problem stringlengths 14 10.4k | solution stringlengths 1 24.1k | answer stringlengths 1 250 ⌀ | problem_is_valid stringclasses 4
values | solution_is_valid stringclasses 3
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values | problem_raw stringlengths 14 10.4k | solution_raw stringlengths 1 24.1k | metadata dict |
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Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be real numbers satisfying the following equations:
$$
\frac{a_{1}}{k^{2}+1}+\frac{a_{2}}{k^{2}+2}+\frac{a_{3}}{k^{2}+3}+\frac{a_{4}}{k^{2}+4}+\frac{a_{5}}{k^{2}+5}=\frac{1}{k^{2}} \text { for } k=1,2,3,4,5
$$
Find the value of $\frac{a_{1}}{37}+\frac{a_{2}}{38}+\frac{a_{3}}{39... | Let $R(x):=\frac{a_{1}}{x^{2}+1}+\frac{a_{2}}{x^{2}+2}+\frac{a_{3}}{x^{2}+3}+\frac{a_{4}}{x^{2}+4}+\frac{a_{5}}{x^{2}+5}$. Then $R( \pm 1)=1$, $R( \pm 2)=\frac{1}{4}, R( \pm 3)=\frac{1}{9}, R( \pm 4)=\frac{1}{16}, R( \pm 5)=\frac{1}{25}$ and $R(6)$ is the value to be found. Let's put $P(x):=\left(x^{2}+1\right)\left(x^... | \frac{187465}{6744582} | Yes | Yes | math-word-problem | Algebra | Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be real numbers satisfying the following equations:
$$
\frac{a_{1}}{k^{2}+1}+\frac{a_{2}}{k^{2}+2}+\frac{a_{3}}{k^{2}+3}+\frac{a_{4}}{k^{2}+4}+\frac{a_{5}}{k^{2}+5}=\frac{1}{k^{2}} \text { for } k=1,2,3,4,5
$$
Find the value of $\frac{a_{1}}{37}+\frac{a_{2}}{38}+\frac{a_{3}}{39... | Let $R(x):=\frac{a_{1}}{x^{2}+1}+\frac{a_{2}}{x^{2}+2}+\frac{a_{3}}{x^{2}+3}+\frac{a_{4}}{x^{2}+4}+\frac{a_{5}}{x^{2}+5}$. Then $R( \pm 1)=1$, $R( \pm 2)=\frac{1}{4}, R( \pm 3)=\frac{1}{9}, R( \pm 4)=\frac{1}{16}, R( \pm 5)=\frac{1}{25}$ and $R(6)$ is the value to be found. Let's put $P(x):=\left(x^{2}+1\right)\left(x^... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2009_sol.jsonl",
"solution_match": "(Solution)",
"tier": "T1",
"year": "2009"
} |
Let three circles $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$, which are non-overlapping and mutually external, be given in the plane. For each point $P$ in the plane, outside the three circles, construct six points $A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3}$ as follows: For each $i=1,2,3, A_{i}, B_{i}$ are distinct points on ... | Let $O_{i}$ be the center and $r_{i}$ the radius of circle $\Gamma_{i}$ for each $i=1,2,3$. Let $P$ be an exceptional point, and let the three corresponding lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ concur at $Q$. Construct the circle with diameter $P Q$. Call the circle $\Gamma$, its center $O$ and its radius $r$.... | proof | Yes | Yes | proof | Geometry | Let three circles $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$, which are non-overlapping and mutually external, be given in the plane. For each point $P$ in the plane, outside the three circles, construct six points $A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3}$ as follows: For each $i=1,2,3, A_{i}, B_{i}$ are distinct points on ... | Let $O_{i}$ be the center and $r_{i}$ the radius of circle $\Gamma_{i}$ for each $i=1,2,3$. Let $P$ be an exceptional point, and let the three corresponding lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ concur at $Q$. Construct the circle with diameter $P Q$. Call the circle $\Gamma$, its center $O$ and its radius $r$.... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2009_sol.jsonl",
"solution_match": "(Solution)",
"tier": "T1",
"year": "2009"
} |
Prove that for any positive integer $k$, there exists an arithmetic sequence
$$
\frac{a_{1}}{b_{1}}, \quad \frac{a_{2}}{b_{2}}, \ldots, \quad \frac{a_{k}}{b_{k}}
$$
of rational numbers, where $a_{i}, b_{i}$ are relatively prime positive integers for each $i=1,2, \ldots, k$, such that the positive integers $a_{1}, b_{... | For $k=1$, there is nothing to prove. Henceforth assume $k \geq 2$.
Let $p_{1}, p_{2}, \ldots, p_{k}$ be $k$ distinct primes such that
$$
k<p_{k}<\cdots<p_{2}<p_{1}
$$
and let $N=p_{1} p_{2} \cdots p_{k}$. By Chinese Remainder Theorem, there exists a positive integer $x$ satisfying
$$
x \equiv-i \quad\left(\bmod p_{... | proof | Yes | Yes | proof | Number Theory | Prove that for any positive integer $k$, there exists an arithmetic sequence
$$
\frac{a_{1}}{b_{1}}, \quad \frac{a_{2}}{b_{2}}, \ldots, \quad \frac{a_{k}}{b_{k}}
$$
of rational numbers, where $a_{i}, b_{i}$ are relatively prime positive integers for each $i=1,2, \ldots, k$, such that the positive integers $a_{1}, b_{... | For $k=1$, there is nothing to prove. Henceforth assume $k \geq 2$.
Let $p_{1}, p_{2}, \ldots, p_{k}$ be $k$ distinct primes such that
$$
k<p_{k}<\cdots<p_{2}<p_{1}
$$
and let $N=p_{1} p_{2} \cdots p_{k}$. By Chinese Remainder Theorem, there exists a positive integer $x$ satisfying
$$
x \equiv-i \quad\left(\bmod p_{... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2009_sol.jsonl",
"solution_match": "(Solution)",
"tier": "T1",
"year": "2009"
} |
Larry and Rob are two robots travelling in one car from Argovia to Zillis. Both robots have control over the steering and steer according to the following algorithm: Larry makes a $90^{\circ}$ left turn after every $\ell$ kilometer driving from start; Rob makes a $90^{\circ}$ right turn after every $r$ kilometer drivin... | Let Zillis be $d$ kilometers away from Argovia, where $d$ is a positive real number. For simplicity, we will position Argovia at $(0,0)$ and Zillis at $(d, 0)$, so that the car starts out facing east. We will investigate how the car moves around in the period of travelling the first $\ell r$ kilometers, the second $\el... | \ell \equiv r \equiv 1 \quad \text { or } \quad \ell \equiv r \equiv 3 \quad(\bmod 4) | Yes | Incomplete | math-word-problem | Number Theory | Larry and Rob are two robots travelling in one car from Argovia to Zillis. Both robots have control over the steering and steer according to the following algorithm: Larry makes a $90^{\circ}$ left turn after every $\ell$ kilometer driving from start; Rob makes a $90^{\circ}$ right turn after every $r$ kilometer drivin... | Let Zillis be $d$ kilometers away from Argovia, where $d$ is a positive real number. For simplicity, we will position Argovia at $(0,0)$ and Zillis at $(d, 0)$, so that the car starts out facing east. We will investigate how the car moves around in the period of travelling the first $\ell r$ kilometers, the second $\el... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2009_sol.jsonl",
"solution_match": "(Solution)",
"tier": "T1",
"year": "2009"
} |
Let $A B C$ be a triangle with $\angle B A C \neq 90^{\circ}$. Let $O$ be the circumcenter of the triangle $A B C$ and let $\Gamma$ be the circumcircle of the triangle $B O C$. Suppose that $\Gamma$ intersects the line segment $A B$ at $P$ different from $B$, and the line segment $A C$ at $Q$ different from $C$. Let $O... | From the assumption that the circle $\Gamma$ intersects both of the line segments $A B$ and $A C$, it follows that the 4 points $N, C, Q, O$ are located on $\Gamma$ in the order of $N, C, Q, O$ or in the order of $N, C, O, Q$. The following argument for the proof of the assertion of the problem is valid in either case.... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with $\angle B A C \neq 90^{\circ}$. Let $O$ be the circumcenter of the triangle $A B C$ and let $\Gamma$ be the circumcircle of the triangle $B O C$. Suppose that $\Gamma$ intersects the line segment $A B$ at $P$ different from $B$, and the line segment $A C$ at $Q$ different from $C$. Let $O... | From the assumption that the circle $\Gamma$ intersects both of the line segments $A B$ and $A C$, it follows that the 4 points $N, C, Q, O$ are located on $\Gamma$ in the order of $N, C, Q, O$ or in the order of $N, C, O, Q$. The following argument for the proof of the assertion of the problem is valid in either case.... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2010_sol.jsonl",
"solution_match": "\nSolution:",
"tier": "T1",
"year": "2010"
} |
For a positive integer $k$, call an integer a pure $k$-th power if it can be represented as $m^{k}$ for some integer $m$. Show that for every positive integer $n$ there exist $n$ distinct positive integers such that their sum is a pure 2009-th power, and their product is a pure 2010-th power. | For the sake of simplicity, let us set $k=2009$.
First of all, choose $n$ distinct positive integers $b_{1}, \cdots, b_{n}$ suitably so that their product is a pure $k+1$-th power (for example, let $b_{i}=i^{k+1}$ for $i=1, \cdots, n$ ). Then we have $b_{1} \cdots b_{n}=t^{k+1}$ for some positive integer $t$. Set $b_{1... | proof | Yes | Yes | proof | Number Theory | For a positive integer $k$, call an integer a pure $k$-th power if it can be represented as $m^{k}$ for some integer $m$. Show that for every positive integer $n$ there exist $n$ distinct positive integers such that their sum is a pure 2009-th power, and their product is a pure 2010-th power. | For the sake of simplicity, let us set $k=2009$.
First of all, choose $n$ distinct positive integers $b_{1}, \cdots, b_{n}$ suitably so that their product is a pure $k+1$-th power (for example, let $b_{i}=i^{k+1}$ for $i=1, \cdots, n$ ). Then we have $b_{1} \cdots b_{n}=t^{k+1}$ for some positive integer $t$. Set $b_{1... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2010_sol.jsonl",
"solution_match": "\nSolution:",
"tier": "T1",
"year": "2010"
} |
Let $n$ be a positive integer. $n$ people take part in a certain party. For any pair of the participants, either the two are acquainted with each other or they are not. What is the maximum possible number of the pairs for which the two are not acquainted but have a common acquaintance among the participants? | When 1 participant, say the person $A$, is mutually acquainted with each of the remaining $n-1$ participants, and if there are no other acquaintance relationships among the participants, then for any pair of participants not involving $A$, the two are not mutual acquaintances, but they have a common acquaintance, namel... | \frac{n^{2}-3 n+2}{2} | Yes | Yes | math-word-problem | Combinatorics | Let $n$ be a positive integer. $n$ people take part in a certain party. For any pair of the participants, either the two are acquainted with each other or they are not. What is the maximum possible number of the pairs for which the two are not acquainted but have a common acquaintance among the participants? | When 1 participant, say the person $A$, is mutually acquainted with each of the remaining $n-1$ participants, and if there are no other acquaintance relationships among the participants, then for any pair of participants not involving $A$, the two are not mutual acquaintances, but they have a common acquaintance, namel... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2010_sol.jsonl",
"solution_match": "\nSolution:",
"tier": "T1",
"year": "2010"
} |
Let $A B C$ be an acute triangle satisfying the condition $A B>B C$ and $A C>B C$. Denote by $O$ and $H$ the circumcenter and the orthocenter, respectively, of the triangle $A B C$. Suppose that the circumcircle of the triangle $A H C$ intersects the line $A B$ at $M$ different from $A$, and that the circumcircle of th... | In the sequel, we denote $\angle B A C=\alpha, \angle C B A=\beta, \angle A C B=\gamma$. Let $O^{\prime}$ be the circumcenter of the triangle $M N H$. The lengths of line segments starting from the point $H$ will be treated as signed quantities.
Let us denote by $M^{\prime}, N^{\prime}$ the point of intersection of $C... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle satisfying the condition $A B>B C$ and $A C>B C$. Denote by $O$ and $H$ the circumcenter and the orthocenter, respectively, of the triangle $A B C$. Suppose that the circumcircle of the triangle $A H C$ intersects the line $A B$ at $M$ different from $A$, and that the circumcircle of th... | In the sequel, we denote $\angle B A C=\alpha, \angle C B A=\beta, \angle A C B=\gamma$. Let $O^{\prime}$ be the circumcenter of the triangle $M N H$. The lengths of line segments starting from the point $H$ will be treated as signed quantities.
Let us denote by $M^{\prime}, N^{\prime}$ the point of intersection of $C... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2010_sol.jsonl",
"solution_match": "\nSolution:",
"tier": "T1",
"year": "2010"
} |
Find all functions $f$ from the set $\mathbf{R}$ of real numbers into $\mathbf{R}$ which satisfy for all $x, y, z \in \mathbf{R}$ the identity
$$
f(f(x)+f(y)+f(z))=f(f(x)-f(y))+f(2 x y+f(z))+2 f(x z-y z) .
$$ | It is clear that if $f$ is a constant function which satisfies the given equation, then the constant must be 0 . Conversely, $f(x)=0$ clearly satisfies the given equation, so, the identically 0 function is a solution. In the sequel, we consider the case where $f$ is not a constant function.
Let $t \in \mathbf{R}$ and ... | f(x)=0 \text{ and } f(x)=x^{2} | Yes | Yes | math-word-problem | Algebra | Find all functions $f$ from the set $\mathbf{R}$ of real numbers into $\mathbf{R}$ which satisfy for all $x, y, z \in \mathbf{R}$ the identity
$$
f(f(x)+f(y)+f(z))=f(f(x)-f(y))+f(2 x y+f(z))+2 f(x z-y z) .
$$ | It is clear that if $f$ is a constant function which satisfies the given equation, then the constant must be 0 . Conversely, $f(x)=0$ clearly satisfies the given equation, so, the identically 0 function is a solution. In the sequel, we consider the case where $f$ is not a constant function.
Let $t \in \mathbf{R}$ and ... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2010_sol.jsonl",
"solution_match": "\nSolution:",
"tier": "T1",
"year": "2010"
} |
Let $A B C$ be an acute triangle with altitudes $A D, B E$ and $C F$, and let $O$ be the center of its circumcircle. Show that the segments $O A, O F, O B, O D, O C, O E$ dissect the triangle $A B C$ into three pairs of triangles that have equal areas. | Let $M$ and $N$ be midpoints of sides $B C$ and $A C$, respectively. Notice that $\angle M O C=\frac{1}{2} \angle B O C=\angle E A B, \angle O M C=90^{\circ}=\angle A E B$, so triangles $O M C$ and $A E B$ are similar and we get $\frac{O M}{A E}=\frac{O C}{A B}$. For triangles $O N A$ and $B D A$ we also have $\frac{O ... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle with altitudes $A D, B E$ and $C F$, and let $O$ be the center of its circumcircle. Show that the segments $O A, O F, O B, O D, O C, O E$ dissect the triangle $A B C$ into three pairs of triangles that have equal areas. | Let $M$ and $N$ be midpoints of sides $B C$ and $A C$, respectively. Notice that $\angle M O C=\frac{1}{2} \angle B O C=\angle E A B, \angle O M C=90^{\circ}=\angle A E B$, so triangles $O M C$ and $A E B$ are similar and we get $\frac{O M}{A E}=\frac{O C}{A B}$. For triangles $O N A$ and $B D A$ we also have $\frac{O ... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2013_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2013"
} |
Determine all positive integers $n$ for which $\frac{n^{2}+1}{[\sqrt{n}]^{2}+2}$ is an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$. | We will show that there are no positive integers $n$ satisfying the condition of the problem.
