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

Latest revision as of 17:56, 25 December 2022

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 you're familiar with inversion, you'll see that the problem's basically asking you to invert the hexagon with respect to the unit circle in the Cartesian Plane using the Inversion Distance Formula. This works because the point in the Cartesian Plane's complex plane equivalent switches places with its conjugate but we can do that in the Cartesian plane too (just reflect a point in the Cartesian plane over the x-axis)! If you're familiar with inversion you can go plot the inverted figure's Cartesian Plane Equivalent. Then simply continue on with the figure shown in the below solution.

Solution 2

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 $R-R_2$ (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 $r\text{cis}\,\theta = a+bi$ 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] unitsize(1.5cm); 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]$ $\Longrightarrow$ $[asy]unitsize(4.5cm); defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype("4 4")); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype("4 4")); draw(Circle((0,0),1),linetype("4 4")); label("$\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 $a+b = \boxed{029}$ . - Th3Numb3rThr33

Solution 3 (Calculus)

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

which rearranges to $z=\left(\dfrac{1}{2\cos{\theta}}\right)(cis{\theta})\implies \frac{1}{z}=2\cos{\theta}cis(-\theta).$

Denote the area of $S$ as $[S].$ Now, dividing the hexagon to 12 equal parts, we have the following integral:

$[S] = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}r^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}(2\cos\theta)^2 d\theta.$

Thankfully, this is a routine computation:

$[S] = 12\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta$

$[S] = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta = 12\left[\frac{1}{2}\sin{2\theta}+\theta\right]_0^{\frac{\pi}{6}}=12\left(\frac{\sqrt{3}}{4}+\frac{\pi}{6}\right)=2\pi+3\sqrt{3}=2\pi + \sqrt{27}$

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

@@ Line 2: / Line 2: @@
 A [[regular polygon|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 <math>R</math> be the region outside the hexagon, and let <math>S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace</math>. Then the area of <math>S</math> has the form <math>a\pi + \sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive integers. Find <math>a + b</math>.
-== Solution ==
+== Solution 1==
-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.
+If you're familiar with inversion, you'll see that the problem's basically asking you to invert the hexagon with respect to the unit circle in the Cartesian Plane using the Inversion Distance Formula. This works because the point in the Cartesian Plane's complex plane equivalent switches places with its conjugate but we can do that in the Cartesian plane too (just reflect a point in the Cartesian plane over the x-axis)! If you're familiar with inversion you can go plot the inverted figure's Cartesian Plane Equivalent. Then simply continue on with the figure shown in the below solution.
-We note that 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>.
+== Solution 2==
+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.
-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>.
+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 (in green below) 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-R_2</math> (in blue below).
-Then <math>S</math> is the union of six unit circles centered at <math>\cis \frac{k\pi}{6}</math>, <math>k = 0,1,2,3,4,5</math>, and the region <math>S_2</math>. That is show below.
+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 = </math>2 . Let <math>r\text{cis}\,\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 an arc of the unit circle centered at <math>(1,0)</math>, after considering the restriction that the side of the hexagon is a segment of length <math>1/\sqrt{3}</math>.
+Including <math>S_2</math>, we find that <math>S</math> is the union of six unit circles centered at <math>\text{cis}\, \frac{k\pi}{6}</math>, <math>k = 0,1,2,3,4,5</math>, as shown below.
 <center><asy>
-picture p;
+unitsize(1.5cm);
-draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60));
+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>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math>\Longrightarrow</math>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<asy>unitsize(4.5cm);
+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></center>
-The area of the regular hexagon is just <math>6 \cdot \frac{(\sqrt{3}^2) \sqrt{3}}{4} = \frac{9}{2}\sqrt{3}</math>. The area of each of the <math>120^{\circ}</math> sectors is <math>6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}</math>. Their sum is <math>2\pi + \sqrt{27}</math>, and <math>a+b = \boxed{029}</math>.
+The area of the regular hexagon is <math>6 \cdot \left( \frac{\left(\sqrt{3}\right)^2 \sqrt{3}}{4} \right) = \frac{9}{2}\sqrt{3}</math>. The total area of the six <math>120^{\circ}</math> sectors is <math>6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}</math>. Their sum is <math>2\pi + \sqrt{27}</math>, and <math>a+b = \boxed{029}</math>.
+- Th3Numb3rThr33
+== Solution 3 (Calculus) ==
+We 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>
+which rearranges to <math>z=\left(\dfrac{1}{2\cos{\theta}}\right)(cis{\theta})\implies \frac{1}{z}=2\cos{\theta}cis(-\theta).</math>
+Denote the area of <math>S</math> as <math>[S].</math> Now, dividing the hexagon to 12 equal parts, we have the following integral:
+<math>[S] = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}r^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}(2\cos\theta)^2 d\theta.</math>
+Thankfully, this is a routine computation:
+<math>[S] = 12\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta</math>
+<math>[S] = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta = 12\left[\frac{1}{2}\sin{2\theta}+\theta\right]_0^{\frac{\pi}{6}}=12\left(\frac{\sqrt{3}}{4}+\frac{\pi}{6}\right)=2\pi+3\sqrt{3}=2\pi + \sqrt{27}</math>
-{{incomplete}}
+<math>a+b = \boxed{029}</math>.
 == See also ==
@@ Line 26: / Line 49: @@
 [[Category:Intermediate Algebra Problems]]
 [[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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