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

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==Solution 3==
 
==Solution 3==
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Call <math>\frac{-1+\sqrt{3}}{2}=q</math>. We have <math>r</math>, <math>qr</math>, and <math>q^2r</math> having to be roots, and since <math>q^3=1</math>, we must choose 2 more roots out of these three such that the condition that <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are all real. There are <math>\fbox{2}</math> ways to do this because of the fundamental theorem of algebra, these being <math>\left(qr,qr^2\right)</math> and <math>\left(r,r\right)</math> because to satisfy FTA if <math>z</math> is a root then <math>\overbar{z}</math> must also be a root.
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Call <math>\frac{-1+\sqrt{3}}{2}=q</math>. We have <math>r</math>, <math>qr</math>, and <math>q^2r</math> having to be roots, and since <math>q^3=1</math>, we must choose 2 more roots out of these three such that the condition that <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are all real. There are <math>\fbox{2}</math> ways to do this because of the fundamental theorem of algebra, these being <math>\left(qr,qr^2\right)</math> and <math>\left(r,r\right)</math> because to satisfy FTA if <math>z</math> is a root then so must its conjugate.
 +
 
 +
~joeythetoey
  
 
==Video Solution by MistyMathMusic==
 
==Video Solution by MistyMathMusic==

Latest revision as of 13:30, 11 October 2024

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

Solution 3

Call $\frac{-1+\sqrt{3}}{2}=q$. We have $r$, $qr$, and $q^2r$ having to be roots, and since $q^3=1$, we must choose 2 more roots out of these three such that the condition that $a$, $b$, $c$, and $d$ are all real. There are $\fbox{2}$ ways to do this because of the fundamental theorem of algebra, these being $\left(qr,qr^2\right)$ and $\left(r,r\right)$ because to satisfy FTA if $z$ is a root then so must its conjugate.

~joeythetoey

Video Solution by MistyMathMusic

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

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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