Question

一个球内切于如图所示的截头正圆锥。截头圆锥的体积是球体积的两倍。截头圆锥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?

$\dfrac32$

$\dfrac{1+\sqrt5}2$

$\sqrt3$

$2$

$\dfrac{3+\sqrt5}2$

答案：E

知识点：圆锥球体积

难度评级：2300

解答：

设上底半径为 $1$ ，下底半径为 $R$ ，内切球半径为 $a$ 。

在截面中，球与两个底面相切，所以截锥的高为 $2a$ 。若以球心为原点，一条斜边连接 $(1,a)$ 与 $(R,-a)$ 。它的方程是 $2ax+(R-1)y-a(R+1)=0.$ 因为这条直线与半径为 $a$ 的圆相切，它到原点的距离为 $a$ 。所以 $\frac{a(R+1)}{\sqrt{4a^2+(R-1)^2}}=a,$ 化简得 $R=a^2$ 。

截锥的体积为 $\frac13\pi(R^2+R+1)(2a)$ $=\frac{2a\pi}{3}(a^4+a^2+1)$ 。

它等于球体积 $\frac{8a^3\pi}{3}$ 的两倍。约去公因子可得 $a^4-3a^2+1=0$ ，所以 $R^2-3R+1=0$ 。

由于下底半径大于上底半径， $R>1$ 。因此 $R=\frac{3+\sqrt5}{2}$ ，正确答案是 E。

Let the top radius be $1$ , the bottom radius be $R$ , and the inscribed sphere radius be $a$ .

In the cross-section, the sphere is tangent to the two bases, so the frustum height is $2a$ . A slanted side joins $(1,a)$ to $(R,-a)$ , if the sphere's center is the origin. Its equation is $2ax+(R-1)y-a(R+1)=0.$ Because this line is tangent to the circle of radius $a$ , its distance from the origin is $a$ . Thus $\frac{a(R+1)}{\sqrt{4a^2+(R-1)^2}}=a,$ which simplifies to $R=a^2$ .

The frustum volume is $\frac13\pi(R^2+R+1)(2a)$ $=\frac{2a\pi}{3}(a^4+a^2+1)$ .

This is twice the sphere volume, $\frac{8a^3\pi}{3}$ . Cancelling gives $a^4-3a^2+1=0$ , so $R^2-3R+1=0$ .

Since the bottom radius is larger than the top radius, $R>1$ . Thus $R=\frac{3+\sqrt5}{2}$ , and the correct answer is E .

2014 AMC 10B 第 23 题

答案：E

难度评级：2300

解答：

其他年份的第 23 题