Question 47
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.
