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focus of circumscribing parabolas

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  • Eisso J. Atzema
    Dear Garcia, Nikolaos, and François, here is an observation that might lead to a more simple construction of the focal points of the two parabolas
    Message 1 of 11 , Sep 22, 2007
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      Dear Garcia, Nikolaos, and François,

      here is an observation that might lead to a more simple construction of
      the focal points of the two parabolas circumscribing a (convex)
      quadrilateral:

      Note that for every triangle self-polar w.r.t. a parabola the three
      lines connecting the midpoints of its sides are tangent lines to the
      parabola. This means that the focus of a parabola lies on the nine-point
      circle of any triangle which is self-polar with respect to that
      parabola. Since the diagonal triangle of any quadrilateral is
      self-polar w.r.t. any conic section circumscribing the quadrilateral, it
      follows that the focal points of both circumscribing parabolas are on
      the nine-point circle of the diagonal triangle. Moreover, in addition to
      the four points on the parabola, we now have four lines tangent to the
      parabola (the three lines connecting the midpoints of the diagonal
      triangle and the line at infinity). Perhaps it is possible to now find
      the focal points as the points of intersection of this nine-point circle
      with a straight line.

      best,
      Eisso

      --

      ========================================
      Eisso J. Atzema, Ph.D.
      Department of Mathematics & Statistics
      University of Maine
      Orono, ME 04469
      Tel.: (207) 581-3928 (office)
      (207) 866-3871 (home)
      Fax.: (207) 581-3902
      E-mail: atzema@...
      ========================================
    • Francois Rideau
      Dear friends and Eisso and Garcia To put it right, I try to find some relationship between the perspector S and the focus F of an ABC- circumparabola . I
      Message 2 of 11 , Sep 22, 2007
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        Dear friends and Eisso and Garcia



        To put it right, I try to find some relationship between the perspector S
        and the focus F of an ABC- circumparabola <P>.

        I don't find it but I notice some interesting facts certainly known for a
        long time.

        As you know it, the locus of S is the inner Steiner ABC- ellipse.

        The locus of F looks like some cubic with lines parallel to the sidelines
        on the centroid as 3 real asymptotes.

        The envelope of line SF is some high degree curve(?)

        The envelope of the axis of <P> looks very nice as the medial triangle
        Steiner deltoid.

        In other words , axis of the ABC-circumparabola are just the Simson lines
        of the medial triangle.

        I would like to have a proof or any reference of this last fact. Any idea?

        Friendly

        Francois


        On 9/22/07, Eisso J. Atzema <atzema@...> wrote:
        >
        > Dear Garcia, Nikolaos, and François,
        >
        > here is an observation that might lead to a more simple construction of
        > the focal points of the two parabolas circumscribing a (convex)
        > quadrilateral:
        >
        > Note that for every triangle self-polar w.r.t. a parabola the three
        > lines connecting the midpoints of its sides are tangent lines to the
        > parabola. This means that the focus of a parabola lies on the nine-point
        > circle of any triangle which is self-polar with respect to that
        > parabola. Since the diagonal triangle of any quadrilateral is
        > self-polar w.r.t. any conic section circumscribing the quadrilateral, it
        > follows that the focal points of both circumscribing parabolas are on
        > the nine-point circle of the diagonal triangle. Moreover, in addition to
        > the four points on the parabola, we now have four lines tangent to the
        > parabola (the three lines connecting the midpoints of the diagonal
        > triangle and the line at infinity). Perhaps it is possible to now find
        > the focal points as the points of intersection of this nine-point circle
        > with a straight line.
        >
        > best,
        > Eisso
        >
        > --
        >
        > ========================================
        > Eisso J. Atzema, Ph.D.
        > Department of Mathematics & Statistics
        > University of Maine
        > Orono, ME 04469
        > Tel.: (207) 581-3928 (office)
        > (207) 866-3871 (home)
        > Fax.: (207) 581-3902
        > E-mail: atzema@... <atzema%40math.umaine.edu>
        > ========================================
        >
        >
        >


