Olympiad problems

2004 All-Russian Olympiad, 1

Tags: geometry , geometric transformation , rotation , analytic geometry , combinatorics , Russia , combinatorial geometry

Each grid point of a cartesian plane is colored with one of three colors, whereby all three colors are used. Show that one can always find a right-angled triangle, whose three vertices have pairwise different colors.

1962 All-Soviet Union Olympiad, 13

Tags: Sequence , algebra , Russia , inequalities

Given are $a_0,a_1, ... , a_n$, satisfying $a_0=a_n = 0$, and $a_{k-1} - 2a_k+a_{k+1}\ge 0$ for $k=0, 1, ... , n-1$. Prove that all the numbers are negative or zero.

2017 Tournament Of Towns, 7

Tags: combinatorics , geometry , Chessboard , Russia

$1\times 2$ dominoes are placed on an $8 \times 8$ chessboard without overlapping. They may partially stick out from the chessboard but the center of each domino must be strictly inside the chessboard (not on its border). Place on the chessboard in such a way: a) at least $40$ dominoes, (3 points) b) at least $41$ dominoes, (3 points) c) more than $41$ dominoes. (6 points) [i](Mikhail Evdokimov)[/i]

2017 Saint Petersburg Mathematical Olympiad, 7

Tags: combinatorics , graph theory , Russia , Petersburg

In a country, some pairs of cities are connected by one-way roads. It turns out that every city has at least two out-going and two in-coming roads assigned to it, and from every city one can travel to any other city by a sequence of roads. Prove that it is possible to delete a cyclic route so that it is still possible to travel from any city to any other city.

2023 All-Russian Olympiad Regional Round, 10.3

Tags: combinatorics , Russia

Given are $50$ distinct sets of positive integers, each of size $30$, such that every $30$ of them have a common element. Prove that all of them have a common element.

2011 All-Russian Olympiad, 4

Tags: geometry , Russia , All Russian Math Olympiad

Perimeter of triangle $ABC$ is $4$. Point $X$ is marked at ray $AB$ and point $Y$ is marked at ray $AC$ such that $AX=AY=1$. Line segments $BC$ and $XY$ intersectat point $M$. Prove that perimeter of one of triangles $ABM$ or $ACM$ is $2$. (V. Shmarov).

2021 All-Russian Olympiad, 1

Tags: combinatorics , All Russian Olympiad , Russia

On a circle there're $1000$ marked points, each colored in one of $k$ colors. It's known that among any $5$ pairwise intersecting segments, endpoints of which are $10$ distinct marked points, there're at least $3$ segments, each of which has its endpoints colored in different colors. Determine the smallest possible value of $k$ for which it's possible.

1962 All-Soviet Union Olympiad, 2

Tags: geometry , Locus , Russia

Given a fixed circle $C$ and a line L through the center $O$ of $C$. Take a variable point $P$ on $L$ and let $K$ be the circle with center $P$ through $O$. Let $T$ be the point where a common tangent to $C$ and $K$ meets $K$. What is the locus of $T$?

2014 All-Russian Olympiad, 2

Tags: function , algebra , Russia , Functional inequality

Given a function $f\colon \mathbb{R}\rightarrow \mathbb{R} $ with $f(x)^2\le f(y)$ for all $x,y\in\mathbb{R} $, $x>y$, prove that $f(x)\in [0,1] $ for all $x\in \mathbb{R}$.

2023 All-Russian Olympiad Regional Round, 9.4

Tags: number theory , Russia

Let $a, b, c$ be positive integers such that no number divides some other number. If $ab-b+1 \mid abc+1$, prove that $c \geq b$.

2023 All-Russian Olympiad Regional Round, 9.8

Tags: geometry , Russia

In an acute triangle $ABC$, let $M$ and $N$ be the midpoints of $AB$ and $AC$ and let $BH$ be its altitude from $B$. Its incircle touches $AC$ at $K$ and the line through $K$ parallel to $MH$ meets $MN$ at $P$. Prove that $AMPK$ has an incircle.

2023 All-Russian Olympiad Regional Round, 9.9

Tags: algebra , Russia , inequalities proposed , Inequality

Find the largest real $m$, such that for all positive real $a, b, c$ with sum $1$, the inequality $\sqrt{\frac{ab} {ab+c}}+\sqrt{\frac{bc} {bc+a}}+\sqrt{\frac{ca} {ca+b}} \geq m$ is satisfied.

