year stringdate 1961-01-01 00:00:00 2025-01-01 00:00:00 β | tier stringclasses 5
values | problem_label stringclasses 119
values | problem_type stringclasses 13
values | exam stringclasses 28
values | problem stringlengths 87 2.77k | solution stringlengths 834 13k | metadata dict | problem_tokens int64 50 903 | solution_tokens int64 500 3.93k |
|---|---|---|---|---|---|---|---|---|---|
2024 | T1 | 4 | null | USAMO | Let $m$ and $n$ be positive integers. A circular necklace contains $m n$ beads, each either red or blue. It turned out that no matter how the necklace was cut into $m$ blocks of $n$ consecutive beads, each block had a distinct number of red beads. Determine, with proof, all possible values of the ordered pair $(m, n)$. | The answer is $m \leq n+1$ only. Construction when $m=n+1$. For concreteness, here is the construction for $n=4$, which obviously generalizes. The beads are listed in reading order as an array with $n+1$ rows and $n$ columns. Four of the blue beads have been labeled $B_{1}, \ldots, B_{n}$ to make them easier to track a... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl",
"solution_match": null
} | 79 | 1,005 |
2024 | T1 | 5 | null | USAMO | Point $D$ is selected inside acute triangle $A B C$ so that $\angle D A C=\angle A C B$ and $\angle B D C=90^{\circ}+\angle B A C$. Point $E$ is chosen on ray $B D$ so that $A E=E C$. Let $M$ be the midpoint of $B C$. Show that line $A B$ is tangent to the circumcircle of triangle $B E M$. | Claim - We have DQCF is cyclic. $$ \begin{aligned} \measuredangle F D C & =-\measuredangle C D B=180^{\circ}-\left(90^{\circ}+\measuredangle C A B\right)=90^{\circ}-\measuredangle C A B \\ & =90^{\circ}-\measuredangle Q C A=\measuredangle F Q C . \end{aligned} $$ To conclude, note that $$ \measuredangle B E M=\measured... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl",
"solution_match": null
} | 98 | 971 |
2024 | T1 | 5 | null | USAMO | Point $D$ is selected inside acute triangle $A B C$ so that $\angle D A C=\angle A C B$ and $\angle B D C=90^{\circ}+\angle B A C$. Point $E$ is chosen on ray $B D$ so that $A E=E C$. Let $M$ be the midpoint of $B C$. Show that line $A B$ is tangent to the circumcircle of triangle $B E M$. | Claim - We have DQCF is cyclic. $$ \begin{aligned} \measuredangle F D C & =-\measuredangle C D B=180^{\circ}-\left(90^{\circ}+\measuredangle C A B\right)=90^{\circ}-\measuredangle C A B \\ & =90^{\circ}-\measuredangle Q C A=\measuredangle F Q C . \end{aligned} $$ To conclude, note that $$ \measuredangle B E M=\measured... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl",
"solution_match": null
} | 98 | 1,130 |
2024 | T1 | 6 | null | USAMO | Let $n>2$ be an integer and let $\ell \in\{1,2, \ldots, n\}$. A collection $A_{1}, \ldots, A_{k}$ of (not necessarily distinct) subsets of $\{1,2, \ldots, n\}$ is called $\ell$-large if $\left|A_{i}\right| \geq \ell$ for all $1 \leq i \leq k$. Find, in terms of $n$ and $\ell$, the largest real number $c$ such that the ... | The answer turns out to be $$ c=\frac{n+\ell^{2}-2 \ell}{n(n-1)} $$ γ Rewriting as a dot product. For $i=1, \ldots, n$ define $\mathbf{v}_{i}$ by $$ \mathbf{v}_{i}[p, q]:=\left\{\begin{array}{ll} \frac{1}{\left|A_{i}\right|} & p \in A_{i} \text { and } q \in A_{i} \\ 0 & \text { otherwise; } \end{array} \quad \mathbf{v... | {
"problem_match": null,
"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl",
"solution_match": null
} | 295 | 1,272 |
2025 | T1 | 2 | null | USAMO | Let \(n > k \geq 1\) be integers. Let \(P(x) \in \mathbb{R}[x]\) be a polynomial of degree \(n\) with no repeated roots and \(P(0) \neq 0\) . Suppose that for any real numbers \(a_0, \ldots , a_k\) such that the polynomial \(a_k x^k + \dots + a_1 x + a_0\) divides \(P(x)\) , the product \(a_0 a_1 \ldots a_k\) is zero. ... | By considering any \(k + 1\) of the roots of \(P\) , we may as well assume WLOG that \(n = k + 1\) . Suppose that \(P(x) = (x + r_{1})\ldots (x + r_{n})\in \mathbb{R}[x]\) has \(P(0)\neq 0\) . Then the problem hypothesis is that each of the \(n\) polynomials (of degree \(n - 1\) ) given by
\[P_{1}(x) = (x + r_{2})(x ... | {
"problem_match": "2. ",
"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl",
"solution_match": "## \\(\\S 1.2\\) USAMO 2025/2, proposed by Carl Schildkraut \n"
} | 135 | 1,179 |
2025 | T1 | 3 | null | USAMO | Alice the architect and Bob the builder play a game. First, Alice chooses two points \(P\) and \(Q\) in the plane and a subset \(S\) of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of eac... | The answer is that Alice wins. Let's define a Bob- set \(V\) to be a set of points in the plane with no three collinear and with all distances at least 1. The point of the problem is to prove the following fact.
Claim β Given a Bob- set \(V\subseteq \mathbb{R}^{2}\) , consider the Bob- graph with vertex set \(V\) def... | {
"problem_match": "3. ",
"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl",
"solution_match": "## \\(\\S 1.3\\) USAMO 2025/3, proposed by Carl Schildkraut \n"
} | 234 | 1,330 |
2025 | T1 | 5 | null | USAMO | Find all positive integers \(k\) such that: for every positive integer \(n\) , the sum
\[\binom{n}{0}^{k} + \binom{n}{1}^{k} + \dots +\binom{n}{n}^{k}\]
is divisible by \(n + 1\) . | The answer is all even \(k\) .
Let's abbreviate \(S(n) := \binom{n}{0}^{k} + \dots + \binom{n}{n}^{k}\) for the sum in the problem.
\(\P\) Proof that even \(k\) is necessary. Choose \(n = 2\) . We need \(3 \mid S(2) = 2 + 2^{k}\) , which requires \(k\) to be even.
Remark. It's actually not much more difficult to ... | {
"problem_match": "5. ",
"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl",
"solution_match": "## \\(\\S 2.2\\) USAMO 2025/5, proposed by John Berman \n"
} | 69 | 1,542 |
2025 | T1 | 6 | null | USAMO | Let \(m\) and \(n\) be positive integers with \(m \geq n\) . There are \(m\) cupcakes of different flavors arranged around a circle and \(n\) people who like cupcakes. Each person assigns a nonnegative real number score to each cupcake, depending on how much they like the cupcake. Suppose that for each person \(P\) , i... | Arbitrarily pick any one person β call her Pip β and her \(n\) arcs. The initial idea is to try to apply Hall's marriage lemma to match the \(n\) people with Pip's arcs (such that each such person is happy with their matched arc). To that end, construct the obvious bipartite graph \(\mathfrak{G}\) between the people an... | {
"problem_match": "6. ",
"resource_path": "USAMO/segmented/en-USAMO-2025-notes.jsonl",
"solution_match": "## \\(\\S 2.3\\) USAMO 2025/6, proposed by Cheng-Yin Chang and Hung-Hsun Yu \n"
} | 160 | 986 |
2014 | T0 | 2 | null | USA_TST | Let $a_{1}, a_{2}, a_{3}, \ldots$ be a sequence of integers, with the property that every consecutive group of $a_{i}$ 's averages to a perfect square. More precisely, for all positive integers $n$ and $k$, the quantity $$ \frac{a_{n}+a_{n+1}+\cdots+a_{n+k-1}}{k} $$ is always the square of an integer. Prove that the ... | Let $\nu_{p}(n)$ denote the largest exponent of $p$ dividing $n$. The problem follows from the following proposition. ## Proposition Let $\left(a_{n}\right)$ be a sequence of integers and let $p$ be a prime. Suppose that every consecutive group of $a_{i}$ 's with length at most $p$ averages to a perfect square. Then $\... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2014.jsonl",
"solution_match": null
} | 123 | 803 |
2014 | T0 | 3 | null | USA_TST | Let $n$ be an even positive integer, and let $G$ be an $n$-vertex (simple) graph with exactly $\frac{n^{2}}{4}$ edges. An unordered pair of distinct vertices $\{x, y\}$ is said to be amicable if they have a common neighbor (there is a vertex $z$ such that $x z$ and $y z$ are both edges). Prove that $G$ has at least $2\... | First, we prove the following lemma. (https://en.wikipedia.org/wiki/Friendship_ paradox). Lemma (On average, your friends are more popular than you) For a vertex $v$, let $a(v)$ denote the average degree of the neighbors of $v$ (setting $a(v)=0$ if $\operatorname{deg} v=0)$. Then $$ \sum_{v} a(v) \geq \sum_{v} \operato... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2014.jsonl",
"solution_match": null
} | 116 | 533 |
2015 | T0 | 1 | null | USA_TST | Let $A B C$ be a scalene triangle with incenter $I$ whose incircle is tangent to $\overline{B C}$, $\overline{C A}, \overline{A B}$ at $D, E, F$, respectively. Denote by $M$ the midpoint of $\overline{B C}$ and let $P$ be a point in the interior of $\triangle A B C$ so that $M D=M P$ and $\angle P A B=\angle P A C$. Le... | First, we claim that $D, P, E$ are collinear. Let $N$ be the midpoint of $\overline{A B}$. It is well-known that the three lines $M N, D E, A I$ are concurrent at a point (see for example problem 6 of USAJMO 2014). Let $P^{\prime}$ be this intersection point, noting that $P^{\prime}$ actually lies on segment $D E$. The... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2015.jsonl",
"solution_match": null
} | 165 | 799 |
2015 | T0 | 2 | null | USA_TST | Prove that for every positive integer $n$, there exists a set $S$ of $n$ positive integers such that for any two distinct $a, b \in S, a-b$ divides $a$ and $b$ but none of the other elements of $S$. | The idea is to look for a sequence $d_{1}, \ldots, d_{n-1}$ of "differences" such that the following two conditions hold. Let $s_{i}=d_{1}+\cdots+d_{i-1}$, and $t_{i, j}=d_{i}+\cdots+d_{j-1}$ for $i \leq j$. (i) No two of the $t_{i, j}$ divide each other. (ii) There exists an integer $a$ satisfying the CRT equivalences... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2015.jsonl",
"solution_match": null
} | 58 | 549 |
2015 | T0 | 5 | null | USA_TST | Fix a positive integer $n$. A tournament on $n$ vertices has all its edges colored by $\chi$ colors, so that any two directed edges $u \rightarrow v$ and $v \rightarrow w$ have different colors. Over all possible tournaments on $n$ vertices, determine the minimum possible value of $\chi$. | The answer is $$ \chi=\left\lceil\log _{2} n\right\rceil $$ First, we prove by induction on $n$ that $\chi \geq \log _{2} n$ for any coloring and any tournament. The base case $n=1$ is obvious. Now given any tournament, consider any used color $c$. Then it should be possible to divide the tournament into two subsets $A... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2015.jsonl",