        [Non-text portions of this message have been removed]
      • Francois Rideau
        Dear Eisso Very nice, you locating the focus on the nine point circle of the diagonal triangle of ABCD. The end of the construction is now clear: the focus is
        Message 3 of 11 , Sep 23, 2007
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          Dear Eisso
          Very nice, you locating the focus on the nine point circle of the diagonal
          triangle of ABCD.
          The end of the construction is now clear: the focus is the isogonal
          conjugate of the point at infinity of the axis wrt the medial triangle of
          <T>, following some Poncelet theorem and as you notice it, this point at
          infinity is also a point at infinity of the nine point conic of ABCD.
          Now if you summarize this construction, all the problem is to construct the
          points at infinity of this nine point conic of ABCD. Newton solved it ,
          (read Dorie!), and I am eager to know your construction. As for me, I have
          an affine construction of these points at infinity for this conic have an
          affine definition..
          Friendly
          Francois
          On 9/22/07, Eisso J. Atzema <atzema@...> wrote:
          >
          > Dear Garcia, Nikolaos, and François,
          >
          > here is an observation that might lead to a more simple construction of
          > the focal points of the two parabolas circumscribing a (convex)
          > quadrilateral:
          >
          > Note that for every triangle self-polar w.r.t. a parabola the three
          > lines connecting the midpoints of its sides are tangent lines to the
          > parabola. This means that the focus of a parabola lies on the nine-point
          > circle of any triangle which is self-polar with respect to that
          > parabola. Since the diagonal triangle of any quadrilateral is
          > self-polar w.r.t. any conic section circumscribing the quadrilateral, it
          > follows that the focal points of both circumscribing parabolas are on
          > the nine-point circle of the diagonal triangle. Moreover, in addition to
          > the four points on the parabola, we now have four lines tangent to the
          > parabola (the three lines connecting the midpoints of the diagonal
          > triangle and the line at infinity). Perhaps it is possible to now find
          > the focal points as the points of intersection of this nine-point circle
          > with a straight line.
          >
          > best,
          > Eisso
          >
          > --
          >
          > ========================================
          > Eisso J. Atzema, Ph.D.
          > Department of Mathematics & Statistics
          > University of Maine
          > Orono, ME 04469
          > Tel.: (207) 581-3928 (office)
          > (207) 866-3871 (home)
          > Fax.: (207) 581-3902
          > E-mail: atzema@... <atzema%40math.umaine.edu>
          > ========================================
          >
          >
          >


          [Non-text portions of this message have been removed]
        • Nikolaos Dergiades
          Dear Francois, ... I think you are right as I can conclude from sketch. Let H be the orthocenter of ABC and A1B1C1 be the medial triangle of ABC. There exists
          Message 4 of 11 , Sep 23, 2007
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            Dear Francois,

            > In other words , axis of the ABC-circumparabola are
            > just the Simson lines
            > of the medial triangle.
            >
            > I would like to have a proof or any reference of
            > this last fact. Any idea?

            I think you are right as I can conclude from sketch.
            Let H be the orthocenter of ABC and
            A1B1C1 be the medial triangle of ABC.
            There exists a unique point D on the circumcircle of
            ABC such that the isogonal conjugate of the tangent
            at D to the circumcircle is the ABC-circumparabola.
            Let P be the mid point of HD.
            The axis of the ABC-circumparabola is the
            Simson line of P wrt the triangle A1B1C1.

