diff --git "a/complete.jsonl" "b/complete.jsonl" new file mode 100644--- /dev/null +++ "b/complete.jsonl" @@ -0,0 +1,174 @@ +{"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)]\n Calculate $f(m,n)$ for all positive integers $m$ and $n$\n which are either both even or both odd.\n Prove that $f(m,n) \\leq \\frac 12 \\max \\{m,n\\}$ for all $m$ and $n$.\n Show that there is no constant $C$\n such that $f(m,n) < C$ for all $m$ and $ n$.", "solution": "In general, we say the \\emph{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\n size(8cm);\n pair A = (0,5);\n pair B = (9,0);\n pair M = midpoint(A--B);\n for (int i=0; i<=5; ++i) {\n draw( (0,i)--(9,i), gray );\n }\n for (int j=0; j<=9; ++j) {\n draw( (j,0)--(j,5), gray );\n }\n dot(\"$M$\", M, dir(50));\n dot(\"$A$\", A, dir(90));\n dot(\"$B$\", B, dir(0));\n dot(\"$C$\", (0,0), dir(180));\n filldraw(A--B--(0,0)--cycle, opacity(0.1)+yellow, black+1.5);\n pair P = (0,2.5);\n pair Q = (9,2.5);\n dot(\"$P$\", P, dir(180));\n dot(\"$Q$\", Q, dir(0));\n draw(P--Q--B, blue+1.5);\n\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\n For 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\n size(8cm);\n fill( (0,4)--(5,0)--(5,4)--cycle, palered );\n draw(box( (0,0), (5,4) ), black);\n fill( (0,3)--(1,3)--(1,3.2)--(0,4)--cycle, gray);\n fill( (1,2)--(2,2)--(2,2.4)--(1.25,3)--(1,3)--cycle, gray);\n fill( (2,1)--(3,1)--(3,1.6)--(2.50,2)--(2,2)--cycle, gray);\n fill( (3,0)--(4,0)--(4,0.8)--(3.75,1)--(3,1)--cycle, gray);\n fill(shift(1,0)*unitsquare, gray);\n fill(shift(0,1)*unitsquare, gray);\n for (int i=1; i<4; ++i) {\n draw( (0,i)--(5,i), gray );\n }\n for (int i=1; i<5; ++i) {\n draw( (i,0)--(i,4), gray );\n }\n draw( (0,4)--(5,0)--(0,0)--cycle, blue+2 );\n\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 \\left[ 1^2 + 2^2 + \\dots + (2k)^2 \\right] = \\frac{4k+1}{12} \\]\nThe area of the $2k$ black polygons sums just below the diagonal to\n\\[ \\sum_{i=1}^{2k} \\left( 1\n - \\frac12 \\cdot \\frac{1}{2k+1} \\cdot \\frac{1}{2k} \\cdot i^2 \\right)\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\\lvert -k + \\frac{20k-1}{12} - \\frac{4k+1}{12} \\right\\rvert\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 $\\ol{BTV}$ meet the circle again at $U_1$,\nso that $AU_1 UB$ is an isosceles trapezoid.\nDefine $U_2$ similarly.\n\npair A = dir(110);\npair B = dir(230);\npair C = dir(310);\npair U = dir(250);\npair U_1 = A*B/U;\npair U_2 = A*C/U;\npair T = extension(B, U_1, C, U_2);\nfilldraw(unitcircle, opacity(0.1)+lightcyan, blue);\nfilldraw(A--B--C--cycle, opacity(0.1)+lightred, red);\ndraw(A--U, deepgreen);\ndraw(C--U_2, deepgreen);\ndraw(B--U_1, deepgreen);\ndraw(B--U_2, red);\ndraw(U_1--U--U_2, red);\n\ndot(\"$A$\", A, dir(A));\ndot(\"$B$\", B, dir(B));\ndot(\"$C$\", C, dir(C));\ndot(\"$U$\", U, dir(U));\ndot(\"$U_1$\", U_1, dir(U_1));\ndot(\"$U_2$\", U_2, dir(U_2));\ndot(\"$T$\", T, dir(T));\n\n/* TSQ Source:\n\nA = dir 110\nB = dir 230\nC = dir 310\nU = dir 250\nU_1 = A*B/U\nU_2 = A*C/U\nT = extension B U_1 C U_2\nunitcircle 0.1 lightcyan / blue\nA--B--C--cycle 0.1 lightred / red\nA--U deepgreen\nC--U_2 deepgreen\nB--U_1 deepgreen\n\n*/\n\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\n The 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. \\]\n Meanwhile, the \\emph{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 = \\left[ -\\frac{n+1}{2}, \\frac{n+1}{2} \\right] \\]\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 \\emph{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)]\n there is no silver matrix for $n = 1997$;\n silver matrices exist for infinitely many values of $n$.", "solution": "\\paragraph{Solution to (a).