"solution_match": null
} | 67 | 552 |
2015 | T0 | 6 | null | USA_TST | Let $A B C$ be a non-equilateral triangle and let $M_{a}, M_{b}, M_{c}$ be the midpoints of the sides $B C, C A, A B$, respectively. Let $S$ be a point lying on the Euler line. Denote by $X, Y, Z$ the second intersections of $M_{a} S, M_{b} S, M_{c} S$ with the nine-point circle. Prove that $A X, B Y, C Z$ are concurre... | We assume now and forever that $A B C$ is scalene since the problem follows by symmetry in the isosceles case. We present four solutions. γ First solution by barycentric coordinates (Evan Chen). Let $A X$ meet $M_{b} M_{c}$ at $D$, and let $X$ reflected over $M_{b} M_{c}^{\prime}$ 's midpoint be $X^{\prime}$. Let $Y^{\... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2015.jsonl",
"solution_match": null
} | 114 | 697 |
2015 | T0 | 6 | null | USA_TST | Let $A B C$ be a non-equilateral triangle and let $M_{a}, M_{b}, M_{c}$ be the midpoints of the sides $B C, C A, A B$, respectively. Let $S$ be a point lying on the Euler line. Denote by $X, Y, Z$ the second intersections of $M_{a} S, M_{b} S, M_{c} S$ with the nine-point circle. Prove that $A X, B Y, C Z$ are concurre... | We assume now and forever that $A B C$ is scalene since the problem follows by symmetry in the isosceles case. We present four solutions. \ Second solution by moving points (Anant Mudgal). Let $H_{a}, H_{b}, H_{c}$ be feet of altitudes, and let $\gamma$ denote the nine-point circle. The main claim is that: Claim - Line... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2015.jsonl",
"solution_match": null
} | 114 | 888 |
2015 | T0 | 6 | null | USA_TST | Let $A B C$ be a non-equilateral triangle and let $M_{a}, M_{b}, M_{c}$ be the midpoints of the sides $B C, C A, A B$, respectively. Let $S$ be a point lying on the Euler line. Denote by $X, Y, Z$ the second intersections of $M_{a} S, M_{b} S, M_{c} S$ with the nine-point circle. Prove that $A X, B Y, C Z$ are concurre... | We assume now and forever that $A B C$ is scalene since the problem follows by symmetry in the isosceles case. We present four solutions. γ Third solution by moving points (Gopal Goel). In this solution, we will instead use barycentric coordinates with resect to $\triangle A B C$ to bound the degrees suitably, and then... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2015.jsonl",
"solution_match": null
} | 114 | 1,263 |
2016 | T0 | 3 | null | USA_TST | Let $p$ be a prime number. Let $\mathbb{F}_{p}$ denote the integers modulo $p$, and let $\mathbb{F}_{p}[x]$ be the set of polynomials with coefficients in $\mathbb{F}_{p}$. Define $\Psi: \mathbb{F}_{p}[x] \rightarrow \mathbb{F}_{p}[x]$ by $$ \Psi\left(\sum_{i=0}^{n} a_{i} x^{i}\right)=\sum_{i=0}^{n} a_{i} x^{x^{i}} . ... | Observe that $\Psi$ is also a linear map of $\mathbb{F}_{p}$ vector spaces, and that $\Psi(x P)=\Psi(P)^{p}$ for any $P \in \mathbb{F}_{p}[x]$. (In particular, $\Psi(1)=x$, not 1 , take caution!) γ Second solution. Here is an alternative (longer but more conceptual) way to finish without Bezout lemma. Let $\beth \subse... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2016.jsonl",
"solution_match": null
} | 184 | 568 |
2016 | T0 | 5 | null | USA_TST | Let $n \geq 4$ be an integer. Find all functions $W:\{1, \ldots, n\}^{2} \rightarrow \mathbb{R}$ such that for every partition $[n]=A \cup B \cup C$ into disjoint sets, $$ \sum_{a \in A} \sum_{b \in B} \sum_{c \in C} W(a, b) W(b, c)=|A||B||C| . $$ | $ Of course, $W(k, k)$ is arbitrary for $k \in[n]$. We claim that $W(a, b)= \pm 1$ for any $a \neq b$, with the sign fixed. (These evidently work.) First, let $X_{a b c}=W(a, b) W(b, c)$ for all distinct $a, b, c$, so the given condition is $$ \sum_{a, b, c \in A \times B \times C} X_{a b c}=|A||B||C| . $$ Consider the... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2016.jsonl",
"solution_match": null
} | 106 | 678 |
2017 | T0 | 2 | null | USA_TST | Let $A B C$ be an acute scalene triangle with circumcenter $O$, and let $T$ be on line $B C$ such that $\angle T A O=90^{\circ}$. The circle with diameter $\overline{A T}$ intersects the circumcircle of $\triangle B O C$ at two points $A_{1}$ and $A_{2}$, where $O A_{1}<O A_{2}$. Points $B_{1}, B_{2}, C_{1}, C_{2}$ are... | Let triangle $A B C$ have circumcircle $\Gamma$. Let $\triangle X Y Z$ be the tangential triangle of $\triangle A B C$ (hence $\Gamma$ is the incircle of $\triangle X Y Z$ ), and denote by $\Omega$ its circumcircle. Suppose the symmedian $\overline{A X}$ meets $\Gamma$ again at $D$, and let $M$ be the midpoint of $\ove... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2017.jsonl",
"solution_match": null
} | 202 | 670 |
2017 | T0 | 3 | null | USA_TST | Let $P, Q \in \mathbb{R}[x]$ be relatively prime nonconstant polynomials. Show that there can be at most three real numbers $\lambda$ such that $P+\lambda Q$ is the square of a polynomial. | This is true even with $\mathbb{R}$ replaced by $\mathbb{C}$, and it will be necessary to work in this generality. γ Second solution using derivatives (by Zack Chroman). We will assume without loss of generality that $\operatorname{deg} P \neq \operatorname{deg} Q$; if not, then one can replace $(P, Q)$ with $(P+c Q, Q... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2017.jsonl",
"solution_match": null
} | 51 | 527 |
2017 | T0 | 4 | null | USA_TST | You are cheating at a trivia contest. For each question, you can peek at each of the $n>1$ other contestant's guesses before writing your own. For each question, after all guesses are submitted, the emcee announces the correct answer. A correct guess is worth 0 points. An incorrect guess is worth -2 points for other co... | We will prove the result with $2^{n-1}$ replaced even by $2^{n-2}+1$. We first make the following reductions. First, change the weights to be $+1,-1,0$ respectively (rather than $0,-2,-1$ ); this clearly has no effect. Also, WLOG that all contestants except you initially have score zero (and that your score exceeds $2^... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2017.jsonl",
"solution_match": null
} | 131 | 1,008 |
2017 | T0 | 6 | null | USA_TST | Prove that there are infinitely many triples $(a, b, p)$ of integers, with $p$ prime and $0<a \leq b<p$, for which $p^{5}$ divides $(a+b)^{p}-a^{p}-b^{p}$. | The key claim is that if $p \equiv 1(\bmod 3)$, then $$ p\left(x^{2}+x y+y^{2}\right)^{2} \text { divides }(x+y)^{p}-x^{p}-y^{p} $$ as polynomials in $x$ and $y$. Since it's known that one can select $a$ and $b$ such that $p^{2} \mid a^{2}+a b+b^{2}$, the conclusion follows. (The theory of quadratic forms tells us we c... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2017.jsonl",
"solution_match": null
} | 57 | 511 |
2018 | T0 | 1 | null | USA_TST | Let $n \geq 2$ be a positive integer, and let $\sigma(n)$ denote the sum of the positive divisors of $n$. Prove that the $n^{\text {th }}$ smallest positive integer relatively prime to $n$ is at least $\sigma(n)$, and determine for which $n$ equality holds. | The equality case is $n=p^{e}$ for $p$ prime and a positive integer $e$. It is easy to check that this works. γ First solution. In what follows, by $[a, b]$ we mean $\{a, a+1, \ldots, b\}$. First, we make the following easy observation. Claim - If $a$ and $d$ are positive integers, then precisely $\varphi(d)$ elements ... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2018.jsonl",
"solution_match": null
} | 72 | 503 |
2018 | T0 | 1 | null | USA_TST | Let $n \geq 2$ be a positive integer, and let $\sigma(n)$ denote the sum of the positive divisors of $n$. Prove that the $n^{\text {th }}$ smallest positive integer relatively prime to $n$ is at least $\sigma(n)$, and determine for which $n$ equality holds. | The equality case is $n=p^{e}$ for $p$ prime and a positive integer $e$. It is easy to check that this works. I Second solution (Ivan Borsenco and Evan Chen). Let $n=p_{1}^{e_{1}} \ldots p_{k}^{e_{k}}$, where $p_{1}<p_{2}<\ldots$. We are going to assume $k \geq 2$, since the $k=1$ case was resolved in the very beginnin... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2018.jsonl",
"solution_match": null
} | 72 | 903 |
2018 | T0 | 2 | null | USA_TST | Find all functions $f: \mathbb{Z}^{2} \rightarrow[0,1]$ such that for any integers $x$ and $y$, $$ f(x, y)=\frac{f(x-1, y)+f(x, y-1)}{2} $$ | We claim that the only functions $f$ are constant functions. (It is easy to see that they work.) I Second solution (random walks, Mark Sellke). We show that if $x+y=x^{\prime}+y^{\prime}$ then $f(x, y)=f\left(x^{\prime}, y^{\prime}\right)$. Let $Z_{n}, Z_{n}^{\prime}$ be random walks starting at $(x, y)$ and $\left(x^{... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2018.jsonl",
"solution_match": null
} | 62 | 621 |
2018 | T0 | 3 | null | USA_TST | At a university dinner, there are 2017 mathematicians who each order two distinct entrΓ©es, with no two mathematicians ordering the same pair of entrΓ©es. The cost of each entrΓ©e is equal to the number of mathematicians who ordered it, and the university pays for each mathematician's less expensive entrΓ©e (ties broken ar... | In graph theoretic terms: we wish to determine the maximum possible value of $$ S(G):=\sum_{e=v w} \min (\operatorname{deg} v, \operatorname{deg} w) $$ across all graphs $G$ with 2017 edges. We claim the answer is $63 \cdot\binom{64}{2}+1=127009$. γ First solution (combinatorial, Evan Chen). First define $L_{k}$ to con... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2018.jsonl",
"solution_match": null
} | 90 | 1,454 |
