2011 USAMO 第 2 题

An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer $m$ from each of the integers at two neighboring vertices and adding 2m to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount $m$ and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.

为正五边形的每个顶点分配一个整数，使得五个整数之和为 2011。单人纸牌游戏的一回合包括从两个相邻顶点的每个整数中减去整数 $m$，并向与前两个顶点中的任何一个都不相邻的相对顶点添加 2m。 (金额 $m$ 和选择的顶点可能会因回合而异。)如果在某个顶点经过一定次数后，该顶点的数字为 2011，而其他四个顶点的数字为 0，则该游戏获胜。证明对于初始整数的任何选择，都恰好有一个顶点可以赢得游戏。