9th Eötvös 1902 Problem 3 The triangle ABC has area k and angle A = θ, and is such that BC is as small as possible. Find AB and AC.
Take AB = c, AC = b, then BC 2 = b 2 + c 2 - 2bc cos θ = (b-c) 2 + 2bc(1 - cos θ). But k = ½bc sin θ, or bc = 2k/sin θ. Hence BC 2 = (b-c) 2 + 4k(1 - cos θ)/sin θ. Since the second term is fixed, we minimise BC by taking b = c. So k = ½bc sin θ, so b = c = √(2k/sin θ). 9th Eötvös 1902 © John Scholes jscholes@kalva.demon.co.uk 29 Oct 2003 Last corrected/updated 29 Oct 03