- Dear friends,

As I have been reworking, after the holidays, the stuff around the

equicenter and the areal center, I have some trouble about - by now-

three related questions:

Two inscribed triangles a1b1c1, a2b2c2 have an equicenter E as

defined and constructed by Thebault. E is the fixed poiny of the

affine transformation sending a1 to a2, b1 to b2, c1 to c2. When

a1b1c1, a2b2c2 are similar, E is its similitude center.

First question: a "choo-choo" menas "a family of inscribed triangles

having a common equicenter" or "a family of inscribed similar

triangles having a common center of similitude"?

Second question: A message stated that the knowledge of E allowed to

build all the others triangles in the family. Is that true for only

inscribed triangles or for inscribed-and-similar triangles? Another

message asserts than the equicenter and the areal center determine

the family. What is true?

Third question: Given a1b1c1, a2b2c2 inscribed, E its equicenter, how

can we construct a3b3c3 also inscribed and having the same equicenter

E wrt a1b1c1 and a2b2c2?

Greetings from a hot Barcelona,

Quim - Dear Quim

As I am responsible but not guilty (as some of our senior ministers like to

remind us) of all the choo-choo fuss, I have to reply of course.

First what is a choo-choo?

It is just a joke to name a uniform linear motion . Here in France, we would

say "teuf-teuf".

If we know the position a(0) for time t = 0 and the position a(1) for time t

= 1, then at time t, the position of the choo-choo is just a(t) = (1 - t)

a(0) + t a(1) = a(0) + t a(0)a(1)

So equation "a(t) = (1 - t) a(0) + t a(1) "means that a(t) is just the

barycenter of points a(0) with mass 1 - t and a(1) with mass t or in other

words that the pair (1 - t, t) is formed by the barycentrics of a(t) in the

affine frame {a(0), a(1)} of the line on which the choo-choo a(t) moves on.

Equation "a(t) = a(0) + t a(0)a(1)" means that the speed of the choo-choo

is the vector a(0)a(1).

Notice that this choo-choo definition is affine, an euclidian structure is

not needed and so I am happy as a bourbakist heir!

The big problem starts with 2 choo-choos a(t) and b(t) going on smoothly on

their tracks L and L'.

What can we say about it?

When I was a little child more than sixty years ago, I was fascinated by

trains going on some parallel tracks with my own train. When I looked at

some passenger of the other train, he seems to stand still or to move

forward or backward.

I did not knew at this time that sixty years after, I should have to

template (or modelize?), "modeliser in french", this situation in geometry.

So we have our 2 choo-choos a(t) and b(t) on their tracks L and L'.

We get a correspondance a(t) <--> b(t) between lines L and L' and this

correspondance is affine.

In fact this is just this correspondence which is interesting and it is

possible to change the definition of a choo-choo to get rid of the time and

so feel like God himself.

With 2 choo-choos many interesting problems arise and one can begin to do

geometry.

The first big one is given positions a(0), b(0), a(1), b(1) is how to

construct b(t) when a(t) is given.

This is not too difficult a task with Cabri or GSP and it is an interesting

problem for young pupils learning geometry.

The situation is already very rich, lines L and L' can be the same or

distinct, parallel or not. It is not very obvious to have a construction

working in all these cases!

But with 3 choo-choos or more, we take it away, "C'est le pied" in french,

for we meet our loving triangles and only God know how Hyacinthists are

triangle addicts.

So we have our 3 choo-choos a(t), b(t), c(t) going on their tracks and they

are entirely determined when we know the 2 triangles a(0)b(0)c(0) and

a(1)b(1)c(1).

I will label La, Lb, Lc their respective tracks and they can form a triangle

ABC ( but this is not necessary) such that La = BC, Lb = CA and Lc = AB. It

is the general situation, interesting enough to forget others at least in

the beginning.

So we get a family of ABC-inscribed triangles a(t)b(t)c(t).

Now, I will explain you what are the equicenter and the areal center.

The most important and known for a long time is the equicenter, maybe it was

Neuberg who gives this name for this point but I am not certain.

If we look at the affine map f : a(0)b(0)c(0) --> a(1)b(1)c(1), i.e f(a(0))

= a(1), f(b(0)) = b(1), f(c(0)) = c(1), then in general, this map has only a

fixed point E and this point has same barycentrics (x:y:z) wrt both

triangles a(0)b(0)c(0) and a(1)b(1)c(1).

But as we are working with choo-choos, it is easy to see that point E has

also the same barycentrics (x:y:z) wrt all triangles a(t)b(t)c(t) of the

family. That's why Neuberg named it equicenter and besides at Neuberg time

aound 1900, mathematicians don't like to use maps nor affine or linear maps

of course. So the barycentric schemas was more suitable.

Now you ask me if given a triangle ABC and a point E, do these data

determine the choo-choos a(t)b(t)c(t) and the answer is not!

