Having a cube, with a point at its center. What are the points that are equidistant from the center point to the cubes vertices? Having a cube, with a point at its center. 
What shape do the points wich are equidistant between the center and the cubes vertices make?
The source of why I had this question is the following photo

What shape is resultant from this composition of equidistant points? 
Thank you very much.
 A: The shape you get is called a truncated octahedron:

The resulting tesselation (tiling) pattern is called a bitruncated cubic honeycomb:

(Both images taken from the Wikipedia article on the truncated octahedron.)
A: The answer to the question from screenshot is a truncated octahedron. If you consider center of the cube and a vertex, the locus of equidistant points is a perpendicular bisector plane. 8 of those planes create an octahedron. However, if you consider centers of other cubes, locus of equidistant points with them will give the initial cube itself. So the answer is the intersection of cube and octahedron, which is a truncated octahedron.
A: Just to add some further notions to the already given answers: Those describe the Voronoi domain resp. Voronoi complex of the body centered cubical (bcc) lattice. - Within crystallography the Voronoi domain also is called Brillouin zone.
A: Within the pic of OP there is the quest for the volume of $D(0)$ as well. $D(0)$ was already being mentioned to be the truncated octahedron. Thus the searched for volume is $$V=8\sqrt 2\ s^3$$
where $s$ is the edge length of that polyhedron. That length in turn can be seen from that very pic to be (in units of the lattice constant $a$) $$s=\frac 12 \sqrt 2\ a$$
Thus the searched for total value would become $$V=4\ a^3$$
