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 2 hours ago
Blue
47.8k870152
asked 2 hours ago
Duero CuadrilleroDuero Cuadrillero
111
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.
$endgroup$
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 2 hours ago
Blue
47.8k870152
asked 2 hours ago
Duero CuadrilleroDuero Cuadrillero
111
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
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 2 hours ago
Blue
47.8k870152
asked 2 hours ago
Duero CuadrilleroDuero Cuadrillero
111
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 2 hours ago
Blue
47.8k870152
asked 2 hours ago
Duero CuadrilleroDuero Cuadrillero
111
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 2 hours ago
Blue
47.8k870152
asked 2 hours ago
Duero CuadrilleroDuero Cuadrillero
111
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 2 hours ago
Blue
47.8k870152
edited 2 hours ago
Blue
47.8k870152
edited 2 hours ago
Blue
47.8k870152
47.8k870152
asked 2 hours ago
Duero CuadrilleroDuero Cuadrillero
111
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
asked 2 hours ago
Duero CuadrilleroDuero Cuadrillero
111
111
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
add a comment |
$begingroup$
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
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 2 hours ago
Vasily MitchVasily Mitch
1,51337
$endgroup$
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 2 hours ago
ArthurArthur
112k7109191
$endgroup$
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.
--- rk
answered 1 hour ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
$endgroup$
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
$$V=4 a^3$$
--- rk
answered 48 mins ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
$endgroup$
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 2 hours ago
Vasily MitchVasily Mitch
1,51337
$endgroup$
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 2 hours ago
Vasily MitchVasily Mitch
1,51337
$endgroup$
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 2 hours ago
Vasily MitchVasily Mitch
1,51337
$endgroup$
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 2 hours ago
Vasily MitchVasily Mitch
1,51337
answered 2 hours ago
Vasily MitchVasily Mitch
1,51337
answered 2 hours ago
Vasily MitchVasily Mitch
1,51337
answered 2 hours ago
Vasily MitchVasily Mitch
1,51337
1,51337
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 2 hours ago
ArthurArthur
112k7109191
$endgroup$
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 2 hours ago
ArthurArthur
112k7109191
$endgroup$
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 2 hours ago
ArthurArthur
112k7109191
$endgroup$
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 2 hours ago
ArthurArthur
112k7109191
edited 2 hours ago
answered 2 hours ago
ArthurArthur
112k7109191
answered 2 hours ago
ArthurArthur
112k7109191
answered 2 hours ago
ArthurArthur
112k7109191
112k7109191
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.
--- rk
answered 1 hour ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
$endgroup$
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.
--- rk
answered 1 hour ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
$endgroup$
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.
--- rk
answered 1 hour ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
$endgroup$
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
answered 1 hour ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
answered 1 hour ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
answered 1 hour ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
answered 1 hour ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
1,54616
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
$$V=4 a^3$$
--- rk
answered 48 mins ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
$endgroup$
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
$$V=4 a^3$$
--- rk
answered 48 mins ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
$endgroup$
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
$$V=4 a^3$$
--- rk
answered 48 mins ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
$endgroup$
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
answered 48 mins ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
answered 48 mins ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
answered 48 mins ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
answered 48 mins ago
Dr. Richard KlitzingDr. Richard Klitzing
1,54616
1,54616
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
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