2014 AMC 12B 第 19 题

Question

如图，一个球内切于一个正圆台。圆台的体积是球体积的两倍。圆台下底半径与上底半径之比是多少？

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?

\dfrac{3}{2}

\dfrac{1+\sqrt{5}}{2}

\sqrt{3}

2

\dfrac{3+\sqrt{5}}{2}

答案：E

知识点：圆锥球体积

难度评级：2220

小提示：

取轴截面，把上底半径规范为

1

，并把下底半径和球半径分别记为

r

和

a

。

Take an axial cross-section, normalize the top radius to

1,

and call the bottom and sphere radii

r

and

a

大提示：

利用球心到斜侧边的距离建立

r

与

a

之间的关系，再比较圆台与球的体积。

Use the distance from the sphere center to a slanted side to relate

r

and

a,

then compare the frustum and sphere volumes

解答：

设上底半径为

1

，下底半径为

r

，球半径为

a

。球与两个底面都相切，所以圆台的高为

2a

。在轴截面中把球心放在

(0,a)

；过

(r,0)

和

(1,2a)

的右侧腰所在直线的方程为

2ax+(r-1)y-2ar=0

。它到

(0,a)

的距离为

a

，所以

(r+1)^2=4a^2+(r-1)^2

，从而

r=a^2

。

圆台体积为

\tfrac13 \pi (r^2 + r + 1)(2a)

。令它等于球体积的两倍

\tfrac43 \pi a^3

，并用

r = a^2

，得到

a^4 - 3a^2 + 1 = 0\text{，}

即

r^2 - 3r + 1 = 0

。

正根为

r = \dfrac{3+\sqrt5}{2}

。

所以正确答案是 E。

Let the top radius be

1,

the bottom radius

r,

and the sphere radius

a.

The sphere touches both bases, so the frustum height is

2a.

In an axial cross-section put the sphere center at

(0,a)

; the right slanted side through

(r,0)

and

(1,2a)

has equation

2ax+(r-1)y-2ar=0.

Its distance from

(0,a)

is

a,

so

(r+1)^2=4a^2+(r-1)^2,

giving

r=a^2.

The frustum volume is

\tfrac13 \pi (r^2 + r + 1)(2a).

Setting it equal to twice the sphere volume

\tfrac43 \pi a^3

and using

r = a^2

yields

a^4 - 3a^2 + 1 = 0,

that is

r^2 - 3r + 1 = 0.

The positive root is

r = \dfrac{3+\sqrt5}{2}.

Thus, the correct answer is E.

2014 AMC 12B 第 19 题

答案：E

难度评级：2220

小提示：

大提示：

解答：

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