Let $m=[\sqrt{n}]$ and $a=n-m^{2}$. We have $m \geq 1$ since $n \geq 1$. From $n^{2}+1=\left(m^{2}+a\right)^{2}+1 \equiv$ $(a-2)^{2}+1\left(\bmod \left(m^{2}+2\right)\right)$, it follows that the condition of the problem is e... | proof | Yes | Yes | math-word-problem | Number Theory | Determine all positive integers $n$ for which $\frac{n^{2}+1}{[\sqrt{n}]^{2}+2}$ is an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$. | We will show that there are no positive integers $n$ satisfying the condition of the problem.
Let $m=[\sqrt{n}]$ and $a=n-m^{2}$. We have $m \geq 1$ since $n \geq 1$. From $n^{2}+1=\left(m^{2}+a\right)^{2}+1 \equiv$ $(a-2)^{2}+1\left(\bmod \left(m^{2}+2\right)\right)$, it follows that the condition of the problem is e... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2013_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2013"
} |
For $2 k$ real numbers $a_{1}, a_{2}, \ldots, a_{k}, b_{1}, b_{2}, \ldots, b_{k}$ define the sequence of numbers $X_{n}$ by
$$
X_{n}=\sum_{i=1}^{k}\left[a_{i} n+b_{i}\right] \quad(n=1,2, \ldots)
$$
If the sequence $X_{n}$ forms an arithmetic progression, show that $\sum_{i=1}^{k} a_{i}$ must be an integer. Here $[r]$... | Let us write $A=\sum_{i=1}^{k} a_{i}$ and $B=\sum_{i=1}^{k} b_{i}$. Summing the corresponding terms of the following inequalities over $i$,
$$
a_{i} n+b_{i}-1<\left[a_{i} n+b_{i}\right] \leq a_{i} n+b_{i}
$$
we obtain $A n+B-k<X_{n}<A n+B$. Now suppose that $\left\{X_{n}\right\}$ is an arithmetic progression with the... | proof | Yes | Yes | proof | Algebra | For $2 k$ real numbers $a_{1}, a_{2}, \ldots, a_{k}, b_{1}, b_{2}, \ldots, b_{k}$ define the sequence of numbers $X_{n}$ by
$$
X_{n}=\sum_{i=1}^{k}\left[a_{i} n+b_{i}\right] \quad(n=1,2, \ldots)
$$
If the sequence $X_{n}$ forms an arithmetic progression, show that $\sum_{i=1}^{k} a_{i}$ must be an integer. Here $[r]$... | Let us write $A=\sum_{i=1}^{k} a_{i}$ and $B=\sum_{i=1}^{k} b_{i}$. Summing the corresponding terms of the following inequalities over $i$,
$$
a_{i} n+b_{i}-1<\left[a_{i} n+b_{i}\right] \leq a_{i} n+b_{i}
$$
we obtain $A n+B-k<X_{n}<A n+B$. Now suppose that $\left\{X_{n}\right\}$ is an arithmetic progression with the... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2013_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2013"
} |
Let $a$ and $b$ be positive integers, and let $A$ and $B$ be finite sets of integers satisfying:
(i) $A$ and $B$ are disjoint;
(ii) if an integer $i$ belongs either to $A$ or to $B$, then $i+a$ belongs to $A$ or $i-b$ belongs to $B$.
Prove that $a|A|=b|B|$. (Here $|X|$ denotes the number of elements in the set $X$.) | Let $A^{*}=\{n-a: n \in A\}$ and $B^{*}=\{n+b: n \in B\}$. Then, by (ii), $A \cup B \subseteq A^{*} \cup B^{*}$ and by (i),
$$
|A \cup B| \leq\left|A^{*} \cup B^{*}\right| \leq\left|A^{*}\right|+\left|B^{*}\right|=|A|+|B|=|A \cup B|
$$
Thus, $A \cup B=A^{*} \cup B^{*}$ and $A^{*}$ and $B^{*}$ have no element in commo... | a|A|=b|B| | Yes | Yes | proof | Combinatorics | Let $a$ and $b$ be positive integers, and let $A$ and $B$ be finite sets of integers satisfying:
(i) $A$ and $B$ are disjoint;
(ii) if an integer $i$ belongs either to $A$ or to $B$, then $i+a$ belongs to $A$ or $i-b$ belongs to $B$.
Prove that $a|A|=b|B|$. (Here $|X|$ denotes the number of elements in the set $X$.) | Let $A^{*}=\{n-a: n \in A\}$ and $B^{*}=\{n+b: n \in B\}$. Then, by (ii), $A \cup B \subseteq A^{*} \cup B^{*}$ and by (i),
$$
|A \cup B| \leq\left|A^{*} \cup B^{*}\right| \leq\left|A^{*}\right|+\left|B^{*}\right|=|A|+|B|=|A \cup B|
$$
Thus, $A \cup B=A^{*} \cup B^{*}$ and $A^{*}$ and $B^{*}$ have no element in commo... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2013_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2013"
} |
Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $A C$ such that $P B$ and $P D$ are tangent to $\omega$. The tangent at $C$ intersects $P D$ at $Q$ and the line $A D$ at $R$. Let $E$ be the second point of intersection between $A Q$ and $\omega$. Prove that $B... | To show $B, E, R$ are collinear, it is equivalent to show the lines $A D, B E, C Q$ are concurrent. Let $C Q$ intersect $A D$ at $R$ and $B E$ intersect $A D$ at $R^{\prime}$. We shall show $R D / R A=R^{\prime} D / R^{\prime} A$ so that $R=R^{\prime}$.
Since $\triangle P A D$ is similar to $\triangle P D C$ and $\tri... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $A C$ such that $P B$ and $P D$ are tangent to $\omega$. The tangent at $C$ intersects $P D$ at $Q$ and the line $A D$ at $R$. Let $E$ be the second point of intersection between $A Q$ and $\omega$. Prove that $B... | To show $B, E, R$ are collinear, it is equivalent to show the lines $A D, B E, C Q$ are concurrent. Let $C Q$ intersect $A D$ at $R$ and $B E$ intersect $A D$ at $R^{\prime}$. We shall show $R D / R A=R^{\prime} D / R^{\prime} A$ so that $R=R^{\prime}$.
Since $\triangle P A D$ is similar to $\triangle P D C$ and $\tri... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2013_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2013"
} |
For a positive integer $m$ denote by $S(m)$ and $P(m)$ the sum and product, respectively, of the digits of $m$. Show that for each positive integer $n$, there exist positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying the following conditions:
$$
S\left(a_{1}\right)<S\left(a_{2}\right)<\cdots<S\left(a_{n}\right)... | Let $k$ be a sufficiently large positive integer. Choose for each $i=2,3, \ldots, n$, $a_{i}$ to be a positive integer among whose digits the number 2 appears exactly $k+i-2$ times and the number 1 appears exactly $2^{k+i-1}-2(k+i-2)$ times, and nothing else. Then, we have $S\left(a_{i}\right)=2^{k+i-1}$ and $P\left(a_... | proof | Yes | Yes | proof | Number Theory | For a positive integer $m$ denote by $S(m)$ and $P(m)$ the sum and product, respectively, of the digits of $m$. Show that for each positive integer $n$, there exist positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying the following conditions:
$$
S\left(a_{1}\right)<S\left(a_{2}\right)<\cdots<S\left(a_{n}\right)... | Let $k$ be a sufficiently large positive integer. Choose for each $i=2,3, \ldots, n$, $a_{i}$ to be a positive integer among whose digits the number 2 appears exactly $k+i-2$ times and the number 1 appears exactly $2^{k+i-1}-2(k+i-2)$ times, and nothing else. Then, we have $S\left(a_{i}\right)=2^{k+i-1}$ and $P\left(a_... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2014_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2014"
} |
Let $S=\{1,2, \ldots, 2014\}$. For each non-empty subset $T \subseteq S$, one of its members is chosen as its representative. Find the number of ways to assign representatives to all non-empty subsets of $S$ so that if a subset $D \subseteq S$ is a disjoint union of non-empty subsets $A, B, C \subseteq S$, then the rep... | Answer: 108 - 2014!.
For any subset $X$ let $r(X)$ denotes the representative of $X$. Suppose that $x_{1}=r(S)$. First, we prove the following fact:
$$
\text { If } x_{1} \in X \text { and } X \subseteq S \text {, then } x_{1}=r(X) \text {. }
$$
If $|X| \leq 2012$, then we can write $S$ as a disjoint union of $X$ and... | 108 \cdot 2014! | Yes | Yes | math-word-problem | Combinatorics | Let $S=\{1,2, \ldots, 2014\}$. For each non-empty subset $T \subseteq S$, one of its members is chosen as its representative. Find the number of ways to assign representatives to all non-empty subsets of $S$ so that if a subset $D \subseteq S$ is a disjoint union of non-empty subsets $A, B, C \subseteq S$, then the rep... | Answer: 108 - 2014!.
For any subset $X$ let $r(X)$ denotes the representative of $X$. Suppose that $x_{1}=r(S)$. First, we prove the following fact:
$$
\text { If } x_{1} \in X \text { and } X \subseteq S \text {, then } x_{1}=r(X) \text {. }
$$
If $|X| \leq 2012$, then we can write $S$ as a disjoint union of $X$ and... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2014_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2014"
} |
Find all positive integers $n$ such that for any integer $k$ there exists an integer $a$ for which $a^{3}+a-k$ is divisible by $n$. (Warut Suksompong, Thailand) | Answer: All integers $n=3^{b}$, where $b$ is a nonnegative integer.
We are looking for integers $n$ such that the set $A=\left\{a^{3}+a \mid a \in \mathbf{Z}\right\}$ is a complete residue system by modulo $n$. Let us call this property by $\left(^{*}\right)$. It is not hard to see that $n=1$ satisfies $\left({ }^{*}\r... | All integers $n=3^{b}$, where $b$ is a nonnegative integer. | Yes | Yes | math-word-problem | Number Theory | Find all positive integers $n$ such that for any integer $k$ there exists an integer $a$ for which $a^{3}+a-k$ is divisible by $n$. (Warut Suksompong, Thailand) | Answer: All integers $n=3^{b}$, where $b$ is a nonnegative integer.
We are looking for integers $n$ such that the set $A=\left\{a^{3}+a \mid a \in \mathbf{Z}\right\}$ is a complete residue system by modulo $n$. Let us call this property by $\left(^{*}\right)$. It is not hard to see that $n=1$ satisfies $\left({ }^{*}\r... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2014_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2014"
} |
Let $n$ and $b$ be positive integers. We say $n$ is $b$-discerning if there exists a set consisting of $n$ different positive integers less than $b$ that has no two different subsets $U$ and $V$ such that the sum of all elements in $U$ equals the sum of all elements in $V$.
(a) Prove that 8 is a 100 -discerning.
(b) Pr... | (a) Take $S=\{3,6,12,24,48,95,96,97\}$, i.e.
$$
S=\left\{3 \cdot 2^{k}: 0 \leq k \leq 5\right\} \cup\left\{3 \cdot 2^{5}-1,3 \cdot 2^{5}+1\right\}
$$
As $k$ ranges between 0 to 5 , the sums obtained from the numbers $3 \cdot 2^{k}$ are $3 t$, where $1 \leq t \leq 63$. These are 63 numbers that are divisible by 3 and ... | proof | Yes | Yes | proof | Combinatorics | Let $n$ and $b$ be positive integers. We say $n$ is $b$-discerning if there exists a set consisting of $n$ different positive integers less than $b$ that has no two different subsets $U$ and $V$ such that the sum of all elements in $U$ equals the sum of all elements in $V$.
(a) Prove that 8 is a 100 -discerning.
(b) Pr... | (a) Take $S=\{3,6,12,24,48,95,96,97\}$, i.e.
$$
S=\left\{3 \cdot 2^{k}: 0 \leq k \leq 5\right\} \cup\left\{3 \cdot 2^{5}-1,3 \cdot 2^{5}+1\right\}
$$
As $k$ ranges between 0 to 5 , the sums obtained from the numbers $3 \cdot 2^{k}$ are $3 t$, where $1 \leq t \leq 63$. These are 63 numbers that are divisible by 3 and ... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2014_sol.jsonl",
"solution_match": "# Solution.",
"tier": "T1",
"year": "2014"
} |
Circles $\omega$ and $\Omega$ meet at points $A$ and $B$. Let $M$ be the midpoint of the $\operatorname{arc} A B$ of circle $\omega$ ( $M$ lies inside $\Omega$ ). A chord $M P$ of circle $\omega$ intersects $\Omega$ at $Q(Q$ lies inside $\omega)$. Let $\ell_{P}$ be the tangent line to $\omega$ at $P$, and let $\ell_{Q}... | Denote $X=A B \cap \ell_{P}, Y=A B \cap \ell_{Q}$, and $Z=\ell_{P} \cap \ell_{Q}$. Without loss of generality we have $A X<B X$. Let $F=M P \cap A B$.
Denote by $R$ the second point of inter... | proof | Yes | Yes | proof | Geometry | Circles $\omega$ and $\Omega$ meet at points $A$ and $B$. Let $M$ be the midpoint of the $\operatorname{arc} A B$ of circle $\omega$ ( $M$ lies inside $\Omega$ ). A chord $M P$ of circle $\omega$ intersects $\Omega$ at $Q(Q$ lies inside $\omega)$. Let $\ell_{P}$ be the tangent line to $\omega$ at $P$, and let $\ell_{Q}... | Denote $X=A B \cap \ell_{P}, Y=A B \cap \ell_{Q}$, and $Z=\ell_{P} \cap \ell_{Q}$. Without loss of generality we have $A X<B X$. Let $F=M P \cap A B$.
Denote by $R$ the second point of inter... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2014_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2014"
} |
Let $A B C$ be a triangle, and let $D$ be a point on side $B C$. A line through $D$ intersects side $A B$ at $X$ and ray $A C$ at $Y$. The circumcircle of triangle $B X D$ intersects the circumcircle $\omega$ of triangle $A B C$ again at point $Z \neq B$. The lines $Z D$ and $Z Y$ intersect $\omega$ again at $V$ and $W... | Suppose $X Y$ intersects $\omega$ at points $P$ and $Q$, where $Q$ lies between $X$ and $Y$. We will show that $V$ and $W$ are the reflections of $A$ and $B$ with respect to the perpendicular bisector of $P Q$. From this, it follows that $A V W B$ is an isosceles trapezoid and hence $A B=V W$.
First, note that
$$
\an... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle, and let $D$ be a point on side $B C$. A line through $D$ intersects side $A B$ at $X$ and ray $A C$ at $Y$. The circumcircle of triangle $B X D$ intersects the circumcircle $\omega$ of triangle $A B C$ again at point $Z \neq B$. The lines $Z D$ and $Z Y$ intersect $\omega$ again at $V$ and $W... | Suppose $X Y$ intersects $\omega$ at points $P$ and $Q$, where $Q$ lies between $X$ and $Y$. We will show that $V$ and $W$ are the reflections of $A$ and $B$ with respect to the perpendicular bisector of $P Q$. From this, it follows that $A V W B$ is an isosceles trapezoid and hence $A B=V W$.
