BMO 1966 Problem 9 The angles of a triangle are A, B, C. Find the smallest possible value of tan A/2 + tan B/2 + tan C/2 and the largest possible value of tan A/2 tan B/2 tan C/2. Solution
: equilateral in both cases, √3, 1/(3√3). Using the familiar tan(a + x) = (tan a + tan x)/(1 - tan a tan x), we have that for a and x in the interval [0, π/4], tan(a + x) + tan(a - x) =2 tan a (1 + tan 2 x)/(1 - tan 2 a tan 2 x) >= 2 tan a, with equality iff x = 0. Hence the minimum value of tan A/2 + tan B/2 + tan C/2 occurs when A = B = C (if any pair of A, B, C are unequal then we can lower the value by replacing the pair by two angles equal to their mean). Similarly, we show that for a and x in the interval (0, π/4), tan(a - x) tan(a + x) < tan 2 a. The lhs = (tan 2 a - tan 2 x)/(1 - tan 2 a tan 2 x) = tan 2 a - tan 2 x (1 - tan 4 a)/(1 - tan 2 a tan 2 x) < tan 2 a. This proves that the maximum value of tan A/2 tan B/2 tan C/2 occurs for A = B = C. For if A ≠ B, then we can increase the value by replacing A and B by their mean (and similarly for any other unequal pair). This does not deal with the degenerate case of a flat triangle. But if the angles are A = 2δ, B = 2ε, C = π - (2δ + 2ε), where δ and ε are small, then tan A/2 tan B/2 tan C/2 = δ ε/(δ + ε) which tends to zero as δ and ε tend to zero. So the flat triangle gives a smaller value. 2nd BMO 1966 © John Scholes jscholes@kalva.demon.co.uk 30 Aug 2000