            Best regards
            Nikos Dergiades



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          • Francois Rideau
            Dear Nikos Nice remark! I go back to Paris in a few day. Better to quit the cider country where I uttered so much blunders! Here I think to find some
            Message 5 of 11 , Sep 24, 2007
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              Dear Nikos
              Nice remark!
              I go back to Paris in a few day. Better to quit the cider country where I
              uttered so much blunders!
              Here I think to find some references on this axis circumparabola
              configuration as we know that nothing is new in Triangle Geometry!
              Now I ask you and everyboby if you know and what you think about the book:
              Divine proportions
              Rational trigonometry to Universal Geometry
              written by NJ Wildberger
              Friendly
              Francois
              On 9/23/07, Nikolaos Dergiades <ndergiades@...> wrote:
              >
              > Dear Francois,
              >
              > > In other words , axis of the ABC-circumparabola are
              > > just the Simson lines
              > > of the medial triangle.
              > >
              > > I would like to have a proof or any reference of
              > > this last fact. Any idea?
              >
              > I think you are right as I can conclude from sketch.
              > Let H be the orthocenter of ABC and
              > A1B1C1 be the medial triangle of ABC.
              > There exists a unique point D on the circumcircle of
              > ABC such that the isogonal conjugate of the tangent
              > at D to the circumcircle is the ABC-circumparabola.
              > Let P be the mid point of HD.
              > The axis of the ABC-circumparabola is the
              > Simson line of P wrt the triangle A1B1C1.
              >
              > Best regards
              > Nikos Dergiades
              >
              >
              >
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              [Non-text portions of this message have been removed]
            • Francois Rideau
              Dear Eisso, Nikos, Francisco Eventually this problem of the construction of the parabolas on 4 points A, B, C, D is fascinating and you learn a lot of geometry
              Message 6 of 11 , Sep 25, 2007
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                Dear Eisso, Nikos, Francisco

                Eventually this problem of the construction of the parabolas on 4 points A,
                B, C, D is fascinating and you learn a lot of geometry solving it.

                Cheer up for your using the diagonal triangle T of the 4 points A, B, C, D
                after noticing it was self conjugate wrt the parabola.

                I just want to make another remark.

                As you told us, the medial triangle of T is tangent to the parabolas, so
                both foci are on its circumcircle that is to say the nine point circle of T
                but orthocenter of the medial triangle that is to say the circumcenter of T
                is on both directrix. (Faure Theorem)

                So both directrix are also easy to draw.



                As I have already said it, the big problem in this construction is to find
                the directions of the axis of the parabolas and it was Newton who solved it
                first without knowing any Desargues theorem and any line at infinity.

                I can elaborate here in this short post but I advise everybody to read Dorie
                book , 100 great problems in Mathematic published by Dover.

                There is a modern way to look at this Newton theorem. In fact he proves
                without noticing it of course that a parabola has a structure of an affine
                line.



                Francisco sends me a very interesting paper on this topic, that is to say
                another « solution » of this construction , named Church's problem as if
                nobody on the Island knew that Isaac had solved it a long time ago!

                I nearly got a headache reading this paper and in fact I gave up in the
                middle for I was not able to read the drawing it.

                But the idea suggested by the writer, a Mr Hyde, is interesting.

                The parabola problem takes place in some plane <P>, so Dr Jekill, sorry, Mr
                Hyde, makes a conic projection from the plane <P> onto another plane <P'>.
                So he solves a new problem that he finds easier in <P'>, ( and maybe he is
                the only one to think it this way), and he projects back the solution in
                <P>.

                What a nightmare and in fact he also uses a making drawing ( epure in
                French) to do the job!

                Clearly Mr Hyde fears nothing nor nobody, even projective geometry!



                In fact this parabola problem is only a special case of the following
                problem in projective geometry that is to say construct the conics on 4
                points A, B, C, D and tangent to a line L.

                Every time you choose an affine chart of the projective plane, that is to
                say that you choose some line of the projective plane as line at infinity,
                then you get a new problem in the affine plane, image of the chart.

                I think that Hyde finds easier to choose line CD as a new line at infinity.

                So he get the new problem, projectively isomorphic to the old:

                To construct hyperbolas on 2 points A and B, knowing the directions of the
                asymptots and also tangent to some line L.

