HMMT 二月 2004 · 冲刺赛 · 第 42 题
HMMT February 2004 — Guts Round — Problem 42
题目详情
- [18] S is a set of complex numbers such that if u, v ∈ S , then uv ∈ S and u + v ∈ S . Suppose that the number N of elements of S with absolute value at most 1 is finite. What is the largest possible value of N ? 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND
解析
- S is a set of complex numbers such that if u, v ∈ S , then uv ∈ S and u + v ∈ S . Suppose that the number N of elements of S with absolute value at most 1 is finite. What is the largest possible value of N ? Solution: 13 First, if S contained some u 6 = 0 with absolute value < 1, then (by the first condition) every power of u would be in S , and S would contain infinitely many different numbers of absolute value < 1. This is a contradiction. Now suppose S contains some number u of absolute value 1 and argument θ . If θ is not an integer multiple of π/ 6, then u has some power v whose argument lies strictly between θ + π/ 3 and θ + π/ 2. Then 2 2 2 2 2 u + v = u (1 + ( v/u ) ) has absolute value between 0 and 1, since ( v/u ) lies on 2 2 the unit circle with angle strictly between 2 π/ 3 and π . But u + v ∈ S , so this is a contradiction. This shows that the only possible elements of S with absolute value ≤ 1 are 0 and the points on the unit circle whose arguments are multiples of π/ 6, giving N ≤ 1+12 = 13. To show that N = 13 is attainable, we need to show that there exists a possible set S containing all these points. Let T be the set of all numbers of the form a + bω , where 2 a, b are integers are ω is a complex cube root of 1. Since ω = − 1 − ω , T is closed under multiplication and addition. Then, if we let S be the set of numbers u such 2 that u ∈ T , S has the required properties, and it contains the 13 complex numbers specified, so we’re in business.