Difference between revisions of "2020 AMC 12B Problems/Problem 17"
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− | Let x_1=r, then x_2=(-1+i√3)/2 r, x_3=〖((-1+i√3)/2)〗^2 r=(-1-i√3)/2 r, x_4=〖((-1+i√3)/2)〗^3 r=r which means x_4 will be back to x_1 | + | Let x_1=r, then <math>x_2=(-1+i√3)/2 r</math>, x_3=〖((-1+i√3)/2)〗^2 r=(-1-i√3)/2 r, x_4=〖((-1+i√3)/2)〗^3 r=r which means x_4 will be back to x_1 |
Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong. | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong. | ||
The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2 | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2 |
Revision as of 01:39, 5 May 2020
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
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
. 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
Let x_1=r, then $x_2=(-1+i√3)/2 r$ (Error compiling LaTeX. Unknown error_msg), x_3=〖((-1+i√3)/2)〗^2 r=(-1-i√3)/2 r, x_4=〖((-1+i√3)/2)〗^3 r=r which means x_4 will be back to x_1 Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong. The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
~Yelong_Li
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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