Distance: sphere related 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 ? 
How do we do here as I have four options as answer : 
(a) $3+\dfrac{\sqrt{30}}{2}$ 
(b) $3+\dfrac{\sqrt{69}}{3}$  
(c) $3+\dfrac{\sqrt{123}}{4}$ 
(d) $\dfrac{52}{9}$
 A: Let the three spheres rest on the plane symmetrically around the origin, with one of them centered at $(h,0,1)$. Here's a picture of the shadows of the spheres in the $xy$-plane.

The first step is to figure out what $h$ is. The next center will lie over the ray $\theta=2\pi/3$, and will therefore have coordinates of the form $(-1/2 h,\sqrt{3}/{2} h,1)$. Because the spheres are tangent and of radius $1$ the distance between these two points must be $2$. So we have
$$\sqrt{(h-(-1/2 h))^2+(0-h\sqrt{3}/2)^2}=2.$$
Solving for $h$, we get $h^2=4/3$.
Now we look at the cross-section $y=0$. We have a circle of radius $1$ centered at $(h,1)$ and a circle of radius $2$ centered at $(k,0)$ for some $k$ and which is tangent to the first circle. Draw a line between the centers of the two circles. This has length $3$, and the total height of the big circle above the $x$-axis is $3$ plus the vertical displacement described by this line segment. The horizontal displacement is $h$, so by the Pythagorean theorem we get the vertical displacement is $\sqrt{9-h^2}=\sqrt{23/3}=\sqrt{69}/{3}$. So the final answer is (b) $3+\sqrt{69}/{3}$.
Here is a picture of the $y=0$ cross-section:

A: Let the centers of your 3 spheres of radius 1 be $C_1, C_2, C_3$. Let the center of the sphere of radius 2 be $C_4$. These 4 points forms a tetrahedron with equilateral triangle base of length 2 ( $C_1C_2 = C_2C_3=C_3C_1=2$) , and a side length of 3 ($C_4C_1 = C_4C_2 = C_4C_3 = 3$). Let $O$ be the center of the equilateral base, then $OC_1 = \frac {2}{3} \sqrt{3} $.
Consider right triangle $OC_1 C_4$, we get that $OC_4 = \sqrt{ C_1C_4 ^2 - O C_1 ^2 } = \sqrt{ 9 - \frac {4}{3} } = \sqrt{ \frac {23}{3} } = \frac {\sqrt{69} }{3}$.
Hence, the distance from the plane to the top is $ 1 + \frac { \sqrt{69} }{3} + 2$.
