## Task 22/70

If one connects the adjacent midpoints of the sides of a regular $n$-gon, a new regular $n$-gon similar to the original one is formed. Let $\Delta A=A-A^{\prime}$ be the difference in the areas of the two $n$-gons.

For which $n$ does $\triangle A=\frac{A}{10}$ hold?
From $\Delta A=A-A^{\prime}\frac{9}{10}$. Furthermore, it holds that: $A=$ $\frac{n}{2} r^{2} \sin \frac{360^{\circ}}{n}$ (where $r$ denotes the circumradius) and

$$
A^{\prime}=\frac{n}{2} r^{\prime 2} \sin \frac{360^{\circ}}{n} \quad \text { so } \quad \frac{A^{\prime}}{A}=\frac{r^{\prime} 2}{r^{2}}>\frac{9}{19}
$$

Since $r^{\prime}=r \cos \frac{360^{\circ}}{n}$, it follows that $\cos ^{2} \frac{360^{\circ}}{2 n}>\frac{9}{10}, \cos \frac{360^{\circ}}{2 n}>0.9487,$ (since $n \geq 3$, $\cos \frac{360^{\circ}}{2 n}>0$), thus $\frac{360^{\circ}}{2 n}<18.5^{\circ}$.

This finally yields $n \geq 10$ (due to the integer nature of $n$).