too complicated to be written here. Click on the link to download a text file. X(6), X(25), X(69), X(74), X(230), X(1495), X(32263), X(40347), X(40348) O1, O2 = bicentric pair PU(4), see below Q1, Q2, Q3 : singular points of the transformation 𝛍 other points below Geometric properties :
 K1166 is K𝛍(X230) mentioned at the bottom of page K1162. K1166 meets the line at infinity at the same points as K478 = pK(X32, X468) and more generally that of pK(Ω, P) where : • Ω is a point on psK(X32 x X5159, X2, X6) passing through X(i) for i = 6, 32, 3162, and the vertices of the medial triangle. • P is a point on a cubic passing through X(i) for i = 4, 66, 468, 1370, 34518, and the vertices of the antimedial triangle. This cubic is the anticomplement of psK(X32 x X5159, X2, X3). K1166 meets the circumcircle at X(74), O1, O2 and more generally at the same points as pK(Ω, P) where : • Ω is a point on K1171 = pK(X32 x X8749, X8749). This cubic is the barycentric product X(74) x K496 hence it must also contain X(40352) = X(74) x X(6), X(40354) = X(74) x X(25), X(40353) = X(74)^2, X(40355) = X(74) x X(1989). • P is a point on K1172 = pK(X8749, X16080). This cubic is the barycentric product X(1494) x K496 hence it must also contain X(74) x X(1494). All these cubics also pass through X(6) and the most remarkable examples are K1169 = pK(X6, X3580), K1170 = pK(X32, X186) and K1172. *** About O1, O2... Recall that these points lie on the orthic axis, on every circle of the pencil generated by the circumcircle and the nine points circle (also containing the orthocentroidal circle, the orthoptic circle of the Steiner inellipse, the polar circle, the tangential circle). O1, O2 are G-Ceva conjugates. Their midpoint is X(468) and their barycentric product is X(25). These points lie on the cubics K397, K489, K490, K491, K492, K493, K496, K533, K535, K816, K817, K818, K1091, K1092, K1164, K1169, K1170, K1172. Remark : if Q is a point on the orthic axis, the cubics pK(X25, Q) and pK(Q, X2) pass through O1, O2. • the cubics pK(X25, Q) belong to a pencil with basis points A, B, C, O1, O2 and the square roots of X(25) i.e. X(20034) and harmonic associates. Examples : K535 = pK(X25, X468), pK(X25, X1990). The isopivot Q' of pK(X25, Q) lies on the circumcircle and its polar conic (C) is a circum-conic of ABC. (C) is a rectangular hyperbola if and only if Q' is one of the three common points of (O) and K176. • the cubics pK(Q, X2) belong to a pencil with basis points A, B, C, O1, O2, G and the midpoints of ABC. Examples : K489 = pK(X3003, X2), pK(X1990, X2).