Question

Let $ABCD$ be a convex quadrilateral with $BC=2$ and $CD=6.$ Suppose that the centroids of $\triangle ABC,$ $\triangle BCD,$ and $\triangle ACD$ form the vertices of an equilateral triangle. What is the maximum possible value of the area of $ABCD?$

$27$

$16\sqrt3$

$12+10\sqrt3$

$9+12\sqrt3$

$30$

Solution:

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

The expression $6\sin C-6\sqrt3\cos C$ has maximum $\sqrt{6^2+(6\sqrt3)^2}=12,$ so the greatest area is $10\sqrt3+12=12+10\sqrt3.$

Thus, C is the correct answer.

2019 AMC 12B Problem 25

Solution:

Problem 25 in Other Years