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Parallelogic Triangles - Loci

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  • Antreas
    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
    Message 1 of 4 , Mar 16, 2013
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      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
    • Francisco Javier
      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
      Message 2 of 4 , Mar 17, 2013
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        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
        >
      • Antreas Hatzipolakis
        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
        Message 3 of 4 , Mar 17, 2013
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          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]
        • Francisco Javier
          Yes, in that case 1. Q007 octic 2. Q009 septic Best regards, Francisco Javier.
          Message 4 of 4 , Mar 17, 2013
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            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]
            >
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