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Gray point X(79) and X(80)

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  • Darij Grinberg <darij_grinberg@web.de>
    Steve Gray posted to the Math Forum the following problem (I don t have the accurate reference, but I saw it mentioned in
    Message 1 of 1 , Feb 5 10:09 AM
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      Steve Gray posted to the Math Forum the following problem (I don't
      have the accurate reference, but I saw it mentioned in
      http://www.pdipas.us.es/r/rbarroso/trianguloscabri/sol/sol39uno.htm
      )

      Here is Gray's theorem:

      Let ABC be a triangle with the incenter I. If A', B', C' are the
      reflections of I in BC, CA, AB, then the lines AA', BB', CC' concur
      at a point J, and the line IJ is parallel to the Euler line of
      triangle ABC.

      Although the fact that the lines AA', BB', CC' concur is a special
      case of the Kariya theorem, it doesn't say anything about the point
      J. With trilinear coordinates, one can see that J is the point X(79)
      in Kimberling's list, having the trilinears

      / 1 1 1 \
      J ( ----------- : ----------- : ----------- )
      \ 1 + 2 cos A 1 + 2 cos B 1 + 2 cos C /

      / sin A/2 sin B/2 sin C/2 \
      = J ( -------- : -------- : -------- ).
      \ sin 3A/2 sin 3B/2 sin 3C/2 /

      I propose to call J the GRAY POINT of triangle ABC.

      But it is interesting to compare J = X(79) with the point J' = X(80),
      having trilinears

      / 1 1 1 \
      J' ( ----------- : ----------- : ----------- ).
      \ 1 - 2 cos A 1 - 2 cos B 1 - 2 cos C /

      In Clark Kimberling's list, we find that J' is the reflection of the
      incenter in the Feuerbach point, and that J' is the inverse of the
      incenter I in the Fuhrmann circle of ABC.

      I have shown that J' is the point of concurrence of the lines AA",
      BB", CC", where A", B", C" are the reflection of the excenters Ia,
      Ib, Ic in BC, CA, AB. By the way, this is a corollary of an analogon
      to the Kariya theorem.

      Darij Grinberg
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