geometry
There are two $d$-dimensional unit spheres centred at $A$ and $B$ in $\mathbb{R}^d$. They are touching at location $I$. How can I find the centre of a third $d$-dimensional unit sphere $C$ that touches the first two spheres? I know that:
I don't know how to proceed from here. I know there can be multiple locations for $C$. In fact the set of possible locations forms a $(d-1)$-sphere. But how to find its radius and centre?
No. It just means that (hyper)cubic lattice sphere packing (where the centers of the spheres are placed in a cubic grid, say spheres with radius $\frac12$ centered at each point with integer cartesian coordinates) is very inefficient in higher dimensions, and the room between the spheres become large enough to fit even larger spheres.
Counterintuitive? Yes, but mostly because we are relatively low-dimensional beings with limited imagination. Paradox or contradiction? No.
A simple approach for producing reasonable lower bounds is to use a face-centered cubic packing or a hexagonal packing (both have the optimal density, $\frac{\pi}{2\sqrt{3}}\approx 74\%$, in the unconstrained space) and to count the number of spheres met by $x^2+y^2+z^2=(20)^2$. Recalling that the optimal packing density in the plane is $\frac{\pi\sqrt{3}}{6}$, in a sphere with radius $20$ it should be possible to pack around
$$ \frac{\pi}{2\sqrt{3}}\cdot 20^3 - \frac{\pi\sqrt{3}}{6}\cdot 4(20)^2 \approx\color{red}{5804}$$ spheres, but not many more. The estimated density is so $\approx 72.5\%$.
Best Answer
From the fact that
$$ \lvert A - B\rvert = \lvert A - C\rvert = \lvert B - C\rvert = 2 $$
you know that $A, B, C$ are vertices of an equilateral triangle of side $2$. It doesn't matter how many dimensions you are working in, three non-collinear points still make the vertices of a triangle and if all three sides are equal then the triangle is equilateral.
Given that $\triangle ABC$ is an equilateral triangle of side $2$, what is the relationship of $C$ to the midpoint of side $AB$?
The point $C$ must be on the perpendicular bisector of $AB$ at a distance $\sqrt3$ from $AB.$
Noting that $I$ is the midpoint of $AB,$ we have $CI = \sqrt3$ and $C$ lies on the $(d-1)$-plane through $I$ orthogonal to the line $AB,$ since that $(d-1)$-plane contains all lines through $I$ perpendicular to $AB$ and all lines through $I$ perpendicular to $AB$ are in that plane.
In summary, the sphere you are looking for has center $I$ and radius $\sqrt3$ and lies in the $(d-1)$-plane through $I$ orthogonal to $AB.$