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X(2), X(3), X(6), X(83), X(98), X(171), X(238), X(385), X(419), X(1429), X(1691), X(2329), X(8290), X(8623), X(9467), X(9468) midpoints of ABC (a^4b^2c^2)^2/a^2 : : = isoconjugate of X(9468) tertiary pivot : a^2(a^4b^2c^2)(b^6+c^6a^6a^2b^2c^2) : : = X6660 x X385 its isoconjugate 1/(b^6+c^6a^6a^2b^2c^2) : : = tX5207 

Denote by H(n) the n^{th} barycentric power of the incenter X(1). For instance H(0) = X(2), H(2) = X(6), H(–1) = X(75). Let F(n) = H(n) × X(385) and G(n) = H(n) ÷ X(385), barycentric product and quotient of H(n) and X(385). Consider now the two pivotal cubics pK1(n) = pK(F(2n + 2), H(n)) and pK2(n) = pK(G(2n + 2), G(n)). Their isopivots are F(n + 2), H(n + 2) and their tertiary pivots are F(n+2) × X(5207), H(n + 2) × X(5207) respectively. These points obviously lie on the curves. It is clear that pK1(0) = K252 and pK2(0) = K354, the isogonal transform of K252. These two families of cubics are connected in many ways. • they are stable under barycentric multiplication (and division) by X(1) : pK1(n) × X(1) = pK1(n + 1) and pK2(n) × X(1) = pK2(n + 1). • pK1(n) is the barycentric product pK2(n) × X(385). • the isotomic transform of pK1(n) is tpK1(n) = pK2(– n – 2). In other words, if n + m = –2 then pK1(n) and pK2(m) are isotomic transforms of each other. In particular, with n = m = –1, we find K862 and K863. • the isogonal transform of pK1(n) is gpK1(n) = pK2(– n). In other words, if n + m = 0 then pK1(n) and pK2(m) are isogonal transforms of each other. In particular, with n = m = 0, we find K252 and K354. • more generally, pK2(n) is the transform of pK1(n) under the isoconjugation with pole H(2n + 2). For instance, if n = m = 1 hence H(4) = X(32), we find K861 and K864. *** The following diagram shows a selection of these cubics. 

× X(1) 
× X(385) 

n = –3 

pK(X75 × tX1933, tX1933) 

n = –2 

pK(tX1691, tX1691) 

n = –1 

n = 0 

n = 1 

n = 2 

pK(X1501 × X385, X6) 

n = 3 

pK(X9233 × X385, X31) 

pK2(n) 
pK1(n) 



Other related cubics : • the GHirst inverses of K354, K356 are K322, K738 respectively. • the anticomplement of K252 is pK(X2, X5207) passing through X(2), X(4), X(69), X(147), X(1031), X(2896). • under the symbolic substitution SS{a –> a^2}, – K251, K767, K774 are transformed into the pK1 cubics K252, K356, pK(X1501 x X385, X6), – K769, K135, K772 are transformed into the pK2 cubics K354, K532, pK(X560 x X385, X292).

