2014 AMC 12B Problem 19

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?

\dfrac{3}{2}

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

\sqrt{3}

2

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

Answer: E

Concepts:cone sphere volume

Difficulty rating: 2220

Small hint:

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

1,

and call the bottom and sphere radii

r

and

a

Big hint:

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

r

and

a,

then compare the frustum and sphere volumes

Solution:

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.