First, note that
$$
\an... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2015_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2015"
} |
Let $S=\{2,3,4, \ldots\}$ denote the set of integers that are greater than or equal to 2 . Does there exist a function $f: S \rightarrow S$ such that
$$
f(a) f(b)=f\left(a^{2} b^{2}\right) \text { for all } a, b \in S \text { with } a \neq b ?
$$ | We prove that there is no such function. For arbitrary elements $a$ and $b$ of $S$, choose an integer $c$ that is greater than both of them. Since $b c>a$ and $c>b$, we have
$$
f\left(a^{4} b^{4} c^{4}\right)=f\left(a^{2}\right) f\left(b^{2} c^{2}\right)=f\left(a^{2}\right) f(b) f(c)
$$
Furthermore, since $a c>b$ and... | proof | Yes | Yes | math-word-problem | Algebra | Let $S=\{2,3,4, \ldots\}$ denote the set of integers that are greater than or equal to 2 . Does there exist a function $f: S \rightarrow S$ such that
$$
f(a) f(b)=f\left(a^{2} b^{2}\right) \text { for all } a, b \in S \text { with } a \neq b ?
$$ | We prove that there is no such function. For arbitrary elements $a$ and $b$ of $S$, choose an integer $c$ that is greater than both of them. Since $b c>a$ and $c>b$, we have
$$
f\left(a^{4} b^{4} c^{4}\right)=f\left(a^{2}\right) f\left(b^{2} c^{2}\right)=f\left(a^{2}\right) f(b) f(c)
$$
Furthermore, since $a c>b$ and... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2015_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2015"
} |
A sequence of real numbers $a_{0}, a_{1}, \ldots$ is said to be good if the following three conditions hold.
(i) The value of $a_{0}$ is a positive integer.
(ii) For each non-negative integer $i$ we have $a_{i+1}=2 a_{i}+1$ or $a_{i+1}=\frac{a_{i}}{a_{i}+2}$.
(iii) There exists a positive integer $k$ such that $a_{k}=2... | Note that
$$
a_{i+1}+1=2\left(a_{i}+1\right) \text { or } a_{i+1}+1=\frac{a_{i}+a_{i}+2}{a_{i}+2}=\frac{2\left(a_{i}+1\right)}{a_{i}+2} .
$$
Hence
$$
\frac{1}{a_{i+1}+1}=\frac{1}{2} \cdot \frac{1}{a_{i}+1} \text { or } \frac{1}{a_{i+1}+1}=\frac{a_{i}+2}{2\left(a_{i}+1\right)}=\frac{1}{2} \cdot \frac{1}{a_{i}+1}+\fra... | 60 | Yes | Yes | math-word-problem | Number Theory | A sequence of real numbers $a_{0}, a_{1}, \ldots$ is said to be good if the following three conditions hold.
(i) The value of $a_{0}$ is a positive integer.
(ii) For each non-negative integer $i$ we have $a_{i+1}=2 a_{i}+1$ or $a_{i+1}=\frac{a_{i}}{a_{i}+2}$.
(iii) There exists a positive integer $k$ such that $a_{k}=2... | Note that
$$
a_{i+1}+1=2\left(a_{i}+1\right) \text { or } a_{i+1}+1=\frac{a_{i}+a_{i}+2}{a_{i}+2}=\frac{2\left(a_{i}+1\right)}{a_{i}+2} .
$$
Hence
$$
\frac{1}{a_{i+1}+1}=\frac{1}{2} \cdot \frac{1}{a_{i}+1} \text { or } \frac{1}{a_{i+1}+1}=\frac{a_{i}+2}{2\left(a_{i}+1\right)}=\frac{1}{2} \cdot \frac{1}{a_{i}+1}+\fra... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2015_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2015"
} |
Let $n$ be a positive integer. Consider $2 n$ distinct lines on the plane, no two of which are parallel. Of the $2 n$ lines, $n$ are colored blue, the other $n$ are colored red. Let $\mathcal{B}$ be the set of all points on the plane that lie on at least one blue line, and $\mathcal{R}$ the set of all points on the pla... | Consider a line $\ell$ on the plane and a point $P$ on it such that $\ell$ is not parallel to any of the $2 n$ lines. Rotate $\ell$ about $P$ counterclockwise until it is parallel to one of the $2 n$ lines. Take note of that line and keep rotating until all the $2 n$ lines are met. The $2 n$ lines are now ordered accor... | proof | Yes | Incomplete | proof | Geometry | Let $n$ be a positive integer. Consider $2 n$ distinct lines on the plane, no two of which are parallel. Of the $2 n$ lines, $n$ are colored blue, the other $n$ are colored red. Let $\mathcal{B}$ be the set of all points on the plane that lie on at least one blue line, and $\mathcal{R}$ the set of all points on the pla... | Consider a line $\ell$ on the plane and a point $P$ on it such that $\ell$ is not parallel to any of the $2 n$ lines. Rotate $\ell$ about $P$ counterclockwise until it is parallel to one of the $2 n$ lines. Take note of that line and keep rotating until all the $2 n$ lines are met. The $2 n$ lines are now ordered accor... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2015_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2015"
} |
Determine all sequences $a_{0}, a_{1}, a_{2}, \ldots$ of positive integers with $a_{0} \geq 2015$ such that for all integers $n \geq 1$ :
(i) $a_{n+2}$ is divisible by $a_{n}$;
(ii) $\left|s_{n+1}-(n+1) a_{n}\right|=1$, where $s_{n+1}=a_{n+1}-a_{n}+a_{n-1}-\cdots+(-1)^{n+1} a_{0}$.
Answer: There are two families of an... | Let $\left\{a_{n}\right\}_{n=0}^{\infty}$ be a sequence of positive integers satisfying the given conditions. We can rewrite (ii) as $s_{n+1}=(n+1) a_{n}+h_{n}$, where $h_{n} \in\{-1,1\}$. Substituting $n$ with $n-1$ yields $s_{n}=n a_{n-1}+h_{n-1}$, where $h_{n-1} \in\{-1,1\}$. Note that $a_{n+1}=s_{n+1}+s_{n}$, there... | a_{n}=c(n+2) n! \text{ for all } n \geq 1 \text{ and } a_{0}=c \pm 1 \text{ for some integer } c \geq 2014 | Yes | Yes | math-word-problem | Number Theory | Determine all sequences $a_{0}, a_{1}, a_{2}, \ldots$ of positive integers with $a_{0} \geq 2015$ such that for all integers $n \geq 1$ :
(i) $a_{n+2}$ is divisible by $a_{n}$;
(ii) $\left|s_{n+1}-(n+1) a_{n}\right|=1$, where $s_{n+1}=a_{n+1}-a_{n}+a_{n-1}-\cdots+(-1)^{n+1} a_{0}$.
Answer: There are two families of an... | Let $\left\{a_{n}\right\}_{n=0}^{\infty}$ be a sequence of positive integers satisfying the given conditions. We can rewrite (ii) as $s_{n+1}=(n+1) a_{n}+h_{n}$, where $h_{n} \in\{-1,1\}$. Substituting $n$ with $n-1$ yields $s_{n}=n a_{n-1}+h_{n-1}$, where $h_{n-1} \in\{-1,1\}$. Note that $a_{n+1}=s_{n+1}+s_{n}$, there... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2015_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2015"
} |
We say that a triangle $A B C$ is great if the following holds: for any point $D$ on the side $B C$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $A B$ and $A C$, respectively, then the reflection of $D$ in the line $P Q$ lies on the circumcircle of the triangle $A B C$.
Prove that triangle ... | For every point $D$ on the side $B C$, let $D^{\prime}$ be the reflection of $D$ in the line $P Q$. We will first prove that if the triangle satisfies the condition then it is isosceles and right-angled at $A$.
Choose $D$ to be the point where the angle bisector from $A$ meets $B C$. Note that $P$ and $Q$ lie on the r... | proof | Yes | Yes | proof | Geometry | We say that a triangle $A B C$ is great if the following holds: for any point $D$ on the side $B C$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $A B$ and $A C$, respectively, then the reflection of $D$ in the line $P Q$ lies on the circumcircle of the triangle $A B C$.
Prove that triangle ... | For every point $D$ on the side $B C$, let $D^{\prime}$ be the reflection of $D$ in the line $P Q$. We will first prove that if the triangle satisfies the condition then it is isosceles and right-angled at $A$.
Choose $D$ to be the point where the angle bisector from $A$ meets $B C$. Note that $P$ and $Q$ lie on the r... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
A positive integer is called fancy if it can be expressed in the form
$$
2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{100}}
$$
where $a_{1}, a_{2}, \ldots, a_{100}$ are non-negative integers that are not necessarily distinct.
Find the smallest positive integer $n$ such that no multiple of $n$ is a fancy number.
Answer: The answe... | Let $k$ be any positive integer less than $2^{101}-1$. Then $k$ can be expressed in binary notation using at most 100 ones, and therefore there exists a positive integer $r$ and non-negative integers $a_{1}, a_{2}, \ldots, a_{r}$ such that $r \leq 100$ and $k=2^{a_{1}}+\cdots+2^{a_{r}}$. Notice that for a positive inte... | 2^{101}-1 | Yes | Yes | math-word-problem | Number Theory | A positive integer is called fancy if it can be expressed in the form
$$
2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{100}}
$$
where $a_{1}, a_{2}, \ldots, a_{100}$ are non-negative integers that are not necessarily distinct.
Find the smallest positive integer $n$ such that no multiple of $n$ is a fancy number.
Answer: The answe... | Let $k$ be any positive integer less than $2^{101}-1$. Then $k$ can be expressed in binary notation using at most 100 ones, and therefore there exists a positive integer $r$ and non-negative integers $a_{1}, a_{2}, \ldots, a_{r}$ such that $r \leq 100$ and $k=2^{a_{1}}+\cdots+2^{a_{r}}$. Notice that for a positive inte... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$... | We present two approaches. The first one introduces an auxiliary point and studies similarities in the figure. The second one reduces the problem to computations involving a particular exradius of a triangle. The second approach has two variants. | proof | Yes | Yes | proof | Geometry | Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$... | We present two approaches. The first one introduces an auxiliary point and studies similarities in the figure. The second one reduces the problem to computations involving a particular exradius of a triangle. The second approach has two variants. | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$... | .
Let the line through $N$ tangent to $\omega$ at point $X \neq E$ intersect $A B$ at point $M^{\prime}$. It suffices to show that $M^{\prime} R \| A C$, since this would yield $M^{\prime}=M... | proof | Yes | Yes | proof | Geometry | Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$... | .
Let the line through $N$ tangent to $\omega$ at point $X \neq E$ intersect $A B$ at point $M^{\prime}$. It suffices to show that $M^{\prime} R \| A C$, since this would yield $M^{\prime}=M... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2016"
} |
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$... | a.
As in Solution 1, we introduce point $M^{\prime}$ and reduce the problem to proving $\frac{P R}{R N}=\frac{P M^{\prime}}{M^{\prime} A}$. Menelaus theorem in triangle $A N P$ with transversal line $F R E$ yields
$$
\frac{P R}{R N} \cdot \frac{N E}{E A} \cdot \frac{A F}{F P}=1
$$
Since $A F=E A$, we have $\frac{F P... | proof | Yes | Yes | proof | Geometry | Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$... | a.
As in Solution 1, we introduce point $M^{\prime}$ and reduce the problem to proving $\frac{P R}{R N}=\frac{P M^{\prime}}{M^{\prime} A}$. Menelaus theorem in triangle $A N P$ with transversal line $F R E$ yields
$$
\frac{P R}{R N} \cdot \frac{N E}{E A} \cdot \frac{A F}{F P}=1
$$
Since $A F=E A$, we have $\frac{F P... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2016"
} |
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$... | b.
As in Solution 1, we introduce point $M^{\prime}$. Let the line through $M^{\prime}$ and parallel to $A N$ intersect $E F$ at $R^{\prime}$. Let $P^{\prime}$ be the intersection of lines $N R^{\prime}$ and $A M$. It suffices to show that $P^{\prime} O \| F E$, since this would yield $P=P^{\prime}$, and then $R=R^{\p... | proof | Yes | Yes | proof | Geometry | Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$... | b.
As in Solution 1, we introduce point $M^{\prime}$. Let the line through $M^{\prime}$ and parallel to $A N$ intersect $E F$ at $R^{\prime}$. Let $P^{\prime}$ be the intersection of lines $N R^{\prime}$ and $A M$. It suffices to show that $P^{\prime} O \| F E$, since this would yield $P=P^{\prime}$, and then $R=R^{\p... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2016"
} |
The country Dreamland consists of 2016 cities. The airline Starways wants to establish some one-way flights between pairs of cities in such a way that each city has exactly one flight out of it. Find the smallest positive integer $k$ such that no matter how Starways establishes its flights, the cities can always be par... | The flights established by Starways yield a directed graph $G$ on 2016 vertices in which each vertex has out-degree equal to 1.
We first show that we need at least 57 groups. For this, suppose that $G$ has a directed cycle of length 57. Then, for any two cities in the cycle, one is reachable from the other using at mo... | 57 | Yes | Yes | math-word-problem | Combinatorics | The country Dreamland consists of 2016 cities. The airline Starways wants to establish some one-way flights between pairs of cities in such a way that each city has exactly one flight out of it. Find the smallest positive integer $k$ such that no matter how Starways establishes its flights, the cities can always be par... | The flights established by Starways yield a directed graph $G$ on 2016 vertices in which each vertex has out-degree equal to 1.
We first show that we need at least 57 groups. For this, suppose that $G$ has a directed cycle of length 57. Then, for any two cities in the cycle, one is reachable from the other using at mo... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
$$
(z+1) f(x+y)=f(x f(z)+y)+f(y f(z)+x)
$$
for all positive real numbers $x, y, z$.
Answer: The only solution is $f(x)=x$ for all positive real numbers $x$. | The identity function $f(x)=x$ clearly satisfies the functional equation. Now, let $f$ be a function satisfying the functional equation. Plugging $x=y=1$ into (3) we get $2 f(f(z)+1)=(z+1)(f(2))$ for all $z \in \mathbb{R}^{+}$. Hence, $f$ is not bounded above.
Lemma. Let $a, b, c$ be positive real numbers. If $c$ is g... | f(x)=x | Yes | Yes | proof | Algebra | Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
$$
(z+1) f(x+y)=f(x f(z)+y)+f(y f(z)+x)
$$
for all positive real numbers $x, y, z$.
Answer: The only solution is $f(x)=x$ for all positive real numbers $x$. | The identity function $f(x)=x$ clearly satisfies the functional equation. Now, let $f$ be a function satisfying the functional equation. Plugging $x=y=1$ into (3) we get $2 f(f(z)+1)=(z+1)(f(2))$ for all $z \in \mathbb{R}^{+}$. Hence, $f$ is not bounded above.
Lemma. Let $a, b, c$ be positive real numbers. If $c$ is g... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
We call a 5-tuple of integers arrangeable if its elements can be labeled $a$, $b, c, d, e$ in some order so that $a-b+c-d+e=29$. Determine all 2017-tuples of integers $n_{1}, n_{2}, \ldots, n_{2017}$ such that if we place them in a circle in clockwise order, then any 5 -tuple of numbers in consecutive positions on the ... | A valid 2017-tuple is $n_{1}=\cdots=n_{2017}=29$. We will show that it is the only solution.
We first replace each number $n_{i}$ in the circle by $m_{i}:=n_{i}-29$. Since the condition $a-b+$ $c-d+e=29$ can be rewritten as $(a-29)-(b-29)+(c-29)-(d-29)+(e-29)=0$, we have that any five consecutive replaced integers in ... | n_{1}=\cdots=n_{2017}=29 | Yes | Yes | math-word-problem | Combinatorics | We call a 5-tuple of integers arrangeable if its elements can be labeled $a$, $b, c, d, e$ in some order so that $a-b+c-d+e=29$. Determine all 2017-tuples of integers $n_{1}, n_{2}, \ldots, n_{2017}$ such that if we place them in a circle in clockwise order, then any 5 -tuple of numbers in consecutive positions on the ... | A valid 2017-tuple is $n_{1}=\cdots=n_{2017}=29$. We will show that it is the only solution.
