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Re: [EMHL] Nagel point
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 Dear Vladimir,
>
I think you are right.
> But here the transversal also cuts the *side* lines, not
> any three
> lines through A, B, C as in Yaglom's book, n'estce
> pas?
Francois' problem is not exactly the same with Yaglom's.
And in my solution the equation of x is now different.
Hence now we must have 8 solutions.
Best regards
Nikos Dergiades
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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr  Dear Francois,
you said to Vladimir
> For your second remark, I suppose you want to say that
No. The general case is that the three lines
> Yaglom don't suppose
> the 3 lines form necessarily a triangle.
form a triangle A*B*C* different from ABC
the transversal is A'B'C',
the points A, A' are on side_line B*C*
the points B, B' are on side_line C*A*
the points C, C' are on side_line A*B*
and we want to have AA' = BB' = CC'.
This is different from your problem and I think
has no 6 solutions but 8.
Best regards
Nikos Dergiades
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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr  Dear Nikos
So if I understand you, you start with a triangle A*B*C* then with any
inscribed triangle ABC and you search a transversal A'B'C' such that AA' =
BB' = CC', that's right?
OK that's an usual choochoo problem and yes there are in general 8
solutions real or not that you can draw following Yaglom euclidian ideas
using 3 similarity centers and all the stuff written in Lalesco, Johnson or
Yaglom books about the "3 similar figures" configuration.
Now let me explain you the affine (choochoo) way to get them.
So you begin to construct any 3 points A", B", C" on B*C*, C*A*, A*B* such
that AA" = BB" = CC" = k where k is any positive given number.
If you are a very very lucky man, maybe the 3 points A", B", C" are on a
same line and you have drawn a line solution, cheer up, but nope you are
unlucky, then what?
You look at the affine map f: ABC > A"B"C" and miracle the invariant lines
of the affine map <f> are the sought transversals.
It seems that we have 8 ways to choose the triangle A"B"C" and maybe that's
here Yaglom make his mistake for if you choose a triangle A"B"C" then the
triangle A"'B"'C"' where segments A"A"', B"B"', C"C"' have middles A, B, C
will lead to the same invariant lines. So there is only 4 choices and maybe
4x2 = 8 solutions for of course we know that an affine map <f> has not
necessarily invariant lines and here there is a lot of affine discussion!
Notice that this method leads to an effective construction of the
transversals solutions if we know to construct the invariant lines of an
affine map given by the images of 3 points not on a same line. Of course I
know to do that with Cabri or GSP and you?
From this point of view, you can say that Yaglom ideas allows to construct
the invariant lines of an affine map but you can also construct these
invariant lines directly even without any euclidian structure except you
will have to use the inner calculator of Cabri or GSP to diagonalize the
linear associated map of the affine map <f>. In fact only the knowledge of
the eigenvalues is needed as I have explained it in a previous post some
months ago.
Friendly
Francois
Friendly
Francois
[Nontext portions of this message have been removed]  Dear Francois,
> So if I understand you, you start with a triangle A*B*C*
Yes.
> then with any
> inscribed triangle ABC and you search a transversal
> A'B'C' such that AA' =
> BB' = CC', that's right?
> Notice that this method leads to an effective construction
No.
> of the
> transversals solutions if we know to construct the
> invariant lines of an
> affine map given by the images of 3 points not on a same
> line. Of course I
> know to do that with Cabri or GSP and you?
A few days ago I managed to construct the affine P' = f(P)
if we know A' = f(A), B' = f(B), C' = f(C).
I had problem to understand in GSP the ordering of points of the makro.
Never mind. I am learning slowly.
By the way if the barycentrics wrt ABC
A1 = (p1 : q1 : r1)
B1 = (p2 : q2 : r2)
C1 = (p3 : q3 : r3)
A2 = (P1 :Q1 : R1)
B2 = (P2 :Q2 : R2)
C2 = (P3 :Q3 : R3)
M = (p : q : r)
N = (x : y : z)
What are the barycentrics of N if f is the affine map
sending A1B1C1 to A2B2C2 and N = f(M)?
Is there a not complicated formula?
