## Task 5 - 261245

Determine all natural numbers $n \geq 3$ for which the following statement holds:

Every plane convex $n$-gon $A_{1} A_{2} \ldots A_{n}$ is completely covered by the areas of the $n$ circles that have the segments $A_{i} A_{i+1}$ as diameters $\left(i=1,2, \ldots, n\right.$; set $A_{n+1}=A_{1}$).

Here, each $n$-gon area and each circle area are understood to include their boundary points.
I. For any triangular area $F=A_{1} A_{2} A_{3}$ and any convex quadrilateral area $F=A_{1} A_{2} A_{3} A_{4}$, the mentioned covering statement holds; this can be proven as follows:

If the statement were false, there would be a point $P$ in $F$ that lies outside each of the three or four mentioned

circles. Since these circles cover the boundary of $F$, $P$ would lie in the interior of $F$. Since $F$ is convex, the angles $\angle A_{i} P A_{i+1}\left(i=1, \ldots, n ; A_{n+1}=A_{1}\right.$, with $n=3$ or $\left.n=4\right)$ would have to satisfy

$$
\sum_{i=1}^{n} \angle A_{i} P A_{i+1}=360^{\circ}
$$

On the other hand, since $P$ lies outside the circles with diameters $A_{i} A_{i+1}$, $\angle A_{i} P A_{i+1}4$ there exists a convex n-sided area $A_{1} A_{2} \ldots A_{n}$ that is not covered by the mentioned circles; this is shown by the following example:

If $A_{1} A_{2} \ldots A_{n}$ is a regular n-gon and $P$ is its center, then (1) (now with $n>4$) holds and thus $\angle A_{i} P A_{i+1}=\frac{1}{n} \cdot 360^{\circ}<90^{\circ}$ for all $i=1, \ldots, n ; A_{n+1}=A_{1}$.

Therefore, $P$ lies outside all circles with diameters $A_{i} A_{i+1}$.

With I. and II: it is proven that the statement mentioned in the problem holds exactly for $n=3$ and $n=4$.

## Adopted from $[5]$