HMMT 二月 2006 · TEAM1 赛 · 第 13 题

HMMT February 2006 — TEAM1 Round — Problem 13

[40] Having lost a game of checkers and my temper, I dash all the pieces to the ground but one. This last checker, which is perfectly circular in shape, remains completely on the 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 of four).

解析

[40] Having lost a game of checkers and my temper, I dash all the pieces to the ground but one. This last checker, which is perfectly circular in shape, remains completely on the 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 of four). 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 the checker 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 parallel to ` . 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 does not lie on any boundary between squares. But, if we were to swap colors within this strip, then the checker would have equal red and black areas, because then it would be 7 colored in a way such that flipping it across d swaps the colors. This shows that, the way it is currently colored, the checker does not have equal red and black areas. This gives us the desired contradiction.