Timeline for An exterior angle theorem for n-dimensional polytopes?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 10, 2010 at 19:52 | vote | accept | user2498 | ||
| Aug 25, 2010 at 6:16 | vote | accept | user2498 | ||
| Oct 10, 2010 at 19:52 | |||||
| Aug 25, 2010 at 3:45 | answer | added | Allen Knutson | timeline score: 3 | |
| Aug 25, 2010 at 0:39 | comment | added | Gerry Myerson | @Jack, your TeX isn't working on my screen, but I get the drift, and of course you are right, if one is willing to accept the notion of the measure of an (exterior) angle being negative. | |
| Aug 24, 2010 at 16:37 | comment | added | Jack Lee | Gerry: For a simple polygon (no self-crossings), if you define the exterior angle measure to be π minus the interior angle measure at the same vertex, with the understanding that the interior angle measure is greater than π at a concave vertex, then the sum of the exterior angle measures will always be 2π. | |
| Aug 24, 2010 at 8:55 | answer | added | Gjergji Zaimi | timeline score: 1 | |
| Aug 24, 2010 at 6:56 | answer | added | Benoît Kloeckner | timeline score: 3 | |
| Aug 24, 2010 at 6:52 | comment | added | Gerry Myerson | I think that for "non-weird" polygons, you want convex polygons. Non-convex polygons, even if they have no self-crossings, can have exterior angles summing to more than $2\pi$. | |
| Aug 24, 2010 at 6:44 | history | asked | user2498 | CC BY-SA 2.5 |