45th Putnam 1984 Problem B4 Find all real valued functions f(x) defined on [0, ∞), such that (1) f is continuous on [0, ∞), (2) f(x) > 0 for x > 0, (3) for all x 0 > 0, the centroid of the region under the curve y = f(x) between 0 and x 0 has y- coordinate equal to the average value of f(x) on [0, x 0 ]. Solution
: f(x) = A x 1 + √2 . The condition is ∫ 0 x 1/2 f 2 (t) dt = 1/x (∫ 0 x f(t) dt) 2 . Put y = ∫ 0 x f(t) dt and differentiate to get 1/2 (y') 2 = 2yy'/x - y 2 /x 2 , or x 2 (y') 2 - 4xyy' + 2y 2 = 0. At this point it is easy to get stuck. The key is to factorize the equation: (x y' - (2 + √2) y ) (x y' - (2 - √2) y ) = 0. Hence y = Ax 2+√2 or Ax 2-√2 . Hence f(x) = Bx 1+√2 or Bx 1-√2 . But f(0) is defined, so f(x) = Bx 1+√2 at least for small x. There remains the possibility that we might switch to Cx 1-√2 at some x 0 . But it is easy to see that if we fix C so that B is continuous, then for x > x 0 , ∫ 0 x f(t) dt will have a non-zero constant term as well as an x 2 - √2 term and hence we will not satisfy the required equation. So the only possible solutions are those given. 45th Putnam 1984 © John Scholes jscholes@kalva.demon.co.uk 7 Jan 2001