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Re: [EMHL] a Morley configuration

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  • Floor van Lamoen
    Dear Steve, Best wishes to you for the new year. ... I know only of one Stammler triangle. What are the other two? About the triangle I came up with. Those are
    Message 1 of 24 , Jan 3, 2001
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      Dear Steve,

      Best wishes to you for the new year.

      [SS]:
      > There are some other equilateral triangles that are parallel to the Morely
      > triangle: the Stammler triangles (3 of them), the den Roussel, triangles and
      > I think there is another floating around. And oh yes, Floor came up with one
      > a year or so ago. Lots of perspectors floating around, methinks.

      I know only of one Stammler triangle. What are the other two?

      About the triangle I came up with. Those are the points of contact of
      the NPC with the three Simson lines that are tangent to the NPC. This
      triangle can be found in a problem proposal by H.D. Drury in
      Mathematical Questions and Solutions from the Educational Times, problem
      17395 (1914).

      Kind regards,
      Floor.
    • xpolakis@otenet.gr
      ... Thread: http://forum.swarthmore.edu/epigone/geom.research/jahclumyax/ APH
      Message 2 of 24 , Jan 4, 2001
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        Floor van Lamoen wrote:

        >On 9 Nov 1998 Den Roussel wrote (as result of discussion) in
        >geometry-research:
        >
        >"The angular trisectors of a triangle cut the circumcircle at 18 points
        >which form 3 sets of 9 parallel lines centered on the circumcenter and
        >parallel to the Morley system."

        Thread:

        http://forum.swarthmore.edu/epigone/geom.research/jahclumyax/

        APH
      • Floor van Lamoen
        Dear all, ... I almost forgot Peter Yff s circumtangential and circumnormal triangles. See TCCT 6.28 and 6.29. Kind regards, Sincerely, Floor.
        Message 3 of 24 , Jan 4, 2001
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          Dear all,

          > [SS]:
          > > There are some other equilateral triangles that are parallel to the Morely
          > > triangle: the Stammler triangles (3 of them), the den Roussel, triangles and
          > > I think there is another floating around. And oh yes, Floor came up with one
          > > a year or so ago. Lots of perspectors floating around, methinks.
          >

          [FvL]:
          > About the triangle I came up with. Those are the points of contact of
          > the NPC with the three Simson lines that are tangent to the NPC. This
          > triangle can be found in a problem proposal by H.D. Drury in
          > Mathematical Questions and Solutions from the Educational Times, problem
          > 17395 (1914).

          I almost forgot Peter Yff's circumtangential and circumnormal triangles.
          See TCCT 6.28 and 6.29.

          Kind regards,
          Sincerely,
          Floor.
        • xpolakis@otenet.gr
          ... .... and of course one can define infinitely many equilateral triangles, but the problem is with the coordinates ! An example is the Equilateral Cevian
          Message 4 of 24 , Jan 4, 2001
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            [FvL]:

            >I almost forgot Peter Yff's circumtangential and circumnormal triangles.
            >See TCCT 6.28 and 6.29.

            .... and of course one can define infinitely many equilateral triangles,
            but the problem is with the coordinates !

            An example is the Equilateral Cevian Triangle (point X_370)

            And here is a related problem:

            For which point(s) P the triangle formed with the circumcenters of PBC,PCA,
            PAB is equilateral?

            APH
          • Steve Sigur
            ... John seems to think that if equilateral triangles are formed by anything having to do with trisection, they are always parallel. This makes the Morley
            Message 5 of 24 , Jan 4, 2001
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              on 1/4/01 1:10 AM, Bernard Gibert wrote:

              > [SS]
              >> Are these triangles parallel to the original Morley one?
              >
              > Yes, they are.


              John seems to think that if equilateral triangles are formed by anything
              having to do with trisection, they are always parallel. This makes the
              Morley inclination a fundamental property of the triangle rather as the
              Brocard angle is.

              Steve
            • Steve Sigur
              ... Hmm.. let me check my pictures.... There are three. Here is the construction. Create the Stammler triangle by drawing tangents to the circumcircle parallel
              Message 6 of 24 , Jan 4, 2001
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                on 1/4/01 2:45 AM, Floor van Lamoen wrote:

                > I know only of one Stammler triangle. What are the other two?

                Hmm.. let me check my pictures....

                There are three.

                Here is the construction. Create the Stammler triangle by drawing tangents
                to the circumcircle parallel to the sides of the Morely triangle. This can
                be done in two ways, one of which is the Stammler triangle, meaning that
                it's vertices are the centers for the Stammler circles. There is a third
                (and fourth I just realize) triangle formed by connecting the contact points
                of the tangents with the circumcircle.

                I think these are also the Peter Yff triangles but I will have to look that
                up.

                I would like to make a list of triangles parallel to Morley.

