# APMO 2007 problem 2 solution

Let $$ABC$$ be an acute angled triangle with $$\angle{BAC}$$ = 60◦ and $$AB$$ > $$AC$$. Let $$I$$ be the incenter, and $$H$$ the orthocenter of the triangle $$ABC$$. Prove that $$2\angle{AHI}$$ = $$3\angle{ABC}$$.

I let $$D,E,F$$ be altitude on $$AB,BC,AC$$ respectively and $$HI$$ meet $$AB$$ at $$X$$. I observed that $$\angle{AHD}$$=$$\angle{ABC}$$. So I assume that the equality hold if $$\angle{DHX}$$=1/2$$\angle{AHD}$$ and to prove that is true I draw $$HZ$$ meet $$AB$$ at $$Z$$ so that $$\triangle{AHZ}$$ is isosceles triangle. But the problem is I don't know how to prove that $$\angle{DHX}=\angle{XHZ}$$. Please help

Let $$C'$$ be on $$AB$$ so that $$AB\bot CC'$$ and let $$\angle CBI = x ={1\over 2}\angle ABC$$.
Then since $$BCHI$$ is cyclic ($$\angle BIC = \angle BHC = 120^{\circ}$$) we have $$\angle IHC = \pi-x$$ so $$\color{red}{\angle IHC' = x}$$.
Since $$\angle BAH = {\pi\over 2}-2x$$ we have $$\color{red}{\angle AHC' = 2x}$$ (observe the triangle $$AHC'$$.) We are done.