Bariz - Problem

Source: Sharygin Geometry Olympiad 2012 - Problem 11

Tags: geometry, circumcircle, geometry unsolved

nsun48

28.04.2012 05:58

Given triangle $ABC$ and point $P$. Points $A', B', C'$ are the projections of $P$ to $BC, CA, AB$. A line passing through $P$ and parallel to $AB$ meets the circumcircle of triangle $PA'B'$ for the second time in point $C_{1}$. Points $A_{1}, B_{1}$ are defined similarly. Prove that a) lines $AA_{1}, BB_{1}, CC_{1}$ concur; b) triangles $ABC$ and $A_{1}B_{1}C_{1}$ are similar.

Clearly $C$ lies on the circumcircle of triangle $PA'B'$ since $\angle PA'C=\angle PB'C=90^{\circ}$. It follows that $CC_1\perp C_1P$ and so $CC_1\perp AB$. Therefore $CC_1$ passes trough the orthocentre $H$ of triangle $ABC$. Similarly, so do $AA_1$ and $BB_1$ and the result follows. Next, using the parallel lines and general angle-chasing, you can show that $\angle C_1PA_1+\angle C_1HA_1=B+180^{\circle}-B=180^{\circle}$, so $P,A_1,C_1,H$ are concyclic. Cyclic reasoning shows that $P,A_1,B_1,C_1,H$ are concyclic. Then angle chasing shows that triangles $ABC$ and $A_1B_1C_1$ are similar.