                Steve
              • jp ehrmann
                Dear Bernard and other Hyacinthists, I ve computed the locus of the centers of the equilateral triangles inscribed in the Steiner circumellipse - a coaxial
                Message 7 of 24 , Jan 4, 2001
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                  Dear Bernard and other Hyacinthists,
                  I've computed the locus of the centers of the equilateral triangles
                  inscribed in the Steiner circumellipse - a coaxial ellipse -
                  In barycentric :
                  a^2 (a^2 + b^2 + c^2) x^2
                  - (2 b^4 + 2 c^4 + b^2 c^2 - c^2 a^2 - a^2 b^2) y z + circular = 0
                  The only remarkable point I've found on this ellipse is the isotomic
                  conjugate of the symedian point, discovered by Bernard in his Morley
                  configuration.
                  May be, somebody can try to find some other ones.
                  Friendly. Jean-Pierre
                • jp ehrmann
                  Dear Antreas and other Hyacinthists, ... PBC,PCA, ... The two Fermat points : for each one the three circumcenters are the vertices of an imperial triangle.
                  Message 8 of 24 , Jan 4, 2001
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                    Dear Antreas and other Hyacinthists,
                    Antreas wrote :
                    > And here is a related problem:
                    >
                    > For which point(s) P the triangle formed with the circumcenters of
                    PBC,PCA,
                    > PAB is equilateral?
                    >
                    The two Fermat points : for each one the three circumcenters are the
                    vertices of an imperial triangle.
                    Friendly. Jean-Pierre
                  • Bernard Gibert
                    Dear Steve and friends, [SS] ... The triangle called gAgBgC in my first Morley configuration message is the circumtangential triangle : that s where the
                    Message 9 of 24 , Jan 5, 2001
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                      Dear Steve and friends,

                      [SS]
                      > Here is the construction. Create the Stammler triangle by drawing tangents
                      > to the circumcircle parallel to the sides of the Morely triangle. This can
                      > be done in two ways, one of which is the Stammler triangle, meaning that
                      > it's vertices are the centers for the Stammler circles. There is a third
                      > (and fourth I just realize) triangle formed by connecting the contact points
                      > of the tangents with the circumcircle.
                      >
                      > I think these are also the Peter Yff triangles but I will have to look that
                      > up.
                      >
                      > I would like to make a list of triangles parallel to Morley.
                      >

                      The triangle called gAgBgC in my first "Morley configuration" message is the
                      circumtangential triangle : that's where the cubic Kjp intersects the
                      circumcircle. The vertices are the contacts (with the circumcircle) of the
                      deltoid which is the envelope of the axes of inscribed parabolas.

                      If you reflect gAgBgC about O, you get another one where the McCay cubic
                      intersects the circumcircle.

                      tAtBtC is analogous to gAgBgC and was obtained from a non-pivotal isotomic
                      cubic, the Kjp isotomic sister although its asymptotes are not concurrent.

                      Another one would be formed by the asymptotes of this cubic, and one more by
                      the asymptotes of the unique pivotal isotomic 60°-cubic.

                      All those triangles are homothetic to the "original" Morley triangle.

                      Best regards

                      Bernard
                    • Bernard Gibert
                      Dear Jean-Pierre and other Hyacinthists, [JP] ... there s at least the symmetric of X76 (= isotomic conjugate of the symedian point) about G which is the point
                      Message 10 of 24 , Jan 5, 2001
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                        Dear Jean-Pierre and other Hyacinthists,
                        [JP]
                        > I've computed the locus of the centers of the equilateral triangles
                        > inscribed in the Steiner circumellipse - a coaxial ellipse -
                        > In barycentric :
                        > a^2 (a^2 + b^2 + c^2) x^2
                        > - (2 b^4 + 2 c^4 + b^2 c^2 - c^2 a^2 - a^2 b^2) y z + circular = 0
                        > The only remarkable point I've found on this ellipse is the isotomic
                        > conjugate of the symedian point, discovered by Bernard in his Morley
                        > configuration.
                        > May be, somebody can try to find some other ones.

                        there's at least the symmetric of X76 (= isotomic
                        conjugate of the symedian point) about G which is the point
                        P=[b^2c^2 - 2a^2(b^2+c^2) : : ]=[u,v,w]

                        That's where everything started because the cubic :
                        ux(y^2 + z^2) + ... + ... = 0 is the Kjp isotomic sister and it passes
                        through the vertices tA, tB, tC of the triangle.

                        Best regards

                        Bernard
                      • xpolakis@otenet.gr
                        Dear Jean-Pierre, ... Another triangle ellipse is Lemoine ellipse. I am not quite sure which is it, since haven t seen the ref. below, but my guess is that
                        Message 11 of 24 , Jan 5, 2001
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                          Dear Jean-Pierre,

                          [J-PE]:
                          >I've computed the locus of the centers of the equilateral triangles
                          >inscribed in the Steiner circumellipse - a coaxial ellipse -
                          >In barycentric :
                          >a^2 (a^2 + b^2 + c^2) x^2
                          >- (2 b^4 + 2 c^4 + b^2 c^2 - c^2 a^2 - a^2 b^2) y z + circular = 0
                          >The only remarkable point I've found on this ellipse is the isotomic
                          >conjugate of the symedian point, discovered by Bernard in his Morley
                          >configuration.
                          >May be, somebody can try to find some other ones.

                          Another triangle ellipse is "Lemoine ellipse."
                          I am not quite sure which is it, since haven't seen the ref. below,
                          but my guess is that this ellipse is the circum-/in-ellipse centered
                          at K, the Lemoine point.

                          PS. Perhaps the locus of the centers of the eq. triangles inscribed in
                          this ellepse is interesting too.

                          References:

                          Cavallaro, Vincenzo G.: On Lemoine's ellipse.
                          Math. Gaz. 34, 266-268 (1950).

                          Toscano, L.: Sur l'ellipse de Lemoine.
                          Bull. Soc. Roy. Sci. Liege 23, 221-229 (1954).


                          Greetings

                          Antreas
                        • Bernard Gibert
                          Dear Antreas, ... The Lemoine ellipse is the in-ellipse with foci G & K. If we call Ka, Kb, Kc the Lemoine points of GBC, GCA, GAB then the points A =BC inter
                          Message 12 of 24 , Jan 5, 2001
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                            Dear Antreas,
                            >
                            > Another triangle ellipse is "Lemoine ellipse."
                            > I am not quite sure which is it, since haven't seen the ref. below,
                            > but my guess is that this ellipse is the circum-/in-ellipse centered
                            > at K, the Lemoine point.
                            >
                            The Lemoine ellipse is the in-ellipse with foci G & K.
                            If we call Ka, Kb, Kc the Lemoine points of GBC, GCA, GAB then the points
                            A'=BC inter GKa, B', C' are the contacts of the ellipse with the sidelines.