We first replace each number $n_{i}$ in the circle by $m_{i}:=n_{i}-29$. Since the condition $a-b+$ $c-d+e=29$ can be rewritten as $(a-29)-(b-29)+(c-29)-(d-29)+(e-29)=0$, we have that any five consecutive replaced integers in ... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$... | . Let $N$ be the midpoint of $A C$. Let $M$ be the intersection point of the circumcircle of triangle $A D Z$ and the segment $A B$. We will show that $M$ is the midpoint of $A B$. To do this, let $D^{\prime}$ the reflection of $D$ with respect to $M$. It suffices to show that $A D B D^{\prime}$ is a parallelogram.
Th... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$... | . Let $N$ be the midpoint of $A C$. Let $M$ be the intersection point of the circumcircle of triangle $A D Z$ and the segment $A B$. We will show that $M$ is the midpoint of $A B$. To do this, let $D^{\prime}$ the reflection of $D$ with respect to $M$. It suffices to show that $A D B D^{\prime}$ is a parallelogram.
Th... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$... | . We proceed directly. As above, we name
$$
\alpha=\angle D A C=\angle A Z N=\angle C Z N=\angle D C B .
$$
Let $L$ be the midpoint of the segment $B C$. Since $M$ and $N$ are midpoints of $A B$ and $A C$, we have that $M N=C L$ and that the segment $M N$ is parallel to $B C$. Thus, $\angle A N M=\angle A C B$.
Ther... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$... | . We proceed directly. As above, we name
$$
\alpha=\angle D A C=\angle A Z N=\angle C Z N=\angle D C B .
$$
Let $L$ be the midpoint of the segment $B C$. Since $M$ and $N$ are midpoints of $A B$ and $A C$, we have that $M N=C L$ and that the segment $M N$ is parallel to $B C$. Thus, $\angle A N M=\angle A C B$.
Ther... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$... | . Let $m$ be the perpendicular bisector of $A D$; thus $m$ passes through the center $O$ of the circumcircle of triangle $A B C$. Since $A D$ is the internal angle bisector of $A$ and $O M$ and $O N$ are perpendicular to $A B$ and $A C$ respectively, we obtain that $O M$ and $O N$ form equal angles with $A D$. This imp... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$... | . Let $m$ be the perpendicular bisector of $A D$; thus $m$ passes through the center $O$ of the circumcircle of triangle $A B C$. Since $A D$ is the internal angle bisector of $A$ and $O M$ and $O N$ are perpendicular to $A B$ and $A C$ respectively, we obtain that $O M$ and $O N$ form equal angles with $A D$. This imp... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"solution_match": "\nSolution 3",
"tier": "T1",
"year": "2017"
} |
Let $A(n)$ denote the number of sequences $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of positive integers for which $a_{1}+\cdots+a_{k}=n$ and each $a_{i}+1$ is a power of two $(i=1,2, \ldots, k)$. Let $B(n)$ denote the number of sequences $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$ of positive integers for which $b_{1}+\... | We say that a sequence $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of positive integers has type $A$ if $a_{i}+1$ is a power of two for $i=1,2, \ldots, k$. We say that a sequence $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$ of positive integer has type $B$ if $b_{j} \geq 2 b_{j+1}$ for $j=1,2, \ldots, m-1$.
Recall that the... | proof | Yes | Yes | proof | Combinatorics | Let $A(n)$ denote the number of sequences $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of positive integers for which $a_{1}+\cdots+a_{k}=n$ and each $a_{i}+1$ is a power of two $(i=1,2, \ldots, k)$. Let $B(n)$ denote the number of sequences $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$ of positive integers for which $b_{1}+\... | We say that a sequence $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of positive integers has type $A$ if $a_{i}+1$ is a power of two for $i=1,2, \ldots, k$. We say that a sequence $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$ of positive integer has type $B$ if $b_{j} \geq 2 b_{j+1}$ for $j=1,2, \ldots, m-1$.
Recall that the... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2017"
} |
Call a rational number $r$ powerful if $r$ can be expressed in the form $\frac{p^{k}}{q}$ for some relatively prime positive integers $p, q$ and some integer $k>1$. Let $a, b, c$ be positive
rational numbers such that $a b c=1$. Suppose there exist positive integers $x, y, z$ such that $a^{x}+b^{y}+c^{z}$ is an integer... | Let $a=\frac{a_{1}}{b_{1}}, b=\frac{a_{2}}{b_{2}}$, where $\operatorname{gcd}\left(a_{1}, b_{1}\right)=\operatorname{gcd}\left(a_{2}, b_{2}\right)=1$. Then $c=\frac{b_{1} b_{2}}{a_{1} a_{2}}$. The condition that $a^{x}+b^{y}+c^{z}$ is an integer becomes
$$
\frac{a_{1}^{x+z} a_{2}^{z} b_{2}^{y}+a_{1}^{z} a_{2}^{y+z} b_... | proof | Yes | Yes | proof | Number Theory | Call a rational number $r$ powerful if $r$ can be expressed in the form $\frac{p^{k}}{q}$ for some relatively prime positive integers $p, q$ and some integer $k>1$. Let $a, b, c$ be positive
rational numbers such that $a b c=1$. Suppose there exist positive integers $x, y, z$ such that $a^{x}+b^{y}+c^{z}$ is an integer... | Let $a=\frac{a_{1}}{b_{1}}, b=\frac{a_{2}}{b_{2}}$, where $\operatorname{gcd}\left(a_{1}, b_{1}\right)=\operatorname{gcd}\left(a_{2}, b_{2}\right)=1$. Then $c=\frac{b_{1} b_{2}}{a_{1} a_{2}}$. The condition that $a^{x}+b^{y}+c^{z}$ is an integer becomes
$$
\frac{a_{1}^{x+z} a_{2}^{z} b_{2}^{y}+a_{1}^{z} a_{2}^{y+z} b_... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2017"
} |
Let $n$ be a positive integer. A pair of $n$-tuples $\left(a_{1}, \ldots, a_{n}\right)$ and $\left(b_{1}, \ldots, b_{n}\right)$ with integer entries is called an exquisite pair if
$$
\left|a_{1} b_{1}+\cdots+a_{n} b_{n}\right| \leq 1
$$
Determine the maximum number of distinct $n$-tuples with integer entries such tha... | First, we construct an example with $n^{2}+n+1 n$-tuples, each two of them forming an exquisite pair. In the following list, $*$ represents any number of zeros as long as the total number of entries is $n$.
- (*)
- $(*, 1, *)$
- $(*,-1, *)$
- $(*, 1, *, 1, *)$
- $(*, 1, *,-1, *)$
For example, for $n=2$ we have the tu... | n^{2}+n+1 | Yes | Incomplete | math-word-problem | Combinatorics | Let $n$ be a positive integer. A pair of $n$-tuples $\left(a_{1}, \ldots, a_{n}\right)$ and $\left(b_{1}, \ldots, b_{n}\right)$ with integer entries is called an exquisite pair if
$$
\left|a_{1} b_{1}+\cdots+a_{n} b_{n}\right| \leq 1
$$
Determine the maximum number of distinct $n$-tuples with integer entries such tha... | First, we construct an example with $n^{2}+n+1 n$-tuples, each two of them forming an exquisite pair. In the following list, $*$ represents any number of zeros as long as the total number of entries is $n$.
- (*)
- $(*, 1, *)$
- $(*,-1, *)$
- $(*, 1, *, 1, *)$
- $(*, 1, *,-1, *)$
For example, for $n=2$ we have the tu... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2017"
} |
Let $H$ be the orthocenter of the triangle $A B C$. Let $M$ and $N$ be the midpoints of the sides $A B$ and $A C$, respectively. Assume that $H$ lies inside the quadrilateral $B M N C$ and that the circumcircles of triangles $B M H$ and $C N H$ are tangent to each other. The line through $H$ parallel to $B C$ intersect... | Lemma 1. In a triangle $A B C$, let $D$ be the intersection of the interior angle bisector at $A$ with the circumcircle of $A B C$, and let $I$ be the incenter of $\triangle A B C$. Then
$$
D I=D B=D C
$$
![](https://cdn.mathpix.com/cropped/2024_11_22_c87ece96a341aa1506ccg-1.jpg?height=637&width=618&top_left_y=938&to... | proof | Yes | Yes | proof | Geometry | Let $H$ be the orthocenter of the triangle $A B C$. Let $M$ and $N$ be the midpoints of the sides $A B$ and $A C$, respectively. Assume that $H$ lies inside the quadrilateral $B M N C$ and that the circumcircles of triangles $B M H$ and $C N H$ are tangent to each other. The line through $H$ parallel to $B C$ intersect... | Lemma 1. In a triangle $A B C$, let $D$ be the intersection of the interior angle bisector at $A$ with the circumcircle of $A B C$, and let $I$ be the incenter of $\triangle A B C$. Then
$$
D I=D B=D C
$$
![](https://cdn.mathpix.com/cropped/2024_11_22_c87ece96a341aa1506ccg-1.jpg?height=637&width=618&top_left_y=938&to... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"solution_match": "# Solution.",
"tier": "T1",
"year": "2018"
} |
Let $H$ be the orthocenter of the triangle $A B C$. Let $M$ and $N$ be the midpoints of the sides $A B$ and $A C$, respectively. Assume that $H$ lies inside the quadrilateral $B M N C$ and that the circumcircles of triangles $B M H$ and $C N H$ are tangent to each other. The line through $H$ parallel to $B C$ intersect... | : According to Solution 1, we have $\angle M H N=180^{\circ}-2 \angle B A C$ and since the point $J$ is the incenter of $\triangle M H N$, we have $\angle M J N=90^{\circ}+\frac{1}{2} \angle M H N=180^{\circ}-\angle B A C$. So the quadrilateral $A M J N$ is cyclic.
According to Solution 1, the point $F$ is the circumc... | proof | Yes | Yes | proof | Geometry | Let $H$ be the orthocenter of the triangle $A B C$. Let $M$ and $N$ be the midpoints of the sides $A B$ and $A C$, respectively. Assume that $H$ lies inside the quadrilateral $B M N C$ and that the circumcircles of triangles $B M H$ and $C N H$ are tangent to each other. The line through $H$ parallel to $B C$ intersect... | : According to Solution 1, we have $\angle M H N=180^{\circ}-2 \angle B A C$ and since the point $J$ is the incenter of $\triangle M H N$, we have $\angle M J N=90^{\circ}+\frac{1}{2} \angle M H N=180^{\circ}-\angle B A C$. So the quadrilateral $A M J N$ is cyclic.
According to Solution 1, the point $F$ is the circumc... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2018"
} |
Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$. | There are two cases: $2 n-1<x<2 n$ and $2 n<x<2 n+1$. Note that $f(2018-x)=-f(x)$ and $g(2018-x)=-g(x)$, that is, a half turn about the point $(1009,0)$ preserves the graphs of $f$ and $g$. So it suffices to consider only the case $2 n-1<x<2 n$.
Let $d(x)=g(x)-f(x)$. We will show that $d(x)>2$ whenever $2 n-1<x<2 n$ a... | proof | Yes | Yes | proof | Algebra | Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$. | There are two cases: $2 n-1<x<2 n$ and $2 n<x<2 n+1$. Note that $f(2018-x)=-f(x)$ and $g(2018-x)=-g(x)$, that is, a half turn about the point $(1009,0)$ preserves the graphs of $f$ and $g$. So it suffices to consider only the case $2 n-1<x<2 n$.
Let $d(x)=g(x)-f(x)$. We will show that $d(x)>2$ whenever $2 n-1<x<2 n$ a... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2018"
} |
Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$. | As in Solution 1, we may assume $2 n-1<x<2 n$ for some $1 \leq n \leq 1009$. Let $d(x)=$ $f(x)-g(x)$, and note that
$$
d(x)=\frac{1}{x}+\sum_{m=1}^{1009} \frac{1}{(x-2 m)(x-2 m+1)}
$$
We split the sum into three parts: the terms before $m=n$, after $m=n$, and the term $m=n$. The first two are
$$
\begin{aligned}
0 & ... | proof | Yes | Yes | proof | Algebra | Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$. | As in Solution 1, we may assume $2 n-1<x<2 n$ for some $1 \leq n \leq 1009$. Let $d(x)=$ $f(x)-g(x)$, and note that
$$
d(x)=\frac{1}{x}+\sum_{m=1}^{1009} \frac{1}{(x-2 m)(x-2 m+1)}
$$
We split the sum into three parts: the terms before $m=n$, after $m=n$, and the term $m=n$. The first two are
$$
\begin{aligned}
0 & ... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2018"
} |
Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$. | First notice that
$$
f(x)-g(x)=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-2}-\cdots-\frac{1}{x-2017}+\frac{1}{x-2018} .
$$
As in Solution 1, we may deal only with the case $2 n<x<2 n+1$. Then $x-2 k+1$ and $x-2 k$ never differ in sign for any integer $k$. Then
$$
\begin{aligned}
-\frac{1}{x-2 k+1}+\frac{1}{x-2 k} & =\frac... | proof | Yes | Yes | proof | Algebra | Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$. | First notice that
$$
f(x)-g(x)=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-2}-\cdots-\frac{1}{x-2017}+\frac{1}{x-2018} .
$$
As in Solution 1, we may deal only with the case $2 n<x<2 n+1$. Then $x-2 k+1$ and $x-2 k$ never differ in sign for any integer $k$. Then
$$
\begin{aligned}
-\frac{1}{x-2 k+1}+\frac{1}{x-2 k} & =\frac... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"solution_match": "# Solution 3",
"tier": "T1",
"year": "2018"
} |
A collection of $n$ squares on the plane is called tri-connected if the following criteria are satisfied:
(i) All the squares are congruent.
(ii) If two squares have a point $P$ in common, then $P$ is a vertex of each of the squares.
(iii) Each square touches exactly three other squares.
How many positive integers $n$... | We will prove that there is no tri-connected collection if $n$ is odd, and that tri-connected collections exist for all even $n \geq 38$. Since there are 501 even numbers in the range from 2018 to 3018, this yields 501 as the answer.
For any two different squares $A$ and $B$, let us write $A \sim B$ to mean that squar... | 501 | Yes | Yes | math-word-problem | Combinatorics | A collection of $n$ squares on the plane is called tri-connected if the following criteria are satisfied:
(i) All the squares are congruent.
(ii) If two squares have a point $P$ in common, then $P$ is a vertex of each of the squares.
(iii) Each square touches exactly three other squares.
How many positive integers $n$... | We will prove that there is no tri-connected collection if $n$ is odd, and that tri-connected collections exist for all even $n \geq 38$. Since there are 501 even numbers in the range from 2018 to 3018, this yields 501 as the answer.
For any two different squares $A$ and $B$, let us write $A \sim B$ to mean that squar... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2018"
} |
Let $A B C$ be an equilateral triangle. From the vertex $A$ we draw a ray towards the interior of the triangle such that the ray reaches one of the sides of the triangle. When the ray reaches a side, it then bounces off following the law of reflection, that is, if it arrives with a directed angle $\alpha$, it leaves wi... | Consider an equilateral triangle $A A_{1} A_{2}$ of side length $m$ and triangulate it with unitary triangles. See the figure. To each of the vertices that remain after the triangulation we can assign a pair of coordinates $(a, b)$ where $a, b$ are non-negative integers, $a$ is the number of edges we travel in the $A A... | n \equiv 1,5 \bmod 6 \text{ with the exception of 5 and 17} | Yes | Yes | math-word-problem | Geometry | Let $A B C$ be an equilateral triangle. From the vertex $A$ we draw a ray towards the interior of the triangle such that the ray reaches one of the sides of the triangle. When the ray reaches a side, it then bounces off following the law of reflection, that is, if it arrives with a directed angle $\alpha$, it leaves wi... | Consider an equilateral triangle $A A_{1} A_{2}$ of side length $m$ and triangulate it with unitary triangles. See the figure. To each of the vertices that remain after the triangulation we can assign a pair of coordinates $(a, b)$ where $a, b$ are non-negative integers, $a$ is the number of edges we travel in the $A A... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2018"
} |
Find all polynomials $P(x)$ with integer coefficients such that for all real numbers $s$ and $t$, if $P(s)$ and $P(t)$ are both integers, then $P(s t)$ is also an integer.
