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Re: [EMHL] Mickel point

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  • Francois Rideau
    Dear Alexey Now I am looking at your proof using angles and I don t understand it! As the triangles AMX and XMC are similar we receive that angle
    Message 1 of 29 , Jan 30, 2008
      Dear Alexey
      Now I am looking at your proof using angles and I don't understand it!

      As the triangles AMX and XMC are similar we receive that angle
      AXC=pi-AMC=AI1C.
      May be you should mean: AXC = pi - (AMC/2) ?

      And the last angle equality : ...= AI1C is also not obvious.

      Friendly
      Francois


      [Non-text portions of this message have been removed]
    • Paris Pamfilos
      Dear Friends, I encountered a trivial property of the parabola, not found in my text books. It gives a very nice description of the parabolas circumscribing a
      Message 2 of 29 , Jan 31, 2008
        Dear Friends, I encountered a trivial property of the parabola, not
        found in my text books. It gives a very nice description of the
        parabolas circumscribing a given triangle.
        Besides, using an appropriate projectivity, generalizes to a nice
        structural description of all conics circumscribing a given triangle and
        tangent to a given line. Here are the remarks:

        Let the vertices of our basic triangle, points {a,b,c}, be on a
        parabola. Consider also points {a'=b+c-a, b'=c+a-b, c'=a+b-c} which are
        the vertices of the anticomplementary triangle.
        The parallels from {a',b',c'} to the axis of the parabola meet the
        opposite sides of the anticomplementary triangle at points lying on the
        parabola.

        This is a special case of the following more general setting that should
        be treated in text-books on conics (may be I have the wrong ones).

        Consider a triangle ABC and a line L . The set of all circumconics of
        ABC tangent to L has the following structure.

        [1] Given the triangle ABC and the arbitrary line L construct its
        trilinear pole P w.r. to ABC and the corresponding precevian triangle
        A0B0C0.
        [2] Consider an arbitrary point Q on L and the lines {QA0,QB0,QC0}
        joining Q to the vertices of A0B0C0. The intersection points
        {A'',B'',C''} of these lines with the opposite sides of triangle A0B0C0
        and points {A,B,C} are six points on a conic c(Q).
        [3] c(Q) passes through Q and is also tangent to L at Q. Thus, it is the
        unique conic passing through the vertices of ABC and tangent to L at its
        point Q.
        [4] The tripole of L with respect to triangle A''B''C'' is a point Q'
        on line QP.
        [5] The tangential triangle A1B1C1 of ABC relative to the conic c(Q) is
        line-perspective to A''B''C'', the perspectrix being line L.
        [6] The locus of perspectors of conics c(Q) is the inscribed conic of
        triangle ABC with perspector P.

        Friendly
        Paris Pamfilos



        Internal Virus Database is out-of-date.
        Checked by AVG Free Edition.
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      • Francois Rideau
        Dear Paris This projective configuration of a conic through 3 points A, B, C and tangent at Q to a line (L) is very beautiful! I am sure there is a nice proof
        Message 3 of 29 , Feb 1, 2008
          Dear Paris

          This projective configuration of a conic through 3 points A, B, C and
          tangent at Q to a line (L) is very beautiful!

          I am sure there is a nice proof in the general case using Desargues and
          Plücker
          involution theorems known for a long time that we can find as an exercise in
          some old book.

          But the idea to send line (L) at infinity and to prove the theorem in this
          special case is very educational.

          Another (educational?) idea should be to send both points B and C at
          infinity that is to say looking at the hyperbola through point A with given
          asymptotic directions and tangent to a line (L) at Q. I am sure that
          elementary affine properties of hyperbola can also do the job.

          Friendly
          Francois


          [Non-text portions of this message have been removed]
        • jpehrmfr
          Dear Eisso, François and Alexei ... such ... in ... angle ... imaginary ... the ... for ... Of course, there is a little problem because there are 4 common
          Message 4 of 29 , Feb 3, 2008
            Dear Eisso, François and Alexei
            > What I think you are saying, François, is that the circular map f
            such
            > that f(A) = A', f(B) = B', f(C) = C' can be represented by complex
            > conjugation w.r.t. to the complex plane if we place the quadrilateral
            in
            > such a way that the Miquel point is at the origin and the (common)
            angle
            > bisectors of AMA', BMB' and CMC' coincide with the real and
            imaginary
            > axes. I think that does not quite answer the concern that I had. In
            > fact, I think you would run into some difficulties too for non-convex
            > cases as it is not immediately evident (at least not to me) which of
            the
            > common angle bisectors should be taken for the real axis and which
            for
            > the complex axis.

