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

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== Problem ==
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==Problem 23==
 
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How many integers <math>n \geq 2</math> are there such that whenever <math>z_1, z_2, ..., z_n</math> are complex numbers such that
 
How many integers <math>n \geq 2</math> are there such that whenever <math>z_1, z_2, ..., z_n</math> are complex numbers such that
  
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<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math>
 
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math>
  
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[[2020 AMC 12B Problems/Problem 23|Solution]]
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== Solution ==
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==Solution==
 
  
 
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.
  
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For <math>n=3</math>, WLOG, we can set <math>z_{1}=1</math>. Notice that now <math>\Re(z_{2}+z{3})=-1</math> and <math>\Im\{z_{2}\}=-\Im\{z_{3}\}</math>. This forces <math>z_{2}</math> and <math>z_{3}</math> to be equal to <math>e^{i\frac{2\pi}{3}}</math> and <math>e^{-i\frac{2\pi}{3}}</math>, meaning that all three are equally spaced along the unit circle.
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For <math>n=3</math>, WLOG, we can set <math>z_{1}=1</math>. Notice that now Re<math>(z_{2}+z_{3})=-1</math> and Im<math>\{z_{2}\}</math> = <math>-</math>Im<math>\{z_{3}\}</math>. This forces <math>z_{2}</math> and <math>z_{3}</math> to be equal to <math>e^{i\frac{2\pi}{3}}</math> and <math>e^{-i\frac{2\pi}{3}}</math>, meaning that all three are equally spaced along the unit circle.
  
 
We can now show that we can construct complex numbers when <math>n\geq 4</math> that do not satisfy the conditions in the problem.
 
We can now show that we can construct complex numbers when <math>n\geq 4</math> that do not satisfy the conditions in the problem.
  
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Suppose that the condition in the problem holds for some <math>n=k</math>. We can now add two points <math>z_{k+1}</math> and <math>z_{k+2}</math> anywhere on the unit circle such that <math>z_{k+1}=-z_{k+2}</math>, which will break the condition. Now that we have shown that <math>n=2</math> and <math>n=3</math> works, by this construction, any <math>n\geq 4</math> does not work, making the answer <math>\boxed{\mathbf(B) 2}</math>.
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Suppose that the condition in the problem holds for some <math>n=k</math>. We can now add two points <math>z_{k+1}</math> and <math>z_{k+2}</math> anywhere on the unit circle such that <math>z_{k+1}=-z_{k+2}</math>, which will break the condition. Now that we have shown that <math>n=2</math> and <math>n=3</math> works, by this construction, any <math>n\geq 4</math> does not work, making the answer <math>\boxed{\textbf{(B)} 2}</math>.
  
 
-Solution by Qqqwerw
 
-Solution by Qqqwerw
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-Minor edit made by HappySharks
  
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==Video Solution==
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== Video Solution by On The Spot STEM ==
 
https://www.youtube.com/watch?v=JOgSOni5HhM
 
https://www.youtube.com/watch?v=JOgSOni5HhM
  

Latest revision as of 16:45, 28 January 2024

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?

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

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

-Solution by Qqqwerw -Minor edit made by HappySharks

Video Solution by 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

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