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?

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

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

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

$\dfrac{52}{9}$

$3 + 2\sqrt{2}$

Solution:

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 $DE = \sqrt{3^2 - \left(\dfrac{2\sqrt{3}}{3}\right)^2} = \sqrt{9 - \dfrac{4}{3}} = \dfrac{\sqrt{69}}{3}.$

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.

2004 AMC 10A Problem 25

Solution:

Problem 25 in Other Years