Baltic Way 2017 · Problem 11

Let $H^{\prime}$ be the reflection of $H$ in $K$. The reflection about the point $K$ sends $Q$ to $P$, and the line $B H$ to the line through $H^{\prime}$ and orthogonal to $A C$. The reflection about the line $A I$ sends $P$ to $C$, and the line through $H^{\prime}$ orthogonal to $A C$ to the line through $H$ orthogonal to $A B$, but this is just $B H$. Since composition of the two reflections sends $B, H$ and $Q$ to the same line, it follows that $B, H$ and $Q$ are collinear.

Let $\alpha=\frac{1}{2} \angle B A C, \beta=\frac{1}{2} \angle C B A$, and $\gamma=\frac{1}{2} \angle A C B$. Clearly then $\alpha+\beta+\gamma=90^{\circ}$, which yields $\angle B I C=180^{\circ}-\beta-\gamma=$ $90^{\circ}+\alpha$. From this we get $\angle C P A=180^{\circ}-\angle B P C=180^{\circ}-\angle B I C=90^{\circ}-\alpha$, so triangle $A P C$ is isosceles.

Now since $A I$ is the anglebisector of $\angle P A C$ it must also be the perpendicular bisector of $C P$. Hence $C K=P K=Q K$ so triangle $K C Q$ is isosceles. Additionally $A I$ bisects $P Q$ so $A I$ is the midline of triangle $P C Q$ parallel to $C Q$. Since $K H$ is perpendicular to $A I$, it is also perpendicular to $C Q$, so we may then conclude by symmetry that $H C Q$ is also isosceles. Moreover $\angle Q C A=\angle I A C=\alpha$, and $\angle A C H=90^{\circ}-2 \alpha$, so $\angle Q C H=\alpha+90^{\circ}-2 \alpha=90^{\circ}-\alpha$, which means that triangles $H C Q$ and $A P C$ are similar. In particular we have $\angle C H Q=2 \alpha$. Since also

$$180^{\circ}-\angle B H C=\angle H C B+\angle C B H=90^{\circ}-2 \beta+90^{\circ}-2 \gamma=2 \alpha$$

it follows that $B, H$, and $Q$ are collinear.

To prove that $Q$ always lies outside of triangle $A B C$ one could do the following: Since $P$ lies on $A B$, angle $C$ is larger than angle $B$ in triangle $A B C$. Thus angle $A D B$ is obtuse, where $D$ is the intersection point between $A I$ and $B C$. As $Q C$ and $A I$ are parallel, angle $Q C B$ is obtuse. Thus $Q$ lies outside of triangle $A B C$.