Answer: $P(x)=x^{n}+k,-x^{n}+k$ for $n$ a non-negative integer and $k$ an integer. | : $P(x)=x^{n}+k,-x^{n}+k$ for $n$ a non-negative integer and $k$ an integer.
Notice that if $P(x)$ is a solution, then so is $P(x)+k$ and $-P(x)+k$ for any integer $k$, so we may assume that the leading coefficient of $P(x)$ is positive and that $P(0)=0$, i.e., we can assume that $P(x)=\sum_{i=1}^{n} a_{i} x^{i}$ with ... | P(x)=x^{n}+k,-x^{n}+k | Yes | Yes | math-word-problem | Algebra | Find all polynomials $P(x)$ with integer coefficients such that for all real numbers $s$ and $t$, if $P(s)$ and $P(t)$ are both integers, then $P(s t)$ is also an integer.
Answer: $P(x)=x^{n}+k,-x^{n}+k$ for $n$ a non-negative integer and $k$ an integer. | : $P(x)=x^{n}+k,-x^{n}+k$ for $n$ a non-negative integer and $k$ an integer.
Notice that if $P(x)$ is a solution, then so is $P(x)+k$ and $-P(x)+k$ for any integer $k$, so we may assume that the leading coefficient of $P(x)$ is positive and that $P(0)=0$, i.e., we can assume that $P(x)=\sum_{i=1}^{n} a_{i} x^{i}$ with ... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2018"
} |
Find all polynomials $P(x)$ with integer coefficients such that for all real numbers $s$ and $t$, if $P(s)$ and $P(t)$ are both integers, then $P(s t)$ is also an integer.
Answer: $P(x)=x^{n}+k,-x^{n}+k$ for $n$ a non-negative integer and $k$ an integer. | : Assume $P(x)=\sum_{i=0}^{n} a_{i} x^{i}$. Consider the following system of equations
$$
\begin{aligned}
& a_{0}=P(0) \\
& a_{n} t^{n}+a_{n-1} t^{n-1}+\cdots+a_{0}=P(t) \\
& 2^{n} a_{n} t^{n}+2^{n-1} a_{n-1} t^{n-1}+\cdots+a_{0}=P(2 t) \\
& \vdots \\
& n^{n} a_{n} t^{n}+n^{n-1} a_{n-1} t^{n-1}+\cdots+a_{0}=P(n t)
\en... | P(x)=x^{n}+k,-x^{n}+k | Yes | Yes | math-word-problem | Algebra | Find all polynomials $P(x)$ with integer coefficients such that for all real numbers $s$ and $t$, if $P(s)$ and $P(t)$ are both integers, then $P(s t)$ is also an integer.
Answer: $P(x)=x^{n}+k,-x^{n}+k$ for $n$ a non-negative integer and $k$ an integer. | : Assume $P(x)=\sum_{i=0}^{n} a_{i} x^{i}$. Consider the following system of equations
$$
\begin{aligned}
& a_{0}=P(0) \\
& a_{n} t^{n}+a_{n-1} t^{n-1}+\cdots+a_{0}=P(t) \\
& 2^{n} a_{n} t^{n}+2^{n-1} a_{n-1} t^{n-1}+\cdots+a_{0}=P(2 t) \\
& \vdots \\
& n^{n} a_{n} t^{n}+n^{n-1} a_{n-1} t^{n-1}+\cdots+a_{0}=P(n t)
\en... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2018"
} |
Let $\mathbb{Z}^{+}$be the set of positive integers. Determine all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ such that $a^{2}+f(a) f(b)$ is divisible by $f(a)+b$ for all positive integers $a$ and $b$.
Answer: The answer is $f(n)=n$ for all positive integers $n$.
Clearly, $f(n)=n$ for all $n \in \mathbb{... | First we perform the following substitutions on the original relation:
1. With $a=b=1$, we find that $f(1)+1 \mid f(1)^{2}+1$, which implies $f(1)=1$.
2. With $a=1$, we find that $b+1 \mid f(b)+1$. In particular, $b \leq f(b)$ for all $b \in \mathbb{Z}^{+}$.
3. With $b=1$, we find that $f(a)+1 \mid a^{2}+f(a)$, and th... | f(n)=n | Yes | Yes | proof | Number Theory | Let $\mathbb{Z}^{+}$be the set of positive integers. Determine all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ such that $a^{2}+f(a) f(b)$ is divisible by $f(a)+b$ for all positive integers $a$ and $b$.
Answer: The answer is $f(n)=n$ for all positive integers $n$.
Clearly, $f(n)=n$ for all $n \in \mathbb{... | First we perform the following substitutions on the original relation:
1. With $a=b=1$, we find that $f(1)+1 \mid f(1)^{2}+1$, which implies $f(1)=1$.
2. With $a=1$, we find that $b+1 \mid f(b)+1$. In particular, $b \leq f(b)$ for all $b \in \mathbb{Z}^{+}$.
3. With $b=1$, we find that $f(a)+1 \mid a^{2}+f(a)$, and th... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $\mathbb{Z}^{+}$be the set of positive integers. Determine all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ such that $a^{2}+f(a) f(b)$ is divisible by $f(a)+b$ for all positive integers $a$ and $b$.
Answer: The answer is $f(n)=n$ for all positive integers $n$.
Clearly, $f(n)=n$ for all $n \in \mathbb{... | : As above, we have relations (1)-(3). In (2) and (3), for $b=2$ we have $3 \mid f(2)+1$ and $f(2)+1 \mid 3$. These imply $f(2)=2$.
Now, using $a=2$ we get $2+b \mid 4+2 f(b)$. Let $f(b)=x$. We have
$$
\begin{aligned}
1+x & \equiv 0 \quad(\bmod b+1) \\
4+2 x & \equiv 0 \quad(\bmod b+2)
\end{aligned}
$$
From the firs... | f(n)=n | Yes | Yes | proof | Number Theory | Let $\mathbb{Z}^{+}$be the set of positive integers. Determine all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ such that $a^{2}+f(a) f(b)$ is divisible by $f(a)+b$ for all positive integers $a$ and $b$.
Answer: The answer is $f(n)=n$ for all positive integers $n$.
Clearly, $f(n)=n$ for all $n \in \mathbb{... | : As above, we have relations (1)-(3). In (2) and (3), for $b=2$ we have $3 \mid f(2)+1$ and $f(2)+1 \mid 3$. These imply $f(2)=2$.
Now, using $a=2$ we get $2+b \mid 4+2 f(b)$. Let $f(b)=x$. We have
$$
\begin{aligned}
1+x & \equiv 0 \quad(\bmod b+1) \\
4+2 x & \equiv 0 \quad(\bmod b+2)
\end{aligned}
$$
From the firs... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2019"
} |
Let $\mathbb{Z}^{+}$be the set of positive integers. Determine all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ such that $a^{2}+f(a) f(b)$ is divisible by $f(a)+b$ for all positive integers $a$ and $b$.
Answer: The answer is $f(n)=n$ for all positive integers $n$.
Clearly, $f(n)=n$ for all $n \in \mathbb{... | : We proceed by induction. As in Solution 1, we have $f(1)=1$. Suppose that $f(n-1)=n-1$ for some integer $n \geq 2$.
With the substitution $a=n$ and $b=n-1$ in the original relation we obtain that $f(n)+$ $n-1 \mid n^{2}+f(n)(n-1)$. Since $f(n)+n-1 \mid(n-1)(f(n)+n-1)$, then $f(n)+n-1 \mid 2 n-1$.
With the substitut... | f(n)=n | Yes | Yes | proof | Number Theory | Let $\mathbb{Z}^{+}$be the set of positive integers. Determine all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ such that $a^{2}+f(a) f(b)$ is divisible by $f(a)+b$ for all positive integers $a$ and $b$.
Answer: The answer is $f(n)=n$ for all positive integers $n$.
Clearly, $f(n)=n$ for all $n \in \mathbb{... | : We proceed by induction. As in Solution 1, we have $f(1)=1$. Suppose that $f(n-1)=n-1$ for some integer $n \geq 2$.
With the substitution $a=n$ and $b=n-1$ in the original relation we obtain that $f(n)+$ $n-1 \mid n^{2}+f(n)(n-1)$. Since $f(n)+n-1 \mid(n-1)(f(n)+n-1)$, then $f(n)+n-1 \mid 2 n-1$.
With the substitut... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"solution_match": "\nSolution 3",
"tier": "T1",
"year": "2019"
} |
Let $m$ be a fixed positive integer. The infinite sequence $\left\{a_{n}\right\}_{n \geq 1}$ is defined in the following way: $a_{1}$ is a positive integer, and for every integer $n \geq 1$ we have
$$
a_{n+1}= \begin{cases}a_{n}^{2}+2^{m} & \text { if } a_{n}<2^{m} \\ a_{n} / 2 & \text { if } a_{n} \geq 2^{m}\end{case... | Suppose that for integers $m$ and $a_{1}$ all the terms of the sequence are integers. For each $i \geq 1$, write the $i$ th term of the sequence as $a_{i}=b_{i} 2^{c_{i}}$ where $b_{i}$ is the largest odd divisor of $a_{i}$ (the "odd part" of $a_{i}$ ) and $c_{i}$ is a nonnegative integer.
Lemma 1. The sequence $b_{1}... | m=2 | Yes | Yes | math-word-problem | Number Theory | Let $m$ be a fixed positive integer. The infinite sequence $\left\{a_{n}\right\}_{n \geq 1}$ is defined in the following way: $a_{1}$ is a positive integer, and for every integer $n \geq 1$ we have
$$
a_{n+1}= \begin{cases}a_{n}^{2}+2^{m} & \text { if } a_{n}<2^{m} \\ a_{n} / 2 & \text { if } a_{n} \geq 2^{m}\end{case... | Suppose that for integers $m$ and $a_{1}$ all the terms of the sequence are integers. For each $i \geq 1$, write the $i$ th term of the sequence as $a_{i}=b_{i} 2^{c_{i}}$ where $b_{i}$ is the largest odd divisor of $a_{i}$ (the "odd part" of $a_{i}$ ) and $c_{i}$ is a nonnegative integer.
Lemma 1. The sequence $b_{1}... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $m$ be a fixed positive integer. The infinite sequence $\left\{a_{n}\right\}_{n \geq 1}$ is defined in the following way: $a_{1}$ is a positive integer, and for every integer $n \geq 1$ we have
$$
a_{n+1}= \begin{cases}a_{n}^{2}+2^{m} & \text { if } a_{n}<2^{m} \\ a_{n} / 2 & \text { if } a_{n} \geq 2^{m}\end{case... | : Let $m$ be a positive integer and suppose that $\left\{a_{n}\right\}$ consists only of positive integers. Call a number small if it is smaller than $2^{m}$ and large otherwise. By the recursion,
after a small number we have a large one and after a large one we successively divide by 2 until we get a small one.
First... | m=2, a_{1}=2^{\ell} \text{ for } \ell \geq 1 | Yes | Yes | math-word-problem | Number Theory | Let $m$ be a fixed positive integer. The infinite sequence $\left\{a_{n}\right\}_{n \geq 1}$ is defined in the following way: $a_{1}$ is a positive integer, and for every integer $n \geq 1$ we have
$$
a_{n+1}= \begin{cases}a_{n}^{2}+2^{m} & \text { if } a_{n}<2^{m} \\ a_{n} / 2 & \text { if } a_{n} \geq 2^{m}\end{case... | : Let $m$ be a positive integer and suppose that $\left\{a_{n}\right\}$ consists only of positive integers. Call a number small if it is smaller than $2^{m}$ and large otherwise. By the recursion,
after a small number we have a large one and after a large one we successively divide by 2 until we get a small one.
First... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2019"
} |
Let $A B C$
be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $B C$. A variable point $P$ is selected in the line segment $A M$. The circumcircles of triangles $B P M$ and $C P M$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $D P$ and $E P$ intersect (a second time... | Let $N$ be the radical center of the circumcircles of triangles $A B C, B M P$ and $C M P$. The pairwise radical axes of these circles are $B D, C E$ and $P M$, and hence they concur at $N$. Now, note that in directed angles:
$$
\angle M C E=\angle M P E=\angle M P Y=\angle M B Y .
$$
![](https://cdn.mathpix.com/crop... | proof | Yes | Yes | proof | Geometry | Let $A B C$
be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $B C$. A variable point $P$ is selected in the line segment $A M$. The circumcircles of triangles $B P M$ and $C P M$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $D P$ and $E P$ intersect (a second time... | Let $N$ be the radical center of the circumcircles of triangles $A B C, B M P$ and $C M P$. The pairwise radical axes of these circles are $B D, C E$ and $P M$, and hence they concur at $N$. Now, note that in directed angles:
$$
\angle M C E=\angle M P E=\angle M P Y=\angle M B Y .
$$
![](https://cdn.mathpix.com/crop... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "# Problem 3.",
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $A B C$
be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $B C$. A variable point $P$ is selected in the line segment $A M$. The circumcircles of triangles $B P M$ and $C P M$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $D P$ and $E P$ intersect (a second time... | . Let the lines $D P$ and $E P$ meet the circumcircle of $A B C$ again at $Q$ and $R$, respectively. Then $\angle D Q C \angle D B C=\angle D P M$, so $Q C \| P M$. Similarly, $R B \| P M$.
No... | proof | Yes | Yes | proof | Geometry | Let $A B C$
be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $B C$. A variable point $P$ is selected in the line segment $A M$. The circumcircles of triangles $B P M$ and $C P M$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $D P$ and $E P$ intersect (a second time... | . Let the lines $D P$ and $E P$ meet the circumcircle of $A B C$ again at $Q$ and $R$, respectively. Then $\angle D Q C \angle D B C=\angle D P M$, so $Q C \| P M$. Similarly, $R B \| P M$.
No... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "# Problem 3.",
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2019"
} |
Consider a $2018 \times 2019$ board with integers in each unit square. Two unit squares are said to be neighbours if they share a common edge. In each turn, you choose some unit squares. Then for each chosen unit square the average of all its neighbours is calculated. Finally, after these calculations are done, the num... | Let $n$ be a positive integer relatively prime to 2 and 3 . We may study the whole process modulo $n$ by replacing divisions by $2,3,4$ with multiplications by the corresponding inverses modulo $n$. If at some point the original process makes all the numbers equal, then the process modulo $n$ will also have all the num... | proof | Yes | Yes | proof | Combinatorics | Consider a $2018 \times 2019$ board with integers in each unit square. Two unit squares are said to be neighbours if they share a common edge. In each turn, you choose some unit squares. Then for each chosen unit square the average of all its neighbours is calculated. Finally, after these calculations are done, the num... | Let $n$ be a positive integer relatively prime to 2 and 3 . We may study the whole process modulo $n$ by replacing divisions by $2,3,4$ with multiplications by the corresponding inverses modulo $n$. If at some point the original process makes all the numbers equal, then the process modulo $n$ will also have all the num... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Determine all the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+f(y)\right)=f(f(x))+f\left(y^{2}\right)+2 f(x y)
$$
for all real number $x$ and $y$.
Answer: The possible functions are $f(x)=0$ for all $x$ and $f(x)=x^{2}$ for all $x$. | By substituting $x=y=0$ in the given equation of the problem, we obtain that $f(0)=0$. Also, by substituting $y=0$, we get $f\left(x^{2}\right)=f(f(x))$ for any $x$.
