Point whose distance from center of a cube is less than that from a family of points Consider a cube $\mathcal{C}$ with side length $\ell$ and let $O$ be its center. Divide the cube in $8$ smaller cubes with one vertex in $O.$ I want to find the family of points such that the nearest point between the vertices of the eight cubes (they are $27$ points) and the centres of their faces (they are $36$ points) is $O:$ in other words, if we call $V$ the family of the vertices and $F$ the family of the centres of the faces we want to determine the set
$$\left\{p\in \mathcal{C}: \min_{q\in V\cup F}|p-q|=|p-O|\right\}$$
I think the answer is another cube centred in $O$ with side length $\ell/8$ but I can’t find a nice proof of this fact (using for instance simmetries to simplify the problem and using calculus the least possible)… any help?
 A: Given two fixed points, the set of all points equidistant from both is the plane perpendicular to the line connecting them, and passing through their midpoint. Now consider any one of the $8$ subcubes, and examine which vertices and face centers can be the closest such point to a $p$ in the set.
Start with the vertex $O'$ opposite $O$, the midplane between them passes though the midpoints of the $6$ edges that do not have either vertex as an endpoint. It is easy to see that the points on this plane are closer to the centers of the faces adjacent to $O$ than $O'$ is, which is also their distance from $O$. So none of the points on that plane are going to be in the set.
But similarly, for the centers of faces adjacent to $O'$, the midplane between them and $O$, the points of the plane are once again closer to the centers of the faces adjacent to $O$, and the same is also true for the other six vertices. In all cases, any point equidistant from them and $O$ will be closer to the center of one of the $O$-adjacent faces. None of these points will be in your set.
Thus it is only the centers of the faces adjacent to $O$ that determine the set. The vertices (other than $O$) and the centers of the far faces can be ignored.
Looking back to the full picture with eight subcubes, by symmetry, each of these nearest face centers to $O$ will have its time to shine as being the $q$ providing the $\min_{q \in V\cup F} |p-q|$ distance, creating a planar face of the resultant shape. There are $12$ such inner face centers, so your shape will have $12$ faces, a dodecahedron. I have not checked to make sure it will be a regular dodecahedron. I'll leave that to you. But the face-to=face diameter is $\frac \ell{2\sqrt 2}$, not $\frac \ell 8$. The interior of that dodecahedron consists of points that are closer to $O$ than to any of the face centers. Per coffeemath's comment, these are automatically in your set. So it is a solid dodecahedron.
