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

22068Re: Locus

Expand Messages
  • Antreas Hatzipolakis
    Mar 25 12:07 AM
    • 0 Attachment


      [APH]:

      Let ABC be a triangle, P = I and A'B'C' the antipedal
      triangle of P.

      Denote:

      Ab = (perpendicular from A' to BC) /\ PB
      Ac = (perpendicular from A' to BC) /\ PC

      Similarly:

      Bc = (perpendicular from B' to CA) /\ PC
      Ba = (perpendicular from B' to CA) /\ PA

      Ca = (perpendicular from C' to AB) /\ PA
      Cb = (perpendicular from C' to AB) /\ PB

      The lines BcCb, CaAc, AbBa are concurrent.

      Point?

      In general, the locus of P such that the lines BcCb, CaAc, AbBa are
      concurrent is Darboux cubic.


       
      [Angel ]:
      For every point P the lines BcCb, CaAc and  AbBa  are concurrent.
      If P = (x: y: z), then the point of concurencia of BcCb, CaAc and  AbBa  is Q=(x (SB y^2 + SC z^2 + SA (y + z)^2):...:...)

      Some pairs {P, Q}:
      {1, 8}, {2, 599}, {3, 3}, {4, 3}, {6, 599}, {8, 2098}, {13, 298},
      {14, 299}, {15, 298}, {16, 299}, {36, 4511}, {56, 2098},  {80, 4511}, ...

      ***************************************************

      Dear Angel

      Whenever the locus is the entire plane, the concurrence  can be
      proved synthetically.

      In our case we have:

      1. BaCa, CbAb, BcAc are concurrent at P.

      2. AbAc, BcBa, CaCb are concurrent, since ABC, A'B'C'
      are orthologic by conctruction, so they concur at the other
      than P orthologic center.

      Now by the Double Perspectivity Theorem we get that
      BcCb, CaAc, AbBa are concurrent.

      APH





    • Show all 234 messages in this topic