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 $OC =$ $\textbf {(A)}\ \sec^2 \theta - \tan \theta \qquad$ $ \textbf {(B)}\ \frac 12 \qquad$ $ \textbf {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad$ $ \textbf {(D)}\ \frac{1}{1+\sin\theta} \qquad$ $ \textbf {(E)}\ \frac{\sin \theta}{\cos^2 \theta}$
Solution
$OA\perp AB$ with $OA=1$, so $OB=\frac1{\cos\theta}$ and $AB=\tan\theta$.
$BC$ bisects $\angle ABO$, so by the angle bisector theorem $OC:CA=OB:AB=\frac1{\sin\theta}$. Since $OC+CA=OA=1$, $OC=\frac{1/\sin\theta}{1+1/\sin\theta}=\frac1{1+\sin\theta}$.
A circle centered at $O$ has radius $2$ and contains the point $A$. Segment $AB$ is tangent to the circle at $A$ and $\angle AOB=30^\circ$. Point $C$ lies on $\overline{OA}$ so that $\overline{BC}$ bisects $\angle ABO$. What is $OC$?
