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values | problem_raw stringlengths 14 10.4k | solution_raw stringlengths 1 24.1k | metadata dict |
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Find all pairs of positive integers $(a, b)$ satisfying the following conditions: (i) $a$ divides $b^{4}+1$, (ii) $b$ divides $a^{4}+1$, (iii) $\lfloor\sqrt{a}\rfloor=\lfloor\sqrt{b}\rfloor$. | Obviously, $\operatorname{gcd}(a, b)=1$, so the problem conditions imply $$ a b \mid(a-b)^{4}+1 $$ since each of $a$ and $b$ divide the right-hand side. We define $$ k \stackrel{\text { def }}{=} \frac{(b-a)^{4}+1}{a b} . $$ Claim (Size estimate) β We must have $k \leq 16$. $$ \begin{aligned} a b & \geq n^{2}\... | (1,1), (1,2) | Yes | Yes | math-word-problem | Number Theory | Find all pairs of positive integers $(a, b)$ satisfying the following conditions: (i) $a$ divides $b^{4}+1$, (ii) $b$ divides $a^{4}+1$, (iii) $\lfloor\sqrt{a}\rfloor=\lfloor\sqrt{b}\rfloor$. | Obviously, $\operatorname{gcd}(a, b)=1$, so the problem conditions imply $$ a b \mid(a-b)^{4}+1 $$ since each of $a$ and $b$ divide the right-hand side. We define $$ k \stackrel{\text { def }}{=} \frac{(b-a)^{4}+1}{a b} . $$ Claim (Size estimate) β We must have $k \leq 16$. $$ \begin{aligned} a b & \geq n^{2}\... | {
"exam": "USA_TSTST",
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Let $\mathbb{N}^{2}$ denote the set of ordered pairs of positive integers. A finite subset $S$ of $\mathbb{N}^{2}$ is stable if whenever $(x, y)$ is in $S$, then so are all points $\left(x^{\prime}, y^{\prime}\right)$ of $\mathbb{N}^{2}$ with both $x^{\prime} \leq x$ and $y^{\prime} \leq y$. Prove that if $S$ is a stab... | Suppose $|S| \geq 2$. For any $p \in S$, let $R(p)$ denote the stable rectangle with upper-right corner $p$. We say such $p$ is pivotal if $p+(1,1) \notin S$ and $|R(p)|$ is even.  Claim -... | proof | Yes | Yes | proof | Combinatorics | Let $\mathbb{N}^{2}$ denote the set of ordered pairs of positive integers. A finite subset $S$ of $\mathbb{N}^{2}$ is stable if whenever $(x, y)$ is in $S$, then so are all points $\left(x^{\prime}, y^{\prime}\right)$ of $\mathbb{N}^{2}$ with both $x^{\prime} \leq x$ and $y^{\prime} \leq y$. Prove that if $S$ is a stab... | Suppose $|S| \geq 2$. For any $p \in S$, let $R(p)$ denote the stable rectangle with upper-right corner $p$. We say such $p$ is pivotal if $p+(1,1) \notin S$ and $|R(p)|$ is even.  Claim -... | {
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Let $A, B, C, D$ be four points such that no three are collinear and $D$ is not the orthocenter of triangle $A B C$. Let $P, Q, R$ be the orthocenters of $\triangle B C D, \triangle C A D$, $\triangle A B D$, respectively. Suppose that lines $A P, B Q, C R$ are pairwise distinct and are concurrent. Show that the four p... | Let $T$ be the concurrency point, and let $H$ be the orthocenter of $\triangle A B C$.  Claim (Key claim) - $T$ is the midpoint of $\overline{A P}, \overline{B Q}, \overline{C R}, \overline{... | proof | Yes | Yes | proof | Geometry | Let $A, B, C, D$ be four points such that no three are collinear and $D$ is not the orthocenter of triangle $A B C$. Let $P, Q, R$ be the orthocenters of $\triangle B C D, \triangle C A D$, $\triangle A B D$, respectively. Suppose that lines $A P, B Q, C R$ are pairwise distinct and are concurrent. Show that the four p... | Let $T$ be the concurrency point, and let $H$ be the orthocenter of $\triangle A B C$.  Claim (Key claim) - $T$ is the midpoint of $\overline{A P}, \overline{B Q}, \overline{C R}, \overline{... | {
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Find all nonconstant polynomials $P(z)$ with complex coefficients for which all complex roots of the polynomials $P(z)$ and $P(z)-1$ have absolute value 1. | The answer is $P(x)$ should be a polynomial of the form $P(x)=\lambda x^{n}-\mu$ where $|\lambda|=|\mu|$ and $\operatorname{Re} \mu=-\frac{1}{2}$. One may check these all work; let's prove they are the only solutions. \l First approach (Evan Chen). We introduce the following notations: $$ \begin{aligned} P(x) & =c_{... | P(x)=\lambda x^{n}-\mu \text{ where } |\lambda|=|\mu| \text{ and } \operatorname{Re} \mu=-\frac{1}{2} | Yes | Yes | math-word-problem | Algebra | Find all nonconstant polynomials $P(z)$ with complex coefficients for which all complex roots of the polynomials $P(z)$ and $P(z)-1$ have absolute value 1. | The answer is $P(x)$ should be a polynomial of the form $P(x)=\lambda x^{n}-\mu$ where $|\lambda|=|\mu|$ and $\operatorname{Re} \mu=-\frac{1}{2}$. One may check these all work; let's prove they are the only solutions. \l First approach (Evan Chen). We introduce the following notations: $$ \begin{aligned} P(x) & =c_{... | {
"exam": "USA_TSTST",
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Find all nonconstant polynomials $P(z)$ with complex coefficients for which all complex roots of the polynomials $P(z)$ and $P(z)-1$ have absolute value 1. | The answer is $P(x)$ should be a polynomial of the form $P(x)=\lambda x^{n}-\mu$ where $|\lambda|=|\mu|$ and $\operatorname{Re} \mu=-\frac{1}{2}$. One may check these all work; let's prove they are the only solutions. II Second approach (from the author). We let $A=P$ and $B=P-1$ to make the notation more symmetric. ... | P(x)=\lambda x^{n}-\mu \text{ where } |\lambda|=|\mu| \text{ and } \operatorname{Re} \mu=-\frac{1}{2} | Yes | Yes | math-word-problem | Algebra | Find all nonconstant polynomials $P(z)$ with complex coefficients for which all complex roots of the polynomials $P(z)$ and $P(z)-1$ have absolute value 1. | The answer is $P(x)$ should be a polynomial of the form $P(x)=\lambda x^{n}-\mu$ where $|\lambda|=|\mu|$ and $\operatorname{Re} \mu=-\frac{1}{2}$. One may check these all work; let's prove they are the only solutions. II Second approach (from the author). We let $A=P$ and $B=P-1$ to make the notation more symmetric. ... | {
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For every positive integer $N$, let $\sigma(N)$ denote the sum of the positive integer divisors of $N$. Find all integers $m \geq n \geq 2$ satisfying $$ \frac{\sigma(m)-1}{m-1}=\frac{\sigma(n)-1}{n-1}=\frac{\sigma(m n)-1}{m n-1} . $$ | The answer is that $m$ and $n$ should be powers of the same prime number. These all work because for a prime power we have $$ \frac{\sigma\left(p^{e}\right)-1}{p^{e}-1}=\frac{\left(1+p+\cdots+p^{e}\right)-1}{p^{e}-1}=\frac{p\left(1+\cdots+p^{e-1}\right)}{p^{e}-1}=\frac{p}{p-1} $$ So we now prove these are the only ... | proof | Yes | Yes | math-word-problem | Number Theory | For every positive integer $N$, let $\sigma(N)$ denote the sum of the positive integer divisors of $N$. Find all integers $m \geq n \geq 2$ satisfying $$ \frac{\sigma(m)-1}{m-1}=\frac{\sigma(n)-1}{n-1}=\frac{\sigma(m n)-1}{m n-1} . $$ | The answer is that $m$ and $n$ should be powers of the same prime number. These all work because for a prime power we have $$ \frac{\sigma\left(p^{e}\right)-1}{p^{e}-1}=\frac{\left(1+p+\cdots+p^{e}\right)-1}{p^{e}-1}=\frac{p\left(1+\cdots+p^{e-1}\right)}{p^{e}-1}=\frac{p}{p-1} $$ So we now prove these are the only ... | {
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Ten million fireflies are glowing in $\mathbb{R}^{3}$ at midnight. Some of the fireflies are friends, and friendship is always mutual. Every second, one firefly moves to a new position so that its distance from each one of its friends is the same as it was before moving. This is the only way that the fireflies ever cha... | In general, we show that when $n \geq 70$, the answer is $f(n)=\left\lfloor\frac{n^{2}}{3}\right\rfloor$. Construction: Choose three pairwise parallel lines $\ell_{A}, \ell_{B}, \ell_{C}$ forming an infinite equilateral triangle prism (with side larger than 1). Split the $n$ fireflies among the lines as equally as po... | \left\lfloor\frac{n^{2}}{3}\right\rfloor | Yes | Incomplete | math-word-problem | Combinatorics | Ten million fireflies are glowing in $\mathbb{R}^{3}$ at midnight. Some of the fireflies are friends, and friendship is always mutual. Every second, one firefly moves to a new position so that its distance from each one of its friends is the same as it was before moving. This is the only way that the fireflies ever cha... | In general, we show that when $n \geq 70$, the answer is $f(n)=\left\lfloor\frac{n^{2}}{3}\right\rfloor$. Construction: Choose three pairwise parallel lines $\ell_{A}, \ell_{B}, \ell_{C}$ forming an infinite equilateral triangle prism (with side larger than 1). Split the $n$ fireflies among the lines as equally as po... | {
"exam": "USA_TSTST",
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Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | γ First solution, angle chasing only (Ankit Bisain). Let lines $B Q$ and $D P$ meet $(A B C D)$ again at $D^{\prime}$ and $B^{\prime}$, respectively.  Then $B B^{\prime} \| P X$ and $D D^{\p... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | γ First solution, angle chasing only (Ankit Bisain). Let lines $B Q$ and $D P$ meet $(A B C D)$ again at $D^{\prime}$ and $B^{\prime}$, respectively.  Then $B B^{\prime} \| P X$ and $D D^{\p... | {
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Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | γ Second solution via isosceles triangles (from contestants). Let $T=\overline{B Q} \cap \overline{D P}$.  Note that $P Q T$ is isosceles because $$ \measuredangle P Q T=\measuredangle Y Q ... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | γ Second solution via isosceles triangles (from contestants). Let $T=\overline{B Q} \cap \overline{D P}$.  Note that $P Q T$ is isosceles because $$ \measuredangle P Q T=\measuredangle Y Q ... | {
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Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | ΰ€¬ Third solution using a parallelogram (from contestants). Let $(B C Y)$ meet $\overline{A B}$ again at $W$ and let $(A D X)$ meet $\overline{C D}$ again at $Z$. Additionally, let $O_{1}$ be the center of $(A D X)$ and $O_{2}$ be the center of $(B C Y)$. . Let $(B C Y)$ meet $\overline{A B}$ again at $W$ and let $(A D X)$ meet $\overline{C D}$ again at $Z$. Additionally, let $O_{1}$ be the center of $(A D X)$ and $O_{2}$ be the center of $(B C Y)$. ![](https://cdn.mathpix.com/cropped/2024_11_19_148cb373ffbe5370b... | {
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Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | ΰ€¬ Fourth solution using congruent circles (from contestants). Let the angle bisector of $\measuredangle B O D$ meet $\overline{X Y}$ at $K$.  Then $(B Q O K)$ is cyclic because $\measuredang... | proof | Yes | Incomplete | proof | Geometry | Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | ΰ€¬ Fourth solution using congruent circles (from contestants). Let the angle bisector of $\measuredangle B O D$ meet $\overline{X Y}$ at $K$.  Then $(B Q O K)$ is cyclic because $\measuredang... | {
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Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | γ Fifth solution by ratio calculation (from contestants). Let $\overline{X Y}$ meet $(A B C D)$ at $X^{\prime}$ and $Y^{\prime}$.  Since $\measuredangle Y^{\prime} B D=\measuredangle P X^{\pr... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | γ Fifth solution by ratio calculation (from contestants). Let $\overline{X Y}$ meet $(A B C D)$ at $X^{\prime}$ and $Y^{\prime}$.  Since $\measuredangle Y^{\prime} B D=\measuredangle P X^{\pr... | {
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Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | I Sixth solution using radical axis (from author). Without loss of generality, assume $\overline{A D} \nVdash \overline{B C}$, as this case holds by continuity. Let $(B C Y)$ meet $\overline{A B}$ again at $W$, let $(A D X)$ meet $\overline{C D}$ again at $Z$, and let $\overline{W Z}$ meet $(A D X)$ and $(B C Y)$ agai... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | I Sixth solution using radical axis (from author). Without loss of generality, assume $\overline{A D} \nVdash \overline{B C}$, as this case holds by continuity. Let $(B C Y)$ meet $\overline{A B}$ again at $W$, let $(A D X)$ meet $\overline{C D}$ again at $Z$, and let $\overline{W Z}$ meet $(A D X)$ and $(B C Y)$ agai... | {
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Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | γ Seventh solution using Cayley-Bacharach (author). Define points $W, Z, R, S$ as in the previous solution.  The quartics $(A D X Z) \cup(B C W Y)$ and $\overline{X Y} \cup \overline{W Z} \cu... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $A B$ and $C D$, respectively. Suppose the circumcircles of $A D X$ and $B C Y$ meet line $X Y$ again at $P$ and $Q$, respectively. Show that $O P=O Q$. | γ Seventh solution using Cayley-Bacharach (author). Define points $W, Z, R, S$ as in the previous solution.  The quartics $(A D X Z) \cup(B C W Y)$ and $\overline{X Y} \cup \overline{W Z} \cu... | {
