Difference between revisions of "2019 AMC 12B Problems/Problem 25"

Latest revision as of 02:38, 10 October 2024

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 the area 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 $10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\frac{1}{2} \sin{\theta} - \frac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}$ , 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 $10\sqrt{3} + 12(1) = \boxed{\textbf{(C) } 12+10\sqrt3}$ .

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) $3MG_1 = AM$ and $3MG_2 = DM$ , we have $\triangle MG_1G_2\sim\triangle MAD$ . This implies that $AD$ is parallel to $G_1G_2$ , and in terms of lengths, $AD = 3G_1G_2$ . (SAS Similarity)

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 $\boxed{\textbf{(C) }12 + 10\sqrt{3}}$ .

Solution 3 (Complex Numbers)

Let $A$ , $B$ , $C$ , and $D$ correspond to the complex numbers $a$ , $b$ , $c$ , and $d$ , respectively. Then, the complex representations of the centroids are $(a+b+c)/3$ , $(b+c+d)/3$ , and $(a+c+d)/3$ . The pairwise distances between the centroids are $\lvert (d-a)/3 \rvert$ , $\lvert (b-a)/3 \rvert$ , and $\lvert (b-d)/3 \rvert$ , all equal. Thus, $\lvert (b-a)/3 \rvert=\lvert (d-a)/3 \rvert=\lvert (b-d)/3 \rvert$ , so $\lvert (b-a) \rvert=\lvert (d-a) \rvert=\lvert (b-d) \rvert$ . Hence, $\triangle DBA$ is equilateral.

By the Law of Cosines, $[ABCD]=[ABD]+[BCD]=\frac{(\sqrt{2^2+6^2-2 \cdot 2 \cdot 6 \cos(\angle BCD)})^2 \cdot \sqrt{3}}{4}+1/2 \cdot 2 \cdot 6 \sin(\angle BCD)$ .

$[ABCD]=10\sqrt{3}+6(\sin{\angle BCD}-\sqrt{3}\cos(\angle BCD))= 10\sqrt{3}+12\sin(\angle BCD-60^{\circ}) \le 12 + 10\sqrt{3}$ . Thus, the maximum possible area of $ABCD$ is $\boxed{\textbf{(C) }12 + 10\sqrt{3}}$ .

~ Leo.Euler

Solution 4 (Homothety)

Let $G_1, G_2$ , and $G_3$ be the centroids of $\triangle ABC, \triangle BCD$ , and $\triangle ACD$ , respectively, and let $X, Y,$ and $Z$ be the midpoints of $\overline{AB}, \overline{BD},$ and $\overline{AD}$ , respectively. Note that $G_1, G_2,$ and $G_3$ are $\frac{2}{3}$ of the way from $C$ to $X, Y,$ and $Z$ , respectively, by a well-known property of centroids. Then a homothety centered at $C$ with ratio $\frac{3}{2}$ maps $G_1, G_2,$ and $G_3$ to $X, Y,$ and $Z$ , respectively, implying that $\triangle XYZ$ is equilateral too. But $\triangle XYZ$ is the medial triangle of $\triangle ABD$ , so $\triangle ABD$ is also equilateral. We may finish with the methods in the solutions above.

~ numberwhiz

While the solutions above have attempted the problem in general, knowing the fact that $\triangle ABD$ is equilateral greatly reduces the effort to find the final answer, hence I propose an alternative after this.

Let $AB = BD = AD = x$ and $\angle BCD = \theta$ . By cosine rule on $\triangle BCD$ : $x^2 = 40 - 24\cos \theta$ Thus, the total area of the quadrilateral is supposedly : $\frac{\sqrt{3}}{4}(x^2) + \frac{1}{2}(2)(6)\sin \theta$ $\implies \frac{\sqrt{3}}{4}(40 - 24\cos \theta) + 6\sin \theta$ $\implies 6(\sin \theta - \sqrt{3}\cos \theta) + 10\sqrt{3} \geq 12 + 10\sqrt{3}$ Where the inequality comes from a common trigonometric identity, $(\sin \theta - \sqrt{3}\cos \theta) \geq \sqrt{1^2 + \big(\sqrt{3}\big)^2} = 2.$

~ SouradipClash_03

Video Solution by MOP 2024

https://youtu.be/c26N2w2MMQE

~r00tsOfUnity

Solution 5

Let $X, Y, Z$ be the centroids of $\triangle ABC, \triangle BCD, \triangle ACD$ respectively, then

, since ,

, since  by midsegment theorem, so

Similarly, ,

So  is an equilateral triangle

Assume , then , the area

$10\sqrt3+6(\sin\alpha-\sqrt3\cos \alpha)=10\sqrt3+12\sin(\alpha-60^\circ)$

The maximal value happens when , and the value is , and the answer is .

$[asy] import graph; size(11.42cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.58, xmax = 9.84, ymin = -7.74, ymax = 8.48; /* image dimensions */ /* draw figures */ draw((0.,0.)--(4.,0.), linewidth(2.)); draw((2.,3.4641016151377544)--(4.,0.), linewidth(2.)); draw((0.,0.)--(2.,3.4641016151377544), linewidth(2.)); draw((4.,0.)--(5.,1.), linewidth(2.)); draw((5.,1.)--(2.,3.4641016151377544), linewidth(2.)); draw((0.,0.)--(5.,1.), linewidth(2.)); draw((2.5,0.5)--(4.,0.), linewidth(2.)); draw((4.,0.)--(3.5,2.232050807568877), linewidth(2.)); draw((2.,3.4641016151377544)--(2.5,0.5), linewidth(2.)); draw((0.,0.)--(3.5,2.232050807568877), linewidth(2.)); draw((0.,0.)--(4.5,0.5), linewidth(2.)); draw((2.,3.4641016151377544)--(4.5,0.5), linewidth(2.)); draw((2.333333333333333,1.4880338717125845)--(3.,0.3333333333333333), linewidth(2.) + linetype("2 2")); draw((3.666666666666666,1.488033871712585)--(3.,0.3333333333333333), linewidth(2.) + linetype("2 2")); /* dots and labels */ dot((0.,0.),dotstyle); label("$A$", (-0.2,-0.18), NE * labelscalefactor); dot((4.,0.),dotstyle); label("$B$", (3.98,-0.3), NE * labelscalefactor); dot((2.,3.4641016151377544),dotstyle); label("$D$", (1.86,3.62), NE * labelscalefactor); dot((5.,1.),dotstyle); label("$C$", (5.06,0.92), NE * labelscalefactor); dot((2.5,0.5),linewidth(4.pt) + dotstyle); label("$E$", (2.2,0.5), NE * labelscalefactor); dot((4.5,0.5),linewidth(4.pt) + dotstyle); label("$F$", (4.52,0.3), NE * labelscalefactor); dot((3.5,2.232050807568877),linewidth(4.pt) + dotstyle); label("$G$", (3.58,2.4), NE * labelscalefactor); dot((2.333333333333333,1.4880338717125845),linewidth(4.pt) + dotstyle); label("$Z$", (2.06,1.44), NE * labelscalefactor); dot((3.,0.3333333333333333),linewidth(4.pt) + dotstyle); label("$X$", (2.92,0.04), NE * labelscalefactor); dot((3.666666666666666,1.488033871712585),linewidth(4.pt) + dotstyle); label("$Y$", (3.76,1.5), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

~szhangmath

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search