12th CanMO 1980 Problem 3 Find the triangle with given angle A and given inradius r with the smallest perimeter.
It is fairly obvious that we take angle B = angle C. But we need to prove it. The perimeter is 2r (cot A/2 + cot B/2 + cot C/2). Both 2r and cot A/2 are fixed, so we have to minimise cot B/2 + cot C/2. Write B/2 = x + y, C/2 = x - y, where x = (B+C)/4, y = (B-C)/4. So we have cot(x + y) + cot(x - y) = (1 - tan x tan y)/(tan x + tan y) + (1 + tan x tan y)/(tan x - tan y) = (2 tan x + 2 tan x tan 2 y)/(tan 2 x - tan 2 y). Now tan x is fixed (because A and hence B+C is fixed), so we both minimise the numerator and maximise the denominator by taking B = C. 12th CanMO 1980 © John Scholes jscholes@kalva.demon.co.uk 21 June 2002