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 $OC?$

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

$\dfrac{1}{2}$

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

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

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

Answer: D

Concepts:angle bisector theorem tangent line trigonometry

Difficulty rating: 1870

Small Hint:

Since $OA = 1$ and $\angle OAB = 90^\circ,$ we have $BA = \tan\theta$ and $OB = \sec\theta$

Big Hint:

The bisector gives $\dfrac{OC}{CA} = \dfrac{OB}{BA},$ and $OC + CA = OA = 1$

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 $OC + CA = OA = 1,$ $\begin{aligned} OC &= \frac{OB}{OB + BA} \\ &= \frac{\sec\theta}{\sec\theta + \tan\theta}. \end{aligned}$

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

Answer: D

Difficulty rating: 1870

Small Hint:

Big Hint:

Solution:

Problem 17 in Other Years