Olympiad problems

2004 All-Russian Olympiad, 3

The polynomials $ P(x)$ and $ Q(x)$ are given. It is known that for a certain polynomial $ R(x, y)$ the identity $ P(x) \minus{} P(y) \equal{} R(x, y) (Q(x) \minus{} Q(y))$ applies. Prove that there is a polynomial $ S(x)$ so that $ P(x) \equal{} S(Q(x)) \quad \forall x.$

2018 Bulgaria EGMO TST, 3

Tags: geometry , circumcircle , Russia

Let be given a semicircle with diameter $AB$ and center $O$, and a line intersecting the semicircle at $C$ and $D$ and the line $AB$ at $M$ ($MB < MA$, $MD < MC$). The circumcircles of the triangles $AOC$ and $DOB$ meet again at $L$. Prove that $\angle MKO$ is right. [i]L. Kuptsov[/i]

2020 Switzerland Team Selection Test, 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$.

2000 All-Russian Olympiad, 7

Tags: geometry , circumcircle , Russia

Let $E$ be a point on the median $CD$ of a triangle $ABC$. The circle $\mathcal S_1$ passing through $E$ and touching $AB$ at $A$ meets the side $AC$ again at $M$. The circle $S_2$ passing through $E$ and touching $AB$ at $B$ meets the side $BC$ at $N$. Prove that the circumcircle of $\triangle CMN$ is tangent to both $\mathcal S_1$ and $\mathcal S_2$.

2007 All-Russian Olympiad, 4

Tags: algorithm , combinatorics , circle , Russia , game , trick

[i]A. Akopyan, A. Akopyan, A. Akopyan, I. Bogdanov[/i] A conjurer Arutyun and his assistant Amayak are going to show following super-trick. A circle is drawn on the board in the room. Spectators mark $2007$ points on this circle, after that Amayak removes one of them. Then Arutyun comes to the room and shows a semicircle, to which the removed point belonged. Explain, how Arutyun and Amayak may show this super-trick.

2021 All-Russian Olympiad, 2

Tags: number theory , Russia , All Russian Olympiad

Let $n$ be a natural number. An integer $a>2$ is called $n$-decomposable, if $a^n-2^n$ is divisible by all the numbers of the form $a^d+2^d$, where $d\neq n$ is a natural divisor of $n$. Find all composite $n\in \mathbb{N}$, for which there's an $n$-decomposable number.

1962 All-Soviet Union Olympiad, 10

Tags: geometry , Russia

In a triangle, $AB=BC$ and $M$ is the midpoint of $AC$. $H$ is chosen on $BC$ so that $MH$ is perpendicular to $BC$. $P$ is the midpoint of $MH$. Prove that $AH$ is perpendicular to $BP$.

2015 Sharygin Geometry Olympiad, 6

Tags: geometry , perpendicular bisector , perpendicular , geometry solved , Sharygin Geometry Olympiad , Russia , median

Lines $b$ and $c$ passing through vertices $B$ and $C$ of triangle $ABC$ are perpendicular to sideline $BC$. The perpendicular bisectors to $AC$ and $AB$ meet $b$ and $c$ at points $P$ and $Q$ respectively. Prove that line $PQ$ is perpendicular to median $AM$ of triangle $ABC$. (D. Prokopenko)

2023 All-Russian Olympiad Regional Round, 10.10

Tags: algebra , Russia , inequalities proposed

Prove that for all positive reals $x, y, z$, the inequality $(x-y)\sqrt{3x^2+y^2}+(y-z)\sqrt{3y^2+z^2}+(z-x)\sqrt{3z^2+x^2} \geq 0$ is satisfied.

2002 All-Russian Olympiad, 3

Tags: inequalities , geometry , circumcircle , Russia

Let O be the circumcenter of a triangle ABC. Points M and N are choosen on the sides AB and BC respectively so that the angle AOC is two times greater than angle MON. Prove that the perimeter of triangle MBN is not less than the lenght of side AC

2000 All-Russian Olympiad, 8

Tags: number theory , greatest common divisor , modular arithmetic , relatively prime , Russia

One hundred natural numbers whose greatest common divisor is $1$ are arranged around a circle. An allowed operation is to add to a number the greatest common divisor of its two neighhbors. Prove that we can make all the numbers pairwise copirme in a finite number of moves.

1962 All-Soviet Union Olympiad, 1

Tags: Russia , geometry

$ABCD$ is any convex quadrilateral. Construct a new quadrilateral as follows. Take $A'$ so that $A$ is the midpoint of $DA'$; similarly, $B'$ so that $B$ is the midpoint of $AB'$; $C'$ so that $C$ is the midpoint of $BC'$; and $D'$ so that $D$ is the midpoint of $CD'$. Show that the area of $A'B'C'D'$ is five times the area of $ABCD$.

