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Difference between revisions of "2019 AMC 12B Problems/Problem 17"

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(→‎Solution 2)
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To be equilateral triangle, we should have
 
To be equilateral triangle, we should have
  
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<cmath>\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(2\pi/3)</cmath>
+
<cmath>\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)</cmath>
  
 
Simplify left side:
 
Simplify left side:

Revision as of 22:11, 15 February 2019

Problem

How many nonzero complex numbers $z$ have the property that $0, z,$ and $z^3,$ when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}$

Solution 1

Convert $z$ and $z^3$ into \[r\text{cis}\theta\] form, giving \[z=r\text{cis}\theta\] and \[z^3=r^3\text{cis}(3\theta)\]. Since the distance from $0$ to $z$ is $r$, the distance from $0$ to $z^3$ must also be $r$, so $r=1$. Now we must find \[2\theta=\pm\frac{\pi}{3}\]the requirements for being an equilateral triangle. From $0 < \theta < \pi/2$, we have \[\theta=\frac{\pi}{6}\] and from $\pi/2 < \theta < \pi$, we see a monotonic increase of $2\theta$, from $\pi$ to $2\pi$, or equivalently, from $-\pi$ to $0$. Hence, there are 2 values that work for $0 < \theta < \pi$. But since the interval $\pi < \theta < 2\pi$ also consists of $2\theta$ going from $0$ to $2\pi$, it also gives us 2 solutions. Our answer is $\boxed{\textbf{(D) 4}}$

Here's a graph of how $z$ and $z^3$ move as $\theta$ increases- https://www.desmos.com/calculator/xtnpzoqkgs

Someone pls help with LaTeX formatting, thanks -FlatSquare , I did, -Dodgers66

Solution 2

To be equilateral triangle, we should have

\[\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)\]

Simplify left side:

\[z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)\]

That is,

\[z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)\]

We have two roots for both equations, therefore the total number of solution for $z$ is $\boxed{\textbf{(D) }4}$

(By Zhen Qin)

See Also

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

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