Sorry, an error occurred while loading the content.

• ## Re: Θέμα: Re: [EMHL] Re:Cev ian Trace Equal Area Points

(10)
• NextPrevious
• Dear Nikos Yes, in fact your proof also gives the general case for on any line there exists always in general 2 isogonal conjugate points. friendly Francois
Message 1 of 10 , Aug 2, 2007
View Source
• 0 Attachment
Dear Nikos
Yes, in fact your proof also gives the general case for on any line there
exists always in general 2 isogonal conjugate points.
friendly
Francois

[Non-text portions of this message have been removed]
• Dear Nikos As I cannot sleep owing to my age, I feel like to begin some summer serial about choo-choos. As this theory is rather complex, today I will only
Message 2 of 10 , Aug 2, 2007
View Source
• 0 Attachment
Dear Nikos
As I cannot sleep owing to my age, I feel like to begin some summer serial
about choo-choos.
As this theory is rather complex, today I will only give you an imperfect
foretaste, just to explain you in general the existence of the equicenter
and areal center and their relationship.
I must tell you that this theory is affine, so no metric is needed at least
at the beginning.
Therefore we start with our beloved reference triangle ABC in the real
affine plane and 2 inscribed triangles a(0)b(0)c(0)and a(1)b(1)c(1).
As in my previous post, I will note S(L,M,N) the signed area defined by the
following equation:
S(L,M, N) = f(LM, LN) where f is some skewsymmetric bilinear form and LM
beeing the vector with origin L and end M.
So S is defined up to a factor but it does not matter for we will use only
area's ratios!
We define 3 reals (a, b, c) by the equations:
a(0)a(1) = a. BC; b(0)b(1) = b. CA; c(0)c(1) = c. AB
Once again, note that a, b, c are not the lengths of the sides of ABC for
there is no metric for the moment.
Choose any point O in the plane and define points U, V, W by :
OU = a(0)a(1); OV = b(0)b(1); OW = c(0)c(1)
Here on this side of Atlantic ocean, we said we have "vectorialized" the
plane in O.
1° I will suppose that these 3 points U, V, W are not on a same line.
2° So O has normalized barycentrics (u, v, w) wrt UVW, that is to say:
x. OU + y. OV + z.OW = 0 with x + y + z = 1.
and I will suppose that : x.y.z <> 0 , i.e none of the 3 reals x, y, z is
equal to 0.

As (1/a) OU + (1/b) OV + (1/c) OW = 0

we have: x = b.c/ (b.c + c.a + a.b); y = c.a/ (b.c + c.a + a.b); x = a.b/ (
b.c + c.a + a.b)
Let P be the point with barycentrics (x,y,z) wrt ABC, then P is the areal
center of the triangles a(0)b(0)c(0) and a(1)b(1)c(1).

In fact S(P,b(0),b(1)) = f(AP, b(0)b(1)), (look at my previous post!)
So S(P, b(0), b(1) = f(y. AB + z. AC, b. CA) = y.b.f(AB, CA) = -(a.b.c)/(b.c+
c.a + a.b) S(A, B, C)
Of course in the same way, we have:
S(P, c(0), c(1)) = f(AP, c(0)c(1)) = f(y. AB + z. AC, c. AB) = z.c.f(AC, AB)
= -(a.b.c)/(b.c + c.a + a.b).S(A, B, C)
and we are done.
From the vectorial equation: x. a(0)a(1) + y. b(0)b(1) + z. c(0)c(1) = 0
we get the affine equation:
x. a(0) + y.b(0) + z. c(0) = x. a(1) + y. b(1) + z. c(1) = E
So the point E is the equicenter and the fixed point of the affine map
sending a(0)b(0)c(0) onto a(1)b(1)c(1)
Note that if g(0) is the affine map sending a(0)b(0)c(0) onto ABC, then we
have g(0)(E) = P as I have said in one of my previous post.
Now you will tell me where are the choo-choos?
A choo-choo is a train going on smoothly on its tracks that is to say a
rectilinear uniform motion (r.u.m)
a(0) and a(1) are the positions at time 0 and time 1 of the choo-choo: t
--> a(t) = (1 -t) a(0) + t . a(1) = a(0) + t . OU
b(0) and b(1) are the positions at time 0 and time 1 of the choo-choo: t
--> b(t) = (1 -t) b(0) + t . b(1) = b(0) + t . OV
c(0) and c(1) are the positions at time 0 and time 1 of the choo-choo: t
--> c(t) = (1 -t) c(0) + t . c(1) = a(0) + t . OW
Notice that the speeds are just OU, OV, OW.
It is easy to see that E = x. a(t) + y. b(t) + z. c(t), E is the equicenter
of the choo-choos, that is the point of view of Neuberg.
In the same way, we also have:
S(P, a(t), a(u))) = S(P, b(t), b(u)) = S(P, c(t), c(u))
so we can also tell that P is the areal center of the choo-choos.

