Team Selection Test for EGMO, 2019

By angle chasing, we have
$$\begin{aligned} \angle BAC + \angle BPC &= 2\angle BAC + \angle ABP + \angle ACP \\ &= 2(\angle BAC + \angle ABQ + \angle ACQ) = 2\angle BQC = 180^{\circ}. \end{aligned}$$
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 $Q' = BQ \cap AP$. (Notice that since the triangle $ABC$ is acute, $Q'$ must be on the line segment $[AP]$.) By bisector theorem, we get
$$\frac{AC}{CP} = \frac{AB}{BP} = \frac{AQ'}{PQ'}$$
and hence $CQ'$ should also be bisector of the angle $ACP$, i.e., $C, Q, Q'$ are collinear. This is possible only if $Q \equiv Q' \in [AP]$.