HMMT 二月 2017 · 冲刺赛 · 第 16 题

HMMT February 2017 — Guts Round — Problem 16

[ 9 ] Let a and b be complex numbers satisfying the two equations 3 2 a − 3 ab = 36 3 2 b − 3 ba = 28 i. Let M be the maximum possible magnitude of a . Find all a such that | a | = M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . February 2017, February 18, 2017 — GUTS ROUND Organization Team Team ID#

解析

[ 9 ] Let a and b be complex numbers satisfying the two equations 3 2 a − 3 ab = 36 3 2 b − 3 ba = 28 i. Let M be the maximum possible magnitude of a . Find all a such that | a | = M . Proposed by: Alexander Katz √ √ 3 3 i 3 3 3 i 3 Answer: 3 , − + , − − 2 2 2 2 Notice that 3 3 2 2 3 ( a − bi ) = a − 3 a bi − 3 ab + b i 3 2 3 2 = ( a − 3 ab ) + ( b − 3 ba ) i = 36 + i (28 i ) = 8 so that a − bi = 2 + i . Additionally 3 3 2 2 3 ( a + bi ) = a + 3 a bi − 3 ab − b i 3 2 3 2 = ( a − 3 ab ) − ( b − 3 ba ) i = 36 − i (28 i ) = 64 ′ ′ ′ It follows that a − bi = 2 ω and a + bi = 4 ω where ω, ω are third roots of unity. So a = ω + 2 ω . From ′ ′ the triangle inequality | a | ≤ | ω | + | 2 ω | = 3, with equality when ω and ω point in the same direction ′ 2 (and thus ω = ω ). It follows that a = 3 , 3 ω, 3 ω , and so √ √ 3 3 i 3 3 3 i 3 a = 3 , − + , − − 2 2 2 2