problem stringlengths 14 10.4k | solution stringlengths 1 24.1k | answer stringlengths 1 250 ⌀ | problem_is_valid stringclasses 4
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values | problem_raw stringlengths 14 10.4k | solution_raw stringlengths 1 24.1k | metadata dict |
|---|---|---|---|---|---|---|---|---|---|
Find all functions \(f: \mathbb{R} \to \mathbb{R}\) such that, for all real numbers \(x\) and \(y\),
\[f(x + y f(x)) + y = x y + f(x + y).\] | .
Let \(P(x,y)\) denote the given relation. If there is an \(a\in \mathbb{R}\) such that \(f(a) = 0\) , then \(P(a,y)\) gives that \(y = a y + f(a + y)\) , and so \(f\) must be linear. Then we can easily check and get that the only linear solutions are \(f(x) = x\) and \(f(x) = 2 - x\) ( \(x\in \mathbb{R}\) ).
No... | f(x) = x \text{ and } f(x) = 2 - x | Yes | Yes | math-word-problem | Algebra | Find all functions \(f: \mathbb{R} \to \mathbb{R}\) such that, for all real numbers \(x\) and \(y\),
\[f(x + y f(x)) + y = x y + f(x + y).\] | .
Let \(P(x,y)\) denote the given relation. If there is an \(a\in \mathbb{R}\) such that \(f(a) = 0\) , then \(P(a,y)\) gives that \(y = a y + f(a + y)\) , and so \(f\) must be linear. Then we can easily check and get that the only linear solutions are \(f(x) = x\) and \(f(x) = 2 - x\) ( \(x\in \mathbb{R}\) ).
No... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution 1",
"tier": "T1",
"year": "2025"
} |
Find all functions \(f: \mathbb{R} \to \mathbb{R}\) such that, for all real numbers \(x\) and \(y\),
\[f(x + y f(x)) + y = x y + f(x + y).\] | .
Let \(P(x,y)\) denote the given relation. Similarly to the first solution, if a root exists ( \(f(a) = 0\) for any \(a\) ), we get that the function is linear and that the two solutions are \(f(x) = x\) and \(f(x) = 2 - x\) . Assertion \(P(x,c - x)\) gives us the following relation:
\[f(x + (c - x)f(x)) = (c - ... | proof | Yes | Yes | math-word-problem | Algebra | Find all functions \(f: \mathbb{R} \to \mathbb{R}\) such that, for all real numbers \(x\) and \(y\),
\[f(x + y f(x)) + y = x y + f(x + y).\] | .
Let \(P(x,y)\) denote the given relation. Similarly to the first solution, if a root exists ( \(f(a) = 0\) for any \(a\) ), we get that the function is linear and that the two solutions are \(f(x) = x\) and \(f(x) = 2 - x\) . Assertion \(P(x,c - x)\) gives us the following relation:
\[f(x + (c - x)f(x)) = (c - ... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution 2",
"tier": "T1",
"year": "2025"
} |
Find all functions \(f: \mathbb{R} \to \mathbb{R}\) such that, for all real numbers \(x\) and \(y\),
\[f(x + y f(x)) + y = x y + f(x + y).\] | .
Let \(P(x,y)\) denote the given relation. As in the previous solutions, if a root exists, then we are done. From \(P(1,y)\) we obtain \(f(1 + yc) = f(1 + y)\) , where we have put \(c = f(1)\) . From the substitution \(P(1 + x,y)\) , we get:
\[f(1 + x + yf(1 + x)) + y = (1 + x)y + f(1 + x + y) \quad (1)\]
Sub... | c^2 = 1 | Yes | Yes | math-word-problem | Algebra | Find all functions \(f: \mathbb{R} \to \mathbb{R}\) such that, for all real numbers \(x\) and \(y\),
\[f(x + y f(x)) + y = x y + f(x + y).\] | .
Let \(P(x,y)\) denote the given relation. As in the previous solutions, if a root exists, then we are done. From \(P(1,y)\) we obtain \(f(1 + yc) = f(1 + y)\) , where we have put \(c = f(1)\) . From the substitution \(P(1 + x,y)\) , we get:
\[f(1 + x + yf(1 + x)) + y = (1 + x)y + f(1 + x + y) \quad (1)\]
Sub... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution 3",
"tier": "T1",
"year": "2025"
} |
Find all functions \(f: \mathbb{R} \to \mathbb{R}\) such that, for all real numbers \(x\) and \(y\),
\[f(x + y f(x)) + y = x y + f(x + y).\] | . (by Stefan Šebez)
Let \(P(x,y)\) denote the given relation. Putting \(P(0,x + y)\) gives us:
\[f((x + y)f(0)) + x + y = 0 + f(x + y)\]
Subtracting this identity from the relation \(P(x,y)\) yields:
\[P(x + yf(x)) - P((x + y)f(0)) = xy + x\]
Suppose that \(f(x)\neq f(0)\) for some \(x\in \mathbf{R}\) (t... | proof | Yes | Yes | math-word-problem | Algebra | Find all functions \(f: \mathbb{R} \to \mathbb{R}\) such that, for all real numbers \(x\) and \(y\),
\[f(x + y f(x)) + y = x y + f(x + y).\] | . (by Stefan Šebez)
Let \(P(x,y)\) denote the given relation. Putting \(P(0,x + y)\) gives us:
\[f((x + y)f(0)) + x + y = 0 + f(x + y)\]
Subtracting this identity from the relation \(P(x,y)\) yields:
\[P(x + yf(x)) - P((x + y)f(0)) = xy + x\]
Suppose that \(f(x)\neq f(0)\) for some \(x\in \mathbf{R}\) (t... | {
"exam": "Balkan_MO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution 4",
"tier": "T1",
"year": "2025"
} |
There are \(n\) cities in a country, where \(n \geqslant 100\) is an integer. Some pairs of cities are connected by direct (two- way) flights. For two cities \(A\) and \(B\) we define:
- a path between \(A\) and \(B\) as a sequence of distinct cities \(A = C_{0}, C_{1}, \ldots , C_{k}, C_{k + 1} = B\) , \(k \geqslan... | Use the obvious graph interpretation. We show that any such graph is one of the following: the full graph \(K_{n}\) , the circular graph \(C_{n}\) , and for \(n\) even, the bipartite graph \(K_{\frac{n}{2}, \frac{n}{2}}\) . First, we show that these graphs satisfy the condition.
- For \(K_{n}\) , we can choose any l... | \frac{n(n - 1)}{2}, n, \frac{n^2}{4} | Yes | Incomplete | math-word-problem | Combinatorics | There are \(n\) cities in a country, where \(n \geqslant 100\) is an integer. Some pairs of cities are connected by direct (two- way) flights. For two cities \(A\) and \(B\) we define:
- a path between \(A\) and \(B\) as a sequence of distinct cities \(A = C_{0}, C_{1}, \ldots , C_{k}, C_{k + 1} = B\) , \(k \geqslan... | Use the obvious graph interpretation. We show that any such graph is one of the following: the full graph \(K_{n}\) , the circular graph \(C_{n}\) , and for \(n\) even, the bipartite graph \(K_{\frac{n}{2}, \frac{n}{2}}\) . First, we show that these graphs satisfy the condition.
- For \(K_{n}\) , we can choose any l... | {
"exam": "Balkan_MO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "Balkan_MO/segmented/en-2025-BMO-type1.jsonl",
"solution_match": "# Solution ",
"tier": "T1",
"year": "2025"
} |
Let $a, b, c$ be positive real numbers. Prove that
$$
\sqrt{a^{3} b+a^{3} c}+\sqrt{b^{3} c+b^{3} a}+\sqrt{c^{3} a+c^{3} b} \geq \frac{4}{3}(a b+b c+c a)
$$ | W.L.O.G. $a \geq b \geq c$.
$$
\begin{gathered}
a \geq b \geq c \Rightarrow a b \geq a c \geq b c \Rightarrow a b+a c \geq a b+b c \geq a c+b c \Rightarrow \sqrt{a b+a c} \geq \sqrt{b c+b a} \geq \sqrt{a c+b c} \\
\sqrt{a^{3} b+a^{3} c}+\sqrt{b^{3} c+b^{3} a}+\sqrt{c^{3} a+c^{3} b}=a \sqrt{a b+a c}+b \sqrt{b c+b a}+c ... | proof | Yes | Yes | proof | Inequalities | Let $a, b, c$ be positive real numbers. Prove that
$$
\sqrt{a^{3} b+a^{3} c}+\sqrt{b^{3} c+b^{3} a}+\sqrt{c^{3} a+c^{3} b} \geq \frac{4}{3}(a b+b c+c a)
$$ | W.L.O.G. $a \geq b \geq c$.
$$
\begin{gathered}
a \geq b \geq c \Rightarrow a b \geq a c \geq b c \Rightarrow a b+a c \geq a b+b c \geq a c+b c \Rightarrow \sqrt{a b+a c} \geq \sqrt{b c+b a} \geq \sqrt{a c+b c} \\
\sqrt{a^{3} b+a^{3} c}+\sqrt{b^{3} c+b^{3} a}+\sqrt{c^{3} a+c^{3} b}=a \sqrt{a b+a c}+b \sqrt{b c+b a}+c ... | {
"exam": "Balkan_Shortlist",
"problem_label": "A1",
"problem_match": "\nA1.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
For all $x, y, z>0$ satisfying $\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y} \leq x+y+z$, prove that
$$
\frac{1}{x^{2}+y+z}+\frac{1}{y^{2}+z+x}+\frac{1}{z^{2}+x+y} \leq 1
$$ | By Cauchy-Schwarz inequality, we have
$$
\left(x^{2}+y+z\right)\left(y^{2}+y z^{2}+z x^{2}\right) \geq(x y+y z+z x)^{2}
$$
and hence we obtain that
$$
\frac{1}{x^{2}+y+z}+\frac{1}{y^{2}+z+x}+\frac{1}{z^{2}+x+y} \leq \frac{2\left(x y^{2}+y z^{2}+z x^{2}\right)+x^{2}+y^{2}+z^{2}}{(x y+y z+z x)^{2}}
$$
Using the condi... | proof | Yes | Yes | proof | Inequalities | For all $x, y, z>0$ satisfying $\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y} \leq x+y+z$, prove that
$$
\frac{1}{x^{2}+y+z}+\frac{1}{y^{2}+z+x}+\frac{1}{z^{2}+x+y} \leq 1
$$ | By Cauchy-Schwarz inequality, we have
$$
\left(x^{2}+y+z\right)\left(y^{2}+y z^{2}+z x^{2}\right) \geq(x y+y z+z x)^{2}
$$
and hence we obtain that
$$
\frac{1}{x^{2}+y+z}+\frac{1}{y^{2}+z+x}+\frac{1}{z^{2}+x+y} \leq \frac{2\left(x y^{2}+y z^{2}+z x^{2}\right)+x^{2}+y^{2}+z^{2}}{(x y+y z+z x)^{2}}
$$
Using the condi... | {
"exam": "Balkan_Shortlist",
"problem_label": "A2",
"problem_match": "\nA2.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2016"
} |
Find all monotonic functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the condition: for every real number $x$ and every natural number $n$
$$
\left|\sum_{i=1}^{n} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<C
$$
where $C>0$ is independent of $x$ and $f^{2}(x)=f(f(x))$. | From the condition of the problem we get $\left|\sum_{i=1}^{n-1} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<C$. Then $\left|n\left(f(x+n+1)-f^{2}(x+n)\right)\right|=\left|\sum_{i=1}^{n} i\left(f(x+i+1)-f^{2}(x+i)\right)-\sum_{i=1}^{n-1} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<2 C$ implying $\left|f(x+n+1)-f^{2}(x+n)\rig... | f(y)=y+1 | Yes | Yes | proof | Algebra | Find all monotonic functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the condition: for every real number $x$ and every natural number $n$
$$
\left|\sum_{i=1}^{n} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<C
$$
where $C>0$ is independent of $x$ and $f^{2}(x)=f(f(x))$. | From the condition of the problem we get $\left|\sum_{i=1}^{n-1} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<C$. Then $\left|n\left(f(x+n+1)-f^{2}(x+n)\right)\right|=\left|\sum_{i=1}^{n} i\left(f(x+i+1)-f^{2}(x+i)\right)-\sum_{i=1}^{n-1} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<2 C$ implying $\left|f(x+n+1)-f^{2}(x+n)\rig... | {
"exam": "Balkan_Shortlist",
"problem_label": "A3",
"problem_match": "\n## A3.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
The positive real numbers $a, b, c$ satisfy the equality $a+b+c=1$. For every natural number $n$ find the minimal possible value of the expression
$$
E=\frac{a^{-n}+b}{1-a}+\frac{b^{-n}+c}{1-b}+\frac{c^{-n}+a}{1-c}
$$ | We transform the first term of the expression $E$ in the following way:
$$
\begin{aligned}
\frac{a^{-n}+b}{1-a}=\frac{1+a^{n} b}{a^{n}(b+c)}=\frac{a^{n+1}+a^{n} b+1-a^{n+1}}{a^{n}(b+c)} & =\frac{a^{n}(a+b)+(1-a)\left(1+a+a^{2}+\ldots+a^{n}\right)}{a^{n}(b+c)} \\
\frac{a^{n}(a+b)}{a^{n}(b+c)}+\frac{(b+c)\left(1+a+a^{2}... | \frac{3^{n+2}+3}{2} | Yes | Incomplete | math-word-problem | Inequalities | The positive real numbers $a, b, c$ satisfy the equality $a+b+c=1$. For every natural number $n$ find the minimal possible value of the expression
$$
E=\frac{a^{-n}+b}{1-a}+\frac{b^{-n}+c}{1-b}+\frac{c^{-n}+a}{1-c}
$$ | We transform the first term of the expression $E$ in the following way:
$$
\begin{aligned}
\frac{a^{-n}+b}{1-a}=\frac{1+a^{n} b}{a^{n}(b+c)}=\frac{a^{n+1}+a^{n} b+1-a^{n+1}}{a^{n}(b+c)} & =\frac{a^{n}(a+b)+(1-a)\left(1+a+a^{2}+\ldots+a^{n}\right)}{a^{n}(b+c)} \\
\frac{a^{n}(a+b)}{a^{n}(b+c)}+\frac{(b+c)\left(1+a+a^{2}... | {
"exam": "Balkan_Shortlist",
"problem_label": "A4",
"problem_match": "\n## A4.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Let $a, b, c$ and $d$ be real numbers such that $a+b+c+d=2$ and $a b+b c+c d+d a+a c+b d=0$.
Find the minimum value and the maximum value of the product $a b c d$. | Let's find the minimum first.
$$
a^{2}+b^{2}+c^{2}+d^{2}=(a+b+c+d)^{2}-2(a b+b c+c d+d a+a c+b d)=4
$$
By AM-GM, $4=a^{2}+b^{2}+c^{2}+d^{2} \geq 4 \sqrt{|a b c d|} \Rightarrow 1 \geq|a b c d| \Rightarrow a b c d \geq-1$.
Note that if $a=b=c=1$ and $d=-1$, then $a b c d=-1$.
We'll find the maximum. We search for $a b ... | \min (a b c d)=-1 \text{ and } \max (a b c d)=\frac{1}{4} | Yes | Yes | math-word-problem | Algebra | Let $a, b, c$ and $d$ be real numbers such that $a+b+c+d=2$ and $a b+b c+c d+d a+a c+b d=0$.
Find the minimum value and the maximum value of the product $a b c d$. | Let's find the minimum first.
$$
a^{2}+b^{2}+c^{2}+d^{2}=(a+b+c+d)^{2}-2(a b+b c+c d+d a+a c+b d)=4
$$
By AM-GM, $4=a^{2}+b^{2}+c^{2}+d^{2} \geq 4 \sqrt{|a b c d|} \Rightarrow 1 \geq|a b c d| \Rightarrow a b c d \geq-1$.
Note that if $a=b=c=1$ and $d=-1$, then $a b c d=-1$.
We'll find the maximum. We search for $a b ... | {
"exam": "Balkan_Shortlist",
"problem_label": "A5",
"problem_match": "\n## A5.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Prove that there is no function from positive real numbers to itself, $f:(0,+\infty) \rightarrow(0,+\infty)$ such that:
$$
f(f(x)+y)=f(x)+3 x+y f(y) \quad \text {,for every } \quad x, y \in(0,+\infty)
$$ | First we prove that $f(x) \geq x$ for all $x>0$.
Indeed, if there is an $a>0$ with $f(a)<a$ then from the initial for $x=a$ and $y=a-f(a)>0$ we get that $3 a+(a-f(a)) f(a-f(a))=0$. This is absurd since $3 a+(a-f(a)) f(a-f(a))>0$.
So we have that
$$
f(x) \geq x \quad \text {,for all } \quad x>0
$$
Then using (1) we ha... | proof | Yes | Yes | proof | Algebra | Prove that there is no function from positive real numbers to itself, $f:(0,+\infty) \rightarrow(0,+\infty)$ such that:
$$
f(f(x)+y)=f(x)+3 x+y f(y) \quad \text {,for every } \quad x, y \in(0,+\infty)
$$ | First we prove that $f(x) \geq x$ for all $x>0$.
Indeed, if there is an $a>0$ with $f(a)<a$ then from the initial for $x=a$ and $y=a-f(a)>0$ we get that $3 a+(a-f(a)) f(a-f(a))=0$. This is absurd since $3 a+(a-f(a)) f(a-f(a))>0$.
