Difference between revisions of "2004 AMC 12A Problems/Problem 22"

The following problem is from both the 2004 AMC 12A #22 and 2004 AMC 10A #25, so both problems redirect to this page.

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 1

Hi there, We draw the three spheres of radius $1$:

And then add the sphere of radius $2$:

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 centers, we get a rectangular 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 $\frac{2\sqrt{3}}{3}$.

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 $\frac{\sqrt{69}}{3} + 1 + 2 \Rightarrow\boxed{\mathrm{(B)}\ 3+\frac{\sqrt{69}}{3}}$

Solution 2

Connect the centers of the spheres. Note that the resulting prism is a tetrahedron with base lengths of 2 and side lengths of 3. Drop a height from the top of the tetrahedron to the centroid of its equilateral triangle base. Using the Pythagorean Theorem, it is easy to see that the circumradius of the base is $\dfrac { 2 \sqrt {3 } } {3}$. We can use PT again to find the height of the tetrahedron given its base's circumradius and it's leg lengths. Finally, we add the distance from the top of the tetrahedron to the top of the sphere of radius 2 and the distance from the bottom of the prism to the ground to get an answer of $\boxed{3 + \dfrac{\sqrt{69}}{3}}$.