            Of course, there is a little problem because there are 4 common points
            of a A-bisector and a A'-bisector, lying on 2 circles going through A
            and A'. We have to choose the right circles through A,A' and B,B' to
            get the fixed points of f.
            So, it is better to use the circles FAA' and FBB' (F = Miquel point)
            because they are the 9P-circles of the orthocentric systems with
            vertices the 4 common points of the A- and A'- bisectors and the 4
            common points of the B- and B'-bisectors; so, there is no more
            ambiguity.
            T = midpoint of the circular arc AFA', T' = midpoint of the circular
            arc BFB'; the fixed points are the common points of the circle with
            center T going through A and the circle with center T' going through B.
            Friendly
            Jean-Pierre
          • Alexey.A.Zaslavsky
            Dear Paris and Francois! It seems that these problems were considered by Lev and Tatyana Emelyanoff. They presented the problems devoted to trypolars at Summer
            Message 5 of 29 , Feb 4, 2008
              Dear Paris and Francois!
              It seems that these problems were considered by Lev and Tatyana Emelyanoff. They presented the problems devoted to trypolars at Summer conference of Towns tournament-2005.
              Also in our with Arsenij Akopyan book next theorem is proved. Given the triangle ABC and the point C0. An arbitrary line passing through C0 intersect AC and BC in B' and A' resp. P is the common point of AA' and BB'. Then
              1. The locus of P is the circumconic touching CC0.
              2. If CP intersect AB in C' then all lines A'C' have the common point B0 and BB0 touches the circumconic.

              Sincerely Alexey

              Dear Paris

              This projective configuration of a conic through 3 points A, B, C and
              tangent at Q to a line (L) is very beautiful!

              I am sure there is a nice proof in the general case using Desargues and
              Plücker
              involution theorems known for a long time that we can find as an exercise in
              some old book.

              But the idea to send line (L) at infinity and to prove the theorem in this
              special case is very educational.

              Another (educational?) idea should be to send both points B and C at
              infinity that is to say looking at the hyperbola through point A with given
              asymptotic directions and tangent to a line (L) at Q. I am sure that
              elementary affine properties of hyperbola can also do the job.

              Friendly
              Francois

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




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            • Francois Rideau
              Dear Alexey Very interesting problem I had had to tackle when I was a young boy more than fifty years ago. I change your notation a little bit to avoid
              Message 6 of 29 , Feb 5, 2008
                Dear Alexey
                Very interesting problem I had had to tackle when I was a young boy more
                than fifty years ago.

                I change your notation a little bit to avoid subscript and I name D your
                point C0.
                I think the locus of P is the ABC-circumconic tangent at A to line AD and
                tangent at B to line BD.
                Let A"B"C" be the ABC-cevian triangle of D.
                Then line B'C' is on the harmonic conjugate of D wrt BB"and line A'C' is on
                the harmonic conjugate of D wrt AA".

                Friendly
                Francois
                On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <zasl@...> wrote:

                > Dear Paris and Francois!
                > It seems that these problems were considered by Lev and Tatyana
                > Emelyanoff. They presented the problems devoted to trypolars at Summer
                > conference of Towns tournament-2005.
                > Also in our with Arsenij Akopyan book next theorem is proved. Given the
                > triangle ABC and the point C0. An arbitrary line passing through C0
                > intersect AC and BC in B' and A' resp. P is the common point of AA' and BB'.
                > Then
                > 1. The locus of P is the circumconic touching CC0.
                > 2. If CP intersect AB in C' then all lines A'C' have the common point B0
                > and BB0 touches the circumconic.
                >
                > Sincerely Alexey
                >
                >
                > Dear Paris
                >
                > This projective configuration of a conic through 3 points A, B, C and
                > tangent at Q to a line (L) is very beautiful!
                >
                > I am sure there is a nice proof in the general case using Desargues and
                > Plücker
                > involution theorems known for a long time that we can find as an exercise
                > in
                > some old book.
                >
                > But the idea to send line (L) at infinity and to prove the theorem in this
                >
                > special case is very educational.
                >
                > Another (educational?) idea should be to send both points B and C at
                > infinity that is to say looking at the hyperbola through point A with
                > given
                > asymptotic directions and tangent to a line (L) at Q. I am sure that
                > elementary affine properties of hyperbola can also do the job.
                >
                > Friendly
                > Francois
                >
                > [Non-text portions of this message have been removed]
                >
                >
                >
                >
                > Antivirus scanning: Symantec Mail Security for SMTP.
                >
                > [Non-text portions of this message have been removed]
                >
                >
                >


                [Non-text portions of this message have been removed]
              • Alexey.A.Zaslavsky
                Dear Francois! Thank you for your correstion. Of course the conic is inscribed on the triangle A0B0C0. Sincerely
                Message 7 of 29 , Feb 5, 2008
                  Dear Francois!
                  Thank you for your correstion. Of course the conic is inscribed on the triangle A0B0C0.

