Question

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

$\sec^2\theta - \tan\theta$

$\dfrac{1}{2}$

$\dfrac{\cos^2\theta}{1 + \sin\theta}$

$\dfrac{1}{1 + \sin\theta}$

$\dfrac{\sin\theta}{\cos^2\theta}$

Solution:

Because $OA = 1$ and $AB$ is tangent at $A,$ angle $OAB$ is right, so $BA = \tan\theta, \qquad OB = \sec\theta.$

Since $BC$ bisects $\angle ABO,$ the angle bisector theorem gives $\dfrac{OC}{CA} = \dfrac{OB}{BA}.$ Using $OC + CA = OA = 1,$ $OC = \frac{OB}{OB + BA} = \frac{\sec\theta}{\sec\theta + \tan\theta}.$

Multiplying numerator and denominator by $\cos\theta$ gives $OC = \dfrac{1}{1 + \sin\theta}.$

Thus, the correct answer is D.

2000 AMC 12 Problem 17

Solution:

Problem 17 in Other Years