Mathematical Reflections 2012, Issue 4 - Problem J237

Problem:
Prove that the diameter of the incircle of a triangle $ABC$ is equal to $\dfrac{AB-BC+CA}{\sqrt{3}}$ if and only if $\angle BAC = 60^{\circ}$.

Proposed by Titu Andreescu.

Solution:
Let $r$ be the radius of the incircle of $\triangle ABC$ and let $\alpha=\angle BAC$. Clearly $0^{\circ} < \alpha < 180^{\circ}$ and $$2r=\dfrac{2AB\cdot CA \sin \alpha}{AB+BC+CA}.$$ Then $$2r=\dfrac{AB-BC+CA}{\sqrt{3}} \iff AB^2+CA^2+2AB\cdot CA - BC^2=2\sqrt{3}AB \cdot CA \sin \alpha.$$ Since $BC^2=AB^2+CA^2-2AB\cdot CA \cos \alpha$, we obtain $$2r=\dfrac{AB-BC+CA}{\sqrt{3}} \iff 1 + \cos \alpha = \sqrt{3} \sin \alpha,$$ i.e. if and only if $\sin (\alpha-30^{\circ})=\dfrac{1}{2},$ which gives $\alpha=60^{\circ}$.