2018 | T0 | 3 | null | USA_TST | At a university dinner, there are 2017 mathematicians who each order two distinct entrΓ©es, with no two mathematicians ordering the same pair of entrΓ©es. The cost of each entrΓ©e is equal to the number of mathematicians who ordered it, and the university pays for each mathematician's less expensive entrΓ©e (ties broken ar... | In graph theoretic terms: we wish to determine the maximum possible value of $$ S(G):=\sum_{e=v w} \min (\operatorname{deg} v, \operatorname{deg} w) $$ across all graphs $G$ with 2017 edges. We claim the answer is $63 \cdot\binom{64}{2}+1=127009$. γI Second solution (algebraic, submitted by contestant James Lin). We gi... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2018.jsonl",
"solution_match": null
} | 90 | 2,329 |
2018 | T0 | 5 | null | USA_TST | Let $A B C D$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\overline{B C}$ and $\overline{C D}$. Rays $M H$ and $N H$ meet $\overline{A D}$ and $\overline{A B}$ at $S$ and $T$, respectively. Prove there exists a poin... | The main claim is that $E$ is the intersection of $(A B C D)$ with the circle with diameter $\overline{A H}$. The following observation can be quickly made without reference to $E$. ## Lemma We have $\angle H S A=\angle H T A=90^{\circ}$. Consequently, quadrilateral $B T S D$ is cyclic. From here it follows that $E$ is... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2018.jsonl",
"solution_match": null
} | 155 | 506 |
2018 | T0 | 5 | null | USA_TST | Let $A B C D$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\overline{B C}$ and $\overline{C D}$. Rays $M H$ and $N H$ meet $\overline{A D}$ and $\overline{A B}$ at $S$ and $T$, respectively. Prove there exists a poin... | The main claim is that $E$ is the intersection of $(A B C D)$ with the circle with diameter $\overline{A H}$. The following observation can be quickly made without reference to $E$. ## Lemma We have $\angle H S A=\angle H T A=90^{\circ}$. Consequently, quadrilateral $B T S D$ is cyclic. From here it follows that $E$ is... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2018.jsonl",
"solution_match": null
} | 155 | 924 |
2018 | T0 | 6 | null | USA_TST | Alice and Bob play a game. First, Alice secretly picks a finite set $S$ of lattice points in the Cartesian plane. Then, for every line $\ell$ in the plane which is horizontal, vertical, or has slope +1 or -1 , she tells Bob the number of points of $S$ that lie on $\ell$. Bob wins if he can then determine the set $S$. P... | Clearly Bob can compute the number $N$ of points. The main claim is that: Claim - Fix $m$ and $n$ as in the problem statement. Among all sets $T \subseteq \mathbb{Z}^{2}$ with $N$ points, the set $S$ is the unique one which maximizes the value of $$ F(T):=\sum_{(x, y) \in T}\left(x^{2}+y^{2}\right)\left(m+n-\left(x^{2}... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2018.jsonl",
"solution_match": null
} | 174 | 863 |
2019 | T0 | 1 | null | USA_TST | Let $A B C$ be a triangle and let $M$ and $N$ denote the midpoints of $\overline{A B}$ and $\overline{A C}$, respectively. Let $X$ be a point such that $\overline{A X}$ is tangent to the circumcircle of triangle $A B C$. Denote by $\omega_{B}$ the circle through $M$ and $B$ tangent to $\overline{M X}$, and by $\omega_{... | γ Third solution by moving points method. Fix triangle $A B C$ and animate $X$ along the tangent at $A$. We let $D$ denote the second intersection point of $\omega_{C}$ with line $\overline{B C}$. Claim - The composed map $X \mapsto D$ is a fractional linear transformation (i.e. a projective map) in terms of a real coo... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2019.jsonl",
"solution_match": null
} | 145 | 656 |
2019 | T0 | 2 | null | USA_TST | Let $\mathbb{Z} / n \mathbb{Z}$ denote the set of integers considered modulo $n$ (hence $\mathbb{Z} / n \mathbb{Z}$ has $n$ elements). Find all positive integers $n$ for which there exists a bijective function $g: \mathbb{Z} / n \mathbb{Z} \rightarrow \mathbb{Z} / n \mathbb{Z}$, such that the 101 functions $$ g(x), \q... | Call a function $g$ valiant if it obeys this condition. We claim the answer is all numbers relatively prime to 101 !. The construction is to just let $g$ be the identity function. Remark (Motivation for both parts). The following solution is dense, and it is easier to think about some small cases first, to motivate the... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2019.jsonl",
"solution_match": null
} | 162 | 1,135 |
2019 | T0 | 2 | null | USA_TST | Let $\mathbb{Z} / n \mathbb{Z}$ denote the set of integers considered modulo $n$ (hence $\mathbb{Z} / n \mathbb{Z}$ has $n$ elements). Find all positive integers $n$ for which there exists a bijective function $g: \mathbb{Z} / n \mathbb{Z} \rightarrow \mathbb{Z} / n \mathbb{Z}$, such that the 101 functions $$ g(x), \q... | Call a function $g$ valiant if it obeys this condition. We claim the answer is all numbers relatively prime to 101 !. The construction is to just let $g$ be the identity function. Remark (Motivation for both parts). The following solution is dense, and it is easier to think about some small cases first, to motivate the... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2019.jsonl",
"solution_match": null
} | 162 | 1,298 |
2019 | T0 | 2 | null | USA_TST | Let $\mathbb{Z} / n \mathbb{Z}$ denote the set of integers considered modulo $n$ (hence $\mathbb{Z} / n \mathbb{Z}$ has $n$ elements). Find all positive integers $n$ for which there exists a bijective function $g: \mathbb{Z} / n \mathbb{Z} \rightarrow \mathbb{Z} / n \mathbb{Z}$, such that the 101 functions $$ g(x), \q... | Call a function $g$ valiant if it obeys this condition. We claim the answer is all numbers relatively prime to 101 !. The construction is to just let $g$ be the identity function. Remark (Motivation for both parts). The following solution is dense, and it is easier to think about some small cases first, to motivate the... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2019.jsonl",
"solution_match": null
} | 162 | 564 |
2019 | T0 | 4 | null | USA_TST | We say a function $f: \mathbb{Z}_{\geq 0} \times \mathbb{Z}_{\geq 0} \rightarrow \mathbb{Z}$ is great if for any nonnegative integers $m$ and $n$, $$ f(m+1, n+1) f(m, n)-f(m+1, n) f(m, n+1)=1 $$ If $A=\left(a_{0}, a_{1}, \ldots\right)$ and $B=\left(b_{0}, b_{1}, \ldots\right)$ are two sequences of integers, we write ... | γ First solution (Nikolai Beluhov). Let $k=a_{0}=b_{0}=c_{0}=d_{0}$. We let $f, g, h$ be great functions for $(A, B),(B, C),(C, D)$ and write the following infinite array: $$ \left[\begin{array}{ccccccc} & \vdots & \vdots & b_{3} & \vdots & \vdots & \\ \cdots & g(2,2) & g(2,1) & b_{2} & f(1,2) & f(2,2) & \cdots \\ \cdo... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2019.jsonl",
"solution_match": null
} | 243 | 572 |
2019 | T0 | 4 | null | USA_TST | We say a function $f: \mathbb{Z}_{\geq 0} \times \mathbb{Z}_{\geq 0} \rightarrow \mathbb{Z}$ is great if for any nonnegative integers $m$ and $n$, $$ f(m+1, n+1) f(m, n)-f(m+1, n) f(m, n+1)=1 $$ If $A=\left(a_{0}, a_{1}, \ldots\right)$ and $B=\left(b_{0}, b_{1}, \ldots\right)$ are two sequences of integers, we write ... | Second solution (Ankan Bhattacharya). We will give an explicit classification of great sequences: ## Lemma The pair $(A, B)$ is great if and only if $a_{0}=b_{0}, a_{0} \mid a_{1} b_{1}+1$, and $a_{n} \mid a_{n-1}+a_{n+1}$ and $b_{n} \mid b_{n-1}+b_{n+1}$ for all $n$. Now, focus on six entries $f(x, y)$ with $x \in\{n-... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2019.jsonl",
"solution_match": null
} | 243 | 795 |
2019 | T0 | 5 | null | USA_TST | Let $n$ be a positive integer. Tasty and Stacy are given a circular necklace with $3 n$ sapphire beads and $3 n$ turquoise beads, such that no three consecutive beads have the same color. They play a cooperative game where they alternate turns removing three consecutive beads, subject to the following conditions: - Ta... | In the necklace, we draw a divider between any two beads of the same color. Unless there are no dividers, this divides the necklace into several zigzags in which the beads in each zigzag alternate. Each zigzag has two endpoints (adjacent to dividers). Observe that the condition about not having three consecutive matchi... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2019.jsonl",
"solution_match": null
} | 159 | 1,049 |
2019 | T0 | 6 | null | USA_TST | Let $A B C$ be a triangle with incenter $I$, and let $D$ be a point on line $B C$ satisfying $\angle A I D=90^{\circ}$. Let the excircle of triangle $A B C$ opposite the vertex $A$ be tangent to $\overline{B C}$ at point $A_{1}$. Define points $B_{1}$ on $\overline{C A}$ and $C_{1}$ on $\overline{A B}$ analogously, usi... | γ First solution using spiral similarity (Ankan Bhattacharya). First, we prove the part of the problem which does not depend on the condition $A B_{1} A_{1} C_{1}$ is cyclic. ## Lemma Let $A B C$ be a triangle and define $I, D, B_{1}, C_{1}$ as in the problem. Moreover, let $M$ denote the midpoint of $\overline{A D}$. ... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2019.jsonl",
"solution_match": null
} | 174 | 870 |
2019 | T0 | 6 | null | USA_TST | Let $A B C$ be a triangle with incenter $I$, and let $D$ be a point on line $B C$ satisfying $\angle A I D=90^{\circ}$. Let the excircle of triangle $A B C$ opposite the vertex $A$ be tangent to $\overline{B C}$ at point $A_{1}$. Define points $B_{1}$ on $\overline{C A}$ and $C_{1}$ on $\overline{A B}$ analogously, usi... | γ Second solution by inversion and mixtilinears (Anant Mudgal). As in the end of the preceding solution, we have $\angle A B_{1} A_{1}=\angle A C_{1} A_{1}=90^{\circ}$ and $I \in \overline{B_{1} C_{1}}$. Let $M$ be the midpoint of minor arc $B C$ and $N$ be the midpoint of $\operatorname{arc} \widehat{B A C}$. Let $L$ ... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2019.jsonl",
"solution_match": null
} | 174 | 762 |
