2019 AMC 12B Problem 25

The differences of the three centroids are $\dfrac{B-D}{3},\ \dfrac{B-A}{3},\ \dfrac{A-D}{3}$

So $\triangle ABD$ is equilateral with side $BD;$ write the area using $\angle BCD$ and maximize

The centroids are $\dfrac{A+B+C}{3},$ $\ \dfrac{B+C+D}{3},$ $\ \dfrac{A+C+D}{3}.$ Their pairwise differences are $\dfrac{A-D}{3},\ \dfrac{B-A}{3},\ \dfrac{B-D}{3},$ so an equilateral centroid triangle forces $AB=BD=DA;$ that is, $\triangle ABD$ is equilateral with side $s=BD.$

Splitting along $BD,$ $$\begin{gathered} [ABCD]=[ABD]+[BCD] \\ =\dfrac{\sqrt3}{4}s^2 \\ {}+\dfrac12\cdot2\cdot6\sin C, \end{gathered}$$ where $C=\angle BCD.$ By the Law of Cosines $s^2=40-24\cos C,$ so $$\begin{gathered} [ABCD]=10\sqrt3 \\ {}-6\sqrt3\cos C+6\sin C. \end{gathered}$$

The expression $6\sin C-6\sqrt3\cos C$ has maximum $\sqrt{6^2+(6\sqrt3)^2}=12,$ so the greatest area is $10\sqrt3+12=12+10\sqrt3.$ Equality occurs at $C=150^\circ;$ constructing $\triangle BCD$ with that angle and placing equilateral $\triangle ABD$ on the opposite side of $\overline{BD}$ produces a convex quadrilateral, so the maximum is attainable.

Thus, C is the correct answer.