Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National olympiad Prove it Mongolia

Let $ABC$ be a triangle with $AB \ne AC$ , and let $H$ be its orthocenter. Let $T$ be a point on the arc $BC$ of the circumcircle of triangle $ABC$ that does not contain point $A$ . Let $l$ be the line through $H$ parallel to $BC$ , and let $l$ intersect lines $TB$ and $TC$ at points $P$ and $Q$ , respectively. Let the circumcircles of triangles $PAB$ and $QAC$ intersect again at point $S$ .
Suppose that $\angle PAQ = 2\angle BAC$ . Prove the following.

(1)
The point $S$ lies on the circumcircle of triangle $BHC$ .
(2)
$\angle PAH = \angle HAQ \text{ or } \angle PAB = \angle BAH.$

Solution

(1) By the cyclic quadrilateral property,
$\angle PSA = \angle PBA = 180^\circ - \angle ABT = \angle ACT = 180^\circ - \angle ACQ = 180^\circ - \angle ASQ,$
so point $S$ lies on line $PQ$ .

From the given condition, we have $\angle PAB + \angle CAQ = \angle BAC$ . Hence, $\angle PSB + \angle CSQ = \angle BAC$ .
Therefore, $\angle BSC = 180^\circ - \angle BAC = \angle BHC$ , and thus $S$ lies on the circum-circle of triangle $BHC$ .

Hence, $BTCH$ is a parallelogram, and under a homothety centered at $T$ with ratio 2, segment $BC$ maps to $PQ$ , so $TB = BP$ , $TC = CQ$ . Since $AB \perp TP$ and $AC \perp TQ$ , point $A$ is the center of the circle through $TPQ$ . As $AH \perp PQ$ , point $H$ is the midpoint of $PQ$ , giving
$\angle PAH = \angle HAQ.$

In this case, point $P$ lies on the circle through triangle $ABS$ . From angle chasing:
$\angle PAB = \angle PSB = \angle SBC = \angle HCB = \angle BAH,$
so
$\angle PAB = \angle BAH.$

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