2014
Problem - 107
Let $z$ be a complex number with $|z|=2014$. Let $P$ be the polygon in the complex plane whose vertices are $z$ and every $w$ such that $\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}$. Then the area enclosed by $P$ can be written in the form $n\sqrt{3}$, where $n$ is an integer. Find the remainder when $n$ is divided by $1000$.
Without loss of generality, let's assume $z=2014$ (because the graph can always rotated so that $z=1024$.). Then the give equation is equivalent to $$\frac{1}{2014+w}=\frac{1}{2014}+\frac{1}{3}\implies w=-1007\cdot(1\pm\sqrt{3}i)$$
Thus $P$ is an isosceles triangle whose area is $$\frac{1}{2}\cdot(2014-(-1007))(2\cdot 1007\sqrt{3})=3021\cdot1007\sqrt{3}$$
Therefore $$n\equiv 3021\cdot 1007\equiv 7\cdot 21\equiv\boxed{147}\pmod {1000}$$