Difference between revisions of "2011 AMC 12B Problems/Problem 6"
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Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer: | Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer: | ||
| − | 1/2 (216°-144°) = 1/2 (72°) <math> =\boxed{36 | + | 1/2 (216°-144°) = 1/2 (72°) <math> =\boxed{36 \textbf{(C)}} </math> |
==See also== | ==See also== | ||
{{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}} | {{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:06, 16 May 2015
Problem
Two tangents to a circle are drawn from a point 
. The points of contact 
and 
divide the circle into arcs with lengths in the ratio 
. What is the degree measure of 
?
Solution
In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).
In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d.
Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.
Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:
1/2 (216°-144°) = 1/2 (72°) 
See also
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
