{"year": "2013", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "USAJMO", "problem": "Are there integers $a$ and $b$ such that $a^{5} b+3$ and $a b^{5}+3$ are both perfect cubes of integers?", "solution": "  No, there do not exist such $a$ and $b$. We prove this in two cases.  - Assume $3 \\mid a b$. WLOG we have $3 \\mid a$, but then $a^{5} b+3 \\equiv 3(\\bmod 9)$, contradiction. - Assume $3 \\nmid a b$. Then $a^{5} b+3$ is a cube not divisible by 3 , so it is $\\pm 1 \\bmod 9$, and we conclude  $$ a^{5} b \\in\\{5,7\\} \\quad(\\bmod 9) $$  Analogously  $$ a b^{5} \\in\\{5,7\\} \\quad(\\bmod 9) $$  We claim however these two equations cannot hold simultaneously. Indeed $(a b)^{6} \\equiv 1$ $(\\bmod 9)$ by Euler's theorem, despite $5 \\cdot 5 \\equiv 7,5 \\cdot 7 \\equiv 8,7 \\cdot 7 \\equiv 4 \\bmod 9$.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2013-notes.jsonl"}}
{"year": "2013", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "USAJMO", "problem": "Each cell of an $m \\times n$ board is filled with some nonnegative integer. Two numbers in the filling are said to be adjacent if their cells share a common side. The filling is called a garden if it satisfies the following two conditions: (i) The difference between any two adjacent numbers is either 0 or 1 . (ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to 0.  Determine the number of distinct gardens in terms of $m$ and $n$.", "solution": "  The numerical answer is $2^{m n}-1$. But we claim much more, by giving an explicit description of all gardens:  Let $S$ be any nonempty subset of the $m n$ cells. Suppose we fill each cell $\\theta$ with the minimum (taxicab) distance from $\\theta$ to some cell in $S$ (in particular, we write 0 if $\\theta \\in S$ ). Then  - This gives a garden, and - All gardens are of this form.  Since there are $2^{m n}-1$ such nonempty subsets $S$, this would finish the problem. An example of a garden with $|S|=3$ is shown below.  $$ \\left[\\begin{array}{llllll} 2 & 1 & 2 & 1 & 0 & 1 \\\\ 1 & 0 & 1 & 2 & 1 & 2 \\\\ 1 & 1 & 2 & 3 & 2 & 3 \\\\ 0 & 1 & 2 & 3 & 3 & 4 \\end{array}\\right] $$  It is actually fairly easy to see that this procedure always gives a garden; so we focus our attention on showing that every garden is of this form.  Given a garden, note first that it has at least one cell with a zero in it - by considering the minimum number across the entire garden. Now let $S$ be the (thus nonempty) set of cells with a zero written in them. We contend that this works, i.e. the following sentence holds:  Claim - If a cell $\\theta$ is labeled $d$, then the minimum distance from that cell to a cell in $S$ is $d$.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2013-notes.jsonl"}}
{"year": "2013", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "USAJMO", "problem": "In triangle $A B C$, points $P, Q, R$ lie on sides $B C, C A, A B$, respectively. Let $\\omega_{A}$, $\\omega_{B}, \\omega_{C}$ denote the circumcircles of triangles $A Q R, B R P, C P Q$, respectively. Given the fact that segment $A P$ intersects $\\omega_{A}, \\omega_{B}, \\omega_{C}$ again at $X, Y, Z$ respectively, prove that $Y X / X Z=B P / P C$.", "solution": "  Let $M$ be the concurrence point of $\\omega_{A}, \\omega_{B}, \\omega_{C}$ (by Miquel's theorem). ![](https://cdn.mathpix.com/cropped/2024_11_19_0a837349e2b78b0909e9g-5.jpg?height=704&width=812&top_left_y=873&top_left_x=622)  Then $M$ is the center of a spiral similarity sending $\\overline{Y Z}$ to $\\overline{B C}$. So it suffices to show that this spiral similarity also sends $X$ to $P$, but  $$ \\measuredangle M X Y=\\measuredangle M X A=\\measuredangle M R A=\\measuredangle M R B=\\measuredangle M P B $$  so this follows.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2013-notes.jsonl"}}