}\nDefine a \\emph{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\n\\paragraph{Solution 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\n M_1 &= \n 1 & 2 \\\\ 3 & 1\n \\\\\n M_2 &= \n {\\color{red}1} & {\\color{red}2} & 4 & 5 \\\\\n {\\color{red}3} & {\\color{red}1} & 6 & 7 \\\\\n 7 & 5 & {\\color{red}1} & {\\color{red}2} \\\\\n 6 & 4 & {\\color{red}3} & {\\color{red}1}\n \\\\\n M_3 &= \n {\\color{red}1} & {\\color{red}2} & {\\color{red}4} & {\\color{red}5} & 8 & 9 & 11 & 12\\\\\n {\\color{red}3} & {\\color{red}1} & {\\color{red}6} & {\\color{red}7} & 10 & 15 & 13 & 14 \\\\\n {\\color{red}7} & {\\color{red}5} & {\\color{red}1} & {\\color{red}2} & 14 & 12 & 8 & 9 \\\\\n {\\color{red}6} & {\\color{red}4} & {\\color{red}3} & {\\color{red}1} & 13 & 11 & 10 & 15 \\\\\n 15 & 9 & 11 & 12 & {\\color{red}1} & {\\color{red}2} & {\\color{red}4}\n & {\\color{red}5} \\\\\n 10 & 8 & 13 & 14 & {\\color{red}3} & {\\color{red}1} & {\\color{red}6}\n & {\\color{red}7} \\\\\n 14 & 12 & 15 & 9 & {\\color{red}7} & {\\color{red}5} & {\\color{red}1}\n & {\\color{red}2} \\\\\n 13 & 11 & 10 & 8 & {\\color{red}6} & {\\color{red}4} & {\\color{red}3}\n & {\\color{red}1} \\\\\n\nThe construction is described recursively as follows.\nLet\n\\[\n M_e' = \\left[\n {c|c}\n {\\color{red}M_{e-1}} & M_{e-1} + (2^e-1) \\\\ \\hline\n M_{e-1} + (2^e-1) & {\\color{red}M_{e-1}} \\\\\n\n \\right].\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\n In fact, it turns out silver matrices exist for all even dimensions.\n A 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\n Clear.\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\n If $e > 0$, then $2t^e \\cdot m > 2e \\cdot m > e+m$.\n If $e=0$ we have $m=n$ and $m = 2m$, contradiction.\n If $e = -1$ we apparently have\n \\[ \\frac{2}{t} \\cdot m = m-1 \\implies\n m = \\frac{t}{t-2} \\]\n so $(t,m) = (3,3)$ or $(t,m) = (4,2)$.\n If $e = -2$ 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}. \\]\n This gives $(t,m) = (2,2)$.\n If $e \\le -3$ 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}. \\]\n However, for $k \\ge 3$ and $t \\ge 2$,\n we always have $2k \\le t^k - 2$,\n with equality only when $(t,k) = (2,3)$;\n this 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) =\n f\\left( \\left\\lfloor \\frac N2 \\right\\rfloor \\right)\n + f\\left( \\left\\lfloor \\frac N2 \\right\\rfloor -1 \\right)\n + f\\left( \\left\\lfloor \\frac N2 \\right\\rfloor -2 \\right)\n + \\dots\n + f(1) + f(0). \\quad (\\bigstar)\n\\]\n\n\\paragraph{Upper 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} - \\half}.\n\\]\n\n\\paragraph{Lower 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]\n If $2 \\mid a+b$ then\n we have $f(2a) + f(2b) \\ge 2 f\\left( a+b \\right)$.\n\n Since $f(2k+1) = f(2k)$, we will only prove the first equation.\n Assume WLOG $a \\ge b$ and use\n $(\\bigstar)$ on all three $f$ expressions here;\n after 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). \\]\n This is true since there are an equal number of terms on each side\n and $f$ is nondecreasing.\n\n For each $1 \\le k < 2^{n-1}$, we have\n \\[ f(2^{n-1} - k) + f(k+1) \\ge 2f(2^{n-2}) \\]\n\n Use the fact that $f(2t+1)=f(2t)$ for all $t$\n and 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\nimport graph; size(10cm);\npen zzttqq = rgb(0.6,0.2,0.); pen aqaqaq = rgb(0.62745,0.62745,0.62745); pen cqcqcq = rgb(0.75294,0.75294,0.75294);\ndraw((2.28154,-9.32038)--(8.62921,-11.79133)--(2.93422,-2.54008)--cycle, linewidth(0.6) + zzttqq);\ndraw((2.28154,-9.32038)--(-2.,-4.)--(-4.,-12.)--cycle, linewidth(0.6) + zzttqq);\n\ndraw((8.62921,-11.79133)--(-2.,-4.), linewidth(0.6));\ndraw((2.93422,-2.54008)--(-4.,-12.), linewidth(0.6));\ndraw((8.62921,-11.79133)--(2.93422,-2.54008), linewidth(0.6));\ndraw((-2.,-4.)--(-4.,-12.), linewidth(0.6));\ndraw((2.28154,-9.32038)--(5.78172,-7.16571), linewidth(0.6) + blue);\ndraw((2.28154,-9.32038)--(-3.,-8.), linewidth(0.6) + blue);\ndraw((2.28154,-9.32038)--(8.62921,-11.79133), linewidth(0.6) + zzttqq);\ndraw((8.62921,-11.79133)--(2.93422,-2.54008), linewidth(0.6) + zzttqq);\ndraw((2.93422,-2.54008)--(2.28154,-9.32038), linewidth(0.6) + zzttqq);\ndraw((2.28154,-9.32038)--(-2.,-4.), linewidth(0.6) + zzttqq);\ndraw((-2.,-4.)--(-4.,-12.), linewidth(0.6) + zzttqq);\ndraw((-4.,-12.)--(2.28154,-9.32038), linewidth(0.6) + zzttqq);\ndraw((2.93422,-2.54008)--(-0.31533,2.73866), linewidth(0.6));\ndraw((-2.,-4.)