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Difference between revisions of "2020 AMC 12B Problems/Problem 17"

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==Solution 1==
 
==Solution 1==
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Let <math>P(x) = x^5+ax^4+bx^3+cx^2+dx+2020</math>. We first notice that <math>\frac{-1+i\sqrt{3}}{2} = e^{i\frac{2\pi}{3}}</math>.  That is because of Euler's Formula : <math>e^{ix} = cos(x) + i \cdot sin(x)</math>. <math>\frac{-1+i\sqrt{3}}{2} = - \frac{1}{2} + i \cdot \frac {sqrt(3)}{2} = cos(120) + i \cdot sin(120) = e^{i \cdot 120(degrees) = e^{i \frac{2}{3} \pi (radians)</math>  .In order <math>r</math> to be a root of <math>P</math>, <math>re^{i\frac{2\pi}{3}}</math> must also be a root of P, meaning that 3 of the roots of <math>P</math> must be <math>r</math>, <math>re^{i\frac{2\pi}{3}}</math>, <math>re^{i\frac{4\pi}{3}}</math>. However, since <math>P</math> is degree 5, there must be two additional roots. Let one of these roots be <math>w</math>, if <math>w</math> is a root, then <math>we^{i\frac{2\pi}{3}}</math> and <math>we^{i\frac{4\pi}{3}}</math> must also be roots. However, <math>P</math> is a fifth degree polynomial, and can therefore only have <math>5</math> roots. This implies that <math>w</math> is either <math>r</math>, <math>re^{i\frac{2\pi}{3}}</math>, or <math>re^{i\frac{4\pi}{3}}</math>. Thus we know that the polynomial <math>P</math> can be written in the form <math>(x-r)^m(x-re^{i\frac{2\pi}{3}})^n(x-re^{i\frac{4\pi}{3}})^p</math>. Moreover, by Vieta's, we know that there is only one possible value for the magnitude of <math>r</math> as <math>||r||^5 = 2020</math>, meaning that the amount of possible polynomials <math>P</math> is equivalent to the possible sets <math>(m,n,p)</math>. In order for the coefficients of the polynomial to all be real, <math>n = p</math> due to <math>re^{i\frac{2\pi}{3}}</math> and <math>re^{i\frac{4\pi}{3}}</math> being conjugates and since <math>m+n+p = 5</math>, (as the polynomial is 5th degree) we have two possible solutions for <math>(m, n, p)</math> which are <math>(1,2,2)</math> and <math>(3,1,1)</math> yielding two possible polynomials. The answer is thus <math>\boxed{\textbf{(C) } 2}</math>.
+
Let <math>P(x) = x^5+ax^4+bx^3+cx^2+dx+2020</math>. We first notice that <math>\frac{-1+i\sqrt{3}}{2} = e^{i\frac{2\pi}{3}}</math>.  That is because of Euler's Formula : <math>e^{ix} = cos(x) + i \cdot sin(x)</math>. <math>\frac{-1+i\sqrt{3}}{2}</math> = <math>-\frac{1}{2} + i \cdot \frac {sqrt(3)}{2}</math> = <math>cos(120) + i \cdot sin(120)</math> = <math>e^{i \cdot 120(degrees)</math> = <math>e^{i \frac{2}{3} \pi (radians)</math>  .In order <math>r</math> to be a root of <math>P</math>, <math>re^{i\frac{2\pi}{3}}</math> must also be a root of P, meaning that 3 of the roots of <math>P</math> must be <math>r</math>, <math>re^{i\frac{2\pi}{3}}</math>, <math>re^{i\frac{4\pi}{3}}</math>. However, since <math>P</math> is degree 5, there must be two additional roots. Let one of these roots be <math>w</math>, if <math>w</math> is a root, then <math>we^{i\frac{2\pi}{3}}</math> and <math>we^{i\frac{4\pi}{3}}</math> must also be roots. However, <math>P</math> is a fifth degree polynomial, and can therefore only have <math>5</math> roots. This implies that <math>w</math> is either <math>r</math>, <math>re^{i\frac{2\pi}{3}}</math>, or <math>re^{i\frac{4\pi}{3}}</math>. Thus we know that the polynomial <math>P</math> can be written in the form <math>(x-r)^m(x-re^{i\frac{2\pi}{3}})^n(x-re^{i\frac{4\pi}{3}})^p</math>. Moreover, by Vieta's, we know that there is only one possible value for the magnitude of <math>r</math> as <math>||r||^5 = 2020</math>, meaning that the amount of possible polynomials <math>P</math> is equivalent to the possible sets <math>(m,n,p)</math>. In order for the coefficients of the polynomial to all be real, <math>n = p</math> due to <math>re^{i\frac{2\pi}{3}}</math> and <math>re^{i\frac{4\pi}{3}}</math> being conjugates and since <math>m+n+p = 5</math>, (as the polynomial is 5th degree) we have two possible solutions for <math>(m, n, p)</math> which are <math>(1,2,2)</math> and <math>(3,1,1)</math> yielding two possible polynomials. The answer is thus <math>\boxed{\textbf{(C) } 2}</math>.
  
