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

Re: [EMHL] Three concurrent lines

Expand Messages
  • 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 1 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 2 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 3 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 4 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 5 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 6 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 7 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 8 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 9 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.