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:

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

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.

2004 AMC 12A Problem 22

Solution:

Problem 22 in Other Years