Timeline for Intersection points of closed curves inscribed in a convex polygon
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2014 at 21:07 | comment | added | Alan Horwitz | OK, thanks Anton. I'll look at it more closely to convince myself. | |
| Oct 25, 2014 at 21:00 | comment | added | Anton Petrunin | yes, yes, sure. | |
| Oct 25, 2014 at 20:46 | comment | added | Alan Horwitz | That appears to be a nice way to prove what I was looking for, though I want to look it over more for now to make sure it does not implicitly assume anywhere what one wants to prove. Is the statement that the boundary contains at least $[\dfrac{n+1}{2}]$ arcs of $C_1$ just based on the fact that $C_1$ is tangent at the $n$ sides of D ? | |
| Oct 25, 2014 at 14:17 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced deprecated tag 'geometry'
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| Oct 24, 2014 at 23:00 | review | Close votes | |||
| Oct 25, 2014 at 13:02 | |||||
| Oct 24, 2014 at 22:49 | comment | added | Anton Petrunin | Look at the union of the rigions bounded by $C_1$ and $C_2$ and note that its boundry contains at least $[\tfrac{n+1}2]$ arcs of $C_1$; here $n$ is the number of sides of the polygon. Each end of such arc is a point of intersection, so we will get at least $2{\cdot}[\tfrac{n+1}2]$ of them. | |
| Oct 24, 2014 at 21:23 | history | edited | Alan Horwitz | CC BY-SA 3.0 |
added 48 characters in body
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| Oct 24, 2014 at 21:11 | review | First posts | |||
| Oct 24, 2014 at 21:16 | |||||
| Oct 24, 2014 at 21:07 | history | asked | Alan Horwitz | CC BY-SA 3.0 |