
Problem 11.2. Given $\triangle A B C$ such that $\angle A B C \geq 60 \mathrm{deg}$ and $\angle B A C \geq 60$ deg. Let $B L(L \in A C)$ be the internal bisector of $\angle A B C$ and $A H, H \in B C$ be the altitude from $A$. Find $\angle A H L$ if $\angle B L C=3 \angle A H L$.


Solution: Denote $\theta=\angle A H L, \alpha=\angle B A C, \beta=\angle A B C$ and $\gamma=$ $\angle A C B$. From the Sine Law for $\triangle A H L$ and $\triangle C H L$ we get $\frac{\sin \theta}{A L}=$ $\frac{\sin \angle A L H}{A H}$ and $\frac{\sin (90 \mathrm{deg}-\theta)}{C L}=\frac{\sin \angle C L H}{C H}$. Since $\frac{C L}{A L}=\frac{B C}{B A}=$ $\frac{\sin \alpha}{\sin \gamma}$ and $\frac{A H}{C H}=\tan \gamma$ (from $\angle A C B3 \theta=\alpha+\frac{\beta}{2} \geq 90 \mathrm{deg}$, so $30 \mathrm{deg} \leq \theta<60 \mathrm{deg}$. Therefore

$$
2 \sin 2 \theta \geq 2 \sin 60 \mathrm{deg}=\sqrt{3}
$$

The inequalities (3) and (4) show that (2) holds only if $\theta=$ $30 \mathrm{deg}$. In this case $\alpha=\beta=\gamma=60 \mathrm{deg}$. (It is obviously true that $90 \mathrm{deg}=\angle B L C=3 \angle A H L$ for any equilateral $\triangle A B C$ ).