Note that we have affine bijections: a(t) <--> b(t) <--> c(t) between the
sides of ABC
In France, they were called in Neuberg time " divisions semblables". Why?
Look at the bijection: b(t) <--> c(t).
In these old times, our forebears think about sides AC and AB as 2 graduate
rules with b(0) and b(1) as graduations 0 and 1 on side AC and c(0) and
c(1) as graduation 0 and 1 on side AB, the point b(t) of graduation t
on side AC corresponding to the point c(t) of graduation t on side AB.
That's look funny now in the bourbakist era . Yeah, they did not know what
were affine or linear maps but they were as good as us if not better in
geometry!
They called the triangles a(t)b(t)c(t) triangles with homologous vertices.
Now I stop this story for the moment and maybe I will able to sleep!
Later, I will tell you the story of the pedal choo-choos with orthopole as
equicenter and areal center on the circumcircle.
Friendly
Francois

[Non-text portions of this message have been removed]
• Dear Francois, and Tuan. Thanks Francois, for your proof that can be written another way as: Let A B C , A B C be two inscribed triangles in ABC, vector BC =
Message 3 of 10 , Aug 4, 2007
View Source
• 0 Attachment
Dear Francois, and Tuan. Thanks Francois,
for your proof that can be written another way as:
Let A'B'C', A"B"C" be two inscribed triangles in ABC,
vector BC = x.A'A", CA = y.B'B", AB = z.C'C" and let
Q = (x : y : z) wrt ABC in barycentrics
E = (x : y : z) wrt A'B'C'.
It is known that
S(Q,B,C) = x.S(A,B,C)/(x+y+z).Hence
S(Q,A',A")=f(A'A",A'Q)=f(BC/x,A'B+BQ)=(1/x).f(BC,BQ)
=(1/x).S(Q,B,C)=S(A,B,C)/(x+y+z).
Similarly S(Q,B',B")=S(Q,C',C")=S(Q,A',A")
Q is the areal point of triangles A'B'C', A"B"C".

Since x.EA' + y.EB' + z.C' = 0 we get
xEA"+yEB"+zEC"-xEA'+yEB'+zEC'=x.A'A"+y.B'B"+z.C'C"
= BC + CA + AB = 0
or x.EA"+y.EB"+z.EC" = 0
or x.EAt+y.EBt+z.ECt = 0 where the triangle
AtBtCt is the choo-choo instance At=(1-t)A'+tA" . . .
Hence E has the same x : y : z barycentrics wrt AtBtCt
and is the equicenter of the choo-choo.
Q is the map of the affine map sending A'B'C'->ABC.

By the way this choo-choo process (Antrea mi gelas)
is very usefull in dynamic geometry.
Forgive me for I am going to add some more lines
and give a long message and may be these thoughts
are trivial to you or may be you will enjoy it
as Tuan likes finding ETC search values.
If we have a geometric construction and find a
triangle center P we can find the ETC search value
of this point working algebraicly with a computer
Mathematics packet or as I often do working with GSP.
One method is to construct the constant triangle
ABC that ETC uses (a, b, c) = (6, 9, 13)
with cartesian coordinates
B(0, 0), C(6, 0) and the circles (B, 13), (C, 9)
give as intersection the vertex A(31/3, 4.sqrt(35)/3).
Working on this triangle ABC and constructing the
point P we get the second cartesian coordinate of
P(ordinate) as the distance from BC that is the
required ETC search value.
But this method is not effective because when
constructing we want always a more convenient figure
that can be changed.
So besides ABC in the same screen I have constructed
an arbitrary triangle A'B'C'.
GSP has the ability to mark the vertices in the
sequence A',A,B',B,C',C and going to edit mode
to create on the screen a button1 that makes the
movement of triangle A'B'C' to triangle ABC.
This movement can be done instantly or slowly by a
number of choo-choo steps.
So working on triangle A'B'C'
(the triangle ABC can be hidden)
free to change the triangle A'B'C' if needed,
I construct the point P e.g. the incenter of A'B'C'
and by a clik on the button1 A'B'C' is transformed
to ABC and P is transformed dynamicaly to the incenter
of ABC and now is easy to see the second cartesian
coordinate of this P that is the required ETC search
value for X(1). Of course the precision is not great
but for me it is a quick evidence if the point is
in ETC. My GSP version has by default greater
precision 3 decimal digits and for X(1) gives on
screen coordinates (5.000, 1.690) but multiplying the
number 1.690 in the calculation option of GSP with
the number 1 000 000 I get the number 1690307.733
So this graphical method gives precision
1.690307733 instead of the correct 1.690308509457.
I created also before some years on the same screen
button2 moving A'B'C' to antimedial triangle of ABC
button3 moving A'B'C' to medial triangle of ABC
button4 moving A'B'C' to excenters triangle of ABC
button5 moving A'B'C' to intouch triangle of ABC
button6 moving A'B'C' to tangential triangle of ABC

Button2 for X(1) gives 4,507487511 instead of the
correct 4.507489358552 which means that X(1)
of a triangle is X(8) of its medial triangle.
(Of course this is not absolutely sure)

Button3 for X(1) gives 3.098899292 instead of the
correct 3.098898934004 which means that X(1)
of a triangle is X(10) of its antimedial triangle.

