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

Revision as of 00:10, 21 May 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 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)

Without loss of generality, let $z = 2014$ . 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 $n \equiv 7\cdot 21\equiv \boxed{147} \pmod{1000}.$

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 == See also ==
 {{AIME box|year=2014|n=II|num-b=9|num-a=11}}
+{{MAA Notice}}

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

Revision as of 00:10, 21 May 2014

Contents

Problem

Solution 1 (long but non-bashy)

Solution 2 (short but a little bashy)

See also