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Difference between revisions of "2000 AMC 12 Problems/Problem 17"

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== Solution ==
 
== Solution ==
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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>.  
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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 19:22, 4 January 2008

Problem

2000 12 AMC-17.png

A circle centered at $O$ has radius $1$ and contains the point $A$. The segment $AB$ is tangent to the circle at $A$ and $\angle AOB = \theta$. If point $C$ lies on $\overline{OA}$ and $\overline{BC}$ bisects $\angle ABO$, then $OC =$

$\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}$

Solution

Since $\overline{AB}$ is tangent to the circle, $\triangle OAB$ is a right triangle. Thus since $OA = 1$, $BA = \tan \theta$ and $OB = \sec \theta$. By the Angle Bisector Theorem, $\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta$. Multiply both sides by $\cos \theta$ to simplify the trigonometric functions. Since $AC + OC = 1$, $1 - OC = OC \sin \theta \Longrightarrow$ $OC = \frac{1}{1+\sin \theta} \Rightarrow \mathrm{(D)}$.

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
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