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

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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}{6}</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>
 
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}{6}</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>
  
Here's a graph of how the points move as <cmath>\theta</cmath> increases: https://www.desmos.com/calculator/xtnpzoqkgs
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Here's a graph of how the points move as <cmath>\theta</cmath> increases- https://www.desmos.com/calculator/xtnpzoqkgs
 
 
  
 
Someone pls help with LaTeX formatting, thanks -FlatSquare
 
Someone pls help with LaTeX formatting, thanks -FlatSquare

Revision as of 16:40, 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 \[\theta=\frac{\pi}{6}\] 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. $\boxed{D}$

Here's a graph of how the points move as

\[\theta\] (Error making remote request. Unexpected URL sent back)

increases- https://www.desmos.com/calculator/xtnpzoqkgs

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

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
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