Button4 for X(1) gives 5.601338330 instead of the
correct 5.601337607766 which means that X(1)
of a triangle is X(164) of its orthic triangle.

Button5 for X(1) gives 0.975427199 instead of the
correct 0.975427635249 which means that X(1)
of a triangle is X(177) of its tangential triangle.

Button6 for X(1) gives -1.326971344 which means that
X(1) of a triangle is not an ETC point of its
intouch triangle.

Francois has Cabri the same ability?

Best regards
Nikos Dergiades

>. . .
> Therefore we start with our beloved reference
> triangle ABC in the real
> affine plane and 2 inscribed triangles
> a(0)b(0)c(0)and a(1)b(1)c(1).
> As in my previous post, I will note S(L,M,N) the
> signed area defined by the
> following equation:
> S(L,M, N) = f(LM, LN) where f is some
> skewsymmetric bilinear form and LM
> beeing the vector with origin L and end M.
> So S is defined up to a factor but it does not
> matter for we will use only
> area's ratios!
> We define 3 reals (a, b, c) by the equations:
> a(0)a(1) = a. BC; b(0)b(1) = b. CA; c(0)c(1) = c. AB
> Once again, note that a, b, c are not the lengths of
> the sides of ABC for
> there is no metric for the moment.
> Choose any point O in the plane and define points U,
> V, W by :
> OU = a(0)a(1); OV = b(0)b(1); OW = c(0)c(1)
> Here on this side of Atlantic ocean, we said we have
> "vectorialized" the
> plane in O.
> 1° I will suppose that these 3 points U, V, W are
> not on a same line.
> 2° So O has normalized barycentrics (u, v, w) wrt
> UVW, that is to say:
> x. OU + y. OV + z.OW = 0 with x + y + z = 1.
> and I will suppose that : x.y.z <> 0 , i.e none of
> the 3 reals x, y, z is
> equal to 0.
>
> As (1/a) OU + (1/b) OV + (1/c) OW = 0
>
> we have: x = b.c/ (b.c + c.a + a.b); y = c.a/ (b.c +
> c.a + a.b); x = a.b/ (
> b.c + c.a + a.b)
> Let P be the point with barycentrics (x,y,z) wrt
> ABC, then P is the areal
> center of the triangles a(0)b(0)c(0) and
> a(1)b(1)c(1).
>
> In fact S(P,b(0),b(1)) = f(AP, b(0)b(1)), (look at
> my previous post!)
> So S(P, b(0), b(1) = f(y. AB + z. AC, b. CA) =
> y.b.f(AB, CA) = -(a.b.c)/(b.c+
> c.a + a.b) S(A, B, C)
> Of course in the same way, we have:
> S(P, c(0), c(1)) = f(AP, c(0)c(1)) = f(y. AB + z.
> AC, c. AB) = z.c.f(AC, AB)
> = -(a.b.c)/(b.c + c.a + a.b).S(A, B, C)
> and we are done.
> From the vectorial equation: x. a(0)a(1) + y.
> b(0)b(1) + z. c(0)c(1) = 0
> we get the affine equation:
> x. a(0) + y.b(0) + z. c(0) = x. a(1) + y. b(1) + z.
> c(1) = E
> So the point E is the equicenter and the fixed
> point of the affine map
> sending a(0)b(0)c(0) onto a(1)b(1)c(1)
> Note that if g(0) is the affine map sending
> a(0)b(0)c(0) onto ABC, then we
> have g(0)(E) = P as I have said in one of my
> previous post.
> Now you will tell me where are the choo-choos?
> A choo-choo is a train going on smoothly on its
> tracks that is to say a
> rectilinear uniform motion (r.u.m)
> a(0) and a(1) are the positions at time 0 and time
> 1 of the choo-choo: t
> --> a(t) = (1 -t) a(0) + t . a(1) = a(0) + t . OU
> b(0) and b(1) are the positions at time 0 and time
> 1 of the choo-choo: t
> --> b(t) = (1 -t) b(0) + t . b(1) = b(0) + t . OV
> c(0) and c(1) are the positions at time 0 and time
> 1 of the choo-choo: t
> --> c(t) = (1 -t) c(0) + t . c(1) = a(0) + t . OW
> Notice that the speeds are just OU, OV, OW.
> It is easy to see that E = x. a(t) + y. b(t) + z.
> c(t), E is the equicenter
> of the choo-choos, that is the point of view of
> Neuberg.
> In the same way, we also have:
> S(P, a(t), a(u))) = S(P, b(t), b(u)) = S(P, c(t),
> c(u))
> so we can also tell that P is the areal center of
> the choo-choos.
> . . .

