A [i]lazy[/i] rook can only move from a square to a vertical or a horizontal neighbour. It follows a path which visits each square of an $8 \times 8$ chessboard exactly once. Prove that the number of such paths starting at a corner square is greater than the number of such paths starting at a diagonal neighbour of a corner square. [i](7 points)[/i]

On a $300 \times 300$ board, several rooks are placed that beat the entire board. Within this case, each rook beats no more than one other rook. At what least $k$, it is possible to state that there is at least one rook in each $k\times k$ square ?

Originally, every square of $8 \times 8$ chessboard contains a rook. One by one, rooks which attack an odd number of others are removed. Find the maximal number of rooks that can be removed. (A rook attacks another rook if they are on the same row or column and there are no other rooks between them.) [i](5 points)[/i]

On a $999\times 999$ board a [i]limp rook[/i] can move in the following way: From any square it can move to any of its adjacent squares, i.e. a square having a common side with it, and every move must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A [i]non-intersecting route[/i] of the limp rook consists of a sequence of pairwise different squares that the limp rook can visit in that order by an admissible sequence of moves. Such a non-intersecting route is called [i]cyclic[/i], if the limp rook can, after reaching the last square of the route, move directly to the first square of the route and start over. How many squares does the longest possible cyclic, non-intersecting route of a limp rook visit? [i]Proposed by Nikolay Beluhov, Bulgaria[/i]

There was a rook at some square of a $10 \times 10{}$ chessboard. At each turn it moved to a square adjacent by side. It visited each square exactly once. Prove that for each main diagonal (the diagonal between the corners of the board) the following statement is true: in the rook’s path there were two consecutive steps at which the rook first stepped away from the diagonal and then returned back to the diagonal. [i]Alexandr Gribalko[/i]

Consider a chessboard $6 \times 6$, made up of $36$ single squares. We want to place $6$ chess rooks on this board, one rook on each square, so that there are no two rooks on the same row, nor two rooks on the same column. Note that, once the rooks have been placed in this way, we have that, for every square where a rook has not been placed, there is a rook in the same row as it and a rook in the same column as it. We will say that such rooks are in line with this square. For each of those $30$ houses without rooks, color it green if the two rooks aligned with that same house are the same distance from it, and color it yellow otherwise. For example, when we place the $6$ rooks ($T$) as below, we have: (a) Is it possible to place the rooks so that there are $30$ green squares? (b) Is it possible to place the rooks so that there are $30$ yellow squares? (c) Is it possible to place the rooks so that there are $15$ green and $15$ yellow squares?

On a $999\times 999$ board a [i]limp rook[/i] can move in the following way: From any square it can move to any of its adjacent squares, i.e. a square having a common side with it, and every move must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A [i]non-intersecting route[/i] of the limp rook consists of a sequence of pairwise different squares that the limp rook can visit in that order by an admissible sequence of moves. Such a non-intersecting route is called [i]cyclic[/i], if the limp rook can, after reaching the last square of the route, move directly to the first square of the route and start over. How many squares does the longest possible cyclic, non-intersecting route of a limp rook visit? [i]Proposed by Nikolay Beluhov, Bulgaria[/i]

There was a rook at some square of a $10 \times 10{}$ chessboard. At each turn it moved to a square adjacent by side. It visited each square exactly once. Prove that for each main diagonal (the diagonal between the corners of the board) the following statement is true: in the rook’s path there were two consecutive steps at which the rook first stepped away from the diagonal and then returned back to the diagonal. [i]Alexandr Gribalko[/i]

Two types of pieces, bishops and rooks, are to be placed on a $10\times 10$ chessboard (without necessarily filling it) such that each piece occupies exactly one square of the board. A bishop $B$ is said to [i]attack[/i] a piece $P$ if $B$ and $P$ are on the same diagonal and there are no pieces between $B$ and $P$ on that diagonal; a rook $R$ is said to attack a piece $P$ if $R$ and $P$ are on the same row or column and there are no pieces between $R$ and $P$ on that row or column. A piece $P$ is [i]chocolate[/i] if no other piece $Q$ attacks $P$. What is the maximum number of chocolate pieces there may be, after placing some pieces on the chessboard? [i]Proposed by José Alejandro Reyes González[/i]

On a $999\times 999$ board a [i]limp rook[/i] can move in the following way: From any square it can move to any of its adjacent squares, i.e. a square having a common side with it, and every move must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A [i]non-intersecting route[/i] of the limp rook consists of a sequence of pairwise different squares that the limp rook can visit in that order by an admissible sequence of moves. Such a non-intersecting route is called [i]cyclic[/i], if the limp rook can, after reaching the last square of the route, move directly to the first square of the route and start over. How many squares does the longest possible cyclic, non-intersecting route of a limp rook visit? [i]Proposed by Nikolay Beluhov, Bulgaria[/i]

A regular hexagon is divided by straight lines parallel to its sides into $6n^2$ equilateral triangles. On them, there are $2n$ rooks, no two of which attack each other (a rook attacks in directions parallel to the sides of the hexagon). Prove that if we color the triangles black and white such that no two adjacent triangles have the same color, there will be as many rooks on the black triangles as on the white ones.