# Please identify this triangle septic

Let $$ABC$$ a triangle in the plane, but $$D$$ a point in (R3) space, such that the angles $$\phi=ADB=BDC=CDA$$ are equal. Let $$E$$ be the footpoint of $$D$$ in $$ABC$$. $$E(\phi)$$ describes a (irreducible) septic $$S$$ (that has 5 asymptotes, for the record).
Points that lie on $$S$$ (checked with ETC): Incenter $$I$$, Circumcenter $$U$$, Orthocenter $$H$$ (isolated), $$A,B,C$$ (double points), 1st (Fermat $$F$$) and 2nd isogonic center $$D$$.
It's near-lying to assume that $$S$$ is the Darboux septic (the only I heard of :-) of some triangle $$XYZ$$ where $$XYZ$$ is the foobar transform of $$ABC$$ but with that septic I'm still skeptic. (SCNR)
Can you help? Gibert's invaluable site went kaput, and I don't even find the original paper to the Darboux septic.

• Bernard Gibert's site has been archived on the Wayback Machine: web.archive.org/web/20190321092028/http://… . Specifically for septics, see web.archive.org/web/20171212225952/http://… . Have you tried checking the isogonal, cyclocevian etc. conjugates of your points, to see whether they lie on a simpler curve? Also, is there an equivalent condition that "stays inside the plane"? – darij grinberg Oct 31 '19 at 21:14
• Ah, the Wayback Machine. Already played with the though of looking there but was too pessimistic. :-) The description already should rule it out - Darboux' septic has triple points. – Hauke Reddmann Nov 1 '19 at 14:39
• For what it's worth, Gilbert's site is alive and well. bernard-gibert.pagesperso-orange.fr – brainjam Jan 9 at 4:25

This is just a back of an envelope calculation so not really an answer, but too long for a comment. I hope it helps: we work in $$(r,s,t)$$ space ($$p$$ and $$q$$ are constants determined by the shape of the triangle). Then your curve is the projection onto the $$(r,s)$$ plane of the algebraic variety $$p_1(r,s,t)=p_2(r,s,t), p_3(r,s,t)=p_4(r,s,t),$$ where $$p_1(r,s,t)=(r^2-r+s^2+t^2)^2((r-p)^2+(s-q)^2+t^2),$$ $$p_2(r,s,t)=((r-1)(r-p)+s(s-q)+t^2)^2(r^2+s^2+t^2),$$ $$p_3(r,s,t)=(r^2-r+s^2+t)^2((r-p)^2+(s-q)^2+t^2)$$ and $$p_4(r,s,t)= ( r^2-rp+s^2-sq+t^2 )^2(( r-1 )^2 +s^2+t^2 ) .$$ Perhaps Gröbner bases can be used to tease out what you want.