                I think this problem is easier for you can take a frame in the affine plane
                with both axis parallel to the known directions of the asymptots. In this
                frame, hyperbolas are homographic graphs.

                I let GSP users to find a metric solution and Cabri users to find an affine
                solution but please, don't send me any making drawing!

                I would like to have some discussion following the position of the tangent
                line in the affine plane.

                Friendly

                Francois


                [Non-text portions of this message have been removed]
              • Eisso J. Atzema
                Dear François, Nikos and Francisco, I d be interested in reading Church s paper. Could anyone mail this paper to me as well? Section V of Book I of Newton s
                Message 7 of 11 , Sep 27, 2007
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                  Dear François, Nikos and Francisco,

                  I'd be interested in reading Church's paper. Could anyone mail this
                  paper to me as well? Section V of Book I of Newton's Principia (the part
                  on fitting conic sections) has been lying on my desk for over a year
                  now, but I still have not had a change to look at all of it. I did look
                  at a few theorems in Section V and it is amazing work -- especially
                  since this Section V predates the rest of the Principia by almost a
                  decade, at which time Newton could not really be considered particularly
                  well-read in mathematics (in fact, he never really was). My own proof of
                  Newton's construction is a few lines, but it uses a lemma that requires
                  proof. The proof of Newton's construction relies heavily on the dual of
                  version of Steiner's Theorem (sometimes referred to as Chasles'
                  Theorem), which certainly was known to Halley (even though he did not
                  grasp the potential of this tool) and (implicitly) to Newton as well (he
                  seems to use something very similar in a few places in Section V, if I
                  remember correctly).

                  The lemma that I need is that for any quadrilateral ABCD with diagonal
                  points E=AC\capBD, F=AD\capBC, G=AB\capCD, the line connecting the
                  midpoints of EF and EG is a tangent line to the exparabola of ABCD.
                  Moreover, the lines connecting the points of intersection of this lines
                  with the sides of ABCD with E are each parallel to the side opposite the
                  side they are on. From here the proof of Newton's result is
                  straightforward. I uploaded a GSP drawing with a sketch of proof (or
                  sketchy proof if you will) of Newton's construction (the file is called
                  parabola.gsp). I leave the proof of the lemma to you. Judging from my
                  own proof, it seems to require the dual of Steiner's Theorem, as well as
                  the harmonic property of quadrilaterals.

