AMC12 2019 B

Answer (C): Let $G_1$, $G_2$, and $G_3$ be the centroids of $\triangle ABC$, $\triangle BCD$, and $\triangle ACD$, respectively. Then with points viewed as position vectors represented by ordered pairs, \[ G_1=\frac13A+\frac13B+\frac13C, \] \[ G_2=\frac13B+\frac13C+\frac13D, \] \[ G_3=\frac13A+\frac13C+\frac13D. \] The directed edges of $\triangle G_1G_2G_3$ are $\overrightarrow{G_1G_2}=\frac13(D-A)$, $\overrightarrow{G_2G_3}=\frac13(A-B)$, and $\overrightarrow{G_3G_1}=\frac13(B-D)$. This triangle is equilateral if and only if $|D-A|=|A-B|=|B-D|$, which in turn is equivalent to the assertion that $\triangle ABD$ is equilateral. Let $\theta$ denote the measure of $\angle BCD$. Because $BC=2$ and $CD=6$, it follows that the area of $\triangle BCD$ is $\frac12\cdot BC\cdot CD\cdot\sin\theta=6\sin\theta$. By the Law of Cosines, \[ BD^2=BC^2+CD^2-2\cdot BC\cdot CD\cdot\cos\theta=40-24\cos\theta. \] The area of the equilateral triangle $\triangle ABD$ is $\frac{\sqrt3}{4}BD^2=10\sqrt3-6\sqrt3\cos\theta$. Adding the areas of $\triangle ABD$ and $\triangle BCD$ gives the area of the quadrilateral $ABCD$: \[ 10\sqrt3-6\sqrt3\cos\theta+6\sin\theta =10\sqrt3+12\left(-\frac{\sqrt3}{2}\cos\theta+\frac12\sin\theta\right) =10\sqrt3+12\sin(\theta-60^\circ). \] This is maximized by taking $\theta=150^\circ$, in which case the area is $10\sqrt3+12$.