Having a cube, with a point at its center. What are the points that are equidistant from the center point to...












2












$begingroup$


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.






geometry







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  • $begingroup$
    What do you mean by shape? Should the connectivity be preserved?
    $endgroup$
    – lightxbulb
    2 hours ago










  • $begingroup$
    They form the vertices of a cube.
    $endgroup$
    – TonyK
    2 hours ago






  • 1




    $begingroup$
    @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.
    $endgroup$
    – MJD
    2 hours ago












  • $begingroup$
    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.
    $endgroup$
    – Arthur
    2 hours ago












  • $begingroup$
    @TonyK Eight of them do, but I think we're after the whole surface . . .
    $endgroup$
    – timtfj
    2 hours ago
















2












$begingroup$


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.






geometry







share|cite|improve this question









New contributor




Duero Cuadrillero is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    What do you mean by shape? Should the connectivity be preserved?
    $endgroup$
    – lightxbulb
    2 hours ago










  • $begingroup$
    They form the vertices of a cube.
    $endgroup$
    – TonyK
    2 hours ago






  • 1




    $begingroup$
    @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.
    $endgroup$
    – MJD
    2 hours ago












  • $begingroup$
    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.
    $endgroup$
    – Arthur
    2 hours ago












  • $begingroup$
    @TonyK Eight of them do, but I think we're after the whole surface . . .
    $endgroup$
    – timtfj
    2 hours ago














2












2








2


1



$begingroup$


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.






geometry







share|cite|improve this question









New contributor




Duero Cuadrillero is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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.






geometry




geometry






share|cite|improve this question









New contributor




Duero Cuadrillero is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Duero Cuadrillero is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Blue

47.8k870152




47.8k870152






New contributor




Duero Cuadrillero is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 hours ago









Duero CuadrilleroDuero Cuadrillero

111




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New contributor




Duero Cuadrillero is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Duero Cuadrillero is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Duero Cuadrillero is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    What do you mean by shape? Should the connectivity be preserved?
    $endgroup$
    – lightxbulb
    2 hours ago










  • $begingroup$
    They form the vertices of a cube.
    $endgroup$
    – TonyK
    2 hours ago






  • 1




    $begingroup$
    @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.
    $endgroup$
    – MJD
    2 hours ago












  • $begingroup$
    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.
    $endgroup$
    – Arthur
    2 hours ago












  • $begingroup$
    @TonyK Eight of them do, but I think we're after the whole surface . . .
    $endgroup$
    – timtfj
    2 hours ago


















  • $begingroup$
    What do you mean by shape? Should the connectivity be preserved?
    $endgroup$
    – lightxbulb
    2 hours ago










  • $begingroup$
    They form the vertices of a cube.
    $endgroup$
    – TonyK
    2 hours ago






  • 1




    $begingroup$
    @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.
    $endgroup$
    – MJD
    2 hours ago












  • $begingroup$
    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.
    $endgroup$
    – Arthur
    2 hours ago












  • $begingroup$
    @TonyK Eight of them do, but I think we're after the whole surface . . .
    $endgroup$
    – timtfj
    2 hours ago
















$begingroup$
What do you mean by shape? Should the connectivity be preserved?
$endgroup$
– lightxbulb
2 hours ago




$begingroup$
What do you mean by shape? Should the connectivity be preserved?
$endgroup$
– lightxbulb
2 hours ago












$begingroup$
They form the vertices of a cube.
$endgroup$
– TonyK
2 hours ago




$begingroup$
They form the vertices of a cube.
$endgroup$
– TonyK
2 hours ago




1




1




$begingroup$
@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.
$endgroup$
– MJD
2 hours ago






$begingroup$
@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.
$endgroup$
– MJD
2 hours ago














$begingroup$
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.
$endgroup$
– Arthur
2 hours ago






$begingroup$
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.
$endgroup$
– Arthur
2 hours ago














$begingroup$
@TonyK Eight of them do, but I think we're after the whole surface . . .
$endgroup$
– timtfj
2 hours ago




$begingroup$
@TonyK Eight of them do, but I think we're after the whole surface . . .
$endgroup$
– timtfj
2 hours ago










4 Answers
4






active

oldest

votes


















5












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    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.)






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      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.



      --- rk






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        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
        $$V=8sqrt 2 s^3$$
        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
        $$V=4 a^3$$
        --- rk






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$


















            5












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$
















              5












              5








              5





              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$



              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.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 2 hours ago









              Vasily MitchVasily Mitch

              1,51337




              1,51337























                  4












                  $begingroup$

                  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.)






                  share|cite|improve this answer











                  $endgroup$


















                    4












                    $begingroup$

                    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.)






                    share|cite|improve this answer











                    $endgroup$
















                      4












                      4








                      4





                      $begingroup$

                      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.)






                      share|cite|improve this answer











                      $endgroup$



                      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.)







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 2 hours ago

























                      answered 2 hours ago









                      ArthurArthur

                      112k7109191




                      112k7109191























                          2












                          $begingroup$

                          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.



                          --- rk






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            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.



                            --- rk






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              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.



                              --- rk






                              share|cite|improve this answer









                              $endgroup$



                              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.



                              --- rk







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 1 hour ago









                              Dr. Richard KlitzingDr. Richard Klitzing

                              1,54616




                              1,54616























                                  0












                                  $begingroup$

                                  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
                                  $$V=8sqrt 2 s^3$$
                                  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
                                  $$V=4 a^3$$
                                  --- rk






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    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
                                    $$V=8sqrt 2 s^3$$
                                    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
                                    $$V=4 a^3$$
                                    --- rk






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      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
                                      $$V=8sqrt 2 s^3$$
                                      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
                                      $$V=4 a^3$$
                                      --- rk






                                      share|cite|improve this answer









                                      $endgroup$



                                      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
                                      $$V=8sqrt 2 s^3$$
                                      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
                                      $$V=4 a^3$$
                                      --- rk







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 48 mins ago









                                      Dr. Richard KlitzingDr. Richard Klitzing

                                      1,54616




                                      1,54616






















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