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Let $a_{1}<a_{2}<a_{3}<a_{4}<\cdots$ be an infinite sequence of real numbers in the interval $(0,1)$. Show that there exists a number that occurs exactly once in the sequence $$ \frac{a_{1}}{1}, \frac{a_{2}}{2}, \frac{a_{3}}{3}, \frac{a_{4}}{4}, \ldots $$ | γ Solution 1 (Merlijn Staps). We argue by contradiction, so suppose that for each $\lambda$ for which the set $S_{\lambda}=\left\{k: a_{k} / k=\lambda\right\}$ is non-empty, it contains at least two elements. Note that $S_{\lambda}$ is always a finite set because $a_{k}=k \lambda$ implies $k<1 / \lambda$. Write $m_{\l... | proof | Yes | Yes | proof | Number Theory | Let $a_{1}<a_{2}<a_{3}<a_{4}<\cdots$ be an infinite sequence of real numbers in the interval $(0,1)$. Show that there exists a number that occurs exactly once in the sequence $$ \frac{a_{1}}{1}, \frac{a_{2}}{2}, \frac{a_{3}}{3}, \frac{a_{4}}{4}, \ldots $$ | γ Solution 1 (Merlijn Staps). We argue by contradiction, so suppose that for each $\lambda$ for which the set $S_{\lambda}=\left\{k: a_{k} / k=\lambda\right\}$ is non-empty, it contains at least two elements. Note that $S_{\lambda}$ is always a finite set because $a_{k}=k \lambda$ implies $k<1 / \lambda$. Write $m_{\l... | {
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Let $a_{1}<a_{2}<a_{3}<a_{4}<\cdots$ be an infinite sequence of real numbers in the interval $(0,1)$. Show that there exists a number that occurs exactly once in the sequence $$ \frac{a_{1}}{1}, \frac{a_{2}}{2}, \frac{a_{3}}{3}, \frac{a_{4}}{4}, \ldots $$ | γ Solution 2 (Sanjana Das). Assume for the sake of contradiction that no number appears exactly once in the sequence. For every $i<j$ with $a_{i} / i=a_{j} / j$, draw an edge between $i$ and $j$, so every $i$ has an edge (and being connected by an edge is a transitive property). Call $i$ good if it has an edge with so... | proof | Yes | Yes | proof | Number Theory | Let $a_{1}<a_{2}<a_{3}<a_{4}<\cdots$ be an infinite sequence of real numbers in the interval $(0,1)$. Show that there exists a number that occurs exactly once in the sequence $$ \frac{a_{1}}{1}, \frac{a_{2}}{2}, \frac{a_{3}}{3}, \frac{a_{4}}{4}, \ldots $$ | γ Solution 2 (Sanjana Das). Assume for the sake of contradiction that no number appears exactly once in the sequence. For every $i<j$ with $a_{i} / i=a_{j} / j$, draw an edge between $i$ and $j$, so every $i$ has an edge (and being connected by an edge is a transitive property). Call $i$ good if it has an edge with so... | {
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Let $a_{1}<a_{2}<a_{3}<a_{4}<\cdots$ be an infinite sequence of real numbers in the interval $(0,1)$. Show that there exists a number that occurs exactly once in the sequence $$ \frac{a_{1}}{1}, \frac{a_{2}}{2}, \frac{a_{3}}{3}, \frac{a_{4}}{4}, \ldots $$ | γ Solution 3 (Gopal Goel). Suppose for sake of contradiction that the problem is false. Call an index $i$ a pin if $$ \frac{a_{j}}{j}=\frac{a_{i}}{i} \Longrightarrow j \geq i $$ ## Lemma There exists $k$ such that if we have $\frac{a_{i}}{i}=\frac{a_{j}}{j}$ with $j>i \geq k$, then $j \leq 1.1 i$. Suppose no such... | proof | Yes | Yes | proof | Number Theory | Let $a_{1}<a_{2}<a_{3}<a_{4}<\cdots$ be an infinite sequence of real numbers in the interval $(0,1)$. Show that there exists a number that occurs exactly once in the sequence $$ \frac{a_{1}}{1}, \frac{a_{2}}{2}, \frac{a_{3}}{3}, \frac{a_{4}}{4}, \ldots $$ | γ Solution 3 (Gopal Goel). Suppose for sake of contradiction that the problem is false. Call an index $i$ a pin if $$ \frac{a_{j}}{j}=\frac{a_{i}}{i} \Longrightarrow j \geq i $$ ## Lemma There exists $k$ such that if we have $\frac{a_{i}}{i}=\frac{a_{j}}{j}$ with $j>i \geq k$, then $j \leq 1.1 i$. Suppose no such... | {
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Find all positive integers $k>1$ for which there exists a positive integer $n$ such that $\binom{n}{k}$ is divisible by $n$, and $\binom{n}{m}$ is not divisible by $n$ for $2 \leq m<k$. | Such an $n$ exists for any $k$. First, suppose $k$ is prime. We choose $n=(k-1)$ !. For $m<k$, it follows from $m!\mid n$ that $$ \begin{aligned} (n-1)(n-2) \cdots(n-m+1) & \equiv(-1)(-2) \cdots(-m+1) \\ & \equiv(-1)^{m-1}(m-1)! \\ & \not \equiv 0 \bmod m! \end{aligned} $$ We see that $\binom{n}{m}$ is not a multip... | proof | Yes | Incomplete | math-word-problem | Combinatorics | Find all positive integers $k>1$ for which there exists a positive integer $n$ such that $\binom{n}{k}$ is divisible by $n$, and $\binom{n}{m}$ is not divisible by $n$ for $2 \leq m<k$. | Such an $n$ exists for any $k$. First, suppose $k$ is prime. We choose $n=(k-1)$ !. For $m<k$, it follows from $m!\mid n$ that $$ \begin{aligned} (n-1)(n-2) \cdots(n-m+1) & \equiv(-1)(-2) \cdots(-m+1) \\ & \equiv(-1)^{m-1}(m-1)! \\ & \not \equiv 0 \bmod m! \end{aligned} $$ We see that $\binom{n}{m}$ is not a multip... | {
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Let $a$ and $b$ be positive integers. Suppose that there are infinitely many pairs of positive integers $(m, n)$ for which $m^{2}+a n+b$ and $n^{2}+a m+b$ are both perfect squares. Prove that $a$ divides $2 b$. | Treating $a$ and $b$ as fixed, we are given that there are infinitely many quadrpules $(m, n, r, s)$ which satisfy the system $$ \begin{aligned} m^{2}+a n+b & =(m+r)^{2} \\ n^{2}+a m+b & =(n+s)^{2} \end{aligned} $$ We say that $(r, s)$ is exceptional if there exists infinitely many $(m, n)$ that satisfy. Claim - If... | proof | Yes | Yes | proof | Number Theory | Let $a$ and $b$ be positive integers. Suppose that there are infinitely many pairs of positive integers $(m, n)$ for which $m^{2}+a n+b$ and $n^{2}+a m+b$ are both perfect squares. Prove that $a$ divides $2 b$. | Treating $a$ and $b$ as fixed, we are given that there are infinitely many quadrpules $(m, n, r, s)$ which satisfy the system $$ \begin{aligned} m^{2}+a n+b & =(m+r)^{2} \\ n^{2}+a m+b & =(n+s)^{2} \end{aligned} $$ We say that $(r, s)$ is exceptional if there exists infinitely many $(m, n)$ that satisfy. Claim - If... | {
"exam": "USA_TSTST",
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Let $T$ be a tree on $n$ vertices with exactly $k$ leaves. Suppose that there exists a subset of at least $\frac{n+k-1}{2}$ vertices of $T$, no two of which are adjacent. Show that the longest path in $T$ contains an even number of edges. | The longest path in $T$ must go between two leaves. The solutions presented here will solve the problem by showing that in the unique 2-coloring of $T$, all leaves are the same color. ## γ Solution 1 (Ankan Bhattacharya, Jeffery Li). ## Lemma If $S$ is an independent set of $T$, then $$ \sum_{v \in S} \operatornam... | proof | Yes | Yes | proof | Combinatorics | Let $T$ be a tree on $n$ vertices with exactly $k$ leaves. Suppose that there exists a subset of at least $\frac{n+k-1}{2}$ vertices of $T$, no two of which are adjacent. Show that the longest path in $T$ contains an even number of edges. | The longest path in $T$ must go between two leaves. The solutions presented here will solve the problem by showing that in the unique 2-coloring of $T$, all leaves are the same color. ## γ Solution 1 (Ankan Bhattacharya, Jeffery Li). ## Lemma If $S$ is an independent set of $T$, then $$ \sum_{v \in S} \operatornam... | {
"exam": "USA_TSTST",
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Let $T$ be a tree on $n$ vertices with exactly $k$ leaves. Suppose that there exists a subset of at least $\frac{n+k-1}{2}$ vertices of $T$, no two of which are adjacent. Show that the longest path in $T$ contains an even number of edges. | The longest path in $T$ must go between two leaves. The solutions presented here will solve the problem by showing that in the unique 2-coloring of $T$, all leaves are the same color. ## γ Solution 2 (Andrew Gu). ## Lemma The vertices of $T$ can be partitioned into $k-1$ paths (i.e. the induced subgraph on each set... | proof | Yes | Yes | proof | Combinatorics | Let $T$ be a tree on $n$ vertices with exactly $k$ leaves. Suppose that there exists a subset of at least $\frac{n+k-1}{2}$ vertices of $T$, no two of which are adjacent. Show that the longest path in $T$ contains an even number of edges. | The longest path in $T$ must go between two leaves. The solutions presented here will solve the problem by showing that in the unique 2-coloring of $T$, all leaves are the same color. ## γ Solution 2 (Andrew Gu). ## Lemma The vertices of $T$ can be partitioned into $k-1$ paths (i.e. the induced subgraph on each set... | {
"exam": "USA_TSTST",
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Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | proof | Yes | Yes | proof | Geometry | Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | {
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Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | proof | Yes | Yes | proof | Geometry | Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | {
"exam": "USA_TSTST",
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Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | proof | Yes | Yes | proof | Geometry | Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | {
"exam": "USA_TSTST",
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Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | proof | Yes | Yes | proof | Geometry | Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | {
"exam": "USA_TSTST",
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Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | proof | Yes | Yes | proof | Geometry | Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | {
"exam": "USA_TSTST",
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Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | proof | Yes | Yes | proof | Geometry | Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | {
"exam": "USA_TSTST",
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Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | proof | Yes | Yes | proof | Geometry | Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | {
"exam": "USA_TSTST",
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Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | proof | Yes | Incomplete | proof | Geometry | Triangles $A B C$ and $D E F$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A, F$, $B, D, C$, and $E$ occur in this order along $\Omega$. Let $\Delta_{A}$ be the triangle formed by lines $A B, A C$, and $E F$, and define triangles $\Delta_{B}, \Delta_{C}, \ldots, \Delta_{F}$ similarly. Furthermore, ... |  Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_{A}$ and $I_{A}$ be the circumcenter and incenter of triangle $\Delta_... | {
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Let $M$ be a finite set of lattice points and $n$ be a positive integer. A mine-avoiding path is a path of lattice points with length $n$, beginning at $(0,0)$ and ending at a point on the line $x+y=n$, that does not contain any point in $M$. Prove that if there exists a mine-avoiding path, then there exist at least $2... | γ Solution 1. We prove the statement by induction on $n$. We use $n=0$ as a base case, where the statement follows from $1 \geq 2^{-|M|}$. For the inductive step, let $n>0$. There exists at least one mine-avoiding path, which must pass through either $(0,1)$ or $(1,0)$. We consider two cases: Case 1: there exist mine... | proof | Yes | Yes | proof | Combinatorics | Let $M$ be a finite set of lattice points and $n$ be a positive integer. A mine-avoiding path is a path of lattice points with length $n$, beginning at $(0,0)$ and ending at a point on the line $x+y=n$, that does not contain any point in $M$. Prove that if there exists a mine-avoiding path, then there exist at least $2... | γ Solution 1. We prove the statement by induction on $n$. We use $n=0$ as a base case, where the statement follows from $1 \geq 2^{-|M|}$. For the inductive step, let $n>0$. There exists at least one mine-avoiding path, which must pass through either $(0,1)$ or $(1,0)$. We consider two cases: Case 1: there exist mine... | {
"exam": "USA_TSTST",
