Loading ...
Sorry, an error occurred while loading the content.

Re: [EMHL] Three concurrent lines

Expand Messages
  • chris.vantienhoven
    ... point of concurs is not in ETC. ... solved. ... Dear Jean-Louis, It s always thrilling when three lines (constructed in the same way) coincide. I checked
    Message 1 of 17 , Feb 11, 2009
    • 0 Attachment
      >"X is the intersection of BC with parallel to AI through O".
      > Francisco Javier Garcia Capitan has verified my conjecture and the
      point of concurs is not in ETC.
      > This problem has a relation with a difficult one. I think it is now
      solved.
      > I must now write a little article on it.
      > Cela me fait beaucoup de plaisir d'avoir de tes nouvelles.
      > Sincerely
      > Jean-Louis

      Dear Jean-Louis, It's always thrilling when three lines (constructed
      in the same way) coincide. I checked also if this point occurs in ETC
      and indeed it is not present. I also made a limited scan of lines on
      which it occurs. Here are the lines I found: * X1.X159 *
      X3.X960.X1158 * X25.X65 * X31.X56.X154.X221 * X40.X197 *
      X64.X71.X198.X220 Besides that it occurs on the
      ABC.X154.X220-hyperbola. After all no bad result. Best regards,
      Chris van Tienhoven


      [Non-text portions of this message have been removed]
    • Moses, Peter J. C.
      Hi Jean-Louis, Chris and all, It is true that the point is not in ETC. It lies on these lines .. {1,159,1486,3220}, {3,960,997,1158,3185,3435},
      Message 2 of 17 , Feb 12, 2009
      • 0 Attachment
        Hi Jean-Louis, Chris and all,

        It is true that the point is not in ETC. It lies on these lines ..

        {1,159,1486,3220},
        {3,960,997,1158,3185,3435},
        {6,1245,1973,2354},
        {20,1610,1633},
        {25,65,1452},
        {31,56,154,221,603,1042,1106,1191,1201,1399,1406,1407,1457,1473,2390},
        {40,197},
        {55,976,1854,2292,3145},
        {63,1619},
        {64,71,198,220,2155,3197},
        {73,2187},
        {100,1265},
        {161,2099},
        {222,1660},
        {859,1780},
        {958,1503},
        {1012,2217},
        {1398,1456},
        {1498,3428},
        {1593,2182},
        {2176,2178,2305}

        It is however X(188) of the tangential triangle.
        It is also X(65) of the excentral triangle of the tangential triangle!

        Best regards,
        Peter.

        ----- Original Message -----
        From: Jean-Louis Ayme
        To: Hyacinthos@yahoogroups.com
        Sent: Wednesday, February 11, 2009 4:30 PM
        Subject: Re: [EMHL] Three concurrent lines


        Dear François,
        sorry... "X is the intersection of BC with parallel to AI through O".
        Francisco Javier Garcia Capitan has verified my conjecture and the point of concurs is not in ETC.
        This problem has a relation with a difficult one. I think it is now solved.
        I must now write a little article on it.
        Cela me fait beaucoup de plaisir d'avoir de tes nouvelles.
        Sincerely
        Jean-Louis

        --- En date de : Mer 11.2.09, Francois Rideau <francois.rideau@...> a écrit :

        De: Francois Rideau <francois.rideau@...>
        Objet: Re: [EMHL] Three concurrent lines
        À: Hyacinthos@yahoogroups.com
        Date: Mercredi 11 Février 2009, 17h19

        Dear Jean-Louis

        You say that X is the meeting point of the parallel to AI with BC but this
        line, parallel to AI, is on which point?

        (Cette droite parallèle à la droite AI passe par quel point?)
        Friendly
        François

        >
        > ____________ _________ _________ _________ _________ _________ _
        > Ne pleurez pas si votre Webmail ferme ! Récupérez votre historique sur
        > Yahoo! Mail ! http://fr.docs yahoo.com/ mail/transfert_ mails.html
        >
        > [Non-text portions of this message have been removed]
        >
        >
        >

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

        __________________________________________________________
        Ne pleurez pas si votre Webmail ferme ! Récupérez votre historique sur Yahoo! Mail ! http://fr.docs.yahoo.com/mail/transfert_mails.html

