2000 AMC 12 第 17 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

17.

一个以

O

为圆心、半径为

1

的圆经过点

A

。线段

AB

在点

A

处与圆相切，且

\angle AOB = \theta

。若点

C

位于

\overline{OA}

上，且

BC

平分

\angle ABO

，则

OC

等于多少？

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}

答案：D

知识点：角平分线定理切线三角学

难度评级：1870

小提示：

因为

OA = 1

且

\angle OAB = 90^\circ

，所以

BA = \tan\theta

，

OB = \sec\theta

。

Since

OA = 1

and

\angle OAB = 90^\circ,

we have

BA = \tan\theta

and

OB = \sec\theta

大提示：

角平分线给出

\dfrac{OC}{CA} = \dfrac{OB}{BA}

，且

OC + CA = OA = 1

。

The bisector gives

\dfrac{OC}{CA} = \dfrac{OB}{BA},

and

OC + CA = OA = 1

解答：

因为

OA = 1

且

AB

在

A

点与圆相切，角

OAB

为直角，所以

BA = \tan\theta, \qquad OB = \sec\theta\text{。}

因为

BC

平分

\angle ABO

，由角平分线定理得

\dfrac{OC}{CA} = \dfrac{OB}{BA}

。又

OC + CA = OA = 1

，

\begin{aligned} OC &= \frac{OB}{OB + BA} \\ &= \frac{\sec\theta}{\sec\theta + \tan\theta} \end{aligned}\text{。}

分子分母同乘

\cos\theta

得到

OC = \dfrac{1}{1 + \sin\theta}

。

因此，正确答案是 D。

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.

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