, 2010

If all numbers written at the vertices of the polygon are equal, then the claim holds trivially. Hence assume that there are both zeros and ones among the numbers at the vertices. We prove by induction that, for every convex polygon, the partition into triangles can be chosen in such a way that Juku writes either $1$ or $2$ to each triangle.

If $n = 3$, then this claim holds since the sum of the numbers at the vertices of a triangle can be neither $0$ nor $3$. If $n = 4$ (Fig. 2), then draw the diagonal that connects the vertices where $0$ and $1$ are written, respectively, or, if such a diagonal does not exist, then an arbitrary diagonal. In both cases, only sums $1$ and $2$ can arise. If $n \ge 5$, then choose two consecutive vertices with different labels and a third vertex $P$ that is not neighbour to either of them (Fig. 3). Irrespective of whether the label of $P$ is $0$ or $1$, we can draw the diagonal from it to one of the two consecutive vertices chosen before so that the labels of its endpoints are different. Now the polygon is divided into two convex polygons with smaller number of vertices so that both $0$ and $1$ occur among their vertex labels. By the induction hypothesis, both polygons can be partitioned into triangles with sum of labels of vertices either $1$ or $2$.

Fig. 2


Fig. 3