Difference between revisions of "2003 AMC 12A Problems/Problem 15"
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A [[semicircle]] of [[diameter]] <math>1</math> sits at the top of a semicircle of diameter <math>2</math>, as shown. The shaded [[area]] inside the smaller semicircle and outside the larger semicircle is called a ''lune''. Determine the area of this lune. | A [[semicircle]] of [[diameter]] <math>1</math> sits at the top of a semicircle of diameter <math>2</math>, as shown. The shaded [[area]] inside the smaller semicircle and outside the larger semicircle is called a ''lune''. Determine the area of this lune. | ||
| − | + | <asy> | |
| + | import graph; | ||
| + | size(150); | ||
| + | defaultpen(fontsize(8)); | ||
| + | pair A=(-2,0), B=(2,0); | ||
| + | filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray); | ||
| + | filldraw(Arc((0,0),2,0,180)--cycle,white); | ||
| + | draw(2*expi(2*pi/6)--2*expi(4*pi/6)); | ||
| + | |||
| + | label("1",(0,sqrt(3)),(0,-1)); | ||
| + | label("2",(0,0),(0,-1)); | ||
| + | </asy> | ||
<math> \mathrm{(A) \ } \frac{1}{6}\pi-\frac{\sqrt{3}}{4}\qquad \mathrm{(B) \ } \frac{\sqrt{3}}{4}-\frac{1}{12}\pi\qquad \mathrm{(C) \ } \frac{\sqrt{3}}{4}-\frac{1}{24}\pi\qquad \mathrm{(D) \ } \frac{\sqrt{3}}{4}+\frac{1}{24}\pi\qquad \mathrm{(E) \ } \frac{\sqrt{3}}{4}+\frac{1}{12}\pi </math> | <math> \mathrm{(A) \ } \frac{1}{6}\pi-\frac{\sqrt{3}}{4}\qquad \mathrm{(B) \ } \frac{\sqrt{3}}{4}-\frac{1}{12}\pi\qquad \mathrm{(C) \ } \frac{\sqrt{3}}{4}-\frac{1}{24}\pi\qquad \mathrm{(D) \ } \frac{\sqrt{3}}{4}+\frac{1}{24}\pi\qquad \mathrm{(E) \ } \frac{\sqrt{3}}{4}+\frac{1}{12}\pi </math> | ||
== Solution == | == Solution == | ||
| − | [ | + | <asy> |
| + | import graph; | ||
| + | size(150); | ||
| + | defaultpen(fontsize(8)); | ||
| + | pair A=(-2,0), B=(2,0); | ||
| + | filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray); | ||
| + | fill(Arc((0,0),2,0,180)--cycle,white); | ||
| + | draw(Arc((0,0),2,0,180)--cycle); | ||
| + | draw((0,0)--2*expi(2*pi/6)--2*expi(2*pi/6*2)--(0,0)); | ||
| + | |||
| + | label("A",(0,2),(0,4)); | ||
| + | label("B",(0,2),(0,-1)); | ||
| + | label("C",(0,sqrt(3)/2),(0,2)); | ||
| + | label("1",(-0.5,sqrt(3)/2),(-1,0)); | ||
| + | label("1",(0.5,sqrt(3)/2),(1,0)); | ||
| + | </asy> | ||
| + | |||
| + | Let <math>[X]</math> denote the area of region <math>X</math> in the figure above. | ||
| − | The shaded area is equal to the area of the smaller semicircle minus the area of a [[sector]] of the larger circle plus the area of a [[triangle]] formed by two [[radius | radii]] of the larger semicircle and the diameter of the smaller semicircle. | + | The shaded area <math>[A]</math> is equal to the area of the smaller semicircle <math>[A+B]</math> minus the area of a [[sector]] of the larger circle <math>[B+C]</math> plus the area of a [[triangle]] formed by two [[radius | radii]] of the larger semicircle and the diameter of the smaller semicircle <math>[C]</math>. |
| − | The area of the smaller semicircle is <math>\frac{1}{2}\pi\cdot(\frac{1}{2})^{2}=\frac{1}{8}\pi</math>. | + | The area of the smaller semicircle is <math>[A+B] = \frac{1}{2}\pi\cdot(\frac{1}{2})^{2}=\frac{1}{8}\pi</math>. |
Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures <math>60^\circ</math>. | Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures <math>60^\circ</math>. | ||
| − | The area of the <math>60^\circ</math> sector of the larger semicircle is <math>\frac{60}{360}\pi\cdot(\frac{2}{2})^{2}=\frac{1}{6}\pi</math>. | + | The area of the <math>60^\circ</math> sector of the larger semicircle is <math>[B+C] = \frac{60}{360}\pi\cdot(\frac{2}{2})^{2}=\frac{1}{6}\pi</math>. |
| − | The area of the triangle is <math>\frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}</math> | + | The area of the triangle is <math>[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}</math>. |
| − | So the shaded area is <math>\frac{1}{8}\pi-\frac{1}{6}\pi+\frac{\sqrt{3}}{4}=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}</math> | + | So the shaded area is <math>[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}</math>. |
== See Also == | == See Also == | ||
Revision as of 11:50, 13 January 2012
- The following problem is from both the 2003 AMC 12A #15 and 2003 AMC 10A #19, so both problems redirect to this page.
