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 |
|---|---|---|---|---|---|---|---|---|---|
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 $a \leq b \leq c$. Expand the inequality $3(a b+b c+c a+c-a-1) \leq$ $(a+b+c)^{2}$ fully to obtain $a^{2}+b^{2}+c^{2}-a b-a c-b c+3 a-3 c+3 \geq 0$. Now fix $\alpha \in \mathbb{R}$ and consider the set
$$
\gamma=\left\{(a, b, c): a^{2}+b^{2}+c^{2}-a b-a c-b c+3 a-3 c+3+\alpha=0\right\} .
$$
Note that $\gamma... | {
"problem_match": "# Problem 2",
"resource_path": "BalticWay/segmented/en-bw12sol.jsonl",
"solution_match": "\nSolution 5"
} | 63 | 798 |
2012 | T3 | 2 | null | BalticWay | Let $a, b, c$ be real numbers. Prove that
$$
a b+b c+c a+\max \{|a-b|,|b-c|,|c-a|\} \leq 1+\frac{1}{3}(a+b+c)^{2} .
$$ | . We start as in Solution 5: construct the quadric
$$
\gamma=\left\{(a, b, c): a^{2}+b^{2}+c^{2}-a b-a c-b c+3 a-3 c+3+\alpha=0\right\} .
$$
Now note that the substitution
$$
\left\{\begin{array}{l}
a=2 x-y+2 z \\
b=2 y+2 z \\
c=-2 x-y+2 z
\end{array}\right.
$$
gives (in the new coordinate system)
$$
\gamma=\left\... | {
"problem_match": "# Problem 2",
"resource_path": "BalticWay/segmented/en-bw12sol.jsonl",
"solution_match": "\nSolution 6"
} | 63 | 612 |
2012 | T3 | 13 | null | BalticWay | Let $A B C$ be an acute triangle, and let $H$ be its orthocentre. Denote by $H_{A}, H_{B}$ and $H_{C}$ the second intersection of the circumcircle with the altitudes from $A, B$ and $C$ respectively. Prove that the area of $\triangle H_{A} H_{B} H_{C}$ does not exceed the area of $\triangle A B C$. | . We know that the points $H_{A}, H_{B}$ and $H_{C}$ are in fact the reflection of $H$ on the sides (Figure 7). Since $A B C$ is acute (i.e. $H$ lies in the interior of $A B C$ ), we have $S_{A H_{C} B H_{A} C H_{B}}=2 S_{A B C}$. We thus have to show that $2 S_{H_{A} H_{B} H_{C}} \leq S_{A H_{C} B H_{A} C H_{B}}$, whi... | {
"problem_match": "# Problem 13",
"resource_path": "BalticWay/segmented/en-bw12sol.jsonl",
"solution_match": "\nSolution 1"
} | 95 | 564 |
2012 | T3 | 15 | null | BalticWay | The circumcentre $O$ of a given cyclic quadrilateral $A B C D$ lies inside the quadrilateral but not on the diagonal $A C$. The diagonals of the quadrilateral intersect at $I$.
The circumcircle of the triangle $A O I$ meets the sides $A D$ and $A B$ at points $P$ and $Q$, respectively; the circumcircle of the triangle... | Assume w.l.o.g. that angle $A B C$ is obtuse (otherwise switch $B$ and $D$, Figure 10). As $A, I, O$ and $P$ are concyclic, we get $\angle Q A I=\angle Q O I$; similarly $\angle R C I=\angle R O I$. Hence
$$
\begin{aligned}
\angle Q O R=\angle Q O I+\angle R O I=\angle Q A I+\angle R C I & =\angle B A C+\angle B C A \... | {
"problem_match": "# Problem 15",
"resource_path": "BalticWay/segmented/en-bw12sol.jsonl",
"solution_match": "\nSolution."
} | 128 | 651 |
2013 | T3 | 2 | null | BalticWay | Let $k$ and $n$ be positive integers and let $x_{1}, x_{2}, \ldots, x_{k}, y_{1}, y_{2}, \ldots, y_{n}$ be distinct integers. A polynomial $P$ with integer coefficients satisfies
$$
P\left(x_{1}\right)=P\left(x_{2}\right)=\ldots=P\left(x_{k}\right)=54
$$
and
$$
P\left(y_{1}\right)=P\left(y_{2}\right)=\ldots=P\left(y... | Letting $Q(x)=P(x)-54$, we see that $Q$ has $k$ zeroes at $x_{1}, \ldots, x_{k}$, while $Q\left(y_{i}\right)=1959$ for $i=1, \ldots, n$. We notice that $1959=3 \cdot 653$, and an easy check shows that 653 is a prime number. As
$$
Q(x)=\prod_{j=1}^{k}\left(x-x_{j}\right) S(x)
$$
and $S(x)$ is a polynomial with integer... | {
"problem_match": "# Problem 2",
"resource_path": "BalticWay/segmented/en-bw13sol.jsonl",
"solution_match": "# Solution\n"
} | 148 | 548 |
2013 | T3 | 3 | null | BalticWay | Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x f(y)+y)+f(-f(x))=f(y f(x)-y)+y \quad \text { for all } x, y \in \mathbb{R}
$$ | Let $f(0)=c$. We make the following substitutions in the initial equation:
1) $x=0, y=0 \Longrightarrow f(0)+f(-c)=f(0) \Longrightarrow f(-c)=0$.
2) $x=0, y=-c \Longrightarrow f(-c)+f(-c)=f\left(c-c^{2}\right)-c \Longrightarrow f\left(c-c^{2}\right)=c$.
3) $x=-c, y=-c \Longrightarrow f(-c)+f(0)=f(c)-c \Longrightarrow ... | {
"problem_match": "# Problem 3",
"resource_path": "BalticWay/segmented/en-bw13sol.jsonl",
"solution_match": "# Solution\n"
} | 80 | 514 |
2013 | T3 | 4 | null | BalticWay | Prove that the following inequality holds for all positive real numbers $x, y, z$ :
$$
\frac{x^{3}}{y^{2}+z^{2}}+\frac{y^{3}}{z^{2}+x^{2}}+\frac{z^{3}}{x^{2}+y^{2}} \geq \frac{x+y+z}{2}
$$ | The inequality is symmetric, so we may assume $x \leq y \leq z$. Then we have
$$
x^{3} \leq y^{3} \leq z^{3} \quad \text { and } \quad \frac{1}{y^{2}+z^{2}} \leq \frac{1}{x^{2}+z^{2}} \leq \frac{1}{x^{2}+y^{2}}
$$
Therefore, by the rearrangement inequality we have:
$$
\begin{gathered}
\frac{x^{3}}{y^{2}+z^{2}}+\frac... | {
"problem_match": "# Problem 4",
"resource_path": "BalticWay/segmented/en-bw13sol.jsonl",
"solution_match": "# Solution\n"
} | 84 | 683 |
2013 | T3 | 5 | null | BalticWay | Numbers 0 and 2013 are written at two opposite vertices of a cube. Some real numbers are to be written at the remaining 6 vertices of the cube. On each edge of the cube the difference between the numbers at its endpoints is written. When is the sum of squares of the numbers written on the edges minimal? | $$
\begin{gathered}
S=\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\left(x_{4}-x_{1}\right)^{2}+\left(x_{4}-x_{2}\right)^{2}+\left(x_{5}-x_{1}\right)^{2}+\left(x_{5}-x_{3}\right)^{2}+\right. \\
\left(x_{6}-x_{2}\right)^{2}+\left(x_{6}-x_{3}\right)^{2}+\left(2013-x_{4}\right)^{2}+\left(2013-x_{5}\right)^{2}+\left(2013-x_{6}\righ... | {
"problem_match": "# Problem 5",
"resource_path": "BalticWay/segmented/en-bw13sol.jsonl",
"solution_match": "# Solution 2"
} | 68 | 902 |
2013 | T3 | 7 | null | BalticWay | A positive integer is written on a blackboard. Players $A$ and $B$ play the following game: in each move one has to choose a divisor $m$ of the number $n$ written on the blackboard for which $1<m<n$ and replace $n$ with $n-m$. Player $A$ makes the first move, players move alternately. The player who can't make a move l... | Firstly note that for a given $n$ exactly one player has a winning strategy. We'll show by induction that $B$ has a winning strategy if $n$ is odd.
First step of the induction is clear. Assume $n$ is odd and $B$ has a winning strategy for all odd integers smaller than $n$. If player $A$ can't make a move, $B$ wins. In ... | {
"problem_match": "# Problem 7",
"resource_path": "BalticWay/segmented/en-bw13sol.jsonl",
"solution_match": "# Solution\n"
} | 104 | 671 |
2013 | T3 | 20 | null | BalticWay | Find all polynomials $f$ with non-negative integer coefficients such that for all primes $p$ and positive integers $n$ there exist a prime $q$ and a positive integer $m$ such that $f\left(p^{n}\right)=q^{m}$. | Notice that among the constant polynomials the only solutions are $P(t)=q^{m}$ where $q$ is a prime and $m$ a positive integer. Assume that
$$
P(t)=a_{k} t^{k}+\cdots a_{0}
$$
where $a_{k} \neq 0$ and $a_{0}, a_{1}, \ldots, a_{k}$ are non-negative integers, is a polynomial that fullfills the conditions.
First conside... | {
"problem_match": "# Problem 20",
"resource_path": "BalticWay/segmented/en-bw13sol.jsonl",
"solution_match": "# Solution\n"
} | 57 | 734 |
2014 | T3 | 1 | null | BalticWay | Show that
$$
\cos \left(56^{\circ}\right) \cdot \cos \left(2 \cdot 56^{\circ}\right) \cdot \cos \left(2^{2} \cdot 56^{\circ}\right) \cdot \ldots \cdot \cos \left(2^{23} \cdot 56^{\circ}\right)=\frac{1}{2^{24}}
$$ | We start by rewriting the expression as follows:
$\cos \left(56^{\circ}\right) \cdot \cos \left(2 \cdot 56^{\circ}\right) \cdot \ldots \cdot \cos \left(2^{23} \cdot 56^{\circ}\right)=\frac{\sin \left(56^{\circ}\right) \cdot \cos \left(56^{\circ}\right) \cdot \cos \left(2 \cdot 56^{\circ}\right) \cdot \ldots \cdot \cos... | {
"problem_match": "# Problem 1",
"resource_path": "BalticWay/segmented/en-bw14sol.jsonl",
"solution_match": "\nSolution."
} | 98 | 681 |
2014 | T3 | 5 | null | BalticWay | Given positive real numbers $a, b, c, d$ that satisfy equalities
$$
a^{2}+d^{2}-a d=b^{2}+c^{2}+b c \quad \text { and } \quad a^{2}+b^{2}=c^{2}+d^{2}
$$
find all possible values of the expression $\frac{a b+c d}{a d+b c}$.
