Given a natural number $k>1$. Find the smallest number $\alpha$ satisfying the following condition. Suppose that the table $(2k + 1) \times (2k + 1)$ is filled with real numbers not exceeding $1$ in absolute value, and the sums of the numbers in all lines are equal to zero. Then you can rearrange the numbers so that each number remains in its row and all the sums over the columns will be at most $\alpha$.

what maximal number of colors can we use in order to color squares of 10 *10 square board so that each row or column contains squares of at most 5 different colors?

$ 2n\plus{}1$ lines are drawn in the plane, in such a way that every 3 lines define a triangle with no right angles. What is the maximal possible number of acute triangles that can be made in this way?

Two ants are moving along the edges of a convex polyhedron. The route of every ant ends in its starting point, so that one ant does not pass through the same point twice along its way. On every face $F$ of the polyhedron are written the number of edges of $F$ belonging to the route of the first ant and the number of edges of $F$ belonging to the route of the second ant. Is there a polyhedron and a pair of routes described as above, such that only one face contains a pair of distinct numbers? [i]Proposed by Nikolai Beluhov[/i]

We are given a bag of sugar, a two-pan balance, and a weight of $1$ gram. How do we obtain $1$ kilogram of sugar in the smallest possible number of weighings?

Let $ X$ be a set of $ k$ vertexes on a plane such that no three of them are collinear. Let $ P$ be the family of all $ {k \choose 2}$ segments that connect each pair of points. Determine $ \tau(P)$.

A polygon $A$ that doesn't intersect itself and has perimeter $p$ is called [b]Rotund[/b] if for each two points $x,y$ on the sides of this polygon which their distance on the plane is less than $1$ their distance on the polygon is at most $\frac{p}{4}$. (Distance on the polygon is the length of smaller path between two points on the polygon) Now we shall prove that we can fit a circle with radius $\frac{1}{4}$ in any rotund polygon. The mathematicians of two planets earth and Tarator have two different approaches to prove the statement. In both approaches by "inner chord" we mean a segment with both endpoints on the polygon, and "diagonal" is an inner chord with vertices of the polygon as the endpoints. [b]Earth approach: Maximal Chord[/b] We know the fact that for every polygon, there exists an inner chord $xy$ with a length of at most 1 such that for any inner chord $x'y'$ with length of at most 1 the distance on the polygon of $x,y$ is more than the distance on the polygon of $x',y'$. This chord is called the [b]maximal chord[/b]. On the rotund polygon $A_0$ there's two different situations for maximal chord: (a) Prove that if the length of the maximal chord is exactly $1$, then a semicircle with diameter maximal chord fits completely inside $A_0$, so we can fit a circle with radius $\frac{1}{4}$ in $A_0$. (b) Prove that if the length of the maximal chord is less than one we still can fit a circle with radius $\frac{1}{4}$ in $A_0$. [b]Tarator approach: Triangulation[/b] Statement 1: For any polygon that the length of all sides is less than one and no circle with radius $\frac{1}{4}$ fits completely inside it, there exists a triangulation of it using diagonals such that no diagonal with length more than $1$ appears in the triangulation. Statement 2: For any polygon that no circle with radius $\frac{1}{4}$ fits completely inside it, can be divided into triangles that their sides are inner chords with length of at most 1. The mathematicians of planet Tarator proved that if the second statement is true, for each rotund polygon there exists a circle with radius $\frac{1}{4}$ that fits completely inside it. (c) Prove that if the second statement is true, then for each rotund polygon there exists a circle with radius $\frac{1}{4}$ that fits completely inside it. They found out that if the first statement is true then the second statement is also true, so they put a bounty of a doogh on proving the first statement. A young earth mathematician named J.N., found a counterexample for statement 1, thus receiving the bounty. (d) Find a 1392-gon that is counterexample for statement 1. But the Tarators are not disappointed and they are still trying to prove the second statement. (e) (Extra points) Prove or disprove the second statement. Time allowed for this problem was 150 minutes.