Furthermore, by letting $y=1$ and simplifying, we get
$$
2 f(x)=f\left(x^{2}+f(1)\right)-f\left(x^{2}\right)-f(1)
$$
from which it follows that $f(-x)=... | f(x)=0 \text{ or } f(x)=x^2 | Yes | Yes | math-word-problem | Algebra | Determine all the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+f(y)\right)=f(f(x))+f\left(y^{2}\right)+2 f(x y)
$$
for all real number $x$ and $y$.
Answer: The possible functions are $f(x)=0$ for all $x$ and $f(x)=x^{2}$ for all $x$. | By substituting $x=y=0$ in the given equation of the problem, we obtain that $f(0)=0$. Also, by substituting $y=0$, we get $f\left(x^{2}\right)=f(f(x))$ for any $x$.
Furthermore, by letting $y=1$ and simplifying, we get
$$
2 f(x)=f\left(x^{2}+f(1)\right)-f\left(x^{2}\right)-f(1)
$$
from which it follows that $f(-x)=... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Determine all the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+f(y)\right)=f(f(x))+f\left(y^{2}\right)+2 f(x y)
$$
for all real number $x$ and $y$.
Answer: The possible functions are $f(x)=0$ for all $x$ and $f(x)=x^{2}$ for all $x$. | After proving that $f(x)>0$ for $x \neq 0$ as in the previous solution, we may also proceed as follows. We claim that $f$ is injective on the positive real numbers. Suppose that $a>b>0$ satisfy $f(a)=f(b)$. Then by setting $x=1 / b$ in (1) we have $f(a / b)=f(1)$. Now, by induction on $n$ and iteratively setting $x=a /... | f(x)=0 \text{ for all } x \text{ and } f(x)=x^{2} \text{ for all } x | Yes | Yes | math-word-problem | Algebra | Determine all the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+f(y)\right)=f(f(x))+f\left(y^{2}\right)+2 f(x y)
$$
for all real number $x$ and $y$.
Answer: The possible functions are $f(x)=0$ for all $x$ and $f(x)=x^{2}$ for all $x$. | After proving that $f(x)>0$ for $x \neq 0$ as in the previous solution, we may also proceed as follows. We claim that $f$ is injective on the positive real numbers. Suppose that $a>b>0$ satisfy $f(a)=f(b)$. Then by setting $x=1 / b$ in (1) we have $f(a / b)=f(1)$. Now, by induction on $n$ and iteratively setting $x=a /... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2019"
} |
Let $\Gamma$ be the circumcircle of $\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent. | From the conditions, we have
Let $P$ be the intersection of $A C$ and $B F$. Then we have
$$
\angle P A E=\angle C B A=\angle B A C=\angle B F C .
$$
This implies $A, P, F, E$ are concyclic.... | proof | Yes | Yes | proof | Geometry | Let $\Gamma$ be the circumcircle of $\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent. | From the conditions, we have
Let $P$ be the intersection of $A C$ and $B F$. Then we have
$$
\angle P A E=\angle C B A=\angle B A C=\angle B F C .
$$
This implies $A, P, F, E$ are concyclic.... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2020"
} |
Let $\Gamma$ be the circumcircle of $\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent. | Let $E^{\prime}$ be any point on the extension of $E A$. From $\angle A E D=\angle E^{\prime} A B=\angle A C D$, points $A, D, C, E$ are concyclic.
Let $P$ be the intersection of $B F$ and $D ... | proof | Yes | Yes | proof | Geometry | Let $\Gamma$ be the circumcircle of $\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent. | Let $E^{\prime}$ be any point on the extension of $E A$. From $\angle A E D=\angle E^{\prime} A B=\angle A C D$, points $A, D, C, E$ are concyclic.
Let $P$ be the intersection of $B F$ and $D ... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2020"
} |
Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+r a_{n+1}}
$$
for every positive integer $n$, then there exists a positive integer $M$ such that $a_{n... | . First, let us assume that $r>2$, and take a positive integer $a \geq 1 /(r-2)$.
Then, if we let $a_{n}=a+\lfloor n / 2\rfloor$ for $n=1,2, \ldots$, the sequence $a_{n}$ satisfies the inequalities
$$
\sqrt{a_{n}^{2}+r a_{n+1}} \geq \sqrt{a_{n}^{2}+r a_{n}} \geq \sqrt{a_{n}^{2}+\left(2+\frac{1}{a}\right) a_{n}} \geq a... | proof | Yes | Yes | proof | Inequalities | Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+r a_{n+1}}
$$
for every positive integer $n$, then there exists a positive integer $M$ such that $a_{n... | . First, let us assume that $r>2$, and take a positive integer $a \geq 1 /(r-2)$.
Then, if we let $a_{n}=a+\lfloor n / 2\rfloor$ for $n=1,2, \ldots$, the sequence $a_{n}$ satisfies the inequalities
$$
\sqrt{a_{n}^{2}+r a_{n+1}} \geq \sqrt{a_{n}^{2}+r a_{n}} \geq \sqrt{a_{n}^{2}+\left(2+\frac{1}{a}\right) a_{n}} \geq a... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2020"
} |
Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+r a_{n+1}}
$$
for every positive integer $n$, then there exists a positive integer $M$ such that $a_{n... | We only give an alternative proof of the assertion $(\dagger)$ in solution 1 . Let $\left\{a_{n}\right\}$ be a sequence satisfying the inequalities given in the problem. We will use the following key observations:
(a) If $a_{n+1} \leq a_{n}$ for some $n \geq 1$, then
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+2 a_{n+1... | proof | Yes | Yes | proof | Inequalities | Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+r a_{n+1}}
$$
for every positive integer $n$, then there exists a positive integer $M$ such that $a_{n... | We only give an alternative proof of the assertion $(\dagger)$ in solution 1 . Let $\left\{a_{n}\right\}$ be a sequence satisfying the inequalities given in the problem. We will use the following key observations:
(a) If $a_{n+1} \leq a_{n}$ for some $n \geq 1$, then
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+2 a_{n+1... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2020"
} |
Determine all positive integers $k$ for which there exist a positive integer $m$ and a set $S$ of positive integers such that any integer $n>m$ can be written as a sum of distinct elements of $S$ in exactly $k$ ways. | We claim that $k=2^{a}$ for all $a \geq 0$.
Let $A=\{1,2,4,8, \ldots\}$ and $B=\mathbb{N} \backslash A$. For any set $T$, let $s(T)$ denote the sum of the elements of $T$. (If $T$ is empty, we let $s(T)=0$.)
We first show that any positive integer $k=2^{a}$ satisfies the desired property. Let $B^{\prime}$ be a subset o... | k=2^{a} | Yes | Yes | math-word-problem | Combinatorics | Determine all positive integers $k$ for which there exist a positive integer $m$ and a set $S$ of positive integers such that any integer $n>m$ can be written as a sum of distinct elements of $S$ in exactly $k$ ways. | We claim that $k=2^{a}$ for all $a \geq 0$.
Let $A=\{1,2,4,8, \ldots\}$ and $B=\mathbb{N} \backslash A$. For any set $T$, let $s(T)$ denote the sum of the elements of $T$. (If $T$ is empty, we let $s(T)=0$.)
We first show that any positive integer $k=2^{a}$ satisfies the desired property. Let $B^{\prime}$ be a subset o... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"solution_match": "# Solution:",
"tier": "T1",
"year": "2020"
} |
Determine all positive integers $k$ for which there exist a positive integer $m$ and a set $S$ of positive integers such that any integer $n>m$ can be written as a sum of distinct elements of $S$ in exactly $k$ ways. | . We give an alternative proof of the first half of the lemma in the Solution 1 above.
Let $s_{1}<s_{2}<\cdots$ be the elements of $S$. For any positive integer $r$, define $A_{r}(x)=\prod_{n=1}^{r}\left(1+x^{s_{n}}\right)$. For each $n$ such that $m \leq n<s_{r+1}$, all $k$ ways of writing $n$ as a sum of elements of ... | proof | Yes | Yes | math-word-problem | Combinatorics | Determine all positive integers $k$ for which there exist a positive integer $m$ and a set $S$ of positive integers such that any integer $n>m$ can be written as a sum of distinct elements of $S$ in exactly $k$ ways. | . We give an alternative proof of the first half of the lemma in the Solution 1 above.
Let $s_{1}<s_{2}<\cdots$ be the elements of $S$. For any positive integer $r$, define $A_{r}(x)=\prod_{n=1}^{r}\left(1+x^{s_{n}}\right)$. For each $n$ such that $m \leq n<s_{r+1}$, all $k$ ways of writing $n$ as a sum of elements of ... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2020"
} |
Let $\mathbb{Z}$ denote the set of all integers. Find all polynomials $P(x)$ with integer coefficients that satisfy the following property:
For any infinite sequence $a_{1}, a_{2}, \ldots$ of integers in which each integer in $\mathbb{Z}$ appears exactly once, there exist indices $i<j$ and an integer $k$ such that $a_{... | Part 1: All polynomials with $\operatorname{deg} P=1$ satisfy the given property.
Suppose $P(x)=c x+d$, and assume without loss of generality that $c>d \geq 0$. Denote $s_{i}=a_{1}+a_{2}+$ $\cdots+a_{i}(\bmod c)$. It suffices to show that there exist indices $i$ and $j$ such that $j-i \geq 2$ and $s_{j}-s_{i} \equiv d$... | proof | Yes | Yes | proof | Number Theory | Let $\mathbb{Z}$ denote the set of all integers. Find all polynomials $P(x)$ with integer coefficients that satisfy the following property:
For any infinite sequence $a_{1}, a_{2}, \ldots$ of integers in which each integer in $\mathbb{Z}$ appears exactly once, there exist indices $i<j$ and an integer $k$ such that $a_{... | Part 1: All polynomials with $\operatorname{deg} P=1$ satisfy the given property.
Suppose $P(x)=c x+d$, and assume without loss of generality that $c>d \geq 0$. Denote $s_{i}=a_{1}+a_{2}+$ $\cdots+a_{i}(\bmod c)$. It suffices to show that there exist indices $i$ and $j$ such that $j-i \geq 2$ and $s_{j}-s_{i} \equiv d$... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"solution_match": "# Solution:",
"tier": "T1",
"year": "2020"
} |
Let $n \geq 3$ be a fixed integer. The number 1 is written $n$ times on a blackboard. Below the blackboard, there are two buckets that are initially empty. A move consists of erasing two of the numbers $a$ and $b$, replacing them with the numbers 1 and $a+b$, then adding one stone to the first bucket and $\operatorname... | The answer is the set of all rational numbers in the interval $[1, n-1)$. First, we show that no other numbers are possible. Clearly the ratio is at least 1, since for every move, at least one stone is added to the second bucket. Note that the number $s$ of stones in the first bucket is always equal to $p-n$, where $p$... | proof | Yes | Incomplete | math-word-problem | Number Theory | Let $n \geq 3$ be a fixed integer. The number 1 is written $n$ times on a blackboard. Below the blackboard, there are two buckets that are initially empty. A move consists of erasing two of the numbers $a$ and $b$, replacing them with the numbers 1 and $a+b$, then adding one stone to the first bucket and $\operatorname... | The answer is the set of all rational numbers in the interval $[1, n-1)$. First, we show that no other numbers are possible. Clearly the ratio is at least 1, since for every move, at least one stone is added to the second bucket. Note that the number $s$ of stones in the first bucket is always equal to $p-n$, where $p$... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"solution_match": "# Solution:",
"tier": "T1",
"year": "2020"
} |
Prove that for each real number $r>2$, there are exactly two or three positive real numbers $x$ satisfying the equation $x^{2}=r\lfloor x\rfloor$.
Note: $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$ | Let $r>2$ be a real number. Let $x$ be a positive real number such that $x^{2}=r\lfloor x\rfloor$ with $\lfloor x\rfloor=k$. Since $x>0$ and $x^{2}=r k$, we also have $k>0$. From $k \leq x<k+1$, we get $k^{2} \leq x^{2}=r k<$ $k^{2}+2 k+1 \leq k^{2}+3 k$, hence $k \leq r<k+3$, or $r-3<k \leq r$. There are at most three... | proof | Yes | Yes | proof | Algebra | Prove that for each real number $r>2$, there are exactly two or three positive real numbers $x$ satisfying the equation $x^{2}=r\lfloor x\rfloor$.
Note: $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$ | Let $r>2$ be a real number. Let $x$ be a positive real number such that $x^{2}=r\lfloor x\rfloor$ with $\lfloor x\rfloor=k$. Since $x>0$ and $x^{2}=r k$, we also have $k>0$. From $k \leq x<k+1$, we get $k^{2} \leq x^{2}=r k<$ $k^{2}+2 k+1 \leq k^{2}+3 k$, hence $k \leq r<k+3$, or $r-3<k \leq r$. There are at most three... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"solution_match": "\nSolution ",
"tier": "T1",
"year": "2021"
} |
For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.
Determine all polynomial $P$ with integer coefficients such that for all positive integers $n, P_{n} \leq 2021$. | There are two possible families of solutions:
- $P(x)=x+d$, for some integer $d \geq-2022$.
- $P(x)=-x+d$, for some integer $d \leq 2022$.
Suppose $P$ satisfies the problem conditions. Clearly $P$ cannot be a constant polynomial. Notice that a polynomial $P$ satifies the conditions if and only if $-P$ also satisfies ... | proof | Yes | Yes | math-word-problem | Number Theory | For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.
Determine all polynomial $P$ with integer coefficients such that for all positive integers $n, P_{n} \leq 2021$. | There are two possible families of solutions:
- $P(x)=x+d$, for some integer $d \geq-2022$.
- $P(x)=-x+d$, for some integer $d \leq 2022$.
Suppose $P$ satisfies the problem conditions. Clearly $P$ cannot be a constant polynomial. Notice that a polynomial $P$ satifies the conditions if and only if $-P$ also satisfies ... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"solution_match": "\nSolution ",
"tier": "T1",
"year": "2021"
} |
For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.
Determine all polynomial $P$ with integer coefficients such that for all positive integers $n, P_{n} \leq 2021$. | If $c \geq 2$, then $P(1)$ and $P(2)$ leave the same remainder upon division by $c$, contradicting the Lemma. Hence $c=1$. | c=1 | Yes | Problem not solved | math-word-problem | Number Theory | For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.
Determine all polynomial $P$ with integer coefficients such that for all positive integers $n, P_{n} \leq 2021$. | If $c \geq 2$, then $P(1)$ and $P(2)$ leave the same remainder upon division by $c$, contradicting the Lemma. Hence $c=1$. | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2021"
} |
For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.
Determine all polynomial $P$ with integer coefficients such that for all positive integers $n, P_{n} \leq 2021$. | Suppose $c \geq 2$. Let $n$ be a positive integer such that $n>2 c M, n\left(1-\frac{3}{2 c}\right)>2022$ and $2 c \mid n$. Notice that for any positive integers $i$ such that $\frac{3 n}{2 c}+i<n, P\left(\frac{3 n}{2 c}+i\right)-P\left(\frac{n}{2 c}+i\right)=n$. Hence, $\left(\frac{n}{2 c}+i, \frac{3 n}{2 c}+i\right)$... | P(x)=x+d \text{ for any } d \geq -2022 | Yes | Yes | math-word-problem | Number Theory | For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.
Determine all polynomial $P$ with integer coefficients such that for all positive integers $n, P_{n} \leq 2021$. | Suppose $c \geq 2$. Let $n$ be a positive integer such that $n>2 c M, n\left(1-\frac{3}{2 c}\right)>2022$ and $2 c \mid n$. Notice that for any positive integers $i$ such that $\frac{3 n}{2 c}+i<n, P\left(\frac{3 n}{2 c}+i\right)-P\left(\frac{n}{2 c}+i\right)=n$. Hence, $\left(\frac{n}{2 c}+i, \frac{3 n}{2 c}+i\right)$... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2021"
} |
Let $A B C D$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\Gamma$ not containing $A$ and $D$. Prove that the e... | Let $L$ be the intersection of the bisectors of $\angle A B C$ and $\angle B C D$. Let $N$ be the $E$-excenter of $\triangle B C E$. Let $\angle B A C=\angle B D C=\alpha, \angle D B C=\beta$ and $\angle A C B=\gamma$.
