- This is the tricircular sextic

Q014 - 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z)

where S=twice the area of ABC.

(tricircular = the circular points are triple points of the curve).

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:

>

> A related locus: Q such that Q and the NPCs of BCQ, CAQ, ABQ are concyclic. This would include X(13), X(14), the bicentric pair PU(5), the circumcircle intercepts of line X(5)X(523), and the point Qi (of ABC) such that I = Qi of the cevian triangle of I.

>

> Randy

>

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

> >

> > A CIRCLE:

> >

> > Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3

> > the NPC centers of IB'C', IC'A', IA'B', resp.

> >

> > The points I, N1,N2,N3 are concyclic.

> >

> > Center of the circle?

> >

> > LOCUS:

> >

> > Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,

> > N2, N3

> > the NPC centers of PB'C', PC'A', PA'B', resp.

> >

> > Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?

> >

> > Antreas

> >

> >

> > On Wed, Apr 17, 2013 at 1:43 AM, Antreas <anopolis72@> wrote:

> >

> > > **

> > >

> > >

> > > Let ABC be a triangle and P a point.

> > >

> > > Which is the locus of P such that the NPC center of PBC

> > > lies on the line AP ?

> > >

> > > Ceva triangle variation:

> > >

> > > Let ABC be a triangle, P a point and A'B'C'

> > > the cevian triangle of P.

> > > Which is the locus of P such that the NPC center

> > > of PB'C' lies on the line APA' ?

> > > (The I is on the locus)

> > >

> > > APH

> > >

> > >

> > >

> > >

> > >

> >

> > http://anopolis72000.blogspot.com/

> >

> >

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

> >

> - Francisco Javier,

While Q014 mentions PU(5), it does not contain them, nor X(13) and X(14), which are on this curve. Also, X(80) which is on Q014, is not on this curve (X(80) and the NPCs of BCX(80), CAX(80), BCX(80) are not concyclic).

By the way, the point Qi on this curve is the point which is the incenter of its anticevian triangle, and has ETC search value 1.999434154060428. Coordinates? Its isogonal conjugate Qi* is the point which is the nine-point center of its pedal triangle (ETC search 1.142779079509848). The line QiQi* passes through X(5).

Randy

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

>

> This is the tricircular sextic

>

> Q014 - 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z)

>

> where S=twice the area of ABC.

>

> (tricircular = the circular points are triple points of the curve).

>

> Francisco Javier.

>

> --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@> wrote:

> >

> > A related locus: Q such that Q and the NPCs of BCQ, CAQ, ABQ are concyclic. This would include X(13), X(14), the bicentric pair PU(5), the circumcircle intercepts of line X(5)X(523), and the point Qi (of ABC) such that I = Qi of the cevian triangle of I.

> >

> > Randy

> >

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

> > >

> > > A CIRCLE:

> > >

> > > Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3

> > > the NPC centers of IB'C', IC'A', IA'B', resp.

> > >

> > > The points I, N1,N2,N3 are concyclic.

> > >

> > > Center of the circle?

> > >

> > > LOCUS:

> > >

> > > Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,

> > > N2, N3

> > > the NPC centers of PB'C', PC'A', PA'B', resp.

> > >

> > > Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?

> > >

> > > Antreas

> > >

> > >

> > > On Wed, Apr 17, 2013 at 1:43 AM, Antreas <anopolis72@> wrote:

> > >

> > > > **

> > > >

> > > >

> > > > Let ABC be a triangle and P a point.

> > > >

> > > > Which is the locus of P such that the NPC center of PBC

> > > > lies on the line AP ?

> > > >

> > > > Ceva triangle variation:

> > > >

> > > > Let ABC be a triangle, P a point and A'B'C'

> > > > the cevian triangle of P.

> > > > Which is the locus of P such that the NPC center

> > > > of PB'C' lies on the line APA' ?

> > > > (The I is on the locus)

> > > >

> > > > APH

> > > >

> > > >

> > > >

> > > >

> > > >

> > >

> > > http://anopolis72000.blogspot.com/

> > >

> > >

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

> > >

> >

> - My apologies. I misread Francisco Javier's post below. As he explained privately, he meant the difference:

LOCUS = SEXTIC(Q014) - SEXTIC(S 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z) )

Question: does this locus contain any ETC centers or bicentric pairs besides X(13), X(14), and PU(5)? Are the circumcircle intercepts of line X(5)X(523) triangle centers or another bicentric pair?

Randy

--- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:

>

> Francisco Javier,

>

> While Q014 mentions PU(5), it does not contain them, nor X(13) and X(14), which are on this curve. Also, X(80) which is on Q014, is not on this curve (X(80) and the NPCs of BCX(80), CAX(80), BCX(80) are not concyclic).

