Having a cube, with a point at its center. What are the points that are equidistant from the center point to...
$begingroup$
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
What shape is resultant from this composition of equidistant points?
Thank you very much.
geometry
edited Jan 18 at 14:15
Blue
47.9k870153
asked Jan 18 at 14:08
Duero CuadrilleroDuero Cuadrillero
262
New contributor
Duero Cuadrillero is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
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
What shape is resultant from this composition of equidistant points?
Thank you very much.
geometry
edited Jan 18 at 14:15
Blue
47.9k870153
asked Jan 18 at 14:08
Duero CuadrilleroDuero Cuadrillero
262
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
Jan 18 at 14:10
$begingroup$
They form the vertices of a cube.
$endgroup$
– TonyK
Jan 18 at 14:18
2
$begingroup$
@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
Jan 18 at 14:25
$begingroup$
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
Jan 18 at 14:34
$begingroup$
@TonyK Eight of them do, but I think we're after the whole surface . . .
$endgroup$
– timtfj
Jan 18 at 14:34
add a comment |
$begingroup$
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
What shape is resultant from this composition of equidistant points?
Thank you very much.
geometry
edited Jan 18 at 14:15
Blue
47.9k870153
asked Jan 18 at 14:08
Duero CuadrilleroDuero Cuadrillero
262
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
What shape is resultant from this composition of equidistant points?
Thank you very much.
geometry
geometry
edited Jan 18 at 14:15
Blue
47.9k870153
asked Jan 18 at 14:08
Duero CuadrilleroDuero Cuadrillero
262
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.
edited Jan 18 at 14:15
Blue
47.9k870153
asked Jan 18 at 14:08
Duero CuadrilleroDuero Cuadrillero
262
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.
edited Jan 18 at 14:15
Blue
47.9k870153
edited Jan 18 at 14:15
Blue
47.9k870153
edited Jan 18 at 14:15
Blue
47.9k870153
47.9k870153
asked Jan 18 at 14:08
Duero CuadrilleroDuero Cuadrillero
262
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 Jan 18 at 14:08
Duero CuadrilleroDuero Cuadrillero
262
asked Jan 18 at 14:08
Duero CuadrilleroDuero Cuadrillero
262
262
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
Jan 18 at 14:10
$begingroup$
They form the vertices of a cube.
$endgroup$
– TonyK
Jan 18 at 14:18
2
$begingroup$
@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
Jan 18 at 14:25
$begingroup$
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
Jan 18 at 14:34
$begingroup$
@TonyK Eight of them do, but I think we're after the whole surface . . .
$endgroup$
– timtfj
Jan 18 at 14:34
add a comment |
$begingroup$
What do you mean by shape? Should the connectivity be preserved?
$endgroup$
– lightxbulb
Jan 18 at 14:10
$begingroup$
They form the vertices of a cube.
$endgroup$
– TonyK
Jan 18 at 14:18
2
$begingroup$
@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
Jan 18 at 14:25
$begingroup$
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
Jan 18 at 14:34
$begingroup$
@TonyK Eight of them do, but I think we're after the whole surface . . .
$endgroup$
– timtfj
Jan 18 at 14:34
$begingroup$
What do you mean by shape? Should the connectivity be preserved?
$endgroup$
– lightxbulb
Jan 18 at 14:10
$begingroup$
What do you mean by shape? Should the connectivity be preserved?
$endgroup$
– lightxbulb
Jan 18 at 14:10
$begingroup$
They form the vertices of a cube.
$endgroup$
– TonyK
Jan 18 at 14:18
$begingroup$
They form the vertices of a cube.
$endgroup$
– TonyK
Jan 18 at 14:18
2
2
$begingroup$
@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
Jan 18 at 14:25
$begingroup$
@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
Jan 18 at 14:25
$begingroup$
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
Jan 18 at 14:34
$begingroup$
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
Jan 18 at 14:34
$begingroup$
@TonyK Eight of them do, but I think we're after the whole surface . . .
$endgroup$
– timtfj
Jan 18 at 14:34
$begingroup$
@TonyK Eight of them do, but I think we're after the whole surface . . .
