AoPS Wiki talk:Problem of the Day/September 9, 2011
Revision as of 16:00, 9 September 2011 by Negativebplusorminus (talk | contribs) (Created page with "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 ma...")
We see that, by Vieta, 
. Thus,
![\[\frac{a}{a+b+c-a}+\frac{b}{a+b+c-b}+\frac{c}{a+b+c-c}=\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\]](http://latex.artofproblemsolving.com/8/9/c/89cd3ae8350ed8c1f303fea82e83f7599102ba6f.png)
which we can manipulate and then put over a common denominator:
![\[=\frac{1}{1-a}-\frac{1-a}{1-a}+\frac{1}{1-b}-\frac{1-b}{1-b}+\frac{1}{1-c}-\frac{1-c}{1-c}=\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}-3\]](http://latex.artofproblemsolving.com/d/3/8/d38cbfb7af5f60d64cd2eae07f2716674137e008.png)
![\[=\frac{3-2(a+b+c)+(ab+ac+bc)}{-(abc-(ab+ac+bc)+a+b+c-1)}-3\]](http://latex.artofproblemsolving.com/7/5/1/751a1a648918506df4b0e69a2de3fbd061cfbf1b.png)
By Vieta, , 
, and 
. Thus,
![\[=\frac{1/2}{3/2}-3=\frac{1}{3}-3=-\frac{8}{3}.\]](http://latex.artofproblemsolving.com/8/d/0/8d09bb5f27cb845d08abaaa93f62cff742ee4d47.png)
Thus, 
.
