
2. A prism is given whose bases are two congruent convex n-gons. The number $v$ of vertices of the prism, the number $s$ of its face diagonals and the number $t$ of its space diagonals form, in a certain order, the first three terms of an arithmetic progression. For which $n$ this holds?

(Remark: By faces of the prism we mean both the bases and the lateral faces. A space diagonal is a segment connecting two vertices which do not lie in the same face.)


Solution. Any $n$-gonal prism has exactly $n$ vertices in each of its bases, thus $v=2 n$. From each vertex there come out $n-3$ diagonals lying in the base and two diagonals lying in the lateral faces; altogether this is $n-1$ face diagonals. From the $n$ vertices there thus come out $2 n(n-1)$ diagonals in total, however, each of them is counted twice; thus $s=n(n-1)$. Similarly, from each vertex there emanate $n-3$ space
diagonals (to all the vertices of the opposite base, except those three to which the given vertex is connected by an edge or a face diagonal); hence $t=2 n(n-3): 2=n(n-3)$.

We are thus looking for those $n \geqslant 3$ for which the numbers

$$
v=2 n, \quad s=n(n-1) \quad \text { and } \quad t=n(n-3)
$$

form, in a suitable order, a triple $x, y, z$ with the property $y-x=z-y$, or $y=\frac{1}{2}(x+z)$. An easy check reveals that for $n=3$ this is not the case (we get the triple $6,6,0$ ), while for $n=4$ we get the triple $8,12,4$ which has the required property ( $\operatorname{as} 8=\frac{1}{2}(4+12)$ ). For any $n \geqslant 5$ we have $n-1>n-3 \geqslant 2$, whence, upon multiplying by $n, s>t \geqslant v$; thus the desired arithmetic mean property must be $t=\frac{1}{2}(v+s)$. This gives the equation

$$
n(n-3)=\frac{2 n+n(n-1)}{2}
$$

whose only relevant root is $n=7$ (the other root $n=0$ has no sense).

Conclusion: the only possible values are $n=4$ and $n=7$.
