2021 Fall AMC 12B Problems/Problem 21

Problem

For real numbers $x$ , let $P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)$ where $i = \sqrt{-1}$ . For how many values of $x$ with $0\leq x<2\pi$ does $P(x)=0?$

$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution 1

Let $a=\cos(x)+i\sin(x)$ . Now $P(a)=1+a-a^2+a^3$ . $P(-1)=-2$ and $P(0)=1$ so there is a real root $a_1$ between $-1$ and $0$ . The other $a$ 's must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex $a$ 's squared is $-\frac{1}{a_1}$ which is greater than $1$ . If $x$ is real number then $a$ must have magnitude of $1$ , but none of the solutions for $a$ have magnitude of $1$ , so the answer is $\boxed{\textbf{(A)}\ 0 }$ ~lopkiloinm

Solution 2

For $\textrm{Im}(P(x))=0$ , we get $\sin(2x)=\sin(x)+\sin(3x)=2\sin(2x)\cos(x)$ So either $\sin(2x)=0$ , i.e. $x\in\{0,\pi\}$ or $\cos(x)=\tfrac 12$ , i.e. $x\in \{\pi/3, 5\pi/3\}$ .

For none of these values do we get $\textrm{Re}(P(x))=0$ .

Therefore, the answer is $\boxed{\textbf{(A) }0}$ .

Solution 3

We have $\begin{align*} P \left( x \right) & = 1 + e^{ix} - e^{i 2x} + e^{i 3x} . \end{align*}$

Denote $y = e^{i x}$ . Hence, this problem asks us to find the number of $y$ with $| y| = 1$ that satisfy $1 + y - y^2 + y^3 = 0 . \hspace{1cm} (1)$

Taking imaginary part of both sides, we have $\begin{align*} 0 & = {\rm Im} \ \left( 1 + y - y^2 + y^3 \right) \\ & = \frac{1}{2i} \left( y - \bar y - y^2 + \bar y^2 + y^3 - \bar y^3 \right) \\ & = \frac{y - \bar y}{2i} \left( 1 - y - \bar y + y^2 + y \bar y + \bar y^2 \right) \\ & = {\rm Im} \ y \left( 1 - \left( y + \bar y \right) + \left( y + \bar y \right)^2 - y \bar y \right) \\ & = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - |y|^2 \right) \\ & = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - 1 \right) \\ & = 2 {\rm Im} \ y \cdot {\rm Re} \ y \left( 2 {\rm Re} \ y - 1 \right) \\ \end{align*}$ The sixth equality follows from the property that $|y| = 1$ .

Therefore, we have either ${\rm Re} \ y = 0$ or ${\rm Im} \ y = 0$ or $2 {\rm Re} \ y - 1 = 0$ .

Case 1: ${\rm Re} \ y = 0$ .

Because $|y| = 1$ , $y = \pm i$ .

However, these solutions fail to satisfy Equation (1).

Therefore, there is no solution in this case.

Case 2: ${\rm Im} \ y = 0$ .

Because $|y| = 1$ , $y = \pm 1$ .

However, these solutions fail to satisfy Equation (1).

Therefore, there is no solution in this case.

Case 3: $2 {\rm Re} \ y - 1 = 0$ .

Because $|y| = 1$ , $y = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i = e^{i \pm \frac{\pi}{3}}$ .

However, these solutions fail to satisfy Equation (1).

Therefore, there is no solution in this case.

All cases above imply that there is no solution in this problem.

Therefore, the answer is $\boxed{\textbf{(A) }0}$ .

~Steven Chen (www.professorchenedu.com)

Solution 4

Let $a=\cos(x)+i\sin(x)$ , so by De Moivre $P(x)=a^3-a^2+a+1$ . The problem essentially asks for the number of real roots of $P$ which lie on the complex unit circle. Let $|r|=1$ be a root of $P$ , and note that we can't have $r^3-r^2+r=0$ , else $P(r)=0$ . Thus, suppose henceforth that $r^3-r^2+r \neq 0$ . We then have $r^3-r^2+r=r^2(r+\tfrac{1}{r}-1)=a^2(2\mathrm{Re}(r)-1)$ , hence the argument of $r^3-r^2+r$ is either the argument of $a^2$ , or the argument of $-a^2$ . Since $r^3-r^2+r=-1$ is real, it follows that $a^2=\pm 1 \implies a \in \{1,i,-1,-i\}$ . Now, we can check all of these values and find that none of them work, yielding an answer of $\boxed{\textbf{(A) }0}$ .

-IAmTheHazard

Solution 5 (Geometry)

$P(x)$ can be written equivalently as $P(x) = 1 + cis(x) - cis(2x) + cis(3x).$ Thus, we aim to find $x$ such that the sum of the vectors $cis(x)$ , $-cis(2x)$ , and $cis(3x)$ is -1. Notice that $cis(x)$ , $-cis(2x)$ , $cis(3x)$ all lie on the unit circle in the complex plane, and the vector $cis(x) + cis(3x)$ is collinear with $-cis(2x).$ Since $|-cis(2x)| = 1$ and we want the three vectors to sum to -1, we either have $-cis(2x) = 1$ and $cis(x) + cis(3x) = -2$ , or $-cis(2x) = -1$ and $cis(x) + cis(3x) = 0.$ If the first condition is true, $cis(x) = cis(3x)=-1.$ This will imply that $x= \pi.$ But then $-cis(2x) = -1$ , which violates the condition. Similarly, we can show that the second condition cannot be met either. Thus $P(x)$ does not have any solutions on the interval $[0, 2\pi].$ Therefore, the answer is $\boxed{\textbf{(A) }0}$ .

-mathy_mathema

Video Solution

https://youtu.be/p7dlJzqyEqQ

~MathProblemSolvingSkills.com

Brute Force Solution

1+cis(x)+cis(3x)=cis(2x) try out all the common values x could be 0,pi/4,pi/3,pi/2,2pi/3,3pi/4,pi,4pi/3,5pi/4,3pi/2,7pi/4,5pi/3,2pi insert them into the equation, and then you find none that work so… let’s move on and pray that our answer is right emilyyunhanq@gmail.com Emily Q

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=vhAc0P09czI

Video Solution by The Power of Logic

https://youtu.be/SHMW3QG4Uu4

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2021 Fall AMC 12B Problems/Problem 21

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5 (Geometry)

Video Solution

Brute Force Solution

Video Solution by Mathematical Dexterity

Video Solution by The Power of Logic

See Also