   too complicated to be written here. Click on the link to download a text file.  X(99), X(115), X(690) X(46457) → X(46464) infinite points of the Steiner ellipses see points below Geometric properties :   K1255 is analogous to K271. These are nodal cubics invariant under an oblique symmetry. Let ABC be a given triangle and t a variable parameter. Let P, Q, R be three points on the sidelines BC, CA, AB of triangle ABC such that : BP/PC = CQ/QA = AR/RB = (1 - t)/t (with directed lengths) i.e. P is the barycenter of (B,t) and (C,1-t), Q and R similarly. K1255 is the locus of the Steiner point M of triangle PQR, as the parameter t varies. K1255 is a nodal cubic with node X(99), passing through X(115) having only one real point at infinity X(690) which is a flex. The asymptote is the trilinear polar of X(9164), a line through X(690) and X(15300). The tangent at X(115) is parallel to this asymptote. The remaining points at infinity of K1255 are always imaginary and lie on the Steiner ellipses. The two imaginary corresponding asymptotes concur at X(2) whose polar conic is the Steiner (circum) ellipse. The nodal tangents at X(99) are the lines passing through T1 and T2, on the circumcircle and on the trilinear polar of X(187). The cubic is invariant under the oblique symmetry with axis the trilinear polar of X(892), a line passing through {2, 99, 111, 115, 126, 148, 543, 574, 620, 671, etc}, and direction the real asymptote. When we swap t and 1-t, we obtain two points M and M' on the curve such that MM' is parallel to the asymptote and the midpoint of MM' lies on the axis. K1255 meets the sidelines of ABC at three triads of real points. The points A1, A2, A3 on BC lie respectively on the trilinear polars of X(99), X(6190), X(6189). The latter two are in fact the axes of the Steiner ellipses. The other six points are defined likewise. K1255 meets the cevian lines again at A0, B0, C0 and their oblique symmetrics are collinear on the cubic and on a line passing through G. Now, if we replace X(99) by its symmetric X(671) with respect to G, we obtain an analogous cubic which is the reflection of K1255 with respect to G. *** Other points on K1255 and on the line passing through X(99) and S on the Steiner ellipse S = X190 : (2 a-b-c) (b-c) (2 a^3-2 a^2 b+b^3-2 a^2 c+2 a b c-b^2 c-b c^2+c^3) : : , SEARCH = -20.43936616040136 S = X668 : (b-c) (-a b-a c+2 b c) (-a^4 b-a^4 c+2 a^3 b c+2 a^2 b^2 c-2 a b^3 c+2 a^2 b c^2-2 a b^2 c^2+b^3 c^2-2 a b c^3+b^2 c^3) : : , SEARCH = -1.015092717986773 S = X670 : (b-c) (b+c) (-a^2 b^2-a^2 c^2+2 b^2 c^2) (-a^6 b^2-a^6 c^2+5 a^4 b^2 c^2-2 a^2 b^4 c^2-2 a^2 b^2 c^4+b^4 c^4) : : , SEARCH = 1.582696450584995 S = X903 : (b-c)^2 (2 a^3-2 a^2 b+b^3-2 a^2 c+2 a b c-b^2 c-b c^2+c^3) : : , SEARCH = 1.452953532417812 S = X3227 : a (b-c)^2 (a^4 b+a^4 c-2 a^3 b c-2 a^2 b^2 c+2 a b^3 c-2 a^2 b c^2+2 a b^2 c^2-b^3 c^2+2 a b c^3-b^2 c^3) : : , SEARCH = 0.1376198560512373 S = X3228 : a^2 (b-c)^2 (b+c)^2 (a^6 b^2+a^6 c^2-5 a^4 b^2 c^2+2 a^2 b^4 c^2+2 a^2 b^2 c^4-b^4 c^4) : : , SEARCH = -0.2098894840508012 