Among four polygons, the second has 4 more sides than the first, the third has 8 more sides than the second, while the fourth has 1 more side than the third. The sum of the number of all diagonals of the first and fourth polygons is equal to the sum of the number of all diagonals of the second and third polygons. How many sides do the polygons have?
The number of sides of the four polygons: $n, n+4, n+12, n+13$. The number of all diagonals of an $n$-sided polygon is $\frac{n(n-3)}{2}$. Therefore, according to the problem,

$$
\frac{n(n-3)}{2}+\frac{(n+13)(n+10)}{2}=\frac{(n+4)(n+1)}{2}+\frac{(n+12)(n+9)}{2}
$$

which simplifies to

$$
n^{2}-3 n+n^{2}+23 n+130=n^{2}+5 n+4+n^{2}+21 n+108
$$

from which

$$
6 n=18, \quad \text { that is } \quad n=3
$$

Thus, the number of sides of the polygons is: $3, 7, 15$, and 16.

Erika Gonda (Miskolc, Zrínyi I. lg. I. o. t.)