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

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

Revision as of 01:56, 8 February 2020


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

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

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AMC 12 Problems and Solutions

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