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2014 AIME II Problems/Problem 10

Revision as of 21:02, 27 March 2014 by Tanuagg13 (talk | contribs) (Created page with "Note that the given equality reduces to <cmath>\frac{1}{w+z} = \frac{w+z}{wz}</cmath> <cmath>wz = {(w+z)}^2</cmath> <cmath>w^2 + wz + z^2 = 0</cmath> <cmath>\frac{w^3 - z^3}{w-z...")
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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

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

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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