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.

Source
Suggested Reading

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.

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