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Re: Sondat
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 Dear Alexei
I give you here a proof of part of the Sondat theorem in case of 2
orthological triangles ABC and A'B'C sharing the same vertice C.
These 2 triangles are clearly perspective, the perspector S beeing the
intersection of lines AA' and BB' and the axis of perspective beeing line CT
where T is the intersection of lines AB and A'B'.
O and O' are the orthology centers, to say:
O is the intersection of the line through A orthogonal to line B'C and the
line through B orthogonal to A'C, then line CO is orthogonal to A'B'.
O' is the intersection of the line through A' orthogonal to line BC and the
line through B' orthogonal to AC, then line CO' is orthogonal to AB.
I will prove here only the first part of Sondat theorem, to say that line
OO' is orthogonal to the axis of perpective CT, the other part beeing to
prove that line OO' is on the perspector S.
I have spent weeks to find this synthetic proof and using another
Sollertinski lemma about the fixed point of an affine map is the most funny
thing in this long story! And I wonder why he never have used his own lemma
to prove Sondat theorem when he has found his own projective intricate proof
of this theorem.
I have already talked here about this Sollertinski lemma on the fixed point
of an affine map: One can find it in RouchéComberousse book and Neuberg
credits Sollertinski on it but the first proof I read of this Sollertinski
lemma is in the Victor Thébault book: "Parmi les plus belles figures de
Géométrie dans l'espace", that is : "Among the most beautiful figures in
Solid Geometry" and by the way there are not a single figure in this book!
Here the Sollertinski lemma:
Let f be an affine map with one fixed point O.
Let L be a line such that line f(L) is not parallel to line L.
Let D be any line in the pencil of lines parallel to L.
Let M = D /\ f(D) be the intersection of lines D and f(D).
Then the locus of M when D moves in the pencil is a line {Delta} through the
fixed point O of the map f.
Of course Sollertinski uses 2 unparallel lines L and L' to get the fixed
point O at the intersection of the 2 lines {Delta} and {Delta'}.
Our Master, the great Coxeter himself uses this Sollertinski construction to
get the fixed point of a similarity in his book: Geometry.
Now I will apply the Sollertinski lemma to prove the first part of the
Sondat theorem.
Let f be the affine orthological map sending ABC on A'B'C, that is: f(A) =
A', f(B) = B' and f(C) = C.
So C is the fixed point of the map f.
I choose L = AB, so f(L) = A'B' and T = L /\ f(L).
I choose for D the line through O parallel to AB, so f(D) is the line
through O' = f(O) parallel to A'B'.
Let H = D /\ f(D) be the intersection of lines D and f(D).
By the Sollertinski lemma, H is on the line CT.
Now D beeing parallel to AB is orthogonal to O'C and f(D) beeing parallel to
A'B' is orthogonal to OC.
So H is the orthocenter of the triangle COO' and line CHT is orthogonal to
line OO' and we are done.
It remains to prove the second part of Sondat theorem, to say that line OO'
is on the perpector S.
I will skip it for the moment for if I know to prove it, I am not still
happy with my proof.
Friendly
François
PS
Have you find my Sondat riddle?
I think it may be useful to prove the Sondat theorem!
[Nontext portions of this message have been removed]  Dear Alexei
>
f(O) = O' is the main result on orthology ! I have proved it in one of my
> Thank you, it is very well. There is only one place which wasn't clear
> immediately: why f(O)=O'. But after sometime I understood it.
>
>
(french) paper! Have you had it? If not, I will send it to you when I come
back to Paris at the end of this week.
I have some idea for acheiving the proof. We must prove that the lines> Af(A), Bf(B) and Of(O) concur. What is the locus of points X such that
Funny! That's what just my idea too and as I find it not enough elementary
> Af(A), Bf(B) and Xf(X) concur?
>
>
, I have skipped it for the moment.
This locus is just the outcome of the following theorem in a projective
plane.
Theorem
Let f be a projective map of the plane and O be any point.
Then the locus of the point M such that the line Mf(M) is on the point O is
a conic {Gamma}.
This conic {Gamma} is on O and on every fixed points of the map f if any.
I recall a projective map has always at least a fixed point, maybe 3 or
infinitely many on a line, of course I suppose f is not the identity map.
Proof
I call g the inverse map of f . If M is a point of the locus, M is on the
line L through O and on the line L' = g(M) through O' = g(M).
