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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.enter image description here

Which will eventually gives a trajectory looking something like this:enter image description here

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 ?

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  • $\begingroup$ HINT: you mention that you already have an important result on $\pi\mathbb{Q}$, maybe you could use that? $\endgroup$ – WNG Sep 3 '16 at 23:27
  • $\begingroup$ Yes I could, but I don't know how to link that result to my question... It's one thing to have a dense trajectory, it's another to have a dense trajectory avoiding a given set $\endgroup$ – E. Joseph Sep 4 '16 at 8:42
  • $\begingroup$ Reading your question, I assumed you had earlier demonstrated that $\frac \theta \pi\notin \mathbb Q$. HINT: Do you know what would happen to such a $\frac \theta \pi\in \mathbb Q$ ? Even more fundamentally, if I give you the start and the first rebound of a trajectory, do you have a formula to determine the second? The $n^{th}$? $\endgroup$ – WNG Sep 4 '16 at 14:02
  • $\begingroup$ Like I said in the question, we don't care for the case $\theta\in \pi\mathbb Q$, because the trajectory is finite so it is not really interesting here. And yes, I know the formula but it's not helping me because the dense set to avoid could be awfully complicated. I think a more theoretical proof is required here. $\endgroup$ – E. Joseph Sep 4 '16 at 14:06
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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 $(x_i)$

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    $\begingroup$ I believe the quantification in (3), (2), and (1) is universal. You have proven a different statement, as if it were existential. It is easy to construct countable sets no two elements of which are $\pi \Bbb{Q}$ commensurable, for instance $\{0, \sqrt{2}\}$. $\endgroup$ – Eric Towers Sep 4 '16 at 19:13
  • $\begingroup$ I had never considered this interpretation of the question. You're probably right, which is a whole other story to demonstrate $\endgroup$ – WNG Sep 4 '16 at 19:19
  • $\begingroup$ Like @EricTowers said, the quantification is universal, otherwise the question is trivial like you said and the demonstration is in the question... Sorry if the way I said it misled you. $\endgroup$ – E. Joseph Sep 4 '16 at 19:23
  • $\begingroup$ OK I added the demonstration for the new interpretation, I used some of the same notations, so I did not delete my first proof. I tried to clarify the two parts in my answer. $\endgroup$ – WNG Sep 4 '16 at 19:55

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