2015 AIME I Problems/Problem 4

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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 (ProblemsAnswer KeyResources)
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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