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

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==Solution==
 
==Solution==
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Let <math>X,Y,Z</math> be the centroids of <math>ABC,BCD,CDA</math>
 
  
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<math>XY=AD/3,ZY=AB/3</math>
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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>.
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<math>XYZ is equilateral therefore ABD is also equilateral</math>
 
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Rotate <math>ACD</math> to <math>AEB</math>
 
  
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Then <math>AEC</math> is also equilateral
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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.
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Which has a larger or equal area than <math>ABCD</math>
 
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<math>CE<=BC+EB =BC+CD =2+6=8</math>
 
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Area of <math>AEC </math> is at most <math>16\sqrt3</math>
 
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Therefore the maximum area for <math>ABCD</math> is therefore <math>16\sqrt3</math>
 
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<math>B</math>
 
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BRUH THIS IS WRONG U TROLL
 
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(ANS IS C)
 
  
 
==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 19:04, 14 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

Set $G_1$, $G_2$, $G_3$ as the centroids of $ABC$, $BCD$, and $CDA$ respectively, while $M$ is the midpoint of line $BC$. $A$, $G_1$, and $M$ are collinear due to the centroid. Likewise, $D$, $G_2$, and $M$ are collinear as well. Because $AG_1 = 3AM$ and $DG_2 = 3DM$, $\triangle MG_1G_2\sim\triangle MAD$. From the similar triangle ratios, we can deduce that $AD = 3G_1G_2$. The similar triangles implies parallel lines, namely $AD$ is parallel to $G_1G_2$.

We can apply the same strategy to the pair of triangles $\triangle BCD$ and $\triangle ACD$. We can conclude that $AB$ is parallel to $G_2G_3$ and $AB = 3G_2G_3$. Because $3G_1G_2 = 3G_2G_3$, $AB = AD$ and the pair of parallel lines preserve the 60 degree angle, meaning $\angle BAD = 60^\circ$. Therefore, $\triangle BAD$ is equilateral.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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