10th Putnam 1950 Problem A1 a and b are positive reals and a > b. Let C be the plane curve r = a - b cos θ. For what values of b/a is C convex?
The curvature is (x'y'' - x''y')/(x' 2 + y' 2 ) 3/2 . So for convexity we require x'y'' ≥ x''y'. Putting k = b/a, c = cos θ, s = sin θ, we have x/a = c - k c 2 , x'/a = - s + 2kcs, x''/a = - c - 2ks 2 + 2kc 2 and y/a = s - ksc, y'/a = c - kc 2 + k s 2 , y''/a = -s + 4ksc. Thus the condition becomes after a little cancellation, 1 + 2k 2 - 3kc ≥ 0. c takes values in the range -1 to 1, so for the condition to be true for all points of the curve we require 1 + 2k 2 - 3k ≥ 0. But 1 + 2k 2 - 3k = (2k - 1)(k - 1). We are given that k  1/2, this becomes a dimple in the surface, so that convexity is broken. For k = 1, the depression in the surface extends as far as the centre (the origin). For k > 0, the curve intersects itself at the origin so that it comprises two ovals one inside the other. 10th Putnam 1950 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002