Online Course Discussion Forum

Q9 for AIME problems of week 1

 
 
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Q9 for AIME problems of week 1
by Henry Zhang - Monday, July 20, 2020, 1:16 PM
 
We used the fact that if x is a root, then 1/2 - x is also a root to solve this problem, but if there is a root r that appears multiple times, do we know if 1/2 - r appears the same number of times such that there is still 1000 root pairs of x and 1/2 - x?
For example (I know these numbers don't actually work for the problem) if 1 was a root with multiplicity 2 would we know that 1/2 - 1 = -1/2 occur two times? Because if 1 was a root with multiplicity 2 and -1/2 only appeared once there wouldn't be root pairs that sum to 1/2 always.
 
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Re: Q9 for AIME problems of week 1
by Dr. Kevin Wang - Thursday, July 23, 2020, 10:20 PM
 

The original question is the following:

Find the sum of the roots of the polynomial $x^{2001} + \left(\dfrac{1}{2} - x\right)^{2001}$.

Because of the complete symmetry between $x$ and $\dfrac{1}{2}-x$, the roots $r$ and $\dfrac{1}{2}-r$ should have the same multiplicity. If you want to be more rigorous, then Vieta's Formulas can be applied to calculate the sum of the roots.

Consider the binomial expansion of the second part. It contains $-x^{2001}$, so the term $x^{2001}$ is canceled, and the highest degree is $2000$.  Then it is a polynomial of degree $2000$, and there are $2000$ roots. Now the coefficient of the term $x^{2000}$ is $\displaystyle\binom{2001}{2000}\frac{1}{2}=\frac{2001}{2}$, and the coefficient of the term $x^{1999}$ is $\displaystyle -\binom{2001}{1999}\frac{1}{2^2}=-\frac{2001\cdot 2000}{8}$. Therefore, by Vieta's formulas, the sum of the roots is

\[ \frac{\dfrac{2001\cdot 2000}{8}}{\dfrac{2001}{2}}=500. \]
Picture of Henry Zhang
Re: Q9 for AIME problems of week 1
by Henry Zhang - Friday, July 24, 2020, 12:30 PM
 

thanks!