year stringdate 1961-01-01 00:00:00 2025-01-01 00:00:00 ⌀ | tier stringclasses 5
values | problem_label stringclasses 119
values | problem_type stringclasses 13
values | exam stringclasses 28
values | problem stringlengths 87 2.77k | solution stringlengths 834 13k | metadata dict | problem_tokens int64 50 903 | solution_tokens int64 500 3.93k |
|---|---|---|---|---|---|---|---|---|---|
2017 | T1 | G1 | Geometry | Balkan_Shortlist | 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
$$
B G \cdot B C=B F \cdot B A
$$
Combining (6) with the problem statement we get
$$
B C^{2}=B A \cdot B F+C E \cdot C A=B G \cdot B C+C E \cdot C A
$$
and from he... | {
"problem_match": "\n## G1",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 84 | 594 |
2017 | T1 | G3 | Geometry | Balkan_Shortlist | 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... | {
"problem_match": "\n## G3",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 178 | 566 |
2017 | T1 | G5 | Geometry | Balkan_Shortlist | 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:... | {
"problem_match": "\n## G5",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 100 | 945 |
2017 | T1 | G6 | Geometry | Balkan_Shortlist | 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,
$$
then $A D \perp B_{A} C_{A}$.
Proof of the lemma. Construct an isosceles triangle $A B E$ outside the triangle $A B C$, so that... | {
"problem_match": "\n## G6",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution."
} | 232 | 740 |
2017 | T1 | G7 | Geometry | Balkan_Shortlist | Let $A B C$ be an acute triangle with $A B \neq A C$ and circumcirle $\Gamma$. The angle bisector of $B A C$ intersects $B C$ and $\Gamma$ at $D$ and $E$ respectively. Circle with diameter $D E$ intersects $\Gamma$ again at $F \neq E$. Point $P$ is on $A F$ such that $P B=P C$ and $X$ and $Y$ are feet of perpendiculars... | WLOG, assume $A B<A C$. Let $M$ be the midpoint of side $B C$ and let the circumcircle of $D F E$ intersect $A F$ again at $K$. Since
$$
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
$$... | {
"problem_match": "\nG7",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 223 | 1,043 |
2017 | T1 | G8 | Geometry | Balkan_Shortlist | Given an acute triangle $\triangle A B C(A C \neq A B)$ and let $(C)$ be its circumcircle. The excircle $\left(C_{1}\right)$ corresponding to the vertex $A$, of center $I_{a}$, tangents to the side $B C$ at the point $D$ and to the extensions of the sides $A B, A C$ at the points $E, Z$ respectively. Let $I$ and $L$ ar... | We have $(e) \perp B C$ and $I_{a} D \perp B C$, so $(e) \| I_{a} D$. Let $T$. $S$ be the midpoints of the segments $H I_{a}, H D$ respectively and $Y$ the point of intersection of the lines $H D, E Z$ Then, $T S \| I_{a} D$. $T S \perp B C$ and $S Y \perp E Z$
The Euler circle ( $\omega$ ) of the triangle EDZ passes t... | {
"problem_match": "\nG8",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 237 | 802 |
2017 | T1 | C2 | Combinatorics | Balkan_Shortlist | Let $n, a, b, c$ be natural numbers. Every point on the coordinate plane with integer coordinates is colored in one of $n$ colors. Prove there exists $c$ triangles whose vertices are colored in the same color, which are pairwise congruent, and which have a side whose lenght is divisible by $a$ and a side whose lenght i... | Let the colors be $d_{1}, d_{2}, d_{3} \ldots, d_{n}$. Look at the coordinates
$$
(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)
$$
for integers $k$ and $r$. By the pigeonhole principle there are two points of the same color. For every pair ( $k, r$ ) we say the color $d... | {
"problem_match": "\nC2",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 80 | 614 |
2017 | T1 | C4 | Combinatorics | Balkan_Shortlist | For any set of points $A_{1}, A_{2}, \ldots, A_{n}$ on the plane, one defines $r\left(A_{1}, A_{2}, \ldots, A_{n}\right)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \geq 3$, there are indices $i, j, k$ such that
$$
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... | {
"problem_match": "\n## C4",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution."
} | 130 | 1,341 |
2017 | T1 | C4 | Combinatorics | Balkan_Shortlist | 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
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=r\left(A_{1}, A_... | The answer turns out to be $\binom{2 n-1}{n}$ if $n$ is odd and $\binom{2 n-1}{n}-2\left(\begin{array}{c}\frac{3 n}{2}-1\end{array}\right)$ if $n$ is even.
Case 1. Suppose $n$ is odd, say $n=2 m+1$. In this case we will show that any distribution of candies is legal. Thus the number of legal distributions is indeed ( $... | {
"problem_match": "\n## C4",
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 130 | 955 |
2018 | T1 | A2 | Algebra | Balkan_Shortlist | Two ants start at the same point in the plane. Each minute they choose whether to walk due north, east, south or west. They each walk 1 meter in the first minute. In each subsequent minute the distance they walk is multiplied by a rational number $q>0$. They meet after a whole number of minutes, but have not taken exac... | Answer: $q=1$.
Let $x_{A}^{(n)}$ (resp. $x_{B}^{(n)}$ ) be the $x$-coordinates of the first (resp. second) ant's position after $n$ minutes. Then $x_{A}^{(n)}-x_{A}^{(n-1)} \in\left\{q^{n},-q^{n}, 0\right\}$, and so $x_{A}^{(n)}, x_{B}^{(n)}$ are given by polynomials in $q$ with coefficients in $\{-1,0,1\}$. So if the ... | {
"problem_match": "\nA2.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution."
} | 91 | 820 |
2018 | T1 | A4 | Algebra | Balkan_Shortlist | Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that the following inequality holds:
$$
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}... | {
"problem_match": "\nA4.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution."
} | 104 | 883 |
2018 | T1 | A5 | Algebra | Balkan_Shortlist | 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... | {
"problem_match": "\nA5.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution."
} | 100 | 637 |
2018 | T1 | A6 | Algebra | Balkan_Shortlist | 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... | {
"problem_match": "\nA6.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution."
} | 126 | 509 |
2018 | T1 | C1 | Combinatorics | Balkan_Shortlist | 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... | {
"problem_match": "\nC1.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 144 | 669 |
2018 | T1 | C2 | Combinatorics | Balkan_Shortlist | 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)\... | {
"problem_match": "\nC2.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution."
} | 133 | 800 |
2018 | T1 | G2 | Geometry | Balkan_Shortlist | 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}... | {
"problem_match": "\nG2.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 108 | 534 |
2018 | T1 | G6 | Geometry | Balkan_Shortlist | 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... | {
"problem_match": "\nG6.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution."
} | 195 | 1,274 |
2018 | T1 | N4 | Number Theory | Balkan_Shortlist | 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 $c \in \mathbb{N}$,... | {
"problem_match": "\nN4.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution."
} | 93 | 587 |
2018 | T1 | N5 | Number Theory | Balkan_Shortlist | 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... | {
"problem_match": "\nN5.",
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"solution_match": "\n## Solution."
} | 55 | 562 |
2019 | T1 | A1b | Algebra | Balkan_Shortlist | 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
$$
\frac{a_{0}+a_{... | {
"problem_match": "\nA1b. ${ }^{4}$",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 131 | 715 |
2019 | T1 | A1b | Algebra | Balkan_Shortlist | Let $a_{0}$ be an arbitrary positive integer. Consider the infinite sequence $\left(a_{n}\right)_{n \geq 1}$, defined inductively as follows: given $a_{0}, a_{1}, \ldots, a_{n-1}$ define the term $a_{n}$ as the smallest positive integer such that $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by $n$. Prove that there exists a... | Firstly, considering $(x, y)=(1,1)$ we get $f(0)=-1$.
Then, setting $y=1$, we see that $-x=f(f(1)-f(x))$, so $f$ must be surjective.