Answer: $\frac{\sqrt{3}}{2}$. | . Let $A_{1} B C_{1}$ be a triangle with $A_{1} B=b, B C_{1}=c$ and $\angle A_{1} B C_{1}=120^{\circ}$, and let $C_{2} D A_{2}$ be another triangle with $C_{2} D=d, D A_{2}=a$ and $\angle C_{2} D A_{2}=60^{\circ}$. By the law of cosines and the assumption $a^{2}+d^{2}-a d=b^{2}+c^{2}+b c$, we have $A_{1} C_{1}=A_{2} C_... | {
"problem_match": "# Problem 5",
"resource_path": "BalticWay/segmented/en-bw14sol.jsonl",
"solution_match": "\nSolution 1"
} | 106 | 547 |
2014 | T3 | 8 | null | BalticWay | Albert and Betty are playing the following game. There are 100 blue balls in a red bowl and 100 red balls in a blue bowl. In each turn a player must make one of the following moves:
a) Take two red balls from the blue bowl and put them in the red bowl.
b) Take two blue balls from the red bowl and put them in the blue... | Betty follows the following strategy. If Albert makes move a), then Betty makes move b) and vice verse. If Albert makes move c) from one bowl, Betty makes move c) from the other bowl. The only exception of this rule is that if Betty can make a winning move, that is, a move where she removes the last blue ball from the ... | {
"problem_match": "# Problem 8",
"resource_path": "BalticWay/segmented/en-bw14sol.jsonl",
"solution_match": "\nSolution."
} | 146 | 575 |
2014 | T3 | 10 | null | BalticWay | In a country there are 100 airports. Super-Air operates direct flights between some pairs of airports (in both directions). The traffic of an airport is the number of airports it has a direct Super-Air connection with. A new company, Concur-Air, establishes a direct flight between two airports if and only if the sum of... | Let $G$ and $G^{\prime}$ be two graphs corresponding to the flights of Super-Air and Concur-Air, respectively. Then the traffic of an airport is simply the degree of a corresponding vertex, and the assertion means that the graph $G$ has a Hamiltonian cycle.
Lemma. Let a graph $H$ has 100 vertices and contains a Hamilt... | {
"problem_match": "# Problem 10",
"resource_path": "BalticWay/segmented/en-bw14sol.jsonl",
"solution_match": "\nSolution."
} | 128 | 611 |
2014 | T3 | 14 | null | BalticWay | Let $A B C D$ be a convex quadrilateral such that the line $B D$ bisects the angle $A B C$. The circumcircle of triangle $A B C$ intersects the sides $A D$ and $C D$ in the points $P$ and $Q$, respectively. The line through $D$ and parallel to $A C$ intersects the lines $B C$ and $B A$ at the points $R$ and $S$, respec... | . Denote by $X^{\prime}$ the image of the point $X$ under some fixed inversion with center $B$. At the beginning of Solution 2 we noticed that the circumcircles of the triangles $A B C$ and $S B R$ are tangent at the point $B$. Therefore, the images of these two circles
under the considered inversion become two paralle... | {
"problem_match": "# Problem 14",
"resource_path": "BalticWay/segmented/en-bw14sol.jsonl",
"solution_match": "\nSolution 3"
} | 123 | 629 |
2014 | T3 | 19 | null | BalticWay | Let $m$ and $n$ be relatively prime positive integers. Determine all possible values of
$$
\operatorname{gcd}\left(2^{m}-2^{n}, 2^{m^{2}+m n+n^{2}}-1\right) .
$$
Answer: 1 and 7 . | Without restriction of generality we may assume that $m \geqslant n$. It is well known that
$$
\operatorname{gcd}\left(2^{p}-1,2^{q}-1\right)=2^{\operatorname{gcd}(p, q)}-1,
$$
SO
$$
\begin{aligned}
\operatorname{gcd}\left(2^{m}-2^{n}, 2^{m^{2}+m n+n^{2}}-1\right) & =\operatorname{gcd}\left(2^{m-n}-1,2^{m^{2}+m n+n^... | {
"problem_match": "# Problem 19",
"resource_path": "BalticWay/segmented/en-bw14sol.jsonl",
"solution_match": "\nSolution."
} | 68 | 543 |
2015 | T3 | 3 | null | BalticWay | Let $n>1$ be an integer. Find all non-constant real polynomials $P(x)$ satisfying, for any real $x$, the identity
$$
P(x) P\left(x^{2}\right) P\left(x^{3}\right) \cdots P\left(x^{n}\right)=P\left(x^{\frac{n(n+1)}{2}}\right) .
$$ | Answer: $P(x)=x^{m}$ if $n$ is even; $P(x)= \pm x^{m}$ if $n$ is odd.
Consider first the case of a monomial $P(x)=a x^{m}$ with $a \neq 0$. Then
$$
a x^{\frac{m n(n+1)}{2}}=P\left(x^{\frac{n(n+1)}{2}}\right)=P(x) P\left(x^{2}\right) P\left(x^{3}\right) \cdots P\left(x^{n}\right)=a x^{m} \cdot a x^{2 m} \cdots a x^{n ... | {
"problem_match": "# Problem 3.",
"resource_path": "BalticWay/segmented/en-bw15sol.jsonl",
"solution_match": "\nSolution."
} | 88 | 616 |
2015 | T3 | 4 | null | BalticWay | A family wears clothes of three colours: red, blue and green, with a separate, identical laundry bin for each colour. At the beginning of the first week, all bins are empty. Each week, the family generates a total of $10 \mathrm{~kg}$ of laundry (the proportion of each colour is subject to variation). The laundry is so... | Answer: $25 \mathrm{~kg}$.
Each week, the accumulation of laundry increases the total amount by $K=10$, after which the washing decreases it by at least one third, because, by the pigeon-hole principle, the bin with the most laundry must contain at least a third of the total. Hence the amount of laundry post-wash afte... | {
"problem_match": "# Problem 4.",
"resource_path": "BalticWay/segmented/en-bw15sol.jsonl",
"solution_match": "# Solution."
} | 129 | 667 |
2015 | T3 | 5 | null | BalticWay | Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfying the equation
$$
|x| f(y)+y f(x)=f(x y)+f\left(x^{2}\right)+f(f(y))
$$
for all real numbers $x$ and $y$. | Answer: all functions $f(x)=c(|x|-x)$, where $c \geq 0$.
Choosing $x=y=0$, we find
$$
f(f(0))=-2 f(0)
$$
Denote $a=f(0)$, so that $f(a)=-2 a$, and choose $y=0$ in the initial equation:
$$
a|x|=a+f\left(x^{2}\right)+f(a)=a+f\left(x^{2}\right)-2 a \quad \Rightarrow \quad f\left(x^{2}\right)=a(|x|+1)
$$
In particular... | {
"problem_match": "# Problem 5.",
"resource_path": "BalticWay/segmented/en-bw15sol.jsonl",
"solution_match": "\nSolution."
} | 65 | 539 |
2015 | T3 | 8 | null | BalticWay | With inspiration drawn from the rectilinear network of streets in New York, the Manhattan distance between two points $(a, b)$ and $(c, d)$ in the plane is defined to be
$$
|a-c|+|b-d| \text {. }
$$
Suppose only two distinct Manhattan distances occur between all pairs of distinct points of some point set. What is the... | Answer: nine.
Let
$$
\left\{\left(x_{1}, y_{1}\right), \ldots,\left(x_{m}, y_{m}\right)\right\}, \quad \text { where } \quad x_{1} \leq \cdots \leq x_{m}
$$
be the set, and suppose $m \geq 10$.
A special case of the Erdős-Szekeres Theorem asserts that a real sequence of length $n^{2}+1$ contains a monotonic subsequ... | {
"problem_match": "# Problem 8.",
"resource_path": "BalticWay/segmented/en-bw15sol.jsonl",
"solution_match": "\nSolution."
} | 87 | 533 |
2015 | T3 | 9 | null | BalticWay | Let $n>2$ be an integer. A deck contains $\frac{n(n-1)}{2}$ cards, numbered
$$
1,2,3, \ldots, \frac{n(n-1)}{2}
$$
Two cards form a magic pair if their numbers are consecutive, or if their numbers are 1 and $\frac{n(n-1)}{2}$.
For which $n$ is it possible to distribute the cards into $n$ stacks in such a manner that,... | . Answer: for all odd $n$.
First assume a stack contains two cards that form a magic pair; say cards number $i$ and $i+1$. Among the cards in this stack and the stack with card number $i+2$ (they might be identical), there are two magic pairs - a contradiction. Hence no stack contains a magic pair.
Each card forms a ... | {
"problem_match": "# Problem 9.",
"resource_path": "BalticWay/segmented/en-bw15sol.jsonl",
"solution_match": "# Solution 1"
} | 120 | 721 |
2015 | T3 | 10 | null | BalticWay | A subset $S$ of $\{1,2, \ldots, n\}$ is called balanced if for every $a \in S$ there exists some $b \in S, b \neq a$, such that $\frac{a+b}{2} \in S$ as well.
(a) Let $k>1$ be an integer and let $n=2^{k}$. Show that every subset $S$ of $\{1,2, \ldots, n\}$ with $|S|>\frac{3 n}{4}$ is balanced.
(b) Does there exist an... | of part (b). Let us introduce the concept of lonely element as an $a \in S$ for which there does not exist a $b \in S$, distinct from $a$, such that $\frac{a+b}{2} \in S$.
We will construct an unbalanced set $S$ with $|S|>\frac{2 n}{3}$ for all $k$. For $n=4$ we can use $S=\{1,2,4\}$ (all elements are lonely), and for... | {
"problem_match": "# Problem 10.",
"resource_path": "BalticWay/segmented/en-bw15sol.jsonl",
"solution_match": "\nSolution 3"
} | 393 | 555 |
2015 | T3 | 14 | null | BalticWay | In the non-isosceles triangle $A B C$ the altitude from $A$ meets side $B C$ in $D$. Let $M$ be the midpoint of $B C$ and let $N$ be the reflection of $M$ in $D$. The circumcircle of the triangle $A M N$ intersects the side $A B$ in $P \neq A$ and the side $A C$ in $Q \neq A$. Prove that $A N, B Q$ and $C P$ are concur... | . Without loss of generality, we assume the order of the points on $B C$ to be $B, M, D, N$, $C$. This implies that $P$ is on the segment $A B$ and $Q$ is on the segment $A C$.
