2013 AMC 12A 第 18 题

Question

六个半径为

1

的球摆放成它们的球心位于边长为

2

的正六边形顶点上。这六个球都内切于一个大球，大球球心是该正六边形的中心。第八个球外切于这六个小球，并内切于大球。这个第八个球的半径是多少？

Six spheres of radius

1

are positioned so that their centers are at the vertices of a regular hexagon of side length

2.

The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?

\sqrt{2}

\dfrac{3}{2}

\dfrac{5}{3}

\sqrt{3}

2

答案：B

知识点：球立体几何勾股定理

难度评级：2100

小提示：

大球半径为

3

；设第八个球半径为

r

，其球心到中心的距离为

x

，且

x + r = 3

。

The large sphere has radius

3;

let the eighth sphere have radius

r

and center at distance

x

from the center, with

x + r = 3

大提示：

一个小球球心、中心

O

，和第八个球心构成直角三角形：

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

A center of a small sphere, the center

O,

and the eighth center form a right triangle:

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

解答：

每个小球球心到中心

O

的距离为

2

，而小球半径为

1

，所以大球半径为

3

。设第八个球半径为

r

，球心

G

到

O

的距离为

x

，则

x + r = 3

。

因为

G

到两个相对的六边形顶点等距，

GO

垂直于通向某个顶点的线，由勾股定理得

\begin{gathered} (r + 1)^2 = 2^2 + x^2 \\ = 4 + (3 - r)^2 \end{gathered}\text{。}

化简得

2r + 1 = 13 - 6r

，所以

r = \tfrac32

。

因此，正确答案是 B。

Each small center is

2

from the center

O,

and the small spheres have radius

1,

so the large sphere has radius

3.

Let the eighth sphere have radius

r

and center

G

at distance

x

from

O;

then

x + r = 3.

Since

G

is equidistant from two opposite hexagon vertices,

GO

is perpendicular to the line to a vertex, and the Pythagorean Theorem gives

\begin{gathered} (r + 1)^2 = 2^2 + x^2 \\ = 4 + (3 - r)^2. \end{gathered}

This simplifies to

2r + 1 = 13 - 6r,

so

r = \tfrac32.

Thus, the correct answer is B.

2013 AMC 12A 第 18 题

答案：B

难度评级：2100

小提示：

大提示：

解答：

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