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

@@ Line 1: / Line 1: @@
 ==Problem==
-==Solution==
+Let <math>ABCD</math> be a convex quadrilateral with <math>BC=2</math> and <math>CD=6.</math> Suppose that the centroids of <math>\triangle ABC,\triangle BCD,</math> and <math>\triangle ACD</math> form the vertices of an equilateral triangle. What is the maximum possible value of the area of <math>ABCD</math>?
+<math>\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30</math>
+==Solution 1 (vectors)==
+Place an origin at <math>A</math>, and assign position vectors of <math>B = \vec{p}</math> and <math>D = \vec{q}</math>. Since <math>AB</math> is not parallel to <math>AD</math>, vectors <math>\vec{p}</math> and <math>\vec{q}</math> are linearly independent, so we can write <math>C = m\vec{p} + n\vec{q}</math> for some constants <math>m</math> and <math>n</math>. Now, recall that the centroid of a triangle <math>\triangle XYZ</math> has position vector <math>\frac{1}{3}\left(\vec{x}+\vec{y}+\vec{z}\right)</math>.
+Thus the centroid of <math>\triangle ABC</math> is <math>g_1 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}n\vec{q}</math>; the centroid of <math>\triangle BCD</math> is <math>g_2 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}(n+1)\vec{q}</math>; and the centroid of <math>\triangle ACD</math> is <math>g_3 = \frac{1}{3}m\vec{p} + \frac{1}{3}(n+1)\vec{q}</math>.
+Hence <math>\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}</math>, <math>\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}</math>, and <math>\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}</math>. For <math>\triangle G_{1}G_{2}G_{3}</math> to be equilateral, we need <math>\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</math>. Further, <math>\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</math>. Hence we have <math>AB = AD = BD</math>, so <math>\triangle ABD</math> is equilateral.
+Now let the side length of <math>\triangle ABD</math> be <math>k</math>, and let <math>\angle BCD = \theta</math>. By the Law of Cosines in <math>\triangle BCD</math>, we have <math>k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}</math>. Since <math>\triangle ABD</math> is equilateral, its area is <math>\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}</math>, while the area of <math>\triangle BCD</math> is <math>\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}</math>. Thus the total area of <math>ABCD</math> is <math>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)}</math>, where in the last step we used the subtraction formula for <math>\sin</math>. Alternatively, we can use calculus to find the local maximum. Observe that <math>\sin{\left(\theta-60^{\circ}\right)}</math> has maximum value <math>1</math> when e.g. <math>\theta = 150^{\circ}</math>, which is a valid configuration, so the maximum area is <math>10\sqrt{3} + 12(1) = \boxed{\textbf{(C) } 12+10\sqrt3}</math>.
+==Solution 2==
+Let <math>G_1</math>, <math>G_2</math>, <math>G_3</math> be the centroids of <math>ABC</math>, <math>BCD</math>, and <math>CDA</math> respectively, and let <math>M</math> be the midpoint of <math>BC</math>. <math>A</math>, <math>G_1</math>, and <math>M</math> are collinear due to well-known properties of the centroid. Likewise, <math>D</math>, <math>G_2</math>, and <math>M</math> are collinear as well. Because (as is also well-known) <math>3MG_1 = AM</math> and <math>3MG_2 = DM</math>, we have <math>\triangle MG_1G_2\sim\triangle MAD</math>. This implies that <math>AD</math> is parallel to <math>G_1G_2</math>, and in terms of lengths, <math>AD = 3G_1G_2</math>. (SAS Similarity)
+We can apply the same argument to the pair of triangles <math>\triangle BCD</math> and <math>\triangle ACD</math>, concluding that <math>AB</math> is parallel to <math>G_2G_3</math> and <math>AB = 3G_2G_3</math>. Because <math>3G_1G_2 = 3G_2G_3</math> (due to the triangle being equilateral), <math>AB = AD</math>, and the pair of parallel lines preserve the <math>60^{\circ}</math> angle, meaning <math>\angle BAD = 60^\circ</math>. Therefore <math>\triangle BAD</math> is equilateral.
+At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows:
+Let <math>BD = 2x</math>, where <math>2 < x < 4</math> due to the Triangle Inequality in <math>\triangle BCD</math>. By breaking the quadrilateral into <math>\triangle ABD</math> and <math>\triangle BCD</math>, we can create an expression for the area of <math>ABCD</math>. We use the formula for the area of an equilateral triangle given its side length to find the area of <math>\triangle ABD</math> and Heron's formula to find the area of <math>\triangle BCD</math>.
+After simplifying,
+<cmath>[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}</cmath>
+Substituting <math>k = x^2 - 10</math>, the expression becomes
+<cmath>[ABCD] = k\sqrt{3} + \sqrt{36 - k^2} + 10\sqrt{3}</cmath>
+We can ignore the <math>10\sqrt{3}</math> for now and focus on <math>k\sqrt{3} + \sqrt{36 - k^2}</math>.
