some finite region with\nan otherwise entirely blue perimeter.\nFor each such line $\\ell$, select one such region,\nand take the next counterclockwise vertex;\nthis is the intersection of two blue lines $v$.\nWe'll say $\\ell$ is the eyelid of $v$.\n\n[Figure omitted]\n\nYou can prove without too much difficulty that every intersection of two blue lines\nhas at most two eyelids.\nSince there are $\\binom k2$ such intersections, we see that\n$ n-k \\le 2 \\binom k2 = k^2 - k$\nso $n \\le k^2$, as required.\n\nIn fact, $k = \\sqrt n$ is ``sharp for greedy algorithms'',\nas illustrated below for $k=3$:\n[Figure omitted]"}
+{"year": 2015, "problem_number": 1, "problem": "We say that a finite set $\\mathcal{S}$ of points in the plane\nis balanced if,\nfor any two different points $A$ and $B$ in $\\mathcal{S}$,\nthere is a point $C$ in $\\mathcal{S}$ such that $AC=BC$.\nWe say that $\\mathcal{S}$ is center-free if for\nany three different points $A$, $B$ and $C$ in $\\mathcal{S}$,\nthere are no points $P$ in $\\mathcal{S}$ such that $PA=PB=PC$.\n\n\\item[(a)] Show that for all integers $n\\ge 3$,\nthere exists a balanced set consisting of $n$ points.\n\\item[(b)] Determine all integers $n\\ge 3$ for which\nthere exists a balanced center-free set consisting of $n$ points.", "solution": "For part (a), take a circle centered at a point $O$,\nand add $n-1$ additional points by adding pairs of points\nseparated by an arc of $60^{\\circ}$ or similar triples.\nAn example for $n = 6$ is shown below.\n[Figure omitted]\n\nFor part (b), the answer is odd $n$, achieved by taking a regular $n$-gon.\nTo show even $n$ fail, note that some point is on the perpendicular bisector of\n$ \\left\\lceil \\frac 1n \\binom n2 \\right\\rceil = \\frac{n}{2} $\npairs of points, which is enough.\n(This is a standard double-counting argument.)\n\nAs an aside, there is a funny joke about this problem.\nThere are two types of people in the world:\n\nThose who solve (b) quickly and then take forever to solve (a),\nthose who solve (a) quickly and then can't solve (b) at all.\n\n(Empirically true when the Taiwan IMO 2014 team was working on it.)"}
+{"year": 2015, "problem_number": 2, "problem": "Find all positive integers $a$, $b$, $c$ such that\neach of $ab-c$, $bc-a$, $ca-b$ is a power of $2$\n(possibly including $2^0=1$).", "solution": "Here is the solution of \\textbf{Telv Cohl},\nwhich is the shortest solution I am aware of.\nWe will prove the only solutions are $(2,2,2)$, $(2,2,3)$,\n$(2,6,11)$ and $(3,5,7)$ and permutations.\n\nWLOG assume $a \\ge b \\ge c > 1$, so $ab-c \\ge ca-b \\ge bc-a$.\nWe consider the following cases:\n\nIf $a$ is even, then\n\nca-b &= \\gcd (ab-c, ca-b) \\le \\gcd (ab-c, a(ca-b)+ab-c) \\\\\n&= \\gcd( ab-c, c(a^2-1) ).\n\nAs $a^2-1$ is odd, we conclude $ca-b \\le c$.\nThis implies $a=b=c=2$.\n\nIf $a$, $b$, $c$ are all odd, then $a > b > c > 1$ follows.\nThen as before\n$ ca-b \\le \\gcd (ab-c, c(a^2-1))\n\\le 2^{\\nu_2(a^2-1)} \\le 2a+2 \\le 3a-b $\nso $c = 3$ and $a = b+2$.\nAs $3a-b = ca-b \\ge 2(bc-a) = 6b-2a$ we then conclude $a=7$ and $b=5$.\n\nIf $a$ is odd and $b$, $c$ are even, then $bc-a=1$\nand hence $bc^2 - b - c = ca - b$.\nThen from the miraculous identity\n$ c^3-b-c = (1-c^2)(ab-c) + a(\\underbrace{bc^2-b-c}_{=ca-b}) + (ca-b) $\nso we conclude $\\gcd(ab-c, ca-b) = \\gcd(ab-c, c^3-b-c)$, in other words\n$ bc^2-b-c = ca-b = \\gcd(ab-c, ca-b) = \\gcd(ab-c, c^3-b-c). $\nWe thus consider two more cases:\n\nIf $c^3-b-c \\neq 0$ then\nthe above implies $|c^3-b-c| \\ge bc^2-b-c$.\nAs $b \\ge c > 1$, we must actually have $b = c$,\nthus $a = c^2-1$.\nFinally $ab-c = c(c^2-2)$ is a power of $2$, hence $b=c=2$, so $a=3$.\n\nIn the second case, assume $c^3-b-c = 0$, hence $c^3-c$.\nFrom $bc-a=1$ we obtain $a=c^4-c^2-1$,\nhence\n$ ca-b = c^5-2c^3 = c^3(c^2-2) $\nis a power of $2$, hence again $c = 2$.\nThus $a=11$ and $b=6$.\n\nThis finishes all cases, so the proof is done."}
+{"year": 2015, "problem_number": 3, "problem": "Let $ABC$ be an acute triangle with $AB > AC$.\nLet $\\Gamma$ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$.\nLet $M$ be the midpoint of $BC$.\nLet $Q$ be the point on $\\Gamma$ such that $\\angle HQA = 90\\dg$\nand let $K$ be the point on $\\Gamma$ such that $\\angle HKQ = 90\\dg$.\nAssume that the points $A$, $B$, $C$, $K$ and $Q$ are all different and lie on $\\Gamma$ in this order.\nProve that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.", "solution": "Let $L$ be on the nine-point circle with $\\angle HML = 90^{\\circ}$.\nThe negative inversion at $H$ swapping $\\Gamma$ and nine-point circle maps\n$ A \\longleftrightarrow F, \\quad\nQ \\longleftrightarrow M, \\quad\nK \\longleftrightarrow L. $\nIn the inverted statement, we want line $ML$ to be tangent to $(AQL)$.\n\n[Figure omitted]\n\n$LM \\parallel AQ$.\n\nBoth are perpendicular to $MHQ$.\n\n$LA = LQ$.\n\nLet $N$ and $T$ be the midpoints of $HQ$ and $AH$,\nand $O$ the circumcenter.\nAs $MT$ is a diameter, we know $LTNM$ is a rectangle,\nso $LT$ passes through $O$.\nSince $LOT \\perp AQ$ and $OA=OQ$, the proof is complete.\n\nTogether these two claims solve the problem."}
+{"year": 2015, "problem_number": 4, "problem": "Triangle $ABC$ has circumcircle $\\Omega$ and circumcenter $O$.\nA circle $\\Gamma$ with center $A$\nintersects the segment $BC$ at points $D$ and $E$,\nsuch that $B$, $D$, $E$, and $C$ are all different\nand lie on line $BC$ in this order.\nLet $F$ and $G$ be the points of intersection of $\\Gamma$ and $\\Omega$,\nsuch that $A$, $F$, $B$, $C$, and $G$ lie on $\\Omega$ in this order.\nLet $K = (BDF) \\cap AB \\neq B$\nand $L = (CGE) \\cap AC \\neq C$\nand assume these points do not lie on line $FG$.\nDefine $X = FK \\cap GL$.\nProve that $X$ lies on the line $AO$.", "solution": "Since $AO \\perp FG$ for obvious reasons,\nwe will only need to show that $XF = XG$,\nor that $\u2220 KFG = \u2220 LGF$.\n\nLet line $FG$ meet $(BDF)$ and $(CGE)$\nagain at $F_2$ and $G_2$.\n[Figure omitted]\n\nQuadrilaterals $FBDF_2$ and $G_2ECG$ are similar,\nactually homothetic through $FG \\cap BC$.\n\nThis is essentially a repeated application\nof being ``anti-parallel'' through $\\angle(FG, BC)$.\nNote the four angle relations\n\n\u2220(FD, FG) = \u2220(BC,GE) = \u2220(G_2C,FG)\n&\\implies FD \\parallel G_2C \\\\\n\u2220(F_2B, FG) = \u2220(BC,FD) = \u2220(GE,FG)\n&\\implies F_2B \\parallel GE \\\\\n\u2220(FB, FG) = \u2220(BC,GC) = \u2220(G_2E,FG)\n&\\implies FB \\parallel G_2E \\\\\n\u2220(F_2D, FG) = \u2220(BC,FB) = \u2220(GC,FG)\n&\\implies F_2D \\parallel GC.\n\nThis gives the desired homotheties.\n\nTo finish the angle chase,\n\n\u2220 GFK = \u2220 F_2BK &= \u2220 F_2BF - \u2220 ABF\n= \u2220 F_2DF - \u2220 ABF \\\\\n&= \u2220 F_2DF - \u2220 GCA\n= \u2220 GCG_2 - \u2220 GCA \\\\\n&= \u2220 LCG_2 = \u2220 LGF\n\nas needed.\n(Here $\u2220 ABF = \u2220 GCA$ since $AF = AG$.)"}
+{"year": 2015, "problem_number": 5, "problem": "Solve the functional equation\n$ f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x) $\nfor $f \\colon \\RR \\to \\RR$.", "solution": "The answers are $f(x) \\equiv x$ and $f(x) \\equiv 2-x$.\nObviously, both of them work.\n\nLet $P(x,y)$ be the given assertion.\nWe also will let $S = \\{t \\mid f(t) = t\\}$\nbe the set of fixed points of $f$.\n\nFrom $P(0,0)$ we get $f(f(0)) = 0$.\nFrom $P(0,f(0))$ we get $2f(0) = f(0)^2$\nand hence $f(0) \\in \\{0,2\\}$.\nFrom $P(x,1)$\nwe find that $x+f(x+1) \\in S$ for all $x$.\n\nWe now solve the case $f(0) = 2$.\n\nIf $f(0) = 2$ then $f(x) \\equiv 2-x$.\n\nLet $t \\in S$ be any fixed point.\nThen $P(0,t)$ gives $2 = 2t$ or $t = 1$;\nso $S = \\{1\\}$.\nBut we also saw $x+f(x+1) \\in S$,\nwhich implies $f(x) \\equiv 2-x$.\n\nHenceforth, assume $f(0) = 0$.\n\nIf $f(0) = 0$ then $f$ is odd.\n\nNote that $P(1,-1) \\implies f(1) + f(-1) = 1 - f(1)$\nand $P(-1,1) \\implies f(-1) + f(-1) = -1 + f(1)$,\ntogether giving $f(1) = 1$ and $f(-1) = -1$.\nTo prove $f$ odd we now obtain more fixed points:\n\nFrom $P(x,0)$ we find that $x+f(x) \\in S$ for all $x \\in \\RR$.\nFrom $P(x-1,1)$ we find that $x-1+f(x) \\in S$ for all $x \\in \\RR$.\nFrom $P(1, f(x)+x-1)$ we find $x+1+f(x) \\in S$ for all $x \\in \\RR$.\n\nFinally $P(x, -1)$ gives $f$ odd.\n\nTo finish from $f$ odd, notice that\n\nP(x,-x) &\\implies f(x) + f(-x^2) = x - xf(x) \\\\\nP(-x,x) &\\implies f(-x) + f(-x^2) = -x + xf(-x)\n\nwhich upon subtracting gives $f(x) \\equiv x$."}
+{"year": 2015, "problem_number": 6, "problem": "The sequence $a_1,a_2,\\dots$ of integers satisfies the conditions:\n[(i)]\n$1\\le a_j\\le2015$ for all $j\\ge1$,\n$k+a_k\\neq \\ell+a_\\ell$ for all $1\\le k<\\ell$.", "solution": "We give two equivalent solutions with different presentations,\none with ``arrows'' and the other by ``juggling''.\n\nFirst solution (arrows).: \nConsider the map\n$ f \\colon k \\mapsto k + a_k. $\nThis map is injective, so if we draw all arrows of the form $k \\mapsto f(k)$\nwe get a partition of $\\NN$ into one or more ascending chains\n(which skip by at most $2015$).\n\nThere are at most $2015$ such chains,\nsince among any $2015$ consecutive points in $\\NN$\nevery chain must have an element.\n\nWe claim we may take $b$ to be the number of such chains,\nand $N$ to be the largest of the start-points of all the chains.\n\nConsider an interval $I = [m+1, n]$.\nWe have that\n$ \\sum_{m<j\\le n} a_j = \\sum_{\\text{chain } c} [ \\min \\left\\{ x > n, x \\in c \\right\\}\n- \\min \\left\\{ x > m, x \\in c \\right\\} ]. $\nThus the upper bound is proved by the calculation\n\n\\sum_{m<j\\le n} (a_j-b)\n&= \\sum_{\\text{chain } c} [ (\\min \\left\\{ x > n, x \\in c \\right\\} - n)\n- (\\min \\left\\{ x > m, x \\in c \\right\\} - m) ] \\\\\n&= \\sum_{\\text{chain } c} [ (\\min \\left\\{ x > n, x \\in c \\right\\} - n) ]\n- \\sum_{\\text{chain } c} [\n\\min \\left\\{ x > m, x \\in c \\right\\} - m ] \\\\\n&\\le (1+2015+2014+\\dots+(2015-(b-2)))-(1+2+\\dots+b) \\\\\n&= (b-1)(2015-b)\n\nfrom above (noting that $n+1$ has to belong to some chain).\nThe lower bound is similar.\n\nSecond solution (juggling).: \nThis solution is essentially the same, but phrased as a juggling problem.\nHere is a solution in this interpretation:\nwe will consider several balls thrown in the air,\nwhich may be at heights $0$, $1$, $2$, \\dots, $2014$.\nThe process is as follows:\n\nInitially, at time $t = 0$, there are no balls in the air.\nThen at each integer time $t$ thereafter,\nif there is a ball at height $0$, it is caught;\notherwise a ball is added to the juggler's hand.\nThis ball (either caught or added) is then thrown to a height of $a_t$.\nImmediately afterwards, all balls have their height decreased by one.\n\nThe condition $a_k + k \\neq \\ell + a_\\ell$ thus ensures that\nno two balls are ever at the same height.\nIn particular, there will never be more than $2016$ balls,\nsince there are only $2015$ possible heights.\n\nWe claim we may set.\n\nb &= \\text{number of balls in entire process} \\\\\nN &= \\text{last moment in time at which a ball is added}.\n\nIndeed, the key fact is that if we let $S_t$ denote\nthe sum of the height of all the balls just after time $t+1/2$, then\n$ S_{t+1} - S_t = a_{t+1} - b $\nAfter all, at each time step $t$, the caught ball is thrown to height $a_t$,\nand then all balls have their height decreased by $1$,\nfrom which the conclusion follows.\nHence the quantity in the problem is exactly equal to\n$ \\left|\\sum_{j=m+1}^n (a_j-b) \\right|\n= \\left| S_m - S_n \\right|. $\nFor a fixed $b$, we easily have the inequalities\n$0 + 1 + \\dots + (b-1) \\le S_t \\le 2014 + 2013 + \\dots + (2015-b)$.\nHence $|S_m - S_n| \\le (b-1)(2015-b) \\le 1007^2$ as desired."}
+{"year": 2016, "problem_number": 1, "problem": "In convex pentagon $ABCDE$ with $\\angle B > 90\\dg$,\nlet $F$ be a point on $AC$ such that $\\angle FBC = 90\\dg$.\nIt is given that $FA=FB$, $DA=DC$, $EA=ED$,\nand rays $AC$ and $AD$ trisect $\\angle BAE$.\nLet $M$ be the midpoint of $CF$.\nLet $X$ be the point such that $AMXE$ is a parallelogram.\nShow that $FX$, $EM$, $BD$ are concurrent.", "solution": "Here is a ``long'' solution which I think\nshows where the ``power'' in the configuration comes from\n(it should be possible to come up with shorter solutions\nby cutting more directly to the desired conclusion).\nThroughout the proof, we let\n$ \\theta = \\angle FAB = \\angle FBA = \\angle DAC = \\angle DCA\n= \\angle EAD = \\angle EDA. $\n\nWe begin by focusing just on $ABCD$ with point $F$,\nignoring for now the points $E$ and $X$\n(and to some extent even point $M$).\nIt turns out this is a very familiar configuration.\n\n[Central lemma]\nThe points $F$ and $C$\nare the incenter and $A$-excenter of $\\triangle DAB$.\nMoreover, $\\triangle DAB$ is isosceles with $DA = DB$.\n\nThe proof uses three observations:\n\nWe already know that $FAC$ is the angle bisector of $\\angle ABD$.\nWe were given $\\angle FBC = 90\\dg$.\nNext, note that $\\triangle AFB \\sim \\triangle ADC$\n(they are similar isosceles triangles).\nFrom this it follows that $AF \\cdot AC = AB \\cdot AD$.\n\nThese three facts, together with $F$ lying inside $\\triangle ABD$,\nare enough to imply the result.\n\n[Figure omitted]\n\nThe point $M$ is the midpoint of arc $\\widehat{BD}$ of $(DAB)$,\nand the center of cyclic quadrilateral $FDCB$.\n\nFact 5.\n\nUsing these observations as the anchor\nfor everything that follows,\nwe now prove several claims about $X$ and $E$ in succession.\n[Figure omitted]\n\nPoint $E$ is the midpoint of arc $\\widehat{AD}$\nin $(ABMD)$, and hence lies on ray $BF$.\n\nThis follows from\n$\\angle EDA = \\theta = \\angle EBA$.\n\nPoints $X$ is the second intersection of ray $ED$\nwith $(BFDC)$.\n\nFirst, $ED \\parallel AC$ already since\n$\\angle AED = 180\\dg - 2\\theta$\nand $\\angle CAE = 2\\theta$.\n\nNow since $DB = DA$, we get $MB = MD = ED = EA$.\nThus, $MX = AE = MB$,\nso $X$ also lies on the circle $(BFDC)$ centered at $M$.\n\nThe quadrilateral $EXMF$ is an isosceles trapezoid.\n\nWe already know $EX \\parallel FM$.\nSince $\\angle EFA = 180\\dg - \\angle AFB = 2\\theta = \\angle FAE$,\nwe have $EF = EA$ as well (and $F \\neq A$).\nAs $EXMA$ was a parallelogram,\nit follows $EXMF$ is an isosceles trapezoid.\n\nThe problem then follows by radical axis theorem\non the three circles $(AEDMB)$, $(BFDXC)$ and $(EXMF)$."}
+{"year": 2016, "problem_number": 2, "problem": "Find all integers $n$ for which each cell of $n \\times n$ table\ncan be filled with one of the letters $I$, $M$ and $O$ in such a way that:\n\n\\item In each row and column, one third of the entries are $I$,\none third are $M$ and one third are $O$; and\n\\item in any diagonal, if the number of entries on the diagonal is a multiple of three,\nthen one third of the entries are $I$, one third are $M$ and one third are $O$.\n\nNote that an $n \\times n$ table has $4n-2$ diagonals.", "solution": "The answer is $n$ divisible by $9$.\n\nFirst we construct $n=9$ and by extension every multiple of $9$.\n$\n{|ccc|ccc|ccc|} \\hline\nI & I & I & M & M & M & O & O & O \\\\\nM & M & M & O & O & O & I & I & I \\\\\nO & O & O & I & I & I & M & M & M \\\\\\hline\nI & I & I & M & M & M & O & O & O \\\\\nM & M & M & O & O & O & I & I & I \\\\\nO & O & O & I & I & I & M & M & M \\\\\\hline\nI & I & I & M & M & M & O & O & O \\\\\nM & M & M & O & O & O & I & I & I \\\\\nO & O & O & I & I & I & M & M & M \\\\\\hline\n\n$\n\nWe now prove $9 \\mid n$ is necessary.\n\nLet $n = 3k$, which divides the given grid into $k^2$ sub-boxes\n(of size $3 \\times 3$ each).\nWe say a multiset of squares $S$ is clean if\nthe letters distribute equally among them;\nnote that unions of clean multisets are clean.\n\nConsider the following clean sets (given to us by problem statement):\n\nAll columns indexed $2 \\pmod 3$,\nAll rows indexed $2 \\pmod 3$, and\nAll $4k-2$ diagonals mentioned in the problem.\n\nTake their union.\nThis covers the center of each box four times,\nand every other cell exactly once.\nWe conclude the set of $k^2$ center squares\nare clean, hence $3 \\mid k^2$ and so $9 \\mid n$,\nas desired.\n\nShown below is the sums over all diagonals only,\nand of the entire union.\n$\n{|ccc|ccc|ccc|} \\hline\n1 & & 1 & 1 & & 1 & 1 & & 1 \\\\\n& 2 & & & 2 & & & 2 & \\\\\n1 & & 1 & 1 & & 1 & 1 & & 1 \\\\\\hline\n1 & & 1 & 1 & & 1 & 1 & & 1 \\\\\n& 2 & & & 2 & & & 2 & \\\\\n1 & & 1 & 1 & & 1 & 1 & & 1 \\\\\\hline\n1 & & 1 & 1 & & 1 & 1 & & 1 \\\\\n& 2 & & & 2 & & & 2 & \\\\\n1 & & 1 & 1 & & 1 & 1 & & 1 \\\\\\hline\n\n\\qquad\n{|ccc|ccc|ccc|} \\hline\n1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\\n1 & 4 & 1 & 1 & 4 & 1 & 1 & 4 & 1 \\\\\n1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\\\hline\n1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\\n1 & 4 & 1 & 1 & 4 & 1 & 1 & 4 & 1 \\\\\n1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\\\hline\n1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\\n1 & 4 & 1 & 1 & 4 & 1 & 1 & 4 & 1 \\\\\n1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\\\hline\n\n$"}
+{"year": 2016, "problem_number": 3, "problem": "Let $P=A_1A_2\\dots A_k$ be a convex polygon in the plane.\nThe vertices $A_1$, $A_2$, \\dots, $A_k$ have integral coordinates\nand lie on a circle. Let $S$ be the area of $P$.\nAn odd positive integer $n$ is given such that\nthe squares of the side lengths of $P$ are integers divisible by $n$.\nProve that $2S$ is an integer divisible by $n$.", "solution": "Solution by Jeck Lim:\nWe will prove the result just for $n = p^e$\nwhere $p$ is an odd prime and $e \\ge 1$.\nThe case $k=3$ is resolved by Heron's formula directly:\nwe have $S = \\frac14\\sqrt{2(a^2b^2 + b^2c^2 + c^2a^2) - a^4-b^4-c^4}$,\nso if $p^e \\mid \\gcd(a^2,b^2,c^2)$ then $p^{2e} \\mid S^2$.\n\nNow we show we can pick a diagonal and induct down on $k$ by using inversion.\n\nLet the polygon be $A_1 A_2 \\dots A_{k+1}$\nand suppose for contradiction that all sides are divisible by $p^e$\nbut no diagonals are.\nLet $O = A_{k+1}$ for notational convenience.\nBy applying inversion around $O$ with radius $1$,\nwe get the ``generalized Ptolemy theorem''\n$\n\\frac{A_1A_2}{OA_1 \\cdot OA_2}\n+ \\frac{A_2A_3}{OA_2 \\cdot OA_3}\n+ \\dots\n+ \\frac{A_{k-1} A_k}{OA_{k-1} \\cdot OA_k}\n= \\frac{A_1 A_k}{OA_1 \\cdot OA_k}\n$\nor, making use of square roots,\n$\n\\sqrt{\\frac{A_1A_2^2}{OA_1^2 \\cdot OA_2^2}}\n+ \\sqrt{\\frac{A_2A_3^2}{OA_2^2 \\cdot OA_3^2}}\n+ \\dots\n+ \\sqrt{\\frac{A_{k-1} A_k^2}{OA_{k-1}^2 \\cdot OA_k^2}}\n= \\sqrt{\\frac{A_1 A_k^2}{OA_1^2 \\cdot OA_k^2}}\n$\nSuppose $\\nu_p$ of all diagonals is strictly less than $e$.\nThen the relation becomes\n$ \\sqrt{q_1} + \\dots + \\sqrt{q_{k-1}} = \\sqrt q $\nwhere $q_i$ are positive rational numbers.\nSince there are no nontrivial relations between square roots\n(see \\href{https://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/}{this link})\nthere is a positive rational number $b$\nsuch that $r_i = \\sqrt{q_i/b}$ and $r = \\sqrt{q/b}$\nare all rational numbers.\nThen\n$ \\sum r_i = r. $\nHowever, the condition implies that $\\nu_p(q_i^2) > \\nu_p(q^2)$ for all $i$\n(check this for $i=1$, $i=k-1$ and $2 \\le i \\le k-2$),\nand hence $\\nu_p(r_i) > \\nu_p(r)$.\nThis is absurd.\n\nI think you basically have to use some Ptolemy-like geometric property,\nand also all correct solutions I know of for $n = p^e$\ndepend on finding a diagonal and inducting down.\n(Actually, the case $k=4$ is pretty motivating;\nPtolemy implies one can cut in two.)"}
+{"year": 2016, "problem_number": 4, "problem": "A set of positive integers is called fragrant\nif it contains at least two elements and each of its elements\nhas a prime factor in common with at least one of the other elements.\nLet $P(n)=n^2+n+1$.\nWhat is the smallest possible positive integer value of $b$ such that\nthere exists a non-negative integer $a$ for which the set\n$ \\{P(a+1),P(a+2),\\dots,P(a+b)\\} $\nis fragrant?", "solution": "The answer is $b = 6$.\n\nFirst, we prove $b \\ge 6$ must hold.\nIt is not hard to prove the following divisibilities by Euclid:\n\n\\gcd(P(n), P(n+1)) &\\mid 1 \\\\\n\\gcd(P(n), P(n+2)) &\\mid 7 \\\\\n\\gcd(P(n), P(n+3)) &\\mid 3 \\\\\n\\gcd(P(n), P(n+4)) &\\mid 19.\n\nNow assume for contradiction $b \\le 5$.\nThen any GCD's among $P(a+1)$, \\dots, $P(a+b)$ must be among $\\{3, 7, 19\\}$.\nConsider a multi-graph on $\\{a+1, \\dots, a+b\\}$ where we join two elements with nontrivial GCD\nand label the edge with the corresponding prime.\nThen we readily see there is at most one edge each of $\\{3, 7, 19\\}$:\nid est at most one edge of gap $2$, $3$, $4$ (and no edges of gap $1$).\n(By the gap of an edge $e = \\{u,v\\}$ we mean $|u - v|$.)\nBut one can see that it's now impossible for every vertex to have nonzero degree, contradiction.\n\nTo construct $b = 6$ we use the Chinese remainder theorem: select $a$ with\n\na+1 & \\equiv 7 \\pmod{19} \\\\\na+5 & \\equiv 11 \\pmod{19} \\\\\na+2 & \\equiv 2 \\pmod{7} \\\\\na+4 & \\equiv 4 \\pmod{7} \\\\\na+3 & \\equiv 1 \\pmod{3} \\\\\na+6 & \\equiv 1 \\pmod{3}\n\nwhich does the trick."}
