MathLabs

Problem 1

Let ABC ABC be an acute-angled triangle with circumcentre O O . Let P P on BC BC be the foot of the altitude from A A . Suppose that ∠BCA≥∠ABC+30∘\angle BCA\ge\angle ABC+30^\circ . Prove that ∠CAB+∠COP<90∘\angle CAB+\angle COP<90^\circ .
Step 3 of 4: Compare OP OP and PC PC
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In plain words

Use acuteness.

OP>YP≥R/2>PCOP>YP\ge R/2>PC
Detailed analysis

O≠Y O\ne Y , otherwise ∠A=90∘\angle A=90^\circ . Hence OP>YP OP>YP . Also PC=YC−YP<R−YP≤R/2 PC=YC-YP<R-YP\le R/2, so OP>PC OP>PC .