- View SourceLet ABC be a triangle, P1,P2 two isogonal conjgate points,

A1B1C1, A2B2C2 the pedal triangles of P1,P2, resp. and H1,H2

the orthocenters of A1B1C1, A2B2C2, resp.

The cubics passing through the eight points A1,B1,C1,A2,B2,C2,H1,H2,

pass also (in general) through a ninth point P*.

1. Which is P* in terms of the h.coordinates of P1 = (x:y:z)?

2. For which P1's the ninth point P* is the common circumcenter

of the triangles A1B1C1, A2B2C2?

APH - View SourceDear Antreas,

I can answer at the second question.

In the pencil of cubics by A1,B1,C1,A2,B2,C2,H1,H2,

there is the degenerated cubic formed by the pedal circle of P1,P2 and by the line H1H2

hence your point P* must be on this cubic too.

If this point is M, the common circumcenter of the triangles A1B1C1, A2B2C2,

then

1) M is on the pedal circle, this circle is a line (Simson line),and your locus is the circumcircle (O) of ABC.

or

2) M is on the line H1H2.

The locus is, in this case, a beautiful isogonal sextic, which contains the points

ABC (triple probably),

the incenter and excenters of ABC,

the Appolonius centers of ABC,

the tangential points, intersection of the tangents to the circumcircle (O) of ABC.

The points of the precevian triangle of ABC (intersections of the parallel to the sides

by the edges).

The cyclic points.

Total: 24 points.

It intersects the sides of ABC and (O) at six points differents of ABC.

May be, Bernard Gibert can give more information on this pretty curve.

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:

>

> Let ABC be a triangle, P1,P2 two isogonal conjgate points,

> A1B1C1, A2B2C2 the pedal triangles of P1,P2, resp. and H1,H2

> the orthocenters of A1B1C1, A2B2C2, resp.

>

> The cubics passing through the eight points A1,B1,C1,A2,B2,C2,H1,H2,

> pass also (in general) through a ninth point P*.

>

> 1. Which is P* in terms of the h.coordinates of P1 = (x:y:z)?

>

> 2. For which P1's the ninth point P* is the common circumcenter

> of the triangles A1B1C1, A2B2C2?

>

> APH

> - View SourceI forgot to say that it is a necessary, but may be not sufficient condition for P1.

Thank's

Fred

--- In Hyacinthos@yahoogroups.com, "fredlangch" <fred.lang@...> wrote:

>

> Dear Antreas,

> I can answer at the second question.

>

> In the pencil of cubics by A1,B1,C1,A2,B2,C2,H1,H2,

> there is the degenerated cubic formed by the pedal circle of P1,P2 and by the line H1H2

> hence your point P* must be on this cubic too.

> If this point is M, the common circumcenter of the triangles A1B1C1, A2B2C2,

> then

> 1) M is on the pedal circle, this circle is a line (Simson line),and your locus is the circumcircle (O) of ABC.

>

> or

>

> 2) M is on the line H1H2.

> The locus is, in this case, a beautiful isogonal sextic, which contains the points

> ABC (triple probably),

> the incenter and excenters of ABC,

> the Appolonius centers of ABC,

> the tangential points, intersection of the tangents to the circumcircle (O) of ABC.

> The points of the precevian triangle of ABC (intersections of the parallel to the sides

> by the edges).

> The cyclic points.

>

> Total: 24 points.

>

> It intersects the sides of ABC and (O) at six points differents of ABC.

>

> May be, Bernard Gibert can give more information on this pretty curve.

>

>

> --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:

> >

> > Let ABC be a triangle, P1,P2 two isogonal conjgate points,

> > A1B1C1, A2B2C2 the pedal triangles of P1,P2, resp. and H1,H2

> > the orthocenters of A1B1C1, A2B2C2, resp.

> >

> > The cubics passing through the eight points A1,B1,C1,A2,B2,C2,H1,H2,

> > pass also (in general) through a ninth point P*.

> >

> > 1. Which is P* in terms of the h.coordinates of P1 = (x:y:z)?

> >

> > 2. For which P1's the ninth point P* is the common circumcenter

> > of the triangles A1B1C1, A2B2C2?

> >

> > APH

> >

> - View Source[APH]
> > Let ABC be a triangle, P1,P2 two isogonal conjgate points,

[Fred]

> > A1B1C1, A2B2C2 the pedal triangles of P1,P2, resp. and H1,H2

> > the orthocenters of A1B1C1, A2B2C2, resp.

> >

> > The cubics passing through the eight points

> > A1,B1,C1,A2,B2,C2,H1,H2,

> > pass also (in general) through a ninth point P*.

> >

> > 1. Which is P* in terms of the h.coordinates of P1 = (x:y:z)?

> >

> > 2. For which P1's the ninth point P* is the common circumcenter

> > of the triangles A1B1C1, A2B2C2?

> I can answer at the second question.

Dear Fred

>

> In the pencil of cubics by A1,B1,C1,A2,B2,C2,H1,H2,

> there is the degenerated cubic formed by the pedal circle

> of P1,P2 and by the line H1H2

> hence your point P* must be on this cubic too.

> If this point is M, the common circumcenter of the

> triangles A1B1C1, A2B2C2,

> then

> 1) M is on the pedal circle, this circle is a line

> (Simson line),and your locus is the circumcircle (O) of ABC.

>

> or

>

> 2) M is on the line H1H2.

> The locus is, in this case, a beautiful isogonal sextic,

> which contains the points

> ABC (triple probably),

> the incenter and excenters of ABC,

> the Appolonius centers of ABC,

> the tangential points, intersection of the tangents

> to the circumcircle (O) of ABC.

> The points of the precevian triangle of ABC (intersections

> of the parallel to the sides

> by the edges).

> The cyclic points.

>

> Total: 24 points.

>

> It intersects the sides of ABC and (O) at six points differents

> of ABC.

>

> May be, Bernard Gibert can give more information on this

> pretty curve.

Nice to hear from you!

If I understand correctly, if P1 lies on the isogonal sextic,

then the pedal triangle A1B1C1 of P1 and the pedal triangle A2B2C2

of P2 [=isogonal conjugate of P1] share the same Euler Line.

We can ask similar questions for other OX lines (other than OH)

of the triangles A1B1C1, A2B2C2.

Brocard axis, for example:

Which is the locus of P1 such that A1B1C1, A2B2C2 share

the same Brocard axis?

APH