HMMT 二月 2006 · TEAM1 赛 · 第 13 题

英文解析

13. [40] Having lost a game of checkers and my temper, I dash all the pieces to the groundbut one. This last checker, which is perfectly circular in shape, remains completely onthe board, and happens to cover equal areas of red and black squares. Prove that the center of this piece must lie on a boundary between two squares (or at a junction offour).
    Solution: Suppose, for the sake of contradiction, that the problem is false. Evidently,
    at least one boundary between adjacent squares must lie within our checker, or else thechecker would exist entirely within one square, meaning it would cover only one color.
    Note also that a checker’s diameter is smaller than the side of any square of the board,
    so there are at most two such boundaries within our checker (one in each direction).
    Let ` be this, or one of these, boundaries. Draw a diameter d of the checker parallelto ` . Presumably, the strip of the checker between ` and d is part red, part black.
    These red and black areas are unequal, however, because the center of the checker doesnot lie on any boundary between squares. But, if we were to swap colors within thisstrip, then the checker would have equal red and black areas, because then it would becolored in a way such that flipping it across d swaps the colors. This shows that, the 7
    way it is currently colored, the checker does not have equal red and black areas. Thisgives us the desired contradiction.