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Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 from Geometry based on Area of Triangle.

## Area of Triangle – AIME 2015

Point B lies on line segment AC with AB =16 and BC =4. Points D and E lie on the same side of line AC forming equilateral triangle ABD and traingle BCE. Let M be the midpoint of AE, and N be the midpoint of CD. The area of triangle BMN is x. Find $x_{2}$.

• is 107
• is 507
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Theory of Equations

Geometry

But try the problem first…

Source

AIME, 2015, Question 4

Geometry Revisited by Coxeter

## Try with Hints

First hint

Let point A be at (0,0).Then B is at (16,0) and C is at (20,0). Due to symmetry, it it allowed to assume D and E are in quadrant 1. By equilateral triangle calculations, point D is at $(8,8\sqrt3)$, and point E is at $(18,2\sqrt3). Second Hint By midpoint formula, M is at $$(9,\sqrt3)$, and N is at (14,4\sqrt3)$$.

Final Step

the distance formula shows that BM=BN=MN=$2\sqrt13)$. Then, by equilateral triangle area formula, x=13\sqrt3 then$x^{2}$=507.