Putnam 1992 Problem B1 Let R be the reals. Let S ⊆ R have n ≥ 2 elements. Let A S = { x ∈ R : x = (s + t)/2 for some s, t ∈ S with s ≠ t}. What is the smallest possible |A S |? Solution
: 2n - 3. Easy. {1, 2, ... , n} gives 2n - 3 averages, namely 1 1/2, 2, 2 1/2, ... , n - 1/2, so it remains to show that we cannot do better. Suppose a 1  a n , so x ∉ A S . Similarly, y = (a n+1 + a n-1 )/2 > (a n + a n-1 )/2, which is the largest element of A S , so y ∉ A S . Hence |A S' | ≥ |A S | + 2. Putnam 1992 © John Scholes jscholes@kalva.demon.co.uk 24 Dec 1998