But the map L > L' = g(M) is a projective map from the pencil of lines
through O onto the pencil of lines through O'.
So the locus of M is a conic through O and O' and clearly also on the fixed
points of f.
Now we apply this theorem to the affine orthological map f sending ABC to
A'B'C.
The fixed points of the map f are the point C of course but also the points
at infinity of the invariant lines of f which are orthogonal for f is
orthological.
So the sought locus is a rectangular hyperbola through A, B, C of which the
asymptots are parallel to the invariant lines of the map f.
So the only thing to do to prove the Sondat theorem is to prove that the
orthology center O is on this rectangular hyperbola!
But what is funny in this little story is that Sollertinski himself has
proved that if you read carefully his proof!
My rectangular hyperbola is just the rectangular hyperbola of the
Sollertinski paper.
But as I don't like his proof, we have to prove in another manner that O is
on the hyperbola.
Friendly
François
PS
I give you (without proof!) the answer of my Sondat riddle:
Given 4 points A, B, A', B', what is the locus of the points C such that the
triangles ABC and A'B'C are orthological.
Answer:
The locus is a line in general. This line is the directrix of the parabola
tangent to the 4 lines AB, A'B', AA', BB'.
This line is also on the 4 orthocenters of the 4 triangles bounded by 3 of
these 4 lines.
This line is orthogonal to the Newton line of these 4 lines AB, A'B', AA',
BB'.
The special cases arise when Vec(AB) = Vec(A'B') and if the parallelogram
ABA'B' is a rectangle or not.
This theorem is useful to draw the special Sondat theorem for orthological
triangles ABC and A'B'C, that's why I have to think about it!
[Nontext portions of this message have been removed]  Dear Alexey
At last, I think I have found a good reason for the line OO' to be on the
perspector S.
And of course, we both have the solution under our noses and we were blind.
What a shame!
So we start with our orthological triangles ABC and A'B'C and their
orthology centers O and O'.
Then triangles ABO and A'B'O' are also orthological but now they share the
same orthology center to say the point C and so they are in perspective.
Yeah!
Now we can see how to prove the Sondat theorem in a synthetic way:
1° Via the lemma on homologies sharing the same center and then having their
axis on a same point, it suffices to prove the Sondat theorem for 2
orthological triangles sharing a same vertice.
2° A synthetic proof of the fact that 2 orthological triangles sharing the
same orthology center are perpective is needed.
So I am eager to have your proof of the 2th point.
Friendly
François
On 9/13/06, Francois Rideau <francois.rideau@...> wrote:
>
> Dear Alexey
> Of course, you are right and I knew this proof too.
> But the only thing I would say about your proof is the third point.
> I want to have another good reason, other than Sollertinski argument, for
> the conic ABCOS to be rectangular.
> The Sollertinski proof for it is intricate and I would find a more simple
> one!
> A rectangular hyperbola has enough affine and metric properties we could
> use!
> Friendly
> François
> PS What do you think of my Sondat riddles?
>
>
> On 9/13/06, alexey <zaslavsky@...> wrote:
> >
> > Dear Francois!
> > I understood how achieve your proof. There are the main steps.
> > 1. The map f has two infinite fixed points x, y which are orthogonal.
> > 2. Consider the rectangular hyperbolaes G=ABCxy and G'=A'B'Cxy. As
> > f(G)=G', f(C)=C, f(x)=x, f(y)=y the forth common point of G and G' is the
> > perspectivity center S.
> > 3. The conic ABCOS (from Sollertinsky proof) pass through the
> > orthocenter H of ABC. So it is a rectangular hyperbola.
> > 4. If S=H(or H') then S=O(O') and the theorem is evident. In other case
> > the recatangular Hyperbolaes G and ABCOS, G' and A'B'CO'S coincide.
> > 5. The lines AA', BB' and OO' concur in S, q.e.d.
> >
> > Sincerely Alexey
> >
>
>
[Nontext portions of this message have been removed]  Dear friends , Alexey, Nikolaos, JeanPierre, Darij, Floor, ...
I am always haunted by a synthetic proof of the Sondat therem on projective
and orthologic triangles.
This subject has been in question very often here with no satisfying answer
if I am not wrong.
I recall that Sollertinski was the first to give such a proof in 1894 in
"L'intermédiaire des mathématiciens", a french periodical which has played
the role of Hyacinthos some 100 years ago.
But Sollertinski proof is too tricky and one don't see clearly the idea
behind it!
