A perfectly elastic particle is projected with velocity v at an elevation $\theta$ . A smooth plane passes through the point of projection and is inclined at an angle $\alpha$  to the horizontal. The particle will return to the point of projection provided $\cot \alpha \cot \left(\theta -\alpha \right)$ is:

 

The velocity parallel to the plane is unaltered by the impacts, so that the distance described parallel to the plane will be zero at the end of a time t given by :

$0=vt\cos (\theta -\alpha )-\frac{\left(g\sin \alpha \right)t^2}{2}\Rightarrow t=\frac{2v\cos (\theta -\alpha )}{g\sin \alpha }$

Also, since the elasticity is perfect, the velocity perpendicular to the plane is just reversed at each impact. The time of flight for each trajectory is thus twice the time in which the velocity $v\sin (\theta -\alpha )$ is destroyed by $g\cos \alpha ,$ and thus $t=\frac{2v\sin (\theta -\alpha )}{g\cos \alpha }$

Clearly the particle will return to the point of projection if the first of these is some multiple, n, of the second, i.e., if $\frac{2v\cos (\theta -\alpha )}{g\sin \alpha }=n\frac{2v\sin (\theta -\alpha )}{\cos \alpha }$

i.e., if $\cot \alpha .\cot (\theta -\alpha )$ is an integer