+By the Cauchy-Schwarz inequality,
+<cmath>\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)</cmath>
+The RHS simplifies to <math>12^2</math>, meaning the maximum value of <math>k\sqrt{3} + \sqrt{36 - k^2}</math> is <math>12</math>.
+Thus the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>.
+==Solution 3 (Complex Numbers)==
+Let <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> correspond to the complex numbers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, respectively. Then, the complex representations of the centroids are <math>(a+b+c)/3</math>, <math>(b+c+d)/3</math>, and <math>(a+c+d)/3</math>. The pairwise distances between the centroids are <math>\lvert (d-a)/3 \rvert</math>, <math>\lvert (b-a)/3 \rvert</math>, and <math>\lvert (b-d)/3 \rvert</math>, all equal. Thus, <math>\lvert (b-a)/3 \rvert=\lvert (d-a)/3 \rvert=\lvert (b-d)/3 \rvert</math>, so <math>\lvert (b-a) \rvert=\lvert (d-a) \rvert=\lvert (b-d) \rvert</math>. Hence, <math>\triangle DBA</math> is equilateral.
+By the Law of Cosines,
+<math>[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)</math>.
+<math>[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}</math>. Thus, the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>.
+~ Leo.Euler
+==Solution 4 (Homothety)==
+Let <math>G_1, G_2</math>, and <math>G_3</math> be the centroids of <math>\triangle ABC, \triangle BCD</math>, and <math>\triangle ACD</math>, respectively, and let <math>X, Y,</math> and <math>Z</math> be the midpoints of <math>\overline{AB}, \overline{BD},</math> and <math>\overline{AD}</math>, respectively. Note that <math>G_1, G_2,</math> and <math>G_3</math> are <math>\frac{2}{3}</math> of the way from <math>C</math> to <math>X, Y,</math> and <math>Z</math>, respectively, by a well-known property of centroids. Then a homothety centered at <math>C</math> with ratio <math>\frac{3}{2}</math> maps <math>G_1, G_2,</math> and <math>G_3</math> to <math>X, Y,</math> and <math>Z</math>, respectively, implying that <math>\triangle XYZ</math> is equilateral too. But <math>\triangle XYZ</math> is the medial triangle of <math>\triangle ABD</math>, so <math>\triangle ABD</math> 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 <math>\triangle ABD</math> is equilateral greatly reduces the effort to find the final answer, hence I propose an alternative after this.
+Let <math>AB = BD = AD = x</math> and <math>\angle BCD = \theta</math>. By cosine rule on <math>\triangle BCD</math> :
+<cmath>x^2 = 40 - 24\cos \theta</cmath>
+Thus, the total area of the quadrilateral is supposedly :
+<cmath>\frac{\sqrt{3}}{4}(x^2) + \frac{1}{2}(2)(6)\sin \theta</cmath>
+<cmath>\implies \frac{\sqrt{3}}{4}(40 - 24\cos \theta) + 6\sin \theta</cmath>
+<cmath>\implies 6(\sin \theta - \sqrt{3}\cos \theta) + 10\sqrt{3} \geq 12 + 10\sqrt{3}</cmath>
+Where the inequality comes from a common trigonometric identity, <math>(\sin \theta - \sqrt{3}\cos \theta) \geq \sqrt{1^2 + \big(\sqrt{3}\big)^2} = 2.</math>
+~ SouradipClash_03
+==Video Solution by MOP 2024==
+https://youtu.be/c26N2w2MMQE
+~r00tsOfUnity
+===Solution 5===
+Let <math>X, Y, Z</math> be the centroids of <math>\triangle ABC, \triangle BCD, \triangle ACD</math> respectively, then
+ <math>XZ//BD</math>, since <math>EX=\frac13 EB, EZ=\frac13 ED</math>,
+ <math>XY//GE</math>, since <math>BX=\frac23BE, BY=\frac23BG, EG//AD</math> by midsegment theorem, so <math>XY//AD</math>
+ Similarly, <math>YZ//AB</math>,
+ So <math>\triangle ABD</math> is an equilateral triangle
+ Assume <math>\alpha=\angle BCD</math>, then <math>BD^2=BC^2+CD^2-2\cdot BC\cdot CD\cos \alpha=40-24\cos\alpha</math>, the area
+ <math>[ABCD]=[ABD]+[BCD]=\frac{\sqrt3}4 BD^2+\frac12\cdot BC\cdot CD\sin\alpha=</math>
+<math>10\sqrt3+6(\sin\alpha-\sqrt3\cos \alpha)=10\sqrt3+12\sin(\alpha-60^\circ)</math>
+ The maximal value happens when <math>\sin(\alpha-60^\circ)=1</math>, and the value is <math>10\sqrt3+12</math>, and the answer is <math>\boxed{\textbf{(C)} 12+10\sqrt3}</math>.
+<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 */
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+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
 ==See Also==
 {{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}}
+{{MAA Notice}}

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
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