Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National Olympiad Prove it Hong Kong

On the sides $AB$ and $AC$ of triangle $ABC$ , there are points $P$ and $Q$ respectively such that $\angle APC = \angle AQB = 45^\circ$ . Let the perpendicular line to side $AB$ through $P$ intersects line $BQ$ at $S$ . Let the perpendicular line to side $AC$ through $Q$ intersects line $CP$ at $R$ . Let $D$ be on side $BC$ such that $AD \perp BC$ . Prove that the lines $PS, AD, QR$ meet at a common point and lines $SR$ and $BC$ are parallel.

Solution

Let $PS$ and $QR$ meet $BC$ at $X$ and $Y$ respectively. Note that $A, P, X, D$ are concyclic since $\angle ADX + \angle APX = 180^\circ$ . Similarly, $A, D, Y, Q$ are concyclic. Also, since
$\angle BPC = 180^\circ - \angle CPA = 180^\circ - \angle AQB = \angle BQC,$

the points $B, C, Q, P$ are concyclic. Thus, we have
$\angle PQY = 90^\circ - \angle AQP = 90^\circ - \angle CBP = \angle PXB.$
This implies $Q, P, X, Y$ are concyclic.
Consider the circles $(APXD)$ , $(ADYQ)$ and $(QPXY)$ . Their pairwise radical axes are $AD, PX, QY$ respectively. Clearly, these lines are not parallel. Therefore, they are concurrent at the radical centre of these circles.

Next, observe that $Q, P, S, R$ are concyclic since
$\angle SQR = 90^\circ - \angle AQS = 90^\circ - \angle RPA = \angle SPR.$
Applying Reim's theorem to $(QPXY)$ and $(QPSR)$ , we obtain $SR \parallel XY$ . (It simply follows from $\angle XSR = \angle PQR = 180^\circ - \angle YXS$ .)

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