+{"year": 2016, "problem_number": 5, "problem": "The equation\n$ (x-1)(x-2)\\dots(x-2016)=(x-1)(x-2)\\dots (x-2016) $\nis written on the board, with $2016$ linear factors on each side.\nWhat is the least possible value of $k$ for which it is possible to\nerase exactly $k$ of these $4032$ linear factors so that at least\none factor remains on each side and the resulting equation\nhas no real solutions?", "solution": "The answer is $2016$.\nObviously this is necessary in order to delete duplicated factors.\nWe now prove it suffices to deleted $2 \\pmod 4$ and $3 \\pmod 4$\nguys from the left-hand side, and $0 \\pmod 4$,\n$1 \\pmod 4$ from the right-hand side.\n\nConsider the $1008$ inequalities\n\n(x-1)(x-4) &< (x-2)(x-3) \\\\\n(x-5)(x-8) &< (x-6)(x-7) \\\\\n(x-9)(x-12) &< (x-10)(x-11) \\\\\n&\\vdots \\\\\n(x-2013)(x-2016) &< (x-2014)(x-2015).\n\nNotice that in all these inequalities, at most one of them\nhas non-positive numbers in it, and we never have both zero.\nIf there is exactly one negative term among the $1008 \\cdot 2 = 2016$ sides,\nit is on the left and we can multiply all together.\nThus the only case that remains is if $x \\in (4m-2, 4m-1)$ for some $m$,\nsay the $m$th inequality.\nIn that case, the two sides of that inequality differ by a factor of at least $9$.\n\nWe have $ \\prod_{k \\ge 0} \\frac{(4k+2)(4k+3)}{(4k+1)(4k+4)} < e. $\n\n[Proof of claim using logarithms]\nIt's equivalent to prove\n$ \\sum_{k \\ge 0} \\log ( 1 + \\frac{2}{(4k+1)(4k+4)} ) < 1. $\nTo this end, we use the deep fact that $\\log(1+t) \\le t$,\nand thus it follows from\n$\\sum_{k \\ge 0} \\frac{1}{(4k+1)(4k+4)} < 1/2$,\nwhich one can obtain for example by noticing\nit's less than $\\frac14\\frac{\\pi^2}{6}$.\n\n[Elementary proof of claim, given by Espen Slettnes]\nTo avoid calculus as above, for each $N \\ge 0$, note the partial product is bounded by\n\n\\prod_{k=0}^N \\frac{(4k+2)(4k+3)}{(4k+1)(4k+4)}\n&=\n\\frac 21 \\cdot ( \\frac 34 \\cdot \\frac 65 )\n\\cdot ( \\frac 78 \\cdot \\frac{10}{9} ) \\cdot \\dots\n\\cdot \\frac{4N+3}{4N+4} \\\\\n&< 2 \\cdot 1 \\cdot 1 \\cdot \\dots \\cdot \\frac{4N+3}{4N+4}\n< 2 < e. \\qedhere\n\nThis solves the problem, because then the factors being multiplied\non by the positive inequalities before the\n$m$th one are both less than $e$, and $e^2 < 9$.\nIn symbols, for $4m-2 < x < 4m-1$ we should have\n$ \\frac{(x-(4m-6))(x-(4m-5))}{(x-(4m-7))(x-(4m-4))}\n\\times \\dots \\times \\frac{(x-2)(x-3)}{(x-1)(x-4)} < e $\nand\n$ \\frac{(x-(4m+2))(x-(4m+3))}{(x-(4m+1))(x-(4m+4))}\n\\times \\dots \\times \\frac{(x-2014)(x-2015)}{(x-2013)(x-2016)} < e $\nbecause the $(k+1)$st term of each left-hand side\nis at most $\\frac{(4k+2)(4k+3)}{(4k+1)(4k+4)}$, for $k \\ge 0$.\nAs $e^2 < 9$, we're okay."}
+{"year": 2016, "problem_number": 6, "problem": "There are $n\\ge 2$ line segments in the plane such that\nevery two segments cross and no three segments meet at a point.\nGeoff has to choose an endpoint of each segment and place a frog\non it facing the other endpoint. Then he will clap his hands $n-1$ times.\nEvery time he claps, each frog will immediately jump forward\nto the next intersection point on its segment.\nFrogs never change the direction of their jumps.\nGeoff wishes to place the frogs in such a way that no two of them\nwill ever occupy the same intersection point at the same time.\n\n[(a)]\nProve that Geoff can always fulfill his wish if $n$ is odd.\nProve that Geoff can never fulfill his wish if $n$ is even.", "solution": "The following solution was communicated to me by Yang Liu.\n\nImagine taking a larger circle $\\omega$ encasing\nall $\\binom{n}{2}$ intersection points.\nDenote by $P_1$, $P_2$, \\dots, $P_{2n}$ the order of the points on $\\omega$\nin clockwise order; we imagine placing the frogs on $P_i$ instead.\nObserve that, in order for every pair of segments to meet,\neach line segment must be of the form $P_i P_{i+n}$.\n[Figure omitted]\nThen:\n[(a)]\nPlace the frogs on $P_1$, $P_3$, \\dots, $P_{2n-1}$.\nA simple parity arguments shows this works.\n\nObserve that we cannot place frogs on consecutive $P_i$,\nso the $n$ frogs must be placed on alternating points.\nBut since we also are supposed to not place frogs on\ndiametrically opposite points,\nfor even $n$ we immediately get a contradiction.\n\nYang says: this is easy to guess if you just do a\nfew small cases and notice that the pairs of\n``violating points'' just forms a large cycle around."}
+{"year": 2017, "problem_number": 1, "problem": "For each integer $a_0 > 1$, define the sequence $a_0$, $a_1$, $a_2$,\n\\dots, by\n$\na_{n+1} =\n\n\\sqrt{a_n} & \\text{if $\\sqrt{a_n}$ is an integer,} \\\\\na_n + 3 & \\text{otherwise}\n\n$\nfor each $n \\ge 0$.\nDetermine all values of $a_0$ for which there is a number $A$\nsuch that $a_n = A$ for infinitely many values of $n$.", "solution": "The answer is $a_0 \\equiv 0 \\pmod 3$ only.\n\nFirst solution.: \nWe first compute the minimal term of any sequence, periodic or not.\n\nLet $c$ be the smallest term in $a_n$.\nThen either $c \\equiv 2 \\pmod 3$ or $c = 3$.\n\nClearly $c \\neq 1, 4$.\nAssume $c \\not\\equiv 2 \\pmod 3$.\nAs $c$ is not itself a square,\nthe next perfect square after $c$ in the sequence\nis one of $( \\left\\lfloor \\sqrt c \\right\\rfloor + 1)^2$,\n$( \\left\\lfloor \\sqrt c \\right\\rfloor + 2)^2$,\nor $( \\left\\lfloor \\sqrt c \\right\\rfloor + 3)^2$.\nSo by minimality we require\n$ c \\le \\left\\lfloor \\sqrt c \\right\\rfloor + 3 \\le \\sqrt c + 3 $\nwhich requires $c \\le 5$.\nSince $c \\neq 1,2,4,5$ we conclude $c = 3$.\n\nNow we split the problem into two cases:\n\nIf $a_0 \\equiv 0 \\pmod 3$, then all terms of the sequence are $0 \\pmod 3$.\nThe smallest term of the sequence is thus $3$ by the lemma\nand we have $ 3 \\to 6 \\to 9 \\to 3 $\nso $A = 3$ works fine.\n\nIf $a_0 \\not\\equiv 0 \\pmod 3$,\nthen no term of the sequence is $0 \\pmod 3$,\nand so in particular $3$ does not appear in the sequence.\nSo the smallest term of the sequence is $2 \\pmod 3$ by lemma.\nBut since no squares are $2 \\pmod 3$,\nthe sequence $a_k$ grows without bound forever after,\nso no such $A$ can exist.\n\nHence the answer is $a_0 \\equiv 0 \\pmod 3$ only.\n\nSecond solution.: \nWe clean up the argument by proving the following lemma.\n\nIf $a_n$ is constant modulo $3$ and not $2 \\pmod 3$,\nthen $a_n$ must eventually cycle in the form\n$(m, m+3, m+6, \\dots, m^2)$,\nwith no squares inside the cycle except $m^2$.\n\nObserve that $a_n$ must eventually hit a square, say $a_k = c^2$;\nthe next term is $a_{k+1} = c$.\nThen it is forever impossible to exceed $c^2$ again,\nby what is essentially discrete intermediate value theorem.\nIndeed, suppose $a_\\ell > c^2$ and take $\\ell > k$ minimal\n(in particular $a_{\\ell} \\neq \\sqrt{a_{\\ell-1}}$).\n\nThus $a_{\\ell-1} \\in \\{c^2-2, c^2-1, c^2\\}$\nand thus for modulo $3$ reasons we have $a_{\\ell-1} = c^2$.\nBut that should imply $a_\\ell = c < c^2$, contradiction.\n\nWe therefore conclude $\\sup \\{a_n, a_{n+1}, \\dots \\}$ is a\ndecreasing integer sequence in $n$.\nIt must eventually stabilize, say at $m^2$.\nNow we can't hit a square between $m$ and $m^2$,\nand so we are done.\n\nNow, we contend that all $a_0 \\equiv 0 \\pmod 3$ work.\nIndeed, for such $a_0$ we have $a_n \\equiv 0 \\pmod 3$ for all $n$,\nso the lemma implies that the problem statement is valid.\n\nNext, we observe that if $a_i \\equiv 2 \\pmod 3$,\nthen the sequence grows without bound afterwards\nsince no squares are $2 \\pmod 3$.\nIn particular, if $a_0 \\equiv 2 \\pmod 3$ the answer is no.\n\nFinally, we claim that if $a_0 \\equiv 1 \\pmod 3$,\nthen eventually some term is $2 \\pmod 3$.\nAssume for contradiction this is not so;\nthen $a_n \\equiv 1 \\pmod 3$ must hold forever,\nand the lemma applies to give us a cycle of the form\n$(m, m+3, \\dots, m^2)$ where $m \\equiv 1 \\pmod 3$.\nIn particular $m \\ge 4$ and\n$ m \\le (m-2)^2 < m^2 $\nbut $(m-2)^2 \\equiv 1 \\pmod 3$ which is a contradiction."}
+{"year": 2017, "problem_number": 2, "problem": "Solve over $\\RR$ the functional equation\n$ f( f(x)f(y) ) + f(x+y) = f(xy). $", "solution": "The only solutions are $f(x) = 0$, $f(x) = x-1$ and $f(x)=1-x$,\nwhich clearly work.\n\nNote that\n\nIf $f$ is a solution, so is $-f$.\nMoreover, if $f(0)=0$ then setting $y=0$ gives $f\\equiv0$.\nSo henceforth we assume $f(0)>0$.\n\nWe have $f(z) = 0 \\iff z =1$.\nAlso, $f(0)=1$ and $f(1)=0$.\n\nFor the forwards direction, if $f(z)=0$ and $z \\neq 1$\none may put $(x,y) = ( z, z(z-1)\\inv )$\n(so that $x+y=xy$) we deduce $f(0) = 0$ which is a contradiction.\n\nFor the reverse, $f(f(0)^2)=0$ by setting $x=y=0$, and use the previous part.\nWe also conclude $f(1) = 0$, $f(0) = 1$.\n\nIf $f$ is injective, we are done.\n\nSetting $y=0$ in the original equation gives $f(f(x)) = 1-f(x)$.\nWe apply this three times on the expression $f^3(x)$:\n$ f(1-f(x)) = f(f(f(x))) = 1 - f(f(x)) = f(x). $\nHence $1-f(x) = x$ or $f(x) = 1-x$.\n\nThe result $f(f(x)) + f(x) = 1$ also implies that surjectivity\nwould solve the problem.\n\n$f$ is injective.\n\nSetting $y=1$ in the original equation gives $f(x+1) = f(x)-1$, and by induction\n\nf(x+n) = f(x)-n.\n\\label{eq:linshift}\n\nAssume now $f(a) = f(b)$.\nBy using \\eqref{eq:linshift} we may shift $a$ and $b$ to be large enough that\nwe may find $x$ and $y$ obeying $x+y=a+1$, $xy=b$.\nSetting these gives\n\nf(f(x)f(y)) &= f(xy) - f(x+y) = f(b) - f(a+1) \\\\\n&= f(b) + 1 - f(a) = 1\n\nfrom which we conclude\n$ f( f(x)f(y) + 1 ) = 0. $\nHence by the first claim\nwe have $f(x)f(y) + 1 = 1$, so $f(x)f(y) = 0$.\nApplying the first claim again gives $1 \\in \\{x,y\\}$.\nBut that implies $a=b$.\n\nJessica Wan points out that for any $a \\neq b$,\nat least one of $a^2 > 4(b-1)$ and $b^2 > 4(a-1)$ is true.\nSo shifting via \\eqref{eq:linshift} is actually unnecessary for this proof.\n\nOne can solve the problem over $\\QQ$ using only \\eqref{eq:linshift} and the easy parts.\nIndeed, that already implies $f(n) = 1-n$ for all $n$.\nNow we induct to show $f(p/q) = 1-p/q$ for all $0 < p < q$ (on $q$).\nBy choosing $x = 1+p/q$, $y = 1+q/p$, we cause $xy = x+y$,\nand hence $0 = f( f(1+p/q)f(1+q/p) )$\nor $1 = f(1+p/q)f(1+q/p)$.\n\nBy induction we compute $f(1+q/p)$ and this gives $f(p/q+1) = f(p/q)-1$."}
+{"year": 2017, "problem_number": 3, "problem": "A hunter and an invisible rabbit play a game in the plane.\nThe rabbit and hunter start at points $A_0 = B_0$.\nIn the $n$th round of the game ($n \\ge 1$), three things occur in order:\n[(i)]\nThe rabbit moves invisibly from $A_{n-1}$ to a point $A_n$\nsuch that $A_{n-1} A_n = 1$.\nThe hunter has a tracking device (e.g.\\ dog)\nwhich reports an approximate location $P_n$ of the rabbit,\nsuch that $P_n A_n \\le 1$.\nThe hunter moves visibly from $B_{n-1}$ to a point $B_n$\nsuch that $B_{n-1} B_n = 1$.", "solution": "No, the hunter cannot.\nWe will show how to increase the distance in the following way:\n\nSuppose the rabbit is at a distance $d \\ge 1$ from the hunter\nat some point in time.\nThen it can increase its distance to at least\n$\\sqrt{d^2+1/2}$ in $4d$ steps\nregardless of what the hunter already knows about the rabbit.\n\nConsider a positive integer $n > d$, to be chosen later.\nLet the hunter start at $B$ and the rabbit at $A$, as shown.\nLet $\\ell$ denote line $AB$.\n\nNow, we may assume the rabbit reveals its location $A$,\nso that all previous information becomes irrelevant.\n\nThe rabbit chooses two points $X$ and $Y$ symmetric about $\\ell$\nsuch that $XY = 2$ and $AX = AY = n$, as shown.\nThe rabbit can then hop to either $X$ or $Y$,\npinging the point $P_n$ on the $\\ell$ each time.\nThis takes $n$ hops.\n[Figure omitted]\n\nNow among all points $H$ the hunter can go to,\n$\\min \\max \\{HX, HY\\}$ is clearly minimized with $H \\in \\ell$ by symmetry.\nSo the hunter moves to a point $H$ such that $BH = n$ as well.\nIn that case the new distance is $HX = HY$.\n\nWe now compute\n\nHX^2 &= 1 + HM^2 = 1 + ( \\sqrt{AX^2-1}-AH )^2 \\\\\n&= 1 + ( \\sqrt{n^2-1}-(n-d) )^2 \\\\\n&\\ge 1 + ( ( n-\\frac1n ) - (n-d) )^2 \\\\\n&= 1 + (d-1/n)^2\n\nwhich exceeds $d^2 + 1/2$ whenever $n \\ge 4d$.\n\nIn particular we can always take $n = 400$ even very crudely;\napplying the lemma $2 \\cdot 100^2$ times,\nthis gives a bound of $400 \\cdot 2 \\cdot 100^2 < 10^9$, as desired.\n\nThe step of revealing the location of the rabbit seems\ncritical because as far as I am aware it is basically\nimpossible to keep track of ping locations in the problem.\n\nReasons to believe the answer is ``no'':\nthe $10^9$ constant,\nand also that ``follow the last ping'' is losing for the hunter.\n\nI think there are roughly two ways you can approach the problem\nonce you recognize the answer.\n\n[(i)]\nTry and control the location of the pings\nAbandon the notion of controlling possible locations,\nand try to increase the distance by a little bit,\nsay from $d$ to $\\sqrt{d^2+\\varepsilon}$.\nThis involves revealing the location of the rabbit\nbefore each iteration of several jumps.\n\nI think it's clear that the difficulty of\nmy approach is realizing that (ii) is possible;\nonce you do, the two-point approach is more or less the only one possible.\n\nMy opinion is that (ii) is not that magical;\nas I said it was the first idea I had.\nBut I am biased, because when I test-solved the problem\nat the IMO it was called ``C5'' and not ``IMO3'';\nthis effectively told me it was unlikely that the official solution\nwas along the lines of (i),\nbecause otherwise it would have been placed much later in the shortlist."}
+{"year": 2017, "problem_number": 4, "problem": "Let $R$ and $S$ be different points on a circle $\\Omega$\nsuch that $RS$ is not a diameter.\nLet $\\ell$ be the tangent line to $\\Omega$ at $R$.\nPoint $T$ is such that $S$ is the midpoint of $RT$.\nPoint $J$ is chosen on minor arc $RS$ of $\\Omega$ so that\nthe circumcircle $\\Gamma$ of triangle $JST$ intersects $\\ell$\nat two distinct points.\nLet $A$ be the common point of $\\Gamma$ and $\\ell$ closer to $R$.\nLine $AJ$ meets $\\Omega$ again at $K$.\nProve that line $KT$ is tangent to $\\Gamma$.", "solution": "First solution (elementary).: \nFirst, note\n$ \u2220 RKA = \u2220 RKJ = \u2220 RSJ = \u2220 TSJ = \u2220 TAJ = \u2220 TAK $\nso $RK \\parallel AT$.\nNow,\n\n$RA$ is tangent at $R$ iff $\\triangle KRS \\sim \\triangle RTA$ (oppositely),\nbecause both equate to $-\u2220 RKS = \u2220 SKR = \u2220 SRA = \u2220 TRA$.\nSimilarly, $TK$ is tangent at $T$\niff $\\triangle KTS \\sim \\triangle ART$.\nThe two similarities are equivalent because $RS = ST$\nthe SAS gives $KR \\cdot TA = RS \\cdot RT = TS \\cdot TR$.\n\n[Figure omitted]\n\nThe problem is actually symmetric with respect to two circles;\n$RA$ is tangent at $R$ if and only if $TK$ at $T$.\n\nSecond solution (inversion).: \nConsider an inversion at $R$ fixing the circumcircle $\\Gamma$ of $TSJA$.\nThen:\n\n$T$ and $S$ swap,\n$A$ and $B$ swap, where $B$ is the second intersection\nof $\\ell$ with $\\Gamma$.\nCircle $\\Omega$ inverts to the line through $T$\nparallel to $RAB$, call it $\\ell$.\n$J^\\ast$ is the second intersection of $\\ell$ with $\\Gamma$.\n$K^\\ast$ is the intersection of $\\ell$ with the circumcircle\nof $RBJ^\\ast$; this implies $RK^\\ast J^\\ast B$ is an isosceles trapezoid.\nIn particular, one reads $RK^\\ast \\parallel AT$ from this,\nhence $RK^\\ast TA$ is a parallelogram.\n\nThus we wish to show the circumcircle of $RSK^\\ast$ is tangent to $\\Gamma$.\nBut that follows from the final parallelogram observed:\n$S$ is the center of the parallelogram since it is the midpoint of the diagonal.\n\nThis also implies $RKTB$ is cyclic, from $K^\\ast SA$ collinear.\nMoreover, quadrilateral $KK^\\ast TS$ is cyclic (by power of a point);\nthis leads to the second official solution to the problem."}
+{"year": 2017, "problem_number": 5, "problem": "Fix $N \\ge 1$. A collection of $N(N+1)$ soccer players of distinct\nheights stand in a row.\nSir Alex Song wishes to remove $N(N-1)$ players from this row\nto obtain a new row of $2N$ players in which the following $N$\nconditions hold: no one stands between the two tallest players,\nno one stands between the third and fourth tallest players, \\dots,\nno one stands between the two shortest players.\nProve that this is possible.", "solution": "Some opening remarks:\n\\textbf{location and height are symmetric to each other},\nif one thinks about this problem as permutation pattern avoidance.\nSo while officially there are multiple solutions,\nthey are basically isomorphic to one another,\nand I am not aware of any solution otherwise.\n\n[Figure omitted]\n\nTake a partition of $N$ groups in order by height:\n$G_1 = \\{1,\\dots,N+1\\}$, $G_2 = \\{N+2, \\dots, 2N+2\\}$, and so on.\nWe will pick two people from each group $G_k$.\n\nScan from the left until we find two people in the same group $G_k$.\nDelete all people scanned and also everyone in $G_k$.\nAll the groups still have at least $N$ people left,\nso we can induct down with the non-deleted people;\nthe chosen pair is to the far left anyways.\n\nThe important bit is to scan by position\nbut group by height,\nand moreover not change the groups as we scan.\nDually, one can have a solution which scans by height\nbut groups by position."}
+{"year": 2017, "problem_number": 6, "problem": "An irreducible lattice point is an ordered pair\nof integers $(x,y)$ satisfying $\\gcd(x,y) = 1$.\nProve that if $S$ is a finite set of irreducible lattice points\nthen there exists a nonconstant\nhomogeneous polynomial $f(x,y)$ with integer coefficients\nsuch that $f(x,y)=1$ for each $(x,y) \\in S$.", "solution": "We present two solutions.\n\nFirst solution (Dan Carmon, Israel).: \nWe prove the result by induction on $|S|$,\nwith the base case being Bezout's Lemma ($n=1$).\nFor the inductive step, suppose we want to add a given pair\n$(a_{m+1},b_{m+1})$ to $\\left\\{ (a_1, \\dots, a_m), (b_1, \\dots, b_m) \\right\\}$.\n\n[Standard]\nBy a suitable linear transformation we may assume\n$ (a_{m+1},b_{m+1}) = (1,0). $\n\n[Outline of proof]\nIt would be sufficient to show there exists a $2 \\times 2$ matrix\n$T = [ u & v \\\\ s & t ]$\nwith integer entries such that $\\det T = 1$ and\n$T \\cdot [ a_{m+1} \\\\ b_{m+1} ]\n= [ 1 \\\\ 0 ]$.\nThen we could apply $T$ to all the ordered pairs in $S$ (viewed as column vectors);\nif $f$ was a polynomial that works on the transformed ordered pairs,\nthen $f(ux+vy, sx+ty)$ works for the original ordered pairs.\n(Here, the condition that $\\det T = 1$ ensures that $T\\inv$ has integer entries,\nand hence that $T$ maps irreducible lattice points to irreducible lattice points.)\n\nTo generate such a matrix $T$, choose\n$T = [ u & v \\\\ -b_{m+1} & a_{m+1} ]$\nwhere $u$ and $v$ are chosen via B\\'{e}zout lemma so that $ua_{m+1} + vb_{m+1} = 1$.\nThis matrix $T$ is rigged so that $\\det T = 1$ and the leftmost column of $T\\inv$ is\n$[ a_{m+1} \\\\ b_{m+1} ]$.\n\nThis transformation claim is not necessary to proceed;\nthe solution below can be rewritten to avoid it with only cosmetic edits.\nHowever, it makes things a bit easier to read.\n\nLet $g(x,y)$ be a polynomial which works on the latter set.\nWe claim we can choose the new polynomial $f$ of the form\n$ f(x,y) = g(x,y)^{M} - C x^{\\deg g \\cdot M-m} \\prod_{i=1}^m (b_i x - a_i y). $\nwhere $C$ and $M$ are integer parameters we may adjust.\n\nSince $f(a_i, b_i) = 1$ by construction we just need\n$ 1 = f(1,0) = g(1,0)^M - C \\prod b_i. $\nIf $\\prod b_i = 0$ we are done,\nsince $b_i = 0 \\implies a_i = \\pm 1$ in that case\nand so $g(1, 0) = \\pm 1$, thus take $M = 2$.\nSo it suffices to prove:\n\nWe have $\\gcd( g(1,0), b_i ) = 1$ when $b_i \\neq 0$.\n\nFix $i$. If $b_i = 0$ then $a_i = \\pm 1$ and $g(\\pm 1,0) = \\pm 1$.\nOtherwise know\n$ 1 = g(a_i, b_i) \\equiv g(a_i, 0) \\pmod{b_i} $\nand since the polynomial is homogeneous with $\\gcd(a_i, b_i) = 1$\nit follows $g(1,0) \\not\\equiv 0 \\pmod{b_i}$ as well.\n\nThen take $M$ a large multiple of $\\varphi(\\prod |b_i|)$ and we're done.\n\nSecond solution (Lagrange).: \nThe main claim is that:\n\nFor every positive integer $N$,\nthere is a homogeneous polynomial $P(x,y)$ such that\n$P(x,y) \\equiv 1 \\pmod N$ whenever $\\gcd(x,y) = 1$.\n\n(This claim is actually implied by the problem.)\n\nFor $N = p^e$ a prime take $(x^{p-1} + y^{p-1})^{\\varphi(N)}$\nwhen $p$ is odd, and $(x^2+xy+y^2)^{\\varphi(N)}$ for $p=2$.\n\nNow, if $N$ is a product of primes,\nwe can collate coefficient by coefficient using the\nChinese remainder theorem.\n\nLet $S = \\left\\{ (a_i, b_i) \\mid i=1, \\dots, m \\right\\}$.\nWe have the natural homogeneous ``Lagrange polynomials''\n$ L_k(x,y) \\coloneq \\prod_{i \\neq k} (b_i x - a_i y) $\nNow let\n$ N \\coloneq \\prod_k L_k(x_k, y_k) $\nand take $P$ as in the claim.\nThen we can take a large power of $P$,\nand for each $i$ subtract an appropriate multiple of $L_i(x,y)$; that is, choose\n$\nf(x,y)\n= P(x,y)^{C} - \\sum_i L_i(x,y) \\cdot Q_i(x,y)\n$\nwhere $C$ is large enough that $C \\deg P > \\max_i \\deg L_i$,\nand $Q_i(x,y)$ is any homogeneous polynomial of degree $C \\deg P - \\deg L_i$\nsuch that $L_k(x_k, y_k) Q_k(x_k, y_k) = \\frac{P(x_k, y_k)^C - 1}{N} \\cdot L_k(x_k, y_k)$\n(which is an integer)."}
