XIX Asian Pacific Mathematics Olympiad

Let $D$ be the intersection point of the lines $AH$ and $BC$. Let $K$ be the intersection point of the circumcircle $O$ of the triangle $ABC$ and the line $AH$. Let the line through $I$ perpendicular to $BC$ meet $BC$ and the minor arc $BC$ of the circumcircle $O$ at $E$ and $N$, respectively. We have $$\angle BIC=180^{\circ }-(\angle IBC+\angle ICB)=180^{\circ }-\frac{1}{2}(\angle ABC+\angle ACB)=90^{\circ }+\frac{1}{2}\angle BAC=120^{\circ }$$ and also $\angle BNC=180^{\circ }-\angle BAC=120^{\circ }=\angle BIC$. Since $IN\perp BC$, the quadrilateral $BICN$ is a kite and thus $IE=EN$.

Now, since $H$ is the orthocenter of the triangle $ABC$, $HD=DK$. Also because $ED\perp IN$ and $ED\perp HK$, we conclude that $IHKN$ is an isosceles trapezoid with $IH=NK$.

Hence $$\angle AHI=180^{\circ }-\angle IHK=180^{\circ }-\angle AKN=\angle ABN.$$ Since $IE=EN$ and $BE\perp IN$, the triangles $IBE$ and $NBE$ are congruent. Therefore $$\angle NBE=\angle IBE=\angle IBC=\angle IBA=\frac{1}{2}\angle ABC$$ and thus $$\angle AHI=\angle ABN=\frac{3}{2}\angle ABC$$

Second solution:

Let $P,Q$ and $R$ be the intersection points of $BH,CH$ and $AH$ with $AC,AB$ and $BC$, respectively. Then we have $\angle IBH=\angle ICH$. Indeed, $$\angle IBH=\angle ABP-\angle ABI=30^{\circ }-\frac{1}{2}\angle ABC$$ and $$\angle ICH=\angle ACI-\angle ACH=\frac{1}{2}\angle ACB-30^{\circ }=30^{\circ }-\frac{1}{2}\angle ABC,$$ because $\angle ABH=\angle ACH=30^{\circ }$ and $\angle ACB+\angle ABC=120^{\circ }$. (Note that $\angle ABP>\angle ABI$ and $\angle ACI>\angle ACH$ because $AB$ is the longest side of the triangle $ABC$ under the given conditions.) Therefore $BIHC$ is a cyclic quadrilateral and thus $$\angle BHI=\angle BCI=\frac{1}{2}\angle ACB$$ On the other hand, $$\angle BHR=90^{\circ }-\angle HBR=90^{\circ }-(\angle ABC-\angle ABH)=120^{\circ }-\angle ABC.$$ Therefore, $$ $$\begin{array}{rl}\angle AHI& =180^{\circ }-\angle BHI-\angle BHR=60^{\circ }-\frac{1}{2}\angle ACB+\angle ABC\\ & =60^{\circ }-\frac{1}{2}(120^{\circ }-\angle ABC)+\angle ABC=\frac{3}{2}\angle ABC.\end{array}$$