Given a point P in the plane of ABC let Pa=P(ABcCb), and define Pb, Pc cyclically. Then1) If P=X(15), PaPbPc is equilateral centered at 2 \Sqrt{3} X(6) + (a^2+b^2+c^2) X(9730);2) If P=X(16), PaPbPc is equilateral centered at 2 \Sqrt{3} X(6) - (a^2+b^2+c^2) X(9730);3) If P lies on the Lemoine axis (line X(187)X(237)), then Pa, Pb, Pc are collinear and lie on line (SA^2 : SB^2 : SC^2) ( barycentrics );4) For all P, lines APa, BPb, CPc concur in P.Note that ABcCb, AcBCa, AbBaC are homothetic to ABC.Regards,Elias M. Hagos
https://www.facebook.com/photo?fbid=959389341574915&set=gm.2961116324131387
Let H=X(4)-Orthocenter of ABC. A 1 B 1 C 1 cevian triangle of H. B1 A , C1 A are reflections of B1, C1 on A. L A , line through B1 A , C1...
Given a point P in the plane of ABC let Pa=P(ABcCb), and define Pb, Pc cyclically. Then
ReplyDelete1) If P=X(15), PaPbPc is equilateral centered at
2 \Sqrt{3} X(6) + (a^2+b^2+c^2) X(9730);
2) If P=X(16), PaPbPc is equilateral centered at
2 \Sqrt{3} X(6) - (a^2+b^2+c^2) X(9730);
3) If P lies on the Lemoine axis (line X(187)X(237)), then Pa, Pb, Pc are collinear and lie on line (SA^2 : SB^2 : SC^2) ( barycentrics );
4) For all P, lines APa, BPb, CPc concur in P.
Note that ABcCb, AcBCa, AbBaC are homothetic to ABC.
Regards,
Elias M. Hagos
https://www.facebook.com/photo?fbid=959389341574915&set=gm.2961116324131387
ReplyDelete