Problem 2. Let $A_{1} A_{2} \ldots A_{n}$ be a regular polygon. Determine the number of subsets $\left\{A_{i}, A_{j}, A_{k}, A_{l}\right\}$ whose elements are the vertices of a trapezoid (a quadrilateral with two parallel and unequal sides).
Solution. If $n$ is odd and $A_{i} A_{j} \| A_{k} A_{l}$, then a diameter of the circumcircle of the polygon passing through one of its vertices will be the perpendicular bisector of the two segments. ... $1 \text{p}$

This can be chosen in $n$ ways, and then the pair of segments in $\binom{(n-1) / 2}{2}$ ways. The segments cannot be congruent, otherwise the quadrilateral determined by them would be a rectangle, so the polygon would contain diametrically opposite points, which is impossible. Therefore, the number sought is $n\binom{(n-1) / 2}{2}=$ $\frac{n(n-1)(n-3)}{8}$

If $n$ is even, pairs of parallel segments are of two types: parallel to a side of the polygon, or perpendicular to a diameter passing through a vertex of the polygon. The number of choices (counting pairs of congruent segments twice) is, in the first case, $\frac{n}{2}\binom{n / 2}{2}$, and in the second case, $\frac{n}{2}\binom{(n-2) / 2}{2}$.

The number of pairs of congruent segments is $\binom{n / 2}{2}$. In conclusion, the number sought is $\frac{n}{2}\binom{n / 2}{2}+\frac{n}{2}\binom{(n-2) / 2}{2}-2\binom{n / 2}{2}=\frac{n(n-2)(n-4)}{8}$ 2p