5. Given a regular 22-gon $M$. Find the number of quadruples of vertices of this 22-gon that are the vertices of trapezoids.
Answer: 990.

Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{22}$ in a circle. Each trapezoid side is defined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{22}$.

Consider a chord connecting two adjacent vertices of the polygon, for example, $K_{6} K_{7}$. There are 10 more chords parallel to it ( $K_{5} K_{8}$, etc.), i.e., we get a set of 11 chords parallel to each other. From these, we can form $C_{11}^{2}=55$ pairs of parallel segments.

We will get similar sets if we consider all chords parallel to $K_{1} K_{2}, \ldots, K_{11} K_{12}$ - a total of 11 such sets.

Now, let's take a chord connecting vertices that are one apart from each other, for example, $K_{6} K_{8}$. There are 9 more chords parallel to it ( $K_{5} K_{9}$, etc.), i.e., we get a set of 10 chords parallel to each other. From these, we can form $C_{10}^{2}=45$ pairs of parallel segments. There are also 11 such sets.

In total, we get $11 \cdot 55 + 11 \cdot 45 = 1100$ quadrilaterals. With this method of counting, rectangles have been counted twice. Note that both diagonals of an inscribed rectangle are diameters, and there are 11 diameters with vertices at the given points, thus there are $C_{11}^{2}=55$ rectangles. Subtracting twice the number of rectangles, we get $1100 - 110 = 990$ quadrilaterals with at least one pair of parallel sides.