+{"year": 2018, "problem_number": 1, "problem": "Let $\\Gamma$ be the circumcircle of acute triangle $ABC$.\nPoints $D$ and $E$ lie on segments $AB$ and $AC$,\nrespectively, such that $AD = AE$.\nThe perpendicular bisectors of $BD$ and $CE$\nintersect the minor arcs $AB$ and $AC$ of $\\Gamma$\nat points $F$ and $G$, respectively.\nProve that the lines $DE$ and $FG$ are parallel.", "solution": "We present a synthetic solution from the IMO shortlist\nas well as a complex numbers approach.\nWe also outline a trig solution (the one I found at IMO),\nand a fourth solution from Derek Liu.\n\nSynthetic solution (from Shortlist).: \nConstruct parallelograms $AXFD$ and $AEGY$,\nnoting that $X$ and $Y$ lie on $\\Gamma$.\nAs $XF \\parallel AB$ we can let $M$\ndenote the midpoint of minor arcs $\\widehat{XF}$ and $\\widehat{AB}$\n(which coincide). Define $N$ similarly.\n\n[Figure omitted]\n\nObserve that $XF = AD = AE = YG$,\nso arcs $\\widehat{XF}$ and $\\widehat{YG}$ have equal measure;\nhence arcs $\\widehat{MF}$ and $\\widehat{NG}$ have equal measure;\ntherefore $MN \\parallel FG$.\n\nSince $MN$ and $DE$ are both perpendicular\nto the $\\angle A$ bisector, we're done.\n\nComplex numbers solution.: \nLet $b$, $c$, $f$, $g$, $a$ be as usual.\nNote that\n\nd-a &= ( 2 \\cdot \\frac{f+a+b-ab\\ol f}{2} -b )-a\n= f - \\frac{ab}{f} \\\\\ne-a &= g - \\frac{ac}{g}\n\nWe are given $AD = AE$ from which one deduces\n\n( \\frac{e-a}{d-a} )^2 &= \\frac cb\n\\implies \\frac{(g^2-ac)^2}{(f^2-ab)^2} = \\frac{g^2 c}{f^2 b} \\\\\n\\implies bc(bg^2-cf^2)a^2 &= g^2f^4c - f^2g^4b = f^2g^2(f^2c-g^2b) \\\\\n\\implies bc \\cdot a^2 &= (fg)^2 \\implies ( -\\frac{fg}{a} )^2 = bc.\n\nSince $\\frac{-fg}{a}$ is the point $X$ on the circle\nwith $AX \\perp FG$,\nwe conclude $FG$ is either parallel or perpendicular\nto the $\\angle A$-bisector; it must the latter\nsince the $\\angle A$-bisector separates the two minor arcs.\n\nTrig solution (outline).: \nLet $\\ell$ denote the $\\angle A$ bisector.\nFix $D$ and $F$.\nWe define the phantom point $G'$ such that $FG' \\perp \\ell$\nand $E'$ on side $AC$ such that $GE'=GC$.\n\n[Converse of the IMO problem]\nWe have $AD = AE'$, so that $E = E'$.\n\nSince $FG' \\perp \\ell$,\none can deduce $\\angle FBD = 1/2 C + x$\nand $\\angle GCA = 1/2 B + x$ for some $x$.\n(One fast way to see this is to note that $FG \\parallel MN$\nwhere $M$ and $N$ are in the first solution.)\nThen $\\angle FAB = 1/2 C -x$ and $\\angle GAC = 1/2 B - x$.\n\nLet $R$ be the circumradius.\nNow, by the law of sines,\n$ BF = 2R \\sin( 1/2 C - x ). $\nFrom there we get\n\nBD &= 2 \\cdot BF \\cos(1/2 C + x)\n= 4R \\cos(1/2 C+x) \\sin (1/2 C-x) \\\\\nDA &= AB - BD = 2R\\sin C\n- 4R\\cos(1/2 C+x) \\sin(1/2 C -x) \\\\\n&= 2R[ \\sin C - 2\\cos( 1/2 C + x ) \\sin ( 1/2 C -x ) ] \\\\\n&= 2R[ \\sin C - ( \\sin C - \\sin 2x ) ]\n= 2R \\sin 2x.\n\nA similar calculation gives $AE' = 2R \\sin 2x$ as needed.\n\nThus, $FG' \\parallel DE$, so $G = G'$ as well.\nThis concludes the proof.\n\nSynthetic solution from Derek Liu.: \nLet lines $FD$ and $GE$ intersect $\\Gamma$ again at $J$ and $K$, respectively.\n[Figure omitted]\nNotice that $\\triangle BFD\\sim\\triangle JAD$; as $FB=FD$, it follows that $AJ=AD$.\nLikewise, $\\triangle CGE\\sim\\triangle KAE$ and $GC=GE$, so $AK=AE$.\nHence,\n$ AK=AE=AD=AJ, $\nso $DEJK$ is cyclic with center $A$.\n\nIt follows that\n$ \u2220 KED=\u2220 KJD=\u2220 KJF=\u2220 KGF, $\nso we're done.\n\nNote that $K$ and $J$ must be distinct for this solution to work.\nSince $G$ and $K$ lie on opposite sides of $AC$, $K$ is on major arc $ABC$.\nAs $AK=AD=AE\\le \\min(AB,AC)$, $K$ lies on minor arc $AB$.\nSimilarly, $J$ lies on minor arc $AC$, so $K\\neq J.$"}
+{"year": 2018, "problem_number": 2, "problem": "Find all integers $n \\geq 3$ for which\nthere exist real numbers $a_1, a_2, \\dots, a_n$ satisfying\n$ a_i a_{i+1} +1 = a_{i+2} $\nfor $i=1,2, \\dots, n$, where indices are taken modulo $n$.", "solution": "The answer is $3 \\mid n$,\nachieved by $(-1,-1,2,-1,-1,2,\\dots)$.\nWe present two solutions.\n\nFirst solution by inequalities.: \nWe compute $a_i a_{i+1} a_{i+2}$ in two ways:\n\na_i a_{i+1} a_{i+2} &= [a_{i+2}-1]a_{i+2} = a_{i+2}^2 - a_{i+2} \\\\\n&= a_i [a_{i+3}-1] = a_i a_{i+3} - a_i.\n\nCyclically summing $a_{i+2}^2 - a_{i+2} = a_i a_{i+3} - a_i$ then gives\n$ \\sum_i a_{i+2}^2 = \\sum_i a_i a_{i+3}\n\\iff \\sum_{\\text{cyc}} ( a_i - a_{i+3} )^2 = 0. $\nThis means for inequality reasons the sequence is $3$-periodic.\nSince the sequence is clearly not $1$-periodic,\nas $x^2 + 1 = x$ has no real solutions.\nThus $3 \\mid n$.\n\nSecond solution by sign counting.: \nExtend $a_n$ to be a periodic sequence.\nThe idea is to look at the signs, and show the sequence\nof the signs must be $--+$ repeated.\nThis takes several steps:\n\nThe pattern $---$ is impossible. Obvious, since the third term should be $ > 1$.\nThe pattern $++$ is impossible. Then the sequence becomes strictly increasing,\nhence may not be periodic.\nZeros are impossible. If $a_1 = 0$, then $a_2 = 0$, $a_3 > 0$, $a_4 > 0$,\nwhich gives the impossible $++$.\nThe pattern $--+-+$ is impossible.\nCompute the terms:\n\na_1 &= -x < 0 \\\\\na_2 &= -y < 0 \\\\\na_3 &= 1 + xy > 1 \\\\\na_4 &= 1 - y(1+xy) < 0 \\\\\na_5 &= 1 + (1+xy)(1-y(1+xy)) < 1.\n\nBut now\n$ a_6 - a_5 = (1 + a_5 a_4) - (1 + a_3 a_4)\n= a_4 (a_5 - a_3) > 0 $\nsince $a_5 > 1 > a_3$.\nThis means we have the impossible $++$ pattern.\nThe infinite alternating pattern $-+-+-+-+\\dots$ is impossible.\nNote that\n$ a_1 a_2 + 1 = a_3 < 0 < a_4 = 1 + a_2 a_3 \\implies a_1 < a_3 $\nsince $a_2 > 0$;\nextending this we get $a_1 < a_3 < a_5 < \\dots$\nwhich contradicts the periodicity.\n\nWe finally collate the logic of sign patterns.\nSince the pattern is not alternating, there must be $--$ somewhere.\nAfterwards must be $+$, and then after that must be two minus signs\n(since one minus sign is impossible by impossibility of $--+-+$\nand $---$ is also forbidden);\nthus we get the periodic $--+$ as desired."}
+{"year": 2018, "problem_number": 3, "problem": "An anti-Pascal triangle is an equilateral triangular array\nof numbers such that, except for the numbers in the bottom row,\neach number is the absolute value of the difference\nof the two numbers immediately below it.\nFor example, the following array is an anti-Pascal triangle\nwith four rows which contains every integer from $1$ to $10$.\n[Figure omitted]\nDoes there exist an anti-Pascal triangle with $2018$ rows\nwhich contains every integer from $1$ to $1+2+\\dots +2018$?", "solution": "The answer is no, there is no anti-Pascal triangle\nwith the required properties.\n\nLet $n = 2018$ and $N = 1+2+\\dots+n$.\nFor every number $d$ not in the bottom row,\ndraw an arrow from $d$ to the larger of the two numbers below it\n(i.e.\\ if $d=a-b$, draw $d \\to a$).\nThis creates an oriented forest (which looks like lightning strikes).\n\nConsider the directed path starting from the top vertex $A$.\nStarting from the first number, it increments by at least $1+2+\\dots+n$,\nsince the increments at each step in the path are distinct;\ntherefore equality must hold\nand thus the path from the top ends at $N = 1+2+\\dots+n$\nwith all the numbers $\\{1,2,\\dots,n\\}$ being close by.\nLet $B$ be that position.\n\n[Figure omitted]\nConsider the two left/right neighbors $X$ and $Y$ of the endpoint $B$.\nAssume that $B$ is to the right of the midpoint of the bottom side,\nand complete the equilateral triangle as shown to an apex $C$.\nConsider the lightning strike from $C$ hitting the bottom at $D$.\nIt travels at least $\\left\\lfloor n/2-1 \\right\\rfloor$ steps, by construction.\nBut the increases must be at least $n+1$, $n+2$, \\dots since $1,2,\\dots,n$\nare close to the $A \\to B$ lightning path.\nThen the number at $D$ is at least\n$ (n+1) + (n+2) + \\dots +\n( n+( \\left\\lfloor n/2-1 \\right\\rfloor ) )\n> 1 + 2 + \\dots + n $\nfor $n \\ge 2018$, contradiction."}
+{"year": 2018, "problem_number": 4, "problem": "A site is any point $(x,y)$ in the plane\nfor which $x,y \\in \\{1, \\dots, 20\\}$.\nInitially, each of the $400$ sites is unoccupied.\nAmy and Ben take turns placing stones on unoccupied sites,\nwith Amy going first;\nAmy has the additional restriction that no two of her stones\nmay be at a distance equal to $\\sqrt5$.\nThey stop once either player cannot move.\nFind the greatest $K$ such that Amy can ensure that\nshe places at least $K$ stones.", "solution": "The answer is $K = 100$.\n\nFirst, we show Amy can always place at least $100$ stones.\nIndeed, treat the problem as a grid with checkerboard coloring.\nThen Amy can choose to always play on one of the $200$ black squares.\nIn this way, she can guarantee half the black squares,\ni.e.\\ she can get $1/2 \\cdot 200 = 100$ stones.\n\nSecond, we show Ben can prevent Amy from placing more than $100$ stones.\nDivide into several $4 \\times 4$ squares and then further partition\neach $4 \\times 4$ squares as shown in the grid below.\n$\n[\n{cccc}\n1 & 2 & 3 & 4 \\\\\n3 & 4 & 1 & 2 \\\\\n2 & 1 & 4 & 3 \\\\\n4 & 3 & 2 & 1\n\n]\n$\nThe squares with each label form $4$-cycles by knight jumps.\nFor each such cycle, whenever Amy plays in the cycle,\nBen plays in the opposite point of the cycle,\npreventing Amy from playing any more stones in that original cycle.\nHence Amy can play at most in $1/4$ of the stones, as desired."}
+{"year": 2018, "problem_number": 5, "problem": "Let $a_1$, $a_2$, \\dots\\ be an infinite sequence of positive integers,\nand $N$ a positive integer.\nSuppose that for all integers $n \\ge N$, the expression\n$ \\frac{a_1}{a_2} + \\frac{a_2}{a_3} + \\dots\n+ \\frac{a_{n-1}}{a_n} + \\frac{a_n}{a_1} $\nis an integer.\nProve that $(a_n)$ is eventually constant.", "solution": "The condition implies that the difference\n$ S(n) = \\frac{a_{n+1} - a_n}{a_1} + \\frac{a_n}{a_{n+1}} $\nis an integer for all $n > N$.\nWe proceed by $p$-adic valuation only henceforth;\nfix a prime $p$.\nThen analyzing the $\\nu_p$, we immediately get that for $n > N$:\n\nIf $\\nu_p(a_n) < \\nu_p(a_{n+1})$, then $\\nu_p(a_{n+1}) = \\nu_p(a_1)$.\nIf $\\nu_p(a_n) = \\nu_p(a_{n+1})$, no conclusion.\nIf $\\nu_p(a_n) > \\nu_p(a_{n+1})$,\nthen $\\nu_p(a_{n+1}) \\ge \\nu_p(a_1)$.\n\nIn other words:\n\nLet $p$ be a prime. Consider the sequence\n$\\nu_p(a_{N+1})$, $\\nu_p(a_{N+2})$, \\dots.\nThen either:\n\nWe have $\\nu_p(a_{N+1}) \\ge \\nu_p(a_{N+2}) \\ge \\dots$\nand so on, i.e.\\ the sequence is weakly decreasing immediately; or\nFor some index $K > N$ we have\n$\\nu_p(a_K) < \\nu_p(a_{K+1}) = \\nu_p(a_{K+2}) = \\dots = \\nu_p(a_1)$,\ni.e.\\ the sequence ``jumps'' to $\\nu_p(a_1)$\nat some point and then stays there forever after.\nNote this requires $\\nu_p(a_1) > 0$.\n\nA cartoon of the situation is drawn below.\n[Figure omitted]\n\nAs only finitely many primes $p$ divide $a_1$,\nafter some time $\\nu_p(a_n)$ is fixed for all such $p \\mid a_1$.\nAfterwards, the sequence satisfies $a_{n+1} \\mid a_n$ for each $n$,\nand thus must be eventually constant.\n\nThis solution is almost completely $p$-adic,\nin the sense that I think a similar result\nholds if one replaces $a_n \\in \\ZZ$\nby $a_n \\in \\ZZ_p$ for any particular prime $p$.\nIn other words, the primes almost do not talk to each other.\n\nThere is one caveat: if $x_n$ is an integer sequence\nsuch that $\\nu_p(x_n)$ is eventually constant for each prime\nthen $x_n$ may not be constant.\nFor example, take $x_n$ to be the $n$th prime!\nThat's why in the first claim (applied to co-finitely many of the primes),\nwe need the stronger non-decreasing condition,\nrather than just eventually constant.\n\nAn alternative approach is to show that, when the fractions $a_n / a_1$\nis written in simplest form for $n = N+1, N+2, \\dots$,\nthe numerator and denominator are both weakly decreasing.\nHence it must eventually be constant; in which case it equals $\\frac11$."}
+{"year": 2018, "problem_number": 6, "problem": "A convex quadrilateral $ABCD$ satisfies $AB \\cdot CD = BC \\cdot DA$.\nPoint $X$ lies inside $ABCD$ so that\n$ \\angle XAB=\\angle XCD \\quad \\text{ and } \\quad \\angle XBC=\\angle XDA. $\nProve that $\\angle BXA + \\angle DXC=180\\dg$.", "solution": "We present two solutions by inversion.\nThe first is the official one.\nThe second is a solution via inversion, completed by USA5 Michael Ren.\n\nOfficial solution by inversion.: \nIn what follows a convex quadrilateral is called\nquasi-harmonic if $AB \\cdot CD = BC \\cdot DA$.\n\nA quasi-harmonic quadrilateral is determined\nup to similarity by its angles.\n\nDo some inequalities.\n\nThis could be expected by degrees of freedom;\na quadrilateral has four degrees of freedom up to similarity;\nthe pseudo-harmonic condition is one\nwhile the measures of angles $\\angle A$, $\\angle B$, $\\angle C$, $\\angle D$\n(summing to $360\\dg$) provide three degrees of freedom.\n(Note that the point $X$ plays no role in this comment.)\n\nPerforming an inversion at $X$, one obtains a\nsecond quasi-harmonic quadrilateral\n$A^\\ast B^\\ast C^\\ast D^\\ast$ which has the same angles\nas the original one, $\\angle D^\\ast = \\angle A$,\n$\\angle A^\\ast = \\angle B$, and so on.\nThus by the claim we obtain similarity\n$ D^\\ast A^\\ast B^\\ast C^\\ast \\sim ABCD. $\nIf one then maps $D^\\ast A^\\ast B^\\ast C^\\ast$,\nonto $ABCD$, the image of $X^\\ast$\nbecomes a point isogonally conjugate to $X$.\nIn other words, $X$ has an isogonal conjugate in $ABCD$.\n\nIt is well-known that this is equivalent to\n$\\angle BXA + \\angle DXC = 180\\dg$,\nfor example by inscribing an ellipse with foci $X$ and $X^\\ast$.\n\nSecond solution: ``rhombus inversion'', by Michael Ren.: \nSince\n$ \\frac{AB}{AD} = \\frac{CB}{CD} $\nand\n$ \\frac{BA}{BC} = \\frac{DA}{DC} $\nit follows that $B$ and $D$ lie on an Apollonian circle $\\omega_{AC}$\nthrough $A$ and $C$,\nwhile $A$ and $C$ lie on an Apollonian circle $\\omega_{BD}$\nthrough $B$ and $D$.\nWe let these two circles intersect at a point $P$ inside $ABCD$.\n\nThe main idea is then to\nperform an inversion about $P$ with radius $1$.\nWe obtain:\n\nThe image of $ABCD$ is a rhombus.\n\nBy the inversion distance formula, we have\n$ \\frac{1}{A'B'} = \\frac{PA}{AB} \\cdot PB = \\frac{PC}{BC} \\cdot PB = \\frac{1}{B'C'} $\nand so $A'B' = B'C'$.\nIn a similar way, we derive $B'C' = C'D' = D'A'$,\nso the image is a rhombus as claimed.\n\nLet us now translate the angle conditions.\nWe were given that $\u2220 XAB = \u2220 XCD$, but\n\n\u2220 XAB &= \u2220 XAP + \u2220 PAB = \u2220 PX'A' + \u2220 A'B'P \\\\\n\u2220 XCD &= \u2220 XCP + \u2220 PCD = \u2220 PX'C' + \u2220 C'D'P\n\\intertext{so subtracting these gives}\n\u2220 A'X'C' &= \u2220 A'B'P + \u2220 PD'C' = \u2220 (A'B', B'P) + \u2220 (PD', C'D') \\\\\n&= \u2220 (A'B', B'P) + \u2220 (PD', A'B') = \u2220 D' P B'. \\tag{1}\n\nsince $A'B' \\parallel C'D'$.\nSimilarly, we obtain\n$ \u2220 B'X'D' = \u2220 A'PC' \\tag{2}. $\nWe now translate the desired condition.\nSince\n\n\u2220 AXB &= \u2220 AXP + \u2220 PXB = \u2220 PA'X' + \u2220 X'B'P \\\\\n\u2220 CXD &= \u2220 CXP + \u2220 PXD = \u2220 PC'X' + \u2220 X'DP'\n\nwe compute\n\n\u2220 AXB + \u2220 CXD &= (\u2220 PA'X' + \u2220 X'B'P) + (\u2220 PC'X' + \u2220 X'D'P) \\\\\n&= -[ ( \u2220 A'X'P + \u2220 X'PA' )\n+ ( \u2220 PX'B' + \u2220 B'PX' ) ] \\\\\n&\\quad- [ ( \u2220 C'X'P + \u2220 X'PC' )\n+ ( \u2220 PX'D' + \u2220 D'PX' ) ] \\\\\n&= [ \u2220 PX'A' + \u2220 BX'P + \u2220 PX'C' + \u2220 D'X'P ] \\\\\n&\\quad+ [ \u2220 A'PX' + \u2220 X'PB' + \u2220 C'PX' + \u2220 X'PD' ] \\\\\n&= \u2220 A'PB' + \u2220 C'PD' + \u2220 B'X'C + \u2220 D'X'A\n\nand we wish to show this is equal to zero, i.e.\\\nthe desired becomes\n$ \u2220 A'PB' + \u2220 C'PD' + \u2220 B'X'C + \u2220 D'X'A = 0. \\tag{3} $\nIn other words, the problem is to show (1) and (2) implies (3).\n\nHenceforth drop apostrophes.\nHere is the inverted diagram (with apostrophes dropped).\n[Figure omitted]\n\nLet $Q$ denote the reflection of $P$\nand let $Y$ denote the second intersection of $(BQC)$ and $(AQD)$.\nThen\n\n-\u2220 AXC &= -\u2220 DPB = \u2220 BQD = \u2220 BQY + \u2220 YQD = \u2220 BCY + \u2220 YAD \\\\\n&= \u2220(BC,CY) + \u2220(YA,AD) = \u2220 YCA = -\u2220 AYC.\n\nHence $XACY$ is concyclic; similarly $XBDY$ is concyclic.\n\n$X \\neq Y$.\n\nTo see this: Work pre-inversion assuming $AB < AC$.\nThen $Q$ was the center of $\\omega_{BD}$.\nIf $T$ was the second intersection of $BA$ with $(QBC)$,\nthen $QB = QD = QT = \\sqrt{QA \\cdot QC}$, by shooting lemma.\nSince $\\angle BAD < 180\\dg$,\nit follows $(QBCY)$ encloses $ABCD$ (pre-inversion).\n(This part is where the hypothesis that\n$ABCD$ is convex with $X$ inside is used.)\n\nFinally, we do an angle chase to finish:\n\n\u2220 DXA &= \u2220 DXY + \u2220 YXA = \u2220 DBY + \u2220 YCA \\\\\n&= \u2220 (DB, YB) + \u2220 (CY, CA) = \u2220 CYB + 90\\dg \\\\\n&= \u2220 CQB + 90\\dg = -\u2220 APB + 90\\dg \\tag{4}.\n\nSimilarly,\n$ \u2220 BXC = \u2220 DPC + 90\\dg. \\tag{5} $\n\nSumming (4) and (5) gives (3).\n\nA difficult part of the problem in many solutions\nis that the conclusion is false in the directed sense,\nif the point $X$ is allowed to lie outside the quadrilateral.\nWe are saved in the first solution because the equivalence\nof the isogonal conjugation requires $X$ inside the quadrilateral.\nOn the other hand, in the second solution,\nthe issue appears in the presence of the second point $Y$."}
+{"year": 2019, "problem_number": 1, "problem": "Solve over $\\ZZ$ the functional equation\n$f(2a) + 2f(b) = f(f(a+b))$.", "solution": "Notice that $f(x) \\equiv 0$ or $f(x) \\equiv 2x+k$ work\nand are clearly the only linear solutions.\nWe now prove all solutions are linear.\n\nLet $P(a,b)$ be the assertion.\n\nFor each $x \\in \\ZZ$ we have $f(2x) = 2f(x) - f(0)$.\n\nCompare $P(0,x)$ and $P(x,0)$.\n\nNow, $P(a,b)$ and $P(0,a+b)$ give\n\nf(f(a+b)) &= f(2a) + 2f(b) = f(0) + 2f(a+b) \\\\\n\\implies [2f(a) - f(0)] + 2f(b) &= f(0) + 2f(a+b) \\\\\n\\implies ( f(a)-f(0) ) + ( f(b)-f(0) )\n&= ( f(a+b)-f(0) ).\n\nThus the map $x \\mapsto f(x) - f(0)$ is additive,\ntherefore linear.\n\nThe same proof works on the functional equation\n$ f(2a) + 2f(b) = g(a+b) $\nwhere $g$ is an arbitrary function (it implies that $f$ is linear)."}
+{"year": 2019, "problem_number": 2, "problem": "In triangle $ABC$ point $A_1$ lies on side $BC$\nand point $B_1$ lies on side $AC$.\nLet $P$ and $Q$ be points on segments $AA_1$ and $BB_1$,\nrespectively, such that $PQ \\parallel AB$.\nPoint $P_1$ is chosen on ray $PB_1$ beyond $B_1$\nsuch that $\\angle PP_1C = \\angle BAC$.\nPoint $Q_1$ is chosen on ray $QA_1$ beyond $A_1$\nsuch that $\\angle CQ_1Q = \\angle CBA$.\nProve that points $P_1$, $Q_1$, $P$, $Q$ are cyclic.", "solution": "We present two solutions.\n\nFirst solution by bary (Evan Chen).: \nLet $PB_1$ and $QA_1$ meet line $AB$ at $X$ and $Y$.\nSince $XY \\parallel PQ$ it is equivalent\nto show $P_1XYQ_1$ is cyclic (Reim's theorem).\n\nNote the angle condition implies $P_1CXA$ and $Q_1CYB$ are cyclic.\n\nLetting $T = PX \\cap QY$ (possibly at infinity),\nit suffices to show that the\nradical axis of $\\triangle CXA$ and $\\triangle CYB$ passes through $T$,\nbecause that would imply $P_1XYQ_1$ is cyclic\n(by power of a point when $T$ is Euclidean,\nand because it is an isosceles trapezoid if $T$ is at infinity).\n\n[Figure omitted]\n\nTo this end we use barycentric coordinates on $\\triangle ABC$.\nWe begin by writing\n$ P = (u+t : s : r), \\quad Q = (t : u+s : r) $\nfrom which it follows that\n$A_1 = (0 : s : r)$ and $B_1 = (t : 0 : r)$.\n\nNext, compute $X = (\n\\det [ u+t & r \\\\ t & r ]\n: \\det [ s & r \\\\ 0 & r ]\n: 0 ) = (u : s : 0)$.\nSimilarly, $Y = (t : u : 0)$.\nSo we have computed all points.\n\nLine $B_1X$ has equation\n$-rs \\cdot x + ru \\cdot y + st \\cdot z = 0$,\nwhile line $C_1 Y$ has equation\n$ru \\cdot x - rt \\cdot y + st \\cdot z = 0$.\n\nLine $B_1X$ is $0 = \\det(B_1, X, -)\n= \\det [ \nt & 0 & r \\\\\nu & s & 0 \\\\\nx & y & z\n]$.\nLine $C_1Y$ is analogous.\n\nThe radical axis $(u+t) y - (u+s) x = 0$.\n\nCircle $(AXC)$ is given by\n$-a^2yz - b^2zx - c^2xy + (x+y+z) \\cdot \\frac{c^2 \\cdot u}{u+s} y = 0$.\nSimilarly, circle $(BYC)$ has equation\n$-a^2yz - b^2zx - c^2xy + (x+y+z) \\cdot \\frac{c^2 \\cdot u}{u+t} x = 0$.\nSubtracting gives the radical axis.\n\nFinally, to see these three lines are concurrent, we now compute\n\n\\det \n-rs & ru & st \\\\\nru & -rt & st \\\\\n-(u+s) & u+t & 0\n\n&= rst [ [u(u+t)-t(u+s)] + [s(u+t)-u(u+s)] ] \\\\\n&= rst [ (u^2-st) + (st-u^2) ] = 0.\n\nThis completes the proof.\n\nSecond official solution by tricky angle chasing.: \nLet lines $AA_1$ and $BB_1$ meet at the circumcircle of $\\triangle ABC$\nagain at points $A_2$ and $B_2$.\nBy Reim's theorem, $PQA_2B_2$ are cyclic.\n[Figure omitted]\n\nThe points $P$, $Q$, $A_2$, $Q_1$ are cyclic.\nSimilarly the points $P$, $Q$, $B_2$, $P_1$ are cyclic.\n\nNote that $CA_1A_2Q_1$ is cyclic since\n$\u2220 CQ_1A_1 = \u2220 CQ_1Q = \u2220 CBA = \u2220 CA_2A = \u2220 CA_2A_1$.\nThen $\u2220 QQ_1A_2 = \u2220 A_1Q_1A_2 = \u2220 A_1 C A_2\n= \u2220 B C A_2 = \u2220 B A A_2 = \u2220 Q P A_2$.\n\nThis claim obviously solves the problem."}
+{"year": 2019, "problem_number": 3, "problem": "A social network has $2019$ users, some pairs of which are friends (friendship is symmetric).\nIf $A$, $B$, $C$ are three users such that $AB$ are friends and $AC$ are friends but $BC$ is not,\nthen the administrator may perform the following operation:\nchange the friendships such that $BC$ are friends, but $AB$ and $AC$ are no longer friends.\n\nInitially, $1009$ users have $1010$ friends and $1010$ users have $1009$ friends.\nProve that the administrator can make a sequence of operations\nsuch that all users have at most $1$ friend.", "solution": "We take the obvious graph formulation\nand call the move a toggle.\n\nLet $G$ be a connected graph.\nThen one can toggle $G$ without disconnecting the graph,\nunless $G$ is a clique, a cycle, or a tree.\n\nAssume $G$ is connected and not a tree, so it has a cycle.\nTake the smallest cycle $C$; by hypothesis $C \\neq G$.\n\nIf $C$ is not a triangle (equivalently, $G$ is triangle-free),\nthen let $b \\notin C$ be a vertex adjacent to $C$, say at $a$.\nTake a vertex $c$ of the cycle adjacent to $a$ (hence not to $b$).\nThen we can toggle $abc$.\n\nNow assume there exists a triangle; let $K$ be the maximal clique.\nBy hypothesis, $K \\neq G$.\nWe take an edge $e = ab$ dangling off the clique,\nwith $a \\in K$ and $b \\notin K$.\nNote some vertex $c$ of $K$ is not adjacent to $b$; now toggle $abc$.\n\nBack to the original problem;\nlet $G_{\\text{imo}}$ be the given graph.\nThe point is that we can apply toggles (by the claim) repeatedly,\nwithout disconnecting the graph, until we get a tree.\nThis is because\n\n$G_{\\text{imo}}$ is connected,\nsince any two vertices which are not adjacent\nhave a common neighbor by pigeonhole\n($1009 + 1009 + 2 > 2019$).\n$G_{\\text{imo}}$ cannot become a cycle,\nbecause it initially has an odd-degree vertex,\nand toggles preserve parity of degree!\n$G_{\\text{imo}}$ is obviously not a clique initially\n(and hence not afterwards).\n\nSo, we can eventually get $G_{\\text{imo}}$ to be a tree.\n\nOnce $G_{\\text{imo}}$ is a tree the problem follows by repeatedly applying\ntoggles arbitrarily until no more are possible;\nthe graph (although now disconnected) remains acyclic\n(in particular having no triangles)\nand therefore can only terminate in the desired situation.\n\nThe above proof in fact shows the following better result:\n\nThe task is possible if and only if\n$G_{\\text{imo}}$ is a connected graph which is not a clique\nand has any vertex of odd degree.\n\nThe ``only if'' follows from the observation\nthat toggles preserve parity of degree.\n\nThus the given condition about the degrees of vertices\nbeing $1009$ and $1010$ is largely a red herring;\nit's a somewhat strange way of masking the correct and more natural\nboth-sufficient-and-necessary condition."}
