Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 12B Problems/Problem 23"

m (→‎Solution 2)
m (→‎Solutions)
Line 8: Line 8:
  
 
== Solutions ==
 
== Solutions ==
−
=== Solution 1 ===
+
 
 
For <math>n=2</math>, we see that if <math>z_{1}+z_{2}=0</math>, then <math>z_{1}=-z_{2}</math>, so they are evenly spaced along the unit circle.
 
For <math>n=2</math>, we see that if <math>z_{1}+z_{2}=0</math>, then <math>z_{1}=-z_{2}</math>, so they are evenly spaced along the unit circle.
  

Revision as of 10:55, 20 June 2022

Problem

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?

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

Solutions

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 $\boxed{\textbf{(B)} 2}$.

-Solution by Qqqwerw

Video Solution

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

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_23&oldid=175097"