2013 All-Russian Olympiad, 3

Tags: geometry , circumcircle , trigonometry , Russia

Squares $CAKL$ and $CBMN$ are constructed on the sides of acute-angled triangle $ABC$, outside of the triangle. Line $CN$ intersects line segment $AK$ at $X$, while line $CL$ intersects line segment $BM$ at $Y$. Point $P$, lying inside triangle $ABC$, is an intersection of the circumcircles of triangles $KXN$ and $LYM$. Point $S$ is the midpoint of $AB$. Prove that angle $\angle ACS=\angle BCP$.

2017 Tournament Of Towns, 6

Tags: combinatorics , Russia

A grasshopper can jump along a checkered strip for $8, 9$ or $10$ cells in any direction. A natural number $n$ is called jumpable if the grasshopper can start from some cell of a strip of length $n$ and visit every cell exactly once. Find at least one non-jumpable number $n > 50$. [i](Egor Bakaev)[/i]

2012 India IMO Training Camp, 1

Tags: geometry , incenter , parallelogram , trigonometry , complex numbers , Russia

A quadrilateral $ABCD$ without parallel sides is circumscribed around a circle with centre $O$. Prove that $O$ is a point of intersection of middle lines of quadrilateral $ABCD$ (i.e. barycentre of points $A,\,B,\,C,\,D$) iff $OA\cdot OC=OB\cdot OD$.

1999 All-Russian Olympiad, 7

Tags: Russia , High school olympiad , algebra , Inequality

Positive numbers $x,y$ satisfy $x^2+y^3 \ge x^3+y^4$. Prove that $x^3+y^3 \le 2$.

1962 All-Soviet Union Olympiad, 3

Tags: Russia , algebra , Sequences

Given integers $a_0,a_1, ... , a_{100}$, satisfying $a_1>a_0$, $a_1>0$, and $a_{r+2}=3 a_{r+1}-2a_r$ for $r=0, 1, ... , 98$. Prove $a_{100}>299$

2021 All-Russian Olympiad, 3

Tags: combinatorics , Russia , All Russian Olympiad

On a line $n+1$ segments are marked such that one of the points of the line is contained in all of them. Prove that one can find $2$ distinct segments $I, J$ which intersect at a segment of length at least $\frac{n-1}{n}d$, where $d$ is the length of the segment $I$.

2012 India IMO Training Camp, 1

Tags: geometry , incenter , parallelogram , trigonometry , complex numbers , Russia

A quadrilateral $ABCD$ without parallel sides is circumscribed around a circle with centre $O$. Prove that $O$ is a point of intersection of middle lines of quadrilateral $ABCD$ (i.e. barycentre of points $A,\,B,\,C,\,D$) iff $OA\cdot OC=OB\cdot OD$.

2023 All-Russian Olympiad Regional Round, 10.9

Tags: combinatorics , Russia

Given is a positive integer $k$. There are $n$ points chosen on a line, such the distance between any two adjacent points is the same. The points are colored in $k$ colors. For each pair of monochromatic points such that there are no points of the same color between them, we record the distance between these two points. If all distances are distinct, find the largest possible $n$.

2023 All-Russian Olympiad Regional Round, 11.8

Tags: geometry , perpendicular bisector , Russia

Given is a triangle $ABC$ with circumcenter $O$. Points $D, E$ are chosen on the angle bisector of $\angle ABC$ such that $EA=EB, DB=DC$. If $P, Q$ are the circumcenters of $(AOE), (COD)$, prove that either the line $PQ$ coincides with $AC$ or $PQCA$ is cyclic.

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.

2022 Dutch IMO TST, 3

Tags: number theory , Russia , All Russian Olympiad

Let $n$ be a natural number. An integer $a>2$ is called $n$-decomposable, if $a^n-2^n$ is divisible by all the numbers of the form $a^d+2^d$, where $d\neq n$ is a natural divisor of $n$. Find all composite $n\in \mathbb{N}$, for which there's an $n$-decomposable number.

1997 All-Russian Olympiad, 4

Tags: geometry , rectangle , induction , combinatorics proposed , combinatorics , Russia

A polygon can be divided into 100 rectangles, but not into 99. Prove that it cannot be divided into 100 triangles. [i]A. Shapovalov[/i]

2013 All-Russian Olympiad, 1

Tags: modular arithmetic , combinatorics proposed , combinatorics , Russia , 2013 , #5 , graph theory

$2n$ real numbers with a positive sum are aligned in a circle. For each of the numbers, we can see there are two sets of $n$ numbers such that this number is on the end. Prove that at least one of the numbers has a positive sum for both of these two sets.