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file with interesting problems

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  • Luís Lopes
    Dear Hyacinthists, I would like to call your attention to the file http://www.lps.ufrj.br/~sergioln/ime/IMEv9b.pdf In it you will find lots of proposed and
    Message 1 of 12 , May 3, 2006
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      Dear Hyacinthists,

      I would like to call your attention to the file

      http://www.lps.ufrj.br/~sergioln/ime/IMEv9b.pdf

      In it you will find lots of proposed and solved problems
      from an admissional exam collection year after year.

      In particular go to pages 202 e 203 where is presented
      in rich details (albeit in portuguese) the solution sent by
      Jean-Pierre Ehrmann (merci encore) to a 1986/1987 question.

      And to page 209 to another Hyacinthos' members contribution.

      Sergio is the author of this huge work and there still is one
      unsolved question for which he is asking for help. There it is:

      ===
      Dados dois circulos externos de raios distintos,
      mostre que o conjunto de secantes que determinam
      em ambos cordas iguais, e' tal que, cada uma dessas
      secantes e' tangente a uma parabola, que se pede identificar.
      ===
      Two external circles with different radii are given.
      Show that the set of secants that determine in both
      circles chords of equal lenghts is such that each of these
      secants is tangent to a parabola. The problem asks to
      identify the parabola as well.

      Last but not least, credits are given at page 3.

      Thank you for your attention.

      Best regards,
      Luis
    • armpist
      ... Dear Luis and all, In Russian edition of Yaglom it is problem #8: equal chords on a line through a given point A. Basically it is revolving one of the two
      Message 2 of 12 , May 3, 2006
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        --- In Hyacinthos@yahoogroups.com, Luís Lopes <qed_texte@...> wrote:
        >
        > Dear Hyacinthists,
        >
        > I would like to call your attention to the file
        >
        > http://www.lps.ufrj.br/~sergioln/ime/IMEv9b.pdf
        >
        >............
        > Two external circles with different radii are given.
        > Show that the set of secants that determine in both
        > circles chords of equal lenghts is such that each of these
        > secants is tangent to a parabola. The problem asks to
        > identify the parabola as well.
        >
        > Last but not least, credits are given at page 3.
        >
        > Thank you for your attention.
        >
        > Best regards,
        > Luis
        >


        Dear Luis and all,

        In Russian edition of Yaglom it is problem #8: equal chords
        on a line through a given point A. Basically it is revolving
        one of the two circles along the circle with diameter O1O2.
        Common chord (when exists, otherwise radical axis) will be a
        tangent to our parabola with focus in midpoint of segment O1O2.

        Radical axis of circles O1,O2 will be a tangent at vertex, so the
        directrix is positioned accordingly.


        Thank you.

        M.T.
      • armpist
        ... the ... Sorry, problem #7(c) in Yaglom book. M.T.
        Message 3 of 12 , May 3, 2006
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          --- In Hyacinthos@yahoogroups.com, "armpist" <armpist@...> wrote:
          >
          > --- In Hyacinthos@yahoogroups.com, Luís Lopes <qed_texte@> wrote:
          > >
          > > Dear Hyacinthists,
          > >
          > > I would like to call your attention to the file
          > >
          > > http://www.lps.ufrj.br/~sergioln/ime/IMEv9b.pdf
          > >
          > >............
          > > Two external circles with different radii are given.
          > > Show that the set of secants that determine in both
          > > circles chords of equal lenghts is such that each of these
          > > secants is tangent to a parabola. The problem asks to
          > > identify the parabola as well.
          > >
          > > Last but not least, credits are given at page 3.
          > >
          > > Thank you for your attention.
          > >
          > > Best regards,
          > > Luis
          > >
          >
          >
          > Dear Luis and all,
          >
          > In Russian edition of Yaglom it is problem #8: equal chords
          > on a line through a given point A. Basically it is revolving
          > one of the two circles along the circle with diameter O1O2.
          > Common chord (when exists, otherwise radical axis) will be a
          > tangent to our parabola with focus in midpoint of segment O1O2.
          >
          > Radical axis of circles O1,O2 will be a tangent at vertex, so
          the
          > directrix is positioned accordingly.
          >
          >
          > Thank you.
          >
          > M.T.
          >


          Sorry, problem #7(c) in Yaglom book.

