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Difference between revisions of "2001 AIME I Problems/Problem 3"

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== Solution 2 ==
 
== Solution 2 ==
  
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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>.
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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>.
  
 
==Solution 3==
 
==Solution 3==

Revision as of 20:28, 18 September 2018

Problem

Find the sum of the roots, real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$, given that there are no multiple roots.

Solution 1

From Vieta's formulas, in a polynomial of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0$, then the sum of the roots is $\frac{-a_{n-1}}{a_n}$.

From the Binomial Theorem, the first term of $\left(\frac 12-x\right)^{2001}$ is $-x^{2001}$, but $x^{2001}+-x^{2001}=0$, so the term with the largest degree is $x^{2000}$. So we need the coefficient of that term, as well as the coefficient of $x^{1999}$.

\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*}

Applying Vieta's formulas, we find that the sum of the roots is $-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}$.

Solution 2

We find that the given equation has a $2000^{\text{th}}$ degree polynomial. Note that there are no multiple roots. Thus, if $\frac{1}{2} - x$ is a root, $x$ is also a root. Thus, we pair up $1000$ pairs of roots that sum to $\frac{1}{2}$ to get a sum of $\boxed{500}$.

Solution 3

Note that if $r$ is a root, then $\frac{1}{2}-r$ is a root and they sum up to $\frac{1}{2}.$ We make the substitution $y=x-\frac{1}{4}$ so \[(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.\] Expanding gives \[2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots\] so by Vieta, the sum of the roots of $y$ is 0. Since $x$ has a degree of 2000, then $x$ has 2000 roots so the sum of the roots is \[2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.\]


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

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