2014 AIME II Problems/Problem 10

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)

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

Notice that \[\frac1{w+z} = \frac{w+z}{wz} \implies 0 = w^2 + wz + z^2 = \frac{w^3-z^3}{w-z}.\] Hence, $w=ze^{2\pi i/3},ze^{4\pi i/3}$, and $P$ is an equilateral triangle with circumradius $2014$. Then, \[[P]=\frac{3}{2}\cdot 2014^2\cdot\sin\frac{\pi}3=3\cdot 1007^2\sqrt3,\] and the answer is $3\cdot 1007^2\equiv 3\cdot 7^2\equiv\boxed{147}\pmod{1000}$.

Solution 4 (Slick)

I find that generally, the trick to these kinds of AIME problems is to interpret the problem geometrically, and that is just what I did here. Looking at the initial equation, this seems like a difficult task, but rearranging yields a nicer equation: \[\frac1{z+w}=\frac1z+\frac1w\] \[\frac w{z+w}=\frac wz+1\] \[\frac w{z+w}=\frac{w+z}z\] \[\frac{w-0}{w-(-z)}=\frac{(-z)-w}{(-z)-0}\] We can interpret the difference of two complex numbers as a vector from one to the other, and we can interpret the quotient as a vector with an angle equal to the angle between the two vectors. Therefore, after labeling the complex numbers with $W$ ($w$), $V$ ($-z$), and $O$ ($0$), we can interpret the above equation to mean that the $\angle OWV=\angle OVW$, and hence triangle $OWV$ is isosceles, so length $OW$ = $OV$. Rearranging the equation \[\frac{w-0}{w-(-z)}=\frac{(-z)-w}{(-z)-0},\] we find that \[(w-0)((-z)-0)=-(w-(-z))^2.\] Taking the magnitude of both sides and using the fact that $OW=OV\implies |w-0|=|(-z)-0|$, we find that \[|w-0|^2=|w-(-z)|^2,\] so length $OW=VW$ and triangle $OWV$ is equilateral. There are only two possible $W$'s for which $OWV$ is equilateral, lying on either side of $OV$. After drawing these points on the circle of radius 2014 centered at the origin, it is easy to see that $z$ and the two $w$'s form an equilateral triangle (this can be verified by simple angle chasing). Drawing a perpendicular bisector of one of the sides and using 30-60-90 triangles shows that the side length of this triangle is $2014\sqrt3$ and hence its area is $\frac{\sqrt3(2014\sqrt3)^2}4=\boxed{147}\sqrt3+1000k\sqrt3,$ for some integer $k$.

~SymbolicPermutation

Notes

This problem is killed by polar coordinates (complex numbers) . Solve using cosine-i-sine.

See also

2014 AIME II (ProblemsAnswer KeyResources)
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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