- 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:

> **

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

>

>

> 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 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

> >

>

>

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

> - 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

> >

>