          M.T.
        • jpehrmfr
          Dear Francois and Hauke ... Neuberg ... are ... the 7 ... another ... Using the group of the cubic - with zero = X(30) -, we have M + M = O+X(74) = X(1138)
          Message 4 of 12 , May 3, 2006
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            Dear Francois and Hauke

            > I don't know if this is useful for finding the points on the
            Neuberg
            > cubic with the quartet property, but may be (I suspect that there
            are
            > only two such points H and X(74)):
            > N(A,B,C) = Neuberg cubic of ABC
            > If M lies on N(A,B,C), then
            > N(A,B,C), N(M,B,C), N(A,M,C), N(A,B,M) have 7 common points :
            > A,B,C,the circular points at infinity, M and another point M'.
            > M' is the 6th common point (apart A,B,C,H,M) of N(A,B,C) with the
            > rectangular hyperbola through A,B,C,M.
            >
            > Note that we know the 9 common points of N(A,B,C) and N(M,B,C) :
            the 7
            > above and the two points P such as PBC is equilateral.
            > It follows that M->M' is a birationnal involutive mapping of the
            > Neuberg cubic and that, if M is a triangle center, M' will be
            another
            > one.

            Using the group of the cubic - with zero = X(30) -, we have
            M + M' = O+X(74) = X(1138)
            Thus the line MM' goes through the isogonal conjugate of X(1138),
            which is X(399).
            Hence we get an easier definition of M' : M' is the second
            intersection of the line X(399)M with the rectangular hyperbola
            going through ABCM.
            I hope that these remarks can lead to the determination of all the
            QP-centers lying on the Neuberg cubic (I strongly suspect that there
            exists only two such points : H and X(74))
            Friendly. Jean-Pierre
          • jpehrmfr
            Dear Francois and Hauke ... Just a little remark : if P(A,B,C) is X(399) of ABC, it follows clearly that, if M lies on the Neuberg cubic, the four lines
            Message 5 of 12 , May 3, 2006
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              Dear Francois and Hauke

              > Using the group of the cubic - with zero = X(30) -, we have
              > M + M' = O+X(74) = X(1138)
              > Thus the line MM' goes through the isogonal conjugate of X(1138),
              > which is X(399).
              > Hence we get an easier definition of M' : M' is the second
              > intersection of the line X(399)M with the rectangular hyperbola
              > going through ABCM.
              > I hope that these remarks can lead to the determination of all the
              > QP-centers lying on the Neuberg cubic (I strongly suspect that there
              > exists only two such points : H and X(74)).
              Just a little remark : if P(A,B,C) is X(399) of ABC, it follows
              clearly that, if M lies on the Neuberg cubic, the four lines
              M-P(A,B,C), A-P(M,B,C), B-P(A,M,B), C-P(A,B,M) concur at a point lying
              on the Neuberg cubic and on the rectangular hyperbola through ABCM.
              Friendly. Jean-Pierre
            • Hauke Reddmann
              ... points, to ... Aargh, bad example, of course for a curve we need ONE equation. Say, the pseudo-roots of AB+AC+AD+BC+BD+CD=0. But first we have to sort out
              Message 6 of 12 , May 4, 2006
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                --- In Hyacinthos@yahoogroups.com, "Francois Rideau"
                <francois.rideau@...> wrote:
                >
                > This relation AB*CD = AC*BD= AD*BC characterizes not one but two
                points, to
                > say the isodynamic points #15 and #16, inverse in the circumcircle.