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Let $M$ be a finite set of lattice points and $n$ be a positive integer. A mine-avoiding path is a path of lattice points with length $n$, beginning at $(0,0)$ and ending at a point on the line $x+y=n$, that does not contain any point in $M$. Prove that if there exists a mine-avoiding path, then there exist at least $2... | ## γ Solution 2. ## Lemma If $|M|<n$, there is more than one mine-avoiding path. - Make the first $i+1$ points $P_{0}, P_{1}, \ldots, P_{i}$. - If $P_{i} \rightarrow P_{i+1}$ is one unit up, go right until $\left(n-y_{i}, y_{i}\right)$. - If $P_{i} \rightarrow P_{i+1}$ is one unit right, go up until $\left(x_{i}, ... | 2^{n-|M|} | Yes | Incomplete | proof | Combinatorics | Let $M$ be a finite set of lattice points and $n$ be a positive integer. A mine-avoiding path is a path of lattice points with length $n$, beginning at $(0,0)$ and ending at a point on the line $x+y=n$, that does not contain any point in $M$. Prove that if there exists a mine-avoiding path, then there exist at least $2... | ## γ Solution 2. ## Lemma If $|M|<n$, there is more than one mine-avoiding path. - Make the first $i+1$ points $P_{0}, P_{1}, \ldots, P_{i}$. - If $P_{i} \rightarrow P_{i+1}$ is one unit up, go right until $\left(n-y_{i}, y_{i}\right)$. - If $P_{i} \rightarrow P_{i+1}$ is one unit right, go up until $\left(x_{i}, ... | {
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Let $A B C$ be a scalene triangle. Points $A_{1}, B_{1}$ and $C_{1}$ are chosen on segments $B C$, $C A$, and $A B$, respectively, such that $\triangle A_{1} B_{1} C_{1}$ and $\triangle A B C$ are similar. Let $A_{2}$ be the unique point on line $B_{1} C_{1}$ such that $A A_{2}=A_{1} A_{2}$. Points $B_{2}$ and $C_{2}$ ... | ΰ€¬ Solution 1 (author). We'll use the following lemma. ## Lemma Suppose that $P Q R S$ is a convex quadrilateral with $\angle P=\angle R$. Then there is a point $T$ on $Q S$ such that $\angle Q P T=\angle S R P, \angle T R Q=\angle R P S$, and $P T=R T$. Before proving the lemma, we will show how it solves the probl... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a scalene triangle. Points $A_{1}, B_{1}$ and $C_{1}$ are chosen on segments $B C$, $C A$, and $A B$, respectively, such that $\triangle A_{1} B_{1} C_{1}$ and $\triangle A B C$ are similar. Let $A_{2}$ be the unique point on line $B_{1} C_{1}$ such that $A A_{2}=A_{1} A_{2}$. Points $B_{2}$ and $C_{2}$ ... | ΰ€¬ Solution 1 (author). We'll use the following lemma. ## Lemma Suppose that $P Q R S$ is a convex quadrilateral with $\angle P=\angle R$. Then there is a point $T$ on $Q S$ such that $\angle Q P T=\angle S R P, \angle T R Q=\angle R P S$, and $P T=R T$. Before proving the lemma, we will show how it solves the probl... | {
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Let $A B C$ be a scalene triangle. Points $A_{1}, B_{1}$ and $C_{1}$ are chosen on segments $B C$, $C A$, and $A B$, respectively, such that $\triangle A_{1} B_{1} C_{1}$ and $\triangle A B C$ are similar. Let $A_{2}$ be the unique point on line $B_{1} C_{1}$ such that $A A_{2}=A_{1} A_{2}$. Points $B_{2}$ and $C_{2}$ ... | Solution 2 (Ankan Bhattacharya). We prove the main claim $\frac{B_{1} A_{2}}{A_{2} C_{1}}=\frac{B A_{1}}{A_{1} C}$. Let $\triangle A_{0} B_{0} C_{0}$ be the medial triangle of $\triangle A B C$. In addition, let $A_{1}^{\prime}$ be the reflection of $A_{1}$ over $\overline{B_{1} C_{1}}$, and let $X$ be the point satis... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a scalene triangle. Points $A_{1}, B_{1}$ and $C_{1}$ are chosen on segments $B C$, $C A$, and $A B$, respectively, such that $\triangle A_{1} B_{1} C_{1}$ and $\triangle A B C$ are similar. Let $A_{2}$ be the unique point on line $B_{1} C_{1}$ such that $A A_{2}=A_{1} A_{2}$. Points $B_{2}$ and $C_{2}$ ... | Solution 2 (Ankan Bhattacharya). We prove the main claim $\frac{B_{1} A_{2}}{A_{2} C_{1}}=\frac{B A_{1}}{A_{1} C}$. Let $\triangle A_{0} B_{0} C_{0}$ be the medial triangle of $\triangle A B C$. In addition, let $A_{1}^{\prime}$ be the reflection of $A_{1}$ over $\overline{B_{1} C_{1}}$, and let $X$ be the point satis... | {
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Let $q=p^{r}$ for a prime number $p$ and positive integer $r$. Let $\zeta=e^{\frac{2 \pi i}{q}}$. Find the least positive integer $n$ such that $$ \sum_{\substack{1 \leq k \leq q \\ \operatorname{gcd}(k, p)=1}} \frac{1}{\left(1-\zeta^{k}\right)^{n}} $$ is not an integer. (The sum is over all $1 \leq k \leq q$ with $p... | Let $S_{q}$ denote the set of primitive $q$ th roots of unity (thus, the sum in question is a sum over $S_{q}$ ). γ Solution 1 (author). Let $\zeta_{p}=e^{2 \pi i / p}$ be a fixed primitive $p$ th root of unity. Observe that the given sum is an integer for all $n \leq 0$ (e.g. because the sum is an integer symmetric ... | z_{q}+1 | Yes | Incomplete | math-word-problem | Number Theory | Let $q=p^{r}$ for a prime number $p$ and positive integer $r$. Let $\zeta=e^{\frac{2 \pi i}{q}}$. Find the least positive integer $n$ such that $$ \sum_{\substack{1 \leq k \leq q \\ \operatorname{gcd}(k, p)=1}} \frac{1}{\left(1-\zeta^{k}\right)^{n}} $$ is not an integer. (The sum is over all $1 \leq k \leq q$ with $p... | Let $S_{q}$ denote the set of primitive $q$ th roots of unity (thus, the sum in question is a sum over $S_{q}$ ). γ Solution 1 (author). Let $\zeta_{p}=e^{2 \pi i / p}$ be a fixed primitive $p$ th root of unity. Observe that the given sum is an integer for all $n \leq 0$ (e.g. because the sum is an integer symmetric ... | {
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Let $q=p^{r}$ for a prime number $p$ and positive integer $r$. Let $\zeta=e^{\frac{2 \pi i}{q}}$. Find the least positive integer $n$ such that $$ \sum_{\substack{1 \leq k \leq q \\ \operatorname{gcd}(k, p)=1}} \frac{1}{\left(1-\zeta^{k}\right)^{n}} $$ is not an integer. (The sum is over all $1 \leq k \leq q$ with $p... | Let $S_{q}$ denote the set of primitive $q$ th roots of unity (thus, the sum in question is a sum over $S_{q}$ ). Solution 2 (Nikolai Beluhov). Suppose that the complex numbers $\frac{1}{1-\omega}$ for $\omega \in S_{q}$ are the roots of $$ P(x)=x^{d}-c_{1} x^{d-1}+c_{2} x^{d-2}-\cdots \pm c_{d} $$ so that $c_{k}$ ... | p-1 | Yes | Incomplete | math-word-problem | Number Theory | Let $q=p^{r}$ for a prime number $p$ and positive integer $r$. Let $\zeta=e^{\frac{2 \pi i}{q}}$. Find the least positive integer $n$ such that $$ \sum_{\substack{1 \leq k \leq q \\ \operatorname{gcd}(k, p)=1}} \frac{1}{\left(1-\zeta^{k}\right)^{n}} $$ is not an integer. (The sum is over all $1 \leq k \leq q$ with $p... | Let $S_{q}$ denote the set of primitive $q$ th roots of unity (thus, the sum in question is a sum over $S_{q}$ ). Solution 2 (Nikolai Beluhov). Suppose that the complex numbers $\frac{1}{1-\omega}$ for $\omega \in S_{q}$ are the roots of $$ P(x)=x^{d}-c_{1} x^{d-1}+c_{2} x^{d-2}-\cdots \pm c_{d} $$ so that $c_{k}$ ... | {
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Let $n$ be a positive integer. Find the smallest positive integer $k$ such that for any set $S$ of $n$ points in the interior of the unit square, there exists a set of $k$ rectangles such that the following hold: - The sides of each rectangle are parallel to the sides of the unit square. - Each point in $S$ is not in ... | We claim the answer is $k=2 n+2$. The lower bound is given by picking $$ S=\left\{\left(s_{1}, s_{1}\right),\left(s_{2}, s_{2}\right), \ldots,\left(s_{n}, s_{n}\right)\right\} $$ for some real numbers $0<s_{1}<s_{2}<\cdots<s_{n}<1$. Consider the $4 n$ points $$ S^{\prime}=S+\{(\varepsilon, 0),(0, \varepsilon),(-\... | 2n+2 | Yes | Yes | math-word-problem | Combinatorics | Let $n$ be a positive integer. Find the smallest positive integer $k$ such that for any set $S$ of $n$ points in the interior of the unit square, there exists a set of $k$ rectangles such that the following hold: - The sides of each rectangle are parallel to the sides of the unit square. - Each point in $S$ is not in ... | We claim the answer is $k=2 n+2$. The lower bound is given by picking $$ S=\left\{\left(s_{1}, s_{1}\right),\left(s_{2}, s_{2}\right), \ldots,\left(s_{n}, s_{n}\right)\right\} $$ for some real numbers $0<s_{1}<s_{2}<\cdots<s_{n}<1$. Consider the $4 n$ points $$ S^{\prime}=S+\{(\varepsilon, 0),(0, \varepsilon),(-\... | {
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Let $A B C$ be a triangle. Let $\theta$ be a fixed angle for which $$ \theta<\frac{1}{2} \min (\angle A, \angle B, \angle C) $$ Points $S_{A}$ and $T_{A}$ lie on segment $B C$ such that $\angle B A S_{A}=\angle T_{A} A C=\theta$. Let $P_{A}$ and $Q_{A}$ be the feet from $B$ and $C$ to $\overline{A S_{A}}$ and $\overl... | We discard the points $S_{A}$ and $T_{A}$ since they are only there to direct the angles correctly in the problem statement. γ First solution, by author. Let $X$ be the projection from $C$ to $A P_{A}, Y$ be the projection from $B$ to $A Q_{A}$.  $$ Points $S_{A}$ and $T_{A}$ lie on segment $B C$ such that $\angle B A S_{A}=\angle T_{A} A C=\theta$. Let $P_{A}$ and $Q_{A}$ be the feet from $B$ and $C$ to $\overline{A S_{A}}$ and $\overl... | We discard the points $S_{A}$ and $T_{A}$ since they are only there to direct the angles correctly in the problem statement. γ First solution, by author. Let $X$ be the projection from $C$ to $A P_{A}, Y$ be the projection from $B$ to $A Q_{A}$. ![](https://cdn.mathpix.com/cropped/2024_11_19_1799270e91f74486a977g-07.... | {
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Let $A B C$ be a triangle. Let $\theta$ be a fixed angle for which $$ \theta<\frac{1}{2} \min (\angle A, \angle B, \angle C) $$ Points $S_{A}$ and $T_{A}$ lie on segment $B C$ such that $\angle B A S_{A}=\angle T_{A} A C=\theta$. Let $P_{A}$ and $Q_{A}$ be the feet from $B$ and $C$ to $\overline{A S_{A}}$ and $\overl... | We discard the points $S_{A}$ and $T_{A}$ since they are only there to direct the angles correctly in the problem statement. γ Second solution via Jacobi, by Maxim Li. Let $D$ be the foot of the $A$-altitude. Note that line $B C$ is the external angle bisector of $\angle P_{A} D Q_{A}$. Claim - $\left(D P_{A} Q_{A}\... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle. Let $\theta$ be a fixed angle for which $$ \theta<\frac{1}{2} \min (\angle A, \angle B, \angle C) $$ Points $S_{A}$ and $T_{A}$ lie on segment $B C$ such that $\angle B A S_{A}=\angle T_{A} A C=\theta$. Let $P_{A}$ and $Q_{A}$ be the feet from $B$ and $C$ to $\overline{A S_{A}}$ and $\overl... | We discard the points $S_{A}$ and $T_{A}$ since they are only there to direct the angles correctly in the problem statement. γ Second solution via Jacobi, by Maxim Li. Let $D$ be the foot of the $A$-altitude. Note that line $B C$ is the external angle bisector of $\angle P_{A} D Q_{A}$. Claim - $\left(D P_{A} Q_{A}\... | {
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Determine all positive integers $N$ for which there exists a strictly increasing sequence of positive integers $s_{0}<s_{1}<s_{2}<\cdots$ satisfying the following properties: - the sequence $s_{1}-s_{0}, s_{2}-s_{1}, s_{3}-s_{2}, \ldots$ is periodic; and - $s_{s_{n}}-s_{s_{n-1}} \leq N<s_{1+s_{n}}-s_{s_{n-1}}$ for all... | γ Answer. All $N$ such that $t^{2} \leq N<t^{2}+t$ for some positive integer $t$. Solution 1 (local). If $t^{2} \leq N<t^{2}+t$ then the sequence $s_{n}=t n+1$ satisfies both conditions. It remains to show that no other values of $N$ work. Define $a_{n}:=s_{n}-s_{n-1}$, and let $p$ be the minimal period of $\left\{a... | t^{2} \leq N < t^{2} + t | Yes | Yes | math-word-problem | Number Theory | Determine all positive integers $N$ for which there exists a strictly increasing sequence of positive integers $s_{0}<s_{1}<s_{2}<\cdots$ satisfying the following properties: - the sequence $s_{1}-s_{0}, s_{2}-s_{1}, s_{3}-s_{2}, \ldots$ is periodic; and - $s_{s_{n}}-s_{s_{n-1}} \leq N<s_{1+s_{n}}-s_{s_{n-1}}$ for all... | γ Answer. All $N$ such that $t^{2} \leq N<t^{2}+t$ for some positive integer $t$. Solution 1 (local). If $t^{2} \leq N<t^{2}+t$ then the sequence $s_{n}=t n+1$ satisfies both conditions. It remains to show that no other values of $N$ work. Define $a_{n}:=s_{n}-s_{n-1}$, and let $p$ be the minimal period of $\left\{a... | {
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Determine all positive integers $N$ for which there exists a strictly increasing sequence of positive integers $s_{0}<s_{1}<s_{2}<\cdots$ satisfying the following properties: - the sequence $s_{1}-s_{0}, s_{2}-s_{1}, s_{3}-s_{2}, \ldots$ is periodic; and - $s_{s_{n}}-s_{s_{n-1}} \leq N<s_{1+s_{n}}-s_{s_{n-1}}$ for all... | γ Answer. All $N$ such that $t^{2} \leq N<t^{2}+t$ for some positive integer $t$. γ Solution 2 (global). Define $\left\{a_{n}\right\}$ and $f$ as in the previous solution. We first show that $s_{i} \not \equiv s_{j}(\bmod p)$ for all $i<j<i+p$. Suppose the contrary, i.e. that $s_{i} \equiv s_{j}$ $(\bmod p)$ for some... | t^{2} \leq N<t^{2}+t | Yes | Yes | math-word-problem | Number Theory | Determine all positive integers $N$ for which there exists a strictly increasing sequence of positive integers $s_{0}<s_{1}<s_{2}<\cdots$ satisfying the following properties: - the sequence $s_{1}-s_{0}, s_{2}-s_{1}, s_{3}-s_{2}, \ldots$ is periodic; and - $s_{s_{n}}-s_{s_{n-1}} \leq N<s_{1+s_{n}}-s_{s_{n-1}}$ for all... | γ Answer. All $N$ such that $t^{2} \leq N<t^{2}+t$ for some positive integer $t$. γ Solution 2 (global). Define $\left\{a_{n}\right\}$ and $f$ as in the previous solution. We first show that $s_{i} \not \equiv s_{j}(\bmod p)$ for all $i<j<i+p$. Suppose the contrary, i.e. that $s_{i} \equiv s_{j}$ $(\bmod p)$ for some... | {
