Difference between revisions of "2000 AMC 12 Problems/Problem 17"
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== Solution == | == Solution == | ||
− | Since <math>OA = 1</math>, <math>BA = \tan \theta</math> and <math>OB = \sec \theta</math>. By the [[Angle Bisector Theorem]], <math>\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta</math>. Multiply both sides by <math>\cos \theta</math> to simplify the trigonometric functions. Since <math>AC + OC = 1</math>, <math>1 - OC = OC \sin \theta \Longrightarrow</math> <math>OC = \frac{1}{1+\sin \theta} \Rightarrow \mathrm{(D)}</math>. | + | Since <math>\overline{AB}</math> is tangent to the circle, <math>\triangle OAB</math> is a [[right triangle]]. Thus since <math>OA = 1</math>, <math>BA = \tan \theta</math> and <math>OB = \sec \theta</math>. By the [[Angle Bisector Theorem]], <math>\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta</math>. Multiply both sides by <math>\cos \theta</math> to simplify the trigonometric functions. Since <math>AC + OC = 1</math>, <math>1 - OC = OC \sin \theta \Longrightarrow</math> <math>OC = \frac{1}{1+\sin \theta} \Rightarrow \mathrm{(D)}</math>. |
== See also == | == See also == |
Revision as of 20:22, 4 January 2008
Problem
A circle centered at has radius
and contains the point
. The segment
is tangent to the circle at
and
. If point
lies on
and
bisects
, then
Solution
Since is tangent to the circle,
is a right triangle. Thus since
,
and
. By the Angle Bisector Theorem,
. Multiply both sides by
to simplify the trigonometric functions. Since
,
.
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 |