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

8899Re: GENERALIZATION (was: 2003 LAST CONJECTURE !)

Expand Messages
  • jpehrmfr
    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
    • Show all 8 messages in this topic