Difference between revisions of "2020 AMC 12B Problems/Problem 17"
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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 r 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>. | + | 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 r 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>. |
-- Murtagh | -- Murtagh |
Revision as of 22:34, 7 February 2020
Problem
How many polynomials of the form , where , , , and are real numbers, have the property that whenever is a root, so is ? (Note that )
Solution
We notice that , so in order for root to exist, must also be a root, meaning that 3 of the roots of the polynomial must be , , . 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 . Moreover, by Vieta's, we know that there is only one possible value for r as . In order for the coefficients of the polynomial to all be real, due to and being conjugates. Since as the polynomial is 5th degree, we have two possible solutions for which are and yielding two possible polynomials. The answer is thus .
-- Murtagh