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

Latest revision as of 12:08, 21 January 2024

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 (Roots of Unity)

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.

@@ Line 33: / Line 33: @@
 and we are done. <math>\blacksquare</math>
-==Solution 2 (short but a little bashy)==
+==Solution 2 (short)==
 Assume <math>z = 2014</math>. Then
 <cmath>\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}</cmath>
@@ Line 47: / Line 47: @@
 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 \boxed{147} \pmod{1000}.</math>
-==Solution 3==
+==Solution 3 (Roots of Unity) ==
 Notice that <cmath>\frac1{w+z} = \frac{w+z}{wz} \implies 0 = w^2 + wz + z^2 = \frac{w^3-z^3}{w-z}.</cmath>
 Hence, <math>w=ze^{2\pi i/3},ze^{4\pi i/3}</math>, and <math>P</math> is an equilateral triangle with circumradius <math>2014</math>. Then, <cmath>[P]=\frac{3}{2}\cdot 2014^2\cdot\sin\frac{\pi}3=3\cdot 1007^2\sqrt3,</cmath>
 and the answer is <math>3\cdot 1007^2\equiv 3\cdot 7^2\equiv\boxed{147}\pmod{1000}</math>.
+==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:
+<cmath>\frac1{z+w}=\frac1z+\frac1w</cmath>
+<cmath>\frac w{z+w}=\frac wz+1</cmath>
+<cmath>\frac w{z+w}=\frac{w+z}z</cmath>
+<cmath>\frac{w-0}{w-(-z)}=\frac{(-z)-w}{(-z)-0}</cmath>
+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 <math>W</math> (<math>w</math>), <math>V</math> (<math>-z</math>), and <math>O</math> (<math>0</math>), we can interpret the above equation to mean that the <math>\angle OWV=\angle OVW</math>, and hence triangle <math>OWV</math> is isosceles, so length <math>OW</math> = <math>OV</math>.  Rearranging the equation
+<cmath>\frac{w-0}{w-(-z)}=\frac{(-z)-w}{(-z)-0},</cmath>
+we find that
+<cmath>(w-0)((-z)-0)=-(w-(-z))^2.</cmath>
+Taking the magnitude of both sides and using the fact that <math>OW=OV\implies |w-0|=|(-z)-0|</math>, we find that
+<cmath>|w-0|^2=|w-(-z)|^2,</cmath>
+so length <math>OW=VW</math> and triangle <math>OWV</math> is equilateral.  There are only two possible <math>W</math>'s for which <math>OWV</math> is equilateral, lying on either side of <math>OV</math>.  After drawing these points on the circle of radius 2014 centered at the origin, it is easy to see that <math>z</math> and the two <math>w</math>'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 <math>2014\sqrt3</math> and hence its area is <math>\frac{\sqrt3(2014\sqrt3)^2}4=\boxed{147}\sqrt3+1000k\sqrt3,</math> for some integer <math>k</math>.
+~SymbolicPermutation
 ==Notes==

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
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