___________________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr
• Dear friends, some typos in my last message Instead of xEA +yEB +zEC -xEA +yEB +zEC =x.A A +y.B B +z.C C = BC + CA + AB = 0 the correct is
Message 4 of 10 , Aug 4, 2007
View Source
• 0 Attachment
Dear friends,
some typos in my last message

Instead of
xEA"+yEB"+zEC"-xEA'+yEB'+zEC'=x.A'A"+y.B'B"+z.C'C"
= BC + CA + AB = 0
the correct is
xEA"+yEB"+zEC"-(xEA'+yEB'+zEC')=
=x(EA"-EA')+y(EB"-EB')+z(EC"-EC')
=x.A'A"+y.B'B"+z.C'C" = BC + CA + AB = 0

and instead of
Q is the map of the affine map sending A'B'C'->ABC.
the correct is
Q is the map of E in the affine map
sending A'B'C'->ABC.

Best regards
Nikos Dergiades

> Dear Francois, and Tuan. Thanks Francois,
> for your proof that can be written another way as:
> Let A'B'C', A"B"C" be two inscribed triangles in
> ABC,
> vector BC = x.A'A", CA = y.B'B", AB = z.C'C" and let
>
> Q = (x : y : z) wrt ABC in barycentrics
> E = (x : y : z) wrt A'B'C'.
> It is known that
> S(Q,B,C) = x.S(A,B,C)/(x+y+z).Hence
> S(Q,A',A")=f(A'A",A'Q)=f(BC/x,A'B+BQ)=(1/x).f(BC,BQ)
> =(1/x).S(Q,B,C)=S(A,B,C)/(x+y+z).
> Similarly S(Q,B',B")=S(Q,C',C")=S(Q,A',A")
> Q is the areal point of triangles A'B'C', A"B"C".
>
> Since x.EA' + y.EB' + z.C' = 0 we get
> xEA"+yEB"+zEC"-xEA'+yEB'+zEC'=x.A'A"+y.B'B"+z.C'C"
> = BC + CA + AB = 0
> or x.EA"+y.EB"+z.EC" = 0
> or x.EAt+y.EBt+z.ECt = 0 where the triangle
> AtBtCt is the choo-choo instance At=(1-t)A'+tA" . .
> .
> Hence E has the same x : y : z barycentrics wrt
> AtBtCt
> and is the equicenter of the choo-choo.
> Q is the map of the affine map sending A'B'C'->ABC.

___________________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr
• Dear Antreas Thanks for your useful remarks. I am not a specialist of search values like Anh Tuan , so I have to think about it before answering. Good news,
Message 5 of 10 , Aug 4, 2007
View Source
• 0 Attachment
Dear Antreas
Thanks for your useful remarks.
I am not a specialist of search values like Anh Tuan , so I have to think
about it before answering.

Good news, one of my sons living in Boston has just offered me GSP and I
begin to play with it!