Let $ABCD$ be a unit square. For any interior points $M,N$ such that the line $MN$ does not contain a vertex of the square, we denote by $s(M,N)$ the least area of the triangles having their vertices in the set of points $\{ A,B,C,D,M,N\}$. Find the least number $k$ such that $s(M,N)\le k$, for all points $M,N$. [i]Dinu Șerbănescu[/i]

Let $ G$ be a graph with $ 2n$ vertexes and $ 2n(n\minus{}1)$ edges.If we color some edge to red,then vertexes,which are connected by this edge,must be colored to red too. But not necessary that all edges from the red vertex are red. Prove that it is possible to color some vertexes and edges in $ G$,such that all red vertexes has exactly $ n$ red edges.

The [i]base factorial[/i] number system is a unique representation for positive integers where the $n$th digit from the right ranges from $0$ to $n$ inclusive and has place value $n!$ for all $n \ge 1.$ For instance, $71$ can be written in base factorial as $2321_{!} = 2 \cdot 4! + 3 \cdot 3! + 2 \cdot 2! + 1 \cdot 1!.$ Let $S_{!}(n)$ be the base $10$ sum of the digits of $n$ when $n$ is written in base factorial. Compute $\sum_{n=1}^{700} S_{!}(n)$ (expressed in base $10$).

$22$ mathematicians are meeting together. Each mathematician has at least $3$ friends (friends are mutual). And each mathematician can pass his or her information to any mathematician through the transfer between friends. Is it possible to divide these $22$ mathematicians into $2$-person groups (that is, two people in each group, a total of $11$ groups), so that the mathematicians in each group are friends? [hide=original wording in Chinese]仃22位数学家一起开会.每位数学家都至少有3个朋友(朋友是相互的).而且每 位数学家都可以通过朋友之间的传递.将门已的资料传给任意一位数学家.问：是否一定可 以将这22位数学家两两分组(即每组两人,共11组),使得每组的数学家都是朋友？[/hide]

The racing tournament has $12$ stages and $ n $ participants. After each stage, all participants, depending on the occupied place $ k $, receive points $ a_k $ (the numbers $ a_k $ are natural and $ a_1> a_2> \dots> a_n $). For what is the smallest $ n $ the tournament organizer can choose the numbers $ a_1 $, $ \dots $, $ a_n $ so that after the penultimate stage for any possible distribution of places at least two participants had a chance to take first place.

A [i]squiggle[/i] is composed of six equilateral triangles with side length $1$ as shown in the figure below. Determine all possible integers $n$ such that an equilateral triangle with side length $n$ can be fully covered with [i]squiggle[/i]s (rotations and reflections of [i]squiggle[/i]s are allowed, overlappings are not). [asy] import graph; size(100); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; draw((0,0)--(0.5,1),linewidth(2pt)); draw((0.5,1)--(1,0),linewidth(2pt)); draw((0,0)--(3,0),linewidth(2pt)); draw((1.5,1)--(2,0),linewidth(2pt)); draw((2,0)--(2.5,1),linewidth(2pt)); draw((0.5,1)--(2.5,1),linewidth(2pt)); draw((1,0)--(2,2),linewidth(2pt)); draw((2,2)--(3,0),linewidth(2pt)); dot((0,0),ds); dot((1,0),ds); dot((0.5,1),ds); dot((2,0),ds); dot((1.5,1),ds); dot((3,0),ds); dot((2.5,1),ds); dot((2,2),ds); clip((-4.28,-10.96)--(-4.28,6.28)--(16.2,6.28)--(16.2,-10.96)--cycle);[/asy]

Two players play a game with a pile of $N$ coins on a table. On a player's turn, if there are $n$ coins, the player can take at most $n/2+1$ coins, and must take at least one coin. The player who grabs the last coin wins. For how many values of $N$ between $1$ and $100$ (inclusive) does the first player have a winning strategy?

Nine people have attended four different meetings sitting around a circular table. Could they have done so in a way that no two people sat next to each other more than once? Justify your answer.

