HMMT 二月 2013 · 冲刺赛 · 第 2 题
HMMT February 2013 — Guts Round — Problem 2
题目详情
- [ 4 ] The real numbers x, y, z satisfy 0 ≤ x ≤ y ≤ z ≤ 4. If their squares form an arithmetic progression with common difference 2, determine the minimum possible value of | x − y | + | y − z | .
解析
- [ 4 ] The real numbers x, y, z satisfy 0 ≤ x ≤ y ≤ z ≤ 4. If their squares form an arithmetic progression with common difference 2, determine the minimum possible value of | x − y | + | y − z | . √ 2 2 z − x 4 Answer: 4 − 2 3 Clearly | x − y | + | y − z | = z − x = = , which is minimized when z = 4 z + x z + x √ √ √ and x = 12. Thus, our answer is 4 − 12 = 4 − 2 3.