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





        ------------------------------------------------------------------------------



        No virus found in this incoming message.
        Checked by AVG - www.avg.com
        Version: 8.0.233 / Virus Database: 270.10.20/1944 - Release Date: 02/10/09 07:20:00


        [Non-text portions of this message have been removed]
      • chris.vantienhoven
        Hi Peter, That s quite a list. Can you tell me how you discovered the relationship with the tangential triangle? Do you check this as a standard procedure? And
        Message 3 of 17 , Feb 12, 2009
        • 0 Attachment
          Hi Peter,

          That's quite a list.
          Can you tell me how you discovered the relationship with the
          tangential triangle?
          Do you check this as a standard procedure? And do you check this with
          data or by observing and construction?
          Best regards,

          Chris van Tienhoven

          --- In Hyacinthos@yahoogroups.com, "Moses, Peter J. C." <mows@...>
          wrote:
          >
          > Hi Jean-Louis, Chris and all,
          >
          > It is true that the point is not in ETC. It lies on these lines ..
          >
          > {1,159,1486,3220},
          > {3,960,997,1158,3185,3435},
          > {6,1245,1973,2354},
          > {20,1610,1633},
          > {25,65,1452},
          >
          {31,56,154,221,603,1042,1106,1191,1201,1399,1406,1407,1457,1473,2390},
          > {40,197},
          > {55,976,1854,2292,3145},
          > {63,1619},
          > {64,71,198,220,2155,3197},
          > {73,2187},
          > {100,1265},
          > {161,2099},
          > {222,1660},
          > {859,1780},
          > {958,1503},
          > {1012,2217},
          > {1398,1456},
          > {1498,3428},
          > {1593,2182},
          > {2176,2178,2305}
          >
          > It is however X(188) of the tangential triangle.
          > It is also X(65) of the excentral triangle of the tangential
          triangle!
          >
          > Best regards,
          > Peter.
          >
        • Moses, Peter J. C.
          Hi Chris, The relationship was found by initially suspecting that it might be something due to the feel of the configuration, then trying it out numerically
          Message 4 of 17 , Feb 12, 2009
          • 0 Attachment
            Hi Chris,

            The relationship was found by initially suspecting that it might be "something" due to the feel of the configuration, then trying it out numerically and finally confirmed algebraically. X(188) was not the "something" I expected though. Then I test it on a drawing in Sketchpad. The algebraic bit is by no means always straight forward.

            The lines are found by a dramatically unsophisticated brute force search with Mathematica against the whole list of ETC points.

            Best regards,
            Peter.



            ----- Original Message -----
            From: chris.vantienhoven
            To: Hyacinthos@yahoogroups.com
            Sent: Thursday, February 12, 2009 5:18 PM
            Subject: [EMHL] Re: Three concurrent lines


            Hi Peter,

            That's quite a list.
            Can you tell me how you discovered the relationship with the
            tangential triangle?
            Do you check this as a standard procedure? And do you check this with
            data or by observing and construction?
            Best regards,

            Chris van Tienhoven

            --- In Hyacinthos@yahoogroups.com, "Moses, Peter J. C." <mows@...>
            wrote:
            >
            > Hi Jean-Louis, Chris and all,
            >
            > It is true that the point is not in ETC. It lies on these lines ..
            >
            > {1,159,1486,3220},
            > {3,960,997,1158,3185,3435},
            > {6,1245,1973,2354},
            > {20,1610,1633},
            > {25,65,1452},
            >
            {31,56,154,221,603,1042,1106,1191,1201,1399,1406,1407,1457,1473,2390},
            > {40,197},
            > {55,976,1854,2292,3145},
            > {63,1619},
            > {64,71,198,220,2155,3197},
            > {73,2187},
            > {100,1265},
            > {161,2099},
            > {222,1660},
            > {859,1780},
            > {958,1503},
            > {1012,2217},
            > {1398,1456},
            > {1498,3428},
            > {1593,2182},
            > {2176,2178,2305}
            >
            > It is however X(188) of the tangential triangle.
            > It is also X(65) of the excentral triangle of the tangential
            triangle!
            >
            > Best regards,
            > Peter.
            >





            ------------------------------------------------------------------------------



            No virus found in this incoming message.
            Checked by AVG - www.avg.com
            Version: 8.0.233 / Virus Database: 270.10.23/1949 - Release Date: 02/11/09 11:13:00


