Difference between revisions of "2008 AIME II Problems/Problem 13"
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== Solution 1== | == 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 <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. | 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 (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_2 | + | 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). |
− | 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\ | + | 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. | 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. | ||
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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>. | 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 | + | == 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> | + | <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> | + | <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> | + | <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> |
<math>a+b = \boxed{029}</math>. | <math>a+b = \boxed{029}</math>. |
Latest revision as of 23:15, 12 September 2020
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 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 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 (in green below) 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 (in blue below).
The equation of the hexagon side in that region is , which is transformed to 2 . Let where ; then , so the equation becomes . Hence the side is sent to an arc of the unit circle centered at , after considering the restriction that the side of the hexagon is a segment of length .
Including , we find that is the union of six unit circles centered at , , as shown below.
The area of the regular hexagon is . The total area of the six sectors is . Their sum is , and . - Th3Numb3rThr33
Solution 3 (Calculus)
We can describe the line parallel to the imaginary axis using polar coordinates as
which rearranges to
Denote the area of as Now, dividing the hexagon to 12 equal parts, we have the following integral:
Thankfully, this is a routine computation:
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.