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A synthetic affine proof of Desargues’ projective theorem

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  • brozolo
    Dear Hyacinthists, here I will give a synthetic affine proof of the projective Desargues theorem using the affine Desargues theorem. Sorry if it s already
    Message 1 of 7 , Sep 9, 2012
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      Dear Hyacinthists, here I will give a synthetic affine proof of the projective Desargues' theorem using the affine Desargues' theorem. Sorry if it's already known.

      The projective Desargues' theorem (pDt) says:
      if AA', BB', CC' are distinct lines concurring in O then L=BC^B'C', M=AC^A'C', N=AB^A'B' are collinear and conversely.

      The affine Desargues' theorem (aDt) says:
      if AA', BB', CC' are distinct concurring (or parallel) lines and BC//B'C' and CA//C'A' then AB//A'B'. Conversely if BC//B'C' and CA//C'A' and AB//A'B' then AA', BB', CC' concur (or are parallel).

      In projective geometry aDt is readily obtained from pDt when L, M, N are on the infinite line. Here instead I'll give a synthetic proof using only affine tools.

      If in pDt, instead of line L, M, N, it's point O that goes to infinity we get a similar theorem, that I will call "Second Affine Desargues' theorem" (aDt2):
      if AA', BB', CC' are parallel lines then L=BC^B'C', M=AC^A'C', N=AB^A'B' are collinear.
      This theorem is a direct consequence of aDt, here's the proof:
      Call P the point of intersection of the parallel lines to AB, A'B' sent from C, C'.
      L, N, P are on a line for aDt in triangles BB'N, CC'P.
      M, N, P are on a line for aDt in triangles AA'N, CC'P.
      Therefore L, M, N are on a line.
      Special cases when one or more of the points P, L, M, N are infinite points can be treated similarly.

      Here's the proof of Desargues projective theorem:
      From B and C send parallel lines to AA' that meet lines A'B', A'C' in B'', C''.
      Call L* the point of intersection of BC and B''C''.
      L*, C', B' are on a line for aDt2 in triangles BCO and B''C''A'.
      Therefore L* is on B'C' and on BC, so L=L*.
      N, L, M are on a line for aDt2 in triangles ABC and A'B''C''.
      Again special cases when one of the points L, M, N is an infinite point can be treated similarly.

      Ciao.
      Marcello Tarquini
    • Nikolaos Dergiades
      Dear friends, If the points A , B , C are not collinear and are on the sides BC, CA, AB of triangle ABC such that BA = CB = AC and the triangles ABC, A B C
      Message 2 of 7 , Sep 24, 2012
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        Dear friends,
        If the points A', B', C' are not collinear and
        are on the sides BC, CA, AB of triangle ABC
        such that BA'= CB' = AC' and the triangles
        ABC, A'B'C' have the same incenter then
        ABC is equilateral. Do you know a proof?

        Best regards
        Nikos Dergiades
      • Nikolaos Dergiades
        Dear friends, does anybody has this book? C. Cocea – 200 de probleme din geometria triunghiului echilateral, Editura ,,Gh. Asachi” Iasi, 1992. Best regards
        Message 3 of 7 , Sep 27, 2012
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          Dear friends,

          does anybody has this book?

          C. Cocea – 200 de probleme din geometria triunghiului echilateral, Editura ,,Gh. Asachi” Iasi, 1992.

          Best regards
          Nikos Dergiades
        • Catalin Barbu
                                  Dear Nikos,  I have the book and I kindly request you to wait for few days in order to scan it, or, if you want, I could
          Message 4 of 7 , Sep 27, 2012
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                                    Dear Nikos,


             I have the book and I kindly request you to wait for few days in order to scan it, or, if you want, I could send you the hard copy of the book.


            Best regards,

            Catalin Barbu





            ________________________________
            From: Nikolaos Dergiades <ndergiades@...>
            To: Hyacinthos@yahoogroups.com
            Sent: Thursday, September 27, 2012 7:10 PM
            Subject: [EMHL] (unknown)


             
            Dear friends,

            does anybody has this book?

            C. Cocea – 200 de probleme din geometria triunghiului echilateral, Editura ,,Gh. Asachi” Iasi, 1992.

            Best regards
            Nikos Dergiades



            [Non-text portions of this message have been removed]
          • Nikolaos Dergiades
            Dear Catalin, Thank you very much A good friend has just sent it to me. Best regards Nikos Dergiades
            Message 5 of 7 , Sep 27, 2012
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              Dear Catalin,
              Thank you very much
              A good friend
              has just sent it to me.

