



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.

K856 is the transform of the Neuberg cubic K001 and the Napoleon cubic K005 under f.

Similarly, K857 is the transform of the Orthocubic K006 under f.

Generalization :

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.

When P = G then Q = X(20), we find Ω = X(6) : K002 and K004 are both transformed into K002.