Now let $(x, y)=(a, 0)$ and $(0, a)$ to get
$$
-1=a+f(-1-f(a)) \quad \text { and } \quad-1=-a+f(f(a)+1)
$$
Since $f$ is surjective, for any real $z$ we may write $z=f(a)+1$ and then add... | {
"problem_match": "\nA1b. ${ }^{4}$",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 131 | 798 |
2019 | T1 | A4 | Algebra | Balkan_Shortlist | Let $a_{i j}, i=1,2, \ldots, m$ and $j=1,2, \ldots, n$, be positive real numbers. Prove that
$$
\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}}}
$$
The equality holds when $\frac{a_{1}}{b_{1}}=\frac{a... | {
"problem_match": "\nA4.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 131 | 1,210 |
2019 | T1 | A5 | Algebra | Balkan_Shortlist | Let $a, b, c$ be positive real numbers, such that $(a b)^{2}+(b c)^{2}+(c a)^{2}=3$. Prove that
$$
\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... | {
"problem_match": "\nA5.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 85 | 967 |
2019 | T1 | G1 | Geometry | Balkan_Shortlist | 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 $... | {
"problem_match": "\nG1.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 90 | 557 |
2019 | T1 | G2 | Geometry | Balkan_Shortlist | 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(... | {
"problem_match": "\nG2.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 118 | 843 |
2019 | T1 | G3 | Geometry | Balkan_Shortlist | 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):
$$
\begin{gathered}
\angle D B F-\angle G C D=180^{\circ}-\angle B F D-\angle B D F-\left(180^{\c... | {
"problem_match": "\nG3.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 138 | 814 |
2019 | T1 | G5 | Geometry | Balkan_Shortlist | Let $A B C(B C>A C)$ be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $A B$ at the point $D$. The circumcircles of triangles $B C D, O C D$ and $A O B$ intersect the ray $C A$ (beyond $A$ ) at the points $Q, P$ and $K$, respectively, such that $P \in(A K)$ and $K... | As $D C$ is tangent to $k$ at $C$ then $\angle O C D=90^{\circ}$. Denote by $X$ the midpoint of $A B$. Then $\angle O X A=90^{\circ}$ because of $O X$ is the perpendicular bisector of the side $A B$. The pentagon $P X O C D$ is inscribed in the circle with diameter $O D$, hence $\angle P X A=$ $\angle P X D=\angle P C ... | {
"problem_match": "\nG5.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 168 | 655 |
2019 | T1 | G6 | Geometry | Balkan_Shortlist | Let $A B C$ be an acute triangle, and $A X, A Y$ two isogonal lines. Also, suppose that $K, S$ are the feet of perpendiculars from $B$ to $A X, A Y$, and $T, L$ are the feet of perpendiculars from $C$ to $A X, A Y$ respectively. Prove that $K L$ and $S T$ intersect on $B C$. | Denote $\phi=\widehat{X A B}=\widehat{Y A C}, \alpha=\widehat{C A X}=\widehat{B A Y}$. Then, because the quadrilaterals ABSK and ACTL are cyclic, we have
$$
\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,... | {
"problem_match": "\nG6.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 95 | 753 |
2019 | T1 | G7 | Geometry | Balkan_Shortlist | 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}$.
Figure 7: G7
Using the fact that $H$ is the incenter of $\triangle D E F$ we get that $D, E^{\prime},... | {
"problem_match": "\nG7.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 141 | 865 |
2019 | T1 | G8 | Geometry | Balkan_Shortlist | 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
Let $L$ be the second point of intersection of $(c)$ and $\left(c_{1}\right)$. First we will prove that the lines $D_{1} H_{2}, D_{2}... | {
"problem_match": "\nG8.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 260 | 1,330 |
2019 | T1 | G9 | Geometry | Balkan_Shortlist | 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 .
$$
Figure 9: G9
We draw the perpendicular line $(v)$ to $H C$ at the point $H$. Let $J$ be the intersection point of lines $(v)$ and $C O$. Then $C J$ is diameter of the circle $(O, O A)$ and
$... | {
"problem_match": "\nG9.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 354 | 534 |
2019 | T1 | TN1 | Number Theory | Balkan_Shortlist | Let $\mathbb{P}$ be the set of all prime numbers. Find all functions $f: \mathbb{P} \rightarrow \mathbb{P}$ such that
$$
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} .
$$
Assume that $f(2) \neq 2$. It follows that $f(2)$ is odd and so... | {
"problem_match": "\nTN1.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 82 | 565 |
2019 | T1 | C1 | Combinatorics | Balkan_Shortlist | 100 couples are invited to a traditional Moldovan dance. The 200 people stand in a line, and then in a step, two of them (not necessarily adjacent) may swap positions. Find the least $C$ such that whatever the initial order, they can arrive at an ordering where everyone is dancing next to their partner in at most $C$ s... | With 100 replaced by $N$, the answer is $C=C(N)=N-1$. Throughout, we will say that the members of a couple have the same.
$N=2$ : We use this as a base case for induction for both bounds. Up to labelling, there is one trivial initial order, and two non-trivial ones, namely
$$
1,1,2,2 ; \quad 1, \sqrt{2,2,1} ; \quad 1,... | {
"problem_match": "\nC1.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 77 | 1,110 |
2019 | T1 | C2b | Combinatorics | Balkan_Shortlist | 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 ... | {
"problem_match": "\nC2b. ${ }^{8}$",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 79 | 851 |
2019 | T1 | C2b | Combinatorics | Balkan_Shortlist | 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... | {
"problem_match": "\nC2b. ${ }^{8}$",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 79 | 1,121 |
2019 | T1 | C4 | Combinatorics | Balkan_Shortlist | 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... | {
"problem_match": "\nC4.",
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 133 | 2,092 |
2021 | T1 | A3 | Algebra | Balkan_Shortlist | Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by Greece | 1. We will show that $f(x)=x$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x+f(x)+f(y))=2 f(x)+y$.
We first show that $f$ is injective. So assume $f(a)=f(b)$. Now $P(1, a)$ and $P(1, b)$ show that
$$
2 f(1)+a=f(1+f(1)+f(a... | {
"problem_match": "\nA3.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 71 | 567 |
2021 | T1 | A3 | Algebra | Balkan_Shortlist | Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by Greece | 2. As in Solution $1, f$ is injective. Furthermore, letting $m=2 f(1)$ we have that the image of $f$ contains $(m, \infty)$. Indeed, if $t>m$, say $t=m+y$ for some $y>0$, then $P(1, y)$ shows that $f(1+f(1)+f(y))=t$.
Let $a, b \in \mathbb{R}$. We will show that $f(a)-a=f(b)-b$. Define $c=2 f(a)-2 f(b)$ and $d=$ $a+f(a... | {
"problem_match": "\nA3.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 71 | 660 |
2021 | T1 | A4 | Algebra | Balkan_Shortlist | Let $f, g$ be functions from the positive integers to the integers. Vlad the impala is jumping around the integer grid. His initial position is $\mathbf{x}_{0}=(0,0)$, and for every $n \geqslant 1$, his jump is
$$
\mathbf{x}_{n}-\mathbf{x}_{n-1}=( \pm f(n), \pm g(n)) \text { or }( \pm g(n), \pm f(n))
$$
with eight po... | 1.
(a) Yes it is always possible. The key idea is the following: Let $b(n)$ be the number of 1 's in the binary expansion of $n=0,1,2, \ldots$.
Lemma: Given a polynomial $f$ with integer coefficients and degree at most $d$, then
$$
\sum_{k=0}^{2^{d+1}-1}(-1)^{b(k)} f(n+k)=f(n)-f(n+1)-f(n+2)+\cdots \pm f\left(n+\left... | {
"problem_match": "\nA4.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 178 | 530 |
2021 | T1 | A4 | Algebra | Balkan_Shortlist | Let $f, g$ be functions from the positive integers to the integers. Vlad the impala is jumping around the integer grid. His initial position is $\mathbf{x}_{0}=(0,0)$, and for every $n \geqslant 1$, his jump is
$$
\mathbf{x}_{n}-\mathbf{x}_{n-1}=( \pm f(n), \pm g(n)) \text { or }( \pm g(n), \pm f(n))
$$
with eight po... | 2.
(a) Given a polynomial $f$ of degree at most $d$ and integers $n, r$, we claim that
$$
\sum_{k=0}^{2^{d+1}-1} \varepsilon_{k} f\left(2^{d} n+r+k\right)=0
$$
for some choice of $\varepsilon_{0}, \varepsilon_{1}, \ldots, \varepsilon_{2^{d+1}-1} \in\{-1,1\}$. (Which are allowed to depend on $d$ and f.)
We proceed b... | {
"problem_match": "\nA4.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\n## Solution"
} | 178 | 617 |
2021 | T1 | A5 | Algebra | Balkan_Shortlist | Find all functions $f: \mathbb{R}^{+} \longrightarrow \mathbb{R}^{+}$such that
$$
f(x f(x+y))=y f(x)+1
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by North Macedonia | 1. We will show that that $f(x)=\frac{1}{x}$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x f(x+y))=y f(x)+1$.
We first show that $f$ is injective. So assume $f\left(x_{1}\right)=f\left(x_{2}\right)$ and take any $x<x_{1},... | {
"problem_match": "\nA5.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 70 | 965 |
2021 | T1 | A6 | Algebra | Balkan_Shortlist | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x y)=f(x) f(y)+f(f(x+y))
$$
holds for all $x, y \in \mathbb{R}$.