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A function $f: \mathbb{N} \rightarrow \mathbb{N}$ has the property that for all positive integers $m$ and $n$, exactly one of the $f(n)$ numbers $$ f(m+1), f(m+2), \ldots, f(m+f(n)) $$ is divisible by $n$. Prove that $f(n)=n$ for infinitely many positive integers $n$. | We start with the following claim: $$ \text { Claim - If } a \mid b \text { then } f(a) \mid f(b) $$ In what follows, let $a \geq 2$ be any positive integer. Because $f(a)$ and $f(2 a)$ are both divisible by $f(a)$, there are $a+1$ consecutive values of $f$ of which at least two divisible by $f(a)$. It follows tha... | proof | Yes | Yes | proof | Number Theory | A function $f: \mathbb{N} \rightarrow \mathbb{N}$ has the property that for all positive integers $m$ and $n$, exactly one of the $f(n)$ numbers $$ f(m+1), f(m+2), \ldots, f(m+f(n)) $$ is divisible by $n$. Prove that $f(n)=n$ for infinitely many positive integers $n$. | We start with the following claim: $$ \text { Claim - If } a \mid b \text { then } f(a) \mid f(b) $$ In what follows, let $a \geq 2$ be any positive integer. Because $f(a)$ and $f(2 a)$ are both divisible by $f(a)$, there are $a+1$ consecutive values of $f$ of which at least two divisible by $f(a)$. It follows tha... | {
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Let $A_{1}, \ldots, A_{2022}$ be the vertices of a regular 2022 -gon in the plane. Alice and Bob play a game. Alice secretly chooses a line and colors all points in the plane on one side of the line blue, and all points on the other side of the line red. Points on the line are colored blue, so every point in the plane ... | The answer is 22 . To prove the lower bound, note that there are $2022 \cdot 2021+2>2^{21}$ possible colorings. If Bob makes less than 22 queries, then he can only output $2^{21}$ possible colorings, which means he is wrong on some coloring. Now we show Bob can always win in 22 queries. A key observation is that the... | 22 | Yes | Yes | math-word-problem | Combinatorics | Let $A_{1}, \ldots, A_{2022}$ be the vertices of a regular 2022 -gon in the plane. Alice and Bob play a game. Alice secretly chooses a line and colors all points in the plane on one side of the line blue, and all points on the other side of the line red. Points on the line are colored blue, so every point in the plane ... | The answer is 22 . To prove the lower bound, note that there are $2022 \cdot 2021+2>2^{21}$ possible colorings. If Bob makes less than 22 queries, then he can only output $2^{21}$ possible colorings, which means he is wrong on some coloring. Now we show Bob can always win in 22 queries. A key observation is that the... | {
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Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $A B C$. The perpendicular bisector of $\overline{A H}$ intersects $\overline{A B}$ and $\overline{A C}$ at $X_{A}$ and $Y_{A}$ respectively. Let $K_{A}$ denote the intersection of the circumcircles of triangles $O X_{A} Y_{... | \ First solution, by author. Let $\odot O X_{A} Y_{A}$ intersects $A B, A C$ again at $U, V$. Then by Reim's theorem $U V C B$ are concyclic. Hence the radical axis of $\odot O X_{A} Y_{A}, \odot O B C$ and $\odot(U V C B)$ are concurrent, i.e. $O K_{A}, B C, U V$ are concurrent, Denote the intersection as $K_{A}^{*}$... | proof | Yes | Yes | proof | Geometry | Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $A B C$. The perpendicular bisector of $\overline{A H}$ intersects $\overline{A B}$ and $\overline{A C}$ at $X_{A}$ and $Y_{A}$ respectively. Let $K_{A}$ denote the intersection of the circumcircles of triangles $O X_{A} Y_{... | \ First solution, by author. Let $\odot O X_{A} Y_{A}$ intersects $A B, A C$ again at $U, V$. Then by Reim's theorem $U V C B$ are concyclic. Hence the radical axis of $\odot O X_{A} Y_{A}, \odot O B C$ and $\odot(U V C B)$ are concurrent, i.e. $O K_{A}, B C, U V$ are concurrent, Denote the intersection as $K_{A}^{*}$... | {
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Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $A B C$. The perpendicular bisector of $\overline{A H}$ intersects $\overline{A B}$ and $\overline{A C}$ at $X_{A}$ and $Y_{A}$ respectively. Let $K_{A}$ denote the intersection of the circumcircles of triangles $O X_{A} Y_{... | I Second solution, from Jeffrey Kwan. Let $O_{A}$ be the circumcenter of $\triangle A X_{A} Y_{A}$. The key claim is that: Claim $-O_{A} X_{A} Y_{A} O$ is cyclic. $$ \frac{A X_{A}}{A B}=\frac{A H / 2}{A D}=\frac{R \cos A}{A D} $$ and so $$ \frac{A O}{A D}=R \cdot \frac{A X_{A}}{A B \cdot R \cos A}=\frac{A X_{A}}{... | proof | Yes | Yes | proof | Geometry | Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $A B C$. The perpendicular bisector of $\overline{A H}$ intersects $\overline{A B}$ and $\overline{A C}$ at $X_{A}$ and $Y_{A}$ respectively. Let $K_{A}$ denote the intersection of the circumcircles of triangles $O X_{A} Y_{... | I Second solution, from Jeffrey Kwan. Let $O_{A}$ be the circumcenter of $\triangle A X_{A} Y_{A}$. The key claim is that: Claim $-O_{A} X_{A} Y_{A} O$ is cyclic. $$ \frac{A X_{A}}{A B}=\frac{A H / 2}{A D}=\frac{R \cos A}{A D} $$ and so $$ \frac{A O}{A D}=R \cdot \frac{A X_{A}}{A B \cdot R \cos A}=\frac{A X_{A}}{... | {
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Let $A B C D$ be a parallelogram. Point $E$ lies on segment $C D$ such that $$ 2 \angle A E B=\angle A D B+\angle A C B, $$ and point $F$ lies on segment $B C$ such that $$ 2 \angle D F A=\angle D C A+\angle D B A . $$ Let $K$ be the circumcenter of triangle $A B D$. Prove that $K E=K F$. | Let the circle through $A, B$, and $E$ intersect $C D$ again at $E^{\prime}$, and let the circle through $D$, $A$, and $F$ intersect $B C$ again at $F^{\prime}$. Now $A B E E^{\prime}$ and $D A F^{\prime} F$ are cyclic quadrilaterals with two parallel sides, so they are isosceles trapezoids. From $K A=K B$, it now fo... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a parallelogram. Point $E$ lies on segment $C D$ such that $$ 2 \angle A E B=\angle A D B+\angle A C B, $$ and point $F$ lies on segment $B C$ such that $$ 2 \angle D F A=\angle D C A+\angle D B A . $$ Let $K$ be the circumcenter of triangle $A B D$. Prove that $K E=K F$. | Let the circle through $A, B$, and $E$ intersect $C D$ again at $E^{\prime}$, and let the circle through $D$, $A$, and $F$ intersect $B C$ again at $F^{\prime}$. Now $A B E E^{\prime}$ and $D A F^{\prime} F$ are cyclic quadrilaterals with two parallel sides, so they are isosceles trapezoids. From $K A=K B$, it now fo... | {
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Find all functions $f: \mathbb{N} \rightarrow \mathbb{Z}$ such that $$ \left\lfloor\frac{f(m n)}{n}\right\rfloor=f(m) $$ for all positive integers $m, n$. | There are two families of functions that work: for each $\alpha \in \mathbb{R}$ the function $f(n)=\lfloor\alpha n\rfloor$, and for each $\alpha \in \mathbb{R}$ the function $f(n)=\lceil\alpha n\rceil-1$. (For irrational $\alpha$ these two functions coincide.) It is straightforward to check that these functions indee... | proof | Yes | Yes | math-word-problem | Number Theory | Find all functions $f: \mathbb{N} \rightarrow \mathbb{Z}$ such that $$ \left\lfloor\frac{f(m n)}{n}\right\rfloor=f(m) $$ for all positive integers $m, n$. | There are two families of functions that work: for each $\alpha \in \mathbb{R}$ the function $f(n)=\lfloor\alpha n\rfloor$, and for each $\alpha \in \mathbb{R}$ the function $f(n)=\lceil\alpha n\rceil-1$. (For irrational $\alpha$ these two functions coincide.) It is straightforward to check that these functions indee... | {
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Let $k>1$ be a fixed positive integer. Prove that if $n$ is a sufficiently large positive integer, there exists a sequence of integers with the following properties: - Each element of the sequence is between 1 and $n$, inclusive. - For any two different contiguous subsequences of the sequence with length between 2 and... | For any positive integer $n$, define an $(n, k)$-good sequence to be a finite sequence of integers each between 1 and $n$ inclusive satisfying the second property in the problem statement. The problems asks to show that, for all sufficiently large integers $n$, there is an $(n, k)$-good sequence of length at least $0... | proof | Yes | Incomplete | proof | Combinatorics | Let $k>1$ be a fixed positive integer. Prove that if $n$ is a sufficiently large positive integer, there exists a sequence of integers with the following properties: - Each element of the sequence is between 1 and $n$, inclusive. - For any two different contiguous subsequences of the sequence with length between 2 and... | For any positive integer $n$, define an $(n, k)$-good sequence to be a finite sequence of integers each between 1 and $n$ inclusive satisfying the second property in the problem statement. The problems asks to show that, for all sufficiently large integers $n$, there is an $(n, k)$-good sequence of length at least $0... | {
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Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$ Prove that $\angle R A S+\angle B A C=\angle B G C$. |  γ Solution 1 using power of a point. From the given condition that $\measuredangle A C R=\measuredangle C G M$, we get that $$ M A^{2}=M C^{2}=M G \cdot M R \Longrightarrow \measuredangle... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$ Prove that $\angle R A S+\angle B A C=\angle B G C$. |  γ Solution 1 using power of a point. From the given condition that $\measuredangle A C R=\measuredangle C G M$, we get that $$ M A^{2}=M C^{2}=M G \cdot M R \Longrightarrow \measuredangle... | {
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Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$ Prove that $\angle R A S+\angle B A C=\angle B G C$. |  ΰ€¬ Solution 2 using similar triangles. As before, $\triangle M G C \sim \triangle M C R$ and $\triangle N G B \sim$ $\triangle N B S$. We obtain $$ \frac{|A C|}{|C R|}=\frac{2|M C|}{|C R|}... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$ Prove that $\angle R A S+\angle B A C=\angle B G C$. |  ΰ€¬ Solution 2 using similar triangles. As before, $\triangle M G C \sim \triangle M C R$ and $\triangle N G B \sim$ $\triangle N B S$. We obtain $$ \frac{|A C|}{|C R|}=\frac{2|M C|}{|C R|}... | {
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Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$ Prove that $\angle R A S+\angle B A C=\angle B G C$. |  I Solution 3 using parallelograms. Let $M$ and $N$ be defined as above. Let $P$ be the reflection of $G$ in $M$ and let $Q$ the reflection of $G$ in $N$. Then $A G C P$ and $A G B Q$ are p... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$ Prove that $\angle R A S+\angle B A C=\angle B G C$. |  I Solution 3 using parallelograms. Let $M$ and $N$ be defined as above. Let $P$ be the reflection of $G$ in $M$ and let $Q$ the reflection of $G$ in $N$. Then $A G C P$ and $A G B Q$ are p... | {
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Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$ Prove that $\angle R A S+\angle B A C=\angle B G C$. |  ΰ€¬ Solution 4 also using parallelograms, by Ankan Bhattacharya. Construct parallelograms $A R C K$ and $A S B L$. Since $$ \measuredangle C A K=\measuredangle A C R=\measuredangle C G B=\m... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$ Prove that $\angle R A S+\angle B A C=\angle B G C$. |  ΰ€¬ Solution 4 also using parallelograms, by Ankan Bhattacharya. Construct parallelograms $A R C K$ and $A S B L$. Since $$ \measuredangle C A K=\measuredangle A C R=\measuredangle C G B=\m... | {
"exam": "USA_TSTST",
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Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$ Prove that $\angle R A S+\angle B A C=\angle B G C$. |  γ Solution 5 using complex numbers, by Milan Haiman. Note that $\angle R A S+\angle B A C=$ $\angle B A S+\angle R A C$. We compute $\angle B A S$ in complex numbers; then $\angle R A C$ w... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$ Prove that $\angle R A S+\angle B A C=\angle B G C$. |  γ Solution 5 using complex numbers, by Milan Haiman. Note that $\angle R A S+\angle B A C=$ $\angle B A S+\angle R A C$. We compute $\angle B A S$ in complex numbers; then $\angle R A C$ w... | {
