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?
$\text {(A)}\ 3 + \frac {\sqrt {30}}{2} \qquad \text {(B)}\ 3 + \frac {\sqrt {69}}{3} \qquad \text {(C)}\ 3 + \frac {\sqrt {123}}{4}\qquad \text {(D)}\ \frac {52}{9}\qquad \text {(E)}\ 3 + 2\sqrt2$
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 $2$. 
[2004 AMC12A-22a.png](https://artofproblemsolving.com/wiki/index.php/File:2004_AMC12A-22a.png)
We now need the vertical height of the centers. If we connect centers, we get a rectangular [pyramid](https://artofproblemsolving.com/wiki/index.php/Pyramid) with an [equilateral triangle](https://artofproblemsolving.com/wiki/index.php/Equilateral_triangle) base. The distance from the vertex of the equilateral triangle to its [centroid](https://artofproblemsolving.com/wiki/index.php/Centroid) can be found by $30\text-60\text-90 \triangle$s to be $\frac{2\sqrt{3}}{3}$.
[2004 AMC12A-22b.png](https://artofproblemsolving.com/wiki/index.php/File:2004_AMC12A-22b.png)
By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/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 $\frac{\sqrt{69}}{3} + 1 + 2 \Rightarrow\boxed{\mathrm{(B)}\ 3+\frac{\sqrt{69}}{3}}$