So we have that
$$
f(x) \geq x \quad \text {,for all } \quad x>0
$$
Then using (1) we ha... | {
"exam": "Balkan_Shortlist",
"problem_label": "A6",
"problem_match": "\n## A6.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Find all integers $n \geq 2$ for which there exist the real numbers $a_{k}, 1 \leq k \leq n$, which are satisfying the following conditions:
$$
\sum_{k=1}^{n} a_{k}=0, \sum_{k=1}^{n} a_{k}^{2}=1 \text { and } \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1), \text { where } b=\max _{1 \leq k \leq n}... | We have: $\left(a_{k}+\frac{1}{\sqrt{n}}\right)^{2}\left(a_{k}-b\right) \leq 0 \Rightarrow\left(a_{k}^{2}+\frac{2}{\sqrt{n}} \cdot a_{k}+\frac{1}{n}\right)\left(a_{k}-b\right) \leq 0 \Rightarrow$
$$
a_{k}^{3} \leq\left(b-\frac{2}{\sqrt{n}}\right) \cdot a_{k}^{2}+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot a_{k... | n \text{ is even} | Yes | Yes | math-word-problem | Algebra | Find all integers $n \geq 2$ for which there exist the real numbers $a_{k}, 1 \leq k \leq n$, which are satisfying the following conditions:
$$
\sum_{k=1}^{n} a_{k}=0, \sum_{k=1}^{n} a_{k}^{2}=1 \text { and } \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1), \text { where } b=\max _{1 \leq k \leq n}... | We have: $\left(a_{k}+\frac{1}{\sqrt{n}}\right)^{2}\left(a_{k}-b\right) \leq 0 \Rightarrow\left(a_{k}^{2}+\frac{2}{\sqrt{n}} \cdot a_{k}+\frac{1}{n}\right)\left(a_{k}-b\right) \leq 0 \Rightarrow$
$$
a_{k}^{3} \leq\left(b-\frac{2}{\sqrt{n}}\right) \cdot a_{k}^{2}+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot a_{k... | {
"exam": "Balkan_Shortlist",
"problem_label": "A7",
"problem_match": "\nA7.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ for which $f(g(n))-g(f(n))$ is independent on $n$ for any $g: \mathbb{Z} \rightarrow \mathbb{Z}$. | First observe that if $f(n)=n$, then $f(g(n))-g(f(n))=0$. Therefore the identity function satisfies the problem condition.
If there is $n_{0}$ with $f\left(n_{0}\right) \neq n_{0}$, consider the characteristic function $g$ that is defined as $g\left(f\left(n_{0}\right)\right)=1$ and $g(n)=0$ for $n \neq f\left(n_{0}\r... | proof | Yes | Yes | math-word-problem | Algebra | Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ for which $f(g(n))-g(f(n))$ is independent on $n$ for any $g: \mathbb{Z} \rightarrow \mathbb{Z}$. | First observe that if $f(n)=n$, then $f(g(n))-g(f(n))=0$. Therefore the identity function satisfies the problem condition.
If there is $n_{0}$ with $f\left(n_{0}\right) \neq n_{0}$, consider the characteristic function $g$ that is defined as $g\left(f\left(n_{0}\right)\right)=1$ and $g(n)=0$ for $n \neq f\left(n_{0}\r... | {
"exam": "Balkan_Shortlist",
"problem_label": "A8",
"problem_match": "\n## A8.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Let positive integers $K$ and $d$ be given. Prove that there exists a positive integer $n$ and a sequence of $K$ positive integers $b_{1}, b_{2}, \ldots, b_{K}$ such that the number $n$ is a $d$-digit palindrome in all number bases $b_{1}, b_{2}, \ldots, b_{K}$. | Let a positive integer $d$ be given. We shall prove that, for each large enough $n$, the number $(n!)^{d-1}$ is a $d$-digit palindrome in all number bases $\frac{n!}{i}-1$ for $1 \leqslant i \leqslant n$. In particular, we shall prove that the digit expansion of $(n!)^{d-1}$ in the base $\frac{n!}{i}-1$ is
$$
\left\la... | proof | Yes | Yes | proof | Number Theory | Let positive integers $K$ and $d$ be given. Prove that there exists a positive integer $n$ and a sequence of $K$ positive integers $b_{1}, b_{2}, \ldots, b_{K}$ such that the number $n$ is a $d$-digit palindrome in all number bases $b_{1}, b_{2}, \ldots, b_{K}$. | Let a positive integer $d$ be given. We shall prove that, for each large enough $n$, the number $(n!)^{d-1}$ is a $d$-digit palindrome in all number bases $\frac{n!}{i}-1$ for $1 \leqslant i \leqslant n$. In particular, we shall prove that the digit expansion of $(n!)^{d-1}$ in the base $\frac{n!}{i}-1$ is
$$
\left\la... | {
"exam": "Balkan_Shortlist",
"problem_label": "C1",
"problem_match": "\n## C1.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
There are 2016 costumers who entered a shop on a particular day. Every customer entered the shop exactly once. (i.e. the customer entered the shop, stayed there for some time and then left the shop without returning back.)
Find the maximal $k$ such that the following holds:
There are $k$ customers such that either all ... | We show that the maximal $k$ is 45 .
First we show that no larger $k$ can be achieved: We break the day at 45 disjoint time intervals and assume that at each time interval there were exactly 45 costumers who stayed in the shop only during that time interval (except in the last interval in which there were only 36 custo... | 45 | Yes | Yes | math-word-problem | Combinatorics | There are 2016 costumers who entered a shop on a particular day. Every customer entered the shop exactly once. (i.e. the customer entered the shop, stayed there for some time and then left the shop without returning back.)
Find the maximal $k$ such that the following holds:
There are $k$ customers such that either all ... | We show that the maximal $k$ is 45 .
First we show that no larger $k$ can be achieved: We break the day at 45 disjoint time intervals and assume that at each time interval there were exactly 45 costumers who stayed in the shop only during that time interval (except in the last interval in which there were only 36 custo... | {
"exam": "Balkan_Shortlist",
"problem_label": "C2",
"problem_match": "\n## C2.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
The plane is divided into unit squares by means of two sets of parallel lines. The unit squares are coloured in 1201 colours so that no rectangle of perimeter 100 contains two squares of the same colour. Show that no rectangle of size $1 \times 1201$ contains two squares of the same colour. | 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 translate of $D$ is called a diamond.
Since any two unit squares that belong to the same diamond also... | proof | Yes | Yes | proof | Combinatorics | The plane is divided into unit squares by means of two sets of parallel lines. The unit squares are coloured in 1201 colours so that no rectangle of perimeter 100 contains two squares of the same colour. Show that no rectangle of size $1 \times 1201$ contains two squares of the same colour. | 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 translate of $D$ is called a diamond.
Since any two unit squares that belong to the same diamond also... | {
"exam": "Balkan_Shortlist",
"problem_label": "C3",
"problem_match": "\n## C3.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
The point $M$ lies on the side $A B$ of the circumscribed quadrilateral $A B C D$. The points $I_{1}, I_{2}$, and $I_{3}$ are the incenters of $\triangle M B C, \triangle M C D$, and $\triangle M D A$. Show that the points $M, I_{1}, I_{2}$, and $I_{3}$ lie on a circle.
![](https://cdn.mathpix.com/cropped/2024_12_07_3d... | Lemma. Let $I$ be the incenter of $\triangle A B C$ and let the points $P$ and $Q$ lie on the lines $A B$ and $A C$. Then the points $A, I, P$, and $Q$ lie on a circle if and only if
$$
\overline{B P}+\overline{C Q}=B C
$$
where $\overline{B P}$ equals $|B P|$ if $P$ lies in the ray $B A \rightarrow$ and $-|B P|$ if ... | proof | Yes | Yes | proof | Geometry | The point $M$ lies on the side $A B$ of the circumscribed quadrilateral $A B C D$. The points $I_{1}, I_{2}$, and $I_{3}$ are the incenters of $\triangle M B C, \triangle M C D$, and $\triangle M D A$. Show that the points $M, I_{1}, I_{2}$, and $I_{3}$ lie on a circle.
![](https://cdn.mathpix.com/cropped/2024_12_07_3d... | Lemma. Let $I$ be the incenter of $\triangle A B C$ and let the points $P$ and $Q$ lie on the lines $A B$ and $A C$. Then the points $A, I, P$, and $Q$ lie on a circle if and only if
$$
\overline{B P}+\overline{C Q}=B C
$$
where $\overline{B P}$ equals $|B P|$ if $P$ lies in the ray $B A \rightarrow$ and $-|B P|$ if ... | {
"exam": "Balkan_Shortlist",
"problem_label": "G1",
"problem_match": "\n## G1.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Let $A B C D$ be a cyclic quadrilateral, with $A B<C D$, whose diagonals intersect at the point $F$ and $A D, B C$ intersect at the point $E$. Let also $K, L$ be the projections of $F$ onto the sides $A D, B C$ respectively, and $M, S, T$ be the midpoints of $E F, C F, D F$. Prove that the second intersection point of ... | Let $N$ be the midpoint of $C D$. We will prove that the circumcircles of the triangles $M K T, M L S$ pass through $N$.
We will prove first 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 of the triangle $E F C$, so it pas... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a cyclic quadrilateral, with $A B<C D$, whose diagonals intersect at the point $F$ and $A D, B C$ intersect at the point $E$. Let also $K, L$ be the projections of $F$ onto the sides $A D, B C$ respectively, and $M, S, T$ be the midpoints of $E F, C F, D F$. Prove that the second intersection point of ... | Let $N$ be the midpoint of $C D$. We will prove that the circumcircles of the triangles $M K T, M L S$ pass through $N$.
We will prove first 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 of the triangle $E F C$, so it pas... | {
"exam": "Balkan_Shortlist",
"problem_label": "G2",
"problem_match": "\nG2.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Given that $A B C$ is a triangle where $A B<A C$. On the half-lines $B A$ and $C A$ we take points $F$ and $E$ respectively such that $B F=C E=B C$. Let $M, N$ and $H$ be the mid-points of the segments $B F, C E$ and $B C$ respectively and $K$ and $O$ be the circumcircles of the triangles $A B C$ and $M N H$ respective... | The circumcenter of the triangle $\triangle M N H$ coincides with the incentre of the triangle $\triangle A B C$ because the triangles $\triangle B M H$ and $\triangle N H C$ are isosceles and therefore the perpendiculars of the $M H, H N$ are also the bisectors of the angles $\angle A B C, \angle A C B$, respectively.... | proof | Yes | Yes | proof | Geometry | Given that $A B C$ is a triangle where $A B<A C$. On the half-lines $B A$ and $C A$ we take points $F$ and $E$ respectively such that $B F=C E=B C$. Let $M, N$ and $H$ be the mid-points of the segments $B F, C E$ and $B C$ respectively and $K$ and $O$ be the circumcircles of the triangles $A B C$ and $M N H$ respective... | The circumcenter of the triangle $\triangle M N H$ coincides with the incentre of the triangle $\triangle A B C$ because the triangles $\triangle B M H$ and $\triangle N H C$ are isosceles and therefore the perpendiculars of the $M H, H N$ are also the bisectors of the angles $\angle A B C, \angle A C B$, respectively.... | {
"exam": "Balkan_Shortlist",
"problem_label": "G3",
"problem_match": "\nG3.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Find all natural numbers $n$ for which $1^{\phi(n)}+2^{\phi(n)}+\ldots+n^{\phi(n)}$ is coprime with $n$. | Consider the given expression $(\bmod p)$ where $p \mid n$ is a prime number. $p|n \Rightarrow p-1| \phi(n)$, thus for any $k$ that is not divisible by $p$, one has $k^{\phi(n)} \equiv 1(\bmod p)$. There are $n-\frac{n}{p}$ numbers among $1,2, \ldots, n$ that are not divisible by $p$. Therefore
$$
1^{\phi(n)}+2^{\phi(... | square-free integers | Yes | Yes | math-word-problem | Number Theory | Find all natural numbers $n$ for which $1^{\phi(n)}+2^{\phi(n)}+\ldots+n^{\phi(n)}$ is coprime with $n$. | Consider the given expression $(\bmod p)$ where $p \mid n$ is a prime number. $p|n \Rightarrow p-1| \phi(n)$, thus for any $k$ that is not divisible by $p$, one has $k^{\phi(n)} \equiv 1(\bmod p)$. There are $n-\frac{n}{p}$ numbers among $1,2, \ldots, n$ that are not divisible by $p$. Therefore
$$
1^{\phi(n)}+2^{\phi(... | {
"exam": "Balkan_Shortlist",
"problem_label": "N1",
"problem_match": "\nN1.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Find all odd natural numbers $n$ such that $d(n)$ is the largest divisor of the number $n$ different from $n$ $(d(n)$ is the number of divisors of the number $n$ including 1 and $n$ ). | From $d(n)\left|n, \frac{n}{d(n)}\right| n$ one obtains $\frac{n}{d(n)} \leq d(n)$.
Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{s}^{\alpha_{s}}$ where $p_{i}, 1 \leq i \leq s$ are prime numbers. The number $n$ is odd from where we get $p_{i}>2,1 \leq i \leq s$. The multiplicativity of the function $d(n)$ imp... | 9 | Incomplete | Yes | math-word-problem | Number Theory | Find all odd natural numbers $n$ such that $d(n)$ is the largest divisor of the number $n$ different from $n$ $(d(n)$ is the number of divisors of the number $n$ including 1 and $n$ ). | From $d(n)\left|n, \frac{n}{d(n)}\right| n$ one obtains $\frac{n}{d(n)} \leq d(n)$.
Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{s}^{\alpha_{s}}$ where $p_{i}, 1 \leq i \leq s$ are prime numbers. The number $n$ is odd from where we get $p_{i}>2,1 \leq i \leq s$. The multiplicativity of the function $d(n)$ imp... | {
"exam": "Balkan_Shortlist",
"problem_label": "N2",
"problem_match": "\nN2.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Find all the integer solutions $(x, y, z)$ of the equation
$$
(x+y+z)^{5}=80 x y z\left(x^{2}+y^{2}+z^{2}\right)
$$ | We directly check the identity
$$
(x+y+z)^{5}-(-x+y+z)^{5}-(x-y+z)^{5}-(x+y-z)^{5}=80 x y z\left(x^{2}+y^{2}+z^{2}\right)
$$
Therefore, if integers $x, y$ and $z$ satisfy the equation from the statement, we then have
$$
(-x+y+z)^{5}+(x-y+z)^{5}+(x+y-z)^{5}=0
$$
By Fermat's theorem at least one of the parenthesis eq... | (x, y, z) \in\{(0, t,-t),(t, 0,-t),(t,-t, 0): t \in \mathbb{Z}\} | Yes | Yes | math-word-problem | Algebra | Find all the integer solutions $(x, y, z)$ of the equation
$$
(x+y+z)^{5}=80 x y z\left(x^{2}+y^{2}+z^{2}\right)
$$ | We directly check the identity
$$
(x+y+z)^{5}-(-x+y+z)^{5}-(x-y+z)^{5}-(x+y-z)^{5}=80 x y z\left(x^{2}+y^{2}+z^{2}\right)
$$
Therefore, if integers $x, y$ and $z$ satisfy the equation from the statement, we then have
$$
(-x+y+z)^{5}+(x-y+z)^{5}+(x+y-z)^{5}=0
$$
By Fermat's theorem at least one of the parenthesis eq... | {
"exam": "Balkan_Shortlist",
"problem_label": "N3",
"problem_match": "\nN3.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.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 for every prime $p>N, p$ divides $2(f(p))!+1$. | From the divisibility relation $p \mid 2(f(p))!+1$ we conclude that:
$$
f(p)<p, \text { for all primes } p>N
$$
In fact, if for some prime number $p$ we have $f(p) \geq p$, then $p \mid(f(p))$ ! and then $p \mid 1$, which is absurd. Now suppose that $\operatorname{deg} f=m>1$. Then $f(x)=x^{m}+Q(x), \operatorname{deg... | f(x)=x-3 | Yes | Yes | proof | Number Theory | Find all monic polynomials $f$ with integer coefficients satisfying the following condition:
There exists a positive integer $N$ such that for every prime $p>N, p$ divides $2(f(p))!+1$. | From the divisibility relation $p \mid 2(f(p))!+1$ we conclude that:
$$
f(p)<p, \text { for all primes } p>N
$$
In fact, if for some prime number $p$ we have $f(p) \geq p$, then $p \mid(f(p))$ ! and then $p \mid 1$, which is absurd. Now suppose that $\operatorname{deg} f=m>1$. Then $f(x)=x^{m}+Q(x), \operatorname{deg... | {
"exam": "Balkan_Shortlist",
"problem_label": "N4",
"problem_match": "\nN4.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
A positive integer $n$ is downhill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \geq a_{k-1} \geq \ldots \geq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and downhill-integervalued if $P(n)$ is an integer for all down... | No, it is not.
A downhill number can always be written as $a-b_{1}-b_{2}-\ldots-b_{9}$, where $a$ is of the form $99 \ldots 99$ and each $b_{i}$ either equals 0 or is of the form $\overline{11 \ldots 11}$.
Let $n$ be a positive integer. The numbers of the form $99 \ldots 99$ yield at most $n$ different remainders upon... | proof | Yes | Yes | proof | Number Theory | A positive integer $n$ is downhill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \geq a_{k-1} \geq \ldots \geq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and downhill-integervalued if $P(n)$ is an integer for all down... | No, it is not.
