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Difference between revisions of "2008 AIME II Problems/Problem 13"

(→‎Solution 2 (Calculus))
(→‎Solution 2 (Calculus))
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One can describe the line parallel to the imaginary axis <math>x=\frac{1}{2}</math> using polar coordinates as <math>r(\theta)=\dfrac{1}{2\cos{\theta}}</math>
 
One can describe the line parallel to the imaginary axis <math>x=\frac{1}{2}</math> using polar coordinates as <math>r(\theta)=\dfrac{1}{2\cos{\theta}}</math>
  
−
so <math>z</math> is equal to <math>z=(\dfrac{1}{2\cos{\theta}})*(cis{\theta}
+
so <math>z</math> is equal to <math>z=(\dfrac{1}{2\cos{\theta}})*(\cis {\theta})
−
)\rightarrow \frac{1}{z}=2\cos{\theta}*cis(-\theta)</math>
+
\rightarrow \frac{1}{z}=2\cos{\theta}*cis(-\theta)</math>
  
 
Dividing the hexagon to 12 equal parts we get that  
 
Dividing the hexagon to 12 equal parts we get that  

Revision as of 15:56, 29 June 2017

Problem

A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$. Then the area of $S$ has the form $a\pi + \sqrt{b}$, where $a$ and $b$ are positive integers. Find $a + b$.

Solution 1

If a point $z = r\text{cis}\,\theta$ is in $R$, then the point $\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)$ is in $S$ (where cis denotes $\text{cis}\, \theta = \cos \theta + i \sin \theta$). Since $R$ is symmetric every $60^{\circ}$ about the origin, it suffices to consider the area of the result of the transformation when $-30 \le \theta \le 30$, and then to multiply by $6$ to account for the entire area.

We note that if the region $S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace$, where $R_2$ is the region (in green below) outside the circle of radius $1/\sqrt{3}$ centered at the origin, then $S_2$ is simply the region inside a circle of radius $\sqrt{3}$ centered at the origin. It now suffices to find what happens to the mapping of the region $R_2 - R$ (in blue below).

The equation of the hexagon side in that region is $x = r \cos \theta = \frac{1}{2}$, which is transformed to $\frac{1}{r} \cos (-\theta) = \frac{1}{r} \cos \theta =$2 . Let $r\cos \theta = a+bi$ where $a,b \in \mathbb{R}$; then $r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}$, so the equation becomes $a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1$. Hence the side is sent to an arc of the unit circle centered at $(1,0)$, after considering the restriction that the side of the hexagon is a segment of length $1/\sqrt{3}$.

Including $S_2$, we find that $S$ is the union of six unit circles centered at $\text{cis}\, \frac{k\pi}{6}$, $k = 0,1,2,3,4,5$, as shown below.

[asy] defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype("4 4"); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i < 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label("$1/\sqrt{3}$",(0,-0.5),W,fontsize(8)); [/asy]     $\Longrightarrow$     [asy] defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype("4 4")); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype("4 4")); draw(Circle((0,0),1),linetype("4 4")); label("$\sqrt{3}$",(0,-0.5),W,fontsize(8)); add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p); [/asy]

The area of the regular hexagon is $6 \cdot \left( \frac{\left(\sqrt{3}\right)^2 \sqrt{3}}{4} \right) = \frac{9}{2}\sqrt{3}$. The total area of the six $120^{\circ}$ sectors is $6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}$. Their sum is $2\pi + \sqrt{27}$, and $a+b = \boxed{029}$.

Solution 2 (Calculus)

One can describe the line parallel to the imaginary axis $x=\frac{1}{2}$ using polar coordinates as $r(\theta)=\dfrac{1}{2\cos{\theta}}$

so $z$ is equal to $z=(\dfrac{1}{2\cos{\theta}})*(\cis {\theta}) \rightarrow \frac{1}{z}=2\cos{\theta}*cis(-\theta)$ (Error compiling LaTeX. Unknown error_msg)

Dividing the hexagon to 12 equal parts we get that

$Area = 12*\int_{0}^{\frac{\pi}{6}}\frac{1}{2}(2\cos\theta)^2 dr$

which is a routine computation:

$Area = 12*\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 dr= 12*\int_{0}^{\frac{\pi}{6}}\cos{2\theta}+1 dr$

$Area = 12*\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 dr=12*[\sin{2\theta}+\theta]_0^{\frac{\pi}{6}}=12*(\frac{\sqrt{3}}{4}+\frac{\pi}{6})=2\pi+3\sqrt{3}=2\pi + \sqrt{27}$

$a+b = \boxed{029}$.

See also

2008 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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