AMC12 2019 12B, Problem 25

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 $ABCD$?

Solution: Place an origin at $A$, and assign position vectors of $B = \vec{p}$ and $D = \vec{q}$. Since $AB$ is not parallel to $AD$, vectors $\vec{p}$ and $\vec{q}$ are linearly independent, so we can write $C = m\vec{p} + n\vec{q}$ for some constants $m$ and $n$. Now, recall that the centroid of a triangle $\triangle XYZ$ has position vector $\frac{1}{3}\left(\vec{x}+\vec{y}+\vec{z}\right)$. Thus the centroid of $\triangle ABC$ is $g_1 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}n\vec{q}$; the centroid of $\triangle BCD$ is $g_2 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}(n+1)\vec{q}$; and the centroid of $\triangle ACD$ is $g_3 = \frac{1}{3}m\vec{p} + \frac{1}{3}(n+1)\vec{q}$. Hence $\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}$, $\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}$, and $\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}$. For $\triangle G_{1}G_{2}G_{3}$ to be equilateral, we need $\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| B = AD$. Further, $\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD$. Hence we have $AB = AD = BD$, so $\triangle ABD$ is equilateral. Now let the side length of $\triangle ABD$ be $k$, and let $\angle BCD = \theta$. By the Law of Cosines in $\triangle BCD$, we have $k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}$. Since $\triangle ABD$ is equilateral, its area is $\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}$, while the area of $\triangle BCD$ is $\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}$. Thus the total area of $ABCD$ is $10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\frac{1}{2} \sin{\theta} - \frac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}$, where in the last step we used the subtraction formula for $\sin$. Alternatively, we can use calculus to find the local maximum. Observe that $\sin{\left(\theta-60^{\circ}\right)}$ has maximum value $1$ when e.g. $\theta = 150^{\circ}$, which is a valid configuration, so the maximum area is $10\sqrt{3} + 12(1) = \fbox{12+10\sqrt3}$.

Answer: 12+10\sqrt3