A downhill number can always be written as $a-b_{1}-b_{2}-\ldots-b_{9}$, where $a$ is of the form $99 \ldots 99$ and each $b_{i}$ either equals 0 or is of the form $\overline{11 \ldots 11}$.
Let $n$ be a positive integer. The numbers of the form $99 \ldots 99$ yield at most $n$ different remainders upon... | {
"exam": "Balkan_Shortlist",
"problem_label": "N5",
"problem_match": "\nN5.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
A positive integer $n$ is downhill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \geq a_{k-1} \geq \ldots \geq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and downhill-integervalued if $P(n)$ is an integer for all down... | First we show that no uphill number is congruent to 10 modulo 11.
To this end, notice that an uphill number can always be written as $b_{1}+b_{2}+\ldots+b_{m}$, where $m \leq 9$, $b_{1} \leq b_{2} \leq \ldots \leq b_{m}$, and each $b_{i}$ is of the form $11 \ldots 11$. Since the remainder of each $b_{i}$ modulo 11 is e... | proof | Yes | Yes | proof | Number Theory | A positive integer $n$ is downhill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \geq a_{k-1} \geq \ldots \geq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and downhill-integervalued if $P(n)$ is an integer for all down... | First we show that no uphill number is congruent to 10 modulo 11.
To this end, notice that an uphill number can always be written as $b_{1}+b_{2}+\ldots+b_{m}$, where $m \leq 9$, $b_{1} \leq b_{2} \leq \ldots \leq b_{m}$, and each $b_{i}$ is of the form $11 \ldots 11$. Since the remainder of each $b_{i}$ modulo 11 is e... | {
"exam": "Balkan_Shortlist",
"problem_label": "N5",
"problem_match": "\nN5.",
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2016"
} |
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$
\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+a^{5}+b^{2}} \leq 1 .
$$ | First we remark that
$$
a^{5}+b^{5} \geq a b\left(a^{3}+b^{3}\right)
$$
Indeed
$$
\begin{aligned}
a^{5}+b^{5} \geq a b\left(a^{3}+b^{3}\right) & \Leftrightarrow a^{5}-a^{4} b-a b^{4}+b^{5} \geq 0 \\
& \Leftrightarrow a^{4}(a-b)-b^{4}(a-b) \geq 0 \\
(a-b)\left(a^{4}-b^{4}\right) \geq 0 & \Leftrightarrow(a-b)^{2}\left... | proof | Yes | Yes | proof | Inequalities | Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$
\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+a^{5}+b^{2}} \leq 1 .
$$ | First we remark that
$$
a^{5}+b^{5} \geq a b\left(a^{3}+b^{3}\right)
$$
Indeed
$$
\begin{aligned}
a^{5}+b^{5} \geq a b\left(a^{3}+b^{3}\right) & \Leftrightarrow a^{5}-a^{4} b-a b^{4}+b^{5} \geq 0 \\
& \Leftrightarrow a^{4}(a-b)-b^{4}(a-b) \geq 0 \\
(a-b)\left(a^{4}-b^{4}\right) \geq 0 & \Leftrightarrow(a-b)^{2}\left... | {
"exam": "Balkan_Shortlist",
"problem_label": "A1",
"problem_match": "\n## A1",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Consider the sequence of rational numbers defned by $x_{1}=\frac{4}{3}$ and $x_{n+1}=\frac{x_{n}^{2}}{x_{n}^{2}-x_{n}+1}, n \geq 1$.
Show that the numerator of the lowest term expression of each sum $\sum_{k=1}^{n} x_{k}$ is a perfect square. | It is easily seen that the $x_{n}$ are all rational numbers greater than 1 . Rewrite the recurrence formula in the form $x_{n}=\frac{1}{x_{n+1}-1}-\frac{1}{x_{n}-1}, n \geq 1$, to get
$$
\sum_{k=1}^{n} x_{k}=\frac{1}{x_{n+1}-1}-\frac{1}{x_{1}-1}=\frac{x_{n}^{2}-x_{n}+1}{x_{n}-1}-3=\frac{\left(x_{n}-2\right)^{2}}{x_{n}... | proof | Yes | Yes | proof | Algebra | Consider the sequence of rational numbers defned by $x_{1}=\frac{4}{3}$ and $x_{n+1}=\frac{x_{n}^{2}}{x_{n}^{2}-x_{n}+1}, n \geq 1$.
Show that the numerator of the lowest term expression of each sum $\sum_{k=1}^{n} x_{k}$ is a perfect square. | It is easily seen that the $x_{n}$ are all rational numbers greater than 1 . Rewrite the recurrence formula in the form $x_{n}=\frac{1}{x_{n+1}-1}-\frac{1}{x_{n}-1}, n \geq 1$, to get
$$
\sum_{k=1}^{n} x_{k}=\frac{1}{x_{n+1}-1}-\frac{1}{x_{1}-1}=\frac{x_{n}^{2}-x_{n}+1}{x_{n}-1}-3=\frac{\left(x_{n}-2\right)^{2}}{x_{n}... | {
"exam": "Balkan_Shortlist",
"problem_label": "A2",
"problem_match": "\n## A2",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Find all the functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:
$$
n+f(m) \mid f(n)+n f(m)
$$
for any $m, n \in \mathbb{N}$ | We will consider 2 cases, whether the range of the functions is infinite or finite or in other words the function take infinite or finite values.
Case 1. The Function has an infinite range. Let's fix a random natural number $n$ and let $m$ be any natural number. Then using (1) we have
$$
n+f(m)\left|f(n)+n f(m)=f(n)-n... | f(n)=n^{2} \text{ or } f(n)=1 | Yes | Yes | proof | Number Theory | Find all the functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:
$$
n+f(m) \mid f(n)+n f(m)
$$
for any $m, n \in \mathbb{N}$ | We will consider 2 cases, whether the range of the functions is infinite or finite or in other words the function take infinite or finite values.
Case 1. The Function has an infinite range. Let's fix a random natural number $n$ and let $m$ be any natural number. Then using (1) we have
$$
n+f(m)\left|f(n)+n f(m)=f(n)-n... | {
"exam": "Balkan_Shortlist",
"problem_label": "A3",
"problem_match": "\n## A3",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Let $M=\left\{(a, b, c) \in \mathbb{R}^{3}: 0<a, b, c<\frac{1}{2}\right.$ with $\left.a+b+c=1\right\}$ and $f: M \rightarrow \mathbb{R}$ given as
$$
f(a, b, c)=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{a b c}
$$
Find the best (real) bounds $\alpha$ and $\beta$ such that
$$
f(M)=\{f(a, b, c):(a, b, c... | Let $\forall(a, b, c) \in M, \alpha \leq f(a, b, c) \leq \beta$ and supose that there are no better bounds, i.e. $\alpha$ is the largest possible and $\beta$ is the smallest possible. Now,
$$
\begin{aligned}
\alpha \leq f(a, b, c) \leq \beta & \Leftrightarrow \alpha a b c \leq 4(a b+b c+c a)-1 \leq \beta a b c \\
& \L... | \alpha=8 \text{ not achievable and } \beta=9 \text{ achievable} | Yes | Yes | math-word-problem | Inequalities | Let $M=\left\{(a, b, c) \in \mathbb{R}^{3}: 0<a, b, c<\frac{1}{2}\right.$ with $\left.a+b+c=1\right\}$ and $f: M \rightarrow \mathbb{R}$ given as
$$
f(a, b, c)=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{a b c}
$$
Find the best (real) bounds $\alpha$ and $\beta$ such that
$$
f(M)=\{f(a, b, c):(a, b, c... | Let $\forall(a, b, c) \in M, \alpha \leq f(a, b, c) \leq \beta$ and supose that there are no better bounds, i.e. $\alpha$ is the largest possible and $\beta$ is the smallest possible. Now,
$$
\begin{aligned}
\alpha \leq f(a, b, c) \leq \beta & \Leftrightarrow \alpha a b c \leq 4(a b+b c+c a)-1 \leq \beta a b c \\
& \L... | {
"exam": "Balkan_Shortlist",
"problem_label": "A4",
"problem_match": "\nA4",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Consider integers $m \geq 2$ and $n \geq 1$. Show that there is a polynomial $P(x)$ of degree equal to $n$ with integer coefficients such that $P(0), P(1), \ldots, P(n)$ are all perfect powers of $m$. | Let $a_{0}, a_{1}, \ldots, a_{n}$ be integers to be chosen later, and consider the polynomial $P(x)=\frac{1}{n!} Q(x)$ where
$$
Q(x)=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} a_{k} \prod_{\substack{0 \leq i \leq n \\ i \neq k}}(x-i) .
$$
Observe that for $l \in\{0,1, \ldots, n\}$ we have
$$
\begin{aligned}
P(l) & =\frac{... | proof | Yes | Yes | proof | Number Theory | Consider integers $m \geq 2$ and $n \geq 1$. Show that there is a polynomial $P(x)$ of degree equal to $n$ with integer coefficients such that $P(0), P(1), \ldots, P(n)$ are all perfect powers of $m$. | Let $a_{0}, a_{1}, \ldots, a_{n}$ be integers to be chosen later, and consider the polynomial $P(x)=\frac{1}{n!} Q(x)$ where
$$
Q(x)=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} a_{k} \prod_{\substack{0 \leq i \leq n \\ i \neq k}}(x-i) .
$$
Observe that for $l \in\{0,1, \ldots, n\}$ we have
$$
\begin{aligned}
P(l) & =\frac{... | {
"exam": "Balkan_Shortlist",
"problem_label": "A5",
"problem_match": "\n## A5",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
$$
f\left(x+y f\left(x^{2}\right)\right)=f(x)+x f(x y)
$$
for all real numbers $x$ and $y$. | Let $P(x, y)$ be the assertion $f\left(x+y f\left(x^{2}\right)\right)=f(x)+x f(x y) . P(1,0)$ yields $f(0)=0$. If there exists $x_{0} \neq 0$ satisfying $f\left(x_{0}^{2}\right)=0$, then considering $P\left(x_{0}, y\right)$, we get $f\left(x_{0} y\right)=0$ for all $y \in \mathbb{R}$. In this case, since $x_{0} \neq 0$... | f(x)=0 \quad \forall x \in \mathbb{R} \text{ and } f(x)=x \quad \forall x \in \mathbb{R} | Yes | Yes | math-word-problem | Algebra | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
$$
f\left(x+y f\left(x^{2}\right)\right)=f(x)+x f(x y)
$$
for all real numbers $x$ and $y$. | Let $P(x, y)$ be the assertion $f\left(x+y f\left(x^{2}\right)\right)=f(x)+x f(x y) . P(1,0)$ yields $f(0)=0$. If there exists $x_{0} \neq 0$ satisfying $f\left(x_{0}^{2}\right)=0$, then considering $P\left(x_{0}, y\right)$, we get $f\left(x_{0} y\right)=0$ for all $y \in \mathbb{R}$. In this case, since $x_{0} \neq 0$... | {
"exam": "Balkan_Shortlist",
"problem_label": "A6",
"problem_match": "\nA6",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Find all pairs $(x, y)$ of positive integers such that
$$
x^{3}+y^{3}=x^{2}+42 x y+y^{2}
$$ | Let $d=(x, y)$ be the greatest common divisor of positive integers $x$ and $y$.
So, $x=a d, y=b d$, where $d \in \mathbb{N},(a, b)=1, a, b \in \mathbb{N}$. We have
$$
\begin{aligned}
x^{3}+y^{3}=x^{2}+42 x y+y^{2} & \Leftrightarrow d^{3}\left(a^{3}+b^{3}\right)=d^{2}\left(a^{2}+42 a b+b^{2}\right) \\
& \Leftrightarrow... | (1,7),(7,1),(22,22) | Yes | Yes | math-word-problem | Algebra | Find all pairs $(x, y)$ of positive integers such that
$$
x^{3}+y^{3}=x^{2}+42 x y+y^{2}
$$ | Let $d=(x, y)$ be the greatest common divisor of positive integers $x$ and $y$.
So, $x=a d, y=b d$, where $d \in \mathbb{N},(a, b)=1, a, b \in \mathbb{N}$. We have
$$
\begin{aligned}
x^{3}+y^{3}=x^{2}+42 x y+y^{2} & \Leftrightarrow d^{3}\left(a^{3}+b^{3}\right)=d^{2}\left(a^{2}+42 a b+b^{2}\right) \\
& \Leftrightarrow... | {
"exam": "Balkan_Shortlist",
"problem_label": "NT1",
"problem_match": "\n## NT1",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Find all functions $f: \mathbf{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that the number $x f(x)+f^{2}(y)+2 x f(y)$ is a perfect square for all positive integers $x, y$. | Let $p$ be a prime number. Then for $x=y=p$ the given condition gives us that the number $f^{2}(p)+3 p f(p)$ is a perfect square. Then, $f^{2}(p)+3 p f(p)=k^{2}$ for some positive integer $k$. Completing the square gives us that $(2 f(p)+3 p)^{2}-9 p^{2}=4 k^{2}$, or
$$
(2 f(p)+3 p-2 k)(2 f(p)+3 p+2 k)=9 p^{2} .
$$
S... | f(x)=x | Yes | Yes | math-word-problem | Number Theory | Find all functions $f: \mathbf{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that the number $x f(x)+f^{2}(y)+2 x f(y)$ is a perfect square for all positive integers $x, y$. | Let $p$ be a prime number. Then for $x=y=p$ the given condition gives us that the number $f^{2}(p)+3 p f(p)$ is a perfect square. Then, $f^{2}(p)+3 p f(p)=k^{2}$ for some positive integer $k$. Completing the square gives us that $(2 f(p)+3 p)^{2}-9 p^{2}=4 k^{2}$, or
$$
(2 f(p)+3 p-2 k)(2 f(p)+3 p+2 k)=9 p^{2} .
$$
S... | {
"exam": "Balkan_Shortlist",
"problem_label": "NT2",
"problem_match": "\n## NT2",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Prove that for all positive integer $n$, there is a positive integer $m$, that $7^{n} \mid 3^{m}+5^{m}-1$. | We prove this by induction on $\boldsymbol{n}$. The case $\boldsymbol{n}=1$ is indeed trivial for $\boldsymbol{m}=1$. Assume that the statement of the problem holds true for $n$, and we have $3^{m}+5^{m}-1=7^{n} l$ for some positive integer $l$ which is not divisible by 7 (if not we are done). Since $3^{6} \equiv 1(\bm... | proof | Yes | Yes | proof | Number Theory | Prove that for all positive integer $n$, there is a positive integer $m$, that $7^{n} \mid 3^{m}+5^{m}-1$. | We prove this by induction on $\boldsymbol{n}$. The case $\boldsymbol{n}=1$ is indeed trivial for $\boldsymbol{m}=1$. Assume that the statement of the problem holds true for $n$, and we have $3^{m}+5^{m}-1=7^{n} l$ for some positive integer $l$ which is not divisible by 7 (if not we are done). Since $3^{6} \equiv 1(\bm... | {
"exam": "Balkan_Shortlist",
"problem_label": "NT3",
"problem_match": "\n## NT3",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Find all pairs of positive integers $(x, y)$, such that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$. | If $y=1$, then $2 x \mid x^{2} \Leftrightarrow x=2 n, n \in \mathbb{N}$. So, the pairs $(x, y)=(2 n, 1), n \in \mathbb{N}$ satisfy the required divizibility.
Let $y>1$ such, that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$. There exist $m \in \mathbb{N}$ such that
$$
x^{2}=m\left(2 x y^{2}-y^{3}+1\right), \text { e.t.... | (x, y) \in\left\{(2 k, 1),(k, 2 k),\left(8 k^{4}-k, 2 k\right) \mid k \in \mathbb{N}\right\} | Yes | Yes | math-word-problem | Number Theory | Find all pairs of positive integers $(x, y)$, such that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$. | If $y=1$, then $2 x \mid x^{2} \Leftrightarrow x=2 n, n \in \mathbb{N}$. So, the pairs $(x, y)=(2 n, 1), n \in \mathbb{N}$ satisfy the required divizibility.
Let $y>1$ such, that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$. There exist $m \in \mathbb{N}$ such that
$$
x^{2}=m\left(2 x y^{2}-y^{3}+1\right), \text { e.t.... | {
"exam": "Balkan_Shortlist",
"problem_label": "NT4",
"problem_match": "\n## NT4",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Given a positive odd integer $n$, show that the arithmetic mean of fractional parts $\left\{\frac{k^{2 n}}{p}\right\}$, $k=1, \ldots, \frac{p-1}{2}$, is the same for infinitely many primes $p$. | We show that the arithmetic mean in question is $\frac{1}{2}$ for infinitely many primes congruent to 1 modulo 4.
Notice that $\left\{\frac{k^{2 n}}{p}\right\}=\frac{r_{k}}{p}$, where $r_{k}$ is the remainder $k^{2 n}$ leaves upon division by $p$. Clearly, the $r_{k}$ are quadratic residues modulo $p$.
If $p$ is prime,... | proof | Yes | Yes | proof | Number Theory | Given a positive odd integer $n$, show that the arithmetic mean of fractional parts $\left\{\frac{k^{2 n}}{p}\right\}$, $k=1, \ldots, \frac{p-1}{2}$, is the same for infinitely many primes $p$. | We show that the arithmetic mean in question is $\frac{1}{2}$ for infinitely many primes congruent to 1 modulo 4.
