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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} = </math>cis <math>(\frac{2\pi}{3})</math>, so in order for every root <math>r</math> to have <math>rcis(\frac{2\pi}{3})</math> as a solution, 3 of the roots of the polynomial must be <math>r</math>, <math>rcis({2\pi}{3})</math>, <math>rcis({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 the three previously determined roots and they have a multiplicity. Moreover, by Vieta's, we know that there is only one possible value for r as <math>r^5 = 2020</math>.Therefore, the polynomial is in the form <math>(x-r)^m(x-rcis({2\pi}{3})^n(x-rcis({4\pi}{3})^p</math>. But in order for the coefficients of the polynomial to all be real, <math>n = p</math> due to <math>rcis{2\pi}{3}</math> and <math>rcis{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{C) 2}</math>.
+
We notice that <math>\frac{1+i\sqrt{3}}{2} = e^(i\frac{2\pi}{3})</math>, so in order for every root <math>r</math> to have <math>rcis(\frac{2\pi}{3})</math> as a solution, 3 of the roots of the polynomial must be <math>r</math>, <math>rcis({2\pi}{3})</math>, <math>rcis({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 the three previously determined roots and they have a multiplicity. Moreover, by Vieta's, we know that there is only one possible value for r as <math>r^5 = 2020</math>.Therefore, the polynomial is in the form <math>(x-r)^m(x-rcis({2\pi}{3})^n(x-rcis({4\pi}{3})^p</math>. But in order for the coefficients of the polynomial to all be real, <math>n = p</math> due to <math>rcis{2\pi}{3}</math> and <math>rcis{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{C) 2}</math>.
  
 
-- Murtagh
 
-- Murtagh

Revision as of 22:25, 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 $i=\sqrt{-1}$)

Solution

We notice that $\frac{1+i\sqrt{3}}{2} = e^(i\frac{2\pi}{3})$, so in order for every root $r$ to have $rcis(\frac{2\pi}{3})$ as a solution, 3 of the roots of the polynomial must be $r$, $rcis({2\pi}{3})$, $rcis({4\pi}{3})$. 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 the three previously determined roots and they have a multiplicity. Moreover, by Vieta's, we know that there is only one possible value for r as $r^5 = 2020$.Therefore, the polynomial is in the form $(x-r)^m(x-rcis({2\pi}{3})^n(x-rcis({4\pi}{3})^p$. But in order for the coefficients of the polynomial to all be real, $n = p$ due to $rcis{2\pi}{3}$ and $rcis{4\pi}{3}$ being conjugates. 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 $\boxed{C) 2}$.

-- Murtagh

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