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

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== Solution ==
 
== Solution ==
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If a point <math>z = r\text{cis}\,\theta</math> is in <math>R</math>, then the point <math>\frac{1}{z} = \frac{1}{r} \text{cis}\, -\theta</math> is in <math>S</math> (where [[cis]] denotes <math>\text{cis}\, \theta = \cos \theta + i \sin \theta</math>). Since <math>R</math> is symmetric about the origin, it suffices to consider the result of the transformation when <math>-30 \le \theta \le 30</math>, and then to multiply by <math>6</math> to account for the entire area.
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If a point <math>z = r\text{cis}\,\theta</math> is in <math>R</math>, then the point <math>\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)</math> is in <math>S</math> (where [[cis]] denotes <math>\text{cis}\, \theta = \cos \theta + i \sin \theta</math>). Since <math>R</math> is symmetric every <math>60^{\circ}</math> about the origin, it suffices to consider the area of the result of the transformation when <math>-30 \le \theta \le 30</math>, and then to multiply by <math>6</math> to account for the entire area.
  
−
We note that if the region <math>S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace</math>, where <math>R_2</math> is the region outside the circle of radius <math>1/2</math> centered at the origin, then <math>S_2</math> is simply the region inside a circle of radius <math>2</math> centered at the origin. It now suffices to find what happens to the mapping of the region <math>R_2 - R</math>.  
+
We note that if the region <math>S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace</math>, where <math>R_2</math> is the region outside the circle of radius <math>1/\sqrt{3}</math> centered at the origin, then <math>S_2</math> is simply the region inside a circle of radius <math>\sqrt{3}</math> centered at the origin. It now suffices to find what happens to the mapping of the region <math>R_2 - R</math>.  
  
 
The equation of the hexagon side in that region is <math>x = r \cos \theta = \frac{1}{2}</math>, which is transformed to <math>\frac{1}{r} \cos -\theta = \frac{1}{r} \cos \theta = \frac 12</math>. Let <math>r\cos \theta = a+bi</math> where <math>a,b \in \mathbb{R}</math>; then <math>r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}</math>, so the equation becomes <math>a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1</math>. Hence the side is sent to a unit circle centered at <math>(1,0)</math>.  
 
The equation of the hexagon side in that region is <math>x = r \cos \theta = \frac{1}{2}</math>, which is transformed to <math>\frac{1}{r} \cos -\theta = \frac{1}{r} \cos \theta = \frac 12</math>. Let <math>r\cos \theta = a+bi</math> where <math>a,b \in \mathbb{R}</math>; then <math>r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}</math>, so the equation becomes <math>a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1</math>. Hence the side is sent to a unit circle centered at <math>(1,0)</math>.  
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<center><asy>
 
<center><asy>
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picture p;
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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));
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draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60));
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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));
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</asy>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math>\Longrightarrow</math>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<asy>
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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(p,arc(1/3^.5,1/3^.5,60,300),linetype("4 4")); */ draw(Circle((0,0),1),linetype("4 4"));
 
add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p);
 
add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p);
 
</asy></center>
 
</asy></center>

Revision as of 13:34, 16 February 2009

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

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 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 $R_2 - R$.

The equation of the hexagon side in that region is $x = r \cos \theta = \frac{1}{2}$, which is transformed to $\frac{1}{r} \cos -\theta = \frac{1}{r} \cos \theta = \frac 12$. 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 a unit circle centered at $(1,0)$.

Then $S$ is the union of six unit circles centered at $\cis \frac{k\pi}{6}$ (Error compiling LaTeX. Unknown error_msg), $k = 0,1,2,3,4,5$, and the region $S_2$. That is show 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)); [/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(p,arc(1/3^.5,1/3^.5,60,300),linetype("4 4")); */ draw(Circle((0,0),1),linetype("4 4")); 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 just $6 \cdot \frac{(\sqrt{3}^2) \sqrt{3}}{4} = \frac{9}{2}\sqrt{3}$. The area of each of the $120^{\circ}$ 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 $a+b = \boxed{029}$.

Template:Incomplete

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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