Notice that $\left\{\frac{k^{2 n}}{p}\right\}=\frac{r_{k}}{p}$, where $r_{k}$ is the remainder $k^{2 n}$ leaves upon division by $p$. Clearly, the $r_{k}$ are quadratic residues modulo $p$.
If $p$ is prime,... | {
"exam": "Balkan_Shortlist",
"problem_label": "NT5",
"problem_match": "\n## NT5",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be an acute triangle. Variable points $E$ and $F$ are on sides $A C$ and $A B$ respectively such that $B C^{2}=B A \cdot B F+C E \cdot C A$. As $E$ and $F$ vary prove that the circumcircle of $A E F$ passes through a fixed point other than $A$. | 1
Let $H$ be the ortocenter of $A B C$ and $K, L, M$ be the feet of perpendiculars respectively from $A, B, C$ to their opposite sides of $A B C$. Also let $D$ be the intersection point of lines $B E$ and $C F$. From power of point we have
$$
B A \cdot B M=B C \cdot B K
$$
and
$$
C A \cdot C L=C B \cdot C K
$$
Add... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle. Variable points $E$ and $F$ are on sides $A C$ and $A B$ respectively such that $B C^{2}=B A \cdot B F+C E \cdot C A$. As $E$ and $F$ vary prove that the circumcircle of $A E F$ passes through a fixed point other than $A$. | 1
Let $H$ be the ortocenter of $A B C$ and $K, L, M$ be the feet of perpendiculars respectively from $A, B, C$ to their opposite sides of $A B C$. Also let $D$ be the intersection point of lines $B E$ and $C F$. From power of point we have
$$
B A \cdot B M=B C \cdot B K
$$
and
$$
C A \cdot C L=C B \cdot C K
$$
Add... | {
"exam": "Balkan_Shortlist",
"problem_label": "G1",
"problem_match": "\n## G1",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be an acute triangle. Variable points $E$ and $F$ are on sides $A C$ and $A B$ respectively such that $B C^{2}=B A \cdot B F+C E \cdot C A$. As $E$ and $F$ vary prove that the circumcircle of $A E F$ passes through a fixed point other than $A$. | 2
Let the $D$ be the intersection of $B E$ and $C F$ and let circumcircle of triangle $C F A$ intersect $B C$ at point $G$. From power of point we have
$$
B G \cdot B C=B F \cdot B A
$$
... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle. Variable points $E$ and $F$ are on sides $A C$ and $A B$ respectively such that $B C^{2}=B A \cdot B F+C E \cdot C A$. As $E$ and $F$ vary prove that the circumcircle of $A E F$ passes through a fixed point other than $A$. | 2
Let the $D$ be the intersection of $B E$ and $C F$ and let circumcircle of triangle $C F A$ intersect $B C$ at point $G$. From power of point we have
$$
B G \cdot B C=B F \cdot B A
$$
... | {
"exam": "Balkan_Shortlist",
"problem_label": "G1",
"problem_match": "\n## G1",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be an acute triangle and $D$ a variable point on side $A C$. Point $E$ is on $B D$ such that $B E=\frac{B C^{2}-C D \cdot C A}{B D}$. As $D$ varies on side $A C$ prove that the circumcircle of $A D E$ passes through a fixed point other than $A$. | Let the circumcircle of triangle $C E D$ intersect $B C$ at point $G$. From power of point we have
$$
B G \cdot B C=B E \cdot B D .
$$
Combining (1) with the problem statement we get
$$
\frac{B C \cdot B C}{B D}=B E=\frac{B C^{2}-C D \cdot C A}{B D}
$$
and from here we get
$$
C D \cdot C A=B C(B C-B G)=B C \cdot C... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle and $D$ a variable point on side $A C$. Point $E$ is on $B D$ such that $B E=\frac{B C^{2}-C D \cdot C A}{B D}$. As $D$ varies on side $A C$ prove that the circumcircle of $A D E$ passes through a fixed point other than $A$. | Let the circumcircle of triangle $C E D$ intersect $B C$ at point $G$. From power of point we have
$$
B G \cdot B C=B E \cdot B D .
$$
Combining (1) with the problem statement we get
$$
\frac{B C \cdot B C}{B D}=B E=\frac{B C^{2}-C D \cdot C A}{B D}
$$
and from here we get
$$
C D \cdot C A=B C(B C-B G)=B C \cdot C... | {
"exam": "Balkan_Shortlist",
"problem_label": "G2",
"problem_match": "\nG2",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be a triangle with $A B<A C$ inscribed into a circle $c$. The tangent of $c$ at the point $C$ meets the parallel from $B$ to $A C$ at the point $D$. The tangent of $c$ at the point $B$ meets the parallel from $C$ to $A B$ at the point $E$ and the tangent of $c$ at the point $C$ at the point $L$. Suppose tha... | We will prove first that the circle $c_{1}$ is tangent to $A B$ at the point $B$. In order to prove this, we have to prove that $\measuredangle B D C=\measuredangle A B C$. Indeed, since $B D \| A C$, we have that $\measuredangle D B C=\measuredangle A C B$. Additionally, $\angle B C D=\measuredangle B A C$ (by chord a... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle with $A B<A C$ inscribed into a circle $c$. The tangent of $c$ at the point $C$ meets the parallel from $B$ to $A C$ at the point $D$. The tangent of $c$ at the point $B$ meets the parallel from $C$ to $A B$ at the point $E$ and the tangent of $c$ at the point $C$ at the point $L$. Suppose tha... | We will prove first that the circle $c_{1}$ is tangent to $A B$ at the point $B$. In order to prove this, we have to prove that $\measuredangle B D C=\measuredangle A B C$. Indeed, since $B D \| A C$, we have that $\measuredangle D B C=\measuredangle A C B$. Additionally, $\angle B C D=\measuredangle B A C$ (by chord a... | {
"exam": "Balkan_Shortlist",
"problem_label": "G3",
"problem_match": "\n## G3",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
The acuteangled triangle $A B C$ with circumcenter $O$ is given. The midpoints of the sides $B C$, $C A$ and $A B$ are $D, E$ and $F$ respectivelly. An arbitrary point $M$ on the side $B C$, different of $D$, is choosen. The straight lines $A M$ and $E F$ intersects at the point $N$ and the straight line $O N$ cut agai... | The straight lines $D O, E O$ and $F O$ are the perpendicular bisectors of the sides $B C, C A$ and $A B$ respectively. It follows that $[O M]$ is the diameter of the circumscribed circle of the triangle $O D M$ and $M P \perp O N$. The point $O$ is the ortocenter of the triangle $D E F$ (see the picture)
Let $O_{1}$ b... | proof | Yes | Yes | proof | Geometry | The acuteangled triangle $A B C$ with circumcenter $O$ is given. The midpoints of the sides $B C$, $C A$ and $A B$ are $D, E$ and $F$ respectivelly. An arbitrary point $M$ on the side $B C$, different of $D$, is choosen. The straight lines $A M$ and $E F$ intersects at the point $N$ and the straight line $O N$ cut agai... | The straight lines $D O, E O$ and $F O$ are the perpendicular bisectors of the sides $B C, C A$ and $A B$ respectively. It follows that $[O M]$ is the diameter of the circumscribed circle of the triangle $O D M$ and $M P \perp O N$. The point $O$ is the ortocenter of the triangle $D E F$ (see the picture)
Let $O_{1}$ b... | {
"exam": "Balkan_Shortlist",
"problem_label": "G4",
"problem_match": "\n## G4",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be an acute angled triangle with ortocenter $H$, centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $A H$ and $A G$ with side $B C$. Rays $M H$ and $D G$ interect $\omega$ again at $P$ and $Q$ respectively. Prove that $P D$ and $Q M$ intersect on $\omega$. | 1
Note that it is enough to prove that $\angle D P A+\angle M Q A=180^{\circ}$.
Without loss of generality assume that $A B<A C$. Let the reflection of $H$ in point $M$ be $H^{\prime}$. Since $B H C H^{\prime}$ is a paralelogram we get
$$
\measuredangle B H^{\prime} C=\angle B H C=180^{\circ}-\measuredangle B A C
$$
... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute angled triangle with ortocenter $H$, centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $A H$ and $A G$ with side $B C$. Rays $M H$ and $D G$ interect $\omega$ again at $P$ and $Q$ respectively. Prove that $P D$ and $Q M$ intersect on $\omega$. | 1
Note that it is enough to prove that $\angle D P A+\angle M Q A=180^{\circ}$.
Without loss of generality assume that $A B<A C$. Let the reflection of $H$ in point $M$ be $H^{\prime}$. Since $B H C H^{\prime}$ is a paralelogram we get
$$
\measuredangle B H^{\prime} C=\angle B H C=180^{\circ}-\measuredangle B A C
$$
... | {
"exam": "Balkan_Shortlist",
"problem_label": "G5",
"problem_match": "\n## G5",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be an acute angled triangle with ortocenter $H$, centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $A H$ and $A G$ with side $B C$. Rays $M H$ and $D G$ interect $\omega$ again at $P$ and $Q$ respectively. Prove that $P D$ and $Q M$ intersect on $\omega$. | 2
We prove that $M, D, P, A$ are concyclic same as in solution 1. Let $P D$ intersect $\omega$ again at $S$ We see that this gives us $\measuredangle S A H^{\prime}=\measuredangle S P H^{\prime}=\measuredangle D P M=\measuredangle D A M$. Combining this with $\measuredangle B A H^{\prime}=\measuredangle C A D$ we get:... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute angled triangle with ortocenter $H$, centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $A H$ and $A G$ with side $B C$. Rays $M H$ and $D G$ interect $\omega$ again at $P$ and $Q$ respectively. Prove that $P D$ and $Q M$ intersect on $\omega$. | 2
We prove that $M, D, P, A$ are concyclic same as in solution 1. Let $P D$ intersect $\omega$ again at $S$ We see that this gives us $\measuredangle S A H^{\prime}=\measuredangle S P H^{\prime}=\measuredangle D P M=\measuredangle D A M$. Combining this with $\measuredangle B A H^{\prime}=\measuredangle C A D$ we get:... | {
"exam": "Balkan_Shortlist",
"problem_label": "G5",
"problem_match": "\n## G5",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Construct outside the acute-angled triangle $A B C$ the isosceles triangles $A B A_{B}, A B B_{A}$, $A C A_{C}, A C C_{A}, B C B_{C}$ and $B C C_{B}$, so that
$$
A B=A B_{A}=B A_{B}, A C=A C_{A}=C A_{C}, B C=B C_{B}=C B_{C}
$$
and
$$
\measuredangle B A B_{A}=\measuredangle A B A_{B}=\measuredangle C A C_{A}=\measure... | Lemma. If $B C D$ is the isoceles triangle which is outside the triangle $A B C$ and has
$$
\measuredangle C B D=\measuredangle B C D=90^{\circ}-\alpha: \stackrel{\text { not }}{=} \beta,
$$
then $A D \perp B_{A} C_{A}$.
Proof of the lemma. Construct an isosceles triangle $A B E$ outside the triangle $A B C$, so that... | proof | Yes | Yes | proof | Geometry | Construct outside the acute-angled triangle $A B C$ the isosceles triangles $A B A_{B}, A B B_{A}$, $A C A_{C}, A C C_{A}, B C B_{C}$ and $B C C_{B}$, so that
$$
A B=A B_{A}=B A_{B}, A C=A C_{A}=C A_{C}, B C=B C_{B}=C B_{C}
$$
and
$$
\measuredangle B A B_{A}=\measuredangle A B A_{B}=\measuredangle C A C_{A}=\measure... | Lemma. If $B C D$ is the isoceles triangle which is outside the triangle $A B C$ and has
$$
\measuredangle C B D=\measuredangle B C D=90^{\circ}-\alpha: \stackrel{\text { not }}{=} \beta,
$$
then $A D \perp B_{A} C_{A}$.
Proof of the lemma. Construct an isosceles triangle $A B E$ outside the triangle $A B C$, so that... | {
"exam": "Balkan_Shortlist",
"problem_label": "G6",
"problem_match": "\n## G6",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2017"
} |
Let $A B C$ be an acute triangle with $A B \neq A C$ and circumcirle $\Gamma$. The angle bisector of $B A C$ intersects $B C$ and $\Gamma$ at $D$ and $E$ respectively. Circle with diameter $D E$ intersects $\Gamma$ again at $F \neq E$. Point $P$ is on $A F$ such that $P B=P C$ and $X$ and $Y$ are feet of perpendiculars... | WLOG, assume $A B<A C$. Let $M$ be the midpoint of side $B C$ and let the circumcircle of $D F E$ intersect $A F$ again at $K$. Since
$$
90^{\circ}+\measuredangle M E D=180^{\circ}-\measuredangle M D E=\measuredangle A B C+\frac{\measuredangle B A C}{2}=\measuredangle A F E=\angle D F E+\angle A F D=90+\angle A F D
$$... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute triangle with $A B \neq A C$ and circumcirle $\Gamma$. The angle bisector of $B A C$ intersects $B C$ and $\Gamma$ at $D$ and $E$ respectively. Circle with diameter $D E$ intersects $\Gamma$ again at $F \neq E$. Point $P$ is on $A F$ such that $P B=P C$ and $X$ and $Y$ are feet of perpendiculars... | WLOG, assume $A B<A C$. Let $M$ be the midpoint of side $B C$ and let the circumcircle of $D F E$ intersect $A F$ again at $K$. Since
$$
90^{\circ}+\measuredangle M E D=180^{\circ}-\measuredangle M D E=\measuredangle A B C+\frac{\measuredangle B A C}{2}=\measuredangle A F E=\angle D F E+\angle A F D=90+\angle A F D
$$... | {
"exam": "Balkan_Shortlist",
"problem_label": "G7",
"problem_match": "\nG7",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Given an acute triangle $\triangle A B C(A C \neq A B)$ and let $(C)$ be its circumcircle. The excircle $\left(C_{1}\right)$ corresponding to the vertex $A$, of center $I_{a}$, tangents to the side $B C$ at the point $D$ and to the extensions of the sides $A B, A C$ at the points $E, Z$ respectively. Let $I$ and $L$ ar... | 
We have $(e) \perp B C$ and $I_{a} D \perp B C$, so $(e) \| I_{a} D$. Let $T$. $S$ be the midpoints of the segments $H I_{a}, H D$ respectively and $Y$ the point of intersection of the lin... | proof | Yes | Yes | proof | Geometry | Given an acute triangle $\triangle A B C(A C \neq A B)$ and let $(C)$ be its circumcircle. The excircle $\left(C_{1}\right)$ corresponding to the vertex $A$, of center $I_{a}$, tangents to the side $B C$ at the point $D$ and to the extensions of the sides $A B, A C$ at the points $E, Z$ respectively. Let $I$ and $L$ ar... | 
We have $(e) \perp B C$ and $I_{a} D \perp B C$, so $(e) \| I_{a} D$. Let $T$. $S$ be the midpoints of the segments $H I_{a}, H D$ respectively and $Y$ the point of intersection of the lin... | {
"exam": "Balkan_Shortlist",
"problem_label": "G8",
"problem_match": "\nG8",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
A grasshopper is sitting at an integer point in the Euclidean plane. Each second it jumps to another integer point in such a way that the jump vector is constant. A hunter that knows neither the starting point of the grasshopper nor the jump vector (but knows that the jump vector for each second is constant) wants to c... | The hunter can catch the grasshopper. Here is the strategy for him. Let $f$ be any bijection between the set of positive integers and the set $\{((x, y),(u, v)): x, y, u, v \in \mathbb{Z}\}$, and denote
$$
f(t)=\left(\left(x_{t}, y_{t}\right),\left(u_{t}, v_{t}\right)\right) .
$$
In the second $t$, the hunter should ... | proof | Yes | Yes | proof | Combinatorics | A grasshopper is sitting at an integer point in the Euclidean plane. Each second it jumps to another integer point in such a way that the jump vector is constant. A hunter that knows neither the starting point of the grasshopper nor the jump vector (but knows that the jump vector for each second is constant) wants to c... | The hunter can catch the grasshopper. Here is the strategy for him. Let $f$ be any bijection between the set of positive integers and the set $\{((x, y),(u, v)): x, y, u, v \in \mathbb{Z}\}$, and denote
$$
f(t)=\left(\left(x_{t}, y_{t}\right),\left(u_{t}, v_{t}\right)\right) .
$$
In the second $t$, the hunter should ... | {
"exam": "Balkan_Shortlist",
"problem_label": "C1",
"problem_match": "\n## C1",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Let $n, a, b, c$ be natural numbers. Every point on the coordinate plane with integer coordinates is colored in one of $n$ colors. Prove there exists $c$ triangles whose vertices are colored in the same color, which are pairwise congruent, and which have a side whose lenght is divisible by $a$ and a side whose lenght i... | Let the colors be $d_{1}, d_{2}, d_{3} \ldots, d_{n}$. Look at the coordinates
$$
(k, 0+(n+1) a b r),(k, a b+(n+1) a b r),(k, 2 a b+(n+1) a b r), \ldots,(k, n a b+(n+1) a b r)
$$
for integers $k$ and $r$. By the pigeonhole principle there are two points of the same color. For every pair ( $k, r$ ) we say the color $d... | proof | Yes | Yes | proof | Combinatorics | Let $n, a, b, c$ be natural numbers. Every point on the coordinate plane with integer coordinates is colored in one of $n$ colors. Prove there exists $c$ triangles whose vertices are colored in the same color, which are pairwise congruent, and which have a side whose lenght is divisible by $a$ and a side whose lenght i... | Let the colors be $d_{1}, d_{2}, d_{3} \ldots, d_{n}$. Look at the coordinates
$$
(k, 0+(n+1) a b r),(k, a b+(n+1) a b r),(k, 2 a b+(n+1) a b r), \ldots,(k, n a b+(n+1) a b r)
$$
for integers $k$ and $r$. By the pigeonhole principle there are two points of the same color. For every pair ( $k, r$ ) we say the color $d... | {
"exam": "Balkan_Shortlist",
"problem_label": "C2",
"problem_match": "\nC2",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Let $n \geq 4$ points in the plane, no three of them are collinear. Prove that the number of parallelograms of area 1 , formed by these points, is at most $\frac{n^{2}-3 n}{4}$. | Fix a direction in the plane. We cannot have three points in the same line parallel to the direction so suppose that in that direction there are $k$ pairs of points, each pair belonging to a parallel line to the fixed direction. Then there are at most $k-1$ parallelograms of area 1 formed by these $k$ pairs of points.
