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Philippine Mathematical Olympiad

Algebra Difficulty 5.3 AIME, harder Prove it Philippines

Problem:
Find all nonempty finite sets XX of real numbers with the following property:
x+∣x∣∈X for all x∈X x + |x| \in X \quad \text{ for all } x \in X

Solution

Solution:
Let X={x1,x2,…,xn}X = \{x_1, x_2, \ldots, x_n\}, n≥1n \geq 1, where x1<x2<⋯<xnx_1 < x_2 < \cdots < x_n.

If xn>0x_n > 0, then xn+∣xn∣=2xn∈Xx_n + |x_n| = 2x_n \in X, which is a contradiction because xn<2xnx_n < 2x_n but xnx_n is the largest element of XX.

The contradiction in the previous paragraph implies that xn≤0x_n \leq 0. If x1<0x_1 < 0, then x1+∣x1∣=x1−x1=0∈Xx_1 + |x_1| = x_1 - x_1 = 0 \in X. Hence, we must have xn=0x_n = 0, so that xi+∣xi∣=xi−xi=0∈Xx_i + |x_i| = x_i - x_i = 0 \in X for any xi∈Xx_i \in X, i=1,2,…,ni = 1, 2, \ldots, n.

Hence, in order for the desired property to be satisfied, XX must be a finite subset of the interval (−∞,0](-\infty, 0] and it must contain 00. On the other hand, such subsets satisfy the said property.

The only nonempty finite sets that satisfy the desired property are those finite subsets of (−∞,0](-\infty, 0] containing 00.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.