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

Set $G_1$ , $G_2$ , $G_3$ as the centroids of $ABC$ , $BCD$ , and $CDA$ respectively, while $M$ is the midpoint of line $BC$ . $A$ , $G_1$ , and $M$ are collinear due to the centroid. Likewise, $D$ , $G_2$ , and $M$ are collinear as well. Because $AG_1 = 3AM$ and $DG_2 = 3DM$ , $\triangle MG_1G_2\sim\triangle MAD$ . From the similar triangle ratios, we can deduce that $AD = 3G_1G_2$ . The similar triangles implies parallel lines, namely $AD$ is parallel to $G_1G_2$ .

We can apply the same strategy to the pair of triangles $\triangle BCD$ and $\triangle ACD$ . We can conclude that $AB$ is parallel to $G_2G_3$ and $AB = 3G_2G_3$ . Because $3G_1G_2 = 3G_2G_3$ , $AB = AD$ and the pair of parallel lines preserve the 60 degree angle, meaning $\angle BAD = 60^\circ$ . Therefore, $\triangle BAD$ is equilateral.

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

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