Best regards
Nikos Dergiades
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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr  Dear Nikos
I am happy to see that you are interested in affine geometry.
I have written many posts on this topic and I feel like that I have already
given here all constructions you are needed for.
Anyway I am ready to help you again.
The only problem is that I am on the verge to quit Paris for Britanny where
I have no direct link with the web. Yes, France is still in the Middle Ages!
So it takes me time to read my mailbox!
There is 2 problems in (affine) geometry:
1° Arithmetic, that is to say calculations very simple in affine geometry
for one uses mostly linear equations or matrix 2x2 or 3x3 which you have to
diagonalize!
2° Implement (affine) macros on your software and that is a little more
difficult!
As for your barycentric question, it is very important though I think it is
badly formulated.
So we start with our beloved triangle ABC, OK!
Then we have a point M with barycentrics (x,y,z) wrt ABC. If you stop your
sentence here, you will have to get some problems in the future except if by
barycentrics of a point M you suppose implicitly that :
x + y + z = 1
I will suppose that in the sequel of my post.
Now why your question is too intricate to begin with?
Only because points A1, B1, C1, A2, B2, C2 are in general not needed.
So I modify slightly your question in this way:
Given 3 points A, B, C not on a same line and 3 other points A'(a1, a2, a3),
B'(b1, b2, b3), C'(c1, c2, c3) given by their barycentrics wrt ABC and at
last a point M(x,y,z), how do you calculate the barycentrics (x',y',z') of
the point M' = f(M) where <f> is the affine map defined by A' = f(A), B' =
f(B), C'= f(C).
The answer is the following:
X is the column vector (x,y,z)
X' is the column vector (x',y',z')
P is the 3x3 matrix of which the column vectors are given by the
barycentrics of A', B', C' wrt ABC, then we have:
X' = P.X
What is amazing is the linearity of this equation and the apparition of this
3x3 matrix!
Of course, there is all a stuff of bourbakist verbiage to explain that using
the vector space extension of the affine plane and you can forget it at
least at the beginning but if you want to be smarter in geometry, you will
have to drink the bourbakist potion for example in the Berger book:
Geometry, translated into english.
Calculation of this matrix P and its (Jordan) reduction is essential to know
the affine map <f>.
Now, I think you can answer yourself to your own original question. You have
to calculate the inverse of some 3x3 matrix. Bon appetit!
Friendly
Francois
[Nontext portions of this message have been removed]  Dear Francois,
thank you very much.
Francisco also sent me a similar message that
simplifies my own formulas.
[Francisco]> if A1B1C1 is the reference triangle, and Z is the matrix
[Francois]
> with {P1, Q1,
> R1},{P2, Q2, R2}, {P3, Q3, R3}} as rows then the image of
> P={x,y,z} is the
> product P.Z, if we suppose that coordinates are normalized
> Then we have a point M with barycentrics (x,y,z) wrt ABC.
So if we put the points with
> If you stop your
> sentence here, you will have to get some problems in the
> future except if by
> barycentrics of a point M you suppose implicitly that :
> x + y + z = 1
> I will suppose that in the sequel of my post.
> Now why your question is too intricate to begin with?
> Only because points A1, B1, C1, A2, B2, C2 are in general
> not needed.
> So I modify slightly your question in this way:
> Given 3 points A, B, C not on a same line and 3 other
> points A'(a1, a2, a3),
> B'(b1, b2, b3), C'(c1, c2, c3) given by their
> barycentrics wrt ABC and at
> last a point M(x,y,z), how do you calculate the
> barycentrics (x',y',z') of
> the point M' = f(M) where <f> is the affine map
> defined by A' = f(A), B' =
> f(B), C'= f(C).
