Difference between revisions of "2021 Fall AMC 12B Problems/Problem 21"
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3 \qquad\textbf{(E)}\ 4</math> | 3 \qquad\textbf{(E)}\ 4</math> | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>a=\cos(x)+i\sin(x)</math>. Now <math>P(a)=1+a-a^2+a^3</math>. <math>P(-1)=-2</math> and <math>P(0)=1</math> so there is a real number <math>a_1</math> between <math>-1</math> and <math>0</math>. The other <math>a</math>'s must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex <math>a</math>'s squared is <math>\frac{1}{a_1}</math> which is greater than <math>1</math>. If <math>x</math> is real number then <math>a</math> must have magnitude of <math>1</math>, but all the solutions for <math>a</math> do not have magnitude of <math>1</math>, so the answer is <math>\boxed{\textbf{(A)}\ 0 }</math> ~lopkiloinm | Let <math>a=\cos(x)+i\sin(x)</math>. Now <math>P(a)=1+a-a^2+a^3</math>. <math>P(-1)=-2</math> and <math>P(0)=1</math> so there is a real number <math>a_1</math> between <math>-1</math> and <math>0</math>. The other <math>a</math>'s must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex <math>a</math>'s squared is <math>\frac{1}{a_1}</math> which is greater than <math>1</math>. If <math>x</math> is real number then <math>a</math> must have magnitude of <math>1</math>, but all the solutions for <math>a</math> do not have magnitude of <math>1</math>, so the answer is <math>\boxed{\textbf{(A)}\ 0 }</math> ~lopkiloinm | ||
+ | |||
+ | == Solution 2 == | ||
+ | We have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | P \left( x \right) | ||
+ | & = 1 + e^{ix} - e^{i 2x} + e^{i 3x} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Denote <math>y = e^{i x}</math>. Hence, this problem asks us to find the number of <math>y</math> with <math>| y| = 1</math> that satisfy | ||
+ | <cmath> | ||
+ | \[ | ||
+ | 1 + y - y^2 + y^3 = 0 . \hspace{1cm} (1) | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Taking imaginary part of both sides, we have | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | The sixth equality follows from the property that <math>|y| = 1</math>. | ||
+ | |||
+ | Therefore, we have either <math>{\rm Re} \ y = 0</math> or <math>{\rm Im} \ y = 0</math> or <math>2 {\rm Re} \ y - 1 = 0</math>. | ||
+ | |||
+ | Case 1: <math>{\rm Re} \ y = 0</math>. | ||
+ | |||
+ | Because <math>|y| = 1</math>, <math>y = \pm i</math>. | ||
+ | |||
+ | However, these solutions fail to satisfy Equation (1). | ||
+ | |||
+ | Therefore, there is no solution in this case. | ||
+ | |||
+ | Case 2: <math>{\rm Im} \ y = 0</math>. | ||
+ | |||
+ | Because <math>|y| = 1</math>, <math>y = \pm 1</math>. | ||
+ | |||
+ | However, these solutions fail to satisfy Equation (1). | ||
+ | |||
+ | Therefore, there is no solution in this case. | ||
+ | |||
+ | Case 3: <math>2 {\rm Re} \ y - 1 = 0</math>. | ||
+ | |||
+ | Because <math>|y| = 1</math>, <math>y = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i = e^{i \pm \frac{\pi}{3}}</math>. | ||
+ | |||
+ | 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 <math>\boxed{\textbf{(A) }0}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) |
Revision as of 23:19, 25 November 2021
Problem
For real numbers , let where . For how many values of with does
Solution 1
Let . Now . and so there is a real number between and . The other 's must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex 's squared is which is greater than . If is real number then must have magnitude of , but all the solutions for do not have magnitude of , so the answer is ~lopkiloinm
Solution 2
We have
Denote . Hence, this problem asks us to find the number of with that satisfy
Taking imaginary part of both sides, we have The sixth equality follows from the property that .
Therefore, we have either or or .
Case 1: .
Because , .
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
Case 2: .
Because , .
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
Case 3: .
Because , .
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 .
~Steven Chen (www.professorchenedu.com)