Mathematical Reflections 2014, Issue 4 - Problem S307

Problem:
Let $ABC$ be a triangle such that $\angle ABC - \angle ACB = 60^{\circ}$. Suppose that the length of the
altitude from $A$ is $\dfrac{1}{4}BC$. Find $\angle ABC$.

 Proposed by Omer Cerrahoglu and Mircea Lascu.

Solution:
Let $\alpha=\angle ABC$ and let $H$ be the foot of the altitude from $A$. Observe that $\alpha>60^{\circ}$, so $\tan \alpha>\sqrt{3}$. We have $$\dfrac{AH}{BH}=\tan \alpha, \qquad \dfrac{AH}{CH}=\tan (\alpha-60^{\circ}).$$ Since $AH=\dfrac{1}{4}BC$, then $$BC=BH+CH=\dfrac{1}{4 \tan \alpha}BC+\dfrac{1}{4 \tan (\alpha-60^{\circ})}BC,$$ which gives $$\dfrac{1}{\tan \alpha}+\dfrac{1}{\tan (\alpha-60^{\circ})}=4.$$ Since $\dfrac{1}{\tan (\alpha)-60^{\circ}}=\dfrac{1+\tan 60^{\circ}\tan \alpha}{\tan \alpha-\tan 60^{\circ}}$, setting $x=\tan \alpha$, we get $$\dfrac{1}{x}+\dfrac{1+\sqrt{3}x}{x-\sqrt{3}}=4 \iff (4-\sqrt{3})x^2-2(2\sqrt{3}+1)x+\sqrt{3}=0.$$ As $x>\sqrt{3}$, then $x=2+\sqrt{3}$, i.e. $\alpha=75^{\circ}$.