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Difference between revisions of "2014 AIME II Problems/Problem 10"
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==Problem==
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Let <math>z</math> be a complex number with <math>|z|=2014</math>. Let <math>P</math> be the polygon in the complex plane whose vertices are <math>z</math> and every <math>w</math> such that <math>\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}</math>. Then the area enclosed by <math>P</math> can be written in the form <math>n\sqrt{3}</math>, where <math>n</math> is an integer. Find the remainder when <math>n</math> is divided by <math>1000</math>.
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==Solution==
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Note that the given equality reduces to
 
Note that the given equality reduces to
  

Revision as of 16:20, 28 March 2014

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

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 $\frac{s^2\sqrt{3}}{4}$, we have

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

We know that $R^2 = 2014^2$. Calculating, we can find that our desired $n$ is $3,042,147$, so our desired answer is $\boxed{147}$.

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