Timeline for Are there infinitely many "generalized triangle vertices"?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 17, 2021 at 17:33 | history | bounty ended | Noah Schweber | ||
| Nov 14, 2021 at 5:23 | comment | added | Blue | "is it at all plausible that being on the Neuberg cubic is a necessary condition?" I've been investigating the Neuberg cubic aspect; conveniently, if $P$ is on the cubic for $\triangle ABC$, then $A$, $B$, $C$ are all on the cubic for $\triangle PBC$ (likewise, $\triangle APC$ and $\triangle ABP$), so this is a not-unreasonable avenue to pursue. It's possible to parameterize the two parts of the cubic as a variable point moves along the Euler line (see the very bottom of Gibert's K001 page), but the algebra is a bit messy. | |
| Nov 13, 2021 at 22:59 | comment | added | Peter Taylor | They're certainly harder to find than I expected, but there's no evidence to suggest that they aren't extremely common even in the rational class that I considered at higher degrees. The comment about the Neuberg cubic was just looking for anything they had in common: I don't see why they would necessarily be on it. It's probably possible to calculate the $\mu$ corresponding to an arbitrary point on the cubic and apply a similar analysis to the one described in the answer, but I don't know when I'll have time to do that. | |
| Nov 13, 2021 at 20:14 | comment | added | Noah Schweber | +1, even though this didn't work out it's quite neat. I'm now starting to wonder if there might actually be only finitely many. You mentioned earlier the Neuberg cubic - is it at all plausible that being on the Neuberg cubic is a necessary condition? | |
| Nov 13, 2021 at 16:34 | history | answered | Peter Taylor | CC BY-SA 4.0 |