Library / Saudi Arabian Mathematical Competitions /113 of 395

Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.2 AIME, harder Prove it Saudi Arabia

Let $ABC$ be a triangle inscribed in the circle $(O)$ and $P$ is a point inside the triangle $ABC$ . Let $D$ be a point on $(O)$ such that $AD \perp AP$ . The line $CD$ cuts the perpendicular bisector of $BC$ at $M$ . The line $AD$ cuts the line passing through $B$ and is perpendicular to $BP$ at $Q$ . Let $N$ be the reflection of $Q$ through $M$ . Prove that $CN \perp CP$ .

Solution

Let $K, L$ be the midpoints of $PQ, PN$ .
Since $PBQA$ is cyclic
$\angle KPB = \angle BAQ = \angle DCB.$
Hence the isosceles triangles $KPB$ and $MCB$ are similar. The line $BM$ cuts $PQ$ at $R$ . We have
$\angle LMC = \angle LMB - \angle BMC = \angle MRQ - \angle BKP = \angle KBM.$
Easily seen $KB = KP = ML$ and $BM = MC$ .
Hence, $\triangle KBM = \triangle LMC$ . We deduce $LC = KM = LP = LN$ .
This, triangle $PCN$ is right at $C$ . We are done. $\square$

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.