2001 AIME I Problems/Problem 3

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 $2000th$ 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 $500$ .

2001 AIME I (Problems • Answer Key • Resources)
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2001 AIME I Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See also