2004 AMC 10A Problem 25

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 sits above the triangle’s centroid; the centroid is

\dfrac{2\sqrt{3}}{3}

from each small center, and the slant distance between centers is

1 + 2 = 3

The three small centers form an equilateral triangle of side

2,

each

1

unit above the plane. Its centroid

D

is at distance

\dfrac{2\sqrt{3}}{3}

from each vertex.

The large sphere’s center

E

sits directly above

D,

and the distance between

E

and a small center is

1 + 2 = 3.

Thus

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

Adding the

1

unit from the plane to

D

and the

2

units from

E

to the top of the large sphere gives

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

Thus, the correct answer is B.

Problem 25 in other years