## Proposed by Romania | 1. We will show that $f(x)=0$ for every $x \in \mathbb{R}$ or $f(x)=x-1$ for every $x \in \mathbb{R}$. It is easy to check that both of these functions work.
We write $P(x, y)$ for the assertion that $f(x y)=f(x) f(y)+f(f(x+y))$. For later use we write $Q(x, y)$ for the assertion that $f(x y)=f(x) f(y)$ and $R(x, y)$ ... | {
"problem_match": "\nA6.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 62 | 2,409 |
2021 | T1 | A6 | Algebra | Balkan_Shortlist | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x y)=f(x) f(y)+f(f(x+y))
$$
holds for all $x, y \in \mathbb{R}$.
## Proposed by Romania | 2 (Sketch). One can prove directly Claims 3 and 4 without the use of Claims 1 and 2. To prove Claim 3 we can make use of $P(x+1, y-1)$ which together with $P(x, y)$ and (1) gives
$$
f(x y+y-x)-c f(x y)=f(y)-c f(x) .
$$
Assuming $c=-1$, then (1) and (3) give that $f(x+2)=f(x)$ for every $x \in \mathbb{R}$. It follows ... | {
"problem_match": "\nA6.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 62 | 505 |
2021 | T1 | C1 | Combinatorics | Balkan_Shortlist | Let $\mathcal{A}_{n}$ be the set of $n$-tuples $x=\left(x_{1}, \ldots, x_{n}\right)$ with $x_{i} \in\{0,1,2\}$. A triple $x, y, z$ of distinct elements of $\mathcal{A}_{n}$ is called good if there is some $i$ such that $\left\{x_{i}, y_{i}, z_{i}\right\}=\{0,1,2\}$. A subset $A$ of $\mathcal{A}_{n}$ is called good if e... | 1. We proceed by induction on $n$, the case $n=1$ being trivial. Let
$$
A_{0}=\left\{\left(x_{1}, \ldots, x_{n}\right) \in A: x_{n} \neq 0\right\}
$$
and define $A_{1}$ and $A_{2}$ similarly.
Since $A$ is good and $A_{0}$ is a subset of $A$, then $A_{0}$ is also good. Therefore, any three of its elements have a coord... | {
"problem_match": "\nC1.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 188 | 551 |
2021 | T1 | C1 | Combinatorics | Balkan_Shortlist | Let $\mathcal{A}_{n}$ be the set of $n$-tuples $x=\left(x_{1}, \ldots, x_{n}\right)$ with $x_{i} \in\{0,1,2\}$. A triple $x, y, z$ of distinct elements of $\mathcal{A}_{n}$ is called good if there is some $i$ such that $\left\{x_{i}, y_{i}, z_{i}\right\}=\{0,1,2\}$. A subset $A$ of $\mathcal{A}_{n}$ is called good if e... | 2. Let
$$
B=\left\{x=\left(x_{1}, \ldots, x_{n}\right) \in \mathcal{A}_{n}: x_{i} \in\{0,1\}\right\}
$$
Let $A$ be a good subset of $\mathcal{A}_{n}$ and define $f: A \times B \rightarrow \mathcal{A}_{n}$ by $f(a, b)=a+b=\left(a_{1}+b_{1}, \ldots, a_{n}+b_{n}\right)$ where the addition is done modulo 3 .
We claim th... | {
"problem_match": "\nC1.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 188 | 1,010 |
2021 | T1 | C2 | Combinatorics | Balkan_Shortlist | Let $K$ and $N>K$ be fixed positive integers. Let $n$ be a positive integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be distinct integers. Suppose that whenever $m_{1}, m_{2}, \ldots, m_{n}$ are integers, not all equal to 0 , such that $\left|m_{i}\right| \leqslant K$ for each $i$, then the sum
$$
\sum_{i=1}^{n} m_{i} a_... | The answer is $n=\left\lfloor\log _{K+1} N\right\rfloor$.
Note first that for $n \leqslant\left\lfloor\log _{K+1} N\right\rfloor$, taking $a_{i}=(K+1)^{i-1}$ works. Indeed let $r$ be maximal such that $m_{r} \neq 0$. Then on the one hand we have
$$
\left|\sum_{i=1}^{n} m_{i} a_{i}\right| \leqslant \sum_{i=1}^{n} K(K+1... | {
"problem_match": "\nC2.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 146 | 562 |
2021 | T1 | C3 | Combinatorics | Balkan_Shortlist | In an exotic country, the National Bank issues coins that can take any value in the interval $[0,1]$. Find the smallest constant $c>0$ such that the following holds, no matter the situation in that country:
Any citizen of the exotic country that has a finite number of coins, with a total value of no more than 1000, ca... | 1. The answer is $c=\frac{1000}{91}=11-\frac{11}{1001}$. Clearly, if $c^{\prime}$ works, so does any $c>c^{\prime}$. First we prove that $c=11-\frac{11}{1001}$ is good.
We start with 100 empty boxes. First, we consider only the coins that individually value more than $\frac{1000}{1001}$. As their sum cannot overpass 1... | {
"problem_match": "\nC3.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 104 | 903 |
2021 | T1 | C4 | Combinatorics | Balkan_Shortlist | A sequence of $2 n+1$ non-negative integers $a_{1}, a_{2}, \ldots, a_{2 n+1}$ is given. There's also a sequence of $2 n+1$ consecutive cells enumerated from 1 to $2 n+1$ from left to right, such that initially the number $a_{i}$ is written on the $i$-th cell, for $i=1,2, \ldots 2 n+1$. Starting from this initial positi... | The answer is: $C_{n} \cdot C_{n}$, where $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$-th Catalan number.
We divide the proof into several steps. First, some terminology: the last (rightmost) $n$ cells will be called the back cells and the front (leftmost) $n$ cells will be called the front cells. The central, $(n+1)... | {
"problem_match": "\nC4.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 363 | 1,652 |
2021 | T1 | C5 | Combinatorics | Balkan_Shortlist | Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel either clears every piece of rubbish from a single pile, or one piece of rubbish from each pile. However, every evening, a demon sneaks into the warehouse and adds one piece of rubbish to each non-empty pile, or... | 1. We will show that he can do so by the morning of day 199 but not earlier.
If we have $n$ piles with at least two pieces of rubbish and $m$ piles with exactly one piece of rubbish, then we define the value of the pile to be
$$
V= \begin{cases}n & m=0 \\ n+\frac{1}{2} & m=1 \\ n+1 & m \geqslant 2\end{cases}
$$
We al... | {
"problem_match": "\nC5.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 104 | 1,025 |
2021 | T1 | C6 | Combinatorics | Balkan_Shortlist | There is a population $P$ of 10000 bacteria, some of which are friends (friendship is mutual), so that each bacterion has at least one friend and if we wish to assign to each bacterion a coloured membrane so that no two friends have the same colour, then there is a way to do it with 2021 colours, but not with 2020 or l... | 1. The answer is affirmative.
We will use the terminology of graph theory. Here the vertices of our main graph $G$ are the bacteria and there is an edge between two precisely when they are friends. The degree $d(v)$ of a vertex $v$ of $G$ is the number of neighbours of $v$. The minimum degree $\delta(G)$ of $G$ is the ... | {
"problem_match": "\nC6.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 215 | 1,117 |
2021 | T1 | G1 | Geometry | Balkan_Shortlist | Let $A B C$ be a triangle with $A B<A C<B C$. On the side $B C$ we consider points $D$ and $E$ such that $B A=B D$ and $C E=C A$. Let $K$ be the circumcenter of triangle $A D E$ and let $F, G$ be the points of intersection of the lines $A D, K C$ and $A E, K B$ respectively. Let $\omega_{1}$ be the circumcircle of tria... | 1. Since the triangles $B A D, K A D$ and $K D E$ are isosceles, then $\angle B A D=\angle B D A$ and $\angle K A D=\angle K D A$ and $\angle K D E=\angle K E D$. Therefore,
$$
\angle B A K=\angle B A D-\angle K A D=\angle B D A-\angle K D A=\angle K D E=\angle K E D=180^{\circ}-\angle B E K .
$$
So the points $B, E,... | {
"problem_match": "\nG1.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 195 | 519 |
2021 | T1 | G1 | Geometry | Balkan_Shortlist | Let $A B C$ be a triangle with $A B<A C<B C$. On the side $B C$ we consider points $D$ and $E$ such that $B A=B D$ and $C E=C A$. Let $K$ be the circumcenter of triangle $A D E$ and let $F, G$ be the points of intersection of the lines $A D, K C$ and $A E, K B$ respectively. Let $\omega_{1}$ be the circumcircle of tria... | 2. Let $M$ the midpoint of $A D$. Then $B M$ is the perpendicular bisector of $A D$, because the triangle $A B D$ is isosceles. $K M$ is also the perpendicular bisector of $A D$, because the point $K$ is the circumcenter of the triangle $A E D$. So points $B, G, K, M$ are collinear and $G M$ is also the perpendicular b... | {
"problem_match": "\nG1.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 195 | 604 |
2021 | T1 | G2 | Geometry | Balkan_Shortlist | Let $I$ and $O$ be the incenter and the circumcenter of a triangle $A B C$, respectively, and let $s_{a}$ be the exterior bisector of angle $\angle B A C$. The line through $I$ perpendicular to $I O$ meets the lines $B C$ and $s_{a}$ at points $P$ and $Q$, respectively. Prove that $I Q=2 I P$.
