1. Determine which pairs of regular polygons have the sizes of their interior angles in the ratio $2: 3$.
1. A regular $n$-sided polygon ($n \geqq 3$) consists of $n$ congruent isosceles triangles, each having an angle of $\frac{1}{n} 360^{\circ}$ at the common vertex. The measure of its interior angle is therefore $\alpha_{n}=180^{\circ}-\frac{1}{n} 360^{\circ}=\frac{n-2}{n} 180^{\circ}$.

The problem leads to the equation for $\mathrm{k}$:

$$
\frac{3(n-2)}{n}=\frac{2(m-2)}{m}
$$

which we can rearrange to:

$$
m n=6 m-4 n \quad \text { or } \quad n=6-4 \frac{m}{n}
$$

From this, it follows that $n<6$. By substituting $n \in\{3,4,5\}$, we find the following three pairs $[n, m]=$ $=[3,4],[4,8],[5,20]$, which satisfy equation (1).

The problem is satisfied by three pairs of regular polygons: the triangle and the square, the square and the octagon, the pentagon and the icosagon.

Full solution is worth 6 points, with 3 points for finding a relationship equivalent to equation (1).