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X(3), X(4), X(5), X(389), X(942) reflections of X(389), X(942) about X(5) CPCC or Hcevian points, see Table 11 their reflections about X(5) 

Let P be a point, PaPbPc its cevian triangle and Ha, Hb, Hc the orthocenters of APbPc, BPcPa, CPaPb. These six points Pa, Pb, Pc, Ha, Hb, Hc lie on a same conic C(P). (Dominik Burek, ADGEOM #424). Barry Wolk (message #436) shows that this is a direct application of Pascal's theorem. Actually, C(P) is the bicevian conic C(P, H÷cP) where H÷cP is the barycentric quotient of the orthocenter H of ABC and the complement cP of P. 

Nikos Dergiades (message #428) observes that C(P) is a circle if and only if P is one of the common points P1, P2 of the circumcircle of ABC and the de Longchamps axis. These points are real when ABC is obtuse angled. In such case, we obtain two circles with same radius and radical axis the Euler line. The corresponding points H÷cP1, H÷cP2 lie on the Steiner ellipse. 

Randy Hutson discovered that a special and interesting configuration arises when P and therefore its isotomic conjugate P' lie on the Lucas cubic K007. Indeed, in this case : • PaPbPc and HaHbHc are homothetic (at the conic center Q) and even congruent, • HaHbHc is perspective to ABC at a point on the Darboux cubic K004. See message #437 by Nikos Dergiades. • The second perspector H÷cP of C(P) lies on K647. When P traverses the Lucas cubic, the center Q of C(P) lies on the cubic K645. Furthermore :
K645 has three real asymptotes concurring at X(5) and passing through the midpoints of ABC. These are the parallels at X(5) to the cevian lines of X(3). K645 is related to orthologic cevian triangles as in Table 7 (see at the bottom of page). 