> The answer is the following:
> X is the column vector (x,y,z)
> X' is the column vector (x',y',z')
> P is the 3x3 matrix of which the column vectors are given
> by the
> barycentrics of A', B', C' wrt ABC, then we
> have:
> X' = P.X
normalized (absolute) barycentrics
as matrix row
A1 = (p1 q1 r1)
B1 = (p2 q2 r2)
C1 = (p3 q3 r3)
M1 the 3x3 matrix of the above rows det(M1)<>0
A2 = (P1 Q1 R1)
B2 = (P2 Q2 R2)
C2 = (P3 Q3 R3)
M2 the 3x3 matrix of the above rows
S1 = (p q r)
S2 = (x y z)
then for the affine mapping f where
f(A1) = A2, f(B1) = B2, f(C1) = C2, f(S1) = S2
we get f = f2 o f1
where f1 sends A1B1C1 to ABC and f2 sends ABC to A2B2C2.
Hence
S2 = S1.(M1^1).M2 or with no normalized barycentrics
S2 = S1.(M1^1).D1.M2.(D2^1)
where D1 is the 3x3 matrix of rows
(p1+q1+r1 0 0)
(0 p2+q2+r2 0)
(0 0 p3+q3+r3)
and D2 is the 3x3 matrix of rows
(P1+Q1+R1 0 0)
(0 P2+Q2+R2 0)
(0 0 P3+Q3+R3)
Thanks again. I hope this formula will be
useful for me.
Best regards
Nikos Dergiades
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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr  Dear Nikos
My matrix P is named matrix of the affine map <f> in the affine frame {A, B,
C}.
I remind you that its columns are the (normalized) barycentrics of A' =
f(A), B' = f(B), C' = f(C) wrt ABC.
If g = Vec(f) is the linear associated map of <f>, then the characteristic
polynomial of P, to say char(P) of degree 3 is given by the following
formula:
Char(P) = (X1)*Char(g)
where Char(g) is the characteristic polynomial of <g> of degree 2
I remind you that:
Char(g) = X²  Trace(g).X + Det(g)
So we have:
Trace(P) = 1 + Trace(g)
and
Det(P) = Det(g)
These formulas are essential in plane affine geometry.
I give you some applications.
If A'B'C' is an inscribed triangle, then the main diagonal of P is made of 0
and then Trace(P) = 0 and so Trace(g) = 1
Moreover if we suppose that points A', B', C' are on a same line, that is to
say that A'B'C' is a transversal line, then Det(P) = Det(g) = 0
In other words, we have:
Char(g) = X² + X and so the spectrum of <g> is {1, 0}
As eigenvalues of <g> are distinct, <g> is diagonalizable and as 1 is not an
eigenvalue of <g> then <f> has a unique fixed point O and a pair of
invariant lines.
I ask you to give me this fixed point O and the invariant lines of the
affine map <f>. Of course the transversal A'B'C' is one member of the pair
of invariant lines. So you just have to give me the other member of the pair
and the point O. As a hint, think about Isaac .
That's a good exercise to begin with and to see how modern linear algebra is
useful even in this modest area of triangle geometry.
Friendly
Francois
[Nontext portions of this message have been removed]  Dear Nikos andv dear friends
Here I give you (without proof !) the euclidian construction of the fixed
point and invariant lines of the affine map f: abc > a'b'c' , given by the
images a', b', c' of 3 points a, b, c not on a same line, based on Yaglom
ideas using the "3 similar figures" configuration.
Of course this construction is rather long and intricate but it has its
inner beauty.
I suppose also we are in a general situation to avoid the numerous special
cases where the following construction don't works.
So we start with a triangle <abc> and 3 other points a', b', c' and we look
at the affine map f: abc > a'b'c'.
First we have the direct similarities:
Sa: bc > b'c' with center <alpha>
Sb: ca > c'a' with center <beta>
Sc: ab > a'b' with center <gamma>
Everybody knows the construction of these centers as circle's intersection.
First we construct line intersections:
A = bc /\ b'c', B = ca /\ c'a', C = ab /\ a'b'
and then the 3 centers:
<alpha> = circle(bcA) /\ circle(b'c'A)
<beta> = circle (caB) /\ circle(c'a'B)
<gamma> = circle(abC) /\ circle(a'b'C)
From now on to understand the sequel of the construction, you have to look
in any book on the "3 similar figures" configuration as the Johnson book or
the Lalesco book or the Yaglom book for example.
As triangle <abc> is inscribed in <ABC>, you look at its Miquel point M =
circle(bcA) /\ circle(caB) /\ circle(abC).
