## Task 4 - 200924

a) Determine all natural numbers $n \geq 3$ for which the following statement is true!

"Every prism that has a convex n-gon as its base has exactly 20n diagonals."

b) Determine, for each prism for which the statement in a) is true, the number of face diagonals and the number of space diagonals!

Hint: An n-gon is called convex if and only if each of its interior angles is less than $180^{\circ}$.
For the "diagonals" of a prism, all line segments between any two of its vertices that are not edges of the prism are referred to as such.

a) For each of the $2n$ vertices of a prism with a convex $n$-gon as its base, it holds that from each vertex, edges extend to exactly three other vertices of the prism. Therefore, it is the endpoint of exactly $2n - 3 - 1 = 2n - 4$ diagonals. (From this vertex, there is no diagonal to the three vertices to which it is connected by an edge, nor to itself.)

Since each diagonal connects exactly two vertices, there are thus exactly $\frac{1}{2} \cdot 2n \cdot (2n - 4) = n \cdot (2n - 4)$ diagonals in such a prism with a convex $n$-gon as its base. For this to be exactly $20n$, it must hold that $2n - 4 = 20$, i.e., $n = 12$.

Since no assumptions were made about the shape of the prism, exactly all prisms whose base is a convex 12-gon satisfy the statement.

b) Space diagonals are those among the diagonals that do not lie in any side face of the prism. This means that they are bounded by one vertex of the base and one vertex of the top face, which do not lie in a common side face of the prism.

For each of the $n$ vertices of the base, it holds that it is connected by an edge to exactly one vertex of the top face and lies in a common side face with exactly two more vertices. Therefore, from each vertex of the base, there are exactly $n - 3$ space diagonals, so there are a total of $n \cdot (n - 3)$ space diagonals, which in the specific case with $n = 12$ means exactly $12 \cdot 9 = 108$ space diagonals.

The remaining $20n = 240$, i.e., 132 (or generally $n \cdot (2n - 4) - n \cdot (n - 3) = n \cdot (n - 1)$), are then the face diagonals.