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

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Therefore the maximum area for <math>ABCD</math> is therefore <math>16\sqrt3</math>
 
Therefore the maximum area for <math>ABCD</math> is therefore <math>16\sqrt3</math>
 
<math>B</math>
 
<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 18:14, 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

Let $X,Y,Z$ be the centroids of $ABC,BCD,CDA$

$XY=AD/3,ZY=AB/3$ $XYZ is equilateral therefore ABD is also equilateral$ Rotate $ACD$ to $AEB$

Then $AEC$ is also equilateral Which has a larger or equal area than $ABCD$

$CE<=BC+EB =BC+CD =2+6=8$ Area of $AEC$ is at most $16\sqrt3$

Therefore the maximum area for $ABCD$ is therefore $16\sqrt3$ $B$

BRUH THIS IS WRONG U TROLL

(ANS IS C)

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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