A cube of edge $2$ with one of the corner unit cubes removed is called a [i]piece[/i]. Prove that if a cube $T$ of edge $2^n$ is divided into $2^{3n}$ unit cubes and one of the unit cubes is removed, then the rest can be cut into [i]pieces[/i].

Given any convex polygon, show that there are three consecutive vertices such that the polygon lies inside the circle through them.

Andy, Bess, Charley and Dick play on a $1000 \times 1000$ board. They make moves in turn: Andy first, then Bess, then Charley and finally Dick, after that Andy moves again and so on. At each move a player must paint several unpainted squares forming $2 \times 1, 1 \times 2, 1 \times 3$, or $3 \times 1$ rectangle. The player that cannot move loses. Prove that some three players can cooperate to make the fourth player lose.

There is a set of $N\geq 3$ points in the plane, such that no three of them are collinear. Three points $A$, $B$, $C$ in the set are said to form a [i]Baltic triangle[/i] if no other point in the set lies on the circumcircle of triangle $ABC$. Assume that there exists at least one Baltic triangle. Show that there exist at least $\displaystyle\frac{N}{3}$ Baltic triangles.

Anton has an isosceles right triangle, which he wants to cut into $9$ triangular parts in the way shown in the picture. What is the largest number of the resulting $9$ parts that can be equilateral triangles? A more formal description of partitioning. Let triangle $ABC$ be given. We choose two points on its sides so that they go in the order $AC_1C_2BA_1A_2CB_1B_2$, and no two coincide. In addition, the segments $C_1A_2$, $A_1B_2$ and $B_1C_2$ must intersect at one point. Then the partition is given by segments $C_1A_2$, $A_1B_2$, $B_1C_2$, $A_1C_2$, $B_1A_2$ and $C_1B_2$. [img]https://cdn.artofproblemsolving.com/attachments/0/5/5dd914b987983216342e23460954d46755d351.png[/img]

Can a paper circle be cut into pieces and then rearranged into a square of the same area, if only a finite number of cuts is allowed and they must be along segments of straight lines or circular arcs? (A Belov)

$30$ segments of lengths$$1,\quad \sqrt{3},\quad \sqrt{5},\quad \sqrt{7},\quad \sqrt{9},\quad \ldots ,\quad \sqrt{59} $$ have been drawn on a blackboard. In each step, two of the segments are deleted and a new segment of length equal to the hypotenuse of the right triangle with legs equal to the two deleted segments is drawn. After $29$ steps only one segment remains. Find the possible values of its length.

All cells of the checkered plane are painted in $5$ colors so that in any figure of the species [img]https://cdn.artofproblemsolving.com/attachments/f/f/49b8d6db20a7e9cca7420e4b51112656e37e81.png[/img] all colors are different. Prove that in any figure of the species $ \begin{tabular}{ | l | c| c | c | r| } \hline & & & &\\ \hline \end{tabular}$, all colors are different..

A $7\times 7$ square is cut into pieces following types: [img]https://cdn.artofproblemsolving.com/attachments/e/d/458b252c719946062b655340cbe8415d1bdaf9.png[/img] Show that exactly one of the pieces is of type (b). [img]https://cdn.artofproblemsolving.com/attachments/4/9/f3dd0e13fed9838969335c82f5fe866edc83e8.png[/img]

A planar polygon $A_1A_2A_3 . . .A_{n-1}A_n$ ($n > 4$) is made of rigid rods that are connected by hinges. Is it possible to bend the polygon (at hinges only!) into a triangle?

Given 2n+3 points in the plane, no three on a line and no four on a circle, prove that it is always possible to find a circle C that goes through three of the given points and splits the other 2n in half, that is, has n on the inside and n on the outside.

Show that there exists an integer $N$ such that for all $n \ge N$ a square can be partitioned into $n$ smaller squares.

A side of an arbitrary triangle has a length greater than $1$. Prove that the given triangle it can be cut into at least $2$ triangles, so that each of them has a side of length equal to $1$.

Determine all natural numbers $n$ for which it is possible to construct a rectangle of sides $15$ and $n$, with pieces congruent to: [asy] unitsize(0.6 cm); draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(1,2)); draw((2,2)--(3,2)); draw((0,0)--(0,2)); draw((1,0)--(1,2)); draw((2,0)--(2,2)); draw((3,0)--(3,2)); draw((5,-0.5)--(6,-0.5)); draw((4,0.5)--(7,0.5)); draw((4,1.5)--(7,1.5)); draw((5,2.5)--(6,2.5)); draw((4,0.5)--(4,1.5)); draw((5,-0.5)--(5,2.5)); draw((6,-0.5)--(6,2.5)); draw((7,0.5)--(7,1.5)); [/asy] The squares of the pieces have side $1$ and the pieces cannot overlap or leave free spaces

Given a finite number of points in the plane as well as many different rays starting at the origin. It is always possible to pair the points with the rays so that they parallell displaced rays starting in respective points do not intersect?

Can one divide a square into finitely many triangles such that no two triangles share a side? (The triangles have pairwise disjoint interiors and their union is the square.)

