**Problem**: (2001 AIME1 - #3) Find the sum of the roots of the polynomial $x^{2001} + \left(\frac{1}{2} - x\right)^{2001}$. [Source: http://www.kalva.demon.co.uk/aime/aime01a.html]

**Solution**: So this isn't really brute force as the post title suggests, but it was a funny play on words so yeah.

We remember back here we used the one and only Vieta's Formulas, and we can do so again.

So let's check the biggest power, $2001$. We get coefficients of 1 from the first term and -1 from the second. But wait, that gives us $1+(-1) = 0$ as the coefficient of $x^{2001}$. What now?

In case you haven't figured out, the polynomial is actually of degree $2000$ instead, so we need to coefficients on $x^{2000}$ and $x^{1999}$. Expanding using the Binomial Theorem, we get

$ 2001C2000 \left(\frac{1}{2}\right)x^{2000} - 2001C1999 \left(\frac{1}{2}\right)^2x^{1999} = \frac{2001}{2}x^{2000}-(2001)(250)x^{1999}$,

which by Vieta's Formulas yields $ \frac{(2001)(250)}{\frac{2001}{2}} = 500 $ as the desired answer. QED.

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Practice Problem #1: Prove Vieta's Formula for the sum of the roots of a polynomial.

Practice Problem #2: (1996 AIME - #5) The roots of $x^3+3x^2+4x-11$ are $a,b,c$. The equation with roots $a+b,b+c,c+a$ is $x^3+rx^2+sx+t$. Find $t$. [Source: http://www.kalva.demon.co.uk/aime/aime96.html]

The sum of the roots of the second polynomial is 2(a+b+c), twice the original sum which was -3. Thus r=6.

ReplyDeleteThe product of the roots is (a+b)(b+c)(c+a) = a^2b + a^2c + b^2a + b^2c + c^2a+c^2b + 2abc or a(ab+ac) + b(ab+bc) + c(ac + bc) + 2(11).

Since ab+ac+bc = 4, we can simplify to get a(4-bc)+b(4-ac)+c(4-ab) + 22 = 4(a+b+c) - 3abc+ 22 = 4(-3) - 3(11) + 22 = -12 - 33 + 22 = -23 = -t, so t = 23.

except abc=-11 (not 11) , so t=43.

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