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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

enter image description here

What shape is resultant from this composition of equidistant points?

Thank you very much.

Blue
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Duero Cuadrillero
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  • What do you mean by shape? Should the connectivity be preserved? – lightxbulb Jan 18 '19 at 14:10
  • They form the vertices of a cube. – TonyK Jan 18 '19 at 14:18
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    @lightxbulb If you look at the original question, you'll see that what's actually being asked for is the volume of the set of all points that are closer to the center of a cube than to any of its vertices. OP wishes to understand the boundary surface of this region. – MJD Jan 18 '19 at 14:25
  • I never expected to learn any math when I watched Assasination Classroom, but seing that problem, and the elegant solution to finding the volume was cool. Actually depicting math in progress (dramaticized though that sequence was) is sadly a rarity in popular culture. – Arthur Jan 18 '19 at 14:34
  • @TonyK Eight of them do, but I think we're after the whole surface . . . – timtfj Jan 18 '19 at 14:34

4 Answers4

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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.

Vasily Mitch
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The shape you get is called a truncated octahedron:

enter image description here

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

enter image description here

(Both images taken from the Wikipedia article on the truncated octahedron.)

Arthur
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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.

Nij
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Dr. Richard Klitzing
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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 $$V=4\ a^3$$

Nij
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Dr. Richard Klitzing
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