... | \frac{n^{2}-3 n}{4} | Yes | Yes | proof | Combinatorics | Let $n \geq 4$ points in the plane, no three of them are collinear. Prove that the number of parallelograms of area 1 , formed by these points, is at most $\frac{n^{2}-3 n}{4}$. | Fix a direction in the plane. We cannot have three points in the same line parallel to the direction so suppose that in that direction there are $k$ pairs of points, each pair belonging to a parallel line to the fixed direction. Then there are at most $k-1$ parallelograms of area 1 formed by these $k$ pairs of points.
... | {
"exam": "Balkan_Shortlist",
"problem_label": "C3",
"problem_match": "\nC3",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
For any set of points $A_{1}, A_{2}, \ldots, A_{n}$ on the plane, one defines $r\left(A_{1}, A_{2}, \ldots, A_{n}\right)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \geq 3$, there are indices $i, j, k$ such that
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=r\left(A_{1}, A_... | We start with a lemma.
Lemma. If the triangle $A B C$ is acute, $r(A, B, C)$ is its circumradius and if it is obtuse. $r(A, B, C)$ is half the length of its longest side.
## Proof.
Let us do the acute case first. The circumcircle contains the vertices, so $r(A, B, C)$ is not greater than the circumradius. Now, let us... | proof | Yes | Incomplete | proof | Geometry | For any set of points $A_{1}, A_{2}, \ldots, A_{n}$ on the plane, one defines $r\left(A_{1}, A_{2}, \ldots, A_{n}\right)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \geq 3$, there are indices $i, j, k$ such that
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=r\left(A_{1}, A_... | We start with a lemma.
Lemma. If the triangle $A B C$ is acute, $r(A, B, C)$ is its circumradius and if it is obtuse. $r(A, B, C)$ is half the length of its longest side.
## Proof.
Let us do the acute case first. The circumcircle contains the vertices, so $r(A, B, C)$ is not greater than the circumradius. Now, let us... | {
"exam": "Balkan_Shortlist",
"problem_label": "C4",
"problem_match": "\n## C4",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2017"
} |
For any set of points $A_{1}, A_{2}, \ldots, A_{n}$ on the plane, one defines $r\left(A_{1}, A_{2}, \ldots, A_{n}\right)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \geq 3$, there are indices $i, j, k$ such that
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=r\left(A_{1}, A_... | The answer turns out to be $\binom{2 n-1}{n}$ if $n$ is odd and $\binom{2 n-1}{n}-2\left(\begin{array}{c}\frac{3 n}{2}-1\end{array}\right)$ if $n$ is even.
Case 1. Suppose $n$ is odd, say $n=2 m+1$. In this case we will show that any distribution of candies is legal. Thus the number of legal distributions is indeed ( $... | proof | Yes | Problem not solved | proof | Geometry | For any set of points $A_{1}, A_{2}, \ldots, A_{n}$ on the plane, one defines $r\left(A_{1}, A_{2}, \ldots, A_{n}\right)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \geq 3$, there are indices $i, j, k$ such that
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=r\left(A_{1}, A_... | The answer turns out to be $\binom{2 n-1}{n}$ if $n$ is odd and $\binom{2 n-1}{n}-2\left(\begin{array}{c}\frac{3 n}{2}-1\end{array}\right)$ if $n$ is even.
Case 1. Suppose $n$ is odd, say $n=2 m+1$. In this case we will show that any distribution of candies is legal. Thus the number of legal distributions is indeed ( $... | {
"exam": "Balkan_Shortlist",
"problem_label": "C4",
"problem_match": "\n## C4",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
What is the least positive integer $k$ such that, in every convex 101 -gon, the sum of any $k$ diagonals is greater than or equal to the sum of the remaining diagonals? | Let $P Q=1$. Consider a convex 101 -gon such that one of its vertices is at $P$ and the remaining 100 vertices are within $\varepsilon$ of $Q$ where $\varepsilon$ is an arbitrarily small positive real. Let $k+l$ equal the total number $\frac{101.98}{2}=4949$ of diagonals. When $k \leq 4851$. the sum of the $k$ shortest... | 4900 | Yes | Yes | math-word-problem | Inequalities | What is the least positive integer $k$ such that, in every convex 101 -gon, the sum of any $k$ diagonals is greater than or equal to the sum of the remaining diagonals? | Let $P Q=1$. Consider a convex 101 -gon such that one of its vertices is at $P$ and the remaining 100 vertices are within $\varepsilon$ of $Q$ where $\varepsilon$ is an arbitrarily small positive real. Let $k+l$ equal the total number $\frac{101.98}{2}=4949$ of diagonals. When $k \leq 4851$. the sum of the $k$ shortest... | {
"exam": "Balkan_Shortlist",
"problem_label": "C6",
"problem_match": "\nC6",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2017"
} |
Let $a, b, c$ be positive real numbers such that $a b c=\frac{2}{3}$. Prove that
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}
$$
(FYR Macedonia) | By the AH mean inequality, we have
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}=\frac{2}{3(a c+b c)}+\frac{2}{3(a b+a c)}+\frac{2}{3(a b+a c)} \geqslant \frac{3}{a b+a c+b c}
$$
so it only remains to prove that $\frac{3}{a b+a c+b c} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}$, or equivalently
$$
3\left(a^{3}+b... | proof | Yes | Yes | proof | Inequalities | Let $a, b, c$ be positive real numbers such that $a b c=\frac{2}{3}$. Prove that
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}
$$
(FYR Macedonia) | By the AH mean inequality, we have
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}=\frac{2}{3(a c+b c)}+\frac{2}{3(a b+a c)}+\frac{2}{3(a b+a c)} \geqslant \frac{3}{a b+a c+b c}
$$
so it only remains to prove that $\frac{3}{a b+a c+b c} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}$, or equivalently
$$
3\left(a^{3}+b... | {
"exam": "Balkan_Shortlist",
"problem_label": "A1",
"problem_match": "\nA1.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Two ants start at the same point in the plane. Each minute they choose whether to walk due north, east, south or west. They each walk 1 meter in the first minute. In each subsequent minute the distance they walk is multiplied by a rational number $q>0$. They meet after a whole number of minutes, but have not taken exac... | 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 | Combinatorics | Two ants start at the same point in the plane. Each minute they choose whether to walk due north, east, south or west. They each walk 1 meter in the first minute. In each subsequent minute the distance they walk is multiplied by a rational number $q>0$. They meet after a whole number of minutes, but have not taken exac... | 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_Shortlist",
"problem_label": "A2",
"problem_match": "\nA2.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Two ants start at the same point in the plane. Each minute they choose whether to walk due north, east, south or west. They each walk 1 meter in the first minute. In each subsequent minute the distance they walk is multiplied by a rational number $q>0$. They meet after a whole number of minutes, but have not taken exac... | 2.
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,... | 2 | Yes | Yes | math-word-problem | Combinatorics | Two ants start at the same point in the plane. Each minute they choose whether to walk due north, east, south or west. They each walk 1 meter in the first minute. In each subsequent minute the distance they walk is multiplied by a rational number $q>0$. They meet after a whole number of minutes, but have not taken exac... | 2.
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,... | {
"exam": "Balkan_Shortlist",
"problem_label": "A2",
"problem_match": "\nA2.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2018"
} |
Show that for every positive integer $n$ we have:
$$
\sum_{k=0}^{n}\left(\frac{2 n+1-k}{k+1}\right)^{k}=\left(\frac{2 n+1}{1}\right)^{0}+\left(\frac{2 n}{2}\right)^{1}+\cdots+\left(\frac{n+1}{n+1}\right)^{n} \leqslant 2^{n}
$$ | We shall prove that
$$
\binom{n}{k} \geqslant\left(\frac{2 n+1-k}{k+1}\right)^{k} \quad \text { for all } \quad k=0,1, \ldots, n
$$
The result will follow immediately, as $\sum_{k=0}^{n}\binom{n}{k}=2^{n}$.
Note that $(*)$ is trivial for $k=0$ and $k=n$. For $0<k<n$, by Hölder's inequality we have
$$
\binom{n}{k}=\l... | proof | Yes | Yes | proof | Inequalities | Show that for every positive integer $n$ we have:
$$
\sum_{k=0}^{n}\left(\frac{2 n+1-k}{k+1}\right)^{k}=\left(\frac{2 n+1}{1}\right)^{0}+\left(\frac{2 n}{2}\right)^{1}+\cdots+\left(\frac{n+1}{n+1}\right)^{n} \leqslant 2^{n}
$$ | We shall prove that
$$
\binom{n}{k} \geqslant\left(\frac{2 n+1-k}{k+1}\right)^{k} \quad \text { for all } \quad k=0,1, \ldots, n
$$
The result will follow immediately, as $\sum_{k=0}^{n}\binom{n}{k}=2^{n}$.
Note that $(*)$ is trivial for $k=0$ and $k=n$. For $0<k<n$, by Hölder's inequality we have
$$
\binom{n}{k}=\l... | {
"exam": "Balkan_Shortlist",
"problem_label": "A3",
"problem_match": "\nA3.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Show that for every positive integer $n$ we have:
$$
\sum_{k=0}^{n}\left(\frac{2 n+1-k}{k+1}\right)^{k}=\left(\frac{2 n+1}{1}\right)^{0}+\left(\frac{2 n}{2}\right)^{1}+\cdots+\left(\frac{n+1}{n+1}\right)^{n} \leqslant 2^{n}
$$ | 2.
As in the previous solution, it is enough to prove (*).
First, we prove that
$$
(n-i+1)(n-k+i)(k+1)^{2} \geqslant i(k-i+1)(2 n+1-k)^{2} \quad \text { for all } i=1,2, \ldots, k
$$
Let us denote the left hand side of the previous inequality with $L$ and the left hand side with $R$. Then
$$
\begin{aligned}
& L=(n+... | proof | Yes | Incomplete | proof | Inequalities | Show that for every positive integer $n$ we have:
$$
\sum_{k=0}^{n}\left(\frac{2 n+1-k}{k+1}\right)^{k}=\left(\frac{2 n+1}{1}\right)^{0}+\left(\frac{2 n}{2}\right)^{1}+\cdots+\left(\frac{n+1}{n+1}\right)^{n} \leqslant 2^{n}
$$ | 2.
As in the previous solution, it is enough to prove (*).
First, we prove that
$$
(n-i+1)(n-k+i)(k+1)^{2} \geqslant i(k-i+1)(2 n+1-k)^{2} \quad \text { for all } i=1,2, \ldots, k
$$
Let us denote the left hand side of the previous inequality with $L$ and the left hand side with $R$. Then
$$
\begin{aligned}
& L=(n+... | {
"exam": "Balkan_Shortlist",
"problem_label": "A3",
"problem_match": "\nA3.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2018"
} |
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that the following inequality holds:
$$
2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant 3(a+b+c+a b+b c+c a)
$$
(Romania) | First, we show that
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a b+b c+c a \quad \text { and } \quad \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a+b+c
$$
By AG inequality, we have
$$
\begin{aligned}
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} & =\frac{1}{3}\left(\frac{a}{b}+\frac{a}{b}+\frac{c}{a}\right)+\frac{1}{3}... | proof | Yes | Yes | proof | Inequalities | Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that the following inequality holds:
$$
2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant 3(a+b+c+a b+b c+c a)
$$
(Romania) | First, we show that
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a b+b c+c a \quad \text { and } \quad \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a+b+c
$$
By AG inequality, we have
$$
\begin{aligned}
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} & =\frac{1}{3}\left(\frac{a}{b}+\frac{a}{b}+\frac{c}{a}\right)+\frac{1}{3}... | {
"exam": "Balkan_Shortlist",
"problem_label": "A4",
"problem_match": "\nA4.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that the following inequality holds:
$$
2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant 3(a+b+c+a b+b c+c a)
$$
(Romania) | 2.
Set $a=x^{3}, b=y^{3}, c=z^{3}$ and denote $T_{p, q, r}=\sum_{s y m} x^{p} y^{q} z^{r}=x^{p} y^{q} z^{r}+y^{p} x^{q} z^{r}+\cdots$. The given inequality is expanded into
$$
4 T_{12,6,0}+2 T_{6,6,6} \geqslant 3 T_{8,5,5}+3 T_{7,7,4}
$$
Applying the Schur inequality on triples $\left(x^{4} y^{2}, y^{4} z^{2}, z^{4}... | proof | Yes | Yes | proof | Inequalities | Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that the following inequality holds:
$$
2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant 3(a+b+c+a b+b c+c a)
$$
(Romania) | 2.
Set $a=x^{3}, b=y^{3}, c=z^{3}$ and denote $T_{p, q, r}=\sum_{s y m} x^{p} y^{q} z^{r}=x^{p} y^{q} z^{r}+y^{p} x^{q} z^{r}+\cdots$. The given inequality is expanded into
$$
4 T_{12,6,0}+2 T_{6,6,6} \geqslant 3 T_{8,5,5}+3 T_{7,7,4}
$$
Applying the Schur inequality on triples $\left(x^{4} y^{2}, y^{4} z^{2}, z^{4}... | {
"exam": "Balkan_Shortlist",
"problem_label": "A4",
"problem_match": "\nA4.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2018"
} |
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a concave function and let $g: \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Given that
$$
f(x+y)+f(x-y)-2 f(x)=g(x) y^{2}
$$
for all $x, y \in \mathbb{R}$, prove that $f$ is a quadratic function.
(Bulgaria) | We plug in the pairs $(a, x),(a, 2 x),(a+x, x)$ and $(a-x, x)$ to get
$$
\begin{aligned}
f(a+x)+f(a-x)-2 f(a) & =g(a) x^{2} \\
f(a+2 x)+f(a-2 x)-2 f(a) & =4 g(a) x^{2} \\
f(a+2 x)+f(a)-2 f(a+x) & =g(a+x) x^{2} \\
f(a-2 x)+f(a)-2 f(a-x) & =g(a-x) x^{2}
\end{aligned}
$$
respectively. Combining these equations in the fo... | proof | Yes | Yes | proof | Algebra | Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a concave function and let $g: \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Given that
$$
f(x+y)+f(x-y)-2 f(x)=g(x) y^{2}
$$
for all $x, y \in \mathbb{R}$, prove that $f$ is a quadratic function.
(Bulgaria) | We plug in the pairs $(a, x),(a, 2 x),(a+x, x)$ and $(a-x, x)$ to get
$$
\begin{aligned}
f(a+x)+f(a-x)-2 f(a) & =g(a) x^{2} \\
f(a+2 x)+f(a-2 x)-2 f(a) & =4 g(a) x^{2} \\
f(a+2 x)+f(a)-2 f(a+x) & =g(a+x) x^{2} \\
f(a-2 x)+f(a)-2 f(a-x) & =g(a-x) x^{2}
\end{aligned}
$$
respectively. Combining these equations in the fo... | {
"exam": "Balkan_Shortlist",
"problem_label": "A5",
"problem_match": "\nA5.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Let $n$ be a positive integer and let $x_{1}, \ldots, x_{n}$ be real numbers. Show that
$$
\sum_{i=1}^{n} x_{i}^{2} \geqslant \frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}+\frac{12\left(\sum_{i=1}^{n} i x_{i}\right)^{2}}{n(n+1)(n+2)(3 n+1)}
$$ | Let $S=\frac{1}{n+1} \sum_{i=1}^{n} x_{i}$, and $y_{i}=x_{i}-S$ for $1 \leqslant i \leqslant n$. Then we have
$$
\sum_{i=1}^{n} i x_{i}=\sum_{i=1}^{n} i y_{i}+\frac{n(n+1)}{2} S
$$
and
$$
\sum_{i=1}^{n} y_{i}^{2}=\sum_{i=1}^{n} x_{i}^{2}-2 S \sum_{i=1}^{n} x_{i}+n S^{2}=\sum_{i=1}^{n} x_{i}^{2}-\frac{1}{n+1}\left(\s... | proof | Yes | Yes | proof | Inequalities | Let $n$ be a positive integer and let $x_{1}, \ldots, x_{n}$ be real numbers. Show that
$$
\sum_{i=1}^{n} x_{i}^{2} \geqslant \frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}+\frac{12\left(\sum_{i=1}^{n} i x_{i}\right)^{2}}{n(n+1)(n+2)(3 n+1)}
$$ | Let $S=\frac{1}{n+1} \sum_{i=1}^{n} x_{i}$, and $y_{i}=x_{i}-S$ for $1 \leqslant i \leqslant n$. Then we have
$$
\sum_{i=1}^{n} i x_{i}=\sum_{i=1}^{n} i y_{i}+\frac{n(n+1)}{2} S
$$
and
$$
\sum_{i=1}^{n} y_{i}^{2}=\sum_{i=1}^{n} x_{i}^{2}-2 S \sum_{i=1}^{n} x_{i}+n S^{2}=\sum_{i=1}^{n} x_{i}^{2}-\frac{1}{n+1}\left(\s... | {
"exam": "Balkan_Shortlist",
"problem_label": "A6",
"problem_match": "\nA6.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Let $N \geqslant 3$ be an odd integer. $N$ tennis players take part in a league. Before the league starts, a committee ranks the players in some order based on perceived skill. During the league, each pair of players plays exactly one match, and each match has one winner. A match is considered an upset if the winner ha... | Answer: $\frac{(N-1)(3 N-1)}{8}$.