+{"year": 2019, "problem_number": 4, "problem": "Solve over positive integers the equation\n$ k! = \\prod_{i=0}^{n-1} (2^n-2^i) = (2^n-1)(2^n-2)(2^n-4) \\dots (2^n-2^{n-1}). $", "solution": "The answer is $(n,k) =(1,1)$ and $(n,k) = (2,3)$ which work.\n\nLet $A = \\prod_i (2^n-2^k)$, and assume $A = k!$ for some $k \\ge 3$.\nRecall by exponent lifting that\n$ \\nu_3(2^t-1) = \n0 & t \\text{ odd} \\\\\n1 + \\nu_3(t/2) & t \\text{ even}.\n$\nThus, we can compute\n\nk > \\nu_2(k!) &= \\nu_2(A) = 1 + 2 + \\dots + (n-1) = \\frac{n(n-1)}{2} \\\\\n\\left\\lfloor \\frac k3 \\right\\rfloor\n\\le \\nu_3(k!) &= \\nu_3(A) = \\left\\lfloor \\frac n2 \\right\\rfloor\n+ \\left\\lfloor \\frac n6 \\right\\rfloor + \\dots < \\frac 34n.\n\nwhere the first inequality follows by Legendre's formula $\\nu_2(k!) = k - s_2(k)$.\n\nIn this way, we get\n$ \\frac 94 n + 3 > k > \\frac{n(n-1)}{2} $\nwhich means $n \\le 6$; a manual check then shows the\nsolutions we claimed earlier are the only ones.\n\nAn amusing corollary of the problem pointed out in the shortlist\nis that the symmetric group $S_k$ cannot be isomorphic to the group $\\GL_n(\\FF_2)$\nunless $(n,k) = (1,1)$ or $(n,k) = (2,3)$, which indeed produce isomorphisms."}
+{"year": 2019, "problem_number": 5, "problem": "Let $n$ be a positive integer.\nHarry has $n$ coins lined up on his desk, which can show either heads or tails.\nHe does the following operation: if there are $k$ coins which show heads and $k > 0$,\nthen he flips the $k$th coin over; otherwise he stops the process.\n(For example, the process starting with $THT$ would be\n$THT \\to HHT \\to HTT \\to TTT$, which takes three steps.)\n\nProve the process will always terminate, and determine the average number of steps\nthis takes over all $2^n$ configurations.", "solution": "The answer is $ E_n = 1/2 (1 + \\dots + n) = \\frac14 n(n+1) $\nwhich is finite.\n\nWe'll represent the operation by a\ndirected graph $G_n$ on vertices $\\{0,1\\}^n$\n(each string points to its successor)\nwith $1$ corresponding to heads and $0$ corresponding to tails.\nFor $b \\in \\{0,1\\}$ we let $\\ol b = 1-b$,\nand denote binary strings as a sequence of $n$ symbols.\n\nThe main claim is that $G_n$\ncan be described explicitly in terms of $G_{n-1}$:\n\nWe take two copies $X$ and $Y$ of $G_{n-1}$.\n\nIn $X$, we take each string of length $n-1$\nand just append a $0$ to it. In symbols,\nwe replace $s_1 \\dots s_{n-1} \\mapsto s_1 \\dots s_{n-1} 0$.\n\nIn $Y$, we toggle every bit, then reverse the order,\nand then append a $1$ to it.\nIn symbols, we replace\n$s_1 \\dots s_{n-1} \\mapsto \\ol s_{n-1} \\ol s_{n-2} \\dots \\ol s_{1} 1$.\n\nFinally, we add one new edge from $Y$ to $X$ by\n$11 \\dots 1 \\to 11\\dots110$.\n\nAn illustration of $G_4$ is given below.\n[Figure omitted]\n\nTo prove this claim, we need only show\nthe arrows of this directed graph remain valid.\nThe graph $X$ is correct as a subgraph of $G_n$,\nsince the extra $0$ makes no difference.\nAs for $Y$, note that if $s = s_1 \\dots s_{n-1}$ had $k$ ones,\nthen the modified string has $(n-1-k)+1 = n-k$ ones, ergo\n$\\ol s_{n-1} \\dots \\ol s_1 1\n\\mapsto \\ol s_{n-1} \\dots \\ol s_{k+1} s_k \\ol s_{k-1} \\dots \\ol s_1 1$\nwhich is what we wanted.\nFinally, the one edge from $Y$ to $X$ is obviously correct.\n\nTo finish, let $E_n$ denote the desired expected value.\nSince $1 \\dots 1$ takes $n$ steps to finish we have\n$ E_n = 1/2 [ E_{n-1} + (E_{n-1}+n) ] $\nbased on cases on whether the chosen string is in $X$ or $Y$ or not.\nBy induction, we have $E_n = 1/2 (1 + \\dots + n) = \\frac14 n(n+1)$,\nas desired.\n\nActually, the following is true:\nif the indices of the $1$'s are $1 \\le i_1 < \\dots < i_\\ell \\le n$,\nthen the number of operations required is\n$ 2(i_1 + \\dots + i_\\ell) - \\ell^2. $\nThis problem also has an interpretation as a Turing machine:\nthe head starts at a position on the tape (the binary string).\nIf it sees a $1$, it changes the cell to a $0$ and moves left;\nif it sees a $0$, it changes the cell to a $1$ and moves right."}
+{"year": 2019, "problem_number": 6, "problem": "Let $ABC$ be a triangle with incenter $I$ and incircle $\\omega$.\nLet $D$, $E$, $F$ denote the tangency points of $\\omega$ with $BC$, $CA$, $AB$.\nThe line through $D$ perpendicular to $EF$ meets $\\omega$ again at $R$ (other than $D$),\nand line $AR$ meets $\\omega$ again at $P$ (other than $R$).\nSuppose the circumcircles of $\\triangle PCE$ and $\\triangle PBF$ meet again at $Q$ (other than $P$).\nProve that lines $DI$ and $PQ$ meet on the external $\\angle A$-bisector.", "solution": "We present five solutions.\n\nFirst solution by complex numbers (Evan Chen, with Yang Liu).: \nWe use complex numbers with $D=x$, $E=y$, $F=z$.\n[Figure omitted]\n\nThen $A = \\frac{2yz}{y+z}$,\n$R = \\frac{-yz}{x}$ and so\n$ P = \\frac{A-R}{1-RA}\n= \\frac{\\frac{2yz}{y+z} + \\frac{yz}{x}}\n{1 + \\frac{yz}{x} \\cdot \\frac{2}{y+z}}\n= \\frac{yz(2x+y+z)}{2yz+x(y+z)}. $\nWe now compute\n\nO_B &= \\det \nP & P \\ol P & 1 \\\\\nF & F \\ol F & 1 \\\\\nB & B \\ol B & 1\n\n\\div \\det \nP & \\ol P & 1 \\\\\nF & \\ol F & 1 \\\\\nB & \\ol B & 1\n\n= \\det \nP & 1 & 1 \\\\\nz & 1 & 1 \\\\\n\\frac{2xz}{x+z} & \\frac{4xz}{(x+z)^2} & 1\n\n\\div \\det \nP & 1/P & 1 \\\\\nz & 1/z & 1 \\\\\n\\frac{2xz}{x+z} & \\frac{2}{x+z} & 1\n\\\\\n&= \\frac{1}{x+z} \\det \nP & 0 & 1 \\\\\nz & 0 & 1 \\\\\n2xz(x+z) & -(x-z)^2 & (x+z)^2\n\n\\div \\det \nP & 1/P & 1 \\\\\nz & 1/z & 1 \\\\\n2xz & 2 & x+z\n\\\\\n&= \\frac{(x-z)^2}{x+z} \\cdot \\frac{P-z}{(x+z)(P/z-z/P)+2z-2x + \\frac{2xz}{P}-2P} \\\\\n&= \\frac{(x-z)^2}{x+z} \\cdot \\frac{P-z}{\n(\\frac xz-1) P - 2(x-z) + (xz-z^2) \\frac 1P } \\\\\n&= \\frac{x-z}{x+z} \\cdot \\frac{P-z}{P/z + z/P - 2}\n= \\frac{x-z}{x+z} \\cdot \\frac{P-z}{\\frac{(P-z)^2}{Pz}}\n= \\frac{x-z}{x+z} \\cdot \\frac{1}{\\frac 1z - \\frac 1P} \\\\\n&= \\frac{x-z}{x+z} \\cdot \\frac{y(2x+y+z)}{y(2x+y+z) - (2yz+xy+xz)}\n= \\frac{x-z}{x+z} \\cdot \\frac{yz(2x+y+z)}{xy+y^2-yz-xz} \\\\\n&= \\frac{x-z}{x+z} \\cdot \\frac{yz(2x+y+z)}{(y-z)(x+y)}.\n\nSimilarly\n$ O_C = \\frac{x-y}{x+y} \\cdot \\frac{yz(2x+y+z)}{(z-y)(x+z)}. $\nTherefore, subtraction gives\n$ O_B-O_C\n=\n\\frac{yz(2x+y+z)}{(x+y)(x+z)(y-z)}\n[ (x-z) + (x-y) ]\n= \\frac{yz(2x+y+z)(2x-y-z)}{(x+y)(x+z)(z-y)}.\n$\nIt remains to compute $T$.\nSince $T \\in ID$ we have $t/x \\in \\RR$\nso $\\ol t = t/x^2$.\nAlso,\n\n\\frac{t - \\frac{2yz}{y+z}}{y+z} \\in i \\RR\n\\implies 0 &= \\frac{t-\\frac{2yz}{y+z}}{y+z}\n+ \\frac{\\frac{t}{x^2}-\\frac{2}{y+z}}{\\frac1y+\\frac1z} \\\\\n&= \\frac{1+\\frac{yz}{x^2}}{y+z} t - \\frac{2yz}{(y+z)^2} - \\frac{2yz}{(y+z)^2} \\\\\n\\implies t &= \\frac{x^2}{x^2+yz} \\cdot \\frac{4yz}{y+z}\n\nThus\n\nP-T &= \\frac{yz(2x+y+z)}{2yz+x(y+z)} - \\frac{4x^2yz}{(x^2+yz)(y+z)} \\\\\n&= yz \\cdot \\frac{(2x+y+z)(x^2+yz)(y+z) - 4x^2(2yz+xy+xz)}\n{(y+z)(x^2+yz)(2yz+xy+xz)} \\\\\n&= -yz \\cdot \\frac{(2x-y-z)(x^2y+x^2z+4xyz+y^2z+yz^2)}\n{(y+z)(x^2+yz)(2yz+xy+xz)}.\n\nThis gives $PT \\perp O_B O_C$ as needed.\n\nSecond solution by tethered moving points, with optimization (Evan Chen).: \nFix $\\triangle DEF$ and $\\omega$, with $B = DD \\cap FF$\nand $C = DD \\cap EE$.\nWe consider a variable point $M$ on $\\omega$\nand let $X$, $Y$ be on $EF$ with\n$CY \\cap \\parallel ME$, $BX \\cap \\parallel MF$.\nWe define $W = CY \\cap BX$.\nAlso, let line $MW$ meet $\\omega$ again at $V$.\n\n[Figure omitted]\n\n[Angle chasing]\nPentagons $CVWXE$ and $BVWYF$ are cyclic.\n\nBy $\u2220 EVW = \u2220 EVM = \u2220 EFM = \u2220 CEM = \u2220 ECW$\nand $\u2220 EXW = \u2220 EFM = \u2220 CEM = \u2220 ECW$.\n\nLet $N = DM \\cap EF$ and $R'$ be the $D$-antipode on $\\omega$.\n\n[Black magic]\nThe points $V$, $N$, $R'$ are collinear.\n\nWe use tethered moving points with $\\triangle DEF$ fixed.\n\nObviously the map $\\omega \\mapsto EF \\mapsto \\omega$ by $M \\mapsto N \\mapsto R'N \\cap \\omega$ is projective.\nAlso, the map $\\omega \\mapsto EF \\mapsto \\omega$ by $M \\mapsto X \\mapsto V$ is also projective\n(the first by projection to the line at infinity at back;\nthe second say by inversion at $E$).\n\nSo it suffices to check for three points.\nWhen $M=E$ we get $N=E$ so $R'N \\cap \\omega = E$, while $W=E$ and thus $V=E$.\nThe case $M=F$ is similar.\nFinally, if $M = R'$, then $W$ is the center of $\\omega$ and so $V = R'N \\cap EF = D$.\n\nWe now address the original problem by specializing $M$:\nchoose it so that $N$ is the midpoint of $EF$.\nLet $M' = DA \\cap (DEF)$.\n\nAfter this specialization, $V=P$ and $W=Q$.\n\nThus $RR'$ and $MM'$ are parallel to $EF$.\nFrom $(EF;PR) = -1 = (EF;N\\infty) \\overset{R'}{=} (EF;NV)$,\nwe derive that $P=V$ and $Q=R$, proving (i).\n\nFinally, the concurrence requested follows by Pascal theorem on $M'MDR'PR$.\n\nThird solution by power of a point linearity (Luke Robitaille).: \n\nLet us define\n$ f(\\bullet) = \\opname{Pow}(\\bullet, (CPE))\n- \\opname{Pow}(\\bullet, (BPF)) $\nwhich is a linear function from the plane to $\\RR$.\n\nDefine $W = BA \\cap PE$, $V = AC \\cap PF$.\nAlso, let $W_1 = ER \\cap AB$, $V_1 = FR \\cap AC$.\nNote that\n$ -1 = (PR;EF) \\overset{E}{=} (WA; W_1F) $\nand similarly $(VA; V_1E) = -1$.\n\nWe have\n\nf(F) &= \\frac{|EF| \\cdot (s-c) \\sin C/2}{\\sin B/2} \\\\\nf(E) &= -\\frac{|EF| \\cdot (s-b) \\sin B/2}{\\sin C/2}.\n\nWe have\n$ f(W) = WF^2 - WB \\cdot WF = WF \\cdot BF $\nwhere lengths are directed.\nNext,\n\nf(F) &= \\frac{AF \\cdot f(W) + FW \\cdot f(A)}{AW} \\\\\n&= \\frac{AF \\cdot WF \\cdot BF + FW \\cdot\n( AE \\cdot AC - AF \\cdot AB )}{AW} \\\\\n&= \\frac{WF(AF \\cdot BF + AF \\cdot AB) + FW \\cdot AE \\cdot AC}{AW} \\\\\n&= \\frac{WF \\cdot AF^2 - WF \\cdot AE \\cdot AC}{AW}\n= \\frac{WF}{AW} \\cdot (AE^2 - AE \\cdot AC) \\\\\n&= \\frac{WF}{AW} \\cdot AE \\cdot CE\n= -\\frac{W_1F}{AW_1} \\cdot AE \\cdot CE.\n\nSince $\\triangle DEF$ is acute,\nthe point $R$ lies inside $\\triangle AEF$.\nThus $W_1$ lies inside segment $AF$\nand the ratio $\\frac{W_1F}{AW_1}$ is positive.\nWe now determine its value: by the ratio lemma\n\n\\frac{|W_1F|}{|AW_1|}\n&= \\frac{|EF| \\sin \\angle W_1 E F}{|AE| \\sin \\angle A E W_1} \\\\\n&= \\frac{|EF| \\sin \\angle REF}{|AE| \\sin \\angle AER} \\\\\n&= \\frac{|EF| \\sin \\angle RDF}{|AE| \\sin \\angle EDR} \\\\\n&= \\frac{|EF| \\sin C/2}{|AE| \\sin B/2}.\n\nAlso, we have $AE \\cdot CE < 0$ since $E$ lies inside $AC$.\nHence\n$ f(F) = -\\frac{|EF| \\sin C/2}{|AE| \\sin B/2}.\n\\cdot AE \\cdot CE\n= |EF| \\cdot \\frac{|CE| \\sin B/2}{\\sin C/2}\n= |EF| \\cdot \\frac{(s-c) \\sin B/2}{\\sin C/2}. $\nThe calculation for $f(E)$ is similar,\n(noting the sign flips since $f$ is anti-symmetric\nin terms of $B$ and $C$).\n\nLet $Z \\in DI$ with $\\angle ZAI = 90\\dg$ be the point requested in the problem now.\nOur goal is to show $f(Z) = 0$.\nWe assume WLOG that $AB < AC$, so $\\frac{ZA}{EF} > 0$.\nThen\n\n|ZA| &= |AI| \\cdot \\tan \\angle AIZ \\\\\n&= |AI| \\cdot \\tan \\angle(AI, DI) \\\\\n&= \\frac{s-a}{\\cos A/2} \\cdot \\tan (BC, EF) \\\\\n&= \\frac{s-a}{\\cos A/2} \\tan (B/2-C/2).\n\nTo this end we compute\n\nf(Z) &= f(A) + [ f(Z) - f(A) ]\n= f(A) + \\frac{ZA}{EF} [ f(E)-f(F) ] \\\\\n&= f(A) - \\frac{ZA}{EF}\n[ \\frac{|EF| \\cdot (s-b) \\sin B/2}{\\sin C/2}\n+ \\frac{|EF| \\cdot (s-c) \\sin C/2}{\\sin B/2} ] \\\\\n&= f(A) - |ZA| [ \\frac{(s-b) \\sin B/2}{\\sin C/2}\n+ \\frac{(s-c) \\sin C/2}{\\sin B/2} ] \\\\\n&= [ b(s-a) - c(s-a) ]\n- |ZA| [ \\frac{(s-b) \\sin B/2}{\\sin C/2}\n+ \\frac{(s-c) \\sin C/2}{\\sin B/2} ] \\\\\n&= (b-c)(s-a) - \\frac{s-a}{\\cos A/2} \\tan (B/2-C/2)\n[ \\frac{(s-b) \\sin B/2}{\\sin C/2}\n+ \\frac{(s-c) \\sin C/2}{\\sin B/2} ].\n\nDividing out,\n\n\\frac{f(Z)}{s-a}\n&= (b-c) - \\frac{1}{\\cos A/2} \\tan (B/2-C/2)\n[ \\frac{r \\cos B/2}{\\sin C/2}\n+ \\frac{r \\cos C/2}{\\sin B/2} ] \\\\\n&= (b-c) - \\frac{r \\tan(B/2-C/2)}{\\cos A/2}\n\\cdot \\frac{\\cos B/2 \\sin B/2 + \\cos C/2 \\sin C/2}\n{\\sin C/2 \\sin B/2} \\\\\n&= (b-c) - \\frac{r \\tan(B/2-C/2)}{\\cos A/2}\n\\cdot \\frac{\\sin B + \\sin C}\n{2\\sin C/2 \\sin B/2} \\\\\n&= (b-c) - \\frac{r \\tan(B/2-C/2)}{\\cos A/2}\n\\cdot \\frac{\\sin(B/2+C/2)\\cos(B/2-C/2)}\n{\\sin C/2 \\sin B/2} \\\\\n&= (b-c) - r \\frac{\\sin(B/2-C/2)}{\\sin B/2 \\sin C/2} \\\\\n&= (b-c) - r(\\cot C/2 - \\cot B/2)\n= (b-c) - ( (s-c) - (s-b) ) = 0.\n\nFourth solution by incircle inversion (USA IMO live stream, led by Andrew Gu).: \nLet $T$ be the intersection of line $DI$ and the external $\\angle A$-bisector.\nAlso, let $G$ be the antipode of $D$ on $\\omega$.\n\nWe perform inversion around $\\omega$, using $\\bullet^\\ast$ for the inverse.\nThen $\\triangle A^\\ast B^\\ast C^\\ast$ is the medial triangle of $\\triangle DEF$,\nand $T^\\ast$ is the foot from $A^\\ast$ on to $DI$.\nIf we denote $Q^\\ast$ as the second intersection\nof $(PC^\\ast E)$ and $(PB^\\ast F)$,\nthen the goal is to show that $Q^\\ast$ lies on $(PIT^\\ast)$.\n\n[Figure omitted]\n\nPoints $Q^\\ast$, $B^\\ast$, $C^\\ast$ are collinear.\n\n$\u2220 PQ^\\ast C^\\ast = \u2220 PEC^\\ast = \u2220 PED = \u2220 PFD\n= \u2220 PFB^\\ast = \u2220 PQ^\\ast B^\\ast$.\n\n[cf Brazil 2011/5]\nPoints $P$, $A^\\ast$, $G$ are collinear.\n\nProject harmonic quadrilateral $PERF$ through $G$,\nnoting $GR \\parallel EF$.\n\nDenote by $M$ the center of parallelogram $DC^\\ast A^\\ast B^\\ast$.\nNote that it is the center of the circle with\ndiameter $DA^\\ast$, which passes through $P$ and $T^\\ast$.\nAlso, $MI \\parallel PA^\\ast G$.\n\nPoints $P$, $M$, $I$, $T^\\ast$ are cyclic.\n\n$\u2220 IT^\\ast P = \u2220 D T^\\ast P = \u2220 DA^\\ast P\n= \u2220 M A^\\ast P = \u2220 A^\\ast P M = \u2220 IMP$.\n\nPoints $P$, $M$, $I$, $Q^\\ast$ are cyclic.\n\n$\u2220 MQ^\\ast P = \u2220 C^\\ast Q^\\ast P = \u2220 C^\\ast E P\n= \u2220 D E P = \u2220 D G P = \u2220 GPI = \u2220 MIP$.\n\nFifth solution by double inversion (Brandon Wang, Luke Robitaille, Michael Ren, Evan Chen).: \nWe outline one final approach.\nAfter inverting about $\\omega$ as in the previous approach,\nwe then apply another inversion around $P$.\nDropping the apostrophes/stars/etc now one can check\nthat the problem we arrive at becomes the following.\n\n[Doubly inverted problem]\nIn $\\triangle PEF$, the $P$-symmedian meets $EF$\nand $(PEF)$ at $K$, $L$.\nLet $D \\in EF$ with $\\angle DPK = 90\\dg$,\nand let $T$ be the foot from $K$ to $DL$.\nDenote by $I$ the reflection of $P$ about $EF$.\nFinally, let $PDNE$ and $PDMF$ be cyclic harmonic quadrilaterals.\nThen lines $EN$, $MF$, $TI$, are concurrent.\n\nThe proof proceeds in three steps.\nSuppose the line through $L$ perpendicular to $EF$\nmeets $EF$ at $W$ and $(PEF)$ at $Z$.\n[Figure omitted]\n\nSince $\u2220 ZEP = \u2220 WLP = \u2220 WDP$,\nit follows $ZE$ is tangent to $(PDNE)$.\nSimilarly, $ZF$ is tangent to $(PDMF)$.\n$\\triangle WTP$ is the orthic triangle of $\\triangle DKL$,\nso $WD$ bisects $\\angle PWT$ and $WTI$ collinear.\n$-1 = E(PN;DZ) = F(PM;DZ) = W(PI;DZ)$, so\n$EN$, $FM$, $WI$ meet on $PZ$."}
+{"year": 2020, "problem_number": 1, "problem": "Consider the convex quadrilateral $ABCD$.\nThe point $P$ is in the interior of $ABCD$.\nThe following ratio equalities hold:\n$\\angle PAD:\\angle PBA:\\angle DPA\n= 1:2:3\n= \\angle CBP:\\angle BAP:\\angle BPC.$\nProve that the following three lines meet in a point:\nthe internal bisectors of angles $\\angle ADP$ and $\\angle PCB$\nand the perpendicular bisector of segment $AB$.", "solution": "Let $O$ denote the circumcenter of $\\triangle PAB$.\nWe claim it is the desired concurrency point.\n[Figure omitted]\nIndeed, $O$ obviously lies on the perpendicular bisector of $AB$.\nNow\n\n\u2220 BCP &= \u2220 CBP + \u2220 BPC \\\\\n&= 2\u2220 BAP = \u2220 BOP\n\nit follows $BOPC$ are cyclic.\nAnd since $OP = OB$, it follows that $O$ is on\nthe bisector of $\\angle PCB$, as needed.\n\nThe angle equality is only used insomuch $\\angle BAP$\nis the average of $\\angle CBP$ and $\\angle BPC$,\ni.e.\\ only $\\frac{1+3}{2} = 2$ matters."}
+{"year": 2020, "problem_number": 2, "problem": "Let $a \\ge b \\ge c \\ge d > 0$ be real numbers satisfying $a+b+c+d=1$.\nProve that\n$ (a+2b+3c+4d) a^a b^b c^c d^d < 1. $", "solution": "By weighted AM-GM we have\n$ a^a b^b c^c d^d \\le \\sum_{\\text{cyc}} \\frac{a}{a+b+c+d} \\cdot a\n= a^2+b^2+c^2+d^2. $\nSo, it is enough to prove that\n$ (a^2+b^2+c^2+d^2)(a+2b+3c+4d) \\le 1 = (a+b+c+d)^3. $\nExpand both sides to get\n$\n{cccc}\n+a^3 &+ b^2a &+ c^2a & +d^2a \\\\\n+2a^2b &+ 2b^3 &+ 2bc^2 & +2d^2b \\\\\n+3a^2c & + 3b^2c & + 3c^3 & + 3d^2c \\\\\n+4a^2d &+ 4b^2d & + 4c^2d & + 4d^3\n\n<\n{cccc}\n+a^3 &+ 3b^2a &+ 3c^2a & +3d^2a \\\\\n+3a^2b &+ b^3 &+ 3bc^2 & +3d^2b \\\\\n+3a^2c &+ 3b^2c &+ c^3 &+ 3d^2c \\\\\n+3a^2d &+ 3b^2d &+ 3c^2d &+ d^3 \\\\\n+6abc &+ 6bcd &+ 6cda &+ 6dab.\n\n$\nIn other words, we need to prove that\n$\n{cccc}\n& && \\\\\n&+ b^3 & & \\\\\n& & +2c^3 & \\\\\n+a^2d &+ b^2d & + c^2d & + 3d^3 \\\\\n\n<\n{cccc}\n&+ 2b^2a &+ 2c^2a & +2d^2a \\\\\n+a^2b & &+ bc^2 & +d^2b \\\\\n&&& \\\\\n&&& \\\\\n+6abc &+ 6bcd &+ 6cda &+ 6dab\n\n$\n\nThis follows since\n\n2b^2a &\\ge b^3 + b^2d \\\\\n2c^2a &\\ge 2c^3 \\\\\n2d^2a &\\ge 2d^3 \\\\\na^2b &\\ge a^2d \\\\\nbc^2 &\\ge c^2d \\\\\nd^2b &\\ge d^3\n\nand $6(abc+bcd+cda+dab) > 0$.\n\nSome students complained this problem was ``unfair'' and they couldn't solve it\nbecause they didn't think an IMO problem would be solved only by expansion.\nI don't agree with this.\nFedor Petrov provides the following motivational comments\nfor why the existence of this solution should not be that surprising:\n\nBetter to think about mathematics.\nYou have to bound from above a product $(a+2b+3c+4d)(a^2+b^2+c^2+d^2)$,\nthe coefficients $1,2,3,4$ are increasing and so play on your side,\nso plausibly $(a+b+c+d)^3$ should majorize this term-wise,\nyou check it and this appears to be true.\n\nHe also gave the \\href{https://aops.com/community/p18331252}{following advice}:\n\nThe general advice is to study mathematics, not olympiad problems in past years.\nIf the IMO problems set the students and their teachers on this path, I am more than satisfied."}
+{"year": 2020, "problem_number": 3, "problem": "There are $4n$ pebbles of weights $1, 2, 3, \\dots, 4n$.\nEach pebble is coloured in one of $n$ colours\nand there are four pebbles of each colour.\nShow that we can arrange the pebbles into two piles\nthe total weights of both piles are the same,\nand each pile contains two pebbles of each colour.", "solution": "The first key idea is the deep fact that\n$ 1+4n = 2+(4n-1) = 3+(4n-2) = \\dots. $\nSo, place all four pebbles of the same colour in a box (hence $n$ boxes).\nFor each $k=1,2,\\dots,2n$\nwe tape a piece of string between pebble $k$ and $4n+1-k$.\nTo solve the problem, it suffices to paint each string\neither blue or green such that each box has two blue strings\nand two green strings\n(where a string between two pebbles in the same box counts double).\n[Figure omitted]\n\nWe can therefore rephrase the problem as follows,\nif we view boxes as vertices and strings as edges:\n\nGiven a $4$-regular multigraph on $n$ vertices\n(where self-loops are allowed and have degree $2$),\none can color the edges blue and green\nsuch that each vertex has two blue and two green edges.\n\nEach connected component of the graph can be decomposed\ninto an Eulerian circuit, since $4$ is even.\nA connected component with $k$ vertices has $2k$\nedges in its Eulerian circuit,\nso we may color the edges in this circuit alternating green and blue.\nThis may be checked to work."}
+{"year": 2020, "problem_number": 4, "problem": "There is an integer $n > 1$.\nThere are $n^2$ stations on a slope of a mountain, all at different altitudes.\nEach of two cable car companies, $A$ and $B$, operates $k$ cable cars;\neach cable car provides a transfer from one of the stations\nto a higher one (with no intermediate stops).\nThe $k$ cable cars of $A$ have $k$ different starting points\nand $k$ different finishing points, and a cable car which starts higher also finishes higher.\nThe same conditions hold for $B$.\nWe say that two stations are linked by a company if one can start from the lower station\nand reach the higher one by using one or more cars of that company\n(no other movements between stations are allowed).\nDetermine the smallest positive integer $k$ for which one can guarantee\nthat there are two stations that are linked by both companies.", "solution": "Answer: $k = n^2 - n + 1$.\n\nWhen $k = n^2-n$,\nthe construction for $n=4$ is shown below which generalizes readily.\n(We draw $A$ in red and $B$ in blue.)\n[Figure omitted]\n\nTo see this is sharp, view $A$ and $B$ as graphs\nwhose connected components are paths (possibly with $0$ edges;\nthe direction of these edges is irrelevant).\nNow, if $k = n^2-n+1$ it follows that $A$ and $B$\neach have exactly $n-1$ connected components.\n\nBut in particular some component of $A$ has at least $n+1$ vertices.\nThis component has two vertices in the same component of $B$, as desired.\n\nThe main foothold for this problem is the hypothesis\nthat the number of stations should be $n^2$ rather than, say, $n$.\nThis gives a big hint towards finding the construction\nwhich in turn shows how the bound can be computed.\n\nOn the other hand, the hypothesis that\n``a cable car which starts higher\nalso finishes higher'' appears to be superfluous."}
