IMO 2001 SL G2

Origin: KOR | Category: Geometry

Problem

In acute triangle ABC with circumcenter O and altitude AP, ∡C \geq∡B + 30◦. Prove that ∡A + ∡COP < 90◦.

Solution

Since \angleOCP = 90◦−\angleA, we are led to showing that \angleOCP > \angleCOP, i.e., OP > CP. By the triangle inequality it suffices to prove CP < 1 2CO. Let CO = R. The law of sines yields CP = AC cos \gamma = 2R sin \beta cos \gamma < 2R sin \beta cos(\beta + 30◦). Finally, we have 2 sin \beta cos(\beta + 30◦) = sin(2\beta + 30◦) −sin 30◦\leq1 2, which completes the proof.