Loading ...
Sorry, an error occurred while loading the content.
 

Re: [EMHL] Nagel point

Expand Messages
  • Nikolaos Dergiades
    Dear Vladimir, ... I think you are right. Francois problem is not exactly the same with Yaglom s. And in my solution the equation of x is now different. Hence
    Message 1 of 24 , Jun 6, 2008
      Dear Vladimir,
      >
      > But here the transversal also cuts the *side* lines, not
      > any three
      > lines through A, B, C as in Yaglom's book, n'est-ce
      > pas?

      I think you are right.
      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



      ___________________________________________________________
      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
    • Nikolaos Dergiades
      Dear Francois, you said to Vladimir ... No. The general case is that the three lines form a triangle A*B*C* different from ABC the transversal is A B C , the
      Message 2 of 24 , Jun 6, 2008
        Dear Francois,
        you said to Vladimir

        > For your second remark, I suppose you want to say that
        > Yaglom don't suppose
        > the 3 lines form necessarily a triangle.

        No. The general case is that the three lines
        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



        ___________________________________________________________
        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
      • Francois Rideau
        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 =
        Message 3 of 24 , Jun 6, 2008
          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 choo-choo 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 (choo-choo) 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


          [Non-text portions of this message have been removed]
        • Nikolaos Dergiades
          Dear Francois, ... Yes. ... No. 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
          Message 4 of 24 , Jun 6, 2008
            Dear Francois,

            > 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?

            Yes.

            > 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?

            No.
            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












            ___________________________________________________________
            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
          • Francois Rideau
            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
            Message 5 of 24 , Jun 7, 2008
              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


              [Non-text portions of this message have been removed]
            • Nikolaos Dergiades
              Dear Francois, thank you very much. Francisco also sent me a similar message that simplifies my own formulas. [Francisco] ... [Francois] ... So if we put the
              Message 6 of 24 , Jun 7, 2008
                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
                > 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

                [Francois]
                > 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

                So if we put the points with
                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















                ___________________________________________________________
                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
              • Francois Rideau
                Dear Nikos My matrix P is named matrix of the affine map in the affine frame {A, B, C}. I remind you that its columns are the (normalized) barycentrics of
                Message 7 of 24 , Jun 7, 2008
                  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) = (X-1)*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


                  [Non-text portions of this message have been removed]
                • Francois Rideau
                  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 --
                  Message 8 of 24 , Jun 8, 2008
                    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


                    [Non-text portions of this message have been removed]
                  • Nikolaos Dergiades
                    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 =
                    Message 9 of 24 , Jun 8, 2008
                      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
                      >
                      >
                      > [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
                    • Francois Rideau
                      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 =
                      Message 10 of 24 , Jun 8, 2008
                        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 choo-choo 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 2th-degree 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,
                        > 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
                        > >
                        > >
                        > > [Non-text portions of this message have been removed]
                        > >
                        > >
                        > > ------------------------------------
                        > >
                        > > Yahoo! Groups Links
                        > >
                        > >
                        > >
                        >
                        >
                        > ___________________________________________________________
                        > �������������� Yahoo!;
                        > ���������� �� ���������� �������� (spam); �� Yahoo! Mail
                        > �������� ��� �������� ������ ��������� ���� ��� �����������
                        > ��������� http://login.yahoo.com/config/mail?.intl=gr
                        >
                        >
                        > ------------------------------------
                        >
                        > Yahoo! Groups Links
                        >
                        >
                        >
                        >


                        [Non-text portions of this message have been removed]
                      • jpehrmfr
                        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
                        Message 11 of 24 , Jun 11, 2008
                          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. Jean-Pierre
                        Your message has been successfully submitted and would be delivered to recipients shortly.