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?
In general these problems are extremely difficult, but we are in luck with the regular hexagon since it tiles the plane under reflection.
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 $a+b$.
Note for experts: While the regular hexagon tiles the plane under reflection, it is not a regularizeable system because the tiling breaks down if we take orientation into account. This means that slightly perturbing the billiard table, even if it preserves the periodicity of a trajectory, may double the period length.