Sphere-packing

This problem can easily be solved using generalised descartes theorem .

The curvature $k=\frac{1}{r}$ of the four spheres is $1$ as all are of radius $1$ . The curvature of the sphere internally tangent to four circles (in this case the unknown sphere) is taken to be negative .

let $k_1$ and $r_1$ be curvature and radius of the bigger sphere.Hence we arrive at the following eqn. $(4-k_1)^2=3(4+k_1^2)$ On solving we get the solution $\sqrt{6}-2$ (ignore the negative soution)

hence the radius is $r_1=\frac{1}{k}=\frac{1}{0.4494}=2.2247$ we get our answer as $\boxed{222}$

Four centers of four sphere form a tetrahedron with each of its edges has the length of $2r$ . We call O is center of the tetrahedron, the distance from to O to the tetrahedron's vertex is $d=\frac{r\sqrt{6}}{2}$ . We get the radius of the containing sphere by adding $r$ to $d$ , that is $R=d+r=r(\frac{\sqrt{6}}{2}+1)$

The radius of the circumsphere of a tetrahedron is a√(3/8), where a is the length of one of its edges. We let a = 2, and add 1, for the final expression for R = 1 + 2√(3/8).