Subfield: Geometry


Question: Let $A B C$ be an acute-angled triangle in which $B C<A B$ and $B C<C A$. Let point $P$ lie on segment $A B$ and point $Q$ lie on segment $A C$ such that $P \neq B, Q \neq C$ and $B Q=B C=C P$. Let $T$ be the circumcentre of triangle $A P Q, H$ the orthocentre of triangle $A B C$, and $S$ the point of intersection of the lines $B Q$ and $C P$. Prove that $T, H$ and $S$ are collinear.

Solution:
<img_3358>

We show that $T$ and $H$ are both on the angle bisector $\ell$ of $\angle B S C$.

We first prove that $H \in \ell$. The altitude $C H$ in triangle $A B C$ is also the altitude in isosceles triangle $P B C$ with $C P=C B$. Therefore, $C H$ is also the angle bisector of $\angle P C B$ and hence also of $\angle S C B$. Analogously, $B H$ is the angle bisector of $\angle S B C$. We conclude that $H$, as the intersection of two angle bisectors of $\triangle B S C$, is also on the third angle bisector, which is $\ell$.

We now prove that $T \in \ell$.

Variant 1. In the isosceles triangles $\triangle B C P$ and $\triangle C B Q$ we see that $\angle B C P=180^{\circ}-2 \angle B$ and $\angle C B Q=180^{\circ}-2 \angle C$. This yields $\angle P S Q=\angle B S C=180^{\circ}-\left(180^{\circ}-2 \angle B\right)-\left(180^{\circ}-2 \angle C\right)=$ $180^{\circ}-2 \angle A$. Furthermore, $\angle P T Q=2 \angle P A Q=2 \angle A$ ( $T$ being circumcentre of $\left.\triangle A P Q\right)$. Now $\angle P T Q+\angle P S Q=180^{\circ}$, so $P T Q S$ is a cyclic quadrilateral. From $P T=T Q$ we then obtain that $\angle P S T=\angle P Q T=\angle Q P T=\angle Q S T$, so $T$ is on the angle bisector $\angle P S Q$, which is also $\ell$.

We conclude that $T, S$ and $H$ are collinear.

Variant 2. Let $R$ be the second intersection of $B Q$ and $\odot A P Q$.

$A P R Q$ is cyclic quadrilateral, so $\angle P R S=\angle A, \angle A P R=\angle B Q C=\angle C$. On the other hand $\angle B P C=\angle B$. Therefore $\angle R P S=180^{\circ}-\angle B-\angle C=\angle A$. Hence, the triangle $P R S$ is isosceles with $S P=S R$; then $\ell$ is the perpendicular bisector of the chord $P R$ in the circle that passes through $T$.

Note that if $B Q$ is tangent to $\odot A P Q$, then $C P$ is also tangent to $\odot A P Q$ and triangle $A B C$ is isosceles, so $T, S$ and $H$ lie on the altitude from $A$.



<img_3681>
<img_3243>

Variant 1. In the isosceles triangles $\triangle B C P$ and $\triangle C B Q$ we see that $\angle B C P=180^{\circ}-2 \angle B$ and $\angle C B Q=180^{\circ}-2 \angle C$. This yields $\angle P S Q=\angle B S C=180^{\circ}-\left(180^{\circ}-2 \angle B\right)-\left(180^{\circ}-2 \angle C\right)=$ $180^{\circ}-2 \angle A$. Furthermore, $\angle P T Q=2 \angle P A Q=2 \angle A$ ( $T$ being circumcentre of $\left.\triangle A P Q\right)$. Now $\angle P T Q+\angle P S Q=180^{\circ}$, so $P T Q S$ is a cyclic quadrilateral. From $P T=T Q$ we then obtain that $\angle P S T=\angle P Q T=\angle Q P T=\angle Q S T$, so $T$ is on the angle bisector $\angle P S Q$, which is also $\ell$.
From above, we see that $\angle P S Q=180^{\circ}-2 \angle A$, so $\angle C S Q=2 \angle A$. From the cyclic quadrilateral $A E H D$ (with $E$ and $D$ feet of the altitudes $C H$ and $B H$ ) we see that $\angle D H C=\angle D A E=\angle A$. Since $B H$ is the perpendicular bisector of $C Q$, we have $\angle D H Q=\angle A$ as well, so $\angle C H Q=2 \angle A$. From $\angle C H Q=2 \angle A=\angle C S Q$, we see $C H S Q$ is a cyclic quadrilateral. This means $\angle Q H S=\angle Q C S$.

Since triangles $P T Q$ and $C H Q$ are both isosceles with apex $2 \angle A$, we get $\triangle P T Q \sim \triangle C H Q$. We see that one can be obtained from the other by a spiral similarity centered at $Q$, so we also obtain $\triangle Q T H \sim \triangle Q P C$. This means that $\angle Q H T=\angle Q C P$. Combining this with $\angle Q H S=\angle Q C S$, we see that $\angle Q H T=\angle Q C P=\angle Q C S=\angle Q H S$. So $\angle Q H T=\angle Q H S$, which means that $T, S$ and $H$ are collinear.
<img_3400>

Let us draw a parallel $f$ to $B C$ through $A$. Let $B^{\prime}=B Q \cap f, C^{\prime}=C P \cap f$. Then $A Q B^{\prime} \sim C Q B$ and $A P C^{\prime} \sim B P C$, therefore both $Q A B^{\prime}$ and $A P C^{\prime}$ will be isosceles.

Also, $B C S \sim B^{\prime} C^{\prime} S$ with respect to the similarity with center $S$, therefore if we take the image of the line $B H$ (which is the perpendicular bisector of the segment $C Q$ ) through this transformation, it will go through the point $B^{\prime}$, and be perpendicular to $A Q$ (as the image of $C Q$ is parallel to $C Q$ ). As $A Q B^{\prime}$ is isosceles, this line is the perpendicular bisector of the segment $A Q$. This means it goes through $T$, the circumcenter of $A P Q$. Similarly on the other side the image of $C H$ also goes through $T$. This means that the image of $H$ with respect to the similarity though $S$ will be $T$, so $T, S, H$ are collinear.