Let $X = \{(x,y) \in \mathbb{R}^2 | y \geq 0, x^2+y^2 = 1\} \cup \{(x,0),-1\leq x\leq 1\} $ be the edge of the closed semicircle with radius 1. a) Let $n>1$ be an integer and $P_1,P_2,\dots,P_n \in X$. Show that there exists a permutation $\sigma \colon \{1,2,\dots,n\}\to \{1,2,\dots,n\}$ such that \[\sum_{j=1}^{n}|P_{\sigma(j+1)}-P_{\sigma(j)}|^2\leq 8\]. Where $\sigma(n+1) = \sigma(1)$. b) Find all sets $\{P_1,P_2,\dots,P_n \} \subset X$ such that for any permutation $\sigma \colon \{1,2,\dots,n\}\to \{1,2,\dots,n\}$, \[\sum_{j=1}^{n}|P_{\sigma(j+1)}-P_{\sigma(j)}|^2 \geq 8\]. Where $\sigma(n+1) = \sigma(1)$.

For an integer $n \geq 5,$ two players play the following game on a regular $n$-gon. Initially, three consecutive vertices are chosen, and one counter is placed on each. A move consists of one player sliding one counter along any number of edges to another vertex of the $n$-gon without jumping over another counter. A move is legal if the area of the triangle formed by the counters is strictly greater after the move than before. The players take turns to make legal moves, and if a player cannot make a legal move, that player loses. For which values of $n$ does the player making the first move have a winning strategy?

In any rectangular game board with black and white squares, call a row $X$ a mix of rows $Y$ and $Z$ whenever each cell in row $X$ has the same colour as either the cell of the same column in row $Y$ or the cell of the same column in row $Z$. Let a natural number $m \ge 3$ be given. In some rectangular board, black and white squares lie in such a way that all the following conditions hold. 1) Among every three rows of the board, one is a mix of two others. 2) For every two rows of the board, their corresponding cells in at least one column have different colours. 3) For every two rows of the board, their corresponding cells in at least one column have equal colours. 4) It is impossible to add a new row with each cell either black or white to the board in a way leaving both conditions 1) and 2) still in force Find all possibilities of what can be the number of rows of the board.

A finite set of [i]axis parallel [/i]cubes in space has the property of each point of the room is located in a maximum of M different cubes. Show that you can divide the amount of cubes in $8 (M - 1) + 1$ subsets (or less) with the property that the cubes in each subset lacks common points. (An axis parallel cube is a cube whose edges are parallel to the coordinate axes.)

Given the point $O$ in the space and $1979$ straight lines $l_1, l_2, ... , l_{1979}$ containing it. Not a pair of lines is orthogonal. Given a point $A_1$ on $l_1$ that doesn't coincide with $O$. Prove that it is possible to choose the points $A_i$ on $l_i$ ($i = 2, 3, ... , 1979$) in so that $1979$ pairs will be orthogonal: $A_1A_3$ and $l_2$, $A_2A_4$ and $l_3$,$ ...$ , $A_{i-1}A_{i+1}$ and $l_i$,$ ...$ , $A_{1977}A_{1979}$ and $l_{1978}$, $A_{1978}A_1$ and $l_{1979}$, $A_{1979}A_2$ and $l_1$

Let us say that a subset $M$ of the plane contains a hole if there exists a disc not contained in $M$, but contained inside some polygon with the boundary lying in $M$. Can the plane be presented as a union of $n$ convex sets such that the union of any $n-1$ from them contains a hole? Proposed by: N.Spivak

Let \(n\) be a positive integer and consider an \(n\times n\) square grid. For \(1\le k\le n\), a [i]python[/i] of length \(k\) is a snake that occupies \(k\) consecutive cells in a single row, and no other cells. Similarly, an [i]anaconda[/i] of length \(k\) is a snake that occupies \(k\) consecutive cells in a single column, and no other cells. The grid contains at least one python or anaconda, and it satisfies the following properties: [list] [*]No cell is occupied by multiple snakes. [*]If a cell in the grid is immediately to the left or immediately to the right of a python, then that cell must be occupied by an anaconda. [*]If a cell in the grid is immediately to above or immediately below an anaconda, then that cell must be occupied by a python. [/list] Prove that the sum of the squares of the lengths of the snakes is at least \(n^2\). [i]Proposed by Linus Tang[/i]

Given positive integer $ n \ge 5 $ and a convex polygon $P$, namely $ A_1A_2...A_n $. No diagonals of $P$ are concurrent. Proof that it is possible to choose a point inside every quadrilateral $ A_iA_jA_kA_l (1\le i<j<k<l\le n) $ not on diagonals of $P$, such that the $ \tbinom{n}{4} $ points chosen are distinct, and any segment connecting these points intersect with some diagonal of P.

Baron Munchausen plays billiards on a table with the shape of an equilateral triangle. He claims to have shot a ball from one of the sides of this table so that it passed through a certain point three times in three different directions and then returned to the original point on the side. Can that be true, assuming that the usual law of reflection holds? (Μ Evdokimov)