Problem:
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 triangles â–³ABD and â–³BCE. Let M be the midpoint of AE, and N be the midpoint of CD. The area of â–³BMN is x. Find x2.
Solution:
Place the triangles on a coordinate grid so that A is at (−16,0),B is at (0,0), and C is at (4,0). Then D is at (−8,83​),E is at (2,23​),M is at (−7,3​), and N is at (−2,43​). Because BM=MN=NB=52​, △BMN is equilateral with area 43​​(BM)2=133​=x. Thus x2=169⋅3=507​.