Harvard-MIT Mathematics Tournament, Team Round A

Solution:

Let $O$ be the center of the $n$-gon. Let us consider two cases, based on the parity of $n$:

- $n$ is odd. In this case, for each diagonal $d$, there is exactly one vertex $D$ of the $n$-gon, such that $d$ is perpendicular to line $O D$; and of course, for each vertex $D$, there is at least one diagonal $d$ perpendicular to $O D$, because $n \geq 5$. The problem of picking a bunch of $d$'s so that no two are parallel is thus transmuted into one of picking a bunch of $d$'s so that none of the corresponding $D$'s are the same. Well, go figure.

- $n$ is even. What can I say? For each diagonal $d$, the perpendicular dropped from $O$ to $d$ either passes through two opposite vertices of the $n$-gon, or else bisects two opposite sides. Conversely, for each line joining opposite vertices or bisecting opposite sides, there is at least one diagonal perpendicular to it, because $n \geq 6$. By reasoning similar to the odd case, we find the answer to be $n$.