Question

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 $1000.$

Answer

Solution:

Multiplying $\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}$ by $zw(z+w)$ gives $zw = (z+w)^2,$ i.e. $z^2 + zw + w^2 = 0.$ Multiplying by $z - w$ yields $z^3 - w^3 = 0,$ so $w = \omega z$ or $w = \omega^2 z,$ where $\omega$ is a primitive cube root of unity (and both indeed satisfy the original equation).

Thus $P$ is the equilateral triangle with vertices $z,$ $\omega z,$ $\omega^2 z,$ inscribed in the circle of radius $2014.$ Its area is $\frac{3\sqrt{3}}{4}\,(2014)^2 = 3 \cdot 1007^2 \sqrt{3},$ so $n = 3 \cdot 1007^2 = 3042147.$

The remainder when $n$ is divided by $1000$ is $147.$

2014 AIME II Problem 10

Solution:

Problem 10 in Other Years