Since $D$ is the midpoint of $M N$ and $A D$ is perpendicular to $M N$, the line $A D$ is the perpendicular bisector of $M N$, which contains ... | {
"problem_match": "# Problem 14.",
"resource_path": "BalticWay/segmented/en-bw15sol.jsonl",
"solution_match": "\nSolution 1"
} | 116 | 697 |
2015 | T3 | 15 | null | BalticWay | In triangle $A B C$, the interior and exterior angle bisectors of $\angle B A C$ intersect the line $B C$ in $D$ and $E$, respectively. Let $F$ be the second point of intersection of the line $A D$ with the circumcircle of the triangle $A B C$. Let $O$ be the circumcentre of the triangle $A B C$ and let $D^{\prime}$ be... | . Again, assume $A B<A C$. We first consider the case $\angle B A C=90^{\circ}$. Define $F^{\prime}$ as in the previous solution. Now $O$ and $D^{\prime}$ lie on $B C$, so $\triangle D^{\prime} F O$ and $\triangle D F O$ are mirror images with respect to $F F^{\prime}$, while $\triangle O F E$ and $\triangle O F^{\prim... | {
"problem_match": "# Problem 15.",
"resource_path": "BalticWay/segmented/en-bw15sol.jsonl",
"solution_match": "\nSolution 3"
} | 122 | 623 |
2015 | T3 | 15 | null | BalticWay | In triangle $A B C$, the interior and exterior angle bisectors of $\angle B A C$ intersect the line $B C$ in $D$ and $E$, respectively. Let $F$ be the second point of intersection of the line $A D$ with the circumcircle of the triangle $A B C$. Let $O$ be the circumcentre of the triangle $A B C$ and let $D^{\prime}$ be... | . We consider the configuration where $C, D, B$ and $E$ are on the line $B C$ in that order. The other configuration can be solved analogously. Let $P$ and $R$ be the feet of the perpendiculars from $D^{\prime}$ and $O$ to the line $A D$, respectively, and let $Q$ and $S$ be the feet of the perpendiculars from $D^{\pri... | {
"problem_match": "# Problem 15.",
"resource_path": "BalticWay/segmented/en-bw15sol.jsonl",
"solution_match": "\nSolution 4"
} | 122 | 886 |
2015 | T3 | 20 | null | BalticWay | For any integer $n \geq 2$, we define $A_{n}$ to be the number of positive integers $m$ with the following property: the distance from $n$ to the nearest non-negative multiple of $m$ is equal to the distance from $n^{3}$ to the nearest non-negative multiple of $m$. Find all integers $n \geq 2$ for which $A_{n}$ is odd.... | For an integer $m$ we consider the distance $d$ from $n$ to the nearest multiple of $m$. Then $m \mid n \pm d$, which means $n \equiv \pm d \bmod m$. So if, for some $m$, the distance from $n$ to the nearest multiple of $m$ is equal to the distance from $n^{3}$ to the nearest multiple of $m$, then $n \equiv \pm n^{3} \... | {
"problem_match": "# Problem 20.",
"resource_path": "BalticWay/segmented/en-bw15sol.jsonl",
"solution_match": "\nSolution."
} | 114 | 1,058 |
2016 | T3 | 2 | null | BalticWay | Prove or disprove the following hypotheses.
a) For all $k \geq 2$, each sequence of $k$ consecutive positive integers contains a number that is not divisible by any prime number less than $k$.
b) For all $k \geq 2$, each sequence of $k$ consecutive positive integers contains a number that is relatively prime to all o... | We give a counterexample to both claims. So neither of them is true.
For a), a counterexample is the sequence $(2,3,4,5,6,7,8,9)$ of eight consecutive integers all of which are divisible by some prime less than 8 .
To construct a counterexample to b), we notice that by the Chinese Remainder Theorem, there exists an i... | {
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw16sol.jsonl",
"solution_match": "\nSolution."
} | 83 | 525 |
2016 | T3 | 5 | null | BalticWay | Let $p>3$ be a prime such that $p \equiv 3(\bmod 4)$. Given a positive integer $a_{0}$, define the sequence $a_{0}, a_{1}, \ldots$ of integers by $a_{n}=a_{n-1}^{2^{n}}$ for all $n=1,2, \ldots$ Prove that it is possible to choose $a_{0}$ such that the subsequence $a_{N}, a_{N+1}, a_{N+2}, \ldots$ is not constant modulo... | Let $p$ be a prime with residue 3 modulo 4 and $p>3$. Then $p-1=u \cdot 2$ where $u>1$ is odd. Choose $a_{0}=2$. The order of 2 modulo $p$ (that is, the smallest positive integer $t$ such that $\left.2^{t} \equiv 1 \bmod p\right)$ is a divisor of $\phi(p)=p-1=u \cdot 2$, but not a divisor of 2 since $1<2^{2}<p$. Hence ... | {
"problem_match": "\n5.",
"resource_path": "BalticWay/segmented/en-bw16sol.jsonl",
"solution_match": "\nSolution."
} | 136 | 505 |
2016 | T3 | 9 | null | BalticWay | Find all quadruples $(a, b, c, d)$ of real numbers that simultaneously satisfy the following equations:
$$
\left\{\begin{aligned}
a^{3}+c^{3} & =2 \\
a^{2} b+c^{2} d & =0 \\
b^{3}+d^{3} & =1 \\
a b^{2}+c d^{2} & =-6
\end{aligned}\right.
$$ | . If $0 \in\{a, b\}$, then one easily gets that $0 \in\{c, d\}$, which contradicts the equation $a b^{2}+c d^{2}=-6$. Similarly, if $0 \in\{c, d\}$, then $0 \in\{a, b\}$ and this contradicts $a b^{2}+c d^{2}=-6$ again. Hence $a, b, c, d \neq 0$.
Let the four equations in the problem be (i), (ii), (iii) and (iv), respe... | {
"problem_match": "\n9.",
"resource_path": "BalticWay/segmented/en-bw16sol.jsonl",
"solution_match": "\nSolution 2"
} | 97 | 590 |
2016 | T3 | 9 | null | BalticWay | Find all quadruples $(a, b, c, d)$ of real numbers that simultaneously satisfy the following equations:
$$
\left\{\begin{aligned}
a^{3}+c^{3} & =2 \\
a^{2} b+c^{2} d & =0 \\
b^{3}+d^{3} & =1 \\
a b^{2}+c d^{2} & =-6
\end{aligned}\right.
$$ | . As in Solution 2, we conclude that $a, b, c, d \neq 0$. The equation $a^{2} b+c^{2} d=0$ yields $a= \pm \sqrt{\frac{-d}{b}} c$. On the other hand, we have $a^{3}+c^{3}=2$ and $a b^{2}+c d^{2}=-6<0$ which implies that $\min \{a, c\}<0<\max \{a, c\}$ and thus $a=-\sqrt{\frac{-d}{b}} c$.
Let $x=-\sqrt{\frac{-d}{b}}$. T... | {
"problem_match": "\n9.",
"resource_path": "BalticWay/segmented/en-bw16sol.jsonl",
"solution_match": "\nSolution 3"
} | 97 | 968 |
2016 | T3 | 19 | null | BalticWay | Consider triangles in the plane where each vertex has integer coordinates. Such a triangle can be legally transformed by moving one vertex parallel to the opposite side to a different point with integer coordinates. Show that if two triangles have the same area, then there exists a series of legal transformations that ... | We will first show that any such triangle can be transformed to a special triangle whose vertices are at $(0,0),(0,1)$ and $(n, 0)$. Since every transformation preserves the triangle's area, triangles with the same area will have the same value for $n$.
Define th $y$-span of a triangle to be the difference between the... | {
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw16sol.jsonl",
"solution_match": "\nSolution."
} | 61 | 525 |
2017 | T3 | 4 | null | BalticWay | A linear form in $k$ variables is an expression of the form $P\left(x_{1}, \ldots, x_{k}\right)=a_{1} x_{1}+\ldots+a_{k} x_{k}$ with real constants $a_{1}, \ldots, a_{k}$. Prove that there exist a positive integer $n$ and linear forms $P_{1}, \ldots, P_{n}$ in 2017 variables such that the equation
$$
x_{1} \cdot x_{2}... | For every $\varepsilon=\left(\varepsilon_{1}, \ldots, \varepsilon_{n}\right) \in\{ \pm 1\}^{2017}$ let
$$
P_{\varepsilon}\left(X_{1}, \ldots, X_{2017}\right)=\varepsilon_{1} X_{1}+\cdots+\varepsilon_{2017} X_{2017}
$$
and $\beta_{\varepsilon}=\varepsilon_{1} \cdots \varepsilon_{2017}$. Consider
$$
\begin{aligned}
g\... | {
"problem_match": "\nProblem 4.",
"resource_path": "BalticWay/segmented/en-bw17sol.jsonl",
"solution_match": "\nSolution 1:"
} | 217 | 1,187 |
2017 | T3 | 4 | null | BalticWay | A linear form in $k$ variables is an expression of the form $P\left(x_{1}, \ldots, x_{k}\right)=a_{1} x_{1}+\ldots+a_{k} x_{k}$ with real constants $a_{1}, \ldots, a_{k}$. Prove that there exist a positive integer $n$ and linear forms $P_{1}, \ldots, P_{n}$ in 2017 variables such that the equation
$$
x_{1} \cdot x_{2}... | We show by induction that for every integer $k \geq 1$ there exist an $n=n_{k}$, real numbers $\lambda_{1}, \ldots, \lambda_{n_{k}}$ and linear forms $P_{k, 1}, \ldots, P_{k, n_{k}}$ in $k$ variables such that
$$
x_{1} \ldots x_{k}=\lambda_{1} P_{k, 1}\left(x_{1}, \ldots, x_{k}\right)^{k}+\cdots+\lambda_{n_{k}} P_{k, ... | {
"problem_match": "\nProblem 4.",
"resource_path": "BalticWay/segmented/en-bw17sol.jsonl",
"solution_match": "\nSolution 2:"
} | 217 | 938 |
2017 | T3 | 11 | null | BalticWay | Let $H$ and $I$ be the orthocentre and incentre, respectively, of an acute angled triangle $A B C$. The circumcircle of the triangle $B C I$ intersects the segment $A B$ at the point $P$ different from $B$. Let $K$ be the projection of $H$ onto $A I$ and $Q$ the reflection of $P$ in $K$. Show that $B, H$ and $Q$ are co... | Let $\alpha=\frac{1}{2} \angle B A C, \beta=\frac{1}{2} \angle C B A$, and $\gamma=\frac{1}{2} \angle A C B$. Clearly then $\alpha+\beta+\gamma=90^{\circ}$, which yields $\angle B I C=180^{\circ}-\beta-\gamma=$ $90^{\circ}+\alpha$. From this we get $\angle C P A=180^{\circ}-\angle B P C=180^{\circ}-\angle B I C=90^{\ci... | {
"problem_match": "\nProblem 11.",
"resource_path": "BalticWay/segmented/en-bw17sol.jsonl",
"solution_match": "\nSolution 2:"
} | 103 | 556 |
2017 | T3 | 19 | null | BalticWay | For an integer $n \geq 1$ let $a(n)$ denote the total number of carries which arise when adding 2017 and $n \cdot 2017$. The first few values are given by $a(1)=1, a(2)=1, a(3)=0$, which can be seen from the following:
| 001 | 001 | 000 |
| :---: | :---: | :---: |
| 2017 | 4034 | 6051 |
| +2017 | +2017 | +2017 |
| $=4... | Let $s(n)$ denote the digit sum of $n$. Then we claim the following.
Lemma. We have
$$
s(n+m)=s(n)+s(m)-9 a(n, m)
$$
where $a(n, m)$ denotes the total number of carries, which arises when adding $n$ and $m$.
Proof: We proceed by induction on the maximal number of digits $k$ of $n$ and $m$.
