## Task 1 - 090811

Investigate whether there are polygons with any of the following properties:

a) The number of diagonals is three times the number of vertices.

b) The number of vertices is three times the number of diagonals.
The formula for the number $z$ of diagonals in a convex n-gon is: $z=\frac{n(n-3)}{2}$.

Justification: From each vertex, there is a diagonal to those vertices that are different from the mentioned vertex and are not adjacent to it. Therefore, exactly $(n-3)$ diagonals emanate from each vertex.

If one adds up these numbers of diagonals formed for all $n$ vertices, each diagonal of the n-gon has been counted exactly twice. Therefore, $2z = n(n-3)$, from which the claimed formula follows.

a) Suppose there is an n-gon of the mentioned type. Then it holds: $z=3n \rightarrow 3n=\frac{n(n-3)}{2} \rightarrow 9n=n^2$ From this, due to $n \neq 0: n=9$.

Conversely: For $n=9$, we get $z=27=3n$. Therefore, all nonagons and only these have property a).

b) Suppose there is an n-gon of the mentioned type. Then it holds:

$$
z=\frac{n}{3} \rightarrow \frac{n}{3}=\frac{n(n-3)}{2} \rightarrow 11n=3n^2
$$

Therefore, there is no n-gon with property b), because the last condition is not satisfied for any positive integer $n$.