On 9/13/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Agreed; I speculate however that the 6 subsidiary incentres lie on a conic when the fourth point is the incentre, and not otherwise.
Presumably somebody has already looked at the 6 subsidiary centroids when the fourth point is the centroid, etc. --- maybe I should wait for RKG's report.
WFL
Or perhaps somebody hasn't ... let S be the fourth point within the original triangle T, the Cevian lines through S partitioning T into six subsidiary triangles. It looks to me as if not only (1) When S is the incentre of T, then the incentres of the six sub-triangles lie on a conic; but also (2) When S is the centroid of T, then the centroids of the six sub-triangles lie on a conic; suggesting analogous results lurking for other special points of T. Possibly these are all special cases of a result in which a general point S of T is put in correspondence with an "analogous" points of arbitrary triangles (including the six). Finally, why should its proposer have thought the "mixed" version mentioned by RKG would work at all? I'm perpetually amazed at my ability to select "random" geometric configurations apparently supporting some pet conjecture, despite the latter eventually proving to be utter wombat's do's. This canard is no exception. If T happens to be right-angled and S is its centroid, then the six incentres do apparently lie on a conic after all! Fred Lunnon