--(-0.31533,2.73866), linewidth(0.6));\ndraw(circle((2.25532,-9.31383), 6.80768), linewidth(0.6) + aqaqaq);\ndraw((0.51354,-5.84245)--(-3.,-8.), linewidth(0.6) + blue);\ndraw((0.51354,-5.84245)--(5.78172,-7.16571), linewidth(0.6) + blue);\ndot(\"$D$\", (-4.,-12.), dir((-76.113, -27.810)));\ndot(\"$C$\", (-2.,-4.), dir((-75.616, 41.734)));\ndot(\"$B$\", (2.93422,-2.54008), dir((1.940, 37.397)));\ndot(\"$A$\", (8.62921,-11.79133), dir((31.751, -33.422)));\ndot(\"$P$\", (2.28154,-9.32038), dir((-11.247, -66.945)));\ndot(\"$M$\", (5.78172,-7.16571), dir((39.728, 27.050)));\ndot(\"$N$\", (-3.,-8.), dir((-82.402, 23.307)));\ndot(\"$E$\", (0.51354,-5.84245), dir((-30.585, 62.531)));\ndot(\"$X$\", (-0.31533,2.73866), dir((-23.973, 36.915)));\n\nLet $M$ and $N$ be the midpoints of $\\ol{AB}$ and $\\ol{CD}$.\n\n Unconditionally, we have $\\dang NEM = \\dang MPN$.\n\n Note that $\\ol{EN}$ is the median of right triangle $\\triangle ECD$, and similarly for $\\ol{EM}$.\n Hence $\\dang NED = \\dang EDN = \\dang BDC$, while $\\dang AEM = \\dang ACB$.\n Since $\\dang DEA = 90\\dg$, by looking at quadrilateral $XDEA$ where $X = \\ol{CD} \\cap \\ol{AB}$,\n we derive that $\\dang NED + \\dang AEM + \\dang DXA = 90\\dg$, so\n \\[ \\dang NEM = \\dang NED + \\dang AEM + 90\\dg = -\\dang DXA = -\\dang NXM = -\\dang NPM \\]\n as 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 $\\ol{ME} \\perp \\ol{CD}$.\nThis implies $\\dang BAE = \\dang CEM = \\dang EDC$ 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 \\left( \\frac{b-1}{2} \\right)^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\n If $y^2 > 7x$, then $0 < y^2-7x < xy^2+y+7$,\n contradiction.\n If $y^2=7x$, let $y = 7k$, so $x = 7k^2$.\n Plugging this back in to the original equation reads\n \\[ 343k^4 + 7k + 7 \\mid 343k^5 + 7k^2 + 7k \\]\n which is always valid, hence these are all solutions.\n If $y^2 < 7x$, then $|y^2-7x| \\le 7x$,\n so $y \\in \\{1,2\\}$.\n\n When $y=1$ we get\n \\[ x+8 \\mid x^2+x+1 \\iff x+8 \\mid 64-8+1=57.\\]\n This has solutions $x=11$ and $x=49$.\n\n When $y=2$\n\n 4x+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\n which 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\n\\paragraph{First 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\n\\paragraph{Second projective proof.}\nLet $N$ be the midpoint of $\\ol{KL}$,\nand let ray $MN$ meet the incircle again at $P$.\n\nNote that line $\\ol{RBS}$ is the polar of $N$.\nBy Brocard's theorem, lines $MK$ and $PL$ should thus\nmeet the polar of $N$, so we conclude $R = \\ol{MK} \\cap \\ol{PL}$.\nAnalogously, $S = \\ol{ML} \\cap \\ol{PK}$.\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\n We have $df(n) = f(dn)$ and $d \\cdot f(ab) = f(a) f(b)$.\n\n Let $P(m,n)$ denote the assertion in the problem statement.\n Off the bat,\n\n $P(1,1)$ implies $f(d) = d^2$.\n $P(n,1)$ implies $f(f(n)) = d^2n$.\n In particular, $f$ is injective.\n $P(1,n)$ implies $f(dn^2) = f(n)^2$.\n\n Then\n\n f(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\n This implies the first claim by taking $(a,b) = (1,n)$.\n Then $df(a) = f(da)$, and so we actually have\n $f(a) f(b) = d f(ab)$.\n\n All values of $f$ are divisible by $d$.\n\n We have\n\n f(n^2) &= \\frac 1d f(n)^2 \\\\\n f(n^3) &= \\frac{f(n^2) f(n)}{d} = \\frac{f(n)^3}{d^2} \\\\\n f(n^4) &= \\frac{f(n^3) f(n)}{d} = \\frac{f(n)^4}{d^3}\n\n and so on,\n which 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\n The IMO problem actually asked for the least value of $f(1998)$.\n But for instruction purposes,\n it is probably better to just find all $f$.\n Since $1998 = 2 \\cdot 3^3 \\cdot 37$,\n this 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\n All points of $S$ lie on a sphere $\\Gamma$ centered at $G$.\n\n Each perpendicular bisector plane passes through $G$.\n So if $A,B \\in S$ it follows $GA = GB$.\n\n Consider any plane passing through three or more points of $S$.\n The points of $S$ in the plane form a regular polygon.\n\n The cross section is a circle because we are intersecting a plane with sphere $\\Gamma$.\n Now if $A$, $B$, $C$ are three adjacent points on this circle,\n by 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\n Every face of $\\Pi$ is an equilateral triangle.\n\n Suppose on the contrary some face $A_1 A_2 \\dots A_n$ has $n > 3$.