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Let $n \geq m \geq 1$ be integers. Prove that $$ \sum_{k=m}^{n}\left(\frac{1}{k^{2}}+\frac{1}{k^{3}}\right) \geq m \cdot\left(\sum_{k=m}^{n} \frac{1}{k^{2}}\right)^{2} $$ | γ First solution (authors). By Cauchy-Schwarz, we have $$ \begin{aligned} \sum_{k=m}^{n} \frac{k+1}{k^{3}} & =\sum_{k=m}^{n} \frac{\left(\frac{1}{k^{2}}\right)^{2}}{\frac{1}{k(k+1)}} \\ & \geq \frac{\left(\frac{1}{m^{2}}+\frac{1}{(m+1)^{2}}+\cdots+\frac{1}{n^{2}}\right)^{2}}{\frac{1}{m(m+1)}+\frac{1}{(m+1)(m+2)}+\cdo... | proof | Yes | Yes | proof | Inequalities | Let $n \geq m \geq 1$ be integers. Prove that $$ \sum_{k=m}^{n}\left(\frac{1}{k^{2}}+\frac{1}{k^{3}}\right) \geq m \cdot\left(\sum_{k=m}^{n} \frac{1}{k^{2}}\right)^{2} $$ | γ First solution (authors). By Cauchy-Schwarz, we have $$ \begin{aligned} \sum_{k=m}^{n} \frac{k+1}{k^{3}} & =\sum_{k=m}^{n} \frac{\left(\frac{1}{k^{2}}\right)^{2}}{\frac{1}{k(k+1)}} \\ & \geq \frac{\left(\frac{1}{m^{2}}+\frac{1}{(m+1)^{2}}+\cdots+\frac{1}{n^{2}}\right)^{2}}{\frac{1}{m(m+1)}+\frac{1}{(m+1)(m+2)}+\cdo... | {
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Let $n \geq m \geq 1$ be integers. Prove that $$ \sum_{k=m}^{n}\left(\frac{1}{k^{2}}+\frac{1}{k^{3}}\right) \geq m \cdot\left(\sum_{k=m}^{n} \frac{1}{k^{2}}\right)^{2} $$ | γ Second approach by inducting down, Luke Robitaille and Carl Schildkraut. Fix $n$; we'll induct downwards on $m$. For the base case of $n=m$ the result is easy, since the left side is $\frac{m+1}{m^{3}}$ and the right side is $\frac{m}{m^{4}}=\frac{1}{m^{3}}$. For the inductive step, suppose we have shown the result... | proof | Yes | Yes | proof | Inequalities | Let $n \geq m \geq 1$ be integers. Prove that $$ \sum_{k=m}^{n}\left(\frac{1}{k^{2}}+\frac{1}{k^{3}}\right) \geq m \cdot\left(\sum_{k=m}^{n} \frac{1}{k^{2}}\right)^{2} $$ | γ Second approach by inducting down, Luke Robitaille and Carl Schildkraut. Fix $n$; we'll induct downwards on $m$. For the base case of $n=m$ the result is easy, since the left side is $\frac{m+1}{m^{3}}$ and the right side is $\frac{m}{m^{4}}=\frac{1}{m^{3}}$. For the inductive step, suppose we have shown the result... | {
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Let $n \geq m \geq 1$ be integers. Prove that $$ \sum_{k=m}^{n}\left(\frac{1}{k^{2}}+\frac{1}{k^{3}}\right) \geq m \cdot\left(\sum_{k=m}^{n} \frac{1}{k^{2}}\right)^{2} $$ | I Third approach by reducing $n \rightarrow \infty$, Michael Ren and Carl Schildkraut. First, we give: Claim (Reduction to $n \rightarrow \infty$ ) - If the problem is true when $n \rightarrow \infty$, it is true for all $n$. However, the region is bounded by a convex curve, and the sequence of points $(0,0)$, $\le... | proof | Yes | Yes | proof | Inequalities | Let $n \geq m \geq 1$ be integers. Prove that $$ \sum_{k=m}^{n}\left(\frac{1}{k^{2}}+\frac{1}{k^{3}}\right) \geq m \cdot\left(\sum_{k=m}^{n} \frac{1}{k^{2}}\right)^{2} $$ | I Third approach by reducing $n \rightarrow \infty$, Michael Ren and Carl Schildkraut. First, we give: Claim (Reduction to $n \rightarrow \infty$ ) - If the problem is true when $n \rightarrow \infty$, it is true for all $n$. However, the region is bounded by a convex curve, and the sequence of points $(0,0)$, $\le... | {
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Let $n \geq m \geq 1$ be integers. Prove that $$ \sum_{k=m}^{n}\left(\frac{1}{k^{2}}+\frac{1}{k^{3}}\right) \geq m \cdot\left(\sum_{k=m}^{n} \frac{1}{k^{2}}\right)^{2} $$ | γ Fourth approach by bashing, Carl Schildkraut. With a bit more work, the third approach can be adapted to avoid the $n \rightarrow \infty$ reduction. Similarly to before, define $$ A=\sum_{k=m}^{n} \frac{1}{k^{2}} \text { and } B=\sum_{k=m}^{n} \frac{1}{k^{3}} $$ we want to show $1+4 m B \geq(2 m A-1)^{2}$. Writing... | proof | Yes | Yes | proof | Inequalities | Let $n \geq m \geq 1$ be integers. Prove that $$ \sum_{k=m}^{n}\left(\frac{1}{k^{2}}+\frac{1}{k^{3}}\right) \geq m \cdot\left(\sum_{k=m}^{n} \frac{1}{k^{2}}\right)^{2} $$ | γ Fourth approach by bashing, Carl Schildkraut. With a bit more work, the third approach can be adapted to avoid the $n \rightarrow \infty$ reduction. Similarly to before, define $$ A=\sum_{k=m}^{n} \frac{1}{k^{2}} \text { and } B=\sum_{k=m}^{n} \frac{1}{k^{3}} $$ we want to show $1+4 m B \geq(2 m A-1)^{2}$. Writing... | {
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Find all positive integers $n$ for which it is possible to color some cells of an infinite grid of unit squares red, such that each rectangle consisting of exactly $n$ cells (and whose edges lie along the lines of the grid) contains an odd number of red cells. | We claim that this is possible for all positive integers $n$. Call a positive integer for which such a coloring is possible good. To show that all positive integers $n$ are good we prove the following: (i) If $n$ is good and $p$ is an odd prime, then $p n$ is good; (ii) For every $k \geq 0$, the number $n=2^{k}$ is g... | proof | Yes | Yes | math-word-problem | Combinatorics | Find all positive integers $n$ for which it is possible to color some cells of an infinite grid of unit squares red, such that each rectangle consisting of exactly $n$ cells (and whose edges lie along the lines of the grid) contains an odd number of red cells. | We claim that this is possible for all positive integers $n$. Call a positive integer for which such a coloring is possible good. To show that all positive integers $n$ are good we prove the following: (i) If $n$ is good and $p$ is an odd prime, then $p n$ is good; (ii) For every $k \geq 0$, the number $n=2^{k}$ is g... | {
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Let $n \geq 3$ be an integer and let $K_{n}$ be the complete graph on $n$ vertices. Each edge of $K_{n}$ is colored either red, green, or blue. Let $A$ denote the number of triangles in $K_{n}$ with all edges of the same color, and let $B$ denote the number of triangles in $K_{n}$ with all edges of different colors. Pr... | $ Consider all unordered pairs of different edges which share exactly one vertex (call these vees for convenience). Assign each vee a charge of +2 if its edge colors are the same, and a charge of -1 otherwise. We compute the total charge in two ways. ## γ Total charge by summing over triangles. Note that - each mon... | B \leq 2 A+\frac{n(n-1)}{3} | Yes | Yes | proof | Combinatorics | Let $n \geq 3$ be an integer and let $K_{n}$ be the complete graph on $n$ vertices. Each edge of $K_{n}$ is colored either red, green, or blue. Let $A$ denote the number of triangles in $K_{n}$ with all edges of the same color, and let $B$ denote the number of triangles in $K_{n}$ with all edges of different colors. Pr... | $ Consider all unordered pairs of different edges which share exactly one vertex (call these vees for convenience). Assign each vee a charge of +2 if its edge colors are the same, and a charge of -1 otherwise. We compute the total charge in two ways. ## γ Total charge by summing over triangles. Note that - each mon... | {
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Suppose $a, b$, and $c$ are three complex numbers with product 1 . Assume that none of $a, b$, and $c$ are real or have absolute value 1 . Define $$ p=(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \quad \text { and } \quad q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} . $$ Given that both $p$ and $q$ are real numb... | We show $(p, q)=(-3,3)$ is the only possible ordered pair. ## γT First solution. $$ \begin{aligned} p+3 & =3+\sum_{\text {cyc }}\left(\frac{x}{y}+\frac{y}{x}\right)=3+\frac{x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)}{x y z} \\ & =\frac{(x+y+z)(x y+y z+z x)}{x y z} \\ q-3 & =-3+\sum_{\text {cyc }} \frac{y^{2}}{z x}=\frac{x^{3... | (-3,3) | Yes | Yes | math-word-problem | Algebra | Suppose $a, b$, and $c$ are three complex numbers with product 1 . Assume that none of $a, b$, and $c$ are real or have absolute value 1 . Define $$ p=(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \quad \text { and } \quad q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} . $$ Given that both $p$ and $q$ are real numb... | We show $(p, q)=(-3,3)$ is the only possible ordered pair. ## γT First solution. $$ \begin{aligned} p+3 & =3+\sum_{\text {cyc }}\left(\frac{x}{y}+\frac{y}{x}\right)=3+\frac{x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)}{x y z} \\ & =\frac{(x+y+z)(x y+y z+z x)}{x y z} \\ q-3 & =-3+\sum_{\text {cyc }} \frac{y^{2}}{z x}=\frac{x^{3... | {
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Suppose $a, b$, and $c$ are three complex numbers with product 1 . Assume that none of $a, b$, and $c$ are real or have absolute value 1 . Define $$ p=(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \quad \text { and } \quad q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} . $$ Given that both $p$ and $q$ are real numb... | We show $(p, q)=(-3,3)$ is the only possible ordered pair. I Second solution, found by contestants. The main idea is to make the substitution $$ x=a+\frac{1}{c}, \quad y=b+\frac{1}{a}, \quad z=c+\frac{1}{b} . $$ Then we can check that $$ \begin{aligned} x+y+z & =p \\ x y+y z+z x & =p+q+3 \\ x y z & =p+2 . \end{ali... | (-3, 3) | Yes | Yes | math-word-problem | Algebra | Suppose $a, b$, and $c$ are three complex numbers with product 1 . Assume that none of $a, b$, and $c$ are real or have absolute value 1 . Define $$ p=(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \quad \text { and } \quad q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} . $$ Given that both $p$ and $q$ are real numb... | We show $(p, q)=(-3,3)$ is the only possible ordered pair. I Second solution, found by contestants. The main idea is to make the substitution $$ x=a+\frac{1}{c}, \quad y=b+\frac{1}{a}, \quad z=c+\frac{1}{b} . $$ Then we can check that $$ \begin{aligned} x+y+z & =p \\ x y+y z+z x & =p+q+3 \\ x y z & =p+2 . \end{ali... | {
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Suppose $a, b$, and $c$ are three complex numbers with product 1 . Assume that none of $a, b$, and $c$ are real or have absolute value 1 . Define $$ p=(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \quad \text { and } \quad q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} . $$ Given that both $p$ and $q$ are real numb... | We show $(p, q)=(-3,3)$ is the only possible ordered pair. γ Third solution by Luke Robitaille and Daniel Zhu. The answer is $p=-3$ and $q=3$. Let's first prove that no other $(p, q)$ work. Let $e_{1}=a+b+c$ and $e_{2}=a^{-1}+b^{-1}+c^{-1}=a b+a c+b c$. Also, let $f=e_{1} e_{2}$. Note that $p=e_{1}+e_{2}$. Our main... | p=-3, q=3 | Yes | Yes | math-word-problem | Algebra | Suppose $a, b$, and $c$ are three complex numbers with product 1 . Assume that none of $a, b$, and $c$ are real or have absolute value 1 . Define $$ p=(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \quad \text { and } \quad q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} . $$ Given that both $p$ and $q$ are real numb... | We show $(p, q)=(-3,3)$ is the only possible ordered pair. γ Third solution by Luke Robitaille and Daniel Zhu. The answer is $p=-3$ and $q=3$. Let's first prove that no other $(p, q)$ work. Let $e_{1}=a+b+c$ and $e_{2}=a^{-1}+b^{-1}+c^{-1}=a b+a c+b c$. Also, let $f=e_{1} e_{2}$. Note that $p=e_{1}+e_{2}$. Our main... | {
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Let $A B C$ be a scalene triangle and let $P$ and $Q$ be two distinct points in its interior. Suppose that the angle bisectors of $\angle P A Q, \angle P B Q$, and $\angle P C Q$ are the altitudes of triangle $A B C$. Prove that the midpoint of $\overline{P Q}$ lies on the Euler line of $A B C$. | γ Solution 1 (Ankit Bisain). Let $H$ be the orthocenter of $A B C$, and construct $P^{\prime}$ using the following claim. Claim - There is a point $P^{\prime}$ for which $$ \measuredangle A P H+\measuredangle A P^{\prime} H=\measuredangle B P H+\measuredangle B P^{\prime} H=\measuredangle C P H+\measuredangle C P^{\... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a scalene triangle and let $P$ and $Q$ be two distinct points in its interior. Suppose that the angle bisectors of $\angle P A Q, \angle P B Q$, and $\angle P C Q$ are the altitudes of triangle $A B C$. Prove that the midpoint of $\overline{P Q}$ lies on the Euler line of $A B C$. | γ Solution 1 (Ankit Bisain). Let $H$ be the orthocenter of $A B C$, and construct $P^{\prime}$ using the following claim. Claim - There is a point $P^{\prime}$ for which $$ \measuredangle A P H+\measuredangle A P^{\prime} H=\measuredangle B P H+\measuredangle B P^{\prime} H=\measuredangle C P H+\measuredangle C P^{\... | {
