- Nice to "meet" old friend Jean-Pierre.... !

Now, I will rewrite the locus problem and cc Bernard, in the case it is not

listed

in the properties of the cubic.

Let ABC be a triangle, P a point and A'B'C' the circumcevian triangle of P.

Let La,Lb,Lc be the parallels to BC,CA,AB through A',B',C', resp. and

Ma,Mb,Mc the reflections of La,Lb,Lc in AA',BB',CC', resp..

Denote:

A* = Ma /\ (circumcircle - A') ie the other than A' intersection of Ma with

the circumcircle.

Similarly B*,C*.

As Francisco found (quoted below), the locus of P such that ABC, A*B*C* are

perspective is Ehrmann strophoid (K025)

APH

On Mon, Mar 11, 2013 at 9:56 AM, Francisco Javier

<garciacapitan@...>wrote:

> **

[Non-text portions of this message have been removed]

>

>

> In this case,

>

> 1. The locus is Ehrmann strophoid (K025)

> 2. The calculation takes a long time. I couldn't get the answer.

>

>

> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...>

> wrote:

> >

> > So only the third is somehow interesting...

> >

> > How about to make it a little complicated?

> >

> > 4. Let A3, B3, C3 be the intersections with circumcircle of the

> reflections

> > in AA', BB', CC' of the parallels to BC, CA, AB through A',B',C', resp.

> >

> > Which is the locus of P such that

> > 1. ABC, A3B3C3

> > 2. ABC, triangle bounded by (A'A3, B'B3, C'C3)

> >

> > are perspectivs?

> >

> > APH

> > On Sun, Mar 10, 2013 at 11:22 PM, Francisco Javier

> > <garciacapitan@...>wrote:

> >

> > > **

>

> > >

> > >

> > > We have perspectivity in the three cases for any P:

> > > 1. The perspector is the isogonal conjugate of P

> > > 2. The perspector is the circumcenter.

> > > 3. The perspector is the isogonal conjugate of the reflection of P on

> the

> > > orthocenter.

> > >

> > > Best regards,

> > >

> > > Francisco Javier.

> > >

> > >

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

> > > >

> > > > Let ABC be a triangle and A'B'C' the circumcevian

> > > > triangle of point P.

> > > > The parallels to BC,CA,AB through A',B',C'

> > > > intersect the circumcircle again at A1,B1,C1,resp.

> > > > The perpendiculars to AA', BB', CC' at A',B',C',

> > > > intersect the circumcircle at A2,B2,C2, resp.

> > > >

> > > > Which is the locus of P such that:

> > > >

> > > > 1. ABC, A1B1C1

> > > >

> > > > 2. ABC, A2B2C2

> > > >

> > > > 3. ABC, triangle bounded by (A1A2, B1B2, C1C2)

> > > >

> > > > are perspective?

> > > >

> > > > APH

>

>

>

- In message #21715 I wrote:

We have perspectivity in the three cases for any P:

1. The perspector is the isogonal conjugate of P

2. The perspector is the circumcenter.

3. The perspector is the isogonal conjugate of the reflection of P on the

orthocenter.

There is a typo here because it should be:

3. The perspector is the isogonal conjugate of the reflection of P on the

CIRCUMCENTER.

(thanks to Angel Montesdeoca for pointing it out)

Best regards,

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:

>

> We have perspectivity in the three cases for any P:

> 1. The perspector is the isogonal conjugate of P

> 2. The perspector is the circumcenter.

> 3. The perspector is the isogonal conjugate of the reflection of P on the orthocenter.

>

> Best regards,

>

> Francisco Javier.

>

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

> >

> > Let ABC be a triangle and A'B'C' the circumcevian

> > triangle of point P.

> > The parallels to BC,CA,AB through A',B',C'

> > intersect the circumcircle again at A1,B1,C1,resp.

> > The perpendiculars to AA', BB', CC' at A',B',C',

> > intersect the circumcircle at A2,B2,C2, resp.

> >

> > Which is the locus of P such that:

> >

> > 1. ABC, A1B1C1

> >

> > 2. ABC, A2B2C2

> >

> > 3. ABC, triangle bounded by (A1A2, B1B2, C1C2)

> >

> > are perspective?

> >

> > APH

> >

>