I was doing a practice exam for maths and I got the following geometry question: A beam of light reflects against two mirrors with an adjustable angle $\alpha$ between them. For which value of $\alpha$ is the doubly reflected beam of light perpendicular to the source beam?
I named the initial angle of incidence $\beta$ (bottom left) and the final angle of reflection $\gamma$ (top right). The total difference between the incident beam and the reflected beam is then equal to $\theta_{total} = \beta + \pi - \alpha + \gamma$ measured anticlockwise from the initial incident beam.
If the incident and reflected beams must be perpendicular, then $\theta_{total} \in \left\{ \frac{\pi}{2}, \frac{3\pi}{2} \right\}$, but since the light beam cannot magically fly back into the mirror, the only option here is $\frac{3\pi}{2}$. Which means that $\beta + \pi - \alpha + \gamma = \frac{3\pi}{2}$. Moreover, using the sum of the interior angles of a triangle, $\gamma = \pi - \alpha - \beta$ and therefore $\theta_{total} = \beta + \pi - \alpha + \left(\pi - \alpha - \beta\right) = 2\cdot{\left(\pi - \alpha\right)}$.
Solving for $\alpha$ when $\theta_{total}=\frac{3\pi}{2}$ gives $\alpha = \frac{\pi}{4}$, but this does not work with the given illustration where it's implicitly assumed that $\alpha > \frac{\pi}{2}$ (all the multiple choice answers were also greater than $\frac{\pi}{2}$).
The correct answer is $135°$ or $\frac{3\pi}{4}$, where is the fault in my reasoning?
EDIT: Why I chose $\theta_{total} = \frac{3\pi}{2}$.
I tried to mentally connect the incident and reflected beam at a point and considered the total angular difference. I figured it would look something a bit like this, hence the $270°$ and not just $90°$.
EDIT 2: The image above actually made me realise I ignore the whole $180°$ traversed when going from the left side of the horizontal to the right before sweeping past $\gamma + \left(\pi - \alpha\right)$. That was the fault in my reasoning.
The correct expression would then be $\theta_{total} = 3\pi - 2\alpha$, in which case I do get the correct value of $\alpha = \frac{3\pi}{4}$.