Inscribed Spheres

Using Pythagoras' Theorem, the distance from the centre of the unit sphere to a vertex of the cube is $\sqrt{3}$ .

Filling in one corner of the cube with progressively smaller tangential spheres, we find the sum of diameters to be $\displaystyle 2 \sum_{n=0}^{\infty} a^{n} = \frac{2}{1-a}$ where $a$ is the ratio between radii.

Equating the former plus a radius 1 and the latter, we can solve for $a$ :

$\frac{2}{1-a} = \sqrt{3}+1 \longleftrightarrow a = 1-\frac{2}{\sqrt{3}+1} = \boxed{2-\sqrt{3}}$

Using this ratio, we can compute the smaller radius since $R = 1$ and $r = aR = a$ , where $r$ and $R$ are the small and big radii respectively.

The distance between the corner of the cube and the spheres' point of tangency is $\sqrt{3}-1$ . This distance can also be expressed as $r(\sqrt{3}+1)$ . So, $r(\sqrt{3}+1) = \sqrt{3}-1$ , which solves to $r=2-\sqrt{3}$ .

Let's consider the corner of the cube which the small-sphere occupies as the origin of the x-y-z coordinate axes. Then, the equation for the large sphere can be written as

$(x-1)^{2}+(y-1)^{2}+(z-1)^{2} = 1$ . (Eq. 1)

Since the distance from each coordinate plane to the center of the small sphere is $r$ , we can write the equation for the small sphere as

$(x-r)^{2}+(y-r)^{2}+(z-r)^{2} = r^{2}$ (Eq. 2)

Now comes the tricky part: finding the point of contact between the small sphere and big sphere.

Since the equation for both circles are symmetric about the line x=y=z, we know that the point of contact lies somewhere along this line.

Furthermore, we know that the "vector" from the center of the small sphere to the point of contact will have magnitude $r$ and direction $<1,1,1>$ . To give this directional vector the correct magnitude, we simply multiply it by $r$ and devide it by $\sqrt{3}$ . We can then add this vectorto the position vector of the center of the small sphere, so that the point of contact between the two spheres can be written as:

$\frac{r(1+\sqrt{3})}{\sqrt{3}}<1,1,1>$

Plugging this point into Eq. 2 won't tell us anything we don't already know. Plugging it into Eq. 1 gives us:

$(\frac{r(1+\sqrt{3})-\sqrt{3}}{\sqrt{3}}-1)^{2} + ditto + ditto = 1$

From there it's a bunch of algebra that I'd rather not type out. Eventually this becomes (but don't take my word for it):

$r^{2}(2+\sqrt{3})-r(3+\sqrt{3})+1 = 0$

Using the quadratic formula and simplifying we get

$r = \frac{3+\sqrt{3} \frac{+}{-} \sqrt{4+2\sqrt{3}}}{(4+2\sqrt{3})}$

Using $+$ in the $+/-$ gives us $r=1$ , so we use the $-$ instead.

Note that $4+2\sqrt{3} = (1+\sqrt{3})^{2}$ , so this can be simplified further to:

$r = \frac{1}{2+\sqrt{3}}$

Rationalizing the denominator we get

$r = 2-\sqrt{3}$

This means $a = 2$ and $b = 3$ , so the answer is $\boxed{a+b = 5}$

The diagonal of cube is 2 (3)^1/2 So, distance b/w centre and corner is root3 Let radius of that circle be R...so distance b/w centre of circle whose radius is R and corner of cube is Rroot3 Hence, Root3=1 (radius of given circle) +R+Rroot3 i.e (root3 -1)/(root3+1)=R Rationalising denominator of L.H.S. we get, R=2-root3 So,a=2 and b=3 hence,a + b=5