2020 AMC 12B 第 23 题

先试着解答 2020 AMC 12B 第 23 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2020 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

23.

有多少个整数 n≥2n \ge 2 具有如下性质:只要复数 z1z_1、z2z_2、…\ldots、znz_n 满足 ∣z1∣=∣z2∣=⋯=∣zn∣=1 |z_1| = |z_2| = \cdots = |z_n| = 1 以及 z1+z2+⋯+zn=0, z_1 + z_2 + \cdots + z_n = 0\text{,} 那么 z1z_1、z2z_2、…\ldots、znz_n 在复平面的单位圆上等间隔分布?
How many integers n≥2n \ge 2 are there such that whenever z1,z_1, z2,z_2, …,\ldots, znz_n are complex numbers such that ∣z1∣=∣z2∣=⋯=∣zn∣=1 |z_1| = |z_2| = \cdots = |z_n| = 1 and z1+z2+⋯+zn=0, z_1 + z_2 + \cdots + z_n = 0, then the numbers z1,z_1, z2,z_2, …,\ldots, znz_n are equally spaced on the unit circle in the complex plane?
11
22
33
44
55
答案:B
知识点:单位根复数反例
小提示:
检查小情形:n=2n = 2 强制两点互为对径,n=3n = 3 强制形成等边三角形。
Check small cases: n=2n = 2 forces antipodal points and n=3n = 3 forces an equilateral triangle
大提示:
对 n≥4n \ge 4 可用一个较小的和为零构型加上一对对径点构造反例。
For n≥4,n \ge 4, build a counterexample by combining a smaller balanced configuration with an antipodal pair
解答:
当 n=2n = 2 时,z1+z2=0z_1 + z_2 = 0 强制 z2=−z1z_2 = -z_1,所以两点等间隔分布。当 n=3n = 3 时,三个和为零的单位向量必须构成等边三角形,所以也等间隔分布。
对每个偶数 n≥4n\ge4,取 n2\frac{n}{2} 对处于一般位置、且不构成正 nn 边形的对径点。对每个奇数 n≥5n\ge5,取一个等边三角形的三个顶点,再加上 n−32\frac{n-3}{2} 对处于一般位置的对径点。每种构造的各数之和都是 00,但它们并不等间隔分布。
因此只有 n=2n = 2 和 n=3n = 3 满足条件,共 22 个值。
所以正确答案是 B。
For n=2,n = 2, z1+z2=0z_1 + z_2 = 0 forces z2=−z1,z_2 = -z_1, which is equally spaced. For n=3,n = 3, three unit vectors summing to zero must form an equilateral triangle, so they are equally spaced.
For every even n≥4,n\ge4, choose n2\frac{n}{2} antipodal pairs at generic angles that do not form a regular nn-gon. For every odd n≥5,n\ge5, choose the vertices of an equilateral triangle together with n−32\frac{n-3}{2} generic antipodal pairs. Each construction has sum 00 but is not equally spaced.
Hence only n=2n = 2 and n=3n = 3 work, giving 22 values.
Thus, the correct answer is B.
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