Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AIME II Problems/Problem 10"
Line 28: Line 28:
  
 
We know that <math>R^2 = 2014^2</math>. Calculating, we can find that our desired <math>n</math> is <math>3,042,147</math>, so our desired answer is <math>\boxed{147}</math>.
 
We know that <math>R^2 = 2014^2</math>. Calculating, we can find that our desired <math>n</math> is <math>3,042,147</math>, so our desired answer is <math>\boxed{147}</math>.
 +
 +
==Solution 2==
 +
Without loss of generality, let <math>z = 2014</math>. Then
 +
<cmath>\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}</cmath>
 +
 +
<cmath>2014w = w(2014 + w) + 2014(2014 + w)</cmath>
 +
 +
<cmath>2014w = 2014w + w^2 + 2014^2 + 2014w</cmath>
 +
 +
<cmath>0 = w^2 + 2014w + 2014^2</cmath>
 +
 +
<cmath>w = \frac{-2014 \pm \sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \pm 1007\sqrt{3}i</cmath>
 +
 +
Thus <math>P</math> is an isosceles triangle with area <math>\frac{1}{2}(2014 - (-1007))(2\cdot 1007\sqrt{3}) = 3021\cdot 1007\sqrt{3}</math> and <math>n \equiv 7\cdot 21\equiv 147 \pmod{1000}</math>

Revision as of 16:25, 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 $\boxed{147}$.

Solution 2

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 $n \equiv 7\cdot 21\equiv 147 \pmod{1000}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_10&oldid=61245"