            [Non-text portions of this message have been removed]
          • chris.vantienhoven
            Hi Peter, ... be something due to the feel of the configuration, then trying it out numerically and finally confirmed algebraically. X(188) was not the
            Message 5 of 17 , Feb 13, 2009
            • 0 Attachment
              Hi Peter,

              > The relationship was found by initially suspecting that it might
              be "something" due to the feel of the configuration, then trying it
              out numerically and finally confirmed algebraically. X(188) was not
              the "something" I expected though. Then I test it on a drawing in
              Sketchpad. The algebraic bit is by no means always straight forward.
              > The lines are found by a dramatically unsophisticated brute force
              search with Mathematica against the whole list of ETC points.


              Very interesting!

              I am not quite familiar with Mathematica.
              Do I understand it right that you have all formulas of all ETC-points
              in Mathematica?
              Then I can understand that it is possible to calculate the whole
              bunch of ETC-points in for example the tangential triangle
              with the outcome if they coincide with the point to be examined or
              not. Is that right?
              I suppose it takes quite of lot of computertime as well for
              calculating in the tangential tringle as well as calculating the
              lines.

              Can you tell me also (if the subject is not too extensive) how you
              algebraically confirm your expectation?
              I find it unbelievable that you calculate the trilinear-coordinate-
              formulas of some point so quick.
              Hope to hear soon.
              Best regards,

              Chris van Tienhoven
            • Nikolaos Dergiades
              Dear Chris and Peter, I am not also familiar with Mathematica. I use Mathcad a less powerfull tool than Mathematica only for calculations not programming. If a
              Message 6 of 17 , Feb 14, 2009
              • 0 Attachment
                Dear Chris and Peter,

                I am not also familiar with Mathematica.
                I use Mathcad a less powerfull tool than Mathematica
                only for calculations not programming.

                If a point P has barycentrics in ABC (x : y : z)
                then the barycentrics of P
                relative to the medial triangle (x' : y' :z') are
                x' = y + z - x
                ...
                relative to the antimedial triangle (x' : y' :z') are
                x' = (y + z)/2
                ...
                relative to the orthic triangle (x' : y' :z') are
                x' = a.a.S_A(y.S_B + z.S_C - x.S_A)
                ...
                relative to the tangential triangle (x' : y' :z') are
                x' = a.a.S_A(a.a.x + y.S_B + z.S_C)
                ...

                When I want to program some calculations
                I work numerically using excel's visual basic
                having all the points of ETC
                with the first and second search values.
                e.g.
                1 1.690 308 509 457 1.690 308 509 457
                2 2.629 368 792 489 1.752 912 528 326
                3 6.782 362 894 196 5.894 950 926 731
                4 -5.676 619 410 927 -6.531 164 268 485
                5 0.552 871 741 635 -0.318 106 670 877
                6 0.992 908 495 066 1.489 362 742 599
                7 0.744 160 979 006 0.793 771 710 940
                8 4.507 489 358 552 1.878 120 566 063
                9 3.571 972 699 230 2.232 482 937 019
                10 3.098 898 934 005 1.784 214 537 760

                These are usefull when you know the barycentrics
                of P in ABC. In some cases it is easy to construct
                and not to calculate. So I work graphically.
                I have constructed in a .gsp file of Sketchpad
                an arbitrary triangle ABC and
                a constant ETC triangle A1B1C1 with coordinates
                B1 (0, 0)
                C1 (6, 0)
                A1 positive coordinates such that distances A1B1 = 13, A1C1 = 9.
                Then I constructed
                a button 1 labeled -> Tr. 6-9-13
                a button 2 labeled -> Medial
                a button 3 labeled -> Antimedial
                a button 4 labeled -> Orthic
                a button 5 labeled -> Tangential
                When I make a double clic on button 1 the triangle
                ABC is moved on triangle A1B1C1 and the point P
                constructed relative to ABC is transformed to the
                analogous point P1 relative to A1B1C1 and hence the
                second coordinate (ordinate) of P1 that is the distance
                of P1 from B1C1 is the first ETC searchnumber.
                Button 2 moves ABC to triangle A2B2C2 that has
                as medial triangle the triangle A1B1C1.
                Button 3 moves ABC to triangle A3B3C3 that has
                as antimedial triangle the triangle A1B1C1.
                Button 4 moves ABC to triangle A4B4C4 that has
                as orthic triangle the triangle A1B1C1.
                Button 5 moves ABC to triangle A5B5C5 that has
                as tangential triangle the triangle A1B1C1.
                The triangles A1B1C1, A2B2C2, . . ., A5B5C5 must
                be hidden. Only the triangle ABC is not hidden
                before the transformation.
                e.g. Construct the orthocenter P of ABC, use button 4
                and see that the second Cartesian coordinate of P4
                is the number 1690.321 when multiplied by 1000
                which means that the searchnumber of this point
                is 1.690321 approximately equal to the first barycentric
                of the incenter and this means that perhaps
                this point is the incenter of the orthic triangle.
                Then it remains to check it algebraically for sure.
                Best regards
                Nikos Dergiades