                  Sincerely Alexey


                  Dear Alexey
                  Very interesting problem I had had to tackle when I was a young boy more
                  than fifty years ago.

                  I change your notation a little bit to avoid subscript and I name D your
                  point C0.
                  I think the locus of P is the ABC-circumconic tangent at A to line AD and
                  tangent at B to line BD.
                  Let A"B"C" be the ABC-cevian triangle of D.
                  Then line B'C' is on the harmonic conjugate of D wrt BB"and line A'C' is on
                  the harmonic conjugate of D wrt AA".

                  Friendly
                  Francois
                  On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <zasl@...> wrote:

                  > Dear Paris and Francois!
                  > It seems that these problems were considered by Lev and Tatyana
                  > Emelyanoff. They presented the problems devoted to trypolars at Summer
                  > conference of Towns tournament-2005.
                  > Also in our with Arsenij Akopyan book next theorem is proved. Given the
                  > triangle ABC and the point C0. An arbitrary line passing through C0
                  > intersect AC and BC in B' and A' resp. P is the common point of AA' and BB'.
                  > Then
                  > 1. The locus of P is the circumconic touching CC0.
                  > 2. If CP intersect AB in C' then all lines A'C' have the common point B0
                  > and BB0 touches the circumconic.
                  >
                  > Sincerely Alexey
                  >
                  >
                  > Dear Paris
                  >
                  > This projective configuration of a conic through 3 points A, B, C and
                  > tangent at Q to a line (L) is very beautiful!
                  >
                  > I am sure there is a nice proof in the general case using Desargues and
                  > Plücker
                  > involution theorems known for a long time that we can find as an exercise
                  > in
                  > some old book.
                  >
                  > But the idea to send line (L) at infinity and to prove the theorem in this
                  >
                  > special case is very educational.
                  >
                  > Another (educational?) idea should be to send both points B and C at
                  > infinity that is to say looking at the hyperbola through point A with
                  > given
                  > asymptotic directions and tangent to a line (L) at Q. I am sure that
                  > elementary affine properties of hyperbola can also do the job.
                  >
                  > Friendly
                  > Francois
                  >
                  > [Non-text portions of this message have been removed]
                  >
                  >
                  >
                  >
                  > Antivirus scanning: Symantec Mail Security for SMTP.
                  >
                  > [Non-text portions of this message have been removed]
                  >
                  >
                  >

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




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                • Francois Rideau
                  Dear Alexey If I name Db the harmonic conjugate of D wrt BB and Da the harmonic conjugate of D wrt AA , then the tangent at C to the circumconic locus of P is
                  Message 8 of 29 , Feb 5, 2008
                    Dear Alexey
                    If I name Db the harmonic conjugate of D wrt BB" and Da the harmonic
                    conjugate of D wrt AA", then the tangent at C to the circumconic locus of P
                    is just the line Cab.

                    1° What is the locus of D in order to the locus of P be a rectangular
                    hyperbola? : ( a line)
                    2° What is the locus of D in order to the locus of P be a parabola?: ( a
                    hyperbola)

                    Friendly
                    Francois


                    On Feb 5, 2008 9:59 AM, Francois Rideau <francois.rideau@...> wrote:

                    > Dear Alexey
                    > Very interesting problem I had had to tackle when I was a young boy more
                    > than fifty years ago.
                    >
                    > I change your notation a little bit to avoid subscript and I name D your
                    > point C0.
                    > I think the locus of P is the ABC-circumconic tangent at A to line AD and
                    > tangent at B to line BD.
                    > Let A"B"C" be the ABC-cevian triangle of D.
                    > Then line B'C' is on the harmonic conjugate of D wrt BB"and line A'C' is
                    > on the harmonic conjugate of D wrt AA".
                    >
                    > Friendly
                    > Francois
                    >
                    > On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <zasl@...> wrote:
                    >
                    > > Dear Paris and Francois!
                    > > It seems that these problems were considered by Lev and Tatyana
                    > > Emelyanoff. They presented the problems devoted to trypolars at Summer
                    > > conference of Towns tournament-2005.
                    > > Also in our with Arsenij Akopyan book next theorem is proved. Given the
                    > > triangle ABC and the point C0. An arbitrary line passing through C0
                    > > intersect AC and BC in B' and A' resp. P is the common point of AA' and BB'.
                    > > Then
                    > > 1. The locus of P is the circumconic touching CC0.
                    > > 2. If CP intersect AB in C' then all lines A'C' have the common point B0
                    > > and BB0 touches the circumconic.
                    > >
                    > > Sincerely Alexey
                    > >
                    > >
                    > > Dear Paris
                    > >
                    > > This projective configuration of a conic through 3 points A, B, C and
                    > > tangent at Q to a line (L) is very beautiful!
                    > >
                    > > I am sure there is a nice proof in the general case using Desargues and
                    > > Plücker
                    > > involution theorems known for a long time that we can find as an
                    > > exercise in
                    > > some old book.
                    > >
                    > > But the idea to send line (L) at infinity and to prove the theorem in
                    > > this
                    > > special case is very educational.
                    > >
                    > > Another (educational?) idea should be to send both points B and C at
                    > > infinity that is to say looking at the hyperbola through point A with
                    > > given
                    > > asymptotic directions and tangent to a line (L) at Q. I am sure that
                    > > elementary affine properties of hyperbola can also do the job.
                    > >
                    > > Friendly
                    > > Francois
                    > >
                    > > [Non-text portions of this message have been removed]
                    > >
                    > >
                    > >
                    > >
                    > > Antivirus scanning: Symantec Mail Security for SMTP.
                    > >
                    > > [Non-text portions of this message have been removed]
                    > >
                    > >
                    > >
                    >
                    >


