2014 AIME II Problem 10

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

Answer: 147

Concepts:complex number roots of unity equilateral triangle

Difficulty rating: 2560

Small hint:

Clear denominators in

\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}

to get

z^2 + zw + w^2 = 0

Big hint:

Multiplying by

z - w

gives

w^3 = z^3,

so the vertices are

z

times the cube roots of unity: an equilateral triangle with circumradius

2014

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