Difference between revisions of "2001 AIME I Problems/Problem 3"
(→Solution 2) |
(→See also) |
||
Line 16: | Line 16: | ||
We find that the given equation has a <math>2000th</math> degree polynomial. Note that there are no multiple roots. Thus, if <math>\frac{1}{2} - x</math> is a root, <math>x</math> is also a root. Thus, we pair up <math>1000</math> pairs of roots that sum to <math>\frac{1}{2}</math> to get a sum of <math>\boxed{500}</math>. | We find that the given equation has a <math>2000th</math> degree polynomial. Note that there are no multiple roots. Thus, if <math>\frac{1}{2} - x</math> is a root, <math>x</math> is also a root. Thus, we pair up <math>1000</math> pairs of roots that sum to <math>\frac{1}{2}</math> to get a sum of <math>\boxed{500}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Note that if <math>r</math> is a root, then <math>\frac{1}{2}-r</math> is a root and they sum up to <math>\frac{1}{2}.</math> We make the substitution <math>y=x-\frac{1}{4}</math> so <cmath>(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.</cmath> Expanding gives <cmath>2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots</cmath> so by Vieta, the sum of the roots of <math>y</math> is 0. Since <math>x</math> has a degree of 2000, then <math>x</math> has 2000 roots so the sum of the roots is <cmath>2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.</cmath> | ||
+ | |||
== See also == | == See also == |
Revision as of 21:27, 18 September 2018
Problem
Find the sum of the roots, real and non-real, of the equation , given that there are no multiple roots.
Solution 1
From Vieta's formulas, in a polynomial of the form , then the sum of the roots is
.
From the Binomial Theorem, the first term of is
, but
, so the term with the largest degree is
. So we need the coefficient of that term, as well as the coefficient of
.
Applying Vieta's formulas, we find that the sum of the roots is .
Solution 2
We find that the given equation has a degree polynomial. Note that there are no multiple roots. Thus, if
is a root,
is also a root. Thus, we pair up
pairs of roots that sum to
to get a sum of
.
Solution 3
Note that if is a root, then
is a root and they sum up to
We make the substitution
so
Expanding gives
so by Vieta, the sum of the roots of
is 0. Since
has a degree of 2000, then
has 2000 roots so the sum of the roots is
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.