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 $(14,4\sqrt{3})$ . 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.

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

Contents

Problem

Solution

Solution 2

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