If both $n$ and $m$ are ... | {
"problem_match": "\nProblem 19.",
"resource_path": "BalticWay/segmented/en-bw17sol.jsonl",
"solution_match": "\nSolution 2:"
} | 237 | 848 |
2017 | T3 | 20 | null | BalticWay | Let $S$ be the set of all ordered pairs $(a, b)$ of integers with $0<2 a<2 b<2017$ such that $a^{2}+b^{2}$ is a multiple of 2017. Prove that
$$
\sum_{(a, b) \in S} a=\frac{1}{2} \sum_{(a, b) \in S} b
$$ | Let $A=\{a:(a, b) \in S\}$ and $B=\{b:(a, b) \in S\}$. The claim is equivalent to
$$
2 \sum_{a \in A} a=\sum_{b \in B} b
$$
Assume that for some $x, y, z \in\{1,2, \ldots, 1008\}$ both, $x^{2}+y^{2}$ and $x^{2}+z^{2}$, are multiples of 2017. By
$$
\left(x^{2}+y^{2}\right)-\left(x^{2}+z^{2}\right)=y^{2}-z^{2}=(y+z)(y... | {
"problem_match": "\nProblem 20.",
"resource_path": "BalticWay/segmented/en-bw17sol.jsonl",
"solution_match": "# Solution\n"
} | 96 | 1,044 |
2018 | T3 | 1 | null | BalticWay | A finite collection of positive real numbers (not necessarily distinct) is balanced if each number is less than the sum of the others. Find all $m \geq 3$ such that every balanced finite collection of $m$ numbers can be split into three parts with the property that the sum of the numbers in each part is less than the s... | Answer: The partition is always possible precisely when $m \neq 4$.
For $m=3$ it is trivially possible, and for $m=4$ the four equal numbers $g, g, g, g$ provide a counter-example. Henceforth, we assume $m \geq 5$.
Among all possible partitions $A \sqcup B \sqcup C=\{1, \ldots, m\}$ such that
$$
S_{A} \leq S_{B} \le... | {
"problem_match": "\n1.",
"resource_path": "BalticWay/segmented/en-bw18sol.jsonl",
"solution_match": "# Solution."
} | 79 | 568 |
2018 | T3 | 3 | null | BalticWay | Let $a, b, c, d$ be positive real numbers such that $a b c d=1$. Prove the inequality
$$
\frac{1}{\sqrt{a+2 b+3 c+10}}+\frac{1}{\sqrt{b+2 c+3 d+10}}+\frac{1}{\sqrt{c+2 d+3 a+10}}+\frac{1}{\sqrt{d+2 a+3 b+10}} \leq 1 .
$$ | Let $x, y, z, t$ be positive numbers such that $a=x^{4}, b=y^{4}, c=z^{4}, d=t^{4}$.
By AM-GM ineguality $x^{4}+y^{4}+z^{4}+1 \geq 4 x y z, y^{4}+z^{4}+1+1 \geq 4 y z$ and $z^{4}+1+1+1 \geq 4 z$. Therefore we have the following estimation for the first fraction
$$
\frac{1}{\sqrt{x^{4}+2 y^{4}+3 z^{4}+10}} \leq \frac{... | {
"problem_match": "\n3.",
"resource_path": "BalticWay/segmented/en-bw18sol.jsonl",
"solution_match": "\nSolution."
} | 114 | 586 |
2018 | T3 | 4 | null | BalticWay | Find all functions $f:[0,+\infty) \rightarrow[0,+\infty)$, such that for any positive integer $n$ and for any non-negative real numbers $x_{1}, \ldots, x_{n}$
$$
f\left(x_{1}^{2}+\cdots+x_{n}^{2}\right)=f\left(x_{1}\right)^{2}+\cdots+f\left(x_{n}\right)^{2} .
$$ | Answer: the functions $f(x)=0$ and $f(x)=x$.
A first observation is that
$$
f(1)=f\left(1^{2}\right)=f(1)^{2}
$$
so that $f(1)$ is either 0 or 1 .
Assume first that $f(1)=0$. For each positive integer $n$, we find
$$
f(n)=f\left(n \cdot 1^{2}\right)=n f(1)^{2}=0 .
$$
Given an arbitrary $x$, find $y$ so that $x^{2... | {
"problem_match": "\n4.",
"resource_path": "BalticWay/segmented/en-bw18sol.jsonl",
"solution_match": "# Solution."
} | 105 | 732 |
2018 | T3 | 7 | null | BalticWay | On a $16 \times 16$ torus as shown all 512 edges are colored red or blue. A coloring is good if every vertex is an endpoint of an even number of red edges. A move consists of switching the color of each of the 4 edges of an arbitrary cell. What is the largest number of good colorings such that none of them can be conve... | Answer: 4. Representatives of the equivalence classes are: all blue, all blue with one longitudinal red ring, all blue with one transversal red ring, all blue with one longitudinal and one transversal red ring.
First, show that these four classes are non equivalent. Consider any ring transversal or longitudinal and co... | {
"problem_match": "\n7.",
"resource_path": "BalticWay/segmented/en-bw18sol.jsonl",
"solution_match": "# Solution."
} | 89 | 854 |
2018 | T3 | 9 | null | BalticWay | Olga and Sasha play a game on an infinite hexagonal grid. They take turns in placing a stone on a free hexagon of their choice. Olga starts the game. Just before the 2018th stone is placed, a new rule comes into play. A stone may now be placed only on those free hexagons having at least two occupied neighbors.
A playe... | Answer: Olga has a winning strategy.
The game cannot go on forever. Draw a large hexagon enclosing all 2017 counters in play after the 2017th move, as in Figure ??. While it will be possible to place future counters in the hexagonal frame at distance 1 from the shaded part (i.e. immediately surrounding it), where $D$ ... | {
"problem_match": "\n9.",
"resource_path": "BalticWay/segmented/en-bw18sol.jsonl",
"solution_match": "# Solution."
} | 200 | 528 |
2020 | T3 | 3 | null | BalticWay | A real sequence $\left(a_{n}\right)_{n=0}^{\infty}$ is defined recursively by $a_{0}=2$ and the recursion formula
$$
a_{n}= \begin{cases}a_{n-1}^{2} & \text { if } a_{n-1}<\sqrt{3} \\ \frac{a_{n-1}^{2}}{3} & \text { if } a_{n-1} \geqslant \sqrt{3}\end{cases}
$$
Another real sequence $\left(b_{n}\right)_{n=1}^{\infty}... | The first step is to prove, using induction, the formula
$$
a_{n}=\frac{2^{2^{n}}}{3^{2^{n}\left(b_{1}+b_{2}+\cdots+b_{n}\right)}} .
$$
The base case $n=0$ is trivial. Assume the formula is valid for $a_{n-1}$, that is,
$$
a_{n-1}=\frac{2^{2^{n-1}}}{3^{2^{n-1}\left(b_{1}+b_{2}+\cdots+b_{n-1}\right)}} .
$$
If now $a... | {
"problem_match": "\nProblem 3.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "\nSolution."
} | 263 | 969 |
2020 | T3 | 5 | null | BalticWay | Find all real numbers $x, y, z$ so that
$$
\begin{aligned}
x^{2} y+y^{2} z+z^{2} & =0 \\
z^{3}+z^{2} y+z y^{3}+x^{2} y & =\frac{1}{4}\left(x^{4}+y^{4}\right)
\end{aligned}
$$ | Answer: $x=y=z=0$.
$y=0 \Longrightarrow z^{2}=0 \Longrightarrow z=0 \Longrightarrow \frac{1}{4} x^{4}=0 \Longrightarrow x=0 . x=y=z=0$ is a solution, so assume that $y \neq 0$. Then $z=0 \Longrightarrow x^{2} y=0 \Longrightarrow x=0 \Longrightarrow \frac{1}{4} y^{4}=0$, which is a contradiction. Hence $z \neq 0$. Now ... | {
"problem_match": "\nProblem 5.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "\nSolution."
} | 87 | 606 |
2020 | T3 | 7 | null | BalticWay | A mason has bricks with dimensions $2 \times 5 \times 8$ and other bricks with dimensions $2 \times 3 \times 7$. She also has a box with dimensions $10 \times 11 \times 14$. The bricks and the box are all rectangular parallelepipeds. The mason wants to pack bricks into the box filling its entire volume and with no bric... | Answer: 24.
Let the number of $2 \times 5 \times 8$ bricks in the box be $x$, and the number of $2 \times 3 \times 7$ bricks $y$. We must figure out the sum $x+y$. The volume of the box is divisible by 7 , and so is the volume of any $2 \times 3 \times 7$ brick. The volume of a $2 \times 5 \times 8$ brick is not divis... | {
"problem_match": "\nProblem 7.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "\nSolution."
} | 104 | 901 |
2020 | T3 | 8 | null | BalticWay | Let $n$ be a given positive integer. A restaurant offers a choice of $n$ starters, $n$ main dishes, $n$ desserts and $n$ wines. A merry company dines at the restaurant, with each guest choosing a starter, a main dish, a dessert and a wine. No two people place exactly the same order. It turns out that there is no collec... | Answer: The maximal number of guests is $n^{4}-n^{3}$.
The possible menus are represented by quadruples
$$
(a, b, c, d), \quad 1 \leqslant a, b, c, d \leqslant n .
$$
Let us count those menus satisfying
$$
a+b+c+d \not \equiv 0 \quad(\bmod n)
$$
The numbers $a, b, c$ may be chosen arbitrarily ( $n$ choices for eac... | {
"problem_match": "\nProblem 8.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "# Solution."
} | 141 | 535 |
2020 | T3 | 9 | null | BalticWay | Each vertex $v$ and each edge $e$ of a graph $G$ are assigned numbers $f(v) \in\{1,2\}$ and $f(e) \in\{1,2,3\}$, respectively. Let $S(v)$ be the sum of numbers assigned to the edges incident to $v$ plus the number $f(v)$. We say that an assignment $f$ is cool if $S(u) \neq S(v)$ for every pair $(u, v)$ of adjacent (i.e... | Let $v_{1}, v_{2}, \ldots, v_{n}$ be any ordering of the vertices of $G$. Initially each vertex assigned number 1 , and each edge assigned number 2. One may imagine that there is a chip lying on each vertex, while two chips are lying on each edge. We are going to refine this assignment so as to get a cool one by perfor... | {
"problem_match": "\nProblem 9.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "# Solution."
} | 136 | 531 |
2020 | T3 | 15 | null | BalticWay | On a plane, Bob chooses 3 points $A_{0}, B_{0}, C_{0}$ (not necessarily distinct) such that $A_{0} B_{0}+B_{0} C_{0}+C_{0} A_{0}=1$. Then he chooses points $A_{1}, B_{1}, C_{1}$ (not necessarily distinct) in such a way that $A_{1} B_{1}=A_{0} B_{0}$ and $B_{1} C_{1}=B_{0} C_{0}$. Next he chooses points $A_{2}, B_{2}, C... | Answer: $\frac{1}{3}$ and 3 .
Denote the lengths $A_{0} B_{0}, B_{0} C_{0}, C_{0} A_{0}$ by $x, y, z$ in non-increasing order. Similarly, denote the lengths $A_{1} B_{1}, B_{1} C_{1}, C_{1} A_{1}$ by $x^{\prime}, y^{\prime}, z^{\prime}$ in non-increasing order, and the lengths $A_{3} B_{3}, B_{3} C_{3}$, $C_{3} A_{3}$... | {
"problem_match": "\nProblem 15.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "# Solution."
} | 262 | 879 |
2020 | T3 | 17 | null | BalticWay | For a prime number $p$ and a positive integer $n$, denote by $f(p, n)$ the largest integer $k$ such that $p^{k} \mid n$ !. Let $p$ be a given prime number and let $m$ and $c$ be given positive integers. Prove that there exist infinitely many positive integers $n$ such that $f(p, n) \equiv c$ $(\bmod m)$. | .