## Proposed by Serbia | Denote by $I_{b}$ and $I_{c}$ the respective excenters opposite to $B$ and $C$. Also denote the midpoint of side $B C$ by $D$, the midpoint of the arc $B A C$ by $M$, and the midpoint of segment $A M$ by $N$. Recall that $M$ is on the perpendicular bisector of $B C$, i.e. on line $O D$. Points $I, O, D, P$ lie on the c... | {
"problem_match": "\nG2.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 98 | 572 |
2021 | T1 | G3 | Geometry | Balkan_Shortlist | Let $A B C$ be a triangle with $A B<A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A=\angle A Y C$ and $X$ lies on the opposite side of $A B$ to $C$ while $Y$ lies on the opposite side of $A C$ to $B$.
Show that, as $X, ... | 2. Let $B^{\prime}$ and $C^{\prime}$ be the points of intersection of the lines $A B$ and $A C$ with $\omega$ respectively and let $\omega_{1}$ be the circumcircle of the triangle $A B^{\prime} C^{\prime}$. Let $\varepsilon$ be the tangent to $\omega_{1}$ at the point $A$. Because $A B<A C$ the lines $B^{\prime} C^{\pr... | {
"problem_match": "\nG3.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 133 | 654 |
2021 | T1 | G4 | Geometry | Balkan_Shortlist | Let $A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$. Let the height from $A$ cut its side $B C$ at $D$. Let $I, I_{B}, I_{C}$ be the incenters of triangles $A B C, A B D, A C D$ respectively. Let also $E_{B}, E_{C}$ be the excenters of $A B C$ with respect to vertices $B$ and $C$ respectively. If $K$ ... | Since $\angle E_{C} B I=90^{\circ}=I C E_{B}$, we conclude that $E_{C} B C E_{B}$ is cyclic. Moreover, we have that
$$
\angle B A I_{B}=\frac{1}{2} \angle B A D=\frac{1}{2} \widehat{C},
$$
so $A I_{B} \perp C I$. Similarly $A I_{C} \perp B I$. Therefore is the orthocenter of triangle $A I_{B} I_{C}$. It follows that
... | {
"problem_match": "\nG4.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 172 | 700 |
2021 | T1 | G5 | Geometry | Balkan_Shortlist | Let $A B C$ be an acute triangle with $A C>A B$ and circumcircle $\Gamma$. The tangent from $A$ to $\Gamma$ intersects $B C$ at $T$. Let $M$ be the midpoint of $B C$ and let $R$ be the reflection of $A$ in $B$. Let $S$ be a point so that $S A B T$ is a parallelogram and finally let $P$ be a point on line $S B$ such tha... | 1. Let $N$ be the midpoint of $B S$ which, as $S A B T$ is a parallelogram, is also the midpoint of $T A$. Using $S T\|A B\| M P$ we get:
$$
\frac{N B}{B P}=\frac{1}{2} \cdot \frac{S B}{B P}=\frac{T B}{2 \cdot B M}=\frac{T B}{B C}
$$
which shows that $T A \| C P$.
Let $\Omega$ be the circle with diameter $O T$. As $... | {
"problem_match": "\nG5.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 155 | 535 |
2021 | T1 | G6 | Geometry | Balkan_Shortlist | Let $A B C$ be an acute triangle such that $A B<A C$. Let $\omega$ be the circumcircle of $A B C$ and assume that the tangent to $\omega$ at $A$ intersects the line $B C$ at $D$. Let $\Omega$ be the circle with center $D$ and radius $A D$. Denote by $E$ the second intersection point of $\omega$ and $\Omega$. Let $M$ be... | 1. Denote by $S$ the intersection point of $\Omega$ and the segment $B C$. Because $D A=D S$, we have $\angle D S A=\angle D A S$. Now using that $D A$ is tangent to $\omega$ we obtain:
$$
\angle B A S=\angle D A S-\angle D A B=\angle D S A-\angle D C A=\angle C A S .
$$
This means that the line $A S$ is the angle bi... | {
"problem_match": "\nG6.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 159 | 832 |
2021 | T1 | G6 | Geometry | Balkan_Shortlist | Let $A B C$ be an acute triangle such that $A B<A C$. Let $\omega$ be the circumcircle of $A B C$ and assume that the tangent to $\omega$ at $A$ intersects the line $B C$ at $D$. Let $\Omega$ be the circle with center $D$ and radius $A D$. Denote by $E$ the second intersection point of $\omega$ and $\Omega$. Let $M$ be... | 2. As in Solution 1, we let $S$ be the intersection of $\Omega$ with $B S$ and obtain that $A S$ is the angle bisector of $\angle B A C$ and that $A E$ and $A Y$ are symmetric with respect to $A S$.
Let $R=\sqrt{(A B)(A C)}$ and let $\Psi$ be the map obtained by first inverting on the circle centered at $A$ of radius $... | {
"problem_match": "\nG6.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 159 | 626 |
2021 | T1 | G7 | Geometry | Balkan_Shortlist | Let $A B C$ be an acute scalene triangle. Its $C$-excircle tangent to the segment $A B$ meets $A B$ at point $M$ and the extension of $B C$ beyond $B$ at point $N$. Analogously, its $B$-excircle tangent to the segment $A C$ meets $A C$ at point $P$ and the extension of $B C$ beyond $C$ at point $Q$. Denote by $A_{1}$ t... | 1. We shall use the standard notations for $A B C$, i.e. $\angle A B C=\beta, B C=a$ etc. We also write $s=\frac{a+b+c}{2}$ for the semiperimeter and $r$ for the inradius.
Let $M N$ intersect the altitude $A D(D$ lies on $B C)$ at the point $L$. We have that $\angle B A D=90^{\circ}-\beta$ and $\angle A M L=\angle B M... | {
"problem_match": "\nG7.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 199 | 755 |
2021 | T1 | G7 | Geometry | Balkan_Shortlist | Let $A B C$ be an acute scalene triangle. Its $C$-excircle tangent to the segment $A B$ meets $A B$ at point $M$ and the extension of $B C$ beyond $B$ at point $N$. Analogously, its $B$-excircle tangent to the segment $A C$ meets $A C$ at point $P$ and the extension of $B C$ beyond $C$ at point $Q$. Denote by $A_{1}$ t... | 2. As in Solution 1, we have that $A_{2}, B_{2}, C_{2}$ belong on the corresponding altitudes with $A A_{2}=B B_{2}=C C_{2}=2 r$. We present an approach with complex numbers (and minimal calculations) which can also complete the proof.
Set the incenter $I$ of the triangle $A B C$ to be the origin. We may assume that $... | {
"problem_match": "\nG7.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 199 | 508 |
2021 | T1 | G8 | Geometry | Balkan_Shortlist | Let $A B C$ be a scalene triangle and let $I$ be its incenter. The projections of $I$ on $B C, C A$ and $A B$ are $D, E$ and $F$ respectively. Let $K$ be the reflection of $D$ over the line $A I$, and let $L$ be the second point of intersection of the circumcircles of the triangles $B F K$ and $C E K$. If $\frac{1}{3} ... | Writing $A E=A F=x, B F=B D=y$ and $C E=C D=z$, the condition $\frac{1}{3} B C=A C-A B$ translates to $y+z=3(z-y)$ giving $z=2 y$, i.e. $C D=2 B D$.
Letting $B^{\prime}$ be the reflection of $B$ on $A I$ we have that $B^{\prime}$ belongs on $A C$ with $B^{\prime} E=B F=$ $B D=\frac{1}{2} C D=\frac{1}{2} C E$ therefore... | {
"problem_match": "\nG8.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 130 | 853 |
2021 | T1 | N1 | Number Theory | Balkan_Shortlist | Let $n \geqslant 3$ be an integer and let
$$
M=\left\{\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}: 1 \leqslant k \leqslant n \text { and } 1 \leqslant a_{1}<\cdots<a_{k} \leqslant n\right\}
$$
be the set of the arithmetic means of the elements of all non-empty subsets of $\{1,2, \ldots, n\}$. Find $\min \{|a-b|: a, b \in M$ w... | We observe that $M$ is composed by rational numbers of the form $a=\frac{x}{k}$, where $1 \leqslant k \leqslant n$. As the arithmetic mean of $1, \ldots, n$ is $\frac{n+1}{2}$, if we look at these rational numbers in their irreducible form, we can say that $1 \leqslant k \leqslant n-1$.
