2015 AIME I Problems/Problem 4

Revision as of 11:22, 21 March 2015 by Brian22 (talk | contribs) (Solution)

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 $\boxed{507}$.

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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