+{"year": 2020, "problem_number": 5, "problem": "A deck of $n > 1$ cards is given.\nA positive integer is written on each card.\nThe deck has the property that the arithmetic mean of the\nnumbers on each pair of cards is also the\ngeometric mean of the numbers on some collection of one or more cards.\nFor which $n$ does it follow that the numbers on the cards are all equal?", "solution": "ik (EST)}\n.}\n\nThe assertion is true for all $n$.\n\n\\bigskip\n\n\\textbf{Setup (boilerplate).}\nSuppose that $a_1$, \\dots, $a_n$ satisfy the required properties\nbut are not all equal.\nLet $d = \\gcd(a_1, \\dots, a_n) > 1$\nthen replace $a_1$, \\dots, $a_n$ by\n$\\frac{a_1}{d}$, \\dots, $\\frac{a_n}{d}$.\nHence without loss of generality we may assume\n$ \\gcd(a_1, a_2, \\dots, a_n) = 1. $\nWLOG we also assume $ a_1 \\ge a_2 \\ge \\dots \\ge a_n. $\n\n\\bigskip\n\n\\textbf{Main proof.}\nAs $a_1 \\ge 2$, let $p$ be a prime divisor of $a_1$.\nLet $k$ be smallest index such that $p \\nmid a_k$ (which must exist).\nIn particular, note that $a_1 \\neq a_k$.\n\nConsider the mean $x = \\frac{a_1+a_k}{2}$; by assumption,\nit equals some geometric mean, hence\n$ \\sqrt[m]{a_{i_1} \\dots a_{i_m}} = \\frac{a_1 + a_k}{2} > a_k. $\nSince the arithmetic mean is an integer not divisible by $p$,\nall the indices $i_1$, $i_2$, \\dots, $i_m$\nmust be at least $k$.\nBut then the GM is at most $a_k$, contradiction.\n\nA similar approach could be attempted by using\nthe smallest numbers rather than the largest ones,\nbut one must then handle the edge case $a_n = 1$\nseparately since no prime divides $1$.\n\nSince $\\frac{27+9}{2} = 18 = \\sqrt[3]{27 \\cdot 27 \\cdot 8}$,\nit is not true that in general the AM of two largest different cards\nis not the GM of other numbers in the sequence\n(say the cards are $27, 27, 9, 8, \\dots$)."}
+{"year": 2020, "problem_number": 6, "problem": "Consider an integer $n > 1$, and a set $\\mathcal S$ of $n$ points\nin the plane such that the distance between any two different points\nin $\\mathcal S$ is at least $1$.\nProve there is a line $\\ell$ separating $\\mathcal S$\nsuch that the distance from any point of $\\mathcal S$ to $\\ell$\nis at least $\\Omega(n^{-1/3})$.\n\n(A line $\\ell$ separates a set of points $S$\nif some segment joining two points in $\\mathcal S$ crosses $\\ell$.)", "solution": "We present the official solution given by the Problem Selection Committee.\n\nLet's suppose that among all projections\nof points in $\\mathcal S$ onto some line $m$,\nthe maximum possible distance between two consecutive projections is $\\delta$.\nWe will prove that $\\delta \\ge \\Omega(n^{-1/3})$,\nsolving the problem.\n\nWe make the following the definitions:\n\nDefine $A$ and $B$ as the two points farthest apart in $\\mathcal S$.\nThis means that all points lie in the intersections\nof the circles centered at $A$ and $B$ with radius $R = AB \\ge 1$.\nWe pick chord $XY$ of $\\odot(B)$\nsuch that $XY \\perp AB$ and the distance\nfrom $A$ to $XY$ is exactly $1/2$.\nWe denote by $\\mathcal T$\nthe smaller region bound by $\\odot(B)$ and chord $XY$.\n\nThe figure is shown below with $\\mathcal T$ drawn in yellow,\nand points of $\\mathcal S$ drawn in blue.\n[Figure omitted]\n\n[Length of $AB$ + Pythagorean theorem]\nWe have $XY < 2\\sqrt{n\\delta}$.\n\nFirst, note that we have $R = AB < (n-1) \\cdot \\delta$,\nsince the $n$ projections of points onto $AB$\nare spaced at most $\\delta$ apart.\nThe Pythagorean theorem gives\n$ XY = 2\\sqrt{R^2 - (R-1/2)^2}\n= 2\\sqrt{R - \\frac14} < 2\\sqrt{n\\delta}. \\qedhere $\n\n[$|\\mathcal T|$ lower bound + narrowness]\nWe have $XY > \\frac{\\sqrt3}{2} ( 1/2 \\delta\\inv - 1 )$.\n\nBecause $\\mathcal T$ is so narrow (has width $1/2$ only),\nthe projections of points in $\\mathcal T$ onto line $XY$\nare spaced at least $\\frac{\\sqrt3}{2}$ apart (more than just $\\delta$).\nThis means\n$ XY > \\frac{\\sqrt3}{2}\n( \\left| \\mathcal T \\right| - 1 ). $\nBut projections of points in $\\mathcal T$\nonto the segment of length $1/2$ are spaced at most $\\delta$ apart,\nso apparently\n$ \\left| \\mathcal T \\right| > 1/2 \\cdot \\delta\\inv. $\nThis implies the result.\n\nCombining these two this implies $\\delta \\ge \\Omega(n^{-1/3})$ as needed.\n\nThe constant $1/3$ in the problem is actually optimal\nand cannot be improved;\nthe constructions give an example showing $\\Theta(n^{-1/3} \\log n)$."}
+{"year": 2021, "problem_number": 1, "problem": "Let $n \\ge 100$ be an integer.\nIvan writes the numbers $n, n+1, \\dots, 2n$ each on different cards.\nHe then shuffles these $n+1$ cards, and divides them into two piles.\nProve that at least one of the piles contains two cards such that\nthe sum of their numbers is a perfect square.", "solution": "We will find three cards $a < b < c$ such that\n\nb+c &= (2k+1)^2 \\\\\nc+a &= (2k)^2 \\\\\na+b &= (2k-1)^2\n\nfor some integer $k$.\nSolving for $a$, $b$, $c$ gives\n\na &= \\frac{(2k)^2+(2k-1)^2-(2k+1)^2}{2} = 2k^2 - 4k \\\\\nb &= \\frac{(2k+1)^2+(2k-1)^2-(2k)^2}{2} = 2k^2 + 1 \\\\\nc &= \\frac{(2k+1)^2+(2k)^2-(2k-1)^2}{2} = 2k^2 + 4k\n\nWe need to show that when $n \\ge 100$, one can find a suitable $k$.\n\nLet\n\nI_k &\\coloneq \\left\\{ n \\in \\ZZ \\mid n \\le a < b < c \\le 2n \\right\\} \\\\\n&= \\{ n \\in \\ZZ \\mid k^2+2k \\le n \\le 2k^2-4k \\}\n\nbe the interval such that when $n \\in I_k$,\nthe problem dies for that choice of $k$.\nIt would be sufficient to show these intervals $I_k$\ncover all the integers $\\ge 100$.\nStarting from $I_9 = \\left\\{ 99 \\le n \\le 126 \\right\\}$,\nwe have\n$ k \\ge 9 \\implies 2k^2 - 4k \\ge (k+1)^2 + 2(k+1) $\nwhich means the right endpoint of $I_k$\nexceeds the left endpoint of $I_{k+1}$.\nHence for $n \\ge 99$ in fact the problem is true.\n\nThe problem turns out to be false for $n = 98$, surprisingly.\nThe counterexample is for one pile to be\n$ \\{98,100,102,\\dots,126\\}\n\\cup \\{129,131,135,\\dots,161 \\}\n\\cup \\{162, 164, \\dots, 196\\}. $"}
+{"year": 2021, "problem_number": 2, "problem": "Show that the inequality\n$\\sum_{i=1}^n \\sum_{j=1}^n \\sqrt{|x_i-x_j|}\n\\le \\sum_{i=1}^n \\sum_{j=1}^n \\sqrt{|x_i+x_j|} $\nholds for all real numbers $x_1$, $x_2$, \\dots, $x_n$.", "solution": "The proof is by induction on $n \\ge 1$ with the base cases $n=1$ and\n$n=2$ being easy to verify by hand.\n\nIn the general situation, consider replacing the tuple $(x_i)_i$\nwith $(x_i+t)_i$ for some parameter $t \\in \\RR$.\nThe inequality becomes\n$\\sum_{i=1}^n \\sum_{j=1}^n \\sqrt{|x_i-x_j|}\n\\le \\sum_{i=1}^n \\sum_{j=1}^n \\sqrt{|x_i+x_j+2t|}. $\nThe left-hand side is independent of $t$.\n\nThe right-hand side, viewed as a function $F(t)$ of $t$,\nis minimized when $2t = -(x_i + x_j)$ for some $i$ and $j$.\n\nSince $F(t)$ is the sum of piecewise concave functions,\nit is hence itself piecewise concave.\nMoreover $F$ increases without bound if $|t| \\to \\infty$.\n\nOn each of the finitely many intervals on which $F(t)$ is\nconcave, the function is minimized at its endpoints.\nHence the minimum value must occur at one of the endpoints.\n\nIf $t = -x_i$ for some $i$, this is the same as shifting all the\nvariables so that $x_i = 0$.\nIn that case, we may apply induction on $n-1$ variables,\ndeleting the variable $x_i$.\n\nIf $t = -\\frac{x_i+x_j}{2}$, then notice\n$ x_i + t = -(x_j + t) $\nso it's the same as shifting all the variables such that $x_i = -x_j$.\nIn that case, we may apply induction on $n-2$ variables,\nafter deleting $x_i$ and $x_j$."}
+{"year": 2021, "problem_number": 3, "problem": "Let $D$ be an interior point of the acute triangle $ABC$\nwith $AB > AC$ so that $\\angle DAB = \\angle CAD$.\nThe point $E$ on the segment $AC$ satisfies $\\angle ADE =\\angle BCD$,\nthe point $F$ on the segment $AB$ satisfies $\\angle FDA =\\angle DBC$,\nand the point $X$ on the line $AC$ satisfies $CX = BX$.\nLet $O_1$ and $O_2$ be the circumcenters of the triangles\n$ADC$ and $EXD$, respectively.\nProve that the lines $BC$, $EF$, and $O_1O_2$ are concurrent.", "solution": "This solution was contributed by Abdullahil Kafi.\n\nQuadrilateral $BCEF$ is cyclic.\n\nLet $D'$ be the isogonal conjugate of the point $D$. The\nangle condition implies quadrilateral $CEDD'$ and $BFDD'$\nare cyclic. By power of point we have $ AE\\cdot AC=AD\\cdot AD'=AF\\cdot AB $\nSo $BCEF$ is cyclic.\n\nLine $ZD$ is tangent to the circles $(BCD)$ and $(DEF)$\nwhere $Z=EF\\cap BC$.\n\nLet $\\angle CAD=\\angle BAD=\\alpha$, $\\angle BCD=\\beta$,\n$\\angle DBC=\\gamma$, $\\angle ACD=\\phi$,\n$\\angle ABD=\\epsilon$.\nFrom $\\triangle ABC$ we have\n$2\\alpha+\\beta+\\gamma+\\phi+\\epsilon=180^\\circ$.\nLet $\\ell$ be a line tangent to $(BCD)$ and $K$ be a\npoint on it in the same side of $AD$ as $C$ and\n$L=AD\\cap BC$. From our labeling we have,\n\n\t \\angle AFE &= \\beta + \\phi \\qquad \\angle BFD =\n\t \\alpha + \\gamma \\qquad \\angle DFE = \\alpha + \\phi\n\t \\qquad \\angle CDL = \\alpha + \\phi\n\nNow $\\angle CDJ = 180^\\circ - \\gamma - \\beta - (\\alpha + \\phi) = \\alpha + \\epsilon$.\nSo $\\angle DFE = \\angle EDK = \\alpha + \\epsilon$, which\nmeans $\\ell$ is also tangent to $(DEF)$. Now by the\nradical center theorem we have $\\ell$ passes through\n$Z$.\n\nLet $M$ be the Miquel point of the cyclic quadrilateral\n$BCEF$. From the Miquel configuration we have $A$, $M$, $Z$\nare collinear and $(AFEM)$, $(ZCEM)$ are cyclic.\n\nPoints $B$, $X$, $M$, $E$ are cyclic.\n\nNotice that $\\angle EMB = 180^\\circ - \\angle AMB -\\angle EMZ$\n$=$ $180^\\circ - 2\\angle ACB = \\angle EXB$.\n\nLet $N$ be the other intersection of circles $(ACD)$ and\n$(DEX)$ and let $R$ be the intersection of $AC$ and $BM$.\n\n[Figure omitted]\n\nPoints $B$, $D$, $M$, $N$ are cyclic.\n\nBy power of point we have\n$\n\\opname{Pow}(R, (ACD)) = RC \\cdot RA = RM \\cdot RB\n= RE \\cdot RX = \\opname{Pow}(R, (DEX)).\n$\nHence $R$ lies on the radical axis of $(ACD)$ and\n$(DEX)$, so $N$, $R$, $D$ are collinear. Also\n$ RN \\cdot RD = RA \\cdot RC = RM \\cdot RB $ So $BDMN$\nis cyclic.\n\nNotice that $(ACD)$, $(BDMN)$, $(DEX)$ are coaxial so their\ncenters are collinear. Now we just need to prove the\ncenters of $(ACD)$, $(BDMN)$ and $Z$ are collinear. To\nprove this, take a circle $\\omega$ with radius $ZD$\ncentered at $Z$. Notice that by power of point\n$ ZC \\cdot ZB = ZD^2 = ZE \\cdot ZF = ZM \\cdot ZA $\nwhich means inversion circle $\\omega$ swaps $(ACD)$ and $(BDMN)$.\nSo the centers of $(ACD)$ and $(BDMN)$ must\nhave to be collinear with the center of inversion circle, as desired."}
+{"year": 2021, "problem_number": 4, "problem": "Let $\\Gamma$ be a circle with center $I$, and $ABCD$ a convex quadrilateral\nsuch that each of the segments $AB$, $BC$, $CD$ and $DA$ is tangent to $\\Gamma$.\nLet $\\Omega$ be the circumcircle of the triangle $AIC$.\nThe extension of $BA$ beyond $A$ meets $\\Omega$ at $X$,\nand the extension of $BC$ beyond $C$ meets $\\Omega$ at $Z$.\nThe extensions of $AD$ and $CD$ beyond $D$ meet $\\Omega$ at $Y$ and $T$, respectively.\nProve that\n$ AD + DT + TX + XA = CD + DY + YZ + ZC. $", "solution": "Let $PQRS$ be the contact points of $\\Gamma$ an $AB$, $BC$,\n$CD$, $DA$.\n\n[Figure omitted]\n\nWe have $\\triangle IQZ \\cong \\triangle IRT$.\nSimilarly, $\\triangle IPX \\cong \\triangle ISY$.\n\nBy considering $(CQIR)$ and $(CITZ)$,\nthere is a spiral similarity similarity\nmapping $\\triangle IQZ$ to $\\triangle IRT$.\nSince $IQ = IR$, it is in fact a congruence.\n\nThis congruence essentially solves the problem.\nFirst, it implies:\n\n$TX = YZ$.\n\nBecause we saw $IX = IY$ and $IT = IZ$.\n\nThen, we can compute\n\nAD + DT + XA\n&= AD + (RT - RD) + (XP-AP) \\\\\n&= (AD-RD-AP) + RT + XP = RT + XP\n\nand\n\nCD + DY + ZC &= CD + (SY-SD) + (ZQ-QC) \\\\\n&= (CD-SD-QC) + SY + ZQ = SY + ZQ\n\nbut $ZQ = RT$ and $XP = SY$, as needed."}
+{"year": 2021, "problem_number": 5, "problem": "Two squirrels, Bushy and Jumpy, have collected $2021$ walnuts for the winter.\nJumpy numbers the walnuts from $1$ through $2021$, and digs $2021$ little holes\nin a circular pattern in the ground around their favourite tree.\nThe next morning Jumpy notices that Bushy had placed one walnut into each hole,\nbut had paid no attention to the numbering.\nUnhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves.\nIn the $k$th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$.\n\nProve that there exists a value of $k$ such that, on the $k$th move,\nJumpy swaps some walnuts $a$ and $b$ such that $a<k<b$.", "solution": "Assume for contradiction no such $k$ exists.\nWe will use a so-called ``threshold trick''.\n\nThis process takes exactly $2021$ steps.\nRight after the $k$th move, we consider a situation where\nwe color walnut $k$ red as well, so at the $k$th step there are $k$ ones.\nFor brevity, a non-red walnut is called black.\nAn example is illustrated below with $2021$ replaced by $6$.\n[Figure omitted]\n\nAt each step, the walnut that becomes red is between two non-red or two red walnuts.\n\nBy definition.\n\nOn the other hand, if there are $2021$ walnuts,\none obtains a parity obstruction to this simplified process:\n\nAfter the first step, there is always a consecutive block of black walnuts positive even length.\n\nAfter the first step, there is a block of $2020$ black walnuts.\n\nThereafter, note that a length $2$ block of black walnuts can never be changed.\nMeanwhile for even lengths at least $4$, if one places a red walnut inside it,\nthe even length block splits into an odd length block and an even length block.\n\nThe statement is true with $2021$ replaced by any odd number, and false for any even number.\n\nThe motivation comes from the following rephrasing of the problem:\n\nStart with all $0$'s and at each step change a $0$ between\ntwo matching numbers from a $0$ to a $1$.\n\nAlthough the coloring (or $0$/$1$) argument may appear to lose information\nat first, I think it should be equivalent to the original process;\nthe ``extra'' information comes down to the choice of\nwhich walnut to color red at each step."}
+{"year": 2021, "problem_number": 6, "problem": "Let $m \\ge 2$ be an integer,\n$A$ a finite set of integers (not necessarily positive)\nand $B_1$, $B_2$, \\dots, $B_m$ subsets of $A$.\nSuppose that, for every $k=1,2,\\dots,m$,\nthe sum of the elements of $B_k$ is $m^k$.\nProve that $A$ contains at least $\\frac{m}{2}$ elements.", "solution": "If $0 \\le X < m^{m+1}$ is a multiple of $m$, then write it in base $m$ as\n$ X = \\sum_{i=1}^m c_i m^i \\qquad c_i \\in \\{0,1,2,\\dots,m-1\\} $\nThen swapping the summation to over $A$ through the $B_i$'s gives\n$ X = \\sum_{i = 1}^n ( \\sum_{b \\in B_i} b ) c_i\n= \\sum_{a \\in A} f_a(X) a\n\\quad\\text{where}\\quad\nf_a(X) \\coloneq \\sum_{i : a \\in B_i} c_i.\n$\n\nEvidently, $0 \\le f_a(X) \\le n(m-1)$ for any $a$ and $X$.\nSo, setting $|A| = n$, the right-hand side of the display takes on at most\n$( n(m-1) + 1 )^n$ distinct values.\nThis means\n$ m^m \\le ( n(m-1) )^n $\nwhich implies $n \\ge m/2$.\n\n[Motivation comments from USJL]\nIn linear algebra terms,\nwe have some $n$-dimensional 0/1 vectors $\\vec{v_1}$, \\dots, $\\vec{v_m}$\nand an $n$-dimensional vector $\\vec a$\nsuch that $\\vec{v_i} \\cdot \\vec a = m^i$ for $i=1, \\dots, m$.\nThe intuition is that if $n$ is too small,\nthen there should be lots of linear dependences between $\\vec{v_i}$.\n\nIn fact, Siegel's lemma is a result that says,\nif there are many more vectors than the dimension of the ambient space,\nthere exist linear dependences whose coefficients are not-too-big integers.\nOn the other hand, any linear dependence between $m$, $m^2$, \\dots, $m^m$\nis going to have coefficients that are pretty big;\nat least one of them needs to exceed $m$.\n\nApplying Siegel's lemma turns out to solve the problem\n(and is roughly equivalent to the solution above).\n\nIn ,\n\\texttt{dgrozev} shows the stronger bound\n$n \\ge (\\frac{2}{3}+\\frac{c}{\\log m} )m$ elements,\nfor some absolute constant $c > 0$."}
+{"year": 2022, "problem_number": 1, "problem": "The Bank of Oslo issues two types of coin: aluminum (denoted $A$) and bronze\n(denoted $B$). Marianne has $n$ aluminum coins and $n$ bronze coins arranged in a\nrow in some arbitrary initial order.\nA chain is any subsequence of consecutive coins of the same type.\nGiven a fixed positive integer $k \\leq 2n$,\nGilberty repeatedly performs the following operation:\nhe identifies the longest chain containing the $k$\\ts{th} coin from the left\nand moves all coins in that chain to the left end of the row.\nFor example, if $n=4$ and $k=4$, the process starting\nfrom the ordering $AABBBABA$ would be\n$AABBBABA \\to BBBAAABA \\to AAABBBBA \\to BBBBAAAA \\to \\dotsb$.\n\nFind all pairs $(n,k)$ with $1 \\leq k \\leq 2n$\nsuch that for every initial ordering,\nat some moment during the process,\nthe leftmost $n$ coins will all be of the same type.", "solution": "Answer: $n \\le k \\le \\left\\lceil \\frac 32 n \\right\\rceil$.\n\nCall a maximal chain a block.\nThen the line can be described as a sequence of blocks: it's one of:\n\n\\underbrace{A\\dots A}_{e_1}\n\\underbrace{B\\dots B}_{e_2}\n\\underbrace{A\\dots A}_{e_3}\n\\dots\n\\underbrace{A\\dots A}_{e_m} & \\text{ for odd $m$} \\\\\n\\underbrace{A\\dots A}_{e_1}\n\\underbrace{B\\dots B}_{e_2}\n\\underbrace{A\\dots A}_{e_3}\n\\dots\n\\underbrace{B\\dots B}_{e_m} & \\text{ for even $m$}\n\nor the same thing with the roles of $A$ and $B$ flipped.\n\nThe main claim is the following:\n\nThe number $m$ of blocks will never increase after an operation.\nMoreover, it stays the same if and only if\n\n$k \\le e_1$; or\n$m$ is even and $e_m \\ge 2n+1-k$.\n\nThis is obvious, just run the operation and see!\n\nThe problem asks for which values of $k$ we always reach $m=2$ eventually;\nwe already know that it's non-increasing.\nWe consider a few cases:\n\nIf $k < n$, then any configuration with $e_1 = n-1$ will never change.\nIf $k > \\left\\lceil 3n/2 \\right\\rceil$,\nthen take $m=4$ and $e_1 = e_2 = \\left\\lfloor n/2 \\right\\rfloor$\nand $e_3 = e_4 = \\left\\lceil n/2 \\right\\rceil$.\nThis configuration retains $m=4$ always:\nthe blocks simply rotate.\n\nConversely, suppose $k \\ge n$ has the property that $m > 2$ stays fixed.\nIf after the first three operations $m$ hasn't changed,\nthen we must have $m \\ge 4$ even, and $e_m, e_{m-1}, e_{m-2} \\ge 2n+1 - k$.\nNow,\n$ n \\ge e_m + e_{m-2} \\ge 2(2n+1-k) \\implies k \\ge \\frac 32 n + 1 $\nso this completes the proof."}
+{"year": 2022, "problem_number": 2, "problem": "Find all functions $f \\colon \\RR^+ \\to \\RR^+$ such that for each $x \\in \\RR^+$,\nthere is exactly one $y \\in \\RR^+$ satisfying $ xf(y)+yf(x) \\leq 2. $", "solution": "The answer is $f(x) \\equiv 1/x$ which obviously works (here $y=x$).\n\nFor the converse, assume we have $f$ such that\neach $x \\in \\RR^+$ has a friend $y$ with $xf(y)+yf(x)\\le2$.\nBy symmetry $y$ is also the friend of $x$.\n\nIn fact every number is its own friend.\n\nAssume for contradiction $a \\neq b$ are friends.\nThen we know that $af(a) + af(a) > 2 \\implies f(a) > \\frac 1a$.\nAnalogously, $f(b) > \\frac 1b$.\nHowever, we then get\n$ 2 \\ge a f(b) + b f(a) > \\frac ab + \\frac ba \\overset{\\text{AMGM}}{\\ge} 2 $\nwhich is impossible.\n\nThe problem condition now simplifies to saying\n$ f(x) \\le \\frac1x \\text{ for all $x$}, \\qquad\nxf(y) + yf(x) > 2 \\text{ for all $x \\neq y$}. $\nIn particular, for any $x>0$ and $\\eps > 0$ we have\n\n2 &< xf(x+\\eps) + (x+\\eps)f(x) \\le \\frac{x}{x+\\eps} + (x+\\eps) f(x) \\\\\n\\implies f(x) &> \\frac{x+2\\eps}{(x+\\eps)^2}\n= \\frac{1}{x + \\frac{\\eps^2}{x+2\\eps}}.\n\nSince this holds for all $\\eps > 0$ this forces $f(x) \\ge \\frac1x$ as well.\nWe're done.\n\nAlternatively, instead of using $x+\\eps$,\nit also works to consider $y = \\frac{1}{f(x)}$.\nFor such a $y$, we would have\n$ xf( \\frac{1}{f(x)} ) + \\frac{1}{f(x)} \\cdot f(x)\n= xf( \\frac{1}{f(x)} ) + 1\n\\leq x \\cdot f(x) + 1 \\leq 1 + 1 = 2 $\nwhich gives a similar contradiction."}
+{"year": 2022, "problem_number": 3, "problem": "Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers.\nProve that there is at most one way (up to rotation and reflection)\nto place the elements of $S$ around the circle such that the product\nof any two neighbors is of the form $x^2+x+k$ for some positive integer $x$.", "solution": "We replace ``positive integer $x$'' with ``nonnegative integer $x$'',\nand say numbers of the form $x^2+x+k$ are good.\nWe could also replace ``nonnegative integer $x$'' with ``integer $x$''\nowing to the obvious map $x \\mapsto 1-x$.\n\nIf $p$ is an odd prime, there are at most two odd primes $q$ and $r$\nless than $p$ for which $pq = x^2+x+k$ and $pr = y^2+y+k$ are good.\n\nMoreover, if the above occurs and $x,y \\ge 0$,\nthen $x+y+1=p$ and $xy \\equiv k \\pmod p$.\n\nThe equation $T^2+T+k \\equiv 0 \\pmod{p}$ has at most two solutions\nmodulo $p$, i.e.\\ at most two solutions in the interval $[0,p-1]$.\nBecause $0 \\le x,y < p$ from $p > \\max(q,r)$ and $k > 0$,\nthe first half follows.\n\nFor the second half,\nVieta also gives $x+y \\equiv -1 \\pmod p$ and $xy \\equiv k \\pmod p$,\nand we know $0 < x+y < 2p$.\n\nIf two such primes do exist as above, then $qr$ is also good (!).\n\nLet $pq = x^2+x+k$ and $pr = y^2+y+k$ for $x,y \\ge 0$ as before.\nFix $\\alpha \\in \\CC$ such that $\\alpha^2 + \\alpha + k = 0$;\nthen for any $n \\in \\ZZ$, we have\n$ n^2 + n + k = \\opname{Norm}(n-\\alpha). $\nHence\n$\npq \\cdot pr = \\opname{Norm}\\Big((x-\\alpha)(y-\\alpha)\\Big)\n= \\opname{Norm}\\Big( (xy-k) - (x+y+1)\\alpha\\Big)\n$\nBut $\\opname{Norm}(p) = p^2$,\nso combining with the second half of the previous claim gives\n$ qr = \\opname{Norm}(\\frac1p(xy-k)-\\alpha) $ as needed.\n\nThese two claims imply the claim directly by induction on $|S|$,\nsince one can now delete the largest element of $S$.\n\nTo show that the condition is not vacuous,\nthe author gave a ring of $385$ primes for $k=41$;\nsee ."}