$endgroup$
– timtfj
Jan 18 at 14:34
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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.
answered Jan 18 at 14:32
Vasily MitchVasily Mitch
1,76638
$endgroup$
add a comment |
$begingroup$
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.)
answered Jan 18 at 14:30
ArthurArthur
113k7110193
$endgroup$
add a comment |
$begingroup$
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.
edited Jan 18 at 21:58
Nij
2,00311223
answered Jan 18 at 15:44
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
add a comment |
$begingroup$
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 $$V=4 a^3$$
edited Jan 18 at 21:58
Nij
2,00311223
answered Jan 18 at 16:01
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 18 at 14:32
Vasily MitchVasily Mitch
1,76638
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 18 at 14:32
Vasily MitchVasily Mitch
1,76638
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 18 at 14:32
Vasily MitchVasily Mitch
1,76638
$endgroup$
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.
answered Jan 18 at 14:32
Vasily MitchVasily Mitch
1,76638
answered Jan 18 at 14:32
Vasily MitchVasily Mitch
1,76638
answered Jan 18 at 14:32
Vasily MitchVasily Mitch
1,76638
answered Jan 18 at 14:32
Vasily MitchVasily Mitch
1,76638
1,76638
add a comment |
add a comment |
$begingroup$
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.)
answered Jan 18 at 14:30
ArthurArthur
113k7110193
$endgroup$
add a comment |
$begingroup$
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.)
answered Jan 18 at 14:30
ArthurArthur
113k7110193
$endgroup$
add a comment |
$begingroup$
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.)
answered Jan 18 at 14:30
ArthurArthur
113k7110193
$endgroup$
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.)
answered Jan 18 at 14:30
ArthurArthur
113k7110193
edited Jan 18 at 14:38
answered Jan 18 at 14:30
ArthurArthur
113k7110193
answered Jan 18 at 14:30
ArthurArthur
113k7110193
answered Jan 18 at 14:30
ArthurArthur
113k7110193
113k7110193
add a comment |
add a comment |
$begingroup$
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.
edited Jan 18 at 21:58
Nij
2,00311223
answered Jan 18 at 15:44
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
add a comment |
$begingroup$
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.
edited Jan 18 at 21:58
Nij
2,00311223
answered Jan 18 at 15:44
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
add a comment |
$begingroup$
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.
edited Jan 18 at 21:58
Nij
2,00311223
answered Jan 18 at 15:44
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
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.
edited Jan 18 at 21:58
Nij
2,00311223
answered Jan 18 at 15:44
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
edited Jan 18 at 21:58
Nij
2,00311223
edited Jan 18 at 21:58
Nij
2,00311223
edited Jan 18 at 21:58
Nij
2,00311223
2,00311223
answered Jan 18 at 15:44
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
answered Jan 18 at 15:44
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
answered Jan 18 at 15:44
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
1,57016
add a comment |
add a comment |
$begingroup$
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 $$V=4 a^3$$
edited Jan 18 at 21:58
Nij
2,00311223
answered Jan 18 at 16:01
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
add a comment |
$begingroup$
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 $$V=4 a^3$$
edited Jan 18 at 21:58
Nij
2,00311223
answered Jan 18 at 16:01
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
add a comment |
$begingroup$
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 $$V=4 a^3$$
edited Jan 18 at 21:58
Nij
2,00311223
answered Jan 18 at 16:01
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
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 $$V=4 a^3$$
edited Jan 18 at 21:58
Nij
2,00311223
answered Jan 18 at 16:01
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
edited Jan 18 at 21:58
Nij
2,00311223
edited Jan 18 at 21:58
Nij
2,00311223
edited Jan 18 at 21:58
Nij
2,00311223
2,00311223
answered Jan 18 at 16:01
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
answered Jan 18 at 16:01
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
answered Jan 18 at 16:01
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
1,57016
add a comment |
add a comment |
Duero Cuadrillero is a new contributor. Be nice, and check out our Code of Conduct.
Duero Cuadrillero is a new contributor. Be nice, and check out our Code of Conduct.
Duero Cuadrillero is a new contributor. Be nice, and check out our Code of Conduct.
Duero Cuadrillero is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
What do you mean by shape? Should the connectivity be preserved?
$endgroup$
– lightxbulb
Jan 18 at 14:10
$begingroup$
They form the vertices of a cube.
$endgroup$
– TonyK
Jan 18 at 14:18
2
$begingroup$
@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
Jan 18 at 14:25
$begingroup$
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
Jan 18 at 14:34
$begingroup$
@TonyK Eight of them do, but I think we're after the whole surface . . .
$endgroup$
– timtfj
Jan 18 at 14:34