57 Given a regular $(2n+1)$-gon. Consider all triangles whose vertices are the vertices of this $(2 n+1)$-gon. Find the number of triangles that contain the center of the regular $(2 n+1)$-gon.
57 First, find the number of triangles with a vertex at $A$ and containing the center $O$ of a regular $2n+1$-sided polygon. Let the vertices to the left of $A$ be $B_{i}(1 \leqslant i \leqslant n)$, and those to the right be $C_{j}(1 \leqslant j \leqslant n)$. Clearly, the other two vertices of such a triangle should be one $B_{i}(1 \leqslant i \leqslant n)$ and one $C_{j}(1 \leqslant j \leqslant n)$. It is not hard to see that $\triangle A B_{i} C_{j}$ contains the center $O$ if and only if $i+j \geqslant n+1$.
When $i=1$, there is only one such triangle $(j=n)$;
When $i=2$, there are two such triangles $(j=n, n-1)$;
When $i=n$, there are $n$ such triangles.
Thus, the total number of $\triangle A B_{i} C_{j}$ containing $O$ with $A$ as a vertex is
$$
1+2+3+\cdots+n=\frac{n(n+1)}{2} .
$$

By similar discussions for each vertex of the $2n+1$-sided polygon, the total number of triangles containing $O$ is
$$
\frac{n(n+1)(2 n+1)}{2} .
$$

However, each such triangle is counted three times, so the total number of triangles sought is
$$
\frac{n(n+1)(2 n+1)}{6} \text {. }
$$