                    [Non-text portions of this message have been removed]
                  • Alexey.A.Zaslavsky
                    Dear Francois! 1 is the partial case of next general fact: if the pencil of conics determined by the points A, B, C, D is considered then the locus of poles of
                    Message 9 of 29 , Feb 6, 2008
                      Dear Francois!
                      1 is the partial case of next general fact: if the pencil of conics determined by the points A, B, C, D is considered then the locus of poles of line AB is the line. This line is the polar of common point of AB and CD wrt any conic of the pencil. In our case it is the common point of AB and the altitude CH.
                      The locus 2 is the hyperbola with asymptotes CA and CB touching AB in its midpoint.

                      Sincerely Alexey

                      Dear Alexey
                      If I name Db the harmonic conjugate of D wrt BB" and Da the harmonic
                      conjugate of D wrt AA", then the tangent at C to the circumconic locus of P
                      is just the line Cab.

                      1° What is the locus of D in order to the locus of P be a rectangular
                      hyperbola? : ( a line)
                      2° What is the locus of D in order to the locus of P be a parabola?: ( a
                      hyperbola)

                      Friendly
                      Francois

                      On Feb 5, 2008 9:59 AM, Francois Rideau <francois.rideau@...> wrote:

                      > Dear Alexey
                      > Very interesting problem I had had to tackle when I was a young boy more
                      > than fifty years ago.
                      >
                      > I change your notation a little bit to avoid subscript and I name D your
                      > point C0.
                      > I think the locus of P is the ABC-circumconic tangent at A to line AD and
                      > tangent at B to line BD.
                      > Let A"B"C" be the ABC-cevian triangle of D.
                      > Then line B'C' is on the harmonic conjugate of D wrt BB"and line A'C' is
                      > on the harmonic conjugate of D wrt AA".
                      >
                      > Friendly
                      > Francois
                      >
                      > On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <zasl@...> wrote:
                      >
                      > > Dear Paris and Francois!
                      > > It seems that these problems were considered by Lev and Tatyana
                      > > Emelyanoff. They presented the problems devoted to trypolars at Summer
                      > > conference of Towns tournament-2005.
                      > > Also in our with Arsenij Akopyan book next theorem is proved. Given the
                      > > triangle ABC and the point C0. An arbitrary line passing through C0
                      > > intersect AC and BC in B' and A' resp. P is the common point of AA' and BB'.
                      > > Then
                      > > 1. The locus of P is the circumconic touching CC0.
                      > > 2. If CP intersect AB in C' then all lines A'C' have the common point B0
                      > > and BB0 touches the circumconic.
                      > >
                      > > Sincerely Alexey
                      > >
                      > >
                      > > Dear Paris
                      > >
                      > > This projective configuration of a conic through 3 points A, B, C and
                      > > tangent at Q to a line (L) is very beautiful!
                      > >
                      > > I am sure there is a nice proof in the general case using Desargues and
                      > > Plücker
                      > > involution theorems known for a long time that we can find as an
                      > > exercise in
                      > > some old book.
                      > >
                      > > But the idea to send line (L) at infinity and to prove the theorem in
                      > > this
                      > > special case is very educational.
                      > >
                      > > Another (educational?) idea should be to send both points B and C at
                      > > infinity that is to say looking at the hyperbola through point A with
                      > > given
                      > > asymptotic directions and tangent to a line (L) at Q. I am sure that
                      > > elementary affine properties of hyperbola can also do the job.
                      > >
                      > > Friendly
                      > > Francois
                      > >
                      > > [Non-text portions of this message have been removed]
                      > >
                      > >
                      > >
                      > >
                      > > Antivirus scanning: Symantec Mail Security for SMTP.
                      > >
                      > > [Non-text portions of this message have been removed]
                      > >
                      > >
                      > >
                      >
                      >

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




                      Antivirus scanning: Symantec Mail Security for SMTP.

