## Special Sondat

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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
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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• 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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