- In Hyacinthos message #7720, I wrote:

>> >> Let Q'a,Q'b,Q'c be the reflections

It is an irreducible sextic. The equation is a

>> >> of Qa,Qb,Qc in BC,CA,AB.

>> >>

>> >> Which are is locus of P such that

>> >>

>> >> (5.2) ABC, Q'aQ'bQ'c are perspective?

>>

>> From the first 6 triangle centers, I have found

>> only X(1) on this locus. The locus seems to

>> meet the Euler line 4 times, so it cannot be a

>> cubic!

simple horror.

Now I will continue with some other loci from

Antreas' message #7714:

>> Let Ha, Ga, (to ask only for two simple points)

For (7.1), the locus is the whole plane, since the

>> be the Orthocenter, Centroid of QaBC

>> and similarly Hb, Gb; Hc,Gc

>>

>> Which are the loci of P such that:

>>

>> (7.1) ABC, HaHbHc

>> (7.2) ABC, GaGbGc

>>

>> are perspective?

quadrilateral BPCHa is a parallelogram, and thus

Ha is the reflection of P in the midpoint of BC,

and similarly for Hb and Hc. It is well-known that

now the perspector of triangles ABC and HaHbHc

will be the complement of P.

The locus in (7.2)... I am seeing that X(4) lies

on the locus, while X(1) and X(3) don't lie, but I

have not tested more points.

Here is a related locus with anticevian instead of

antipedal triangle: If Pa, Pb, Pc are the vertices

of the anticevian triangle of a point P, and Ga,

Gb, Gc are the centroids of triangles BPaC, CPbA,

APcB, then the triangles ABC and GaGbGc are

perspective if and only if the point P lies on a

degenerate cubic consisting of the three medians

of triangle ABC.

Sincerely,

Darij Grinberg - Dear Paul

>[APH]: Let ABC be a triangle P a point and PaPbPc the pedal

Probably this variation is interesting:

> triangle of P.

>

> Let Oa, Ob, Oc be the circumcenters of the triangles

> APbPc, BPcPa, CPaPb, resp.

>

> Which is the locus of P such that PaPbPc, OaObOc

> are perspective?

>

> [PY]: This is the Neuberg cubic.

>

> [PY] This is already in Bernard's CTC, K001, attributed to Jean-Pierre.

Let PaPbPc, P*aP*bP*c be the pedal triangles of the

isogonal points P,P*.

And let Oa,Ob,Oc / Qa, Qb, Qc be the circumcenters of the

triangles APbPc, BPcPa, CPaPb / AP*bP*c, BP*cP*a, CP*aP*b

Which is the locus of P such that the triangles

1. PaPbPc, QaQbQc

2. P*aP*bP*c, OaObOc

are perspective?

Greetings from Athens

Antreas

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