                > Hi Peter,
                >
                > > The relationship was found by initially suspecting
                > that it might
                > be "something" due to the feel of the
                > configuration, then trying it
                > out numerically and finally confirmed algebraically.
                > X(188) was not
                > the "something" I expected though. Then I test
                > it on a drawing in
                > Sketchpad. The algebraic bit is by no means always
                > straight forward.
                > > The lines are found by a dramatically unsophisticated
                > brute force
                > search with Mathematica against the whole list of ETC
                > points.
                >
                >
                > Very interesting!
                >
                > I am not quite familiar with Mathematica.
                > Do I understand it right that you have all formulas of all
                > ETC-points
                > in Mathematica?
                > Then I can understand that it is possible to calculate the
                > whole
                > bunch of ETC-points in for example the tangential triangle
                > with the outcome if they coincide with the point to be
                > examined or
                > not. Is that right?
                > I suppose it takes quite of lot of computertime as well for
                >
                > calculating in the tangential tringle as well as
                > calculating the
                > lines.
                >
                > Can you tell me also (if the subject is not too extensive)
                > how you
                > algebraically confirm your expectation?
                > I find it unbelievable that you calculate the
                > trilinear-coordinate-
                > formulas of some point so quick.
                > Hope to hear soon.
                > Best regards,
                >
                > Chris van Tienhoven
                >




                ___________________________________________________________
                Χρησιμοποιείτε Yahoo!;
                Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
                διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
                μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr
              • Nikolaos Dergiades
                Sorry, in my previous message I wrote ... I wanted to say the first trillinear. Best regards Nikos Dergiades
                Message 7 of 17 , Feb 14, 2009
                • 0 Attachment
                  Sorry, in my previous message I wrote

                  > is 1.690321 approximately equal to the first barycentric

                  I wanted to say the first trillinear.

                  Best regards
                  Nikos Dergiades




                  ___________________________________________________________
                  Χρησιμοποιείτε Yahoo!;
                  Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
                  διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
                  μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr
                • Jeffrey Brooks
                  Dear friends, I have just recently installed MATLAB from MathWorks and am trying to get familiar with it. Anyone use this program? Seems the base element in
                  Message 8 of 17 , Feb 14, 2009
                  • 0 Attachment
                    Dear friends,

                    I have just recently installed MATLAB from MathWorks and am trying to
                    get familiar with it. Anyone use this program?
                    Seems the base element in the program is the matrix. Should I be
                    surprised? Since much of TCCT focuses on use of matrices, I thought
                    this might be of some interest.

                    Sincerely, Jeff
                  • chris.vantienhoven
                    Dear Nikos, Peter and all, ... Nikos, This really helps me! Now I understand how to do systematic calculations as well in the basic triangle as well in the
                    Message 9 of 17 , Feb 16, 2009
                    • 0 Attachment
                      Dear Nikos, Peter and all,

                      > I use Mathcad a less powerfull tool than Mathematica
                      > only for calculations not programming.
                      > Then I constructed
                      > a button 1 labeled -> Tr. 6-9-13
                      > a button 2 labeled -> Medial
                      > a button 3 labeled -> Antimedial
                      > a button 4 labeled -> Orthic
                      > a button 5 labeled -> Tangential
                      > Construct the orthocenter P of ABC, use button 4
                      > and see that the second Cartesian coordinate of P4
                      > is the number 1690.321 when multiplied by 1000
                      > which means that the searchnumber of this point
                      > is 1.690321 approximately equal to the first barycentric
                      > of the incenter and this means that perhaps
                      > this point is the incenter of the orthic triangle.
                      > Then it remains to check it algebraically for sure.

