2019/20 AHSMC Part 1 Problem 16

Consider the circle centred at $O$ that is circumscribed to $\triangle ABC$. Let $EBC$ be the isosceles triangle containing $O$ and inscribed in this circle. Since $\angle BEC = \angle BAC = 60^\circ$, $\triangle EBC$ is in fact equilateral. Let $EF \perp BC$, $OG \perp AD$.

By simple calculations one obtains $$EF = 7\sqrt{3},\quad OE = OA = \frac{14\sqrt{3}}{3},\quad OG = FD = 5,\quad GA = \frac{11\sqrt{3}}{3},\quad GD = OF = \frac{7\sqrt{3}}{3}.$$ Hence $AD = 6\sqrt{3}$ and the area of $\triangle ABC$ is equal to $42\sqrt{3}$. The answer is (C).

Alternative solution: Let $\theta = \angle BAD$, and $x = AD$. In $\triangle ABD$, $\tan \theta = \dfrac{BD}{x} = \dfrac{12}{x}$ and in $\triangle ADC$, $\tan\bigl(\dfrac{\pi}{3} - \theta\bigr) = \dfrac{DC}{x} = \dfrac{2}{x}$. Hence $$\sqrt{3} = \tan \frac{\pi}{3} = \tan\bigl(\theta + (\pi/3 - \theta)\bigr) = \frac{\tan\theta + \tan(\pi/3 - \theta)}{1 - \tan\theta \tan(\pi/3 - \theta)} = \frac{\frac{12}{x} + \frac{2}{x}}{1 - \frac{24}{x^2}} = \frac{14x}{x^2 - 24}.$$ Solving for $x$ one obtains $x = 6\sqrt{3}$ and thus the area of $\triangle ABC$ is equal to $42\sqrt{3}$.