                            (This is from Brocard & Lemoyne)

                            May I add :

                            Its center is the midpoint of GK ie [b^2 + c^2 + 4a^2 : : ]

                            Its perspector is [1/(2b^2 + 2c^2 - a^2) : : ], isotomic of
                            [2b^2 + 2c^2 - a^2 : : ], those three points not in ETC.

                            The in-ellipse centered at K is called "ellipse K" in Brocard & Lemoyne, the
                            contacts being the feet of the altitudes therefore the perspector is H.

                            The circum-ellipse centered at K is not mentionned in the book. Its
                            perspector is O.

                            Best regards

                            Bernard
                          • John Conway
                            ... I think that s a garbled version of something I told Steve some time ago, which in turn was a garbled version of something I once proved, but couldn t
                            Message 13 of 24 , Jan 5, 2001
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                              On Thu, 4 Jan 2001, Steve Sigur wrote:

                              > John seems to think that if equilateral triangles are formed by anything
                              > having to do with trisection, they are always parallel. This makes the
                              > Morley inclination a fundamental property of the triangle rather as the
                              > Brocard angle is.

                              I think that's a garbled version of something I told Steve some
                              time ago, which in turn was a garbled version of something I once
                              proved, but couldn't quite remember!

                              Let me say something about what's going on here. I use
                              [l] to denote the angle of inclination of the line l with
                              some arbitrary fixed "horizontal" direction. Then there
                              are three meanings to "the mean of [a],[b],[c]"; for since
                              these quantities are only defined modulo 2pi, their mean
                              is only defined modulo pi/3. Now we can obviously find
                              the inclinations of the edges of the Morley triangle from
                              the angles in that figure (which are all known), and it
                              turns out that they are in the three mean directions. It's
                              even easier for the den Roussel triangle, while any of the
                              proofs that the Steiner deltoid "works" will show the same
                              for the triangles formed by either its cusps or its contact
                              points with the NPC.

                              When I first heard of Stammler's triangle I immediately
                              guess it would be true there too (because of the "forgotten
                              principle" Steve was talking about), and rapidly found a
                              short algebraic proof, but later heard that Stammler had
                              already verified it another way.

                              Regards, JHC
                            • Richard Guy
                              These ramblings are mainly talking to myself -- there s always the delete key. Andrew Bremner and I are interested in those rational triangles which have
                              Message 14 of 24 , Jan 7, 2001
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                                These ramblings are mainly talking to myself -- there's always the
                                delete key. Andrew Bremner and I are interested in those rational
                                triangles which have Malfatti circles of rational radius.

                                First of all, perhaps a Hyacinthian can put me right about *the* (?)
                                construction. Coxeter, Coxeter & Greitzer, Johnson, Altschiller-Court,
                                none of them have Malfattri in their Index. Eves gives some
                                interesting history, including that the construction is due to
                                Steiner, who knew that there were 32 solutions, but does not give
                                the construction.

                                Here's the best I can do. It relies heavily on trigonometry,
                                it can be turned into a Euclidean construction, but I don't
                                have a neat one.

                                I don't even know know why it depends on the quarter-angles.

                                Let the tangents of these be u, v, w. note that u, v, w
                                are not independent. The following formulas are useful when
                                doing manipulations:

                                a : b : c = sin A : sin B : sin C = 4u(1-uu)/(1+uu)^2 : :

                                u + v + w + vw + wu + uv = 1 + uvw

                                (1 + u)(1 + v)(1 + w) = 2(1 + uvw)

                                (1 + u)(1 - v)(1 - w) = 2(u + vw)

                                (1 - u)(1 - vw) = (1 + u)(v + w)

                                Divide edge BC in the ratio cot B/4 : cot C/4, i.e. w : v.
                                This is most expeditiously and Euclideanly done by drawing the
                                incircle of BIC, where I is the incentre of ABC, and X
                                is where the incircle of BIC touches BC.

                                This point X is on the radical axis (transverse common tangent)
                                of the two M-circles which touch BC, and the tans from it
                                to these circles are all equal, equal in fact to (1 + u)/2.
                                [I think I've normalized it so that the inradius is 1.]
                                Why? Because it works out! With centre X and
                                radius (1 + u)/2 draw a circle, which cuts BC in the two
                                points of contact at the Malfatti b- and c-circles. This
                                last step is Euclidean, and there must be a neat construction.
                                [Hyacinthians: Help!]

                                Note that the (direct) common tan to two touching circles
                                has length = geom mean of their diameters, so that we could
                                start from three equations

                                rb cot B/2 + 2 root (rb.rc) + rc cot C/2 = a, etc. (***)

                                where ra, rb, rc are the three Malfatti radii, but I find
                                them hard to solve, and giving at most 8, perhaps only 4,
                                solutions.

                                I said it works out. It does if you know the answer, i.e.,
                                that ra = (1+v)(1+w) / 2(1+u), etc. and
                                2 root (rb.rc) = 1 + u. [remember I've taken the inradius as 1.]

                                Let me convince at least myself of this.

