too complicated to be written here. Click on the link to download a text file. X(3), X(5), X(1499)
 K271 is the solution of a question raised in a MathLinks thread (http://www.mathlinks.ro/Forum/viewtopic.php?t=17597) and solved by Jean-Pierre Ehrmann. 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. K271 is the locus of the circumcenter M of triangle PQR, as the parameter t varies. K271 is a nodal cubic with node O, passing through the nine point center N = X(5) having only one real point at infinity X(1499) which is a flex. The asymptote is the line through X(1499) and S on the Euler line defined by OS = -1/3 ON. The tangent at N is parallel to this asymptote. The nodal tangents are the lines : (a^2b^2 + c^4 - 2b^2c^2)x + cyclic = 0 and (a^2c^2 + b^4 - 2b^2c^2)x + cyclic = 0. The cubic is invariant under the oblique symmetry with axis the Euler line 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 Euler line. The radical axis of two corresponding (distinct) circles centered at M and M' is the line GK for any t. Notice that the centroid of PQR is G for any t. This shows that the locus of the orthocenter of PQR is another nodal cubic with node H which is the anticomplement of K271. See the related cubic K272.