Difference between revisions of "2004 AMC 10A Problems/Problem 21"
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So, the shaded region is <math>3\theta+3\pi</math>. This means that the unshaded region is <math>9\pi-(3\theta+3\pi)</math>. | So, the shaded region is <math>3\theta+3\pi</math>. This means that the unshaded region is <math>9\pi-(3\theta+3\pi)</math>. | ||
− | Also, the shaded region is <math>\frac{8}{13}</math> of the unshaded region. Hence, we can now make an equation and solve for theta | + | Also, the shaded region is <math>\frac{8}{13}</math> of the unshaded region. Hence, we can now make an equation and solve for <math>\theta</math>. |
Latest revision as of 23:02, 25 January 2021
Contents
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 1
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
Solution 2
As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of .
.
So, the shaded region is . This means that the unshaded region is .
Also, the shaded region is of the unshaded region. Hence, we can now make an equation and solve for .
.
Simplifying, we get
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 |
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