Nowadays we have made some progress, above all in the definition of an
orthological affine map which is central in such a synthetic proof.
In message 10325, Alexey gives a synthetic proof , criticized by Darij in
message 10328 and by Nikolaos in message 10336
In this message, Nikolaos talks about a JeanPierre email where he refers to
some Mitrea's paper.
So please, JeanPierre, can you give me the reference of this paper, thanks
in advance!
I summarize briefly the notations so everybody can understand what I say
about the state of this problem.
ABC and A'B'C' are orthologic(al) triangles with O and O' as orthology (or
orthological centers), that is:
lines AO, BO, CO are respectively orthogonal to lines B'C', C'A', A'B'
and
lines A'O', B'O', C'O' are respectively orthogonal to lines BC, CA, AB.
If f is the affine map such that f(A) = A', f(B) = B', f(C) = C', then
Vec(f), the associated linear map (a.l.m) of the affine map f is symmetric
or self adjoint, that is Vec(f) is diagonalizable in an orthonormal frame.
One says that such an f is an orthologic map.
Then we have the important relation: f(O) = O'.
Practically, that is to say:
There exists a cartesian orthonormal frame of the plane such that in these
coordinates, f looks like:
M(x,y) > f(M)( a.x + b, c.x + d) where and b are the eigenvalues of
Vec(f).
In general, one can choose b and d to be null, that is to say the origin of
the frame is a fixed point of the map f.
but we have to be careful for an orthologic map may have no fixed points,
think about a translation or to the map:
M(x,y) > f(M)(x+1, 2y) (in an orthonormal frame!).
It is important to notice that an orthological map transforms every triangle
LMN of the plane in an orthologic triangle f(L)f(M)f(N).
The Sondat theorem states that if ABC and A'B'C' are orthologic and
perspective wrt some perspector S, then line OO' connecting the orthology
centers is on the perspector S and orthogonal to the axis L of perspective.
(That is to say there exists a homology of center S and axis L sending A on
A', B on B', C on C'.)
Note that one supposes that O and O' are distinct in order to the Sondat
theorem keeps some sense.
This Sondat theorem is very useful in many geometrical configurations and
so it is very important to have such a synthetic proof .
Most of the synthetic proofs already known, even the first Sollertinski
proof, reduce to prove the Sondat theorem in case of 2 orthological
triangles ABC and A'B'C, sharing the same vertice C.
For example Sollertinski proves line SO is orthogonal to L and as O and O'
play a symmetrical role, SO' is also orthogonal to the axis L and we are
done. But his proof is very intricate, using the projective properties of
some rectangular hyperbola.
This rectangular hyperbola plays a great role in the proof. It is the conic
{Gamma} on A, B, C, S, O and it is not obvious such a conic is a rectangular
hyperbola. In fact its asymptotic directions are just the eigenspaces of
Vec(f), which are orthogonal for Vec(f) is symmetric.
My own synthetic proof begins as all the previous ones and reduces to the
case of proving the Sondat theorem for 2 orthological triangles ABC and
A'B'C sharing the same vertice C.
But instead to prove that line SO is orthogonal to the axis L as
Sollertinski, I prove that line OO' is orthogonal to the axis L. This fact
is very very easy to prove using that f is a orthologic map, (a theorem
Sollertinski don't knew at his time) and another lemma of Sollertinki on the
fixed point of an affine map.
So it remains to prove that line OO' is on the perspector S.
It is easy to see that this is equivalent to prove that O is on the
rectangular hyperbola through ABC with the asymptotic directions the
eigenspaces of Vec(f) and for such a proof, some projective trick ( à la
Sollertinski) is needed. So why to get tired, we might as well keep the
first whole Sollertinki proof.
But I have seen another manner to prove this last point without using this
damned hyperbola but for that I need to look at the case where Sondat
theorem has no meaning, that is the case where ABC and A'B'C' are orthologic
but their orthology centers O and O' are the same.
This configuration was also lenghty studied here.
In this case O = O', one proves that the triangle ABC and A'B'C' are also
perspective!
I remember everybody, even Floor, want to have a synthetic proof of this
theorem.
But I need a synthetic proof independant of the Sondat theorem, of course
for I use this result to prove the Sondat theorem.
I remember to have sent to Floor such an analytical and very short proof of
this theorem (independant of Sondat theorem) but I am eager to have a
synthetic and short proof of this case. So if anybody knows such a proof, I
would be very happy to know it!
