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Re: [EMHL] Lemoine query

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  • Dick Tahta
    Antreas: Many thanks for your note on Soddy circles. I see now that their centres (a +/- ra::) are Clark Kimberling s X176, X175. It is intriguing that
    Message 1 of 13 , May 1, 2000
      Antreas:

      Many thanks for your note on Soddy circles. I see now that their centres
      (a +/- ra::) are Clark Kimberling's X176, X175. It is intriguing that
      these were being discussed by Lemoine, some decades before Soddy's 1936
      article. Coxeter ascribes the result to Steiner (1826) and Adrian Oldknow
      has also mentioned a Philip Beecroft (1842). The radius of the inner Soddy
      circle is t where 1/t=2/r+1/sa+1/sb+1/sc; Lemoine gives this in the form
      you quoted, namely t=delta/(2s+4R+r).

      I am interested in Lemoine's "continuous transformation", say the the one
      in A where the sides a,b,c transform to a,-b,-c. (This is also mentioned
      briefly in Johnson's Advanced Euclidean Geometry, para 300.) The
      original circles with centres A,B,C and radii sa,sb,sc respectively
      transform (in Lemoine's sense) into circles with radii s,sc,sb. Lemoine
      then deduces that the circle touching these is the transform of the inner
      Soddy centre (a+ra:b+rb:c+rc), namely (a+r:b+rc:c+rb). I note that
      extraversion would yield the same new three circles but that the inner
      Soddy centre extraverts to (-a+r:b-rc:c-rb), which is the harmonic of the
      Lemoine transform of the outer Soddy centre.

      best wishes, Dick
    • xpolakis@otenet.gr
      ... You are welcome, Dick. ... Coxeter - Greitzer (Geometry Revisited, p. 114, Exercise #4) write parenthetically: (They are sometimes called Soddy s circles
      Message 2 of 13 , May 1, 2000
        Dick Tahta wrote:

        >Antreas:
        >
        >Many thanks for your note on Soddy circles. I see now that their centres

        You are welcome, Dick.

        >(a +/- ra::) are Clark Kimberling's X176, X175. It is intriguing that
        >these were being discussed by Lemoine, some decades before Soddy's 1936
        >article. Coxeter ascribes the result to Steiner (1826) and Adrian Oldknow
        >has also mentioned a Philip Beecroft (1842). The radius of the inner Soddy

        Coxeter - Greitzer (Geometry Revisited, p. 114, Exercise #4) write
        parenthetically:
        "(They are sometimes called Soddy's circles ... although they were described
        by Steiner as long as 1826 in the first volume of Crelle's Journal fu"r
        Mathematik, p. 274.)"

        Steiner's paper (Einige geometrische Sa"tze) is reprinted in the first volume
        of his _Gesammelte Werke_ (A Chelsea 1971 reprint, p. 3ff)

        Oldknow (p. 326) writes, refering to Coxeter (Introduction to Geometry):
        "In fact these had already discovered by Descartes in the 17th century".

        Coxeter (Introduction..., p. 14) writes:
        "In a letter of November 1643 to Princess Elizabeth of Bohemia, Rene Descartes
        developed a formula relating the radii of four mutually tangent circles".

        Coxeter continues:
        "This Descartes circle theorem was rediscovered in 1842 by an English amateur,
        Philip Beecroft, who observed that the four circles E_i determine
        anotherset of four circles Hi, mutually tangent at the same six points
        ...."

        >circle is t where 1/t=2/r+1/sa+1/sb+1/sc; Lemoine gives this in the form
        >you quoted, namely t=delta/(2s+4R+r).

