- Jun 30, 2014---------- 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:

For which P's the O, P, Rp are collinear?Aq = The antipode of Q in the Circumcircle of BQCAp = The antipode of P in the Circumcircle of BPCLet ABC be a triangle and P,Q two isogonal conjugate points.Denote:

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

Antreas

--- Antreas P. Hatzipolakis wrote :APHFor P = H, the Rh is not on the OH line.For P = K (symmedian point), the Rk is on the OK line (is it?)Rp = the radical center of AApAq, BBpBq, CCpCq,Aq = AQ /\ (Circumcircle of BQC) - QAp = AP /\ (Circumcircle of BPC) - PLet ABC be a triangle and P,Q two isogonal conjugate points.Denote:

Bp = BP /\ (Circumcircle of CPA) - P

Bq = BQ /\ (Circumcircle of CQA) - Q

Cp = CP /\ (Circumcircle of APB) - P

Cq = CQ /\ (Circumcircle of AQB) - Q(H is simple, but is Rh interesting?)For which P's the O,P,Rp are collinear?.__,._

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