2001 All-Russian Olympiad, 4

Tags: number theory , relatively prime , Russia , Divisors , MONT

Find all odd positive integers $ n > 1$ such that if $ a$ and $ b$ are relatively prime divisors of $ n$, then $ a\plus{}b\minus{}1$ divides $ n$.

2022 All-Russian Olympiad, 1

Tags: number theory , All russia mo , All Russian Olympiad , Russia , Russian , number theory solved , Divisors

We call the $main$ $divisors$ of a composite number $n$ the two largest of its natural divisors other than $n$. Composite numbers $a$ and $b$ are such that the main divisors of $a$ and $b$ coincide. Prove that $a=b$.

2004 All-Russian Olympiad, 1

Tags: geometry , geometric transformation , rotation , analytic geometry , combinatorics , Russia , combinatorial geometry

Each grid point of a cartesian plane is colored with one of three colors, whereby all three colors are used. Show that one can always find a right-angled triangle, whose three vertices have pairwise different colors.

2013 All-Russian Olympiad, 4

Tags: geometry , Russia

Inside the inscribed quadrilateral $ABCD$ are marked points $P$ and $Q$, such that $\angle PDC + \angle PCB,$ $\angle PAB + \angle PBC,$ $\angle QCD + \angle QDA$ and $\angle QBA + \angle QAD$ are all equal to $90^\circ$. Prove that the line $PQ$ has equal angles with lines $AD$ and $BC$. [i]A. Pastor[/i]

2023 All-Russian Olympiad Regional Round, 11.10

Tags: combinatorics , Russia , graph theory , spanning tree , ARO

Given is a simple connected graph with $2n$ vertices. Prove that its vertices can be colored with two colors so that if there are $k$ edges connecting vertices with different colors and $m$ edges connecting vertices with the same color, then $k-m \geq n$.

1962 All-Soviet Union Olympiad, 12

Tags: number theory , Russia , algebra , Divisibility

Given unequal integers $x, y, z$ prove that $(x-y)^5 + (y-z)^5 + (z-x)^5$ is divisible by $5(x-y)(y- z)(z-x)$.

2023 All-Russian Olympiad Regional Round, 10.5

Tags: geometry , Russia

In a triangle $ABC$, let $BD$ be its altitude and let $H$ be its orthocenter. The perpendicular bisector of of $HD$ meets $(BCD)$ at $P, Q$. Prove that $\angle APB+\angle AQB=180^{o}$

Kvant 2023, M2739

Tags: geometry , Russia

In an acute triangle $ABC$, let $M$ and $N$ be the midpoints of $AB$ and $AC$ and let $BH$ be its altitude from $B$. Its incircle touches $AC$ at $K$ and the line through $K$ parallel to $MH$ meets $MN$ at $P$. Prove that $AMPK$ has an incircle.

2023 All-Russian Olympiad Regional Round, 11.4

Tags: number theory , Russia

We write pairs of integers on a blackboard. Initially, the pair $(1,2)$ is written. On a move, if $(a, b)$ is on the blackboard, we can add $(-a, -b)$ or $(-b, a+b)$. In addition, if $(a, b)$ and $(c, d)$ are written on the blackboard, we can add $(a+c, b+d)$. Can we reach $(2022, 2023)$?

1962 All-Soviet Union Olympiad, 9

Tags: number theory , Russia

Given is a number with $1998$ digits which is divisible by $9$. Let $x$ be the sum of its digits, let $y$ be the sum of the digits of $x$, and $z$ the sum of the digits of $y$. Find $z$.

2009 All-Russian Olympiad, 4

Tags: modular arithmetic , combinatorics , Russia

There are n cups arranged on the circle. Under one of cups is hiden a coin. For every move, it is allowed to choose 4 cups and verify if the coin lies under these cups. After that, the cups are returned into its former places and the coin moves to one of two neigbor cups. What is the minimal number of moves we need in order to eventually find where the coin is?

1962 All-Soviet Union Olympiad, 6

Tags: Russia , geometry

Given the lengths $AB$ and $BC$ and the fact that the medians to those two sides are perpendicular, construct the triangle $ABC$.

2021 All-Russian Olympiad, 4

Tags: combinatorics , Russia , All Russian Olympiad

Given a natural number $n>4$ and $2n+4$ cards numbered with $1, 2, \dots, 2n+4$. On the card with number $m$ a real number $a_m$ is written such that $\lfloor a_{m}\rfloor=m$. Prove that it's possible to choose $4$ cards in such a way that the sum of the numbers on the first two cards differs from the sum of the numbers on the two remaining cards by less than $$\frac{1}{n-\sqrt{\frac{n}{2}}}$$.