- View SourceDear 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

[Non-text portions of this message have been removed] - View SourceDear 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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