2004 AMC 12A Problems/Problem 22

Problem

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?

$\text {(A)} 3 + \frac {\sqrt {30}}{2} \qquad \text {(B)} 3 + \frac {\sqrt {69}}{3} \qquad \text {(C)} 3 + \frac {\sqrt {123}}{4}\qquad \text {(D)} \frac {52}{9}\qquad \text {(E)}3 + 2\sqrt2$

Solution

The height from the center of the bottom sphere to the plane is $1$ , and from the center of the top sphere to the tip is $2$ . We now need the vertical height of the centers. If we connect the centers, we get a triangular pyramid with an equilateral triangle base. The distance from the vertex of the equilateral triangle to its centroid can be found by $30-60-90 \triangle$ s to be $\frac{2}{\sqrt{3}}$ .

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By the Pythagorean Theorem, we have $\left(\frac{2}{\sqrt{3}}\right)^2 + h^2 = 3^2 \Longrightarrow h = \frac{\sqrt{69}}{3}$ . Adding the heights up, we get $\frac{\sqrt{69}}{3} + 1 + 2 = \frac{\sqrt{69} + 9}{3} \Rightarrow \mathrm{(B)}$ .

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2004 AMC 12A Problems/Problem 22

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