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NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Geometry Difficulty 4.8 AIME Prove it Romania

The measure of the angle A^\hat{A} of the acute triangle ABCABC is 60∘60^\circ, and HI=HBHI = HB, where II and HH are the incenter and the orthocenter of the triangle ABCABC. Find the measure of the angle B^\hat{B}.

Solution

We have m(∠BIC)≡m(∠BHC)=120∘m(\angle BIC) \equiv m(\angle BHC) = 120^\circ, hence BB, HH, II, CC are situated on a circle.

If m(∠B)>60∘m(\angle B) > 60^\circ then m(∠HBI)=m(∠ABI)−m(∠ABH)=12m(∠B)−30∘m(\angle HBI) = m(\angle ABI) - m(\angle ABH) = \frac{1}{2}m(\angle B) - 30^\circ.

But m(∠HBI)=m(∠HIB)=m(∠HCB)=90∘−m(∠B)m(\angle HBI) = m(\angle HIB) = m(\angle HCB) = 90^\circ - m(\angle B), hence m(∠B)=m(\angle B) =

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