Difference between revisions of "2008 AIME II Problems/Problem 13"
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== Solution == | == Solution == | ||
− | 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 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/ | + | 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> | ||
− | picture p; | + | 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);draw(p,arc(1/3^.5,1/3^.5,-60,60)); | + | 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> <math>\Longrightarrow</math> <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); | 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 14:34, 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 be the region outside the hexagon, and let . Then the area of has the form , where and are positive integers. Find .
Solution
If a point is in , then the point is in (where cis denotes ). Since is symmetric every about the origin, it suffices to consider the area of the result of the transformation when , and then to multiply by to account for the entire area.
We note that if the region , where is the region outside the circle of radius centered at the origin, then is simply the region inside a circle of radius centered at the origin. It now suffices to find what happens to the mapping of the region .
The equation of the hexagon side in that region is , which is transformed to . Let where ; then , so the equation becomes . Hence the side is sent to a unit circle centered at .
Then is the union of six unit circles centered at $\cis \frac{k\pi}{6}$ (Error compiling LaTeX. ! Undefined control sequence.), , and the region . That is show below.
The area of the regular hexagon is just . The area of each of the sectors is . Their sum is , and .
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 |