2000 AMC 12 Problem 17

Attempt Problem 17 of the 2000 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 12 solutions, or check the answer key.

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17.

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.

Problem 17 in other years