\n Let $B$ be any vertex adjacent to $A_1$ in $\\Pi$ other than $A_2$ or $A_n$.\n Consider the plane determined by $\\triangle A_1 A_3 B$.\n This is supposed to be a regular polygon, but arc $A_1 A_3$ is longer than arc $A_1 B$,\n and by construction there are no points inside these arcs.\n This 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 \\left( \\sum_{1 \\le i \\le n} x_i \\right)^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\n\\paragraph{First 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]\n Let $f(x) = x^3 - x^4$.\n If $u + v \\le \\frac 34$, then $f(u) + f(v) \\le f(0) + f(u+v)$.\n\n Note 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\n If $u+v\\le \\frac 34$ 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\n\\paragraph{Second solution by AM-GM (Nairit Sarkar).}\nWrite\n\n \\text{LHS}\n &\\le \\left( \\sum_{1 \\le k \\le n} x_k^2 \\right)\n \\left( \\sum_{1 \\le i < j \\le n} x_i x_j \\right)\n = \\half \\left( \\sum_{1 \\le k \\le n} x_k^2 \\right)\n \\left( \\sum_{1 \\le i < j \\le n} 2 x_i x_j \\right) \\\\\n &\\le \\half \\left( \\frac{\\sum_k x_k^2 + 2 \\sum_{i 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 $\\ol{PTQ} \\perp \\ol{AB}$, it follows $\\ol{LW}$ is a diameter of $\\Omega$.\nLet $O$ be its center.\n\nsize(10cm);\npen xfqqff = rgb(0.49803,0.,1.); pen qqffff = rgb(0.,1.,1.); pen cqcqcq = rgb(0.75294,0.75294,0.75294);\npair O = (0.,0.), M = (-0.54799,0.83648), P = (-0.16957,0.25884), Q = (-0.16861,-0.43170), A = (-0.97449,-0.22439), B = (0.97512,-0.22166), C = (-0.84251,0.10388), D = (0.50379,0.10577), T = (-0.16936,0.10483), L = (-0.00139,0.99999);\ndraw(circle(O, 1.), linewidth(0.6));\ndraw(circle(P, 0.69055), linewidth(0.6) + xfqqff);\ndraw(circle(Q, 0.53653), linewidth(0.6) + xfqqff);\ndraw(O--M, linewidth(0.6) + red);\ndraw(M--A, linewidth(0.6) + qqffff);\ndraw(M--B, linewidth(0.6) + qqffff);\ndraw(O--(-0.36380,-0.93147), linewidth(0.6) + red);\ndraw(P--Q, linewidth(0.6));\ndraw(M--(0.00139,-0.99999), linewidth(0.6) + green);\ndraw((0.00139,-0.99999)--L, linewidth(0.6));\ndraw(L--(-0.36380,-0.93147), linewidth(0.6) + green);\ndraw(M--(-0.36380,-0.93147), linewidth(0.6));\ndraw(A--B, linewidth(0.6));\ndot(\"$O$\", O, dir((1.868, -2.897)));\ndot(\"$N$\", (-0.36380,-0.93147), dir((-5.700, -12.058)));\ndot(\"$M$\", M, dir((-8.656, 10.478)));\ndot(\"$P$\", P, dir((-2.789, 6.521)));\ndot(\"$Q$\", Q, dir((3.600, -6.558)));\ndot(\"$A$\", A, dir((-15.478, -5.915)));\ndot(\"$B$\", B, dir((4.982, -4.507)));\ndot(\"$C$\", C, dir((-8.023, -3.199)));\ndot(\"$D$\", D, dir((1.442, 0.454)));\ndot(\"$T$\", T, dir((-7.133, 1.989)));\ndot(\"$W$\", (0.00139,-0.99999), dir((-2.113, -15.052)));\ndot(\"$L$\", L, dir((1.046, 1.812)));\ndot(\"$E$\", midpoint(A--B), dir(-45));\n\n $MNTQ$ is cyclic.\n\n By Reim: $\\dang TQM = \\dang LWM = \\dang LNM = \\dang TNM$.\n\nLet $E$ be the midpoint of $\\ol{AB}$.\n\n $OEMN$ is cyclic.\n\n By radical axis, the lines $MM$, $NN$, $AEB$ meet at a point $R$.\n Then $OEMN$ is on the circle with diameter $\\ol{OR}$.\n\n $MTE$ are collinear.\n\n $\\dang NMT = \\dang TQN = \\dang LON = \\dang NOE = \\dang NME$.\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 $\\ol{AB}$; 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 $f(x) = -\\half x^2+1$\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) = -\\half s^2 + (c + 1). \\]\nThus, the given functional equation can be rewritten as\n\\[ Q(x,s) : f(x-s)=-\\half s^2 + sx + f(x) - c. \\]\n\n [Main claim]\n We can find a function $g \\colon \\RR \\to \\RR$ such that\n \\[ f(x-z) = zx + f(x) + g(z). \\qquad (\\spadesuit). \\]\n\n If $z \\neq 0$,\n the idea is to fix a nonzero value $s_0 \\in S$ (it exists)\n and then choose $x_0$ such that $- \\half s_0^2 + s_0 x_0 - c = z$.\n Then, $Q(x_0, s)$ gives an pair $(u,v)$ with $u-v = z$.\n\n But now for any $x$, using $Q(x+v,u)$ and $Q(x,-v)$ gives\n\n f(x-z)-f(x) &= f(x-u+v)-f(x)\n = f(x+v)-f(x) + u(x+v) - \\half u^2 + c \\\\\n &= -vx-\\half s^2-c + u(x+v) - \\half u^2 + c \\\\\n &= -vx-\\half v^2 + u(x+v) - \\half u^2 = zx + g(z)\n\n where $g(z) = -\\half(u^2+v^2)$ depends only on $z$.