HMMT 十一月 2017 · GEN 赛 · 第 5 题

HMMT November 2017 — GEN Round — Problem 5

Given that a, b, c are integers with abc = 60, and that complex number ω 6 = 1 satisfies ω = 1, find the ∣ ∣ 2 ∣ ∣ minimum possible value of a + bω + cω .

解析

Given that a, b, c are integers with abc = 60, and that complex number ω 6 = 1 satisfies ω = 1, find the ∣ ∣ 2 ∣ ∣ minimum possible value of a + bω + cω . Proposed by: Ashwin Sah √ Answer: 3 3 2 Since ω = 1, and ω 6 = 1, ω is a third root of unity. For any complex number z , | z | = z · z ¯ . Letting 2 2 z = a + bω + cω , we find that ¯ z = a + cω + bω , and 2 2 2 2 2 2 2 | z | = a + abω + acω + abω + b + bcω + acω + bcω + c 2 2 2 2 = ( a + b + c ) + ( ab + bc + ca )( ω ) + ( ab + bc + ca )( ω ) 2 2 2 = ( a + b + c ) − ( ab + bc + ca ) 1 2 2 2 = (( a − b ) + ( b − c ) + ( c − a ) ) , 2 3 2 where we have used the fact that ω = 1 and that ω + ω = − 1. This quantity is minimized when a, b, and c are as close to each other as possible, making a = 3 , b = 4 , c = 5 the optimal choice, giving 2 | z | = 3. (A smaller value of | z | requires two of a, b, c to be equal and the third differing from them by √ at most 2, which is impossible.) So | z | = 3. min