Here is an old unanswered Hyacinthos message:

APH:

Let P be a point on the plane of triangle ABC.

The parallels to AB, AC through P intersect BC at Ab, Ac resp.

Let A1,A2 be the traces of the circumcenter,incenter of PAbAc on AbAc.

Similarly we define the points B1,B2; C1,C2.

Which are the loci of P such that (1) A1B1C1 (2) A2B2C2 be in perspective with ABC?

---

I found that the loci are the following cubics:

(1) K044 - Euler Central Cubic

(2) K033 - Spieker Central Cubic

Changing circumcenter, incenter by an arbitrary point Q=(u:v:w), the locus is the cubic

(cyclic sum) (u(v+w)y z((u+v-w)y -(u-v+w)z))) = 0

Central cubic with center the complement of Q.

--- In Hyacinthos@yahoogroups.com, xpolakis@... wrote:

>

> Let P be a point on the plane of triangle ABC.

> The parallels to AB, AC through P intersect BC at Ab, Ac resp.

> Let A1,A2 be the traces of the circumcenter,incenter of PAbAc on AbAc.

> Similarly we define the points B1,B2; C1,C2.

>

> Which are the loci of P such that (1) A1B1C1 (2) A2B2C2 be in perspective

> with ABC?

>

> I found that the loci are these cubics:

>

> xcos(B+u) + zcosu ycos(C+v) + xcosv zcos(A+w) + ycosw

> ----------------- * ------------------ * ------------------ = 1

> xcos(C-u) + ycosu ycos(A-v) + zcosv zcos(B-w) + xcosw

>

> where u := B-C, v := C-A, w := A-B for the (1), and

> 2u := B-C, 2v := C-A, 2w := A-B for the (2).

>

> But maybe these are Darboux cubics of some triangles of the reference

> triangle ABC...

>

> Antreas

>