I have a little problem about this late configuration:
We start with a triangle ABC and any point O.
Prove there are infinitely many triangle A'B'C' orthologic with ABC such
that O is a common orthology center.
As we know that ABC and A'B'C' are perspective, what is the locus of the
perspector S (easy, we can find it in this post!), and the envelope of the
axis of perspective (perspectrix?) L, (a little more difficult!)
Friendly
François.
[Nontext portions of this message have been removed]  Dear Francois
> I am always haunted by a synthetic proof of the Sondat therem on
projective
> and orthologic triangles.
satisfying answer
> This subject has been in question very often here with no
> if I am not wrong.
1894 in
> I recall that Sollertinski was the first to give such a proof in
> "L'intermédiaire des mathématiciens", a french periodical which
has played
> the role of Hyacinthos some 100 years ago.
idea
> But Sollertinski proof is too tricky and one don't see clearly the
> behind it!
of an
> Nowadays we have made some progress, above all in the definition
> orthological affine map which is central in such a synthetic proof.
Darij in
> In message 10325, Alexey gives a synthetic proof , criticized by
> message 10328 and by Nikolaos in message 10336
refers to
> In this message, Nikolaos talks about a JeanPierre email where he
> some Mitrea's paper.
paper, thanks
>
> So please, JeanPierre, can you give me the reference of this
> in advance!
Mitrea & Mitrea : "A generalization of a theorem of Euler"
AMM 101 vol 1 (1994) pages 5558
Unfortunately, I'haven't this article here but probably one of the
Hyacinthos members can upload the article in Hyacinthos files (the
paper is very short)
Friendly JeanPierre  Dear friends
Really, I am an orthologic addict!
I just find in an old book written by Balliccioni: "Coordonnées
Barycentriques et Géométrie", a theorem about orthology and I wonder if you,
Hyacinthists, know it. As it looks like Sondat theorem but it is different,
maybe it can be used to prove Sondat theorem. It's up to you to appreciate!
So here is the Balliccioni theorem (whitout proof of course!):
Let ABC and A'1B'1C'1 two orthological triangles with orthology centers O1
and O'.
That is to say:
lines AO1, BO1, CO1 are respectively orthogonal to B'1C'1, C'1A'1, A'1B'1
and
lines A'1O', B'1O', C'1O' are respectively orthogonal to BC, CA, AB.
Now one chooses A'2 on line A'1O', B'2 on line B'1O', C'2 on line C'1O'.
Then it is clear that triangles ABC and A'2B'2C'2 are still orthologic with
orthology centers O2 and O', (the same previous O' !).
On the other way triangles A'1B'1C'1 and A'2B'2C'2 are perspective with
perspector O' and with axis of perspective L (perspectrix? please tell me
the english name of this line!).
Then line O1O2 is orthogonal to the axis L.
Funny as this theorem has the same perfume as Sondat's one but it is not the
same.
I don't give you the proof for it is too long to write and you know my
lazyness and above all to let you think about it and maybe find a better
synthetic proof than Balliccioni and why not use this result to prove Sondat
theorem?
Friendly
François
PS
Balliccioni is the same guy who also is the coauthor of "Le premier livre du
tétraèdre", i;e: "The first book on tetraedra".
[Nontext portions of this message have been removed]  On 9/14/06, Francois Rideau <francois.rideau@...> wrote:
>
[Nontext portions of this message have been removed]
> Dear Friends
> I want to add something to my previous post.
> First:
> About the case where ABC and A'B'C' are orthologic with the same
> orthology center O = O';
> I will formulate the theorem in this way:
> Theorem
> Let ABC and A'B'C' be two orthologic triangles sharing the same orthology
> centers, (i.e: O = O'), then:
> 1° ABC and A'B'C' are perspective wrt some perspector S.
> 2° Line OS is orthogonal to the axis of perspective of ABC and A'B'C'.
> (In some way, this second assertion looks like the Sondat theorem for this
> special case.)
> So I want a synthetic proof of this theorem of course independant of the
> general Sondat theorem.
>
> Second:
> About the little problem at the end of my previous post, I suppose of
> course that O is different from the orthocenter H of ABC. Then what can be
> said if O = H?
> Third:
> I correct a little typo:
>
> Practically, that is to say:
> > There exists a cartesian orthonormal frame of the plane such that in
> > these coordinates, f looks like:
> > M(x,y) > f(M)( a.x + b, c.x + d) where and b are the eigenvalues of
> > Vec(f).