Suppose the players are ranked $1,2, \ldots, N=2 n+1$, where 1 is the highest ranking.
For $k \leqslant n$, the player ranked $k$ could have beaten at most $k-1$ players with a higher ranking. Thus the top $n$ players could have made at most $\sum_{k=1}^{n}(k-1)=\frac{n(n-1)}{2}$ upset... | \frac{(N-1)(3 N-1)}{8} | Yes | Yes | math-word-problem | Combinatorics | Let $N \geqslant 3$ be an odd integer. $N$ tennis players take part in a league. Before the league starts, a committee ranks the players in some order based on perceived skill. During the league, each pair of players plays exactly one match, and each match has one winner. A match is considered an upset if the winner ha... | Answer: $\frac{(N-1)(3 N-1)}{8}$.
Suppose the players are ranked $1,2, \ldots, N=2 n+1$, where 1 is the highest ranking.
For $k \leqslant n$, the player ranked $k$ could have beaten at most $k-1$ players with a higher ranking. Thus the top $n$ players could have made at most $\sum_{k=1}^{n}(k-1)=\frac{n(n-1)}{2}$ upset... | {
"exam": "Balkan_Shortlist",
"problem_label": "C1",
"problem_match": "\nC1.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Let $N \geqslant 3$ be an odd integer. $N$ tennis players take part in a league. Before the league starts, a committee ranks the players in some order based on perceived skill. During the league, each pair of players plays exactly one match, and each match has one winner. A match is considered an upset if the winner ha... | 2.
Write $N=2 n+1$. We only prove the upper bound.
Consider a tournament $\mathbb{T}$ with correct final ranking, but where not everyone won $n$ matches. Let $A$ be the worst-ranked player with the maximal number of wins, and let $B$ be the best-ranked player with minimal wins. Clearly, $A$ was ranked above $B$.
Assum... | 2 | Yes | Yes | math-word-problem | Combinatorics | Let $N \geqslant 3$ be an odd integer. $N$ tennis players take part in a league. Before the league starts, a committee ranks the players in some order based on perceived skill. During the league, each pair of players plays exactly one match, and each match has one winner. A match is considered an upset if the winner ha... | 2.
Write $N=2 n+1$. We only prove the upper bound.
Consider a tournament $\mathbb{T}$ with correct final ranking, but where not everyone won $n$ matches. Let $A$ be the worst-ranked player with the maximal number of wins, and let $B$ be the best-ranked player with minimal wins. Clearly, $A$ was ranked above $B$.
Assum... | {
"exam": "Balkan_Shortlist",
"problem_label": "C1",
"problem_match": "\nC1.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"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, they choose a pile with an even number of coins and move half of the coins of this pile to the other pile. The game ends if a player cannot move, or if we reach a previously reached position. In ... | 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, they choose a pile with an even number of coins and move half of the coins of this pile to the other pile. The game ends if a player cannot move, or if we reach a previously reached position. In ... | 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_Shortlist",
"problem_label": "C2",
"problem_match": "\nC2.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
An open necklace can contain rubies, emeralds and sapphires. At every step we can perform any of the following operations:
$\left(1^{\circ}\right)$ We can replace two consecutive rubies with an emerald and a sapphire, where the emerald is on the left of the sapphire.
$\left(2^{\circ}\right)$ We can replace three consec... | For each precious stone on the necklace, we define its value as $(-1)^{r} \cdot s$, where $r$ denotes the number of emeralds and sapphires preceding it, and $s$ equals $-2,1$ or -1 for a ruby, emerald or sapphire, respectively.
The value of the necklace is equal to the sum of the values of its precious stones. We claim... | proof | Yes | Yes | proof | Combinatorics | An open necklace can contain rubies, emeralds and sapphires. At every step we can perform any of the following operations:
$\left(1^{\circ}\right)$ We can replace two consecutive rubies with an emerald and a sapphire, where the emerald is on the left of the sapphire.
$\left(2^{\circ}\right)$ We can replace three consec... | For each precious stone on the necklace, we define its value as $(-1)^{r} \cdot s$, where $r$ denotes the number of emeralds and sapphires preceding it, and $s$ equals $-2,1$ or -1 for a ruby, emerald or sapphire, respectively.
The value of the necklace is equal to the sum of the values of its precious stones. We claim... | {
"exam": "Balkan_Shortlist",
"problem_label": "C3",
"problem_match": "\nC3.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
An open necklace can contain rubies, emeralds and sapphires. At every step we can perform any of the following operations:
$\left(1^{\circ}\right)$ We can replace two consecutive rubies with an emerald and a sapphire, where the emerald is on the left of the sapphire.
$\left(2^{\circ}\right)$ We can replace three consec... | 2.
Write $a, b$ and $c$ respectively for a ruby, emerald and sapphire. Each necklace corresponds to an element of a group $G$ containing elements $a, b, c$. If we impose the conditions $a^{2}=b c, b^{3}=c a, c^{2}=1$ and $a b c=1$, the allowed operations will preserve this element. In this group we have $c=a b$ (since... | proof | Yes | Yes | proof | Combinatorics | An open necklace can contain rubies, emeralds and sapphires. At every step we can perform any of the following operations:
$\left(1^{\circ}\right)$ We can replace two consecutive rubies with an emerald and a sapphire, where the emerald is on the left of the sapphire.
$\left(2^{\circ}\right)$ We can replace three consec... | 2.
Write $a, b$ and $c$ respectively for a ruby, emerald and sapphire. Each necklace corresponds to an element of a group $G$ containing elements $a, b, c$. If we impose the conditions $a^{2}=b c, b^{3}=c a, c^{2}=1$ and $a b c=1$, the allowed operations will preserve this element. In this group we have $c=a b$ (since... | {
"exam": "Balkan_Shortlist",
"problem_label": "C3",
"problem_match": "\nC3.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2018"
} |
In an acute triangle $A B C$, the midpoint of the side $B C$ is $M$ and the centers of the excircles relative to $M$ of the triangles $A M B$ and $A M C$ are $D$ and $E$ respectively. The circumcircle of the triangle $A B D$ meets line $B C$ at $B$ and $F$. The circumcircle of the triangle $A C E$ meets line $B C$ at $... | We have $\varangle A D B=90^{\circ}-\frac{1}{2} \varangle A M B$ and $\varangle A E C=90^{\circ}-\frac{1}{2} \varangle A M C$.
Let the circles $A D B$ and $A E C$ respectively meet the line $A M$ again at points $P$ and $P^{\prime}$. Note that $M$ lies outside the circles $A B D$ and $A C E$ because $\varangle A D B+\v... | B F=C G | Yes | Yes | proof | Geometry | In an acute triangle $A B C$, the midpoint of the side $B C$ is $M$ and the centers of the excircles relative to $M$ of the triangles $A M B$ and $A M C$ are $D$ and $E$ respectively. The circumcircle of the triangle $A B D$ meets line $B C$ at $B$ and $F$. The circumcircle of the triangle $A C E$ meets line $B C$ at $... | We have $\varangle A D B=90^{\circ}-\frac{1}{2} \varangle A M B$ and $\varangle A E C=90^{\circ}-\frac{1}{2} \varangle A M C$.
Let the circles $A D B$ and $A E C$ respectively meet the line $A M$ again at points $P$ and $P^{\prime}$. Note that $M$ lies outside the circles $A B D$ and $A C E$ because $\varangle A D B+\v... | {
"exam": "Balkan_Shortlist",
"problem_label": "G1",
"problem_match": "\nG1.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Let $A B C$ be a triangle inscribed in circle $\Gamma$ with center $O$ and let $H$ its orthocenter and $K$ be the midpoint of $O H$. The tangent of $\Gamma$ at $B$ meets the perpendicular bisector of $A C$ meets at $L$ and the tangent of $\Gamma$ at $C$ meets the perpendicular bisector of $A B$ at $M$. Prove that $A K ... | The polar of $L$ with respect to $\Gamma$ is the line $\ell_{B}$ through $B$ parallel to $A C$, and the polar of $M$ with respect to $\Gamma$ is the line $\ell_{C}$ through $C$ parallel to $A B$. Therefore the pole of the line $L M$ is the intersection $A^{\prime}$ of $\ell_{B}$ and $\ell_{C}$. It follows that $O A^{\p... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle inscribed in circle $\Gamma$ with center $O$ and let $H$ its orthocenter and $K$ be the midpoint of $O H$. The tangent of $\Gamma$ at $B$ meets the perpendicular bisector of $A C$ meets at $L$ and the tangent of $\Gamma$ at $C$ meets the perpendicular bisector of $A B$ at $M$. Prove that $A K ... | The polar of $L$ with respect to $\Gamma$ is the line $\ell_{B}$ through $B$ parallel to $A C$, and the polar of $M$ with respect to $\Gamma$ is the line $\ell_{C}$ through $C$ parallel to $A B$. Therefore the pole of the line $L M$ is the intersection $A^{\prime}$ of $\ell_{B}$ and $\ell_{C}$. It follows that $O A^{\p... | {
"exam": "Balkan_Shortlist",
"problem_label": "G2",
"problem_match": "\nG2.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Let $A B C$ be a triangle inscribed in circle $\Gamma$ with center $O$ and let $H$ its orthocenter and $K$ be the midpoint of $O H$. The tangent of $\Gamma$ at $B$ meets the perpendicular bisector of $A C$ meets at $L$ and the tangent of $\Gamma$ at $C$ meets the perpendicular bisector of $A B$ at $M$. Prove that $A K ... | 2.
We introduce the complex plane such that $\Gamma$ is the unit cycle. Also, let the lower-case letters denote complex numbers corresponding to the points denoted by capital letters. First, note that $o=0, \bar{a}=1 / a, \bar{b}=1 / b$ and $\bar{c}=1 / c$.
Since $B L \perp B O$, we have
$$
\frac{b-l}{\bar{b}-\bar{l}... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a triangle inscribed in circle $\Gamma$ with center $O$ and let $H$ its orthocenter and $K$ be the midpoint of $O H$. The tangent of $\Gamma$ at $B$ meets the perpendicular bisector of $A C$ meets at $L$ and the tangent of $\Gamma$ at $C$ meets the perpendicular bisector of $A B$ at $M$. Prove that $A K ... | 2.
We introduce the complex plane such that $\Gamma$ is the unit cycle. Also, let the lower-case letters denote complex numbers corresponding to the points denoted by capital letters. First, note that $o=0, \bar{a}=1 / a, \bar{b}=1 / b$ and $\bar{c}=1 / c$.
Since $B L \perp B O$, we have
$$
\frac{b-l}{\bar{b}-\bar{l}... | {
"exam": "Balkan_Shortlist",
"problem_label": "G2",
"problem_match": "\nG2.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution",
"tier": "T1",
"year": "2018"
} |
Let $P$ be a point inside a triangle $A B C$ and let $a, b, c$ be the side lengths and $p$ the semi-perimeter of the triangle. Find the maximum value of
$$
\min \left(\frac{P A}{p-a}, \frac{P B}{p-b}, \frac{P C}{p-c}\right)
$$
over all possible choices of triangle $A B C$ and point $P$. | If $A B C$ is an equilateral triangle and $P$ its center, then $\frac{P A}{p-a}=\frac{P B}{p-b}=\frac{P C}{p-c}=\frac{2}{\sqrt{3}}$.
We shall prove that $\frac{2}{\sqrt{3}}$ is the required value. Suppose without loss of generality that $\varangle A P B \geqslant 120^{\circ}$. Then
$$
A B^{2} \geqslant P A^{2}+P B^{2}... | \frac{2}{\sqrt{3}} | Yes | Yes | math-word-problem | Geometry | Let $P$ be a point inside a triangle $A B C$ and let $a, b, c$ be the side lengths and $p$ the semi-perimeter of the triangle. Find the maximum value of
$$
\min \left(\frac{P A}{p-a}, \frac{P B}{p-b}, \frac{P C}{p-c}\right)
$$
over all possible choices of triangle $A B C$ and point $P$. | If $A B C$ is an equilateral triangle and $P$ its center, then $\frac{P A}{p-a}=\frac{P B}{p-b}=\frac{P C}{p-c}=\frac{2}{\sqrt{3}}$.
We shall prove that $\frac{2}{\sqrt{3}}$ is the required value. Suppose without loss of generality that $\varangle A P B \geqslant 120^{\circ}$. Then
$$
A B^{2} \geqslant P A^{2}+P B^{2}... | {
"exam": "Balkan_Shortlist",
"problem_label": "G3",
"problem_match": "\nG3.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
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 point $H$ is the foot of the perpendicular from $M$ to $A B$. Given that $\varangle M H C=\varangle M H D$, prove that $A B$ is a diameter of $k$... | 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 point $H$ is the foot of the perpendicular from $M$ to $A B$. Given that $\varangle M H C=\varangle M H D$, prove that $A B$ is a diameter of $k$... | 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_Shortlist",
"problem_label": "G4",
"problem_match": "\nG4.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Let $A B C$ be an acute-angled triangle with $A B<A C<B C$ and let $D$ be an arbitrary point on the extension of $B C$ beyond $C$. The circle $\gamma(A, A D)$ intersects the rays $A C$, $A B, C B$ at points $E, F, G$, respectively. The circumcircle $\omega_{1}$ of triangle $A F G$ intersects the lines $F E, B C, G E, D... | From $\varangle F A H=\varangle F G H=\varangle F G D=\frac{1}{2} \varangle F A D=90^{\circ}-\varangle A F D$ we deduce that $A H \perp$ $D F$. Similarly, $\varangle D A I=180^{\circ}-\varangle D E I=180^{\circ}-\varangle D E F=\varangle D G F=\frac{1}{2} \varangle D A F$, so we also have $A I \perp D F$. Therefore, po... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be an acute-angled triangle with $A B<A C<B C$ and let $D$ be an arbitrary point on the extension of $B C$ beyond $C$. The circle $\gamma(A, A D)$ intersects the rays $A C$, $A B, C B$ at points $E, F, G$, respectively. The circumcircle $\omega_{1}$ of triangle $A F G$ intersects the lines $F E, B C, G E, D... | From $\varangle F A H=\varangle F G H=\varangle F G D=\frac{1}{2} \varangle F A D=90^{\circ}-\varangle A F D$ we deduce that $A H \perp$ $D F$. Similarly, $\varangle D A I=180^{\circ}-\varangle D E I=180^{\circ}-\varangle D E F=\varangle D G F=\frac{1}{2} \varangle D A F$, so we also have $A I \perp D F$. Therefore, po... | {
"exam": "Balkan_Shortlist",
"problem_label": "G5",
"problem_match": "\nG5.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
In a triangle $A B C$ with $A B=A C, \omega$ is the circumcircle and $O$ its center. Let $D$ be a point on the extension of $B A$ beyond $A$. The circumcircle $\omega_{1}$ of triangle $O A D$ intersects the line $A C$ and the circle $\omega$ again at points $E$ and $G$, respectively. Point $H$ is such that $D A E H$ is... | We first observe that $\varangle D O E=\varangle D A E=2 \varangle A B C=\varangle B O A$ and hence $\varangle D O B=$ $\varangle E O A$, which together with $O B=O A$ and $\varangle O B D=\varangle B A O=\varangle O A E$ gives us $\triangle O B D \cong \triangle O A E$. Therefore $B D=A E$.
Next, $O G=O A$ implies $\v... | proof | Yes | Yes | proof | Geometry | In a triangle $A B C$ with $A B=A C, \omega$ is the circumcircle and $O$ its center. Let $D$ be a point on the extension of $B A$ beyond $A$. The circumcircle $\omega_{1}$ of triangle $O A D$ intersects the line $A C$ and the circle $\omega$ again at points $E$ and $G$, respectively. Point $H$ is such that $D A E H$ is... | We first observe that $\varangle D O E=\varangle D A E=2 \varangle A B C=\varangle B O A$ and hence $\varangle D O B=$ $\varangle E O A$, which together with $O B=O A$ and $\varangle O B D=\varangle B A O=\varangle O A E$ gives us $\triangle O B D \cong \triangle O A E$. Therefore $B D=A E$.
