2014 dimensions?

Using Pythagoras' Theorem, the distance from the centre of the unit hypersphere to a vertex of the hypercube is $\sqrt{2014}$ .

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 adjacent radii.

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

$\frac{2}{1-a} = \sqrt{2014}+1 \longleftrightarrow a = 1-\frac{2}{\sqrt{2014}+1} = \boxed{\frac{2015-\sqrt{8056}}{2013}}$

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.

Generally, $\frac{2}{1-a} = \sqrt{n}+1$ for $n$ dimensions.

Note: in this solution hypercube refers to the hypercube in $2014$ dimensions.

Let the radius of an inscribed hypersphere in a hypercube be $r$ . By repeated Pythagoras' Theorem, the distance between the center of that hypercube to one of the corners is $\underbrace{\sqrt{r^2+r^2+\cdots +r^2}}_{2014\text{ }r^2\text{'s}}=r\sqrt{2014}$

Thus, the distance between the center of the hypercube in the problem to the corner is $\sqrt{2014}$ .

Call the center of the hypercube $O_1$ and the corner $A$ . Call the intersection of $O_1A$ to the hypersphere $X$ . Note that $OX=1$ because the hypersphere is a unit hypersphere. Now call the center of the smaller inscribed hypersphere $O_2$ and its radius $r$ . Note that $XO_2=r$ because the line connecting the centers of any two tangent hyperspheres goes through their tangency point. Also, by the lemma we proved above, $O_2A=r\sqrt{2014}$ .

But wait: $O_1A=\sqrt{2014}$ and $O_1A=O_1X+XO_2+O_2A=1+r+r\sqrt{2014}$

Thus, $\sqrt{2014}=1+r+r\sqrt{2014}$

Solving the linear system of equations gives $r=\boxed{\dfrac{2015-\sqrt{8056}}{2013}}$ so our answer is $2015+8056+2013=\boxed{12084}$

Consider the same problem on 2-space.

Here, you'd be solving this: $\sqrt{2} = 1 + r + \sqrt{2} r$

because $\sqrt{2}$ is the distance from the corner of the square to the center, $1$ is the radius of the circle and $\sqrt{2} r$ is the distance between the center of the second circle and the corner of the square.

Similar analogy in 2014-space follows where you solve:

$\sqrt{2014} = 1 + r + \sqrt{2014} r \\ \implies \boxed{r = \frac{2015-2 \sqrt{2014}}{2013}}$

The distance between the corner of the cube and the spheres' point of tangency is $\sqrt{2014}-1$ . This distance can also be expressed as $r(\sqrt{2014}+1)$ . So, $r(\sqrt{2014}+1) = \sqrt{2014}-1$ , which solves to $r=\frac{2015-\sqrt{8056}}{2013}$ . So, $a+b+c=12084$ .