Four balls and a cone covering them

Let the z-axis be the axis of symmetry of the assembly. The centers of the three balls are separated by $2R$ , and the three centers form an equilateral triangle. The distance from the central axis to one of the vertices is $(2/3) \dfrac{\sqrt{3}}{2} (2 R) = 2 R / \sqrt{3}$ . We can take the centers of the three lower balls to be

$C_1 = ( 2 R/\sqrt{3}, 0, R)$

$C_2 = ( - R/\sqrt{3}, R , R)$

$C_3 = ( - R/\sqrt{3}, -R, R)$

One can verify the distances between any two of these centers is $2 R$ .

Now the fourth ball has its center at $C_4 = (0, 0, z )$ such the distance between $C_4$ and any of the first three centers is $2R$ . This gives

$(z - R)^2 + 4/3 R^2 = 4 R^2$ , so that $z = R (1 + \sqrt{\dfrac{8}{3}} )$

Now we can find a common tangent to one of the lower balls and the top ball, let's say the first ball and the fourth ball. The tangent is parallel to the line segment connecting the two centers, and is a distance R from it.

In the x-z plane, the equation of the line segment connecting $C_1$ and $C_4$ is

$z = R - ( \sqrt{8/3} * \sqrt{3} / 2 ) (x - 2 R/\sqrt{3} ) = R - \sqrt{2} (x - 2 R / \sqrt{3} )$

Thus,

$(z - R) + \sqrt{2} (x - 2R/\sqrt{3} ) = 0$

which is of the form $n \cdot ( r - r_0) = 0$

To generate the line that is parallel to this line but R units away in the direction of the normal, we add $R n / |n |$ to $r_0$ . The normal vector to this line is $n = (\sqrt{2}, 1)$ . Therefore, the equation of the tangent line is

$( \sqrt{2}, 1) . ( r - r_0 ) = R \sqrt{3}$

Thus the required equation is,

$(z - R) + \sqrt{2} (x - 2R/\sqrt{3}) = R \sqrt{3}$

The z-intercept is the height of the cone, which is

$h = R (1 + \sqrt{3} + 2 \sqrt{\dfrac{2}{3}} )$

And the radius of the cone is the x-intercept,

$r = 2 R / \sqrt{3} + \dfrac{1}{\sqrt{2}} ( R (1 + \sqrt{3} ) )$

which becomes, after simplification,

$r = R ( \dfrac{2}{ \sqrt{3}} + \dfrac{1}{\sqrt{2}} + \sqrt{\dfrac{3}{2}} )$

Substituting $R = 10$ , gives us, $h = 43.65$ and $r = 30.87$ and this makes the answer $43.65 + 30.87 = \boxed{74.52}$