, 2018

Solution:

A rotation by $120^{\circ}$ about the center of the circle will take $A B E$ to $B C D$, so $B C D$ has area $3$. Let $A D = x$, $B D = y$, and observe that $\angle A D C = \angle C D B = 60^{\circ}$. By Ptolemy's Theorem, $C D = x + y$. We have
$$\begin{aligned} & 4 = [A C D] = \frac{1}{2} A D \cdot C D \cdot \sin 60^{\circ} = \frac{\sqrt{3}}{4} x(x + y) \\ & 3 = [B C D] = \frac{1}{2} B D \cdot C D \cdot \sin 60^{\circ} = \frac{\sqrt{3}}{4} y(x + y) \end{aligned}$$

By dividing these equations find $x : y = 4 : 3$. Let $x = 4 t$, $y = 3 t$. Substitute this into the first equation to get $1 = \frac{\sqrt{3}}{4} \cdot 7 t^{2}$. By the Law of Cosines,
$$A B^{2} = x^{2} + x y + y^{2} = 37 t^{2}$$

The area of $A B C$ is then
$$\frac{A B^{2} \sqrt{3}}{4} = \frac{37}{7}$$