Suppose you have a random point inside a square and you set the point moving in a random direction. When the point hits a side of the square, it bounces off the side like a billiard ball, i.e. angle in $=$ angle out.

Will the point always end up in a fixed pattern, i.e. will it always come to a position it has been at before, moving in the same direction it did at that earlier time?

I suspect the answer is No, based on computer simulations, but then computers have limited accuracy. It is clear that there are many cases where a fixed pattern does occur. One example is if the initial angle of movement $\theta$ is a multiple of $45^\circ$. In fact, it seems to me that if $\theta = \text{arctan}(\frac{1}{n})$, where $n$ is a positive integer, a fixed pattern will occur.

But will a fixed pattern occur for any starting point and angle? If not, why not?

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    $\begingroup$ Often with these problems about reflection you can instead ask an equivalent question about a straight line traveling through an infinite grid of squares $\endgroup$ – Zubin Mukerjee Sep 26 '18 at 20:45
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    $\begingroup$ If the gradient of the initial trajectory is rational, it will repeat. Otherwise, not. $\endgroup$ – TonyK Sep 26 '18 at 20:48
  • $\begingroup$ @TonyK: Sounds good. Could you show this is the case in an answer? $\endgroup$ – Jens Sep 26 '18 at 20:51
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    $\begingroup$ A fact that is closely related: any irrational number $r$ will have integer multiples arbitrarily close to integers, but none that exactly equal integers (you can use this fact to show that irrational slope lines will never reach a fixed repeating sequence of crossing through squares, because the line will eventually always reach a point where it's closer to a corner than it ever was before) $\endgroup$ – Zubin Mukerjee Sep 26 '18 at 20:51

A fixed pattern occurs if and only if the slope of the first direction is rational. Otherwise, it doesn't occur, and moreover, you can prove that the trajectory is dense in the square (which means that it approaches every point of the square with every precision you can choose).

  • $\begingroup$ Thanks. But could you explain why this is the case? $\endgroup$ – Jens Sep 26 '18 at 20:58
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    $\begingroup$ @Jens Going off of Zubin's comment, look at the infinite unit grid and say your line has equation $y=ax+b$ (vertical lines are trivial). A necessary (and sufficient) condition for a pattern to occur is that there exists some positive integer $k$ such that $kb$ is also an integer. This is only possible when $b$ is rational. $\endgroup$ – N.Bach Sep 26 '18 at 21:09
  • $\begingroup$ @N Bach: Got it. So simple. Thanks! $\endgroup$ – Jens Sep 26 '18 at 21:16

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