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Croatian Mathematical Competitions

Geometry Difficulty 5.8 AIME, harder Prove it Croatia

Let $ABC$ be an isosceles triangle with a base $\overline{AB}$ . A point $P$ is chosen on the segment $\overline{AC}$ and a point $Q$ is chosen on the segment $\overline{BC}$ such that $|AP| + |BQ| = |PQ|$ . The line parallel with the line $BC$ which passes through the midpoint of the segment $\overline{PQ}$ intersects the segment $\overline{AB}$ in the point $N$ . Circumcircle of the triangle $PNQ$ intersects the line $AC$ in the points $P$ and $K$ , and the line $BC$ in the points $Q$ and $L$ . If the point $R$ is the intersection of the lines $PL$ and $QK$ , prove that the line $PQ$ is perpendicular to the line $CR$ . (Stipe Vidak)

Solution

Let $S$ be the midpoint of the segment $\overline{PQ}$ and let $M$ be the point on the segment $\overline{AB}$ such that $MP \parallel BC$ .

We have $\angle PMA = \angle CBA = \angle CAB = \angle PAM$ , so the triangle $PAM$ is isosceles and $|PA| = |PM|$ . The quadrilateral $PMBQ$ is a trapezium with the midline $\overline{SN}$ , so we have
$|SN| = \frac{|PM| + |QB|}{2} = \frac{|AP| + |QB|}{2} = \frac{|PQ|}{2}.$
Hence $S$ is the circumcentre of the triangle $PQN$ and the segment $\overline{PQ}$ is its diameter. Thales' theorem implies $QK \perp CP$ and $PL \perp CQ$ , so the point $R$ is the orthocentre of the triangle $CPQ$ . From this we conclude that $CR \perp PQ$ .

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