Difference between revisions of "2008 AMC 12B Problems/Problem 9"
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The [[Trigonometric_identities#Half_Angle_Identities  halfangle formula]] says that  The [[Trigonometric_identities#Half_Angle_Identities  halfangle formula]] says that  
−  <math>\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}</math>. The law of cosines tells us <math>AC = \sqrt{5^2+5^22  +  <math>\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}</math>. The law of cosines tells us <math>AC = \sqrt{5^2+5^22\cdot 5\cdot 5\cdot \frac{4}{5}} = \sqrt{5050\frac{4}{5}} = \sqrt{10}</math>, which is answer choice <math>\boxed{\text{A}}</math>. 
===Solution 2===  ===Solution 2=== 
Revision as of 00:23, 15 August 2011
Problem 9
Points and are on a circle of radius and . Point is the midpoint of the minor arc . What is the length of the line segment ?
Solutions
Solution 1
Let be the angle that subtends the arc . By the law of cosines, implies .
The halfangle formula says that . The law of cosines tells us , which is answer choice .
Solution 2

Figure 1 
Define as the midpoint of line segment , and the center of the circle. Then , , and are collinear, and since is the midpoint of , and so . Since , , and so .
See Also
2008 AMC 12B (Problems • Answer Key • Resources)  
Preceded by Problem 8 
Followed by Problem 10 
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 