        I found the formula in the Greek book:
        I. Panakis: Plane Trigonometry. Athens 1973, vol. II, p. 457, exercise #2

        In Edward Brisse's very good geometry web page, located at:

        http://pages.infinit.net/spqrsncf/math1.htm

        the Soddy Line is called "Calegari Line":

        <quote>
        17.Droite de Calegari
        Les cercles de centres A, B, C, sommets d'un triangle et
        mutuellement tangents 2 à 2, sont tangents a 2 cercles
        de centre T et T'. Ces centres sont les points de
        Calegari et TT' est la droite de Calegari.
        TT' coupe la droite d'Euler en K tel que GK/GH=-2
        </quote>

        The author refers to:
        The american mathematical monthly, vol. 75, 1968, p.5-15
        The problem of Apollonius, H.S.M. Coxeter

        The american mathematical monthly, 1964, p.176-179
        Soddy's circles and the De Longchamps point of a triangle, A. Vandeghen

        He also has a link to the web page:
        http://www.maths.uq.oz.au/~bab/calegari.html

        but it is dead.

        >
        >I am interested in Lemoine's "continuous transformation", say the the one

        George Kapetis, the author of "Triangle Geometry" vol. I (in Greek), told me
        once (in telephon) that the second volume of his book will contain (among
        other things) "Lemoine Transformation".

        >in A where the sides a,b,c transform to a,-b,-c. (This is also mentioned
        >briefly in Johnson's Advanced Euclidean Geometry, para 300.) The
        >original circles with centres A,B,C and radii sa,sb,sc respectively
        >transform (in Lemoine's sense) into circles with radii s,sc,sb. Lemoine
        >then deduces that the circle touching these is the transform of the inner
        >Soddy centre (a+ra:b+rb:c+rc), namely (a+r:b+rc:c+rb). I note that
        >extraversion would yield the same new three circles but that the inner
        >Soddy centre extraverts to (-a+r:b-rc:c-rb), which is the harmonic of the
        >Lemoine transform of the outer Soddy centre.
        >
        >best wishes, Dick


        Greetings from Athens

        Antreas
      • Ignacio Larrosa Cañestro
        ... From: Dick Tahta To: Hyacinthos@egroups.com Cc: Hyacinthos@egroups.com Sent: Monday, May 01, 2000 2:45 PM Subject: Re: [EMHL] Lemoine query Antreas: Many
        Message 3 of 13 , May 1, 2000
          ----- Original Message -----
          From: Dick Tahta
          To: Hyacinthos@egroups.com
          Cc: Hyacinthos@egroups.com
          Sent: Monday, May 01, 2000 2:45 PM
          Subject: Re: [EMHL] Lemoine query


          Antreas:

          Many thanks for your note on Soddy circles. I see now that their centres
          (a +/- ra::) are Clark Kimberling's X176, X175. It is intriguing that
          these were being discussed by Lemoine, some decades before Soddy's 1936
          -----------------

          In ETC, at X(176), it says:

          "If X is a point not between A and B, we make a detour of magnitude |AX| +
          |XB| - |AB| if we walk from A to B via X; then a point has the equal detour
          property if the magnitues of the three detours, A to B via X, B to C via X,
          and C to A via X, are equal; X(176) is the only such point unless ABC has an
          angle greater than 2*arcsin(4/5), and then X(175) also has the equal detour
          property. "

          But I think that its only certainly if the two sides, e.g. b and c, of that
          angle, e.g. A, are equals. If they aren't equals, the limit angle is
          alfa=2*arctan(u/(2u^2 + 3u + 2)) + pi/2, where u^2=b/c. If u=1, b=c,
          alfa=2*arctan(1/7) + pi/2= 2*arcsin(4/5), but alfa slowly go to pi/2 when u
          go to 0 or infinite. For that angle alfa, the outer Soddy circle is a
          straight line, tangent to the
          circles with centres A,B,C and radii sa,sb,sc. Or am I mistaken?


          Best regards,

          Ignacio Larrosa Cañestro
          A Coruña (España)
          ilarrosa@...
        • Steve Sigur
          Hello all, A while ago I misunderstood something that John Conway told me. I thought he said that almost all equilateral triangles were parallel to the Morley
          Message 4 of 13 , May 2, 2000
            Hello all,

            A while ago I misunderstood something that John Conway told me. I thought he
            said that almost all equilateral triangles were parallel to the Morley
            triangle created from trisectors.

