Let $a$ and $b$ be two positive integers satifying $gcd(a, b) = 1$. Consider a pawn standing on the grid point $(x, y)$. A step of type A consists of moving the pawn to one of the following grid points: $(x+a, y+a),(x+a,y-a), (x-a, y + a)$ or $(x - a, y - a)$. A step of type B consists of moving the pawn to $(x + b,y + b),(x + b,y - b), (x - b,y + b)$ or $(x - b,y - b)$. Now put a pawn on $(0, 0)$. You can make a (nite) number of steps, alternatingly of type A and type B, starting with a step of type A. You can make an even or odd number of steps, i.e., the last step could be of either type A or type B. Determine the set of all grid points $(x,y)$ that you can reach with such a series of steps.

Let $k$ be a fixed positive integer. For each $n = 1, 2,...,$ we will call [i]configuration [/i] of order $n$ any set of $kn$ points of the plane, which does not contain $3$ collinear, colored with $k$ given colors, so that there are $n$ points of each color. Determine all positive integers $n$ with the following property: in each configuration of order $n$, it is possible to select three points of each color, such that the $k$ triangles with vertices of the same color that are determined are disjoint in pairs.

The radius $r$ of a circle with center at the origin is an odd integer. There is a point ($p^m, q^n$) on the circle, with $p,q$ prime numbers and $m,n$ positive integers. Determine $r$.

$A_0B, A_0C$ rays that satisfy $\angle BA_0C=14^{\circ}$. You are to place points $A_1, A_2, ...$ by the following rules. [b]Rules[/b] (1) On the first move, place $A_1$ on any point on $A_0B$(except $A_0$). (2) On the $n>1$th move, place $A_n$ on $A_0B$ iff $A_{n-1}$ is on $A_0C$, and place $A_n$ on $A_0C$ iff $A_{n-1}$ is one $A_0B$. $A_n$ must be place on the point that satisfies $A_{n-2}A_n{n-1}=A_{n-1}A_n$. All the points must be placed in different locations. What is the maximum number of points that can be placed?

World Cup in America introduced a new point system. For a victory $3$ points are given, for a draw $1$ point and for defeat $0$ points. In the preliminary games, the teams are divided into groups of $4$ teams. In groups, teams play with each other, once, then in accordance with the points scored $a,b,c$ and $d$ ($a>b>c>d$) teams take the first, second, third and fourth place in their groups. Give all possible options for the distribution points $a,b,c$ and $d$

There are $n$ points in a plane. Prove that there exist a point $O$ (not necessarily from the given $n$) such that on each side of an arbitrary line, through $O$, lie at least $\frac{n}{3}$ points (including the points on the line).

Prove that from any five points in the plane it is possible to choose three points that are not vertices of an acute triangle.

There are $2017$ points on the plane, no three of them are collinear. Some pairs of the points are connected by $n$ segments. Find the smallest value of $n$ so that there always exists two disjoint segments in any case.

Let $k$ be a positive integer. Find all positive integers $n$, such that it is possible to mark $n$ points on the sides of a triangle (different from its vertices) and connect some of them with a line in such a way that the following conditions are satisfied: 1) there is at least $1$ marked point on each side, 2) for each pair of points $X$ and $Y$ marked on different sides, on the third side there exist exactly $k$ marked points which are connected to both $X$ and $Y$ and exactly k points which are connected to neither $X$ nor $Y$

On the lines $AB$ are located $2009$ different points that do not belong to the segment $[AB]$. Prove that the sum of the distances from point $A$ to these points is not equal to the sum of the distances from point $B$ to these points.

In a circle with a radius of $1$, $64$ mutually different points are selected. Prove that $10$ mutually different points can be selected from them, which lie in a circle with a radius $\frac12$.

Given $n$ points (in sequence)$ A_1, A_2, ... , A_n$ on a line. All the segments $A_1A_2$, $A_2A_3$,$ ...$, $A_{n-1}A_n$ are shorter than $1$. We need to mark $(k-1)$ points so that the difference of every two segments, with the ends in the marked points, is shorter than $1$. Prove that it is possible a) for $k=3$, b) for every $k$ less than $(n-1)$.

