Chinese Mathematical Olympiad

Notice that if a convex $m$-gon whose vertex set is contained in $P$ has exactly two acute angles, they must be at consecutive vertices; for otherwise there would be two disjoint pairs of sides that take up more than half of the circle each.

Now assume that the last vertex, clockwise, of these four vertices that make up two acute angles is fixed; this reduces the total number of regular $m$-gons $2n+1$ times and we will later multiply by this factor.

Suppose that the larger arc that the first and the last of these four vertices make contains $k$ points, and the other arc contains $2n-1-k$ points. For each $k$, the vertices of the $m$-polygon on the smaller arc may be arranged in $\left(\genfrac{}{}{0ex}{}{2n-1-k}{m-4}\right)$ ways, and the two vertices on the larger arc may be arranged in $(k-n-1{)}^2$ ways (so that the two angles cut off more than half of the circle).

The total number of polygons given by $k$ is thus $(k-n-1{)}^2\times (\genfrac{}{}{0ex}{}{2n-1-k}{m-4})$. Summation over all $k$ and a change of variable shows that the total number of polygons (divided by a factor of $2n+1$) is ${\sum }_{k\ge 0}{k}^2\times \left(\genfrac{}{}{0ex}{}{n-k-2}{m-4}\right)$.

This can be proven to be exactly $\left(\genfrac{}{}{0ex}{}{n}{m-1}\right)+\left(\genfrac{}{}{0ex}{}{n+1}{m-1}\right)$ by double induction on $n>m$ and $m>4$. The base cases $n=m+1$ and $m=5$ are readily calculated. The induction step is $$\begin{array}{rl}& \sum _{k\ge 0}{k}^2\times \left(\genfrac{}{}{0ex}{}{n-k-2}{m-4}\right)\\ & =\sum _{k\ge 0}{k}^2\times \left(\genfrac{}{}{0ex}{}{(n-1)-k-2}{m-4}\right)+\sum _{k\ge 0}{k}^2\times \left(\genfrac{}{}{0ex}{}{(n-1)-k-2}{m-4-2}\right)\\ & =\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right)+\left(\genfrac{}{}{0ex}{}{n}{m-1}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{m-2}\right)+\left(\genfrac{}{}{0ex}{}{n}{m-2}\right)\\ & =\left(\genfrac{}{}{0ex}{}{n}{m-1}\right)+\left(\genfrac{}{}{0ex}{}{n+1}{m-1}\right).\end{array}$$