We start by noting that
$$
f(p, n)=\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^{2}}\right\rfloor+\left\lfloor\frac{n}{p^{3}}\right\rfloor+\ldots
$$
which is the well-known Legendre Formula. Now, if we choose
$$
n=p^{a_{1}}+p^{a_{2}}+\cdots+p^{a_{k}}
$$
for positive integers $a_{1}>a_{2}>\cdots>a_... | {
"problem_match": "\nProblem 17.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "\nSolution 1"
} | 95 | 602 |
2020 | T3 | 17 | null | BalticWay | For a prime number $p$ and a positive integer $n$, denote by $f(p, n)$ the largest integer $k$ such that $p^{k} \mid n$ !. Let $p$ be a given prime number and let $m$ and $c$ be given positive integers. Prove that there exist infinitely many positive integers $n$ such that $f(p, n) \equiv c$ $(\bmod m)$. | .
We denote $v_{p}(n)$ for the largest power of $p$ dividing $n$.
We start with a lemma.
Lemma. For any prime $q$ and modulus $m^{\prime}$ not divisible by $q$, there exists infinitely many powers $q^{n}$ of $q$ such that $v_{p}\left(q^{n} !\right) \equiv 1\left(\bmod m^{\prime}\right)$.
Proof. Define $a_{k}=v_{q}\... | {
"problem_match": "\nProblem 17.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "# Solution 2"
} | 95 | 1,024 |
2020 | T3 | 18 | null | BalticWay | Let $n \geqslant 1$ be a positive integer. We say that an integer $k$ is a fan of $n$ if $0 \leqslant k \leqslant n-1$ and there exist integers $x, y, z \in \mathbb{Z}$ such that
$$
\begin{aligned}
x^{2}+y^{2}+z^{2} & \equiv 0 \quad(\bmod n) ; \\
x y z & \equiv k \quad(\bmod n) .
\end{aligned}
$$
Let $f(n)$ be the nu... | Answer: $f(2020)=f(4) \cdot f(5) \cdot f(101)=1 \cdot 1 \cdot 101=101$.
To prove our claim we show that $f$ is multiplicative, that is, $f(r s)=f(r) f(s)$ for coprime numbers $r, s \in \mathbb{N}$, and that
(i) $f(4)=1$,
(ii) $f(5)=1$,
(iii) $f(101)=101$.
The multiplicative property follows from the Chinese Remain... | {
"problem_match": "\nProblem 18.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "\nSolution."
} | 151 | 533 |
2020 | T3 | 19 | null | BalticWay | Denote by $d(n)$ the number of positive divisors of a positive integer $n$. Prove that there are infinitely many positive integers $n$ such that $|\sqrt{3} \cdot d(n)|$ divides $n$. | .
Based on the solution by the Finnish team: Instead of finding infinitely many solutions with $d(n)=2^{3}$, we prove that every power $2^{k}, k \geqslant 0$, has at least one solution with $d(n)=2^{k}$.
Let $k \geqslant 0$ be given, and let $p_{1}, p_{2}, \ldots$ be the sequence of all prime numbers. We then write t... | {
"problem_match": "\nProblem 19.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "# Solution 2"
} | 50 | 846 |
2020 | T3 | 19 | null | BalticWay | Denote by $d(n)$ the number of positive divisors of a positive integer $n$. Prove that there are infinitely many positive integers $n$ such that $|\sqrt{3} \cdot d(n)|$ divides $n$. | .
Based on the solution by the Norwegian team: In this third solution, instead of letting $d(n)$ be a power of 2 , we prove that there a infinitely many solutions with $n=2^{k}$.
Consider the sequence $a_{i}=\lfloor i \sqrt{3}\rfloor$ for $i \geqslant 1$. If $k \geqslant 3$, then $(k+1) \sqrt{3}<2(k+1) \leqslant 2^{k... | {
"problem_match": "\nProblem 19.",
"resource_path": "BalticWay/segmented/en-bw20sol.jsonl",
"solution_match": "\nSolution 3"
} | 50 | 1,025 |
2021 | T3 | 1 | null | BalticWay | Let $n$ be a positive integer. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the equation
$$
(f(x))^{n} f(x+y)=(f(x))^{n+1}+x^{n} f(y)
$$
for all $x, y \in \mathbb{R}$. | The functions we are looking for are $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=0$ and $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=x$. For $n$ even $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=-x$ is also a solution.
Throughout the solution, $P\left(x_{0}, y_{0}\right)$ will denote the substitution of $x_{0}$ and $y_{... | {
"problem_match": "\nProblem 1.",
"resource_path": "BalticWay/segmented/en-bw21sol.jsonl",
"solution_match": "\nSolution."
} | 79 | 874 |
2021 | T3 | 4 | null | BalticWay | Let $\Gamma$ be a circle in the plane and $S$ be a point on $\Gamma$. Mario and Luigi drive around the circle $\Gamma$ with their go-karts. They both start at $S$ at the same time. They both drive for exactly 6 minutes at constant speed counterclockwise around the track. During these 6 minutes, Luigi makes exactly one ... | . Without loss of generality, we assume that $\Gamma$ is the unit circle and $S=(1,0)$. Three points are marked with bananas:
(i) After 45 seconds, Luigi has passed through an arc with a subtended angle of $45^{\circ}$ and is at the point $\left(\sqrt{2} / 2, \sqrt{2} / 2\right.$ ), whereas Mario has passed through an... | {
"problem_match": "\nProblem 4.",
"resource_path": "BalticWay/segmented/en-bw21sol.jsonl",
"solution_match": "\nSolution 1"
} | 167 | 1,209 |
2021 | T3 | 7 | null | BalticWay | Let $n>2$ be an integer. Anna, Edda and Magni play a game on a hexagonal board tiled with regular hexagons, with $n$ tiles on each side. The figure shows a board with 5 tiles on each side. The central tile is marked.
## Baltic Way
Reykjavík, November 11th - 15th
Solutions
The game begins with a stone on a tile in o... | We colour the board in three colours in such a way that no neighbouring tiles are of the same colour. We can give each hexagon a coordinate using $\overrightarrow{e_{1}}=(1,0)$ and $\overrightarrow{e_{2}}=\left(\cos \left(120^{\circ}, \sin \left(120^{\circ}\right)\right)=\right.$ $\left(\frac{-1}{2}, \frac{\sqrt{3}}{2}... | {
"problem_match": "\nProblem 7.",
"resource_path": "BalticWay/segmented/en-bw21sol.jsonl",
"solution_match": "\nSolution."
} | 330 | 563 |
2021 | T3 | 8 | null | BalticWay | We are given a collection of $2^{2^{k}}$ coins, where $k$ is a non-negative integer. Exactly one coin is fake. We have an unlimited number of service dogs. One dog is sick but we do not know which one. A test consists of three steps: select some coins from the collection of all coins; choose a service dog; the dog smel... | Number the coins by $2^{k}$-digit binary numbers from $\overbrace{00 \ldots 0}^{\text {length } 2^{k}}$ to $\overbrace{11 \ldots 1}^{\text {length } 2^{k}}$. Let $A_{i}$ be the set of coins which have 0 in $i$-th position of the binary number. The first $2^{k}$ tests we perform with the help of $2^{k}$ different dogs. ... | {
"problem_match": "\nProblem 8.",
"resource_path": "BalticWay/segmented/en-bw21sol.jsonl",
"solution_match": "\nSolution."
} | 144 | 533 |
2021 | T3 | 9 | null | BalticWay | We are given 2021 points on a plane, no three of which are collinear. Among any 5 of these points, at least 4 lie on the same circle. Is it necessarily true that at least 2020 of the points lie on the same circle? | The answer is positive.
Let us first prove a lemma that if 4 points $A, B, C, D$ all lie on circle $\Gamma$ and some two points $X, Y$ do not lie on $\Gamma$, then these 6 points are pairs of intersections of three circles, circle $\Gamma$ and two other circles. Indeed, according to the problem statement there are 4 p... | {
"problem_match": "\nProblem 9.",
"resource_path": "BalticWay/segmented/en-bw21sol.jsonl",
"solution_match": "\nSolution."
} | 60 | 674 |
2021 | T3 | 10 | null | BalticWay | John has a string of paper where $n$ real numbers $a_{i} \in[0,1]$, for all $i \in\{1, \ldots, n\}$, are written in a row. Show that for any given $k<n$, he can cut the string of paper into $k$ non-empty pieces, between adjacent numbers, in such a way that the sum of the numbers on each piece does not differ from any o... | . Denote the sums on each piece by
$$
\begin{aligned}
& S_{1}=a_{1}+a_{2}+\ldots+a_{m_{1}}, \\
& S_{2}=a_{m_{1}+1}+a_{m_{1}+2}+\ldots+a_{m_{2}}, \\
& \quad \ldots \\
& S_{k}=a_{m_{k-1}+1}+\ldots+a_{m_{k}} .
\end{aligned}
$$
By abuse of notation $S_{i}$ will both denote the set of numbers enclosed by cuts and its sum,... | {
"problem_match": "\nProblem 10.",
"resource_path": "BalticWay/segmented/en-bw21sol.jsonl",
"solution_match": "\nSolution 1"
} | 126 | 1,185 |
2021 | T3 | 10 | null | BalticWay | John has a string of paper where $n$ real numbers $a_{i} \in[0,1]$, for all $i \in\{1, \ldots, n\}$, are written in a row. Show that for any given $k<n$, he can cut the string of paper into $k$ non-empty pieces, between adjacent numbers, in such a way that the sum of the numbers on each piece does not differ from any o... | . This problem can be solved by finding a certain graph having a directed path of length $k$. For real $x$ let $\left(V_{x}, E_{x}\right)$ be a directed graph having vertices $V_{x}=\{0,1, \ldots, n\}$. If $i, j \in V_{x}$ we have a directed edge $(i, j) \in V_{x}$ iff $i \leq j$ and $\sum_{l=i+1}^{j} \in[x, x+1]$.
Su... | {
"problem_match": "\nProblem 10.",
"resource_path": "BalticWay/segmented/en-bw21sol.jsonl",
"solution_match": "\nSolution 2"
} | 126 | 754 |
2021 | T3 | 11 | null | BalticWay | A point $P$ lies inside a triangle $A B C$. The points $K$ and $L$ are the projections of $P$ onto $A B$ and $A C$, respectively. The point $M$ lies on the line $B C$ so that $K M=L M$, and the point $P^{\prime}$ is symmetric to $P$ with respect to $M$. Prove that $\angle B A P=\angle P^{\prime} A C$.
## Baltic Way
R... | For points $X, Y, Z, X \neq Y$ and $Z \neq Y$ let rot $X Y Z$ denote the rotation that takes rotates line $X Y$ to line $Z Y$ modulo half turns. We consider two rotations equivalent one of them is a composition of some translation and the other rotation. It is clear that this is indeed an equivalence relation (as the E... | {
"problem_match": "\nProblem 11.",
"resource_path": "BalticWay/segmented/en-bw21sol.jsonl",
"solution_match": "\nSolution."