                      [Non-text portions of this message have been removed]
                    • Francois Rideau
                      Dear Alexey Cheer up! Thanks to your problem, it was true nostalgia for me! Friendly Francois ... [Non-text portions of this message have been removed]
                      Message 10 of 29 , Feb 6, 2008
                        Dear Alexey
                        Cheer up!
                        Thanks to your problem, it was true nostalgia for me!
                        Friendly
                        Francois

                        On Feb 6, 2008 11:07 AM, Alexey.A.Zaslavsky <zasl@...> wrote:

                        > Dear Francois!
                        > 1 is the partial case of next general fact: if the pencil of conics
                        > determined by the points A, B, C, D is considered then the locus of poles of
                        > line AB is the line. This line is the polar of common point of AB and CD wrt
                        > any conic of the pencil. In our case it is the common point of AB and the
                        > altitude CH.
                        > The locus 2 is the hyperbola with asymptotes CA and CB touching AB in its
                        > midpoint.
                        >
                        > Sincerely Alexey
                        >
                        >
                        > Dear Alexey
                        > If I name Db the harmonic conjugate of D wrt BB" and Da the harmonic
                        > conjugate of D wrt AA", then the tangent at C to the circumconic locus of
                        > P
                        > is just the line Cab.
                        >
                        > 1° What is the locus of D in order to the locus of P be a rectangular
                        > hyperbola? : ( a line)
                        > 2° What is the locus of D in order to the locus of P be a parabola?: ( a
                        > hyperbola)
                        >
                        > Friendly
                        > Francois
                        >
                        > On Feb 5, 2008 9:59 AM, Francois Rideau <francois.rideau@...<francois.rideau%40gmail.com>>
                        > wrote:
                        >
                        > > Dear Alexey
                        > > Very interesting problem I had had to tackle when I was a young boy more
                        >
                        > > than fifty years ago.
                        > >
                        > > I change your notation a little bit to avoid subscript and I name D your
                        >
                        > > point C0.
                        > > I think the locus of P is the ABC-circumconic tangent at A to line AD
                        > and
                        > > tangent at B to line BD.
                        > > Let A"B"C" be the ABC-cevian triangle of D.
                        > > Then line B'C' is on the harmonic conjugate of D wrt BB"and line A'C' is
                        >
                        > > on the harmonic conjugate of D wrt AA".
                        > >
                        > > Friendly
                        > > Francois
                        > >
                        > > On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <zasl@...<zasl%40cemi.rssi.ru>>
                        > wrote:
                        > >
                        > > > Dear Paris and Francois!
                        > > > It seems that these problems were considered by Lev and Tatyana
                        > > > Emelyanoff. They presented the problems devoted to trypolars at Summer
                        >
                        > > > conference of Towns tournament-2005.
                        > > > Also in our with Arsenij Akopyan book next theorem is proved. Given
                        > the
                        > > > triangle ABC and the point C0. An arbitrary line passing through C0
                        > > > intersect AC and BC in B' and A' resp. P is the common point of AA'
                        > and BB'.
                        > > > Then
                        > > > 1. The locus of P is the circumconic touching CC0.
                        > > > 2. If CP intersect AB in C' then all lines A'C' have the common point
                        > B0
                        > > > and BB0 touches the circumconic.
                        > > >
                        > > > Sincerely Alexey
                        > > >
                        > > >
                        > > > Dear Paris
                        > > >
                        > > > This projective configuration of a conic through 3 points A, B, C and
                        > > > tangent at Q to a line (L) is very beautiful!
                        > > >
                        > > > I am sure there is a nice proof in the general case using Desargues
                        > and
                        > > > Plücker
                        > > > involution theorems known for a long time that we can find as an
                        > > > exercise in
                        > > > some old book.
                        > > >
                        > > > But the idea to send line (L) at infinity and to prove the theorem in
                        > > > this
                        > > > special case is very educational.
                        > > >
                        > > > Another (educational?) idea should be to send both points B and C at
                        > > > infinity that is to say looking at the hyperbola through point A with
                        > > > given
                        > > > asymptotic directions and tangent to a line (L) at Q. I am sure that
                        > > > elementary affine properties of hyperbola can also do the job.
                        > > >
                        > > > Friendly
                        > > > Francois
                        > > >
                        > > > [Non-text portions of this message have been removed]
                        > > >
                        > > >
                        > > >
                        > > >
                        > > > Antivirus scanning: Symantec Mail Security for SMTP.
                        > > >
                        > > > [Non-text portions of this message have been removed]
                        > > >
                        > > >
                        > > >
                        > >
                        > >
                        >
                        > [Non-text portions of this message have been removed]
                        >
                        >
                        >
                        >
                        > Antivirus scanning: Symantec Mail Security for SMTP.
                        >
                        > [Non-text portions of this message have been removed]
                        >
                        >
                        >