                      Nikos,

                      This really helps me!
                      Now I understand how to do systematic calculations as well in the
                      basic triangle as well in the tangential triangle etc.
                      I think I can use parts of your system.
                      I think it is very interesting to hear sytematics from each other.
                      We always can learn from each other.

                      I myself work also with MS-Excel and VBA (visual Basic for
                      Applications). I also work with Cabri to draw triangles.
                      I work with a standard triangle (I noticed that I often draw the same
                      form to start with and I use this as a standard form) in Cabri.
                      I store all my data in 2 Excel-sheets:
                      SHEET 1: contains cartesian coordinates of points I found
                      SHEET 2: contains coefficients of lines, circles, conics and cubics I
                      encountered.
                      When I find a new point I temporarily place the cartesian coordinates
                      as input (found in Cabri) in sheet 2. Instantly I see then at which
                      lines, circles, conics or cubics it occurs.
                      When this is interesting enough I register this point in sheet 1.
                      Of course I first check in the KimberlingTriangle if the point is
                      registered in ETC. If so it gets the ETC-X-nr. If not I give it a P-
                      number of my own.
                      When I find a new line, circle, conic or cubic I place the
                      coefficients at the top of sheet 1 in special cells. Instantly I see
                      then which points occur at this curve.
                      When this is interesting enough I register the coefficients of this
                      curve in sheet 2. In case it is a cubic I copy the coefficients right
                      away from Cabri. In other cases the coefficients are calculated in
                      Excel.
                      I also have a special procedure for new non-ETC-points.
                      I made an Excel-macro to check on which possible lines it occurs.
                      This takes about 2 minutes computertime.
                      I also made another Excel-macro to check on which possible
                      circumscribed conics it occurs. This takes another 2 minutes
                      computertime.
                      Of course it is possible to check other coincidences in this way.
                      After this procedure I know much more about this new point.
                      As a matter of fact in one year I found more than 400 non-ETC-
                      registered points some of which are very interesting.
                      Right now I have an Excel-database of about 1.200 points (800 ETC,
                      400 non-ETC). Most of them I encountered during researchtime.
                      In Sheet 2 I registered so far about 130 lines, 60 circles, 200 other
                      conics and 45 cubics.
                      All of them only registered when they came along during research.
                      It is not my aim to register "all there is". It is my aim to find
                      connections and I try to find an overall explanation for all these
                      astonishing regularities in Triangle Geometry.

                      What I would like to learn now is how experienced "colleague's" make
                      algebraic calculations. Doing it on a piece of paper takes an awful
                      lot of time. Is there a quicker/automated way to substite in formulae
                      etc? Are there tips and trics? Is there software that helps? Who can
                      help me?

                      Best regards,

                      Chris van Tienhoven
                    • Moses, Peter J. C.
                      Hi Chris & Nikos, Yes, I have all the ETC barcentric coordinates in a Mathematica file. In this particular case, the way X(188) of the tangential triangle was
                      Message 10 of 17 , Feb 16, 2009
                      • 0 Attachment
                        Hi Chris & Nikos,

                        Yes, I have all the ETC barcentric coordinates in a Mathematica file.

                        In this particular case, the way X(188) of the tangential triangle was found is that once we know the barycentric coordinates of the point P wrt ABC, we can then ask what is that P of the intouch triangle, which ends up at X(188), i.e. P wrt ABC is X(188) wrt tangential. This is done firstly numerically against Clark's {6,9,13} triangle, and then using the same formula but this time symbolically. Sometimes this involves square roots, and needs careful pruning is needed to eek out the answer. The formula is a general one that can in principle find some point wrt some triangle. The answer can be somewhat formidable though. Here is the M code I made up ...