} | 205 | 773 |
2021 | T3 | 13 | null | BalticWay | Let $D$ be the foot of the $A$-altitude of an acute triangle $A B C$. The internal bisector of the angle $D A C$ intersects $B C$ at $K$. Let $L$ be the projection of $K$ onto $A C$. Let $M$ be the intersection point of $B L$ and $A D$. Let $P$ be the intersection point of $M C$ and $D L$. Prove that $P K \perp A B$. | . Let $X$ be a point on $B C$ such that $L X \perp A B$, as seen in figure 6. It is enough to prove that
because then $P K \| L X$ and $L X \perp A B$.
$$
\frac{D P}{P L}=\frac{D K}{K X}
$$
Applying Menelaos for triangle $B D L$ and transversal $M P C$ we get
$$
\frac{D P}{P L} \cdot \frac{L M}{M B} \cdot \frac{B C... | {
"problem_match": "\nProblem 13.",
"resource_path": "BalticWay/segmented/en-bw21sol.jsonl",
"solution_match": "\nSolution 1"
} | 108 | 564 |
2021 | T3 | 15 | null | BalticWay | For which positive integers $n \geq 4$ does there exist a convex $n$-gon with side lengths $1,2, \ldots, n$ (in some order) and with all of its sides tangent to the same circle? | It exists if $n=4 k$ or $n=4 k+1$ where $k$ is a positive integer.
Let us consider $n$-gon $P_{1} P_{2} \ldots P_{n}$. Tangent points of the inscribed circle divide each of its sides in two segments. Lengths of these segments that has a common vertex $P_{i}$ are equal. Denote the length of tangent segments that origin... | {
"problem_match": "\nProblem 15.",
"resource_path": "BalticWay/segmented/en-bw21sol.jsonl",
"solution_match": "\nSolution."
} | 53 | 1,548 |
2023 | T3 | 1 | null | BalticWay | Find all strictly increasing sequences $1=a_{1}<a_{2}<a_{3}<\cdots$ of positive integers satisfying
$$
3\left(a_{1}+a_{2}+\cdots+a_{n}\right)=a_{n+1}+a_{n+2}+\cdots+a_{2 n}
$$
for all positive integers $n$. | The strictly increasing sequence $\left(a_{n}\right)$ with $a_{n}=2 n-1$ for all $n \in \mathbb{Z}^{+}$ satisfies $a_{1}=1$ and solves the given equation, since $1+3+\cdots+(2 n-1)=n^{2}$ and $(2 n+1)+(2 n+3)+\cdots+(4 n-1)=(2 n)^{2}-n^{2}=3 n^{2}$ for all $n \in \mathbb{Z}^{+}$.
We claim that no other sequence is sui... | {
"problem_match": "\nProblem 1:",
"resource_path": "BalticWay/segmented/en-bw23sol.jsonl",
"solution_match": "\nSolution:"
} | 81 | 1,164 |
2023 | T3 | 2 | null | BalticWay | Let $a_{1}, a_{2}, \ldots, a_{2023}$ be positive real numbers with
$$
a_{1}+a_{2}^{2}+a_{3}^{3}+\cdots+a_{2023}^{2023}=2023
$$
Show that
$$
a_{1}^{2023}+a_{2}^{2022}+\cdots+a_{2022}^{2}+a_{2023}>1+\frac{1}{2023} .
$$ | Let us prove that conversely, the condition
$$
a_{1}^{2023}+a_{2}^{2022}+\cdots+a_{2023} \leq 1+\frac{1}{2023}
$$
implies that
$$
S:=a_{1}+a_{2}^{2}+\cdots+a_{2023}^{2023}<2023 .
$$
This is trivial if all $a_{i}$ are less than 1 . So suppose that there is an $i$ with $a_{i} \geq 1$, clearly it is unique and $a_{i}<... | {
"problem_match": "\nProblem 2:",
"resource_path": "BalticWay/segmented/en-bw23sol.jsonl",
"solution_match": "\nSolution:"
} | 132 | 672 |
2023 | T3 | 4 | null | BalticWay | Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy
$$
f(f(x)+y)+x f(y)=f(x y+y)+f(x)
$$
for all real numbers $x$ and $y$. | Let $P(x, y)$ denote the assertion of the given functional equation.
Claim 1: $f(0)=0$.
Proof. Note that $P(0, y)$ and $P(x, 0)$ gives us the following:
$$
\begin{aligned}
f(y+f(0)) & =f(y)+f(0) \\
f(f(x))+x f(0) & =f(0)+f(x) .
\end{aligned}
$$
Consider the first expression. Plugging $y=-f(0)$ in it yields
$$
f(-f... | {
"problem_match": "\nProblem 4:",
"resource_path": "BalticWay/segmented/en-bw23sol.jsonl",
"solution_match": "\nSolution:"
} | 57 | 1,060 |
2023 | T3 | 5 | null | BalticWay | Find the smallest positive real number $\alpha$, such that
$$
\frac{x+y}{2} \geq \alpha \sqrt{x y}+(1-\alpha) \sqrt{\frac{x^{2}+y^{2}}{2}}
$$
for all positive real numbers $x$ and $y$. | Let us prove that $\alpha=\frac{1}{2}$ works. Then the following inequality should hold for all positive real numbers $x$ and $y$ :
$$
\begin{aligned}
& \frac{x+y}{2} \geq \frac{1}{2} \sqrt{x y}+\frac{1}{2} \sqrt{\frac{x^{2}+y^{2}}{2}} \\
\Longleftrightarrow & (x+y)^{2} \geq x y+\frac{x^{2}+y^{2}}{2}+2 \sqrt{x y \cdot... | {
"problem_match": "\nProblem 5:",
"resource_path": "BalticWay/segmented/en-bw23sol.jsonl",
"solution_match": "\nSolution:"
} | 66 | 959 |
2023 | T3 | 8 | null | BalticWay | In the city of Flensburg there is a single, infinitely long, street with houses numbered 2, 3, .. The police in Flensburg is trying to catch a thief who every night moves from the house where she is currently hiding to one of its neighbouring houses.
To taunt the local law enforcement the thief reveals every morning t... | We will prove that the police are always able to catch the thief in finite time.
Let $h_{i}$ denote the house the thief stays at the $i$-th night and $p_{i}$ denote the greatest prime divisor of $h_{i}$.
The police knows that she stays at different neighbouring houses every night, so $h_{i+1}-h_{i}=1$ for all non-neg... | {
"problem_match": "\nProblem 8:",
"resource_path": "BalticWay/segmented/en-bw23sol.jsonl",
"solution_match": "\nSolution:"
} | 121 | 1,131 |
2023 | T3 | 10 | null | BalticWay | On a circle, $n \geq 3$ points are marked. Each marked point is coloured red, green or blue. In one step, one can erase two neighbouring marked points of different colours and mark a new point between the locations of the erased points with the third colour. In a final state, all marked points have the same colour whic... | Answer: All even numbers $n$ greater than 2 .
We show first that required initial states are impossible for odd $n$. Note that if one colour is missing then the numbers of marked points of existing two colours have different parities, i.e., the difference of these numbers is odd. Each step keeps the parity of the diff... | {
"problem_match": "\nProblem 10:",
"resource_path": "BalticWay/segmented/en-bw23sol.jsonl",
"solution_match": "\nSolution:"
} | 125 | 1,057 |
2023 | T3 | 14 | null | BalticWay | Let $A B C$ be a triangle with centroid $G$. Let $D, E$ and $F$ be the circumcentres of $B C G, C A G$ and $A B G$, respectively. Let $X$ be the intersection of the perpendiculars from $E$ to $A B$ and from $F$ to $A C$. Prove that $D X$ bisects the segment $E F$. | In all three solutions we will prove that the $D$ median coincides with the perpendicular bisector of the segment $B C$. Thus the solutions con- sist of two parts, proving that $X$ lies on the perpendicular bisector of $\mathrm{BC}$ and proving that the midpoint of $E F$ lies on the perpendicular bisector of $B C$.
Th... | {
"problem_match": "\nProblem 14:",
"resource_path": "BalticWay/segmented/en-bw23sol.jsonl",
"solution_match": "# Solution:"
} | 93 | 1,604 |
2023 | T3 | 15 | null | BalticWay | Let $\omega_{1}$ and $\omega_{2}$ be circles with no common points, such that neither circle lies inside the other. Points $M$ and $N$ are chosen on the circles $\omega_{1}$ and $\omega_{2}$, respectively, such that the tangent to the circle $\omega_{1}$ at $M$ and the tangent to the circle $\omega_{2}$ at $N$ intersec... | Since $M P N$ is an isosceles triangle, we have $\angle P M A=\angle P M N=\angle M N P=\angle B N P$. By tangent and chord theorem, $\angle M C A=$ $\angle P M A=\angle B N P=\angle B D N$.
Since $\angle M C P=\angle M N P$, the quadrilateral $C M P N$ is cyclic. Analogously, from $\angle P D N=\angle P M N$, we get ... | {
"problem_match": "\nProblem 15:",
"resource_path": "BalticWay/segmented/en-bw23sol.jsonl",
"solution_match": "\nSolution:"
} | 197 | 690 |
2023 | T3 | 18 | null | BalticWay | Let $p>7$ be a prime number and let $A$ be a subset of $\{0,1, \ldots, p-1\}$ consisting of at least $\frac{p-1}{2}$ elements. Show that for each integer $r$, there exist (not necessarily distinct) numbers $a, b, c, d \in A$ such that
$$
a b-c d \equiv r \quad(\bmod p)
$$ | Let $P$ be the set of residues modulo of possible products $a b$, for $a, b \in A$. Clearly, we have $|P| \geq \frac{p-1}{2}$, since we get $|A|$ different products by fixing an arbitrary $0 \neq a \in A$ and let run $b$ through $A$. If $|P| \geq \frac{p+1}{2}$, then $|r+P| \geq \frac{p+1}{2}$, too. Hence, $|P|+|r+P| \... | {
"problem_match": "\nProblem 18:",
"resource_path": "BalticWay/segmented/en-bw23sol.jsonl",
"solution_match": "\nSolution:"
} | 97 | 1,244 |
2023 | T3 | 20 | null | BalticWay | Let $n$ be a positive integer. A German set in an $n \times n$ square grid is a set of $n$ cells which contains exactly one cell in each row and column. Given a labelling of the cells with the integers from 1 to $n^{2}$ using each integer exactly once, we say that an integer is a German product if it is the product of ... | (a) No, there is no such labelling.
On the contrary, we show that for every labelling there exist two German products whose difference is not divisible by 65 . Suppose that an $8 \times 8$ square grid is labelled with the numbers $1,2, \ldots, 64$ such that no number is used twice.
We can construct a German product t... | {
"problem_match": "\nProblem 20:",
"resource_path": "BalticWay/segmented/en-bw23sol.jsonl",
"solution_match": "# Solution:"
} | 191 | 2,602 |
2024 | T3 | 1 | null | BalticWay | Let $\alpha$ be a non-zero real number. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
x f(x+y)=(x+\alpha y) f(x)+x f(y)
$$
for all $x, y \in \mathbb{R}$.