On the circumference of a circle are $21$ points. Prove that among the arcs which join any two of these points, at least $100$ of them must subtend an angle at the centre of the circle not exceeding $120^o$ . ( A . F . Sidorenko)

There are $5$ points in the plane no three of which are collinear. We draw all the segments whose vertices are these points. What is the minimum number of new points made by the intersection of the drawn segments? $\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5$

On the plane $n+1$ points are marked, no three of which lie on one straight line. For what natural $k$ can they be connected by segments so that for any $n$ marked points there are exactly $k$ segments with ends at these points?

$ n > 2 $ points are chosen on a circle and each of them is connected to every other by a segment. Is it possible to draw all of these segments in one sequence, i.e. so that the end of the first segment is the beginning of the second, the end of the second - the beginning of the third, etc., and so that the end of the last segment is the beginning of the first?

Let $k$ be a positive integer. Find all positive integers $n$, such that it is possible to mark $n$ points on the sides of a triangle (different from its vertices) and connect some of them with a line in such a way that the following conditions are satisfied: 1) there is at least $1$ marked point on each side, 2) for each pair of points $X$ and $Y$ marked on different sides, on the third side there exist exactly $k$ marked points which are connected to both $X$ and $Y$ and exactly k points which are connected to neither $X$ nor $Y$

The radius $r$ of a circle with center at the origin is an odd integer. There is a point ($p^m, q^n$) on the circle, with $p,q$ prime numbers and $m,n$ positive integers. Determine $r$.

A circle is divided into $n$ equal arcs by $n$ points. Assume that, no matter how we color the $n$ points in two colors, there always exists an axis of symmetry of the set of points such that any two of the $n$ points which are symmetric with respect to that axis have the same color. Find all possible values of $n$.

Let $A$ be a countable set, some of its countable subsets are selected such that; the intersection of any two selected subsets has at most one element. Find the smallest $k$ for which one can ensure that we can color elements of $A$ with $k$ colors such that each selected subsets exactly contain one element of one of the colors and an infinite number of elements of each of the other colors.

Let $n$ be a positive integer. A square $ABCD$ is divided into $n^2$ identical small squares by drawing $(n-1)$ equally spaced lines parallel to the side $AB$ and another $(n- 1)$ equally spaced lines parallel to $BC$, thus giving rise to $(n+1)^2$ intersection points. The points $A, C$ are coloured red and the points $B, D$ are coloured blue. The rest of the intersection points are coloured either red or blue. Prove that the number of small squares having exactly $3$ vertices of the same colour is even.

Let $a$ and $b$ be two positive integers satifying $gcd(a, b) = 1$. Consider a pawn standing on the grid point $(x, y)$. A step of type A consists of moving the pawn to one of the following grid points: $(x+a, y+a),(x+a,y-a), (x-a, y + a)$ or $(x - a, y - a)$. A step of type B consists of moving the pawn to $(x + b,y + b),(x + b,y - b), (x - b,y + b)$ or $(x - b,y - b)$. Now put a pawn on $(0, 0)$. You can make a (nite) number of steps, alternatingly of type A and type B, starting with a step of type A. You can make an even or odd number of steps, i.e., the last step could be of either type A or type B. Determine the set of all grid points $(x,y)$ that you can reach with such a series of steps.

On the plane $n$ points are chosen so that exactly $m$ of them lie on one straight line and no three points not included in these $m$ points lie on one straight line. What is the number of all lines, each of which contains at least two of these points?

At a football tournament, each team plays with each of the remaining teams, winning three points for the win, one point for the draw score and zero points for the defeat. At the end of the tournament it turned out that the sum of the winning points of all teams was $50$. (a) How many teams participated in this tournament? (b) How big is the difference between the team with the highest number and the number of points won?

Assuming that each point of a straight line is painted red or blue, arbitrarily, show that it is always possible to choose three points $A, B$ and $C$ in such a way straight, that are painted the same color and that: $$\frac{AB}{1}=\frac{BC}{2}=\frac{AC}{3}.$$