Question

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?

$\dfrac32$

$\dfrac{1+\sqrt5}2$

$\sqrt3$

$2$

$\dfrac{3+\sqrt5}2$

Solution:

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 $2a$ . The right triangle formed by the side, a radius to the tangency point, and the base radii gives $R=a^2$ , as in the official diagram.

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 $R^2-3R+1=0$ .

Thus $R=\frac{3+\sqrt5}{2}$ , and the correct answer is E .

2014 AMC 10B Problem 23

Solution:

Problem 23 in Other Years