Problem
A semicircle of diameter 
sits at the top of a semicircle of diameter 
, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.
![[asy] import graph; size(150); defaultpen(fontsize(8)); pair A=(-2,0), B=(2,0); filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray); filldraw(Arc((0,0),2,0,180)--cycle,white); draw(2*expi(2*pi/6)--2*expi(4*pi/6)); label("1",(0,sqrt(3)),(0,-1)); label("2",(0,0),(0,-1)); [/asy]](http://latex.artofproblemsolving.com/e/e/8/ee832dfa1925af1fe40d6af62243894f0614e222.png)
Solution
![[asy] import graph; size(150); defaultpen(fontsize(8)); pair A=(-2,0), B=(2,0); filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray); fill(Arc((0,0),2,0,180)--cycle,white); draw(Arc((0,0),2,0,180)--cycle); draw((0,0)--2*expi(2*pi/6)--2*expi(2*pi/6*2)--(0,0)); label("A",(0,2),(0,4)); label("B",(0,2),(0,-1)); label("C",(0,sqrt(3)/2),(0,2)); label("1",(-0.5,sqrt(3)/2),(-1,0)); label("1",(0.5,sqrt(3)/2),(1,0)); [/asy]](http://latex.artofproblemsolving.com/c/7/d/c7dc6582f0dd4a6d448e272a4a65408d00646000.png)
Let ![$[X]$](http://latex.artofproblemsolving.com/f/6/3/f631f7a53d834de52b718ed6b496689e9658eadf.png)
denote the area of region 
in the figure above.
The shaded area ![$[A]$](http://latex.artofproblemsolving.com/5/9/7/5974998abb6495b7e7cb9c32143687050121fe86.png)
is equal to the area of the smaller semicircle ![$[A+B]$](http://latex.artofproblemsolving.com/a/e/1/ae1d8ff00ee5641906bcafa9a9ec1e98cb41498a.png)
minus the area of a sector of the larger circle ![$[B+C]$](http://latex.artofproblemsolving.com/d/7/7/d77dc2c402c80073c4eb779f64c4b3c5fbc1d9c1.png)
plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle ![$[C]$](http://latex.artofproblemsolving.com/8/e/b/8eb0d5cccb9df6a659750553ad358f3f7c446956.png)
.
The area of the smaller semicircle is ![$[A+B] = \frac{1}{2}\pi\cdot(\frac{1}{2})^{2}=\frac{1}{8}\pi$](http://latex.artofproblemsolving.com/b/b/7/bb737be2a96ad1fba215df3a97727aebb88ddebb.png)
.
Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures 
.
The area of the sector of the larger semicircle is ![$[B+C] = \frac{60}{360}\pi\cdot(\frac{2}{2})^{2}=\frac{1}{6}\pi$](http://latex.artofproblemsolving.com/e/7/5/e75d6a4b3ab5416fffaeb9294fdc4101475d2934.png)
.
The area of the triangle is ![$[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$](http://latex.artofproblemsolving.com/1/e/3/1e3b9d2aae4c54ac73fe5f02af335976a89ee9db.png)
.
So the shaded area is ![$[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}$](http://latex.artofproblemsolving.com/8/5/1/851075ddcae6c08caa75052db7fd1cbfec7b9ca9.png)
.
See Also
| 2003 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2003 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
