2008 AMC 12A 第 13 题

Question

点

A

和

B

在以

O

为圆心的圆上，且

\angle AOB = 60^\circ

。第二个圆内切于第一个圆，并与

OA

和

OB

都相切。小圆面积与大圆面积之比是多少？

Points

A

and

B

lie on a circle centered at

O,

and

\angle AOB = 60^\circ.

A second circle is internally tangent to the first and tangent to both

OA

and

OB.

What is the ratio of the area of the smaller circle to that of the larger circle?

\dfrac{1}{16}

\dfrac{1}{9}

\dfrac{1}{8}

\dfrac{1}{6}

\dfrac{1}{4}

答案：B

知识点：相切圆特殊直角三角形面积比

难度评级：1620

小提示：

小圆圆心在

\angle AOB

的角平分线上。

The center of the small circle lies on the bisector of

\angle AOB

大提示：

向

OA

作半径垂线，会得到一个

30

-

60

-

90

三角形，其中

OE = 2r

。

Dropping a radius to

OA

makes a

30

-

60

-

90

triangle in which

OE = 2r

解答：

设小圆和大圆半径分别为

r

和

R

，小圆圆心为

E

。由对称性，

E

在

\angle AOB

的角平分线上，所以

OE

与

OA

成

30^\circ

。

作半径

ED

垂直于

OA

，得到

30

-

60

-

90

三角形，所以

OE = 2 \cdot ED = 2r

。两圆内切还给出

OE = R - r

。

于是

R - r = 2r

，所以

R = 3r

，

\tfrac{r}{R} = \tfrac{1}{3}

。于是面积之比为

\left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}\text{。}

所以正确答案是 B。

Let

r

and

R

be the radii of the smaller and larger circles, and let

E

be the center of the smaller circle. By symmetry

E

lies on the bisector of

\angle AOB,

so

OE

makes a

30^\circ

angle with

OA.

Dropping the radius

ED

perpendicular to

OA

gives a

30

-

60

-

90

triangle with

OE = 2 \cdot ED = 2r.

Since the circles are internally tangent,

OE = R - r.

Then

R - r = 2r,

so

R = 3r

and

\tfrac{r}{R} = \tfrac{1}{3}.

The ratio of areas is

\left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}.

Thus, B is the correct answer.

2008 AMC 12A 第 13 题

答案：B

难度评级：1620

小提示：

大提示：

解答：

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