23. How many trapezoids are there with vertices from the vertices of a regular $n$-sided polygon?
23. When $n=2 m+1$ is odd, any line connecting two vertices of a regular $(2 m+1)$-gon is either a side of the regular $(2 m+1)$-gon or parallel to one of its sides. Additionally, no four vertices of a regular $(2 m+1)$-gon form the vertices of a rectangle. Therefore, any trapezoid that meets the conditions must have one base parallel to a side of the regular $(2 m+1)$-gon. The number of such trapezoids is $C_{m}^{2}$. Hence, when $n$ is odd, the number of trapezoids that meet the conditions is $(2 m+1) C_{m}^{2}=n C_{\frac{n-1}{2}}^{2}$; when $n=2 m$ is even, let the set of rectangles and trapezoids formed by the vertices of a regular $2 m$-gon be $S$. The quadrilaterals in $S$ can be divided into two categories: (1) those with one pair of opposite sides parallel or coincident with a side of the regular $2 m$-gon, of which there are $m C_{m}^{2}$; (2) those with one pair of opposite sides perpendicular to the diameter (the longest diagonal) of the regular $2 m$-gon, of which there are $m C_{m-1}^{2}$. The total number of quadrilaterals in these two categories is $m C_{m}^{2}+m C_{m-1}^{2}$. Each rectangle in $S$ is determined by two diameters (i.e., the diagonals of the rectangle) of the regular $2 m$-gon, so the number of rectangles in $S$ is $C_{m}^{2}$. Since each rectangle is counted twice (as each rectangle has two pairs of parallel sides), the number of trapezoids that meet the conditions when $n=2 m$ is even is $m\left(C_{m}^{2}+C_{m-1}^{2}\right)-2 C_{m}^{2}=m(m-1)(m-2)=6 C_{m}^{3}=6 C_{\frac{n}{2}}^{3}$.