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too complicated to be written here. Click on the link to download a text file.

X(1), X(3), X(30), X(110), X(191), X(484), X(2475)

other points below

K853 is the only cubic of the pencil described in the page K725 that passes through X(1).

(K) is pK(X6, X2475), a cubic of the Euler pencil, meeting the circumcircle again at Q1, Q2, Q3. The orthocenter of this triangle T is X(2475), the pivot of (K), hence its Euler line is that of ABC. K853 is the locus of foci of conics inscribed in T whose center lies on the Euler line. Hence K853 is K187 for T, i.e. K853 is an isogonal focal nK with respect to T. For instance, {X1, X191}, {X3, X2475}, {X30, X110} are pairs of isogonal conjugates with respect to T.

K853 must meet the sidelines of T at three collinear points R1, R2, R3 which lie on the perpendicular bisector of X(110)X(2475).

K853 can be seen also as :

• the locus of contacts of tangents drawn through X(110) to the circles passing through X(3) and X(2475),

• the locus of point M from which the segments X(1)X(3) and X(191)X(2475) are seen under equal or supplementary angles. More precisely, for any point M on K853, the sum of the directed line angles (MX1, MX3) + (MX191, MX2475) = 0 (mod. π).

K853 and the Neuberg cubic K001 share the same points at infinity with the same tangents at these points. Indeed, both are circular with same focus X(110), same real asymptote namely the line X(30), X(74). They must meet again at three finite collinear points which are X(1), X(3), X(484).

K853 and (K) meet at Q1, Q2, Q3, X(1), X(3), X(191), X(2475) and four other points which are the foci of the conic inscribed in T whose center is the midpoint X(5499) of X(3)X(2475).

At last, the intersection Y of the line X(30)X(110) with the perpendicular bisector of X(3)X(2475) lies on K853. Its isogonal conjugate X with respect to T is the intersection with the real asymptote and the tangential of X(110). These two points are rather complicated with SEARCH = 4.62510874861319 and 1.66911873606704 respectively.