The difference in the number of sides of two regular polygons is 10, the difference in the number of their diagonals is 15 less than the degree measure of an interior angle of the smaller polygon. How many sides do these polygons have?
Let the number of sides of one polygon be $x$, then the other's is $x+10$. We know that the number of diagonals in an $n$-sided polygon is $\frac{n(n-3)}{2}$, and the measure of an angle in a regular $n$-sided polygon is $\frac{180^{\circ}(n-2)}{n}$.

From the given conditions:

$$
\frac{(10+x)(7+x)}{2}-\frac{x(x-3)}{2}+15=\frac{180(x-2)}{x}
$$

From this:

$$
\begin{gathered}
20 x^{2}-260 x+720=0 \\
x^{2}-13 x+36=0
\end{gathered}
$$

The number of sides of the polygons:

![](https://cdn.mathpix.com/cropped/2024_05_02_3444b15a60521c522034g-1.jpg?height=219&width=675&top_left_y=655&top_left_x=714)

Thus, two pairs of polygons satisfy the requirements of the problem.

Zsigmond Ildikó (Bp. VIII., Ságvári lg. I. o. t.)