Problem 988

If a polygon is divided into parts by several diagonals, these parts can be colored in two colors such that parts sharing a common side are of different colors. This can be done as follows. We will sequentially draw the diagonals. Each diagonal divides the polygon into two parts. In one of them, we preserve the coloring, and the other we recolor, replacing white with black and black with white everywhere. By performing this operation for all necessary diagonals, we obtain the required coloring. Since in our case an odd number of triangles meet at each vertex, with such a coloring all sides of the polygon will belong to triangles of one color, for example, black (see figure). Let the number of sides of white triangles be denoted by $m$. It is clear that $m$ is divisible by 3. Since each side of a white triangle is also a side of a black triangle, and all sides of the polygon are sides of black triangles, the number of sides of black triangles is $n+m$. Therefore, $n+m$ is divisible by 3, and since $m$ is divisible by 3, $n$ is also divisible by 3.

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