L(t,P) = Trilinear Polar of triangle t = (ABC) with respect to point P. But t is constructed by errecting isosceli triangles, similar to u=(GHI), on the sides of another triangle s = (DEF). Find the envelope of L(t,P) as the isosceles changes its altitude.
A problem not related to the well known Kiepert Parabola and having a locus which may possess cusps, depending on the location of P and the shape of triangle DEF. Kiepert's Parabola of DEF results as the envelope of trilinear polars LP with respect to DEF, where P is the center of perspective of triangles DEF and ABC (hence Lp is their perspective axis).
![[alogo]](alogo.png){kind=small}
Trilinear polar envelope ![[0_0]](Trilinear_Polar_Envelope_files/Trilinear_Polar_Envelope_0_0_0.png)
![[0_1]](Trilinear_Polar_Envelope_files/Trilinear_Polar_Envelope_0_0_1.png)
![[0_2]](Trilinear_Polar_Envelope_files/Trilinear_Polar_Envelope_0_0_2.png)