Next, $O G=O A$ implies $\v... | {
"exam": "Balkan_Shortlist",
"problem_label": "G6",
"problem_match": "\nG6.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
For positive integers $m$ and $n$, let $d(m, n)$ be the number of distinct primes that divide both $m$ and $n$. For instance, $d(60,126)=d\left(2^{2} \times 3 \times 5,2 \times 3^{2} \times 7\right)=2$. Does there exist a sequence $\left(a_{n}\right)$ of positive integers such that:
(i) $a_{1} \geqslant 2018^{2018}$;
(... | Such a sequence does exist.
Let $p_{1}<p_{2}<p_{3}<\ldots$ be the usual list of primes, and $q_{1}<q_{2}<\ldots, r_{1}<r_{2}<\ldots$ be disjoint sequences of primes greater than $2018^{2018}$. For example, let $q_{i} \equiv 1$ and $r_{i} \equiv 3$ modulo 4. Then, if $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots$, whe... | proof | Yes | Yes | proof | Number Theory | For positive integers $m$ and $n$, let $d(m, n)$ be the number of distinct primes that divide both $m$ and $n$. For instance, $d(60,126)=d\left(2^{2} \times 3 \times 5,2 \times 3^{2} \times 7\right)=2$. Does there exist a sequence $\left(a_{n}\right)$ of positive integers such that:
(i) $a_{1} \geqslant 2018^{2018}$;
(... | Such a sequence does exist.
Let $p_{1}<p_{2}<p_{3}<\ldots$ be the usual list of primes, and $q_{1}<q_{2}<\ldots, r_{1}<r_{2}<\ldots$ be disjoint sequences of primes greater than $2018^{2018}$. For example, let $q_{i} \equiv 1$ and $r_{i} \equiv 3$ modulo 4. Then, if $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots$, whe... | {
"exam": "Balkan_Shortlist",
"problem_label": "N1",
"problem_match": "\nN1.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that
$$
n!+f(m)!\mid f(n)!+f(m!)
$$
for all $m, n \in \mathbb{N}$. | Answer: $f(n)=n$ for all $n \in \mathbb{N}$.
Taking $m=n=1$ in $(*)$ yields $1+f(1)!\mid f(1)!+f(1)$ and hence $1+f(1)!\mid f(1)-1$. Since $|f(1)-1|<f(1)!+1$, this implies $f(1)=1$.
For $m=1$ in $(*)$ we have $n!+1 \mid f(n)!+1$, which implies $n!\leqslant f(n)$ !, i.e. $f(n) \geqslant n$.
On the other hand, taking $(m... | f(n)=n | Yes | Yes | proof | Number Theory | Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that
$$
n!+f(m)!\mid f(n)!+f(m!)
$$
for all $m, n \in \mathbb{N}$. | Answer: $f(n)=n$ for all $n \in \mathbb{N}$.
Taking $m=n=1$ in $(*)$ yields $1+f(1)!\mid f(1)!+f(1)$ and hence $1+f(1)!\mid f(1)-1$. Since $|f(1)-1|<f(1)!+1$, this implies $f(1)=1$.
For $m=1$ in $(*)$ we have $n!+1 \mid f(n)!+1$, which implies $n!\leqslant f(n)$ !, i.e. $f(n) \geqslant n$.
On the other hand, taking $(m... | {
"exam": "Balkan_Shortlist",
"problem_label": "N2",
"problem_match": "\nN2.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$. | 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 $11^{p}+17^{p}$. | 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_Shortlist",
"problem_label": "N3",
"problem_match": "\nN3.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Let $P(x)=a_{d} x^{d}+\cdots+a_{1} x+a_{0}$ be a non-constant polynomial with nonnegative integer coefficients having $d$ rational roots. Prove that
$$
\operatorname{lcm}(P(m), P(m+1), \ldots, P(n)) \geqslant m\binom{n}{m}
$$
for all positive integers $n>m$. | Let $x_{i}=-\frac{p_{i}}{q_{i}}(1 \leqslant i \leqslant d)$ be the roots of $P(x)$, where $p_{i}, q_{i} \in \mathbb{N}$ and $\operatorname{gcd}\left(p_{i}, q_{i}\right)=1$. By Gauss' lemma, we have $P(x)=c\left(q_{1} x+p_{1}\right)\left(q_{2} x+p_{2}\right) \cdots\left(q_{d} x+p_{d}\right)$ for some $c \in \mathbb{N}$,... | proof | Yes | Yes | proof | Number Theory | Let $P(x)=a_{d} x^{d}+\cdots+a_{1} x+a_{0}$ be a non-constant polynomial with nonnegative integer coefficients having $d$ rational roots. Prove that
$$
\operatorname{lcm}(P(m), P(m+1), \ldots, P(n)) \geqslant m\binom{n}{m}
$$
for all positive integers $n>m$. | Let $x_{i}=-\frac{p_{i}}{q_{i}}(1 \leqslant i \leqslant d)$ be the roots of $P(x)$, where $p_{i}, q_{i} \in \mathbb{N}$ and $\operatorname{gcd}\left(p_{i}, q_{i}\right)=1$. By Gauss' lemma, we have $P(x)=c\left(q_{1} x+p_{1}\right)\left(q_{2} x+p_{2}\right) \cdots\left(q_{d} x+p_{d}\right)$ for some $c \in \mathbb{N}$,... | {
"exam": "Balkan_Shortlist",
"problem_label": "N4",
"problem_match": "\nN4.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Let $x$ and $y$ be positive integers. If for each positive integer $n$ we have that
$$
(n y)^{2}+1 \mid x^{\varphi(n)}-1
$$
prove that $x=1$. | Let us take $n=3^{k}$ and suppose that $p$ is a prime divisor of $\left(3^{k} y\right)^{2}+1$ such that $p \equiv 2$ $(\bmod 3)$.
Since $p$ divides $x^{\varphi(n)}-1=x^{2 \cdot 3^{k-1}}-1$, the order of $x$ modulo $p$ divides both $p-1$ and $2 \cdot 3^{k-1}$, but $\operatorname{gcd}\left(p-1,2 \cdot 3^{k-1}\right) \mid... | proof | Yes | Yes | proof | Number Theory | Let $x$ and $y$ be positive integers. If for each positive integer $n$ we have that
$$
(n y)^{2}+1 \mid x^{\varphi(n)}-1
$$
prove that $x=1$. | Let us take $n=3^{k}$ and suppose that $p$ is a prime divisor of $\left(3^{k} y\right)^{2}+1$ such that $p \equiv 2$ $(\bmod 3)$.
Since $p$ divides $x^{\varphi(n)}-1=x^{2 \cdot 3^{k-1}}-1$, the order of $x$ modulo $p$ divides both $p-1$ and $2 \cdot 3^{k-1}$, but $\operatorname{gcd}\left(p-1,2 \cdot 3^{k-1}\right) \mid... | {
"exam": "Balkan_Shortlist",
"problem_label": "N5",
"problem_match": "\nN5.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution.",
"tier": "T1",
"year": "2018"
} |
Let $a_{0}$ be an arbitrary positive integer. Consider the infinite sequence $\left(a_{n}\right)_{n \geq 1}$, defined inductively as follows: given $a_{0}, a_{1}, \ldots, a_{n-1}$ define the term $a_{n}$ as the smallest positive integer such that $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by $n$. Prove that there exists a... | Define $b_{n}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}$ for every positive integer $n$. According to condition, $b_{n}$ is a positive integer for every positive integer $n$.
Since $a_{n+1}$ is the smallest positive integer such that $\frac{a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}}{n+1}$ is a positive integer and
$$
\frac{a_{0}+a_{... | proof | Yes | Yes | proof | Number Theory | Let $a_{0}$ be an arbitrary positive integer. Consider the infinite sequence $\left(a_{n}\right)_{n \geq 1}$, defined inductively as follows: given $a_{0}, a_{1}, \ldots, a_{n-1}$ define the term $a_{n}$ as the smallest positive integer such that $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by $n$. Prove that there exists a... | Define $b_{n}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}$ for every positive integer $n$. According to condition, $b_{n}$ is a positive integer for every positive integer $n$.
Since $a_{n+1}$ is the smallest positive integer such that $\frac{a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}}{n+1}$ is a positive integer and
$$
\frac{a_{0}+a_{... | {
"exam": "Balkan_Shortlist",
"problem_label": "A1b",
"problem_match": "\nA1b. ${ }^{4}$",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $a_{0}$ be an arbitrary positive integer. Consider the infinite sequence $\left(a_{n}\right)_{n \geq 1}$, defined inductively as follows: given $a_{0}, a_{1}, \ldots, a_{n-1}$ define the term $a_{n}$ as the smallest positive integer such that $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by $n$. Prove that there exists a... | Firstly, considering $(x, y)=(1,1)$ we get $f(0)=-1$.
Then, setting $y=1$, we see that $-x=f(f(1)-f(x))$, so $f$ must be surjective.
Now let $(x, y)=(a, 0)$ and $(0, a)$ to get
$$
-1=a+f(-1-f(a)) \quad \text { and } \quad-1=-a+f(f(a)+1)
$$
Since $f$ is surjective, for any real $z$ we may write $z=f(a)+1$ and then add... | proof | Yes | Incomplete | proof | Number Theory | Let $a_{0}$ be an arbitrary positive integer. Consider the infinite sequence $\left(a_{n}\right)_{n \geq 1}$, defined inductively as follows: given $a_{0}, a_{1}, \ldots, a_{n-1}$ define the term $a_{n}$ as the smallest positive integer such that $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by $n$. Prove that there exists a... | Firstly, considering $(x, y)=(1,1)$ we get $f(0)=-1$.
Then, setting $y=1$, we see that $-x=f(f(1)-f(x))$, so $f$ must be surjective.
Now let $(x, y)=(a, 0)$ and $(0, a)$ to get
$$
-1=a+f(-1-f(a)) \quad \text { and } \quad-1=-a+f(f(a)+1)
$$
Since $f$ is surjective, for any real $z$ we may write $z=f(a)+1$ and then add... | {
"exam": "Balkan_Shortlist",
"problem_label": "A1b",
"problem_match": "\nA1b. ${ }^{4}$",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $a, b, c$ be real numbers such that $0 \leq a \leq b \leq c$. Prove that if
$$
a+b+c=a b+b c+c a>0,
$$
then $\sqrt{b c}(a+1) \geq 2$. When does the equality hold? | 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 $0 \leq a \leq b \leq c$. Prove that if
$$
a+b+c=a b+b c+c a>0,
$$
then $\sqrt{b c}(a+1) \geq 2$. When does the equality hold? | 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_Shortlist",
"problem_label": "A3",
"problem_match": "\nA3.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $a_{i j}, i=1,2, \ldots, m$ and $j=1,2, \ldots, n$, be positive real numbers. Prove that
$$
\sum_{i=1}^{m}\left(\sum_{j=1}^{n} \frac{1}{a_{i j}}\right)^{-1} \leq\left(\sum_{j=1}^{n}\left(\sum_{i=1}^{m} a_{i j}\right)^{-1}\right)^{-1} .
$$
When does the equality hold? | We will use the following
Lemma. If $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ are positive real numbers then
$$
\frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{b_{j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}+b_{j}}}
$$
The equality holds when $\frac{a_{1}}{b_{1}}=\frac{a... | proof | Yes | Yes | proof | Inequalities | Let $a_{i j}, i=1,2, \ldots, m$ and $j=1,2, \ldots, n$, be positive real numbers. Prove that
$$
\sum_{i=1}^{m}\left(\sum_{j=1}^{n} \frac{1}{a_{i j}}\right)^{-1} \leq\left(\sum_{j=1}^{n}\left(\sum_{i=1}^{m} a_{i j}\right)^{-1}\right)^{-1} .
$$
When does the equality hold? | We will use the following
Lemma. If $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ are positive real numbers then
$$
\frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{b_{j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}+b_{j}}}
$$
The equality holds when $\frac{a_{1}}{b_{1}}=\frac{a... | {
"exam": "Balkan_Shortlist",
"problem_label": "A4",
"problem_match": "\nA4.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $a, b, c$ be positive real numbers, such that $(a b)^{2}+(b c)^{2}+(c a)^{2}=3$. Prove that
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right) \geq 1 .
$$ | The inequality is equivalent with
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right) \geq 1 \Leftrightarrow\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq(a+1)(b+1)(c+1) .
$$
Thus:
$$
\begin{gathered}
\prod_{c y c}\left(a^{3}+1\right)-\prod_{c y c}(a+1)=\sum_{c y c} a^{3}+\sum_{c... | proof | Yes | Yes | proof | Inequalities | Let $a, b, c$ be positive real numbers, such that $(a b)^{2}+(b c)^{2}+(c a)^{2}=3$. Prove that
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right) \geq 1 .
$$ | The inequality is equivalent with
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right) \geq 1 \Leftrightarrow\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq(a+1)(b+1)(c+1) .
$$
Thus:
$$
\begin{gathered}
\prod_{c y c}\left(a^{3}+1\right)-\prod_{c y c}(a+1)=\sum_{c y c} a^{3}+\sum_{c... | {
"exam": "Balkan_Shortlist",
"problem_label": "A5",
"problem_match": "\nA5.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $A B C D$ be a square of center $O$ and let $M$ be the symmetric of the point $B$ with respect to the point $A$. Let $E$ be the intersection of $C M$ and $B D$, and let $S$ be the intersection of $M O$ and $A E$. Show that $S O$ is the angle bisector of $\angle E S B$. | We have
$$
\left\{\begin{array}{l}
D C \equiv D A \\
\angle E D C \equiv \angle E D A \quad \Rightarrow \triangle D E C \equiv \triangle D E A \Rightarrow \angle D A E \equiv \angle D C E(*) . \\
D E \equiv D E
\end{array}\right.
$$
Let $C M \cap A D=\{P\}$, then follows $\triangle C D P \equiv \triangle B A P$ and $... | proof | Yes | Yes | proof | Geometry | Let $A B C D$ be a square of center $O$ and let $M$ be the symmetric of the point $B$ with respect to the point $A$. Let $E$ be the intersection of $C M$ and $B D$, and let $S$ be the intersection of $M O$ and $A E$. Show that $S O$ is the angle bisector of $\angle E S B$. | We have
$$
\left\{\begin{array}{l}
D C \equiv D A \\
\angle E D C \equiv \angle E D A \quad \Rightarrow \triangle D E C \equiv \triangle D E A \Rightarrow \angle D A E \equiv \angle D C E(*) . \\
D E \equiv D E
\end{array}\right.
