2004 AMC 12B 第 19 题

Question

一个截锥有水平底面，两个底面半径分别为

18

和

2

。一个球与截锥的上底面、下底面和侧面都相切。这个球的半径是多少？

A truncated cone has horizontal bases with radii

18

and

2.

A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?

6

4\sqrt{5}

9

10

6\sqrt{3}

知识点：圆锥球切线

取过轴的截面：它是一个内切圆的等腰梯形。

Take the cross-section through the axis: an isosceles trapezoid with an inscribed circle

切线长关系给出斜边长

18 + 2 = 20

；作高得到一个直角三角形。

Tangent lengths give a slant side of

18 + 2 = 20;

drop a height to form a right triangle

轴截面是梯形

ABCD

，平行边长为

2

和

18

，且有一个内切圆（球的大圆）。从

B

和

C

引出的切线段长度相等，所以斜边

BC = 18 + 2 = 20

。从

C

向下底作垂线，得到一个水平直角边为

18 - 2 = 16

的直角三角形，所以高为

\sqrt{20^2 - 16^2} = 12

。球的半径是高的一半，即

6

。

因此正确答案是 A。

The axial cross-section is a trapezoid

ABCD

with parallel sides

2

and

18

and an inscribed circle (a great circle of the sphere). By equal tangent lengths from

B

and

C,

the slant side

BC = 18 + 2 = 20.

Dropping a perpendicular from

C

to the bottom base gives a right triangle with horizontal leg

18 - 2 = 16,

so the height is

\sqrt{20^2 - 16^2} = 12.

The sphere’s radius is half the height,

6.

Thus, the correct answer is A.

其他年份的第 19 题