Consider a hypercube of side length 2 in a 2014-dimensional space. Inscribe a unit hypersphere in the hypercube. Inscribe a smaller hypersphere tangent to the unit hypersphere and the sides of the hypercube.

The radius of the smaller sphere can be written as $\frac{a-\sqrt{b}}{c}$ where $a$ , $b$ and $c$ are coprime positive integers. Find $a+b+c$ .

Based on this problem .

The answer is 12084.

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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.