Number of diagonals that cut the center of a regular polygon

Question

An $n$-sided regular polygon ($n \geq 3$) has, as we know, $\dfrac{n(n-3)}{2}$ diagonals. If $n$ is odd, then none of those diagonals pass through the center of it. If $n$ is even, then $\dfrac{n}{2}$ diagonal passes through its center.

That's the thing: how to prove it?

All I've found was: drawing a diagonal that passes through the center of a $n$-sided polygon, the other points are separated in two equal groups (because of its symmetry) of $p$ points. So the total of points is $p + p + 2=2p+2=2(p+1)$, which is even. But this doesn't really feel like proof to me, since it uses arguments based on nothing.

Answer 1

Another way to see this is to connect all the vertices to the center. The regular polygon would be divided into $n$ equal parts, and form an angle of $\dfrac {2\pi}n$ at the center.

For a diagonal to pass through the center, we need to show that the vertices involved, when connected to the center, forms one straight line. The angle formed at the center would then be $\pi$. This must also be equal to $k\left(\dfrac {2\pi}n\right)$ for some integer $k$.

Now it is clear that for odd $n$, $\dfrac {2k\pi}n \ne\pi$, or $2k \ne n$ for any $k$, so no diagonals satisfy this property.

For even $n$, the unique $k$ that satisfy this is $k=\dfrac n2$. Hence for each vertex, there is one unique "opposite" vertex that forms a diagonal passing through the center. Since we have $n$ vertices, we can draw $n$ diagonals. However we have double-counted, since each diagonal is formed by $2$ vertices. Hence the number of diagonals passing through the center in this case is precisely $\dfrac n2$.

Answer 2

Welcome to MSE. I think you have a proof, of half of it at least, though you might want to state it a bit more carefully.

Consider the line joining a vertex to the center. If it passes, through a second vertex, then every vertex lies on a diagonal passing through the center, by symmetry. Since there are two vertices on each such diagonal, there must be an even number of vertices.

What we haven't shown is that, in the case that $n$ is even, these $\frac n2$ diagonals actually exist. The only way I can see to do that is by construction, but that's easy.