> >
>
> Of course, you must read:
> Practically, that is to say:
> There exists a cartesian orthonormal frame of the plane such that in these
> coordinates, f looks like:
> M(x,y) > f(M)( a.x + b, c.y + d) where and b are the eigenvalues of
> Vec(f).
> .
> Four:
> About the synthetic proof of the Sondat theorem, my philosophy to get a
> neat proof is to divide it in two parts:
> 1° Prove that the line OO' connecting the orthology centers is orthogonal
> to the axis of perspective L.
> That's the metric part of the theorem
> 2° Prove that the line OO' is on the perspector S, that's the projective
> part of the theorem.
>
> Now I have a synthetic proof of 1° and as for 2°, I feel like one cannot
> escape to use some projective trick à la Sollertinski to prove it.
> And I feel the same about the proof of the case where O = O', we cannot
> escape to use a projective trick à la Sollertinski and in fact I know it but
> it is easier to use it in this special case than in the general one where
> the trick is well hidden.
> Friendly
> François.
>
 Dear JeanPierre
Don't worry. A Hyacinthist just send me the Mitrea's paper and I am reading
it.
Effectively, the paper proves the Sondat theorem in another way I should
have thought.
I will criticize it when I will finish to understand it.
Friendly
François
[Nontext portions of this message have been removed]  Dear friends
Finally I have read the Mitrea's proof of Sondat theorem and can give my
opinion about it.
First, this is a paper from the Mitrea family, Dorina and Marius, cheer up!
Second , maybe I have already read it in the past and as my memory is not
good , I mistake Rumanian with Russian for I also forgotten their name.
Third, the paper is very short and the proof correct but hard to read for
the Mitrea did not have given to the points some suggestive names invariant
by permutations of the vertices.
Fourth, their proof is calculative. One cannot say it is very synthetic
though their calculus proceed from an astutely geometric idea that I should
have thought of. If S is the perspector of the orthologic triangles ABC and
A'B'C' with orthology centers O and O' and axis of perspective L = abc where
a = BC /\ B'C', b = CA /\ C'A' and c = AB /\ A'B', they notice line SO' is
orthogonal to the axis L if and only if the triangles SB'C' and Acb are
orthologic.
Fifth, to prove that two triangles are orthologic, they use an equation that
I knew of course for I have also used it very often in my own paper on
orthology.That's why I begin to worry about my brain.
Six, when you think about their short but not very aesthetically calculus,
it amounts to prove that some affine map is orthologic by using coordinates
wrt a general frame not orthonormal, here to say the frame: Vec(SA),
Vec(SB).
To sum up, the Mitrea's proof is correct and was new at this time. Cheer up!
However, I feel like one can slightly improve it but it will remain
calculative.
I also notice a serious mistake at the beginning of the paper: They say that
orthology is an equivalence relation! Of course not, orthology is not
transitive.
They also assert without proof that 2 triangles ABC and A'B'C' can be
"moved" in the plane such that they become orthologic and perspective. Why
not but I don't know if it is true.
Friendly
François
On 9/13/06, Francois Rideau <francois.rideau@...> wrote:
>
> Dear Alexey
> At last, I think I have found a good reason for the line OO' to be on the
> perspector S.
> And of course, we both have the solution under our noses and we were
> blind. What a shame!
> So we start with our orthological triangles ABC and A'B'C and their
> orthology centers O and O'.
> Then triangles ABO and A'B'O' are also orthological but now they share the
> same orthology center to say the point C and so they are in perspective.
> Yeah!
> Now we can see how to prove the Sondat theorem in a synthetic way:
> 1° Via the lemma on homologies sharing the same center and then having
> their axis on a same point, it suffices to prove the Sondat theorem for 2
> orthological triangles sharing a same vertice.
> 2° A synthetic proof of the fact that 2 orthological triangles sharing the
> same orthology center are perpective is needed.
>
> So I am eager to have your proof of the 2th point.
> Friendly
> François
>
>
>
> On 9/13/06, Francois Rideau <francois.rideau@...> wrote:
> >
> > Dear Alexey
> > Of course, you are right and I knew this proof too.
> > But the only thing I would say about your proof is the third point.
> > I want to have another good reason, other than Sollertinski argument,
> > for the conic ABCOS to be rectangular.
> > The Sollertinski proof for it is intricate and I would find a more
> > simple one!
