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
values | question_type stringclasses 4
values | problem_type stringclasses 8
values | problem_raw stringlengths 14 10.4k | solution_raw stringlengths 1 24.1k | metadata dict |
|---|---|---|---|---|---|---|---|---|---|
Let \(A B C\) be an acute triangle inscribed in a circle \(\Gamma\) . Let \(A_{1}\) be the orthogonal projection of \(A\) onto \(B C\) so that \(A A_{1}\) is an altitude. Let \(B_{1}\) and \(C_{1}\) be the orthogonal projections of \(A_{1}\) onto \(A B\) and \(A C\) , respectively. Point \(P\) is such that quadrilatera... | First notice that, since angles \(\angle A A_{1}B_{1}\) and \(\angle A A_{1}C_{1}\) are both right, the points \(B_{1}\) and \(C_{1}\) lie on the circle with \(A A_{1}\) as a diameter. Therefore, \(A C_{1} = A A_{1}\sin \angle A A_{1}C_{1} = A A_{1}\sin (90^{\circ}-\) \(\angle A_{1}A C) = A A_{1}\sin \angle C\) , simil... | proof | Yes | Yes | proof | Geometry | Let \(A B C\) be an acute triangle inscribed in a circle \(\Gamma\) . Let \(A_{1}\) be the orthogonal projection of \(A\) onto \(B C\) so that \(A A_{1}\) is an altitude. Let \(B_{1}\) and \(C_{1}\) be the orthogonal projections of \(A_{1}\) onto \(A B\) and \(A C\) , respectively. Point \(P\) is such that quadrilatera... | First notice that, since angles \(\angle A A_{1}B_{1}\) and \(\angle A A_{1}C_{1}\) are both right, the points \(B_{1}\) and \(C_{1}\) lie on the circle with \(A A_{1}\) as a diameter. Therefore, \(A C_{1} = A A_{1}\sin \angle A A_{1}C_{1} = A A_{1}\sin (90^{\circ}-\) \(\angle A_{1}A C) = A A_{1}\sin \angle C\) , simil... | {
"exam": "APMO",
"problem_label": "1",
"problem_match": "# Problem 1 ",
"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl",
"solution_match": "# Solution \n\n",
"tier": "T1",
"year": "2025"
} |
Let \(\alpha\) and \(\beta\) be positive real numbers. Emerald makes a trip in the coordinate plane, starting off from the origin \((0,0)\) . Each minute she moves one unit up or one unit to the right, restricting herself to the region \(|x - y|< 2025\) , in the coordinate plane. By the time she visits a point \((x,y)\... | Let \((x_{n},y_{n})\) be the point that Emerald visits after \(n\) minutes. Then \((x_{n + 1},y_{n + 1})\in \{(x_{n}+\) \(1,y_{n}),(x_{n},y_{n} + 1)\}\) . Either way, \(x_{n + 1} + y_{n + 1} = x_{n} + y_{n} + 1\) , and since \(x_{0} + y_{0} = 0 + 0 = 0\) \(x_{n} + y_{n} = n\)
The \(n\) - th number would be then
\... | \alpha + \beta = 2 | Yes | Yes | math-word-problem | Combinatorics | Let \(\alpha\) and \(\beta\) be positive real numbers. Emerald makes a trip in the coordinate plane, starting off from the origin \((0,0)\) . Each minute she moves one unit up or one unit to the right, restricting herself to the region \(|x - y|< 2025\) , in the coordinate plane. By the time she visits a point \((x,y)\... | Let \((x_{n},y_{n})\) be the point that Emerald visits after \(n\) minutes. Then \((x_{n + 1},y_{n + 1})\in \{(x_{n}+\) \(1,y_{n}),(x_{n},y_{n} + 1)\}\) . Either way, \(x_{n + 1} + y_{n + 1} = x_{n} + y_{n} + 1\) , and since \(x_{0} + y_{0} = 0 + 0 = 0\) \(x_{n} + y_{n} = n\)
The \(n\) - th number would be then
\... | {
"exam": "APMO",
"problem_label": "2",
"problem_match": "# Problem 2 ",
"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl",
"solution_match": "# Solution \n\n",
"tier": "T1",
"year": "2025"
} |
Let \(P(x)\) be a non- constant polynomial with integer coefficients such that \(P(0) \neq 0\) . Let \(a_{1}, a_{2}, a_{3}, \ldots\) be an infinite sequence of integers such that \(P(i - j)\) divides \(a_{i} - a_{j}\) for all distinct positive integers \(i, j\) . Prove that the sequence \(a_{1}, a_{2}, a_{3}, \ldots\) ... | Let \(a_{0} = P(0) \neq 0\) be the independent coefficient, i.e., the constant term of \(P(x)\) . Then there are infinitely many primes \(p\) such that \(p\) divides \(P(k)\) but \(p\) does not divide \(k\) . In fact, since \(P(k) - a_{0}\) is a multiple of \(k\) , \(\gcd (P(k), k) = \gcd (k, a_{0}) \leq a_{0}\) is bou... | proof | Yes | Yes | proof | Number Theory | Let \(P(x)\) be a non- constant polynomial with integer coefficients such that \(P(0) \neq 0\) . Let \(a_{1}, a_{2}, a_{3}, \ldots\) be an infinite sequence of integers such that \(P(i - j)\) divides \(a_{i} - a_{j}\) for all distinct positive integers \(i, j\) . Prove that the sequence \(a_{1}, a_{2}, a_{3}, \ldots\) ... | Let \(a_{0} = P(0) \neq 0\) be the independent coefficient, i.e., the constant term of \(P(x)\) . Then there are infinitely many primes \(p\) such that \(p\) divides \(P(k)\) but \(p\) does not divide \(k\) . In fact, since \(P(k) - a_{0}\) is a multiple of \(k\) , \(\gcd (P(k), k) = \gcd (k, a_{0}) \leq a_{0}\) is bou... | {
"exam": "APMO",
"problem_label": "3",
"problem_match": "# Problem 3 ",
"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl",
"solution_match": "# Solution \n\n",
"tier": "T1",
"year": "2025"
} |
Let \(n \geq 3\) be an integer. There are \(n\) cells on a circle, and each cell is assigned either 0 or 1. There is a rooster on one of these cells, and it repeats the following operations:
If the rooster is on a cell assigned 0, it changes the assigned number to 1 and moves to the next cell counterclockwise. If th... | Reformulate the problem as a \(n\) - string of numbers in \(\{0,1\}\) and a position at which the action described in the problem is performed, and add 1 or 2 modulo \(n\) to the position according to the action. Say that a lap is complete for each time the position resets to 0 or 1. We will prove that the statement cl... | proof | Yes | Incomplete | proof | Combinatorics | Let \(n \geq 3\) be an integer. There are \(n\) cells on a circle, and each cell is assigned either 0 or 1. There is a rooster on one of these cells, and it repeats the following operations:
If the rooster is on a cell assigned 0, it changes the assigned number to 1 and moves to the next cell counterclockwise. If th... | Reformulate the problem as a \(n\) - string of numbers in \(\{0,1\}\) and a position at which the action described in the problem is performed, and add 1 or 2 modulo \(n\) to the position according to the action. Say that a lap is complete for each time the position resets to 0 or 1. We will prove that the statement cl... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "# Problem 4 ",
"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2025"
} |
Let \(n \geq 3\) be an integer. There are \(n\) cells on a circle, and each cell is assigned either 0 or 1. There is a rooster on one of these cells, and it repeats the following operations:
If the rooster is on a cell assigned 0, it changes the assigned number to 1 and moves to the next cell counterclockwise. If th... | Define positions, laps, stoppings, and bypassing as in Solution 1. This other pair of lemmata also solves the problem.
Lemma 3. There is a position and a lap in which the rooster stops twice and bypasses once (in some order) in the next three laps.
Proof. There is a position \(j\) the rooster stops for infinitely... | proof | Yes | Yes | proof | Combinatorics | Let \(n \geq 3\) be an integer. There are \(n\) cells on a circle, and each cell is assigned either 0 or 1. There is a rooster on one of these cells, and it repeats the following operations:
If the rooster is on a cell assigned 0, it changes the assigned number to 1 and moves to the next cell counterclockwise. If th... | Define positions, laps, stoppings, and bypassing as in Solution 1. This other pair of lemmata also solves the problem.
Lemma 3. There is a position and a lap in which the rooster stops twice and bypasses once (in some order) in the next three laps.
Proof. There is a position \(j\) the rooster stops for infinitely... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "# Problem 4 ",
"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2025"
} |
Let \(n \geq 3\) be an integer. There are \(n\) cells on a circle, and each cell is assigned either 0 or 1. There is a rooster on one of these cells, and it repeats the following operations:
If the rooster is on a cell assigned 0, it changes the assigned number to 1 and moves to the next cell counterclockwise. If th... | Let us reformulate the problem in terms of Graphs: we have a directed graph \(G\) with \(V = \{v_{1}, v_{2}, \ldots , v_{n}\}\) representing positions and \(E = \{v_{i} \to v_{i + 1}, v_{i} \to v_{i + 2} \mid 1 \le i \le n\}\) representing moves. Indices are taken mod \(n\) . Note that each vertex has in- degree and ou... | proof | Yes | Incomplete | proof | Combinatorics | Let \(n \geq 3\) be an integer. There are \(n\) cells on a circle, and each cell is assigned either 0 or 1. There is a rooster on one of these cells, and it repeats the following operations:
If the rooster is on a cell assigned 0, it changes the assigned number to 1 and moves to the next cell counterclockwise. If th... | Let us reformulate the problem in terms of Graphs: we have a directed graph \(G\) with \(V = \{v_{1}, v_{2}, \ldots , v_{n}\}\) representing positions and \(E = \{v_{i} \to v_{i + 1}, v_{i} \to v_{i + 2} \mid 1 \le i \le n\}\) representing moves. Indices are taken mod \(n\) . Note that each vertex has in- degree and ou... | {
"exam": "APMO",
"problem_label": "4",
"problem_match": "# Problem 4 ",
"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl",
"solution_match": "# Solution 3",
"tier": "T1",
"year": "2025"
} |
Consider an infinite sequence \(a_{1},a_{2},\ldots\) of positive integers such that
\[100!(a_{m} + a_{m + 1} + \cdot \cdot \cdot +a_{n})\quad \mathrm{is~a~multiple~of}\quad a_{n - m + 1}a_{n + m}\]
for all positive integers \(m,n\) such that \(m\leq n\)
Prove that the sequence is either bounded or linear.
O... | Let \(c = 100!\) . Suppose that \(n\geq m + 2\) . Then \(a_{m + n} = a_{(m + 1) + (n - 1)}\) divides both \(c(a_{m} + a_{m + 1}+\) \(\cdot \cdot \cdot +a_{n - 1} + a_{n})\) and \(c(a_{m + 1} + \cdot \cdot \cdot +a_{n - 1})\) , so it also divides the difference \(c(a_{m} + a_{n})\) . Notice that if \(n = m + 1\) then \(... | proof | Yes | Incomplete | proof | Number Theory | Consider an infinite sequence \(a_{1},a_{2},\ldots\) of positive integers such that
\[100!(a_{m} + a_{m + 1} + \cdot \cdot \cdot +a_{n})\quad \mathrm{is~a~multiple~of}\quad a_{n - m + 1}a_{n + m}\]
for all positive integers \(m,n\) such that \(m\leq n\)
Prove that the sequence is either bounded or linear.
O... | Let \(c = 100!\) . Suppose that \(n\geq m + 2\) . Then \(a_{m + n} = a_{(m + 1) + (n - 1)}\) divides both \(c(a_{m} + a_{m + 1}+\) \(\cdot \cdot \cdot +a_{n - 1} + a_{n})\) and \(c(a_{m + 1} + \cdot \cdot \cdot +a_{n - 1})\) , so it also divides the difference \(c(a_{m} + a_{n})\) . Notice that if \(n = m + 1\) then \(... | {
"exam": "APMO",
"problem_label": "5",
"problem_match": "# Problem 5 ",
"resource_path": "APMO/segmented/en-apmo2025_sol.jsonl",
"solution_match": "# Solution \n\n",
"tier": "T1",
"year": "2025"
} |
An acute-angled scalene triangle $A B C$ is given, with $A C>B C$. Let $O$ be its circumcentre, $H$ its orthocentre, and $F$ the foot of the altitude from $C$. Let $P$ be the point (other than $A$ ) on the line $A B$ such that $A F=P F$, and $M$ be the midpoint of $A C$. We denote the intersection of $P H$ and $B C$ by... | It is enough to show that $\mathrm{O} F \perp F X$.
Let $\mathrm{OE} \perp \mathrm{AB}$, then it is trivial that :
$$
C \mathrm{H}=2 \mathrm{OE} .
$$
Since from the hypothesis we have $\mathrm{P} F=\mathrm{A} F$ then we take $\mathrm{PB}=\mathrm{P} F-\mathrm{B} F$ or
$$
\mathrm{PB}=\mathrm{A} F-\mathrm{B} F
$$
Also... | proof | Yes | Yes | proof | Geometry | An acute-angled scalene triangle $A B C$ is given, with $A C>B C$. Let $O$ be its circumcentre, $H$ its orthocentre, and $F$ the foot of the altitude from $C$. Let $P$ be the point (other than $A$ ) on the line $A B$ such that $A F=P F$, and $M$ be the midpoint of $A C$. We denote the intersection of $P H$ and $B C$ by... | It is enough to show that $\mathrm{O} F \perp F X$.
Let $\mathrm{OE} \perp \mathrm{AB}$, then it is trivial that :
$$
C \mathrm{H}=2 \mathrm{OE} .
$$
Since from the hypothesis we have $\mathrm{P} F=\mathrm{A} F$ then we take $\mathrm{PB}=\mathrm{P} F-\mathrm{B} F$ or
$$
\mathrm{PB}=\mathrm{A} F-\mathrm{B} F
$$
Also... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "Balkan_MO/segmented/en-2008-BMO-type1.jsonl",
"solution_match": "# Solution:",
"tier": "T1",
"year": "2008"
} |
Does there exist a sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ of positive real numbers satisfying both of the following conditions:
(i) $\sum_{i=1}^{n} a_{i} \leq n^{2}$, for every positive integer $n$;
(ii) $\sum_{i=1}^{n} \frac{1}{a_{i}} \leq 2008$, for every positive integer $n$ ? | The answer is no.
It is enough to show that
if $\sum_{i=1}^{n} a_{i} \leq n^{2}$ for any $n$, then $\sum_{i=2}^{2^{n}} \frac{1}{a_{i}}>\frac{n}{4}$. (or any other precise estimate)
For this, we use that $\sum_{i=2^{k}+1}^{2^{k+1}} a_{i} \sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{a_{i}} \geq 2^{2 k}$ for any $k \geq 0$ by the ... | proof | Yes | Yes | proof | Inequalities | Does there exist a sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ of positive real numbers satisfying both of the following conditions:
(i) $\sum_{i=1}^{n} a_{i} \leq n^{2}$, for every positive integer $n$;
(ii) $\sum_{i=1}^{n} \frac{1}{a_{i}} \leq 2008$, for every positive integer $n$ ? | The answer is no.
It is enough to show that
if $\sum_{i=1}^{n} a_{i} \leq n^{2}$ for any $n$, then $\sum_{i=2}^{2^{n}} \frac{1}{a_{i}}>\frac{n}{4}$. (or any other precise estimate)
For this, we use that $\sum_{i=2^{k}+1}^{2^{k+1}} a_{i} \sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{a_{i}} \geq 2^{2 k}$ for any $k \geq 0$ by the ... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "# Problem 2",
"resource_path": "Balkan_MO/segmented/en-2008-BMO-type1.jsonl",
"solution_match": "# Solution.",
"tier": "T1",
"year": "2008"
} |
Let $n$ be a positive integer. The rectangle $A B C D$ with side lengths $A B=90 n+1$ and $B C=90 n+5$ is partitioned into unit squares with sides parallel to the sides of $A B C D$. Let $S$ be the set of all points which are vertices of these unit squares. Prove that the number of lines which pass through at least two... | Denote $90 n+1=m$. We investigate the number of the lines modulo 4 consecutively reducing different types of lines.
The vertical and horizontal lines are
$(m+5)+(m+1)=2(m+3)$ which is divisible to 4.
Moreover, every line which makes an acute angle to the axe $O x$ (i.e. that line has a positive angular coefficient) cor... | proof | Yes | Yes | proof | Combinatorics | Let $n$ be a positive integer. The rectangle $A B C D$ with side lengths $A B=90 n+1$ and $B C=90 n+5$ is partitioned into unit squares with sides parallel to the sides of $A B C D$. Let $S$ be the set of all points which are vertices of these unit squares. Prove that the number of lines which pass through at least two... | Denote $90 n+1=m$. We investigate the number of the lines modulo 4 consecutively reducing different types of lines.
The vertical and horizontal lines are
$(m+5)+(m+1)=2(m+3)$ which is divisible to 4.
Moreover, every line which makes an acute angle to the axe $O x$ (i.e. that line has a positive angular coefficient) cor... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "# Problem 3",
"resource_path": "Balkan_MO/segmented/en-2008-BMO-type1.jsonl",
"solution_match": "# Solution.",
"tier": "T1",
"year": "2008"
} |
Let $c$ be a positive integer. The sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ is defined by $a_{1}=c$, and $a_{n+1}=a_{n}^{2}+a_{n}+c^{3}$, for every positive integer $n$. Find all values of $c$ for which there exist some integers $k \geq 1$ and $m \geq 2$, such that $a_{k}^{2}+c^{3}$ is the $m^{\text {th }}$ power... | First, notice:
$$
a_{n+1}^{2}+c^{3}=\left(a_{n}^{2}+a_{n}+c^{3}\right)^{2}+c^{3}=\left(a_{n}^{2}+c^{3}\right)\left(a_{n}^{2}+2 a_{n}+1+c^{3}\right)
$$
We first prove that $a_{n}^{2}+c^{3}$ and $a_{n}^{2}+2 a_{n}+1+c^{3}$ are coprime.
