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Difference between revisions of "2019 AMC 12B Problems/Problem 25"
(Added a new solution (using vectors), and cleaned up the old solution)
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<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>
 
<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==
+
==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>.
  
−
Set <math>G_1</math>, <math>G_2</math>, <math>G_3</math> as the centroids of <math>ABC</math>, <math>BCD</math>, and <math>CDA</math> respectively, while <math>M</math> is the midpoint of line <math>BC</math>. <math>A</math>, <math>G_1</math>, and <math>M</math> are collinear due to the centroid. Likewise, <math>D</math>, <math>G_2</math>, and <math>M</math> are collinear as well. Because <math>AG_1 = 3AM</math> and <math>DG_2 = 3DM</math>,  <math>\triangle MG_1G_2\sim\triangle MAD</math>. From the similar triangle ratios, we can deduce that <math>AD = 3G_1G_2</math>. The similar triangles implies parallel lines, namely <math>AD</math> is parallel to <math>G_1G_2</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>.  
  
−
We can apply the same strategy to the pair of triangles <math>\triangle BCD</math> and <math>\triangle ACD</math>. We can conclude 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>, <math>AB = AD</math> and the pair of parallel lines preserve the 60 degree angle, meaning <math>\angle BAD = 60^\circ</math>. Therefore, <math>\triangle BAD</math> is equilateral.
+
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.
  
−
Set <math>BD = 2x</math> where <math>2 < x < 4</math> due to the triangle inequality. By breaking the quadrilateral into <math>[ABD]</math> and <math>[BCD]</math>, we can create an expression for the area of <math>[ABCD]</math>. We will use the formula for the area of an equilateral triangle given its side length to find the area of <math>[ABD]</math> and Heron's formula to find the area of <math>[BCD]</math>.
+
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\sin{\left(\theta-60^{\circ}\right)}</math>, where in the last step we used the subtraction formula for <math>\sin</math>. 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>AG_1 = 3AM</math> and <math>DG_2 = 3DM</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>.
 +
 
 +
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,
 
After simplifying,
  
−
<math>[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}</math>
+
<cmath>[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}</cmath>
  
−
Substitute <math>k = x^2 - 10</math> and then the expression becomes
+
Substituting <math>k = x^2 - 10</math>, the expression becomes
  
−
<math>[ABCD] = k\sqrt{3} + \sqrt{36 - k^2} + 10\sqrt{3}</math>
+
<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>.
 
We can ignore the <math>10\sqrt{3}</math> for now and focus on <math>k\sqrt{3} + \sqrt{36 - k^2}</math>.
Line 25: Line 36:
 
By the Cauchy-Schwarz Inequality,
 
By the Cauchy-Schwarz Inequality,
  
−
<math>(k\sqrt 3 + \sqrt{36-k^2})^2 \leq ((\sqrt{3})^2+1^2)((k)^2 + (\sqrt{36-k^2})^2).</math>
+
<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>.
 
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>.
  
−
Finally, the maximum value of the area of <math>[ABCD]</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>.
+
Thus the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>.
  
 
==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}}

Revision as of 00:14, 19 February 2019

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 $10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}$, where in the last step we used the subtraction formula for $\sin$. 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
Last Problem
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