# Difference between revisions of "2014 AIME II Problems/Problem 10"

## Problem

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 $1000$.

## Solution 1 (long but non-bashy)

Note that the given equality reduces to

$$\frac{1}{w+z} = \frac{w+z}{wz}$$ $$wz = {(w+z)}^2$$ $$w^2 + wz + z^2 = 0$$ $$\frac{w^3 - z^3}{w-z} = 0$$ $$w^3 = z^3, w \neq z$$

Now, let $w = r_w e^{i \theta_w}$ and likewise for $z$. Consider circle $O$ with the origin as the center and radius 2014 on the complex plane. It is clear that $z$ must be one of the points on this circle, as $|z| = 2014$.

By DeMoivre's Theorem, the complex modulus of $w$ is cubed when $w$ is cubed. Thus $w$ must lie on $O$, since its the cube of its modulus, and thus its modulus, must be equal to $z$'s modulus.

Again, by DeMoivre's Theorem, $\theta_w$ is tripled when $w$ is cubed and likewise for $z$. For $w$, $z$, and the origin to lie on the same line, $3 \theta_w$ must be some multiple of 360 degrees apart from $3 \theta_z$ , so $\theta_w$ must differ from $\theta_z$ by some multiple of 120 degrees.

Now, without loss of generality, assume that $z$ is on the real axis. (The circle can be rotated to put $z$ in any other location.) Then there are precisely two possible distinct locations for $w$; one is obtained by going 120 degrees clockwise from $z$ about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points.

Let the two possible locations for $w$ be $W_1$ and $W_2$ and the location of $z$ be point $Z$. Note that by symmetry, $W_1W_2Z$ is equilateral, say, with side length $x$. We know that the circumradius of this equilateral triangle is $2014$, so using the formula $\frac{abc}{4R} = [ABC]$ and that the area of an equilateral triangle with side length $s$ is $\frac{s^2\sqrt{3}}{4}$, so we have

$$\frac{x^3}{4R} = \frac{x^2\sqrt{3}}{4}$$ $$x = R \sqrt{3}$$ $$\frac{x^2\sqrt{3}}{4} = \frac{3R^2 \sqrt{3}}{4}$$

Since we're concerned with the non-radical part of this expression and $R = 2014$,

$$\frac{3R^2}{4} \equiv 3 \cdot 1007^2 \equiv 3 \cdot 7^2 \equiv \boxed{147} \pmod{1000}$$

and we are done. $\blacksquare$

## Solution 2 (short but a little bashy)

Assume $z = 2014$. Then $$\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}$$

$$2014w = w(2014 + w) + 2014(2014 + w)$$

$$2014w = 2014w + w^2 + 2014^2 + 2014w$$

$$0 = w^2 + 2014w + 2014^2$$

$$w = \frac{-2014 \pm \sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \pm 1007\sqrt{3}i$$

Thus $P$ is an isosceles triangle with area $\frac{1}{2}(2014 - (-1007))(2\cdot 1007\sqrt{3}) = 3021\cdot 1007\sqrt{3}$ and $n \equiv 7\cdot 21\equiv \boxed{147} \pmod{1000}.$

## Solution 3 (easy)

Our equation can be simplified like the following. $$\frac{1}{w+z} = \frac{w+z}{wz}$$ $$wz = {(w+z)}^2$$ $$w^2 + wz + z^2 = 0$$ We recognize this as the Law of Cosines with angle $120$ degrees. Our polygon is an equilateral triangle, say $ABC$, with center $O$ at the origin and $AB=AC=BC=2014$. The area of $ABC$ is $3*[ABO]=3*(1007*1007\sqrt{3})=3*1007^2*\sqrt{3}=3042147\sqrt{3}$. Thus, the answer is \boxed{147}.

Solution by TheUltimate123 (Eric Shen)