                  best,
                  Eisso






                  Francois Rideau wrote:
                  >
                  > Dear Eisso, Nikos, Francisco
                  >
                  > Eventually this problem of the construction of the parabolas on 4
                  > points A,
                  > B, C, D is fascinating and you learn a lot of geometry solving it.
                  >
                  > Cheer up for your using the diagonal triangle T of the 4 points A, B, C, D
                  > after noticing it was self conjugate wrt the parabola.
                  >
                  > I just want to make another remark.
                  >
                  > As you told us, the medial triangle of T is tangent to the parabolas, so
                  > both foci are on its circumcircle that is to say the nine point circle
                  > of T
                  > but orthocenter of the medial triangle that is to say the circumcenter
                  > of T
                  > is on both directrix. (Faure Theorem)
                  >
                  > So both directrix are also easy to draw.
                  >
                  > As I have already said it, the big problem in this construction is to find
                  > the directions of the axis of the parabolas and it was Newton who
                  > solved it
                  > first without knowing any Desargues theorem and any line at infinity.
                  >
                  > I can elaborate here in this short post but I advise everybody to read
                  > Dorie
                  > book , 100 great problems in Mathematic published by Dover.
                  >
                  > There is a modern way to look at this Newton theorem. In fact he proves
                  > without noticing it of course that a parabola has a structure of an affine
                  > line.
                  >
                  > Francisco sends me a very interesting paper on this topic, that is to say
                  > another « solution » of this construction , named Church's problem as if
                  > nobody on the Island knew that Isaac had solved it a long time ago!
                  >
                  > I nearly got a headache reading this paper and in fact I gave up in the
                  > middle for I was not able to read the drawing it.
                  >
                  > But the idea suggested by the writer, a Mr Hyde, is interesting.
                  >
                  > The parabola problem takes place in some plane <P>, so Dr Jekill,
                  > sorry, Mr
                  > Hyde, makes a conic projection from the plane <P> onto another plane <P'>.
                  > So he solves a new problem that he finds easier in <P'>, ( and maybe he is
                  > the only one to think it this way), and he projects back the solution in
                  > <P>.
                  >
                  > What a nightmare and in fact he also uses a making drawing ( epure in
                  > French) to do the job!
                  >
                  > Clearly Mr Hyde fears nothing nor nobody, even projective geometry!
                  >
                  > In fact this parabola problem is only a special case of the following
                  > problem in projective geometry that is to say construct the conics on 4
                  > points A, B, C, D and tangent to a line L.
                  >
                  > Every time you choose an affine chart of the projective plane, that is to
                  > say that you choose some line of the projective plane as line at infinity,
                  > then you get a new problem in the affine plane, image of the chart.
                  >
                  > I think that Hyde finds easier to choose line CD as a new line at
                  > infinity.
                  >
                  > So he get the new problem, projectively isomorphic to the old:
                  >
                  > To construct hyperbolas on 2 points A and B, knowing the directions of the
                  > asymptots and also tangent to some line L.
                  >
                  > I think this problem is easier for you can take a frame in the affine
                  > plane
                  > with both axis parallel to the known directions of the asymptots. In this
                  > frame, hyperbolas are homographic graphs.
                  >
                  > I let GSP users to find a metric solution and Cabri users to find an
                  > affine
                  > solution but please, don't send me any making drawing!
                  >
                  > I would like to have some discussion following the position of the tangent
                  > line in the affine plane.
                  >
                  > Friendly
                  >
                  > Francois
                  >
                  > [Non-text portions of this message have been removed]
                  >
                  >


                  --

                  ========================================
                  Eisso J. Atzema, Ph.D.
                  Department of Mathematics & Statistics
                  University of Maine
                  Orono, ME 04469
                  Tel.: (207) 581-3928 (office)
                  (207) 866-3871 (home)
                  Fax.: (207) 581-3902
                  E-mail: atzema@...
                  ========================================
                • Jeff
                  Dear Francois, ... I find Epure (I don t know the plural in French) especially helpful, don t you? Sincerely, Jeff
                  Message 8 of 11 , Sep 28, 2007
                  • 0 Attachment
                    Dear Francois,

                    > What a nightmare and in fact he also uses a making drawing ( epure in
                    > French) to do the job!

                    I find 'Epure (I don't know the plural in French) especially helpful,
                    don't you?

                    Sincerely, Jeff
                  • Eisso J. Atzema
                    Dear François, Nikos and Francisco, I tried to post this message earlier, but never showed up. I d be interested in reading Church s paper. Could anyone mail
                    Message 9 of 11 , Sep 28, 2007
                    • 0 Attachment
                      Dear François, Nikos and Francisco,

                      I tried to post this message earlier, but never showed up. I'd be
                      interested in reading Church's paper. Could anyone mail this
                      paper to me as well? Section V of Book I of Newton's Principia (the part
                      on fitting conic sections) has been lying on my desk for over a year
                      now, but I still have not had a change to look at all of it. I did look
                      at a few theorems in Section V and it is amazing work -- especially
                      since this Section V predates the rest of the Principia by almost a
                      decade, at which time Newton could not really be considered particularly
                      well-read in mathematics (in fact, he never really was). My own proof of
                      Newton's construction is a few lines, but it uses a lemma that requires
                      proof. The proof of Newton's construction relies heavily on the dual of
                      version of Steiner's Theorem (sometimes referred to as Chasles'
                      Theorem), which certainly was known to Halley (even though he did not
                      grasp the potential of this tool) and (implicitly) to Newton as well (he
                      seems to use something very similar in a few places in Section V, if I
                      remember correctly).