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Let $A B C$ be a scalene triangle and let $P$ and $Q$ be two distinct points in its interior. Suppose that the angle bisectors of $\angle P A Q, \angle P B Q$, and $\angle P C Q$ are the altitudes of triangle $A B C$. Prove that the midpoint of $\overline{P Q}$ lies on the Euler line of $A B C$. | Solution 2 using complex numbers (Carl Schildkraut and Milan Haiman). Let $(A B C)$ be the unit circle in the complex plane, and let $A=a, B=b, C=c$ such that $|a|=|b|=|c|=1$. Let $P=p$ and $Q=q$, and $O=0$ and $H=h=a+b+c$ be the circumcenter and orthocenter of $A B C$ respectively. The first step is to translate the... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a scalene triangle and let $P$ and $Q$ be two distinct points in its interior. Suppose that the angle bisectors of $\angle P A Q, \angle P B Q$, and $\angle P C Q$ are the altitudes of triangle $A B C$. Prove that the midpoint of $\overline{P Q}$ lies on the Euler line of $A B C$. | Solution 2 using complex numbers (Carl Schildkraut and Milan Haiman). Let $(A B C)$ be the unit circle in the complex plane, and let $A=a, B=b, C=c$ such that $|a|=|b|=|c|=1$. Let $P=p$ and $Q=q$, and $O=0$ and $H=h=a+b+c$ be the circumcenter and orthocenter of $A B C$ respectively. The first step is to translate the... | {
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Let $A B C$ be a scalene triangle and let $P$ and $Q$ be two distinct points in its interior. Suppose that the angle bisectors of $\angle P A Q, \angle P B Q$, and $\angle P C Q$ are the altitudes of triangle $A B C$. Prove that the midpoint of $\overline{P Q}$ lies on the Euler line of $A B C$. | γ Solution 3 also using complex numbers (Michael Ren). We use complex numbers as in the previous solution. The angle conditions imply that $\frac{(a-p)(a-q)}{(b-c)^{2}}, \frac{(b-p)(b-q)}{(c-a)^{2}}$, and $\frac{(c-p)(c-q)}{(a-b)^{2}}$ are real numbers. Take a linear combination of these with real coefficients $X$, $Y... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a scalene triangle and let $P$ and $Q$ be two distinct points in its interior. Suppose that the angle bisectors of $\angle P A Q, \angle P B Q$, and $\angle P C Q$ are the altitudes of triangle $A B C$. Prove that the midpoint of $\overline{P Q}$ lies on the Euler line of $A B C$. | γ Solution 3 also using complex numbers (Michael Ren). We use complex numbers as in the previous solution. The angle conditions imply that $\frac{(a-p)(a-q)}{(b-c)^{2}}, \frac{(b-p)(b-q)}{(c-a)^{2}}$, and $\frac{(c-p)(c-q)}{(a-b)^{2}}$ are real numbers. Take a linear combination of these with real coefficients $X$, $Y... | {
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The Bank of Pittsburgh issues coins that have a heads side and a tails side. Vera has a row of 2023 such coins alternately tails-up and heads-up, with the leftmost coin tails-up. In a move, Vera may flip over one of the coins in the row, subject to the following rules: - On the first move, Vera may flip over any of th... | The answer is 4044 . In general, replacing 2023 with $4 n+3$, the answer is $8 n+4$. Bound. Observe that the first and last coins must be flipped, and so every coin is flipped at least once. Then, the $2 n+1$ even-indexed coins must be flipped at least twice, so they are flipped at least $4 n+2$ times. The $2 n+2$ ... | 4044 | Yes | Yes | math-word-problem | Combinatorics | The Bank of Pittsburgh issues coins that have a heads side and a tails side. Vera has a row of 2023 such coins alternately tails-up and heads-up, with the leftmost coin tails-up. In a move, Vera may flip over one of the coins in the row, subject to the following rules: - On the first move, Vera may flip over any of th... | The answer is 4044 . In general, replacing 2023 with $4 n+3$, the answer is $8 n+4$. Bound. Observe that the first and last coins must be flipped, and so every coin is flipped at least once. Then, the $2 n+1$ even-indexed coins must be flipped at least twice, so they are flipped at least $4 n+2$ times. The $2 n+2$ ... | {
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Let $A B C$ be an equilateral triangle with side length 1. Points $A_{1}$ and $A_{2}$ are chosen on side $B C$, points $B_{1}$ and $B_{2}$ are chosen on side $C A$, and points $C_{1}$ and $C_{2}$ are chosen on side $A B$ such that $B A_{1}<B A_{2}, C B_{1}<C B_{2}$, and $A C_{1}<A C_{2}$. Suppose that the three line se... | The only possible value of the common perimeter, denoted $p$, is 1 . γ Synthetic approach (from author). We prove the converse of the problem first: Claim ( $p=1$ implies concurrence) - Suppose the six points are chosen so that triangles $A B_{2} C_{1}, B C_{2} A_{1}, C A_{2} B_{1}$ all have perimeter 1. Then lines $... | 1 | Yes | Yes | math-word-problem | Geometry | Let $A B C$ be an equilateral triangle with side length 1. Points $A_{1}$ and $A_{2}$ are chosen on side $B C$, points $B_{1}$ and $B_{2}$ are chosen on side $C A$, and points $C_{1}$ and $C_{2}$ are chosen on side $A B$ such that $B A_{1}<B A_{2}, C B_{1}<C B_{2}$, and $A C_{1}<A C_{2}$. Suppose that the three line se... | The only possible value of the common perimeter, denoted $p$, is 1 . γ Synthetic approach (from author). We prove the converse of the problem first: Claim ( $p=1$ implies concurrence) - Suppose the six points are chosen so that triangles $A B_{2} C_{1}, B C_{2} A_{1}, C A_{2} B_{1}$ all have perimeter 1. Then lines $... | {
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Let $A B C$ be an equilateral triangle with side length 1. Points $A_{1}$ and $A_{2}$ are chosen on side $B C$, points $B_{1}$ and $B_{2}$ are chosen on side $C A$, and points $C_{1}$ and $C_{2}$ are chosen on side $A B$ such that $B A_{1}<B A_{2}, C B_{1}<C B_{2}$, and $A C_{1}<A C_{2}$. Suppose that the three line se... | The only possible value of the common perimeter, denoted $p$, is 1 . γ Barycentric solution (by Carl, Krit, Milan). We show that, if the common perimeter is 1 , then the lines concur. To do this, we use barycentric coordinates. Let $A=(1: 0: 0)$, $B=(0: 1: 0)$, and $C=(0: 0: 1)$. Let $A_{1}=\left(0: 1-a_{1}: a_{1}\ri... | 1 | Yes | Incomplete | math-word-problem | Geometry | Let $A B C$ be an equilateral triangle with side length 1. Points $A_{1}$ and $A_{2}$ are chosen on side $B C$, points $B_{1}$ and $B_{2}$ are chosen on side $C A$, and points $C_{1}$ and $C_{2}$ are chosen on side $A B$ such that $B A_{1}<B A_{2}, C B_{1}<C B_{2}$, and $A C_{1}<A C_{2}$. Suppose that the three line se... | The only possible value of the common perimeter, denoted $p$, is 1 . γ Barycentric solution (by Carl, Krit, Milan). We show that, if the common perimeter is 1 , then the lines concur. To do this, we use barycentric coordinates. Let $A=(1: 0: 0)$, $B=(0: 1: 0)$, and $C=(0: 0: 1)$. Let $A_{1}=\left(0: 1-a_{1}: a_{1}\ri... | {
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"year": "2023"
} |
Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e}... | Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e}... | p^{e\left(a-p^{\nu_{p}(a)}\right)} | Yes | Yes | math-word-problem | Number Theory | Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e}... | Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e}... | {
"exam": "USA_TSTST",
"problem_label": "9",
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2023.jsonl",
"solution_match": null,
"tier": "T0",
"year": "2023"
} |
Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e}... | Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e}... | p^{e\left(a-p^{\nu_{p}(a)}\right)} | Yes | Incomplete | math-word-problem | Number Theory | Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e}... | Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e}... | {
"exam": "USA_TSTST",
"problem_label": "9",
"problem_match": null,
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2023.jsonl",
"solution_match": null,
"tier": "T0",
"year": "2023"
} |
For every ordered pair of integers \((i,j)\) , not necessarily positive, we wish to select a point \(P_{i,j}\) in the Cartesian plane whose coordinates lie inside the unit square defined by
\[i< x< i + 1,\qquad j< y< j + 1.\]
Find all real numbers \(c > 0\) for which it's possible to choose these points such that... | Answer. \(c\geq 4\)
Proof \(c< 4\) is not possible. Let \(n\) be an arbitrary positive integer. We take an \(n\times n\) subgrid of unit squares (i.e. \(P_{i,j}\) for \(1\leq i,j\leq n\) ), and compute a lower bound on the average of all possible quadrilaterals from this subgrid.
Consider the average length of th... | c \geq 4 | Yes | Yes | math-word-problem | Geometry | For every ordered pair of integers \((i,j)\) , not necessarily positive, we wish to select a point \(P_{i,j}\) in the Cartesian plane whose coordinates lie inside the unit square defined by
\[i< x< i + 1,\qquad j< y< j + 1.\]
Find all real numbers \(c > 0\) for which it's possible to choose these points such that... | Answer. \(c\geq 4\)
Proof \(c< 4\) is not possible. Let \(n\) be an arbitrary positive integer. We take an \(n\times n\) subgrid of unit squares (i.e. \(P_{i,j}\) for \(1\leq i,j\leq n\) ), and compute a lower bound on the average of all possible quadrilaterals from this subgrid.
Consider the average length of th... | {
"exam": "USA_TSTST",
"problem_label": "1",
"problem_match": "1. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl",
"solution_match": "## \\(\\S 1.1\\) TSTST 2024/1, proposed by Karthik Vedula \n",
"tier": "T0",
"year": "2024"
} |
Let \(p\) be an odd prime number. Suppose \(P\) and \(Q\) are polynomials with integer coefficients such that \(P(0) = Q(0) = 1\) , there is no nonconstant polynomial dividing both \(P\) and \(Q\) , and
\[1 + \frac{x}{1 + \frac{2x}{1 + \frac{\cdots}{1 + (p - 1)x}}} = \frac{P(x)}{Q(x)}.\]
Show that all coefficient... | \(\P\) Solution 1. We first make some general observations about rational functions represented through continued fractions.
Claim β Let \(a_{1}, a_{2}, \ldots\) , be a sequence of nonzero integers. Define the sequence of polynomials \(P_{1}(x) = 1\) , \(P_{2}(x) = 1 + a_{1}x\) , and
\[P_{k + 1}(x) = P_{k}(x) + a... | proof | Yes | Incomplete | proof | Number Theory | Let \(p\) be an odd prime number. Suppose \(P\) and \(Q\) are polynomials with integer coefficients such that \(P(0) = Q(0) = 1\) , there is no nonconstant polynomial dividing both \(P\) and \(Q\) , and
\[1 + \frac{x}{1 + \frac{2x}{1 + \frac{\cdots}{1 + (p - 1)x}}} = \frac{P(x)}{Q(x)}.\]
Show that all coefficient... | \(\P\) Solution 1. We first make some general observations about rational functions represented through continued fractions.
Claim β Let \(a_{1}, a_{2}, \ldots\) , be a sequence of nonzero integers. Define the sequence of polynomials \(P_{1}(x) = 1\) , \(P_{2}(x) = 1 + a_{1}x\) , and
\[P_{k + 1}(x) = P_{k}(x) + a... | {
"exam": "USA_TSTST",
"problem_label": "2",
"problem_match": "2. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl",
"solution_match": "## \\(\\S 1.2\\) TSTST 2024/2, proposed by Andrew Gu \n",
"tier": "T0",
"year": "2024"
} |
Let \(A = \{a_{1},\ldots ,a_{2024}\}\) be a set of 2024 pairwise distinct real numbers. Assume that there exist positive integers \(b_{1},b_{2},\ldots ,b_{2024}\) such that
\[a_{1}b_{1} + a_{2}b_{2} + \cdot \cdot \cdot +a_{2024}b_{2024} = 0.\]
Prove that one can choose \(a_{2025},a_{2026},a_{2027},\ldots\) such t... | It will be convenient to use 0- based indexing here, i.e. \(A = \{a_{0},\ldots ,a_{2023}\}\) and so on. Let \(m = \sum_{i = 0}^{2023}b_{i}\) . By appending \(b_{i} - 1\) copies of \(a_{i}\) for each \(i\) , we may extend the sequence to \(a_{0},\ldots ,a_{m - 1}\) such that \(a_{0} + \cdot \cdot \cdot +a_{m - 1} = 0\) ... | proof | Yes | Incomplete | proof | Algebra | Let \(A = \{a_{1},\ldots ,a_{2024}\}\) be a set of 2024 pairwise distinct real numbers. Assume that there exist positive integers \(b_{1},b_{2},\ldots ,b_{2024}\) such that
\[a_{1}b_{1} + a_{2}b_{2} + \cdot \cdot \cdot +a_{2024}b_{2024} = 0.\]
Prove that one can choose \(a_{2025},a_{2026},a_{2027},\ldots\) such t... | It will be convenient to use 0- based indexing here, i.e. \(A = \{a_{0},\ldots ,a_{2023}\}\) and so on. Let \(m = \sum_{i = 0}^{2023}b_{i}\) . By appending \(b_{i} - 1\) copies of \(a_{i}\) for each \(i\) , we may extend the sequence to \(a_{0},\ldots ,a_{m - 1}\) such that \(a_{0} + \cdot \cdot \cdot +a_{m - 1} = 0\) ... | {
"exam": "USA_TSTST",
"problem_label": "3",
"problem_match": "3. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl",
"solution_match": "## \\(\\S 1.3\\) TSTST 2024/3, proposed by Daniel Zhu \n",
"tier": "T0",
"year": "2024"
} |
Let \(ABCD\) be a quadrilateral inscribed in a circle with center \(O\) and \(E\) be the intersection of segments \(AC\) and \(BD\) . Let \(\omega_{1}\) be the circumcircle of \(ADE\) and \(\omega_{2}\) be the circumcircle of \(BCE\) . The tangent to \(\omega_{1}\) at \(A\) and the tangent to \(\omega_{2}\) at \(C\) me... | \(\P\) Solution 1. Let \(R = \overline{AD} \cap \overline{BC}\) (possibly at infinity, but we'll see it's an Euclidean point later).
