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Brocard-like points

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  • Antreas P. Hatzipolakis
    Let ABC be a triangle, P a point and A B C the pedal triangle of P. AaAbAc := The Orthic Triangle of AB C (ie Ab = the Orth. Proj. of B on AC , Ac = the
    Message 1 of 1 , Sep 1, 2006
      Let ABC be a triangle, P a point and A'B'C'
      the pedal triangle of P.

      AaAbAc := The Orthic Triangle of AB'C'
      (ie Ab = the Orth. Proj. of B' on AC',
      Ac = the orth. proj. of C' on AB')

      BaBbBc := The Orthic Triangle of A'BC'
      (ie Bc = the orth. proj. of C' on BA',
      Ba = the orth. proj. of A' on BC')

      CaCbCc := The Orthic Triangle of A'B'C
      (ie Ca = the orth. proj. of A' on CB',
      Cb = the orth. proj. of B' on CA')

      For which points P

      (1) The Circumcircle of AcBaCb is centered at P?

      (2) The Circumcircle of AbBcCa is centered at P?

      If P = (x:y:z) in Trilinears, then

      (1) ==> y^2 + (zsinA)^2 = z^2 + (xsinB)^2 = x^2 + (ysinC)^2

      ==> The squared Trilinears of P are

      ((sinAsinC)^2 + (cosC)^2 : (sinBsinA)^2 + (cosA)^2 :
      : (sinCsinB)^2 + (cosB)^2)


      (2) ==> z^2 + (ysinA)^2 = x^2 + (zsinB)^2 = y^2 + (xsinC)^2

      ==> The squared Trilinears of P are

      ((sinAsinB)^2 + (cosB)^2 : (sinBsinC)^2 + (cosC)^2 :
      : (sinCsinA)^2 + (cosA)^2)


      Equality of the altitudes ie
      for which points P

      (1) AcC' = BaA' = CbB'

      (2) AbB' = BcC' = CaA'


      (1) ==> P = (1 + cosAcosC - cosC : 1 + cosBcosA - cosA :
      : 1 + cosCcosB - cosB)

      (2) ==> P = (1 + cosAcosB - cosB : 1 + cosBcosC - cosC :
      : 1 + cosCcosA - cosA)


      Antreas



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