+{"year": 2022, "problem_number": 4, "problem": "Let $ABCDE$ be a convex pentagon such that $BC=DE$.\nAssume that there is a point $T$ inside $ABCDE$\nwith $TB=TD$, $TC=TE$ and $\\angle ABT = \\angle TEA$.\nLet line $AB$ intersect lines $CD$ and $CT$ at points $P$ and $Q$, respectively.\nAssume that the points $P$, $B$, $A$, $Q$ occur on their line in that order.\nLet line $AE$ intersect $CD$ and $DT$ at points $R$ and $S$, respectively.\nAssume that the points $R$, $E$, $A$, $S$ occur on their line in that order.\nProve that the points $P$, $S$, $Q$, $R$ lie on a circle.", "solution": "The conditions imply\n$ \\triangle BTC \\cong \\triangle DTE,\n\\qquad\\text{and}\\qquad\n\\triangle BTY \\overset{-}{\\sim} \\triangle ETX. $\nDefine $K = CT \\cap AE$, $L = DT \\cap AB$,\n$X = BT \\cap AE$, $Y = ET \\cap BY$.\n\n[Figure omitted]\n\n[Main claim]\nWe have\n$ \\triangle BTQ \\overset{-}{\\sim} \\triangle ETS,\n\\qquad\\text{and}\\qquad\nBY:YL:LQ = EX:XK:KS. $\nIn other words, $TBYLQ \\overset{-}{\\sim} TEXKS$.\n\nWe know $\\triangle BTY \\overset{-}{\\sim} \\triangle ETX$.\nAlso, $\u2220 BTL = \u2220 BTD = \u2220 CTE = \u2220 KTE$\nand $\u2220 BTQ = \u2220 BTC = \u2220 DTE = \u2220 STE$.\n\nIt follows from the claim that:\n\n$TL/TQ = TK/TS$, ergo $TL \\cdot TS = TK \\cdot TQ$,\nso $KLSQ$ is cyclic; and\n$TC/TK = TE/TK = TB/TL = TD/TL$, so $KL \\parallel PCDR$.\n\nWith these two bullets, we're done by Reim theorem."}
+{"year": 2022, "problem_number": 5, "problem": "Find all triples $(a,b,p)$ of positive integers with $p$ prime and\n$ a^p=b!+p. $", "solution": "The answer is $(2,2,2)$ and $(3,4,3)$ only, which work.\n\nIn what follows we assume $a \\ge 2$.\n\nWe have $b \\le 2p-2$, and hence $a < p^2$.\n\nFor the first half, assume first for contradiction that $b \\ge 2p$.\nThen $b!+p \\equiv p \\pmod{p^2}$, so $\\nu_p(b!+p)=1$,\nbut $\\nu_p(a^p)=1$ never occurs.\n\nWe can also rule out $b = 2p-1$ since that would give\n$ (2p-1)!+p = p [ (p-1)! (p+1)(p+2)\\dots(2p-1) + 1 ] $\nBy Wilson theorem the inner bracket is $(-1)^2+1 \\equiv 2 \\pmod p$\nexactly, contradiction for $p > 2$.\nAnd when $p=2$, $3!+2=8$ is not a perfect square.\n\nThe second half follows as $a^p \\le (2p-2)!+p < p^{2p}$.\n(Here we used the crude estimate\n$(2p-2)! = \\prod_{k=1}^{p-1}k \\cdot (2p-1-k) < (p(p-1))^{p-1}$).\n\nWe have $a \\ge p$, and hence $b \\ge p$.\n\nFor the first half, assume for contradiction that $p > a$.\nThen\n$ b! + p = a^p \\ge a^{p-1} + p \\ge a^a + p > a! + p \\implies b > a. $\nThen taking modulo $a$ now gives $0 \\equiv 0 + p \\pmod{a}$,\nwhich is obviously impossible.\n\nThe second half follows from $b! = a^p-p \\ge p! - p > (p-1)!$.\n\nWe have $a=p$ exactly.\n\nWe know $p \\ge b$ hence $p \\mid b!+p$, so let $a = pk$ for $k < p$.\nThen $k \\mid b!$ yet $k \\nmid a^p-p$, contradiction.\n\nLet's get the small $p$ out of the way:\n\nFor $p=2$, checking $2 \\le b \\le 3$ gives $(a,b)=(2,2)$ only.\nFor $p=3$, checking $3 \\le b \\le 5$ gives $(a,b)=(3,4)$ only.\n\nOnce $p \\ge 5$, if $b! = p^p - p = p(p^{p-1}-1)$\nthen applying Zsigmondy gets a prime factor $q \\equiv 1 \\pmod{p-1}$\nwhich divides $p^{p-1}-1$.\nYet $q \\le b \\le 2p-2$ and $q \\neq p$, contradiction."}
+{"year": 2022, "problem_number": 6, "problem": "Let $n$ be a positive integer.\nA Nordic square is an $n \\times n$ board\ncontaining all the integers from $1$ to $n^2$\nso that each cell contains exactly one number.\nAn uphill path is a sequence of one or more cells such that:\n\nthe first cell in the sequence is a valley,\nmeaning the number written is less than all its orthogonal neighbors,\n\neach subsequent cell in the sequence is orthogonally\nadjacent to the previous cell, and\n\nthe numbers written in the cells in the sequence are in increasing order.", "solution": "(SRB)}\n.}\n\nAnswer: $2n^2-2n+1$.\n\nBound.: \nThe lower bound is the ``obvious'' one:\n\nFor any it generalizes to any $3 \\mid n$,\nand then to any $n$ by deleting the last $n \\bmod 3$ rows\nand either/both of the leftmost/rightmost column.\n\n[Figure omitted]\n\nPlace $1$ anywhere in $T$ and then place all the small numbers at most $|T|$\nadjacent to previously placed numbers (example above).\nThen place the remaining numbers outside $T$ arbitrarily.\n\nBy construction, as $1$ is the only valley, any uphill path must start from $1$.\nAnd by construction, it may only reach a given pair of terminal cells in one\nway, i.e.\\ the downhill paths we mentioned are the only one.\nEnd proof."}
+{"year": 2023, "problem_number": 1, "problem": "Determine all composite integers $n>1$ that satisfy the following property:\nif $d_1 < d_2 < \\dots < d_k$ are all the positive divisors of $n$ with\nthen $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \\leq i \\leq k - 2$.", "solution": "The answer is prime powers.\n\nVerification that these work.: \nWhen $n = p^e$, we get $d_i = p^{i-1}$.\nThe $i$\\ts{th} relationship reads $ p^{i-1} \\mid p^i + p^{i+1} $\nwhich is obviously true.\n\nProof that these are the only answers.: \nConversely, suppose $n$ has at least two distinct prime divisors.\nLet $p < q$ denote the two smallest ones,\nand let $p^e$ be the largest power of $p$ which both divides $n$\nand is less than $q$, hence $e \\ge 1$.\nThen the smallest factors of $n$ are $1$, $p$, \\dots, $p^e$, $q$.\nSo we are supposed to have\n$ \\frac{n}{q} \\mid \\frac{n}{p^e} + \\frac{n}{p^{e-1}}\n= \\frac{(p+1)n}{p^e} $\nwhich means that the ratio\n$ \\frac{q(p+1)}{p^e} $\nneeds to be an integer, which is obviously not possible."}
+{"year": 2023, "problem_number": 2, "problem": "Let $ABC$ be an acute-angled triangle with $AB < AC$.\nLet $\\Omega$ be the circumcircle of $ABC$.\nLet $S$ be the midpoint of the arc $CB$ of $\\Omega$ containing $A$.\nThe perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\\Omega$ again at $E \\neq A$.\nThe line through $D$ parallel to $BC$ meets line $BE$ at $L$.\nDenote the circumcircle of triangle $BDL$ by $\\omega$.\nLet $\\omega$ meet $\\Omega$ again at $P \\neq B$.\nProve that the line tangent to $\\omega$ at $P$ meets line $BS$\non the internal angle bisector of $\\angle BAC$.", "solution": "We have $LPS$ collinear.\n\nBecause $\u2220 LPB = \u2220 LDB = \u2220 CBD = \u2220 CBS = \u2220 SCB = \u2220 SPB$.\n\nLet $F$ be the antipode of $A$, so $AMFS$ is a rectangle.\n\nWe have $PDF$ collinear. (This lets us erase $L$.)\n\nBecause $\u2220 SPD = \u2220 LPD = \u2220 LBD = \u2220 SBE = \u2220 FCS = \u2220 FPS$.\n\nLet us define $X = AM \\cap BS$ and complete chord $PXQ$.\nWe aim to show that $PXQ$ is tangent to $(PDLB)$.\n\n[Figure omitted]\n\n[Main projective claim]\nWe have $XP = XA$.\n\nIntroduce $Y = PDF \\cap AM$.\nNote that\n$ -1 = (SM;EF) \\overset{A}{=} (S,X;D,AF \\cap ES) \\overset{F}{=} (\\infty X;YA) $\nwhere $\\infty = AM \\cap SF$ is at infinity (because $AMSF$ is a rectangle).\nThus, $XY = XA$.\n[Figure omitted]\nSince $\\triangle APY$ is also right, we get $XP = XA$.\n\n[Alternative proof of claim without harmonic bundles,\nfrom Solution 9 of the marking scheme]\nWith $Y = PDF \\cap AM$ defined as before, note that\n$AE \\parallel SM$ and $AM \\parallel SF$ (as $AMFS$ is a rectangle)\ngives respectively the similar triangles\n$ \\triangle AXD \\sim \\triangle MXS, \\qquad \\triangle XDY \\sim \\triangle SDF. $\nFrom this we conclude\n$ \\frac{AX}{XD} = \\frac{AX+XM}{XD+SX} = \\frac{AM}{SD} = \\frac{SF}{SD} = \\frac{XY}{XD}. $\nSo $AX = XY$ and as before we conclude $XP = XA$.\n\nFrom $XP = XA$, we conclude that $\\arc{PM}$ and $\\arc{AQ}$ have the same measure.\nSince $\\arc{AS}$ and $\\arc{EM}$ have the same measure,\nit follows $\\arc{PE}$ and $\\arc{SQ}$ have the same measure.\nThe desired tangency then follows from\n$ \u2220 QPL = \u2220 QPS = \u2220 PQE = \u2220 PFE = \u2220 PDL. $\n\n[Logical ordering]\nThis solution is split into two phases:\nthe ``synthetic phase'' where we do a bunch of angle chasing, and the\n``projective phase'' where we use cross-ratios because I like projective.\nFor logical readability (so we write in only one logical direction),\nthe projective phase is squeezed in two halves of the synthetic phase,\nbut during an actual solve it's expected to complete\nthe whole synthetic phase first (i.e.\\ to reduce the problem to show $XP=XA$).\n\nThere are quite a multitude of approaches for this problem;\nthe marking scheme for this problem at the actual IMO had 13 different solutions."}
+{"year": 2023, "problem_number": 3, "problem": "For each integer $k\\geq 2$, determine all infinite sequences of positive integers\n$a_1$, $a_2$, \\dots\\ for which there exists a polynomial $P$ of the form\n$ P(x)=x^k+c_{k-1}x^{k-1}+\\dots + c_1 x+c_0, $\nwhere $c_0$, $c_1$, \\dots, $c_{k-1}$ are non-negative integers, such that\n$ P(a_n)=a_{n+1}a_{n+2}\\dotsm a_{n+k} $\nfor every integer $n\\geq 1$.", "solution": "The answer is $a_n$ being an arithmetic progression.\nIndeed, if $a_n = d(n-1) + a_1$ for $d \\ge 0$ and $n \\ge 1$, then\n$ a_{n+1} a_{n+2} \\dots a_{n+k} = (a_n+d)(a_n+2d)\\dots(a_n+kd) $\nso we can just take $P(x) = (x+d)(x+2d) \\dots (x+kd)$.\n\nThe converse direction takes a few parts.\n\nEither $a_1 < a_2 < \\dotsb$ or the sequence is constant.\n\nNote that\n\nP(a_{n-1}) &= a_{n}a_{n+1}\\dotsm a_{n+k-1} \\\\\nP(a_n) &= a_{n+1}a_{n+2}\\dotsm a_{n+k} \\\\\n\\implies a_{n+k} &= \\frac{P(a_n)}{P(a_{n-1})} \\cdot a_n.\n\nNow the polynomial $P$ is strictly increasing over $\\NN$.\n\nSo assume for contradiction there's an index $n$ such that $a_n < a_{n-1}$.\nThen in fact the above equation shows $a_{n+k} < a_n < a_{n-1}$.\nThen there's an index $\\ell \\in [n+1,n+k]$ such that\n$a_\\ell < a_{\\ell-1}$, and also $a_\\ell < a_n$.\nContinuing in this way, we can an infinite descending subsequence of $(a_n)$,\nbut that's impossible because we assumed integers.\n\nHence we have $a_1 \\le a_2 \\le \\dotsb$.\nNow similarly, if $a_n = a_{n-1}$ for any index $n$, then $a_{n+k} = a_n$,\nergo $a_{n-1} = a_n = a_{n+1} = \\dots = a_{n+k}$.\nSo the sequence is eventually constant, and then by downwards induction,\nit is fully constant.\n\nThere exists a constant $C$ (depending only $P$, $k$)\nsuch that we have $a_{n+1} \\leq a_n + C$.\n\nLet $C$ be a constant such that $P(x) < x^k + Cx^{k-1}$ for all $x \\in \\NN$\n(for example $C = c_0 + c_1 + \\dots + c_{k-1} + 1$ works).\nWe have\n\na_{n+k} &= \\frac{P(a_n)}{a_{n+1} a_{n+2} \\dots a_{n+k-1}} \\\\\n&< \\frac{P(a_n)}{(a_n+1)(a_n+2)\\dots(a_n+k-1)} \\\\\n&< \\frac{a_n^k + C \\cdot a_n^{k-1}}{(a_n+1)(a_n+2)\\dots(a_n+k-1)} \\\\\n&< a_n + C + 1. \\qedhere\n\nAssume henceforth $a_n$ is nonconstant, and hence unbounded.\nFor each index $n$ and term $a_n$ in the sequence,\nconsider the associated differences\n$d_1 = a_{n+1} - a_n$, $d_2 = a_{n+2} - a_{n+1}$, \\dots, $d_k = a_{n+k}-a_{n+k-1}$,\nwhich we denote by\n$ \\Delta(n) \\coloneq (d_1, \\dots, d_k).$\nThis $\\Delta$ can only take up to $C^k$ different values.\nSo in particular, some tuple $(d_1, \\dots, d_n)$\nmust appear infinitely often as $\\Delta(n)$; for that tuple, we obtain\n$ P(a_N) = (a_N+d_1)(a_N+d_1+d_2) \\dots (a_N+d_1+\\dots+d_k) $\nfor infinitely many $N$.\nBut because of that, we actually must have\n$ P(X) = (X+d_1)(X+d_1+d_2) \\dots (X+d_1+\\dots+d_k). $\n\nHowever, this also means that exactly one output to $\\Delta$\noccurs infinitely often (because that output is determined by $P$).\nConsequently, it follows that $\\Delta$ is eventually constant.\nFor this to happen, $a_n$ must eventually coincide with an arithmetic\nprogression of some common difference $d$,\nand $P(X) = (X+d)(X+2d) \\dots (X+kd)$.\nFinally, this implies by downwards induction that $a_n$ is\nan arithmetic progression on all inputs."}
+{"year": 2023, "problem_number": 4, "problem": "Let $x_1$, $x_2$, \\dots, $x_{2023}$ be pairwise different positive real numbers such that\n$ a_n = \\sqrt{(x_1+x_2+\\dots+x_n)\n(\\frac{1}{x_1}+\\frac{1}{x_2}+\\dots+\\frac{1}{x_n})} $\nis an integer for every $n=1,2,\\dots,2023$. Prove that $a_{2023} \\geq 3034$.", "solution": "Note that $a_{n+1} > \\sqrt{\\sum_1^n x_i \\sum_1^n \\frac{1}{x_i}} = a_n$ for all $n$,\nso that $a_{n+1} \\geq a_n + 1$.\nObserve $a_1 = 1$.\nWe are going to prove that $ a_{2m+1} \\geq 3m+1 \\qquad \\text{for all } m \\geq 0 $\nby induction on $m$, with the base case being clear.\n\nWe now present two variations of the induction.\nThe first shorter solution compares $a_{n+2}$ directly to $a_n$,\nshowing it increases by at least $3$.\nThen we give a longer approach that compares $a_{n+1}$ to $a_n$,\nand shows it cannot increase by $1$ twice in a row.\n\nInduct-by-two solution.: \nLet $u = \\sqrt{\\frac{x_{n+1}}{x_{n+2}}} \\neq 1$.\nNote that by using Cauchy-Schwarz with three terms:\n\na_{n+2}^2 &= \\Bigg[ (x_1+\\dots+x_n)+x_{n+1}+x_{n+2} \\Bigg]\n\\Bigg[ (\\frac{1}{x_1}+\\dots+\\frac{1}{x_n})\n+\\frac{1}{x_{n+2}} + \\frac{1}{x_{n+1}} \\Bigg] \\\\\n&\\geq ( \\sqrt{ (x_1+\\dots+x_n)(\\frac{1}{x_1}+\\dots+\\frac{1}{x_n})}\n+ \\sqrt{\\frac{x_{n+1}}{x_{n+2}}} + \\sqrt{\\frac{x_{n+2}}{x_{n+1}}} )^2 \\\\\n&= ( a_n + u + \\frac 1u )^2. \\\\\n\\implies a_{n+2} &\\ge a_n + u + \\frac 1u > a_n + 2\n\nwhere the last equality $u + \\frac 1u > 2$ is by AM-GM, strict as $u \\neq 1$.\nIt follows that $a_{n+2} \\geq a_n + 3$, completing the proof.\n\nInduct-by-one solution.: \nThe main claim is:\n\nIt's impossible to have\n$a_n = c$, $a_{n+1} = c+1$, $a_{n+2} = c+2$ for any $c$ and $n$.\n\nLet $p = x_{n+1}$ and $q = x_{n+2}$ for brevity.\nLet $s = \\sum_1^n x_i$ and $t = \\sum_1^n \\frac{1}{x_n}$, so $c^2 = a_n^2 = st$.\n\nFrom $a_n = c$ and $a_{n+1} = c+1$ we have\n\n(c+1)^2 &= a_{n+1}^2 = (p+s)( \\frac 1p+t ) \\\\\n&= st + pt + \\frac1ps + 1 = c^2 + pt + \\frac1ps + 1 \\\\\n&\\overset{\\text{AM-GM}}{\\geq} c^2 + 2\\sqrt{st} + 1 = c^2 + 2\\sqrt{c^2} + 1 = (c+1)^2.\n\nHence, equality must hold in the AM-GM we must have exactly\n$ p t = \\frac 1p s = c. $\nIf we repeat the argument again on $a_{n+1}=c+1$ and $a_{n+2}=c+2$, then\n$ q ( \\frac 1p + t ) = \\frac 1q ( p + s ) = c + 1. $\nHowever this forces $\\frac pq = \\frac qp = 1$ which is impossible."}
+{"year": 2023, "problem_number": 5, "problem": "Let $n$ be a positive integer.\nA Japanese triangle consists of $1 + 2 + \\dots + n$ circles arranged in an\nequilateral triangular shape such that for each $1 \\le i \\le n$,\nthe $i$\\ts{th} row contains exactly $i$ circles, exactly one of which is colored red.\nA ninja path in a Japanese triangle is a sequence of $n$ circles\nobtained by starting in the top row, then repeatedly going from a circle to\none of the two circles immediately below it and finishing in the bottom row.\nHere is an example of a Japanese triangle with $n = 6$,\nalong with a ninja path in that triangle containing two red circles.\n[Figure omitted]\nIn terms of $n$, find the greatest $k$ such that in each Japanese triangle\nthere is a ninja path containing at least $k$ red circles.", "solution": "The answer is\n$ k = \\left\\lfloor \\log_2(n) \\right\\rfloor + 1. $\n\nConstruction.: \nIt suffices to find a Japanese triangle for $n = 2^e-1$\nwith the property that at most $e$ red circles in any ninja path.\nThe construction shown below for $e=4$ obviously generalizes,\nand works because in each of the sets $\\{1\\}$, $\\{2,3\\}$, $\\{4,5,6,7\\}$,\n\\dots, $\\{2^{e-1},\\dots,2^e-1\\}$, at most one red circle can be taken.\n(These sets are colored in different shades of red for visual clarity).\n\n[Figure omitted]\n\nBound.: \nConversely, we show that in any Japanese triangle,\none can find a ninja path containing at least\n$ k = \\left\\lfloor \\log_2(n) \\right\\rfloor + 1. $\nThe following short solution was posted at ,\napparently first found by the team leader for Iran.\n\nWe construct a rooted binary tree $T_1$ on the set of all circles as follows.\nFor each row, other than the bottom row:\n\nConnect the red circle to both circles under it;\nWhite circles to the left of the red circle in its row are connected to the left;\nWhite circles to the right of the red circle in its row are connected to the right.\n\nThe circles in the bottom row are all leaves of this tree.\nFor example, the $n=6$ construction in the beginning gives the tree\nshown on the left half of the figure below:\n[Figure omitted]\n\nNow focus on only the red circles, as shown in the right half of the figure.\nWe build a new rooted tree $T_2$ where each red circle is joined to the\nred circle below it if there was a path of (zero or more)\nwhite circles in $T_1$ between them.\nThen each red circle has at most $2$ direct descendants in $T_2$.\nHence the depth of the new tree $T_2$ exceeds $\\log_2(n)$, which produces the desired path.\n\nAnother recursive proof of bound, communicated by Helio Ng.: \nWe give another proof that $\\lfloor \\log_2 n\\rfloor + 1$ is always achievable.\nDefine $f(i, j)$ to be the maximum number of red circles contained in\nthe portion of a ninja path from $(1, 1)$ to $(i, j)$,\nincluding the endpoints $(1, 1)$ and $(i, j)$.\n(If $(i,j)$ is not a valid circle in the triangle, define $f(i, j)=0$ for convenience.)\nAn example is shown below with the values of $f(i,j)$ drawn in the circles.\n\n[Figure omitted]\n\nWe have that\n$ f(i, j) = \\max \\left\\{f(i-1,j-1), f(i,j-1) \\right\\} +\n\n1 & \\text{if $(i,j)$ is red} \\\\\n0 & \\text{otherwise}\n$\nsince every ninja path passing through $(i, j)$ also passes through\neither $(i-1,j-1)$ or $(i,j-1)$. Now consider the quantity\n$S_j = f(0, j) + \\dots + f(j, j)$. We obtain the following recurrence:\n\n$S_{j+1} \\geq S_j + \\left\\lceil \\frac{S_j}{j} \\right\\rceil + 1$.\n\nConsider a maximal element $f(m, j)$ of $ \\{f(0, j), \\dots, f(j, j) \\}$.\nWe perform the following manipulations:\n\nS_{j+1}\n&= \\sum_{i=0}^{j+1} \\max \\left\\{f(i-1,j), f(i,j) \\right\\} + \\sum_{i=0}^{j+1}\n\n1 & \\text{if $(i,j+1)$ is red} \\\\\n0 & \\text{otherwise}\n\\\\\n&= \\sum_{i=0}^{m} \\max \\left\\{f(i-1,j), f(i,j) \\right\\} +\n\\sum_{i=m+1}^{j} \\max \\left\\{f(i-1,j), f(i,j) \\right\\} + 1 \\\\\n&\\geq \\sum_{i=0}^{m} f(i,j) + \\sum_{i=m+1}^{j} f(i-1,j) + 1 \\\\\n&= S_j + f(m, j) + 1 \\\\\n&\\geq S_j + \\left\\lceil \\frac{S_j}{j} \\right\\rceil + 1\n\nwhere the last inequality is due to Pigeonhole.\n\nThis is actually enough to solve the problem.\nWrite $n = 2^c + r$, where $0 \\leq r \\leq 2^c - 1$.\n\n$S_n \\geq cn + 2r + 1$. In particular, $\\left\\lceil \\frac{S_n}{n} \\right\\rceil \\geq c + 1$.\n\nFirst note that $S_n \\geq cn + 2r + 1$ implies\n$\\left\\lceil \\frac{S_n}{n} \\right\\rceil \\geq c + 1$ because\n$ \\left\\lceil \\frac{S_n}{n} \\right\\rceil \\geq \\left\\lceil \\frac{cn + 2r + 1}{n} \\right\\rceil\n= c + \\left\\lceil \\frac{2r + 1}{n} \\right\\rceil = c + 1.\n$\nWe proceed by induction on $n$.\nThe base case $n = 1$ is clearly true as $S_1 = 1$.\nAssuming that the claim holds for some $n=j$, we have\n\nS_{j+1} &\\geq S_j + \\left\\lceil \\frac{S_j}{j} \\right\\rceil + 1 \\\\\n&\\geq cj + 2r + 1 + c + 1 + 1 \\\\\n&= c(j+1) + 2(r+1) + 1\n\nso the claim is proved for $n=j+1$ if $j+1$ is not a power of $2$.\nIf $j+1 = 2^{c+1}$, then by writing\n$c(j+1) + 2(r+1) + 1 = c(j+1) + (j+1) + 1 = (c+2)(j+1) + 1$, the claim is also proved.\n\nNow $\\left\\lceil \\frac{S_n}{n} \\right\\rceil \\geq c + 1$ implies the existence of\nsome ninja path containing at least $c+1$ red circles, and we are done."}
