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

Revision as of 00:55, 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 the points 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 (Z^3-Z)/(Z-0)=cis(pi/3) or cis(2*pi/3) simplify left side Z^2-1=cis(pi/3) or Z^2-1=cis(2*pi/3) That is Z^2=1+cis(pi/3) or Z^2=1+cis(2*pi/3) we have two roots for both equations, therefore the total number of solution for Z is 4

(By Zhen Qin)

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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Difference between revisions of "2019 AMC 12B Problems/Problem 17"

Revision as of 00:55, 15 February 2019

Contents

Problem

Solution 1

Solution 2

See Also

@@ Line 20: / Line 20: @@
 (Z^3-Z)/(Z-0)=cis(pi/3)   or  cis(2*pi/3)
 simplify left side
-Z^2-1=cis(pi/3) or cis(2*pi/3)
+Z^2-1=cis(pi/3) or Z^2-1=cis(2*pi/3)
-Z^2=1+cis(pi/3) or 1+cis(2*pi/3)
+That is Z^2=1+cis(pi/3) or Z^2=1+cis(2*pi/3)
 we have two roots for both equations, therefore the total number of solution for Z is 4