HMMT 二月 2004 · 冲刺赛 · 第 42 题

英文解析

42. 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 numbersof absolute value < 1. This is a contradiction. Now suppose S contains some numberu of absolute value 1 and argument θ . If θ is not an integer multiple of π/ 6, then uhas 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 acontradiction.
    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 Scontaining all these points. Let T be the set of all numbers of the form a + bω , wherea, b are integers are ω is a complex cube root of 1. Since ω = − 1 − ω , T is closed 2
    under multiplication and addition. Then, if we let S be the set of numbers u suchthat u ∈ T , S has the required properties, and it contains the 13 complex numbers 2
    specified, so we’re in business.