2004 AMC 12A Problem 22

Question

Three mutually tangent spheres of radius

1

rest on a horizontal plane. A sphere of radius

2

rests on them. What is the distance from the plane to the top of the larger sphere?

3 + \dfrac{\sqrt{30}}{2}

3 + \dfrac{\sqrt{69}}{3}

3 + \dfrac{\sqrt{123}}{4}

\dfrac{52}{9}

3 + 2\sqrt{2}

The three small centers form an equilateral triangle of side

2

at height

1

The big center lies above the triangle’s centroid; use the distance between centers

1 + 2 = 3

to find its height

Let the centers of the three unit spheres be

A,

B,

C,

forming an equilateral triangle of side

2

at height

1

above the plane, and let

E

be the center of the large sphere directly above the centroid

D

of

\triangle ABC.

The distance from a vertex to the centroid is

AD = \tfrac{2\sqrt3}{3},

and

AE = 1 + 2 = 3,

so

\begin{aligned} DE &= \sqrt{3^2 - \left(\tfrac{2\sqrt3}{3}\right)^2} \\ &= \sqrt{9 - \tfrac{4}{3}} \\ &= \dfrac{\sqrt{69}}{3}. \end{aligned}

Since

D

is

1

unit above the plane and the top of the large sphere is

2

units above

E,

the total height is

1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}.

Thus, the correct answer is B.

Problem 22 in other years