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

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Let <math>x_1=r</math>, then <math>x_2=(-1+i\sqrt{3})/2 r</math>, 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
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Let <math>x_1=r</math>, then <cmath>x_2=\frac{-1+i\sqrt{3}}{2} r,</cmath> <cmath>x_3=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 r =\left( \frac{-1-i\sqrt{3}}{2} \right) r,</cmath> <cmath>x_4=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 r=r,</cmath> which means <math>x_4</math> is the same as <math>x_1</math>.
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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.
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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
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Now we have 3 different roots of the polynomial, <math>x_1</math>, <math>x_2</math>, and <math>x_3</math>. Next, we will prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there is one root <math>x_4=p</math> which is different from the three roots we already know, then there must be two other roots, <cmath>x_5=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 p =\left( \frac{-1-i\sqrt{3}}{2} \right) p,</cmath> <cmath>x_6=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 p=p,</cmath> different from all known roots. We get 6 different roots for the polynomial, which contradicts the limit of 5 roots. Therefore the assumption of a different root is wrong, thus the roots must be chosen from <math>x_1</math>, <math>x_2</math>, and <math>x_3</math>.
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The polynomial then can be written like <math>f(x)=(x-x_1^m (x-x_2)^n (x-x_3)^q</math>, where <math>m</math>, <math>n</math>, and <math>q</math> are non-negative integers and <math>m+n+q=5</math>. Since <math>a</math>, <math>b</math>, <math>c</math> and <math>d</math> are real numbers, then <math>n</math> must be equal to <math>q</math>. Therefore <math>(m,n,q)</math> can only be <math>(1,2,2)</math> or <math>(3,1,1)</math>, so the answer is <math>\boxed{\mathbf{(C)} 2}</math>.
  
 
~Yelong_Li
 
~Yelong_Li
 +
 +
Edit and <math>\LaTeX</math> by DandelionDreams
  
 
==Video Solution==
 
==Video Solution==

Revision as of 00:33, 24 January 2021

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 $\frac{-1+i\sqrt{3}}{2} = e^{2\pi i / 3}$. That is because of Euler's Formula : $e^{ix} = \cos(x) + i \cdot \sin(x)$. $\frac{-1+i\sqrt{3}}{2}$ = $-\frac{1}{2} + i \cdot \frac {\sqrt{3}}{2}$ = $\cos(120^\circ) + i \cdot \sin(120^\circ) = e^{ 2\pi i / 3}$.

In order $r$ to be a root of $P$, $re^{2\pi i / 3}$ 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 $we^{2\pi i / 3}$ and $we^{4\pi i / 3}$ must also be roots. However, $P$ is a fifth degree polynomial, and can therefore only have $5$ roots. This implies that $w$ is either $r$, $re^{2\pi i / 3}$, or $re^{4\pi i / 3}$. Thus we know that the polynomial $P$ can be written in the form $(x-r)^m(x-re^{2\pi i / 3})^n(x-re^{4\pi i / 3})^p$. 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 $re^{2\pi i / 3}$ and $re^{4 \pi i / 3}$ 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

Solution 2

Let $x_1=r$, then \[x_2=\frac{-1+i\sqrt{3}}{2} r,\] \[x_3=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 r =\left( \frac{-1-i\sqrt{3}}{2} \right) r,\] \[x_4=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 r=r,\] which means $x_4$ is the same as $x_1$.

Now we have 3 different roots of the polynomial, $x_1$, $x_2$, and $x_3$. Next, we will prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there is one root $x_4=p$ which is different from the three roots we already know, then there must be two other roots, \[x_5=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 p =\left( \frac{-1-i\sqrt{3}}{2} \right) p,\] \[x_6=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 p=p,\] different from all known roots. We get 6 different roots for the polynomial, which contradicts the limit of 5 roots. Therefore the assumption of a different root is wrong, thus the roots must be chosen from $x_1$, $x_2$, and $x_3$.

The polynomial then can be written like $f(x)=(x-x_1^m (x-x_2)^n (x-x_3)^q$, where $m$, $n$, and $q$ are non-negative integers and $m+n+q=5$. Since $a$, $b$, $c$ and $d$ are real numbers, then $n$ must be equal to $q$. Therefore $(m,n,q)$ can only be $(1,2,2)$ or $(3,1,1)$, so the answer is $\boxed{\mathbf{(C)} 2}$.

~Yelong_Li

Edit and $\LaTeX$ by DandelionDreams

Video Solution

Link: https://www.youtube.com/watch?v=8V5l5jeQjNg

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AMC 12 Problems and Solutions

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