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16519Re: Simson lines

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  • jpehrmfr
    Jul 1, 2008
      Dear Alxei
      I wrote :
      > Here is a better presentation of a generalization :
      > Starting with a point J in the plane of ABC, the perpendicular
      lines
      > through J to JA, JB, JC intersect BC, CA, AB respectively at A',
      B', C'.
      > It is well known that A',B',C' are on a same line L(J).
      > For any point M,
      > Ma is the point of the line AM such as <MJA = <AJMa
      > Mb is the point of the line BM such as <MJB = <BJMb
      > Mc is the point of the line CM such as <MJC = <CJMc
      > Then Ma,Mb,Mc are collinear if and only if M lies on L(J)
      > In this case, the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the
      > inscribed conic with focus J
      > This can be proved with projective tools using the fact that M->Ma
      is
      > the harmonic homology with center A and axis the line JA'
      > * The harmonic homology (I don't know if this English name is
      correct)
      > with center O, axis L maps M to his harmonic conjugate wrt O and OM
      > inter L
      >
      >
      > I think that, if we consider three harmonic homologies with centers
      > A,B,C, the locus of M with three colinear images is generally a
      cubic,
      > even when the three axis concur. In the particular case of three
      axis
      > La,Lb,Lc going through a point J and such as there exists an
      involution
      > swapping (JA,La), (JB,Lb),(JC,Lc), then the cubic degenerates in
      three
      > lines :
      > the fixed lines of the involution and the line through La inter BC,
      Lb
      > inter CA, Lc inter AB

      Now, if we consider a line L with trilinear pole P and intersecting
      the sidelines of ABC at A',B',C', we get :
      Suppose that M is a point of L; the harmonic homology with center A,
      axis JA' maps M to Ma,.., similarly Mb,Mc
      Then
      Ma, Mb, Mc are collinear
      the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the inscribed conic
      with perspector the cevian product J*P.
      Looking at that, I've realized something which should be interesting
      for Bernard :
      For any point J, the cevian product J*o(J), where o(J) is the
      orthocorrespondant of J, is the perspector of the inscribed conic
      with focus J.
      Friendly. Jean-Pierre
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