Of course it was the first question I had to solve and for that I looked at

the point S having barycentrics (x:y:z) wrt triangle ABC, the same

barycentrics point E has wrt every triangle a(t)b(t)c(t).

And miracle, this point S is just the areal center, that is to say points

a(t), b(t), c(t) has the same areal speed wrt point S.

If we look at the signed area, we have the following equations:

Area(S, a(t), a(u)) = Area(S, b(t), b(u)) = Area(S, c(t), c(u)) = (u -t)

Area(S,a(0), a(1)) = (t - u) Area(S, b(0), b(1)) = (t-u) Area(S, c(0), c(1).

So the common areal speed is Area(S, a(0), a(1)) = Area(S, b(0), b(1)) =

Area(S,c(0),c(1))

Notice that both points E and S have affine definitions, no metric structure

is needed and so their constructions is also affine.

Nikos constuction of S is affine as mine and Sollertinski construction of E

is affine.

If g(t) is the affine map g(t): ABC --> a(t)b(t)c(t), then g(t)(S) = E for

every time t.

It is easy to see that this last equation determines entirely the

choo-choos!

In other words the data, A, B, C, S, E determines the choo-choos with the

little restriction that the areal center S cannot be on the sides of the

triangle ABC.

I have found an affine construction of points b(t) and c(t) with the data A,

B, C, S, E and a(t) on line BC.

It is my main Cabri or GSP tool in choo-choo theory.

So you can start with ABC-inscribed triangles a(0)b(0)c(0) and a(1)b(1)c(1)

and construct point S and E.

Notice you can first construct S with Nikos construction and then get E by

g(t)(S) = E or construct E by Sollertinski construction and then get S.

I have also in one of my previous post given another construction of E, more

natural and very different from Sollertinski.

Of course, even if a metric structure is not needed for the definitions,

such a structure gives birth to many interesting choo-choos.

I give you 2 examples:

1° Choos-choos where triangles a(t)b(t)c(t) are directly similar between

themselves.

As you said it, equicenter E is just the common center of similarity and in

this case, areal center S is the isogonal conjugate of equicenter E wrt

triangle ABC.

2° Pedal choo-choos.

You have a first choo-choo m(t) going on its line L and you look at the

pedal triangles a(t)b(t)c(t) of point m(t) wrt triangle ABC.

In fact, I think these pedal choo-choos were the first studied by Neuberg.

He proves that the equicenter was the orthopole of line L wrt triangle ABC.

As for me, I proved that the areal center S was on the ABC-circumcircle the

point of which the Simson line is parallel to line L.

That's all I can say for the moment but the theory is very very rich and

there are many other interesting results.

Friendly

Francois

PS Of course, I can send you some of my tools.

On 8/29/07, Quim Castellsaguer <qcastell@...> wrote:

>

> Dear friends,

> As I have been reworking, after the holidays, the stuff around the

> equicenter and the areal center, I have some trouble about - by now-

> three related questions:

> Two inscribed triangles a1b1c1, a2b2c2 have an equicenter E as

> defined and constructed by Thebault. E is the fixed poiny of the

> affine transformation sending a1 to a2, b1 to b2, c1 to c2. When

> a1b1c1, a2b2c2 are similar, E is its similitude center.

> First question: a "choo-choo" menas "a family of inscribed triangles

> having a common equicenter" or "a family of inscribed similar

> triangles having a common center of similitude"?

> Second question: A message stated that the knowledge of E allowed to

> build all the others triangles in the family. Is that true for only

> inscribed triangles or for inscribed-and-similar triangles? Another

> message asserts than the equicenter and the areal center determine

> the family. What is true?

> Third question: Given a1b1c1, a2b2c2 inscribed, E its equicenter, how

> can we construct a3b3c3 also inscribed and having the same equicenter

> E wrt a1b1c1 and a2b2c2?

>

> Greetings from a hot Barcelona,

> Quim

>

>

>

[Non-text portions of this message have been removed] - Dear Quim,

> Two inscribed triangles a1b1c1, a2b2c2 have an equicenter E as

I'm not sure about who first defined the term equicenter, but I found

> defined and constructed by Thebault. E is the fixed point of the

> affine transformation sending a1 to a2, b1 to b2, c1 to c2. When

> a1b1c1, a2b2c2 are similar, E is its similitude center.

something online that suggests a person named E. Duporcq coined the

term.

http://links.jstor.org/sici?sici=0002-9890%28194606%2F07%2953%3A6%

3C324%3ACPCAS%3E2.0.CO%3B2-7&size=LARGE&origin=JSTOR-enlargePage

This link will probably not work after the message is posted and the

lines will have to be concatenated to make it link properly to JSTOR.

I can send a screenshot of the introductory materials if anyone is

interested. I want to get my hands on this entire three page paper.

Friendly, Jeff