2014 IMO Problems/Problem 2
Let be an integer. Consider an chessboard consisting of unit squares. A configuration of rooks on this board is if every row and every column contains exactly one rook. Find the greatest positive integer such that, for each peaceful configuration of rooks, there is a square which does not contain a rook on any of its squares.
We claim the answer is , where is the ceiling function of ; i.e., the least integer greater than or equal to . Notice that .
First, we shall show that each chessboard with a peaceful configuration of rooks contains a valid square. Consider firstly the rook on the top row of the board. Because , there exists a set of consecutive columns, one of which contains rook . Consider the squares in . Of them, only one contains the rook on one of its squares. Furthermore, each of the other rooks in can only make up to of the squares have it on one of its squares. Therefore, because by the Pigeonhole Principle there exists a square in not covered by any rook in . Clearly, that square cannot be covered by any rook outside of , and so it is a valid choice.
It remains to show that there exists a chessboard without a square. Indeed, such a board exists. First, place a rook at the upper-left corner of the board. Next, place a rook 1 space to the right and spaces down of the first rook. Then, place a rook 1 space to the right and spaces down of the second rook, and so on, until the placement of a new rook will be under the lower boundary of the board. In that case, place a rook in the same unoccupied column and in the first unoccupied row, and continue placement of subsequent rooks. This arrangement of rooks is clearly peaceful, and because it has the property that any square in the board will be occupied by at least one rook, completing the proof.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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