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Question on conics

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  • Giovanni Artico
    It is known that if we have two common points of two conics and other sufficient elements of each one, we can construct the other two common points; the same
    Message 1 of 15 , May 12 12:48 PM
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      It is known that if we have two common points of two conics and other
      sufficient elements of each one, we can construct the other two common
      points; the same is true for the common tangents. My question is: is it
      true that if we know two common tangents (and all other elements of
      each conic) we can construct the common points? And if this is true,
      how can this be accomplished ?
      Thanks for all illuminations

      Giovanni Artico
    • maedu@hotmail.com
      To construct a conic it is enough to know 5 data about it: 5 points, or 5 tangents, or combinations of these. If You know 2 tangents and 3 other points, You
      Message 2 of 15 , May 12 12:52 PM
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        To construct a conic it is enough to know 5 data about it: 5 points, or 5 tangents, or combinations of these. If You know 2 tangents and 3 other points, You can construct the conic.

        Martin


        From: Giovanni Artico
        Sent: Monday, May 12, 2008 9:48 PM
        To: Hyacinthos@yahoogroups.com
        Subject: [EMHL] Question on conics


        It is known that if we have two common points of two conics and other
        sufficient elements of each one, we can construct the other two common
        points; the same is true for the common tangents. My question is: is it
        true that if we know two common tangents (and all other elements of
        each conic) we can construct the common points? And if this is true,
        how can this be accomplished ?
        Thanks for all illuminations

        Giovanni Artico





        [Non-text portions of this message have been removed]
      • Alexey.A.Zaslavsky
        Dear Giovanni! I don t understand the problem. What are the given elements of conics and which means can we use for construction? Sincerely
        Message 3 of 15 , May 12 11:10 PM
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          Dear Giovanni!
          I don't understand the problem. What are the given elements of conics and which means can we use for construction?

          Sincerely Alexey


          It is known that if we have two common points of two conics and other
          sufficient elements of each one, we can construct the other two common
          points; the same is true for the common tangents. My question is: is it
          true that if we know two common tangents (and all other elements of
          each conic) we can construct the common points? And if this is true,
          how can this be accomplished ?
          Thanks for all illuminations

          Giovanni Artico




          Antivirus scanning: Symantec Mail Security for SMTP.

          [Non-text portions of this message have been removed]
        • Martin Acosta
          R. Cuppens in Faire de la géométrie supérieure en jouant avec Cabri géomètre II writes: the problem of finding the common tangents to two conics is the
          Message 4 of 15 , May 12 11:35 PM
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            R. Cuppens in 'Faire de la géométrie supérieure en jouant avec Cabri géomètre II' writes:

            "the problem of finding the common tangents to two conics is the dual problem of finding the common points. If You have to construct the common points A, B, C, D of conic 1 and the porar conic of conic 2: the tangents to conic 1 by these points are the common tangents of conics 1 and 2".

            I think it wold not be difficult to inverse this construction to find the common points having the common tangents.

            Martin


            To: Hyacinthos@yahoogroups.comFrom: zasl@...: Tue, 13 May 2008 10:10:01 +0400Subject: Re: [EMHL] Question on conics




            Dear Giovanni! I don't understand the problem. What are the given elements of conics and which means can we use for construction? Sincerely Alexey It is known that if we have two common points of two conics and other sufficient elements of each one, we can construct the other two common points; the same is true for the common tangents. My question is: is it true that if we know two common tangents (and all other elements of each conic) we can construct the common points? And if this is true, how can this be accomplished ? Thanks for all illuminations Giovanni Artico Antivirus scanning: Symantec Mail Security for SMTP.[Non-text portions of this message have been removed]






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          • jpehrmfr
            Dear Martin and Giovanni ... There is a little problem : there are 4 conics going through 3 given points and touching 2 given lines. Of course, these 4 conics
            Message 5 of 15 , May 12 11:54 PM
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              Dear Martin and Giovanni
              > To construct a conic it is enough to know 5 data about it: 5 points,
              > or 5 tangents, or combinations of these. If You know 2 tangents and 3
              > other points, You can construct the conic.

              There is a little problem : there are 4 conics going through 3 given
              points and touching 2 given lines.
              Of course, these 4 conics are constructible : if the three common
              points are A,B,C and the common tangents are L1, L2 with trilinear
              poles P1, P2 (wrt ABC), the perspectors of the 4 conics are the common
              points of the inscribed conics with perspectors P1 and P2 and these
              common points are constructible (because the degenerate conics of the
              pencil generated these two inscribed conics are constructible pair of
              lines intersecting respectively at the vertices of the precevian
              triangle of L1 inter L2)
              But now, if you want to construct the common points of two conics
              knowing three points and to tangents for each conic, you have 16 pairs
              of such conics and the complete construction (which is possible) will
              be a bit tedious.
              friendly. Jean-Pierre
            • Giovanni Artico
              ... and which means can we use for construction? ... Example: two hyperbolae; I know the foci and vertices of each hyperbola; I know also the four common
              Message 6 of 15 , May 13 12:36 AM
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                --- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@...>
                wrote:
                >
                > Dear Giovanni!
                > I don't understand the problem. What are the given elements of conics
                and which means can we use for construction?
                >