\n\nNow, let\n\\[ h(x) \\coloneq \\half x^2 + f(x) - (2c+1), \\]\nso $h(0) = 0$.\n\n The function $h$ is additive.\n\n We just need to rewrite $(\\spadesuit)$.\n Letting $x=z$ in $(\\spadesuit)$,\n we find that actually $g(x)=f(0)-x^2-f(x)$.\n Using 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) - \\half x^2 + (2c+1) \\mid x \\in \\RR \\right\\}. \\]\nThus, we derive\n\\[ h\\left( h(x)-\\half x^2+(2c+1) \\right) = -c\n \\qquad \\forall x \\in \\RR. \\qquad(\\heartsuit) \\]\nWe can take the following two instances of $\\heartsuit$:\n\n h\\left( h(2x)-2x^2+(2c+1) \\right) &= -c \\\\\n h\\left( 2h(x)-x^2+2(2c+1) \\right) &= -2c.\n\nNow subtracting these and using $2h(x)=h(2x)$ gives\n\\[ c = h\\left( -x^2 - (2c+1) \\right). \\]\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": ""} +{"year": 2000, "problem_number": 2, "problem": "Let $a$, $b$, $c$ be positive real numbers with $abc = 1$.\nShow that\n\\[\n \\left( a - 1 + \\frac 1b \\right)\n \\left( b - 1 + \\frac 1c \\right)\n \\left( c - 1 + \\frac 1a \\right)\n \\le 1.\n\\]", "solution": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"year": 2000, "problem_number": 6, "problem": "Let $\\ol{AH_1}$, $\\ol{BH_2}$, and $\\ol{CH_3}$ 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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"year": 2001, "problem_number": 5, "problem": "Let $ABC$ be a triangle.\nLet $\\ol{AP}$ bisect $\\angle BAC$ and let $\\ol{BQ}$ bisect $\\angle ABC$,\nwith $P$ on $\\ol{BC}$ and $Q$ on $\\ol{AC}$.\nIf $AB + BP = AQ + QB$ and $\\angle BAC = 60\\dg$,\nwhat are the angles of the triangle?", "solution": ""} +{"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": ""} +{"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 \\left(t_1 + t_2 + \\dots + t_n\\right)\n \\left( \\frac{1}{t_1} + \\frac{1}{t_2} + \\dots + \\frac{1}{t_n} \\right). \\]\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": ""} +{"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": ""} +{"year": 2004, "problem_number": 6, "problem": "We call a positive integer \\emph{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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"year": 2005, "problem_number": 5, "problem": "Let $ABCD$ be a fixed convex quadrilateral\nwith $BC=DA$ and $\\ol{BC} \\nparallel \\ol{DA}$.\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": ""} +{"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": ""} +{"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": ""} +{"year": 2006, "problem_number": 2, "problem": "Let $P$ be a regular $2006$-gon.\nA diagonal is called \\emph{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 \\emph{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": ""} +{"year": 2006, "problem_number": 3, "problem": "Determine the least real number $M$ such that the inequality\n\\[ \\left\\lvert ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2) \\right\\rvert\n \\leq M\\left( a^2+b^2+c^2 \\right)^2 \\]\nholds for all real numbers $a$, $b$ and $c$.", "solution": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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)]\n Prove that for any real numbers $x_1 \\le \\dots \\le x_n$ we have\n \\[\n \\max \\left\\{ \\left\\lvert x_i - a_i \\right\\rvert :\n 1 \\le i \\le n \\right\\}\n \\ge \\half d.\n \\]\n Moreover, show that there exists some\n choice of $x_1 \\le \\dots \\le x_n$ which achieves equality.", "solution": ""} +{"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": ""} +{"year": 2007, "problem_number": 3, "problem": "In a mathematical competition some competitors are (mutual) friends.\nCall a group of competitors a \\emph{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": ""} +{"year": 2007, "problem_number": 4, "problem": "In triangle $ABC$ the bisector of $\\angle BCA$\nmeets the circumcircle again at $R$,\nthe perpendicular bisector of $\\ol{BC}$ at $P$,\nand the perpendicular bisector of $\\ol{AC}$ at $Q$.\nThe midpoint of $\\ol{BC}$ is $K$ and the midpoint of $\\ol{AC}$ is $L$.