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Special Sondat

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  • Francois Rideau
    Dear friends I have noticed that the Nikos configuration in message 6466 was simply the figure of the special Sondat theorem that is to say 2 orthologic
    Message 1 of 2 , Oct 1, 2006
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      Dear friends
      I have noticed that the Nikos configuration in message 6466 was simply the
      figure of the special Sondat theorem that is to say 2 orthologic triangles
      ABC and A'B'C' with the same orthology center O.
      Then :
      1°ABC and A'B'C' are perspective wrt some perspector S
      2°Line SO is orthogonal to the perspectrix.
      Nikos only prove the 1st point and I have said I don't like his proof very
      much.
      In the next message n°6467, Darij noticed that ABC and A'B'C' are conjugate
      wrt some circle of center O and that implies the 1st point (and also the 2st
      as we shall see).
      But Darij regretted this proof was not valid if the circle is not real.
      I don't agree with him for polarity is in fact defined wrt a quadratic form
      before beeing defined wrt the conic possibly vanishing the form.
      So Darij proof is valid in all cases.
      Now if we come back to the general situation of 2 triangles ABC and A'B'C'
      conjugate wrt some conic {Gamma}, then it is well known that ABC and A'B'C'
      are perspective with perspector S and perspectrix L. Moreover L is the polar
      line of S wrt conic {Gamma}.
      In the special Sondat configuration, as {Gamma} is a circle of center O and
      L is the polar line of S wrt circle {Gamma}, then line SO is orthogonal to L
      and we are done with the 2st point.
      So I think Darij proof is The Synthetic Solution of the special Sondat
      Theorem.
      Friendly
      François


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    • Francois Rideau
      Dear friends Now the special Sondat theorem has been proved thanks to Darij, it remains to prove the general Sondat theorem always in a synthetic way. That is
      Message 2 of 2 , Oct 1, 2006
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        Dear friends
        Now the special Sondat theorem has been proved thanks to Darij, it remains
        to prove the general Sondat theorem always in a synthetic way.
        That is to say:
        Let ABC and A'B'C', 2 triangles both orthologic and perspective with
        perspector S, perspectrix L and orthology centers O and O'.
        Then:
        1° S, O, O' are on a same line.
        2° This line is orthogonal to L
        Suppose the 1st point has been proved and let h' be the dilation of center
        S, sending O' on O, that is to say h'(O') = O, and A'B'C' on A"B"C"
        Let h be the homology of center S and axis L, sending ABC on A'B'C'.
        Then h" = h'.h is a homology of center S and axis L', sending ABC on A"B"C".
        As we can consider the dilation h' as a homology of center S and axis the
        line at infinity, that implies that lines L and L' are parallel. So as
        triangles ABC and A"B"C" are orthologic with same orthology center O, then
        line SO is orthogonal to axis L', thanks to Darij theorem and then
        orthogonal to axis L and we are done.

        So it remains to prove the 1st point:
        Let f be the affine orthologic map sending ABC on A'B'C'. We know that
        Vec(f) the linear associated map of f is symmetric and then has 2 orthogonal
        eigenspaces, so defining 2 orthogonal directions D1 and D2 in the affine
        plane.
        If ABC and A'B'C' are 2 orthologic triangles (not necessarily perpective),
        we know that the orthology center O is on the rectangular hyperbola (H)
        through A, B, C, D1, D2, that is to say the rectangular hyperbola through A,
        B, C and D1 and D2 as asymptotic directions. For a proof, look at the
        message 14172 I have sent to Jean-Pierre.
        Now suppose ABC and A'B'C' are also perspective with perspector S.
        Then the locus of the point M of the plane such that S, M, f(M) are on a
        same line is in general a conic through S, S' (defined by f(S') = S) and the
        fixed points of f.
        As D1 and D2 are such fixed points, this locus is the rectangular hyperbola
        through D1 and D2 and also through A, B, C for lines AA', BB', CC' are on S.
        So this locus is just our previous and damned rectangular hyperbola (H) ans
        as we know that O is on it, we are done!

        It is interesring to compare this proof with the first one of Sollertinski.
        He also uses the dilation trick but to do h(A') = A in order to the product
        homology h" = h'.h have a fixed point A and so to reason on 2 orthologic
        triangles ABC and AB"C" sharing one vertice A.
        However that may be, all affine, euclidian and projective machinery is
        needed to prove Sondat theorem and any elementary solution if any would hide
        the true nature of this beautiful theorem!
        Friendly
        François


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