HMMT 二月 2002 · 冲刺赛 · 第 36 题

HMMT February 2002 — Guts Round — Problem 36

[ ± 6] Find the set consisting of all real values of x such that the three numbers 2 , 2 , 2 form a non-constant arithmetic progression (in that order). 6

解析

Find the set consisting of all real values of x such that the three numbers 2 , 2 , 2 form a non-constant arithmetic progression (in that order). Solution: The empty set, Ø . Trivially, x = 0 , 1 yield constant arithmetic progressions; we show that there are no other possibilities. If these numbers do form a progression, then, by the AM-GM (arithmetic mean-geometric mean) inequality, √ 2 3 3 x x x x x 2 · 2 = 2 + 2 ≥ 2 2 · 2 2 3 x ( x + x ) / 2 2 3 ⇒ 2 ≥ 2 ⇒ x ≥ ( x + x ) / 2 2 3 2 ⇒ x ( x − 1) = x − 2 x + x ≤ 0 . 3 2 x x Assuming x 6 = 0 , 1, we can divide by ( x − 1) > 0 and obtain x < 0. However, then 2 , 2 are 2 x less than 1, while 2 is more than 1, so the given sequence cannot possibly be an arithmetic progression.