                                BC = a = cot B/2 + cot C/2 = (1 - vv)2v + (1 - ww)/2w =
                                (v + w)(1 - vw) / 2vw = (1 + u)(v + w)^2 / 2vw( 1 - u)

                                BI = cosec B/2 = (1 + vv)/2v, CI = (1 + ww)/2w, perim of BIC =

                                [w(1 + vv) + v(1 + ww) + (v + w)(1 - vw)] / 2vw = (v + w) / vw.

                                and lengths of tans to incircle of BIC from B and C are

                                BX = (v + w)/2vw - (1 + vv)/2v and CX = (v + w)/2vw - (1 + ww)/2w
                                = (1 - vw)/2v = (1 - vw)/2w

                                whose sum checks out to be a. The claim is that the tans from B
                                and C to their respective M-circles have lengths

                                BX - (1 + u)/2 and CX - (1 + u)/2, i.e.
                                BX = (1 - vw)/2v - (1 + u)/2
                                = (v + w)(1 + u)/2v(1 - u) - (1 + u)/2 (v. formulas above)
                                = (1 + u)(v + w - v + uv)/2v(1-u)
                                = (1 + u)(w + uv)/2v(1 - u)
                                = (1 + u)(1 - u)(1 - v)(1 + w)/4v(1 - u) (v. above),
                                which, when multiplied by tan B/2 = 2v/(1 - vv) becomes
                                (1 + u)(1 + w)/2(1 + v) = rb, and similarly CX - (1 + u)/2.
                                multiplied by tan C/2, gives (1 + u)(1 + v)/2(1+w) = rc.

                                Moreover, these quantities satisy the three equations of which
                                we quoted the first above.

                                How the construction carries over to the other 31 cases I shudder
                                to think, but luckily we have Conway's extraversion:

                                (I use the numbers ..., 3, 2, +, 0, -, = (=-2), ...)

                                22= -
                                |
                                0=2 --2 =02 +
                                / \ / \ / \
                                03- +2- 2+- 30- -
                                | | | |
                                +=+ 0-+ -0+ =++ +
                                / \ / \ / \ / \
                                -30 020 ++0 200 3-0 -
                                | | | | |
                                2=0 +-0 000 -+0 =20 +
                                \ / \ / \ / \ /
                                -2+ 0++ +0+ 2-+ -
                                | | | |
                                2-- +0- 0+- -2- +
                                / \ / \ / \ / \
                                =22 -+2 002 +-2 2=2 -
                                | | |
                                20= 11= 02= +
                                \ / \ /
                                -03 0-3 -

                                where a node, abc, means a triangle with angles

                                a pi + A, b pi + B, c pi + C
                                or a pi - A, b pi - B, c pi - C
                                ^
                                according as a + b + c = 0 or 2 (see signs on right ... | ).

                                We are working mod 4, so 4 x 4 of the hexagons of the honeycomb
                                pattern roll up into a torus, where we identify 22=, =22 & 2=2,
                                and also 0=2 with 02=, =02 with 20=, 03- with 0-3,
                                30- with -03, -30 with 3-0 and 2=0 with =20,

                                So tan A/4 has the value

                                u -u -1/u 1/u (1+u)/(1-u) (1-u)/(1+u) (u-1)/(u+1) (u+1)/(u-1)

                                according as a =

                                0+ 0- 2+ 2- 1+ 1- -1+ or 3+ -1- or 3-
                                -2+ -2-

                                These form a group and similarly for the values of tan B/4
                                and tan C/4. One can think of taking arbitrary integers for
                                a and b, but only distinguishing them mod 4, and then c
                                has two choices, making a + b + c = 0 or 2. 4 x 4 x 2 = 32.

                                I've worked out all the details for the triangle (169,125,84)
                                for which u = 1/2, v = 1/5, w = 1/8. The radii, tangents,
                                common tangents, are all rational, of course, in all 32 cases.
                                I've drawn some of the pictures, but haven't completely
                                satisfied myself about the signs of the lengths -- the circles
                                lie in the `interior' angles (or their vertically opposites)
                                when a, b, c are even, and in the `exterior' angles when
                                a, b, c are odd.

                                I could fax the calculations to anyone interested, but you'll
                                learn much more if you do it yourself.

                                [An aside for Hyacinthians: with encouragement from John Conway
                                I drew a particular triangle and put in the 3 x 32 Malfatti
                                centres. Apart from the obvious fact that they lie 16 on each
                                member of the 3 pairs of angle bisectors, I didn't discern any
                                other collinearities. Has any Hyacinthian or triangle centre
                                buff looked at the 32 radical centres of the 32 triads of
                                Malfatti circles?]

                                I'm mainly interested in rationality, and have a vague
                                classification of triangles into:

                                1-rational: only the edges (and cosines of the angles) rational.

                                2-rational: also the sines, area, inradii, circumradius,

                                3-rational: the cosines of the one-third angles are rational,
                                the sines are rational multiples of root 3,
                                the sides of the Morley triangles are rational.

                                4-rational: the tangents, but not necessarily the sines and
                                cosines, of the quarter angles are rational.
                                Automatically 2-rational. Malfatti radii rational.

                                Is it possible to have one radius rational, but not all?
                                Yes, the triangle (6,5,5) has u = 1/3, v = w = (root 5) - 2
                                and Malfatti radii 6 - 2(root 5), 1 and 1.
                                The triangle (8,5,5) has u = 1/2, v = w = (root 10) - 3
                                and Malfatti radii 14 - 4(root 10), 1 and 1.