                        [Non-text portions of this message have been removed]
                      • Paris Pamfilos
                        Dear Alexey and Francois I was in leave last week and could not participate. I ll read your letters soon and think about. Thank you for your interest. Paris
                        Message 11 of 29 , Feb 9, 2008
                          Dear Alexey and Francois
                          I was in leave last week and could not participate.
                          I'll read your letters soon and think about.
                          Thank you for your interest.
                          Paris


                          -----Original Message-----
                          From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On
                          Behalf Of Francois Rideau
                          Sent: Wednesday, February 06, 2008 2:02 PM
                          To: Hyacinthos@yahoogroups.com
                          Subject: [SPAM] Re: [EMHL] Circumconics



                          Dear Alexey
                          Cheer up!
                          Thanks to your problem, it was true nostalgia for me!
                          Friendly
                          Francois

                          On Feb 6, 2008 11:07 AM, Alexey.A.Zaslavsky <HYPERLINK
                          "mailto:zasl%40cemi.rssi.ru"zasl@....-ru> wrote:

                          > Dear Francois!
                          > 1 is the partial case of next general fact: if the pencil of conics
                          > determined by the points A, B, C, D is considered then the locus of
                          poles of
                          > line AB is the line. This line is the polar of common point of AB and
                          CD wrt
                          > any conic of the pencil. In our case it is the common point of AB and
                          the
                          > altitude CH.
                          > The locus 2 is the hyperbola with asymptotes CA and CB touching AB in
                          its
                          > midpoint.
                          >
                          > Sincerely Alexey
                          >
                          >
                          > Dear Alexey
                          > If I name Db the harmonic conjugate of D wrt BB" and Da the harmonic
                          > conjugate of D wrt AA", then the tangent at C to the circumconic locus
                          of
                          > P
                          > is just the line Cab.
                          >
                          > 1° What is the locus of D in order to the locus of P be a rectangular
                          > hyperbola? : ( a line)
                          > 2° What is the locus of D in order to the locus of P be a parabola?: (
                          a
                          > hyperbola)
                          >
                          > Friendly
                          > Francois
                          >
                          > On Feb 5, 2008 9:59 AM, Francois Rideau <HYPERLINK
                          "mailto:francois.rideau%40gmail.com"francois.rideau@...<francois.
                          rideau%-40gmail.com>->
                          > wrote:
                          >
                          > > Dear Alexey
                          > > Very interesting problem I had had to tackle when I was a young boy
                          more
                          >
                          > > than fifty years ago.
                          > >
                          > > I change your notation a little bit to avoid subscript and I name D
                          your
                          >
                          > > point C0.
                          > > I think the locus of P is the ABC-circumconic tangent at A to line
                          AD
                          > and
                          > > tangent at B to line BD.
                          > > Let A"B"C" be the ABC-cevian triangle of D.
                          > > Then line B'C' is on the harmonic conjugate of D wrt BB"and line
                          A'C' is
                          >
                          > > on the harmonic conjugate of D wrt AA".
                          > >
                          > > Friendly
                          > > Francois
                          > >
                          > > On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <HYPERLINK
                          "mailto:zasl%40cemi.rssi.ru"zasl@....-ru<zasl%40cemi.-rssi.ru>>
                          > wrote:
                          > >
                          > > > Dear Paris and Francois!
                          > > > It seems that these problems were considered by Lev and Tatyana
                          > > > Emelyanoff. They presented the problems devoted to trypolars at
                          Summer
                          >
                          > > > conference of Towns tournament-2005.
                          > > > Also in our with Arsenij Akopyan book next theorem is proved.
                          Given
                          > the
                          > > > triangle ABC and the point C0. An arbitrary line passing through
                          C0
                          > > > intersect AC and BC in B' and A' resp. P is the common point of
                          AA'
                          > and BB'.
                          > > > Then
                          > > > 1. The locus of P is the circumconic touching CC0.
                          > > > 2. If CP intersect AB in C' then all lines A'C' have the common
                          point
                          > B0
                          > > > and BB0 touches the circumconic.
                          > > >
                          > > > Sincerely Alexey
                          > > >
                          > > >
                          > > > Dear Paris
                          > > >
                          > > > This projective configuration of a conic through 3 points A, B, C
                          and
                          > > > tangent at Q to a line (L) is very beautiful!
                          > > >
                          > > > I am sure there is a nice proof in the general case using
                          Desargues
                          > and
                          > > > Plücker
                          > > > involution theorems known for a long time that we can find as an
                          > > > exercise in
                          > > > some old book.
                          > > >
                          > > > But the idea to send line (L) at infinity and to prove the theorem
                          in
                          > > > this
                          > > > special case is very educational.
                          > > >
                          > > > Another (educational?-) idea should be to send both points B and C
                          at
                          > > > infinity that is to say looking at the hyperbola through point A
                          with
                          > > > given
                          > > > asymptotic directions and tangent to a line (L) at Q. I am sure
                          that
                          > > > elementary affine properties of hyperbola can also do the job.
                          > > >
                          > > > Friendly
                          > > > Francois
                          > > >
                          > > > [Non-text portions of this message have been removed]
                          > > >
                          > > >
                          > > >
                          > > >
                          > > > Antivirus scanning: Symantec Mail Security for SMTP.
                          > > >
                          > > > [Non-text portions of this message have been removed]
                          > > >
                          > > >
                          > > >
                          > >
                          > >
                          >
                          > [Non-text portions of this message have been removed]
                          >
                          >
                          >
                          >
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                          >
                          >
                          >

