Difference between revisions of "2020 AMC 12B Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
− | 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^ | + | 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>. |
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+ | 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 | ~Murtagh | ||
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− | Let <math>x_1=r</math>, then < | + | 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, | + | |
− | The polynomial then can be written like f(x)= | + | 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>. |
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+ | 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 | ~Yelong_Li | ||
+ | |||
+ | Edit and <math>\LaTeX</math> by DandelionDreams | ||
==Video Solution== | ==Video Solution== |
Revision as of 00:33, 24 January 2021
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 1
Let . We first notice that . That is because of Euler's Formula : . = = .
In order to be a root of , must also be a root of P, meaning that 3 of the roots of must be , , . However, since is degree 5, there must be two additional roots. Let one of these roots be , if is a root, then and must also be roots. However, is a fifth degree polynomial, and can therefore only have roots. This implies that is either , , or . Thus we know that the polynomial can be written in the form . Moreover, by Vieta's, we know that there is only one possible value for the magnitude of as , meaning that the amount of possible polynomials is equivalent to the possible sets . In order for the coefficients of the polynomial to all be real, due to and being conjugates and 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
Solution 2
Let , then which means is the same as .
Now we have 3 different roots of the polynomial, , , and . 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 which is different from the three roots we already know, then there must be two other roots, 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 , , and .
The polynomial then can be written like , where , , and are non-negative integers and . Since , , and are real numbers, then must be equal to . Therefore can only be or , so the answer is .
~Yelong_Li
Edit and by DandelionDreams
Video Solution
Link: 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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