2020 | T0 | 1 | null | USA_TST | Choose positive integers $b_{1}, b_{2}, \ldots$ satisfying $$ 1=\frac{b_{1}}{1^{2}}>\frac{b_{2}}{2^{2}}>\frac{b_{3}}{3^{2}}>\frac{b_{4}}{4^{2}}>\cdots $$ and let $r$ denote the largest real number satisfying $\frac{b_{n}}{n^{2}} \geq r$ for all positive integers $n$. What are the possible values of $r$ across all pos... | The answer is $0 \leq r \leq 1 / 2$. Obviously $r \geq 0$. In one direction, we show that Claim (Greedy bound) - For all integers $n$, we have $$ \frac{b_{n}}{n^{2}} \leq \frac{1}{2}+\frac{1}{2 n} $$ $$ \begin{aligned} b_{n} & <n^{2} \frac{b_{n-1}}{(n-1)^{2}} \leq n^{2}\left(\frac{1}{2}+\frac{1}{2(n-1)}\right)=\frac{n^... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2020.jsonl",
"solution_match": null
} | 134 | 662 |
2020 | T0 | 2 | null | USA_TST | Two circles $\Gamma_{1}$ and $\Gamma_{2}$ have common external tangents $\ell_{1}$ and $\ell_{2}$ meeting at $T$. Suppose $\ell_{1}$ touches $\Gamma_{1}$ at $A$ and $\ell_{2}$ touches $\Gamma_{2}$ at $B$. A circle $\Omega$ through $A$ and $B$ intersects $\Gamma_{1}$ again at $C$ and $\Gamma_{2}$ again at $D$, such that... | γ Fourth solution by untethered moving points. Fix $\ell_{1}, \ell_{2}, T, \Gamma_{1}$ and $\Gamma_{2}$, and let $\Gamma_{1}$ and $\Gamma_{2}$ meet at $U$ and $V$. By the radical axis theorem, $X$ lies on $U V$. Thus we instead treat $X$ as a variable point on line $U V$ and let $C=A X \cap \Gamma_{1}$, $D=B X \cap \Ga... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2020.jsonl",
"solution_match": null
} | 162 | 512 |
2020 | T0 | 3 | null | USA_TST | Let $\alpha \geq 1$ be a real number. Hephaestus and Poseidon play a turn-based game on an infinite grid of unit squares. Before the game starts, Poseidon chooses a finite number of cells to be flooded. Hephaestus is building a levee, which is a subset of unit edges of the grid, called walls, forming a connected, non-s... | We show that if $\alpha>2$ then Hephaestus wins, but when $\alpha=2$ (and hence $\alpha \leq 2$ ) Hephaestus cannot contain even a single-cell flood initially. Strategy for $\alpha>2$ : Impose $\mathbb{Z}^{2}$ coordinates on the cells. Adding more flooded cells does not make our task easier, so let us assume that initi... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2020.jsonl",
"solution_match": null
} | 243 | 1,361 |
2020 | T0 | 5 | null | USA_TST | Find all integers $n \geq 2$ for which there exists an integer $m$ and a polynomial $P(x)$ with integer coefficients satisfying the following three conditions: - $m>1$ and $\operatorname{gcd}(m, n)=1$; - the numbers $P(0), P^{2}(0), \ldots, P^{m-1}(0)$ are not divisible by $n$; and - $P^{m}(0)$ is divisible by $n$. H... | The answer is that this is possible if and only if there exists primes $p^{\prime}<p$ such that $p \mid n$ and $p^{\prime} \nmid n$. (Equivalently, the radical $\operatorname{rad}(n)$ must not be the product of the first several primes.) For a polynomial $P$, and an integer $N$, we introduce the notation $$ \operatorna... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2020.jsonl",
"solution_match": null
} | 149 | 1,040 |
2020 | T0 | 6 | null | USA_TST | Let $P_{1} P_{2} \ldots P_{100}$ be a cyclic 100-gon, and let $P_{i}=P_{i+100}$ for all $i$. Define $Q_{i}$ as the intersection of diagonals $\overline{P_{i-2} P_{i+1}}$ and $\overline{P_{i-1} P_{i+2}}$ for all integers $i$. Suppose there exists a point $P$ satisfying $\overline{P P_{i}} \perp \overline{P_{i-1} P_{i+1}... | Claim - The points $K_{\bullet}$ are concyclic say with circumcircle $\gamma$. Let $E_{i}$ be the second intersection of line $\overline{P_{i-1} K_{i} P_{i+1}}$ with $\gamma$; then it follows that the perpendiculars to $\overline{P_{i-1} P_{i+1}}$ at $E_{i}$ all concur at a point $E$, which is the reflection of $P$ acr... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2020.jsonl",
"solution_match": null
} | 174 | 2,286 |
2021 | T0 | 2 | null | USA_TST | Points $A, V_{1}, V_{2}, B, U_{2}, U_{1}$ lie fixed on a circle $\Gamma$, in that order, and such that $B U_{2}>A U_{1}>B V_{2}>A V_{1}$. Let $X$ be a variable point on the $\operatorname{arc} V_{1} V_{2}$ of $\Gamma$ not containing $A$ or $B$. Line $X A$ meets line $U_{1} V_{1}$ at $C$, while line $X B$ meets line $U_... | For brevity, we let $\ell_{i}$ denote line $U_{i} V_{i}$ for $i=1,2$. We first give an explicit description of the fixed point $K$. Let $E$ and $F$ be points on $\Gamma$ such that $\overline{A E} \| \ell_{1}$ and $\overline{B F} \| \ell_{2}$. The problem conditions imply that $E$ lies between $U_{1}$ and $A$ while $F$ ... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2021.jsonl",
"solution_match": null
} | 174 | 564 |
2021 | T0 | 2 | null | USA_TST | Points $A, V_{1}, V_{2}, B, U_{2}, U_{1}$ lie fixed on a circle $\Gamma$, in that order, and such that $B U_{2}>A U_{1}>B V_{2}>A V_{1}$. Let $X$ be a variable point on the $\operatorname{arc} V_{1} V_{2}$ of $\Gamma$ not containing $A$ or $B$. Line $X A$ meets line $U_{1} V_{1}$ at $C$, while line $X B$ meets line $U_... | For brevity, we let $\ell_{i}$ denote line $U_{i} V_{i}$ for $i=1,2$. We first give an explicit description of the fixed point $K$. Let $E$ and $F$ be points on $\Gamma$ such that $\overline{A E} \| \ell_{1}$ and $\overline{B F} \| \ell_{2}$. The problem conditions imply that $E$ lies between $U_{1}$ and $A$ while $F$ ... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2021.jsonl",
"solution_match": null
} | 174 | 715 |
2021 | T0 | 3 | null | USA_TST | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the inequality $$ f(y)-\left(\frac{z-y}{z-x} f(x)+\frac{y-x}{z-x} f(z)\right) \leq f\left(\frac{x+z}{2}\right)-\frac{f(x)+f(z)}{2} $$ for all real numbers $x<y<z$. | Answer: all functions of the form $f(y)=a y^{2}+b y+c$, where $a, b, c$ are constants with $a \leq 0$. If $I=(x, z)$ is an interval, we say that a real number $\alpha$ is a supergradient of $f$ at $y \in I$ if we always have $$ f(t) \leq f(y)+\alpha(t-y) $$ for every $t \in I$. (This inequality may be familiar as the s... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2021.jsonl",
"solution_match": null
} | 96 | 1,457 |
2023 | T0 | 2 | null | USA_TST | Let $A B C$ be an acute triangle. Let $M$ be the midpoint of side $B C$, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. Suppose that the common external tangents to the circumcircles of triangles $B M E$ and $C M F$ intersect at a point $K$, and that $K$ lies on the circumcircle of $A ... | I Synthetic solution due to Andrew Gu (Harvard 2026). Define $O_{1}$ and $O_{2}$ as the circumcenters of $(B M E)$ and $(C M F)$. Let $T$ be the point on $(A B C)$ such that $\overline{A T} \perp \overline{B C}$. Denote by $L$ the midpoint of minor arc $\overparen{B C}$. We are going to ignore the condition that $K$ li... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2023.jsonl",
"solution_match": null
} | 114 | 989 |
2023 | T0 | 2 | null | USA_TST | Let $A B C$ be an acute triangle. Let $M$ be the midpoint of side $B C$, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. Suppose that the common external tangents to the circumcircles of triangles $B M E$ and $C M F$ intersect at a point $K$, and that $K$ lies on the circumcircle of $A ... | γ Complex numbers approach with Apollonian circles, by Carl Schildkraut. We use complex numbers. As in the first approach, we will ignore the hypothesis that $K$ lies on ( $A B C$ ). Let $Q:=(A H) \cap(A B C) \cap(A E F) \neq A$ be the Miquel point of $B F E C$ again. Construct the point $T$ on $(A B C)$ for which $A T... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2023.jsonl",
"solution_match": null
} | 114 | 1,440 |
2023 | T0 | 3 | null | USA_TST | Consider pairs $(f, g)$ of functions from the set of nonnegative integers to itself such that - $f(0) \geq f(1) \geq f(2) \geq \cdots \geq f(300) \geq 0 ;$ - $f(0)+f(1)+f(2)+\cdots+f(300) \leq 300$; - for any 20 nonnegative integers $n_{1}, n_{2}, \ldots, n_{20}$, not necessarily distinct, we have $$ g\left(n_{1}+n_{... | Replace $300=\frac{24 \cdot 25}{2}$ with $\frac{s(s+1)}{2}$ where $s=24$, and 20 with $k$. The answer is $115440=$ $\frac{k s(k s+1)}{2}$. Equality is achieved at $f(n)=\max (s-n, 0)$ and $g(n)=\max (k s-n, 0)$. To prove $$ g\left(n_{1}+\cdots+n_{k}\right) \leq f\left(n_{1}\right)+\cdots+f\left(n_{k}\right) $$ write it... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2023.jsonl",
"solution_match": null
} | 223 | 1,743 |
2023 | T0 | 5 | null | USA_TST | Let $m$ and $n$ be fixed positive integers. Tsvety and Freyja play a game on an infinite grid of unit square cells. Tsvety has secretly written a real number inside of each cell so that the sum of the numbers within every rectangle of size either $m \times n$ or $n \times m$ is zero. Freyja wants to learn all of these ... | The answer is the following: - If $\operatorname{gcd}(m, n)>1$, then Freyja cannot win. - If $\operatorname{gcd}(m, n)=1$, then Freyja can win in a minimum of $(m-1)^{2}+(n-1)^{2}$ questions. First, we dispose of the case where $\operatorname{gcd}(m, n)>1$. Write $d=\operatorname{gcd}(m, n)$. The idea is that any label... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2023.jsonl",
"solution_match": null
} | 201 | 2,599 |
2023 | T0 | 6 | null | USA_TST | Fix a function $f: \mathbb{N} \rightarrow \mathbb{N}$ and for any $m, n \in \mathbb{N}$ define $$ \Delta(m, n)=\underbrace{f(f(\ldots f}_{f(n) \text { times }}(m) \ldots))-\underbrace{f(f(\ldots f}_{f(m) \text { times }}(n) \ldots)) . $$ Suppose $\Delta(m, n) \neq 0$ for any distinct $m, n \in \mathbb{N}$. Show that ... | Suppose for the sake of contradiction that $|\Delta(m, n)| \leq N$ for all $m, n$. Note that $f$ is injective, as $$ f(m)=f(n) \Longrightarrow \Delta(m, n)=0 \Longrightarrow m=n $$ as desired. Let $G$ be the "arrow graph" of $f$, which is the directed graph with vertex set $\mathbb{N}$ and edges $n \rightarrow f(n)$. T... | {
"problem_match": null,
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2023.jsonl",
"solution_match": null
} | 170 | 1,176 |
2024 | T0 | 1 | null | USA_TST | Find the smallest constant \(C > 1\) such that the following statement holds: for every integer \(n\geq 2\) and sequence of non-integer positive real numbers \(a_{1}\) , \(a_{2}\) , ..., \(a_{n}\) satisfying
\[\frac{1}{a_{1}} +\frac{1}{a_{2}} +\dots +\frac{1}{a_{n}} = 1,\]
it's possible to choose positive integer... | Answer. The answer is \(C = \frac{3}{2}\) .
