- Here is how, from a property of H, we get some, more or

less complicated, generalizations....

Let ABC be a triangle and A'B'C' the pedal/cevian triangle

of H (orthic triangle)

Denote:

A" = AA' /\ B'C', B" = BB' /\ C'A', C" = CC' /\ A'B'

La = the redical axis (common chord) of the circles ((B", B"A'),

(C", C"A')) (ie the circles centered at B",C" with radii B"A',

C"A', resp.)

Lb = the redical axis (common chord) of the circles ((C", C"B'),

(A", A"B'))

Lc = the redical axis (common chord) of the circles ((A", A"C'),

(B", B"C'))

The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic.

The parallelogic center (ABC) y(*) (La,Lb,Lc) is the center

of the pedal/cevian circle of H (ie N)

(*) I like the "y" that sometimes our Spanish friends use instead

of the English "and"!

[for two reasons: 1. The word is monogrammatic, that is, of one

letter only 2. "y' is named in Latin as "y Graeca" = "Greek y"]

Which is the other parallelogic centers?

Gerneralizations:

Given that for H, its pedal and cevian triangle are same,

and also that the triangle bounded by the radical axes could be 0

(ie to be concurrent the lines), we have these possible loci:

Let ABC be a triangle P a point and A'B'C' the (1) cevian or (2)

the pedal triangle of P.

Denote:

A" = AA' /\ B'C', B" = BB' /\ C'A', C" = CC' /\ A'B'

La = the redical axis (common chord) of the circles ((B", B"A'),

(C", C"A')) (ie the circles centered at B",C" with radii B"A', C"A',

resp.)

Lb = the redical axis (common chord) of the circles ((C", C"B'),

(A", A"B'))

Lc = the redical axis (common chord) of the circles ((A", A"C'),

(B", B"C'))

Which is the locus of P such that:

i. La,Lb,Lc are concurrent?

ii. The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

APH - I only give the cevian case (for the pedal case, the calculations delayed too much)

i. La,Lb,Lc are concurrent?

The sextic Q058

ii. The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

The cubic K117

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

>

> Here is how, from a property of H, we get some, more or

> less complicated, generalizations....

>

> Let ABC be a triangle and A'B'C' the pedal/cevian triangle

> of H (orthic triangle)

>

> Denote:

>

> A" = AA' /\ B'C', B" = BB' /\ C'A', C" = CC' /\ A'B'

>

> La = the redical axis (common chord) of the circles ((B", B"A'),

> (C", C"A')) (ie the circles centered at B",C" with radii B"A',

> C"A', resp.)

>

> Lb = the redical axis (common chord) of the circles ((C", C"B'),

> (A", A"B'))

>

> Lc = the redical axis (common chord) of the circles ((A", A"C'),

> (B", B"C'))

>

> The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic.

>

> The parallelogic center (ABC) y(*) (La,Lb,Lc) is the center

> of the pedal/cevian circle of H (ie N)

>

> (*) I like the "y" that sometimes our Spanish friends use instead

> of the English "and"!

> [for two reasons: 1. The word is monogrammatic, that is, of one

> letter only 2. "y' is named in Latin as "y Graeca" = "Greek y"]

>

> Which is the other parallelogic centers?

>

> Gerneralizations:

>

> Given that for H, its pedal and cevian triangle are same,

> and also that the triangle bounded by the radical axes could be 0

> (ie to be concurrent the lines), we have these possible loci:

>

> Let ABC be a triangle P a point and A'B'C' the (1) cevian or (2)

> the pedal triangle of P.

>

> Denote:

>

> A" = AA' /\ B'C', B" = BB' /\ C'A', C" = CC' /\ A'B'

>

> La = the redical axis (common chord) of the circles ((B", B"A'),

> (C", C"A')) (ie the circles centered at B",C" with radii B"A', C"A',

> resp.)

>

> Lb = the redical axis (common chord) of the circles ((C", C"B'),

> (A", A"B'))

>

> Lc = the redical axis (common chord) of the circles ((A", A"C'),

> (B", B"C'))

>

> Which is the locus of P such that:

>

> i. La,Lb,Lc are concurrent?

>

> ii. The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

>

> APH

> - Dear Francisco,

I realize that the pedal version is complicated since AA',BB', CC' are not

concurrent in general!

But we can make it simpler if we replace the intersections of the pedal

triangle sides with AA',BB', CC' by the lines AP, BP, CP.

That is:

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

Denote:

A" = AP /\ B'C', B" = BP /\ C'A', C" = CP /\ A'B'

La = the redical axis (common chord) of the circles ((B", B"A'),

(C", C"A')) (ie the circles centered at B",C" with radii B"A', C"A', resp.)

Lb = the redical axis (common chord) of the circles ((C", C"B'), A", A"B'))

Lc = the redical axis (common chord) of the circles ((A", A"C'), (B", B"C'))

Which is the locus of P such that:

i. La,Lb,Lc are concurrent?

ii. The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

APH

On Sun, Mar 17, 2013 at 5:45 PM, Francisco Javier

<garciacapitan@...>wrote:

> **

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

>

>

> I only give the cevian case (for the pedal case, the calculations delayed

> too much)

>

> i. La,Lb,Lc are concurrent?

>

> The sextic Q058

>

>

> ii. The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

>

> The cubic K117

>

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

> >

> > Here is how, from a property of H, we get some, more or

> > less complicated, generalizations....

> >

> > Let ABC be a triangle and A'B'C' the pedal/cevian triangle

> > of H (orthic triangle)

> >

> > Denote:

> >

> > A" = AA' /\ B'C', B" = BB' /\ C'A', C" = CC' /\ A'B'

> >

> > La = the redical axis (common chord) of the circles ((B", B"A'),

> > (C", C"A')) (ie the circles centered at B",C" with radii B"A',

> > C"A', resp.)

