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2008 AMC 12B Problems/Problem 9

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Problem 9

Points A and B are on a circle of radius 5 and AB = 6. Point C is the midpoint of the minor arc AB. What is the length of the line segment AC?

\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \q...

Solution

Trig Solution:

Let \alpha be the angle that subtends the arc AB. By the law of cosines, 6^2=5^2+5^2-2*5*5cos(\alpha)

\alpha = cos^{-1}(7/25)

The half-angle formula says that cos(\alpha/2) = \frac{\sqrt{1+cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5} AC = \sqrt{5^2+5^2-2*5*5*\frac{4}{5}}

AC = \sqrt{50-50\frac{4}{5}}

AC = \sqrt{10}, which is answer choice A.

Other Solution

Define D as the midpoint of AB, and R the center of the circle. R, C, and D are collinear, and since D is the midpoint of AB, m\angle RDA=90\deg, and so RD=\sqrt{5^2-3^2}=4. Since RD=4, CD=5-4=1, and so AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow A

See Also

2008 AMC 12B (ProblemsResources)
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
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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