We prove by induction that $4 c^{3}+1$ is coprime with $2 a_{n}+1$, for every $n \ge... | c=s^{2}-1, s \geq 2, s \in \mathbb{N} | Yes | Yes | math-word-problem | Number Theory | Let $c$ be a positive integer. The sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ is defined by $a_{1}=c$, and $a_{n+1}=a_{n}^{2}+a_{n}+c^{3}$, for every positive integer $n$. Find all values of $c$ for which there exist some integers $k \geq 1$ and $m \geq 2$, such that $a_{k}^{2}+c^{3}$ is the $m^{\text {th }}$ power... | First, notice:
$$
a_{n+1}^{2}+c^{3}=\left(a_{n}^{2}+a_{n}+c^{3}\right)^{2}+c^{3}=\left(a_{n}^{2}+c^{3}\right)\left(a_{n}^{2}+2 a_{n}+1+c^{3}\right)
$$
We first prove that $a_{n}^{2}+c^{3}$ and $a_{n}^{2}+2 a_{n}+1+c^{3}$ are coprime.
We prove by induction that $4 c^{3}+1$ is coprime with $2 a_{n}+1$, for every $n \ge... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "# Problem 4",
"resource_path": "Balkan_MO/segmented/en-2008-BMO-type1.jsonl",
"solution_match": "# Solution.",
"tier": "T1",
"year": "2008"
} |
We start by observing that $z$ must be even, so $z^{2}=3^{x}-5^{y} \equiv(-1)^{x}-1(\bmod 4)$ is divisible by 4 , which implies that $x$ is even, say $x=2 t$. Then our equation can be rewritten as $\left(3^{t}-z\right)\left(3^{t}+z\right)=5^{y}$, which means that both $3^{t}-z=5^{k}$ and $3^{t}+z=5^{y-k}$ for some nonn... | We start by observing that $f$ is injective. From the known identity
$$
\left(a^{2}+2 b^{2}\right)\left(c^{2}+2 d^{2}\right)=(a c \pm 2 b d)^{2}+2(a d \mp b c)^{2}
$$
we obtain $f(a c+2 b d)^{2}+2 f(a d-b c)^{2}=f(a c-2 b d)^{2}+2 f(a d+b c)^{2}$, assuming that the arguments are positive integers. Specially, for $b=c... | (x, y, z)=(2,1,2) | Yes | Incomplete | proof | Number Theory | We start by observing that $z$ must be even, so $z^{2}=3^{x}-5^{y} \equiv(-1)^{x}-1(\bmod 4)$ is divisible by 4 , which implies that $x$ is even, say $x=2 t$. Then our equation can be rewritten as $\left(3^{t}-z\right)\left(3^{t}+z\right)=5^{y}$, which means that both $3^{t}-z=5^{k}$ and $3^{t}+z=5^{y-k}$ for some nonn... | We start by observing that $f$ is injective. From the known identity
$$
\left(a^{2}+2 b^{2}\right)\left(c^{2}+2 d^{2}\right)=(a c \pm 2 b d)^{2}+2(a d \mp b c)^{2}
$$
we obtain $f(a c+2 b d)^{2}+2 f(a d-b c)^{2}=f(a c-2 b d)^{2}+2 f(a d+b c)^{2}$, assuming that the arguments are positive integers. Specially, for $b=c... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\n1.",
"resource_path": "Balkan_MO/segmented/en-2009-BMO-type2.jsonl",
"solution_match": "\n4.",
"tier": "T1",
"year": "2009"
} |
In a triangle $A B C$, points $M$ and $N$ on the sides $A B$ and $A C$ respectively are such that $M N \| B C$. Let $B N$ and $C M$ intersect at point $P$. The circumcircles of triangles $B M P$ and $C N P$ intersect at two distinct points $P$ and $Q$. Prove that $\angle B A Q=\angle C A P$.
(Moldova) | Since the quadrilaterals $B M P Q$ and $C N P Q$ are cyclic, we have $\angle B Q N=\angle B Q P+$ $\angle P Q N=\angle A M C+\angle M C A=180^{\circ}-$ $\angle C A B$, so $A B Q N$ is cyclic as well. Hence $\frac{\sin \angle B A Q}{\sin \angle N A Q}=\frac{B Q}{N Q}$. Moreover, triangles $M B Q$ and $C N Q$ are similar... | proof | Yes | Yes | proof | Geometry | In a triangle $A B C$, points $M$ and $N$ on the sides $A B$ and $A C$ respectively are such that $M N \| B C$. Let $B N$ and $C M$ intersect at point $P$. The circumcircles of triangles $B M P$ and $C N P$ intersect at two distinct points $P$ and $Q$. Prove that $\angle B A Q=\angle C A P$.
(Moldova) | Since the quadrilaterals $B M P Q$ and $C N P Q$ are cyclic, we have $\angle B Q N=\angle B Q P+$ $\angle P Q N=\angle A M C+\angle M C A=180^{\circ}-$ $\angle C A B$, so $A B Q N$ is cyclic as well. Hence $\frac{\sin \angle B A Q}{\sin \angle N A Q}=\frac{B Q}{N Q}$. Moreover, triangles $M B Q$ and $C N Q$ are similar... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\n2.",
"resource_path": "Balkan_MO/segmented/en-2009-BMO-type2.jsonl",
"solution_match": "\n2.",
"tier": "T1",
"year": "2009"
} |
A $9 \times 12$ rectangle is divided into unit squares. The centers of all the unit squares, except the four corner squares and the eight squares adjacent (by side) to them, are colored red. Is it possible to numerate the red centers by $C_{1}, C_{2}, \ldots, C_{96}$ so that the following two conditions are fulfilled:
... | Place the given rectangle into the coordinate plane so that the center of the square at the intersection of $i$-th column and $j$-th row has the coordinates $(i, j)$. Suppose that a desired numeration of the red points exists; it corresponds to a path, i.e. a closed poligonal line consisting of 96 segments of length $\... | proof | Yes | Yes | proof | Geometry | A $9 \times 12$ rectangle is divided into unit squares. The centers of all the unit squares, except the four corner squares and the eight squares adjacent (by side) to them, are colored red. Is it possible to numerate the red centers by $C_{1}, C_{2}, \ldots, C_{96}$ so that the following two conditions are fulfilled:
... | Place the given rectangle into the coordinate plane so that the center of the square at the intersection of $i$-th column and $j$-th row has the coordinates $(i, j)$. Suppose that a desired numeration of the red points exists; it corresponds to a path, i.e. a closed poligonal line consisting of 96 segments of length $\... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\n3.",
"resource_path": "Balkan_MO/segmented/en-2009-BMO-type2.jsonl",
"solution_match": "\n3.",
"tier": "T1",
"year": "2009"
} |
Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfying
$$
f\left(f(m)^{2}+2 f(n)^{2}\right)=m^{2}+2 n^{2} \quad \text { for all } m, n \in \mathbb{N} . \quad \text { (Bulgaria) }
$$
Time allowed: 270 minutes.
Each problem is worth 10 points.
## SOLUTIONS | We start by observing that $f$ is injective. From the known identity
$$
\left(a^{2}+2 b^{2}\right)\left(c^{2}+2 d^{2}\right)=(a c \pm 2 b d)^{2}+2(a d \mp b c)^{2}
$$
we obtain $f(a c+2 b d)^{2}+2 f(a d-b c)^{2}=f(a c-2 b d)^{2}+2 f(a d+b c)^{2}$, assuming that the arguments are positive integers. Specially, for $b=c... | f(n)=n | Yes | Yes | proof | Algebra | Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfying
$$
f\left(f(m)^{2}+2 f(n)^{2}\right)=m^{2}+2 n^{2} \quad \text { for all } m, n \in \mathbb{N} . \quad \text { (Bulgaria) }
$$
Time allowed: 270 minutes.
Each problem is worth 10 points.
## SOLUTIONS | We start by observing that $f$ is injective. From the known identity
$$
\left(a^{2}+2 b^{2}\right)\left(c^{2}+2 d^{2}\right)=(a c \pm 2 b d)^{2}+2(a d \mp b c)^{2}
$$
we obtain $f(a c+2 b d)^{2}+2 f(a d-b c)^{2}=f(a c-2 b d)^{2}+2 f(a d+b c)^{2}$, assuming that the arguments are positive integers. Specially, for $b=c... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\n4.",
"resource_path": "Balkan_MO/segmented/en-2009-BMO-type2.jsonl",
"solution_match": "\n4.",
"tier": "T1",
"year": "2009"
} |
The left-hand side is equal to
$$
\frac{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-a^{3} b^{2} c-b^{3} c^{2} a-c^{3} a^{2} b}{(a+b)(b+c)(c+a)}
$$
so it is enough to show that $a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3} \geq a^{3} b^{2} c+b^{3} c^{2} a+c^{3} a^{2} b$. The AM-GM inequality gives us $a^{3} b^{3}+a^{3} b^{3}+a^{3} c^{... | There are $n+1-\varphi(n)$ nonnegative integers not coprime with $n$, and whenever $r$ is among them, so is $n-r$. This gives us the formula $f(n)=\frac{1}{2} n(n+1-\varphi(n))$. Suppose that $f(n)=f(n+p)$. We observe first that $n$ and $n+p$ divide $2 f(n)<n(n+p)$, so $n$ and $n+p$ are not coprime, which implies that ... | not found | Yes | Incomplete | proof | Inequalities | The left-hand side is equal to
$$
\frac{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-a^{3} b^{2} c-b^{3} c^{2} a-c^{3} a^{2} b}{(a+b)(b+c)(c+a)}
$$
so it is enough to show that $a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3} \geq a^{3} b^{2} c+b^{3} c^{2} a+c^{3} a^{2} b$. The AM-GM inequality gives us $a^{3} b^{3}+a^{3} b^{3}+a^{3} c^{... | There are $n+1-\varphi(n)$ nonnegative integers not coprime with $n$, and whenever $r$ is among them, so is $n-r$. This gives us the formula $f(n)=\frac{1}{2} n(n+1-\varphi(n))$. Suppose that $f(n)=f(n+p)$. We observe first that $n$ and $n+p$ divide $2 f(n)<n(n+p)$, so $n$ and $n+p$ are not coprime, which implies that ... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\n1.",
"resource_path": "Balkan_MO/segmented/en-2010-BMO-type2.jsonl",
"solution_match": "\n4.",
"tier": "T1",
"year": "2010"
} |
Let $A B C$ be an acute-angled triangle with orthocenter $H$ and let $M$ be the midpoint of $A C$. The foot of the altitude from $C$ is $C_{1}$. Point $H_{1}$ is symmetric to $H$ in $A B$. The projections of $C_{1}$ on lines $A H_{1}, A C$ and $B C$ are $P, Q$ and $R$ respectively. If $M_{1}$ is the circumcenter of tri... | We shall use the following simple statement.
Lemma. Let $A_{1} A_{2} A_{3} A_{4}$ be a convex cyclic quadrilateral whose diagonals are orthogonal and meet at $X$. If $B_{i}$ is the midpoint of side $A_{i} A_{i+1}$ and $X_{i}$ the projection of $X$ on this side $\left(A_{5}=A_{1}\right)$, then the eight points $B_{i}, ... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute-angled triangle with orthocenter $H$ and let $M$ be the midpoint of $A C$. The foot of the altitude from $C$ is $C_{1}$. Point $H_{1}$ is symmetric to $H$ in $A B$. The projections of $C_{1}$ on lines $A H_{1}, A C$ and $B C$ are $P, Q$ and $R$ respectively. If $M_{1}$ is the circumcenter of tri... | We shall use the following simple statement.
Lemma. Let $A_{1} A_{2} A_{3} A_{4}$ be a convex cyclic quadrilateral whose diagonals are orthogonal and meet at $X$. If $B_{i}$ is the midpoint of side $A_{i} A_{i+1}$ and $X_{i}$ the projection of $X$ on this side $\left(A_{5}=A_{1}\right)$, then the eight points $B_{i}, ... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\n2.",
"resource_path": "Balkan_MO/segmented/en-2010-BMO-type2.jsonl",
"solution_match": "\n2.",
"tier": "T1",
"year": "2010"
} |
We define a $w$-strip as the set of all points in the plane that are between or on two parallel lines on a mutual distance $w$. Let $S$ be a set of $n$ points in the plane such that any three points from $S$ can be covered by a 1 -strip. Show that the entire set $S$ can be covered by a 2 -strip.
(Romania) | Of all triangles with the vertices in $S$, consider one with a maximum area, say $\triangle A B C$. Let $A^{\prime}, B^{\prime}, C^{\prime}$ be the points symmetric to $A, B, C$ with respect to the midpoints of $B C, C A, A B$, respectively. We claim that all points from $S$ must lie inside or on the boundary of $\tria... | proof | Yes | Yes | proof | Geometry | We define a $w$-strip as the set of all points in the plane that are between or on two parallel lines on a mutual distance $w$. Let $S$ be a set of $n$ points in the plane such that any three points from $S$ can be covered by a 1 -strip. Show that the entire set $S$ can be covered by a 2 -strip.
(Romania) | Of all triangles with the vertices in $S$, consider one with a maximum area, say $\triangle A B C$. Let $A^{\prime}, B^{\prime}, C^{\prime}$ be the points symmetric to $A, B, C$ with respect to the midpoints of $B C, C A, A B$, respectively. We claim that all points from $S$ must lie inside or on the boundary of $\tria... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\n3.",
"resource_path": "Balkan_MO/segmented/en-2010-BMO-type2.jsonl",
"solution_match": "\n3.",
"tier": "T1",
"year": "2010"
} |
For every integer $n \geq 2$, denote by $f(n)$ the sum of positive integers not exceeding $n$ that are not coprime to $n$. Prove that $f(n+p) \neq f(n)$ for any such $n$ and any prime number $p$.
(Turkey)
Time allowed: 270 minutes.
Each problem is worth 10 points.
## SOLUTIONS | There are $n+1-\varphi(n)$ nonnegative integers not coprime with $n$, and whenever $r$ is among them, so is $n-r$. This gives us the formula $f(n)=\frac{1}{2} n(n+1-\varphi(n))$. Suppose that $f(n)=f(n+p)$. We observe first that $n$ and $n+p$ divide $2 f(n)<n(n+p)$, so $n$ and $n+p$ are not coprime, which implies that ... | proof | Yes | Yes | proof | Number Theory | For every integer $n \geq 2$, denote by $f(n)$ the sum of positive integers not exceeding $n$ that are not coprime to $n$. Prove that $f(n+p) \neq f(n)$ for any such $n$ and any prime number $p$.
(Turkey)
Time allowed: 270 minutes.
Each problem is worth 10 points.
## SOLUTIONS | There are $n+1-\varphi(n)$ nonnegative integers not coprime with $n$, and whenever $r$ is among them, so is $n-r$. This gives us the formula $f(n)=\frac{1}{2} n(n+1-\varphi(n))$. Suppose that $f(n)=f(n+p)$. We observe first that $n$ and $n+p$ divide $2 f(n)<n(n+p)$, so $n$ and $n+p$ are not coprime, which implies that ... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\n4.",
"resource_path": "Balkan_MO/segmented/en-2010-BMO-type2.jsonl",
"solution_match": "\n4.",
"tier": "T1",
"year": "2010"
} |
Let $A B C D$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $A B$ and $C D$ are $F$ and $G$ respectively, and $\ell$ is the line through $G$ parallel to $A B$. The feet of the perpendiculars from $E$ onto the lines $\ell$ and $C D$ are $H$ and $K$, respectively. Pr... | The points $E, K, H, G$ are on the circle of diameter $G E$, so
$$
\angle E H K=\angle E G K
$$
Also, from $\angle D C A=\angle D B A$ and $\frac{C E}{C D}=\frac{B E}{B A}$ it follows
$$
\frac{C E}{C G}=\frac{2 C E}{C D}=\frac{2 B E}{B A}=\frac{B E}{B F},
$$
therefore $\triangle C G E \sim \triangle B F E$. In part... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $A B$ and $C D$ are $F$ and $G$ respectively, and $\ell$ is the line through $G$ parallel to $A B$. The feet of the perpendiculars from $E$ onto the lines $\ell$ and $C D$ are $H$ and $K$, respectively. Pr... | The points $E, K, H, G$ are on the circle of diameter $G E$, so
$$
\angle E H K=\angle E G K
$$
Also, from $\angle D C A=\angle D B A$ and $\frac{C E}{C D}=\frac{B E}{B A}$ it follows
$$
\frac{C E}{C G}=\frac{2 C E}{C D}=\frac{2 B E}{B A}=\frac{B E}{B F},
$$
therefore $\triangle C G E \sim \triangle B F E$. In part... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "# PROBLEM 1",
"resource_path": "Balkan_MO/segmented/en-2011-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2011"
} |
Given real numbers $x, y, z$ such that $x+y+z=0$, show that
$$
\frac{x(x+2)}{2 x^{2}+1}+\frac{y(y+2)}{2 y^{2}+1}+\frac{z(z+2)}{2 z^{2}+1} \geq 0 .
$$
When does equality hold? | The inequality is clear if $x y z=0$, in which case equality holds if and only if $x=y=z=0$.