                      The lemma that I need is that for any quadrilateral ABCD with diagonal
                      points E=AC\capBD, F=AD\capBC, G=AB\capCD, the line connecting the
                      midpoints of EF and EG is a tangent line to the exparabola of ABCD.
                      Moreover, the lines connecting the points of intersection of this lines
                      with the sides of ABCD with E are each parallel to the side opposite the
                      side they are on. From here the proof of Newton's result is
                      straightforward. I uploaded a GSP drawing with a sketch of proof (or
                      sketchy proof if you will) of Newton's construction (the file is called
                      parabola.gsp). I leave the proof of the lemma to you. Judging from my
                      own proof, it seems to require the dual of Steiner's Theorem, as well as
                      the harmonic property of quadrilaterals.

                      best,
                      Eisso






                      Francois Rideau wrote:
                      >
                      > Dear Eisso, Nikos, Francisco
                      >
                      > Eventually this problem of the construction of the parabolas on 4
                      > points A,
                      > B, C, D is fascinating and you learn a lot of geometry solving it.
                      >
                      > Cheer up for your using the diagonal triangle T of the 4 points A, B, C, D
                      > after noticing it was self conjugate wrt the parabola.
                      >
                      > I just want to make another remark.
                      >
                      > As you told us, the medial triangle of T is tangent to the parabolas, so
                      > both foci are on its circumcircle that is to say the nine point circle
                      > of T
                      > but orthocenter of the medial triangle that is to say the circumcenter
                      > of T
                      > is on both directrix. (Faure Theorem)
                      >
                      > So both directrix are also easy to draw.
                      >
                      > As I have already said it, the big problem in this construction is to find
                      > the directions of the axis of the parabolas and it was Newton who
                      > solved it
                      > first without knowing any Desargues theorem and any line at infinity.
                      >
                      > I can elaborate here in this short post but I advise everybody to read
                      > Dorie
                      > book , 100 great problems in Mathematic published by Dover.
                      >
                      > There is a modern way to look at this Newton theorem. In fact he proves
                      > without noticing it of course that a parabola has a structure of an affine
                      > line.
                      >
                      > Francisco sends me a very interesting paper on this topic, that is to say
                      > another « solution » of this construction , named Church's problem as if
                      > nobody on the Island knew that Isaac had solved it a long time ago!
                      >
                      > I nearly got a headache reading this paper and in fact I gave up in the
                      > middle for I was not able to read the drawing it.
                      >
                      > But the idea suggested by the writer, a Mr Hyde, is interesting.
                      >
                      > The parabola problem takes place in some plane <P>, so Dr Jekill,
                      > sorry, Mr
                      > Hyde, makes a conic projection from the plane <P> onto another plane <P'>.
                      > So he solves a new problem that he finds easier in <P'>, ( and maybe he is
                      > the only one to think it this way), and he projects back the solution in
                      > <P>.
                      >
                      > What a nightmare and in fact he also uses a making drawing ( epure in
                      > French) to do the job!
                      >
                      > Clearly Mr Hyde fears nothing nor nobody, even projective geometry!
                      >
                      > In fact this parabola problem is only a special case of the following
                      > problem in projective geometry that is to say construct the conics on 4
                      > points A, B, C, D and tangent to a line L.
                      >
                      > Every time you choose an affine chart of the projective plane, that is to
                      > say that you choose some line of the projective plane as line at infinity,
                      > then you get a new problem in the affine plane, image of the chart.
                      >
                      > I think that Hyde finds easier to choose line CD as a new line at
                      > infinity.
                      >
                      > So he get the new problem, projectively isomorphic to the old:
                      >
                      > To construct hyperbolas on 2 points A and B, knowing the directions of the
                      > asymptots and also tangent to some line L.
                      >
                      > I think this problem is easier for you can take a frame in the affine
                      > plane
                      > with both axis parallel to the known directions of the asymptots. In this
                      > frame, hyperbolas are homographic graphs.
                      >
                      > I let GSP users to find a metric solution and Cabri users to find an
                      > affine
                      > solution but please, don't send me any making drawing!
                      >
                      > I would like to have some discussion following the position of the tangent
                      > line in the affine plane.
                      >
                      > Friendly
                      >
                      > Francois
                      >
                      > [Non-text portions of this message have been removed]
                      >
                      >


