8/11/2023

Today I continued the math book I was doing, and I did problems 1-9 in Chapter 2 :)

My favorite problem was an AIME problem: 

My Solution: 

Note that, by the Binomial Theorem, $$x^{2001}+(\frac{1}{2}-x{)}^{2001}$$ $$=x^{2001}+(-x^{2001}+(\genfrac{}{}{0ex}{}{2001}{1})x^{2000}{\left(\frac{1}{2}\right)}^1-(\genfrac{}{}{0ex}{}{2001}{2})x^{1999}{\left(\frac{1}{2}\right)}^2+\cdots +(\genfrac{}{}{0ex}{}{2001}{2001})x^0{\left(\frac{1}{2}\right)}^{2001})$$ The $x^{2001}$ terms cancel, and we know that this whole expression is $0$, so $$=(\genfrac{}{}{0ex}{}{2001}{1})x^{2000}{\left(\frac{1}{2}\right)}^1-(\genfrac{}{}{0ex}{}{2001}{2})x^{1999}{\left(\frac{1}{2}\right)}^2+\cdots +(\genfrac{}{}{0ex}{}{2001}{2001})x^0{\left(\frac{1}{2}\right)}^{2001}=0$$ Now, we have to find the sum of roots of this expression. For this, we can use Vieta's formula for the sum of roots: $$\sum \text{roots}=-\frac{-(\genfrac{}{}{0ex}{}{2001}{2})\cdot \frac{1}{4}}{(\genfrac{}{}{0ex}{}{2001}{1})\cdot \frac{1}{2}}=\boxed{{\displaystyle 500}}.$$

Cool solution by the author:

Note that when you plug in $x=\frac{1}{2}-r$, then you get the same equation as when you plug in $x=r$. 

This implies that the solutions to this equation come in pairs: if $r$ is a solution, then $\frac{1}{2}-r$ is also a solution. Since there are pairs of $(r,\frac{1}{2}-r)$, the sum of all of the roots would be $\frac{1}{2}$ summed up several times. It would be summed for each pair, and since we have a polynomial of degree $2000$, there are $$\frac{2000}{2}=1000$$ pairs. The sum of roots is thus $\frac{1}{2}\cdot 1000=\boxed{{\displaystyle 500}}.$

I think the author's solution was a cool and slick way that was applicable in this problem, but in my opinion, the best way to approach these types of problems with a foolproof approach is the Vieta's formulas.