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

(Solution)
m (Solution)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
Convert z 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 0 to z is r, the distance from 0 to z^3 must also be r, so r=1. Now we must find <cmath>\text{cis}(2\theta)=60</cmath>. From 0 < theta < pi/2, we have <cmath>\theta=\frac{\pi}{2}</cmath> and from pi/2 < theta < pi, we see a monotonic decrease of <cmath>\text{cis}(2\theta)</cmath>, from 180 to 0. Hence, there are 2 values that work for 0 < theta < pi. But since the interval pi < theta < 2pi is identical, because 3theta=theta 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^{\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>
  
 
-FlatSquare
 
-FlatSquare
  
 
Someone pls help with LaTeX formatting, thanks
 
Someone pls help with LaTeX formatting, thanks
 +
, I did, -Dodgers66
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=16|num-a=18}}
 
{{AMC12 box|year=2019|ab=B|num-b=16|num-a=18}}

Revision as of 15:53, 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 $180^{\circ}$ to $0^{\circ}$. 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

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
Invalid username
Login to AoPS