IMO 1986 SL 7

Origin: FRA

Problem

Let real numbers x1, x2, . . . , xn satisfy 0 < x1 < x2 < \cdot \cdot \cdot < xn < 1 and set x0 = 0, xn+1 = 1. Suppose that these numbers satisfy the following system of equations: n+1  j=0, j̸=i xi −xj = 0 where i = 1, 2, . . ., n. (1) Prove that xn+1−i = 1 −xi for i = 1, 2, . . . , n.

Solution

Let P(x) = (x −x0)(x −x1) \cdot \cdot \cdot (x −xn)(x −xn+1). Then P ′(x) = n+1  j=0 P(x) x −xj and P ′′(x) = n+1  j=0  k̸=j P(x) (x −xj)(x −xk) . Therefore P ′′(xi) = 2P ′(xi)  j̸=i (xi −xj) for i = 0, 1, . . ., n + 1, and the given condition implies P ′′(xi) = 0 for i = 1, 2, . . . , n. Consequently,

x(x −1)P ′′(x) = (n + 2)(n + 1)P(x). (1) It is easy to observe that there is a unique monic polynomial of degree n+2 satisfying differential equation (1). On the other hand, the polynomial Q(x) = (−1)nP(1 −x) also satisfies this equation, is monic, and deg Q = n + 2. Therefore (−1)nP(1 −x) = P(x), and the result follows.