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

(→‎Solution 5)
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~ numberwhiz
 
~ 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==
 
==Video Solution by MOP 2024==
Line 63: Line 75:
  
 
===Solution 5===
 
===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>
 
<asy>
Line 109: Line 141:
 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  
 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  
 
</asy>
 
</asy>
 +
 +
 +
~szhangmath
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}}
 
{{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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 

$XZ//BD$, since $EX=\frac13 EB, EZ=\frac13 ED$,

$XY//GE$, since $BX=\frac23BE, BY=\frac23BG, EG//AD$ by midsegment theorem, so $XY//AD$

Similarly, $YZ//AB$,

So $\triangle ABD$ is an equilateral triangle

Assume $\alpha=\angle BCD$, then $BD^2=BC^2+CD^2-2\cdot BC\cdot CD\cos \alpha=40-24\cos\alpha$, the area

$[ABCD]=[ABD]+[BCD]=\frac{\sqrt3}4 BD^2+\frac12\cdot BC\cdot CD\sin\alpha=$

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


The maximal value happens when $\sin(\alpha-60^\circ)=1$, and the value is $10\sqrt3+12$, and the answer is $\boxed{\textbf{(C)} 12+10\sqrt3}$.

[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

See Also

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

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