HMMT February

Algebra Difficulty 4.8 AIME Find the answer

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 $|x-y|+|y-z|$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.