30th IMO 1989 shortlist

The real numbers a₀, a₁, ... , a_n satisfy a₀ = a_n = 0, a_k = c + ∑_i=k^n-1 a_i-k(a_i + a_i+1). Show that c <= 1/(4n).

Solution

This is awkward, because it is not easy to get an expression for any a_i in terms of c (apart from i = 0, n-1, n). The key is not to try!

Put s_k = a₀ + a₁ + ... + a_k. Then we have s_n = s_n-1 = ∑_k=0..n-1 (c + ∑_i=k..n-1 a_i-k(a_i + a_i+1) ) = nc + ∑_i=0..n-1 (a_i + a_i+1) ∑_k=0..i a_k. (As usual, inverting the order of summation requires a little care, but is basically straightforward. Just check the low and high values.). So s_n = nc + ∑_i=0..n-1 (a_i + a_i+1) s_i = (s₂ - s₀)s₁ + (s₃ - s₁)s₂ + (s₄ - s₂)s₃ + ... + (s_n - s_n-2)s_n-1 = s_ns_n-1 (we have s₀ = 0 and the other terms cancel in pairs) = s_n².

So nc = s_n - s_n² = ¼ - (s_n - ½)² ≤ ¼. Hence c ≤ 1/4n.

30th IMO shortlist 1989

© John Scholes
jscholes@kalva.demon.co.uk
28 Dec 2002
Last corrected/updated 28 Dec 02