2020 AMC 12B Problems/Problem 23

Problem 23

How many integers $n \geq 2$ are there such that whenever $z_1, z_2, ..., z_n$ are complex numbers such that

$|z_1| = |z_2| = ... = |z_n| = 1 \text{ and } z_1 + z_2 + ... + z_n = 0,$ then the numbers $z_1, z_2, ..., z_n$ are equally spaced on the unit circle in the complex plane?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5$

Solution

For $n=2$ , we see that if $z_{1}+z_{2}=0$ , then $z_{1}=-z_{2}$ , so they are evenly spaced along the unit circle.

For $n=3$ , WLOG, we can set $z_{1}=1$ . Notice that now $\Re(z_{2}+z{3})=-1$ and $\Im\{z_{2}\}=-\Im\{z_{3}\}$ . This forces $z_{2}$ and $z_{3}$ to be equal to $e^{i\frac{2\pi}{3}}$ and $e^{-i\frac{2\pi}{3}}$ , meaning that all three are equally spaced along the unit circle.

We can now show that we can construct complex numbers when $n\geq 4$ that do not satisfy the conditions in the problem.

Suppose that the condition in the problem holds for some $n=k$ . We can now add two points $z_{k+1}$ and $z_{k+2}$ anywhere on the unit circle such that $z_{k+1}=-z_{k+2}$ , which will break the condition. Now that we have shown that $n=2$ and $n=3$ works, by this construction, any $n\geq 4$ does not work, making the answer $\boxed{\mathbf(B) 2}$ .

-Solution by Qqqwerw

Video Solution

On The Spot STEM: https://www.youtube.com/watch?v=JOgSOni5HhM

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2020 AMC 12B Problems/Problem 23

Contents

Problem 23

Solution

Video Solution

See Also