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

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