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Six NPC Centers

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  • Antreas
    Let ABC be a triangle. The cevians of N intersect the NPC at six points: The AN at A1 (near to A) and A2 The BN at B1 (near to B) and B2 The CN at C1 (near to
    Message 1 of 4 , Mar 12, 2013
      Let ABC be a triangle. The cevians of N intersect the
      NPC at six points:
      The AN at A1 (near to A) and A2
      The BN at B1 (near to B) and B2
      The CN at C1 (near to C) and C2

      Denote:

      Nab, Nac = The NPC centers of A1OB2, A1OC2

      Nbc, Nba = The NPC centers of B1OC2, B1OA2

      Nca, Ncb = The NPC centers of C1OA2, C1OB2

      The lines NabNba, NbcNcb, NcaNac are concurrent at the
      midpoint of ON.

      We can ask a number of questions related to the configuration.

      For example, for which points P (instead of O) the lines
      NabNba, NbcNcb, NcaNac are concurrenrt?
      The six NPC centers (Nab, Nac, Nbc, Nba, Nca, Ncb) are on
      a conic (and for which ones the conic is a circle)?

      APH
    • Antreas
      More Six NPC centers: Let ABC be a triangle, P a point, A B C the cevian triangle of P and Ha,Hb,Hc the orthocenters of AB C , BC A , CA B , resp. Denote:
      Message 2 of 4 , Mar 22, 2013
        More Six NPC centers:

        Let ABC be a triangle, P a point, A'B'C'
        the cevian triangle of P and Ha,Hb,Hc
        the orthocenters of AB'C', BC'A', CA'B', resp.

        Denote:

        Nab, Nac = The NPC centers of PA'Hb, PA'Hc, resp.

        Nbc, Nba = The NPC centers of PB'Hc, PB'Ha, resp.

        Nca, Ncb = The NPC centers of PC'Ha, PC'Hb, resp.

        For which points the six centers lie on a conic and
        the lines NabNba, NbcNcb, NcaNac are concurrent?

        The same question for circumcenters Oab, Oac etc
        instead of NPCs centers.

        Antreas

        --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
        >
        > Let ABC be a triangle. The cevians of N intersect the
        > NPC at six points:
        > The AN at A1 (near to A) and A2
        > The BN at B1 (near to B) and B2
        > The CN at C1 (near to C) and C2
        >
        > Denote:
        >
        > Nab, Nac = The NPC centers of A1OB2, A1OC2
        >
        > Nbc, Nba = The NPC centers of B1OC2, B1OA2
        >
        > Nca, Ncb = The NPC centers of C1OA2, C1OB2
        >
        > The lines NabNba, NbcNcb, NcaNac are concurrent at the
        > midpoint of ON.
        >
        > We can ask a number of questions related to the configuration.
        >
        > For example, for which points P (instead of O) the lines
        > NabNba, NbcNcb, NcaNac are concurrenrt?
        > The six NPC centers (Nab, Nac, Nbc, Nba, Nca, Ncb) are on
        > a conic (and for which ones the conic is a circle)?
        >
        > APH
      • Antreas
        http://anthrakitis.blogspot.gr/2013/03/six-npcs-n-of-medial-g-of-orthic.html Let ABC be a triangle and A B C the cevian triangle of P. Denote: Nab = The NPC
        Message 3 of 4 , Mar 27, 2013
          http://anthrakitis.blogspot.gr/2013/03/six-npcs-n-of-medial-g-of-orthic.html

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

          Denote:

          Nab = The NPC center of AA'B'

          Nac = The NPC center of AA'C'

          Nbc = The NPC center of BB'C'

          Nba = The NPC center of BB'A'

          Nca = The NPC center of CC'A'

          Ncb = The NPC center of CC'B'

          1. P = G (A'B'C' = medial triangle)

          The lines NbaNca, NcbNab, NacNbc are concurrent at the NPC center
          of A'B'C'.

          2. P = H (A'B'C' = orthic triangle)

          2.1.The triangles ABC and bounded by the lines (NbcNcb, NcaNac, NabNba) are homothetic
          The homothetic center is G'(centroid) of A'B'C'
          2.2. The homothetic center of the medial triangle and (NbcNcb, NcaNac, NabNba) is the infinite point of the line GG'.

          APH
        • Antreas
          [APH] ... The radical axes of ((Nab),(Nac)), ((Nbc),(Nba)),((Nca),(Ncb)) are concurrent. Which is the point of concurrence? APH
          Message 4 of 4 , Apr 1, 2013
            [APH]
            > Let ABC be a triangle and A'B'C' the cevian triangle of P.
            >
            > Denote:
            >
            > Nab = The NPC center of AA'B'
            >
            > Nac = The NPC center of AA'C'
            >
            > Nbc = The NPC center of BB'C'
            >
            > Nba = The NPC center of BB'A'
            >
            > Nca = The NPC center of CC'A'
            >
            > Ncb = The NPC center of CC'B'
            >
            > 2. P = H (A'B'C' = orthic triangle)

            The radical axes of ((Nab),(Nac)), ((Nbc),(Nba)),((Nca),(Ncb))
            are concurrent.

            Which is the point of concurrence?

            APH
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