Henceforth assume $x y z \neq 0$ and rewrite the inequality as
$$
\frac{(2 x+1)^{2}}{2 x^{2}+1}+\frac{(2 y+1)^{2}}{2 y^{2}+1}+\frac{(2 z+1)^{2}}{2 z^{2}+1} \geq 3 .
$$
Notice that (exactly) one of the products $x y, y z, z x... | proof | Yes | Yes | proof | Inequalities | Given real numbers $x, y, z$ such that $x+y+z=0$, show that
$$
\frac{x(x+2)}{2 x^{2}+1}+\frac{y(y+2)}{2 y^{2}+1}+\frac{z(z+2)}{2 z^{2}+1} \geq 0 .
$$
When does equality hold? | The inequality is clear if $x y z=0$, in which case equality holds if and only if $x=y=z=0$.
Henceforth assume $x y z \neq 0$ and rewrite the inequality as
$$
\frac{(2 x+1)^{2}}{2 x^{2}+1}+\frac{(2 y+1)^{2}}{2 y^{2}+1}+\frac{(2 z+1)^{2}}{2 z^{2}+1} \geq 3 .
$$
Notice that (exactly) one of the products $x y, y z, z x... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "# PROBLEM 2",
"resource_path": "Balkan_MO/segmented/en-2011-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2011"
} |
Let $S$ be a finite set of positive integers which has the following property: if $x$ is a member of $S$, then so are all positive divisors of $x$. A non-empty subset $T$ of $S$ is good if whenever $x, y \in T$ and $x<y$, the ratio $y / x$ is a power of a prime number. A non-empty subset $T$ of $S$ is bad if whenever $... | Notice first that a bad subset of $S$ contains at most one element from a good one, to deduce that a partition of $S$ into bad subsets has at least as many members as a maximal good subset.
Notice further that the elements of a good subset of $S$ must be among the terms of a geometric sequence whose ratio is a prime: ... | proof | Yes | Yes | proof | Combinatorics | Let $S$ be a finite set of positive integers which has the following property: if $x$ is a member of $S$, then so are all positive divisors of $x$. A non-empty subset $T$ of $S$ is good if whenever $x, y \in T$ and $x<y$, the ratio $y / x$ is a power of a prime number. A non-empty subset $T$ of $S$ is bad if whenever $... | Notice first that a bad subset of $S$ contains at most one element from a good one, to deduce that a partition of $S$ into bad subsets has at least as many members as a maximal good subset.
Notice further that the elements of a good subset of $S$ must be among the terms of a geometric sequence whose ratio is a prime: ... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "# PROBLEM 3",
"resource_path": "Balkan_MO/segmented/en-2011-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2011"
} |
Let $A B C D E F$ be a convex hexagon of area 1 , whose opposite sides are parallel. The lines $A B, C D$ and $E F$ meet in pairs to determine the vertices of a triangle. Similarly, the lines $B C, D E$ and $F A$ meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these tw... | Unless otherwise stated, throughout the proof indices take on values from 0 to 5 and are reduced modulo 6 . Label the vertices of the hexagon in circular order, $A_{0}, A_{1}, \cdots, A_{5}$, and let the lines of support of the alternate sides $A_{i} A_{i+1}$ and $A_{i+2} A_{i+3}$ meet at $B_{i}$. To show that the area... | proof | Yes | Yes | proof | Geometry | Let $A B C D E F$ be a convex hexagon of area 1 , whose opposite sides are parallel. The lines $A B, C D$ and $E F$ meet in pairs to determine the vertices of a triangle. Similarly, the lines $B C, D E$ and $F A$ meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these tw... | Unless otherwise stated, throughout the proof indices take on values from 0 to 5 and are reduced modulo 6 . Label the vertices of the hexagon in circular order, $A_{0}, A_{1}, \cdots, A_{5}$, and let the lines of support of the alternate sides $A_{i} A_{i+1}$ and $A_{i+2} A_{i+3}$ meet at $B_{i}$. To show that the area... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "# PROBLEM 4",
"resource_path": "Balkan_MO/segmented/en-2011-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2011"
} |
\quad$ Let $A, B$ and $C$ be points lying on a circle $\Gamma$ with centre $O$. Assume that $\angle A B C>90$. Let $D$ be the point of intersection of the line $A B$ with the line perpendicular to $A C$ at $C$. Let $l$ be the line through $D$ which is perpendicular to $A O$. Let $E$ be the point of intersection of $l$ ... | Let $\ell \cap A O=\{K\}$ and $G$ be the other end point of the diameter of $\Gamma$ through $A$. Then $D, C, G$ are collinear. Moreover, $E$ is the orthocenter of triangle $A D G$. Therefore $G E \perp A D$ and $G, E, B$ are collinear.
![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-1.jpg?height=1... | proof | Yes | Yes | proof | Geometry | \quad$ Let $A, B$ and $C$ be points lying on a circle $\Gamma$ with centre $O$. Assume that $\angle A B C>90$. Let $D$ be the point of intersection of the line $A B$ with the line perpendicular to $A C$ at $C$. Let $l$ be the line through $D$ which is perpendicular to $A O$. Let $E$ be the point of intersection of $l$ ... | Let $\ell \cap A O=\{K\}$ and $G$ be the other end point of the diameter of $\Gamma$ through $A$. Then $D, C, G$ are collinear. Moreover, $E$ is the orthocenter of triangle $A D G$. Therefore $G E \perp A D$ and $G, E, B$ are collinear.
![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-1.jpg?height=1... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\n1. ",
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2012"
} |
Prove that
$$
\sum_{c y c}(x+y) \sqrt{(z+x)(z+y)} \geq 4(x y+y z+z x),
$$
for all positive real numbers $x, y$ and $z$. | We will obtain the inequality by adding the inequalities
$$
(x+y) \sqrt{(z+x)(z+y)} \geq 2 x y+y z+z x
$$
for cyclic permutation of $x, y, z$.
Squaring both sides of this inequality we obtain
$$
(x+y)^{2}(z+x)(z+y) \geq 4 x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+4 x y^{2} z+4 x^{2} y z+2 x y z^{2}
$$
which is equivalent... | proof | Yes | Yes | proof | Inequalities | Prove that
$$
\sum_{c y c}(x+y) \sqrt{(z+x)(z+y)} \geq 4(x y+y z+z x),
$$
for all positive real numbers $x, y$ and $z$. | We will obtain the inequality by adding the inequalities
$$
(x+y) \sqrt{(z+x)(z+y)} \geq 2 x y+y z+z x
$$
for cyclic permutation of $x, y, z$.
Squaring both sides of this inequality we obtain
$$
(x+y)^{2}(z+x)(z+y) \geq 4 x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+4 x y^{2} z+4 x^{2} y z+2 x y z^{2}
$$
which is equivalent... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\n2. ",
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"solution_match": "\nSolution 1.",
"tier": "T1",
"year": "2012"
} |
Prove that
$$
\sum_{c y c}(x+y) \sqrt{(z+x)(z+y)} \geq 4(x y+y z+z x),
$$
for all positive real numbers $x, y$ and $z$. | For positive real numbers $x, y, z$ there exists a triangle with the side lengths $\sqrt{x+y}, \sqrt{y+z}, \sqrt{z+x}$ and the area $K=\sqrt{x y+y z+z x} / 2$.
The existence of the triangle is clear by simple checking of the triangle inequality. To prove the area formula, we have
$$
K=\frac{1}{2} \sqrt{x+y} \sqrt{z+x... | proof | Yes | Yes | proof | Inequalities | Prove that
$$
\sum_{c y c}(x+y) \sqrt{(z+x)(z+y)} \geq 4(x y+y z+z x),
$$
for all positive real numbers $x, y$ and $z$. | For positive real numbers $x, y, z$ there exists a triangle with the side lengths $\sqrt{x+y}, \sqrt{y+z}, \sqrt{z+x}$ and the area $K=\sqrt{x y+y z+z x} / 2$.
The existence of the triangle is clear by simple checking of the triangle inequality. To prove the area formula, we have
$$
K=\frac{1}{2} \sqrt{x+y} \sqrt{z+x... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\n2. ",
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"solution_match": "\nSolution 2.",
"tier": "T1",
"year": "2012"
} |
Let $n$ be a positive integer. Let $P_{n}=\left\{2^{n}, 2^{n-1} \cdot 3,2^{n-2} \cdot 3^{2}, \ldots, 3^{n}\right\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with ... | Let $\alpha=3 / 2$ so $1+\alpha>\alpha^{2}$.
Given $y$, we construct $Y$ algorithmically. Let $Y=\varnothing$ and of course $S_{\varnothing}=0$. For $i=0$ to $m$, perform the following operation:
$$
\text { If } S_{Y}+2^{i} 3^{m-i} \leq y \text {, then replace } Y \text { by } Y \cup\left\{2^{i} 3^{m-i}\right\}
$$
Wh... | proof | Yes | Yes | proof | Number Theory | Let $n$ be a positive integer. Let $P_{n}=\left\{2^{n}, 2^{n-1} \cdot 3,2^{n-2} \cdot 3^{2}, \ldots, 3^{n}\right\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with ... | Let $\alpha=3 / 2$ so $1+\alpha>\alpha^{2}$.
Given $y$, we construct $Y$ algorithmically. Let $Y=\varnothing$ and of course $S_{\varnothing}=0$. For $i=0$ to $m$, perform the following operation:
$$
\text { If } S_{Y}+2^{i} 3^{m-i} \leq y \text {, then replace } Y \text { by } Y \cup\left\{2^{i} 3^{m-i}\right\}
$$
Wh... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\n3. ",
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"solution_match": "\nSolution 1.",
"tier": "T1",
"year": "2012"
} |
Let $n$ be a positive integer. Let $P_{n}=\left\{2^{n}, 2^{n-1} \cdot 3,2^{n-2} \cdot 3^{2}, \ldots, 3^{n}\right\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with ... | Note that $3^{m+1}-2^{m+1}=(3-2)\left(3^{m}+3^{m-1} \cdot 2+\cdots+3 \cdot 2^{m-1}+2^{m}\right)=S_{P_{m}}$. Dividing every element of $P_{m}$ by $2^{m}$ gives us the following equivalent problem:
Let $m$ be a positive integer, $a=3 / 2$, and $Q_{m}=\left\{1, a, a^{2}, \ldots, a^{m}\right\}$. Show that for any real num... | proof | Yes | Yes | proof | Number Theory | Let $n$ be a positive integer. Let $P_{n}=\left\{2^{n}, 2^{n-1} \cdot 3,2^{n-2} \cdot 3^{2}, \ldots, 3^{n}\right\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with ... | Note that $3^{m+1}-2^{m+1}=(3-2)\left(3^{m}+3^{m-1} \cdot 2+\cdots+3 \cdot 2^{m-1}+2^{m}\right)=S_{P_{m}}$. Dividing every element of $P_{m}$ by $2^{m}$ gives us the following equivalent problem:
Let $m$ be a positive integer, $a=3 / 2$, and $Q_{m}=\left\{1, a, a^{2}, \ldots, a^{m}\right\}$. Show that for any real num... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\n3. ",
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"solution_match": "\nSolution 2.",
"tier": "T1",
"year": "2012"
} |
\quad$ Let $\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$such that the following conditions both hold:
(i) $f(n!)=f(n)$ ! for every positive integer $n$,
(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers. | There are three such functions: the constant functions 1, 2 and the identity function $\mathrm{id}_{\mathbf{Z}^{+}}$. These functions clearly satisfy the conditions in the hypothesis. Let us prove that there are only ones.
Consider such a function $f$ and suppose that it has a fixed point $a \geq 3$, that is $f(a)=a$.... | proof | Yes | Yes | math-word-problem | Number Theory | \quad$ Let $\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$such that the following conditions both hold:
(i) $f(n!)=f(n)$ ! for every positive integer $n$,
(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers. | There are three such functions: the constant functions 1, 2 and the identity function $\mathrm{id}_{\mathbf{Z}^{+}}$. These functions clearly satisfy the conditions in the hypothesis. Let us prove that there are only ones.
Consider such a function $f$ and suppose that it has a fixed point $a \geq 3$, that is $f(a)=a$.... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\n4. ",
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"solution_match": "\nSolution 1.",
"tier": "T1",
"year": "2012"
} |
\quad$ Let $\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$such that the following conditions both hold:
(i) $f(n!)=f(n)$ ! for every positive integer $n$,
(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers. | Note first that if $f\left(n_{0}\right)=n_{0}$, then $m-n_{0} \mid f(m)-m$ for all $m \in \mathbf{Z}^{+}$. If $f\left(n_{0}\right)=n_{0}$ for infinitely many $n_{0} \in \mathbf{Z}^{+}$, then $f(m)-m$ has infinitely many divisors, hence $f(m)=m$ for all $m \in \mathbf{Z}^{+}$. On the other hand, if $f\left(n_{0}\right)=... | f \equiv 1, f \equiv 2, f=\mathrm{id}_{\mathbf{z}^{+}} | Yes | Yes | math-word-problem | Number Theory | \quad$ Let $\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$such that the following conditions both hold:
(i) $f(n!)=f(n)$ ! for every positive integer $n$,
(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers. | Note first that if $f\left(n_{0}\right)=n_{0}$, then $m-n_{0} \mid f(m)-m$ for all $m \in \mathbf{Z}^{+}$. If $f\left(n_{0}\right)=n_{0}$ for infinitely many $n_{0} \in \mathbf{Z}^{+}$, then $f(m)-m$ has infinitely many divisors, hence $f(m)=m$ for all $m \in \mathbf{Z}^{+}$. On the other hand, if $f\left(n_{0}\right)=... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\n4. ",
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"solution_match": "\nSolution 2.",
"tier": "T1",
"year": "2012"
} |
We denote the angles of the triangle by $\alpha, \beta$ and $\gamma$ as usual. Since $\angle K M P=90^{\circ}-\frac{\beta}{2}$, it suffices to prove that $\angle K L P=90^{\circ}+\frac{\beta}{2}$, which is equivalent to $\angle K L C=\frac{\beta}{2}$.
Let $I$ be the incenter of triangle $A B C$ and let $D$ be the tange... | Consider the graph $\mathcal{G}$ whose vertices are the contestants, where there is an edge between two contestants if and only if they are not friends.
Lemma. There is a vertex in graph $\mathcal{G}$ with degree at most 2 .
Proof. Suppose that each vertex has a degree at least three. Consider the longest induced path... | proof | Yes | Incomplete | proof | Geometry | We denote the angles of the triangle by $\alpha, \beta$ and $\gamma$ as usual. Since $\angle K M P=90^{\circ}-\frac{\beta}{2}$, it suffices to prove that $\angle K L P=90^{\circ}+\frac{\beta}{2}$, which is equivalent to $\angle K L C=\frac{\beta}{2}$.
Let $I$ be the incenter of triangle $A B C$ and let $D$ be the tange... | Consider the graph $\mathcal{G}$ whose vertices are the contestants, where there is an edge between two contestants if and only if they are not friends.
Lemma. There is a vertex in graph $\mathcal{G}$ with degree at most 2 .
Proof. Suppose that each vertex has a degree at least three. Consider the longest induced path... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\n1.",
"resource_path": "Balkan_MO/segmented/en-2013-BMO-type2.jsonl",
"solution_match": "\n4.",
"tier": "T1",
"year": "2013"
} |
Determine all positive integers $x, y$ and $z$ such that
$$
x^{5}+4^{y}=2013^{z}
$$ | Reducing modulo 11 yields $x^{5}+4^{y} \equiv 0(\bmod 11)$, where $x^{5} \equiv \pm 1(\bmod 11)$, so we also have $4^{y} \equiv \pm 1(\bmod 11)$. Congruence $4^{y} \equiv-1(\bmod 11)$ does not hold for any $y$, whereas $4^{y} \equiv 1(\bmod 11)$ holds if and only if $5 \mid y$.
Setting $t=4^{y / 5}$, the equation becom... | not found | Yes | Yes | math-word-problem | Number Theory | Determine all positive integers $x, y$ and $z$ such that
$$
x^{5}+4^{y}=2013^{z}
$$ | Reducing modulo 11 yields $x^{5}+4^{y} \equiv 0(\bmod 11)$, where $x^{5} \equiv \pm 1(\bmod 11)$, so we also have $4^{y} \equiv \pm 1(\bmod 11)$. Congruence $4^{y} \equiv-1(\bmod 11)$ does not hold for any $y$, whereas $4^{y} \equiv 1(\bmod 11)$ holds if and only if $5 \mid y$.
Setting $t=4^{y / 5}$, the equation becom... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\n2.",
"resource_path": "Balkan_MO/segmented/en-2013-BMO-type2.jsonl",
"solution_match": "\n2.",
"tier": "T1",
"year": "2013"
} |
Let $S$ be the set of positive real numbers. Find all functions $f: S^{3} \rightarrow S$ such that, for all positive real numbers $x, y, z$ and $k$, the following three conditions are satisfied:
(i) $x f(x, y, z)=z f(z, y, x)$;
(ii) $f\left(x, k y, k^{2} z\right)=k f(x, y, z)$;
(iii) $f(1, k, k+1)=k+1$.
(United Kingdom... | It follows from the properties of function $f$ that, for all $x, y, z, a, b>0$, $f\left(a^{2} x, a b y, b^{2} z\right)=b f\left(a^{2} x, a y, z\right)=b \cdot \frac{z}{a^{2} x} f\left(z, a y, a^{2} x\right)=\frac{b z}{a x} f(z, y, x)=\frac{b}{a} f(x, y, z)$.
We shall choose $a$ and $b$ in such a way that the triple $\... | \frac{y+\sqrt{y^{2}+4 x z}}{2 x} | Yes | Yes | math-word-problem | Algebra | Let $S$ be the set of positive real numbers. Find all functions $f: S^{3} \rightarrow S$ such that, for all positive real numbers $x, y, z$ and $k$, the following three conditions are satisfied:
(i) $x f(x, y, z)=z f(z, y, x)$;
(ii) $f\left(x, k y, k^{2} z\right)=k f(x, y, z)$;
(iii) $f(1, k, k+1)=k+1$.