> > A rectangular hyperbola has enough affine and metric properties we could
> > use!
> > Friendly
> > François
> > PS What do you think of my Sondat riddles?
> >
> >
> > On 9/13/06, alexey < zaslavsky@...> wrote:
> > >
> > > Dear Francois!
> > > I understood how achieve your proof. There are the main steps.
> > > 1. The map f has two infinite fixed points x, y which are orthogonal.
> > > 2. Consider the rectangular hyperbolaes G=ABCxy and G'=A'B'Cxy. As
> > > f(G)=G', f(C)=C, f(x)=x, f(y)=y the forth common point of G and G' is the
> > > perspectivity center S.
> > > 3. The conic ABCOS (from Sollertinsky proof) pass through the
> > > orthocenter H of ABC. So it is a rectangular hyperbola.
> > > 4. If S=H(or H') then S=O(O') and the theorem is evident. In other
> > > case the recatangular Hyperbolaes G and ABCOS, G' and A'B'CO'S coincide.
> > > 5. The lines AA', BB' and OO' concur in S, q.e.d.
> > >
> > > Sincerely Alexey
> > >
> >
> >
>
[Nontext portions of this message have been removed]  Dear friends
About the Mitrea's assertion on 2 triangles which can be moved to become
both perspective and orthologic, of course it is true.
When we have 2 orthologic triangles ABC and A'B'C' with orthology centers O
and O', (directed) angles (of lines) of triangle A'B'C' can be read at the
point O.
For we have the following equalities between directed angles of lines:
(OB, OC) = (A'C', A'B')
(OC, OA) = (B'A', B'C')
(OA, OB) = (C'B', C'A')
I recall that angles (OB, OC), (OC, OA), (OA, OB) are just the angular
coordinates of O wrt ABC.
They well define the point O except in one case, when O is on the
ABCcircumcircle.
And this case arises when ABC and A'B'C' are indirectly similar.
Of course an indirect similarity is orthologic but has also other
properties, for example it is also parallelogic, (see my paper on
isogonology).
So if you know the class of similarity of A'B'C' orthological with ABC, the
orthology center O is well defined, except in the case of A'B'C' indirectly
similar with ABC where you can choose for the orthology center O any point
on the ABCcircumcircle.
Now if h is any dilation and A" = h(A'), B" = h(B'), C" = h(C'), ABC and
A"B"C" are still orthologic with the same orthology center O, the other
orthology center beeing h(O').
As the group of dilations is transitive on the plane, we can as well choose
O = O'.
And we find a problem I gave in one of my previous mail:
Proof there are infinitely many triangles A'B'C' orthologic with ABC and
sharing the same orthology center O = O'
and we are done. This is just a special case of the Sondat theorem if one
wants.
Friendly
François
[Nontext portions of this message have been removed]  Dear friends
Now back in Paris, it's easier for me to chat!
I am always on the Sondat stuff and Mark Tudosi pointed out to me something
new on Sondat that I think important to understand this figure.
I recall here the (temporary) notations and I say temporary for maybe after
what I will tell you, we will have to modify some of them.
(I remember how the Mitrea's paper was hard to read with its bad notations!)
Let ABC and A'B'C' 2 triangles perspective and orthologic.
1°S is the perspector, so lines AA', BB', CC' are on S.
2°L is the perspectrix, so intersections a = BC /\ B'C', b = CA /\C'A', c =
AB /\ A'B' are on L.
O and O' are the orthology centers. Then:
1°lines AO, BO, CO are respectively orthogonal to lines B'C', C'A', A'B'.
2°lines A'O', B'O', C'O' are respectively orthogonal to lines BC, CA, AB.
Sondat states that S, O, O' are on a same line orthogonal to the perspectrix
L.
Now Mark attracts my attention on the points:
A" = AO /\ A'O', B" = BO /\ B'O', C" = CO /\ C'O'
Why?
For the triangles ABC, A'B'C', A"B"C" are perspective and orthologic by
pairs!!
For you thinking a little about it, I let you find their perspectors,
perspectrix and orthology centers. They are all already drawn on the figure.
I also notice triangles AA'A", BB'B", CC'C" are also perspective and
orthologic by pairs and I let you find their perspectors, perspectrix and
orthology centers, also all already drawn on the figure.
So I begin to wonder if there is not some structure on this Sondat
configuration seen as a whole.
It looks desmic for Mark but I think not. However if you look at the
different homologies and different orthologic maps, there is some structure
on them.