              Best regards
              Nikos Dergiades


              --- Στις Πέμ., 27/09/12, ο/η Catalin Barbu <kafka_mate@...> έγραψε:

              > Από: Catalin Barbu <kafka_mate@...>
              > Θέμα: Re: [EMHL] (unknown)
              > Προς: "Hyacinthos@yahoogroups.com" <Hyacinthos@yahoogroups.com>
              > Ημερομηνία: Πέμπτη, 27 Σεπτέμβριος 2012, 21:05
              >                        
              > Dear Nikos,
              >
              >
              >  I have the book and I kindly request you to wait for few
              > days in order to scan it, or, if you want, I could send you
              > the hard copy of the book.
              >
              >
              > Best regards,
              >
              > Catalin Barbu
              >
              >
              >
              >
              >
              > ________________________________
              > From: Nikolaos Dergiades <ndergiades@...>
              > To: Hyacinthos@yahoogroups.com
              >
              > Sent: Thursday, September 27, 2012 7:10 PM
              > Subject: [EMHL] (unknown)
              >
              >
              >  
              > Dear friends,
              >
              > does anybody has this book?
              >
              > C. Cocea – 200 de probleme din geometria triunghiului
              > echilateral, Editura ,,Gh. Asachi” Iasi, 1992.
              >
              > Best regards
              > Nikos Dergiades
              >
              >
              >
              > [Non-text portions of this message have been removed]
              >
              >
              >
              > ------------------------------------
              >
              > Yahoo! Groups Links
              >
              >
              >     Hyacinthos-fullfeatured@yahoogroups.com
              >
              >
            • Nikolaos Dergiades
              Dear friends, Let the points A1, A2 on BC are equidistant from the mid point of BC and BA1 = CA2 = t. Similarly we take the points B1, B2 on CA and the points
              Message 6 of 7 , Oct 4, 2012
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                Dear friends,
                Let the points A1, A2 on BC are equidistant
                from the mid point of BC and BA1 = CA2 = t.
                Similarly we take the points B1, B2 on CA
                and the points C1, C2 on AB with the same t.
                The points A1,A2,B1,B2,C1,C2 are on a conic
                and the locus of the center of this conic
                as t changes is a quartic that is the
                anticomplement of the quartic
                P(ayz)^2+Q(bzx)^2+R(cxy)^2-PQRxyz(x+y+z)=0
                where P = b-c, Q = c-a, R = a-b
                that is the isotomic of the conic
                P(ax)^2 + Q(by)^2 + R(cz)^2 - PQR(xy + yz + zx) = 0.
                I think that the conic is a hyperbola with
                center not in ETC and passes through the points
                X(8), X(69), X(1330), X(1654), X(2891), X(3436).

                Best regards
                Nikos Degiades
              • rhutson2
                Dear Nikos, The center of this hyperbola is the anticomplement of the center of hyperbola {A,B,C,X(1),X(6)} (the isogonal conjugate of the Nagel line and the
                Message 7 of 7 , Oct 4, 2012
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                  Dear Nikos,

                  The center of this hyperbola is the anticomplement of the center of hyperbola {A,B,C,X(1),X(6)} (the isogonal conjugate of the Nagel line and the isotomic conjugate of line X(10)X(75)).

                  The center of hyperbola {A,B,C,X(1),X(6)} is also:
                  Perspector of ABC and tangential triangle of circumconic centered at X(649);
                  Center of circumconic that is locus of trilinear poles of lines passing through X(649);
                  X(2)-Ceva conjugate of X(649).

                  Hyperbola {A,B,C,X(1),X(6)} also passes through X(34), X(56), X(58), X(86), X(87), X(106), X(269), X(292), X(870), X(937), X(939), X(977), X(979), X(996), X(998), X(1027), X(1120), X(1126), X(1167), X(1220), X(1222), X(1256), X(1474), X(2297), X(2424), X(2665), X(3226), X(3445).

                  Best regards,
                  Randy Hutson

                  --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...> wrote:
                  >
                  > Dear friends,
                  > Let the points A1, A2 on BC are equidistant
                  > from the mid point of BC and BA1 = CA2 = t.
                  > Similarly we take the points B1, B2 on CA
                  > and the points C1, C2 on AB with the same t.
                  > The points A1,A2,B1,B2,C1,C2 are on a conic
                  > and the locus of the center of this conic
                  > as t changes is a quartic that is the
                  > anticomplement of the quartic
                  > P(ayz)^2+Q(bzx)^2+R(cxy)^2-PQRxyz(x+y+z)=0
                  > where P = b-c, Q = c-a, R = a-b
                  > that is the isotomic of the conic
                  > P(ax)^2 + Q(by)^2 + R(cz)^2 - PQR(xy + yz + zx) = 0.
                  > I think that the conic is a hyperbola with
                  > center not in ETC and passes through the points
                  > X(8), X(69), X(1330), X(1654), X(2891), X(3436).
                  >
                  > Best regards
                  > Nikos Degiades
                  >
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