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2019 AMC 12B Problems/Problem 25

Revision as of 16:06, 23 January 2020 by Zeric (talk | contribs) (→‎Solution 1 (vectors))

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 1 (vectors)

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 $\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| \Rightarrow AB = AD$. 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 $10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\sin{\theta} \frac{1}{2} - \frac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}$, 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 $10\sqrt{3} + 12(1) = \boxed{\textbf{(C) } 12+10\sqrt3}$.

Solution 2

Let $G_1$, $G_2$, $G_3$ be the centroids of $ABC$, $BCD$, and $CDA$ respectively, and let $M$ be the midpoint of $BC$. $A$, $G_1$, and $M$ are collinear due to well-known properties of the centroid. Likewise, $D$, $G_2$, and $M$ are collinear as well. Because (as is also well-known) $AG_1 = 3AM$ and $DG_2 = 3DM$, we have $\triangle MG_1G_2\sim\triangle MAD$. This implies that $AD$ is parallel to $G_1G_2$, and in terms of lengths, $AD = 3G_1G_2$.

We can apply the same argument to the pair of triangles $\triangle BCD$ and $\triangle ACD$, concluding that $AB$ is parallel to $G_2G_3$ and $AB = 3G_2G_3$. Because $3G_1G_2 = 3G_2G_3$ (due to the triangle being equilateral), $AB = AD$, and the pair of parallel lines preserve the $60^{\circ}$ angle, meaning $\angle BAD = 60^\circ$. Therefore $\triangle BAD$ is equilateral.

At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows:

Let $BD = 2x$, where $2 < x < 4$ due to the Triangle Inequality in $\triangle BCD$. By breaking the quadrilateral into $\triangle ABD$ and $\triangle BCD$, we can create an expression for the area of $ABCD$. We use the formula for the area of an equilateral triangle given its side length to find the area of $\triangle ABD$ and Heron's formula to find the area of $\triangle BCD$.

After simplifying,

\[[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}\]

Substituting $k = x^2 - 10$, the expression becomes

\[[ABCD] = k\sqrt{3} + \sqrt{36 - k^2} + 10\sqrt{3}\]

We can ignore the $10\sqrt{3}$ for now and focus on $k\sqrt{3} + \sqrt{36 - k^2}$.

By the Cauchy-Schwarz inequality,

\[\left(k\sqrt 3 + \sqrt{36-k^2}\right)^2 \leq \left(\left(\sqrt{3}\right)^2+1^2\right)\left(\left(k\right)^2 + \left(\sqrt{36-k^2}\right)^2\right)\]

The RHS simplifies to $12^2$, meaning the maximum value of $k\sqrt{3} + \sqrt{36 - k^2}$ is $12$.

Thus the maximum possible area of $ABCD$ is $\boxed{\textbf{(C) }12 + 10\sqrt{3}}$.

See Also

2019 AMC 12B (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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All AMC 12 Problems and Solutions

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