A non-zero difference $|a-b|$ w... | {
"problem_match": "\nN1.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 154 | 664 |
2021 | T1 | N2 | Number Theory | Balkan_Shortlist | Denote by $\ell(n)$ the largest prime divisor of $n$. Let $a_{n+1}=a_{n}+\ell\left(a_{n}\right)$ be a recursively defined sequence of integers with $a_{1}=2$. Determine all natural numbers $m$ such that there exists some $i \in \mathbb{N}$ with $a_{i}=m^{2}$. | We will show that all such numbers are exactly the prime numbers.
Let $p_{1}, p_{2}, \ldots$ be the sequence of prime numbers. We will prove the following:
Claim: Assume $a_{n}=p_{i} p_{i+1}$. Then for each $k=1,2, \ldots, p_{i+2}-p_{i}$ we have that $a_{n+k}=$ $\left(p_{i}+k\right) p_{i+1}$.
Proof. By induction on $k$... | {
"problem_match": "\nN2.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 85 | 804 |
2021 | T1 | N5 | Number Theory | Balkan_Shortlist | A natural number $n$ is given. Determine all $(n-1)$-tuples of nonnegative integers $a_{1}, a_{2}, \ldots, a_{n-1}$ such that
$$
\left[\frac{m}{2^{n}-1}\right]+\left[\frac{2 m+a_{1}}{2^{n}-1}\right]+\left[\frac{2^{2} m+a_{2}}{2^{n}-1}\right]+\left[\frac{2^{3} m+a_{3}}{2^{n}-1}\right]+\cdots+\left[\frac{2^{n-1} m+a_{n-... | 1. We will show that there is a unique such $n$-tuple: $a_{k}=2^{n-1}+2^{k-1}-1$ for $k=1, \ldots, n-1$.
Write $N=2^{n}-1$ and $f_{k}(x)=\left[\frac{2^{k} x+a_{k}}{N}\right]$ for $k=0,1, \ldots, n-1$, where $a_{0}=0$. Since
$$
\sum_{k=0}^{n-1} f_{k}(m)-\sum_{k=0}^{n-1} f_{k}(m-1)=1
$$
for each $m \in \mathbb{Z}$, the... | {
"problem_match": "\nN5.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 176 | 1,110 |
2021 | T1 | N5 | Number Theory | Balkan_Shortlist | A natural number $n$ is given. Determine all $(n-1)$-tuples of nonnegative integers $a_{1}, a_{2}, \ldots, a_{n-1}$ such that
$$
\left[\frac{m}{2^{n}-1}\right]+\left[\frac{2 m+a_{1}}{2^{n}-1}\right]+\left[\frac{2^{2} m+a_{2}}{2^{n}-1}\right]+\left[\frac{2^{3} m+a_{3}}{2^{n}-1}\right]+\cdots+\left[\frac{2^{n-1} m+a_{n-... | 2. We will use the identity
$$
[x]+\left[x+\frac{1}{N}\right]+\left[x+\frac{2}{N}\right]+\cdots+\left[x+\frac{N-1}{N}\right]=[N x]
$$
which holds for every $x \in \mathbb{R}$ and every $N \in \mathbb{N}$. (One can check this by noting that the difference between the two sides of the identity is periodic with period $... | {
"problem_match": "\nN5.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution"
} | 176 | 2,220 |
2021 | T1 | N6 | Number Theory | Balkan_Shortlist | Let $a, b$ and $c$ be positive integers satisfying the equation $(a, b)+[a, b]=2021^{c}$. If $|a-b|$ is a prime number, prove that the number $(a+b)^{2}+4$ is composite.
## Proposed by Serbia | We write $p=|a-b|$ and assume for contradiction that $q=(a+b)^{2}+4$ is a prime number.
Since $(a, b) \mid[a, b]$, we have that $(a, b) \mid 2021^{c}$. As $(a, b)$ also divides $p=|a-b|$, it follows that $(a, b) \in\{1,43,47\}$. We will consider all 3 cases separately:
(1) If $(a, b)=1$, then $1+a b=2021^{c}$, and the... | {
"problem_match": "\nN6.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 66 | 844 |
2021 | T1 | N7 | Number Theory | Balkan_Shortlist | A super-integer triangle is defined to be a triangle whose lengths of all sides and at least one height are positive integers. We will deem certain positive integer numbers to be good with the condition that if the lengths of two sides of a super-integer triangle are two (not necessarily different) good numbers, then t... | Evidently, all right-angle triangles with integer sides are super-integer triangles. We will use the following notation $(a, b, c\{h\})$ to denote a super-integer triangle whose sides are $a$, $b$ and $c$ and the height of integer length is $h$. The height will be written in curly brackets next to the corresponding sid... | {
"problem_match": "\nN7.",
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"solution_match": "\nSolution."
} | 99 | 1,996 |
2000 | T3 | 6 | null | BalticWay | Fredek runs a private hotel. He claims that whenever $n \geqslant 3$ guests visit the hotel, it is possible to select two guests who have equally many acquaintances among the other guests, and who also have a common acquaintance or a common unknown among the guests. For which values of $n$ is Fredek right?
(Acquaintan... | Answer: Fredek is right for all $n \neq 4$.
Suppose that any two guests of Fredek having the same number of acquaintances have neither a common acquaintance nor a common unknown. From the set $\mathcal{K}$ of Fredek's guests choose any two guests $A$ and $B$ having the same number of acquaintances (the existence of su... | {
"problem_match": "\n6.",
"resource_path": "BalticWay/segmented/en-bw00sol.jsonl",
"solution_match": "\n6."
} | 81 | 616 |
2001 | T3 | 5 | null | BalticWay | Let 2001 given points on a circle be colored either red or green. In one step all points are recolored simultaneously in the following way: If both direct neighbors of a point $P$ have the same color as $P$, then the color of $P$ remains unchanged, otherwise $P$ obtains the other color. Starting with the first coloring... | Answer: no.
Let the points be denoted by $1,2, \ldots, 2001$ such that $i, j$ are neighbors if $|i-j|=1$ or $\{i, j\}=\{1,2001\}$. We say that $k$ points form a monochromatic segment of length $k$ if the points are consecutive on the circle and if
they all have the same color. For a coloring $F$ let $d(F)$ be the maxi... | {
"problem_match": "\n5.",
"resource_path": "BalticWay/segmented/en-bw01sol.jsonl",
"solution_match": "\n5."
} | 168 | 778 |
2001 | T3 | 11 | null | BalticWay | The real-valued function $f$ is defined for all positive integers. For any integers $a>1, b>1$ with $d=\operatorname{gcd}(a, b)$, we have
$$
f(a b)=f(d) \cdot\left(f\left(\frac{a}{d}\right)+f\left(\frac{b}{d}\right)\right),
$$
Determine all possible values of $f(2001)$. | Answer: 0 and $\frac{1}{2}$.
Obviously the constant functions $f(n)=0$ and $f(n)=\frac{1}{2}$ provide solutions.
We show that there are no other solutions. Assume $f(2001) \neq 0$. Since $2001=3 \cdot 667$ and $\operatorname{gcd}(3,667)=1$, then
$$
f(2001)=f(1) \cdot(f(3)+f(667)),
$$
and $f(1) \neq 0$. Since $\oper... | {
"problem_match": "\n11.",
"resource_path": "BalticWay/segmented/en-bw01sol.jsonl",
"solution_match": "\n11."
} | 100 | 626 |
2001 | T3 | 15 | null | BalticWay | Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of positive real numbers satisfying $i \cdot a_{i}^{2} \geqslant(i+1) \cdot a_{i-1} a_{i+1}$ for $i=1,2, \ldots$ Furthermore, let $x$ and $y$ be positive reals, and let $b_{i}=x a_{i}+y a_{i-1}$ for $i=1,2, \ldots$ Prove that the inequality $i \cdot b_{i}^{2}>(i+1) \cdot ... | Let $i \geqslant 2$. We are given the inequalities
$$
(i-1) \cdot a_{i-1}^{2} \geqslant i \cdot a_{i} a_{i-2}
$$
and
$$
i \cdot a_{i}^{2} \geqslant(i+1) \cdot a_{i+1} a_{i-1} .
$$
Multiplying both sides of (6) by $x^{2}$, we obtain
$$
i \cdot x^{2} \cdot a_{i}^{2} \geqslant(i+1) \cdot x^{2} \cdot a_{i+1} a_{i-1}
$... | {
"problem_match": "\n15.",
"resource_path": "BalticWay/segmented/en-bw01sol.jsonl",
"solution_match": "\n15."
