# Trajectories on a circular billiards table

This question is related to The case of Captain America's shield: a variation of Alhazen's Billard problem, but more focused. Let the unit disc in the plane be our billiards table, and let $C$ be the point $(c,0)$, where $c$ is a fixed real number between $0$ and $1$. How many different trajectories starting from $C$ return to $C$ after exactly $n$ bounces, where $n$ is a given positive integer?

Bouncing off either of the two points on the $x$-axis are always solutions, and reflecting a solution in the $x$-axis always give another solution, so it's enough to consider only those trajectories which begin by entering the upper half of the disc. Let the number of such trajectories be $f(n,c)$. I'm pretty sure $f(n,c)$ depends on $c$ in general (see below), so let's consider most particularly $f(n,0)=\lim_{c\to0} f(n,c)$ and $f(n,1)=\lim_{c\to 1} f(n,c)$.

With rapidly decreasing confidence I think we have the following values for small $n$.

$$f(1,c)=0,\\f(2,c)=1,\\f(3,c)=1,\\f(4,c)=3,\\f(5,0)=2,\,f(5,1)\approx 5.$$

For small $n$ it helps to think of matters this way: If $n$ is even then a trajectory returns to $C$ after $n$ bounces if and only if after $n/2$ bounces the trajectory is moving vertically, while if $n$ is odd then a trajectory returns to $C$ after $n$ bounces if and only if the $(n+1)/2$th bounce is at one of the two points on the $x$-axis. EDIT: This is bogus. See comments.

Any ideas?

• $c=1$ seems to be a very special case. Because of rotational symmetry of the circle, and the fact that angle of incidence equals angle of reflection, a path returning after $n-1$ bounces must actually be periodic with period $n$ (depending on whether you count the starting point as a bounce; I do.) The vertices must be equally spaced around the circle, and we get $f(n,1)=\phi(n)$ (or $\phi(n+1)$, depending.) This says $f(5,1)=4$ (or $2$) maybe you can prove me wrong.) – Tad Aug 22 '15 at 18:37
• The $n-1$ above should read $n$, sorry. – Tad Aug 22 '15 at 18:43
• @Tad Good point. I'm struggling to decide whether the cases $c=1$ and $c\approx 1$ are really the same, but already what you say shows that my paragraph starting "For small $n$..." is bogus, at least the "only if" parts. – Sean Eberhard Aug 22 '15 at 22:53
• Here's another idea that your comment suggests: Consider rotating a regular pentagon inscribed in the unit circle. For every position for which this regular pentagon passes through $C$ we get a valid trajectory which returns in $5$ bounces. For $c<\cos(2\pi/10)$ we get $0$ trajectories in this way, but for $\cos(2\pi/10)<c<1$ we get $2$ essentially different trajectories. Similarly for a regular pentagram, with two other transition values of $c$ ($c=\cos(2\pi/5)$ and whatever the distance from $0$ to a vertex of the inner pentagon is). This doesn't seem to be the whole story though. – Sean Eberhard Aug 22 '15 at 23:04