                Example: two hyperbolae; I know the foci and vertices of each
                hyperbola; I know also the four common tangents; by ruler and compass
                is it possible to construct the four intersection points? If yes, how ?
                Sincerely

                Giovanni
              • jpehrmfr
                Dear Hyacinthists, I wrote ... given ... Here is a possible - and quite classical - construction of the conics going through A,B,C and touching the lines L1
                Message 7 of 15 , May 13 5:08 AM
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                  Dear Hyacinthists,
                  I wrote
                  > There is a little problem : there are 4 conics going through 3
                  given
                  > points and touching 2 given lines.
                  > Of course, these 4 conics are constructible.

                  Here is a possible - and quite classical - construction of the conics
                  going through A,B,C and touching the lines L1 and L2. (at most 4 real
                  solutions)
                  C1,C2 = common points of AB with L1,L2
                  B1,B2 = common points of AC with L1,L2
                  The radical axis of the circles (O) = ABC and C1C2C intersects AB at
                  Oc; the circle centered at Oc and orthogonal to (O) intersects AB at
                  P,P'.
                  Same way, the radical axis of the circles (O) and B1B2B intersects AC
                  at Ob; the circle centered at Ob and orthogonal to (O) intersects AC
                  at Q,Q'.
                  Now take L = PQ or PQ' or P'Q or P'Q'
                  If L intersects L1 and L2 at D1 and D2, the conic ABCD1D2 is a
                  solution

                  This due to the fact that
                  - P, P' are the fixed points of the involution on the line AB
                  swapping (A,B) and (C1,C2)
                  - Q, Q' are the fixed points of the involution on the line AC
                  swapping (A,C) and (B1,B2)
                  - As the line D1D2 (twice) is member of the pencil generated by the
                  required conic and the degenerated conic L1 union L2, the common
                  point of D1D2 and AB - or AC - must be a fixed point of the
                  involution i - ot i' -

                  Friendly. Jean-Pierre
                • Martin Acosta
                  Knowing the foci and vertices You can construct as many points of the conic as You want. But I think that common tangents don t give information about common
                  Message 8 of 15 , May 13 5:22 AM
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                    Knowing the foci and vertices You can construct as many points of the conic as You want. But I think that common tangents don't give information about common points. So my previous message is wrong: if You know how to construct the common points, You cant construct common tangents using the polar conic; but You can't construct common points having the common tangents, I think.

                    Martin


                    To: Hyacinthos@yahoogroups.comFrom: zeroprof@...: Tue, 13 May 2008 07:36:55 +0000Subject: Re: [EMHL] Question on conics




                    --- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@...> wrote:>> Dear Giovanni!> I don't understand the problem. What are the given elements of conics and which means can we use for construction?> Example: two hyperbolae; I know the foci and vertices of each hyperbola; I know also the four common tangents; by ruler and compass is it possible to construct the four intersection points? If yes, how ?SincerelyGiovanni






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                  • jpehrmfr
                    Dear Martin ... conic as You want. But I think that common tangents don t give information about common points. The diagonal triangle of the complete
                    Message 9 of 15 , May 13 8:23 AM
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                      Dear Martin

                      > Knowing the foci and vertices You can construct as many points of the
                      conic as You want. But I think that common tangents don't give
                      information about common points.

                      The diagonal triangle of the complete quadrilateral with sidelines the
                      4 common tangents = the diagonal triangle of the quadrangle with
                      vertices the 4 common points = the only triangle selfpolar wrt each
                      conic
                      Friendly. Jean-Pierre
                    • Giovanni Artico
                      ... the ... Dear Jean-Pierre, I apologize, but to a not expert geometer as me this sounds a bit cryptic. I guess it means that the common points can be
                      Message 10 of 15 , May 13 1:11 PM
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                        --- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-pierre.ehrmann@...>
                        wrote:
                        > The diagonal triangle of the complete quadrilateral with sidelines
                        the
                        > 4 common tangents = the diagonal triangle of the quadrangle with
                        > vertices the 4 common points = the only triangle selfpolar wrt each
                        > conic
                        > Friendly. Jean-Pierre
                        >

                        Dear Jean-Pierre,
                        I apologize, but to a not expert geometer as me this sounds a bit
                        cryptic. I guess it means that the common points can be constructed,
                        and in the construction is involved a selfpolar triangle: but could you
                        briefly outline the principal steps of the construction? Many thanks in
                        advance.

                        Giovanni
                      • jpehrmfr
                        Dear Giovanni [JPE] ... [GA] ... you ... No, you cannot find the common points if you know only the common tangents (because there are infinitely many conics
                        Message 11 of 15 , May 14 12:20 AM
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                          Dear Giovanni
                          [JPE]
                          > > The diagonal triangle of the complete quadrilateral with sidelines
                          > the
                          > > 4 common tangents = the diagonal triangle of the quadrangle with
                          > > vertices the 4 common points = the only triangle selfpolar wrt each
                          > > conic
                          [GA]
                          > I guess it means that the common points can be constructed,
                          > and in the construction is involved a selfpolar triangle: but could
                          you
                          > briefly outline the principal steps of the construction?