I have also something to say about equicenters and areal centers.
In f

On 8/4/07, Nikolaos Dergiades <ndergiades@...> wrote:
>
> Dear Francois, and Tuan. Thanks Francois,
> for your proof that can be written another way as:
> Let A'B'C', A"B"C" be two inscribed triangles in ABC,
> vector BC = x.A'A", CA = y.B'B", AB = z.C'C" and let
> Q = (x : y : z) wrt ABC in barycentrics
> E = (x : y : z) wrt A'B'C'.
> It is known that
> S(Q,B,C) = x.S(A,B,C)/(x+y+z).Hence
> S(Q,A',A")=f(A'A",A'Q)=f(BC/x,A'B+BQ)=(1/x).f(BC,BQ)
> =(1/x).S(Q,B,C)=S(A,B,C)/(x+y+z).
> Similarly S(Q,B',B")=S(Q,C',C")=S(Q,A',A")
> Q is the areal point of triangles A'B'C', A"B"C".
>
> Since x.EA' + y.EB' + z.C' = 0 we get
> xEA"+yEB"+zEC"-xEA'+yEB'+zEC'=x.A'A"+y.B'B"+z.C'C"
> = BC + CA + AB = 0
> or x.EA"+y.EB"+z.EC" = 0
> or x.EAt+y.EBt+z.ECt = 0 where the triangle
> AtBtCt is the choo-choo instance At=(1-t)A'+tA" . . .
> Hence E has the same x : y : z barycentrics wrt AtBtCt
> and is the equicenter of the choo-choo.
> Q is the map of the affine map sending A'B'C'->ABC.
>
> By the way this choo-choo process (Antrea mi gelas)
> is very usefull in dynamic geometry.
> Forgive me for I am going to add some more lines
> and give a long message and may be these thoughts
> are trivial to you or may be you will enjoy it
> as Tuan likes finding ETC search values.
> If we have a geometric construction and find a
> triangle center P we can find the ETC search value
> of this point working algebraicly with a computer
> Mathematics packet or as I often do working with GSP.
> One method is to construct the constant triangle
> ABC that ETC uses (a, b, c) = (6, 9, 13)
> with cartesian coordinates
> B(0, 0), C(6, 0) and the circles (B, 13), (C, 9)
> give as intersection the vertex A(31/3, 4.sqrt(35)/3).
> Working on this triangle ABC and constructing the
> point P we get the second cartesian coordinate of
> P(ordinate) as the distance from BC that is the
> required ETC search value.
> But this method is not effective because when
> constructing we want always a more convenient figure
> that can be changed.
> So besides ABC in the same screen I have constructed
> an arbitrary triangle A'B'C'.
> GSP has the ability to mark the vertices in the
> sequence A',A,B',B,C',C and going to edit mode
> to create on the screen a button1 that makes the
> movement of triangle A'B'C' to triangle ABC.
> This movement can be done instantly or slowly by a
> number of choo-choo steps.
> So working on triangle A'B'C'
> (the triangle ABC can be hidden)
> free to change the triangle A'B'C' if needed,
> I construct the point P e.g. the incenter of A'B'C'
> and by a clik on the button1 A'B'C' is transformed
> to ABC and P is transformed dynamicaly to the incenter
> of ABC and now is easy to see the second cartesian
> coordinate of this P that is the required ETC search
> value for X(1). Of course the precision is not great
> but for me it is a quick evidence if the point is
> in ETC. My GSP version has by default greater
> precision 3 decimal digits and for X(1) gives on
> screen coordinates (5.000, 1.690) but multiplying the
> number 1.690 in the calculation option of GSP with
> the number 1 000 000 I get the number 1690307.733
> So this graphical method gives precision
> 1.690307733 instead of the correct 1.690308509457.
> I created also before some years on the same screen
> button2 moving A'B'C' to antimedial triangle of ABC
> button3 moving A'B'C' to medial triangle of ABC
> button4 moving A'B'C' to excenters triangle of ABC
> button5 moving A'B'C' to intouch triangle of ABC
> button6 moving A'B'C' to tangential triangle of ABC
>
> Button2 for X(1) gives 4,507487511 instead of the
> correct 4.507489358552 which means that X(1)
> of a triangle is X(8) of its medial triangle.
> (Of course this is not absolutely sure)
>
> Button3 for X(1) gives 3.098899292 instead of the
> correct 3.098898934004 which means that X(1)
> of a triangle is X(10) of its antimedial triangle.
>
> Button4 for X(1) gives 5.601338330 instead of the
> correct 5.601337607766 which means that X(1)
