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Difference between revisions of "2015 AIME I Problems/Problem 4"

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==Solution==
 
==Solution==
 
Let point <math>A</math> be at <math>(0,0)</math>. Then, <math>B</math> is at <math>(16,0)</math>, and <math>C</math> is at <math>(20,0)</math>. Due to symmetry, it is allowed to assume <math>D</math> and <math>E</math> are in quadrant 1. By equilateral triangle calculations, Point <math>D</math> is at <math>(8,8\sqrt{3})</math>, and Point <math>E</math> is at <math>(18,2\sqrt{3})</math>. By Midpoint Formula, <math>M</math> is at <math>(9,\sqrt{3})</math>, and <math>N</math> is at <math>(14,4\sqrt{3})</math>. The distance formula shows that <math>BM=BN=MN=2\sqrt{13}</math>. Therefore, by equilateral triangle area formula, <math>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>.
 
Let point <math>A</math> be at <math>(0,0)</math>. Then, <math>B</math> is at <math>(16,0)</math>, and <math>C</math> is at <math>(20,0)</math>. Due to symmetry, it is allowed to assume <math>D</math> and <math>E</math> are in quadrant 1. By equilateral triangle calculations, Point <math>D</math> is at <math>(8,8\sqrt{3})</math>, and Point <math>E</math> is at <math>(18,2\sqrt{3})</math>. By Midpoint Formula, <math>M</math> is at <math>(9,\sqrt{3})</math>, and <math>N</math> is at <math>(14,4\sqrt{3})</math>. The distance formula shows that <math>BM=BN=MN=2\sqrt{13}</math>. Therefore, by equilateral triangle area formula, <math>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>.
 +
==Solution 2==
 +
Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle <math>BMN</math> to solve for its area.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2015|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2015|n=I|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:54, 23 March 2015

Problem

Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$. Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$. Let $M$ be the midpoint of $\overline{AE}$, and $N$ be the midpoint of $\overline{CD}$. The area of $\triangle BMN$ is $x$. Find $x^2$.

Solution

Let point $A$ be at $(0,0)$. Then, $B$ is at $(16,0)$, and $C$ is at $(20,0)$. Due to symmetry, it is allowed to assume $D$ and $E$ are in quadrant 1. By equilateral triangle calculations, Point $D$ is at $(8,8\sqrt{3})$, and Point $E$ is at $(18,2\sqrt{3})$. By Midpoint Formula, $M$ is at $(9,\sqrt{3})$, and $N$ is at $(14,4\sqrt{3})$. The distance formula shows that $BM=BN=MN=2\sqrt{13}$. Therefore, by equilateral triangle area formula, $x=13\sqrt{3}$, so $x^2$ is $\boxed{507}$.

Solution 2

Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle $BMN$ to solve for its area.

See Also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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