Apparent existence of a semi-regular polyhedron, but that I cannot find in any table.
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I propose the existence of a semi-regular polyhedron with one square, one hexagon and two triangles at each vertex. The sum of angles at reach vertex is $330°$ and therefore the external angle is $30°$, which divides $720°$. That would imply $frac{720}{30} = 24$ vertices, from which can be easily calculated that there are $48$ sides and $26$ faces ($4$ hexagons, $6$ squares and $16$ triangles). That´s fine, but I cannot see any sign of this supposed polyhedron in lists of the $13$ Archimedean solids, etc. What condition does this polyhedron violate? Why does it not exist?
geometry
edited Dec 1 '18 at 2:50
timtfj
1,248318
asked Nov 30 '18 at 23:45
Bryon HallBryon Hall
261
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add a comment |
$begingroup$
I propose the existence of a semi-regular polyhedron with one square, one hexagon and two triangles at each vertex. The sum of angles at reach vertex is $330°$ and therefore the external angle is $30°$, which divides $720°$. That would imply $frac{720}{30} = 24$ vertices, from which can be easily calculated that there are $48$ sides and $26$ faces ($4$ hexagons, $6$ squares and $16$ triangles). That´s fine, but I cannot see any sign of this supposed polyhedron in lists of the $13$ Archimedean solids, etc. What condition does this polyhedron violate? Why does it not exist?
geometry
edited Dec 1 '18 at 2:50
timtfj
1,248318
asked Nov 30 '18 at 23:45
Bryon HallBryon Hall
261
$endgroup$
$begingroup$
I am a bit confused. If you go around a vertex, in what order do the faces appear in your suggested construction?
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– fedja
Dec 1 '18 at 1:38
$begingroup$
In other words, can you draw the planar graph corresponding to your suggestion? I tried and I couldn't make all vertices look the same (remember that semi-regularity requires a group of isometries transitive on the vertices, not just the same unordered collections of faces at each vertex) but, perhaps, I'm misunderstanding your suggestion.
$endgroup$
– fedja
Dec 1 '18 at 1:49
1
$begingroup$
Have you tried making one (or part of one) out of card? Does it seem to fit together?
$endgroup$
– timtfj
Dec 1 '18 at 1:57
add a comment |
$begingroup$
I propose the existence of a semi-regular polyhedron with one square, one hexagon and two triangles at each vertex. The sum of angles at reach vertex is $330°$ and therefore the external angle is $30°$, which divides $720°$. That would imply $frac{720}{30} = 24$ vertices, from which can be easily calculated that there are $48$ sides and $26$ faces ($4$ hexagons, $6$ squares and $16$ triangles). That´s fine, but I cannot see any sign of this supposed polyhedron in lists of the $13$ Archimedean solids, etc. What condition does this polyhedron violate? Why does it not exist?
geometry
edited Dec 1 '18 at 2:50
timtfj
1,248318
asked Nov 30 '18 at 23:45
Bryon HallBryon Hall
261
$endgroup$
I propose the existence of a semi-regular polyhedron with one square, one hexagon and two triangles at each vertex. The sum of angles at reach vertex is $330°$ and therefore the external angle is $30°$, which divides $720°$. That would imply $frac{720}{30} = 24$ vertices, from which can be easily calculated that there are $48$ sides and $26$ faces ($4$ hexagons, $6$ squares and $16$ triangles). That´s fine, but I cannot see any sign of this supposed polyhedron in lists of the $13$ Archimedean solids, etc. What condition does this polyhedron violate? Why does it not exist?
geometry
geometry
edited Dec 1 '18 at 2:50
timtfj
1,248318
asked Nov 30 '18 at 23:45
Bryon HallBryon Hall
261
edited Dec 1 '18 at 2:50
timtfj
1,248318
asked Nov 30 '18 at 23:45
Bryon HallBryon Hall
261
edited Dec 1 '18 at 2:50
timtfj
1,248318
edited Dec 1 '18 at 2:50
timtfj
1,248318
edited Dec 1 '18 at 2:50
timtfj
1,248318
1,248318
asked Nov 30 '18 at 23:45
Bryon HallBryon Hall
261
asked Nov 30 '18 at 23:45
Bryon HallBryon Hall
261
asked Nov 30 '18 at 23:45
Bryon HallBryon Hall
261
261
$begingroup$
I am a bit confused. If you go around a vertex, in what order do the faces appear in your suggested construction?