> of a triangle is X(164) of its orthic triangle.
>
> Button5 for X(1) gives 0.975427199 instead of the
> correct 0.975427635249 which means that X(1)
> of a triangle is X(177) of its tangential triangle.
>
> Button6 for X(1) gives -1.326971344 which means that
> X(1) of a triangle is not an ETC point of its
> intouch triangle.
>
> Francois has Cabri the same ability?
>
> Best regards
> Nikos Dergiades
>
>
> >. . .
> > Therefore we start with our beloved reference
> > triangle ABC in the real
> > affine plane and 2 inscribed triangles
> > a(0)b(0)c(0)and a(1)b(1)c(1).
> > As in my previous post, I will note S(L,M,N) the
> > signed area defined by the
> > following equation:
> > S(L,M, N) = f(LM, LN) where f is some
> > skewsymmetric bilinear form and LM
> > beeing the vector with origin L and end M.
> > So S is defined up to a factor but it does not
> > matter for we will use only
> > area's ratios!
> > We define 3 reals (a, b, c) by the equations:
> > a(0)a(1) = a. BC; b(0)b(1) = b. CA; c(0)c(1) = c. AB
> > Once again, note that a, b, c are not the lengths of
> > the sides of ABC for
> > there is no metric for the moment.
> > Choose any point O in the plane and define points U,
> > V, W by :
> > OU = a(0)a(1); OV = b(0)b(1); OW = c(0)c(1)
> > Here on this side of Atlantic ocean, we said we have
> > "vectorialized" the
> > plane in O.
> > 1� I will suppose that these 3 points U, V, W are
> > not on a same line.
> > 2� So O has normalized barycentrics (u, v, w) wrt
> > UVW, that is to say:
> > x. OU + y. OV + z.OW = 0 with x + y + z = 1.
> > and I will suppose that : x.y.z <> 0 , i.e none of
> > the 3 reals x, y, z is
> > equal to 0.
> >
> > As (1/a) OU + (1/b) OV + (1/c) OW = 0
> >
> > we have: x = b.c/ (b.c + c.a + a.b); y = c.a/ (b.c +
> > c.a + a.b); x = a.b/ (
> > b.c + c.a + a.b)
> > Let P be the point with barycentrics (x,y,z) wrt
> > ABC, then P is the areal
> > center of the triangles a(0)b(0)c(0) and
> > a(1)b(1)c(1).
> >
> > In fact S(P,b(0),b(1)) = f(AP, b(0)b(1)), (look at
> > my previous post!)
> > So S(P, b(0), b(1) = f(y. AB + z. AC, b. CA) =
> > y.b.f(AB, CA) = -(a.b.c)/(b.c+
> > c.a + a.b) S(A, B, C)
> > Of course in the same way, we have:
> > S(P, c(0), c(1)) = f(AP, c(0)c(1)) = f(y. AB + z.
> > AC, c. AB) = z.c.f(AC, AB)
> > = -(a.b.c)/(b.c + c.a + a.b).S(A, B, C)
> > and we are done.
> > From the vectorial equation: x. a(0)a(1) + y.
> > b(0)b(1) + z. c(0)c(1) = 0
> > we get the affine equation:
> > x. a(0) + y.b(0) + z. c(0) = x. a(1) + y. b(1) + z.
> > c(1) = E
> > So the point E is the equicenter and the fixed
> > point of the affine map
> > sending a(0)b(0)c(0) onto a(1)b(1)c(1)
> > Note that if g(0) is the affine map sending
> > a(0)b(0)c(0) onto ABC, then we
> > have g(0)(E) = P as I have said in one of my
> > previous post.
> > Now you will tell me where are the choo-choos?
> > A choo-choo is a train going on smoothly on its
> > tracks that is to say a
> > rectilinear uniform motion (r.u.m)
> > a(0) and a(1) are the positions at time 0 and time
> > 1 of the choo-choo: t
> > --> a(t) = (1 -t) a(0) + t . a(1) = a(0) + t . OU
> > b(0) and b(1) are the positions at time 0 and time
> > 1 of the choo-choo: t
> > --> b(t) = (1 -t) b(0) + t . b(1) = b(0) + t . OV
> > c(0) and c(1) are the positions at time 0 and time
> > 1 of the choo-choo: t
> > --> c(t) = (1 -t) c(0) + t . c(1) = a(0) + t . OW
> > Notice that the speeds are just OU, OV, OW.
> > It is easy to see that E = x. a(t) + y. b(t) + z.
> > c(t), E is the equicenter
> > of the choo-choos, that is the point of view of
> > Neuberg.
> > In the same way, we also have:
> > S(P, a(t), a(u))) = S(P, b(t), b(u)) = S(P, c(t),
> > c(u))
> > so we can also tell that P is the areal center of
> > the choo-choos.
> > . . .
>
>
>
>
> ___________________________________________________________
> �������������� Yahoo!;
> ���������� �� ���������� �������� (spam); �� Yahoo! Mail
> �������� ��� �������� ������ ��������� ���� ��� �����������
> ��������� http://login.yahoo.com/config/mail?.intl=gr
>
>
>
>
> Yahoo! Groups Links
>
>
>
>

