Timeline for Is this bounded?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Dec 6, 2012 at 14:26 | history | edited | Palt | CC BY-SA 3.0 |
added 1 characters in body
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| Dec 6, 2012 at 14:24 | comment | added | Palt | Squark, $v_{m+1}$ is also a vertex in the whole graph, sorry for unclearness. In an equilateral polygon, the smallest angle would be an interior angle of the triangle that has $v_{m+1}$ as a vertex, so it would be less than the one you expressed. | |
| Dec 6, 2012 at 14:20 | answer | added | Ben Barber | timeline score: 5 | |
| Dec 6, 2012 at 13:56 | comment | added | Vanessa | Consider an equilateral polygon. $alpha_m = (m - 2) \pi / m$, thus $m^2 alpha_m = m(m - 2) \pi$ which is unbounded | |
| Dec 6, 2012 at 12:51 | history | asked | Palt | CC BY-SA 3.0 |