Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution by Mathematical Dexterity)
(sol 4)
Line 82: Line 82:
  
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
 +
 +
== Solution 4 ==
 +
Let <math>a=\cos(x)+i\sin(x)</math>, so by De Moivre <math>P(x)=a^3-a^2+a+1</math>. The problem essentially asks for the number of real roots of <math>P</math> which lie on the complex unit circle.
 +
Let <math>|r|=1</math> be a root of <math>P</math>, and note that we can't have <math>r^3-r^2+r=0</math>, else <math>P(r)=0</math>. Thus, suppose henceforth that <math>r^3-r^2+r \neq 0</math>. We then have <math>r^3-r^2+r=r^2(r+\tfrac{1}{r}-1)=a^2(2\mathrm{Re}(r)-1)</math>, hence the argument of <math>r^3-r^2+r</math> is either the argument of <math>a^2</math>, or the argument of <math>-a^2</math>. Since <math>r^3-r^2+r=-1</math> is real, it follows that a^2=\pm 1 \implies a \in \{1,i,-1,-i\}<math>. Now, we can check all of these values and find that none of them work, yielding an answer of </math>\boxed{\textbf{(A) }0}$.
 +
 +
-IAmTheHazard
  
 
==Video Solution by Mathematical Dexterity==
 
==Video Solution by Mathematical Dexterity==

Revision as of 12:05, 2 August 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)

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

Video Solution by Mathematical Dexterity

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

Video Solution by The Power of Logic

https://youtu.be/SHMW3QG4Uu4

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_21&oldid=176566"