Difference between revisions of "2008 AMC 12B Problems/Problem 9"
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− | 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, <math>m\angle RDA=90\deg</math>, and so <math>RD=\sqrt{5^2-3^2}=4</math>. Since <math>RD=4</math>, <math>CD=5-4=1</math>, and so <math>AC=sqrt{3^2+1^2}=\sqrt{10} \rightarrow A</math> | + | 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, <math>m\angle RDA=90\deg</math>, and so <math>RD=\sqrt{5^2-3^2}=4</math>. Since <math>RD=4</math>, <math>CD=5-4=1</math>, and so <math>AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow A</math> |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=7|num-a=9}} | {{AMC12 box|year=2008|ab=B|num-b=7|num-a=9}} |
Revision as of 20:58, 30 November 2008
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 ?
Solution
Trig Solution:
Let be the angle that subtends the arc AB. By the law of cosines,
The half-angle formula says that
, 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, , and so . Since , , and so
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 |