## Task A-4.5. (4 points)

For which $n \in \mathbb{N}$ do there exist an angle $\alpha$ and a convex $n$-gon with angles $\alpha, 2 \alpha, \ldots, n \alpha$?
## Solution.

The sum of the angles of a convex $n$-gon is $(n-2) \cdot 180^{\circ}$.

On the other hand, the sum of the angles $\alpha, 2 \alpha, \ldots, n \alpha$ is

$$
(1+2+\ldots+n) \cdot \alpha=\frac{n(n+1)}{2} \cdot \alpha
$$

From the equality $\frac{n(n+1)}{2} \cdot \alpha=(n-2) \cdot 180^{\circ}$, it follows that $\alpha=\frac{2(n-2)}{n(n+1)} \cdot 180^{\circ}$.

For the polygon to be convex, all its angles must be less than a straight angle, i.e., $n \alpha < 180^{\circ}$.

This gives us the condition $\frac{2(n-2)}{n+1} < 1$, from which it follows that $n < 5$.

Thus, the only possible solutions are $n=3$ and $n=4$.

It is easy to see that these are indeed solutions:

For $n=3$, from $\alpha + 2 \alpha + 3 \alpha = 180^{\circ}$, it follows that $\alpha = 30^{\circ}$, so the angles of the triangle are $30^{\circ}, 60^{\circ}$, and $90^{\circ}$.

For $n=4$, from $\alpha + 2 \alpha + 3 \alpha + 4 \alpha = 360^{\circ}$, it follows that $\alpha = 36^{\circ}$, so the condition is satisfied by any quadrilateral with angles $36^{\circ}, 72^{\circ}$, $108^{\circ}$, and $144^{\circ}$.