Lower bound. Note that if \(a_{1} = \frac{4n - 3}{2n - 1}\) and \(a_{i} = \frac{4n - 3}{2n - 1}\) for \(i > 1\) , then we must have \(b_{1} \in \{1,2\}\) and \(b_{i} \in \{2n - 2,2n - 1\}\) for \(i > 1\) . If we take \(b_{1} = 2\) then we obtain
\[\frac{1}{b_{1}} +\frac{1... | {
"problem_match": "1. ",
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl",
"solution_match": "## \\(\\S 1.1\\) USA TST 2024/1, proposed by Merlijn Staps \n"
} | 218 | 1,296 |
2024 | T0 | 2 | null | USA_TST | Let \(A B C\) be a triangle with incenter \(I\) . Let segment \(A I\) intersect the incircle of triangle \(A B C\) at point \(D\) . Suppose that line \(B D\) is perpendicular to line \(A C\) . Let \(P\) be a point such that \(\angle B P A = \angle P A I = 90^{\circ}\) . Point \(Q\) lies on segment \(B D\) such that the... | We show several approaches.
First solution, by author.
Claim β We have \(B P = B Q\) .
Proof. For readability, we split the proof into three unconditional parts.
- We translate the condition \(\overline{B D} \perp \overline{A C}\) .... | {
"problem_match": "2. ",
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl",
"solution_match": "## \\(\\S 1.2\\) USA TST 2024/2, proposed by Luke Robitaille \n"
} | 164 | 1,970 |
2024 | T0 | 3 | null | USA_TST | Let \(n > k\geq 1\) be integers and let \(p\) be a prime dividing \(\textstyle{\binom{n}{k}}\) . Prove that the \(k\) -element subsets of \(\{1,\ldots ,n\}\) can be split into \(p\) classes of equal size, such that any two subsets with the same sum of elements belong to the same class. | ## Problem statem
Let \(\sigma (S)\) denote the sum of the elements of \(S\) , so that
\[P(x):= \sum_{S\subseteq \{1,\ldots ,n\}}x^{\sigma (S)}\]
is the generating function for the sums of \(k\) - element subsets of \(\{1, \ldots , n\}\) .
By Legendre's formula,
\[\nu_{p}\left(\binom{n}{k}\right) = \sum_{r = ... | {
"problem_match": "3. ",
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl",
"solution_match": "## \\(\\S 1.3\\) USA TST 2024/3, proposed by Ankan Bhattacharya \n"
} | 87 | 1,717 |
2024 | T0 | 4 | null | USA_TST | Find all integers \(n\geq 2\) for which there exists a sequence of \(2n\) pairwise distinct points \((P_{1},\ldots ,P_{n},Q_{1},\ldots ,Q_{n})\) in the plane satisfying the following four conditions:
(i) no three of the \(2n\) points are collinear;
(ii) \(P_{i}P_{i + 1}\geq 1\) for all \(i = 1,2,\ldots ,n\) , whe... | Find alln\) .
Answer. Even integers only.
Proof that even \(n\) work. If we ignore the conditions that the points are pairwise distinct and form no collinear triples, we may take
\[P_{2i + 1} = (0.51,0), \quad P_{2i} = (-0.51,0), \quad Q_{2i + 1} = (0,0.51), \quad Q_{2i} = (0, -0.51).\]
The distances \(P_{i}P_{... | {
"problem_match": "4. ",
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl",
"solution_match": "## \\(\\S 2.1\\) USA TST 2024/4, proposed by Ray Li \n"
} | 228 | 1,173 |
2024 | T0 | 5 | null | USA_TST | Suppose \(a_{1}< a_{2}< \dots < a_{2024}\) is an arithmetic sequence of positive integers, and \(b_{1}< b_{2}< \dots < b_{2024}\) is a geometric sequence of positive integers. Find the maximum possible number of integers that could appear in both sequences, over all possible choices of the two sequences. | Answer. 11 terms.
Construction. Let \(a_{i} = i\) and \(b_{i} = 2^{i - 1}\)
Bound. We show a \(\nu_{p}\) - based approach communicated by Derek Liu, which seems to be the shortest one. At first, we completely ignore the geometric sequence \(b_{i}\) and focus only on the arithmetic sequence.
Claim β Let \(p\) be a... | {
"problem_match": "5. ",
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2024.jsonl",
"solution_match": "## \\(\\S 2.2\\) USA TST 2024/5, proposed by Ray Li \n"
} | 82 | 1,280 |
2025 | T0 | 1 | null | IMO | Let \(n\) be a positive integer. Ana and Banana play a game. Banana thinks of a function \(f\colon \mathbb{Z}\to \mathbb{Z}\) and a prime number \(p\) . He tells Ana that \(f\) is nonconstant, \(p< 100\) and \(f(x + p) = f(x)\) for all integers \(x\) . Ana's goal is to determine the value of \(p\) . She writes down \(n... | The answer is \(n = 83 + 89 - 1 = 171\)
In general, if Ana has to decide between periods from the set \(\mathcal{P}:= \{p_{1} > p_{2} > \dots >p_{r}\}\) of pairwise distinct relatively prime positive integers for \(r\geq 3\) , the answer is \(p_{2} + p_{3} - 1\)
## Bound
Suppose for the sake of contradiction that... | {
"problem_match": "Problem 1. ",
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl",
"solution_match": "## \\(\\S 1.1\\) Solution to TST 1, by Anthony Wang \n"
} | 179 | 2,086 |
2025 | T0 | 2 | null | IMO | Let \(a_{1},a_{2},\ldots\) and \(b_{1},b_{2},\ldots\) be sequences of real numbers for which \(a_{1} > b_{1}\) and
\[a_{n + 1} = a_{n}^{2} - 2b_{n\] \[b_{n + 1} = b_{n}^{2} - 2a_{n\]
for all positive integers \(n\) . Prove that \(a_{1},a_{2},\ldots\) is eventually increasing (that is, there exists a positive inte... | Let \(r\) , \(s\) , and \(t\) be the complex roots of the polynomial \(p(\lambda) = \lambda^{3} - a_{1}\lambda^{2} + b_{1}\lambda - 1\) . By Vieta's formulas,
\[a_{1} = r + s + t\] \[b_{1} = 1 / r + 1 / s + 1 / t\] \[1 = rst.\]
Claim β For every positive integer \(n\) ,
\[a_{n} = r^{2^{n - 1}} + s^{2^{n - 1}} + t... | {
"problem_match": "Problem 2. ",
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl",
"solution_match": "## \\(\\S 1.2\\) Solution to TST 2, by Holden Mui \n"
} | 160 | 749 |
2025 | T0 | 3 | null | IMO | Let \(A_{1}A_{2}\dots A_{2025}\) be a convex 2025- gon, and let \(A_{i} = A_{i + 2025}\) for all integers \(i\) . Distinct points \(P\) and \(Q\) lie in its interior such that \(\angle A_{i - 1}A_{i}P = \angle QA_{i}A_{i + 1}\) for all \(i\) . Define points \(P_{i}^{j}\) and \(Q_{i}^{j}\) for integers \(i\) and positiv... | Let \(n = 2025\) . Let \(\mathcal{P}_{i}\) and \(\mathcal{Q}_{i}\) denote the polygons \(P_{1}^{i}\cdot \cdot \cdot P_{n}^{i}\) and \(Q_{1}^{i}\cdot \cdot \cdot Q_{n}^{i}\) . In this notation, \(\mathcal{P} = \mathcal{P}_{n}\) , \(\mathcal{Q} = \mathcal{Q}_{n}\) , and \(\mathcal{P}_{1} = \mathcal{Q}_{1} = A_{1}\cdot \c... | {
"problem_match": "Problem 3. ",
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl",
"solution_match": "## \\(\\S 1.3\\) Solution to TST 3, by Ruben Carpenter \n"
} | 456 | 2,536 |
2025 | T0 | 4 | null | IMO | Let \(ABC\) be a triangle, and let \(X\) , \(Y\) , and \(Z\) be collinear points such that \(AY = AZ\) , \(BZ = BX\) , and \(CX = CY\) . Points \(X'\) , \(Y'\) , and \(Z'\) are the reflections of \(X\) , \(Y\) , and \(Z\) over \(BC\) , \(CA\) , and \(AB\) , respectively. Prove that if \(XY'Z'\) is a nondegenerate trian... | ## \(\P\) Solution 1 (Pitchayut Saengrungkonga)
Let \(S\) denote the circumcenter of \(\triangle X^{\prime}Y^{\prime}Z^{\prime}\) . Observe that \(A Y = A Z = A Y^{\prime} = A Z^{\prime}\) , so \(Y Z Y^{\prime}Z^{\prime}\) is cyclic and \... | {
"problem_match": "Problem 4. ",
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl",
"solution_match": "## \\(\\S 2.1\\) Solution to TST 4, by Michael Ren \n"
} | 127 | 1,362 |
2025 | T0 | 5 | null | IMO | A pond has 2025 lily pads arranged in a circle. Two frogs, Alice and Bob, begin on different lily pads. A frog jump is a jump which travels 2, 3, or 5 positions clockwise. Alice and Bob each make a series of frog jumps, and each frog ends on the same lily pad that it started from. Given that each lily pad is the destin... | Let \(\pi :\mathbb{Z} / 2025\mathbb{Z}\to \mathbb{Z} / 2025\mathbb{Z}\) be the permutation where the jump with source \(i\) has destination \(\pi (i)\) . We know that \(\pi\) has exactly two cycles, corresponding to the paths of each frog.