Then M is on the circle circle(<alpha><beta><gamma>), named in every book
the similitude circle.
Line aM cuts again the similitude circle in U.
Line bM cuts again the similtude circle in V.
Line cM cuts again the similitude circle in W.
These 3 points U, V, W are named in every book the invariable points. Why?
In fact if you do again the same construction from the other inscribed
triangle <a'b'c'>, you will get the same points U, V, W.
Now first miracle, lines <alpha>U, <beta>V, <gamma>W are on a same point O,
named the director point in every book and this director point E is just the
fixed point of the affine map <f>.
Then we have 3 circles: circle(<beta><gamma>O), circle(<gamma><alpha>O),
circle(<alpha><beta>O), named associated circles by Lalesco but they have no
names in Johnson or Yaglom.
Suppose that circle (<beta><gamma>O) has intersections {u, u'} with line
BC. Then the invariant lines of <f> are lines Ou and Ou'.
In this case the other associated circles: circle(<gamma><alpha>O) and
circle( <alpha><beta>O) have also respective intersections {v, v'} and {w,
w'} with lines CA and AB and even it means some swapping, points {u, v, w}
are on an invariant line of <f> and points {u', v', w'} are on the other
invariant line.
I think you can create a nice GSP macro with all this stuff.
Friendly
Francois
PS
Of course as a damned bourbakist, I don't like this construction very much
but it works in general.
Friendly
Francois
[Nontext portions of this message have been removed]  Dear Francois,
I can't understand your previous message?
Since A = bc /\ b'c'
the points A, b, c and A, b', c' are collinear.
Why you say
<alpha> = circle(bcA) /\ circle(b'c'A)?
Also you say
"As triangle <abc> is inscribed in <ABC>"
I think that <ABC> is inscribed in <abc>.
Best regards
Nikos Dergiades
 Στις Κυρ., 08/06/08, ο/η Francois Rideau <francois.rideau@...> έγραψε:
> Από: Francois Rideau <francois.rideau@...>
___________________________________________________________
> Θέμα: Re: [EMHL] Nagel point>Affine bar
> Προς: "hyacinthos" <hyacinthos@yahoogroups.com>
> Ημερομηνία: Κυριακή, 8 Ιούνιος 2008, 11:41
> Dear Nikos andv dear friends
>
> Here I give you (without proof !) the euclidian
> construction of the fixed
> point and invariant lines of the affine map f: abc >
> a'b'c' , given by the
> images a', b', c' of 3 points a, b, c not on a
> same line, based on Yaglom
> ideas using the "3 similar figures"
> configuration.
>
> Of course this construction is rather long and intricate
> but it has its
> inner beauty.
>
> I suppose also we are in a general situation to avoid the
> numerous special
> cases where the following construction don't works.
>
> So we start with a triangle <abc> and 3 other points
> a', b', c' and we look
> at the affine map f: abc > a'b'c'.
>
> First we have the direct similarities:
> Sa: bc > b'c' with center <alpha>
> Sb: ca > c'a' with center <beta>
> Sc: ab > a'b' with center <gamma>
>
> Everybody knows the construction of these centers as
> circle's intersection.
> First we construct line intersections:
> A = bc /\ b'c', B = ca /\ c'a', C
> = ab /\ a'b'
> and then the 3 centers:
> <alpha> = circle(bcA) /\ circle(b'c'A)
> <beta> = circle (caB) /\
> circle(c'a'B)
> <gamma> = circle(abC) /\ circle(a'b'C)
>
> From now on to understand the sequel of the construction,
> you have to look
> in any book on the "3 similar figures"
> configuration as the Johnson book or
> the Lalesco book or the Yaglom book for example.
>
> As triangle <abc> is inscribed in <ABC>, you
> look at its Miquel point M =
> circle(bcA) /\ circle(caB) /\ circle(abC).
> Then M is on the circle
> circle(<alpha><beta><gamma>), named in
> every book
> the similitude circle.
> Line aM cuts again the similitude circle in U.
> Line bM cuts again the similtude circle in V.
> Line cM cuts again the similitude circle in W.