(United Kingdom... | It follows from the properties of function $f$ that, for all $x, y, z, a, b>0$, $f\left(a^{2} x, a b y, b^{2} z\right)=b f\left(a^{2} x, a y, z\right)=b \cdot \frac{z}{a^{2} x} f\left(z, a y, a^{2} x\right)=\frac{b z}{a x} f(z, y, x)=\frac{b}{a} f(x, y, z)$.
We shall choose $a$ and $b$ in such a way that the triple $\... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\n3.",
"resource_path": "Balkan_MO/segmented/en-2013-BMO-type2.jsonl",
"solution_match": "\n3.",
"tier": "T1",
"year": "2013"
} |
In a mathematical competition some competitors are friends; friendship is always mutual, that is to say that when $A$ is a friend of $B$, then also $B$ is a friend of $A$. We say that $n \geq 3$ different competitors $A_{1}, A_{2}, \ldots, A_{n}$ form a weakly-friendly cycle if $A_{i}$ is not a friend of $A_{i+1}$ for ... | Consider the graph $\mathcal{G}$ whose vertices are the contestants, where there is an edge between two contestants if and only if they are not friends.
Lemma. There is a vertex in graph $\mathcal{G}$ with degree at most 2 .
Proof. Suppose that each vertex has a degree at least three. Consider the longest induced path... | proof | Yes | Yes | proof | Combinatorics | In a mathematical competition some competitors are friends; friendship is always mutual, that is to say that when $A$ is a friend of $B$, then also $B$ is a friend of $A$. We say that $n \geq 3$ different competitors $A_{1}, A_{2}, \ldots, A_{n}$ form a weakly-friendly cycle if $A_{i}$ is not a friend of $A_{i+1}$ for ... | Consider the graph $\mathcal{G}$ whose vertices are the contestants, where there is an edge between two contestants if and only if they are not friends.
Lemma. There is a vertex in graph $\mathcal{G}$ with degree at most 2 .
Proof. Suppose that each vertex has a degree at least three. Consider the longest induced path... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\n4.",
"resource_path": "Balkan_MO/segmented/en-2013-BMO-type2.jsonl",
"solution_match": "\n4.",
"tier": "T1",
"year": "2013"
} |
Let $x, y$ and $z$ be positive real numbers such that $x y+y z+z x=3 x y z$. Prove that
$$
x^{2} y+y^{2} z+z^{2} x \geq 2(x+y+z)-3
$$
and determine when equality holds. | The given condition can be rearranged to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$. Using this, we obtain:
$$
\begin{aligned}
x^{2} y+y^{2} z+z^{2} x-2(x+y+z)+3 & =x^{2} y-2 x+\frac{1}{y}+y^{2} z-2 y+\frac{1}{z}+z^{2} x-2 x+\frac{1}{x}= \\
& =y\left(x-\frac{1}{y}\right)^{2}+z\left(y-\frac{1}{z}\right)^{2}+x\left(z-\frac... | proof | Yes | Yes | proof | Inequalities | Let $x, y$ and $z$ be positive real numbers such that $x y+y z+z x=3 x y z$. Prove that
$$
x^{2} y+y^{2} z+z^{2} x \geq 2(x+y+z)-3
$$
and determine when equality holds. | The given condition can be rearranged to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$. Using this, we obtain:
$$
\begin{aligned}
x^{2} y+y^{2} z+z^{2} x-2(x+y+z)+3 & =x^{2} y-2 x+\frac{1}{y}+y^{2} z-2 y+\frac{1}{z}+z^{2} x-2 x+\frac{1}{x}= \\
& =y\left(x-\frac{1}{y}\right)^{2}+z\left(y-\frac{1}{z}\right)^{2}+x\left(z-\frac... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "Balkan_MO/segmented/en-2014-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2014"
} |
A special number is a positive integer $n$ for which there exist positive integers $a, b, c$ and $d$ with
$$
n=\frac{a^{3}+2 b^{3}}{c^{3}+2 d^{3}}
$$
Prove that:
(a) there are infinitely many special numbers;
(b) 2014 is not a special number. | (a) Every perfect cube $k^{3}$ of a positive integer is special because we can write
$$
k^{3}=k^{3} \frac{a^{3}+2 b^{3}}{a^{3}+2 b^{3}}=\frac{(k a)^{3}+2(k b)^{3}}{a^{3}+2 b^{3}}
$$
for some positive integers $a, b$.
(b) Observe that $2014=2.19 .53$. If 2014 is special, then we have,
$$
x^{3}+2 y^{3}=2014\left(u^{3}... | proof | Yes | Yes | proof | Number Theory | A special number is a positive integer $n$ for which there exist positive integers $a, b, c$ and $d$ with
$$
n=\frac{a^{3}+2 b^{3}}{c^{3}+2 d^{3}}
$$
Prove that:
(a) there are infinitely many special numbers;
(b) 2014 is not a special number. | (a) Every perfect cube $k^{3}$ of a positive integer is special because we can write
$$
k^{3}=k^{3} \frac{a^{3}+2 b^{3}}{a^{3}+2 b^{3}}=\frac{(k a)^{3}+2(k b)^{3}}{a^{3}+2 b^{3}}
$$
for some positive integers $a, b$.
(b) Observe that $2014=2.19 .53$. If 2014 is special, then we have,
$$
x^{3}+2 y^{3}=2014\left(u^{3}... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2014-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2014"
} |
Let $A B C D$ be a trapezium inscribed in a circle $\Gamma$ with diameter $A B$. Let $E$ be the intersection point of the diagonals $A C$ and $B D$. The circle with center $B$ and radius $B E$ meets $\Gamma$ at the points $K$ and $L$, where $K$ is on the same side of $A B$ as $C$. The line perpendicular to $B D$ at $E$... | Since $A B \| C D$, we have that $A B C D$ is isosceles trapezium. Let $O$ be the center of $k$ and $E M$ meets $A B$ at point $Q$. Then, from the right angled triangle $B E Q$, we have $B E^{2}=B O \cdot B Q$. Since $B E=B K$, we get $B K^{2}=B O \cdot B Q$ (1). Suppose that $K L$ meets $A B$ at $P$. Then, from the ri... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a trapezium inscribed in a circle $\Gamma$ with diameter $A B$. Let $E$ be the intersection point of the diagonals $A C$ and $B D$. The circle with center $B$ and radius $B E$ meets $\Gamma$ at the points $K$ and $L$, where $K$ is on the same side of $A B$ as $C$. The line perpendicular to $B D$ at $E$... | Since $A B \| C D$, we have that $A B C D$ is isosceles trapezium. Let $O$ be the center of $k$ and $E M$ meets $A B$ at point $Q$. Then, from the right angled triangle $B E Q$, we have $B E^{2}=B O \cdot B Q$. Since $B E=B K$, we get $B K^{2}=B O \cdot B Q$ (1). Suppose that $K L$ meets $A B$ at $P$. Then, from the ri... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "Balkan_MO/segmented/en-2014-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2014"
} |
Let $n$ be a positive integer. A regular hexagon with side length $n$ is divided into equilateral triangles with side length 1 by lines parallel to its sides.
Find the number of regular hexagons all of whose vertices are among the vertices of the equilateral triangles. | By a lattice hexagon we will mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we will call the enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ of sid... | \left(\frac{n(n+1)}{2}\right)^{2} | Yes | Yes | math-word-problem | Combinatorics | Let $n$ be a positive integer. A regular hexagon with side length $n$ is divided into equilateral triangles with side length 1 by lines parallel to its sides.
Find the number of regular hexagons all of whose vertices are among the vertices of the equilateral triangles. | By a lattice hexagon we will mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we will call the enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ of sid... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "Balkan_MO/segmented/en-2014-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2014"
} |
Let $a, b$ and $c$ be positive real numbers. Prove that
$$
a^{3} b^{6}+b^{3} c^{6}+c^{3} a^{6}+3 a^{3} b^{3} c^{3} \geq a b c\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\right)+a^{2} b^{2} c^{2}\left(a^{3}+b^{3}+c^{3}\right)
$$ | After dividing both sides of the given inequality by $a^{3} b^{3} c^{3}$ it becomes
$$
\left(\frac{b}{c}\right)^{3}+\left(\frac{c}{a}\right)^{3}+\left(\frac{a}{b}\right)^{3}+3 \geq\left(\frac{a}{c} \cdot \frac{b}{c}+\frac{b}{a} \cdot \frac{c}{a}+\frac{c}{b} \cdot \frac{a}{b}\right)+\left(\frac{a}{b} \cdot \frac{a}{c}+... | proof | Yes | Yes | proof | Inequalities | Let $a, b$ and $c$ be positive real numbers. Prove that
$$
a^{3} b^{6}+b^{3} c^{6}+c^{3} a^{6}+3 a^{3} b^{3} c^{3} \geq a b c\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\right)+a^{2} b^{2} c^{2}\left(a^{3}+b^{3}+c^{3}\right)
$$ | After dividing both sides of the given inequality by $a^{3} b^{3} c^{3}$ it becomes
$$
\left(\frac{b}{c}\right)^{3}+\left(\frac{c}{a}\right)^{3}+\left(\frac{a}{b}\right)^{3}+3 \geq\left(\frac{a}{c} \cdot \frac{b}{c}+\frac{b}{a} \cdot \frac{c}{a}+\frac{c}{b} \cdot \frac{a}{b}\right)+\left(\frac{a}{b} \cdot \frac{a}{c}+... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "Balkan_MO/segmented/en-2015-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2015"
} |
Let $A B C$ be a scalene triangle with incentre $I$ and circumcircle ( $\omega$ ). The lines $A I, B I, C I$ intersect $(\omega)$ for the second time at the points $D, E, F$, respectively. The lines through $I$ parallel to the sides $B C, A C, A B$ intersect the lines $E F, D F, D E$ at the points $K, L, M$, respective... | First we will prove that $K A$ is tangent to $(\omega)$.
Indeed, it is a well-known fact that $F A=F B=F I$ and $E A=E C=E I$, so $F E$ is the perpendicular bisector of $A I$. It follows that $K A=K I$ and
$$
\angle K A F=\angle K I F=\angle F C B=\angle F E B=\angle F E A,
$$
so $K A$ is tangent to $(\omega)$. Simil... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a scalene triangle with incentre $I$ and circumcircle ( $\omega$ ). The lines $A I, B I, C I$ intersect $(\omega)$ for the second time at the points $D, E, F$, respectively. The lines through $I$ parallel to the sides $B C, A C, A B$ intersect the lines $E F, D F, D E$ at the points $K, L, M$, respective... | First we will prove that $K A$ is tangent to $(\omega)$.
Indeed, it is a well-known fact that $F A=F B=F I$ and $E A=E C=E I$, so $F E$ is the perpendicular bisector of $A I$. It follows that $K A=K I$ and
$$
\angle K A F=\angle K I F=\angle F C B=\angle F E B=\angle F E A,
$$
so $K A$ is tangent to $(\omega)$. Simil... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2015-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2015"
} |
A jury of 3366 film critics are judging the Oscars. Each critic makes a single vote for his favourite actor, and a single vote for his favourite actress. It turns out that for every integer $n \in\{1,2, \ldots, 100\}$ there is an actor or actress who has been voted for exactly $n$ times. Show that there are two critics... | Let us assume that every critic votes for a different pair of actor and actress. We'll arrive at a contradiction proving the required result. Indeed:
Call the vote of each critic, i.e his choice for the pair of an actor and an actress, as a double-vote, and call as a single-vote each one of the two choices he makes, i... | proof | Yes | Yes | proof | Combinatorics | A jury of 3366 film critics are judging the Oscars. Each critic makes a single vote for his favourite actor, and a single vote for his favourite actress. It turns out that for every integer $n \in\{1,2, \ldots, 100\}$ there is an actor or actress who has been voted for exactly $n$ times. Show that there are two critics... | Let us assume that every critic votes for a different pair of actor and actress. We'll arrive at a contradiction proving the required result. Indeed:
Call the vote of each critic, i.e his choice for the pair of an actor and an actress, as a double-vote, and call as a single-vote each one of the two choices he makes, i... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "Balkan_MO/segmented/en-2015-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2015"
} |
Prove that among any 20 consecutive positive integers there exists an integer $d$ such that for each positive integer $n$ we have the inequality
$$
n \sqrt{d}\{n \sqrt{d}\}>\frac{5}{2}
$$
where $\{x\}$ denotes the fractional part of the real number $x$. The fractional part of a real number $x$ is $x$ minus the greate... | Among the given numbers there is a number of the form $20 k+15=5(4 k+3)$. We shall prove that $d=5(4 k+3)$ satisfies the statement's condition. Since $d \equiv-1(\bmod 4)$, it follows that $d$ is not a perfect square, and thus for any $n \in \mathbb{N}$ there exists $a \in \mathbb{N}$ such that $a+1>n \sqrt{d}>a$, that... | proof | Yes | Yes | proof | Number Theory | Prove that among any 20 consecutive positive integers there exists an integer $d$ such that for each positive integer $n$ we have the inequality
$$
n \sqrt{d}\{n \sqrt{d}\}>\frac{5}{2}
$$
where $\{x\}$ denotes the fractional part of the real number $x$. The fractional part of a real number $x$ is $x$ minus the greate... | Among the given numbers there is a number of the form $20 k+15=5(4 k+3)$. We shall prove that $d=5(4 k+3)$ satisfies the statement's condition. Since $d \equiv-1(\bmod 4)$, it follows that $d$ is not a perfect square, and thus for any $n \in \mathbb{N}$ there exists $a \in \mathbb{N}$ such that $a+1>n \sqrt{d}>a$, that... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "Balkan_MO/segmented/en-2015-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2015"
} |
Find all injective functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for every real number $x$ and every positive integer $n$,
$$
\left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))\right|<2016
$$ | From the condition of the problem we get
$$
\left|\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\right|<2016
$$
Then
$$
\begin{aligned}
& |n(f(x+n+1)-f(f(x+n)))| \\
= & \left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))-\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\right| \\
< & 2 \cdot 2016=4032
\end{aligned}
$$
implying
$$
|f(x+n+1)-f... | f(y)=y+1 | Yes | Yes | math-word-problem | Algebra | Find all injective functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for every real number $x$ and every positive integer $n$,
$$
\left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))\right|<2016
$$ | From the condition of the problem we get
$$
\left|\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\right|<2016
$$
Then
$$
\begin{aligned}
& |n(f(x+n+1)-f(f(x+n)))| \\
= & \left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))-\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\right| \\
< & 2 \cdot 2016=4032
\end{aligned}
$$
implying
$$
|f(x+n+1)-f... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "# Problem 1.",
"resource_path": "Balkan_MO/segmented/en-2016-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Let $A B C D$ be a cyclic quadrilateral with $A B<C D$. The diagonals intersect at the point $F$ and lines $A D$ and $B C$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $A D$ and $B C$ respectively, and let $M, S$ and $T$ be the midpoints of $E F, C F$ and $D F$ respectivel... | Let $N$ be the midpoint of $C D$. We will prove that the circumcircles of the triangles $M K T$ and $M L S$ pass through $N$. (1)
First will prove that the circumcircle of $M L S$ passes through $N$.
Let $Q$ be the midpoint of $E C$. Note that the circumcircle of $M L S$ is the Euler circle (2) of the triangle $E F C$,... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a cyclic quadrilateral with $A B<C D$. The diagonals intersect at the point $F$ and lines $A D$ and $B C$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $A D$ and $B C$ respectively, and let $M, S$ and $T$ be the midpoints of $E F, C F$ and $D F$ respectivel... | Let $N$ be the midpoint of $C D$. We will prove that the circumcircles of the triangles $M K T$ and $M L S$ pass through $N$. (1)
First will prove that the circumcircle of $M L S$ passes through $N$.
Let $Q$ be the midpoint of $E C$. Note that the circumcircle of $M L S$ is the Euler circle (2) of the triangle $E F C$,... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "# Problem 2.",
"resource_path": "Balkan_MO/segmented/en-2016-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Find all monic polynomials $f$ with integer coefficients satisfying the following condition: there exists a positive integer $N$ such that $p$ divides $2(f(p)!)+1$ for every prime $p>N$ for which $f(p)$ is a positive integer.
Note: A monic polynomial has leading coefficient equal to 1. | If $f$ is a constant polynomial then it's obvious that the condition cannot hold for
$$
p \geq 5 \text { since } f(p)=1
$$
From the divisibility relation $p \mid 2(f(p))$ ! +1 we conclude that:
$$
f(p)<p, \text { for all primes } p>N \quad(*)
$$
In fact, if for some prime number $p$ we have $f(p) \geq p$, then $p \... | f(x)=x-3 | Yes | Yes | math-word-problem | Number Theory | Find all monic polynomials $f$ with integer coefficients satisfying the following condition: there exists a positive integer $N$ such that $p$ divides $2(f(p)!)+1$ for every prime $p>N$ for which $f(p)$ is a positive integer.
Note: A monic polynomial has leading coefficient equal to 1. | If $f$ is a constant polynomial then it's obvious that the condition cannot hold for
$$
p \geq 5 \text { since } f(p)=1
$$
From the divisibility relation $p \mid 2(f(p))$ ! +1 we conclude that:
$$
f(p)<p, \text { for all primes } p>N \quad(*)
$$
In fact, if for some prime number $p$ we have $f(p) \geq p$, then $p \... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "# Problem 3.",
"resource_path": "Balkan_MO/segmented/en-2016-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
The plane is divided into unit squares by two sets of parallel lines, forming an infinite grid. Each unit square is coloured with one of 1201 colours so that no rectangle with perimeter 100 contains two squares of the same colour. Show that no rectangle of size $1 \times 1201$ or $1201 \times 1$ contains two squares of... | Let the centers of the unit squares be the integer points in the plane, and denote each unit square by the coordinates of its center.
