## Task $3 / 83$

For which natural numbers $n$ does the area $A_{2 n}$ of the regular $2 n$-gon equal twice the area $A_{n}$ of the regular $n$-gon with the same circumradius?
Let $r$ be the circumradius. Then for the area $A_{k}$ of the regular $k$-gon,

$$
A_{k}=\frac{k}{2} r^{2} \sin \frac{2 \pi}{k}
$$

and for the ratio $\frac{A_{2 n}}{A_{n}}=2$

$$
\frac{A_{2 n}}{A_{n}}=2=\frac{\frac{2 n}{2} r^{2} \sin \frac{2 \pi}{2 n}}{\frac{n}{2} r^{2} \sin \frac{2 \pi}{n}}=2 \frac{\sin \frac{\pi}{n}}{\sin \frac{2 \pi}{n}}=\frac{2 \sin \frac{\pi}{n}}{2 \sin \frac{\pi}{n} \cos \frac{\pi}{n}}=\frac{1}{\cos \frac{\pi}{n}}
$$

thus $\cos \frac{\pi}{n}=\frac{1}{2}$ holds for $n \in \mathbb{N}$. Therefore, $n=3$ is the only solution. Indeed,

$$
A_{3}=\frac{3}{2} r^{2} \sin \frac{2 \pi}{3} \quad ; \quad A_{6}=\frac{6}{2} r^{2} \sin \frac{2 \pi}{6}=2 A_{3}
$$

since $\sin \frac{2 \pi}{6}=\sin \frac{\pi}{3}=\sin \frac{2 \pi}{3}$.