74th Romanian Mathematical Olympiad

The projection of the line $CP$ on the plane $(BAD)$ is the line $OC$, hence the angle of the line $CP$ with the plane $(BAD)$ is $\angle OCQ$. From $CN\perp MN$, $CN\perp VO$ and $MN\cap VO=\{O\}$ follows $CN\perp (VMN)$. So, the projection of the line $CP$ on the plane $(VMN)$ is the line $NQ$, hence the angle of the line $CP$ with the plane $(VMN)$ is $\angle CQN$. If $\angle OCQ=45^{\circ }$, then $CQ=\sqrt{2}\cdot OC$ and, since $CN=\frac{\sqrt{2}}{2}\cdot OC$, the leg $CN$ of the right triangle $CQN$ is half of the hypotenuse $CQ$, therefore $\angle CQN=30^{\circ }$. Conversely, if $\angle CQN=30^{\circ }$, then $CQ=2CN=\sqrt{2}\cdot OC$, so $\angle OCQ=45^{\circ }$.