2000 AMC 12 Problems/Problem 17

Problem

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 $AC + OC = 1$ , $1 - OC = OC \sin \theta (from$ AC \sec \theta = OC \tan \theta $) \Longrightarrow$ $OC = \frac{1}{1+\sin \theta} \Rightarrow \mathrm{(D)}$ .

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
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2000 AMC 12 Problems/Problem 17

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