BdMO 2016 National Junior Question 8. 
$\triangle ABC$ 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 -

 A: 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, $\angle CPQ=90^\circ+30^\circ=120^\circ$.
A: Let us consider $\triangle APB$:
$$\angle ABP = 180° - (\angle APB + \angle PAB)$$
$$\angle ABP = 70°$$
Let us consider $\triangle BPQ:$
Let $\alpha=\angle BPQ$
$$ \dfrac{\sin{\alpha}}{BQ}= \dfrac{\sin{70°}}{QP}$$
$$ \sin{\alpha}= \dfrac{\sin{70°}}{QP}BQ$$
$$ \sin{\alpha}= 6\dfrac{\sin{70°}}{QP}$$
Now let us find $QP$
Let us consider $\triangle APQ:$
$$QP^2 = AP^2 + AQ^2 - 2\cdot AP\cdot AQ \cdot \cos(\angle PAQ)$$
Now let us find $AP \text{ and } AQ$
Let us consider $\triangle BPC:$
$$ \dfrac{\sin{\angle PBC}}{PC}= \dfrac{\sin{\angle PCB}}{BP}= \dfrac{\sin{\angle BPC}}{BC}$$
$$ BP = \dfrac{\sin{\angle PCB}}{\sin{\angle BPC}}\cdot BC$$
$$ BP = 11.82$$
$$ PC = \dfrac{\sin{\angle PBC}}{\sin{\angle PCB}}\cdot PB$$
$$ BP = 2.08$$
Let us consider $\triangle ABP\text{ to find } AP$
$$ \dfrac{\sin{\angle BAP}}{BP}= \dfrac{\sin{\angle ABP}}{AP}$$
$$ AP= \dfrac{\sin{\angle ABP}}{\sin{\angle BAP}}BP$$
$$ AP= 32.48$$
$P \in [AC] \Rightarrow AC = AP + PC \Leftrightarrow AC = 34.56$
Since $\triangle ABC \text{ is isosceles, then } AB = AC = 34.56$
$Q \in [AB] \Rightarrow AQ = AB - BQ \Leftrightarrow AQ = 28.56$
$$QP^2 = AP^2 + AQ^2 - 2 AP \cdot AQ \cdot \cos(\angle QAP)$$
$$QP = 11.28$$
Now let us consider $\triangle BPQ$
$$ \dfrac{\sin{\angle \alpha}}{BQ}= \dfrac{\sin{\angle QBP}}{QP}$$
$$ \sin{\alpha}= \dfrac{\sin{\angle QBP}}{QP}BQ$$
$$ \sin{\alpha}= \dfrac{1}{2} \Rightarrow \alpha = 30°$$
$$\angle CPQ = \angle CPB + \angle BPQ$$
$$\angle CPQ = 90° + 30°$$
$$\angle CPQ = 120°$$
A: Since $\angle BPC = 90^{\circ}$, it follows that $P$ lies on the circle whose diameter is $BC$. As such, let $D$ be the midpoint of $BC$, and let $E$ be a point inside $\Delta ABC$ such that $\Delta BDE$ is equilateral. Then $E$ also lies on the circle with diameter $BC$, so quadrilateral $BEPC$ is cyclic. Thus, $\angle BEP$ and $\angle PCB$ are supplementary, and since $\angle PCB = 80^{\circ}$, it follows that $\angle BEP = 100^{\circ}$. Now, $BC = 12$, so $BE = BD = \frac{1}{2}BC = 6$. Thus, $BE = 6 = QB$, so $\triangle QBE$ is isosceles. Since
$$\angle QBE = \angle QBC - \angle EBC = 80^{\circ} - 60^{\circ} = 20^{\circ}$$
it follows that $\angle QEB = 80^{\circ}$. Thus, $\angle QEB$ and $\angle BEP$ are supplementary, so $E$ lies on line $QP$. As such, we have
$$\angle CPQ = \angle CPE$$
and $\angle CPE = 120^{\circ}$, either by noting that it subtends an arc of $240^{\circ}$, or by noting that it is supplementary to angle $\angle EBC = 60^{\circ}$.
