Problem (Algebra):
If $\frac{(x-2013)(y-2014)}{(x-2013)^{2}+(y-2014)^{2}}=-\frac{1}{2}$, what is the value of $x+y$ ?

Solution:
Let $a=x-2013$ and let $b=y-2014$.

The given equation becomes $\frac{a b}{a^{2}+b^{2}}=-\frac{1}{2}$, which is equivalent to $2 a b=-a^{2}-b^{2}$ and $a^{2}+2 a b+b^{2}=0$.

This is equivalent to $(a+b)^{2}=0$ which is equivalent to $a+b=0$.

Since $a=x-2013$ and $b=y-2014$, then $x-2013+y-2014=0$ or $x+y=4027$.

Final answer: 4027