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2019 AMC 12B Problems/Problem 17: Difference between revisions

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==Solution 1==
==Solution 1==


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>
Convert <math>z</math> and <math>z^3</math> into modulus-argument (polar) form, giving <math>z=r\text{cis}(\theta)</math> for some <math>r</math> and <math>\theta</math>. Thus, by De Moivre's Theorem, <math>z^3=r^3\text{cis}(3\theta)</math>. Since the distance from <math>0</math> to <math>z</math> is <math>r</math>, and the triangle is equilateral, the distance from <math>0</math> to <math>z^3</math> must also be <math>r</math>, so <math>r^3=r</math>, giving <math>r=1</math>. (We know <math>r \neq 0</math> since the problem statement specifies that <math>z</math> must be nonzero.)


Here's a graph of how <math>z</math> and <math>z^3</math> move as <math>\theta</math> increases- https://www.desmos.com/calculator/xtnpzoqkgs
Now, to get from <math>z</math> to <math>z^3</math>, which should be a rotation of <math>120^{\circ}</math> if the triangle is equilateral, we multiply by <math>z^2 = r^2\text{cis}(2\theta)</math>, again using De Moivre's Theorem. Thus we require <math>2\theta=\pm\frac{\pi}{3} + 2\pi k</math> (where <math>k</math> can be any integer). If <math>0 < \theta < \frac{\pi}{2}</math>, we must have <math>\theta=\frac{\pi}{6}</math>, while if <math>\frac{\pi}{2} \leq \theta < \pi</math>, we must have <math>\theta = \frac{5\pi}{6}</math>. Hence there are <math>2</math> values that work for <math>0 < \theta < \pi</math>. By symmetry, the interval <math>\pi \leq \theta < 2\pi</math> will also give <math>2</math> solutions. The answer is thus <math>2 + 2 = \boxed{\textbf{(D) }4}</math>.
 
''Note'': Here's a graph showing how <math>z</math> and <math>z^3</math> move as <math>\theta</math> increases: https://www.desmos.com/calculator/xtnpzoqkgs.


==Solution 2==
==Solution 2==


To be equilateral triangle, we should have
For the triangle to be equilateral, the vector from <math>z</math> to <math>z^3</math>, i.e <math>z^3 - z</math>, must be a <math>60^{\circ}</math> rotation of the vector from <math>0</math> to <math>z</math>, i.e. just <math>z</math>. Thus we must have


<cmath>\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)</cmath>
<cmath>\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)</cmath>


Simplify left side:
Simplifying gives
 
<cmath>z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)</cmath>
<cmath>z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)</cmath>
 
so
That is,
 
<cmath>z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)</cmath>
<cmath>z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)</cmath>


We have two roots for both equations, therefore the total number of solution for <math>z</math> is <math>\boxed{\textbf{(D) }4}</math>
Since any nonzero complex number will have two square roots, each equation gives two solutions. Thus, as before, the total number of possible values of <math>z</math> is <math>\boxed{\textbf{(D) }4}</math>.
 
(By Zhen Qin)


==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}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 22:31, 18 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 modulus-argument (polar) form, giving $z=r\text{cis}(\theta)$ for some $r$ and $\theta$. Thus, by De Moivre's Theorem, $z^3=r^3\text{cis}(3\theta)$. Since the distance from $0$ to $z$ is $r$, and the triangle is equilateral, the distance from $0$ to $z^3$ must also be $r$, so $r^3=r$, giving $r=1$. (We know $r \neq 0$ since the problem statement specifies that $z$ must be nonzero.)

Now, to get from $z$ to $z^3$, which should be a rotation of $120^{\circ}$ if the triangle is equilateral, we multiply by $z^2 = r^2\text{cis}(2\theta)$, again using De Moivre's Theorem. Thus we require $2\theta=\pm\frac{\pi}{3} + 2\pi k$ (where $k$ can be any integer). If $0 < \theta < \frac{\pi}{2}$, we must have $\theta=\frac{\pi}{6}$, while if $\frac{\pi}{2} \leq \theta < \pi$, we must have $\theta = \frac{5\pi}{6}$. Hence there are $2$ values that work for $0 < \theta < \pi$. By symmetry, the interval $\pi \leq \theta < 2\pi$ will also give $2$ solutions. The answer is thus $2 + 2 = \boxed{\textbf{(D) }4}$.

Note: Here's a graph showing how $z$ and $z^3$ move as $\theta$ increases: https://www.desmos.com/calculator/xtnpzoqkgs.

Solution 2

For the triangle to be equilateral, the vector from $z$ to $z^3$, i.e $z^3 - z$, must be a $60^{\circ}$ rotation of the vector from $0$ to $z$, i.e. just $z$. Thus we must have

\[\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)\]

Simplifying gives \[z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)\] so \[z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)\]

Since any nonzero complex number will have two square roots, each equation gives two solutions. Thus, as before, the total number of possible values of $z$ is $\boxed{\textbf{(D) }4}$.

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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