2004 AMC 12A 第 19 题

Question

圆

A

、

B

和

C

两两外切，并且都内切于圆

D

。圆

B

和

C

全等。圆

A

的半径为

1

，并且经过圆

D

的圆心。圆

B

的半径是多少？

Circles

A,

B,

and

C

are externally tangent to each other and internally tangent to circle

D.

Circles

B

and

C

are congruent. Circle

A

has radius

1

and passes through the center of

D.

What is the radius of circle

B?

\dfrac{2}{3}

\dfrac{\sqrt{3}}{2}

\dfrac{7}{8}

\dfrac{8}{9}

\dfrac{1 + \sqrt{3}}{3}

答案：D

知识点：相切圆勾股定理方程组

难度评级：1920

小提示：

因为圆

A

经过圆

D

的圆心且与圆

D

内切，所以大圆的半径为

2

。

Since

A

passes through the center of

D

and is internally tangent, circle

D

has radius

2

大提示：

把圆

D

的圆心放在原点；利用各圆心与对称轴形成的直角三角形。

Place the center of

D

at the origin; use right triangles formed by the centers and the axis of symmetry

解答：

圆

A

半径为

1

，经过圆

D

的圆心并与圆

D

内切，所以圆

D

的半径为

2

。

将圆

D

的圆心放在原点，圆

A

的圆心在

(-1, 0)

。由对称性，

B

的圆心为

(x, y)

，半径为

r

而

C

是它关于水平轴的镜像，所以两圆在该轴上相切，且

y = r

。

与圆

D

内切给出

x^2 + y^2 = (2 - r)^2

，与圆

A

外切给出

(x + 1)^2 + y^2 = (1 + r)^2

。

两式相减并使用

y = r

得到

x = \tfrac23

且

r = \tfrac89

。因此圆

B

的半径是

\tfrac89

。

所以正确答案是 D。

Circle

A

has radius

1

and passes through the center of

D

while being internally tangent to

D,

so

D

has radius

2.

Place the center of

D

at the origin, with

A

centered at

(-1, 0).

By symmetry,

B

has center

(x, y)

and radius

r,

with

C

its mirror image across the horizontal axis, so the two congruent circles touch on that axis and

y = r.

Internal tangency to

D

gives

x^2 + y^2 = (2 - r)^2,

and external tangency to

A

gives

(x + 1)^2 + y^2 = (1 + r)^2.

Subtracting and using

y = r

yields

x = \tfrac23

and

r = \tfrac89.

The radius of circle

B

is

\tfrac89.

Thus, the correct answer is D.

2004 AMC 12A 第 19 题

答案：D

难度评级：1920

小提示：

大提示：

解答：

其他年份的第 19 题