                                [all calculations done by aged hand, so probably rife with error]

                                Triangles with u = 1/3 have
                                a : b : c = 12(pp+qq) : 12pp+7pq-12qq : 25pq
                                and with
                                p = q = 1 24 : 7 : 25 Pythagorean -- I claim that a Pythagorean
                                triangle can't have any rational Malfatti radii. A P-triangle
                                is 2-rational, but never 4-rational.
                                p = 2, q = 1 6 : 5 : 5 -- see above
                                p = 3, q = 1 40 : 39 : 25 w = 3-root10, v = (5root10 - 13)/9
                                p = 3, q = 2 26 : 17 : 25 w = (root13 - 3)/2, v = (5root13 - 17)/6
                                etc. -- Presumably v and w are never rational ?

                                Is it possible to have rational Malfatti radii without the
                                triangle's being 4-rational? I suspect not. Proof, if true?
                                The possibilities for exception are where the quadrisectors
                                coincide with edges of the triangle, e.g. if there's an
                                angle pi/3, pi/2 or 2 pi/3. But the tans of a quarter of
                                each of these is irrational. So what?

                                But I want to start again from the three circles, and find
                                lots of maverick solutions. Pity I can't draw the circles --
                                you'll just have to sketch along.

                                1. The three circles coincide. They must be incircles (I
                                include excircles whenever I say this) Any three tangents
                                serve as edges of the triangle -- 4 (rational) solutions
                                whenever the triangle is 2-rational (or 4-rational).

                                2. Two circles coincide, the third touching (externally
                                if you want real solutions). The coincident ones must be
                                incircles, the edges being the transverse common tangent
                                and the two direct ones. The triangle is isosceles, and
                                we have the incircle and symmetrical excircle. Which is
                                the double circle gives 2 solutions.

                                3. Three circles distinct, but all touching at same point.
                                Two on one side of transverse common tan, one on the other,
                                in order to have real direct common tans. There are two
                                pairs of such and four choices for the 2 edges you take
                                with the transverse common tan. [Can't take both members
                                of a pair, else one circle touches only one edge.] One
                                circle is an incircle (4 possibilities) and there are 3
                                ways of choosing the point of contact for the other two
                                circles. E.g., the tans from A and B to the incircle of
                                (169,125,84) are 20 and 64 and there are circles touching
                                AB at the same point, but outside the triangle. One of
                                radius 15 will touch AC and one of radius 768/5 will
                                touch CB. 12 solutions?

                                4. Three circles distinct, one pair touching externally,
                                but each touching the third internally. There's only one
                                real pair of direct common tans, but three transverse
                                common tans. We can't choose all three of these, since
                                they concur. However, it is worth noting that they
                                form a point triangle, and one can in general visualize
                                3 circles, each touching two edges of a finite triangle
                                at infinity.

                                If you choose two of the transverse common tans (3 ways)
                                and one of the direct common tans (2 ways) you get 6
                                solutions? (or more?). They don't seem to come out rational
                                for a 4-rational triangle, and it may not be possible to find
                                a rational solution of this kind. Two of the three
                                equations (***) degenerate in that the square root term
                                disappears; e.g.,
                                rb cot B/2 + rc cot C/2 = a
                                ra cot A/2 + rb cot B/2 = c
                                rc cot C/2 + 2 root(rc.ra) + ra cot A/2 = b
                                General solution, anyone?

                                If you choose both the direct common tangentss, and one
                                transverse one, then the outside circle doesn't touch
                                two sides.

                                5. Three distinct circles touching externally in pairs.
                                Three pairs of direct common tans and three transverse
                                common tans (radical axes, concurrent). As before,
                                we can't choose all three of these last.

                                If we choose two of them (3 ways), then 4 of the choices
                                of the direct common tangent make one circle an incircle
                                and leaves one circle touching only one side. The other
                                2 choices have each circle touching just two sides. The
                                three equations are again in the same shape as in section 4.
                                How many solutions are there? 6? or more? Can they
                                be rational? Not, I suspect, with the above sense of
                                `4-rational' triangle.

                                If we choose one transverse common tan (3 ways), then there
                                are 4 choices for the direct common tans, leading to 12,
                                possibly 24 solutions of equations of shape

                                tc = a + tb + 2 root(rb.rc)
                                tc = b + ta
                                tb + 2 root(ra.rb) = c + ta

                                where ta, tb, tc are the lengths of the tans from A, B, C to
                                the repective Malfattis. E.g., ta = ra cot A/4, for some
                                suitable value of A (k pi + or - A).

                                If the edges are chosen (in one of 8 ways) from the 3 pairs
                                of direct common tans, then we have the classical case.
                                32 solutions.

                                Comments welcome, if anyone's still awake! R.
                              • xpolakis@otenet.gr
                                ... Two or three words before read the rest: A year ago or so, John Conway and I had discussed the Steiner s synthetic construction, as it is found in the
                                Message 15 of 24 , Jan 7, 2001
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                                  Richard Guy wrote:

                                  >These ramblings are mainly talking to myself -- there's always the
                                  >delete key. Andrew Bremner and I are interested in those rational
                                  >triangles which have Malfatti circles of rational radius.
                                  >
                                  >First of all, perhaps a Hyacinthian can put me right about *the* (?)
                                  >construction. Coxeter, Coxeter & Greitzer, Johnson, Altschiller-Court,
                                  >none of them have Malfattri in their Index. Eves gives some
                                  >interesting history, including that the construction is due to
                                  >Steiner, who knew that there were 32 solutions, but does not give
                                  >the construction.

                                  Two or three words before read the rest:

                                  A year ago or so, John Conway and I had discussed the Steiner's synthetic
                                  construction, as it is found in the French book: F.G.-M. _Exercices de
                                  geometrie..._,, and, If I remember correctly, we had CC-ied you.

