4th USAMO 1975 Problem 3 A polynomial p(x) of degree n satisfies p(0) = 0, p(1) = 1/2, p(2) = 2/3, ... , p(n) = n/(n+1). Find p(n+1).
The polynomial (x + 1)p(x) - x has degree n+1 and zeros at 0, 1, 2, ... , n, so it must be k x (x - 1)(x - 2) ... (x - n). Also it has value 1 at x = -1, so k (n+1)! (-1) n+1 . Hence (n+2) p(n+1) - (n+1) = (-1) n+1 . So for n odd, p(n+1) = 1, and for n even, p(n+1) = n/(n+2). 4th USAMO 1975 © John Scholes jscholes@kalva.demon.co.uk 11 Jul 2002