We have the following:
$$
\begin{array}{r}
\angle C B L=\frac{1}{2} \angle A B C=90^{\circ}-\frac{1}... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\Gamma$ not containing $A$ and $D$. Prove that the e... | Let $L$ be the intersection of the bisectors of $\angle A B C$ and $\angle B C D$. Let $N$ be the $E$-excenter of $\triangle B C E$. Let $\angle B A C=\angle B D C=\alpha, \angle D B C=\beta$ and $\angle A C B=\gamma$.
We have the following:
$$
\begin{array}{r}
\angle C B L=\frac{1}{2} \angle A B C=90^{\circ}-\frac{1}... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2021"
} |
Let $A B C D$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\Gamma$ not containing $A$ and $D$. Prove that the e... | Denote by $N$ the excenter of triangle $B C E$ opposite $E$. Since $B L$ bisects $\angle A B C$, we have $\angle C B L=$ $\frac{\angle A B C}{2}$. Since $M$ is the midpoint of arc $B C$, we have $\angle M B C=\frac{1}{2}(\angle M B C+\angle M C B)$ It follows by angle chasing that
$$
\begin{aligned}
\angle M B L & =\a... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\Gamma$ not containing $A$ and $D$. Prove that the e... | Denote by $N$ the excenter of triangle $B C E$ opposite $E$. Since $B L$ bisects $\angle A B C$, we have $\angle C B L=$ $\frac{\angle A B C}{2}$. Since $M$ is the midpoint of arc $B C$, we have $\angle M B C=\frac{1}{2}(\angle M B C+\angle M C B)$ It follows by angle chasing that
$$
\begin{aligned}
\angle M B L & =\a... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2021"
} |
Given a $32 \times 32$ table, we put a mouse (facing up) at the bottom left cell and a piece of cheese at several other cells. The mouse then starts moving. It moves forward except that when it reaches a piece of cheese, it eats a part of it, turns right, and continues moving forward. We say that a subset of cells cont... | (a) For the sake of contradiction, assume a good subset consisting of 888 cells exists. We call those cheese-cells and the other ones gap-cells. Observe that since each cheese-cell is visited once, each gap-cell is visited at most twice (once vertically and once horizontally). Define a finite sequence $s$ whose $i$-th ... | proof | Yes | Yes | proof | Combinatorics | Given a $32 \times 32$ table, we put a mouse (facing up) at the bottom left cell and a piece of cheese at several other cells. The mouse then starts moving. It moves forward except that when it reaches a piece of cheese, it eats a part of it, turns right, and continues moving forward. We say that a subset of cells cont... | (a) For the sake of contradiction, assume a good subset consisting of 888 cells exists. We call those cheese-cells and the other ones gap-cells. Observe that since each cheese-cell is visited once, each gap-cell is visited at most twice (once vertically and once horizontally). Define a finite sequence $s$ whose $i$-th ... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"solution_match": "# Solution.",
"tier": "T1",
"year": "2021"
} |
Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(f(a)-b)+b f(2 a)$ is a perfect square for all integers $a$ and $b$. | .
There are two families of functions which satisfy the condition:
(1) $f(n)= \begin{cases}0 & \text { if } n \text { is even, and } \\ \text { any perfect square } & \text { if } n \text { is odd }\end{cases}$
(2) $f(n)=n^{2}$, for every integer $n$.
It is straightforward to verify that the two families of functions... | proof | Yes | Yes | math-word-problem | Algebra | Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(f(a)-b)+b f(2 a)$ is a perfect square for all integers $a$ and $b$. | .
There are two families of functions which satisfy the condition:
(1) $f(n)= \begin{cases}0 & \text { if } n \text { is even, and } \\ \text { any perfect square } & \text { if } n \text { is odd }\end{cases}$
(2) $f(n)=n^{2}$, for every integer $n$.
It is straightforward to verify that the two families of functions... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2021"
} |
Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$. | .1
By inspection, we see that the pairs $(a, b)$ with $a=b$ are solutions, and so too are the pairs $(a, 1)$. We will see that these are the only solutions.
- Case 1. Consider the case $b<a$. Since $b-1$ is a multiple of $a-1$, it follows that $b=1$. This yields the second set of solutions described above.
- Case 2. ... | proof | Yes | Yes | math-word-problem | Number Theory | Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$. | .1
By inspection, we see that the pairs $(a, b)$ with $a=b$ are solutions, and so too are the pairs $(a, 1)$. We will see that these are the only solutions.
- Case 1. Consider the case $b<a$. Since $b-1$ is a multiple of $a-1$, it follows that $b=1$. This yields the second set of solutions described above.
- Case 2. ... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2022"
} |
Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$. | .2
We will start by showing that there are positive integers $x, c, d$ such that $a=x^{2} c d$ and $b=x^{3} c$. Let $g=\operatorname{gcd}(a, b)$ so that $a=g d$ and $b=g x$ for some coprime $d$ and $x$. Then, $b^{2} \mid a^{3}$ is equivalent to $g^{2} x^{2} \mid g^{3} d^{3}$, which is equivalent to $x^{2} \mid g d^{3}... | a=b \text{ or } b=1 | Yes | Yes | math-word-problem | Number Theory | Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$. | .2
We will start by showing that there are positive integers $x, c, d$ such that $a=x^{2} c d$ and $b=x^{3} c$. Let $g=\operatorname{gcd}(a, b)$ so that $a=g d$ and $b=g x$ for some coprime $d$ and $x$. Then, $b^{2} \mid a^{3}$ is equivalent to $g^{2} x^{2} \mid g^{3} d^{3}$, which is equivalent to $x^{2} \mid g d^{3}... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2022"
} |
Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$. | .3
All answers are $(n, n)$ and $(n, 1)$ where $n$ is any positive integer. They all clearly work.
To show that these are all solutions, note that we can easily eliminate the case $a=1$ or $b=1$. Thus, assume that $a, b \neq 1$ and $a \neq b$. By the second divisibility, we see that $a-1 \mid b-a$. However, $\operator... | (n, n) \text{ and } (n, 1) | Yes | Yes | math-word-problem | Number Theory | Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$. | .3
All answers are $(n, n)$ and $(n, 1)$ where $n$ is any positive integer. They all clearly work.
To show that these are all solutions, note that we can easily eliminate the case $a=1$ or $b=1$. Thus, assume that $a, b \neq 1$ and $a \neq b$. By the second divisibility, we see that $a-1 \mid b-a$. However, $\operator... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2022"
} |
Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\triangle A C D$ and the circumcircle of $\triangle B D E$. Prove that as $D$ varies, t... | .1
Let the line $E F$ intersect the line $B C$ at $P$ and the circumcircle of $\triangle A C D$ at $G$ distinct from $F$. We will prove that $P$ is the fixed point.
First, notice that $\triangle B E D$ is isosceles with $E B=E D$. This implies $\angle E B C=\angle E D P$.
Then, $\angle D A G=\angle D F G=\angle E B C=... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\triangle A C D$ and the circumcircle of $\triangle B D E$. Prove that as $D$ varies, t... | .1
Let the line $E F$ intersect the line $B C$ at $P$ and the circumcircle of $\triangle A C D$ at $G$ distinct from $F$. We will prove that $P$ is the fixed point.
First, notice that $\triangle B E D$ is isosceles with $E B=E D$. This implies $\angle E B C=\angle E D P$.
Then, $\angle D A G=\angle D F G=\angle E B C=... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2022"
} |
Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\triangle A C D$ and the circumcircle of $\triangle B D E$. Prove that as $D$ varies, t... | .2
Set up a coordinate system where $B C$ is along the positive $x$-axis, $B A$ is along the positive $y$-axis, and $B$ is the origin. Take $A=(0, a), B=(0,0), C=(c, 0), D=(-d, 0)$ where $a, b, c, d>0$. Then $E=\left(-\frac{d}{2}, \frac{a}{2}\right)$. The general equation of a circle is
$$
x^{2}+y^{2}+2 f x+2 g y+h=0... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\triangle A C D$ and the circumcircle of $\triangle B D E$. Prove that as $D$ varies, t... | .2
Set up a coordinate system where $B C$ is along the positive $x$-axis, $B A$ is along the positive $y$-axis, and $B$ is the origin. Take $A=(0, a), B=(0,0), C=(c, 0), D=(-d, 0)$ where $a, b, c, d>0$. Then $E=\left(-\frac{d}{2}, \frac{a}{2}\right)$. The general equation of a circle is
$$
x^{2}+y^{2}+2 f x+2 g y+h=0... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2022"
} |
Find all positive integers $k<202$ for which there exists a positive integer $n$ such that
$$
\left\{\frac{n}{202}\right\}+\left\{\frac{2 n}{202}\right\}+\cdots+\left\{\frac{k n}{202}\right\}=\frac{k}{2}
$$
where $\{x\}$ denote the fractional part of $x$.
Note: $\{x\}$ denotes the real number $k$ with $0 \leq k<1$ su... | Denote the equation in the problem statement as $\left(^{*}\right)$, and note that it is equivalent to the condition that the average of the remainders when dividing $n, 2 n, \ldots, k n$ by 202 is 101 . Since $\left\{\frac{i n}{202}\right\}$ is invariant in each residue class modulo 202 for each $1 \leq i \leq k$, it ... | k \in\{1,100,101,201\} | Yes | Yes | math-word-problem | Number Theory | Find all positive integers $k<202$ for which there exists a positive integer $n$ such that
$$
\left\{\frac{n}{202}\right\}+\left\{\frac{2 n}{202}\right\}+\cdots+\left\{\frac{k n}{202}\right\}=\frac{k}{2}
$$
where $\{x\}$ denote the fractional part of $x$.
Note: $\{x\}$ denotes the real number $k$ with $0 \leq k<1$ su... | Denote the equation in the problem statement as $\left(^{*}\right)$, and note that it is equivalent to the condition that the average of the remainders when dividing $n, 2 n, \ldots, k n$ by 202 is 101 . Since $\left\{\frac{i n}{202}\right\}$ is invariant in each residue class modulo 202 for each $1 \leq i \leq k$, it ... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"solution_match": "# Solution\n\n",
"tier": "T1",
"year": "2022"
} |
Let $n$ and $k$ be positive integers. Cathy is playing the following game. There are $n$ marbles and $k$ boxes, with the marbles labelled 1 to $n$. Initially, all marbles are placed inside one box. Each turn, Cathy chooses a box and then moves the marbles with the smallest label, say $i$, to either any empty box or the... | We claim Cathy can win if and only if $n \leq 2^{k-1}$.
First, note that each non-empty box always contains a consecutive sequence of labeled marbles. This is true since Cathy is always either removing from or placing in the lowest marble in a box. As a consequence, every move made is reversible.
Next, we prove by in... | n \leq 2^{k-1} | Yes | Yes | math-word-problem | Combinatorics | Let $n$ and $k$ be positive integers. Cathy is playing the following game. There are $n$ marbles and $k$ boxes, with the marbles labelled 1 to $n$. Initially, all marbles are placed inside one box. Each turn, Cathy chooses a box and then moves the marbles with the smallest label, say $i$, to either any empty box or the... | We claim Cathy can win if and only if $n \leq 2^{k-1}$.
First, note that each non-empty box always contains a consecutive sequence of labeled marbles. This is true since Cathy is always either removing from or placing in the lowest marble in a box. As a consequence, every move made is reversible.
Next, we prove by in... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"solution_match": "# Solution\n\n",
"tier": "T1",
"year": "2022"
} |
Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved. | The minimum value is $-\frac{1}{8}$. There are eight equality cases in total. The first one is
$$
\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right) .
$$
Cyclic shifting all the entries give three more quadruples. Moreover, flipp... | -\frac{1}{8} | Yes | Yes | math-word-problem | Algebra | Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved. | The minimum value is $-\frac{1}{8}$. There are eight equality cases in total. The first one is
$$
\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right) .
$$
Cyclic shifting all the entries give three more quadruples. Moreover, flipp... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"solution_match": "# Solution\n\n",
"tier": "T1",
"year": "2022"
} |
Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved. | .1
Since the expression is cyclic, we could WLOG $a=\max \{a, b, c, d\}$. Let
$$
S(a, b, c, d)=(a-b)(b-c)(c-d)(d-a)
$$
Note that we have given $(a, b, c, d)$ such that $S(a, b, c, d)=-\frac{1}{8}$. Therefore, to prove that $S(a, b, c, d) \geq$ $-\frac{1}{8}$, we just need to consider the case where $S(a, b, c, d)<0$... | \left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}\right) | Yes | Yes | math-word-problem | Algebra | Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved. | .1
Since the expression is cyclic, we could WLOG $a=\max \{a, b, c, d\}$. Let
$$
S(a, b, c, d)=(a-b)(b-c)(c-d)(d-a)
$$
Note that we have given $(a, b, c, d)$ such that $S(a, b, c, d)=-\frac{1}{8}$. Therefore, to prove that $S(a, b, c, d) \geq$ $-\frac{1}{8}$, we just need to consider the case where $S(a, b, c, d)<0$... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"solution_match": "# Solution 5",
"tier": "T1",
"year": "2022"
} |
Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved. | .2
The minimum value is $-\frac{1}{8}$. There are eight equality cases in total. The first one is
$$
\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right) .
$$
Cyclic shifting all the entries give three more quadruples. Moreover, f... | -\frac{1}{8} | Yes | Yes | math-word-problem | Algebra | Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved. | .2
The minimum value is $-\frac{1}{8}$. There are eight equality cases in total. The first one is
$$
\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right) .
$$
Cyclic shifting all the entries give three more quadruples. Moreover, f... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"solution_match": "# Solution 5",
"tier": "T1",
"year": "2022"
} |
Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.
Show that it is possible to arrange these squares in a way s... | Set aside the squares with sidelengths $n-3, n-2, n-1$, and $n$ and suppose we can split the remaining squares into two sets $A$ and $B$ such that the sum of the sidelengths of the squares in $A$ is 1 or 2 units larger than the sum of the sidelengths of the squares in $B$.
String the squares of each set $A, B$ along tw... | proof | Yes | Yes | proof | Combinatorics | Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.
Show that it is possible to arrange these squares in a way s... | Set aside the squares with sidelengths $n-3, n-2, n-1$, and $n$ and suppose we can split the remaining squares into two sets $A$ and $B$ such that the sum of the sidelengths of the squares in $A$ is 1 or 2 units larger than the sum of the sidelengths of the squares in $B$.
String the squares of each set $A, B$ along tw... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2023"
} |
Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.
Show that it is possible to arrange these squares in a way s... | Solve the problem by induction in $n$. Construct examples for $n=5,6,7,8,9,10$ (one can use the constructions from the previous solution, for instance). For $n>10$, set aside the six larger squares and arrange them in the following fashion:
. For $n>10$, set aside the six larger squares and arrange them in the following fashion:
![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-02.jpg?hei... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2023"
} |
Find all integers $n$ satisfying $n \geq 2$ and $\frac{\sigma(n)}{p(n)-1}=n$, in which $\sigma(n)$ denotes the sum of all positive divisors of $n$, and $p(n)$ denotes the largest prime divisor of $n$.