Suppose that the frogs complete a total of \(\ell\) laps around the circle, so... | {
"problem_match": "Problem 5. ",
"resource_path": "USA_TST/segmented/en-sols-TST-IMO-2025.jsonl",
"solution_match": "## \\(\\S 2.2\\) Solution to TST 5, by Linus Tang \n"
} | 115 | 2,161 |
2011 | T0 | 1 | null | USA_TSTST | Find all real-valued functions $f$ defined on pairs of real numbers, having the following property: for all real numbers $a, b, c$, the median of $f(a, b), f(b, c), f(c, a)$ equals the median of $a, b, c$. (The median of three real numbers, not necessarily distinct, is the number that is in the middle when the three nu... | We prove the following main claim, from which repeated applications can deduce the problem. Claim - Let $a<b<c$ be arbitrary. On $\{a, b, c\}^{2}, f$ takes one of the following two forms, where the column indicates the $x$-value and the row indicates the $y$-value. | $f$ | $a$ | $b$ | $c$ | | ---: | ---: | ---: | ---: ... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2011.jsonl",
"solution_match": null
} | 94 | 509 |
2012 | T0 | 3 | null | USA_TSTST | Let $\mathbb{N}$ be the set of positive integers. Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying the following two conditions: (a) $f(m)$ and $f(n)$ are relatively prime whenever $m$ and $n$ are relatively prime. (b) $n \leq f(n) \leq n+2012$ for all $n$. Prove that for any natural number $n$ and ... | γ Second solution with a grid. Fix $n$ and $p$, and assume for contradiction $p \nmid n$. Claim - There exists a large integer $N$ with $f(N)=N$, that also satisfies $N \equiv 1$ $(\bmod n)$ and $N \equiv 0(\bmod p)$. it to fill in the following table: | | $N+1$ | $N+2$ | $\ldots$ | $N+2012$ | | :---: | :---: | :---: |... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2012.jsonl",
"solution_match": null
} | 128 | 680 |
2012 | T0 | 8 | null | USA_TSTST | Let $n$ be a positive integer. Consider a triangular array of nonnegative integers as follows:  Call such a triangular array stable if for every $0 \leq i<j<k \leq n$ we have $$ a_{i, j}+a_{j... | Let $n$ be a positive integer. Consider a triangular array of nonnegative integers as follows: $$ \begin{array}{clllll} \text { Row 1: } & & & a_{0,1} \\ \text { Row 2: } & & a_{0,2} & a_{1,2} & \\ & \vdots & \vdots & & \ddots & \\ \text { Row } n-1: & a_{0, n-1} & a_{1, n-1} & \ldots & a_{n-2, n-1} \\ \text { Row } n ... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2012.jsonl",
"solution_match": null
} | 218 | 1,407 |
2013 | T0 | 2 | null | USA_TSTST | A finite sequence of integers $a_{1}, a_{2}, \ldots, a_{n}$ is called regular if there exists a real number $x$ satisfying $$ \lfloor k x\rfloor=a_{k} \quad \text { for } 1 \leq k \leq n $$ Given a regular sequence $a_{1}, a_{2}, \ldots, a_{n}$, for $1 \leq k \leq n$ we say that the term $a_{k}$ is forced if the foll... | The answer is 985 . WLOG, by shifting $a_{1}=0$ (clearly $a_{1}$ isn't forced). Now, we construct regular sequences inductively using the following procedure. Start with the inequality $$ \frac{0}{1} \leq x<\frac{1}{1} $$ Then for each $k=2,3, \ldots, 1000$ we perform the following procedure. If there is no fraction of... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2013.jsonl",
"solution_match": null
} | 179 | 520 |
2013 | T0 | 3 | null | USA_TSTST | Divide the plane into an infinite square grid by drawing all the lines $x=m$ and $y=n$ for $m, n \in \mathbb{Z}$. Next, if a square's upper-right corner has both coordinates even, color it black; otherwise, color it white (in this way, exactly $1 / 4$ of the squares are black and no two black squares are adjacent). Let... | We re-color the white squares as red, blue, or green according as to whether they have a black square directly to the left/right, above/below, or neither, as shown below. Finally, we fix time zero to be a moment just before the laser passes a horizontal (WLOG) lattice line (not necessarily a wall). Shown below is an ex... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2013.jsonl",
"solution_match": null
} | 165 | 615 |
2013 | T0 | 6 | null | USA_TSTST | Let $\mathbb{N}$ be the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ that satisfy the equation $$ f^{a b c-a}(a b c)+f^{a b c-b}(a b c)+f^{a b c-c}(a b c)=a+b+c $$ for all $a, b, c \geq 2$. (Here $f^{k}$ means $f$ applied $k$ times.) | The answer is $f(n)=n-1$ for $n \geq 3$ with $f(1)$ and $f(2)$ arbitrary; check these work. ## Lemma We have $f^{t^{2}-t}\left(t^{2}\right)=t$ for all $t$. $$ f^{t^{9}-t^{3}}\left(t^{9}\right)=t^{3} \quad \text { and } \quad f^{t^{3}-t}\left(t^{3}\right)=t \Longrightarrow f^{t^{9}-t}\left(t^{9}\right)=t $$ so $k=1$ and... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2013.jsonl",
"solution_match": null
} | 109 | 728 |
2013 | T0 | 9 | null | USA_TSTST | Let $r$ be a rational number in the interval $[-1,1]$ and let $\theta=\cos ^{-1} r$. Call a subset $S$ of the plane good if $S$ is unchanged upon rotation by $\theta$ around any point of $S$ (in both clockwise and counterclockwise directions). Determine all values of $r$ satisfying the following property: The midpoint ... | The answer is that $r$ has this property if and only if $r=\frac{4 n-1}{4 n}$ for some integer $n$. $$ \omega=e^{i \theta}=\frac{a}{b} \pm \frac{\sqrt{b^{2}-a^{2}}}{b} i $$ This means we may work with complex multiplication in the usual way; the rotation of $z$ through center $c$ is given by $z \mapsto \omega(z-c)+c$. ... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2013.jsonl",
"solution_match": null
} | 97 | 1,464 |
2014 | T0 | 6 | null | USA_TSTST | Suppose we have distinct positive integers $a, b, c, d$ and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence $$ \begin{gathered} c a-d b \\ c a^{2}-d b^{2} \\ c a^{3}-d b^{3} \\ c a^{4}-d b^{4} \end{gathered} $$ and looks at the highest power of $p$ that d... | By orders, the indices of terms divisible by $p$ is an arithmetic subsequence of $\mathbb{N}$ : say they are $\kappa, \kappa+\lambda, \kappa+2 \lambda, \ldots$, where $\lambda$ is the order of $a / b$. That means we want $$ \nu_{p}\left(c a^{\kappa+n \lambda}-d b^{\kappa+n \lambda}\right)=\nu_{p}\left(\left(\frac{a^{\l... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2014.jsonl",
"solution_match": null
} | 186 | 835 |
2015 | T0 | 4 | null | USA_TSTST | Let $x, y, z$ be real numbers (not necessarily positive) such that $x^{4}+y^{4}+z^{4}+x y z=4$. Prove that $x \leq 2$ and $$ \sqrt{2-x} \geq \frac{y+z}{2} $$ | We prove that the condition $x^{4}+y^{4}+z^{4}+x y z=4$ implies $$ \sqrt{2-x} \geq \frac{y+z}{2} $$ We first prove the easy part. Claim - We have $x \leq 2$. $$ \begin{aligned} 5=x^{4}+y^{4}+\left(z^{4}+1\right)+x y z & =\frac{3 x^{4}}{4}+\left(\frac{x^{4}}{4}+y^{4}\right)+\left(z^{4}+1\right)+x y z \\ & \geq \frac{3 x... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2015.jsonl",
"solution_match": null
} | 72 | 665 |
2015 | T0 | 6 | null | USA_TSTST | A Nim-style game is defined as follows. Two positive integers $k$ and $n$ are specified, along with a finite set $S$ of $k$-tuples of integers (not necessarily positive). At the start of the game, the $k$-tuple $(~ n, 0,0, \ldots, 0)$ is written on the blackboard. A legal move consists of erasing the tuple $\left(a_{1}... | Now, the "game" is played as follows. The mechanics are controlled by the turn counters $A$ and $B$. Observe the game starts with Alice playing Init. Thereafter, we say that the game is - In the main part if $A+B=1$, and no one has played Init a second time. - In the death part otherwise. Observe that in the main state... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2015.jsonl",
"solution_match": null
} | 273 | 840 |
2016 | T0 | 1 | null | USA_TSTST | Let $A=A(x, y)$ and $B=B(x, y)$ be two-variable polynomials with real coefficients. Suppose that $A(x, y) / B(x, y)$ is a polynomial in $x$ for infinitely many values of $y$, and a polynomial in $y$ for infinitely many values of $x$. Prove that $B$ divides $A$, meaning there exists a third polynomial $C$ with real coef... | This is essentially an application of the division algorithm, but the details require significant care. First, we claim that $A / B$ can be written as a polynomial in $x$ whose coefficients are rational functions in $y$. To see this, use the division algorithm to get $$ A=Q \cdot B+R \quad Q, R \in(\mathbb{R}(y))[x] $$... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2016.jsonl",
"solution_match": null
} | 100 | 599 |
2016 | T0 | 2 | null | USA_TSTST | Let $A B C$ be a scalene triangle with orthocenter $H$ and circumcenter $O$ and denote by $M, N$ the midpoints of $\overline{A H}, \overline{B C}$. Suppose the circle $\gamma$ with diameter $\overline{A H}$ meets the circumcircle of $A B C$ at $G \neq A$, and meets line $\overline{A N}$ at $Q \neq A$. The tangent to $\... | γ Alternate solution (by proposer). Let $L$ be diametrically opposite $A$ on the circumcircle. Denote by $\triangle D E F$ the orthic triangle. Let $X=\overline{A H} \cap \overline{E F}$. Finally, let $T$ be the second intersection of (MFDNE) and (MBC). We begin with a few easy observations. First, points $H, G, N, L$ ... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2016.jsonl",
"solution_match": null
} | 152 | 611 |
2016 | T0 | 3 | null | USA_TSTST | Decide whether or not there exists a nonconstant polynomial $Q(x)$ with integer coefficients with the following property: for every positive integer $n>2$, the numbers $$ Q(0), Q(1), Q(2), \ldots, Q(n-1) $$ produce at most $0.499 n$ distinct residues when taken modulo $n$. | We claim that $$ Q(x)=420\left(x^{2}-1\right)^{2} $$ works. Clearly, it suffices to prove the result when $n=4$ and when $n$ is an odd prime $p$. The case $n=4$ is trivial, so assume now $n=p$ is an odd prime. First, we prove the following easy claim. Claim - For any odd prime $p$, there are at least $\frac{1}{2}(p-3)$... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2016.jsonl",