\nProve that the triangles $RPK$ and $RQL$ have the same area.", "solution": ""} +{"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": ""} +{"year": 2007, "problem_number": 6, "problem": "Let $n$ be a positive integer.\nConsider\n\\[ S = \\left\\{ (x,y,z) \\mid\n x,y,z \\in \\{ 0, 1, \\dots, n\\}, \\;\n x+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": ""} +{"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 $\\ol{BC}$ 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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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 $\\ol{BP}$, $\\ol{CQ}$, $\\ol{PQ}$.\nSuppose that $\\ol{PQ}$ is tangent to the circumcircle of $\\triangle KLM$.\nProve that $OP = OQ$.", "solution": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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 $\\ol{IF}$.\nProve that $\\ol{DG}$ and $\\ol{EI}$ intersect on $\\Gamma$.", "solution": ""} +{"year": 2010, "problem_number": 3, "problem": "Find all functions $g \\colon \\ZZ_{>0} \\to \\ZZ_{>0}$ such that\n\\[ \\left( g(m)+n \\right)\\left( g(n)+m \\right) \\]\nis always a perfect square.", "solution": ""} +{"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": ""} +{"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\n Choose a non-empty box $B_j$, $1\\leq j \\leq 5$,\n remove one coin from $B_j$ and add two coins to $B_{j+1}$;\n Choose a non-empty box $B_k$, $1\\leq k \\leq 4$,\n remove one coin from $B_k$ and swap the contents\n (possibly empty) of the boxes $B_{k+1}$ and $B_{k+2}$.", "solution": ""} +{"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\\[\n a_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\\[\n a_n =\n a_{\\ell} + a_{n - \\ell} \\text{ for all $n \\ge N$}.\n\\]", "solution": ""} +{"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": ""} +{"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 \\emph{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 \\emph{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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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 $\\ol{ST}$.", "solution": ""} +{"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": ""} +{"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)]\n If $n \\ge 2^k$, then $B$ can guarantee a win.\n For all sufficiently large $k$,\n there exists an integer $n \\ge (1.99)^k$\n such that $B$ cannot guarantee a win.", "solution": ""} +{"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": ""} +{"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 $M = \\ol{AL} \\cap \\ol{BK}$.\nProve that $MK = ML$.", "solution": ""} +{"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": ""} +{"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} = \\left( 1 + \\frac{1}{m_1} \\right) \\left( 1 + \\frac{1}{m_2} \\right)\n \\dots \\left( 1 + \\frac{1}{m_k} \\right). \\]", "solution": ""} +{"year": 2013, "problem_number": 2, "problem": "A configuration of $4027$ points in the plane is called\n\\emph{Colombian} 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 \\emph{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": ""} +{"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": ""} +{"year": 2013, "problem_number": 4, "problem": "Let $ABC$ be an acute triangle with orthocenter $H$,\nand let $W$ be a point on the side $\\ol{BC}$, 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 $\\ol{WX}$ is a diameter of $\\omega_1$.\nSimilarly, $\\omega_2$ is the circumcircle of triangle $CWM$\nand $Y$ is a point such that $\\ol{WY}$ is a diameter of $\\omega_2$.\nShow that the points $X$, $Y$, and $H$ are collinear.", "solution": ""} +{"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": ""} +{"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 \\emph{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": ""} +{"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": ""} +{"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 \\emph{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": ""} +{"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 $\\ol{BD}$.