We drew all the diagonals in three polygons with different numbers of sides. The second polygon has 3 more sides and 3 times as many diagonals as the first one. The third one has 7 times as many diagonals as the second one. How many more sides does the third polygon have compared to the second one?
Let the number of sides in the second polygon be $m$, in the third $m+x$, then in the first polygon it is $m-3$ (> $>$, since there are no diagonals in a triangle) and we need to determine $x$.

It is known that the number of diagonals in an $n$-sided polygon is $n(n-3) / 2$, so according to our data,

$$
\begin{gathered}
\frac{m(m-3)}{2}=3 \cdot \frac{(m-3)(m-6)}{2} \\
\frac{(m+x)(m+x-3)}{2}=7 \cdot \frac{m(m-3)}{2}
\end{gathered}
$$

From the first equation, $m=3(m-6), m=9$ (since $m-3 \neq 0$), and thus from the second equation, $x=12$ (only the positive root can be used).

Haeffner Erika (Veszprém, Lovassy L. Gymn., I. o. t.) Mátrai István (Szombathely, Nagy Lajos Gymn., I. o. t.)