Circumscribed Sphere

This is an application of the Descartes Circle Theorem for 3-dimensional space. We have, $3\left (\sum_{i=1}^{5} k_i^2\right)=\left(\sum_{i=1}^{5}k_i\right)^2$ where $k_i$ for $1\le i\le 5$ denotes the curvature of the spheres. In this case $k_1=k_2=k_3=k_4=\dfrac{1}{10}$ and $k_5=-\dfrac{1}{R}$ . So, we have, $\begin{aligned}3\left(\frac{1}{10^2}+\frac{1}{10^2}+\frac{1}{10^2}+\frac{1}{10^2}+\frac{1}{R^2}\right) &=\left(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}-\frac{1}{R}\right)^2 \\ \\ \implies 3\left(\frac{4}{100}+\frac{1}{R^2} \right) &=\left(\frac{4}{10}-\frac{1}{R}\right)^2 \\ \\ \implies \frac{3}{R^2}+\frac{3}{25}&=\frac{1}{R^2}-\frac{4}{5R}+\frac{4}{25}\end{aligned}$ Solving for $R$ in the above equation, we get $R=10+5\sqrt{6}\implies a+b+c=\boxed{21}$

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Note: For $n$ dimensions, the Soddy-Gosset Theorem states that the maximum number of mutually tangent spheres is $n+2$ and the relationship of their curvatures is $n\left (\sum_{i=1}^{n+2} k_i^2\right)=\left(\sum_{i=1}^{n+2}k_i\right)^2$

This problem is the 3D analogous to this problem. Here, the centers of the four spheres are the vertices of a regular tetrahedron with edge length of 20 and the center $G$ of the outer sphere coincides with the centroid of the tetrahedron. Using the labelling seen on the figure, $GD$ is the circumradius of the tetrahedron. The formula for the circumradius $r$ of a regular tetrahedron of edge length $a$ is $r=\dfrac{\sqrt{6}}{4}a$ , hence $GD=\frac{\sqrt{6}}{4}BC=\frac{\sqrt{6}}{4}\times 20=5\sqrt{6}$ Therefore, for the radius $R$ of the big sphere we have $R=ED+DG=10+5\sqrt{6}$ For the answer, $a=10$ , $b=5$ , $c=6$ , thus, $a+b+c=\boxed{21}$ .

Four centers of spheres are tetrahedron $A(a,a,a), B(-a,-a,a),C(-a,a-a),D(a,-a,-a)$ . Center of tetrahedron $O(0,0,0)$ .

$\mid{AB}\mid= a\sqrt{2}=20$ and $a=5 \sqrt{2}$ .

$R=\mid OA+10 \frac{OA}{\mid{OA}\mid} \mid$

Wolframalpha give answer $R=5 \sqrt{6} + 10$ .

Bonus.

For inner sphere $r=5 \sqrt{6} - 10$ .

$\frac{R}{r}=5+2\sqrt{6}$

When we join the centers of the four spheres with straight lines. a regular tetrahedron is formed with the four centers being the vertices. Therefore the edge length of the tetrahedron so formed is $a = 2r =20$ , where $r$ is the radius of the spheres. The distance between the centroid and the vertex of a regular tetrahedron is given by $\sqrt{\frac 38}a$ ( reference ). Since the outer sphere, which touches the four spheres, shares the same centroid with the tetrahedron, its radius is then

$R = r + \sqrt{\frac 38}a = r + \sqrt{\frac 32}r = 10 + 5\sqrt 6$

Therefore $a+b+c = 10+5+6 = \boxed{21}$ .