Timeline for The Symmetry of a Soccer Ball
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 30, 2010 at 19:10 | comment | added | Gordon Williams | A follow up to Tomaz Pisanski's question: What if we allow skew regular hexagons, i.e., non-planar regular hexagons? An example of this sort of thing is the Petrie polygon of the cube. Note that we can't have non-planar skew regular pentagons because of parity. | |
| Mar 29, 2010 at 20:56 | comment | added | Tomaž Pisanski | I do not see where the conditions (2) and (3) are used in the proof. If you relax the conditions of the problem, an infinite hexagonal lattice would be the third solution. Are there any other solutionos possible, if (2a) more than three polygons may meet at a vertex, (3a) P is topologically a closed surface (4) P may self-intersect? | |
| Mar 29, 2010 at 20:17 | comment | added | Anton Petrunin | I added something, hope it is better now. | |
| Mar 29, 2010 at 20:16 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
added 178 characters in body
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| Mar 29, 2010 at 19:10 | comment | added | Sergei Ivanov | What is missing in Anton's answer is the fact that all vertices must have the same type. This can be proved as follows: let $[pq]$ be a common edge of faces $A$ and $B$. Since $A$ and $B$ are regular polygons, the angle between their remaining edges at $p$ is the same as at $q$. Hence the face meeting $A$ and $B$ at $p$ has the same angles as the one meeting $A$ and $B$ at $q$. Therefore $p$ and $q$ are of the same type. This is so for every pair of adjacent vertices and hence for all vertices. | |
| Mar 29, 2010 at 19:10 | vote | accept | Bill Kronholm | ||
| Mar 29, 2010 at 19:01 | comment | added | Sergei Ivanov | Anton, you are too brief. I'm afraid that those who can understand this proof do not need it. | |
| Mar 29, 2010 at 18:59 | vote | accept | Bill Kronholm | ||
| Mar 29, 2010 at 19:00 | |||||
| Mar 29, 2010 at 18:53 | comment | added | David Eppstein | That's assuming all vertices look the same, but it's also possible to have polyhedra in which all faces are hexagons or pentagons and all vertices have degree three but e.g. some vertices have three hexagons while others have one pentagon and two hexagons. | |
| Mar 29, 2010 at 18:45 | history | answered | Anton Petrunin | CC BY-SA 2.5 |