                                  A trigonometrical solution of the problem can be found in:
                                  H. Do"rrie: _100 Great Problems of Elementary Mathematics.._(Dover), p. 147ff

                                  I am waiting a paper from a Greek mathematician (G. Kapetis) on the
                                  solution of the problem [by "Lemoine Transformation", but I am not quite sure
                                  what is it!] for publication in the forthcoming geometry journal
                                  _Forum Geometricorum_

                                  Antreas
                                • Dick Klingens
                                  Hello Richard, Antreas and other Hyacinthers, There is a Malfatti reference in Julian Coolidge, A Treatise on the Circle and the Sphere (Chelsea Publishing
                                  Message 16 of 24 , Jan 8, 2001
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                                    Hello Richard, Antreas and other Hyacinthers,

                                    There is a Malfatti reference in
                                    Julian Coolidge, A Treatise on the Circle and the Sphere (Chelsea Publishing
                                    Company, 1971)
                                    illustrating Lemoine's simplicity test of geometrical constructions
                                    and in
                                    George E. Martin, Geometric Constructions (Springer, 1998)
                                    with (a part of) Malfatti's equations.
                                    Steiner's construction is given as a problem (p. 96).

                                    Regards,
                                    Dick

                                    [:]-----Original Message-----
                                    [:]From: Richard Guy [mailto:rkg@...]
                                    [:]Sent: maandag 8 januari 2001 0:49
                                    [:]To: Andrew Bremner; Mike Guy; Math Fun; Hyacinthos@egroups.com
                                    [:]Subject: [EMHL] Malfatti musings.
                                    [:]
                                    [:]
                                    [:]These ramblings are mainly talking to myself -- there's always the
                                    [:]delete key. Andrew Bremner and I are interested in those rational
                                    [:]triangles which have Malfatti circles of rational radius.
                                    [:]
                                    [:]First of all, perhaps a Hyacinthian can put me right about *the* (?)
                                    [:]construction. Coxeter, Coxeter & Greitzer, Johnson, Altschiller-Court,
                                    [:]none of them have Malfattri in their Index. Eves gives some
                                    [:]interesting history, including that the construction is due to
                                    [:]Steiner, who knew that there were 32 solutions, but does not give
                                    [:]the construction.
                                    [:]
                                    [:]Here's the best I can do. It relies heavily on trigonometry,
                                    [:]it can be turned into a Euclidean construction, but I don't
                                    [:]have a neat one.
                                    [:]
                                    [:]I don't even know know why it depends on the quarter-angles.
                                    [:]
                                    [:]Let the tangents of these be u, v, w. note that u, v, w
                                    [:]are not independent. The following formulas are useful when
                                    [:]doing manipulations:
                                    [:]
                                    [:] a : b : c = sin A : sin B : sin C = 4u(1-uu)/(1+uu)^2 : :
                                    [:]
                                    [:] u + v + w + vw + wu + uv = 1 + uvw
                                    [:]
                                    [:] (1 + u)(1 + v)(1 + w) = 2(1 + uvw)
                                    [:]
                                    [:] (1 + u)(1 - v)(1 - w) = 2(u + vw)
                                    [:]
                                    [:] (1 - u)(1 - vw) = (1 + u)(v + w)
                                    [:]
                                    [:]Divide edge BC in the ratio cot B/4 : cot C/4, i.e. w : v.
                                    [:]This is most expeditiously and Euclideanly done by drawing the
                                    [:]incircle of BIC, where I is the incentre of ABC, and X
                                    [:]is where the incircle of BIC touches BC.
                                    [:]
                                    [:]This point X is on the radical axis (transverse common tangent)
                                    [:]of the two M-circles which touch BC, and the tans from it
                                    [:]to these circles are all equal, equal in fact to (1 + u)/2.
                                    [:][I think I've normalized it so that the inradius is 1.]
                                    [:]Why? Because it works out! With centre X and
                                    [:]radius (1 + u)/2 draw a circle, which cuts BC in the two
                                    [:]points of contact at the Malfatti b- and c-circles. This
                                    [:]last step is Euclidean, and there must be a neat construction.
                                    [:][Hyacinthians: Help!]
                                    [:]
                                    [:]Note that the (direct) common tan to two touching circles
                                    [:]has length = geom mean of their diameters, so that we could
                                    [:]start from three equations
                                    [:]
                                    [:]rb cot B/2 + 2 root (rb.rc) + rc cot C/2 = a, etc. (***)
                                    [:]
                                    [:]where ra, rb, rc are the three Malfatti radii, but I find
                                    [:]them hard to solve, and giving at most 8, perhaps only 4,
                                    [:]solutions.
                                    [:]
                                    [:]I said it works out. It does if you know the answer, i.e.,
                                    [:]that ra = (1+v)(1+w) / 2(1+u), etc. and
                                    [:]2 root (rb.rc) = 1 + u. [remember I've taken the inradius as 1.]
                                    [:]
                                    [:]Let me convince at least myself of this.
                                    [:]
                                    [:]BC = a = cot B/2 + cot C/2 = (1 - vv)2v + (1 - ww)/2w =
                                    [:] (v + w)(1 - vw) / 2vw = (1 + u)(v + w)^2 / 2vw( 1 - u)
                                    [:]
                                    [:]BI = cosec B/2 = (1 + vv)/2v, CI = (1 + ww)/2w, perim of BIC =
                                    [:]
                                    [:][w(1 + vv) + v(1 + ww) + (v + w)(1 - vw)] / 2vw = (v + w) / vw.
                                    [:]
                                    [:]and lengths of tans to incircle of BIC from B and C are
                                    [:]
                                    [:]BX = (v + w)/2vw - (1 + vv)/2v and CX = (v + w)/2vw - (1 + ww)/2w