Consider the set $D$ of all unit squares $(x, y)$ such that $|x|+|y| \leq 24$. Any integer translate of $D$ is called a diamond.
Since any two unit squares that belong to the same diamon... | proof | Yes | Yes | proof | Combinatorics | The plane is divided into unit squares by two sets of parallel lines, forming an infinite grid. Each unit square is coloured with one of 1201 colours so that no rectangle with perimeter 100 contains two squares of the same colour. Show that no rectangle of size $1 \times 1201$ or $1201 \times 1$ contains two squares of... | Let the centers of the unit squares be the integer points in the plane, and denote each unit square by the coordinates of its center.
Consider the set $D$ of all unit squares $(x, y)$ such that $|x|+|y| \leq 24$. Any integer translate of $D$ is called a diamond.
Since any two unit squares that belong to the same diamon... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "# Problem 4.",
"resource_path": "Balkan_MO/segmented/en-2016-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Find all ordered pairs $(x, y)$ of positive integers such that:
$$
x^{3}+y^{3}=x^{2}+42 x y+y^{2} \text {. }
$$ | Possible initial thoughts about this equation might include:
(i) I can factorise the sum of cubes on the left.
(ii) How can I use the 42?
(iii) The left is cubic and the right is quadratic, so if $x$ or $y$ is very large there will be no solutions.
The first two might lead us to rewrite the equation as $(x+y)\left(x^{... | (22,22),(1,7),(7,1) | Yes | Yes | math-word-problem | Algebra | Find all ordered pairs $(x, y)$ of positive integers such that:
$$
x^{3}+y^{3}=x^{2}+42 x y+y^{2} \text {. }
$$ | Possible initial thoughts about this equation might include:
(i) I can factorise the sum of cubes on the left.
(ii) How can I use the 42?
(iii) The left is cubic and the right is quadratic, so if $x$ or $y$ is very large there will be no solutions.
The first two might lead us to rewrite the equation as $(x+y)\left(x^{... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\n1. ",
"resource_path": "Balkan_MO/segmented/en-2017-BMO-type3.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be an acute triangle with with $A B<A C$ and let $\Gamma$ be its circumcircle. Let the tangents to $\Gamma$ at $B$ and $C$ be $t_{B}$ and $t_{C}$ respectively and let their point of intersection be $L$. The line through $B$ parallel to $A C$ intersects $t_{C}$ at $D$. The line through $C$ parallel to $A B$ ... | How we attack this problem depends on how much triangle geometry we can effortlessly recall - a good knowledge of some standard results helps a great deal.
We might instantly note that $A L$ is a symmedian of $A B C$, and so divides the line $B C$ in the ratio $c^{2}: b^{2}$. Now the plan is to show that $S T$ also di... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle with with $A B<A C$ and let $\Gamma$ be its circumcircle. Let the tangents to $\Gamma$ at $B$ and $C$ be $t_{B}$ and $t_{C}$ respectively and let their point of intersection be $L$. The line through $B$ parallel to $A C$ intersects $t_{C}$ at $D$. The line through $C$ parallel to $A B$ ... | How we attack this problem depends on how much triangle geometry we can effortlessly recall - a good knowledge of some standard results helps a great deal.
We might instantly note that $A L$ is a symmedian of $A B C$, and so divides the line $B C$ in the ratio $c^{2}: b^{2}$. Now the plan is to show that $S T$ also di... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\n2. ",
"resource_path": "Balkan_MO/segmented/en-2017-BMO-type3.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2017"
} |
Let $\mathbb{N}$ be the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:
$$
n+f(m) \text { divides } f(n)+n f(m)
$$ | The striking thing about this problem is that the relation concerns divisibility rather than equality. How can we exploit this? We are given that $n+f(m) \mid f(n)+n f(m)$ but we can certainly add or subtract multiples of the left hand side from the right hand side and preserve the divisibility. This leads to a key ide... | f(n)=n^2 \text{ or } f(n)=1 | Yes | Yes | math-word-problem | Number Theory | Let $\mathbb{N}$ be the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:
$$
n+f(m) \text { divides } f(n)+n f(m)
$$ | The striking thing about this problem is that the relation concerns divisibility rather than equality. How can we exploit this? We are given that $n+f(m) \mid f(n)+n f(m)$ but we can certainly add or subtract multiples of the left hand side from the right hand side and preserve the divisibility. This leads to a key ide... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\n3. ",
"resource_path": "Balkan_MO/segmented/en-2017-BMO-type3.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2017"
} |
There are $n>2$ students sitting at a round table. Initially each student has exactly one candy. At each step, each student chooses one of the following operations:
(a) Pass one candy to the student on their left or the student on their right.
(b) Divide all their candies into two, possibly empty, sets and pass one set... | One possible initial reaction to this problem is that there is rather too much movement of caramels ${ }^{2}$ at each step to keep track of easily. This leads to the question: 'How little can I do in, say, two steps?' If every student passes all their caramels left on one step using (b), and all their caramels right on... | not found | Yes | Incomplete | math-word-problem | Combinatorics | There are $n>2$ students sitting at a round table. Initially each student has exactly one candy. At each step, each student chooses one of the following operations:
(a) Pass one candy to the student on their left or the student on their right.
(b) Divide all their candies into two, possibly empty, sets and pass one set... | One possible initial reaction to this problem is that there is rather too much movement of caramels ${ }^{2}$ at each step to keep track of easily. This leads to the question: 'How little can I do in, say, two steps?' If every student passes all their caramels left on one step using (b), and all their caramels right on... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\n4. ",
"resource_path": "Balkan_MO/segmented/en-2017-BMO-type3.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2017"
} |
A quadrilateral $A B C D$ is inscribed in a circle $k$, where $A B>C D$ and $A B$ is not parallel to $C D$. Point $M$ is the intersection of the diagonals $A C$ and $B D$ and the perpendicular from $M$ to $A B$ intersects the segment $A B$ at the point $E$. If $E M$ bisects the angle $C E D$, prove that $A B$ is a diam... | Let the line through $M$ parallel to $A B$ meet the segments $A D, D H, B C, C H$ at points $K, P, L, Q$, respectively. Triangle $H P Q$ is isosceles, so $M P=M Q$. Now from
$$
\frac{M P}{B H}=\frac{D M}{D B}=\frac{K M}{A B} \quad \text { and } \quad \frac{M Q}{A H}=\frac{C M}{C A}=\frac{M L}{A B}
$$
we obtain $A H /... | proof | Yes | Yes | proof | Geometry | A quadrilateral $A B C D$ is inscribed in a circle $k$, where $A B>C D$ and $A B$ is not parallel to $C D$. Point $M$ is the intersection of the diagonals $A C$ and $B D$ and the perpendicular from $M$ to $A B$ intersects the segment $A B$ at the point $E$. If $E M$ bisects the angle $C E D$, prove that $A B$ is a diam... | Let the line through $M$ parallel to $A B$ meet the segments $A D, D H, B C, C H$ at points $K, P, L, Q$, respectively. Triangle $H P Q$ is isosceles, so $M P=M Q$. Now from
$$
\frac{M P}{B H}=\frac{D M}{D B}=\frac{K M}{A B} \quad \text { and } \quad \frac{M Q}{A H}=\frac{C M}{C A}=\frac{M L}{A B}
$$
we obtain $A H /... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\n1. ",
"resource_path": "Balkan_MO/segmented/en-2018-BMO-type3.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2018"
} |
Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the p... | Answer: $q=1$.
Let $x_{A}^{(n)}$ (resp. $x_{B}^{(n)}$ ) be the $x$-coordinates of the first (resp. second) ant's position after $n$ minutes. Then $x_{A}^{(n)}-x_{A}^{(n-1)} \in\left\{q^{n},-q^{n}, 0\right\}$, and so $x_{A}^{(n)}, x_{B}^{(n)}$ are given by polynomials in $q$ with coefficients in $\{-1,0,1\}$. So if the ... | q=1 | Yes | Yes | math-word-problem | Number Theory | Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the p... | Answer: $q=1$.
Let $x_{A}^{(n)}$ (resp. $x_{B}^{(n)}$ ) be the $x$-coordinates of the first (resp. second) ant's position after $n$ minutes. Then $x_{A}^{(n)}-x_{A}^{(n-1)} \in\left\{q^{n},-q^{n}, 0\right\}$, and so $x_{A}^{(n)}, x_{B}^{(n)}$ are given by polynomials in $q$ with coefficients in $\{-1,0,1\}$. So if the ... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\n2. ",
"resource_path": "Balkan_MO/segmented/en-2018-BMO-type3.jsonl",
"solution_match": "\nSolution 1.",
"tier": "T1",
"year": "2018"
} |
Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the p... | Consider the ants' positions $\alpha_{k}$ and $\beta_{k}$ after $k$ steps in the complex plane, assuming that their initial positions are at the origin and that all steps are parallel to one of the axes. We have $\alpha_{k+1}-\alpha_{k}=a_{k} q^{k}$ and $\beta_{k+1}-\beta_{k}=b_{k} q^{k}$ with $a_{k}, b_{k} \in\{1,-1, ... | q=1 | Yes | Yes | math-word-problem | Number Theory | Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the p... | Consider the ants' positions $\alpha_{k}$ and $\beta_{k}$ after $k$ steps in the complex plane, assuming that their initial positions are at the origin and that all steps are parallel to one of the axes. We have $\alpha_{k+1}-\alpha_{k}=a_{k} q^{k}$ and $\beta_{k+1}-\beta_{k}=b_{k} q^{k}$ with $a_{k}, b_{k} \in\{1,-1, ... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\n2. ",
"resource_path": "Balkan_MO/segmented/en-2018-BMO-type3.jsonl",
"solution_match": "\nSolution 2.",
"tier": "T1",
"year": "2018"
} |
Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of the coins of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins.
Det... | By $v_{2}(n)$ we denote the largest nonnegative integer $r$ such that $2^{r} \mid n$.
A position $(a, b)$ (i.e. two piles of sizes $a$ and $b$ ) is said to be $k$-happy if $v_{2}(a)=v_{2}(b)=k$ for some integer $k \geqslant 0$, and $k$-unhappy if $\min \left\{v_{2}(a), v_{2}(b)\right\}=k<\max \left\{v_{2}(a), v_{2}(b)\... | proof | Yes | Yes | math-word-problem | Combinatorics | Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of the coins of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins.
Det... | By $v_{2}(n)$ we denote the largest nonnegative integer $r$ such that $2^{r} \mid n$.
A position $(a, b)$ (i.e. two piles of sizes $a$ and $b$ ) is said to be $k$-happy if $v_{2}(a)=v_{2}(b)=k$ for some integer $k \geqslant 0$, and $k$-unhappy if $\min \left\{v_{2}(a), v_{2}(b)\right\}=k<\max \left\{v_{2}(a), v_{2}(b)\... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\n3. ",
"resource_path": "Balkan_MO/segmented/en-2018-BMO-type3.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2018"
} |
Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$.
Time allowed: 4 hours and 30 minutes.
Each problem is worth 10 points. | Answer: $(p, q)=(3,3)$.
For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$.
Observe that $N=11^{p}+17^{p} \equiv 4(\bmod 8)$, so $8 \nmid 3 p^{q-1}+1>4$. Consider an odd prime divisor $r$ of $3 p^{q-1}+1$. Obviously, $r \notin\{3,11,17\}$. There exists $b$ such that $17 b \equiv 1$ $(\bmod ... | (p, q)=(3,3) | Yes | Yes | math-word-problem | Number Theory | Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$.
Time allowed: 4 hours and 30 minutes.
Each problem is worth 10 points. | Answer: $(p, q)=(3,3)$.
For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$.
Observe that $N=11^{p}+17^{p} \equiv 4(\bmod 8)$, so $8 \nmid 3 p^{q-1}+1>4$. Consider an odd prime divisor $r$ of $3 p^{q-1}+1$. Obviously, $r \notin\{3,11,17\}$. There exists $b$ such that $17 b \equiv 1$ $(\bmod ... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\n4. ",
"resource_path": "Balkan_MO/segmented/en-2018-BMO-type3.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2018"
} |
Let $\mathbb{P}$ be the set of all prime numbers. Find all functions $f: \mathbb{P} \rightarrow \mathbb{P}$ such that
$$
f(p)^{f(q)}+q^{p}=f(q)^{f(p)}+p^{q}
$$
holds for all $p, q \in \mathbb{P}$.
Proposed by Albania | Obviously, the identical function $f(p)=p$ for all $p \in \mathbb{P}$ is a solution. We will show that this is the only one.
First we will show that $f(2)=2$. Taking $q=2$ and $p$ any odd prime number, we have
$$
f(p)^{f(2)}+2^{p}=f(2)^{f(p)}+p^{2}
$$
Assume that $f(2) \neq 2$. It follows that $f(2)$ is odd and so $... | f(p)=p | Yes | Yes | proof | Number Theory | Let $\mathbb{P}$ be the set of all prime numbers. Find all functions $f: \mathbb{P} \rightarrow \mathbb{P}$ such that
$$
f(p)^{f(q)}+q^{p}=f(q)^{f(p)}+p^{q}
$$
holds for all $p, q \in \mathbb{P}$.
Proposed by Albania | Obviously, the identical function $f(p)=p$ for all $p \in \mathbb{P}$ is a solution. We will show that this is the only one.
First we will show that $f(2)=2$. Taking $q=2$ and $p$ any odd prime number, we have
$$
f(p)^{f(2)}+2^{p}=f(2)^{f(p)}+p^{2}
$$
Assume that $f(2) \neq 2$. It follows that $f(2)$ is odd and so $... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "# Problem 1.",
"resource_path": "Balkan_MO/segmented/en-2019-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $a, b, c$ be real numbers, such that $0 \leq a \leq b \leq c$ and $a+b+c=a b+b c+c a>0$. Prove that $\sqrt{b c}(a+1) \geq 2$. Find all triples $(a, b, c)$ for which equality holds.
## Proposed by Romania | Let $a+b+c=a b+b c+c a=k$. Since $(a+b+c)^{2} \geq 3(a b+b c+c a)$, we get that $k^{2} \geq 3 k$. Since $k>0$, we obtain that $k \geq 3$.
We have $b c \geq c a \geq a b$, so from the above relation we deduce that $b c \geq 1$.
By AM-GM, $b+c \geq 2 \sqrt{b c}$ and consequently $b+c \geq 2$. The equality holds iff $b=c... | proof | Yes | Yes | proof | Inequalities | Let $a, b, c$ be real numbers, such that $0 \leq a \leq b \leq c$ and $a+b+c=a b+b c+c a>0$. Prove that $\sqrt{b c}(a+1) \geq 2$. Find all triples $(a, b, c)$ for which equality holds.
## Proposed by Romania | Let $a+b+c=a b+b c+c a=k$. Since $(a+b+c)^{2} \geq 3(a b+b c+c a)$, we get that $k^{2} \geq 3 k$. Since $k>0$, we obtain that $k \geq 3$.
We have $b c \geq c a \geq a b$, so from the above relation we deduce that $b c \geq 1$.
By AM-GM, $b+c \geq 2 \sqrt{b c}$ and consequently $b+c \geq 2$. The equality holds iff $b=c... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "# Problem 2.",
"resource_path": "Balkan_MO/segmented/en-2019-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $A B C$ be an acute scalene triangle. Let $X$ and $Y$ be two distinct interior points of the segment $B C$ such that $\angle C A X=\angle Y A B$. Suppose that:
1) $K$ and $S$ are the feet of perpendiculars from $B$ to the lines $A X$ and $A Y$ respectively;
2) $T$ and $L$ are the feet of perpendiculars from $C$ to... | Denote $\phi=\widehat{X A B}=\widehat{Y A C}, \alpha=\widehat{C A X}=\widehat{B A Y}$. Then, because the quadrilaterals ABSK and ACTL are cyclic, we have
$$
\widehat{B S K}+\widehat{B A K}=180^{\circ}=\widehat{B S K}+\phi=\widehat{L A C}+\widehat{L T C}=\widehat{L T C}+\phi
$$
so, due to the 90-degree angles formed, ... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute scalene triangle. Let $X$ and $Y$ be two distinct interior points of the segment $B C$ such that $\angle C A X=\angle Y A B$. Suppose that:
1) $K$ and $S$ are the feet of perpendiculars from $B$ to the lines $A X$ and $A Y$ respectively;
2) $T$ and $L$ are the feet of perpendiculars from $C$ to... | Denote $\phi=\widehat{X A B}=\widehat{Y A C}, \alpha=\widehat{C A X}=\widehat{B A Y}$. Then, because the quadrilaterals ABSK and ACTL are cyclic, we have
$$
\widehat{B S K}+\widehat{B A K}=180^{\circ}=\widehat{B S K}+\phi=\widehat{L A C}+\widehat{L T C}=\widehat{L T C}+\phi
$$
so, due to the 90-degree angles formed, ... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "# Problem 3.",
"resource_path": "Balkan_MO/segmented/en-2019-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
A grid consists of all points of the form $(m, n)$ where $m$ and $n$ are integers with $|m| \leqslant 2019$, $|n| \leqslant 2019$ and $|m|+|n|<4038$. We call the points $(m, n)$ of the grid with either $|m|=2019$ or $|n|=2019$ the boundary points. The four lines $x= \pm 2019$ and $y= \pm 2019$ are called boundary lines... | Anna does not have a winning strategy. We will provide a winning strategy for Bob. It is enough to describe his strategy for the deletions on the line $y=2019$.