[Non-text portions of this message have been removed]
• Dear Antreas Sorry, I send my post too soon! I said from the 2 points equicenter and areal center, the first is the more important for we all know importance
Message 6 of 10 , Aug 4, 2007
View Source
• 0 Attachment
Dear Antreas
Sorry, I send my post too soon!
I said from the 2 points equicenter and areal center, the first is the more
important for we all know importance of the fixed points of a map in Math.
Remember the configuration of 2 inscribed triangles abc and a'b'c' in the
reference triangle ABC.
Equicenter E is the fixed point of the affine map f : abc --> a'b'c'.

My first idea was to recover E by the Sollertinski construction and then to
get the areal center S as g(E) where g is the affine map g: abc --> ABC

We can also get the areal center S by your construction as intersection of
the 3 equal area axis then get E = h(S) where h is the affine map: g^<-1>:
ABC --> abc.
So we get a different construction of the fixed point E.

But if we look at the general situation of 2 triangles abc and a'b'c',
intersections: A = bc/\b'c', B = ca /\c'a', C = ab/\a'b'
might not exist and so we cannot apply this construction.

So it seems that the Sollertinski construction only remains!
But no, my sketch is still valid for the areal center is not needed to get
E.
We start with any point O, we construct the vectors OU, OV, OW as I said in
my previous post.
In general points U, V, W are not in a same line and O has normalized
barycentrics (x,y,z) wrt UVW.
So O = x.U + y.V + z.W
As E = x.a + y.b + z. c, then E = h(O) where h is the affine map: UVW -->
abc.
I think this construction really new though I think we can find it in germ
in old books like Rouche-Comberousse (~ 1900).
The case where U, V, W are on a same line is also very interesting but I
have no room here to explain it in detail.
This is the case where the affine map f: abc --> a'b'c' may have no fixed
points or infinitely many on a same line.
And it is easy to reduce f to some "normal form" in this case.

Nevertheless areal center plays a central role in choo-choos along with
equicenter!
Friendly
Francois

[Non-text portions of this message have been removed]
• Dear Nikos Of course, your proof is better and one can effectively define directly on the drawing the barycentrics (x:y:z) of the areal center. In fact the
Message 7 of 10 , Aug 4, 2007
View Source
• 0 Attachment
Dear Nikos
Of course, your proof is better and one can effectively define directly on
the drawing the barycentrics (x:y:z) of the areal center. In fact the reals
(a:b:c) defined by A'A" = a. BC; B'B" = b.CA; C'C" = c.AB are the "affine"
speeds of the 3 choo-choos, that's why I feel like to call R(a:b:c) the
speed center of the choo-choos. Of course, R is the isotomic conjugate of
the areal center and it plays also a role in the theory.
Friendly
Francois

On 8/4/07, Nikolaos Dergiades <ndergiades@...> wrote:
>
> Dear friends,
> some typos in my last message
>
> Instead of
> xEA"+yEB"+zEC"-xEA'+yEB'+zEC'=x.A'A"+y.B'B"+z.C'C"
> = BC + CA + AB = 0
> the correct is
> xEA"+yEB"+zEC"-(xEA'+yEB'+zEC')=
> =x(EA"-EA')+y(EB"-EB')+z(EC"-EC')
> =x.A'A"+y.B'B"+z.C'C" = BC + CA + AB = 0
>
> and instead of
> Q is the map of the affine map sending A'B'C'->ABC.
> the correct is
> Q is the map of E in the affine map
> sending A'B'C'->ABC.
>
> Best regards
> Nikos Dergiades
>
>
>
>
> > Dear Francois, and Tuan. Thanks Francois,
> > for your proof that can be written another way as:
> > Let A'B'C', A"B"C" be two inscribed triangles in
> > ABC,
> > vector BC = x.A'A", CA = y.B'B", AB = z.C'C" and let
> >
> > Q = (x : y : z) wrt ABC in barycentrics
> > E = (x : y : z) wrt A'B'C'.
> > It is known that
> > S(Q,B,C) = x.S(A,B,C)/(x+y+z).Hence
> > S(Q,A',A")=f(A'A",A'Q)=f(BC/x,A'B+BQ)=(1/x).f(BC,BQ)
> > =(1/x).S(Q,B,C)=S(A,B,C)/(x+y+z).
> > Similarly S(Q,B',B")=S(Q,C',C")=S(Q,A',A")
> > Q is the areal point of triangles A'B'C', A"B"C".
> >
> > Since x.EA' + y.EB' + z.C' = 0 we get
> > xEA"+yEB"+zEC"-xEA'+yEB'+zEC'=x.A'A"+y.B'B"+z.C'C"
> > = BC + CA + AB = 0
> > or x.EA"+y.EB"+z.EC" = 0
> > or x.EAt+y.EBt+z.ECt = 0 where the triangle
> > AtBtCt is the choo-choo instance At=(1-t)A'+tA" . .
> > .
> > Hence E has the same x : y : z barycentrics wrt
> > AtBtCt
> > and is the equicenter of the choo-choo.
> > Q is the map of the affine map sending A'B'C'->ABC.
>
>
>
>
> ___________________________________________________________
> �������������� Yahoo!;
> ���������� �� ���������� �������� (spam); �� Yahoo! Mail
> �������� ��� �������� ������ ��������� ���� ��� �����������
> ��������� http://login.yahoo.com/config/mail?.intl=gr
>
>
>
>
> Yahoo! Groups Links
>
>
>
>

[Non-text portions of this message have been removed]
• Dear Francois, You answered to Antreas, I think by mistake. I am not Antreas. An interesting case is when abc, a b c are similar. If the triangle abc is
Message 8 of 10 , Aug 4, 2007
View Source
• 0 Attachment
Dear Francois,
You answered to Antreas, I think by mistake.
I am not Antreas.