                                    [:] = (1 - vw)/2v = (1 - vw)/2w
                                    [:]
                                    [:]whose sum checks out to be a. The claim is that the tans from B
                                    [:]and C to their respective M-circles have lengths
                                    [:]
                                    [:] BX - (1 + u)/2 and CX - (1 + u)/2, i.e.
                                    [:]BX = (1 - vw)/2v - (1 + u)/2
                                    [:] = (v + w)(1 + u)/2v(1 - u) - (1 + u)/2 (v. formulas above)
                                    [:] = (1 + u)(v + w - v + uv)/2v(1-u)
                                    [:] = (1 + u)(w + uv)/2v(1 - u)
                                    [:] = (1 + u)(1 - u)(1 - v)(1 + w)/4v(1 - u) (v. above),
                                    [:]which, when multiplied by tan B/2 = 2v/(1 - vv) becomes
                                    [:] (1 + u)(1 + w)/2(1 + v) = rb, and similarly CX - (1 + u)/2.
                                    [:]multiplied by tan C/2, gives (1 + u)(1 + v)/2(1+w) = rc.
                                    [:]
                                    [:]Moreover, these quantities satisy the three equations of which
                                    [:]we quoted the first above.
                                    [:]
                                    [:]How the construction carries over to the other 31 cases I shudder
                                    [:]to think, but luckily we have Conway's extraversion:
                                    [:]
                                    [:](I use the numbers ..., 3, 2, +, 0, -, = (=-2), ...)
                                    [:]
                                    [:] 22= -
                                    [:] |
                                    [:] 0=2 --2 =02 +
                                    [:] / \ / \ / \
                                    [:] 03- +2- 2+- 30- -
                                    [:] | | | |
                                    [:] +=+ 0-+ -0+ =++ +
                                    [:] / \ / \ / \ / \
                                    [:] -30 020 ++0 200 3-0 -
                                    [:] | | | | |
                                    [:] 2=0 +-0 000 -+0 =20 +
                                    [:] \ / \ / \ / \ /
                                    [:] -2+ 0++ +0+ 2-+ -
                                    [:] | | | |
                                    [:] 2-- +0- 0+- -2- +
                                    [:] / \ / \ / \ / \
                                    [:] =22 -+2 002 +-2 2=2 -
                                    [:] | | |
                                    [:] 20= 11= 02= +
                                    [:] \ / \ /
                                    [:] -03 0-3 -
                                    [:]
                                    [:]where a node, abc, means a triangle with angles
                                    [:]
                                    [:] a pi + A, b pi + B, c pi + C
                                    [:] or a pi - A, b pi - B, c pi - C
                                    [:] ^
                                    [:]according as a + b + c = 0 or 2 (see signs on right ... | ).
                                    [:]
                                    [:]We are working mod 4, so 4 x 4 of the hexagons of the honeycomb
                                    [:]pattern roll up into a torus, where we identify 22=, =22 & 2=2,
                                    [:]and also 0=2 with 02=, =02 with 20=, 03- with 0-3,
                                    [:]30- with -03, -30 with 3-0 and 2=0 with =20,
                                    [:]
                                    [:]So tan A/4 has the value
                                    [:]
                                    [:]u -u -1/u 1/u (1+u)/(1-u) (1-u)/(1+u) (u-1)/(u+1) (u+1)/(u-1)
                                    [:]
                                    [:]according as a =
                                    [:]
                                    [:]0+ 0- 2+ 2- 1+ 1- -1+ or 3+ -1- or 3-
                                    [:] -2+ -2-
                                    [:]
                                    [:]These form a group and similarly for the values of tan B/4
                                    [:]and tan C/4. One can think of taking arbitrary integers for
                                    [:]a and b, but only distinguishing them mod 4, and then c
                                    [:]has two choices, making a + b + c = 0 or 2. 4 x 4 x 2 = 32.
                                    [:]
                                    [:]I've worked out all the details for the triangle (169,125,84)
                                    [:]for which u = 1/2, v = 1/5, w = 1/8. The radii, tangents,
                                    [:]common tangents, are all rational, of course, in all 32 cases.
                                    [:]I've drawn some of the pictures, but haven't completely
                                    [:]satisfied myself about the signs of the lengths -- the circles
                                    [:]lie in the `interior' angles (or their vertically opposites)
                                    [:]when a, b, c are even, and in the `exterior' angles when
                                    [:]a, b, c are odd.
                                    [:]
                                    [:]I could fax the calculations to anyone interested, but you'll
                                    [:]learn much more if you do it yourself.
                                    [:]
                                    [:][An aside for Hyacinthians: with encouragement from John Conway
                                    [:]I drew a particular triangle and put in the 3 x 32 Malfatti
                                    [:]centres. Apart from the obvious fact that they lie 16 on each
                                    [:]member of the 3 pairs of angle bisectors, I didn't discern any
                                    [:]other collinearities. Has any Hyacinthian or triangle centre
                                    [:]buff looked at the 32 radical centres of the 32 triads of
                                    [:]Malfatti circles?]
                                    [:]
                                    [:]I'm mainly interested in rationality, and have a vague
                                    [:]classification of triangles into:
                                    [:]
                                    [:]1-rational: only the edges (and cosines of the angles) rational.
                                    [:]
                                    [:]2-rational: also the sines, area, inradii, circumradius,
                                    [:]
                                    [:]3-rational: the cosines of the one-third angles are rational,
                                    [:] the sines are rational multiples of root 3,
                                    [:] the sides of the Morley triangles are rational.
                                    [:]
                                    [:]4-rational: the tangents, but not necessarily the sines and
                                    [:] cosines, of the quarter angles are rational.
                                    [:] Automatically 2-rational. Malfatti radii rational.
                                    [:]
                                    [:]Is it possible to have one radius rational, but not all?
                                    [:]Yes, the triangle (6,5,5) has u = 1/3, v = w = (root 5) - 2