Claim β \(ACRP\) is an isosceles trapezoid with \(\overline{AC} \parallel \overline{PR}\) . Consequen... | proof | Yes | Incomplete | proof | Geometry | Let \(ABCD\) be a quadrilateral inscribed in a circle with center \(O\) and \(E\) be the intersection of segments \(AC\) and \(BD\) . Let \(\omega_{1}\) be the circumcircle of \(ADE\) and \(\omega_{2}\) be the circumcircle of \(BCE\) . The tangent to \(\omega_{1}\) at \(A\) and the tangent to \(\omega_{2}\) at \(C\) me... | \(\P\) Solution 1. Let \(R = \overline{AD} \cap \overline{BC}\) (possibly at infinity, but we'll see it's an Euclidean point later).
Claim β \(ACRP\) is an isosceles trapezoid with \(\overline{AC} \parallel \overline{PR}\) . Consequen... | {
"exam": "USA_TSTST",
"problem_label": "4",
"problem_match": "4. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl",
"solution_match": "## \\(\\S 2.1\\) TSTST 2024/4, proposed by Merlijn Staps \n",
"tier": "T0",
"year": "2024"
} |
For a positive integer \(k\) , let \(s(k)\) denote the number of 1s in the binary representation of \(k\) . Prove that for any positive integer \(n\) ,
\[\sum_{i = 1}^{n}(-1)^{s(3i)} > 0.\] | \(\P\) Solution 1. Given a set of positive integers \(S\) , define
\[f(S) = \sum_{k\in S}(-1)^{s(k)}.\]
We also define
\[S_{\mathrm{even}} = \{k\in S\mid k\mathrm{~is~even}\}\] \[S_{\mathrm{odd}} = \{k\in S\mid k\mathrm{~is~odd}\}\]
and apply functions on sets pointwise, e.g.
\[\frac{S - 1}{2} = \left\{\... | proof | Yes | Yes | proof | Number Theory | For a positive integer \(k\) , let \(s(k)\) denote the number of 1s in the binary representation of \(k\) . Prove that for any positive integer \(n\) ,
\[\sum_{i = 1}^{n}(-1)^{s(3i)} > 0.\] | \(\P\) Solution 1. Given a set of positive integers \(S\) , define
\[f(S) = \sum_{k\in S}(-1)^{s(k)}.\]
We also define
\[S_{\mathrm{even}} = \{k\in S\mid k\mathrm{~is~even}\}\] \[S_{\mathrm{odd}} = \{k\in S\mid k\mathrm{~is~odd}\}\]
and apply functions on sets pointwise, e.g.
\[\frac{S - 1}{2} = \left\{\... | {
"exam": "USA_TSTST",
"problem_label": "5",
"problem_match": "5. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl",
"solution_match": "## \\(\\S 2.2\\) TSTST 2024/5, proposed by Holden Mui \n",
"tier": "T0",
"year": "2024"
} |
Determine whether there exists a function \(f\colon \mathbb{Z}_{>0}\to \mathbb{Z}_{>0}\) such that for all positive integers \(m\) and \(n\) ,
\[f(m + n f(m)) = f(n)^{m} + 2024!\cdot m.\] | The answer is no. Let \(P(m,n)\) denote the given FE.
\(\P\) Solution 1 (Gopal Goel). Suppose there was a function \(f\) , and let \(r = f(1)\) . Note that \(P(1,n)\) gives
\[f(1 + r n) = f(n) + 2024!.\]
Iterating this result gives
\[f(1 + r + \dots +r^{k}) = r + k\cdot 2024!\]
for all \(k\in \mathbb{Z}_... | proof | Yes | Yes | proof | Number Theory | Determine whether there exists a function \(f\colon \mathbb{Z}_{>0}\to \mathbb{Z}_{>0}\) such that for all positive integers \(m\) and \(n\) ,
\[f(m + n f(m)) = f(n)^{m} + 2024!\cdot m.\] | The answer is no. Let \(P(m,n)\) denote the given FE.
\(\P\) Solution 1 (Gopal Goel). Suppose there was a function \(f\) , and let \(r = f(1)\) . Note that \(P(1,n)\) gives
\[f(1 + r n) = f(n) + 2024!.\]
Iterating this result gives
\[f(1 + r + \dots +r^{k}) = r + k\cdot 2024!\]
for all \(k\in \mathbb{Z}_... | {
"exam": "USA_TSTST",
"problem_label": "6",
"problem_match": "6. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl",
"solution_match": "## \\(\\S 2.3\\) TSTST 2024/6, proposed by Jaedon Whyte \n",
"tier": "T0",
"year": "2024"
} |
An infinite sequence \(a_{1},a_{2},a_{3},\ldots\) of real numbers satisfies
\[a_{2n - 1} + a_{2n} > a_{2n + 1} + a_{2n + 2}\qquad \mathrm{and}\qquad a_{2n} + a_{2n + 1}< a_{2n + 2} + a_{2n + 3}\]
for every positive integer \(n\) . Prove that there exists a real number \(C\) such that \(a_{n}a_{n + 1}< C\) for eve... | Aer \(n\) .
It suffices to solve the problem for sufficiently large \(n\) . Let \(d_{n} = (- 1)^{n - 1}(a_{n + 2} - a_{n})\) . The assertion simply says that \(d_{1}, d_{2}, \ldots\) is strictly increasing.
We consider the following cases.
- Suppose that \(d_{k} > 0\) for some \(k\) . Then,
\[a_{2n + 1} = a... | proof | Yes | Yes | proof | Inequalities | An infinite sequence \(a_{1},a_{2},a_{3},\ldots\) of real numbers satisfies
\[a_{2n - 1} + a_{2n} > a_{2n + 1} + a_{2n + 2}\qquad \mathrm{and}\qquad a_{2n} + a_{2n + 1}< a_{2n + 2} + a_{2n + 3}\]
for every positive integer \(n\) . Prove that there exists a real number \(C\) such that \(a_{n}a_{n + 1}< C\) for eve... | Aer \(n\) .
It suffices to solve the problem for sufficiently large \(n\) . Let \(d_{n} = (- 1)^{n - 1}(a_{n + 2} - a_{n})\) . The assertion simply says that \(d_{1}, d_{2}, \ldots\) is strictly increasing.
We consider the following cases.
- Suppose that \(d_{k} > 0\) for some \(k\) . Then,
\[a_{2n + 1} = a... | {
"exam": "USA_TSTST",
"problem_label": "7",
"problem_match": "7. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl",
"solution_match": "## \\(\\S 3.1\\) TSTST 2024/7, proposed by Merlijn Staps \n",
"tier": "T0",
"year": "2024"
} |
Let \(ABC\) be a scalene triangle, and let \(D\) be a point on side \(BC\) satisfying \(\angle BAD = \angle DAC\) . Suppose that \(X\) and \(Y\) are points inside \(ABC\) such that triangles \(ABX\) and \(ACY\) are similar and quadrilaterals \(ACDX\) and \(ABDY\) are cyclic. Let lines \(BX\) and \(CY\) meet at \(S\) an... | \(\P\) Solution by characterizing \(X\) and \(Y\) . We first state an important property of \(X\) and \(Y\) .
Claim β Points \(X\) and \(Y\) are isogonal conjugates with respect to \(\triangle ABC\) .
Here are two proofs of the cla... | proof | Yes | Incomplete | proof | Geometry | Let \(ABC\) be a scalene triangle, and let \(D\) be a point on side \(BC\) satisfying \(\angle BAD = \angle DAC\) . Suppose that \(X\) and \(Y\) are points inside \(ABC\) such that triangles \(ABX\) and \(ACY\) are similar and quadrilaterals \(ACDX\) and \(ABDY\) are cyclic. Let lines \(BX\) and \(CY\) meet at \(S\) an... | \(\P\) Solution by characterizing \(X\) and \(Y\) . We first state an important property of \(X\) and \(Y\) .
Claim β Points \(X\) and \(Y\) are isogonal conjugates with respect to \(\triangle ABC\) .
Here are two proofs of the cla... | {
"exam": "USA_TSTST",
"problem_label": "8",
"problem_match": "8. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl",
"solution_match": "## \\(\\S 3.2\\) TSTST 2024/8, proposed by Michael Ren \n",
"tier": "T0",
"year": "2024"
} |
Let \(n \geq 2\) be a fixed integer. The cells of an \(n \times n\) table are filled with the integers from 1 to \(n^2\) with each number appearing exactly once. Let \(N\) be the number of unordered quadruples of cells on this board which form an axis-aligned rectangle, with the two smaller integers being on opposite v... | L
The largest possible value of \(N\) is \(\frac{1}{12} n^2 (n^2 - 1)\) . Call these rectangles wobbly. We defer the construction until the proof is complete, since the proof suggests the construction.
Proof of bound. Call a triple of integers \((a, b, c)\) an elbow if \(a\) and \(b\) are in the same row, \(b\) a... | \frac{1}{12} n^2 (n^2 - 1) | Yes | Yes | math-word-problem | Combinatorics | Let \(n \geq 2\) be a fixed integer. The cells of an \(n \times n\) table are filled with the integers from 1 to \(n^2\) with each number appearing exactly once. Let \(N\) be the number of unordered quadruples of cells on this board which form an axis-aligned rectangle, with the two smaller integers being on opposite v... | L
The largest possible value of \(N\) is \(\frac{1}{12} n^2 (n^2 - 1)\) . Call these rectangles wobbly. We defer the construction until the proof is complete, since the proof suggests the construction.
Proof of bound. Call a triple of integers \((a, b, c)\) an elbow if \(a\) and \(b\) are in the same row, \(b\) a... | {
"exam": "USA_TSTST",
"problem_label": "9",
"problem_match": "9. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2024.jsonl",
"solution_match": "## Β§3.3 TSTST 2024/9 \n",
"tier": "T0",
"year": "2024"
} |
In a finite group of people, some pairs are friends (friendship is mutual). Each person \(p\) has a list \(f_{1}(p), f_{2}(p), \ldots , f_{d(p)}(p)\) of their friends, where \(d(p)\) is the number of distinct friends \(p\) has. Additionally, any two people are connected by a series of friendships. Each person also has ... | Given a person \(p\) , let \(F(p)\) be the set of friends of \(p\) . Choose a person \(p\) with the most friends. Note that for each friend \(q\) of \(p\) , \(p\) receives a water balloon from \(q\) once out of every \(d(q)\) turns. Since \(p\) always receives 1 water balloon, we must have
\[\sum_{q\in F(p)}\frac{1}... | proof | Yes | Yes | proof | Combinatorics | In a finite group of people, some pairs are friends (friendship is mutual). Each person \(p\) has a list \(f_{1}(p), f_{2}(p), \ldots , f_{d(p)}(p)\) of their friends, where \(d(p)\) is the number of distinct friends \(p\) has. Additionally, any two people are connected by a series of friendships. Each person also has ... | Given a person \(p\) , let \(F(p)\) be the set of friends of \(p\) . Choose a person \(p\) with the most friends. Note that for each friend \(q\) of \(p\) , \(p\) receives a water balloon from \(q\) once out of every \(d(q)\) turns. Since \(p\) always receives 1 water balloon, we must have
\[\sum_{q\in F(p)}\frac{1}... | {
"exam": "IMO",
"problem_label": "1",
"problem_match": "Problem 1. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl",
"solution_match": "## \\(\\S 1.1\\) Solution to TSTST 1, by Milan Haiman \n",
"tier": "T0",
"year": "2025"
} |
Find all sets \(S \subseteq \mathbb{Z}\) for which there exists a function \(f: \mathbb{R} \to \mathbb{Z}\) such that
- \(f(x - y) - 2f(x) + f(x + y) \geq -1\) for all \(x, y \in \mathbb{R}\) , and
- \(S = \{f(z) \mid z \in \mathbb{R}\}\) . | The answer is \(\{a\}\) , \(\{a,a + 1\}\) , \(\{a,a + 1,a + 2,\ldots \}\) , and \(\mathbb{Z}\) , for arbitrary \(a\in \mathbb{Z}\) . For constructions, it is not hard to show that if \(g\colon \mathbb{R}\to \mathbb{R}\) is a convex function, then \(\lfloor g\rfloor\) satisfies the functional equation. Thus \(f(x) = a\)... | proof | Yes | Yes | math-word-problem | Algebra | Find all sets \(S \subseteq \mathbb{Z}\) for which there exists a function \(f: \mathbb{R} \to \mathbb{Z}\) such that
- \(f(x - y) - 2f(x) + f(x + y) \geq -1\) for all \(x, y \in \mathbb{R}\) , and
- \(S = \{f(z) \mid z \in \mathbb{R}\}\) . | The answer is \(\{a\}\) , \(\{a,a + 1\}\) , \(\{a,a + 1,a + 2,\ldots \}\) , and \(\mathbb{Z}\) , for arbitrary \(a\in \mathbb{Z}\) . For constructions, it is not hard to show that if \(g\colon \mathbb{R}\to \mathbb{R}\) is a convex function, then \(\lfloor g\rfloor\) satisfies the functional equation. Thus \(f(x) = a\)... | {
"exam": "IMO",
"problem_label": "2",
"problem_match": "Problem 2. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl",
"solution_match": "## \\(\\S 1.2\\) Solution to TSTST 2, by Daniel Zhu \n",
"tier": "T0",
"year": "2025"
} |
Let \(a_{1}, a_{2}, r\) , and \(s\) be positive integers with \(r\) and \(s\) odd. The sequence \(a_{1}, a_{2}, a_{3}, \ldots\) is defined by
\[a_{n + 2} = r a_{n + 1} + s a_{n}\]
for all \(n \geq 1\) . Determine the maximum possible number of integers \(1 \leq \ell \leq 2025\) such that \(a_{\ell}\) divides \(a_... | Answer 1350.
