Billiard problem variation

Question

Suppose a square table. You are shooting from a corner. A ball ends its path upon landing in a corner. At what angles can you shoot a ball from your corner such that the ball will never come back to your corner?

Edit: seemingly my question would be how can you prove that if the trajectory is periodic, how can you prove that it will encounter another "corner" point before it returns to itself?

Does this answer your question? The forever moving billiard ball Or this https://plus.maths.org/content/billiards-donuts — Ethan Bolker, Oct 14 '23 at 19:42
It seems slightly different, a closed loop is formed for $\tan(\theta) \in \mathbb{Q}$ but the solution set allows for angles like $\frac{\pi}{4}, \frac{\pi}{2}, 0$. So seemingly the answer would be all angles. — redrobinyum, Oct 14 '23 at 19:57

score 0 · Answer 1 · answered Oct 14 '23 at 19:42

0

Imagine replicating your square billiard table over the whole plane. Then "bounces" and "crossing into the next billiard table" become more or less equivalent (there's some flipping-over to consider).

So this question becomes "which rays from the origin contain no point of the integer grid?"

You should be able to turn that into a statment about angles, I believe.

answered Oct 14 '23 at 19:42

John Hughes

93,729

If the ray does contain a lattice point, the first such point will represent one of the other corners. The ball ends there and never returns to you corner. – Daniel Mathias Oct 14 '23 at 20:04
How can you prove that the lattice point the ray encounters does not map to the starting point? – redrobinyum Oct 14 '23 at 20:30

Billiard problem variation

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