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                          [Non-text portions of this message have been removed]
                        • Paris Pamfilos
                          Dear Alexey, 1) Could I have some concrete reference on Lev and Tatyana s presentation? 2) The theorem you proved with Arsenij is the old and classical way of
                          Message 12 of 29 , Feb 14, 2008
                            Dear Alexey,
                            1) Could I have some concrete reference on Lev and Tatyana's
                            presentation?
                            2) The theorem you proved with Arsenij is the old and classical way of
                            "Maclaurin's generation of conics"
                            and have already pictures in my gallery (HYPERLINK
                            "http://www.math.uoc.gr/~pamfilos/eGallery/problems/Maclaurin.html"http:
                            //www.math.uoc.gr/~pamfilos/eGallery/problems/Maclaurin.html)
                            3) This gives of course a direct proof of the proposition I posted.
                            4) I prefer though the reduction to parabola and the stress of this
                            property of anticomplementary
                            and more general the precevian triangle. They clarify greatly I think
                            the structure of the
                            circumconics tangent to a given line, and especially the circumparabolas
                            of a triangle.
                            Sincerely Paris



                            -----Original Message-----
                            From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On
                            Behalf Of Alexey.A.Zaslavsky
                            Sent: Tuesday, February 05, 2008 7:29 AM
                            To: Hyacinthos@yahoogroups.com
                            Subject: [SPAM] Re: [EMHL] Circumconics



                            Dear Paris and Francois!
                            It seems that these problems were considered by Lev and Tatyana
                            Emelyanoff. They presented the problems devoted to trypolars at Summer
                            conference of Towns tournament-2005.
                            Also in our with Arsenij Akopyan book next theorem is proved. Given the
                            triangle ABC and the point C0. An arbitrary line passing through C0
                            intersect AC and BC in B' and A' resp. P is the common point of AA' and
                            BB'. Then
                            1. The locus of P is the circumconic touching CC0.
                            2. If CP intersect AB in C' then all lines A'C' have the common point B0
                            and BB0 touches the circumconic.

                            Sincerely Alexey

                            Dear Paris

                            This projective configuration of a conic through 3 points A, B, C and
                            tangent at Q to a line (L) is very beautiful!

                            I am sure there is a nice proof in the general case using Desargues and
                            Plücker
                            involution theorems known for a long time that we can find as an
                            exercise in
                            some old book.

                            But the idea to send line (L) at infinity and to prove the theorem in
                            this
                            special case is very educational.

                            Another (educational?-) idea should be to send both points B and C at
                            infinity that is to say looking at the hyperbola through point A with
                            given
                            asymptotic directions and tangent to a line (L) at Q. I am sure that
                            elementary affine properties of hyperbola can also do the job.