"solution_match": null
} | 78 | 975 |
2016 | T0 | 6 | null | USA_TSTST | Let $A B C$ be a triangle with incenter $I$, and whose incircle is tangent to $\overline{B C}, \overline{C A}$, $\overline{A B}$ at $D, E, F$, respectively. Let $K$ be the foot of the altitude from $D$ to $\overline{E F}$. Suppose that the circumcircle of $\triangle A I B$ meets the incircle at two distinct points $C_{... | \I First solution (Allen Liu). Let $X, Y, Z$ be midpoints of $E F, F D, D E$, and let $G$ be the Gergonne point. By radical axis on $(A E I F),(D E F),(A I C)$ we see that $B_{1}$, $X, B_{2}$ are collinear. Likewise, $B_{1}, Z, B_{2}$ are collinear, so lines $B_{1} B_{2}$ and $X Z$ coincide. Similarly, lines $C_{1} C_{... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2016.jsonl",
"solution_match": null
} | 191 | 844 |
2016 | T0 | 6 | null | USA_TSTST | Let $A B C$ be a triangle with incenter $I$, and whose incircle is tangent to $\overline{B C}, \overline{C A}$, $\overline{A B}$ at $D, E, F$, respectively. Let $K$ be the foot of the altitude from $D$ to $\overline{E F}$. Suppose that the circumcircle of $\triangle A I B$ meets the incircle at two distinct points $C_{... | γ Second solution (Evan Chen). As before, we just have to prove $G$ lies on the radical axis. Construct parallelograms $G P F Q, G R D S, G T U E$ such that $P, R \in D F, S, T \in D E$, $Q, U \in E F$. As $F G$ bisects $P Q$ and is isogonal to $F Z$, we find $P Q E D$, hence $P Q R U$, is cyclic. Repeating the same lo... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2016.jsonl",
"solution_match": null
} | 191 | 736 |
2017 | T0 | 1 | null | USA_TSTST | Let $A B C$ be a triangle with circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. Assume that $A B \neq A C$ and $\angle A \neq 90^{\circ}$. Let $M$ and $N$ be the midpoints of $\overline{A B}$ and $\overline{A C}$, respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle... | γ Third solution (complex numbers). Let $A B C$ be the unit circle. We first compute $P$ as the midpoint of $A$ and $\overline{A A} \cap \overline{B C}$ : $$ \begin{aligned} p & =\frac{1}{2}\left(a+\frac{a^{2}(b+c)-b c \cdot 2 a}{a^{2}-b c}\right) \\ & =\frac{a\left(a^{2}-b c\right)+a^{2}(b+c)-2 a b c}{2\left(a^{2}-b c... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2017.jsonl",
"solution_match": null
} | 212 | 537 |
2017 | T0 | 2 | null | USA_TSTST | Ana and Banana are playing a game. First Ana picks a word, which is defined to be a nonempty sequence of capital English letters. Then Banana picks a nonnegative integer $k$ and challenges Ana to supply a word with exactly $k$ subsequences which are equal to Ana's word. Ana wins if she is able to supply such a word, ot... | First we introduce some notation. Define a block of letters to be a maximal contiguous subsequence of consecutive letters. For example, the word $A A B B B C A A A$ has four blocks, namely $A A, B B B, C, A A A$. Throughout the solution, we fix the word $A$ that Ana picks, and introduce the following notation for its $... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2017.jsonl",
"solution_match": null
} | 142 | 862 |
2017 | T0 | 3 | null | USA_TSTST | Consider solutions to the equation $$ x^{2}-c x+1=\frac{f(x)}{g(x)} $$ where $f$ and $g$ are nonzero polynomials with nonnegative real coefficients. For each $c>0$, determine the minimum possible degree of $f$, or show that no such $f, g$ exist. | Indeed, one simply takes $x=1$ to get $f(1) / g(1) \leq 0$, impossible. For $c<2$, let $c=2 \cos \theta$, where $0<\theta<\pi$. We claim that $f$ exists and has minimum degree equal to $n$, where $n$ is defined as the smallest integer satisfying $\sin n \theta \leq 0$. In other words $$ n=\left\lceil\frac{\pi}{\arccos ... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2017.jsonl",
"solution_match": null
} | 73 | 1,502 |
2017 | T0 | 6 | null | USA_TSTST | A sequence of positive integers $\left(a_{n}\right)_{n \geq 1}$ is of Fibonacci type if it satisfies the recursive relation $a_{n+2}=a_{n+1}+a_{n}$ for all $n \geq 1$. Is it possible to partition the set of positive integers into an infinite number of Fibonacci type sequences? | γ First solution (Kevin Sun). We are going to appeal to the so-called Zeckendorf theorem: Theorem (Zeckendorf) Every positive integer can be uniquely expressed as the sum of nonconsecutive Fibonacci numbers. This means every positive integer has a Zeckendorf ("Fibonacci-binary") representation where we put 1 in the $i$... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2017.jsonl",
"solution_match": null
} | 78 | 895 |
2017 | T0 | 6 | null | USA_TSTST | A sequence of positive integers $\left(a_{n}\right)_{n \geq 1}$ is of Fibonacci type if it satisfies the recursive relation $a_{n+2}=a_{n+1}+a_{n}$ for all $n \geq 1$. Is it possible to partition the set of positive integers into an infinite number of Fibonacci type sequences? | \I Second solution. Call an infinite set of integers $S$ sandwiched if there exist increasing sequences $\left\{a_{i}\right\}_{i=0}^{\infty},\left\{b_{i}\right\}_{i=0}^{\infty}$ such that the following are true: - $a_{i}+a_{i+1}=a_{i+2}$ and $b_{i}+b_{i+1}=b_{i+2}$. - The intervals $\left[a_{i}+1, b_{i}-1\right]$ are d... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2017.jsonl",
"solution_match": null
} | 78 | 1,108 |
2017 | T0 | 6 | null | USA_TSTST | A sequence of positive integers $\left(a_{n}\right)_{n \geq 1}$ is of Fibonacci type if it satisfies the recursive relation $a_{n+2}=a_{n+1}+a_{n}$ for all $n \geq 1$. Is it possible to partition the set of positive integers into an infinite number of Fibonacci type sequences? | \l Third solution. We add Fibonacci-type sequences one-by-one. At each step, let $x$ be the smallest number that has not been used in any previous sequence. We generate a new Fibonacci-type sequence as follows. Set $a_{0}=x$ and for $i \geq 1$, set $$ a_{i}=\left\lfloor\varphi a_{i-1}+\frac{1}{2}\right\rfloor $$ Equiva... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2017.jsonl",
"solution_match": null
} | 78 | 926 |
2017 | T0 | 6 | null | USA_TSTST | A sequence of positive integers $\left(a_{n}\right)_{n \geq 1}$ is of Fibonacci type if it satisfies the recursive relation $a_{n+2}=a_{n+1}+a_{n}$ for all $n \geq 1$. Is it possible to partition the set of positive integers into an infinite number of Fibonacci type sequences? | I Fourth solution (Mark Sellke). For later reference let $$ f_{1}=0, f_{2}=1, f_{3}=1, \ldots $$ denote the ordinary Fibonacci numbers. We will denote the Fibonacci-like sequences by $F^{i}$ and the elements with subscripts; hence $F_{1}^{2}$ is the first element of the second sequence. Our construction amounts to just... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2017.jsonl",
"solution_match": null
} | 78 | 1,464 |
2018 | T0 | 1 | null | USA_TSTST | As usual, let $\mathbb{Z}[x]$ denote the set of single-variable polynomials in $x$ with integer coefficients. Find all functions $\theta: \mathbb{Z}[x] \rightarrow \mathbb{Z}$ such that for any polynomials $p, q \in$ $\mathbb{Z}[x]$, - $\theta(p+1)=\theta(p)+1$, and - if $\theta(p) \neq 0$ then $\theta(p)$ divides $\t... | The answer is $\theta: p \mapsto p(c)$, for each choice of $c \in \mathbb{Z}$. Obviously these work, so we prove these are the only ones. In what follows, $x \in \mathbb{Z}[x]$ is the identity polynomial, and $c=\theta(x)$. II Second solution. First, we settle the case $\operatorname{deg} p=0$. In that case, from the s... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2018.jsonl",
"solution_match": null
} | 113 | 862 |
2018 | T0 | 3 | null | USA_TSTST | Let $A B C$ be an acute triangle with incenter $I$, circumcenter $O$, and circumcircle $\Gamma$. Let $M$ be the midpoint of $\overline{A B}$. Ray $A I$ meets $\overline{B C}$ at $D$. Denote by $\omega$ and $\gamma$ the circumcircles of $\triangle B I C$ and $\triangle B A D$, respectively. Line $M O$ meets $\omega$ at ... | Henceforth assume $\angle A \neq 60^{\circ}$; we prove the concurrence. Let $L$ denote the center of $\omega$, which is the midpoint of minor arc $B C$. Claim - Let $K$ be the point on $\omega$ such that $\overline{K L} \| \overline{A B}$ and $\overline{K C} \| \overline{A L}$. Then $\overline{K A}$ is tangent to $\gam... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2018.jsonl",
"solution_match": null
} | 212 | 554 |
2018 | T0 | 3 | null | USA_TSTST | Let $A B C$ be an acute triangle with incenter $I$, circumcenter $O$, and circumcircle $\Gamma$. Let $M$ be the midpoint of $\overline{A B}$. Ray $A I$ meets $\overline{B C}$ at $D$. Denote by $\omega$ and $\gamma$ the circumcircles of $\triangle B I C$ and $\triangle B A D$, respectively. Line $M O$ meets $\omega$ at ... | Henceforth assume $\angle A \neq 60^{\circ}$; we prove the concurrence. Let $L$ denote the center of $\omega$, which is the midpoint of minor arc $B C$. Claim - Let $K$ be the point on $\omega$ such that $\overline{K L} \| \overline{A B}$ and $\overline{K C} \| \overline{A L}$. Then $\overline{K A}$ is tangent to $\gam... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2018.jsonl",
"solution_match": null
} | 212 | 597 |
2018 | T0 | 3 | null | USA_TSTST | Let $A B C$ be an acute triangle with incenter $I$, circumcenter $O$, and circumcircle $\Gamma$. Let $M$ be the midpoint of $\overline{A B}$. Ray $A I$ meets $\overline{B C}$ at $D$. Denote by $\omega$ and $\gamma$ the circumcircles of $\triangle B I C$ and $\triangle B A D$, respectively. Line $M O$ meets $\omega$ at ... | Henceforth assume $\angle A \neq 60^{\circ}$; we prove the concurrence. Let $L$ denote the center of $\omega$, which is the midpoint of minor arc $B C$. Claim - Let $K$ be the point on $\omega$ such that $\overline{K L} \| \overline{A B}$ and $\overline{K C} \| \overline{A L}$. Then $\overline{K A}$ is tangent to $\gam... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2018.jsonl",
