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

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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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Consider <math>n=4</math>. We can set <math>z_{1}</math> and <math>z_{3}</math> arbitrarily and let <math>z_{2}=-z_{1}</math> and <math>z_{4}=-z_{3}</math>. We constructed two pairs of points that cancel each other out. For example, the complex numbers <math>1,-1,e^{i\frac{\pi}{4}},e^{-i\frac{3\pi}{4}}</math>
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Suppose that the problem condition 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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-Solution by Qqqwerw
  
 
==See Also==
 
==See Also==

Revision as of 00:05, 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 problem condition 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

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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