$endgroup$
– fedja
Dec 1 '18 at 1:38
$begingroup$
In other words, can you draw the planar graph corresponding to your suggestion? I tried and I couldn't make all vertices look the same (remember that semi-regularity requires a group of isometries transitive on the vertices, not just the same unordered collections of faces at each vertex) but, perhaps, I'm misunderstanding your suggestion.
$endgroup$
– fedja
Dec 1 '18 at 1:49
1
$begingroup$
Have you tried making one (or part of one) out of card? Does it seem to fit together?
$endgroup$
– timtfj
Dec 1 '18 at 1:57
add a comment |
$begingroup$
I am a bit confused. If you go around a vertex, in what order do the faces appear in your suggested construction?
$endgroup$
– fedja
Dec 1 '18 at 1:38
$begingroup$
In other words, can you draw the planar graph corresponding to your suggestion? I tried and I couldn't make all vertices look the same (remember that semi-regularity requires a group of isometries transitive on the vertices, not just the same unordered collections of faces at each vertex) but, perhaps, I'm misunderstanding your suggestion.
$endgroup$
– fedja
Dec 1 '18 at 1:49
1
$begingroup$
Have you tried making one (or part of one) out of card? Does it seem to fit together?
$endgroup$
– timtfj
Dec 1 '18 at 1:57
$begingroup$
I am a bit confused. If you go around a vertex, in what order do the faces appear in your suggested construction?
$endgroup$
– fedja
Dec 1 '18 at 1:38
$begingroup$
I am a bit confused. If you go around a vertex, in what order do the faces appear in your suggested construction?
$endgroup$
– fedja
Dec 1 '18 at 1:38
$begingroup$
In other words, can you draw the planar graph corresponding to your suggestion? I tried and I couldn't make all vertices look the same (remember that semi-regularity requires a group of isometries transitive on the vertices, not just the same unordered collections of faces at each vertex) but, perhaps, I'm misunderstanding your suggestion.
$endgroup$
– fedja
Dec 1 '18 at 1:49
$begingroup$
In other words, can you draw the planar graph corresponding to your suggestion? I tried and I couldn't make all vertices look the same (remember that semi-regularity requires a group of isometries transitive on the vertices, not just the same unordered collections of faces at each vertex) but, perhaps, I'm misunderstanding your suggestion.
$endgroup$
– fedja
Dec 1 '18 at 1:49
1
1
$begingroup$
Have you tried making one (or part of one) out of card? Does it seem to fit together?
$endgroup$
– timtfj
Dec 1 '18 at 1:57
$begingroup$
Have you tried making one (or part of one) out of card? Does it seem to fit together?
$endgroup$
– timtfj
Dec 1 '18 at 1:57
add a comment |
1 Answer
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Did you try to build them?
As mentioned by @fedja in the comment there are 2 possibilities for the (to be used unique) vertex configuration: either 6-3-4-3 or 6-3-3-4 (cyclically each). For sure, in the second case it might be allowed to use the mirror copy as well.
Consider 6-3-4-3 for the first try. So start with one hexagon. Then you have to attach triangles to each side. Into each gap you would have to insert a square. But then the tips of those triangles already would contain a partial configuration 4-3-4, which is not compatible with the assumed configuration.
Now consider the other possibility 6-3-3-4. Then you'd have to start with a hexagon again. Because 6=2*3 and 3 is odd, you would have to attach alternatingly triangles and squares to that hexagon, and will have to insert further triangles inbetween those attached faces. But then again at the tips of the hexagon-attached triangles you'd get a partial configuration 3-3-3, which is not compatible with the assumed configuration.
--- rk
answered Dec 1 '18 at 9:17
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
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add a comment |
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$begingroup$
Did you try to build them?
As mentioned by @fedja in the comment there are 2 possibilities for the (to be used unique) vertex configuration: either 6-3-4-3 or 6-3-3-4 (cyclically each). For sure, in the second case it might be allowed to use the mirror copy as well.
Consider 6-3-4-3 for the first try. So start with one hexagon. Then you have to attach triangles to each side. Into each gap you would have to insert a square. But then the tips of those triangles already would contain a partial configuration 4-3-4, which is not compatible with the assumed configuration.
