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

解答：

设上底半径为 $1$ 下底半径为 $r$ ，球半径为 $a$ 。球与两个底面相切，所以圆台高为 $2a$ ，在侧面截面中应用勾股定理得到 $(0,a)$ 。 $(r,0)$ $(1,2a)$ $2ax+(r-1)y-2ar=0$ 。 $(0,a)$ $a$ ， $(r+1)^2=4a^2+(r-1)^2$ ， $r=a^2$ 。

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 $r = a^2$ ，得到即 $r^2 - 3r + 1 = 0$ 。 $a^4 - 3a^2 + 1 = 0,$

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

解答：

其他年份的第 19 题