Kissing Spheres

$\text{Four spheres radius r touch one another, so their centers must be at 2r.} \\ \text{So the four centers are the four vertices of a TETRAHEDRON side 2r.}\\ Circumradius ~~ of~ a~ tetrahedron~R=\dfrac{\sqrt 6 }{4}*side=\dfrac{\sqrt 6 }{4}*2r=\dfrac{\sqrt 6 *r}{2}.\\\therefore~1=radius~ of~ big~ sphere=R+r=\dfrac{\sqrt 6 *r}{2}+r~~\implies r=\dfrac{1}{ \dfrac{\sqrt 6 }{2} + 1 } \\r=\dfrac{2}{\sqrt 6 +2} = \sqrt 6 - 2=\sqrt n-m.~~~~~m+n= ~~~~~~~~~~~~\large \color{#D61F06}{ \boxed{8} }$

Apply the Soddy-Gosset theorem for three dimensions and 4 spheres,

$n(\displaystyle\sum\limits_{i=1}^{n+2} k_i^2)=(\displaystyle\sum\limits_{i=1}^{n+2} k_i)^2$

We can solve for the curvatures for the 4 inner spheres by equating all of them, expressing them as $K_s$ and setting the larger sphere’s curvature to -1.

$3({K_s}^2+{K_s}^2+{K_s}^2+{K_s}^2+(-1)^2)=({K_s}+{K_s}^2+{K_s}^2+{K_s}^2-1)^2$

$3(4{K_s}^2+(-1)^2)=(4{K_s}-1)^2$

$3+12{K_s}^2=16{K_s}-8{K_s}+1$

$4{K_s}^2-8{K_s}-2=0$

Now applying the quadratic formula we can get

$\frac{8\pm\sqrt{64+32}}{8}$

Which simplifies to

$\frac{2\pm\sqrt{6}}{2}$

Now since curvature is defined as the reciprocal of the radius, with is parity dependent on whether the n-spheres are externally or internally tangent to each other.

The Inner smaller spheres are all externally tangent to each other so their curvature must be positive, allowing for us to rule out the negative solution of our quadratic.

Solving for the reciprocal of $K_s$ .

$\frac{1}{K_s}=\frac{2}{2+\sqrt{6}}$

$R_s=\frac{2}{2+\sqrt{6}}\cdot\frac{2-\sqrt{6}}{2-\sqrt{6}}$

$\frac{4-2\sqrt{6}}{-2}=-2+\sqrt{6}$

The radius of the inner circles therefore equals $-2+\sqrt{6}$ and our answer of $m+n$ is $2+6=\boxed{8}$ .