Problem 12. Let $m \geq 5$ and $n$ be positive integers and $M$ be a regular ( $2 n+1$ gon. Find the number of convex $m$-gons with vertices among the vertices of $M$ and having at least one acute angle.

## S O L U T I O N S

## Bulgarian Mathematical Competitions 2003 
12. It is easy to observe that there are at most two acute angles in every convex $m$-gon. Moreover, if there are two acute angles then they are located at one and the same side.

Fix $l=0,1, \ldots, n-1$ and let $A$ and $B$ be two vertices of $M$ such that there are $l$ vertices on the arc $\widehat{A B}$. Consider the following expression

$$
\binom{l}{m-2}+\binom{n}{m-2}-\binom{n-l-1}{m-2}-(l+1)\binom{n-l-1}{m-3}
$$

We count in it all $m$-gons having two acute angles to the side $A B$ and all $m$-gons having one acute angle to the right of $A B$.

It is easy to see that summation on $l=0,1, \ldots, n-1$ and then multiplication by $2 n+1$ counts every $m$-gon exactly ones.

Now using the identity $\sum_{s=0}^{k}\binom{s}{t}=\binom{k+1}{t+1}$ we obtain

$$
\begin{aligned}
& \sum_{l=0}^{n-1}\left[\binom{l}{m-2}+\binom{n}{m-2}-\binom{n-l-1}{m-2}-(l+1)\binom{n-l-1}{m-3}\right] \\
= & \sum_{l=0}^{n-1}\binom{l}{m-2}+n\binom{n}{m-2}-\sum_{s=0}^{n-1}\binom{s}{m-2}-(n-s) \sum_{s=0}^{n-1}\binom{s}{m-3} \\
= & n\binom{n}{m-2}-n \sum_{s=0}^{n-1}\binom{s}{m-3}+(s+1) \sum_{s=0}^{n-1}\binom{s}{m-3}-\sum_{s=0}^{n-1}\binom{s}{m-3} \\
= & (m-2) \sum_{s=0}^{n-1}\binom{s+1}{m-2}-\binom{n}{m-2}=(m-2)\binom{n+1}{m-1}-\binom{n}{m-2} \\
= & \frac{m n-2 n-1}{m-1}\binom{n}{m-2} .
\end{aligned}
$$

Therefore the answer is $\frac{(2 n+1)(m n-2 n-1)}{m-1}\binom{n}{m-2}$.