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

Revision as of 02:35, 20 February 2021

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 $3\cdot 1007^2\equiv 3\cdot 7^2\equiv\boxed{147}\pmod{1000}$ .

Notes

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

@@ Line 4: / Line 4: @@
 ==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
@@ Line 35: / 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 50: / Line 48: @@
 ==Solution 3==
-Our equation can be simplified like the following.
+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>
-<cmath>\frac{1}{w+z} = \frac{w+z}{wz}</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>
-<cmath>wz = {(w+z)}^2</cmath>
+and the answer is <math>3\cdot 1007^2\equiv 3\cdot 7^2\equiv\boxed{147}\pmod{1000}</math>.
-<cmath>w^2 + wz + z^2 = 0</cmath>
-We recognize this as the Law of Cosines with angle <math>120</math> degrees.
-Our polygon is an equilateral triangle, say <math>ABC</math>, with center <math>O</math> at the origin and <math>AO=BO=CO=2014</math>. The area of <math>ABC</math> is <math>3*[ABO]=3*(1007*1007\sqrt{3})=3*1007^2*\sqrt{3}=3042147\sqrt{3}</math>. Thus, the answer is <math>\boxed{147}</math>.
--Solution by TheUltimate123
+==Notes==
+This problem is killed by polar coordinates (complex numbers) . Solve using cosine-i-sine.
 == See also ==
 {{AIME box|year=2014|n=II|num-b=9|num-a=11}}
 {{MAA Notice}}

2014 AIME II (Problems • Answer Key • Resources)
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