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19146Re: A Circle From Reflections In Incenter Cevians And Symmetries Of Vertices

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  • jpehrmfr
    Aug 1, 2010
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      Dear Quang Tuan Bui
      > Given triangle ABC with incenter I. Select X is variable point on line BC. We do some reflections and symmetries as following:
      > By dicrection BAC
      > X1 = reflection of X in BI
      > X2 = symmetry of B in X1
      > X3 = reflection of X2 in AI
      > X4 = symmetry of A in X3
      > X5 = reflection of X4 in CI
      > X6 = symmetry of C in X5
      >
      > There is only one X on BC such that X=X6. We name this point as Ab.
      > Do the same procedures but with direction CAB, we get another point on BC and name it as Ac
      >
      > It is easy to get barycentrics of X1, X2... X6 from barycentrics of X. If we calculate intersection point Z of two lines XX1 and X4X6 then Z is always on a line. By choose X=B and X=C and do procedures as above, we can easy to construct this line. Intersection of this line with BC is Ab or Ac depend on direction we choosen. So Ab, Ac are easy to construct by ruler and compass.
      >
      > We can do the same procedures to get another similar points on lines CA, AB and we have total six points: Ab, Ac, Bc, Ba, Ca, Cb.
      >
      > It is interesting that these six points are concyclic on one circle with center T.
      > Barycentrics of T:
      > a*(3*a*(a^2 - b^2 - c^2) - 2*(b + c)*(a - b + c)*(a + b - c) + 4*a*b*c) : :
      > Search value: +3.3876599710368

      We have IT = IO/3 Friendly. Jean-Pierre
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