Suppose I have a square with sides $a$, $b$, $c$, $d$. Now I shoot a ray from the vertex of the square formed by the sides, let's say $a$ and $b$ at angle $50^{\circ}$ from $a$. If I let the ray bounce off the sides of the square continuously, how many times would the ray intersect itself through exactly the same point in the square? Could you generalise this result to any angle $\alpha$?
Bouncing ray inside a square
analytic geometryeuclidean-geometrygeometry
Related Solutions
In a possible attempt to explain a), let us focus solely on a single angle, say angle $A$. Similarly, draw tangent lines extending from the two adjacent sides, namely $AB$ and $AD$. Assuming $A\neq180$ (which we can, because it would cause $ABCD$ to be a triangle), $AB$ and $AD$ are not parallel.
This means that they meet at $A$ and continue, getting further apart as they go. If $A \lt 180$, meaning $ABCD$ is convex, $AB$ and $AD$ continue away from the shape, not intersecting any sides.
However, if $A \gt 180$, $AB$ and $AD$ enter the interior or $ABCD$ after intersecting at $A$. As the lines are infinite and the quadrilateral is not, the lines must at some point leave the shape. As two lines can only meet at a single point, and will not intersect themselves, they must leave the shape through one of the other two sides (Note Pasch's Theorem).
As both $AB$ and $AD$ are equally dependent on the angle of $A$, it is not possible for only one of the two lines to split one of the other sides.
When the ball rebounds on an edge, reflect the entire board on that edge. Then the path of the ball is the straight line extension to the reflected board. Thus when the ball lands in a corner of the board, the number of horizontal reflections $m$ and the number of vertical reflections $n$ must satisfy $297 m = 165 n$. Thus $m=5, n=9$. The ball lands in a corner after $14$ rebounds.
Best Answer
Look at this figure, think about the symmetry of reflections at each wall:
You will get termination whenever $\theta = \tan^{-1} a/b$ where $(a,b) \in \mathbb{Z}$ and the number of reflections is $a+b$.
Conversely, you will never get termination if the $\theta = \tan^{-1} c$ where $c \notin \mathbb{Q}$, and thus the ray will intersect itself an infinite number of times.