APMO 2016

For every point $D$ on the side $BC$, let $D^{\prime }$ be the reflection of $D$ in the line $PQ$. We will first prove that if the triangle satisfies the condition then it is isosceles and right-angled at $A$.

Choose $D$ to be the point where the angle bisector from $A$ meets $BC$. Note that $P$ and $Q$ lie on the rays $AB$ and $AC$ respectively. Furthermore, $P$ and $Q$ are reflections of each other in the line $AD$, from which it follows that $PQ\perp AD$. Therefore, $D^{\prime }$ lies on the line $AD$ and we may deduce that either $D^{\prime }=A$ or $D^{\prime }$ is the second point of the angle bisector at $A$ and the circumcircle of $ABC$. However, since $APDQ$ is a cyclic quadrilateral, the segment $PQ$ intersects the segment $AD$. Therefore, $D^{\prime }$ lies on the ray $DA$ and therefore $D^{\prime }=A$. By angle chasing we obtain $$\angle P{D}^{\prime }Q=\angle PDQ=180^{\circ }-\angle BAC$$ and since $D^{\prime }=A$ we also know $\angle P{D}^{\prime }Q=\angle BAC$. This implies that $\angle BAC=90^{\circ }$.

Now we choose $D$ to be the midpoint of $BC$. Since $\angle BAC=90^{\circ }$, we can deduce that $DQP$ is the medial triangle of triangle $ABC$. Therefore, $PQ\parallel BC$ from which it follows that $D{D}^{\prime }\perp BC$. But the distance from $D^{\prime }$ to $BC$ is equal to both the circumradius of triangle $ABC$ and to the distance from $A$ to $BC$. This can only happen if $A=D^{\prime }$. This implies that $ABC$ is isosceles and right-angled at $A$.



We will now prove that if $ABC$ is isosceles and right-angled at $A$ then the required property in the problem holds. Let $D$ be any point on side $BC$. Then $D^{\prime }P=DP$ and we also have $DP=BP$. Hence, $D^{\prime }P=BP$ and similarly $D^{\prime }Q=CQ$. Note that $APDQ{D}^{\prime }$ is cyclic with diameter $PQ$. Therefore, $\angle AP{D}^{\prime }=\angle AQ{D}^{\prime }$, from which we obtain $\angle BP{D}^{\prime }=\angle CQ{D}^{\prime }$. So triangles $D^{\prime }PB$ and $D^{\prime }QC$ are similar. It follows that $\angle P{D}^{\prime }Q=\angle P{D}^{\prime }C+\angle C{D}^{\prime }Q=\angle P{D}^{\prime }C+\angle B{D}^{\prime }P=\angle B{D}^{\prime }C$ and $\frac{D^{\prime }P}{D^{\prime }Q}=\frac{D^{\prime }B}{D^{\prime }C}$. So we also obtain that triangles $D^{\prime }PQ$ and $D^{\prime }BC$ are similar. But since $DPQ$ and $D^{\prime }PQ$ are congruent, we may deduce that $\angle B{D}^{\prime }C=\angle P{D}^{\prime }Q=\angle PDQ=90^{\circ }$. Therefore, $D^{\prime }$ lies on the circle with diameter $BC$, which is the circumcircle of triangle $ABC$.