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

m (→‎Solution)
(→‎Solution: fixed a few formatting issues and a few errors regarding the use of cis(theta) to mean the argument of a complex number, switched all degrees to radians, and fixed an incorrect assertion.)
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==Solution==
 
==Solution==
  
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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>
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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>2\theta=\pm\frac{\pi}{3}</cmath>the requirements for being an equilateral triangle. 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 increase of <math>2\theta</math>, from <math>\pi</math> to <math>2\pi</math>, or equivalently, from <math>-\pi</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> also consists of <math>2\theta</math> going from <math>0</math> to <math>2\pi</math>, it also gives us 2 solutions. Our answer is <math>\boxed{\textbf{(D)  4}}</math>
  
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Here's a graph of how the points move as <cmath>\theta</cmath> increases- https://www.desmos.com/calculator/xtnpzoqkgs
+
Here's a graph of how the points move as <math>\theta</math> 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 16:08, 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 \[2\theta=\pm\frac{\pi}{3}\]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

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
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