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

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

@@ Line 4: / Line 4: @@
 <math>\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4</math>
-==Solution==
+==Solution 1==
-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 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-re^{i\frac{2\pi}{3}})^n(x-re^{i\frac{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>re^{i\frac{2\pi}{3}}</math> and <math>re^{i\frac{2\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>\textbf{(C) } 2</math>.
+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^{2\pi i / 3}</math>.  That is because of Euler's Formula : <math>e^{ix} = \cos(x) + i \cdot \sin(x)</math>. <math>\frac{-1+i\sqrt{3}}{2}</math> = <math>-\frac{1}{2} + i \cdot \frac {\sqrt{3}}{2}</math> = <math>\cos(120^\circ) + i \cdot \sin(120^\circ) = e^{ 2\pi i / 3}</math>.
--- Murtagh
+In order <math>r</math> to be a root of <math>P</math>, <math>re^{2\pi i / 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^{2\pi i / 3}</math> and <math>we^{4\pi i / 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^{2\pi i / 3}</math>, or <math>re^{4\pi i / 3}</math>. Thus we know that the polynomial <math>P</math> can be written in the form <math>(x-r)^m(x-re^{2\pi i / 3})^n(x-re^{4\pi i / 3})^p</math>. Moreover, by Vieta's, we know that there is only one possible value for the magnitude of <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^{2\pi i / 3}</math> and <math>re^{4 \pi i / 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
+==Solution 2==
+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>.
+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>.
+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
+==Solution 3==
+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==
+https://www.youtube.com/watch?v=8V5l5jeQjNg
+==See Also==
+{{AMC12 box|year=2020|ab=B|num-b=16|num-a=18}}
+{{MAA Notice}}

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

Page

Toolbox

Search