} | 168 | 555 |
2002 | T3 | 2 | null | BalticWay | Let $a, b, c, d$ be real numbers such that
$$
\begin{aligned}
a+b+c+d & =-2 \\
a b+a c+a d+b c+b d+c d & =0
\end{aligned}
$$
Prove that at least one of the numbers $a, b, c, d$ is not greater than -1 . | We can assume that $a$ is the least among $a, b, c, d$ (or one of the least, if some of them are equal), there are $n>0$ negative numbers among $a, b, c, d$, and the sum of the positive ones is $x$.
Then we obtain
$$
-2=a+b+c+d \geqslant n a+x .
$$
Squaring we get
$$
4=a^{2}+b^{2}+c^{2}+d^{2}
$$
which implies
$$
... | {
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw02sol.jsonl",
"solution_match": "\nSolution."
} | 80 | 889 |
2002 | T3 | 3 | null | BalticWay | Find all sequences $a_{0} \leqslant a_{1} \leqslant a_{2} \leqslant \ldots$ of real numbers such that
$$
a_{m^{2}+n^{2}}=a_{m}^{2}+a_{n}^{2}
$$
for all integers $m, n \geqslant 0$.
Answer: $a_{n} \equiv 0, a_{n} \equiv \frac{1}{2}$ and $a_{n}=n$. | Denoting $f(n)=a_{n}$ we have
$$
f\left(m^{2}+n^{2}\right)=f^{2}(m)+f^{2}(n) .
$$
Substituting $m=n=0$ into (7) we get $f(0)=2 f^{2}(0)$, hence either $f(0)=\frac{1}{2}$ or $f(0)=0$. We consider these cases separately.
(1) If $f(0)=\frac{1}{2}$ then substituting $m=1$ and $n=0$ into (7) we obtain $f(1)=f^{2}(1)+\fra... | {
"problem_match": "\n3.",
"resource_path": "BalticWay/segmented/en-bw02sol.jsonl",
"solution_match": "\nSolution."
} | 121 | 971 |
2002 | T3 | 4 | null | BalticWay | Let $n$ be a positive integer. Prove that
$$
\sum_{i=1}^{n} x_{i}\left(1-x_{i}\right)^{2} \leqslant\left(1-\frac{1}{n}\right)^{2}
$$
for all nonnegative real numbers $x_{1}, x_{2}, \ldots, x_{n}$ such that $x_{1}+x_{2}+\cdots+x_{n}=1$. | Expanding the expressions at both sides we obtain the equivalent inequality
$$
-\sum_{i} x_{i}^{3}+2 \sum_{i} x_{i}^{2}-\frac{2}{n}+\frac{1}{n^{2}} \geqslant 0
$$
It is easy to check that the left hand side is equal to
$$
\sum_{i}\left(2-\frac{2}{n}-x_{i}\right)\left(x_{i}-\frac{1}{n}\right)^{2}
$$
and hence is non... | {
"problem_match": "\n4.",
"resource_path": "BalticWay/segmented/en-bw02sol.jsonl",
"solution_match": "\nSolution."
} | 108 | 741 |
2002 | T3 | 6 | null | BalticWay | The following solitaire game is played on an $m \times n$ rectangular board, $m, n \geqslant 2$, divided into unit squares. First, a rook is placed on some square. At each move, the rook can be moved an arbitrary number of squares horizontally or vertically, with the extra condition that each move has to be made in the... | First, consider any row that is not the row where the rook starts from. The rook has to visit all the squares of that row exactly once, and on its tour around the board, every time it visits this row, exactly two squares get visited. Hence, $m$ must be even; a similar argument for the columns shows that $n$ must also b... | {
"problem_match": "\n6.",
"resource_path": "BalticWay/segmented/en-bw02sol.jsonl",
"solution_match": "\nSolution."
} | 192 | 739 |
2002 | T3 | 7 | null | BalticWay | We draw $n$ convex quadrilaterals in the plane. They divide the plane into regions (one of the regions is infinite). Determine the maximal possible number of these regions.
Answer: The maximal number of regions is $4 n^{2}-4 n+2$. | One quadrilateral produces two regions. Suppose we have drawn $k$ quadrilaterals $Q_{1}, \ldots, Q_{k}$ and produced $a_{k}$ regions. We draw another quadrilateral $Q_{k+1}$ and try to evaluate the number of regions $a_{k+1}$ now produced. Our task is to make $a_{k+1}$ as large as possible. Note that in a maximal confi... | {
"problem_match": "\n7.",
"resource_path": "BalticWay/segmented/en-bw02sol.jsonl",
"solution_match": "\nSolution."
} | 55 | 529 |
2002 | T3 | 8 | null | BalticWay | Let $P$ be a set of $n \geqslant 3$ points in the plane, no three of which are on a line. How many possibilities are there to choose a set $T$ of $\left(\begin{array}{c}n-1 \\ 2\end{array}\right)$ triangles, whose vertices are all in $P$, such that each triangle in $T$ has a side that is not a side of any other triangl... | For a fixed point $x \in P$, let $T_{x}$ be the set of all triangles with vertices in $P$ which have $x$ as a vertex. Clearly, $\left|T_{x}\right|=\left(\begin{array}{c}n-1 \\ 2\end{array}\right)$, and each triangle in $T_{x}$ has a side which is not a side of any other triangle in $T_{x}$. For any $x, y \in P$ such th... | {
"problem_match": "\n8.",
"resource_path": "BalticWay/segmented/en-bw02sol.jsonl",
"solution_match": "\nSolution."
} | 130 | 880 |
2002 | T3 | 9 | null | BalticWay | Two magicians show the following trick. The first magician goes out of the room. The second magician takes a deck of 100 cards labelled by numbers $1,2, \ldots, 100$ and asks three spectators to choose in turn one card each. The second magician sees what card each spectator has taken. Then he adds one more card from th... | We will identify ourselves with the second magician. Then we need to choose a card in such a manner that another magician will be able to understand which of the 4 cards we have chosen and what information it gives about the order of the other cards. We will reach these two goals independently.
Let $a, b, c$ be remain... | {
"problem_match": "\n9.",
"resource_path": "BalticWay/segmented/en-bw02sol.jsonl",
"solution_match": "\nSolution."
} | 148 | 1,869 |
2002 | T3 | 16 | null | BalticWay | Find all nonnegative integers $m$ such that
$$
a_{m}=\left(2^{2 m+1}\right)^{2}+1
$$
is divisible by at most two different primes.
Answer: $m=0,1,2$ are the only solutions. | Obviously $m=0,1,2$ are solutions as $a_{0}=5, a_{1}=65=5 \cdot 13$, and $a_{2}=1025=25 \cdot 41$. We show that these are the only solutions.
Assume that $m \geqslant 3$ and that $a_{m}$ contains at most two different prime factors. Clearly, $a_{m}=4^{2 m+1}+1$ is divisible by 5 , and
$$
a_{m}=\left(2^{2 m+1}+2^{m+1}... | {
"problem_match": "\n16.",
"resource_path": "BalticWay/segmented/en-bw02sol.jsonl",
"solution_match": "\nSolution."
} | 62 | 567 |
2002 | T3 | 17 | null | BalticWay | Show that the sequence
$$
\left(\begin{array}{l}
2002 \\
2002
\end{array}\right),\left(\begin{array}{c}
2003 \\
2002
\end{array}\right),\left(\begin{array}{l}
2004 \\
2002
\end{array}\right), \ldots
$$
considered modulo 2002, is periodic. | Define
$$
x_{n}^{k}=\left(\begin{array}{l}
n \\
k
\end{array}\right)
$$
and note that
$$
x_{n+1}^{k}-x_{n}^{k}=\left(\begin{array}{c}
n+1 \\
k
\end{array}\right)-\left(\begin{array}{l}
n \\
k
\end{array}\right)=\left(\begin{array}{c}
n \\
k-1
\end{array}\right)=x_{n}^{k-1}
$$
Let $m$ be any positive integer. We wil... | {
"problem_match": "\n17.",
"resource_path": "BalticWay/segmented/en-bw02sol.jsonl",
"solution_match": "\nSolution."
} | 103 | 544 |
2002 | T3 | 20 | null | BalticWay | Does there exist an infinite non-constant arithmetic progression, each term of which is of the form $a^{b}$, where $a$ and $b$ are positive integers with $b \geqslant 2$ ?
Answer: no. | For an arithmetic progression $a_{1}, a_{2}, \ldots$ with difference $d$ the following holds:
$$
\begin{aligned}
S_{n} & =\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n+1}}=\frac{1}{a_{1}}+\frac{1}{a_{1}+d}+\ldots+\frac{1}{a_{1}+n d} \geqslant \\
& \geqslant \frac{1}{m}\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{... | {
"problem_match": "\n20.",
"resource_path": "BalticWay/segmented/en-bw02sol.jsonl",
"solution_match": "\nSolution."
