We double the number of sides of a polygon. Which are the polygons for which the sum of the angles of the new polygon is an integer multiple of the sum of the angles of the original polygon?
Solution. Let the number of sides of the original polygon be $m$, then the new polygon has $2m$ sides. The sum of the interior angles of an $n$-sided polygon is $(n-2) \cdot 180^{\circ}$, so in the original polygon it is $(m-2) \cdot 180^{\circ}$, while in the new polygon it is $(2m-2) \cdot 180^{\circ}$.

According to the problem, $(2m-2) \cdot 180^{\circ}$ is an integer multiple of $(m-2) \cdot 180^{\circ}$, that is, $\frac{(2m-2) \cdot 180^{\circ}}{(m-2) \cdot 180^{\circ}} \in \mathbb{Z}$.

$$
\frac{(2m-2) \cdot 180^{\circ}}{(m-2) \cdot 180^{\circ}}=\frac{2m-2}{m-2}=\frac{2(m-2)+2}{m-2}=2+\frac{2}{m-2}
$$

thus $\frac{2}{m-2} \in \mathbb{Z}$, meaning that $m-2$ is a divisor of 2. The divisors of 2 are: $-2; -1; 1; 2$. Since $m \geq 3$, then $m-2 \geq 1$, so $m-2=1$ or $m-2=2$. We have obtained two possible values for $m$.

Therefore, triangles and quadrilaterals are the polygons that meet the requirements of the problem.