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

Revision as of 21:11, 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)

@@ Line 18: / Line 18: @@
 To be equilateral triangle, we should have
-<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:

2019 AMC 12B (Problems • Answer Key • Resources)
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Difference between revisions of "2019 AMC 12B Problems/Problem 17"

Revision as of 21:11, 15 February 2019

Contents

Problem

Solution 1

Solution 2

See Also