                          No, you cannot find the common points if you know only the common
                          tangents (because there are infinitely many conics with 4 given
                          tangents).
                          I just wanted to say that there exists is a very close relation between
                          the complete quadrilateral with sidelines the 4 common tangents, the
                          quadrangle with vertices the 4 common points and the triangle selfpolar
                          wrt both conics.
                          Of course, if you know one common point M and the 4 common tangents,
                          the remark above gives an immediate construction (with ruler only) of
                          the 3 other common points : they are the vertices of the precevian
                          triangle of M wrt to the diagonal triangle of the complete quadrileral
                          with sidelines the 4 common tangents.
                          Same way, if you know the 4 common point and one common tangent, you
                          can construct the 3 other common tangents.
                          Friendly. Jean-Pierre
                        • Giovanni Artico
                          ... The question is a bit different: if I know four common tangents AND another point on each conic, can I construct the four intersection points? (in this
                          Message 12 of 15 , May 14 9:15 AM
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                            > No, you cannot find the common points if you know only the common
                            > tangents (because there are infinitely many conics with 4 given
                            > tangents).

                            The question is a bit different: if I know four common tangents AND
                            another point on each conic, can I construct the four intersection
                            points? (in this case there are not infinitely many conics).
                            The problem could be restated : if I know five points of each conic AND
                            four of them are common, can I costruct the four common tangents?
                            I don't have so far an answer to this question.
                            Many thanks for your help Jean-Pierre

                            Giovanni
                          • jpehrmfr
                            Dear Giovanni ... AND ... Yes, of course. Here is a possible solution (there is probably a better solution): 1)Find the diagonal triangle A B C of the
                            Message 13 of 15 , May 15 8:35 AM
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                              Dear Giovanni

                              > The question is a bit different: if I know four common tangents AND
                              > another point on each conic, can I construct the four intersection
                              > points? (in this case there are not infinitely many conics).
                              > The problem could be restated : if I know five points of each conic
                              AND
                              > four of them are common, can I costruct the four common tangents?
                              > I don't have so far an answer to this question.
                              > Many thanks for your help Jean-Pierre

                              Yes, of course.
                              Here is a possible solution (there is probably a better solution):
                              1)Find the diagonal triangle A'B'C' of the quadrangle with vertices
                              the 4 common points
                              2) Find the centers W,W' of the two conics
                              3) If the line WW' intersects B'C' at S, find the points U1, U2 on
                              the line B'C' with midpoint S and harmonic conjugates wrt (B',C')
                              These points are opposite vertices of the complete quadrilateral with
                              sidelines the 4 common tangents
                              Do the same with the lines WW' and C'A' to get the 4 common tangents.

                              Friendly. Jean-Pierre
                            • Giovanni Artico
                              Merci beaucoup Jean-Pierre Giovanni
                              Message 14 of 15 , May 15 12:03 PM
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                                Merci beaucoup Jean-Pierre

                                Giovanni
                              • jpehrmfr
                                Dear Giovanni [Giovanni] ... AND ... intersection ... conic ... [JP] ... with ... tangents. I ve just used two simple facts : - U1,U2 are harmonic conjugates
                                Message 15 of 15 , May 15 2:34 PM
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                                  Dear Giovanni
                                  [Giovanni]
                                  > > The question is a bit different: if I know four common tangents
                                  AND
                                  > > another point on each conic, can I construct the four
                                  intersection
                                  > > points? (in this case there are not infinitely many conics).
                                  > > The problem could be restated : if I know five points of each
                                  conic
                                  > AND
                                  > > four of them are common, can I costruct the four common tangents?
                                  > > I don't have so far an answer to this question.
                                  [JP]
                                  > Yes, of course.
                                  > Here is a possible solution (there is probably a better solution):
                                  > 1)Find the diagonal triangle A'B'C' of the quadrangle with vertices
                                  > the 4 common points
                                  > 2) Find the centers W,W' of the two conics
                                  > 3) If the line WW' intersects B'C' at S, find the points U1, U2 on
                                  > the line B'C' with midpoint S and harmonic conjugates wrt (B',C')
                                  > These points are opposite vertices of the complete quadrilateral
                                  with
                                  > sidelines the 4 common tangents
                                  > Do the same with the lines WW' and C'A' to get the 4 common
                                  tangents.

                                  I've just used two simple facts :
                                  - U1,U2 are harmonic conjugates wrt B', C'
                                  - their midpoint lies on the Newton line of the complete
                                  quadrilateral with vertices the four common tangents and this Newton
                                  line is the line WW'
                                  In fact, there is probably something better if, instead of the
                                  centers of the conics, we use the poles of a well choosen line wrt
                                  the conics.
                                  Friendly. Jean-Pierre
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