Japan Mathematical Olympiad

Question

Consider an equilateral triangle ABC with each side measuring 10. A circle passing through A touches the side BC (excluding the endpoints) at point X, and intersects sides AB and AC at points D and E respectively, both of which are different from point A. Given that BX > CX and AD + AE = 13, find the length of segment BX. FIGURE_0

Accepted Answer

5 + 10 Let x = BX. Since CX = 10 - x and BX > CX, we have 5 < x < 10. By the power of a point theorem, x^2 + (10-x)^2 = BX^2 + CX^2 = BD BA + CE CA = 10(BD + CE) holds. Also, from the condition AD + AE = 13, we find BD + CE = (10 - AD) + (10 - AE) = 20 - (AD + AE) = 7 and thus x^2 + (10 - x)^2 = 70 Since 5 < x < 10, this yields x = 5 + 10.