Competition problem — Geometry — Maths Olympiad Prep

Library / /49 of 189

Geometry Difficulty 5.2 AIME, harder Prove it Ukraine

In the acute triangle $ABC$ there are altitudes $BP$ and $CQ$ , point $T$ is the intersection of altitudes of $\triangle PAQ$ . It turns out that $\angle CTB = 90^\circ$ . Find the value of $\angle BAC$ .

Solution

The statement of the problem yields $\angle BQC = \angle BTC = \angle BPC = 90^\circ$ , hence, points $Q, T, P$ lie on the circle with diameter $BC$ , and in this exact order: $B, Q, T, P, C$ (Fig. 6), since $\triangle ABC$ is acute. Then,
$\angle QTP = 180^\circ - \angle ABP = 180^\circ - (90^\circ - \angle BAC) = 90^\circ + \angle BAC.$
On the other hand, $QT \perp AC$ , $PT \perp AB$ , $\angle QTP = 180^\circ - \angle BAC$ . Hence,
$90^\circ + \angle BAC = 180^\circ - \angle BAC \Rightarrow \angle BAC = 45^\circ.$

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 reproduced verbatim; metadata (topic, difficulty) added by this project.