"solution_match": null
} | 212 | 1,237 |
2018 | T0 | 4 | null | USA_TSTST | For an integer $n>0$, denote by $\mathcal{F}(n)$ the set of integers $m>0$ for which the polynomial $p(x)=x^{2}+m x+n$ has an integer root. (a) Let $S$ denote the set of integers $n>0$ for which $\mathcal{F}(n)$ contains two consecutive integers. Show that $S$ is infinite but $$ \sum_{n \in S} \frac{1}{n} \leq 1 $$ (... | We prove the following. Claim - The set $S$ is given explicitly by $S=\{x(x+1) y(y+1) \mid x, y>0\}$. $$ \begin{aligned} m^{2}-4 n & =p^{2} \\ (m+1)^{2}-4 n & =q^{2} \end{aligned} $$ Subtraction gives $2 m+1=q^{2}-p^{2}$, so $p$ and $q$ are different parities. We can thus let $q-p=2 x+1, q+p=2 y+1$, where $y \geq x \ge... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2018.jsonl",
"solution_match": null
} | 143 | 1,033 |
2018 | T0 | 5 | null | USA_TSTST | Let $A B C$ be an acute triangle with circumcircle $\omega$, and let $H$ be the foot of the altitude from $A$ to $\overline{B C}$. Let $P$ and $Q$ be the points on $\omega$ with $P A=P H$ and $Q A=Q H$. The tangent to $\omega$ at $P$ intersects lines $A C$ and $A B$ at $E_{1}$ and $F_{1}$ respectively; the tangent to $... | Let $O$ be the center of $\omega$, and let $M=\overline{P Q} \cap \overline{A B}$ and $N=\overline{P Q} \cap \overline{A C}$ be the midpoints of $\overline{A B}$ and $\overline{A C}$ respectively. Refer to the diagram below. The main idea is to prove two key claims involving $O$, which imply the result: (i) quadrilater... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2018.jsonl",
"solution_match": null
} | 193 | 2,391 |
2018 | T0 | 6 | null | USA_TSTST | Let $S=\{1, \ldots, 100\}$, and for every positive integer $n$ define $$ T_{n}=\left\{\left(a_{1}, \ldots, a_{n}\right) \in S^{n} \mid a_{1}+\cdots+a_{n} \equiv 0 \quad(\bmod 100)\right\} $$ Determine which $n$ have the following property: if we color any 75 elements of $S$ red, then at least half of the $n$-tuples i... | We claim this holds exactly for $n$ even. \I First solution by generating functions. Define $$ R(x)=\sum_{s \text { red }} x^{s}, \quad B(x)=\sum_{s \text { blue }} x^{s} . $$ (Here "blue" means "not-red", as always.) Then, the number of tuples in $T_{n}$ with exactly $k$ red coordinates is exactly equal to $$ \binom{n... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2018.jsonl",
"solution_match": null
} | 142 | 703 |
2018 | T0 | 6 | null | USA_TSTST | Let $S=\{1, \ldots, 100\}$, and for every positive integer $n$ define $$ T_{n}=\left\{\left(a_{1}, \ldots, a_{n}\right) \in S^{n} \mid a_{1}+\cdots+a_{n} \equiv 0 \quad(\bmod 100)\right\} $$ Determine which $n$ have the following property: if we color any 75 elements of $S$ red, then at least half of the $n$-tuples i... | We claim this holds exactly for $n$ even. γ Second solution by strengthened induction and random coloring. We again prove that $n$ even work. Let us define $$ T_{n}(a)=\left\{\left(a_{1}, \ldots, a_{n}\right) \in S^{n} \mid a_{1}+\cdots+a_{n} \equiv a \quad(\bmod 100)\right\} $$ Also, call an $n$-tuple good if it has a... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2018.jsonl",
"solution_match": null
} | 142 | 1,133 |
2018 | T0 | 7 | null | USA_TSTST | Let $n$ be a positive integer. A frog starts on the number line at 0 . Suppose it makes a finite sequence of hops, subject to two conditions: - The frog visits only points in $\left\{1,2, \ldots, 2^{n}-1\right\}$, each at most once. - The length of each hop is in $\left\{2^{0}, 2^{1}, 2^{2}, \ldots\right\}$. (The hops... | We claim the answer is $\frac{4^{n}-1}{3}$. We first prove the bound. First notice that the hop sizes are in $\left\{2^{0}, 2^{1}, \ldots, 2^{n-1}\right\}$, since the frog must stay within bounds the whole time. Let $a_{i}$ be the number of hops of size $2^{i}$ the frog makes, for $0 \leq i \leq n-1$. Claim - For any $... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2018.jsonl",
"solution_match": null
} | 148 | 1,270 |
2018 | T0 | 9 | null | USA_TSTST | Show that there is an absolute constant $c<1$ with the following property: whenever $\mathcal{P}$ is a polygon with area 1 in the plane, one can translate it by a distance of $\frac{1}{100}$ in some direction to obtain a polygon $\mathcal{Q}$, for which the intersection of the interiors of $\mathcal{P}$ and $\mathcal{Q... | Suppose $\mathcal{P}$ is a polygon of area 1 , and $\varepsilon>0$ is a constant, such that for any translate $\mathcal{Q}=\mathcal{P}+v$, where $v$ has length exactly $\frac{1}{100}$, the intersection of $\mathcal{P}$ and $\mathcal{Q}$ has area at least $1-\varepsilon$. The problem asks us to prove a lower bound on $\... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2018.jsonl",
"solution_match": null
} | 95 | 619 |
2019 | T0 | 1 | null | USA_TSTST | Find all binary operations $\diamond: \mathbb{R}_{>0} \times \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ (meaning $\diamond$ takes pairs of positive real numbers to positive real numbers) such that for any real numbers $a, b, c>0$, - the equation $a \diamond(b \diamond c)=(a \diamond b) \cdot c$ holds; and - if $a \g... | The answer is only multiplication and division, which both obviously work. γ First solution using Cauchy FE. We prove: Claim - We have $a \diamond b=a f(b)$ where $f$ is some involutive and totally multiplicative function. (In fact, this classifies all functions satisfying the first condition completely.) - Note that f... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2019.jsonl",
"solution_match": null
} | 116 | 570 |
2019 | T0 | 1 | null | USA_TSTST | Find all binary operations $\diamond: \mathbb{R}_{>0} \times \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ (meaning $\diamond$ takes pairs of positive real numbers to positive real numbers) such that for any real numbers $a, b, c>0$, - the equation $a \diamond(b \diamond c)=(a \diamond b) \cdot c$ holds; and - if $a \g... | The answer is only multiplication and division, which both obviously work. γ Second solution manually. As before we arrive at $a \diamond b=a f(b)$, with $f$ an involutive and totally multiplicative function. We prove that: Claim - For any $a>0$, we have $f(a) \in\{1 / a, a\}$. Assume that $a b>1$; we show $a=b$. Note ... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2019.jsonl",
"solution_match": null
} | 116 | 588 |
2019 | T0 | 2 | null | USA_TSTST | Let $A B C$ be an acute triangle with circumcircle $\Omega$ and orthocenter $H$. Points $D$ and $E$ lie on segments $A B$ and $A C$ respectively, such that $A D=A E$. The lines through $B$ and $C$ parallel to $\overline{D E}$ intersect $\Omega$ again at $P$ and $Q$, respectively. Denote by $\omega$ the circumcircle of ... | Claim - Points $L, D, P$ are collinear. $$ \begin{aligned} \measuredangle C L D & =\measuredangle D H L=\measuredangle D H A+\measuredangle A H L=\measuredangle D E A+\measuredangle A H C \\ & =\measuredangle A D E+\measuredangle C B A=\measuredangle A B P+\measuredangle C B A=\measuredangle C B P=\measuredangle C L P ... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2019.jsonl",
"solution_match": null
} | 157 | 1,921 |
2019 | T0 | 4 | null | USA_TSTST | Consider coins with positive real denominations not exceeding 1. Find the smallest $C>0$ such that the following holds: if we are given any 100 such coins with total value 50 , then we can always split them into two stacks of 50 coins each such that the absolute difference between the total values of the two stacks is ... | The answer is $C=\frac{50}{51}$. The lower bound is obtained if we have 51 coins of value $\frac{1}{51}$ and 49 coins of value 1. (Alternatively, 51 coins of value $1-\frac{\varepsilon}{51}$ and 49 coins of value $\frac{\varepsilon}{49}$ works fine for $\varepsilon>0$.) We now present two (similar) proofs that this $C=... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2019.jsonl",
"solution_match": null
} | 78 | 620 |
2019 | T0 | 5 | null | USA_TSTST | Let $A B C$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. A line through $H$ intersects segments $A B$ and $A C$ at $E$ and $F$, respectively. Let $K$ be the circumcenter of $\triangle A E F$, and suppose line $A K$ intersects $\Gamma$ again at a point $D$. Prove that line $H K$ and the line thro... | II Second solution (Ankan Bhattacharya). We let $D^{\prime}$ be the second intersection of $\overline{E F}$ with $(B H C)$ and redefine $D$ as the reflection of $D^{\prime}$ across $\overline{B C}$. We will first prove that this point $D$ coincides with the point $D$ given in the problem statement. The idea is that: Cl... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2019.jsonl",
"solution_match": null
} | 112 | 643 |
2019 | T0 | 5 | null | USA_TSTST | Let $A B C$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. A line through $H$ intersects segments $A B$ and $A C$ at $E$ and $F$, respectively. Let $K$ be the circumcenter of $\triangle A E F$, and suppose line $A K$ intersects $\Gamma$ again at a point $D$. Prove that line $H K$ and the line thro... | I Fourth solution, complex numbers with spiral similarity (Evan Chen). First if $\overline{A D} \perp \overline{B C}$ there is nothing to prove, so we assume this is not the case. Let $W$ be the antipode of $D$. Let $S$ denote the second intersection of $(A E F)$ and $(A B C)$. Consider the spiral similarity sending $\... | {
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2019.jsonl",
"solution_match": null
} | 112 | 1,652 |
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