+{"year": 2023, "problem_number": 6, "problem": "Let $ABC$ be an equilateral triangle.\nLet $A_1$, $B_1$, $C_1$ be interior points of $ABC$\nsuch that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and\n$ \\angle BA_1C + \\angle CB_1A + \\angle AC_1B = 480\\dg. $\nLet $A_2 = BC_1 \\cap CB_1$, $B_2 = CA_1 \\cap AC_1$,\n$C_2 = AB_1 \\cap BA_1$.\nProve that if triangle $A_1B_1C_1$ is scalene,\nthen the circumcircles of triangles $AA_1A_2$, $BB_1B_2$, and $CC_1C_2$\nall pass through two common points.", "solution": "This is the second official solution from the marking scheme,\nalso communicated to me by Michael Ren.\nDefine $O$ as the center of $ABC$ and set the angles\n\n\\alpha &\\coloneq \\angle A_1CB = \\angle CBA_1 \\\\\n\\beta &\\coloneq \\angle ACB_1 = \\angle B_1AC \\\\\n\\gamma &\\coloneq \\angle C_1AB = \\angle C_1BA\n\nso that\n$ \\alpha + \\beta + \\gamma = 30\\dg. $\nIn particular, $\\max(\\alpha,\\beta,\\gamma) < 30\\dg$,\nso it follows that $A_1$ lies inside $\\triangle OBC$, and similarly for the others.\nThis means for example that $C_1$ lies between $B$ and $A_2$, and so on.\nTherefore the polygon $A_2C_1B_2A_1C_2B_1$ is convex.\n\n[Figure omitted]\n\nWe start by providing the ``interpretation'' for the $480\\dg$ angle in the statement:\n\nPoint $A_1$ is the circumcenter of $\\triangle A_2BC$, and similarly for the others.\n\nWe have $\\angle BA_1C = 180\\dg - 2\\alpha$, and\n\n\\angle BA_2C &= 180\\dg - \\angle CBC_1 - \\angle B_1CB \\\\\n&= 180\\dg - ( 60\\dg - \\gamma ) - ( 60\\dg - \\beta ) \\\\\n&= 60\\dg + \\beta + \\gamma = 90\\dg - \\alpha = 1/2 \\angle BA_1 C.\n\nSince $A_1$ lies inside $\\triangle BA_2C$, it follows $A_1$ is exactly the circumcenter.\n\nQuadrilateral $B_2C_1B_1C$ can be inscribed in a circle, say $\\gamma_a$.\nCircles $\\gamma_b$ and $\\gamma_c$ can be defined similarly.\nFinally, these three circles are pairwise distinct.\n\nUsing directed angles now, we have\n$ \u2220 B_2B_1C_2 = 180\\dg-\u2220 AB_1B_2\n= 180\\dg - 2 \u2220 ACB = 180\\dg-2(60\\dg-\\alpha) = 60\\dg+2\\alpha. $\nBy the same token, $\u2220 B_2C_1C_2 = 60\\dg+2\\alpha$.\nThis establishes the existence of $\\gamma_a$.\n\nThe proof for $\\gamma_b$ and $\\gamma_c$ is the same.\nFinally, to show the three circles are distinct, it would be enough to\nverify that the convex hexagon $A_2C_1B_2A_1C_2B_1$ is not cyclic.\n\nAssume for contradiction it was cyclic. Then\n$ 360\\dg = \\angle C_2A_1B_2 + \\angle B_2C_1A_2 + \\angle A_2B_1C_2\n= \\angle BA_1C + \\angle CB_1A + \\angle AC_1B = 480\\dg $\nwhich is absurd.\nThis contradiction eliminates the degenerate case, so the three circles are distinct.\n\nFor the remainder of the solution, let $\\opname{Pow}(P, \\omega)$\ndenote the power of a point $P$ with respect to a circle $\\omega$.\n\nLet line $AA_1$ meet $\\gamma_b$ and $\\gamma_c$ again at $X$ and $Y$,\nand set $k_a \\coloneq \\frac{AX}{AY}$.\nConsider the locus of all points $P$ such that\n$ \\mathcal C_a \\coloneq \\Big\\{ \\text{points } P \\text{ in the plane satisfying }\n\\opname{Pow}(P, \\gamma_b) = k_a \\opname{Pow}(P, \\gamma_c) \\Big\\}. $\nWe recall the coaxiality lemma\\footnote{We quickly outline a proof of this lemma:\nin the Cartesian coordinate system, the expression $\\opname{Pow}((x,y), \\omega)$\nis an expression of the form $x^2 + y^2 + {\\bullet} x + {\\bullet} y + {\\bullet}$\nfor some constants ${\\bullet}$ whose value does not matter.\nSubstituting this into the equation\n$\\frac{k_a \\opname{Pow}(P, \\gamma_c) - \\opname{Pow}(P, \\gamma_b)}{k_a-1} = 0$\ngives the equation of a circle provided $k_a \\neq 1$,\nand when $k_a = 1$, one instead recovers the radical axis.},\nwhich states that (given $\\gamma_b$ and $\\gamma_c$ are not concentric)\nthe locus $\\mathcal C_a$ must be either a circle (if $k_a \\neq 1$) or a line (if $k_a=1$).\n\nOn the other hand, $A_1$, $A_2$, and $A$ all obviously lie on $\\mathcal C_a$.\n(For $A_1$ and $A_2$, the powers are both zero, and for the point $A$,\nwe have $\\opname{Pow}(P, \\gamma_b) = AX \\cdot AA_1$\nand $\\opname{Pow}(P, \\gamma_c) = AY \\cdot AA_1$.)\nSo $\\mathcal C_a$ must be exactly the circumcircle of $\\triangle AA_1A_2$\nfrom the problem statement.\n\nWe turn to evaluating $k_a$ more carefully.\nFirst, note that\n$ \\angle A_1XB_1 = \\angle A_1B_2B_1 = \\angle CB_2B_1\n= 90\\dg - \\angle B_2AC = 90\\dg - (60\\dg-\\gamma) = 30\\dg + \\gamma. $\nNow using the law of sines, we derive\n\n\\frac{AX}{AB_1} &= \\frac{\\sin \\angle AB_1X}{\\sin \\angle AXB_1}\n= \\frac{\\sin(\\angle A_1XB_1 - \\angle XAB_1)}{\\sin \\angle A_1XB_1} \\\\\n&= \\frac{\\sin( (30\\dg+\\gamma)-(30\\dg-\\beta) )}{\\sin (30\\dg+\\gamma)}\n= \\frac{\\sin(\\beta+\\gamma)}{\\sin (30\\dg+\\gamma)}.\n\nSimilarly, $AY = AC_1 \\cdot \\frac{\\sin(\\beta+\\gamma)}{\\sin(30\\dg+\\beta)}$, so\n$ k_a = \\frac{AX}{AY}\n= \\frac{AB_1}{AC_1} \\cdot \\frac{\\sin(30\\dg+\\beta)}{\\sin(30\\dg+\\gamma)}. $\nNow define analogous constants $k_b$ and $k_c$\nand circles $\\mathcal C_b$ and $\\mathcal C_c$.\nOwing to the symmetry of the expressions, we have the key relation\n$ k_a k_b k_c = 1. $\n\nIn summary, the three circles in the problem statement may be described as\n\n\\mathcal C_a &= (AA_1A_2)\n= \\left\\{ \\text{points $P$ such that } \\opname{Pow}(P, \\gamma_b) = k_a \\opname{Pow}(P, \\gamma_c) \\right\\} \\\\\n\\mathcal C_b &= (BB_1B_2)\n= \\left\\{ \\text{points $P$ such that } \\opname{Pow}(P, \\gamma_c) = k_b \\opname{Pow}(P, \\gamma_a) \\right\\} \\\\\n\\mathcal C_c &= (CC_1C_2)\n= \\left\\{ \\text{points $P$ such that } \\opname{Pow}(P, \\gamma_a) = k_c \\opname{Pow}(P, \\gamma_b) \\right\\}.\n\nSince $k_a$, $k_b$, $k_c$ have product $1$,\nit follows that any point on at least two of the circles\nmust lie on the third circle as well.\nThe convexity of hexagon $A_2C_1B_2A_1C_2B_1$ mentioned earlier ensures these\nany two of these circles do intersect at two different points, completing the solution."}
+{"year": 2024, "problem_number": 1, "problem": "Find all real numbers $\\alpha$ so that, for every positive integer $n$, the integer\n$ \\left\\lfloor \\alpha \\right\\rfloor + \\left\\lfloor 2 \\alpha \\right\\rfloor\n+ \\left\\lfloor 3 \\alpha \\right\\rfloor + \\dots + \\left\\lfloor n \\alpha \\right\\rfloor $\nis divisible by $n$.", "solution": "The answer is that $\\alpha$ must be an even integer.\nLet $S(n, \\alpha)$ denote the sum in question.\n\nAnalysis for $\\alpha$ an integer.: \nIf $\\alpha$ is an integer, then the sum equals\n$ S(n, \\alpha) = (1+2+\\dots+n) \\alpha = \\frac{n(n+1)}{2} \\cdot \\alpha $\nwhich is obviously a multiple of $n$ if $2 \\mid \\alpha$;\nmeanwhile, if $\\alpha$ is an odd integer then $n = 2$ gives a counterexample.\n\nMain case.: \nSuppose $\\alpha$ is not an integer; we show the desired condition can never be true.\nNote that replacing $\\alpha$ with $\\alpha \\pm 2$ changes by\n$ S(n \\pm 2, \\alpha) - S(n, \\alpha) = 2(1+2+\\dots+n) = n(n+1) \\equiv 0 \\pmod n $\nfor every $n$.\nThus, by shifting appropriately we may assume $-1 < \\alpha < 1$ and $\\alpha \\notin \\ZZ$.\n\nIf $0 < \\alpha < 1$,\nthen let $m \\ge 2$ be the smallest integer such that $m \\alpha \\ge 1$.\nThen\n$ S(m, \\alpha) = \\underbrace{0 + \\dots + 0}_{m-1\\text{ terms}} + 1 = 1 $\nis not a multiple of $m$.\n\nIf $-1 < \\alpha < 0$,\nthen let $m \\ge 2$ be the smallest integer such that $m \\alpha \\le -1$.\nThen\n$ S(m, \\alpha) = \\underbrace{(-1) + \\dots + (-1)}_{m-1\\text{ terms}} + 0 = -(m-1) $\nis not a multiple of $m$."}
+{"year": 2024, "problem_number": 2, "problem": "For which pairs of positive integers $(a,b)$ is the sequence\n$ \\gcd(a^n+b, b^n+a) \\qquad n = 1, 2, \\dotsc $\neventually constant?", "solution": "The answer is $(a,b)=(1,1)$ only, which obviously works since the sequence is always $2$.\n\nConversely, assume the sequence\n$ x_n \\coloneq \\gcd(a^n+b, b^n+a) $ is eventually constant.\nThe main crux of the other direction is to consider\n$ M \\coloneq ab+1. $\n\n[Motivation]\nThe reason to consider the number is the same technique used in IMO 2005/4,\nnamely the idea to consider ``$n = -1$''.\nThe point is that the two rational numbers\n$ \\frac 1a + b = \\frac{ab+1}{b}, \\qquad \\frac 1b + a = \\frac{ab+1}{a} $\nhave a large common factor: we could write ``$x_{-1} = ab + 1$'', loosely speaking.\n\nNow, the sequence is really only defined for $n \\ge 1$,\nso one should instead take $n \\equiv -1 \\pmod{\\varphi(M)}$\n--- and this is exactly what we do.\n\nObviously $\\gcd(a,M) = \\gcd(b,M) = 1$.\nLet $n$ be a sufficiently large multiple of $\\varphi(M)$\nso that $ x_{n-1} = x_n = x_{n+1} = \\dotsb. $\nWe consider the first three terms;\nthe first one is the ``key'' one that gets the bulk of the work,\nand the rest is bookkeeping and extraction.\n\nConsider $x_{n-1}$.\nNote that\n$ a (a^{n-1} + b) = a^n + ab \\equiv 1 + (-1) \\equiv 0 \\pmod M $\nand similarly $b (b^n + a) \\equiv 0 \\pmod M$.\nHence $M \\mid x_{n-1}$.\n\nConsider $x_n$, which is now known to be divisible by $M$. Note that\n\n0 &\\equiv a^n + b \\equiv 1 + b \\pmod M \\\\\n0 &\\equiv b^n + a \\equiv 1 + a \\pmod M.\n\nSo $a \\equiv b \\equiv -1 \\pmod M$.\n\nConsider $x_{n+1}$, which is now known to be divisible by $M$. Note that\n$ 0 \\equiv a^{n+1} + b \\equiv b^{n+1} + a \\equiv a + b \\pmod M. $\nWe knew $a \\equiv b \\equiv -1 \\pmod M$,\nhence this means $0 \\equiv 2 \\pmod M$, so $M = 2$.\n\nFrom $M = 2$ we then conclude $a = b = 1$, as desired.\n\n[No alternate solutions known]\nAt the time nobody seems to know any solution not depending critically\non $M = ab+1$ (or prime numbers dividing $M$, etc.).\nThey vary in execution once some term of the form $x_{k\\varphi(n)-1}$ is taken,\nbut avoiding the key idea altogether does not currently seem possible.\n\nA good example to consider for ruling out candidate ideas is $(a,b) = (18,9)$."}
+{"year": 2024, "problem_number": 3, "problem": "Let $a_1$, $a_2$, $a_3$, \\dots\\ be an infinite sequence of positive integers,\nand let $N$ be a positive integer.\nSuppose that, for each $n > N$,\nthe number $a_n$ is equal to the number of times $a_{n-1}$ appears\nin the list $(a_1, a_2, \\dots, a_{n-1})$.\nProve that at least one of the sequences $a_1$, $a_3$, $a_5$, \\dots\\\nand $a_2$, $a_4$, $a_6$, \\dots\\ is eventually periodic.", "solution": "We present the solution from ``gigamilkmen'tgeg''\nin ,\nwith some adaptation from the first shortlist official solution as well.\nSet $M \\coloneq \\max(a_1, \\dots, a_N)$.\n\nSetup.: \nWe will visualize the entire process as follows.\nWe draw a stack of towers labeled $1$, $2$, \\dots, each initially empty.\nFor $i=1,2,\\dots$, we imagine the term $a_i$ as adding a block $B_i$ to tower $a_i$.\n\nThen there are $N$ initial blocks placed, colored red.\nThe rest of the blocks are colored yellow:\nif the last block $B_i$ was added to a tower that then reaches height $a_{i+1}$,\nthe next block $B_{i+1}$ is added to tower $a_{i+1}$.\nWe'll say $B_i$ contributes to the tower containing $B_{i+1}$.\n\nIn other words, the yellow blocks $B_i$ for $i > N$\nare given coordinates $B_i = (a_i, a_{i+1})$ for $i>N$.\nNote in particular that in towers $M+1$, $M+2$, \\dots, the blocks are all yellow.\n\n[Figure omitted]\n\nWe let $h_\\ell$ denote the height of the $\\ell$\\ts{th} tower at a given time $n$.\n(This is an abuse of notation and we should write $h_\\ell(n)$ at time $n$,\nbut $n$ will always be clear from context.)\n\nUp to alternating up and down.: \nWe start with two independent easy observations:\nthe set of numbers that occur infinitely often is downwards closed,\nand consecutive terms cannot both be huge.\n\n[Figure omitted]\n\nIf the $(k+1)$\\ts{st} tower grows arbitrarily high, so does tower $k$.\nIn fact, there exists a constant $C$ such that $h_{k} \\ge h_{k+1} - C$ at all times.\n\nSuppose $B_n$ is a yellow block in tower $k+1$.\nThen with at most finitely many exceptions, $B_{n-1}$ is a yellow block at height $k+1$,\nand the block $B_r$ right below $B_{n-1}$ is also yellow;\nthen $B_{r+1}$ is in tower $k$.\nHence, with at most finitely many exceptions, the map\n$ B_n \\mapsto B_{n-1} \\mapsto B_r \\mapsto B_{r+1} $\nprovides an injective map taking each yellow block in tower $k+1$\nto a yellow block in tower $k$.\n(The figure above shows $B_{32} \\to B_{31} \\to B_{19} \\to B_{20}$ as an example.)\n\nIf $a_n > M$ then $a_{n+1} \\le M$.\n\nAssume for contradiction there's a first moment where $a_n > M$ and $a_{n+1} > M$,\nmeaning the block $B_n$ was added to an all-yellow tower past $M$\nthat has height exceeding $M$.\n(This is the X'ed out region in the figure above.)\nIn $B_n$'s tower, every (yellow) block (including $B_n$)\nwas contributed by a block placed in different towers at height $a_n > M$.\nSo before $B_n$, there were already $a_{n+1} > M$ towers of height more than $M$.\nThis contradicts minimality of $n$.\n\nIt follows that the set of indices with $a_n \\le M$\nhas arithmetic density at least half, so certainly\nat least some of the numbers must occur infinitely often.\nOf the numbers in $\\{1,2,\\dots,M\\}$,\ndefine $L$ such that towers $1$ through $L$ grow unbounded\nbut towers $L+1$ through $M$ do not.\nThen we can pick a larger threshold $N' > N$ such that\n\nTowers $1$ through $L$ have height greater than $(M,N)$;\nTowers $L+1$ through $M$ will receive no further blocks;\n$a_{N'} \\le L$.\n\nAfter this threshold, the following statement is true:\n\n[Alternating small and big]\nThe terms $a_{N'}$, $a_{N' + 2}$, $a_{N' + 4}$, \\dots\\ are all at most $L$ while\nthe terms $a_{N' + 1}$, $a_{N' + 3}$, $a_{N' + 5}$, \\dots\\ are all greater than $M$.\n\nAutomaton for $n \\equiv N' \\pmod 2$.: \nFrom now on we always assume $n > N'$.\nWhen $n \\equiv N' \\pmod 2$, i.e., when $a_n$ is small, we define the state\n$ S(n) = (h_1, h_2, \\dots, h_L; a_n). $\nFor example, in the figure below, we illustrate how\n$ S(34) = (9,11;a_{34}=1) \\longrightarrow S(36) = (9,12;a_{36}=2) $\n[Figure omitted]\n\nThe final element $a_n$ simply reminds us which tower was most recently incremented.\nAt this point we can give a complete description of how to move from $S(n)$ to $S(n+2)$:\n\nThe intermediate block $B_{n+1}$ is placed in the tower\ncorresponding to the height $a_{n+1}$ of $B_n$;\nThat tower will have height $a_{n+2}$ equal to the number of towers\nwith height at least $a_{n+1}$; that is, it equals the cardinality of the set\n$ \\{ i \\colon h_i \\ge h_{a_n} \\} $\nWe increment $h_{a_{n+2}}$ by $1$ and update $a_n$.\n\nFor example, the illustrated $S(34) \\to S(36)$\ncorresponds to the block $B_{34}$ at height $h_1$ in tower $1$\ngiving the block $B_{35}$ at height $2$ in tower $h_1$,\nthen block $B_{36}$ at height $h_2 + 1$ being placed in tower $2$.\n\nPigeonhole periodicity argument.: \nBecause only the relative heights matter in the automata above,\nif we instead define\n$ T(n) = (h_1-h_2, h_2-h_3, \\dots, h_{L-1}-h_L; a_n). $\nthen $T(n+2)$ can be determined from just $T(n)$.\n\nSo it would be sufficient to show $T(n)$ only takes on finitely many values\nto show that $T(n)$ (and hence $a_n$) is eventually periodic.\n\nSince we have the bound $h_{k+1} \\le h_k + C$,\nwe are done upon proving the following lower bound:\n\nFor every $1 \\le \\ell < L$ and $n > N'$,\nwe have $h_\\ell \\le h_{\\ell+1} + C \\cdot (L-1)$.\n\nAssume for contradiction that there is some moment $n > N'$ such that\n$ h_\\ell > h_{\\ell+1} + C \\cdot (L-1) $\nand WLOG assume that $h_\\ell$ was just updated at the moment $n$.\nTogether with $h_{k+1} \\le h_k + C$ for all $k$ and triangle inequality, we conclude\n$ \\min(h_1, \\dots, h_\\ell) > q \\coloneq \\max(h_{\\ell+1}, \\dots, h_L). $\nWe find that the blocks now in fact alternate between being placed\namong the first $\\ell$ towers and in towers with indices greater than $q$ thereafter.\nHence the heights $h_{\\ell+1}$, \\dots, $h_L$ never grow after this moment.\nThis contradicts the definition of $L$.\n\nIn fact, it can be shown that the period is actually exactly $L$,\nmeaning the periodic part will be exactly a permutation of $(1,2,\\dots,L)$.\nFor any $L$, it turns out there is indeed a permutation achieving that periodic part."}
+{"year": 2024, "problem_number": 4, "problem": "Let triangle $ABC$ with incenter $I$ satisfying $AB < AC < BC$.\nLet $X$ be a point on line $BC$, different from $C$,\nsuch that the line through $X$ and parallel to $AC$ is tangent to the incircle.\nSimilarly, let $Y$ be a point on line $BC$, different from $B$,\nsuch that the line through $Y$ and parallel to $AB$ is tangent to the incircle.\nLine $AI$ intersects the circumcircle of triangle $ABC$ again at $P$.\nLet $K$ and $L$ be the midpoints of $AC$ and $AB$, respectively.\nProve that $\\angle KIL + \\angle YPX = 180^{\\circ}$.", "solution": "Let $T$ be the reflection of $A$ over $I$, the most important point to add\nsince it gets rid of $K$ and $L$ as follows.\n\nWe have $\\angle KIL = \\angle BTC$,\nand lines $TX$ and $TY$ are tangent to the incircle.\n\nThe first part is true since $\\triangle BTC$ is the image of $\\triangle KIL$\nunder a homothety of ratio $2$.\nThe second part is true because lines $AB$, $AC$, $TX$, $TY$\ndetermine a rhombus with center $I$.\n\nWe thus delete $K$ and $L$ from the picture altogether; they aren't needed anymore.\n\n[Figure omitted]\n\nWe have $BXPT$ and $CYPT$ are cyclic.\n\n$\u2220 TYC = \u2220 TYB = \u2220 ABC = \u2220 APC = \u2220 TPC$ and similarly.\n(Some people call this Reim's theorem.)\n\nTo finish, observe that\n$ \u2220 CTB = \u2220 CTP + \u2220 PTB = \u2220 CYP + \u2220 PXB\n= \u2220 XYP + \u2220 XYP = \u2220 XPY $\nas desired.\n(The length conditions $AC > AB > BC$ ensure that $B$, $X$, $Y$, $C$ are collinear\nin that order, and that $T$ lies on the opposite side of $BC$ as $A$.\nHence the directed equality $\u2220 CTB = \u2220 XPY$\ntranslates to the undirected $\\angle BTC + \\angle XPY = 180\\dg$.)"}
+{"year": 2024, "problem_number": 5, "problem": "Turbo the snail is in the top row of a grid with $2024$ rows and $2023$ columns\nand wants to get to the bottom row.\nHowever, there are $2022$ hidden monsters, one in every row except the first and last,\nwith no two monsters in the same column.\n\nTurbo makes a series of attempts to go from the first row to the last row.\nOn each attempt, he chooses to start on any cell in the first row,\nthen repeatedly moves to an orthogonal neighbor.\n(He is allowed to return to a previously visited cell.)\nIf Turbo reaches a cell with a monster,\nhis attempt ends and he is transported back to the first row to start a new attempt.\nThe monsters do not move between attempts, and Turbo remembers whether or not each cell\nhe has visited contains a monster.\nIf he reaches any cell in the last row, his attempt ends and Turbo wins.\n\nFind the smallest integer $n$ such that Turbo has a strategy which guarantees\nbeing able to reach the bottom row in at most $n$ attempts,\nregardless of how the monsters are placed.", "solution": "Surprisingly the answer is $n = 3$ for any grid size $s \\times (s-1)$ when $s \\ge 4$.\nWe prove this in that generality.\n\nProof that at least three attempts are needed.: \nWhen Turbo first moves into the second row, Turbo could encounter a monster $M_1$ right away.\nThen on the next attempt, Turbo must enter the third row in different column as $M_1$,\nand again could encounter a monster $M_2$ right after doing so.\nThis means no strategy can guarantee fewer than three attempts.\n\nStrategy with three attempts.: \nOn the first attempt, we have Turbo walk through the entire second row\nuntil he finds the monster $M_1$ in it.\nThen we get two possible cases.\n\n\\subparagraph{Case where $M_1$ is not on the edge.}\nIn the first case, if that monster $M_1$ is not on the edge of the row,\nthen Turbo can trace two paths below it as shown below.\nAt least one of these paths works, hence three attempts is sufficient.\n\n[Figure omitted]\n\n\\subparagraph{Case where $M_1$ is on the edge.}\nWLOG, $M_1$ is in the leftmost cell.\nThen Turbo follows the green staircase pattern shown in the left figure below.\nIf the staircase is free of monsters, then Turbo wins on the second attempt.\nOtherwise, if a monster $M_2$ is encountered on the staircase,\nTurbo has found a safe path to the left of $M_2$;\nthen Turbo can use this to reach the column $M_1$ is in, and escape from there.\nThis is shown in purple in the center and right figure\n(there are two slightly different cases depending on whether $M_2$\nwas encountered going east or south).\n[Figure omitted]\nThus the problem is solved in three attempts, as promised.\n\nExtended remark: all working strategies look similar to this.: \nAs far as we know, all working strategies are variations of the above.\nIn fact, we will try to give a description of the space of possible strategies,\nalthough this needs a bit of notation.\n\nFor simplicity, we only use $s$ even in the figures below.\nWe define the happy triangle as the following cells:\n\n\\item All $s-1$ cells in the first row (which has no monsters).\n\\item The center $s-3$ cells in the second row.\n\\item The center $s-5$ cells in the third row.\n\\item \\dots\n\\item The center cell in the $\\frac s2$\\textsuperscript{th} row.\n\nFor $s=12$, the happy triangle is the region shaded in the thick border below.\n[Figure omitted]\n\nGiven a cell, define a shoulder to be the cell directly northwest or northeast of it.\nHence there are two shoulders of cells outside the first and last column,\nand one shoulder otherwise.\n\nThen solutions roughly must distinguish between these two cases:\n\n\\item \\textbf{Inside happy triangle:}\nIf the first monster $M_1$ is found in the happy triangle,\nand there is a safe path found by Turbo to the two shoulders\n(marked $\\bigstar$ in the figure),\nthen one can finish in two more moves by considering the two paths from $\\bigstar$\nthat cut under the monster $M_1$; one of them must work.\nThis slightly generalizes the easier case in the solution above\n(which focuses only on the case where $M_1$ is in the first row).\n[Figure omitted]\n\n\\item \\textbf{Outside happy triangle:}\nNow suppose the first monster $M_1$ is outside the happy triangle.\nOf the two shoulders, take the one closer to the center\n(if in the center column, either one works; if only one shoulder, use it).\nIf there is a safe path to that shoulder,\nthen one can take a staircase pattern towards the center, as shown in the figure.\nIn that case, the choice of shoulder and position guarantees the staircase\nreaches the bottom row, so that if no monster is along this path, the algorithm ends.\nOtherwise, if one encounters a second monster along the staircase,\nthen one can use the third trial to cut under the monster $M_1$.\n[Figure omitted]\n\nWe now prove the following proposition:\nin any valid strategy for Turbo,\nin the case where Turbo first encounters a monster upon leaving the happy triangle,\nthe second path must outline the same staircase shape.\n\nThe monsters pre-commit to choosing their pattern to be\neither a NW-SE diagonal or NE-SW diagonal, with a single one-column gap;\nsee figure below for an example.\nNote that this forces any valid path for Turbo to pass through the particular gap.\n\n[Figure omitted]\n\nWe may assume without loss of generality that Turbo first encounters a monster $M_1$\nwhen Turbo first leaves the happy triangle, and that this forces an NW-SE configuration.\n\n[Figure omitted]\n\nThen the following is true:\n\nThe strategy of Turbo on the second path must\nvisit every cell in ``slightly raised diagonal'' marked with\n$\\clubsuit$ in the figure above\nin order from top to bottom, until it encounters a second Monster $M_2$\n(or reaches the bottom row and wins anyway).\nIt's both okay and irrelevant if Turbo visits other cells above this diagonal,\nbut the marked cells must be visited from top to bottom in that order.\n\nIf Turbo tries to sidestep by visiting the cell southeast of $M_1$\n(marked {X} in the Figure),\nthen Turbo clearly cannot finish after this (for $s$ large enough).\nMeanwhile, suppose Turbo tries to ``skip'' one of the $\\clubsuit$,\nsay in column $C$, then the gap could equally well be in the column to the left of $C$.\nThis proves the proposition.\n\n[Memories of safe cells are important, not just monster cells]\nHere is one additional observation that one can deduce from this.\nWe say a set $\\mathcal S$ of revealed monsters is called obviously winnable if,\nbased on only the positions of the monsters\n(and not the moves or algorithm that were used to obtain them),\none can identify a guaranteed winning path for Turbo using only $\\mathcal S$.\nFor example, two monsters in adjacent columns which are not diagonally\nadjacent is obviously winnable.\n\nThen no strategy can guarantee obtaining an obviously winnable set in $2$ moves\n(or even $k$ moves for any constant $k$, if $s$ is large enough in terms of $k$).\nSo any valid strategy must also use the memory of identified safe cells\nthat do not follow just from the revealed monster positions."}