                            Friendly
                            Francois

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                            [Non-text portions of this message have been removed]
                          • Francois Rideau
                            Dear Alexey and Paris As for the circumparabolas of a triangle ABC, it is also useful to transform them by isotomy. By isotomy wrt ABC, such a ABC-
                            Message 13 of 29 , Feb 15, 2008
                              Dear Alexey and Paris
                              As for the circumparabolas of a triangle ABC, it is also useful to
                              transform them by isotomy.
                              By isotomy wrt ABC, such a ABC- circumparabola (P) is transformed in a line
                              (L) tangent to the ABC-Steiner circumellipse (S).
                              For example to get the parabolas through 4 points A, B, C, D, you look at
                              the ABC-isotomic conjugate D' of D.
                              We construct lines (L1) and (L2) through D' tangent to the ellipse (S) and
                              we get the 2 parabolas (P1) and (P2) on A, B, C, D transformed by
                              ABC-isotomy from the lines (L1) and (L2).
                              Of course there is some discussion to do!
                              It is interesting to compare this affine solution with the original
                              Newton's euclidian construction.
                              Friendly
                              Francois


                              [Non-text portions of this message have been removed]
                            • Alexey.A.Zaslavsky
                              Dear Paris! The Emelyanov s paper was published (in Russsian) in Matematicheskoe prosveschenije , 2002, N 6. May be it also was published in FG, but I amn t
                              Message 14 of 29 , Feb 17, 2008
                                Dear Paris!
                                The Emelyanov's paper was published (in Russsian) in "Matematicheskoe prosveschenije", 2002, N 6. May be it also was published in FG, but I amn't sure.

                                Sincerely Alexey

                                Dear Alexey,
                                1) Could I have some concrete reference on Lev and Tatyana's
                                presentation?
                                2) The theorem you proved with Arsenij is the old and classical way of
                                "Maclaurin's generation of conics"
                                and have already pictures in my gallery (HYPERLINK
                                "http://www.math.uoc.gr/~pamfilos/eGallery/problems/Maclaurin.html"http:
                                //www.math.uoc.gr/~pamfilos/eGallery/problems/Maclaurin.html)
                                3) This gives of course a direct proof of the proposition I posted.
                                4) I prefer though the reduction to parabola and the stress of this
                                property of anticomplementary
                                and more general the precevian triangle. They clarify greatly I think
                                the structure of the
                                circumconics tangent to a given line, and especially the circumparabolas
                                of a triangle.
                                Sincerely Paris



                                -----Original Message-----
                                From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On
                                Behalf Of Alexey.A.Zaslavsky
                                Sent: Tuesday, February 05, 2008 7:29 AM
                                To: Hyacinthos@yahoogroups.com
                                Subject: [SPAM] Re: [EMHL] Circumconics

                                Dear Paris and Francois!
                                It seems that these problems were considered by Lev and Tatyana
                                Emelyanoff. They presented the problems devoted to trypolars at Summer
                                conference of Towns tournament-2005.
                                Also in our with Arsenij Akopyan book next theorem is proved. Given the
                                triangle ABC and the point C0. An arbitrary line passing through C0
                                intersect AC and BC in B' and A' resp. P is the common point of AA' and
                                BB'. Then
                                1. The locus of P is the circumconic touching CC0.
                                2. If CP intersect AB in C' then all lines A'C' have the common point B0
                                and BB0 touches the circumconic.

                                Sincerely Alexey

                                Dear Paris

                                This projective configuration of a conic through 3 points A, B, C and
                                tangent at Q to a line (L) is very beautiful!

                                I am sure there is a nice proof in the general case using Desargues and
                                Plücker
                                involution theorems known for a long time that we can find as an
                                exercise in
                                some old book.

                                But the idea to send line (L) at infinity and to prove the theorem in
                                this
                                special case is very educational.

                                Another (educational?-) idea should be to send both points B and C at
                                infinity that is to say looking at the hyperbola through point A with
                                given
                                asymptotic directions and tangent to a line (L) at Q. I am sure that
                                elementary affine properties of hyperbola can also do the job.

                                Friendly
                                Francois

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                              • Paris Pamfilos
                                Dear Francois thank you for the nice remarks on the alternative description of circumparabolas. Dear Alexei thank you for the reference. Is Matematicheskoe
                                Message 15 of 29 , Feb 18, 2008
                                  Dear Francois
                                  thank you for the nice remarks on the alternative description
                                  of circumparabolas.
                                  Dear Alexei
                                  thank you for the reference.
                                  Is "Matematicheskoe prosveschenije" electronically accessible?
                                  Do you know?
                                  Best regards to all
                                  Paris

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                                • Alexey.A.Zaslavsky
                                  Dear Paris! Yes in www.mccme.ru you can find the numbers of MP but I think that there is only russian text. Sincerely
                                  Message 16 of 29 , Feb 19, 2008
                                    Dear Paris!
                                    Yes in www.mccme.ru you can find the numbers of "MP" but I think that there is only russian text.

                                    Sincerely Alexey

                                    Dear Francois
                                    thank you for the nice remarks on the alternative description
                                    of circumparabolas.
                                    Dear Alexei
                                    thank you for the reference.
                                    Is "Matematicheskoe prosveschenije" electronically accessible?
                                    Do you know?
                                    Best regards to all
                                    Paris

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