Think of a semicircle with a diameter AB. Now consider a light beam going off of point A and travelling towards the arc of the semicircle.
The ray does not bounce off straight lines tangent to the arc of the semi-circle.
Instead, the ray bounces off the arc of the semi-circle as if the arc was instead a straight line parallel to the diameter. The ray goes towards the diameter, and then bounces from the diameter back towards the arc.
If the light beam exits point A with an angle theta, hits the arc four times and ends up perfectly at point B, what angle would theta be?
Be aware that the sketch is not to scale! Because of symmetry, the second bounce off of the diameter would be in the center of the semicircle and all colored angles would be the same.
Best Answer
You are asking for the interior angle $\angle CAE$ of an isosceles triangle $ACE$ with $AC = CE$ whose base $AE$ is on diameter $AB$ and opposite vertex $C$ on semicircle $AB$, such that another similar triangle $\triangle EDO \sim \triangle ACE$ with $D$ on semicircle $AB$ has $O$ at the midpoint of $AB$.
As such, let $\theta = \angle CAE$ and let $x = AE$. Assume without loss of generality that $AO = 1$, thus $EO = 1-x$. But $\triangle AOC \cong \triangle EDO$ because $\angle CAO = \angle CAE = \theta$ and $AO = CO = 1$, hence $\angle ACO = \angle CAO = \theta$. It follows that $AC = 1-x$. But by similarity, $$\frac{AC}{AE} = \frac{DE}{EO},$$ or $$\frac{1-x}{x} = \frac{1}{1-x},$$ hence $$x = \frac{3-\sqrt{5}}{2}$$ and $$\theta = \cos^{-1} \frac{1-x}{2} = \cos^{-1} \frac{\sqrt{5}-1}{4} \approx 1.2566370614359172954$$ radians.