Trajectory of light rays in a mirror polygon

Question

Given a general polygon and we are given a ray of light bouncing between the sides of the polygon where each side is a mirror. they hit at points $P_1,P_2...$, we define $\alpha_i$ to be the smaller of the angles between $P_iP_{i+1}$ and the side $P_i$ is on. Is it possible that $\alpha_1>\alpha_i$ for all $i>1$?
We assume the trajectory never touches a corner of the polygon.

We can notice the light ray can not have a periodic path since it would have the angle $\alpha_1$ again.
But in general, paths of light inside mirror polygons need not be periodic.

Answer 1

Consider this arrangement of $8$ mirrors (the thick black lines). The three horizontal mirrors are on $y=0,y=1,y=2$ and the three vertical mirrors are on $x=0,y=2,x=\beta $, where $\beta >2$ is irrational.

The red arrow shows a light ray making any angle $\alpha_1>45^o$ with mirror $L$. The direction of $L$ can be adjusted at will. The crucial thing is that the ray is at $45^o$ to the horizontal.

The ray then enters and is trapped in the blue region. The path is then aperiodic and then the ray will never escape back along the red path. Therefore all further $\alpha_i =45^o$.

PROOF

The ray first hits a mirror at $(3,0)$. Between this hit and subsequent hits on a horizontal mirror the ray will travel an integer distance in a vertical direction. It will of course travel the same integer distance in a horizontal direction.

When this distance is $D-3$, the $x$-coordinate of the hitting point will be $D$ if $D\le \beta$ but $2\beta -D$ if $\beta <D\le 2\beta$.

Therefore the $x$-coordinate of any hitting point will be either $D-2n\beta $ or $2n\beta -D$, for some integer $n$. It will never again be $D$.

Answer 2

In a square, it may not be possible, since $\alpha_i = 90-\alpha_{i-1}$ OR $\alpha_i = \alpha_{i-1}$.

In a equilateral triangle also it may not be possible, $\alpha_i = 120-\alpha_{i-1}$ OR $\alpha_i = 60+\alpha_{i-1}$.

In a general regular $n$-gon, $\alpha_i = \ell \frac{360}{n} + \alpha_{i-1}$ OR $\alpha_i = 180 - \ell \frac{360}{n} - \alpha_{i-1}$ for some $0 \leq \ell \leq n-1$.

Since the possibilities are finitely many, the angles will repeat. Although not sure if initial angle $\alpha_1$ will be part of the repeating angles in $\{\alpha_i\}$.

Based on this intuition: I suggest you try to construct polygon with irrational angles and rational angles or co-prime angles. In such a case you will atleast have aperiodic angles $\{\alpha_i\}$. May be try a triangle first.

Cheers man !