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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 K131 = pK(X31, X171) if and only if : ∑ a (b-c) (a^2-b c) (b^2 + b c + c^2) q r u^2 = 0. (E) (E) can be construed as follows : • X2, X292, P^2 ÷ Ω are collinear, • Ω, Ω x X292, P^2 are collinear, • P, P*, P* x X292 are collinear, 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 K131. See the related Table 66 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). Note that X(335) is the isotomic conjugate of X(239). (1) : K323(n) = pK(X1^(2n+1), X1^n x X239) (2) : K251(n) = pK(X1^(2n+1) x X239, X1^n) (3) : K768(n) = pK(X1^(2n) x X335, X1^(n+1) x X335) (4) : K769(n) = pK(X1^(2n+1) x X335, X1^n x X335) (5) : K770(n) = pK(X1^(2n) x X239, X1^(n+1)) (6) : K132(n) = pK(X1^(2n+1), X1^n x X894) (7) : K1002(n) = pK(X1^(2n), X1^n x X4645) Remarks : • these equations clearly show that the cubics above are weak for any n. It follows that the symbolic substitution SS{a -> a^2} transforms each cubic into a strong pK equivalent to K020 as in Table 66. • when Ω = X2 (resp. P = X2), the complement (resp. anticomplement) of pK(Ω, P) is another pK equivalent to K131. *** Additional data by Peter Moses The following table shows other cubics passing through at least 10 ETC centers. All of them are simple barycentric products, see column 3. |
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