The equilateral triangles ABF and CAG are constructed in the exterior of a right-angled triangle ABC with ∠C = 90. Let M be the midpoint of BC. Given that MF = 11 and MG = 7, find the length of BC.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to my brother-in-law Quan Nguyễn)

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Let x = \frac{BC}{2}, AB = b and AC = c.

Applying the law of cosines, we have

MG = x + c - 2xc cos(∠ACB + 60) or MG = x + c - 2xc cos150, and MF = x + b - 2xb cos(∠ABC + 60)

Expanding those two equations with MG = 7 and MF = 11, cos(∠ABC + 60) = cos∠ABCcos60 - sin∠ABCsin60,

cos60 = \frac{1}{2}, sin60 = \frac{\sqrt{3}}{2}, cos150 = - \frac{\sqrt{3}}{2} and observe the Pythagoreans theorem, we have

49 = c + x + xc \sqrt{3}

121 = b - x + xc \sqrt{3}

b = c + 4x

Solving for x, we obtain x = 6 or BC = 12.