Timeline for Incenter-of-mass of a polygon
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 23 at 6:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 24 at 4:16 | comment | added | Don Hatch | @GerryMyerson by "side-triple" I mean a subset of the set of n sides, of cardinality 3. If there are n sides, then there are (n choose 3) side triples, so a quadrilateral has (4 choose 3) = 4 of them, and a pentagon has (5 choose 3) = 10 of them. But if you pick a particular triangulation of the dual to work with, that means you're selecting a particular n-2 of those (2 of them for a quad, 3 of them for a pentagon). | |
| Jul 24 at 2:14 | comment | added | Gerry Myerson | It's not clear to me what a "side-triple" is. It's not clear why a quadrilateral has four of them, but a pentagon has only three. | |
| Jul 24 at 2:04 | history | edited | Don Hatch | CC BY-SA 4.0 |
fix typo
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| Jul 24 at 0:11 | answer | added | Don Hatch | timeline score: 2 | |
| Jul 22 at 8:45 | comment | added | Don Hatch | I figured it out. Given P, let Q be the cyclic polygon whose vertices are the outward unit normals of the sides of P. Then the weights are proportional to the triangle areas in an arbitrary triangulation of Q, weighting the incenters of the corresponding side-triples of P. The result is magically independent of the triangulation chosen. I'll write up an answer with pictures soon, and maybe a proof of triangulation-independence. | |
| Jul 19 at 20:32 | history | edited | Don Hatch | CC BY-SA 4.0 |
further clarify the ansatz
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| Jul 19 at 18:35 | history | edited | Don Hatch | CC BY-SA 4.0 |
minor clarification
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| Jul 19 at 4:10 | history | asked | Don Hatch | CC BY-SA 4.0 |