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

(Undo revision 230966 by Elpianista227 (talk))
(Tag: Undo)
Line 73: Line 73:
 
<cmath>(a + bi)^3 = (a^3 - 3ab^2) + i(3a^2b - b)</cmath>
 
<cmath>(a + bi)^3 = (a^3 - 3ab^2) + i(3a^2b - b)</cmath>
 
Now, we let <math>z = z^3</math> and equate real and complex parts. It then follows that :  
 
Now, we let <math>z = z^3</math> and equate real and complex parts. It then follows that :  
−
<cmath>a^3 - 3ab^2, 3a^2b - b = b</cmath>
+
<cmath>a^3 - 3ab^2 = a, 3a^2b - b = b</cmath>
 
from which solving gives  
 
from which solving gives  
−
<math></math>a = \pm\frac{\sqrt6}{3}, b = \pm\frac{<math>i</math>}{3}<math></math>
+
<cmath>a^2 - 3b^2 = 1 (1), 3a^2 - 1 = 1 (2)</cmath>
 +
<cmath>3a^2 = 2 -> a^2 = \frac{2}{3}</cmath>
 +
and
 +
<cmath>\frac{2}{3} - 3b^2 = 1</cmath>
 +
Focusing on (1) then gives
 +
<cmath>b^2 = \frac{1}{9}</cmath>
 +
Now, finding the square roots gives
 +
<cmath>a = \pm\frac{\sqrt6}{3}, b = \pm\frac{i}{3}</cmath>
 
And so there are <math>\boxed{(D)4}</math> possible solutions.  
 
And so there are <math>\boxed{(D)4}</math> possible solutions.  
 
~elpianista227
 
~elpianista227

Revision as of 10:15, 30 October 2024

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 $60^{\circ}$ if the triangle is equilateral, we multiply by $z^2 = \text{cis}(2\theta)$, again using De Moivre's Theorem and $r=1$. 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 (Quick Look)

As before, $r=1$. Represent $z$ in polar form. By De Moivre's Theorem, $z^3=\text{cis}(3\theta)$. To form an equilateral triangle, their difference in angle must be $\frac{\pi}{3}$, so \[\frac{\text{cis}(3\theta)}{\text{cis}(\theta)}=\text{cis}(2\theta)=\text{cis}(\pm\frac{\pi}{3})\] From the polar form of $z$, we know that $0\geq\theta\leq2\pi$, so $\text{cis}(2\theta)$ cycles in a circle twice. By contrast, $\pm\frac{\pi}{3}$ represent $2$ fixed, distinct points. Thus, $\text{cis}(2\theta)$ intersects these points twice each$\implies\boxed{\textbf{(D) }4}$

Visual: https://www.desmos.com/calculator/rnpxzns0jn 


To be more rigorous, you can find the $4$ solutions. $\text{cis}(2\theta)$ cycles twice, so $2\theta=\pm\frac{\pi}{3}+2n\pi$, where $n=0,1$. Then, $\theta=\frac{\pi}{6}$, $\frac{7\pi}{6}$, $\frac{5\pi}{6}$, $\frac{11\pi}{6}$. Substitute those values into $z$ and check that they are valid. $\implies\boxed{\textbf{(D) }4}$

(Solution by BJHHar)

Solution 3

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 $\boxed{\textbf{(D) }4}$.

Solution 4 (Quick and Easy)

Since the complex numbers $0,z,$ and $z^3$ form an equilateral triangle in the complex plane, we note that either $z^3$ is a 60 degrees counterclockwise rotation about the origin from $z$ or $z$ is a 60 degrees counterclockwise rotation about the origin from $z^3$.

Therefore, we note that either $z^3 = z \text{cis} 60^\circ{}$ or $z \text{cis}(-60^\circ{}) = z^3$

The first equation in $z$ (meaning $z^3 = z \text{cis} 60^\circ{}$) gives us: $z^2 = cis 60^\circ{}$, which gives 2 solutions in $z$.

The second equation in $z$ (which is $z \text{cis} (-60^\circ{}) = z^3$) gives us $z^2 = \text{cis} (-60^\circ{})$, which must give another 2 solutions in $z$.

Therefore, there are $\boxed{(D) 4}$ solutions for $z$. (Professor-Mom)

Note: The motivation for this method came from an older AIME problem, namely https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8.

Solution 5

Let the $z = re^{i\theta}$, so $z^3 = r^3e^{3i\theta}$. To have an equilateral triangle, we must have $|z^3| = |z|$, so $r^3 = r$, so $r= 1$.

Note that the angle between $z^3$ and $z$ is $3\theta - \theta = 2\theta$. Then, by the Law of Cosines,

\[|z^3 - z|^2 = |z|^2 + |z^3|^2 - 2|z||z^3|\cos 2\theta.\]

Since we have an equilateral triangle, it must be that $|z^3 - z| =|z| = |z^3| = r = 1$. Hence, \[1 = 2 - 2\cos 2\theta = 2 - 2(1-2\sin^2\theta) = 4\sin^2\theta,\] \[\sin^2\theta = \frac{1}{4}\] \[\sin\theta = \pm\frac{1}{2}\] \[\theta = \pm\frac{\pi}{6}, \pm\frac{5\pi}{6}.\]

These $4$ values of $\theta$ correspond to $\boxed{(D)4}$ distinct values of $z$. ~ brainfertilzer

Solution 6 (Quick, rectangular form)

Let $z = a + bi$, where $a$ and $b$ are real numbers. Then, since 0, $z$, and $z^3$ are vertices of an equilateral triangle in the complex plane, expanding $z^3$ gives the following : \[(a + bi)^3 = (a^3 - 3ab^2) + i(3a^2b - b)\] Now, we let $z = z^3$ and equate real and complex parts. It then follows that : \[a^3 - 3ab^2 = a, 3a^2b - b = b\] from which solving gives \[a^2 - 3b^2 = 1 (1), 3a^2 - 1 = 1 (2)\] \[3a^2 = 2 -> a^2 = \frac{2}{3}\] and \[\frac{2}{3} - 3b^2 = 1\] Focusing on (1) then gives \[b^2 = \frac{1}{9}\] Now, finding the square roots gives \[a = \pm\frac{\sqrt6}{3}, b = \pm\frac{i}{3}\] And so there are $\boxed{(D)4}$ possible solutions. ~elpianista227

Video Solution

For those who prefer a video: https://www.youtube.com/watch?v=uBL80yd1ihc

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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