\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": ""} +{"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 $\\ol{AP}$ and $\\ol{AQ}$,\nrespectively, such that $P$ is the midpoint of $\\ol{AM}$\nand $Q$ is the midpoint of $\\ol{AN}$.\nProve that $\\ol{BM}$ and $\\ol{CN}$ meet on the\ncircumcircle of $\\triangle ABC$.", "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": ""} +{"year": 2014, "problem_number": 6, "problem": "A set of lines in the plane is in \\emph{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 \\emph{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": ""} +{"year": 2015, "problem_number": 1, "problem": "We say that a finite set $\\mathcal{S}$ of points in the plane\nis \\emph{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 \\emph{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 $n\\ge 3$,\n there exists a balanced set consisting of $n$ points.\n\\item[(b)] Determine all integers $n\\ge 3$ for which\n there exists a balanced center-free set consisting of $n$ points.", "solution": ""} +{"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": ""} +{"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 $\\ol{BC}$.\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": ""} +{"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 $K = (BDF) \\cap \\ol{AB} \\neq B$\nand $L = (CGE) \\cap \\ol{AC} \\neq C$\nand assume these points do not lie on line $FG$.\nDefine $X = \\ol{FK} \\cap \\ol{GL}$.\nProve that $X$ lies on the line $AO$.", "solution": ""} +{"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": ""} +{"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": ""} +{"year": 2016, "problem_number": 1, "problem": "In convex pentagon $ABCDE$ with $\\angle B > 90\\dg$,\nlet $F$ be a point on $\\ol{AC}$ such that $\\angle FBC = 90\\dg$.\nIt is given that $FA=FB$, $DA=DC$, $EA=ED$,\nand rays $\\ol{AC}$ and $\\ol{AD}$ trisect $\\angle BAE$.\nLet $M$ be the midpoint of $\\ol{CF}$.\nLet $X$ be the point such that $AMXE$ is a parallelogram.\nShow that $\\ol{FX}$, $\\ol{EM}$, $\\ol{BD}$ are concurrent.", "solution": ""} +{"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": ""} +{"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": ""} +{"year": 2016, "problem_number": 4, "problem": "A set of positive integers is called \\emph{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": ""} +{"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": ""} +{"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)]\n Prove that Geoff can always fulfill his wish if $n$ is odd.\n Prove that Geoff can never fulfill his wish if $n$ is even.", "solution": ""} +{"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\\[\n a_{n+1} =\n\n \\sqrt{a_n} & \\text{if $\\sqrt{a_n}$ is an integer,} \\\\\n a_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": ""} +{"year": 2017, "problem_number": 2, "problem": "Solve over $\\RR$ the functional equation\n\\[ f\\left( f(x)f(y) \\right) + f(x+y) = f(xy). \\]", "solution": ""} +{"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)]\n The rabbit moves invisibly from $A_{n-1}$ to a point $A_n$\n such that $A_{n-1} A_n = 1$.\n The hunter has a tracking device (e.g.\\ dog)\n which reports an approximate location $P_n$ of the rabbit,\n such that $P_n A_n \\le 1$.\n The hunter moves visibly from $B_{n-1}$ to a point $B_n$\n such that $B_{n-1} B_n = 1$.", "solution": ""} +{"year": 2017, "problem_number": 4, "problem": "Let $R$ and $S$ be different points on a circle $\\Omega$\nsuch that $\\ol{RS}$ is not a diameter.\nLet $\\ell$ be the tangent line to $\\Omega$ at $R$.\nPoint $T$ is such that $S$ is the midpoint of $\\ol{RT}$.