Consider a $d-$dimensional hypercube $Q$ of side length $l ∈ R, i. e. |x_i − y_i | ≤ l$ for all $x, y ∈ Q$ and all $i ∈ [d]$. Note that $Q$ has $2^d$ corners. We fill $Q$ with ($L_2-)$hyperballs the following way:
- We place 2d hyperballs of radius $\frac{l}{4}$ close to the $2^d$ corners of $Q$, such that their distance to the center of $Q$ is maximal while still being completely contained in $Q$.
- We place an additional single hyperball in the center of $Q$ such that its radius is maximal with the property that it intersects with none of the other hyperballs’ intereriors.
What is the minimal dimension d such that the the surface of the central hyperball peeks trough the surface of Q?
I was thinking about computing the volume, but things got very complicated, as i could not found the lower bound for the dimension $d$.
I would appreciate any ideas that could make it easier for me to solve the task.
We may as well set $l=1$ and talk about the unit hypercube. The hyperspheres in the corners then had diameter $\frac 12$. Consider the body diagonal of the hypercube. It goes through the centers of two of the corner hyperspheres, the center of the center hypersphere, and two of the points of tangency between the center hypersphere and corner hyperspheres.
The length of the body diagonal is $\sqrt d$. The distance from the corner of the hypercube to the center of a corner hypersphere is $\sqrt{\frac d{16}}=\frac {\sqrt d}4$. The distance from the corner of the hypercube to a tangency point is then $\frac {\sqrt d+1}4$. The radius of the central hypersphere is then $\frac {\sqrt d}2-\frac{\sqrt d+1}4$. The central hypersphere touches the face of the hypercube when this is $\frac 12$ $$\frac {\sqrt d}2-\frac{\sqrt d+1}4=\frac 12\\\frac {\sqrt d}4=\frac 34\\d=9$$