The relationships between the 6 homologies are easy to find and rather
surprising but those between the 6 affine orthologic maps are maybe more
difficult to discover for one always talks about orthological triangles and
never about orthologic maps and I think the later is the proper concept on
orthology!
I remember the Mitrea thank orthology is an equivalence relation and of
course it is not for it would imply that the product of 2 orthologic maps is
still orthologic and it is false for the product of 2 linear symmetric maps
is not always symmetric. But in the case of Sondat figure, this is true. So
if you look at the fixed points and invariant lines of the 6 orthologic
maps, they are the same and all these maps are commuting!
For hyacinthists having Cabri or GSP, they can use a macro giving the fixed
point and invariant lines of an affine map, given by the images of 3
distinct points.
Now I come to another point totally different about Sondat.
The first time you learn Sondat, you will have some difficulty to draw the
configuration. Try to draw 2 triangles ABC and A'B'C' both perspective and
orthologic from the void! That's not so easy and that's why I come across
some of the riddles I have sent you.
So I start with a triangle ABC and 2 other points A' and B' and I try to
construct another point C' in order to ABC and A'B'C' are perspective and
orthologic.
Then it is relatively easy to get C'. I let you find the construction.
So we get (in general) a map C > C' but I am not able to see what is this
map. I only know that if C moves on a line, then C' will move on some
hyperbola, so the degree of this map is rather high!
I say in general for of course they are exceptions I let you find. I only
say they are exceptions for formulating Sondat, the most known beeing O =
O'but you can have also S = O or O'.
Friendly
François
[Nontext portions of this message have been removed]  Dear Francois!
This configuration was also lenghty studied here.
In this case O = O', one proves that the triangle ABC and A'B'C' are also perspective!
I remember everybody, even Floor, want to have a synthetic proof of this theorem.
But I need a synthetic proof independant of the Sondat theorem, of course for I use this result to prove the Sondat theorem.
You have this proof! It can be easy obtained using next generalization of Sollertinsky lemma. Let the projective map f be given. Consider a pensil of lines d passing through the fixed point O and the respective pensil of lines f(d) passing through O'=f(P). Then the locus of points d^f(d) is the conic passing through O and O'. If the line OO' is fixed line of f, this conic is degenerated.
Now let the triangle ABC and the point O be fixed. Consider the set of triangles A'B'C' such that O is the common orthology center of ABC and A'B'C'. From this lemma the common point of AA' and BB' lies on the conic passing through A and B. Also it os easy to note that this conic pass through O, C and the orthocenter H of ABC. So CC' intersect the conic in the same point.
(It is interesting that this theorem can be used in Klein model of Lobachevsky geometry for proving that the altitudes of triangle are in the pensil.)
Also I found the answers to your riddles.
1. The locus of the perspective centers is the rectangular hyperbola ABCO and the envelop of perspective axis is the parabola inscribed to ABC. If we use the polar transformation wrt some circle with center O the triangle ABC is transformed to orthologic triangle A'B'C'. So this parabola is the image of rectangular hyperbola A'B'C'O. The special case is O=H. Then the triangles ABC and A'B'C' are homothetic with center H and the perspective axis is infinite.
2. Given 4 points A, B, A', B', what is the locus of the points C such that the triangles ABC and A'B'C are orthological.
As the scalar products (CA,CB') are (CB,CA') equal the locus of C is the line. This line is the radical axis of two circles with diameters AB' and A'B or the Aubert line of the quadrilateral formed by lines AA', BB', AB, A'B'. The orthology centers by Sollertinsky lemma lies on two conics. One of these conics pass through A and B, and the other through A' and B'. Also both conics pass through the common point of AA' and BB', two common points of circles with diameters AB' and A'B (may be not real) and the infinite point of the line passing through the centers of these circles. The special case is when AA'B'B is the rectangle. Then the locus of C (and of orthology centers) is whole plane.
Sincerely Alexey
Antivirus scanning: Symantec Mail Security for SMTP.
[Nontext portions of this message have been removed]  Dear Alexey
I come back to my idea of a bilogic map.
Given 4 points A,B,A',B' and S = AA' /\ BB', I say that given any point C on
the plane there exists "in general" a unique point C' such that triangles
ABC and A'B'C' are bilogic, i.e both orthologic and perspective.