\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": ""} +{"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": ""} +{"year": 2017, "problem_number": 6, "problem": "An \\emph{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\n\\emph{homogeneous} polynomial $f(x,y)$ with integer coefficients\nsuch that $f(x,y)=1$ for each $(x,y) \\in S$.", "solution": ""} +{"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 $\\ol{BD}$ and $\\ol{CE}$\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": ""} +{"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": ""} +{"year": 2018, "problem_number": 3, "problem": "An \\emph{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\n[scale = 0.8]\n\\node at (1.5,2.58) {$4$};\n\\node at (1,1.72) {$2$};\n\\node at (2,1.72) {$6$};\n\\node at (0.5,0.86) {$5$};\n\\node at (1.5,0.86) {$7$};\n\\node at (2.5,0.86) {$1$};\n\\node at (0,0) {$8$};\n\\node at (1,0) {$3$};\n\\node at (2,0) {$10$};\n\\node at (3,0) {$9$};\n\nDoes there exist an anti-Pascal triangle with $2018$ rows\nwhich contains every integer from $1$ to $1+2+\\dots +2018$?", "solution": ""} +{"year": 2018, "problem_number": 4, "problem": "A \\emph{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": ""} +{"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": ""} +{"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": ""} +{"year": 2019, "problem_number": 1, "problem": "Solve over $\\ZZ$ the functional equation\n$f(2a) + 2f(b) = f(f(a+b))$.", "solution": ""} +{"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 $\\ol{PQ} \\parallel \\ol{AB}$.\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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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 $\\ol{BC}$, $\\ol{CA}$, $\\ol{AB}$.\nThe line through $D$ perpendicular to $\\ol{EF}$ 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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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 $a1$ 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": ""} +{"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": ""} +{"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": ""} +{"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 \\left(\\frac{1}{x_1}+\\frac{1}{x_2}+\\dots+\\frac{1}{x_n}\\right)} \\]\nis an integer for every $n=1,2,\\dots,2023$. Prove that $a_{2023} \\geq 3034$.", "solution": ""} +{"year": 2023, "problem_number": 5, "problem": "Let $n$ be a positive integer.\nA \\emph{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 \\emph{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\n size(4cm);\n pair X = dir(240); pair Y = dir(0);\n path c = scale(0.5)*unitcircle;\n int[] t = {0,0,2,2,3,0};\n for (int i=0; i<=5; ++i) {\n for (int j=0; j<=i; ++j) {\n filldraw(shift(i*X+j*Y)*c, (t[i]==j) ? lightred : white);\n draw(shift(i*X+j*Y)*c);\n }\n }\n draw((0,0)--(X+Y)--(2*X+Y)--(3*X+2*Y)--(4*X+2*Y)--(5*X+2*Y),linewidth(1.5));\n path q = (3,-3sqrt(3))--(-3,-3sqrt(3));\n draw(q,Arrows(TeXHead, 1));\n label(\"$n = 6$\", q, S);\n\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": ""} +{"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 $A_2 = \\ol{BC_1} \\cap \\ol{CB_1}$, $B_2 = \\ol{CA_1} \\cap \\ol{AC_1}$,\n$C_2 = \\ol{AB_1} \\cap \\ol{BA_1}$.\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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"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": ""} +{"year": 2024, "problem_number": 6, "problem": "A function $f \\colon \\QQ \\to \\QQ$ is called \\emph{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": ""} +{"year": 2025, "problem_number": 1, "problem": "A line in the plane is called \\emph{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\n for all positive integers $a$ and $b$ with $a+b\\le n+1$,\n the point $(a,b)$ lies on at least one of the lines; and\n exactly $k$ of the $n$ lines are sunny.", "solution": ""} +{"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": ""} +{"year": 2025, "problem_number": 3, "problem": "A function $f \\colon \\NN \\to \\NN$ is said to be \\emph{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": ""} +{"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": ""} +{"year": 2025, "problem_number": 5, "problem": "Alice and Bazza are playing the \\emph{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\n If $n$ is odd, Alice chooses a nonnegative real number $x_n$ such that\n \\[ x_1 + x_2 + \\cdots + x_n \\le \\lambda n. \\]\n If $n$ 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": ""} +{"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": ""} \ No newline at end of file