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

(Created page with "==Problem== Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on...")
 
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==Problem==
 
==Problem==
 
Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>.
 
Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>.
 +
 +
==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,<math>8\sqrt{3}</math>), and Point E is at (18,<math>2\sqrt{3}</math>). By Midpoint Formula, M is at (9,<math>\sqrt{3}</math>), and N is at (14,<math>4\sqrt{3}</math>). Distance formula shows that BM=BN=MN=<math>2\sqrt{13}</math>. Therefore, by equilateral triangle area formula, x=13<math>\sqrt{3}</math>, so <math>x^2</math> is 507.
  
 
==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 21:58, 20 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,$4\sqrt{3}$). Distance formula shows that BM=BN=MN=$2\sqrt{13}$. Therefore, by equilateral triangle area formula, x=13$\sqrt{3}$, so $x^2$ is 507.

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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