\(\P\) Solution We first provide the upper bound. We start by dividing out any common factors of \(a_{1}\) and \(a_{2}\) from the whole sequence. Note that since \(r\) and \(s\) are odd, and \(a_{1}\) and \(a_{2}\) cannot both be divisible by 2, the sequence \(a_{1}, a_{2}, a_{3}, \ldots\) (mod 2) must ... | 1350 | Yes | Yes | math-word-problem | Number Theory | Let \(a_{1}, a_{2}, r\) , and \(s\) be positive integers with \(r\) and \(s\) odd. The sequence \(a_{1}, a_{2}, a_{3}, \ldots\) is defined by
\[a_{n + 2} = r a_{n + 1} + s a_{n}\]
for all \(n \geq 1\) . Determine the maximum possible number of integers \(1 \leq \ell \leq 2025\) such that \(a_{\ell}\) divides \(a_... | Answer 1350.
\(\P\) Solution We first provide the upper bound. We start by dividing out any common factors of \(a_{1}\) and \(a_{2}\) from the whole sequence. Note that since \(r\) and \(s\) are odd, and \(a_{1}\) and \(a_{2}\) cannot both be divisible by 2, the sequence \(a_{1}, a_{2}, a_{3}, \ldots\) (mod 2) must ... | {
"exam": "IMO",
"problem_label": "3",
"problem_match": "Problem 3. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl",
"solution_match": "## Β§1.3 Solution to TSTST 3, by Carlos Rodriguez, Albert Wang, Kevin Wu, Isaac Zhu, Nathan Cho \n",
"tier": "T0",
"year": "2025"
} |
Let \(n \geq 2\) be a positive integer. Let \(a_{1}, a_{2}, \ldots , a_{n}\) be a sequence of positive integers such that
\[\gcd (a_{1},a_{2}),\gcd (a_{2},a_{3}),\ldots ,\gcd (a_{n - 1},a_{n})\]
is a strictly increasing sequence. Find, in terms of \(n\) , the maximum possible value of
\[\frac{1}{a_{1}} +\frac{... | We claim the maximum possible value is 2. To see that this is achievable, let the sequence \((a_{i})_{i = 1}^{n}\) be 1, 2, 4, ..., \(2^{n - 2}\) , \(2^{n - 2}\) . Then \(\gcd (a_{i},a_{i + 1}) = 2^{i - 1}\) , which is an increasing sequence, and it is easy to check that \(\sum \frac{1}{a_{i}} = 2\) . We now show this ... | 2 | Yes | Yes | math-word-problem | Number Theory | Let \(n \geq 2\) be a positive integer. Let \(a_{1}, a_{2}, \ldots , a_{n}\) be a sequence of positive integers such that
\[\gcd (a_{1},a_{2}),\gcd (a_{2},a_{3}),\ldots ,\gcd (a_{n - 1},a_{n})\]
is a strictly increasing sequence. Find, in terms of \(n\) , the maximum possible value of
\[\frac{1}{a_{1}} +\frac{... | We claim the maximum possible value is 2. To see that this is achievable, let the sequence \((a_{i})_{i = 1}^{n}\) be 1, 2, 4, ..., \(2^{n - 2}\) , \(2^{n - 2}\) . Then \(\gcd (a_{i},a_{i + 1}) = 2^{i - 1}\) , which is an increasing sequence, and it is easy to check that \(\sum \frac{1}{a_{i}} = 2\) . We now show this ... | {
"exam": "IMO",
"problem_label": "4",
"problem_match": "Problem 4. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl",
"solution_match": "## \\(\\S 2.1\\) Solution to TSTST 4, by Maxim Li \n",
"tier": "T0",
"year": "2025"
} |
A tetrahedron \(ABCD\) is said to be angelic if it has nonzero volume and satisfies
\[\angle BAC + \angle CAD + \angle DAB = \angle ABC + \angle CBD + \angle DBA,\] \[\angle ACB + \angle BCD + \angle DCA = \angle ADB + \angle BDC + \angle CDA.\]
Across all angelic tetrahedrons, what is the maximum number of disti... | 
We claim the maximum cardinality is \(\boxed{4}\) . This is attained by taking a non- square rectangle (or parallelogram) \(A C B D\) and folding it along diagonal \(A B\) , creating edges \(A B\) and \(C D\) in the process. Here, \(A C ... | 4 | Yes | Yes | math-word-problem | Geometry | A tetrahedron \(ABCD\) is said to be angelic if it has nonzero volume and satisfies
\[\angle BAC + \angle CAD + \angle DAB = \angle ABC + \angle CBD + \angle DBA,\] \[\angle ACB + \angle BCD + \angle DCA = \angle ADB + \angle BDC + \angle CDA.\]
Across all angelic tetrahedrons, what is the maximum number of disti... | 
We claim the maximum cardinality is \(\boxed{4}\) . This is attained by taking a non- square rectangle (or parallelogram) \(A C B D\) and folding it along diagonal \(A B\) , creating edges \(A B\) and \(C D\) in the process. Here, \(A C ... | {
"exam": "IMO",
"problem_label": "5",
"problem_match": "Problem 5. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl",
"solution_match": "## \\(\\S 2.2\\) Solution to TSTST 5, by Karthik Vedula \n",
"tier": "T0",
"year": "2025"
} |
Alice and Bob play a game on \(n\) vertices labelled \(1, 2, \ldots , n\) . They take turns adding edges \(\{i, j\}\) , with Alice going first. Neither player is allowed to make a move that creates a cycle, and the game ends after \(n - 1\) total turns.
Let the weight of the edge \(\{i, j\}\) be \(|i - j|\) , and le... | \(\P\) Solution Let \(k = \lceil \frac{n - 1}{2}\rceil\) . The answer is
\[\frac{1}{2} (k + 1)(2n - k - 2) = (n - k - 1) + (n - k) + \dots +(n - 1).\]
When \(n = 1\) , this is clear.
Consider now when \(n\geq 2\) . Note that the game consists of \(n - 1\) moves, with Alice making \(k\) moves and Bob making \(n... | \frac{1}{2} (k + 1)(2n - k - 2) | Yes | Yes | math-word-problem | Combinatorics | Alice and Bob play a game on \(n\) vertices labelled \(1, 2, \ldots , n\) . They take turns adding edges \(\{i, j\}\) , with Alice going first. Neither player is allowed to make a move that creates a cycle, and the game ends after \(n - 1\) total turns.
Let the weight of the edge \(\{i, j\}\) be \(|i - j|\) , and le... | \(\P\) Solution Let \(k = \lceil \frac{n - 1}{2}\rceil\) . The answer is
\[\frac{1}{2} (k + 1)(2n - k - 2) = (n - k - 1) + (n - k) + \dots +(n - 1).\]
When \(n = 1\) , this is clear.
Consider now when \(n\geq 2\) . Note that the game consists of \(n - 1\) moves, with Alice making \(k\) moves and Bob making \(n... | {
"exam": "IMO",
"problem_label": "6",
"problem_match": "Problem 6. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl",
"solution_match": "## \\(\\S 2.3\\) Solution to TSTST 6, by Max Lu, Kevin Wu \n",
"tier": "T0",
"year": "2025"
} |
For a positive real number \(c\) , the sequence \(a_{1}, a_{2}, \ldots\) of real numbers is defined as follows. Let \(a_{1} = c\) , and for \(n \geq 2\) , let
\[a_{n} = \sum_{i = 1}^{n - 1}(a_{i})^{n - i + 1}.\]
Find all positive real numbers \(c\) such that \(a_{i} > a_{i + 1}\) for all positive integers \(i\) . | \(\P\) Solution (author) The answer is \(c< \frac{\sqrt{5} - 1}{2}\)
To show this is necessary, note that \(a_{2} = c^{2}\) and \(a_{3} = c^{3} + c^{4}\) , so if the sequence is decreasing, we have \(c^{2} > c^{3} + c^{4}\) , implying \(c< \frac{\sqrt{5} - 1}{2}\)
In the other direction, suppose \(c\) is a positi... | c< \frac{\sqrt{5} - 1}{2} | Yes | Yes | math-word-problem | Algebra | For a positive real number \(c\) , the sequence \(a_{1}, a_{2}, \ldots\) of real numbers is defined as follows. Let \(a_{1} = c\) , and for \(n \geq 2\) , let
\[a_{n} = \sum_{i = 1}^{n - 1}(a_{i})^{n - i + 1}.\]
Find all positive real numbers \(c\) such that \(a_{i} > a_{i + 1}\) for all positive integers \(i\) . | \(\P\) Solution (author) The answer is \(c< \frac{\sqrt{5} - 1}{2}\)
To show this is necessary, note that \(a_{2} = c^{2}\) and \(a_{3} = c^{3} + c^{4}\) , so if the sequence is decreasing, we have \(c^{2} > c^{3} + c^{4}\) , implying \(c< \frac{\sqrt{5} - 1}{2}\)
In the other direction, suppose \(c\) is a positi... | {
"exam": "IMO",
"problem_label": "7",
"problem_match": "Problem 7. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl",
"solution_match": "## \\(\\S 3.1\\) Solution to TSTST 7, by Luke Robitaille \n",
"tier": "T0",
"year": "2025"
} |
Find all polynomials \(f\) with integer coefficients such that for all positive integers \(n\) ,
\[n \text{ divides } \frac{f(f(\ldots(f(0)) \ldots) - 1}{n + 1 f^{\mathrm{s}}}\] | There are three families.
- \(f(x) = x + 1\) .
- \(f(x) = x(x - 1)g(x) + 1\) for any polynomial \(g(x)\) (i.e., any \(f(x)\) such that \(f(0) = f(1) = 1\) ).
- \(f(x) = x(x - 1)(x + 1)g(x) + (2x^2 - 1)\) for any polynomial \(g(x)\) (i.e., any \(f(x)\) such that \(f(0) = -1\) , \(f(-1) = f(1) = 1\) ).
These ... | proof | Incomplete | Yes | math-word-problem | Number Theory | Find all polynomials \(f\) with integer coefficients such that for all positive integers \(n\) ,
\[n \text{ divides } \frac{f(f(\ldots(f(0)) \ldots) - 1}{n + 1 f^{\mathrm{s}}}\] | There are three families.
- \(f(x) = x + 1\) .
- \(f(x) = x(x - 1)g(x) + 1\) for any polynomial \(g(x)\) (i.e., any \(f(x)\) such that \(f(0) = f(1) = 1\) ).
- \(f(x) = x(x - 1)(x + 1)g(x) + (2x^2 - 1)\) for any polynomial \(g(x)\) (i.e., any \(f(x)\) such that \(f(0) = -1\) , \(f(-1) = f(1) = 1\) ).
These ... | {
"exam": "IMO",
"problem_label": "8",
"problem_match": "Problem 8. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl",
"solution_match": "## \\(\\S 3.2\\) Solution to TSTST 8, by Pitchayut Saengrungkonga \n",
"tier": "T0",
"year": "2025"
} |
Let acute triangle \(ABC\) have orthocenter \(H\) . Let \(B_{1}\) , \(C_{1}\) , \(B_{2}\) , and \(C_{2}\) be collinear points which lie on lines \(AB\) , \(AC\) , \(BH\) , and \(CH\) , respectively. Let \(\omega_{B}\) and \(\omega_{C}\) be the circumcircles of triangles \(BB_{1}B_{2}\) and \(CC_{1}C_{2}\) , respectivel... | \(\P\) Solution (author) The first important step is to introduce \(N\) , the circumcenter of \(H B_{2}C_{2}\) .
Claim β Lines \(N B_{2}\) and \(N C_{2}\) are tangent to \((B B_{1}B_{2})\) and \((C C_{1}C_{2})\) , respectively.
Proof. This follows from chasing
\[\angle N B_{2}B = \angle N B_{2}H = 90^{\circ} -... | proof | Yes | Incomplete | proof | Geometry | Let acute triangle \(ABC\) have orthocenter \(H\) . Let \(B_{1}\) , \(C_{1}\) , \(B_{2}\) , and \(C_{2}\) be collinear points which lie on lines \(AB\) , \(AC\) , \(BH\) , and \(CH\) , respectively. Let \(\omega_{B}\) and \(\omega_{C}\) be the circumcircles of triangles \(BB_{1}B_{2}\) and \(CC_{1}C_{2}\) , respectivel... | \(\P\) Solution (author) The first important step is to introduce \(N\) , the circumcenter of \(H B_{2}C_{2}\) .
Claim β Lines \(N B_{2}\) and \(N C_{2}\) are tangent to \((B B_{1}B_{2})\) and \((C C_{1}C_{2})\) , respectively.
Proof. This follows from chasing
\[\angle N B_{2}B = \angle N B_{2}H = 90^{\circ} -... | {
"exam": "IMO",
"problem_label": "9",
"problem_match": "Problem 9. ",
"resource_path": "USA_TSTST/segmented/en-sols-TSTST-2025.jsonl",
"solution_match": "## \\(\\S 3.3\\) Solution to TSTST 9, by Ruben Carpenter \n",
"tier": "T0",
"year": "2025"
} |
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