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

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==Solution==
 
==Solution==
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We notice that <math>\frac{-1+i\sqrt{3}}{2} = e^{i\frac{2\pi}{3}}</math>, so in order for root <math>r</math> to exist, <math>re^{i\frac{2\pi}{3}}</math> must also be a root, meaning that 3 of the roots of the polynomial must be <math>r</math>, <math>re^{i\frac{2\pi}{3}}</math>, <math>re^{i\frac{4\pi}{3}}</math>. However, since the polynomial is degree 5, there must be two additional roots, but in order for each of these roots to reach eachother by multiplying by 120 degrees, there must be three of them, but this clearly cannot be the case. This means that the polynomial is degree 5 with only the three previously determined roots 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 <math>r</math> as <math>r^5 = 2020</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. 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>.
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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>, so 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> and thus 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 <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
 
-- Murtagh

Revision as of 23:52, 7 February 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

Let $P(x) = x^5+ax^4+bx^3+cx^2+dx+2020$. We first notice that $\frac{-1+i\sqrt{3}}{2} = e^{i\frac{2\pi}{3}}$, so 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 $re^{i\frac{4\pi}{3}}$ and thus 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 $r$ 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

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