>
> These 3 points U, V, W are named in every book the
> invariable points. Why?
> In fact if you do again the same construction from the
> other inscribed
> triangle <a'b'c'>, you will get the same
> points U, V, W.
>
> Now first miracle, lines <alpha>U, <beta>V,
> <gamma>W are on a same point O,
> named the director point in every book and this director
> point E is just the
> fixed point of the affine map <f>.
>
> Then we have 3 circles: circle(<beta><gamma>O),
> circle(<gamma><alpha>O),
> circle(<alpha><beta>O), named associated
> circles by Lalesco but they have no
> names in Johnson or Yaglom.
>
> Suppose that circle (<beta><gamma>O) has
> intersections {u, u'} with line
> BC. Then the invariant lines of <f> are lines Ou and
> Ou'.
> In this case the other associated circles:
> circle(<gamma><alpha>O) and
> circle( <alpha><beta>O) have also respective
> intersections {v, v'} and {w,
> w'} with lines CA and AB and even it means some
> swapping, points {u, v, w}
> are on an invariant line of <f> and points {u',
> v', w'} are on the other
> invariant line.
>
> I think you can create a nice GSP macro with all this
> stuff.
>
> Friendly
> Francois
>
> PS
> Of course as a damned bourbakist, I don't like this
> construction very much
> but it works in general.
> Friendly
> Francois
>
>
> [Nontext portions of this message have been removed]
>
>
> 
>
> Yahoo! Groups Links
>
>
>
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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr  Dear Nikos
Sorry, I send my message when I was not quite awake and I mixed up my poor
points a, b, c, a', b', c'!
Of course, I was wrong
So we have:
A = bb'/\cc'
B = cc'/\ aa'
C = aa'/\bb'
In other words triangles abc and a'b'c' are inscribed in ABC as I said
further in my previous post.
That is the starting figure of the choochoo theory!
I hope that I don't make any other mistake.
Besides you are lucky for I send you my sketch as other Hyacinthists have
nothing to see!
I think this "3 similar figures" configuration is essential in plane
geometry and I don't understand why it was not teached at a regular level.
After all, it just explains how to make the product of 2 direct similarities
of the plane.
I am sure that you can make a nice GSP macro with all this stuff but it's
just a stylistic composition.
I feel like that I explained some months ago how to create the same macro in
an affine way without using any circles.
It is based on the calculation of the spectrum of the associated linear map
of the affine map <f>, using the inner calculator of the software.
That's just a 2thdegree equation to solve!
This method gives the invariant lines at first and then the fixed point as
their intersection.
Another way is first to draw the fixed point by the Sollertinski
construction for example and then the invariant lines.
Suppose you know the fixed point O. Let L be any line and L' = f(L) its
image by <f>.
If m is on L, then m' = f(m) is on L' and we know that the envelope of the
line mm' is a parabola. You get the fixed lines when line mm' is on O. In
other words, the invariant lines are the tangents to the parabola through O,
a well known construction.
Friendly
Francois
PS
A question for Antreas:
Have I the right to attach a sketch in my posts to Hyacinthos?
2008/6/8 Nikolaos Dergiades <ndergiades@...>:
> Dear Francois,
[Nontext portions of this message have been removed]
> I can't understand your previous message?
>
> Since A = bc /\ b'c'
> the points A, b, c and A, b', c' are collinear.
> Why you say
> <alpha> = circle(bcA) /\ circle(b'c'A)?
> Also you say
> "As triangle <abc> is inscribed in <ABC>"
> I think that <ABC> is inscribed in <abc>.
>
> Best regards
> Nikos Dergiades
>
>
>
>
>  ���� ���., 08/06/08, �/� Francois Rideau <francois.rideau@...>
> ������:
>
> > ���: Francois Rideau <francois.rideau@...>
> > ����: Re: [EMHL] Nagel point>Affine bar
> > ����: "hyacinthos" <hyacinthos@yahoogroups.com>
> > ����������: �������, 8 ������� 2008, 11:41
> > Dear Nikos andv dear friends
> >
> > Here I give you (without proof !) the euclidian
> > construction of the fixed
> > point and invariant lines of the affine map f: abc >
> > a'b'c' , given by the
> > images a', b', c' of 3 points a, b, c not on a
> > same line, based on Yaglom
> > ideas using the "3 similar figures"
> > configuration.