 
~Murtagh (edited by totoro.selsel)
 
~Murtagh (edited by totoro.selsel)

Revision as of 22:35, 18 August 2020

Problem

How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$, where $a$, $b$, $c$, and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$? (Note that $i=\sqrt{-1}$)

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

Solution 1

Let $P(x) = x^5+ax^4+bx^3+cx^2+dx+2020$. We first notice that $\frac{-1+i\sqrt{3}}{2} = e^{i\frac{2\pi}{3}}$. That is because of Euler's Formula : $e^{ix} = cos(x) + i \cdot sin(x)$. $\frac{-1+i\sqrt{3}}{2}$ = $-\frac{1}{2} + i \cdot \frac {sqrt(3)}{2}$ = $cos(120) + i \cdot sin(120)$ = $e^{i \cdot 120(degrees)$ (Error compiling LaTeX. Unknown error_msg) = $e^{i \frac{2}{3} \pi (radians)$ (Error compiling LaTeX. Unknown error_msg) .In order $r$ to be a root of $P$, $re^{i\frac{2\pi}{3}}$ must also be a root of P, meaning that 3 of the roots of $P$ must be $r$, $re^{i\frac{2\pi}{3}}$, $re^{i\frac{4\pi}{3}}$. However, since $P$ is degree 5, there must be two additional roots. Let one of these roots be $w$, if $w$ is a root, then $we^{i\frac{2\pi}{3}}$ and $we^{i\frac{4\pi}{3}}$ must also be roots. However, $P$ is a fifth degree polynomial, and can therefore only have $5$ roots. This implies that $w$ is either $r$, $re^{i\frac{2\pi}{3}}$, or $re^{i\frac{4\pi}{3}}$. Thus we know that the polynomial $P$ can be written in the form $(x-r)^m(x-re^{i\frac{2\pi}{3}})^n(x-re^{i\frac{4\pi}{3}})^p$. Moreover, by Vieta's, we know that there is only one possible value for the magnitude of $r$ as $||r||^5 = 2020$, meaning that the amount of possible polynomials $P$ is equivalent to the possible sets $(m,n,p)$. In order for the coefficients of the polynomial to all be real, $n = p$ due to $re^{i\frac{2\pi}{3}}$ and $re^{i\frac{4\pi}{3}}$ being conjugates and since $m+n+p = 5$, (as the polynomial is 5th degree) we have two possible solutions for $(m, n, p)$ which are $(1,2,2)$ and $(3,1,1)$ yielding two possible polynomials. The answer is thus $\boxed{\textbf{(C) } 2}$.

~Murtagh (edited by totoro.selsel)

Solution 2

Let $x_1=r$, then $x_2=(-1+i\sqrt{3})/2 r$, x_3=〖((-1+i√3)/2)〗^2 r=(-1-i√3)/2 r, x_4=〖((-1+i√3)/2)〗^3 r=r which means x_4 will be back to x_1 Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2, and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of the different root is wrong. The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2

~Yelong_Li

See Also

2020 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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