The construction of such a C' from A,B,A',B' and C is quite easy:
The lines through A' orthogonal to BC and through B'orthogonal to CA are
cutting in O', (one of the orthology center). So C' must be on the line
through O' and orthogonal to AB and also on the line CS in order to ABC and
A'B'C' are bilogic.
So we get a map f: C > C', a priori defined on the whole plane but it is
easy to see that it is not defined
on H, the orthocenter of triangle ABS where there is some "blowing up".
As this construction of C' seems rather complex, I give you another one in
which the role of the orthocenter H is obvious:
Let K be the intersection of the line through A' parallel to AH with the
line through B' parallel to BH, then C' is the intersection of line CS with
the line through K parallel to CH.
With such a construction the blowing up of f in H is rather obvious.
I feel like this map is important to understand Sondat theorem and that's
why I was asking for!
Friendly
François
On 9/19/06, Francois Rideau <francois.rideau@...> wrote:
>
> Dear Alexey
> I don't understand your answer very well.
> What I say is the following:
> Given four points A, B, A', B' and S = AA' /\ BB', then for each point C
> of the plane, there is "in general" a unique point C' such that triangles
> ABC and A'B'C' are bilogic.
> So we get a (bilogic?) map: f: C > C',defined in fact on the whole
> plane minus the orthocenter of triangle ABS and not on a hyperbola!
> That is this map f I was talking about!
> Friendly
> François
>
>
> On 9/19/06, Alexey.A.Zaslavsky <zasl@...> wrote:
> >
> > Dear Francois!
> >
> >
> > What do you think of my map C > C' in order to triangles ABC and
> > A'B'C'are bilogic?
> > Is this map known?
> >
> > The line CC' pass through the perspective center which is the common
> > point of AA' and BB'. This center lies on two hyperbolaes which are the
> > locus of C and C'. So this map is the projection of one hyperbola to the
> > other.
> >
> > Sincerely Alexey
> >
> > Antivirus scanning: Symantec Mail Security for SMTP.
> >
> >
>
[Nontext portions of this message have been removed]  Dear friends
In my previous mail, by bilogic triangles I mean triangles both orthologic
and perspective, of course.
Excuse me to have not defined this word!
Besides I am not sure this terminology was universally adopted and why not
biologic triangles for the next time. As for me I prefer loving triangles.
So if you know a better word, let me know about it!
Friendly
François
[Nontext portions of this message have been removed]  Dear Francois!
Given 4 points A,B,A',B' and S = AA' /\ BB', I say that given any point C on the plane there exists "in general" a unique point C' such that triangles ABC and A'B'C' are bilogic, i.e both orthologic and perspective.
I feel like this map is important to understand Sondat theorem and that's why I was asking for!
I think that this map is fractionaly quadratic. My sketchpad displays that it transforms the lines to the conics passing through S. But I doubt that there is another simple definition.
Sincerely Alexey
Antivirus scanning: Symantec Mail Security for SMTP.
[Nontext portions of this message have been removed]  Dear friends and Alexey
Searching always a new synthetic proof of Sondat theorem, I have found it is
a simple consequence of Balliccioni theorem.
So here is the Balliccioni theorem (without proof of course!):
Let ABC and A'1B'1C'1 two orthological triangles with orthology centers O1
and O'.
That is to say:
lines AO1, BO1, CO1 are respectively orthogonal to B'1C'1, C'1A'1, A'1B'1
and
lines A'1O', B'1O', C'1O' are respectively orthogonal to BC, CA, AB.
Now one chooses A'2 on line A'1O', B'2 on line B'1O', C'2 on line C'1O'.
Then it is clear that triangles ABC and A'2B'2C'2 are still orthologic with
orthology centers O2 and O', (the same previous O' !).
On the other way triangles A'1B'1C'1 and A'2B'2C'2 are perspective with
perspector O' and with axis of perspective L (perspectrix? please tell me
the english name of this line!).
Then line O1O2 is orthogonal to the axis L.
Balliccioni proof is very cleaver, using circles, power wrt circle, radical
axis ans so one but he often uses circles with imaginary radius and I don't
like that. It is clear that one can improve it but maybe , there are other
proofs entirely different.
I would be happy to get your advices.
Friendly
François
[Nontext portions of this message have been removed]  Dear Hyacinthists,
an article written by myself and intitled
"Le théorème de Sondat, a new purely synthetic proof"
has been published at
http://www.campusoei.org/oim/revistaoim/ (n° 27)
Sincerely
JeanLouis Ayme
jeanlouisayme@...

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