2015 AIME I Problems/Problem 4

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.

2015 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2015 AIME I Problems/Problem 4

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