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Let pK(Ω = p : q : r, P = u : v : w) be a pivotal cubic with isopivot P*, the barycentric quotient Ω ÷ P. We suppose Ω ≠ P^2 in order to discard pKs decomposed into the cevian lines of P. This cubic is anharmonically equivalent to K020 = pK(X6, X384) if and only if : ∑ a^2 (b^2 - c^2) (a^4 - b^2 c^2) [(b^2 + c^2)^2 - b^2 c^2] q r u^2 = 0. (E) (E) can be construed as follows : • X2, X3114, P^2 ÷ Ω are collinear, • Ω, Ω x X3114, P^2 are collinear, • P, P*, P* x X3114 are collinear, Recall that X3114 = t X3094, X3407 = g X3094, X3314 = t X3407. X x Y denotes the barycentric product of X and Y. For a given pole Ω, (E) shows that P must lie on a diagonal conic and for a given pivot P, (E) shows that Ω must lie on a circum-conic. *** The following table shows a large selection of cubics equivalent to K020. W denotes a Weak cubic otherwise it is Strong. See the related Table 67 and also Table 68. |
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Notes : each number refers to cubics with equations of the same type and with the same color in the table. Furthermore, Kxxxx(n+1) = X(1) x Kxxxx(n) with exceptions K354(n) and K1023(n) for which Kxxxx(n-1) = X(1) x Kxxxx(n). Note that X(1916) is the isotomic conjugate of X(385). (1) : K020(n) = pK(X1^(2n+2), X1^n x X384) (2) : K128(n) = pK(X1^(2n+2), X1^n x X385) (3) : K739(n) = pK(X1^(2n) x X385, X1^(n+2)) (4) : K252(n) = pK(X1^(2n+2) x X385, X1^n) (5) : K322(n) = pK(X1^(2n) x X1916, X1^(n+2) x X1916) (6) : K354(n) = pK(X1^(2-2n) x X1916, X1^(-n) x X1916) (7) : K1012(n) = pK(X1^(2n) x X3094, X1^n) (8) : K1013(n) = pK(X1^(2-2n) x X3407, X1^(-n) x X3407) Remarks : • these equations clearly show that the cubics above are always strong for n even and weak for n odd. • when Ω = X2 (resp. P = X2), the complement (resp. anticomplement) of pK(Ω, P) is another pK equivalent to K020. *** Additional data by Peter Moses The following table shows other cubics passing through at least 10 ETC centers. Most of them are simple barycentric products, see column 3. |
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note 1 : these two cubics are isotomic transforms from one another.
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