An interesting case is when
abc, a'b'c' are similar.
If the triangle abc is changed then
the areal point is constant.
Best regards
Nikos Dergiades

> Dear Antreas
> Sorry, I send my post too soon!
> I said from the 2 points equicenter and areal
> center, the first is the more
> important for we all know importance of the fixed
> points of a map in Math.
> Remember the configuration of 2 inscribed triangles
> abc and a'b'c' in the
> reference triangle ABC.
> Equicenter E is the fixed point of the affine map f
> : abc --> a'b'c'.
>
> My first idea was to recover E by the Sollertinski
> construction and then to
> get the areal center S as g(E) where g is the affine
> map g: abc --> ABC
>
> We can also get the areal center S by your
> construction as intersection of
> the 3 equal area axis then get E = h(S) where h is
> the affine map: g^<-1>:
> ABC --> abc.
> So we get a different construction of the fixed
> point E.
>
> But if we look at the general situation of 2
> triangles abc and a'b'c',
> intersections: A = bc/\b'c', B = ca /\c'a', C =
> ab/\a'b'
> might not exist and so we cannot apply this
> construction.
>
> So it seems that the Sollertinski construction only
> remains!
> But no, my sketch is still valid for the areal
> center is not needed to get
> E.
> We start with any point O, we construct the vectors
> OU, OV, OW as I said in
> my previous post.
> In general points U, V, W are not in a same line and
> O has normalized
> barycentrics (x,y,z) wrt UVW.
> So O = x.U + y.V + z.W
> As E = x.a + y.b + z. c, then E = h(O) where h is
> the affine map: UVW -->
> abc.
> I think this construction really new though I think
> we can find it in germ
> in old books like Rouche-Comberousse (~ 1900).
> The case where U, V, W are on a same line is also
> very interesting but I
> have no room here to explain it in detail.
> This is the case where the affine map f: abc -->
> a'b'c' may have no fixed
> points or infinitely many on a same line.
> And it is easy to reduce f to some "normal form" in
> this case.
>
> Nevertheless areal center plays a central role in
> choo-choos along with
> equicenter!
> Friendly
> Francois
>
>
> [Non-text portions of this message have been
> removed]
>
>
>
>
> Yahoo! Groups Links
>
>
>
>
>

___________________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr
• Dear Nikos and (always understood of course !) Dear Antreas ... constant areal center . If we look at 2 inscribed directly similar triangles abc and a b c
Message 9 of 10 , Aug 5, 2007
View Source
• 0 Attachment
Dear Nikos and (always understood of course !) Dear Antreas

>
> An interesting case is when
> abc, a'b'c' are similar.
> If the triangle abc is changed then
> the areal point is constant.
> Best regards
> Nikos Dergiades
>
> Yes , this case is very important but I don't understand what you mean by
"constant areal center".
If we look at 2 inscribed directly similar triangles abc and a'b'c' in ABC,
then of course all triangles a(t)b(t)c(t) in the choo-choos are directly
similar to abc, (where a(t) = (1-t) . a + t . b and so on...)
This configuration was intensively studied in France with the pivot theory
(théorie du pivot).
The equicenter was named pivot of the inscribed triangles directly similar
between themselves.
If E is the equicenter, then the pedal triangle of E wrt ABC is a triangle
in the choo-choos, i.e a triangle with homologous vertices.
But what about the areal center S in this case?
It is the same for every pair of homologous triangles in the choo-choos.
Maybe that's what you mean?
S is simply the isogonal conjugate of E wrt ABC !

In the other way, if the equicenter E and the areal center S are isogonally
conjugate wrt ABC, then E is a pivot for all homologous triangles in the
choo-choos directly similar between themselves..
That explain why some configurations like Brocard points take place
naturally in choo-choos theory!

One of the most celebrated configuration, studied for example in
Rouche-Comberousse or F.G-M is that one of 3 directly similar triangles.
It amounts to study the product of 2 direct similarities.
This configuration is intimately linked with choo-choo theory and some of
its properties generalize very well.

I give you a little problem to think about on the sandy beaches during your
summer holidays:

Study the case of 2 inscribed triangles indirectly similar. What can be said
about the equicenter E and the areal center S?

Friendly
Francois

[Non-text portions of this message have been removed]
• ... a(t) = (1-t) .a + t . a of course! [Non-text portions of this message have been removed]
Message 10 of 10 , Aug 5, 2007
View Source
• 0 Attachment
I correct a little typo:

> >
> > Yes , this case is very important but I don't understand what you mean
> by "constant areal center".
> If we look at 2 inscribed directly similar triangles abc and a'b'c' in
> ABC, then of course all triangles a(t)b(t)c(t) in the choo-choos are
> directly similar to abc, (where a(t) = (1-t) . a + t . b and so on...)

a(t) = (1-t) .a + t . a'
of course!

[Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.