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.
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)
• 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.
note 1 : these two cubics are isotomic transforms from one another.