            I made a bunch of drawings and, sure enough, triangles such as the Napoleon
            triangle were (or seemed to be) parallel to the Morley triangle. In addition
            Floor came up with a pair of triangles with this property.

            Later John told me that this was not true and only triangles (such as the
            den Roussel and Stammler triangles) that have to do with trisectors are
            parallel to Morley.

            So I went back to my pictures and darned if it does not seem that the
            Napoleon triangle is parallel to the Morley triangle.

            I think they are parallel and, along with my students, am trying to prove or
            disprove this. But we have had no success either way.

            Does anyone out there know about this or have any ideas that might help.

            If true we have another homothety, potentially important.

            Steve
          • Richard Guy
            If you ve found one homothety, you ve found 18. The sides of the Morley triangles make angles (B-C)/3, (C-A)/3, (A-B)/3 with the resp sides of the original
            Message 5 of 13 , May 2, 2000
              If you've found one homothety, you've found 18. The sides of the Morley
              triangles make angles (B-C)/3, (C-A)/3, (A-B)/3 with the resp sides of
              the original triangle, but I don't think those of the `Napoleon'
              triangle do. R.

              On Tue, 2 May 2000, Steve Sigur wrote:

              > Hello all,
              >
              > A while ago I misunderstood something that John Conway told me. I thought he
              > said that almost all equilateral triangles were parallel to the Morley
              > triangle created from trisectors.
              >
              > I made a bunch of drawings and, sure enough, triangles such as the Napoleon
              > triangle were (or seemed to be) parallel to the Morley triangle. In addition
              > Floor came up with a pair of triangles with this property.
              >
              > Later John told me that this was not true and only triangles (such as the
              > den Roussel and Stammler triangles) that have to do with trisectors are
              > parallel to Morley.
              >
              > So I went back to my pictures and darned if it does not seem that the
              > Napoleon triangle is parallel to the Morley triangle.
              >
              > I think they are parallel and, along with my students, am trying to prove or
              > disprove this. But we have had no success either way.
              >
              > Does anyone out there know about this or have any ideas that might help.
              >
              > If true we have another homothety, potentially important.
              >
              > Steve
              >
              >
              >
              >
              >
              > ------------------------------------------------------------------------
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              > http://click.egroups.com/1/3013/3/_/672428/_/957277896/
              > ------------------------------------------------------------------------
              >
              >
              >
            • Dick Tahta
              Antreas recently wrote that ... I have remained curious about this transformation and about some inconsistencies in Lemoine s use of it. And, also, about the
              Message 6 of 13 , May 6, 2000
                Antreas recently wrote that
                >George Kapetis, the author of "Triangle Geometry" vol. I (in Greek), told me
                >once (in telephon) that the second volume of his book will contain (among
                >other things) "Lemoine Transformation".

                I have remained curious about this transformation and about some
                inconsistencies in Lemoine's use of it. And, also, about the link with
                extraversion. To recap: Lemoine's "transformation continue en A" maps
                a,b,c to a,-b,-c and s,sa,sb,sc to -sa,-s,sc,sb respectively, and so on
                (whereas Conway's extraversion (in A) maps a,b,c to -a,b,c ...).

                In the paper I have quoted recently, Lemoine transforms various points by
                applying the above mapping to their trilinear coordinates. For example,
                the inner Soddy centre {(a+ra/a)::} maps to {(a+r)/a:(b+rc)/b:(c+rb)/c}.
                But in the case of the Nagel point {sa/a::} which should - according to the
                above - map to {s/a:sc/b:sb/c}, Lemoine gives {-sa/a:sc/b:sb/c}. This is,
                of course, the more preferable Nagel extraversion, but the inconsistency is
                puzzling.