} | 52 | 841 |
2003 | T3 | 7 | null | BalticWay | Let $X$ be a subset of $\{1,2,3, \ldots, 10000\}$ with the following property: If $a, b \in X, a \neq b$, then $a \cdot b \notin X$. What is the maximal number of elements in $X$ ?
Answer: 9901. | If $X=\{100,101,102, \ldots, 9999,10000\}$, then for any two selected $a$ and $b, a \neq b$, $a \cdot b \geq 100 \cdot 101>10000$, so $a \cdot b \notin X$. So $X$ may have 9901 elements.
Suppose that $x_{1}<x_{2}<\cdots<x_{k}$ are all elements of $X$ that are less than 100. If there are none of them, no more than 9901... | {
"problem_match": "\n7.",
"resource_path": "BalticWay/segmented/en-bw03sol.jsonl",
"solution_match": "\nSolution:"
} | 78 | 503 |
2003 | T3 | 18 | null | BalticWay | Every integer is coloured with exactly one of the colours BLUE, GREEN, RED, YELLOW. Can this be done in such a way that if $a, b, c, d$ are not all 0 and have the same colour, then $3 a-2 b \neq 2 c-3 d$ ?
Answer: Yes. | A colouring with the required property can be defined as follows. For a non-zero integer $k$ let $k^{*}$ be the integer uniquely defined by $k=5^{m} \cdot k^{*}$, where $m$ is a nonnegative integer and $5 \nmid k^{*}$. We also define $0^{*}=0$. Two non-zero integers $k_{1}, k_{2}$ receive the same colour if and only if... | {
"problem_match": "\n18.",
"resource_path": "BalticWay/segmented/en-bw03sol.jsonl",
"solution_match": "\nSolution:"
} | 70 | 534 |
2004 | T3 | 14 | null | BalticWay | We say that a pile is a set of four or more nuts. Two persons play the following game. They start with one pile of $n \geq 4$ nuts. During a move a player takes one of the piles that they have and split it into two non-empty subsets (these sets are not necessarily piles, they can contain an arbitrary number of nuts). I... | Let $n=4 k+r$, where $0 \leq r \leq 3$. We will prove the above answer by induction on $k$; clearly it holds for $k=1$. We are also going to need the following useful fact:
If at some point there are exactly two piles with $4 s+1$ and $4 t+1$ nuts, $s+t \leq k$, then the second player to move from that point wins.
Th... | {
"problem_match": "\n14.",
"resource_path": "BalticWay/segmented/en-bw04sol.jsonl",
"solution_match": "\nSolution:"
} | 136 | 792 |
2005 | T3 | 1 | null | BalticWay | Let $a_{0}$ be a positive integer. Define the sequence $a_{n}, n \geq 0$, as follows: If
$$
a_{n}=\sum_{i=0}^{j} c_{i} 10^{i}
$$
where $c_{i}$ are integers with $0 \leq c_{i} \leq 9$, then
$$
a_{n+1}=c_{0}^{2005}+c_{1}^{2005}+\cdots+c_{j}^{2005} .
$$
Is it possible to choose $a_{0}$ so that all the terms in the seq... | It is clear that there exists a smallest positive integer $k$ such that
$$
10^{k}>(k+1) \cdot 9^{2005} .
$$
We will show that there exists a positive integer $N$ such that $a_{n}$ consists of less than $k+1$ decimal digits for all $n \geq N$. Let $a_{i}$ be a positive integer which consists of exactly $j+1$ digits, t... | {
"problem_match": "\n1.",
"resource_path": "BalticWay/segmented/en-bw05sol.jsonl",
"solution_match": "\nSolution:"
} | 163 | 551 |
2005 | T3 | 6 | null | BalticWay | Let $K$ and $N$ be positive integers with $1 \leq K \leq N$. A deck of $N$ different playing cards is shuffled by repeating the operation of reversing the order of the $K$ topmost cards and moving these to the bottom of the deck. Prove that the deck will be back in its initial order after a number of operations not gre... | Let $N=q \cdot K+r, 0 \leq r<K$, and let us number the cards $1,2, \ldots, N$, starting from the one at the bottom of the deck. First we find out how the cards $1,2, \ldots K$ are moving in the deck.
If $i \leq r$ then the card $i$ is moving along the cycle
$$
\begin{aligned}
& i \rightarrow K+i \rightarrow 2 K+i \ri... | {
"problem_match": "\n6.",
"resource_path": "BalticWay/segmented/en-bw05sol.jsonl",
"solution_match": "\nSolution:"
} | 94 | 558 |
2006 | T3 | 18 | null | BalticWay | For a positive integer $n$ let $a_{n}$ denote the last digit of $n^{\left(n^{n}\right)}$. Prove that the sequence $\left(a_{n}\right)$ is periodic and determine the length of the minimal period. | Let $b_{n}$ and $c_{n}$ denote the last digit of $n$ and $n^{n}$, respectively. Obviously, if $b_{n}=0,1,5,6$, then $c_{n}=0,1,5,6$ and $a_{n}=0,1,5,6$, respectively.
If $b_{n}=9$, then $n^{n} \equiv 1(\bmod 2)$ and consequently $a_{n}=9$. If $b_{n}=4$, then $n^{n} \equiv 0$ $(\bmod 2)$ and consequently $a_{n}=6$.
If... | {
"problem_match": "\n18.",
"resource_path": "BalticWay/segmented/en-bw06sol.jsonl",
"solution_match": "\nSolution:"
} | 54 | 554 |
2006 | T3 | 19 | null | BalticWay | Does there exist a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers such that the sum of every $n$ consecutive elements is divisible by $n^{2}$ for every positive integer $n$ ?
Answer: Yes. One such sequence begins 1, 3, 5, 55, 561, 851, 63253, 110055,... | We will show that whenever we have positive integers $a_{1}, \ldots, a_{k}$ such that $n^{2} \mid a_{i+1}+\cdots+a_{i+n}$ for every $n \leq k$ and $i \leq k-n$, then it is possible to choose $a_{k+1}$ such that $n^{2} \mid a_{i+1}+\cdots+a_{i+n}$ for every $n \leq k+1$ and $i \leq k+1-n$. This directly implies the posi... | {
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw06sol.jsonl",
"solution_match": "\nSolution:"
} | 98 | 725 |
2008 | T3 | 3 | null | BalticWay | Does there exist an angle $\alpha \in(0, \pi / 2)$ such that $\sin \alpha, \cos \alpha, \tan \alpha$ and $\cot \alpha$, taken in some order, are consecutive terms of an arithmetic progression?
Answer: No. | Suppose that there is an $x$ such that $0<x<\frac{\pi}{2}$ and $\sin x, \cos x, \tan x, \cot x$ in some order are consecutive terms of an arithmetic progression.
Suppose $x \leq \frac{\pi}{4}$. Then $\sin x \leq \sin \frac{\pi}{4}=\cos \frac{\pi}{4} \leq \cos x<1 \leq \cot x$ and $\sin x<\frac{\sin x}{\cos x}=\tan x \... | {
"problem_match": "\nProblem 3.",
"resource_path": "BalticWay/segmented/en-bw08sol.jsonl",
"solution_match": "\nSolution:"
} | 58 | 543 |
2008 | T3 | 5 | null | BalticWay | Suppose that Romeo and Juliet each have a regular tetrahedron to the vertices of which some positive real numbers are assigned. They associate each edge of their tetrahedra with the product of the two numbers assigned to its end points. Then they write on each face of their tetrahedra the sum of the three numbers assoc... | Let us prove that this conclusion can in fact be drawn. For this purpose we denote the numbers assigned to the vertices of Romeo's tetrahedron by $r_{1}, r_{2}, r_{3}, r_{4}$ and the numbers assigned to the vertices of Juliette's tetrahedron by $j_{1}, j_{2}, j_{3}, j_{4}$ in such a way that
$$
\begin{aligned}
& r_{2}... | {
"problem_match": "\nProblem 5.",
"resource_path": "BalticWay/segmented/en-bw08sol.jsonl",
"solution_match": "\nSolution:"
} | 144 | 725 |
2008 | T3 | 9 | null | BalticWay | Suppose that the positive integers $a$ and $b$ satisfy the equation
$$
a^{b}-b^{a}=1008 .
