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22481Re: Radical Center - OP line

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  • Antreas Hatzipolakis
    Jun 30, 2014
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      ---------- Forwarded message ----------
      From: Antreas P. Hatzipolakis
      Date: Tue, Jul 1, 2014 at 3:04 AM
      Subject: [EGML] Re: Radical Center - OP line
      To: Anopolis@yahoogroups.com


       

      A variation:

      Let ABC be a triangle and P,Q two isogonal conjugate points.

      Denote:

      Ap = The antipode of P in the Circumcircle of BPC
      Aq = The antipode of Q in the Circumcircle of BQC

      Bp = The antipode of P in the Circumcircle of CPA
      Bq = The antipode of Q in the Circumcircle of CQA

      Cp = The antipode of P in the Circumcircle of APB
      Cq = The antipode of Q in the Circumcircle of AQB

      Rp = the radical center of AApAq, BBpBq, CCpCq,

      For which P's the O, P, Rp are collinear?

      Antreas


      --- Antreas P. Hatzipolakis  wrote :


      Let ABC be a triangle and P,Q two isogonal conjugate points.

      Denote:

      Ap = AP /\ (Circumcircle of BPC)  - P
      Aq = AQ /\ (Circumcircle of BQC) - Q

      Bp = BP /\ (Circumcircle of CPA)  - P
      Bq = BQ /\ (Circumcircle of CQA) - Q

      Cp = CP /\ (Circumcircle of APB)  - P
      Cq = CQ /\ (Circumcircle of AQB) - Q

      Rp = the radical center of AApAq, BBpBq, CCpCq,

      For P = K (symmedian point), the Rk is on the OK line (is it?)

      For P = H, the Rh is not on the OH line.
      (H is simple, but is Rh interesting?)

      For which P's the O,P,Rp are collinear?

      APH












      .

      __,._


      --
      http://anopolis72000.blogspot.com/
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