TITLE: BdMO 2016 National Junior Question 8.
QUESTION [10 upvotes]: $\triangle ABC$ is isosceles with $AB = AC$ ans $\angle A = 20^{\circ}$ and $BC = 12$.  $BP \perp AC$ ans $Q$ is a point on $AB$ such that $QB=6$. Find $\angle CPQ$.  

I was not able to solve this problem. Tried it many times. It can be solved very easily if I use trigonometric functions and calculator. But I need to solve it in Olympiad Math's way. The Diagram is something like this -

REPLY [8 votes]: A short solution from the construction suggested by @Fimpellizieri in the comments.
Let $X\in AP$ such that $\Delta CBX$ is isosceles, and $Y\in AB$ such that $XY\bot AB$. Then $\angle XBY=\angle CBA-\angle CBX=80^\circ-20^\circ=60^\circ$, and, hence, $|BY|=12\cdot\cos 60^\circ=6$, i.e. $Y=Q$.

Now consider the quadrilateral $BQXP$. Since $\angle BQX=\angle BPX=90^\circ$, all four points lay on the same (circumscribed) circle. From the inscribed angles theorem it follows that $\angle BPQ=\angle BXQ$, but the latter is $90^\circ-60^\circ=30^\circ$. Finally, $\angle CPQ=90^\circ+30^\circ=120^\circ$.