> >
> > Of course this construction is rather long and intricate
> > but it has its
> > inner beauty.
> >
> > I suppose also we are in a general situation to avoid the
> > numerous special
> > cases where the following construction don't works.
> >
> > So we start with a triangle <abc> and 3 other points
> > a', b', c' and we look
> > at the affine map f: abc > a'b'c'.
> >
> > First we have the direct similarities:
> > Sa: bc > b'c' with center <alpha>
> > Sb: ca > c'a' with center <beta>
> > Sc: ab > a'b' with center <gamma>
> >
> > Everybody knows the construction of these centers as
> > circle's intersection.
> > First we construct line intersections:
> > A = bc /\ b'c', B = ca /\ c'a', C
> > = ab /\ a'b'
> > and then the 3 centers:
> > <alpha> = circle(bcA) /\ circle(b'c'A)
> > <beta> = circle (caB) /\
> > circle(c'a'B)
> > <gamma> = circle(abC) /\ circle(a'b'C)
> >
> > From now on to understand the sequel of the construction,
> > you have to look
> > in any book on the "3 similar figures"
> > configuration as the Johnson book or
> > the Lalesco book or the Yaglom book for example.
> >
> > As triangle <abc> is inscribed in <ABC>, you
> > look at its Miquel point M =
> > circle(bcA) /\ circle(caB) /\ circle(abC).
> > Then M is on the circle
> > circle(<alpha><beta><gamma>), named in
> > every book
> > the similitude circle.
> > Line aM cuts again the similitude circle in U.
> > Line bM cuts again the similtude circle in V.
> > Line cM cuts again the similitude circle in W.
> >
> > These 3 points U, V, W are named in every book the
> > invariable points. Why?
> > In fact if you do again the same construction from the
> > other inscribed
> > triangle <a'b'c'>, you will get the same
> > points U, V, W.
> >
> > Now first miracle, lines <alpha>U, <beta>V,
> > <gamma>W are on a same point O,
> > named the director point in every book and this director
> > point E is just the
> > fixed point of the affine map <f>.
> >
> > Then we have 3 circles: circle(<beta><gamma>O),
> > circle(<gamma><alpha>O),
> > circle(<alpha><beta>O), named associated
> > circles by Lalesco but they have no
> > names in Johnson or Yaglom.
> >
> > Suppose that circle (<beta><gamma>O) has
> > intersections {u, u'} with line
> > BC. Then the invariant lines of <f> are lines Ou and
> > Ou'.
> > In this case the other associated circles:
> > circle(<gamma><alpha>O) and
> > circle( <alpha><beta>O) have also respective
> > intersections {v, v'} and {w,
> > w'} with lines CA and AB and even it means some
> > swapping, points {u, v, w}
> > are on an invariant line of <f> and points {u',
> > v', w'} are on the other
> > invariant line.
> >
> > I think you can create a nice GSP macro with all this
> > stuff.
> >
> > Friendly
> > Francois
> >
> > PS
> > Of course as a damned bourbakist, I don't like this
> > construction very much
> > but it works in general.
> > Friendly
> > Francois
> >
> >
> > [Nontext portions of this message have been removed]
> >
> >
> > 
> >
> > Yahoo! Groups Links
> >
> >
> >
>
>
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>
 Dear Francois and Nikolaos
using the so interesting Francois ideas, we can get the following.
UVW is a triangle inscribed in ABC; we want to construct  using only
affine tools  the transversal lines U'V'W' such as
BC/UU' : CA/VV' : AB/WW' = p : q : r (signed distances)
Let P = p:q:r in barycentrics and f the affine mapping ABC > UVW
Then the solutions are the tangents from f(P) to the inscribed conic
with center the midpoint of Pf(P)
Of course, if we take for P the incenter or an excenter, we get the 8
transversal lines (not necessarily real) such as
UU' = VV' = WW' (here the distances are not signed)
Friendly. JeanPierre
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