## Task 4 - 130934

In a plane, regular n-gons (with a uniform number of vertices) are to be laid around a vertex so that the sum of the sizes of the interior angles lying at this vertex is $360^{\circ}$.

Give all natural numbers $n$ for which this is possible; in each case, also give the total number of n-gons required!
Let $n \geq 3$. Then the interior angles of a regular $n$-gon are each $\frac{n-2}{n} \cdot 180^{\circ}$. For $n=3 ; 4 ; 5$ and 6, this results in interior angles of $60^{\circ}, 90^{\circ}, 108^{\circ}$, and $120^{\circ}$.

Obviously, since $\frac{n-2}{n}=1-\frac{2}{n}$, the size of the interior angles is a strictly monotonically increasing function in $n$, so for all $n > 6$, the interior angles of a regular $n$-gon are greater than $120^{\circ} = \frac{1}{3} \cdot 360^{\circ}$, but because $\frac{n-2}{n} < 1$, no number of regular $n$-gons can be arranged without overlap at a vertex. Similarly, for $n=5$, $\frac{360^{\circ}}{108^{\circ}}=\frac{10}{3} \notin \mathbb{Z}$. Thus, only $n=3 ; 4$ and 6 remain, where 6, 4, and 3 regular $n$-gons, respectively, can be arranged without overlap at a vertex.