too complicated to be written here. Click on the link to download a text file. X(1), X(9), X(55), X(1155), X(1156), X(2291), X(4845), X(5526), X(10426) cevian triangle of X(1156) X(15732) = X(55)-crossconjugate of X(1155) X(15734) = X(55)-crossconjugate of X(1156) X(15733) on the line at infinity Geometric properties :
 The Pelletier strophoid K040 and K806 generate a pencil of circular circum-cubics that contains the three pKs K949, K950, K951. This pencil is stable under isogonal conjugation and the isogonal transform of K949 is K950. Hence, for any P = X(i) on K949, the isogonal conjugate P* of P lies on K950. K950 is the pK with pivot X(1156) and isopivot X(55). The singular focus is X(15747), the inverse of the focus X(15746) of K949 in the circumcircle. Note that K950 is also the barycentric product of K949 by Z = X(8) x X(1156), on the lines {2,664}, {9,100}, etc. In other words, for any P = X(i) on K949, the barycentric product Q = X(j) = P x Z lies on K950.
 P = X(i) for i = 1 7 57 527 1155 1156 1323 3321 10427 15727 15728 15729 15730 ? Q = X(j) for j = 4845 1156 2291 9 55 ? 1 1155 15733 ? 10426 15732 5526 15734 P* = X(k) for k = 1 55 9 2291 1156 1155 4845 ? 10426 15732 15733 ? 15734 5526
 Remark : more generally, the isogonal transform of any pK(Ω, P) is its barycentric product by the isogonal conjugate of Ω.