   too complicated to be written here. Click on the link to download a text file.  X(2), X(25), X(612), X(614), X(1184), X(3162), X(7386), X(15487), X(39951), X(40124), X(40125), X(40174), X(40175), X(40176), X(40177), X(40178), X(40179), X(40180), X(40181), X(40182), X(40183), X(40184), X(40185), X(40186), X(40191), X(40192), X(40193), X(40194), X(40195) vertices of the medial triangle extraversions of X(612), X(614) A', B', C' : see below squared roots of X(1184), on the Wallace hyperbola Geometric properties :   The cubic pK(X6, P) meets the circum-circle (O) of ABC at A, B, C and three other points P1, P2, P3. See the paper How pivotal cubics intersect the circumcircle. The antipodes on (O) of P1, P2, P3 are three points denoted by Q1, Q2, Q3. There is a unique cubic nK0(X6, R) passing through these three points. With P = p:q:r, the first barycentric of R is (SO^2 - b^2c^2) p : a^2 SC q : a^2 SB r, with SO = (a^2+b^2+c^2) / 2, and the other coordinates are obtained cyclically. The mapping f : P → R is of the 1st degree, without singular points, having three fixed points F1, F2, F3 which are the common points (apart G) of the Wallace and Jerabek-Thomson rectangular hyperbolas. Hence G is the orthocenter of triangle F1F2F3 whose circum-center Ω is the reflection of X(351) about O. Note that F1F2F3 is always acutangle since its polar circle (with center G) has radius – (a^2+b^2+c^2) / 9 < 0. These points F1, F2, F3 lie on the circle with center Ω passing through X(110) and X(111). They also lie on nK0(X669, X69) and on nK(X187, X69, X110), a circular cubic passing through X(690), X(2396), with singular focus X(18860). f(X3) = X6 corresponding to the cubics K003 = pK(X6, X3) and K024 = nK0(X6, X6). Here the triangles P1P2P3 and Q1Q2Q3 are the CircumNormal and CircumTangential triangles. See Table 25. f(X20) = X2 corresponding to the cubics K004 = pK(X6, X20) and K082 = nK0(X6, X2). Here the triangles P1P2P3 and Q1Q2Q3 are the antipodal triangle of ABC and ABC itself. f transforms a line into another line and, in particular, the line at infinity into the orthic axis, the Euler line into the line GK. *** f transforms ABC into the triangle A'B'C' inscribed in K1162 and perspective at X(25) to ABC. A' = SO^2 - b^2c^2 : b^2 SC : c^2 SB, with. B' and C' defined cyclically. A'B'C' is perspective to • the cevian triangle of M on K169 and then the perspector N lies on K1162. In particular, with the medial triangle the perspector is X(40179) and with the symmedial triangle KaKbKc it is X(1184). Hence, A' is the intersection of the lines A, X(25) and Ka, X(1184). The transformations M → N and N → M are examined below. The following list gives pairs {M on K169, N on K1162} = {X(i), X(j)} for these {i, j}: {1,614}, {2,40179}, {6,1184}, {20,7386}, {25,25}, {64,40124}, {69,2}, {159,40125}, {200,612}, {269,40180}, {1763,15487}, {2138,3162}, {2139,40186}, {7097,40176}, {13575,40185}, {17742,40181}, {40187,39951}, {40188,40184}, {40189,40182}, {40190,40192}. Note that, if M1 and M2 are two isogonal conjugates on K169 (hence collinear with X(69)), then the corresponding perspectors N1 and N2 are collinear with X(7386). • the anticevian triangle of M for M on pK(X1184, X20) passing through X(20), X(25). • the circumcevian triangle of M for M on a quartic passing through X(25), X(1661). • the pedal triangle of M for M on a cubic passing through X(3), X(30), X(1498). • the fourth Brocard triangle at a point on the lines {3767,5094}, {5354,37077}. A'B'C' is orthologic to • the pedal triangle of X(69) at X(25) and X(40196), more generally to the pedal triangle of every point on the line X(69), X(4232). The orthology centers lie on a hyperbola passing through X(6), X(25) and a line passing through X(40196). • the antipedal triangle of X(25) at X(25) and a complicated point with SEARCH = 445.5824518181439. *** Other properties of K1162 A classical property of non singular cubics (Salmon) is fruitfully adapted to K1162. Recall that, if S is a point on the cubic with tangential T, then the polar conic C(T) of T meets the cubic at T (twice), S and three other points S1, S2, S3 such that the cubic is a pK with pivot S, isopivot T in triangle S1S2S3. It must pass through the cevians T1, T2, T3 of S in this triangle. T1T2T3 is the diagonal triangle of S, S1, S2, S3. The following table shows a selection of remarkable triangles. pivot S isopivot T triangle S1S2S3 X(2) X(1184) ABC X(1184) X(40125) MaMbMc, medial triangle X(7386) X(40124) A'B'C' X(612) X(40179) extraversions of X(612) X(614) X(7386) extraversions of X(614) X(25) X(40177) X(1184)-isoconjugates of A', B', C' X(40179) X(40195) X(2)-Ceva conjugates of A', B', C' Ro X(2) RaRbRc, anticevian triangle of Ro   Ro is the squared root of X(1184) which is inside ABC. Ro = a Sqrt[a^2 + (b - c)^2] Sqrt[a^2 + (b + c)^2] : : , SEARCH = 1.18156606472281.    The transformations 𝛍 : M → N and 𝛎 : N → M Both are quadratic with fixed points A, B, C, X(25).Each has three singular points which lie on K169 for 𝛍 and on K1162 for 𝛎. If P is not one of these fixed points, the locus of S such that P, S, 𝛍(S) – resp. 𝛎(S) – are collinear is a circum-cubic passing through X(25) and the three corresponding singular points. In each case, we obtain a net of cubics denoted by K𝛍(P) and K𝛎(P) respectively. K𝛍(G) = K169 and K𝛎(G) = K1162. For S on the line GK, each cubic is a K0 and each net contains two other pKs obtained for S = X(1611) and S = barycentric product of X(193) and X(5254). K𝛍(X1611) = pK(X32, X6353) = K1164 and K𝛍(X193 x X5254) = pK(X1196, X4) = K1165. The corresponding cubics K𝛎 are complicated and less interesting. K𝛍(S on GK) passes through X(6) and X(69) in addition to the points already mentioned and these cubics form a pencil. This pencil contains K1163 = K𝛍(X5913), a circular cubic also passing through X(111), X(2393), X(5913) with singular focus F on the line X(3), X(126). Another remarkable member is K1166. In the same manner, K𝛎(S on GK) passes through X(2) and X(1184).  