2006 AMC 12B 第 15 题

Question

圆心为

O

和

P

的两个圆半径分别为

2

和

4

，并且外切。点

A

和

B

在圆心为

O

的圆上，点

C

和

D

在圆心为

P

的圆上，使得

AD

和

BC

是两个圆的公外切线。六边形

AOBCPD

的面积是多少？

Circles with centers

O

and

P

have radii

2

and

4,

respectively, and are externally tangent. Points

A

and

B

are on the circle centered at

O,

and points

C

and

D

are on the circle centered at

P,

such that

AD

and

BC

are common external tangents to the circles. What is the area of hexagon

AOBCPD?

18\sqrt{3}

24\sqrt{2}

36

24\sqrt{3}

32\sqrt{2}

答案：B

知识点：相切圆梯形勾股定理

难度评级：1680

小提示：

两圆圆心距离

OP = 2 + 4 = 6

，到切点的半径垂直于切线。

The centers are

OP = 2 + 4 = 6

apart, and the radii to the tangent points are perpendicular to the tangent

大提示：

AOPD

是直角梯形；用勾股定理求它的斜边，再把面积加倍

AOPD

is a right trapezoid; find its slant side with the Pythagorean theorem, then double the area

解答：

两圆外切，所以

OP = 2 + 4 = 6

。在四边形

AOPD

中，

OA = 2

和

PD = 4

都垂直于切线

AD

，因此它是直角梯形。

过

O

作平行于

AD

的直线，可形成一个直角三角形，斜边为

OP = 6

，一条直角边为

PD - OA = 2

，所以

AD = \sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt2

。

梯形

AOPD

的面积为

\frac{1}{2}(2 + 4)(4\sqrt2) = 12\sqrt2\text{。}

由对称性，六边形

AOBCPD

由两个这样的梯形组成，所以面积为

2 \cdot 12\sqrt2 = 24\sqrt2

。

因此，正确答案是 B。

The circles are externally tangent, so

OP = 2 + 4 = 6.

In quadrilateral

AOPD,

both

OA = 2

and

PD = 4

are perpendicular to the tangent line

AD,

making it a right trapezoid.

Drawing the line through

O

parallel to

AD

creates a right triangle with hypotenuse

OP = 6

and one leg

PD - OA = 2,

so

AD = \sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt2.

The trapezoid

AOPD

has area

\frac{1}{2}(2 + 4)(4\sqrt2) = 12\sqrt2.

By symmetry the hexagon

AOBCPD

is made of two such trapezoids, so its area is

2 \cdot 12\sqrt2 = 24\sqrt2.

Thus, the correct answer is B.

2006 AMC 12B 第 15 题

答案：B

难度评级：1680

小提示：

大提示：

解答：

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