Timeline for Half the pentagon!
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2016 at 12:06 | history | edited | Beastly Gerbil | CC BY-SA 3.0 |
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| Oct 24, 2016 at 11:32 | comment | added | Beastly Gerbil | @MOehm yeah this is flawed | |
| Oct 24, 2016 at 11:22 | comment | added | oleslaw | @Beastly Gerbil I believe the counterexample is indeed a counter to your theory as actually none of the vertices passed to your algorithm provides two equal halves. | |
| Oct 24, 2016 at 11:18 | comment | added | M Oehm | But even a horizontal line through the rightmost vertex of the inner pentagon won't create two halves of equal area. | |
| Oct 24, 2016 at 11:14 | history | edited | Beastly Gerbil | CC BY-SA 3.0 |
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| Oct 24, 2016 at 11:07 | comment | added | Beastly Gerbil | @oleslaw I wouldn't have drawn that red line though, I'd have drawn sort of horizontally | |
| Oct 24, 2016 at 10:31 | comment | added | Rosie F | @oleslaw Good counterexample. Selection of the vertex to start from in Beastly Gerbil's method won't always work, because for any candidate vertex to start from, you could just move all the vertices so that they, and the dividing line, are in oleslaw's counterexample diagram. Any method must work for any positions of the vertices that make a convex pentagon. | |
| Oct 24, 2016 at 10:20 | comment | added | oleslaw | How about this pentagon: i.sstatic.net/i9Hv6.png ? I believe I used your algorithm, but the halves are certainly not equal areawise. Maybe selection of the vertex to start from is important? | |
| Oct 24, 2016 at 10:17 | history | edited | Beastly Gerbil | CC BY-SA 3.0 |
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| Oct 24, 2016 at 10:15 | comment | added | Beastly Gerbil | @oleslaw well I was using an irregular pentagon in the pic and it has to be convex | |
| Oct 24, 2016 at 10:11 | comment | added | oleslaw | This wouldn't work for a "less perfect" pentagon. At least with this amount of details. | |
| Oct 24, 2016 at 10:08 | history | answered | Beastly Gerbil | CC BY-SA 3.0 |
