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X(2), X(97), X(323), X(2981), X(3218), X(3219), X(6151)
midpoints of ABC
S1 = X(14918) = X(53) ÷ X(1989)
S2 = X(14919) = isogonal conjugate of X(1990)
S3 = X(14920) = X(1990) ÷ X(1989)
S4 = X(14921) = X(395) ÷ X(1989)
S5 = X(14922) = X(396) ÷ X(1989)
K856 = pK(X323, X2) is the isogonal transform of K095. It is obviously invariant under X2-Ceva conjugation hence, for any point M on K856, the center (resp. perspector) of the circum-conic with perspector (resp. center) M is another point M' on K856 and M, M', X(323) are collinear.
It is also the barycentric quotient of :
• K095 by X(1989),
• K261a by X(13),
• K261b by X(14).
Indeed, X(323) coincide with the barycentric quotients X(15) ÷ X(14) and X(16) ÷ X(13).
Let us consider the mapping f : u : v : w –> (v^2 / b^2 – w^2 / c^2) / (SB v – SC w) : : . For M different of the in/excenters and the orthocenter of ABC, let (H) be the diagonal rectangular hyperbola passing through M and the in/excenters. Denote by P the pole of the line HM in (H) then f(M) is the barycentric quotient X(3) ÷ P. If M' is the second point of the line HM on (H) then obviously f(M) = f(M'). The transformation 𝝋 : M –> M' is a third degree involution with singular points H and the in/excenters. It fixes the Orthocubic K006, it swaps K001 and K005, also K002 and K004. 𝝋 contracts the points on a line through H and one in/excenter into this same in/excenter and the points on the diagonal hyperbola passing through H into H.
Note that f contracts all the points of K003 into G.
Let P, Q be two points on the Euler inverses in the circle with diameter OH. f transforms the two cubics of the Euler pencil pK(X6, P) and pK(X6, Q) into the same cubic pK(Ω, X2) where Ω is a point of the line GK.
In particular, these two cubics coincide when
• P = O then Ω = G hence pK(Ω, X2) is the union of the medians of ABC,
• P = H then Ω = X(394) hence pK(Ω, X2) is K857.