                      --

                      ========================================
                      Eisso J. Atzema, Ph.D.
                      Department of Mathematics & Statistics
                      University of Maine
                      Orono, ME 04469
                      Tel.: (207) 581-3928 (office)
                      (207) 866-3871 (home)
                      Fax.: (207) 581-3902
                      E-mail: atzema@...
                      ========================================
                    • Francois Rideau
                      ... I don t see exactly what you mean here.Can you be more explicit? ... As for the sides of the medial triangle of the diagonal triangle of ABCD are ... . ...
                      Message 10 of 11 , Oct 1, 2007
                      • 0 Attachment
                        >
                        > Dear Eisso
                        >
                        > Moreover, the lines connecting the points of intersection of this lines
                        > with the sides of ABCD with E are each parallel to the side opposite the
                        > side they are on.
                        >






                        I don't see exactly what you mean here.Can you be more explicit?
                        >



                        As for the sides of the medial triangle of the diagonal triangle of ABCD are
                        > tangent to the parabola, it is just a consequence of the fact that for a
                        > point M having polar line L wrt a parabola <P>, the line homothetic of L in
                        > the dilation of center M and ratio 1/2 is tangent to the parabola.
                        >



                        .
                        > Friendly
                        >

                        Francois


                        >


                        [Non-text portions of this message have been removed]
                      • Eisso J. Atzema
                        Dear François, I hoped the drawing that I uploaded would clear that up. Anyway, what I mean is that if the line through the midpoints of EFG intersects AD in
                        Message 11 of 11 , Oct 1, 2007
                        • 0 Attachment
                          Dear François,

                          I hoped the drawing that I uploaded would clear that up. Anyway, what I
                          mean is that if the line through the midpoints of EFG intersects AD in
                          F*, BC in F**, AB in G* and CD in G** then EF* is parallel to BC, EF**
                          is parallel to AD, EG* is parallel to CD and G** is parallel to AB.

                          Eisso

                          Francois Rideau wrote:
                          >
                          > >
                          > > Dear Eisso
                          > >
                          > > Moreover, the lines connecting the points of intersection of this lines
                          > > with the sides of ABCD with E are each parallel to the side opposite the
                          > > side they are on.
                          > >
                          >
                          > I don't see exactly what you mean here.Can you be more explicit?
                          > >
                          >
                          > As for the sides of the medial triangle of the diagonal triangle of
                          > ABCD are
                          > > tangent to the parabola, it is just a consequence of the fact that for a
                          > > point M having polar line L wrt a parabola <P>, the line homothetic
                          > of L in
                          > > the dilation of center M and ratio 1/2 is tangent to the parabola.
                          > >
                          >
                          > .
                          > > Friendly
                          > >
                          >
                          > Francois
                          >
                          > >
                          >
                          > [Non-text portions of this message have been removed]
                          >
                          >

                          --

                          ========================================
                          Eisso J. Atzema, Ph.D.
                          Department of Mathematics & Statistics
                          University of Maine
                          Orono, ME 04469
                          Tel.: (207) 581-3928 (office)
                          (207) 866-3871 (home)
                          Fax.: (207) 581-3902
                          E-mail: atzema@...
                          ========================================
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