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

Revision as of 14:54, 14 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

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 $\text{cis}(2\theta)=60$ . From $0 < \theta < \pi/2$ , we have $\theta=\frac{\pi}{2}$ and from $\pi/2 < \theta < \pi$ , we see a monotonic decrease of $\text{cis}(2\theta)$ , from $180$ to $0$ . Hence, there are 2 values that work for $0 < \theta < \pi$ . But since the interval $\pi < \theta < 2\pi$ is identical, because $3\theta=\theta$ at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. $\boxed{D}$

-FlatSquare

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

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

@@ Line 7: / Line 7: @@
 ==Solution==
-Convert <math>z</math> and <math>z^3</math> into <cmath>r\text{cis}\theta</cmath> form, giving <cmath>z=r\text{cis}\theta</cmath> and <cmath>z^3=r^3\text{cis}(3\theta)</cmath>. Since the distance from <math>0</math> to <math>z</math> is <math>r</math>, the distance from <math>0</math> to <math>z^3</math> must also be <math>r</math>, so <math>r=1</math>. Now we must find <cmath>\text{cis}(2\theta)=60</cmath>. From <math>0 < \theta < \pi/2</math>, we have <cmath>\theta=\frac{\pi}{2}</cmath> and from <math>\pi/2 < \theta < \pi</math>, we see a monotonic decrease of <cmath>\text{cis}(2\theta)</cmath>, from <math>180^{\circ}</math> to <math>0^{\circ}</math>. Hence, there are 2 values that work for <math>0 < \theta < \pi</math>. But since the interval <math>\pi < \theta < 2\pi</math> is identical, because <math>3\theta=\theta</math> at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. <math>\boxed{D}</math>
+Convert <math>z</math> and <math>z^3</math> into <cmath>r\text{cis}\theta</cmath> form, giving <cmath>z=r\text{cis}\theta</cmath> and <cmath>z^3=r^3\text{cis}(3\theta)</cmath>. Since the distance from <math>0</math> to <math>z</math> is <math>r</math>, the distance from <math>0</math> to <math>z^3</math> must also be <math>r</math>, so <math>r=1</math>. Now we must find <cmath>\text{cis}(2\theta)=60</cmath>. From <math>0 < \theta < \pi/2</math>, we have <cmath>\theta=\frac{\pi}{2}</cmath> and from <math>\pi/2 < \theta < \pi</math>, we see a monotonic decrease of <cmath>\text{cis}(2\theta)</cmath>, from <math>180</math> to <math>0</math>. Hence, there are 2 values that work for <math>0 < \theta < \pi</math>. But since the interval <math>\pi < \theta < 2\pi</math> is identical, because <math>3\theta=\theta</math> at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. <math>\boxed{D}</math>
 -FlatSquare

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

Revision as of 14:54, 14 February 2019

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