2011 AMC 12A 第 11 题

Question

圆

A

、

B

和

C

的半径都为

1

。圆

A

和圆

B

共有一个切点。圆

C

与

\overline{AB}

的中点相切。位于圆

C

内部、但在圆

A

和圆

B

外部的面积是多少？

Circles

A,

B,

and

C

each have radius

1.

Circles

A

and

B

share one point of tangency. Circle

C

has a point of tangency with the midpoint of

\overline{AB}.

What is the area inside circle

C

but outside circle

A

and circle

B?

3 - \dfrac{\pi}{2}

\dfrac{\pi}{2}

2

\dfrac{3\pi}{4}

1 + \dfrac{\pi}{2}

答案：C

知识点：圆面积相切圆面积分割

难度评级：1540

小提示：

设圆心为

A(-1,0)

，

B(1,0)

，和

C(0,1)

；从

C

到

A, B

的距离都是

\sqrt2

Place the centers at

A(-1,0),

B(1,0),

and

C(0,1);

the distance from

C

to each of

A, B

is

\sqrt2

大提示：

所求面积是圆

C

的面积减去圆

C

分别与圆

A

和圆

B

的两个透镜形重叠区域

The wanted area is the area of

C

minus the two lens-shaped overlaps of

C

with

A

and with

B

解答：

设

A = (-1, 0)

，

B = (1, 0)

，则它们的切点是原点，也就是

\overline{AB}

的中点。因此

C = (0, 1)

，因为圆

C

经过原点。

从

C

到

A

（以及到

B

）的距离为

\sqrt2

。两个圆心距离为

\sqrt2

的单位圆重叠成一个透镜形，其面积为

\begin{aligned} 2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) \\ {}- \tfrac{\sqrt2}{2}\sqrt{4 - 2} \\ = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1 \end{aligned}\text{。}

圆

A

和圆

B

只在原点相交，所以两个透镜形区域不重叠。所求面积为

\pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2\text{。}

因此，正确答案是 C。

Place

A = (-1, 0),

B = (1, 0),

so their tangency point is the origin, the midpoint of

\overline{AB}.

Then

C = (0, 1),

since

C

passes through the origin.

The distance from

C

to

A

(and to

B

) is

\sqrt2.

Two unit circles whose centers are

\sqrt2

apart overlap in a lens of area

\begin{aligned} 2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) \\ {}- \tfrac{\sqrt2}{2}\sqrt{4 - 2} \\ = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1. \end{aligned}

Circles

A

and

B

meet only at the origin, so the two lenses do not overlap. The wanted area is

\pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2.

Thus, the correct answer is C.

2011 AMC 12A 第 11 题

答案：C

难度评级：1540

小提示：

大提示：

解答：

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