Timeline for Reordering vertices of a polygon
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 3, 2018 at 4:16 | comment | added | Wlod AA | @JosephO'Rourke, the very first characteristic and essential assumption of the QUESTION reads: strictly convex internal angles only in three vertices v1,v2,v3 (resp. v′1,v′2,v′3).--yes, it says three (not ten or whatever). | |
| May 31, 2018 at 11:05 | comment | added | Joseph O'Rourke | @user389604: Apologies for not understanding your question. I thought the correspondence between $Q$ and $Q'$ implied the same ordering of their vertices. | |
| May 31, 2018 at 8:18 | comment | added | user101163 | The distance between some pair of vertices of $Q''$ can be greater than the distance between corresponding vertices of $Q'$, but not greater than the corresponding vertices of $Q$. I am not trying to expand $Q'$ in order to make it convex, but rather to move its vertices inside $Q'$ (fixing three $v_i'$) in order to reorder them | |
| May 31, 2018 at 8:15 | comment | added | user101163 | Thank you for your answer, but I don't see how the result you cite could be applied to my case. Vertices of polygons $Q$ and $Q'$ can have different order in the sense that if $\overline{xy}$ is a side of $Q$ then $\overline{x'y'}$ can be a diagonal of $Q'$. I am searching for another polygon $Q''$ entirely contained in $Q'$ and which shares with it three vertices $v_i'=v_i''$, $i=1,2,3$ (the ones of $Q'$ with strictly convex internal angles). | |
| May 31, 2018 at 1:02 | history | answered | Joseph O'Rourke | CC BY-SA 4.0 |