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

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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>.  
 
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 1 (long but non-bashy commentors note: this solution made me lose brain cells, way too wordy for something so simple)==
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==Solution 1 (long but non-bashy)
 +
 
 +
*Commenter's note: this solution made me lose brain cells, as it is so wordy for something so simple.
  
 
Note that the given equality reduces to
 
Note that the given equality reduces to

Revision as of 20:52, 18 February 2018

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)

  • Commenter's note: this solution made me lose brain cells, as it is so wordy for something so simple.

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

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 $AO=BO=CO=2014$. 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)

See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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