Answer: $f(x)=c x^{2}$ for any real constant $c$ if $\alpha=2 ; f(x)=0$ otherwise. | Let $P(x, y)$ denote the assertion of the given functional equation. Note that $P(1,0)$ is $f(1)=f(1)+f(0)$ which implies
$$
f(0)=0
$$
Applying this result to $P(x,-x)$ and $P(-x, x)$ where $x \neq 0$ we get:
$$
\begin{aligned}
& 0=(1-\alpha) x f(x)+x f(-x) \\
& 0=(\alpha-1) x f(-x)-x f(x)
\end{aligned}
$$
By addin... | {
"problem_match": "\n1.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "\nSolution 1:"
} | 101 | 675 |
2024 | T3 | 2 | null | BalticWay | Let $\mathbb{R}^{+}$be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
$$
\frac{f(a)}{1+a+c a}+\frac{f(b)}{1+b+a b}+\frac{f(c)}{1+c+b c}=1
$$
for all $a, b, c \in \mathbb{R}^{+}$that satisfy $a b c=1$.
Answer: $f(x)=k x+1-k$ where $k$ is any real number... | Note that $\frac{1}{1+a+c a}=b c \cdot \frac{1}{1+c+b c}$ since $a b c=1$. Similarly,
$$
\frac{1}{1+b+a b}=a c \cdot \frac{1}{1+a+c a}=c \cdot \frac{1}{1+c+b c}
$$
So the initial equality becomes $\frac{b c f(a)+c f(b)+f(c)}{1+c+b c}=1$ which yields
$$
b c f\left(\frac{1}{b c}\right)+c f(b)+f(c)=1+c+b c
$$
Taking $... | {
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "\nSolution:"
} | 153 | 564 |
2024 | T3 | 3 | null | BalticWay | Positive real numbers $a_{1}, a_{2}, \ldots, a_{2024}$ are written on the blackboard. A move consists of choosing two numbers $x$ and $y$ on the blackboard, erasing them and writing the number $\frac{x^{2}+6 x y+y^{2}}{x+y}$ on the blackboard. After 2023 moves, only one number $c$ will remain on the blackboard. Prove t... | Note that by GM-HM we have
$$
\frac{x^{2}+6 x y+y^{2}}{x+y}=x+y+\frac{4 x y}{x+y}=x+y+2 \cdot \frac{2}{\frac{1}{x}+\frac{1}{y}} \leq x+y+2 \sqrt{x y}=(\sqrt{x}+\sqrt{y})^{2}
$$
which means that
$$
\sqrt{\frac{x^{2}+6 x y+y^{2}}{x+y}} \leq \sqrt{x}+\sqrt{y}
$$
Therefore after each move the sum of square roots of all... | {
"problem_match": "\n3.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "\nSolution:"
} | 136 | 615 |
2024 | T3 | 5 | null | BalticWay | Find all positive real numbers $\lambda$ such that every sequence $a_{1}, a_{2}, \ldots$ of positive real numbers satisfying
$$
a_{n+1}=\lambda \cdot \frac{a_{1}+a_{2}+\ldots+a_{n}}{n}
$$
for all $n \geq 2024^{2024}$ is bounded.
Remark: A sequence $a_{1}, a_{2}, \ldots$ of positive real numbers is bounded if there ex... | First we will show that for all $\lambda>1$ every such sequence is unbounded. Note that $a_{n}=\lambda \cdot \frac{a_{1}+a_{2}+\ldots+a_{n-1}}{n-1}$ implies
$$
\frac{a_{n}(n-1)}{\lambda}=a_{1}+a_{2}+\ldots+a_{n-1}
$$
for all $n>2024^{2024}$. Therefore
$$
\begin{aligned}
a_{n+1} & =\lambda \cdot \frac{a_{1}+a_{2}+\ld... | {
"problem_match": "\n5.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "\nSolution:"
} | 156 | 877 |
2024 | T3 | 6 | null | BalticWay | A labyrinth is a system of 2024 caves and 2023 non-intersecting (bidirectional) corridors, each of which connects exactly two caves, where each pair of caves is connected through some sequence of corridors. Initially, Erik is standing in a corridor connecting some two caves. In a move, he can walk through one of the ca... | Throughout the solution, we denote a corridor directly connecting caves $a$ and $b$ by $a b$.
First we show that Erik can reverse his moves. Indeed, consider three caves $a, b, c$ such that $a b$ and $b c$ are corridors, and assume that Erik stands in the corridor $a b$. He can then perform the moves $a b \rightarrow b... | {
"problem_match": "\n6.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "\nSolution:"
} | 247 | 697 |
2024 | T3 | 10 | null | BalticWay | A frog is located on a unit square of an infinite grid oriented according to the cardinal directions. The frog makes moves consisting of jumping either one or two squares in the direction it is facing, and then turning according to the following rules:
(i) If the frog jumps one square, it then turns $90^{\circ}$ to the... | (a) We color the grid with 5 colors so that the color of a square is determined by the expression $2 x+y$ modulo 5 , where $(x, y)$ are the coordinates of the square (we assume that the side length of the square is 1 ; see Fig. 9). Without loss of generality, let the color of the target square be 0 , and let the frog b... | {
"problem_match": "\n10.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "# Solution:"
} | 147 | 1,334 |
2024 | T3 | 10 | null | BalticWay | A frog is located on a unit square of an infinite grid oriented according to the cardinal directions. The frog makes moves consisting of jumping either one or two squares in the direction it is facing, and then turning according to the following rules:
(i) If the frog jumps one square, it then turns $90^{\circ}$ to the... | Denote the circumradius of $A B C D$ by $r$ and the circumcircle of triangle $B O D$ by $\omega$. Let $T=A C \cap B D$, let $O T$ meet $\omega$ again at $S$, and let $O E$ be a diameter of $\omega$ (Fig. 11). We see that $A C \| O E$ as $A C \perp B D$ and $B D \perp O E$. Furthermore, note that
$$
\measuredangle D S ... | {
"problem_match": "\n10.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "\nSolution:"
} | 147 | 511 |
2024 | T3 | 16 | null | BalticWay | Determine all composite positive integers $n$ such that, for each positive divisor $d$ of $n$, there are integers $k \geq 0$ and $m \geq 2$ such that $d=k^{m}+1$.
Answer: 10. | Call a positive integer $n$ powerless if, for each positive divisor $d$ of $n$, there are integers $k \geq 0$ and $m \geq 2$ such that $d=k^{m}+1$. The solution is composed of proofs of three claims.
Claim 1: If $n$ is powerless, then each positive divisor $d$ of $n$ can be written as $k^{2}+1$ for some integer $k$.
Pr... | {
"problem_match": "\n16.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "\nSolution:"
} | 61 | 1,229 |
2024 | T3 | 16 | null | BalticWay | Determine all composite positive integers $n$ such that, for each positive divisor $d$ of $n$, there are integers $k \geq 0$ and $m \geq 2$ such that $d=k^{m}+1$.
Answer: 10. | Assume that there exists a prime $p<d$ such that $p \nmid a b c d$. Then, since $p-1 \mid d$ ! and $p \nmid d$, by Fermat's little theorem $d^{d!} \equiv\left(d^{p-1}\right)^{\frac{d!}{p-1}} \equiv 1(\bmod p)$. By the same argument $a^{a!} \equiv b^{b!} \equiv c^{c!} \equiv 1$ $(\bmod p)$, and therefore $a^{a!}+b^{b!}-... | {
"problem_match": "\n16.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "\nSolution:"
} | 61 | 909 |
2024 | T3 | 19 | null | BalticWay | Does there exist a positive integer $N$ which is divisible by at least 2024 distinct primes and whose positive divisors $1=d_{1}<d_{2}<\ldots<d_{k}=N$ are such that the number
$$
\frac{d_{2}}{d_{1}}+\frac{d_{3}}{d_{2}}+\ldots+\frac{d_{k}}{d_{k-1}}
$$
is an integer?
Answer: Yes. | For arbitrary positive integer $N$, we will write $f(N)=\frac{d_{2}}{d_{1}}+\frac{d_{3}}{d_{2}}+\ldots+\frac{d_{k}}{d_{k-1}}$ where $1=d_{1}<d_{2}<\ldots<d_{k}=N$ are all positive divisors of $N$. Let us prove by induction that for any positive integer $M$ there is a positive integer $N$ with exactly $M$ different prim... | {
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "\nSolution:"
} | 105 | 502 |
2024 | T3 | 20 | null | BalticWay | Positive integers $a, b$ and $c$ satisfy the system of equations
$$
\left\{\begin{aligned}
(a b-1)^{2} & =c\left(a^{2}+b^{2}\right)+a b+1 \\
a^{2}+b^{2} & =c^{2}+a b
\end{aligned}\right.
$$
(a) Prove that $c+1$ is a perfect square.
(b) Find all such triples $(a, b, c)$.
Answer: (b) $a=b=c=3$. | (a) Substituting $a^{2}+b^{2}$ from the second equation to the first one gives
$$
(a b-1)^{2}=c\left(c^{2}+a b\right)+a b+1
$$
Rearranging terms in the obtained equation gives
$$
(a b)^{2}-(c+3) a b-c^{3}=0
$$
which we can consider as a quadratic equation in $a b$. Its discriminant is
$$
D=(c+3)^{2}+4 c^{3}=4 c^{3... | {
"problem_match": "\n20.",
"resource_path": "BalticWay/segmented/en-bw24sol.jsonl",
"solution_match": "# Solution 2:"
} | 125 | 1,271 |
1992 | T3 | 11 | null | BalticWay | Let $\mathbb{Q}^{+}$denote the set of positive rational numbers. Show that there exists one and only one function $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$satisfying the following conditions:
(i) If $0<q<\frac{1}{2}$ then $f(q)=1+f\left(\frac{q}{1-2 q}\right)$.
(ii) If $1<q \leq 2$ then $f(q)=1+f(q-1)$.
(iii) $... | By condition (iii) we have $f(1)=1$. Applying condition (iii) to each of (i) and (ii) gives two new conditions $\left(i^{\prime}\right)$ and $\left(i i^{\prime}\right)$ taking care of $q>2$ and $\frac{1}{2} \leq q<1$ respectively. Now, for any rational number $\frac{a}{b} \neq 1$ we can use $(i),\left(i^{\prime}\right)... | {
"problem_match": "\n11.",
"resource_path": "BalticWay/segmented/en-bw92sol.jsonl",
"solution_match": "\nSolution."
} | 159 | 637 |
1993 | T3 | 19 | null | BalticWay | A convex quadrangle $A B C D$ is inscribed in a circle with the centre $O$. The angles $\angle A O B, \angle B O C, \angle C O D$ and $\angle D O A$, taken in some order, are of the same size as the angles of quadrangle $A B C D$. Prove that $A B C D$ is a square. | As the quadrangle $A B C D$ is inscribed in a circle, we have $\angle A B C+\angle C D A=\angle B C D+\angle D A B=$ $180^{\circ}$. It suffices to show that if each of these angles is equal to $90^{\circ}$, then each of the angles $A O B, B O C$, $C O D$ and $D O A$ is also equal to $90^{\circ}$ and thus $A B C D$ is a... | {
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw93sol.jsonl",
"solution_match": "\nSolution."