                One explanation might be the reminder that Lemoine transformation (as,
                indeed, extraversion) is extrinsic, ie dependent on the coordinate system.
                Thus if a point with barycentric coordinates [x:y:z] maps to [x':y':z']
                then the trilinear version {x/a:y/b:z/c} maps to {x'/a:-y'/b:-z'/c}. A
                barycentric version of this is [-x':y:'z']. I note that when applied to
                barycentrics the Lemoine transformation is equivalent to extraversion; but
                that when applied to trilinears (as originally intended) it yields the
                harmonic of the (barycentric) extraversion.

                There is, I think, a further difficulty in the case of Lemoine's
                transformation of the Soddy centres. These are defined as centres of
                circles touching the three circles with centres A,B,C and radii sa,sb,sc.
                Lemoine maps the latter into circles with centres A,B,C and radii -s,sc,sb
                (and extraversion yields the same circles). He then assumes that the Soddy
                centres map into the centres of these circles. Now drawing a figure (the
                simplest case takes ABC equilateral) verifies that the x-coordinate of the
                inner Soddy centre must be negative. But, according to Lemoine, the inner
                Soddy centre maps to a point with (trilinear) cooordinates
                {(a+r)/a:(b+rc)/b:(c+rb)/c} whose components are all positive. The
                harmonic of this yields the appropriate negative coordinate: extraversion
                rules OK?

                Dick Tahta
              • Dick Tahta
                A brief historical note. In an article (Proc.Edin.Math.Soc., 1893, pp 86-105) J S Mackay invokes the Lemoine transformation (a,b,c maps to a,-b,-c) and gives a
                Message 7 of 13 , May 21, 2000
                  A brief historical note.

                  In an article (Proc.Edin.Math.Soc., 1893, pp 86-105) J S Mackay invokes the
                  Lemoine transformation (a,b,c maps to a,-b,-c) and gives a table of the
                  corresponding maps of A,B,C,R,S, sx, rx,hx. He then comments:

                  "The greater part of this table is given in the Ladies and Gentleman's
                  Diary for 1871, p 93, and it is due either to the editor of the Diary W S B
                  Woolhouse, or to one of his correspondents, W B G (William Bywater
                  Grove?). No demonstration however is offered of the law of transformation
                  thus enunciated."

                  I haven't seen the Diary reference, but assume that like Mackay's article
                  it is concerned with various triangle formulae. I guess Lemoine was the
                  first to transform the co-ordinates of points. And I assume that the
                  Lemoine transformation is the trilinear form of extraversion.


                  Dick Tahta
                • Steve Sigur
                  ... Dick, This looks like extraversion to me too. Johnson s triangle geometry book has a little section on transformations of formulae which was probably what
                  Message 8 of 13 , May 21, 2000
                    on 5/21/2000 6:39 AM, Dick Tahta wrote:

                    > A brief historical note.
                    >
                    > In an article (Proc.Edin.Math.Soc., 1893, pp 86-105) J S Mackay invokes the
                    > Lemoine transformation (a,b,c maps to a,-b,-c) and gives a table of the
                    > corresponding maps of A,B,C,R,S, sx, rx,hx. He then comments:
                    >
                    > "The greater part of this table is given in the Ladies and Gentleman's
                    > Diary for 1871, p 93, and it is due either to the editor of the Diary W S B
                    > Woolhouse, or to one of his correspondents, W B G (William Bywater
                    > Grove?). No demonstration however is offered of the law of transformation
                    > thus enunciated."
                    >
                    > I haven't seen the Diary reference, but assume that like Mackay's article
                    > it is concerned with various triangle formulae. I guess Lemoine was the
                    > first to transform the co-ordinates of points. And I assume that the
                    > Lemoine transformation is the trilinear form of extraversion.
                    >
                    >
                    > Dick Tahta


                    Dick,

                    This looks like extraversion to me too.

                    Johnson's triangle geometry book has a little section on transformations of
                    formulae which was probably what got Conway thinking about it.

                    STeve
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