Bob starts by deleting $(0,2019)$ and $(-1,2019)$. Once Anna completes her turn, he deletes the next two available points on the left if Anna decreased her $x... | proof | Yes | Yes | math-word-problem | Combinatorics | A grid consists of all points of the form $(m, n)$ where $m$ and $n$ are integers with $|m| \leqslant 2019$, $|n| \leqslant 2019$ and $|m|+|n|<4038$. We call the points $(m, n)$ of the grid with either $|m|=2019$ or $|n|=2019$ the boundary points. The four lines $x= \pm 2019$ and $y= \pm 2019$ are called boundary lines... | Anna does not have a winning strategy. We will provide a winning strategy for Bob. It is enough to describe his strategy for the deletions on the line $y=2019$.
Bob starts by deleting $(0,2019)$ and $(-1,2019)$. Once Anna completes her turn, he deletes the next two available points on the left if Anna decreased her $x... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "# Problem 4.",
"resource_path": "Balkan_MO/segmented/en-2019-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies o... | . We will first prove that $C$ is the midpoint of the segment $B E$. From the angle equalities
- $\angle B C D=\angle A C B=\angle C B A=\angle E B A$
- $\angle B D C=\angle B A D+\angle D B A=\angle B A D+\angle D A E=\angle B A E$
we can conclude that the triangles $\triangle A B E$ and $\triangle D C B$ are similar... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies o... | . We will first prove that $C$ is the midpoint of the segment $B E$. From the angle equalities
- $\angle B C D=\angle A C B=\angle C B A=\angle E B A$
- $\angle B D C=\angle B A D+\angle D B A=\angle B A D+\angle D A E=\angle B A E$
we can conclude that the triangles $\triangle A B E$ and $\triangle D C B$ are similar... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2020"
} |
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies o... | . Like in the previous solution, we will first prove that $C$ is the midpoint of the segment $B E$. Let $F$ be the midpoint of the segment $A B$. Because $\angle E A C=\angle A B D=\angle F C D$, we have that $C F \| A E$. This implies that $C F$ is a midline in triangle $\triangle B A E$, so $C$ is indeed the midpoint... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies o... | . Like in the previous solution, we will first prove that $C$ is the midpoint of the segment $B E$. Let $F$ be the midpoint of the segment $A B$. Because $\angle E A C=\angle A B D=\angle F C D$, we have that $C F \| A E$. This implies that $C F$ is a midline in triangle $\triangle B A E$, so $C$ is indeed the midpoint... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2020"
} |
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies o... | . Like in the previous solutions, establish that $C$ is the midpoint of the segment $B E$.
Let $S$ be the intersection between $B D$ and $A E$. We will first show that $S$ also lies on $C O$. Because
$$
\angle B D C=\angle B A D+\angle D B A=\angle B A D+\angle D A E=\angle B A E
$$
(just like in solution 1), we obta... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies o... | . Like in the previous solutions, establish that $C$ is the midpoint of the segment $B E$.
Let $S$ be the intersection between $B D$ and $A E$. We will first show that $S$ also lies on $C O$. Because
$$
\angle B D C=\angle B A D+\angle D B A=\angle B A D+\angle D A E=\angle B A E
$$
(just like in solution 1), we obta... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"solution_match": "\nSolution 3",
"tier": "T1",
"year": "2020"
} |
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies o... | . Like in the previous solutions, establish that $C$ is the midpoint of the segment $B E$.
Let $F$ be the midpoint of the segment $A B$. Then both $F$ and $C$ lie on the circle of diameter $B O$. Because the quadrilateral $B F D C$ is cyclic, it means that $D$ also lies on that circle.
Let $K$ be the midpoint of $B O$... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies o... | . Like in the previous solutions, establish that $C$ is the midpoint of the segment $B E$.
Let $F$ be the midpoint of the segment $A B$. Then both $F$ and $C$ lie on the circle of diameter $B O$. Because the quadrilateral $B F D C$ is cyclic, it means that $D$ also lies on that circle.
Let $K$ be the midpoint of $B O$... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"solution_match": "\nSolution 4",
"tier": "T1",
"year": "2020"
} |
Denote $\mathbb{Z}_{>0}=\{1,2,3, \ldots\}$ the set of all positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that, for each positive integer $n$,
i) $\sum_{k=1}^{n} f(k)$ is a perfect square, and
ii) $f(n)$ divides $n^{3}$. | Induct on $n$ to show that $f(n)=n^{3}$ for all positive integers $n$. It is readily checked that this $f$ satisfies the conditions in the statement. The base case, $n=1$, is clear.
Let $n \geqslant 2$ and assume that $f(m)=m^{3}$ for all positive integers $m<n$. Then $\sum_{k=1}^{n-1} f(k)=\frac{n^{2}(n-1)^{2}}{4}$, ... | f(n)=n^{3} | Yes | Yes | math-word-problem | Number Theory | Denote $\mathbb{Z}_{>0}=\{1,2,3, \ldots\}$ the set of all positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that, for each positive integer $n$,
i) $\sum_{k=1}^{n} f(k)$ is a perfect square, and
ii) $f(n)$ divides $n^{3}$. | Induct on $n$ to show that $f(n)=n^{3}$ for all positive integers $n$. It is readily checked that this $f$ satisfies the conditions in the statement. The base case, $n=1$, is clear.
Let $n \geqslant 2$ and assume that $f(m)=m^{3}$ for all positive integers $m<n$. Then $\sum_{k=1}^{n-1} f(k)=\frac{n^{2}(n-1)^{2}}{4}$, ... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "## 2020 BMO, Problem 2",
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2020"
} |
Let $k$ be a positive integer. Determine the least integer $n \geqslant k+1$ for which the game below can be played indefinitely:
Consider $n$ boxes, labelled $b_{1}, b_{2}, \ldots, b_{n}$. For each index $i$, box $b_{i}$ contains initially exactly $i$ coins. At each step, the following three substeps are performed in... | The required minimum is $n=2^{k}+k-1$.
In this case the game can be played indefinitely by choosing the last $k+1$ boxes, $b_{2^{k}-1}, b_{2^{k}}, \ldots, b_{2^{k}+k-1}$, at each step: At step $r$, if box $b_{2^{k}+i-1}$ has exactly $m_{i}$ coins, then $\left\lceil m_{i} / 2\right\rceil$ coins are removed from that box... | 2^{k}+k-1 | Yes | Yes | math-word-problem | Combinatorics | Let $k$ be a positive integer. Determine the least integer $n \geqslant k+1$ for which the game below can be played indefinitely:
Consider $n$ boxes, labelled $b_{1}, b_{2}, \ldots, b_{n}$. For each index $i$, box $b_{i}$ contains initially exactly $i$ coins. At each step, the following three substeps are performed in... | The required minimum is $n=2^{k}+k-1$.
In this case the game can be played indefinitely by choosing the last $k+1$ boxes, $b_{2^{k}-1}, b_{2^{k}}, \ldots, b_{2^{k}+k-1}$, at each step: At step $r$, if box $b_{2^{k}+i-1}$ has exactly $m_{i}$ coins, then $\left\lceil m_{i} / 2\right\rceil$ coins are removed from that box... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "## 2020 BMO, Problem 3",
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2020"
} |
Let $a_{1}=2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_{n}$ that has more positive divisors than $a_{n}$. Prove that $2 a_{n+1}=3 a_{n}$ only for finitely many indices $n$. | Begin with a mere remark on the terms of the sequence under consideration.
Lemma 1. Each $a_{n}$ is minimal amongst all positive integers having the same number of positive divisors as $a_{n}$.
Proof. Suppose, if possible, that for some $n$, some positive integer $b<a_{n}$ has as many positive divisors as $a_{n}$. The... | proof | Yes | Yes | proof | Number Theory | Let $a_{1}=2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_{n}$ that has more positive divisors than $a_{n}$. Prove that $2 a_{n+1}=3 a_{n}$ only for finitely many indices $n$. | Begin with a mere remark on the terms of the sequence under consideration.
Lemma 1. Each $a_{n}$ is minimal amongst all positive integers having the same number of positive divisors as $a_{n}$.
Proof. Suppose, if possible, that for some $n$, some positive integer $b<a_{n}$ has as many positive divisors as $a_{n}$. The... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "## 2020 BMO, Problem 4",
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2020"
} |
Let $A B C$ be a triangle with $A B<A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A=\angle A Y C$. Suppose also that $X$ and $C$ lie on opposite sides of the line $A B$ and that $Y$ and $B$ lie on opposite sides of the ... | . Extend $X A$ and $Y A$ to meet $\omega$ again at $X^{\prime}$ and $Y^{\prime}$ respectively. We then have that:
$$
\angle Y^{\prime} Y C=\angle A Y C=\angle B X A=\angle B X X^{\prime} .
$$
so $B C X^{\prime} Y^{\prime}$ is an isosceles trapezium and hence $X^{\prime} Y^{\prime} \| B C$.
![](https://cdn.mathpix.com... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with $A B<A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A=\angle A Y C$. Suppose also that $X$ and $C$ lie on opposite sides of the line $A B$ and that $Y$ and $B$ lie on opposite sides of the ... | . Extend $X A$ and $Y A$ to meet $\omega$ again at $X^{\prime}$ and $Y^{\prime}$ respectively. We then have that:
$$
\angle Y^{\prime} Y C=\angle A Y C=\angle B X A=\angle B X X^{\prime} .
$$
so $B C X^{\prime} Y^{\prime}$ is an isosceles trapezium and hence $X^{\prime} Y^{\prime} \| B C$.
![](https://cdn.mathpix.com... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "## BMO 2021 - Problem 1",
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2021"
} |
Let $A B C$ be a triangle with $A B<A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A=\angle A Y C$. Suppose also that $X$ and $C$ lie on opposite sides of the line $A B$ and that $Y$ and $B$ lie on opposite sides of the ... | . Let $B^{\prime}$ and $C^{\prime}$ be the points of intersection of the lines $A B$ and $A C$ with $\omega$ respectively and let $\omega_{1}$ be the circumcircle of the triangle $A B^{\prime} C^{\prime}$. Let $\varepsilon$ be the tangent to $\omega_{1}$ at the point $A$. Because $A B<A C$ the lines $B^{\prime} C^{\pri... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with $A B<A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A=\angle A Y C$. Suppose also that $X$ and $C$ lie on opposite sides of the line $A B$ and that $Y$ and $B$ lie on opposite sides of the ... | . Let $B^{\prime}$ and $C^{\prime}$ be the points of intersection of the lines $A B$ and $A C$ with $\omega$ respectively and let $\omega_{1}$ be the circumcircle of the triangle $A B^{\prime} C^{\prime}$. Let $\varepsilon$ be the tangent to $\omega_{1}$ at the point $A$. Because $A B<A C$ the lines $B^{\prime} C^{\pri... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "## BMO 2021 - Problem 1",
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2021"
} |
Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$ such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in(0,+\infty)$. | . We will show that $f(x)=x$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x+f(x)+f(y))=2 f(x)+y$.
We first show that $f$ is injective. So assume $f(a)=f(b)$. Now $P(1, a)$ and $P(1, b)$ show that
$$
2 f(1)+a=f(1+f(1)+f(a)... | f(x)=x | Yes | Yes | math-word-problem | Algebra | Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$ such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in(0,+\infty)$. | . We will show that $f(x)=x$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x+f(x)+f(y))=2 f(x)+y$.
We first show that $f$ is injective. So assume $f(a)=f(b)$. Now $P(1, a)$ and $P(1, b)$ show that
$$
2 f(1)+a=f(1+f(1)+f(a)... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "## BMO 2021 - Problem 2",
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2021"
} |
Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$ such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in(0,+\infty)$. | . As in Solution 1, $f$ is injective. Furthermore, letting $m=2 f(1)$ we have that the image of $f$ contains $(m, \infty)$. Indeed, if $t>m$, say $t=m+y$ for some $y>0$, then $P(1, y)$ shows that $f(1+f(1)+f(y))=t$.
Let $a, b \in \mathbb{R}$. We will show that $f(a)-a=f(b)-b$. Define $c=2 f(a)-2 f(b)$ and $d=a+f(a)-b-... | f(x)=x | Yes | Yes | math-word-problem | Algebra | Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$ such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in(0,+\infty)$. | . As in Solution 1, $f$ is injective. Furthermore, letting $m=2 f(1)$ we have that the image of $f$ contains $(m, \infty)$. Indeed, if $t>m$, say $t=m+y$ for some $y>0$, then $P(1, y)$ shows that $f(1+f(1)+f(y))=t$.
Let $a, b \in \mathbb{R}$. We will show that $f(a)-a=f(b)-b$. Define $c=2 f(a)-2 f(b)$ and $d=a+f(a)-b-... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "## BMO 2021 - Problem 2",
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2021"
} |
Let $a, b$ and $c$ be positive integers satisfying the equation
$$
(a, b)+[a, b]=2021^{c} .
$$
If $|a-b|$ is a prime number, prove that the number $(a+b)^{2}+4$ is composite.
Here, $(a, b)$ denotes the greatest common divisor of $a$ and $b$, and $[a, b]$ denotes the least common multiple of $a$ and $b$. | We write $p=|a-b|$ and assume for contradiction that $q=(a+b)^{2}+4$ is a prime number.
Since $(a, b) \mid[a, b]$, we have that $(a, b) \mid 2021^{c}$. As $(a, b)$ also divides $p=|a-b|$, it follows that $(a, b) \in\{1,43,47\}$. We will consider all 3 cases separately:
(1) If $(a, b)=1$, then $1+a b=2021^{c}$, and the... | proof | Yes | Yes | proof | Number Theory | Let $a, b$ and $c$ be positive integers satisfying the equation
$$
(a, b)+[a, b]=2021^{c} .
$$
If $|a-b|$ is a prime number, prove that the number $(a+b)^{2}+4$ is composite.
Here, $(a, b)$ denotes the greatest common divisor of $a$ and $b$, and $[a, b]$ denotes the least common multiple of $a$ and $b$. | We write $p=|a-b|$ and assume for contradiction that $q=(a+b)^{2}+4$ is a prime number.
Since $(a, b) \mid[a, b]$, we have that $(a, b) \mid 2021^{c}$. As $(a, b)$ also divides $p=|a-b|$, it follows that $(a, b) \in\{1,43,47\}$. We will consider all 3 cases separately:
(1) If $(a, b)=1$, then $1+a b=2021^{c}$, and the... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "## BMO 2021 - Problem 3",
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2021"
} |
Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel performs exactly one of the following moves:
(a) He clears every piece of rubbish from a single pile.
(b) He clears one piece of rubbish from each pile.
However, every evening, a demon sneaks into the warehouse... | . We will show that he can do so by the morning of day 199 but not earlier.If we have $n$ piles with at least two pieces of rubbish and $m$ piles with exactly one piece of rubbish, then we define the value of the pile to be
$$
V= \begin{cases}n & m=0 \\ n+\frac{1}{2} & m=1 \\ n+1 & m \geqslant 2\end{cases}
$$
We also... | 199 | Yes | Yes | math-word-problem | Logic and Puzzles | Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel performs exactly one of the following moves:
(a) He clears every piece of rubbish from a single pile.
(b) He clears one piece of rubbish from each pile.
However, every evening, a demon sneaks into the warehouse... | . We will show that he can do so by the morning of day 199 but not earlier.If we have $n$ piles with at least two pieces of rubbish and $m$ piles with exactly one piece of rubbish, then we define the value of the pile to be
$$
V= \begin{cases}n & m=0 \\ n+\frac{1}{2} & m=1 \\ n+1 & m \geqslant 2\end{cases}
$$
We also... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "## BMO 2021 - Problem 4",
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2021"
} |
Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel performs exactly one of the following moves:
(a) He clears every piece of rubbish from a single pile.
(b) He clears one piece of rubbish from each pile.
However, every evening, a demon sneaks into the warehouse... | .
Define Angel's score $S_{A}$ to be $S_{A}=2 n+m-1$. The Angel can clear the rubbish in at most $\max \left\{S_{A}, 1\right\}$ days. The proof is by induction on $(n, m)$ in lexicographic order.
Angel's strategy is the same as in Solution 1 and in each of cases (ii)-(iv) one needs to check that $S_{A}$ reduces by at... | not found | Yes | Yes | math-word-problem | Logic and Puzzles | Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel performs exactly one of the following moves:
(a) He clears every piece of rubbish from a single pile.
(b) He clears one piece of rubbish from each pile.
However, every evening, a demon sneaks into the warehouse... | .
Define Angel's score $S_{A}$ to be $S_{A}=2 n+m-1$. The Angel can clear the rubbish in at most $\max \left\{S_{A}, 1\right\}$ days. The proof is by induction on $(n, m)$ in lexicographic order.
Angel's strategy is the same as in Solution 1 and in each of cases (ii)-(iv) one needs to check that $S_{A}$ reduces by at... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "## BMO 2021 - Problem 4",
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2021"
} |
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ para... | . Firstly observe that $O A X B$ is cyclic, with diameter $O X$, and $Y$ also lies on this circle since $O Y \perp X C$. Hence:
$$
\angle A Z C=\angle X A B=\angle A B X=\angle A Y X
$$
and so $C Y A Z$ is cyclic.
![](https://cdn.mathpix.com/cropped/2024_12_10_b5da7837303016ab5616g-1.jpg?height=1121&width=1112&top_le... | proof | Yes | Yes | proof | Geometry | Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ para... | . Firstly observe that $O A X B$ is cyclic, with diameter $O X$, and $Y$ also lies on this circle since $O Y \perp X C$. Hence:
$$
\angle A Z C=\angle X A B=\angle A B X=\angle A Y X
$$
and so $C Y A Z$ is cyclic.