> >

> > Lb = the redical axis (common chord) of the circles ((C", C"B'),

> > (A", A"B'))

> >

> > Lc = the redical axis (common chord) of the circles ((A", A"C'),

> > (B", B"C'))

> >

> > The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic.

> >

> > The parallelogic center (ABC) y(*) (La,Lb,Lc) is the center

> > of the pedal/cevian circle of H (ie N)

> >

> > (*) I like the "y" that sometimes our Spanish friends use instead

> > of the English "and"!

> > [for two reasons: 1. The word is monogrammatic, that is, of one

> > letter only 2. "y' is named in Latin as "y Graeca" = "Greek y"]

> >

> > Which is the other parallelogic centers?

> >

> > Gerneralizations:

> >

> > Given that for H, its pedal and cevian triangle are same,

> > and also that the triangle bounded by the radical axes could be 0

> > (ie to be concurrent the lines), we have these possible loci:

> >

> > Let ABC be a triangle P a point and A'B'C' the (1) cevian or (2)

> > the pedal triangle of P.

> >

> > Denote:

> >

> > A" = AA' /\ B'C', B" = BB' /\ C'A', C" = CC' /\ A'B'

> >

> > La = the redical axis (common chord) of the circles ((B", B"A'),

> > (C", C"A')) (ie the circles centered at B",C" with radii B"A', C"A',

> > resp.)

> >

> > Lb = the redical axis (common chord) of the circles ((C", C"B'),

> > (A", A"B'))

> >

> > Lc = the redical axis (common chord) of the circles ((A", A"C'),

> > (B", B"C'))

> >

> > Which is the locus of P such that:

> >

> > i. La,Lb,Lc are concurrent?

> >

> > ii. The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

> >

> > APH

> >

>

> __.

>

- Yes, in that case

1. Q007 octic

2. Q009 septic

Best regards,

Francisco Javier.

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

>

> Dear Francisco,

>

> I realize that the pedal version is complicated since AA',BB', CC' are not

> concurrent in general!

> But we can make it simpler if we replace the intersections of the pedal

> triangle sides with AA',BB', CC' by the lines AP, BP, CP.

>

> That is:

>

> Let ABC be a triangle P a point and A'B'C' the pedal triangle of P.

>

> Denote:

>

> A" = AP /\ B'C', B" = BP /\ C'A', C" = CP /\ A'B'

>

> La = the redical axis (common chord) of the circles ((B", B"A'),

> (C", C"A')) (ie the circles centered at B",C" with radii B"A', C"A', resp.)

>

> Lb = the redical axis (common chord) of the circles ((C", C"B'), A", A"B'))

>

> Lc = the redical axis (common chord) of the circles ((A", A"C'), (B", B"C'))

>

> Which is the locus of P such that:

>

> i. La,Lb,Lc are concurrent?

>

> ii. The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

>

> APH

>

>

> On Sun, Mar 17, 2013 at 5:45 PM, Francisco Javier

> <garciacapitan@...>wrote:

>

> > **

> >

> >

> > I only give the cevian case (for the pedal case, the calculations delayed

> > too much)

> >

> > i. La,Lb,Lc are concurrent?

> >

> > The sextic Q058

> >

> >

> > ii. The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

> >

> > The cubic K117

> >

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

> > >

> > > Here is how, from a property of H, we get some, more or

> > > less complicated, generalizations....

> > >

> > > Let ABC be a triangle and A'B'C' the pedal/cevian triangle

> > > of H (orthic triangle)

> > >

> > > Denote:

> > >

> > > A" = AA' /\ B'C', B" = BB' /\ C'A', C" = CC' /\ A'B'

> > >

> > > La = the redical axis (common chord) of the circles ((B", B"A'),

> > > (C", C"A')) (ie the circles centered at B",C" with radii B"A',

> > > C"A', resp.)

> > >

> > > Lb = the redical axis (common chord) of the circles ((C", C"B'),

> > > (A", A"B'))

> > >

> > > Lc = the redical axis (common chord) of the circles ((A", A"C'),

> > > (B", B"C'))

> > >

> > > The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic.

> > >

> > > The parallelogic center (ABC) y(*) (La,Lb,Lc) is the center

> > > of the pedal/cevian circle of H (ie N)

> > >

> > > (*) I like the "y" that sometimes our Spanish friends use instead

> > > of the English "and"!

> > > [for two reasons: 1. The word is monogrammatic, that is, of one

> > > letter only 2. "y' is named in Latin as "y Graeca" = "Greek y"]

> > >

> > > Which is the other parallelogic centers?

> > >

> > > Gerneralizations:

> > >

> > > Given that for H, its pedal and cevian triangle are same,

> > > and also that the triangle bounded by the radical axes could be 0

> > > (ie to be concurrent the lines), we have these possible loci:

> > >

> > > Let ABC be a triangle P a point and A'B'C' the (1) cevian or (2)

> > > the pedal triangle of P.

> > >

> > > Denote:

> > >

> > > A" = AA' /\ B'C', B" = BB' /\ C'A', C" = CC' /\ A'B'

> > >

> > > La = the redical axis (common chord) of the circles ((B", B"A'),

> > > (C", C"A')) (ie the circles centered at B",C" with radii B"A', C"A',

> > > resp.)

> > >

> > > Lb = the redical axis (common chord) of the circles ((C", C"B'),

> > > (A", A"B'))

> > >

> > > Lc = the redical axis (common chord) of the circles ((A", A"C'),

> > > (B", B"C'))

> > >

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

> > >

> > > i. La,Lb,Lc are concurrent?

> > >

> > > ii. The triangles ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

> > >

> > > APH

> > >

> >

> > __.

> >

>

>

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

>