AoPS Wiki talk:Problem of the Day/September 9, 2011

We see that, by Vieta, $a+b+c=1$ . Thus, $\frac{a}{a+b+c-a}+\frac{b}{a+b+c-b}+\frac{c}{a+b+c-c}=\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}$ which we can manipulate and then put over a common denominator: $=\frac{1}{1-a}-\frac{1-a}{1-a}+\frac{1}{1-b}-\frac{1-b}{1-b}+\frac{1}{1-c}-\frac{1-c}{1-c}=\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}-3$ $=\frac{3-2(a+b+c)+(ab+ac+bc)}{-(abc-(ab+ac+bc)+a+b+c-1)}-3$ By Vieta, $a+b+c=1$ , $ab+ac+bc=-\frac{1}{2}$ , and $abc=-2$ . Thus, $=\frac{1/2}{3/2}-3=\frac{1}{3}-3=-\frac{8}{3}.$ Thus, $p+q=8+3=\boxed{11}$ .

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AoPS Wiki talk:Problem of the Day/September 9, 2011