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?

  • $\begingroup$ You have no idea how long I have been waiting for someone to ask a polygonal billiards question. (It's my area of research) $\endgroup$ – Alex Becker Apr 5 '14 at 7:38
  • $\begingroup$ 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). $\endgroup$ – Hagen von Eitzen Apr 5 '14 at 7:39

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.

  • $\begingroup$ Similar arguments work for rectangles and equilateral triangles. $\endgroup$ – Hagen von Eitzen Apr 5 '14 at 7:40
  • $\begingroup$ @HagenvonEitzen As well as the right and 30-60-90 triangles. In fact all those cases are nicer because (as I addressed in my edit) the hexagon is not regularizable, so there is a disconnect between the algebraic and geometric picture. $\endgroup$ – Alex Becker Apr 5 '14 at 7:44
  • $\begingroup$ Thank-you! I'm actually trying to study and periodic orbits on all shapes that tessellate the plane and see what interesting connections I can find. This was a really helpful insight! $\endgroup$ – illysial Apr 5 '14 at 7:47
  • $\begingroup$ @illysial Tessellating the plane on its own isn't very helpful--the key thing that makes these cases easy is the fact that they do so using only reflections. $\endgroup$ – Alex Becker Apr 5 '14 at 7:48
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    $\begingroup$ @illysial They all behave basically the same way as the hexagon. The non-regularizability issue results in certain complications for the last 4 in your list, but determining period lengths doesn't change. $\endgroup$ – Alex Becker Apr 5 '14 at 8:01

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