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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 21"

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==Video Solution by Mathematical Dexterity==
 
==Video Solution by Mathematical Dexterity==
 
https://www.youtube.com/watch?v=vhAc0P09czI
 
https://www.youtube.com/watch?v=vhAc0P09czI
 +
 +
==Video Solution by The Power of Logic==
 +
https://youtu.be/SHMW3QG4Uu4

Revision as of 17:55, 13 June 2022

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)

Video Solution by Mathematical Dexterity

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

Video Solution by The Power of Logic

https://youtu.be/SHMW3QG4Uu4

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