$$
Prove that $a$ and $b$ are congruent modulo 1008. | Observe that $1008=2^{4} \cdot 3^{2} \cdot 7$. First we show that $a$ and $b$ cannot both be even. For suppose the largest of them were equal to $2 x$ and the smallest of them equal to $2 y$, where $x \geq y \geq 1$. Then
$$
\pm 1008=(2 x)^{2 y}-(2 y)^{2 x}
$$
so that $2^{2 y}$ divides 1008 . It follows that $y \leq ... | {
"problem_match": "\nProblem 9.",
"resource_path": "BalticWay/segmented/en-bw08sol.jsonl",
"solution_match": "\nSolution:"
} | 54 | 889 |
2008 | T3 | 12 | null | BalticWay | In a school class with $3 n$ children, any two children make a common present to exactly one other child. Prove that for all odd $n$ it is possible that the following holds:
For any three children $A, B$ and $C$ in the class, if $A$ and $B$ make a present to $C$ then $A$ and $C$ make a present to $B$. | Assume there exists a set $\mathscr{S}$ of sets of three children such that any set of two children is a subset of exactly one member of $\mathscr{S}$, and assume that the children $A$ and $B$ make a common present to $C$ if and only if $\{A, B, C\} \in \mathscr{S}$. Then it is true that any two children $A$ and $B$ ma... | {
"problem_match": "\nProblem 12.",
"resource_path": "BalticWay/segmented/en-bw08sol.jsonl",
"solution_match": "\nSolution:"
} | 88 | 896 |
2010 | T3 | 3 | null | BalticWay | Let $x_{1}, x_{2}, \ldots, x_{n}(n \geq 2)$ be real numbers greater than 1 . Suppose that $\left|x_{i}-x_{i+1}\right|<1$ for $i=1,2, \ldots, n-1$. Prove that
$$
\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{3}}+\ldots+\frac{x_{n-1}}{x_{n}}+\frac{x_{n}}{x_{1}}<2 n-1
$$ | The proof is by induction on $n$.
We establish first the base case $n=2$. Suppose that $x_{1}>1, x_{2}>1,\left|x_{1}-x_{2}\right|<1$ and moreover $x_{1} \leq x_{2}$. Then
$$
\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}} \leq 1+\frac{x_{2}}{x_{1}}<1+\frac{x_{1}+1}{x_{1}}=2+\frac{1}{x_{1}}<2+1=2 \cdot 2-1 .
$$
Now we procee... | {
"problem_match": "\nProblem 3.",
"resource_path": "BalticWay/segmented/en-bw10sol.jsonl",
"solution_match": "\nSolution."
} | 130 | 705 |
2010 | T3 | 7 | null | BalticWay | There are some cities in a country; one of them is the capital. For any two cities $A$ and $B$ there is a direct flight from $A$ to $B$ and a direct flight from $B$ to $A$, both having the same price. Suppose that all round trips with exactly one landing in every city have the same total cost. Prove that all round trip... | Let $C$ be the capital and $C_{1}, C_{2}, \ldots, C_{n}$ be the remaining cities. Denote by $d(x, y)$ the price of the connection between the cities $x$ and $y$, and let $\sigma$ be the total price of a round trip going exactly once through each city.
Now consider a round trip missing the capital and visiting every ot... | {
"problem_match": "\nProblem 7.",
"resource_path": "BalticWay/segmented/en-bw10sol.jsonl",
"solution_match": "\nSolution."
} | 98 | 605 |
2010 | T3 | 10 | null | BalticWay | Let $n$ be an integer with $n \geq 3$. Consider all dissections of a convex $n$-gon into triangles by $n-3$ non-intersecting diagonals, and all colourings of the triangles with black and white so that triangles with a common side are always of a different colour. Find the least possible number of black triangles. | . The answer is $\left\lfloor\frac{n-1}{3}\right\rfloor$.
Let $f(n)$ denote the minimum number of black triangles in an $n$-gon. It is clear that $f(3)=0$ and that $f(n)$ is at least 1 for $n=4,5,6$. It is easy to see that for $n=4,5,6$ there is a coloring with only one black triangle, so $f(n)=1$ for $n=4,5,6$.
Firs... | {
"problem_match": "\nProblem 10.",
"resource_path": "BalticWay/segmented/en-bw10sol.jsonl",
"solution_match": "\nSolution 1"
} | 77 | 769 |
2010 | T3 | 10 | null | BalticWay | Let $n$ be an integer with $n \geq 3$. Consider all dissections of a convex $n$-gon into triangles by $n-3$ non-intersecting diagonals, and all colourings of the triangles with black and white so that triangles with a common side are always of a different colour. Find the least possible number of black triangles. | . Call two triangles neighbours if they have a common side. Let the dissections of convex $n$-gons together with appropriate colourings be called $n$-colourings.
Observe that all triangles of an arbitrary $n$-colouring can be listed, starting with an arbitrary triangle and always continuing the list by a triangle that... | {
"problem_match": "\nProblem 10.",
"resource_path": "BalticWay/segmented/en-bw10sol.jsonl",
"solution_match": "\nSolution 2"
} | 77 | 650 |
2010 | T3 | 19 | null | BalticWay | For which $k$ do there exist $k$ pairwise distinct primes $p_{1}, p_{2}, \ldots, p_{k}$ such that
$$
p_{1}^{2}+p_{2}^{2}+\cdots+p_{k}^{2}=2010 ?
$$ | We show that it is possible only if $k=7$.
The 15 smallest prime squares are:
$$
4,9,25,49,121,169,289,361,529,841,961,1369,1681,1849,2209
$$
Since $2209>2010$ we see that $k \leq 14$.
Now we note that $p^{2} \equiv 1 \bmod 8$ if $p$ is an odd prime. We also have that $2010 \equiv 2 \bmod 8$. If all the primes are ... | {
"problem_match": "\nProblem 19.",
"resource_path": "BalticWay/segmented/en-bw10sol.jsonl",
"solution_match": "\nSolution."
} | 67 | 764 |
2011 | T3 | A-1 | Algebra | BalticWay | The real numbers $x_{1}, \ldots, x_{2011}$ satisfy
$$
x_{1}+x_{2}=2 x_{1}^{\prime}, \quad x_{2}+x_{3}=2 x_{2}^{\prime}, \quad \ldots, \quad x_{2011}+x_{1}=2 x_{2011}^{\prime}
$$
where $x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{2011}^{\prime}$ is a permutation of $x_{1}, x_{2}, \ldots, x_{2011}$. Prove that $x_{1}=x_... | For convenience we call $x_{2011}$ also $x_{0}$. Let $k$ be the largest of the numbers $x_{1}, \ldots, x_{2011}$, and consider an equation $x_{n-1}+x_{n}=2 k$, where $1 \leq n \leq 2011$. Hence we get $2 \max \left(x_{n-1}, x_{n}\right) \geq x_{n-1}+x_{n}=2 k$, so either $x_{n-1}$ or $x_{n}$, say $x_{n-1}$, satisfies $... | {
"problem_match": "\n## A-1 DEN\n",
"resource_path": "BalticWay/segmented/en-bw11sol.jsonl",
"solution_match": "\nSolution 1"
} | 176 | 502 |
2011 | T3 | G-4 | Geometry | BalticWay | The incircle of a triangle $A B C$ touches the sides $B C, C A, A B$ at $D, E, F$, respectively. Let $G$ be a point on the incircle such that $F G$ is a diameter. The lines $E G$ and $F D$ intersect at $H$. Prove that $C H \| A B$. | We work in the opposite direction. Suppose that $H^{\prime}$ is the point where $D F$ intersect the line through $C$ parallel to $A B$. We need to show that $H^{\prime}=H$. For this purpose it suffices to prove that $E, G, H^{\prime}$ are collinear, which reduces to showing that if $G^{\prime} \neq E$ is the common poi... | {
"problem_match": "\n## G-4 POL\n",
"resource_path": "BalticWay/segmented/en-bw11sol.jsonl",
"solution_match": "\nSolution."
} | 81 | 620 |
2012 | T3 | 2 | null | BalticWay | Let $a, b, c$ be real numbers. Prove that
$$
a b+b c+c a+\max \{|a-b|,|b-c|,|c-a|\} \leq 1+\frac{1}{3}(a+b+c)^{2} .
$$ | . Assume $a \leq b \leq c$ and take $c=a+x, b=a+y$, where $x \geq y \geq 0$. The inequality $3(a b+b c+c a+c-a-1) \leq(a+b+c)^{2}$ then reduces to
$$
x^{2}-x y+y^{2}+3 \geq 3 x \text {. }
$$
The latter inequality is equivalent to the inequality
$$
\left(\frac{x}{2}-y\right)^{2}+\frac{3}{4} x^{2}-3 x+3 \geq 0
$$
whi... | {
"problem_match": "# Problem 2",
"resource_path": "BalticWay/segmented/en-bw12sol.jsonl",
"solution_match": "\nSolution 3"
} | 63 | 503 |
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