Difference between revisions of "2001 AIME I Problems/Problem 3"
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== Problem == | == Problem == | ||
− | Find the sum of the | + | Find the sum of the [[root]]s, real and non-real, of the equation <math>x^{2001}+\left(\frac 12-x\right)^{2001}=0</math>, given that there are no multiple roots. |
+ | |||
+ | == Solution 1 == | ||
+ | From [[Vieta's formulas]], in a [[polynomial]] of the form <math>a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0</math>, then the sum of the roots is <math>\frac{-a_{n-1}}{a_n}</math>. | ||
+ | |||
+ | From the [[Binomial Theorem]], the first term of <math>\left(\frac 12-x\right)^{2001}</math> is <math>-x^{2001}</math>, but <math>x^{2001}+-x^{2001}=0</math>, so the term with the largest degree is <math>x^{2000}</math>. So we need the coefficient of that term, as well as the coefficient of <math>x^{1999}</math>. | ||
+ | |||
+ | <cmath>\begin{align*}\binom{2001}{1} \cdot (-x)^{2000} \cdot \left(\frac{1}{2}\right)^1&=\frac{2001x^{2000}}{2}\\ | ||
+ | \binom{2001}{2} \cdot (-x)^{1999} \cdot \left(\frac{1}{2}\right)^2 &=\frac{-x^{1999}*2001*2000}{8}=-2001 \cdot 250x^{1999} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Applying Vieta's formulas, we find that the sum of the roots is <math>-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | We find that the given equation has a <math>2000^{\text{th}}</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>. | ||
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+ | ==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> | ||
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== See also == | == See also == | ||
{{AIME box|year=2001|n=I|num-b=2|num-a=4}} | {{AIME box|year=2001|n=I|num-b=2|num-a=4}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:28, 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.