Timeline for Fair cutting of the plane with lines
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2021 at 9:05 | comment | added | Yaakov Baruch | Please see my other answer, where "wiggling" is only used to simplify the problem enough to make it solvable by easy cartesian coordinate computations. | |
| Mar 25, 2021 at 7:57 | comment | added | Yaakov Baruch | Indeed for the contraction idea to work it needs to happen withing a compact family of pentagons (then an accumulation point would exist and it would have to have min ear=max ear). But so far I don't see how to define such compact family that would be clearly closed under equalizations. | |
| Mar 25, 2021 at 0:36 | comment | added | Pierre PC | @YaakovBaruch I can see it is a strict contraction, but it does not have to converge, does it? It could look something like $[m-1/n,M+1/n]$ with $m<M$. | |
| Mar 24, 2021 at 23:56 | comment | added | Yaakov Baruch | The rotation (with ensuing shift) can be pushed to the point where the ear being reduced can be reduced as far as 0. Therefore there is an intermediate wiggling that equalizes the 2 adjacent ears. Then, a sequence of 4 equalizations clockwise is a necessarily a strict contraction of the compact interval [min ear, max ear] (unless min=max already) and therefore its iteration must converge to a unique fixed point, which can only be when min=max. | |
| Mar 24, 2021 at 21:51 | comment | added | Pierre PC | @YaakovBaruch Oh I see, very nice! I don't see an easy argument to see that the procedure converges. Side note: we can get rid of the degenerate ears by starting with ears left and right from a non-degenerate one (at least one of them must exist) and then working our way around the star. | |
| Mar 24, 2021 at 21:15 | comment | added | Yaakov Baruch | When you rotate clockwise you reduce the ear on the right and increase the one on the left. The shift does not change that because it is limited by having to keep the intersection with the old edge inside it (or else the pentagon area itself would be changed). I hope this clarifies. | |
| Mar 24, 2021 at 21:06 | comment | added | Pierre PC | @YaakovBaruch I don't understand why your rotating-and-shifting procedure could not decrease the area of both ears. The sum of the areas of the two ears does not have to be preserved, does it? | |
| Mar 24, 2021 at 20:40 | comment | added | Yaakov Baruch | I think what I meant back then was this: if we rotate an edge of the pentagon a bit around its midpoint, and then also parallel-shift it so that the area of the corresponding ear (and therefore of the pentagon too) is preserved, we increase the area of one of the 2 adjacent ears and decrease the area of the other. Always chose so that you increase the smaller and decrease the larger one, so that the smallest never decreases. Applying this procedure infinitely many times should lead to all 5 ears having equal area (then it's easy to see that such area is less than the pentagon's). | |
| S Mar 24, 2021 at 20:01 | history | answered | Pierre PC | CC BY-SA 4.0 | |
| S Mar 24, 2021 at 20:01 | history | made wiki | Post Made Community Wiki by Pierre PC |