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

## Check the Answer

But try the problem first…

Answer: is 507.

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.

## Other useful links

- https://www.cheenta.com/number-and-series-aime-2015/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s