On a table are given $1999$ red candies and $6661$ yellow candies. The candies are indistinguishable due to the same packing. A gourmet applies the following procedure as long as it is possible: (i) He picks any of the remaining candies, notes its color, eats it and goes to (ii). (ii) He picks any of the remaining candies, and notes its color: if it is the same as the color of the last eaten candy, eats it and goes to (ii); otherwise returns it upon repacking and goes to (i). Prove that all the candies will be eaten and find the probability that the last eaten candy will be red.

Is it possible to divide all non-empty subsets of a set of 10 elements into triples so that in each triple, two of the subsets do not intersect and in their union give the third? [i]Proposed by Vladislav Frank[/i]

A cent, a stuiver ($5$ cent coin), a dubbeltje ($10$ cent coin), a kwartje ($25$ cent coin), a gulden ($100$ cent coin) and a rijksdaalder ($250$ cent coin) are divided among four children in such a way that each of them receives at least one of the six coins. How many such distributions are there?

A football is covered by some polygonal pieces of leather which are sewed up by three different colors threads. It features as follows: i) any edge of a polygonal piece of leather is sewed up with an equal-length edge of another polygonal piece of leather by a certain color thread; ii) each node on the ball is vertex to exactly three polygons, and the three threads joint at the node are of different colors. Show that we can assign to each node on the ball a complex number (not equal to $1$), such that the product of the numbers assigned to the vertices of any polygonal face is equal to $1$.

Let $n \ge 6$ be an integer and $F$ be the system of the $3$-element subsets of the set $\{1, 2,...,n \}$ satisfying the following condition: for every $1 \le i < j \le n$ there is at least $ \lfloor \frac{1}{3} n \rfloor -1$ subsets $A\in F$ such that $i, j \in A$. Prove that for some integer $m \ge 1$ exist the mutually disjoint subsets $A_1, A_2 , ... , A_m \in F $ also, that $|A_1\cup A_2 \cup ... \cup A_m |\ge n-5 $ (Poland) PS. just in case my translation does not make sense, I leave the original in Slovak, in case someone understands something else

We say that a prism is [i]binary[/i] if there exists a labelling of the vertices of the prism with integers from the set $\{-1,1\}$ such that the product of the numbers assigned to the vertices of each face (base or lateral face) is equal to $-1$. a) Prove that any [i]binary[/i] prism has the number of total vertices divisible by 8; b) Prove that any prism with 2000 vertices is [i]binary[/i].

Let $n$ and $d$ be positive integers satisfying $d<\dfrac{n}{2}$. There are $n$ boys and $n$ girls in a school. Each boy has at most $d$ girlfriends and each girl has at most $d$ boyfriends. Prove that one can introduce some of them to make each boy have exactly $2d$ girlfriends and each girl have exactly $2d$ boyfriends. (I think we assume if a girl has a boyfriend, she is his girlfriend as well and vice versa) (proposed by B. Batbaysgalan, folklore).

An $n\times n$ board is coloured in $n$ colours such that the main diagonal (from top-left to bottom-right) is coloured in the first colour; the two adjacent diagonals are coloured in the second colour; the two next diagonals (one from above and one from below) are coloured in the third colour, etc; the two corners (top-right and bottom-left) are coloured in the $n$-th colour. It happens that it is possible to place on the board $n$ rooks, no two attacking each other and such that no two rooks stand on cells of the same colour. Prove that $n=0\pmod{4}$ or $n=1\pmod{4}$.

Can the numbers $1,2,...,121$ be written in the cells of an $11\times 11$ board in such a way that any two consecutive numbers are in adjacent cells (sharing a side), and all perfect squares are in the same column?

Simon draws some line segments on the face of a regular polygon, dissecting it into exactly $2021$ triangles, such that no two drawn line segments are collinear, and no two triangles share a pair of vertices. Simon then assigns each drawn line segment and each side of the polygon with one of three colours. Prove that there is some triangle in the dissection with a pair of identically-coloured sides.