Competition problem — Geometry — Maths Olympiad Prep

Library / /21 of 155

Geometry Difficulty 5.0 AIME, harder Prove it Saudi Arabia

Let $ABC$ be a triangle, the circle having $BC$ as diameter cuts $AB$ , $AC$ at $F$ , $E$ respectively. Let $P$ be a point on this circle. Let $C_{0}$ , $B_{0}$ be the projections of $P$ upon the sides $AB$ , $AC$ respectively. Let $H$ be the orthocenter of the triangle $AB_{0}C_{0}$ . Show that $\angle EHF = 90^{\circ}$ .

Solution

Let $B' E'$ , $C' F'$ be the altitudes of $\triangle AB'C'$ and concur at $H$ .
We have that $AB' P C'$ is inscribed in the circle with diameter $AP$ , so $AP$ passes through the circumcenter of $\triangle AB'C'$ . Then $AP$ , $AH$ reflect each other by the angle bisector of $\angle B'AC'$ . From that, easy to prove that $\triangle AHF' \sim \triangle APC'$ .

But $BE \parallel C'F'$ , as Thales's Theorem, we have $\frac{AE}{AF'} = \frac{AB}{AC'}$ , implies that
$\frac{HE}{PB} = \frac{AE}{AB} = \frac{EF}{BC} .$
Similarly, we get $\frac{HF}{PC} = \frac{EF}{BC}$ , thus $\triangle HEF \sim \triangle PBC$ . Then we get
$\angle EHF = \angle BPC = 90^{\circ} .$ $\square$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.