Answer: $n=6$. | Let $n=p_{1}^{\alpha_{1}} \cdot \ldots \cdot p_{k}^{\alpha_{k}}$ be the prime factorization of $n$ with $p_{1}<\ldots<p_{k}$, so that $p(n)=p_{k}$ and $\sigma(n)=\left(1+p_{1}+\cdots+p_{1}^{\alpha_{1}}\right) \ldots\left(1+p_{k}+\cdots+p_{k}^{\alpha_{k}}\right)$. Hence
$p_{k}-1=\frac{\sigma(n)}{n}=\prod_{i=1}^{k}\left(... | 6 | Yes | Yes | math-word-problem | Number Theory | Find all integers $n$ satisfying $n \geq 2$ and $\frac{\sigma(n)}{p(n)-1}=n$, in which $\sigma(n)$ denotes the sum of all positive divisors of $n$, and $p(n)$ denotes the largest prime divisor of $n$.
Answer: $n=6$. | Let $n=p_{1}^{\alpha_{1}} \cdot \ldots \cdot p_{k}^{\alpha_{k}}$ be the prime factorization of $n$ with $p_{1}<\ldots<p_{k}$, so that $p(n)=p_{k}$ and $\sigma(n)=\left(1+p_{1}+\cdots+p_{1}^{\alpha_{1}}\right) \ldots\left(1+p_{k}+\cdots+p_{k}^{\alpha_{k}}\right)$. Hence
$p_{k}-1=\frac{\sigma(n)}{n}=\prod_{i=1}^{k}\left(... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "# Problem 2",
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"solution_match": "# Solution\n\n",
"tier": "T1",
"year": "2023"
} |
Let $A B C D$ be a parallelogram. Let $W, X, Y$, and $Z$ be points on sides $A B, B C, C D$, and $D A$, respectively, such that the incenters of triangles $A W Z, B X W, C Y X$ and $D Z Y$ form a parallelogram. Prove that $W X Y Z$ is a parallelogram. | Let the four incenters be $I_{1}, I_{2}, I_{3}$, and $I_{4}$ with inradii $r_{1}, r_{2}, r_{3}$, and $r_{4}$ respectively (in the order given in the question). Without loss of generality, let $I_{1}$ be closer to $A B$ than $I_{2}$. Let the acute angle between $I_{1} I_{2}$ and $A B$ (and hence also the angle between $... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a parallelogram. Let $W, X, Y$, and $Z$ be points on sides $A B, B C, C D$, and $D A$, respectively, such that the incenters of triangles $A W Z, B X W, C Y X$ and $D Z Y$ form a parallelogram. Prove that $W X Y Z$ is a parallelogram. | Let the four incenters be $I_{1}, I_{2}, I_{3}$, and $I_{4}$ with inradii $r_{1}, r_{2}, r_{3}$, and $r_{4}$ respectively (in the order given in the question). Without loss of generality, let $I_{1}$ be closer to $A B$ than $I_{2}$. Let the acute angle between $I_{1} I_{2}$ and $A B$ (and hence also the angle between $... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "# Problem 3",
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"solution_match": "# Solution\n\n",
"tier": "T1",
"year": "2023"
} |
Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that
$$
f((c+1) x+f(y))=f(x+2 y)+2 c x \quad \text { for all } x, y \in \mathbb{R}_{>0}
$$
Answer: $f(x)=2 x$ for all $x>0$. | We first prove that $f(x) \geq 2 x$ for all $x>0$. Suppose, for the sake of contradiction, that $f(y)<2 y$ for some positive $y$. Choose $x$ such that $f((c+1) x+f(y))$ and $f(x+2 y)$ cancel out, that is,
$$
(c+1) x+f(y)=x+2 y \Longleftrightarrow x=\frac{2 y-f(y)}{c}
$$
Notice that $x>0$ because $2 y-f(y)>0$. Then $2... | f(x)=2x | Yes | Yes | math-word-problem | Algebra | Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that
$$
f((c+1) x+f(y))=f(x+2 y)+2 c x \quad \text { for all } x, y \in \mathbb{R}_{>0}
$$
Answer: $f(x)=2 x$ for all $x>0$. | We first prove that $f(x) \geq 2 x$ for all $x>0$. Suppose, for the sake of contradiction, that $f(y)<2 y$ for some positive $y$. Choose $x$ such that $f((c+1) x+f(y))$ and $f(x+2 y)$ cancel out, that is,
$$
(c+1) x+f(y)=x+2 y \Longleftrightarrow x=\frac{2 y-f(y)}{c}
$$
Notice that $x>0$ because $2 y-f(y)>0$. Then $2... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "# Problem 4",
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2023"
} |
Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that
$$
f((c+1) x+f(y))=f(x+2 y)+2 c x \quad \text { for all } x, y \in \mathbb{R}_{>0}
$$
Answer: $f(x)=2 x$ for all $x>0$. | After proving that $f(y) \geq 2 y$ for all $y>0$, one can define $g(x)=f(x)-2 x, g: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{\geq 0}$, and our goal is proving that $g(x)=0$ for all $x>0$. The problem is now rewritten as
$$
\begin{aligned}
& g((c+1) x+g(y)+2 y)+2((c+1) x+g(y)+2 y)=g(x+2 y)+2(x+2 y)+2 c x \\
\Longleftrig... | f(x)=2x | Yes | Yes | math-word-problem | Algebra | Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that
$$
f((c+1) x+f(y))=f(x+2 y)+2 c x \quad \text { for all } x, y \in \mathbb{R}_{>0}
$$
Answer: $f(x)=2 x$ for all $x>0$. | After proving that $f(y) \geq 2 y$ for all $y>0$, one can define $g(x)=f(x)-2 x, g: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{\geq 0}$, and our goal is proving that $g(x)=0$ for all $x>0$. The problem is now rewritten as
$$
\begin{aligned}
& g((c+1) x+g(y)+2 y)+2((c+1) x+g(y)+2 y)=g(x+2 y)+2(x+2 y)+2 c x \\
\Longleftrig... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "# Problem 4",
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2023"
} |
There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2 n-1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:
First, he chooses an endp... | Draw a circle that encloses all the intersection points between line segments and extend all line segments until they meet the circle, and then move Tony and all his friends to the circle. Number the intersection points with the circle from 1 to $2 n$ anticlockwise, starting from Tony (Tony has number 1). We will prove... | proof | Yes | Incomplete | proof | Combinatorics | There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2 n-1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:
First, he chooses an endp... | Draw a circle that encloses all the intersection points between line segments and extend all line segments until they meet the circle, and then move Tony and all his friends to the circle. Number the intersection points with the circle from 1 to $2 n$ anticlockwise, starting from Tony (Tony has number 1). We will prove... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2023"
} |
There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2 n-1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:
First, he chooses an endp... | First part: at most n friends can receive a present.
Similarly to the first solution, consider a circle that encompasses all line segments, extend the lines, and use the endpoints of the chords instead of the line segments, and prove that each chord connects vertices $k$ and $k+n$. We also consider, even in the first p... | proof | Yes | Yes | proof | Combinatorics | There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2 n-1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:
First, he chooses an endp... | First part: at most n friends can receive a present.
Similarly to the first solution, consider a circle that encompasses all line segments, extend the lines, and use the endpoints of the chords instead of the line segments, and prove that each chord connects vertices $k$ and $k+n$. We also consider, even in the first p... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2023"
} |
Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$... | 
Let $\ell$ be the radical axis of circles $B Q X$ and $C P X$. Since $X$ and $Y$ are on $\ell$, it is sufficient to show that $A$ is on $\ell$. Let line $A X$ intersect segments $B C$ and $D... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$... | 
Let $\ell$ be the radical axis of circles $B Q X$ and $C P X$. Since $X$ and $Y$ are on $\ell$, it is sufficient to show that $A$ is on $\ell$. Let line $A X$ intersect segments $B C$ and $D... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2024"
} |
Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$... | 
Let circle $B Q X$ intersect line $A B$ at a point $S$ which is different from $B$. Then $\angle D E X=$ $\angle X Q C=\angle B S X$, thus $S$ is on circle $D E X$. Similarly, let circle $C ... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$... | 
Let circle $B Q X$ intersect line $A B$ at a point $S$ which is different from $B$. Then $\angle D E X=$ $\angle X Q C=\angle B S X$, thus $S$ is on circle $D E X$. Similarly, let circle $C ... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2024"
} |
Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$... | Consider the (direct) homothety that takes triangle $A D E$ to triangle $A B C$, and let $Y^{\prime}$ be the image of $Y$ under this homothety; in other words, let $Y^{\prime}$ be the intersection of the line parallel to $B Y$ through $D$ and the line parallel to $C Y$ through $E$.
 homothety that takes triangle $A D E$ to triangle $A B C$, and let $Y^{\prime}$ be the image of $Y$ under this homothety; in other words, let $Y^{\prime}$ be the intersection of the line parallel to $B Y$ through $D$ and the line parallel to $C Y$ through $E$.
![](https://cdn.mathpix.com/cropped/2... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl",
"solution_match": "# Solution 3",
"tier": "T1",
"year": "2024"
} |
Consider a $100 \times 100$ table, and identify the cell in row $a$ and column $b, 1 \leq a, b \leq 100$, with the ordered pair $(a, b)$. Let $k$ be an integer such that $51 \leq k \leq 99$. A $k$-knight is a piece that moves one cell vertically or horizontally and $k$ cells to the other direction; that is, it moves fr... | Cell $(x, y)$ is directly reachable from another cell if and only if $x-k \geq 1$ or $x+k \leq 100$ or $y-k \geq 1$ or $y+k \leq 100$, that is, $x \geq k+1$ or $x \leq 100-k$ or $y \geq k+1$ or $y \leq 100-k(*)$. Therefore the cells $(x, y)$ for which $101-k \leq x \leq k$ and $101-k \leq y \leq k$ are unreachable. Let... | L(k)=\left\{\begin{array}{ll}100^{2}-(2 k-100)^{2} & \text { if } k \text { is even } \\ \frac{100^{2}-(2 k-100)^{2}}{2} & \text { if } k \text { is odd }\end{array}\right.} | Yes | Yes | math-word-problem | Combinatorics | Consider a $100 \times 100$ table, and identify the cell in row $a$ and column $b, 1 \leq a, b \leq 100$, with the ordered pair $(a, b)$. Let $k$ be an integer such that $51 \leq k \leq 99$. A $k$-knight is a piece that moves one cell vertically or horizontally and $k$ cells to the other direction; that is, it moves fr... | Cell $(x, y)$ is directly reachable from another cell if and only if $x-k \geq 1$ or $x+k \leq 100$ or $y-k \geq 1$ or $y+k \leq 100$, that is, $x \geq k+1$ or $x \leq 100-k$ or $y \geq k+1$ or $y \leq 100-k(*)$. Therefore the cells $(x, y)$ for which $101-k \leq x \leq k$ and $101-k \leq y \leq k$ are unreachable. Let... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "# Problem 2",
"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl",
"solution_match": "# Solution\n\n",
"tier": "T1",
"year": "2024"
} |
Let $n$ be a positive integer and $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers. Prove that
$$
\sum_{i=1}^{n} \frac{1}{2^{i}}\left(\frac{2}{1+a_{i}}\right)^{2^{i}} \geq \frac{2}{1+a_{1} a_{2} \ldots a_{n}}-\frac{1}{2^{n}}
$$ | We first prove the following lemma:
Lemma 1. For $k$ positive integer and $x, y>0$,
$$
\left(\frac{2}{1+x}\right)^{2^{k}}+\left(\frac{2}{1+y}\right)^{2^{k}} \geq 2\left(\frac{2}{1+x y}\right)^{2^{k-1}}
$$
The proof goes by induction. For $k=1$, we have
$$
\left(\frac{2}{1+x}\right)^{2}+\left(\frac{2}{1+y}\right)^{2}... | proof | Yes | Yes | proof | Inequalities | Let $n$ be a positive integer and $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers. Prove that
$$
\sum_{i=1}^{n} \frac{1}{2^{i}}\left(\frac{2}{1+a_{i}}\right)^{2^{i}} \geq \frac{2}{1+a_{1} a_{2} \ldots a_{n}}-\frac{1}{2^{n}}
$$ | We first prove the following lemma:
Lemma 1. For $k$ positive integer and $x, y>0$,
$$
\left(\frac{2}{1+x}\right)^{2^{k}}+\left(\frac{2}{1+y}\right)^{2^{k}} \geq 2\left(\frac{2}{1+x y}\right)^{2^{k-1}}
$$
The proof goes by induction. For $k=1$, we have
$$
\left(\frac{2}{1+x}\right)^{2}+\left(\frac{2}{1+y}\right)^{2}... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "# Problem 3",
"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl",
"solution_match": "# Solution\n\n",
"tier": "T1",
"year": "2024"
} |
Prove that for every positive integer $t$ there is a unique permutation $a_{0}, a_{1}, \ldots, a_{t-1}$ of $0,1, \ldots, t-$ 1 such that, for every $0 \leq i \leq t-1$, the binomial coefficient $\binom{t+i}{2 a_{i}}$ is odd and $2 a_{i} \neq t+i$. | We constantly make use of Kummer's theorem which, in particular, implies that $\binom{n}{k}$ is odd if and only if $k$ and $n-k$ have ones in different positions in binary. In other words, if $S(x)$ is the set of positions of the digits 1 of $x$ in binary (in which the digit multiplied by $2^{i}$ is in position $i),\bi... | proof | Yes | Incomplete | proof | Combinatorics | Prove that for every positive integer $t$ there is a unique permutation $a_{0}, a_{1}, \ldots, a_{t-1}$ of $0,1, \ldots, t-$ 1 such that, for every $0 \leq i \leq t-1$, the binomial coefficient $\binom{t+i}{2 a_{i}}$ is odd and $2 a_{i} \neq t+i$. | We constantly make use of Kummer's theorem which, in particular, implies that $\binom{n}{k}$ is odd if and only if $k$ and $n-k$ have ones in different positions in binary. In other words, if $S(x)$ is the set of positions of the digits 1 of $x$ in binary (in which the digit multiplied by $2^{i}$ is in position $i),\bi... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "# Problem 4",
"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl",
"solution_match": "# Solution\n\n",
"tier": "T1",
"year": "2024"
} |
Line $\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \neq Q$, while circumcircles of the ... | We start with the following lemma.
Lemma 1. Points $M, N, P, Q$ are concyclic.
Point $M$ is the Miquel point of lines $A P=A B, P S=\ell, A S=A D$, and $B R=B C$, and point $N$ is the Miquel point of lines $C Q=C D, R C=B C, Q R=\ell$, and $D S=A D$. Both points $M$ and $N$ are on the circumcircle of the triangle deter... | proof | Yes | Yes | proof | Geometry | Line $\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \neq Q$, while circumcircles of the ... | We start with the following lemma.
Lemma 1. Points $M, N, P, Q$ are concyclic.
Point $M$ is the Miquel point of lines $A P=A B, P S=\ell, A S=A D$, and $B R=B C$, and point $N$ is the Miquel point of lines $C Q=C D, R C=B C, Q R=\ell$, and $D S=A D$. Both points $M$ and $N$ are on the circumcircle of the triangle deter... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2024"
} |
Line $\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \neq Q$, while circumcircles of the ... | Barycentric coordinates are a viable way to solve the problem, but even the solution we have found had some clever computations. Here is an outline of this solution.
Lemma 2. Denote by $\operatorname{pow}_{\omega} X$ the power of point $X$ with respect to circle $\omega$. Let $\Gamma_{1}$ and $\Gamma_{2}$ be circles w... | proof | Yes | Yes | proof | Geometry | Line $\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \neq Q$, while circumcircles of the ... | Barycentric coordinates are a viable way to solve the problem, but even the solution we have found had some clever computations. Here is an outline of this solution.
Lemma 2. Denote by $\operatorname{pow}_{\omega} X$ the power of point $X$ with respect to circle $\omega$. Let $\Gamma_{1}$ and $\Gamma_{2}$ be circles w... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2024"
} |
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