2000 IMO Shortlist S09

A nonempty set $A$ of real numbers is called a $B_3$ -set if the conditions $a_1, a_2, a_3, a_4, a_5, a_6 \in A$ and $a_1 + a_2 + a_3 = a_4 + a_5 + a_6$ imply that the sequences $(a_1, a_2, a_3)$ and $(a_4, a_5, a_6)$ are identical up to a permutation. Let $A = \{a_0 = 0 < a_1 < a_2 < \cdots \}$ , $B = \{b_0 = 0 < b_1 < b_2 < \cdots \}$ 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 $\{|x-y|\mid x, y \in X \}.$ Prove that if $A$ is a $B_3$ -set, then $A = B.$

如果条件 $a_1, a_2, a_3, a_4, a_5, a_6 \in A$ 和 $a_1 + a_2 + a_3 = a_4 + a_5 + a_6$ 意味着序列 $(a_1, a_2, a_3)$ 和 $(a_4, a_5, a_6)$ 相同，则实数的非空集合 $A$ 称为 $B_3$ -set到一个排列。设 $A = \{a_0 = 0 < a_1 < a_2 < \cdots \}$ ， $B = \{b_0 = 0 < b_1 < b_2 < \cdots \}$ 为无限实数序列，$D(A) = D(B),$ 其中，对于一组 $X$ 实数，$D(X)$ 表示差集 $\{|x-y|\mid x, y \in X \}.$ 证明如果 $A$ 是 $B_3$ 集合，则 $A = B.$