                        XnBaryTriangle[{b1_,b2_,b3_},{{p1_,p2_,p3_},{q1_,q2_,q3_},{r1_,r2_,r3_}}]:=Module[{siderules},siderules={a->Sqrt[(p1+p2+p3)^2 (((-q2-q3) r1+q1 (r2+r3))^2 SA+((-q1-q3) r2+q2 (r1+r3))^2 SB+(q3 (r1+r2)-(q1+q2) r3)^2 SC)],b->Sqrt[(q1+q2+q3)^2 (((p2+p3) r1-p1 (r2+r3))^2 SA+((p1+p3) r2-p2 (r1+r3))^2 SB+(-p3 (r1+r2)+(p1+p2) r3)^2 SC)],c->Sqrt[(r1+r2+r3)^2 (((-p2-p3) q1+p1 (q2+q3))^2 SA+((-p1-p3) q2+p2 (q1+q3))^2 SB+(p3 (q1+q2)-(p1+p2) q3)^2 SC)]};{{p1,q1,r1},{p2,q2,r2},{p3,q3,r3}}.{(b1/.siderules)/(p1+p2+p3),(b2/.siderules)/(q1+q2+q3),(b3/.siderules)/(r1+r2+r3)}]

                        So if we want say X(9) of the intouch triangle we would do something like ....

                        XnBaryTriangle[{a (-a+b+c),b (a-b+c),(a+b-c) c},{{0,1/(a-b+c),1/(a+b-c)},{1/(-a+b+c),0,1/(a+b-c)},{1/(-a+b+c),1/(a-b+c),0}}];

                        This gives a wretched answer .. but we can ask for the coordinates of test triangle like the {6,9,13}. In this case we get {5.0108402139564704446,1.5503054676667462861}. The y-coordinate (First Trilinear), 1.5503..., is not in the ETC SEARCH page, so we are probably out of luck there, i.e. X(9) of the intouch is not X(something in ETC) of ABC. If we now feel brave, we could contemplate carefully whittling down the symbolic answer, for example to
                        (a+b-c) (Cos[A/2]-Cos[B/2]+Cos[C/2]) Csc[B/2]+(a-b+c) (Cos[A/2]+Cos[B/2]-Cos[C/2]) Csc[C/2] :: .. on lines {1, 166} & {7, 177}

                        There are times to be careful .. for example suppose we want X(1) of the orthic triangle, we might be tempted by
                        XnBaryTriangle[{a,b,c},{{0,a^2+b^2-c^2,a^2-b^2+c^2},{a^2+b^2-c^2,0,-a^2+b^2+c^2},{a^2-b^2+c^2,-a^2+b^2+c^2,0}}]
                        This would lead us to the B vertex. In this case we need to ask for
                        XnBaryTriangle[{a,b,-c},{{0,a^2+b^2-c^2,a^2-b^2+c^2},{a^2+b^2-c^2,0,-a^2+b^2+c^2},{a^2-b^2+c^2,-a^2+b^2+c^2,0}}] to find X(4).

                        Clark's TCCT mentions coordinate transfomation on pages 205 - 209, which might also be of interest.

                        Best regards,
                        Peter.


                        ----- Original Message -----
                        From: chris.vantienhoven
                        To: Hyacinthos@yahoogroups.com
                        Sent: Friday, February 13, 2009 10:04 AM
                        Subject: [EMHL] Re: Three concurrent lines


                        Hi Peter,

                        > The relationship was found by initially suspecting that it might
                        be "something" due to the feel of the configuration, then trying it
                        out numerically and finally confirmed algebraically. X(188) was not
                        the "something" I expected though. Then I test it on a drawing in
                        Sketchpad. The algebraic bit is by no means always straight forward.
                        > The lines are found by a dramatically unsophisticated brute force
                        search with Mathematica against the whole list of ETC points.

                        Very interesting!

                        I am not quite familiar with Mathematica.
                        Do I understand it right that you have all formulas of all ETC-points
                        in Mathematica?
                        Then I can understand that it is possible to calculate the whole
                        bunch of ETC-points in for example the tangential triangle
                        with the outcome if they coincide with the point to be examined or
                        not. Is that right?
                        I suppose it takes quite of lot of computertime as well for
                        calculating in the tangential tringle as well as calculating the
                        lines.

                        Can you tell me also (if the subject is not too extensive) how you
                        algebraically confirm your expectation?
                        I find it unbelievable that you calculate the trilinear-coordinate-
                        formulas of some point so quick.
                        Hope to hear soon.
                        Best regards,

                        Chris van Tienhoven





                        ------------------------------------------------------------------------------



                        No virus found in this incoming message.
                        Checked by AVG - www.avg.com
                        Version: 8.0.237 / Virus Database: 270.10.23/1951 - Release Date: 02/13/09 06:51:00


                        [Non-text portions of this message have been removed]
                      Your message has been successfully submitted and would be delivered to recipients shortly.