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)**

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

Applying the law of cosines, we have

MG² = x² + c² - 2xc cos(∠ACB + 60°) or MG² = x² + c² - 2xc cos150°, and MF² = x² + b² - 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 Pythagorean’s theorem, we have

49 = c² + x² + xc \sqrt{3}

121 = b² - x² + xc \sqrt{3}

b² = c² + 4x²

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