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 | + | 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^{i\frac{2\pi}{3}}</math>, so in order <math>r</math> to be a root of <math>P</math>, <math>re^{i\frac{2\pi}{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^{i\frac{2\pi}{3}}</math> and <math>we^{i\frac{4\pi}{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^{i\frac{2\pi}{3}}</math>, or <math>re^{i\frac{4\pi}{3}}</math> and thus the polynomial <math>P</math> can be written 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 <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^{i\frac{2\pi}{3}}</math> and <math>re^{i\frac{4\pi}{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 |
Revision as of 22:52, 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
Let . We first notice that
, so 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
and thus the polynomial
can be written in the form
. Moreover, by Vieta's, we know that there is only one possible value for
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