+{"year": 2024, "problem_number": 6, "problem": "A function $f \\colon \\QQ \\to \\QQ$ is called aquaesulian\nif the following property holds: for every $x,y \\in \\mathbb{Q}$,\n$ f(x+f(y)) = f(x) + y \\quad \\text{or} \\quad f(f(x)+y) = x + f(y). $\nShow that there exists an integer $c$ such that for any aquaesulian function $f$\nthere are at most $c$ different rational numbers of the\nform $f(r) + f(-r)$ for some rational number $r$,\nand find the smallest possible value of $c$.", "solution": "We will prove that\n$ \\left\\{ f(x) + f(-x) \\mid x \\in \\QQ \\right\\} $\ncontains at most $2$ elements\nand give an example where there are indeed $2$ elements.\n\nWe fix the notation $x \\to y$ to mean that $f(x+f(y)) = f(x)+y$.\nSo the problem statement means that either $x \\to y$ or $y \\to x$ for all $x$, $y$.\nIn particular, we always have $x \\to x$, and hence\n$ f(x+f(x)) = x+f(x) $\nfor every $x$.\n\nConstruction.: \nThe function\n$ f(x) = \\left\\lfloor 2x \\right\\rfloor - x $\ncan be seen to satisfy the problem conditions.\nMoreover, $f(0)+f(0) = 0$ but $f(1/3)+f(-1/3) = -1$.\n\nHere is how I (Evan) found the construction.\nLet $h(x) \\coloneq x+f(x)$, and let $S \\coloneq h(\\QQ) = \\{h(x) \\mid x \\in \\QQ\\}$.\nHence $f$ is the identity on all of $S$.\nIf we rewrite the problem condition in terms of $h$ instead of $f$,\nit asserts that at least one of the equations\n\nh(x+h(y)-y) &= h(x)+h(y) \\\\\nh(y+h(x)-x) &= h(x)+h(y)\n\nis true.\nIn particular, $S$ is closed under addition.\n\nNow, the two trivial solutions for $h$ are $h(x) = 2x$ and $h(x) = 0$.\nTo get a nontrivial construction, we must also have $S \\neq \\{0\\}$ and $S \\neq \\QQ$.\nSo a natural guess is to take $S = \\ZZ$.\nAnd indeed $h(x) = \\left\\lfloor 2x \\right\\rfloor$ works fine.\n\nThis construction is far from unique. For example,\n$f(x) = 2\\left\\lfloor x \\right\\rfloor - x = \\left\\lfloor x \\right\\rfloor - \\{x\\}$\nseems to have been more popular to find.\n\nProof (communicated by Abel George Mathew).: \nWe start by proving:\n\n$f$ is injective.\n\nSuppose $f(a) = f(b)$. WLOG $a \\to b$. Then\n$ f(a)+a = f(a+f(a)) = f(a+f(b)) = f(a)+b \\implies a=b. \\qedhere. $\n\nSuppose $s \\to r$.\nThen either $f(r) + f(-r) = 0$ or $f(f(s)) = s+f(r)+f(-r)$.\n\nTake the given statement with $x = s+f(r)$ and $y = -r$; then\n\nx + f(y) &= s + f(r) + f(-r) \\\\\ny + f(x) &= f(s+f(r)) - r = f(s).\n\nBecause $f$ is injective, if $x \\to y$ then $f(r) + f(-r) = 0$.\nMeanwhile, if $y \\to x$ then indeed $f(f(s)) = s + f(r) + f(-r)$.\n\nFinally, suppose $a$ and $b$ are different numbers for which\n$f(a)+f(-a)$ and $f(b)+f(-b)$ are both nonzero.\nAgain, WLOG $a \\to b$.\nThen\n$ f(a) + f(-a) \\overset{a \\to a}{=} f(f(a))-a \\overset{a \\to b}{=} f(b) + f(-b). $\nThis shows at most two values can occur.\n\nThe above solution works equally well for $f \\colon \\RR \\to \\RR$.\nBut the choice of $\\QQ$ permits some additional alternate solutions.\n\nAfter showing $f$ injective,\na common lemma proved is that $-f(-f(x)) = x$, i.e.\\ $f$ is an involution.\nThis provides some alternative paths for solutions."}
+{"year": 2025, "problem_number": 1, "problem": "A line in the plane is called sunny\nif it is not parallel to any of the $x$\u2013axis, the $y$\u2013axis, or the line $x+y=0$.\n\nLet $n \\ge 3$ be a given integer.\nDetermine all nonnegative integers $k$ such that there exist $n$ distinct lines\nin the plane satisfying both of the following:\n\nfor all positive integers $a$ and $b$ with $a+b\\le n+1$,\nthe point $(a,b)$ lies on at least one of the lines; and\nexactly $k$ of the $n$ lines are sunny.", "solution": "The answer is $0$, $1$, or $3$ sunny lines.\n\nIn what follows, we draw the grid as equilateral instead of a right triangle;\nthis has no effect on the problem statement but is more symmetric.\n\nWe say a long line is one of the three lines at the edge of the grid,\ni.e.\\ one of the (non-sunny) lines passing through $n$ points.\nThe main claim is the following.\n\nIf $n \\ge 4$, any set of $n$ lines must have at least one long line.\n\nConsider the $3(n-1)$ points on the outer edge of the grid.\nIf there was no long line, each of the $n$ lines passes through at most two such points.\nSo we obtain $2n \\ge 3(n-1)$, which forces $n \\le 3$.\n\nHence, by induction we may repeatedly delete a long line without changing the number\nof sunny lines until $n = 3$ (and vice-versa: given a construction for smaller $n$\nwe can increase $n$ by one and add a long line).\n\nWe now classify all the ways to cover\nthe $1+2+3=6$ points in an $n=3$ grid with $3$ lines.\n[Figure omitted]\n\nIf there is a long line\n(say, the red one in the figure),\nthe remaining $1+2=3$ points (circled in blue) are covered with two lines.\nOne of the lines passes through $2$ points and must not be sunny;\nthe other line may or may not be sunny.\nHence in this case the possible counts of sunny lines are $0$ or $1$.\n\nIf there is no long line, each of the three lines passes through at most $2$ points.\nBut there are $6$ total lines, so in fact each line must pass through\nexactly two points.\nThe only way to do this is depicted in the figure in the right.\nIn this case there are $3$ sunny lines.\n\nThis proves that $0$, $1$, $3$ are the only possible answers.\n\nThe concept of a sunny line is not that important to the problem.\nThe proof above essentially classifies all the ways to cover\nthe $1+2+\\dots+n$ points with exactly $n$ lines.\nNamely, one should repeatedly take a long line and decrease $n$ until $n=3$,\nand then pick one of the finitely many cases for $n=3$.\nThe count of sunny lines just happens to be whatever is possible for $n=3$,\nsince long lines are not sunny."}
+{"year": 2025, "problem_number": 2, "problem": "Let $\\Omega$ and $\\Gamma$ be circles with centres $M$ and $N$, respectively,\nsuch that the radius of $\\Omega$ is less than the radius of $\\Gamma$.\nSuppose $\\Omega$ and $\\Gamma$ intersect at two distinct points $A$ and $B$.\nLine $MN$ intersects $\\Omega$ at $C$ and $\\Gamma$ at $D$,\nso that $C$, $M$, $N$, $D$ lie on $MN$ in that order.\nLet $P$ be the circumcenter of triangle $ACD$.\nLine $AP$ meets $\\Omega$ again at $E \\neq A$ and meets $\\Gamma$ again at $F \\neq A$.\nLet $H$ be the orthocenter of triangle $PMN$.\n\nProve that the line through $H$ parallel to $AP$ is tangent\nto the circumcircle of triangle $BEF$.", "solution": "Throughout the solution, we define\n\n\\alpha &\\coloneq \u2220 DCA = \u2220 BCD \\implies \u2220 PAD = \u2220 CAB = 90\\dg - \\alpha \\\\\n\\beta &\\coloneq \u2220 ADC = \u2220 CDB \\implies \u2220 CAP = \u2220 BAD = 90\\dg - \\beta.\n\nIgnore the points $H$, $M$, $N$ for now and focus on the remaining ones.\n\nWe have $CE \\parallel AD$ and $DF \\parallel AC$.\n\n$\u2220 AEC = \u2220 ABC = \u2220 CAB = 90\\dg - \\alpha$.\n\nHence, if we let $A' \\coloneq CE \\cap DF$, we have a parallelogram $ACA'D$.\nNote in particular that $BA' \\parallel CD$.\n\n[Figure omitted]\n\nNext, let $T$ denote the circumcenter of $\\triangle A'EF$.\n(This will be the tangency point later in the problem.)\n\nPoint $T$ also lies on $BA'$ and is also the arc midpoint of $\\widehat{EF}$ on $(BEF)$.\n\nWe compute the angles of $\\triangle A'EF$:\n\n\u2220 FEA' &= \u2220 AEC = \u2220 ABC = \u2220 CAB = 90\\dg - \\alpha \\\\\n\u2220 A'FE &= \u2220 DFA = \u2220 DBA = \u2220 BAD = 90\\dg - \\beta \\\\\n\u2220 EA'F &= \\alpha + \\beta.\n\nThen, since $T$ is the circumcenter, it follows that:\n$ \u2220 EA'T = \u2220 90\\dg - \u2220 A'FE = \\beta = \u2220 A'CD = \u2220 CA'B. $\nThis shows that $T$ lies on $BA'$.\n\nAlso, we have $\u2220 ETF = 2 \u2220 EA'F = 2(\\alpha+\\beta)$ and\n\n\u2220 EBF &= \u2220 EBA + \u2220 ABF = \u2220 ECA + \u2220 ADF \\\\\n&= \u2220 A'CA + \u2220 ADA' = (\\alpha+\\beta) + (\\alpha+\\beta) = 2(\\alpha+\\beta)\n\nwhich proves that $T$ also lies on $AB'$.\n\nWe then bring $M$ and $N$ into the picture as follows:\n\nPoint $T$ lies on both lines $ME$ and $NF$.\n\nTo show that $F$, $T$, $N$ are collinear,\nnote that $\\triangle FEA' \\sim \\triangle FAD$ via a homothety at $F$.\nThis homothety maps $T$ to $N$.\n\nWe now deal with point $H$ using two claims.\n\nWe have $MH \\parallel AD$ and $NH \\parallel AC$.\n\nNote that $MH \\perp PN$,\nbut $PN$ is the perpendicular bisector of $AD$,\nso in fact $MH \\parallel AD$.\nSimilarly, $NH \\parallel AC$.\n\nLines $MH$ and $NH$ bisect $\\angle NMT$ and $\\angle MNT$.\nIn fact, point $H$ is the incenter of $\\triangle TMN$,\nand $\u2220 NTH = \u2220 HTM = 90\\dg-(\\alpha+\\beta)$.\n\nHence, $\u2220 HMN = \u2220 A'CD = \u2220 ADC = \\beta$.\nBut $\u2220 TMN = \u2220 CME = 2 \u2220 CAE = -2(90\\dg-2\\beta) = 2\\beta$.\nThat proves $MH$ bisects $\\angle NMT$; the other one is similar.\n\nTo show that $H$ is an incenter (rather than an excenter)\nand get the last angle equality, we need to temporarily undirect our angles.\nAssume WLOG that $\\triangle ACD$ is directed counterclockwise.\nThe problem condition that $C$ and $D$ are the farther intersections of line $MN$\nmean that $\\angle NHM = \\angle CAD > 90\\dg$.\nWe are also promised $C$, $M$, $N$, $D$ are collinear in that order.\nHence the reflections of line $MN$ over lines $MH$ and $NH$, which meet at $T$,\nshould meet at a point for which $T$ lies on the same side as $H$.\nIn other words, $\\triangle MTN$ is oriented counterclockwise and contains $H$.\n\nWorking with undirected $\\alpha = \\angle DCA$ and $\\beta = \\angle ADC$ with $\\alpha + \\beta < 90\\dg$,\n$ \\angle NTH = \\angle HTM = 1/2 \\angle NTM = 1/2(180\\dg-2(\\alpha+\\beta))\n= 90\\dg - (\\alpha + \\beta). $\nThis matches the claim and finishes the result.\n\nNow\n$ \u2220 NFA = 90\\dg - \u2220 ADF = 90\\dg-(\\alpha+\\beta) = \u2220 NTH $\nso $HT \\parallel AP$.\nAnd since $TE = TF$, we have the tangency requested too now, as desired.\n\nThere are many other ways to describe the point $T$.\nFor example, $AMTN$ is a parallelogram and $MBTN$ is an isosceles trapezoid.\nIn coordination, we joked that it was impossible to write a false conjecture."}
+{"year": 2025, "problem_number": 3, "problem": "A function $f \\colon \\NN \\to \\NN$ is said to be bonza if\n$ f(a)\\quad\\text{divides}\\quad b^a-f(b)^{f(a)} $\nfor all positive integers $a$ and $b$.\n\nDetermine the smallest real constant $c$ such that $f(n) \\leq cn$\nfor all bonza functions $f$ and all positive integers $n$.", "solution": "The answer is $c=4$.\n\nLet $P(a,b)$ denote the given statement $f(a) \\mid b^a - f(b)^{f(a)}$.\n\nWe have $f(n) \\mid n^n$ for all $n$.\n\nTake $P(n,n)$.\n\nUnless $f = \\id$, we have $f(p) = 1$ for all odd primes $p$.\n\nConsider any prime $q$ with $f(q) > 1$.\nThen $f(q)$ is a power of $q$, and for each $n$ we get\n$ P(q,n) \\implies q \\mid f(q) \\mid n^q - f(n)^{f(q)}. $\nFermat's little theorem now gives\n$n^q \\equiv n \\pmod q$ and $f(n)^{f(q)} \\equiv f(n) \\pmod q$\n(since $f(q)$ is a power of $q$), and therefore $q \\mid n - f(n)$.\nHence, unless $f$ is the identity function,\nonly finitely many $q$ could have $f(q) > 1$.\n\nNow let $p$ be any odd prime,\nand let $q$ be a large prime such that $q \\not\\equiv 1 \\pmod p$\n(possible for all $p > 2$, say by Dirichlet).\nThen\n$ P(p,q) \\implies f(p) \\mid q^p - 1^p. $\nThe RHS is $q^p - 1 \\equiv q - 1 \\not\\equiv 0 \\pmod p$, so $f(p) = 1$.\n\nWe have $f(n) \\mid 2^\\infty$ for all $n$.\n\nIf $p \\mid f(n)$ is odd then $P(n,p)$ gives $p \\mid f(n) \\mid p^n-1^n$,\ncontradiction.\n\n(In particular, we now know $f(n) = 1$ for all odd $n$, though we don't use this.)\n\nWe have $f(n) \\le 2^{\\nu_2(n)+2}$ for all $n$.\n\nConsider $P(n,5) \\implies f(n) \\mid 5^n - 1^n$.\nIt's well-known that $\\nu_2(5^n-1) = \\nu_2(n)+2$ for all $n$.\n\nThis immediately shows $f(n) \\le 4n$ for all $n$, hence $c=4$ in the problem statement works.\n\nFor the construction, the simplest one seems to be\n$\nf(n) = \n1 & n \\text{ odd} \\\\\n16 & n = 4 \\\\\n2 & n \\text{ even}, n \\neq 4\n\n$\nwhich is easily checked to work and has $f(4) = 16$.\n\nWith a little more case analysis we can classify all functions $f$.\nThe two trivial solutions are $f(n) = n$ and $f(n) = 1$;\nthe others are described by writing $f(n) = 2^{e(n)}$ for any function $e$ satisfying\n\n$e(n) = 0$ for odd $n$;\n$1 \\le e(2) \\le 2$;\n$1 \\le e(n) \\le \\nu_2(n)+2$ for even $n > 2$.\n\nThis basically means that there are almost no additional constraints\nbeyond what is suggested by the latter two claims."}
+{"year": 2025, "problem_number": 4, "problem": "An infinite sequence $a_1$, $a_2$, \\dots\\ consists of positive integers\nhas each of which has at least three proper divisors.\nSuppose that for each $n\\geq 1$,\n$a_{n+1}$ is the sum of the three largest proper divisors of $a_n$.\nDetermine all possible values of $a_1$.", "solution": "The answer is $a_1 = 12^e \\cdot 6 \\cdot \\ell$\nfor any $e, \\ell \\ge 0$ with $\\gcd(\\ell, 10) = 1$.\n\nLet $\\mathbf{S}$ denote the set of positive integers with at least three divisors.\nFor $x \\in \\mathbf{S}$, let $\\psi(x)$ denote the sum of the three largest ones,\nso that $\\psi(a_n) = a_{n+1}$.\n\nProof that all such $a_1$ work.: \nLet $x = 12^e \\cdot 6 \\cdot \\ell \\in \\mathbf{S}$ with $\\gcd(\\ell, 10) = 1$.\nAs $\\frac12+\\frac13+\\frac14 = \\frac{13}{12}$, we get\n$ \\psi(x) = \nx & e = 0 \\\\\n\\frac{13}{12} x & e > 0\n$\nso by induction on the value of $e$ we see that $\\psi(x) \\in \\mathbf{S}$\n(the base $e=0$ coming from $\\psi$ fixing $x$).\n\nProof that all $a_1$ are of this form.: \nIn what follows $x$ is always an element of $\\mathbf{S}$,\nnot necessarily an element of the sequence.\n\nLet $x \\in \\mathbf{S}$. If $2 \\mid \\psi(x)$ then $2 \\mid x$.\n\nIf $x$ is odd then every divisor of $x$ is odd,\nso $f(x)$ is the sum of three odd numbers.\n\nLet $x \\in \\mathbf{S}$. If $6 \\mid \\psi(x)$ then $6 \\mid x$.\n\nWe consider only $x$ even because of the previous claim.\nWe prove the contrapositive that $3 \\nmid x \\implies 3 \\nmid \\psi(x)$.\n\nIf $4 \\mid x$, then letting $d$ be the third largest proper divisor of $x$,\n$ \\psi(x) = \\frac x2 + \\frac x4 + d = \\frac 34 x + d \\equiv d \\not\\equiv 0 \\pmod 3. $\nOtherwise, let $p \\mid x$ be the smallest prime dividing $x$, with $p > 3$.\nIf the third smallest nontrivial divisor of $x$ is $2p$, then\n$ \\psi(x) = \\frac x2 + \\frac xp + \\frac{x}{2p} = \\frac{3}{2p} x + \\frac x2\n\\equiv \\frac x2 \\not\\equiv 0 \\pmod 3. $\nIf the third smallest nontrivial divisor of $x$ is instead an odd prime $q$, then\n$ \\psi(x) = \\frac x2 + \\frac xp + \\frac{x}{q} \\equiv 1+0+0 \\equiv 1 \\pmod 2. $\n\nTo tie these two claims into the problem, we assert:\n\nEvery $a_i$ must be divisible by $6$.\n\nThe idea is to combine the previous two claims\n(which have no dependence on the sequence) with a size argument.\n\nFor odd $x \\in \\mathbf{S}$ note that $\\psi(x) < ( \\frac13+\\frac15+\\frac17 ) x < x$\nand $\\psi(x)$ is still odd.\nSo if any $a_i$ is odd the sequence is strictly decreasing and that's impossible.\nHence, we may assume $a_1$, $a_2$, \\dots\\ are all even.\n\nIf $x \\in \\mathbf{S}$ is even but $3 \\nmid x$ then $\\psi(x) < ( 1/2+\\frac14+\\frac15 ) x < x$\nand $\\psi(x)$ is still not a multiple of $3$.\nSo if any $a_i$ is not divisible by $3$ the sequence is again strictly decreasing.\n\\qedhere\n\nOn the other hand, if $x$ is a multiple of $6$, we have the following formula for $\\psi(x)$:\n$ \\psi(x) = \n\\frac{13}{12}x & 4 \\mid x \\\\\n\\frac{31}{30}x & 4 \\nmid x \\text{ but } 5 \\mid x \\\\\nx & 4 \\nmid x \\text{ and } 5 \\nmid x.\n$\nLooking back on our sequence of $a_i$ (which are all multiples of $6$),\nthe center case cannot happen with our $a_i$, because $\\frac{31}{30}x$\nis odd when $x \\equiv 2 \\pmod 4$.\nHence in actuality\n$ a_{n+1} = \\frac{13}{12} a_n \\quad\\text{or}\\quad a_{n+1} = a_n $\nfor every $n$.\n\nLet $T$ be the smallest index such that $a_T = a_{T+1} = a_{T+2} = \\dotsb$\n(it must exist because we cannot multiply by $\\frac{13}{12}$ forever).\nThen we can exactly describe the sequence by\n$ a_n = a_1 \\cdot ( \\frac{13}{12} )^{\\min(n,T)-1}. $\nHence $a_1 = ( \\frac{12}{13} )^{T-1} a_T$,\nand since $a_T$ is a multiple of $6$ not divisible by $4$ or $5$,\nit follows $a_1$ has the required form."}
+{"year": 2025, "problem_number": 5, "problem": "Alice and Bazza are playing the inekoalaty game,\na two\u2011player game whose rules depend on a positive real number $\\lambda$ which is known to both players.\nOn the $n$th turn of the game (starting with $n=1$) the following happens:\n\nIf $n$ is odd, Alice chooses a nonnegative real number $x_n$ such that\n$ x_1 + x_2 + \\cdots + x_n \\le \\lambda n. $\nIf $n$ is even, Bazza chooses a nonnegative real number $x_n$ such that\n$ x_1^2 + x_2^2 + \\cdots + x_n^2 \\le n. $\n\nIf a player cannot choose a suitable $x_n$, the game ends and the other player wins.\nIf the game goes on forever, neither player wins.\nAll chosen numbers are known to both players.\n\nDetermine all values of $\\lambda$ for which Alice has a winning strategy\nand all those for which Bazza has a winning strategy.", "solution": "The answer is that Alice has a winning strategy for $\\lambda > 1/\\sqrt2$,\nand Bazza has a winning strategy for $\\lambda < 1/\\sqrt2$.\n(Neither player can guarantee winning for $\\lambda = 1/\\sqrt2$.)\n\nWe divide the proof into two parts.\n\nAlice's strategy when $\\lambda \\ge 1/\\sqrt2$.: \nConsider the strategy where Alice always plays $x_{2i+1} = 0$ for $i=0, \\dots, k-1$.\n\nIn this situation, when $n = 2k+1$ we have\n\n\\sum_1^{2k} x_i &= 0 + x_2 + 0 + x_4 + \\dots + 0 + x_{2k} \\\\\n&\\le k \\cdot \\sqrt{\\frac{x_2^2 + \\dots + x_{2k}^2}{k}}\n= \\sqrt{2} \\cdot k < \\lambda \\cdot (2k+1)\n\nand so the choices for $x_{2k+1}$ are\n$ x_{2k+1} \\in [0, \\lambda \\cdot (2k+1) - \\sqrt{2} k] $\nwhich is nonempty.\nHence Alice can't ever lose with this strategy.\n\nBut suppose further $\\lambda > \\frac{1}{\\sqrt 2}$; we show Alice can win.\nChoose $k$ large enough that\n$ \\sqrt{2} \\cdot k < \\lambda \\cdot (2k+1) - \\sqrt{2k+2}. $\nThen on the $(2k+1)$st turn, Alice can (after playing $0$ on all earlier turns)\nplay a number greater than $\\sqrt{2k+2}$ and cause Bob to lose.\n\nBob strategy when $\\lambda \\le 1/\\sqrt2$.: \nConsider the strategy where Bob always plays $x_{2i+2} = \\sqrt{2 - x_{2i+1}^2}$\nfor all $i = 0, \\dots, k-1$ (i.e.\\ that is, the largest possible value Bob can play).\n\nTo analyze Bob's choices on each of his turns, we first need to estimate $x_{2k+1}$.\nWe do this by writing\n\n\\lambda \\cdot (2k+1) &\\ge x_1 + x_2 + \\dots + x_{2k+1} \\\\\n&= (x_1 + \\sqrt{2-x_1^2}) + (x_3 + \\sqrt{2-x_3^2}) \\\\\n&\\qquad+ \\dots + (x_{2k-1} + \\sqrt{2-x_{2k}^2}) + x_{2k+1} \\\\\n&\\ge {\\underbrace{\\sqrt{2} + \\dots + \\sqrt{2}}_{k \\text{ times}}} + x_{2k+1}\n= \\sqrt 2 \\cdot k + x_{2k+1}\n\nwhere we have used the fact that $t + \\sqrt{2-t^2} \\ge 2$ for all $t \\ge 0$.\nThis means that\n$ x_{2k+1} \\le \\lambda \\cdot (2k+1) - \\sqrt 2 k < \\sqrt 2. $\nAnd $x_1^2 + x_2^2 + \\dots + x_{2k}^2 + x_{2k+1}^2 = (2 + \\dots + 2) + x_{2k+1}^2$,\nBob can indeed choose $x_{2k+2} = \\sqrt{2-x_{2k+1}^2}$ and always has a move.\n\nBut suppose further $\\lambda < 1/\\sqrt2$.\nThen the above calculation also shows that Alice couldn't\nhave made a valid choice for large enough $k$,\nsince $\\lambda \\cdot (2k+1) - \\sqrt 2 k < 0$ for large $k$.\n\nIn the strategies above,\nwe saw that Alice prefers to always play $0$\nand Bob prefers to always play as large as possible.\nOne could consider what happens in the opposite case:\n\nIf Alice tries to always play the largest number possible,\nher strategy still wins for $\\lambda > 1$.\nIf Bob tries to always play $0$,\nAlice can win no matter the value for $\\lambda > 0$."}
+{"year": 2025, "problem_number": 6, "problem": "Consider a $2025\\times2025$ grid of unit squares.\nMatilda wishes to place on the grid some rectangular tiles,\npossibly of different sizes,\nsuch that each side of every tile lies on a grid line\nand every unit square is covered by at most one tile.\n\nDetermine the minimum number of tiles Matilda needs to place\nso that each row and each column of the grid has exactly one unit square\nthat is not covered by any tile.", "solution": "The answer is $2112 = 2025 + 2 \\cdot 45 - 3$.\nIn general, the answer turns out to be $\\left\\lceil n + 2 \\sqrt n - 3 \\right\\rceil$,\nbut when $n$ is not a perfect square the solution is more complicated.\n\nConstruction.: \nTo be added.\n\nBound.: \nTo be added.\n\nIronically, the construction obtaining the answer\nin the floor pattern at Sunshine Coast airport,\nthe closest airport to the site of the exam.\nSee \\href{https://i0.wp.com/havewheelchairwilltravel.net/wp-content/uploads/2019/10/IMG_6815.jpg}{this image}\nor \\href{https://stea.com.au/wp-content/uploads/2020/12/sunshine-coast-airport-design-stea-architectural-projects-20.jpg}{this image}."}