$$
Let $C M \cap A D=\{P\}$, then follows $\triangle C D P \equiv \triangle B A P$ and $... | {
"exam": "Balkan_Shortlist",
"problem_label": "G1",
"problem_match": "\nG1.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let be a triangle $\triangle A B C$ with $m(\angle A B C)=75^{\circ}$ and $m(\angle A C B)=45^{\circ}$. The angle bisector of $\angle C A B$ intersects $C B$ at the point $D$. We consider the point $E \in(A B)$, such that $D E=D C$. Let $P$ be the intersection of the lines $A D$ and $C E$. Prove that $P$ is the midpoin... | Let $P^{\prime}$ be the midpoint of the segment $A D$. We will prove that $P^{\prime}=P$. Let $F \in A C$ such that $D F \perp A C$. The triangle $C D F$ is isosceles with $F D=F C$ and the triangle $D P^{\prime} F$ is equilateral as $m(\angle A D F)=60^{\circ}$. Thus, the triangle $F C P^{\prime}$ is isosceles $\left(... | proof | Yes | Yes | proof | Geometry | Let be a triangle $\triangle A B C$ with $m(\angle A B C)=75^{\circ}$ and $m(\angle A C B)=45^{\circ}$. The angle bisector of $\angle C A B$ intersects $C B$ at the point $D$. We consider the point $E \in(A B)$, such that $D E=D C$. Let $P$ be the intersection of the lines $A D$ and $C E$. Prove that $P$ is the midpoin... | Let $P^{\prime}$ be the midpoint of the segment $A D$. We will prove that $P^{\prime}=P$. Let $F \in A C$ such that $D F \perp A C$. The triangle $C D F$ is isosceles with $F D=F C$ and the triangle $D P^{\prime} F$ is equilateral as $m(\angle A D F)=60^{\circ}$. Thus, the triangle $F C P^{\prime}$ is isosceles $\left(... | {
"exam": "Balkan_Shortlist",
"problem_label": "G2",
"problem_match": "\nG2.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $A B C$ be a scalene and acute triangle, with circumcentre $O$. Let $\omega$ be the circle with centre $A$, tangent to $B C$ at $D$. Suppose there are two points $F$ and $G$ on $\omega$ such that $F G \perp A O, \angle B F D=\angle D G C$ and the couples of points $(B, F)$ and $(C, G)$ are in different halfplanes w... | Consider any two points $F, G$ on $\omega$ such that $\angle B F D=\angle D G C$. Exploiting the isosceles triangles $\triangle A F G, \triangle A F D$, and $\triangle A D G$, we deduce (using directed angles throughout):
$$
\begin{gathered}
\angle D B F-\angle G C D=180^{\circ}-\angle B F D-\angle B D F-\left(180^{\c... | proof | Yes | Yes | proof | Geometry | Let $A B C$ be a scalene and acute triangle, with circumcentre $O$. Let $\omega$ be the circle with centre $A$, tangent to $B C$ at $D$. Suppose there are two points $F$ and $G$ on $\omega$ such that $F G \perp A O, \angle B F D=\angle D G C$ and the couples of points $(B, F)$ and $(C, G)$ are in different halfplanes w... | Consider any two points $F, G$ on $\omega$ such that $\angle B F D=\angle D G C$. Exploiting the isosceles triangles $\triangle A F G, \triangle A F D$, and $\triangle A D G$, we deduce (using directed angles throughout):
$$
\begin{gathered}
\angle D B F-\angle G C D=180^{\circ}-\angle B F D-\angle B D F-\left(180^{\c... | {
"exam": "Balkan_Shortlist",
"problem_label": "G3",
"problem_match": "\nG3.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Given an acute triangle $A B C$, let $M$ be the midpoint of $B C$ and $H$ the orthocentre. Let $\Gamma$ be the circle with diameter $H M$, and let $X, Y$ be distinct points on $\Gamma$ such that $A X, A Y$ are tangent to $\Gamma$. Prove that $B X Y C$ is cyclic. | Let $D$ be the foot of the altitude from $A$ to $B C$, which also lies on $\Gamma$. Let $O$ be the circumcentre of $\triangle A B C$. Since $\angle H D M=90^{\circ}$, note that rays $H D$ and $H M$ meet the circumcircle at points which are reflections in $O M$. Then, since $\angle B A D=\angle O A C$, we recover the we... | proof | Yes | Yes | proof | Geometry | Given an acute triangle $A B C$, let $M$ be the midpoint of $B C$ and $H$ the orthocentre. Let $\Gamma$ be the circle with diameter $H M$, and let $X, Y$ be distinct points on $\Gamma$ such that $A X, A Y$ are tangent to $\Gamma$. Prove that $B X Y C$ is cyclic. | Let $D$ be the foot of the altitude from $A$ to $B C$, which also lies on $\Gamma$. Let $O$ be the circumcentre of $\triangle A B C$. Since $\angle H D M=90^{\circ}$, note that rays $H D$ and $H M$ meet the circumcircle at points which are reflections in $O M$. Then, since $\angle B A D=\angle O A C$, we recover the we... | {
"exam": "Balkan_Shortlist",
"problem_label": "G4",
"problem_match": "\nG4.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $A B C(B C>A C)$ be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $A B$ at the point $D$. The circumcircles of triangles $B C D, O C D$ and $A O B$ intersect the ray $C A$ (beyond $A$ ) at the points $Q, P$ and $K$, respectively, such that $P \in(A K)$ and $K... | As $D C$ is tangent to $k$ at $C$ then $\angle O C D=90^{\circ}$. Denote by $X$ the midpoint of $A B$. Then $\angle O X A=90^{\circ}$ because of $O X$ is the perpendicular bisector of the side $A B$. The pentagon $P X O C D$ is inscribed in the circle with diameter $O D$, hence $\angle P X A=$ $\angle P X D=\angle P C ... | proof | Yes | Yes | proof | Geometry | Let $A B C(B C>A C)$ be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $A B$ at the point $D$. The circumcircles of triangles $B C D, O C D$ and $A O B$ intersect the ray $C A$ (beyond $A$ ) at the points $Q, P$ and $K$, respectively, such that $P \in(A K)$ and $K... | As $D C$ is tangent to $k$ at $C$ then $\angle O C D=90^{\circ}$. Denote by $X$ the midpoint of $A B$. Then $\angle O X A=90^{\circ}$ because of $O X$ is the perpendicular bisector of the side $A B$. The pentagon $P X O C D$ is inscribed in the circle with diameter $O D$, hence $\angle P X A=$ $\angle P X D=\angle P C ... | {
"exam": "Balkan_Shortlist",
"problem_label": "G5",
"problem_match": "\nG5.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $A B C$ be an acute triangle, and $A X, A Y$ two isogonal lines. Also, suppose that $K, S$ are the feet of perpendiculars from $B$ to $A X, A Y$, and $T, L$ are the feet of perpendiculars from $C$ to $A X, A Y$ respectively. Prove that $K L$ and $S T$ intersect on $B C$. | 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 triangle, and $A X, A Y$ two isogonal lines. Also, suppose that $K, S$ are the feet of perpendiculars from $B$ to $A X, A Y$, and $T, L$ are the feet of perpendiculars from $C$ to $A X, A Y$ respectively. Prove that $K L$ and $S T$ intersect on $B C$. | 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_Shortlist",
"problem_label": "G6",
"problem_match": "\nG6.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $A D, B E$, and $C F$ denote the altitudes of triangle $\triangle A B C$. Points $E^{\prime}$ and $F^{\prime}$ are the reflections of $E$ and $F$ over $A D$, respectively. The lines $B F^{\prime}$ and $C E^{\prime}$ intersect at $X$, while the lines $B E^{\prime}$ and $C F^{\prime}$ intersect at the point $Y$. Prov... | We will prove that the desired point of concurrency is the midpoint of $B C$. Assume that $\triangle A B C$ is acute. Let $(A B C)^{5}$ intersect $(A E F)$ at the point $Y^{\prime}$; we will prove that $Y=Y^{\prime}$.
^{5}$ intersect $(A E F)$ at the point $Y^{\prime}$; we will prove that $Y=Y^{\prime}$.
![](https://cdn.mathpix.com/cropped/2024_12_07_82afe765b60b274413c4g-24.jpg?height=1623&width=1529&top... | {
"exam": "Balkan_Shortlist",
"problem_label": "G7",
"problem_match": "\nG7.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Given an acute triangle $A B C,(c)$ is circumcircle with center $O$ and $H$ the orthocenter of the triangle $A B C$. The line $A O$ intersects $(c)$ at the point $D$. Let $D_{1}, D_{2}$ and $H_{2}, H_{3}$ be the symmetrical points of the points $D$ and $H$ with respect to the lines $A B, A C$ respectively. Let $\left(c... | It is well known that the symmetrical points $H_{1}, H_{2}, H_{3}$ of $H$ with respect the sides $B C, A B, A C$ of the triangle $A B C$ respectively lie on the circle (c).
Figure 8: G8
Le... | proof | Yes | Yes | proof | Geometry | Given an acute triangle $A B C,(c)$ is circumcircle with center $O$ and $H$ the orthocenter of the triangle $A B C$. The line $A O$ intersects $(c)$ at the point $D$. Let $D_{1}, D_{2}$ and $H_{2}, H_{3}$ be the symmetrical points of the points $D$ and $H$ with respect to the lines $A B, A C$ respectively. Let $\left(c... | It is well known that the symmetrical points $H_{1}, H_{2}, H_{3}$ of $H$ with respect the sides $B C, A B, A C$ of the triangle $A B C$ respectively lie on the circle (c).
Figure 8: G8
Le... | {
"exam": "Balkan_Shortlist",
"problem_label": "G8",
"problem_match": "\nG8.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Given semicircle (c) with diameter $A B$ and center $O$. On the (c) we take point $C$ such that the tangent at the $C$ intersects the line $A B$ at the point $E$. The perpendicular line from $C$ to $A B$ intersects the diameter $A B$ at the point $D$. On the (c) we get the points $H, Z$ such that $C D=C H=C Z$. The lin... | Since $C H=C Z$ we have $O C \perp H Z$. So from the cyclic quadrilateral $S O D I$ we get
$$
C S \cdot C O=C I \cdot C D .
$$
Figure 9: G9
We draw the perpendicular line $(v)$ to $H C$ ... | proof | Yes | Yes | proof | Geometry | Given semicircle (c) with diameter $A B$ and center $O$. On the (c) we take point $C$ such that the tangent at the $C$ intersects the line $A B$ at the point $E$. The perpendicular line from $C$ to $A B$ intersects the diameter $A B$ at the point $D$. On the (c) we get the points $H, Z$ such that $C D=C H=C Z$. The lin... | Since $C H=C Z$ we have $O C \perp H Z$. So from the cyclic quadrilateral $S O D I$ we get
$$
C S \cdot C O=C I \cdot C D .
$$
Figure 9: G9
We draw the perpendicular line $(v)$ to $H C$ ... | {
"exam": "Balkan_Shortlist",
"problem_label": "G9",
"problem_match": "\nG9.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
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 $p, q \in \mathbb{P}$. | 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 | math-word-problem | 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 $p, q \in \mathbb{P}$. | 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_Shortlist",
"problem_label": "TN1",
"problem_match": "\nTN1.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $S \subset\{1, \ldots, n\}$ be a nonempty set, where $n$ is a positive integer. We denote by $s$ the greatest common divisor of the elements of the set $S$. We assume that $s \neq 1$ and let $d$ be its smallest divisor greater than 1 . Let $T \subset\{1, \ldots, n\}$ be a set such that $S \subset T$ and $|T| \geq 1... | Let $t$ be the greatest common divisor of the elements in $T$. Due to the fact that $S \subset T$, we immediately get that $t / s$. Let us assume for the sake of contradiction that $t \neq 1$. From the previous observation we get that $t \geq d$.
By taking into account that $|T| \geq 1+\left[\frac{n}{d}\right]$, we in... | proof | Yes | Yes | proof | Number Theory | Let $S \subset\{1, \ldots, n\}$ be a nonempty set, where $n$ is a positive integer. We denote by $s$ the greatest common divisor of the elements of the set $S$. We assume that $s \neq 1$ and let $d$ be its smallest divisor greater than 1 . Let $T \subset\{1, \ldots, n\}$ be a set such that $S \subset T$ and $|T| \geq 1... | Let $t$ be the greatest common divisor of the elements in $T$. Due to the fact that $S \subset T$, we immediately get that $t / s$. Let us assume for the sake of contradiction that $t \neq 1$. From the previous observation we get that $t \geq d$.
By taking into account that $|T| \geq 1+\left[\frac{n}{d}\right]$, we in... | {
"exam": "Balkan_Shortlist",
"problem_label": "TN2",
"problem_match": "\nTN2.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Let $n(n \geq 1)$ be a positive integer and $U=\{1, \ldots, n\}$. Let $S$ be a nonempty subset of $U$ and let $d(d \neq 1)$ be the smallest common divisor of all elements of the set $S$. Find the smallest positive integer $k$ such that for any subset $T$ of $U$, consisting of $k$ elements, with $S \subset T$, the great... | We will show that $k_{\min }=1+\left[\frac{n}{d}\right]$ (here [.] denotes the integer part).
Obviously, the number of elements of $S$ is not grater than $\left[\frac{n}{d}\right]$, i.e. $|S| \leq\left[\frac{n}{d}\right]$, and $S \neq U$.
If $S \subset T$ and the greatest common divisor of elements of $T$ is equal to ... | 1+\left[\frac{n}{d}\right] | Yes | Yes | math-word-problem | Number Theory | Let $n(n \geq 1)$ be a positive integer and $U=\{1, \ldots, n\}$. Let $S$ be a nonempty subset of $U$ and let $d(d \neq 1)$ be the smallest common divisor of all elements of the set $S$. Find the smallest positive integer $k$ such that for any subset $T$ of $U$, consisting of $k$ elements, with $S \subset T$, the great... | We will show that $k_{\min }=1+\left[\frac{n}{d}\right]$ (here [.] denotes the integer part).
Obviously, the number of elements of $S$ is not grater than $\left[\frac{n}{d}\right]$, i.e. $|S| \leq\left[\frac{n}{d}\right]$, and $S \neq U$.
If $S \subset T$ and the greatest common divisor of elements of $T$ is equal to ... | {
"exam": "Balkan_Shortlist",
"problem_label": "TN2b",
"problem_match": "\nTN2b. ${ }^{7}$",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
100 couples are invited to a traditional Moldovan dance. The 200 people stand in a line, and then in a step, two of them (not necessarily adjacent) may swap positions. Find the least $C$ such that whatever the initial order, they can arrive at an ordering where everyone is dancing next to their partner in at most $C$ s... | With 100 replaced by $N$, the answer is $C=C(N)=N-1$. Throughout, we will say that the members of a couple have the same.
$N=2$ : We use this as a base case for induction for both bounds. Up to labelling, there is one trivial initial order, and two non-trivial ones, namely
$$
1,1,2,2 ; \quad 1, \sqrt{2,2,1} ; \quad 1,... | N-1 | Yes | Yes | math-word-problem | Combinatorics | 100 couples are invited to a traditional Moldovan dance. The 200 people stand in a line, and then in a step, two of them (not necessarily adjacent) may swap positions. Find the least $C$ such that whatever the initial order, they can arrive at an ordering where everyone is dancing next to their partner in at most $C$ s... | With 100 replaced by $N$, the answer is $C=C(N)=N-1$. Throughout, we will say that the members of a couple have the same.
$N=2$ : We use this as a base case for induction for both bounds. Up to labelling, there is one trivial initial order, and two non-trivial ones, namely
$$
1,1,2,2 ; \quad 1, \sqrt{2,2,1} ; \quad 1,... | {
"exam": "Balkan_Shortlist",
"problem_label": "C1",
"problem_match": "\nC1.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
An $5 \times 5$ array must be completed with all numbers $\{1,2, \ldots, 25\}$, one number in each cell. Find the maximal positive integer $k$, such that for any completion of the array there is a $2 \times 2$ square (subarray), whose numbers have a sum not less than $k$. | We will prove that $k_{\max }=45$.
We number the columns and the rows and we select all possible $3^{2}=9$ choices of an odd column with an odd row.
Collecting all such pairs of an odd column with an odd row, we double count some squares. Indeed, we take some $3^{2}$ squares 5 times, some 12 squares 3 times and there ... | 45 | Yes | Yes | math-word-problem | Combinatorics | An $5 \times 5$ array must be completed with all numbers $\{1,2, \ldots, 25\}$, one number in each cell. Find the maximal positive integer $k$, such that for any completion of the array there is a $2 \times 2$ square (subarray), whose numbers have a sum not less than $k$. | We will prove that $k_{\max }=45$.
We number the columns and the rows and we select all possible $3^{2}=9$ choices of an odd column with an odd row.
Collecting all such pairs of an odd column with an odd row, we double count some squares. Indeed, we take some $3^{2}$ squares 5 times, some 12 squares 3 times and there ... | {
"exam": "Balkan_Shortlist",
"problem_label": "C2b",
"problem_match": "\nC2b. ${ }^{8}$",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
An $5 \times 5$ array must be completed with all numbers $\{1,2, \ldots, 25\}$, one number in each cell. Find the maximal positive integer $k$, such that for any completion of the array there is a $2 \times 2$ square (subarray), whose numbers have a sum not less than $k$. | 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 step, he deletes the next two available points on the left if Anna decreased her $x... | not found | Yes | Incomplete | math-word-problem | Combinatorics | An $5 \times 5$ array must be completed with all numbers $\{1,2, \ldots, 25\}$, one number in each cell. Find the maximal positive integer $k$, such that for any completion of the array there is a $2 \times 2$ square (subarray), whose numbers have a sum not less than $k$. | 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 step, he deletes the next two available points on the left if Anna decreased her $x... | {
"exam": "Balkan_Shortlist",
"problem_label": "C2b",
"problem_match": "\nC2b. ${ }^{8}$",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
A town-planner has built an isolated city whose road network consists of 2 N roundabouts, each connecting exactly three roads. A series of tunnels and bridges ensure that all roads in the town meet only at roundabouts. All roads are two-way, and each roundabout is oriented clockwise.
Vlad has recently passed his drivi... | $N$ odd. In fact, the number of trajectories has the same parity as $N$.
The setting is a (multi)graph where every vertex has degree three. Each vertex has an orientation, an ordering of its incident edges. We call Vlad's possible paths trajectories, and a complete trajectory if he traverses every edge in both directio... | N \text{ odd} | Yes | Incomplete | math-word-problem | Combinatorics | A town-planner has built an isolated city whose road network consists of 2 N roundabouts, each connecting exactly three roads. A series of tunnels and bridges ensure that all roads in the town meet only at roundabouts. All roads are two-way, and each roundabout is oriented clockwise.
Vlad has recently passed his drivi... | $N$ odd. In fact, the number of trajectories has the same parity as $N$.
The setting is a (multi)graph where every vertex has degree three. Each vertex has an orientation, an ordering of its incident edges. We call Vlad's possible paths trajectories, and a complete trajectory if he traverses every edge in both directio... | {
"exam": "Balkan_Shortlist",
"problem_label": "C4",
"problem_match": "\nC4.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2019"
} |
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}$ and $g: \mathbb{R}^{+} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+y^{2}\right)=g(x y)
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by Greece | Given any $u \geqslant 2$, take $a, b \in \mathbb{R}^{+}$such that $a+b=u$ and $a b=1$. This is possible as the equation $x^{2}-u x+1$ for $u \geqslant 2$ has two positive real solutions. (Discriminant is $u^{2}-4 \geqslant 0$, sum and product of solutions are positive.) Now taking $x=\sqrt{a}, y=\sqrt{b}$ we get $f(u)... | proof | Yes | Yes | math-word-problem | Algebra | Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}$ and $g: \mathbb{R}^{+} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+y^{2}\right)=g(x y)
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by Greece | Given any $u \geqslant 2$, take $a, b \in \mathbb{R}^{+}$such that $a+b=u$ and $a b=1$. This is possible as the equation $x^{2}-u x+1$ for $u \geqslant 2$ has two positive real solutions. (Discriminant is $u^{2}-4 \geqslant 0$, sum and product of solutions are positive.) Now taking $x=\sqrt{a}, y=\sqrt{b}$ we get $f(u)... | {
"exam": "Balkan_Shortlist",
"problem_label": "A1",
"problem_match": "\nA1.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution.",
"tier": "T1",
"year": "2021"
} |
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