Affine stuff

      [Darij] Subject: An affine generalization of the Schiffler point

      It is well-known that if I is the incenter and Ia, Ib, Ic are the excenters of a triangle ABC, then     

      (1) the Euler lines of triangles BIC, CIA, AIB concur at the Schiffler point X(21).      

      (2) the Euler lines of triangles BIaC, CIbA, AIcB concur at the point X(100).

      These both properties can be generalized as follows:

      Let P be a point with homogeneous barycentrics ( x : y : z ), and let PaPbPc be its anticevian triangle. Call Ma, Mb, Mc the midpoints of PPa, PPb, PPc, call Ga, Gb, Gc the centroids of the triangles BPC, CPA, APB, and Ha, Hb, Hc the centroids of the triangles BPaC, CPbA, APcB.

      Then,

      (1) the lines MaGa, MbGb, McGc concur at the point with homogeneous barycentrics

            / x(y+z-x)   y(z+x-y)   z(x+y-z) \

           (  --------    :    --------   :   --------   ).

            \    y+z            z+x          x+y     /

      (2) the lines MaHa, MbHb, McHc concur at the point with homogeneous barycentrics

            /  x      y       z  \

           (  ---  :  ---  :  ---  ).

            \ y-z   z-x    x-y /

      These affine theorems indeed generalize the Schiffler results, because:

      In the case P = I, we can easily show that Ma, Mb, Mc are the circumcenters of triangles BIC, CIA, AIB, but also the circumcenters of triangles BIaC, CIbA, AIcB simultaneously. Then we infer that the lines MaGa, MbGb, McGc are the Euler lines of triangles BIC, CIA, AIB, while the lines MaHa, MbHb, McHc are the Euler lines of triangles BIaC, CIbA, AIcB.

      By the way, for every P, if Na, Nb, Nc are the midpoints of segments BC, CA, AB, then the lines MaNa, MbNb, McNc also concur, and the point of concurrence has homogeneous barycentrics

                                       ( x(y+z-x) : y(z+x-y) : z(x+y-z) ).

      For P = K, this is the circumcenter of triangle ABC. 

      Darij Grinberg

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           In Hyacinthos messages #6374, #6376, #6380, Nikolaos and me gave some theorems on triangles and isotomic conjugates.

      If ABC is a triangle and D is a point, call A' the isotomic conjugate of A with respect to triangle BCD,

      B' the isotomic conjugate of B with respect to triangle CDA, C' the isotomic conjugate of C with respect to triangle DAB, D' the isotomic conjugate of D with respect to triangle ABC.

      Then the lines AA', BB', CC' and DD' concur.

      The point of concurrence of AA', BB', CC' and DD' will be denoted by P. I call it the ISOTOMION of the point D with respect to triangle ABC. So we have a new affine point mapping related to a triangle. As Nikolaos Dergiades computed in message #6376, for a point D with homogeneous barycentrics ( x : y : z ), the point P has barycentrics

            / (2x+y+z)(y+z)   (2y+z+x)(z+x)   (2z+x+y)(x+y) \

      P (  -------------      :       -------------       :        -------------  ).

           \        x                              y                              z       /

      Now, it follows that the locus of all P so that the isotomion lies on a given line l will be generally a quartic. What I found interesting is: If l is the line at infinity, our quartic factors into the line at infinity and the three sidelines of the antimedial (= dilated = superior) triangle.

      BTW, for every point P on the line at infinity, the isotomion of P is just P itself.