Question

有多少个整数 $n \ge 2$ 具有如下性质：对任意复数 $z_1, z_2, \ldots, z_n$ ，若且则 $z_1, z_2, \ldots, z_n$ 在复平面的单位圆上等间隔分布？ $|z_1| = |z_2| = \cdots = |z_n| = 1$ $z_1 + z_2 + \cdots + z_n = 0,$

How many integers $n \ge 2$ are there such that whenever $z_1, z_2, \ldots, z_n$ are complex numbers such that $|z_1| = |z_2| = \cdots = |z_n| = 1$ and $z_1 + z_2 + \cdots + z_n = 0,$ then the numbers $z_1, z_2, \ldots, z_n$ are equally spaced on the unit circle in the complex plane?

$1$

$2$

$3$

$4$

$5$

答案：B

知识点：单位根复数反例

难度评级：2100

解答：

当 $n = 2$ 时， $z_1 + z_2 = 0$ 强制 $z_2 = -z_1$ 所以两点等间隔分布。当 $n = 3$ 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 $n\ge4$ 都存在反例。例如，取对径点对 $n/2$ 再加上任意其他和为零的集合： $n$ 时取两对方向不同的对径点， $n\ge5$ 时取一个等边三角形加一对对径点。 $(n-3)/2$ $0$

这些数之和为零，但并非等间隔分布。因此只有 $n = 2$ 和 $n = 3$ 满足条件，共 $2$ 个值。

所以正确答案是 B。

For $n = 2,$ $z_1 + z_2 = 0$ forces $z_2 = -z_1,$ which is equally spaced. For $n = 3,$ three unit vectors summing to zero must form an equilateral triangle, so they are equally spaced.

For every even $n\ge4,$ choose $n/2$ antipodal pairs at generic angles that do not form a regular $n$ -gon. For every odd $n\ge5,$ choose the vertices of an equilateral triangle together with $(n-3)/2$ generic antipodal pairs. Each construction has sum $0$ but is not equally spaced.

Hence only $n = 2$ and $n = 3$ work, giving $2$ values.

Thus, the correct answer is B.

2020 AMC 12B 第 23 题

答案：B

难度评级：2100

解答：

其他年份的第 23 题