2019 AMC 12B Problems/Problem 25

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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.

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
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Problem 24
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