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>. |
− | + | 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== | ==Solution 2== | ||
− | 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>. |
+ | |||
+ | 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 | ||
+ | |||
+ | ==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== | ==See Also== |
Latest revision as of 13:30, 11 October 2024
Contents
[hide]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
Solution 3
Call . We have
,
, and
having to be roots, and since
, we must choose 2 more roots out of these three such that the condition that
,
,
, and
are all real. There are
ways to do this because of the fundamental theorem of algebra, these being
and
because to satisfy FTA if
is a root then so must its conjugate.
~joeythetoey
Video Solution by MistyMathMusic
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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