About trajectories on a circular billiard

Question

Let consider a circular billiard.

You start by putting the ball on the edge off the table. The trajectory then is quite simple (see the picture), with equal angles at each rebound.

Which will eventually gives a trajectory looking something like this:

Let's call the angle between the center of the billiard, the starting position and the first rebound $\theta$.

We already know that if $\frac \theta \pi\notin \mathbb Q$ then the trajectory is dense on the edge of the billiard (that we call $\mathcal C$). We assume that $\frac \theta \pi\notin \mathbb Q$ from now on.

My three questions are:

(1) If $(x_i)_{i=1}^n$ is a finite family on $\mathcal C$, does it exist a dense trajectory which avoid every $x_i$ ?

(2) If $(x_i)_{i=1}^\infty$ is a countable family on $\mathcal C$ which is discrete on $\mathcal C$, does it exist a dense trajectory which avoid every $x_i$ ?

(3) If $(x_i)_{i=1}^\infty$ is a countable family on $\mathcal C$, does it exist a dense trajectory which avoid every $x_i$ ?

We obviously have $$(3)\Rightarrow (2)\Rightarrow (1).$$

I believe that (1) is true, I think (2) is true too, and I don't know about $(3)$ (but I would like to think that it is !).

What are your thoughts on the subjects ? How would your prove such a result ?

Answer 1

1. Every trajectory avoids a dense family of $(x_i)$, and this set was stated in your question, because it is $\pi \mathbb Q$. This is why I insisted to know what you had already demonstrated.

For every trajectory, let's set $x_0$ as the origin of the circle, and associate each $x_i$ with its polar angle.

Thus, $x_0=0$, $x_1 \in [0,2\pi[$, and more importantly, because the billiard is round :

$\forall i \in \mathbb{N} \quad x_i=ix_1$ mod $2\pi$

This equality yields that the trajectory will come back to its starting point iff $x_1/\pi \in \mathbb Q$, that we note $x_1 \in \pi\mathbb Q$ (because $\exists i,n \in \mathbb N^2$ so that $ix_1=2n\pi$ iff $\frac {x_i}{\pi}=\frac{2n}{i} \in \mathbb Q$)

Also, we know that $\mathbb{Q}$ and thus $\pi\mathbb Q$ is dense in $\mathbb R$ and thus in $[0,2\pi[$, so its image by mod $2\pi$ is dense.

Thus every non-repeating trajectory avoids the family $\pi\mathbb Q$ mod $2\pi$ which is a denombrable, dense family on $\mathcal C$, which demonstrates $(3)$

By the way, every repeating trajectory has a finite set of values, so they also avoid a dense set of points, for example they all avoid the points $\mathbb Q$ mod $2\pi$

2. Every countable family is avoided by a dense trajectory:

As some comments point out, I might have misunderstood the question. So, if the point is to demonstrate that for any countable family $(x_i)$, a non-repeating suite exists, the approach would be to exhibit such a trajectory $(y)$ using the following argument :

Let $(y)$ be a trajectory, it will encounter an $x_i$ iff

$\exists k,i,n \in \mathbb N^3, ny_1$ mod $2\pi=x_i $ mod $2\pi\iff y_1=\frac{x_i+2k\pi}{n}$

As $\mathbb N^3$ is countable, the set of all $y_1$ for which $(y)$ encounters an $x_i$ is countable. As $[0,2\pi[$ is not, we can always find a $\hat y$ that is neitherin this set, nor in $\pi\mathbb{Q}$.

Then the $(y)$ trajectory defined by $y_0=0$ and $y_1=\hat y$ is non-repeating, and will avoid all $(x_i)$