Loading ...
Sorry, an error occurred while loading the content.

2004 FIRST CONJECTURE !

Expand Messages
  • Antreas P. Hatzipolakis
    -- Let A B C be the orthic and A B C the medial triangles of ABC. Denote: Ab, Ac = the reflections of A in CC , BB , resp. Bc, Ba = the reflections of B in
    Message 1 of 8 , Dec 31, 2003
    • 0 Attachment
      --
      Let A'B'C' be the orthic and A"B"C" the medial
      triangles of ABC.

      Denote:

      Ab, Ac = the reflections of A" in CC', BB', resp.
      Bc, Ba = the reflections of B" in AA', CC', resp.
      Ca, Cb = the reflections of C" in BB', AA', resp.

      CONJECTURE:

      The Nine Point Circles of the triangles
      A"AbAc, B"BcBa, C"CaCb concur.


      HAPPY NEW YEAR 2004 !

      Greetings from Athens

      Antreas
    • jpehrmfr
      Dear Antreas ... In both cases, your assertions are true and the locus of the common point of the three NP-circles is the line joining H to X(143) = the
      Message 2 of 8 , Jan 1, 2004
      • 0 Attachment
        Dear Antreas
        > Let ABC be a triangle, and A'B'C' its Orthic Triangle.
        >
        > 1. Let Pa, Pb, Pc be points on AH, BH, CH, resp. such that:
        >
        > APa / AH = BPb / BH = CPc / CH = t
        >
        > Denote
        >
        > Ab, Ac = the reflections of Pa in CC', BB', resp.
        > Bc, Ba = the reflections of Pb in AA', CC', resp.
        > Ca, Cb = the reflections of Pc in BB', AA', resp.
        >
        > The Nine Point Circles of the Triangles
        > PaAbAc, PbBcBa, PcCaCb are concurrent (??)
        >
        > Which is the locus of the point P of concurrence,
        > as t varies?
        >
        > [If t = 0 (ie Pa = A, Pb = B, Pc = C), then P = X(1986) (BW)]
        >
        > 2. Let Pa, Pb, Pc be points on A'H, B'H, C'H, resp. such that:
        >
        > A'Pa / A'H = B'Pb / B'H = C'Pc / C'H = t'
        >
        > Denote
        >
        > Ab, Ac = the reflections of Pa in CC', BB', resp.
        > Bc, Ba = the reflections of Pb in AA', CC', resp.
        > Ca, Cb = the reflections of Pc in BB', AA', resp.
        >
        > The Nine Point Circles of the Triangles
        > PaAbAc, PbBcBa, PcCaCb are concurrent (??)
        >
        > Which is the locus of the point P of concurrence,
        > as t' varies?
        >
        > [If t' = 0 (ie Pa = A', Pb = B', Pc = C'), then P = X(1112) (JPE) ]

        In both cases, your assertions are true and the locus of the common
        point of the three NP-circles is the line joining H to X(143) = the
        NP-center of the orthic triangle

        Friendly. Jean-Pierre
      • jpehrmfr
        Dear Antreas ... They go through X(974) Happy and peaceful 2004 to every Hyacinthist Friendly. Jean-Pierre
        Message 3 of 8 , Jan 1, 2004
        • 0 Attachment
          Dear Antreas
          > Let A'B'C' be the orthic and A"B"C" the medial
          > triangles of ABC.
          >
          > Denote:
          >
          > Ab, Ac = the reflections of A" in CC', BB', resp.
          > Bc, Ba = the reflections of B" in AA', CC', resp.
          > Ca, Cb = the reflections of C" in BB', AA', resp.
          >
          > CONJECTURE:
          >
          > The Nine Point Circles of the triangles
          > A"AbAc, B"BcBa, C"CaCb concur.

          They go through X(974)
          Happy and peaceful 2004 to every Hyacinthist
          Friendly. Jean-Pierre
        Your message has been successfully submitted and would be delivered to recipients shortly.