2019 AMC 12B Problems/Problem 25

Problem

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 $ABCD$ ?

$\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30$

Solution

Let $X,Y,Z$ be the centroids of $ABC,BCD,CDA$

$XY=AD/3,ZY=AB/3$ $XYZ is equilateral therefore ABD is also equilateral$ Rotate $ACD$ to $AEB$

Then $AEC$ is also equilateral Which has a larger or equal area than $ABCD$

$CE<=BC+EB =BC+CD =2+6=8$ Area of $AEC$ is at most $16\sqrt3$

Therefore the maximum area for $ABCD$ is $16\sqrt3$

2019 AMC 12B (Problems • Answer Key • Resources)
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2019 AMC 12B Problems/Problem 25

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