# Counting Periodic Orbits on a regular Hexagon

An orbit on a polygon is a path that a "billiards ball" (a point) would follow if it obeyed Snell's law of reflection (the angle of incidence is equal to the angle of reflection). A periodic orbit is an orbit that returns to the same point with the same initial angle (we count the smallest such number so as to avoid repeats). In this case, the period of a periodic orbit is the number of edge-hitting events.

My question is this:

Is anyone aware of a (possibly classical) result with proof that tells us what the period of a periodic orbit on a regular hexagon would be given either the initial angle or a vector that defines the angle?

• You have no idea how long I have been waiting for someone to ask a polygonal billiards question. (It's my area of research) – Alex Becker Apr 5 '14 at 7:38
• I would suggest to define the period rather as number of edge-hitting events, not the number of edges hit (edges may be hit several times during one period). – Hagen von Eitzen Apr 5 '14 at 7:39

Let's let the bottom of the (unit) hexagon lie along the $x$-axis. Let $v$ be the initial vector of the billiard ball, and write $$v=a\begin{pmatrix}\sqrt{3}/2\\1/2\end{pmatrix}+b\begin{pmatrix}0\\1\end{pmatrix}$$ These vectors are chosen because they are a basis for the lattice of hexagons in the plane. The trajectory is periodic iff $a/b$ is rational, in which case we can scale $v$ such that $a,b\in\mathbb Z$ and $\gcd(a,b)=1$, and the period length is then just $\|v\|$. The number of edges hit is $a+b$.