Difference between revisions of "2002 AMC 12A Problems/Problem 5"
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==Solution== | ==Solution== | ||
The outer circle has radius <math>1+1+1=3</math>, and thus area <math>9\pi</math>. The little circles have area <math>\pi</math> each; since there are 7, their total area is <math>7\pi</math>. Thus, our answer is <math>9\pi-7\pi=\boxed{2\pi\Rightarrow \textbf{(C)}}</math>. | The outer circle has radius <math>1+1+1=3</math>, and thus area <math>9\pi</math>. The little circles have area <math>\pi</math> each; since there are 7, their total area is <math>7\pi</math>. Thus, our answer is <math>9\pi-7\pi=\boxed{2\pi\Rightarrow \textbf{(C)}}</math>. | ||
| + | |||
| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/VSPM5_pTK34?si=EIQDi3PF6I0P7wMR | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
==See Also== | ==See Also== | ||
Latest revision as of 10:40, 21 July 2024
- The following problem is from both the 2002 AMC 12A #5 and 2002 AMC 10A #5, so both problems redirect to this page.
Problem
Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
![[asy] unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]](http://latex.artofproblemsolving.com/e/4/2/e42bf637d6aea8385012d9ddf6ed3b031156bddb.png)
Solution
The outer circle has radius 
, and thus area 
. The little circles have area 
each; since there are 7, their total area is 
. Thus, our answer is 
.
Video Solution by Daily Dose of Math
https://youtu.be/VSPM5_pTK34?si=EIQDi3PF6I0P7wMR
~Thesmartgreekmathdude
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 | |
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
