   too complicated to be written here. Click on the link to download a text file.  X(4), X(194), X(14251) infinite points of K003 A', B', C' : vertices of the ITB triangle foci of the inconic with center the midpoint Q = X(32448) of X(3), X(194) U intersection of BC and the parallel at X(194) to the A-cevian line of O, V and W similarly Geometric properties :   The ITB triangle A'B'C' is defined at K1098. K1101 is a stelloid inscribed in both ABC and ITB triangles. See K1098, another stelloid. Its radial center X is the image of X(194) under h(X5, 1/3). X = a^2 (a^4 b^2-3 a^2 b^4+2 b^6+a^4 c^2-7 a^2 b^2 c^2-b^4 c^2-3 a^2 c^4-b^2 c^4+2 c^6) : : , SEARCH = -0.174814769448192. X is now X(32447) in ETC (2019-05-08). K1101 meets K003 at three points on the line at infinity and six finite points on the rectangular circum-hyperbola passing through X(694), X(3224). K1101 meets K1098 at three points on the line at infinity and six finite points on the rectangular hyperbola passing through X(110), X(194), X(2574), X(2575) and A', B', C'. K1101 meets K444 at A, B, C, A', B', C' on the circumcircle and the three remaining common points are X(4), X(194), X(14251). K1101 meets K1100 at A, B, C, A', B', C' on the circumcircle and three remaining points on the Euler line, one of them being X(4). Since K1101 is spK(X3, Q) as in CL055, its isogonal transform is spK(X194, Q) and both cubics generate a pencil containing pK(X6, Q) hence all cubics pass through the foci of the inconic with center the midpoint Q of X(3), X(194). Q is actually the nine points center of the ITB triangle. Note that Q is now X(32448) in ETC (2019-05-08). 