| {"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}. \\]"} | |
| {"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."} | |
| {"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."} | |
| {"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 ."} | |
| {"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."} | |
| {"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\\]"} | |