![](https://cdn.mathpix.com/cropped/2024_12_10_b5da7837303016ab5616g-1.jpg?height=1121&width=1112&top_le... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2022"
} |
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ para... | . Let $M$ be the midpoint of $A C$. We have $\angle C A Z=\angle C B A$ and $\angle Z C A=$ $\angle B A C$ so the triangles $C A Z$ and $A B C$ are similar. The line $C Y X$ is the $C$-symmedian of triangle $A B C$, and $Z M$ is the corresponding median in triangle $C A Z$, hence by isogonality $\angle A Z M=\angle A C... | proof | Yes | Yes | proof | Geometry | Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ para... | . Let $M$ be the midpoint of $A C$. We have $\angle C A Z=\angle C B A$ and $\angle Z C A=$ $\angle B A C$ so the triangles $C A Z$ and $A B C$ are similar. The line $C Y X$ is the $C$-symmedian of triangle $A B C$, and $Z M$ is the corresponding median in triangle $C A Z$, hence by isogonality $\angle A Z M=\angle A C... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2022"
} |
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ para... | . As in Solution 2 we have that $C X$ is the $A$-symmedian of triangle $A B C$ and that triangle $A B C$ is similar to triangle $C A Z$.
Let $f$ be the spiral similarity which maps $A C$ onto $A B$ and let $g$ be the reflection on the perpendicular bisector of $A B$. Note that $f$ is a rotation about $A$ by an angle o... | proof | Yes | Yes | proof | Geometry | Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ para... | . As in Solution 2 we have that $C X$ is the $A$-symmedian of triangle $A B C$ and that triangle $A B C$ is similar to triangle $C A Z$.
Let $f$ be the spiral similarity which maps $A C$ onto $A B$ and let $g$ be the reflection on the perpendicular bisector of $A B$. Note that $f$ is a rotation about $A$ by an angle o... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"solution_match": "\nSolution 3",
"tier": "T1",
"year": "2022"
} |
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ para... | . Let $E=A B \cap C X$ and $F=A W \cap C Z$. We have $(C, W ; X, E)=-1$. Projecting from the line $C X$ onto the line $C Z$ from $A$ we get that $(C, F ; Z, \infty)=-1$. Thus $Z$ is the midpoint of $C F$. Since also $Y$ is the midpoint of $C W$, we get that $Z Y$ bisects $C A$. | proof | Yes | Yes | proof | Geometry | Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ para... | . Let $E=A B \cap C X$ and $F=A W \cap C Z$. We have $(C, W ; X, E)=-1$. Projecting from the line $C X$ onto the line $C Z$ from $A$ we get that $(C, F ; Z, \infty)=-1$. Thus $Z$ is the midpoint of $C F$. Since also $Y$ is the midpoint of $C W$, we get that $Z Y$ bisects $C A$. | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"solution_match": "\nSolution 4",
"tier": "T1",
"year": "2022"
} |
Problem. Let $a, b$ and $n$ be positive integers with $a>b$ such that all of the following hold:
(i) $a^{2021}$ divides $n$,
(ii) $b^{2021}$ divides $n$,
(iii) 2022 divides $a-b$.
Prove that there is a subset $T$ of the set of positive divisors of the number $n$ such that the sum of the elements of $T$ is divisible by... | If $1011 \mid a$, then $1011^{2021} \mid n$ and we can take $T=\left\{1011,1011^{2}\right\}$. So we can assume that $3 \nmid a$ or $337 \nmid a$.
We continue with the following claim:
Claim. If $k$ is a positive integer, then $a^{k} b^{2021-k} \mid n$.
Proof of the Claim. We have that $n^{2021}=n^{k} \cdot n^{2021-k}$... | proof | Yes | Yes | proof | Number Theory | Problem. Let $a, b$ and $n$ be positive integers with $a>b$ such that all of the following hold:
(i) $a^{2021}$ divides $n$,
(ii) $b^{2021}$ divides $n$,
(iii) 2022 divides $a-b$.
Prove that there is a subset $T$ of the set of positive divisors of the number $n$ such that the sum of the elements of $T$ is divisible by... | If $1011 \mid a$, then $1011^{2021} \mid n$ and we can take $T=\left\{1011,1011^{2}\right\}$. So we can assume that $3 \nmid a$ or $337 \nmid a$.
We continue with the following claim:
Claim. If $k$ is a positive integer, then $a^{k} b^{2021-k} \mid n$.
Proof of the Claim. We have that $n^{2021}=n^{k} \cdot n^{2021-k}$... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "# Problem 2",
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2022"
} |
Problem. Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that
$$
f\left(y(f(x))^{3}+x\right)=x^{3} f(y)+f(x)
$$
for all $x, y>0$. | . Setting $y=\frac{t}{f(x)^{3}}$ we get
$$
f(x+t)=x^{3} f\left(\frac{t}{f(x)^{3}}\right)+f(x)
$$
for every $x, t>0$.
From (1) it is immediate that $f$ is increasing.
Claim. $f(1)=1$
Proof of Claim. Let $c=f(1)$. If $c<1$, taking $x=1$ and $y=\frac{1}{1-c^{3}}$ we have $y-y c^{3}=1$, so $y f(1)^{3}+1=y$ and $f\left(y ... | f(x)=x | Yes | Yes | math-word-problem | Algebra | Problem. Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that
$$
f\left(y(f(x))^{3}+x\right)=x^{3} f(y)+f(x)
$$
for all $x, y>0$. | . Setting $y=\frac{t}{f(x)^{3}}$ we get
$$
f(x+t)=x^{3} f\left(\frac{t}{f(x)^{3}}\right)+f(x)
$$
for every $x, t>0$.
From (1) it is immediate that $f$ is increasing.
Claim. $f(1)=1$
Proof of Claim. Let $c=f(1)$. If $c<1$, taking $x=1$ and $y=\frac{1}{1-c^{3}}$ we have $y-y c^{3}=1$, so $y f(1)^{3}+1=y$ and $f\left(y ... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "# Problem 3",
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2022"
} |
Problem. Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that
$$
f\left(y(f(x))^{3}+x\right)=x^{3} f(y)+f(x)
$$
for all $x, y>0$. | . We can also derive a contradiction in the case $c>1$ as follows:
Since $f$ is strictly increasing then
$$
f(y)+f(1)=f\left(y f(1)^{3}+1\right)>f\left(y f(1)^{3}\right) \Longrightarrow f\left(c^{3} y\right)<f(y)+c
$$
for every $y>0$. So by induction we get $f\left(c^{3 n}\right)<(n+1) c$ for every $n \in \mathbb{N}$... | proof | Yes | Yes | math-word-problem | Algebra | Problem. Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that
$$
f\left(y(f(x))^{3}+x\right)=x^{3} f(y)+f(x)
$$
for all $x, y>0$. | . We can also derive a contradiction in the case $c>1$ as follows:
Since $f$ is strictly increasing then
$$
f(y)+f(1)=f\left(y f(1)^{3}+1\right)>f\left(y f(1)^{3}\right) \Longrightarrow f\left(c^{3} y\right)<f(y)+c
$$
for every $y>0$. So by induction we get $f\left(c^{3 n}\right)<(n+1) c$ for every $n \in \mathbb{N}$... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "# Problem 3",
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2022"
} |
Problem. Consider an $n \times n$ grid consisting of $n^{2}$ unit cells, where $n \geqslant 3$ is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Th... | . Let $G$ be the graph whose vertices are all $(n+1)^{2}$ vertices of the grid and where two vertices are adjacent if and only if they are adjacent in the grid and moreover the two cells in either side of the corresponding edge have different colours.
The connnected components of $G$, excluding the isolated vertices, ... | \frac{(n+1)^{2}}{4}+1 | Yes | Yes | math-word-problem | Combinatorics | Problem. Consider an $n \times n$ grid consisting of $n^{2}$ unit cells, where $n \geqslant 3$ is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Th... | . Let $G$ be the graph whose vertices are all $(n+1)^{2}$ vertices of the grid and where two vertices are adjacent if and only if they are adjacent in the grid and moreover the two cells in either side of the corresponding edge have different colours.
The connnected components of $G$, excluding the isolated vertices, ... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "# Problem 4",
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2022"
} |
Problem. Consider an $n \times n$ grid consisting of $n^{2}$ unit cells, where $n \geqslant 3$ is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Th... | . Consider an $n \times m$ grid with $n, m \geqslant 3$ being odd. We say that a column is of 'Type A' if, when partitioned into its monochromatic pieces, the first and last piece have the same colour with each one containing at least two cells. Otherwise we say that that it is of 'Type B'.
It is enough to show that t... | not found | Yes | Incomplete | math-word-problem | Combinatorics | Problem. Consider an $n \times n$ grid consisting of $n^{2}$ unit cells, where $n \geqslant 3$ is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Th... | . Consider an $n \times m$ grid with $n, m \geqslant 3$ being odd. We say that a column is of 'Type A' if, when partitioned into its monochromatic pieces, the first and last piece have the same colour with each one containing at least two cells. Otherwise we say that that it is of 'Type B'.
It is enough to show that t... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "# Problem 4",
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2022"
} |
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,
$$
x f(x+f(y))=(y-x) f(f(x))
$$ | Answer: For any real $c, f(x)=c-x$ for all $x \in \mathbb{R}$ and $f(x)=0$ for all $x \in \mathbb{R}$.
Let $P(x, y)$ denote the proposition that $x$ and $y$ satisfy the given equation. $P(0,1)$ gives us $f(f(0))=0$.
From $P(x, x)$ we get that $x f(x+f(x))=0$ for all $x \in \mathbb{R}$, which together with $f(f(0))=$ ... | proof | Yes | Yes | math-word-problem | Algebra | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,
$$
x f(x+f(y))=(y-x) f(f(x))
$$ | Answer: For any real $c, f(x)=c-x$ for all $x \in \mathbb{R}$ and $f(x)=0$ for all $x \in \mathbb{R}$.
Let $P(x, y)$ denote the proposition that $x$ and $y$ satisfy the given equation. $P(0,1)$ gives us $f(f(0))=0$.
From $P(x, x)$ we get that $x f(x+f(x))=0$ for all $x \in \mathbb{R}$, which together with $f(f(0))=$ ... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "Balkan_MO/segmented/en-2023-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2023"
} |
In triangle $A B C$, the incircle touches sides $B C, C A, A B$ at $D, E, F$ respectively. Assume there exists a point $X$ on the line $E F$ such that
$$
\angle X B C=\angle X C B=45^{\circ} .
$$
Let $M$ be the midpoint of the arc $B C$ on the circumcircle of $A B C$ not containing $A$. Prove that the line $M D$ pass... | . We first state a well-known lemma.
Lemma: In triangle $A B C$, let $D, E, F$ be the points of tangency of the incircle to the sides $B C, C A, A B$ and let $I$ be the incenter. Then the intersection of $E F$ and $B I$ lies on the circle of diameter $B C$.
Proof: Let $S$ be the intersection of $E F$ and $B I . \angle... | proof | Yes | Yes | proof | Geometry | In triangle $A B C$, the incircle touches sides $B C, C A, A B$ at $D, E, F$ respectively. Assume there exists a point $X$ on the line $E F$ such that
$$
\angle X B C=\angle X C B=45^{\circ} .
$$
Let $M$ be the midpoint of the arc $B C$ on the circumcircle of $A B C$ not containing $A$. Prove that the line $M D$ pass... | . We first state a well-known lemma.
Lemma: In triangle $A B C$, let $D, E, F$ be the points of tangency of the incircle to the sides $B C, C A, A B$ and let $I$ be the incenter. Then the intersection of $E F$ and $B I$ lies on the circle of diameter $B C$.
Proof: Let $S$ be the intersection of $E F$ and $B I . \angle... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2023-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2023"
} |
In triangle $A B C$, the incircle touches sides $B C, C A, A B$ at $D, E, F$ respectively. Assume there exists a point $X$ on the line $E F$ such that
$$
\angle X B C=\angle X C B=45^{\circ} .
$$
Let $M$ be the midpoint of the arc $B C$ on the circumcircle of $A B C$ not containing $A$. Prove that the line $M D$ pass... | .
Let $I$ be the incenter of $\triangle A B C$ and let $K$ be the foot of the perpendicular from $D$ to $E F$. We begin by proving that $B K X C$ is cyclic, which can be done in two ways:
First Way. Note that $\angle K F D=90^{\circ}-\frac{\angle C}{2}$ and $\angle K E D=90^{\circ}-\frac{\angle B}{2}$, so by using $K... | proof | Yes | Yes | proof | Geometry | In triangle $A B C$, the incircle touches sides $B C, C A, A B$ at $D, E, F$ respectively. Assume there exists a point $X$ on the line $E F$ such that
$$
\angle X B C=\angle X C B=45^{\circ} .
$$
Let $M$ be the midpoint of the arc $B C$ on the circumcircle of $A B C$ not containing $A$. Prove that the line $M D$ pass... | .
Let $I$ be the incenter of $\triangle A B C$ and let $K$ be the foot of the perpendicular from $D$ to $E F$. We begin by proving that $B K X C$ is cyclic, which can be done in two ways:
First Way. Note that $\angle K F D=90^{\circ}-\frac{\angle C}{2}$ and $\angle K E D=90^{\circ}-\frac{\angle B}{2}$, so by using $K... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2023-BMO-type1.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2023"
} |
For each positive integer $n$, denote by $\omega(n)$ the number of distinct prime divisors of $n$ (for example, $\omega(1)=0$ and $\omega(12)=2$ ). Find all polynomials $P(x)$ with integer coefficients, such that whenever $n$ is a positive integer satisfying $\omega(n)>2023^{2023}$, then $P(n)$ is also a positive integ... | Answer: All polynomials of the form $f(x)=x^{m}$ for some $m \in \mathbb{Z}^{+}$and $f(x)=c$ for some $c \in \mathbb{Z}^{+}$with $\omega(c) \leq 2023^{2023}+1$.
First of all we prove the following (well-known) Lemma. Lemma: Let $f(x)$ be a nonconstant polynomial with integer coefficients. Then, the number of primes $p... | proof | Yes | Yes | proof | Number Theory | For each positive integer $n$, denote by $\omega(n)$ the number of distinct prime divisors of $n$ (for example, $\omega(1)=0$ and $\omega(12)=2$ ). Find all polynomials $P(x)$ with integer coefficients, such that whenever $n$ is a positive integer satisfying $\omega(n)>2023^{2023}$, then $P(n)$ is also a positive integ... | Answer: All polynomials of the form $f(x)=x^{m}$ for some $m \in \mathbb{Z}^{+}$and $f(x)=c$ for some $c \in \mathbb{Z}^{+}$with $\omega(c) \leq 2023^{2023}+1$.
First of all we prove the following (well-known) Lemma. Lemma: Let $f(x)$ be a nonconstant polynomial with integer coefficients. Then, the number of primes $p... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "Balkan_MO/segmented/en-2023-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2023"
} |
Find the greatest integer $k \leq 2023$ for which the following holds: whenever Alice colours exactly $k$ numbers of the set $\{1,2, \ldots, 2023\}$ in red, Bob can colour some of the remaining uncoloured numbers in blue, such that the sum of the red numbers is the same as the sum of the blue numbers. | Answer: 592.
For $k \geq 593$, Alice can color the greatest 593 numbers $1431,1432, \ldots, 2023$ and any other $(k-593)$ numbers so that their sum $s$ would satisfy
$$
s \geq \frac{2023 \cdot 2024}{2}-\frac{1430 \cdot 1431}{2}>\frac{1}{2} \cdot\left(\frac{2023 \cdot 2024}{2}\right),
$$
thus anyhow Bob chooses his nu... | 592 | Yes | Yes | math-word-problem | Combinatorics | Find the greatest integer $k \leq 2023$ for which the following holds: whenever Alice colours exactly $k$ numbers of the set $\{1,2, \ldots, 2023\}$ in red, Bob can colour some of the remaining uncoloured numbers in blue, such that the sum of the red numbers is the same as the sum of the blue numbers. | Answer: 592.
