The ratio of the interior angles of two regular (convex) polygons is 5:7. How many vertices do the two polygons have?
Let the number of sides of the first polygon be $n$, one of its interior angles $\alpha$, the number of sides of the second polygon $m$, and one of its interior angles $\beta$. Then, from $\alpha: \beta=5: 7$,

$$
\alpha=\frac{5}{7} \beta<\frac{5}{7} \cdot 180^{\circ}=\left(128 \frac{4}{7}\right)^{\circ}
$$

The central angle of a regular polygon with $n$ sides is $\frac{360^{\circ}}{n}$, so its interior angle is

$$
\alpha=\left(180^{\circ}-\frac{360^{\circ}}{n}\right)=180^{\circ}\left(1-\frac{2}{n}\right)
$$

$n$ is an integer and at least 3, but at most 6, because for $n \geq 7$, $\alpha \geq\left(128 \frac{4}{7}\right)^{\circ}$ would result, which contradicts (1). Thus, we only need to consider the values $n=3$, 4, 5, and 6:

| $n$ | $\alpha$ | $\beta=\frac{7}{5} \alpha$ | $m=\frac{360^{\circ}}{180^{\circ}-\beta}$ |
| :---: | :---: | :---: | :---: |
| 3 | $60^{\circ}$ | $84^{\circ}$ | not an integer |
| 4 | $90^{\circ}$ | $126^{\circ}$ | not an integer |
| 5 | $108^{\circ}$ | $\left(151 \frac{1}{5}\right)^{\circ}$ | not an integer |
| 6 | $120^{\circ}$ | $168^{\circ}$ | 30 |

We see that only in one case do we get integer values for the sides, so the only solution is when one polygon has 6 sides and the other has 30 sides.