Conjecture is true. X_6(PQR) = (SA+SB+SC) X + S (H - X(3)) ,where H is the orthocenter, X(3) the circumcenter of ABC and S is twice the area of ABC.Elias M. Hagos
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...
Conjecture is true. X_6(PQR) = (SA+SB+SC) X + S (H - X(3)) ,
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Elias M. Hagos