>

> By the way, the point Qi on this curve is the point which is the incenter of its anticevian triangle, and has ETC search value 1.999434154060428. Coordinates? Its isogonal conjugate Qi* is the point which is the nine-point center of its pedal triangle (ETC search 1.142779079509848). The line QiQi* passes through X(5).

>

> Randy

>

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

> >

> > This is the tricircular sextic

> >

> > Q014 - 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z)

> >

> > where S=twice the area of ABC.

> >

> > (tricircular = the circular points are triple points of the curve).

> >

> > Francisco Javier.

> >

> > --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@> wrote:

> > >

> > > A related locus: Q such that Q and the NPCs of BCQ, CAQ, ABQ are concyclic. This would include X(13), X(14), the bicentric pair PU(5), the circumcircle intercepts of line X(5)X(523), and the point Qi (of ABC) such that I = Qi of the cevian triangle of I.

> > >

> > > Randy

> > >

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

> > > >

> > > > A CIRCLE:

> > > >

> > > > Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3

> > > > the NPC centers of IB'C', IC'A', IA'B', resp.

> > > >

> > > > The points I, N1,N2,N3 are concyclic.

> > > >

> > > > Center of the circle?

> > > >

> > > > LOCUS:

> > > >

> > > > Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,

> > > > N2, N3

> > > > the NPC centers of PB'C', PC'A', PA'B', resp.

> > > >

> > > > Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?

> > > >

> > > > Antreas

> > > >

> > > >

> > > > On Wed, Apr 17, 2013 at 1:43 AM, Antreas <anopolis72@> wrote:

> > > >

> > > > > **

> > > > >

> > > > >

> > > > > Let ABC be a triangle and P a point.

> > > > >

> > > > > Which is the locus of P such that the NPC center of PBC

> > > > > lies on the line AP ?

> > > > >

> > > > > Ceva triangle variation:

> > > > >

> > > > > Let ABC be a triangle, P a point and A'B'C'

> > > > > the cevian triangle of P.

> > > > > Which is the locus of P such that the NPC center

> > > > > of PB'C' lies on the line APA' ?

> > > > > (The I is on the locus)

> > > > >

> > > > > APH

> > > > >

> > > > >

> > > > >

> > > > >

> > > > >

> > > >

> > > > http://anopolis72000.blogspot.com/

> > > >

> > > >

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

> > > >

> > >

> >

> - [APH Hyacinthos 21971]:

A CIRCLE:

Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3

the NPC centers of IB'C', IC'A', IA'B', resp.

The points I, N1,N2,N3 are concyclic.

Center of the circle?

[César Lozada]:

Yes, they are concyclic.

The center X of the circle has trilinears:

2*cos(A)+4*sin(3*A/2)*cos(B/2-C/2)+ cos(B-C)+2 : :

[....]

****************************2. Stathis Koutras proved it synthetically in the FB group "Romantics of Geometry" #2653. How about the extraversions?[APH Hyacinthos 21968]

Let Ia, Ib, Ic be the excenters of ABC.

Denote:

A1B1C1, A2B2C2, A3B3C3 = the cevian triangles of Ia, Ib, Ic, resp.

I1 = the center of the circle (Ia, N11, N12, N13), where

N11 = the NPC center of IaB1C1,

N12 = the NPC center of IaC1A1,

N13 = the NPC center of IaA1B1.

Similarly I2, I3.************What properties have the circles (I1), (I2), (I3) ?Are the triangles IaIbIc, I1I2I3 perspective ?(Are IaI1, IbI2, IcI3 parallels?)APH - [APH Hyacinthos 21971]:

A CIRCLE:

Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3

the NPC centers of IB'C', IC'A', IA'B', resp.

The points I, N1,N2,N3 are concyclic.

Center of the circle?

[César Lozada]:

Yes, they are concyclic.

The center X of the circle has trilinears:

2*cos(A)+4*sin(3*A/2)*cos(B/2- C/2)+ cos(B-C)+2 : :

[....]

****************************2. Stathis Koutras proved it synthetically in the FB group "Romantics of Geometry" #2653. How about the extraversions?[APH Hyacinthos 21968]

Let Ia, Ib, Ic be the excenters of ABC.

Denote:

A1B1C1, A2B2C2, A3B3C3 = the cevian triangles of Ia, Ib, Ic, resp.

I1 = the center of the circle (Ia, N11, N12, N13), where

N11 = the NPC center of IaB1C1,

N12 = the NPC center of IaC1A1,

N13 = the NPC center of IaA1B1.

Similarly I2, I3.************What properties have the circles (I1), (I2), (I3) ?Are the triangles IaIbIc, I1I2I3 perspective ?(Are IaI1, IbI2, IcI3 parallels to II0 = X1X5453 ?If so then they concur at the infinity point of X1X5453 )APH