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 x0 > 0, the centroid of the region under the curve y = f(x) between 0 and x0 has y-coordinate equal to the average value of f(x) on [0, x0].
Answer: f(x) = A x1 + √2.
The condition is ∫0x 1/2 f2(t) dt = 1/x (∫0x f(t) dt)2. Put y = ∫0x f(t) dt and differentiate to get 1/2 (y')2 = 2yy'/x - y2/x2, or x2(y')2 - 4xyy' + 2y2 = 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 = Ax2+√2 or Ax2-√2. Hence f(x) = Bx1+√2 or Bx1-√2. But f(0) is defined, so f(x) = Bx1+√2 at least for small x. There remains the possibility that we might switch to Cx1-√2 at some x0. But it is easy to see that if we fix C so that B is continuous, then for x > x0, ∫0x f(t) dt will have a non-zero constant term as well as an x2 - √2 term and hence we will not satisfy the required equation. So the only possible solutions are those given.
45th Putnam 1984
© John Scholes
7 Jan 2001