Timeline for Conjecture: Given any five points, we can always draw a pair of non-intersecting circles whose diameter endpoints are four of those points
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 29, 2023 at 3:30 | vote | accept | Dan | ||
| Nov 29, 2023 at 2:56 | answer | added | Iosif Pinelis | timeline score: 24 | |
| Nov 29, 2023 at 2:43 | comment | added | Alex Ravsky | @JackHuizenga I think to understand the condition it can be helpful to read the answer from the MSE thread linked by OP. But although I did this, I think that some pieces of the puzzle are still missed. | |
| Nov 28, 2023 at 22:16 | comment | added | Jack Huizenga | If five points are a counterexample, then any subset of four points are a counterexample to the 4 points and 2 circles version of the problem. So perhaps it is worth better understanding the condition on 4 points which makes them a counterexample. | |
| Nov 28, 2023 at 8:42 | comment | added | RavenclawPrefect | Note that since we permit tangent circles, any counterexample will be robust to epsilon perturbations, so we can safely assume any "generic" conditions we like. | |
| Nov 28, 2023 at 0:08 | history | asked | Dan | CC BY-SA 4.0 |