{"year": "2013", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "USAJMO", "problem": "Let $f(n)$ be the number of ways to write $n$ as a sum of powers of 2 , where we keep track of the order of the summation. For example, $f(4)=6$ because 4 can be written as $4,2+2,2+1+1,1+2+1,1+1+2$, and $1+1+1+1$. Find the smallest $n$ greater than 2013 for which $f(n)$ is odd.", "solution": "  The answer is 2047. For convenience, we agree that $f(0)=1$. Then by considering cases on the first number in the representation, we derive the recurrence  $$ f(n)=\\sum_{k=0}^{\\left\\lfloor\\log _{2} n\\right\\rfloor} f\\left(n-2^{k}\\right) $$  We wish to understand the parity of $f$. The first few values are  $$ \\begin{aligned} & f(0)=1 \\\\ & f(1)=1 \\\\ & f(2)=2 \\\\ & f(3)=3 \\\\ & f(4)=6 \\\\ & f(5)=10 \\\\ & f(6)=18 \\\\ & f(7)=31 . \\end{aligned} $$  Inspired by the data we make the key claim that Claim - $f(n)$ is odd if and only if $n+1$ is a power of 2.   This only takes a few cases:  - If $n=2^{k}$, then $(\\odot)$ has exactly two repetitive terms on the right-hand side, namely 0 and $2^{k}-1$. - If $n=2^{k}+2^{\\ell}-1$, then $(\\odot)$ has exactly two repetitive terms on the right-hand side, namely $2^{\\ell+1}-1$ and $2^{\\ell}-1$. - If $n=2^{k}-1$, then $(\\bigcirc)$ has exactly one repetitive terms on the right-hand side, namely $2^{k-1}-1$. - For other $n$, there are no repetitive terms at all on the right-hand side of $(\\Omega)$.  Thus the induction checks out. So the final answer to the problem is 2047.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2013-notes.jsonl"}}
{"year": "2013", "tier": "T3", "problem_label": "5", "problem_type": null, "exam": "USAJMO", "problem": "Quadrilateral $X A B Y$ is inscribed in the semicircle $\\omega$ with diameter $\\overline{X Y}$. Segments $A Y$ and $B X$ meet at $P$. Point $Z$ is the foot of the perpendicular from $P$ to line $\\overline{X Y}$. Point $C$ lies on $\\omega$ such that line $X C$ is perpendicular to line $A Z$. Let $Q$ be the intersection of segments $A Y$ and $X C$. Prove that  $$ \\frac{B Y}{X P}+\\frac{C Y}{X Q}=\\frac{A Y}{A X} $$", "solution": "  Let $\\beta=\\angle Y X P$ and $\\alpha=\\angle P Y X$ and set $X Y=1$. We do not direct angles in the following solution. ![](https://cdn.mathpix.com/cropped/2024_11_19_0a837349e2b78b0909e9g-7.jpg?height=552&width=1012&top_left_y=1046&top_left_x=525)  Observe that  $$ \\angle A Z X=\\angle A P X=\\alpha+\\beta $$  since $A P Z X$ is cyclic. In particular, $\\angle C X Y=90^{\\circ}-(\\alpha+\\beta)$. It is immediate that  $$ B Y=\\sin \\beta, \\quad C Y=\\cos (\\alpha+\\beta), \\quad A Y=\\cos \\alpha, \\quad A X=\\sin \\alpha $$  The Law of Sines on $\\triangle X P Y$ gives $X P=X Y \\frac{\\sin \\alpha}{\\sin (\\alpha+\\beta)}$, and on $\\triangle X Q Y$ gives $X Q=$ $X Y \\frac{\\sin \\alpha}{\\sin (90+\\beta)}=\\frac{\\sin \\alpha}{\\cos \\beta}$. So, the given is equivalent to  $$ \\frac{\\sin \\beta}{\\frac{\\sin \\alpha}{\\sin (\\alpha+\\beta)}}+\\frac{\\cos (\\alpha+\\beta)}{\\frac{\\sin \\alpha}{\\cos \\beta}}=\\frac{\\cos \\alpha}{\\sin \\alpha} $$  which is equivalent to $\\cos \\alpha=\\cos \\beta \\cos (\\alpha+\\beta)+\\sin \\beta \\sin (\\alpha+\\beta)$. This is obvious, because the right-hand side is just $\\cos ((\\alpha+\\beta)-\\beta)$.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2013-notes.jsonl"}}
{"year": "2013", "tier": "T3", "problem_label": "6", "problem_type": null, "exam": "USAJMO", "problem": "Find all real numbers $x, y, z \\geq 1$ satisfying  $$ \\min (\\sqrt{x+x y z}, \\sqrt{y+x y z}, \\sqrt{z+x y z})=\\sqrt{x-1}+\\sqrt{y-1}+\\sqrt{z-1} . $$", "solution": "  Set $x=1+a, y=1+b, z=1+c$ which eliminates the $x, y, z \\geq 1$ condition. Assume without loss of generality that $a \\leq b \\leq c$. Then the given equation rewrites as  $$ \\sqrt{(1+a)(1+(1+b)(1+c))}=\\sqrt{a}+\\sqrt{b}+\\sqrt{c} $$  In fact, we are going to prove the left-hand side always exceeds the right-hand side, and then determine the equality cases. We have:  $$ \\begin{aligned} (1+a)(1+(1+b)(1+c)) & =(a+1)(1+(b+1)(1+c)) \\\\ & \\leq(a+1)\\left(1+(\\sqrt{b}+\\sqrt{c})^{2}\\right) \\\\ & \\leq(\\sqrt{a}+(\\sqrt{b}+\\sqrt{c}))^{2} \\end{aligned} $$  by two applications of Cauchy-Schwarz. Equality holds if $b c=1$ and $1 / a=\\sqrt{b}+\\sqrt{c}$. Letting $c=t^{2}$ for $t \\geq 1$, we recover $b=t^{-2} \\leq t^{2}$ and $a=\\frac{1}{t+1 / t} \\leq t^{2}$.   $$ (x, y, z)=\\left(1+\\left(\\frac{t}{t^{2}+1}\\right)^{2}, 1+\\frac{1}{t^{2}}, 1+t^{2}\\right) $$  and permutations, for any $t>0$.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2013-notes.jsonl"}}