                Aargh, bad example, of course for a curve we need ONE equation.
                Say, the pseudo-roots of AB+AC+AD+BC+BD+CD=0.
                But first we have to sort out a definitory problem.
                What do you here associate with the entity "#13"?
                a) the 120-120-120 Steiner triangulation point of ABC
                (which means that #13 doesn't exist for the Kimberling triangle),
                b) the point with trilinears sec(@-pi/6):... ?
                If the latter holds (and comparable problems arise for
                #175/#176 etc.) the QP can be described analytically,
                and any references to geometric definitions like a)
                are irrelevant. (I.e., I think that at least #13-#16
                of the Neuberg cubic do have the QP too because I stick
                to definition b).)

                Hauke
              • jpehrmfr
                Dear Hauke ... There exists allways two points verifying a) but b) is better because of tha ambiguity when these points are outside ABC. If you want a
                Message 7 of 12 , May 4, 2006
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                  Dear Hauke

                  > But first we have to sort out a definitory problem.
                  > What do you here associate with the entity "#13"?
                  > a) the 120-120-120 Steiner triangulation point of ABC
                  > (which means that #13 doesn't exist for the Kimberling triangle),
                  > b) the point with trilinears sec(@-pi/6):... ?
                  > If the latter holds (and comparable problems arise for
                  > #175/#176 etc.) the QP can be described analytically,
                  > and any references to geometric definitions like a)
                  > are irrelevant. (I.e., I think that at least #13-#16
                  > of the Neuberg cubic do have the QP too because I stick
                  > to definition b).)

                  There exists allways two points verifying a) but b) is better
                  because of tha ambiguity when these points are outside ABC.
                  If you want a geometric definition without ambiguity, I suggest you
                  either
                  c) to draw externally three equilteral triangles A'BC, B'CA, C'AB;
                  A'B'C' and ABC will be perspective at #13.
                  or
                  d) to begin with #15 and #16 : they are the common points of the
                  three Appolonian circles and lye on the line OK (K=symedian point)
                  Now #15 is the one lying on the segment OK or equivalently inside
                  the circumcircle or equivalently the only point with pedal triangle
                  equilateral with the same orientation than ABC. Now, #13 is the
                  isogonal conjugate of #15.
                  In any case, none of these 4 points has the QP-property (a sketch
                  shows that immediately)
                  Friendly. Jean-Pierre
                • jpehrmfr
                  Dear Hauke [HR] ... [JPE] ... I m sorry. This was not very clear; I was meaning that there allways exist two points M for wgich the oriented angles of the
                  Message 8 of 12 , May 4, 2006
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                    Dear Hauke
                    [HR]
                    > > But first we have to sort out a definitory problem.
                    > > What do you here associate with the entity "#13"?
                    > > a) the 120-120-120 Steiner triangulation point of ABC
                    > > (which means that #13 doesn't exist for the Kimberling triangle),

                    [JPE]
                    > There exists allways two points verifying a)
                    I'm sorry. This was not very clear; I was meaning that there allways
                    exist two points M for wgich the oriented angles of the cevian lines
                    verify
                    <AMB = <BMC = <CMA = 120 or -120 modulo 180
                    In any case, for the Kimberling triangle 6,9,13, each angle of ABC is
                    <120; so #13 is inside ABC and there is no p^roblem with the Steiner
                    characterization.
                    Friendly. Jean-Pierre
                  • Hauke Reddmann
                    ... I can t think geometrically! :-) From the definition of #15/16 (AB*CD=AC*BD=AD*BC [1]) the QP property follows almost by default...uh-oh, I begin to see
                    Message 9 of 12 , May 5, 2006
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                      --- In Hyacinthos@yahoogroups.com, "jpehrmfr" <Jean-
                      Pierre.Ehrmann.70@...> wrote:
                      >
                      > In any case, none of these 4 points has the QP-property (a sketch
                      > shows that immediately)

                      I can't think geometrically! :-)
                      From the "definition" of #15/16 (AB*CD=AC*BD=AD*BC [1]) the
                      QP property follows almost by default...uh-oh, I begin
                      to see what's cooking. [1] defines a {set} of points,
                      not a single one. Thus if D is #15 of ABC, A might be
                      #16 of BCD. Same trouble with #13/14.

                      Hauke
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