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Algebra Difficulty 5.0 AIME Prove it United States

Problem:

The real numbers $x, y, z$ satisfy $0 \leq x \leq y \leq z \leq 4$ . If their squares form an arithmetic progression with common difference $2$ , determine the minimum possible value of $|x-y| + |y-z|$ .

Solution

Solution:

$|x-y| + |y-z| = z - x = \frac{z^2 - x^2}{z + x} = \frac{4}{z + x}$ , which is minimized when $z = 4$ and $x = \sqrt{12}$ . Thus, our answer is $4 - \sqrt{12} = 4 - 2\sqrt{3}$ .

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