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Three questions about equicenters

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  • Quim Castellsaguer
    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
    Message 1 of 3 , Aug 29, 2007
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      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
    • Francois Rideau
      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
      Message 2 of 3 , Aug 31, 2007
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        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]
      • Jeff
        Dear Quim, ... I m not sure about who first defined the term equicenter, but I found something online that suggests a person named E. Duporcq coined the term.
        Message 3 of 3 , Aug 31, 2007
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          Dear Quim,

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

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