} | 84 | 575 |
1994 | T3 | 2 | null | BalticWay | Let $a_{1}, a_{2}, \ldots, a_{9}$ be any non-negative numbers such that $a_{1}=a_{9}=0$ and at least one of the numbers is non-zero. Prove that for some $i, 2 \leq i \leq 8$, the inequality $a_{i-1}+a_{i+1}<2 a_{i}$ holds. Will the statement remain true if we change the number 2 in the last inequality to $1.9 ?$ | Suppose we have the opposite inequality $a_{i-1}+a_{i+1} \geq 2 a_{i}$ for all $i=2, \ldots, 8$. Let $a_{k}=\max _{1 \leq i \leq 9} a_{i}$. Then we have $a_{k-1}=a_{k+1}=a_{k}, a_{k-2}=a_{k-1}=a_{k}$, etc. Finally we get $a_{1}=a_{k}$, a contradiction.
Suppose now $a_{i-1}+a_{i+1} \geq 1.9 a_{i}$, i.e., $a_{i+1} \geq ... | {
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution."
} | 113 | 770 |
1994 | T3 | 10 | null | BalticWay | How many positive integers satisfy the following three conditions:
(i) All digits of the number are from the set $\{1,2,3,4,5\}$;
(ii) The absolute value of the difference between any two consecutive digits is 1 ;
(iii) The integer has 1994 digits? | Consider all positive integers with $2 n$ digits satisfying conditions $(i)$ and (ii) of the problem. Let the number of such integers beginning with $1,2,3,4$ and 5 be $a_{n}, b_{n}, c_{n}, d_{n}$ and $e_{n}$, respectively. Then, for $n=1$ we have $a_{1}=1$ (integer 12), $b_{1}=2$ (integers 21 and 23), $c_{1}=2$ (integ... | {
"problem_match": "\n10.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution."
} | 65 | 776 |
1994 | T3 | 12 | null | BalticWay | The inscribed circle of the triangle $A_{1} A_{2} A_{3}$ touches the sides $A_{2} A_{3}, A_{3} A_{1}$ and $A_{1} A_{2}$ at points $S_{1}, S_{2}, S_{3}$, respectively. Let $O_{1}, O_{2}, O_{3}$ be the centres of the inscribed circles of triangles $A_{1} S_{2} S_{3}, A_{2} S_{3} S_{1}$ and $A_{3} S_{1} S_{2}$, respective... | We shall prove that the lines $S_{1} O_{1}, S_{2} O_{2}, S_{3} O_{3}$ are the bisectors of the angles of the triangle $S_{1} S_{2} S_{3}$. Let $O$ and $r$ be the centre and radius of the inscribed circle $C$ of the triangle $A_{1} A_{2} A_{3}$. Further, let $P_{1}$ and $H_{1}$ be the points where the inscribed circle o... | {
"problem_match": "\n12.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution."
} | 172 | 592 |
1994 | T3 | 19 | null | BalticWay | The Wonder Island Intelligence Service has 16 spies in Tartu. Each of them watches on some of his colleagues. It is known that if spy $A$ watches on spy $B$ then $B$ does not watch on $A$. Moreover, any 10 spies can be numbered in such a way that the first spy watches on the second, the second watches on the third, ..,... | We call two spies $A$ and $B$ neutral to each other if neither $A$ watches on $B$ nor $B$ watches on $A$.
Denote the spies $A_{1}, A_{2}, \ldots, A_{16}$. Let $a_{i}, b_{i}$ and $c_{i}$ denote the number of spies that watch on $A_{i}$, the number of that are watched by $A_{i}$ and the number of spies neutral to $A_{i}... | {
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution."
} | 106 | 519 |
1994 | T3 | 20 | null | BalticWay | An equilateral triangle is divided into 9000000 congruent equilateral triangles by lines parallel to its sides. Each vertex of the small triangles is coloured in one of three colours. Prove that there exist three points of the same colour being the vertices of a triangle with its sides parallel to the sides of the orig... | Consider the side $A B$ of the big triangle $A B C$ as "horizontal" and suppose the statement of the problem does not hold. The side $A B$ contains 3001 vertices $A=A_{0}, A_{1}, \ldots, A_{3000}=B$ of 3 colours. Hence, there are at least 1001 vertices of one colour, e.g., red. For any two red vertices $A_{k}$ and $A_{... | {
"problem_match": "\n20.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution."
} | 70 | 593 |
1996 | T3 | 3 | null | BalticWay | Let $A B C D$ be a unit square and let $P$ and $Q$ be points in the plane such that $Q$ is the circumcentre of triangle $B P C$ and $D$ is the circumcentre of triangle $P Q A$. Find all possible values of the length of segment $P Q$. | As $Q$ is the circumcentre of triangle $B P C$, we have $|P Q|=|Q C|$ and $Q$ lies on the perpendicular bisector $s$ of $B C$. On the other hand, as $D$ is the circumcentre of triangle $P Q A, Q$ lies on the circle centred at $D$ and passing through $A$. Thus $Q$ must be one of the two intersection points $Q_{1}$ and $... | {
"problem_match": "\n3.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution."
} | 69 | 602 |
1996 | T3 | 18 | null | BalticWay | The jury of an olympiad has 30 members in the beginning. Each member of the jury thinks that some of his colleagues are competent, while all the others are not, and these opinions do not change. At the beginning of every session a voting takes place, and those members who are not competent in the opinion of more than o... | First we note that if nobody is excluded in some session, then the situation becomes stable and nobody can be excluded in any later session.
We use induction to prove the slightly more general claim that if the jury has $2 n$ members, $n \geq 2$, then after at most $n$ sessions nobody will be excluded anymore. For $n=... | {
"problem_match": "\n18.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution."
} | 115 | 645 |
1997 | T3 | 4 | null | BalticWay | Prove that the arithmetic mean $a$ of $x_{1}, \ldots, x_{n}$ satisfies
$$
\left(x_{1}-a\right)^{2}+\cdots+\left(x_{n}-a\right)^{2} \leqslant \frac{1}{2}\left(\left|x_{1}-a\right|+\cdots+\left|x_{n}-a\right|\right)^{2} .
$$ | Denote $y_{i}=x_{i}-a$. Then $y_{1}+y_{2}+\cdots+y_{n}=0$. We can assume $y_{1} \leqslant y_{2} \leqslant \cdots \leqslant y_{k} \leqslant 0 \leqslant y_{k+1} \leqslant \cdots \leqslant y_{n}$. Let $y_{1}+y_{2}+\cdots+y_{k}=-z$, then $y_{k+1}+\cdots+y_{n}=z$ and
$$
\begin{aligned}
y_{1}^{2}+y_{2}^{2}+\cdots+y_{n}^{2} ... | {
"problem_match": "\n4.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n4."
} | 99 | 553 |
1997 | T3 | 13 | null | BalticWay | Five distinct points $A, B, C, D$ and $E$ lie on a line with
$$
|A B|=|B C|=|C D|=|D E| \text {. }
$$
The point $F$ lies outside the line. Let $G$ be the circumcentre of triangle $A D F$ and $H$ be the circumcentre of triangle $B E F$. Show that lines $G H$ and $F C$ are perpendicular.
. Then $O, G$ and $G^{\prime}$ lie on the perpendicular bisector of the segment $D F$, while $O, H$ and $H^{\prime}$ lie on the perpendicular bisector of the segment $B F$. Moreover, $G$ and... | {
"problem_match": "\n13.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n13."
} | 172 | 942 |
1997 | T3 | 16 | null | BalticWay | On a $5 \times 5$ chessboard, two players play the following game. The first player places a knight on some square. Then the players alternately move the knight according to the rules of chess, starting with the second player. It is not allowed to move the knight to a square that has been visited previously. The player... | Answer: the first player has a winning strategy.
Divide all the squares of the board except one in pairs so that the squares of each pair are accessible from each other by one move of the knight (see Figure 10 where the squares of each pair are marked with the same number, and the remaining square is marked by $X$ ). ... | {
"problem_match": "\n16.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n16."
} | 84 | 699 |
1997 | T3 | 18 | null | BalticWay | a) Prove the existence of two infinite sets $A$ and $B$, not necessarily disjoint, of non-negative integers such that each non-negative integer $n$ is uniquely representable in the form $n=a+b$ with $a \in A, b \in B$.
b) Prove that for each such pair $(A, B)$, either $A$ or $B$ contains only multiples of some integer... | a) Let $A$ be the set of non-negative integers whose only non-zero decimal digits are in even positions counted from the right, and $B$ the set of nonnegative integers whose only non-zero decimal digits are in odd positions counted from the right. It is obvious that $A$ and $B$ have the required property.
b) Since the... | {
"problem_match": "\n18.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n18."
} | 93 | 751 |
1997 | T3 | 19 | null | BalticWay | In a forest each of $n$ animals $(n \geqslant 3)$ lives in its own cave, and there is exactly one separate path between any two of these caves. Before the election for King of the Forest some of the animals make an election campaign. Each campaign-making animal visits each of the other caves exactly once,
uses only the... | Answer: b) 4 .
a) As each campaign-making animal uses exactly $n$ paths and the total number of paths is $\frac{n(n-1)}{2}$, the number of campaign-making animals cannot exceed $\frac{n-1}{2}$. Labeling the caves by integers $0,1,2, \ldots, n-1$, we can construct $\frac{n-1}{2}$ non-intersecting campaign routes as fol... | {
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n19."
} | 168 | 577 |
1997 | T3 | 20 | null | BalticWay | Twelve cards lie in a row. The cards are of three kinds: with both sides white, both sides black, or with a white and a black side. Initially, nine of the twelve cards have a black side up. The cards 1-6 are turned, and subsequently four of the twelve cards have a black side up. Now cards $4-9$ are turned, and six card... | Answer: there are 9 cards with one black and one white side and 3 cards with both sides white.
Divide the cards into four types according to the table below.
| Type | Initially up | Initially down |
| :---: | :---: | :---: |
| $A$ | black | white |
| $B$ | white | black |
| $C$ | white | white |
| $D$ | black | black... | {
"problem_match": "\n20.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n20."
} | 126 | 742 |
1998 | T3 | 2 | null | BalticWay | A triple of positive integers $(a, b, c)$ is called quasi-Pythagorean if there exists a triangle with lengths of the sides $a, b, c$ and the angle opposite to the side $c$ equal to $120^{\circ}$. Prove that if $(a, b, c)$ is a quasi-Pythagorean triple then $c$ has a prime divisor greater than 5 . | By the cosine law, a triple of positive integers $(a, b, c)$ is quasi-Pythagorean if and only if
$$
c^{2}=a^{2}+a b+b^{2}
$$
If a triple $(a, b, c)$ with a common divisor $d>1$ satisfies (1), then so does the reduced triple $\left(\frac{a}{d}, \frac{b}{d}, \frac{c}{d}\right)$. Hence it suffices to prove that in every... | {
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n2."
} | 89 | 754 |
1998 | T3 | 5 | null | BalticWay | Let $a$ be an odd digit and $b$ an even digit. Prove that for every positive integer $n$ there exists a positive integer, divisible by $2^{n}$, whose decimal representation contains no digits other than $a$ and $b$. | If $b=0$, then $N=10^{n} a$ meets the demands. For the sequel, suppose $b \neq 0$.
Let $n$ be fixed. We prove that if $1 \leqslant k \leqslant n$, then we can find a positive integer $m_{k}<5^{k}$ such that the last $k$ digits of $m_{k} 2^{n}$ are all $a$ or $b$. Clearly, for $k=1$ we can find $m_{1}$ with $1 \leqslan... | {
"problem_match": "\n5.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n5."
} | 55 | 1,013 |
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