HMMT-Nov 2018 gen, Problem 8

Equilateral triangle $A B C$ has circumcircle $\Omega$. Points $D$ and $E$ are chosen on minor arcs $A B$ and $A C$ of $\Omega$ respectively such that $B C=D E$. Given that triangle $A B E$ has area 3 and triangle $A C D$ has area 4, find the area of triangle $A B C$.

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 $x=4 t, y=3$. 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 \[ \frac{A B^{2} \sqrt{3}}{4}=\frac{37}{7} \] $\fbox{\frac{37}{7}}$.

Answer: \frac{37}{7}