Difference between revisions of "2004 AMC 10A Problems/Problem 21"
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<math>\dfrac{2\theta}{2\pi}*\pi +\dfrac{2(\pi-\theta)}{2\pi}*(4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi=\dfrac{24\pi}{7}</math> | <math>\dfrac{2\theta}{2\pi}*\pi +\dfrac{2(\pi-\theta)}{2\pi}*(4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi=\dfrac{24\pi}{7}</math> | ||
| − | Thus <math>3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow \mathrm{(B) \ }</math> | + | Thus <math>3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow\boxed{\mathrm{(B)}\ \frac{\pi}{7}}</math> |
== See also == | == See also == | ||
Revision as of 11:50, 21 July 2014
Problem
Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is 
of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: 
radians is 
degrees.)
Solution
Let the area of the shaded region be 
, the area of the unshaded region be 
, and the acute angle that is formed by the two lines be 
. We can set up two equations between and :
Thus 
, and 
, and thus 
.
Now we can make a formula for the area of the shaded region in terms of :
Thus 
See also
| 2004 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