Now consider the other possibility 6-3-3-4. Then you'd have to start with a hexagon again. Because 6=2*3 and 3 is odd, you would have to attach alternatingly triangles and squares to that hexagon, and will have to insert further triangles inbetween those attached faces. But then again at the tips of the hexagon-attached triangles you'd get a partial configuration 3-3-3, which is not compatible with the assumed configuration.
--- rk
answered Dec 1 '18 at 9:17
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
$endgroup$
add a comment |
$begingroup$
Did you try to build them?
As mentioned by @fedja in the comment there are 2 possibilities for the (to be used unique) vertex configuration: either 6-3-4-3 or 6-3-3-4 (cyclically each). For sure, in the second case it might be allowed to use the mirror copy as well.
Consider 6-3-4-3 for the first try. So start with one hexagon. Then you have to attach triangles to each side. Into each gap you would have to insert a square. But then the tips of those triangles already would contain a partial configuration 4-3-4, which is not compatible with the assumed configuration.
Now consider the other possibility 6-3-3-4. Then you'd have to start with a hexagon again. Because 6=2*3 and 3 is odd, you would have to attach alternatingly triangles and squares to that hexagon, and will have to insert further triangles inbetween those attached faces. But then again at the tips of the hexagon-attached triangles you'd get a partial configuration 3-3-3, which is not compatible with the assumed configuration.
--- rk
answered Dec 1 '18 at 9:17
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
$endgroup$
add a comment |
$begingroup$
Did you try to build them?
As mentioned by @fedja in the comment there are 2 possibilities for the (to be used unique) vertex configuration: either 6-3-4-3 or 6-3-3-4 (cyclically each). For sure, in the second case it might be allowed to use the mirror copy as well.
Consider 6-3-4-3 for the first try. So start with one hexagon. Then you have to attach triangles to each side. Into each gap you would have to insert a square. But then the tips of those triangles already would contain a partial configuration 4-3-4, which is not compatible with the assumed configuration.
Now consider the other possibility 6-3-3-4. Then you'd have to start with a hexagon again. Because 6=2*3 and 3 is odd, you would have to attach alternatingly triangles and squares to that hexagon, and will have to insert further triangles inbetween those attached faces. But then again at the tips of the hexagon-attached triangles you'd get a partial configuration 3-3-3, which is not compatible with the assumed configuration.
--- rk
answered Dec 1 '18 at 9:17
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
$endgroup$
Did you try to build them?
As mentioned by @fedja in the comment there are 2 possibilities for the (to be used unique) vertex configuration: either 6-3-4-3 or 6-3-3-4 (cyclically each). For sure, in the second case it might be allowed to use the mirror copy as well.
Consider 6-3-4-3 for the first try. So start with one hexagon. Then you have to attach triangles to each side. Into each gap you would have to insert a square. But then the tips of those triangles already would contain a partial configuration 4-3-4, which is not compatible with the assumed configuration.
Now consider the other possibility 6-3-3-4. Then you'd have to start with a hexagon again. Because 6=2*3 and 3 is odd, you would have to attach alternatingly triangles and squares to that hexagon, and will have to insert further triangles inbetween those attached faces. But then again at the tips of the hexagon-attached triangles you'd get a partial configuration 3-3-3, which is not compatible with the assumed configuration.
--- rk
answered Dec 1 '18 at 9:17
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
answered Dec 1 '18 at 9:17
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
answered Dec 1 '18 at 9:17
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
answered Dec 1 '18 at 9:17
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
1,50616
add a comment |
add a comment |
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$begingroup$
I am a bit confused. If you go around a vertex, in what order do the faces appear in your suggested construction?
$endgroup$
– fedja
Dec 1 '18 at 1:38
$begingroup$
In other words, can you draw the planar graph corresponding to your suggestion? I tried and I couldn't make all vertices look the same (remember that semi-regularity requires a group of isometries transitive on the vertices, not just the same unordered collections of faces at each vertex) but, perhaps, I'm misunderstanding your suggestion.
$endgroup$
– fedja
Dec 1 '18 at 1:49
1
$begingroup$
Have you tried making one (or part of one) out of card? Does it seem to fit together?
$endgroup$
– timtfj
Dec 1 '18 at 1:57