A larger sphere A, having a radius $R$ is snugly fitted in a cube (i.e. sphere A touches all six faces of the cube). Further, a small sphere B is snugly fitted in the corner of cube (i.e. sphere B touches sphere A & three orthogonal faces meeting at the same vertex). Further, a smaller sphere C, having a radius $r$, is snugly fitted in the same corner of the cube (i.e. sphere C touches sphere B & three orthogonal faces meeting at the same vertex).
How to find out the radius $r$ (of smaller sphere C) in terms of the radius $R$ (of larger sphere A)? I have tried this using vector analysis but I did not find the correct answer. Any help or pointing in the right direction is appreciated.
Note: None of the spheres touches any of 12 edges of the cube.
Unit vector equally inclined with three orthogonal axes X, Y & Z is given as $$ \hat{r}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$$
angle of inclination of the vector with three axes is $$=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\approx 54.73^{o}$$ but further if angle of inclination of this vector with each of three orthogonal faces is $\alpha \space$ then I have $$\sin\alpha =\frac{R_{B}}{R\sqrt{3}-R-R_{B}}$$ Similarly, for radius $r$ $$\sin\alpha =\frac{r}{R_{B}\sqrt{3}-R_{B}-r}$$
Further, how I can determine angle of inclination $\alpha$ with the orthogonal planes?
Extend the problem to get infinitely many spheres, each fitting snugly in the space between the previous one and the corner. From the scaling symmetry (around the bottom left vertex) of the problem the radii of the spheres decrease in geometric series. So the distance from the far-side of $A$ to the corner, which we denote by $d$, should be
$$ d = \frac{2R}{1 - r} $$
where $r$ is the ratio of the radii. We know that $d = R + \sqrt{3}R$. So we have
$$ r = (2- \sqrt {3}) $$
and the radius of $C$ being $r^2 R$ which you can compute yourself.