IMO 2000 SL A6

Origin: IRE | Category: Algebra

Problem

A nonempty set A of real numbers is called a B3-set if the conditions a1, a2, a3, a4, a5, a6 \inA and a1+a2+a3 = a4+a5+a6 imply that the sequences (a1, a2, a3) and (a4, a5, a6) are identical up to a permutation. Let A = {a0 = 0 < a1 < a2 < \cdot \cdot \cdot }, B = {b0 = 0 < b1 < b2 < \cdot \cdot \cdot } be infinite sequences of real numbers with D(A) = D(B), where, for a set X of real numbers, D(X) denotes the difference set {|x −y| | x, y \inX}. Prove that if A is a B3-set, then A = B.

Solution

Since D(A) = D(B), we can define f(i) > g(i) \geq0 that satisfy bi −bi−1 = af(i) −ag(i) for all i. The number bi+1 −bi−1 \inD(B) = D(A) can be written in the form au −av, u > v \geq0. Then bi+1 −bi−1 = bi+1 −bi + bi −bi−1 implies

af(i+1) + af(i) + av = ag(i+1) + ag(i) + au, so the B3 property of A implies that (f(i + 1), f(i), v) and (g(i + 1), g(i), u) coincide up to a permutation. It follows that either f(i + 1) = g(i) or f(i) = g(i + 1). Hence if we define R = {i \inN0 | f(i + 1) = g(i)} and S = {i \inN0 | f(i) = g(i + 1)} it holds that R \cupS = N0. Lemma. If i \inR , then also i + 1 \inR. Proof. Suppose to the contrary that i \inR and i + 1 \inS, i.e., g(i) = f(i+ 1) = g(i+ 2). There are integers x and y such that bi+2 −bi−1 = ax −ay. Then ax −ay = af(i+2) −ag(i+2) + af(i+1) −ag(i+1) + af(i) − ag(i) = af(i+2) + af(i) −ag(i+1) −ag(i), so by the B3 property (x, g(i + 1), g(i)) and (y, f(i + 2), f(i)) coincide up to a permutation. But this is impossible, since f(i+2), f(i) > g(i+2) = g(i) = f(i+1) > g(i+1). This proves the lemma. Therefore if i \inR ̸= \emptyset, then it follows that every j > i belongs to R. Consequently g(i) = f(i + 1) > g(i + 1) = f(i + 2) > g(i + 2) = f(i + 3) > \cdot \cdot \cdot is an infinite decreasing sequence of nonnegative integers, which is impossible. Hence S = N0, i.e., bi+1 −bi = af(i+1) −af(i) for all i \inN0. Thus f(0) = g(1) < f(1) < f(2) < \cdot \cdot \cdot , implying f(i) \geqi. On the other hand, for any i there exist j, k such that af(i)−ai = bj −bk = af(j)−af(k), so by the B3 property i \in{f(i), f(k)} is a value of f. Hence we must have f(i) = i for all i, which finally gives A = B.