Team Selection Test for EGMO 2019

By angle chasing, we have $$\begin{array}{rl}\angle BAC+\angle BPC& =2\angle BAC+\angle ABP+\angle ACP\\ & =2(\angle BAC+\angle ABQ+\angle ACQ)=2\angle BQC=180^{\circ }.\end{array}$$ Let $R$ be a point on $[PD]$ such that $PD=DR$. Since $BD=DC$ and $PD=DR$, we conclude that $BPCR$ is a parallelogram. Therefore $\angle BRC=\angle BPC=180^{\circ }-\angle BAC$ which means that $BACR$ is a cyclic quadrilateral. Since $BD=DC$, the areas of the triangles $ABR$ and $ACR$ are equal. We also have $\angle ABR+\angle ACR=180^{\circ }$ and hence we get $$AB\cdot BR=AC\cdot CR$$ which shows that $$\frac{AB}{AC}=\frac{CR}{BR}=\frac{BP}{CP}.$$

Let $Q^{\prime }=BQ\cap AP$. (Notice that since the triangle $ABC$ is acute, $Q^{\prime }$ must be on the line segment $[AP]$.) By bisector theorem, we get $$\frac{AC}{CP}=\frac{AB}{BP}=\frac{A{Q}^{\prime }}{P{Q}^{\prime }}$$ and hence $C{Q}^{\prime }$ should also be bisector of the angle $ACP$, i.e., $C,Q,Q^{\prime }$ are collinear. This is possible only if $Q\equiv Q^{\prime }\in [AP]$.