Timeline for A pentagon that can measure the first 7 integer distances
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 24, 2022 at 0:17 | comment | added | noedne | @Bass You're right, that only holds for convex pentagons. | |
| Jul 23, 2022 at 23:40 | comment | added | Bass | The proof isn't quite convincing though: having three vertices on the same line doesn't necessarily make a pentagon degenerate. | |
| Jul 22, 2022 at 22:25 | vote | accept | Dmitry Kamenetsky | ||
| Jul 22, 2022 at 16:39 | comment | added | Bass | Here's a degenerate pentagon that can count up to 9: i.sstatic.net/slrz7.png | |
| Jul 22, 2022 at 14:31 | comment | added | noedne | @DmitryKamenetsky No because the segment of length 1 together with the 3 vertices not on that segment form 3 triangles with 3 distinct pairs of segments. Each pair can have at most 1 integral length by the triangle inequality (unless we permit degenerate pentagons), so we can have at most 10 - 3 = 7 integral lengths. | |
| Jul 22, 2022 at 14:27 | comment | added | Dmitry Kamenetsky | Very nice work! Do you think it's possible to add distance 8? | |
| Jul 22, 2022 at 14:20 | history | answered | noedne | CC BY-SA 4.0 |