For $k \geq 593$, Alice can color the greatest 593 numbers $1431,1432, \ldots, 2023$ and any other $(k-593)$ numbers so that their sum $s$ would satisfy
$$
s \geq \frac{2023 \cdot 2024}{2}-\frac{1430 \cdot 1431}{2}>\frac{1}{2} \cdot\left(\frac{2023 \cdot 2024}{2}\right),
$$
thus anyhow Bob chooses his nu... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "Balkan_MO/segmented/en-2023-BMO-type1.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2023"
} |
Let $A B C$ be an acute-angled triangle with $A C>A B$ and let $D$ be the foot of the $A$-angle bisector on $B C$. The reflections of lines $A B$ and $A C$ in line $B C$ meet $A C$ and $A B$ at points $E$ and $F$ respectively. A line through $D$ meets $A C$ and $A B$ at $G$ and $H$ respectively such that $G$ lies stric... | . Let $X$ and $Y$ lie on the tangent to the circumcircle of $\triangle E D G$ on the opposite side to $D$ as shown in the figure below. Regarding diagram dependency, the acute condition with $A C>A B$ ensures $E$ lies on extension of $C A$ beyond $A$, and $F$ lies on extension of $A B$ beyond $B$. The condition on $\el... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute-angled triangle with $A C>A B$ and let $D$ be the foot of the $A$-angle bisector on $B C$. The reflections of lines $A B$ and $A C$ in line $B C$ meet $A C$ and $A B$ at points $E$ and $F$ respectively. A line through $D$ meets $A C$ and $A B$ at $G$ and $H$ respectively such that $G$ lies stric... | . Let $X$ and $Y$ lie on the tangent to the circumcircle of $\triangle E D G$ on the opposite side to $D$ as shown in the figure below. Regarding diagram dependency, the acute condition with $A C>A B$ ensures $E$ lies on extension of $C A$ beyond $A$, and $F$ lies on extension of $A B$ beyond $B$. The condition on $\el... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2024"
} |
Let $n \geq k \geq 3$ be integers. Show that for every integer sequence $1 \leq a_{1}<a_{2}<\ldots<$ $a_{k} \leq n$ one can choose non-negative integers $b_{1}, b_{2}, \ldots, b_{k}$, satisfying the following conditions:
(i) $0 \leq b_{i} \leq n$ for each $1 \leq i \leq k$,
(ii) all the positive $b_{i}$ are distinct,
(... | . Let the resulting progression be $A n s:=\left\{a_{k}-(k-1), a_{k}-(k-2), \ldots, a_{k}\right\}$ and $a_{t}$ be the largest number not belonging to $A n s$. Clearly the set $A n s \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ has cardinality $t$; let its members be $c_{1}>c_{2}>\cdots>c_{t}$. Define $b_{j}:=c... | proof | Yes | Yes | proof | Combinatorics | Let $n \geq k \geq 3$ be integers. Show that for every integer sequence $1 \leq a_{1}<a_{2}<\ldots<$ $a_{k} \leq n$ one can choose non-negative integers $b_{1}, b_{2}, \ldots, b_{k}$, satisfying the following conditions:
(i) $0 \leq b_{i} \leq n$ for each $1 \leq i \leq k$,
(ii) all the positive $b_{i}$ are distinct,
(... | . Let the resulting progression be $A n s:=\left\{a_{k}-(k-1), a_{k}-(k-2), \ldots, a_{k}\right\}$ and $a_{t}$ be the largest number not belonging to $A n s$. Clearly the set $A n s \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ has cardinality $t$; let its members be $c_{1}>c_{2}>\cdots>c_{t}$. Define $b_{j}:=c... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2024"
} |
Let $n \geq k \geq 3$ be integers. Show that for every integer sequence $1 \leq a_{1}<a_{2}<\ldots<$ $a_{k} \leq n$ one can choose non-negative integers $b_{1}, b_{2}, \ldots, b_{k}$, satisfying the following conditions:
(i) $0 \leq b_{i} \leq n$ for each $1 \leq i \leq k$,
(ii) all the positive $b_{i}$ are distinct,
(... | . Let the resulting progression be $A n s:=\left\{a_{k}-(k-1), a_{k}-(k-2), \ldots, a_{k}\right\}$.
We proceed with the following reduction. Let $\delta$ be the smallest $b$ we used before (in the beginning it is $n$ ). While $a_{1} \notin A n s$ we map $a_{1}$ to the largest element $q$ of $A n s \backslash\left\{a_{1... | proof | Yes | Yes | proof | Combinatorics | Let $n \geq k \geq 3$ be integers. Show that for every integer sequence $1 \leq a_{1}<a_{2}<\ldots<$ $a_{k} \leq n$ one can choose non-negative integers $b_{1}, b_{2}, \ldots, b_{k}$, satisfying the following conditions:
(i) $0 \leq b_{i} \leq n$ for each $1 \leq i \leq k$,
(ii) all the positive $b_{i}$ are distinct,
(... | . Let the resulting progression be $A n s:=\left\{a_{k}-(k-1), a_{k}-(k-2), \ldots, a_{k}\right\}$.
We proceed with the following reduction. Let $\delta$ be the smallest $b$ we used before (in the beginning it is $n$ ). While $a_{1} \notin A n s$ we map $a_{1}$ to the largest element $q$ of $A n s \backslash\left\{a_{1... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2024"
} |
Let $a$ and $b$ be distinct positive integers such that $3^{a}+2$ is divisible by $3^{b}+2$. Prove that $a>b^{2}$ 。 | . Obviously we have $a>b$. Let $a=b q+r$, where $0 \leq r<b$. Then
$$
3^{a} \equiv 3^{b q+r} \equiv(-2)^{q} \cdot 3^{r} \equiv-2 \quad\left(\bmod 3^{b}+2\right)
$$
So $3^{b}+2$ divides $A=(-2)^{q} .3^{r}+2$ and it follows that
$$
\left|(-2)^{q} \cdot 3^{r}+2\right| \geq 3^{b}+2 \text { or }(-2)^{q} \cdot 3^{r}+2=0
$... | proof | Yes | Yes | proof | Number Theory | Let $a$ and $b$ be distinct positive integers such that $3^{a}+2$ is divisible by $3^{b}+2$. Prove that $a>b^{2}$ 。 | . Obviously we have $a>b$. Let $a=b q+r$, where $0 \leq r<b$. Then
$$
3^{a} \equiv 3^{b q+r} \equiv(-2)^{q} \cdot 3^{r} \equiv-2 \quad\left(\bmod 3^{b}+2\right)
$$
So $3^{b}+2$ divides $A=(-2)^{q} .3^{r}+2$ and it follows that
$$
\left|(-2)^{q} \cdot 3^{r}+2\right| \geq 3^{b}+2 \text { or }(-2)^{q} \cdot 3^{r}+2=0
$... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2024"
} |
Let $a$ and $b$ be distinct positive integers such that $3^{a}+2$ is divisible by $3^{b}+2$. Prove that $a>b^{2}$ 。 | . $D=a-b$, and we shall show $D>b^{2}-b$. We have $3^{b}+2 \mid 3^{a}+2$, so $3^{b}+2 \mid 3^{D}-1$. Let $D=b q+r$ where $r<b$. First suppose that $r \neq 0$. We have
$$
1 \equiv 3^{D} \equiv 3^{b q+r} \equiv(-2)^{q+1} 3^{r-b} \quad\left(\bmod 3^{b}+2\right) \Longrightarrow 3^{b-r} \equiv(-2)^{q+1} \quad\left(\bmod 3^... | proof | Yes | Yes | proof | Number Theory | Let $a$ and $b$ be distinct positive integers such that $3^{a}+2$ is divisible by $3^{b}+2$. Prove that $a>b^{2}$ 。 | . $D=a-b$, and we shall show $D>b^{2}-b$. We have $3^{b}+2 \mid 3^{a}+2$, so $3^{b}+2 \mid 3^{D}-1$. Let $D=b q+r$ where $r<b$. First suppose that $r \neq 0$. We have
$$
1 \equiv 3^{D} \equiv 3^{b q+r} \equiv(-2)^{q+1} 3^{r-b} \quad\left(\bmod 3^{b}+2\right) \Longrightarrow 3^{b-r} \equiv(-2)^{q+1} \quad\left(\bmod 3^... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2024"
} |
Let $\mathbb{R}^{+}=(0, \infty)$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0)=0$ which satisfy the equality
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
for all real numbers $x>y>0$. | . Assume that $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$and the polynomial $P$ with non-negative coefficients and $P(0)=0$ satisfy the conditions of the problem. For positive reals with $x>y$, we shall write $Q(x, y)$ for the relation:
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
1. Step 1. $f(x) \geq x$. Assume that this is no... | f(x)=x \text{ and } P(y)=y | Yes | Yes | math-word-problem | Algebra | Let $\mathbb{R}^{+}=(0, \infty)$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0)=0$ which satisfy the equality
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
for all real numbers $x>y>0$. | . Assume that $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$and the polynomial $P$ with non-negative coefficients and $P(0)=0$ satisfy the conditions of the problem. For positive reals with $x>y$, we shall write $Q(x, y)$ for the relation:
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
1. Step 1. $f(x) \geq x$. Assume that this is no... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"solution_match": "\nSolution 1",
"tier": "T1",
"year": "2024"
} |
Let $\mathbb{R}^{+}=(0, \infty)$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0)=0$ which satisfy the equality
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
for all real numbers $x>y>0$. | . Assume that the function $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$and the polynomial with non-negative coefficients $P(y)=y P_{1}(y)$ satisfy the given equation. Fix $x=x_{0}>0$ and note that:
$$
f\left(f\left(x_{0}+y\right)+P(y)\right)=f\left(x_{0}+y-y\right)+2 y=f\left(x_{0}\right)+2 y
$$
Assume that $g=0$. ... | f(x)=x \text{ and } P(y)=y | Yes | Incomplete | math-word-problem | Algebra | Let $\mathbb{R}^{+}=(0, \infty)$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0)=0$ which satisfy the equality
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
for all real numbers $x>y>0$. | . Assume that the function $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$and the polynomial with non-negative coefficients $P(y)=y P_{1}(y)$ satisfy the given equation. Fix $x=x_{0}>0$ and note that:
$$
f\left(f\left(x_{0}+y\right)+P(y)\right)=f\left(x_{0}+y-y\right)+2 y=f\left(x_{0}\right)+2 y
$$
Assume that $g=0$. ... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"solution_match": "\nSolution 2",
"tier": "T1",
"year": "2024"
} |
An integer \(n > 1\) is called good if there exists a permutation \(a_{1},a_{2},a_{3},\ldots ,a_{n}\) of the numbers \(1,2,3,\ldots ,n\) , such that:
- \(a_{i}\) and \(a_{i + 1}\) have different parities for every \(1 \leqslant i \leqslant n - 1\) ;
- the sum \(a_{1} + a_{2} + \dots + a_{k}\) is a quadratic resid... | We will split the problem into two parts - the first one proving there are infinitely many numbers that are not good, and the second part proving there are infinitely many good numbers.
## Infinitely many numbers are not good
## Proof #1
We will show that all numbers \(n = 4^{m}\) with \(m \in \mathbb{Z}^{+}\)... | proof | Yes | Incomplete | proof | Number Theory | An integer \(n > 1\) is called good if there exists a permutation \(a_{1},a_{2},a_{3},\ldots ,a_{n}\) of the numbers \(1,2,3,\ldots ,n\) , such that:
- \(a_{i}\) and \(a_{i + 1}\) have different parities for every \(1 \leqslant i \leqslant n - 1\) ;
- the sum \(a_{1} + a_{2} + \dots + a_{k}\) is a quadratic resid... | We will split the problem into two parts - the first one proving there are infinitely many numbers that are not good, and the second part proving there are infinitely many good numbers.
## Infinitely many numbers are not good
## Proof #1
We will show that all numbers \(n = 4^{m}\) with \(m \in \mathbb{Z}^{+}\)... | {
"exam": "Balkan_MO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution ",
"tier": "T1",
"year": "2025"
} |
Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \(H\) and let \(D\) be an arbitrary interior point on side \(BC\) . Suppose \(E\) and \(F\) are points on the segments \(AB\) and \(AC\) respectively such that the quadrilaterals \(ABDF\) and \(ACDE\) are cyclic, and let \(BF\) and \(CE\) intersect at ... | 
We have \(\angle PCD = \angle ECD = \angle EAD\) and \(\angle PBD = \angle FBD = \angle FAD\) , therefore
\[\angle BPC = 180^{\circ} - \angle EAD - \angle FAD = 180^{\circ} - \angle BAC = \angle BHC,\]
meaning that \(BHPC\) is cycl... | proof | Yes | Yes | proof | Geometry | Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \(H\) and let \(D\) be an arbitrary interior point on side \(BC\) . Suppose \(E\) and \(F\) are points on the segments \(AB\) and \(AC\) respectively such that the quadrilaterals \(ABDF\) and \(ACDE\) are cyclic, and let \(BF\) and \(CE\) intersect at ... | 
We have \(\angle PCD = \angle ECD = \angle EAD\) and \(\angle PBD = \angle FBD = \angle FAD\) , therefore
\[\angle BPC = 180^{\circ} - \angle EAD - \angle FAD = 180^{\circ} - \angle BAC = \angle BHC,\]
meaning that \(BHPC\) is cycl... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2025"
} |
Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \(H\) and let \(D\) be an arbitrary interior point on side \(BC\) . Suppose \(E\) and \(F\) are points on the segments \(AB\) and \(AC\) respectively such that the quadrilaterals \(ABDF\) and \(ACDE\) are cyclic, and let \(BF\) and \(CE\) intersect at ... | As in Solution 1, we introduce the point \(Y\) and prove that the quadrilateral \(BHPC\) is cyclic and that points \(Y, D, P\) are collinear. Define point \(Z\) as the second intersection of \((CXH)\) with the line \(DX\) .
Since \(\angle HZD = \angle HZX = \angle HCX = \angle HCP = \angle HYP = \angle HYD\) , we ge... | proof | Yes | Yes | proof | Geometry | Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \(H\) and let \(D\) be an arbitrary interior point on side \(BC\) . Suppose \(E\) and \(F\) are points on the segments \(AB\) and \(AC\) respectively such that the quadrilaterals \(ABDF\) and \(ACDE\) are cyclic, and let \(BF\) and \(CE\) intersect at ... | As in Solution 1, we introduce the point \(Y\) and prove that the quadrilateral \(BHPC\) is cyclic and that points \(Y, D, P\) are collinear. Define point \(Z\) as the second intersection of \((CXH)\) with the line \(DX\) .
Since \(\angle HZD = \angle HZX = \angle HCX = \angle HCP = \angle HYP = \angle HYD\) , we ge... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2025"
} |
Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \(H\) and let \(D\) be an arbitrary interior point on side \(BC\) . Suppose \(E\) and \(F\) are points on the segments \(AB\) and \(AC\) respectively such that the quadrilaterals \(ABDF\) and \(ACDE\) are cyclic, and let \(BF\) and \(CE\) intersect at ... | Let \(H_{A}\) and \(H_{C}\) be the feet of the altitudes from \(A\) and \(C\) , respectively. Also, let \(B A\) and \(B H\) meet \(C L\) at \(Z\) and \(T\) respectively.
We prove that \(B C P H\) is cyclic as in Solution 1. \(C H_{A}H C_{A}\) and \(C D E A\) are cyclic, so \(\angle B H_{A}H_{C} = \angle B D E = \ang... | proof | Yes | Yes | proof | Geometry | Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \(H\) and let \(D\) be an arbitrary interior point on side \(BC\) . Suppose \(E\) and \(F\) are points on the segments \(AB\) and \(AC\) respectively such that the quadrilaterals \(ABDF\) and \(ACDE\) are cyclic, and let \(BF\) and \(CE\) intersect at ... | Let \(H_{A}\) and \(H_{C}\) be the feet of the altitudes from \(A\) and \(C\) , respectively. Also, let \(B A\) and \(B H\) meet \(C L\) at \(Z\) and \(T\) respectively.
We prove that \(B C P H\) is cyclic as in Solution 1. \(C H_{A}H C_{A}\) and \(C D E A\) are cyclic, so \(\angle B H_{A}H_{C} = \angle B D E = \ang... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution 3",
"tier": "T1",
"year": "2025"
} |
Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \(H\) and let \(D\) be an arbitrary interior point on side \(BC\) . Suppose \(E\) and \(F\) are points on the segments \(AB\) and \(AC\) respectively such that the quadrilaterals \(ABDF\) and \(ACDE\) are cyclic, and let \(BF\) and \(CE\) intersect at ... | Same as in Solution 3, we prove that \(H_{C}H_{A}\parallel E D\parallel l\) , where \(l\) is the tangent to ( \(B H P C\) ) at \(C\) . Now define \(L\) as the intersection of \(A H\) and \(D X\) . Apply Desargues's theorem on triangles \(\triangle B H_{A}H_{C}\) and \(\triangle X L C\) . Since \(L H_{A}\) , \(C H_{C}\)... | proof | Yes | Yes | proof | Geometry | Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \(H\) and let \(D\) be an arbitrary interior point on side \(BC\) . Suppose \(E\) and \(F\) are points on the segments \(AB\) and \(AC\) respectively such that the quadrilaterals \(ABDF\) and \(ACDE\) are cyclic, and let \(BF\) and \(CE\) intersect at ... | Same as in Solution 3, we prove that \(H_{C}H_{A}\parallel E D\parallel l\) , where \(l\) is the tangent to ( \(B H P C\) ) at \(C\) . Now define \(L\) as the intersection of \(A H\) and \(D X\) . Apply Desargues's theorem on triangles \(\triangle B H_{A}H_{C}\) and \(\triangle X L C\) . Since \(L H_{A}\) , \(C H_{C}\)... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution 4",
"tier": "T1",
"year": "2025"
} |
Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \(H\) and let \(D\) be an arbitrary interior point on side \(BC\) . Suppose \(E\) and \(F\) are points on the segments \(AB\) and \(AC\) respectively such that the quadrilaterals \(ABDF\) and \(ACDE\) are cyclic, and let \(BF\) and \(CE\) intersect at ... | Let \(H_{A},H_{B},H_{C}\) to be the feet of the altitudes. We again obtain that \(B H C P\) is cyclic and that \(D E\parallel L C\) . We want to prove \(L C,A H\) and \(D X\) are concurrent. Applying Trigonometric Ceva's theorem to \(\triangle A E D\) , we need to prove:
\[\frac{\sin\angle A B H}{\sin\angle H B C}\c... | proof | Yes | Yes | proof | Geometry | Let \(\triangle ABC\) be an acute- angled triangle with orthocentre \(H\) and let \(D\) be an arbitrary interior point on side \(BC\) . Suppose \(E\) and \(F\) are points on the segments \(AB\) and \(AC\) respectively such that the quadrilaterals \(ABDF\) and \(ACDE\) are cyclic, and let \(BF\) and \(CE\) intersect at ... | Let \(H_{A},H_{B},H_{C}\) to be the feet of the altitudes. We again obtain that \(B H C P\) is cyclic and that \(D E\parallel L C\) . We want to prove \(L C,A H\) and \(D X\) are concurrent. Applying Trigonometric Ceva's theorem to \(\triangle A E D\) , we need to prove:
\[\frac{\sin\angle A B H}{\sin\angle H B C}\c... | {
"exam": "Balkan_MO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution 5",
"tier": "T1",
"year": "2025"
} |
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