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The Thomson cubic K002 and the Lemoine cubic K009 are both circumcubics passing through the vertices of the medial triangle, X(3) and X(4). The tangent at X(3) passes through X(64) except for K009 which has a node at X(3). They generate a pencil of cubics through these same 8 points and meeting the Euler line again at Q. All these cubics are psK(Ω, X2, X3) with pseudopole Ω on the line X(6)X(25) or, equivalently, psK(X6 x Q, X2, X3) where X6 x Q is a barycentric product. There is only one pK namely the Thomson cubic. See PseudoPivotal Cubics and Poristic Triangles for further details. See below for geometric properties of these cubics. The following table shows a selection of these cubics according to the point Q on the Euler line. 



The pencil contains seven other nodal cubics : • three are decomposed into a sideline of ABC and a conic passing through X(3), X(4) and the remaining vertices of ABC and the medial triangle. • four (not always real) with complicated corresponding points Q on the Euler line. *** 

Properties of K = psK(X6 x Q, X2, X3) with Q on the Euler line 

The cubic K = psK(X6 x Q, X2, X3) meets : • the circumcircle again at the same points O1, O2, O3 as pK(X6, Q), a member of the Euler pencil, the remaining common points being obviously X(3), X(4) and Q. • the line at infinity at the same points as pK(X6, aQ), another member of the Euler pencil, where aQ denotes the anticomplement of Q. The remaining common points lie on the Jerabek hyperbola namely A, B, C, X(3), X(4) and the isogonal conjugate Q4 of aQ. K also passes through : • Q1 = GCeva(X4 x Q), on the complement (H1) of the rectangular circumhyperbola passing through X(20). • Q2 = GCeva(X3 x Q), on the complement (H2) of the Jerabek hyperbola. Q2 is the tangential of Q. • Q3 = X1073 x Q, on the line X(3)X(64), the tangent at X(3) to K. Q3 is therefore the tangential of X(3). Note the triads of collinear points on K : X(3), Q2, Q4  X(4), Q1, Q4  Q, Q1, Q3.
See also Table 54 for a generalization. 