                                    [:]and Malfatti radii 6 - 2(root 5), 1 and 1.
                                    [:] The triangle (8,5,5) has u = 1/2, v = w = (root 10) - 3
                                    [:]and Malfatti radii 14 - 4(root 10), 1 and 1.
                                    [:]
                                    [:][all calculations done by aged hand, so probably rife with error]
                                    [:]
                                    [:]Triangles with u = 1/3 have
                                    [:] a : b : c = 12(pp+qq) : 12pp+7pq-12qq : 25pq
                                    [:]and with
                                    [:]p = q = 1 24 : 7 : 25 Pythagorean -- I claim that a Pythagorean
                                    [:]triangle can't have any rational Malfatti radii. A P-triangle
                                    [:]is 2-rational, but never 4-rational.
                                    [:]p = 2, q = 1 6 : 5 : 5 -- see above
                                    [:]p = 3, q = 1 40 : 39 : 25 w = 3-root10, v = (5root10 - 13)/9
                                    [:]p = 3, q = 2 26 : 17 : 25 w = (root13 - 3)/2, v = (5root13 - 17)/6
                                    [:]etc. -- Presumably v and w are never rational ?
                                    [:]
                                    [:]Is it possible to have rational Malfatti radii without the
                                    [:]triangle's being 4-rational? I suspect not. Proof, if true?
                                    [:]The possibilities for exception are where the quadrisectors
                                    [:]coincide with edges of the triangle, e.g. if there's an
                                    [:]angle pi/3, pi/2 or 2 pi/3. But the tans of a quarter of
                                    [:]each of these is irrational. So what?
                                    [:]
                                    [:]But I want to start again from the three circles, and find
                                    [:]lots of maverick solutions. Pity I can't draw the circles --
                                    [:]you'll just have to sketch along.
                                    [:]
                                    [:]1. The three circles coincide. They must be incircles (I
                                    [:]include excircles whenever I say this) Any three tangents
                                    [:]serve as edges of the triangle -- 4 (rational) solutions
                                    [:]whenever the triangle is 2-rational (or 4-rational).
                                    [:]
                                    [:]2. Two circles coincide, the third touching (externally
                                    [:]if you want real solutions). The coincident ones must be
                                    [:]incircles, the edges being the transverse common tangent
                                    [:]and the two direct ones. The triangle is isosceles, and
                                    [:]we have the incircle and symmetrical excircle. Which is
                                    [:]the double circle gives 2 solutions.
                                    [:]
                                    [:]3. Three circles distinct, but all touching at same point.
                                    [:]Two on one side of transverse common tan, one on the other,
                                    [:]in order to have real direct common tans. There are two
                                    [:]pairs of such and four choices for the 2 edges you take
                                    [:]with the transverse common tan. [Can't take both members
                                    [:]of a pair, else one circle touches only one edge.] One
                                    [:]circle is an incircle (4 possibilities) and there are 3
                                    [:]ways of choosing the point of contact for the other two
                                    [:]circles. E.g., the tans from A and B to the incircle of
                                    [:](169,125,84) are 20 and 64 and there are circles touching
                                    [:]AB at the same point, but outside the triangle. One of
                                    [:]radius 15 will touch AC and one of radius 768/5 will
                                    [:]touch CB. 12 solutions?
                                    [:]
                                    [:]4. Three circles distinct, one pair touching externally,
                                    [:]but each touching the third internally. There's only one
                                    [:]real pair of direct common tans, but three transverse
                                    [:]common tans. We can't choose all three of these, since
                                    [:]they concur. However, it is worth noting that they
                                    [:]form a point triangle, and one can in general visualize
                                    [:]3 circles, each touching two edges of a finite triangle
                                    [:]at infinity.
                                    [:]
                                    [:]If you choose two of the transverse common tans (3 ways)
                                    [:]and one of the direct common tans (2 ways) you get 6
                                    [:]solutions? (or more?). They don't seem to come out rational
                                    [:]for a 4-rational triangle, and it may not be possible to find
                                    [:]a rational solution of this kind. Two of the three
                                    [:]equations (***) degenerate in that the square root term
                                    [:]disappears; e.g.,
                                    [:] rb cot B/2 + rc cot C/2 = a
                                    [:] ra cot A/2 + rb cot B/2 = c
                                    [:] rc cot C/2 + 2 root(rc.ra) + ra cot A/2 = b
                                    [:]General solution, anyone?
                                    [:]
                                    [:]If you choose both the direct common tangentss, and one
                                    [:]transverse one, then the outside circle doesn't touch
                                    [:]two sides.
                                    [:]
                                    [:]5. Three distinct circles touching externally in pairs.
                                    [:]Three pairs of direct common tans and three transverse
                                    [:]common tans (radical axes, concurrent). As before,
                                    [:]we can't choose all three of these last.
                                    [:]
                                    [:]If we choose two of them (3 ways), then 4 of the choices
                                    [:]of the direct common tangent make one circle an incircle
                                    [:]and leaves one circle touching only one side. The other
                                    [:]2 choices have each circle touching just two sides. The
                                    [:]three equations are again in the same shape as in section 4.
                                    [:]How many solutions are there? 6? or more? Can they
                                    [:]be rational? Not, I suspect, with the above sense of
                                    [:]`4-rational' triangle.
                                    [:]
                                    [:]If we choose one transverse common tan (3 ways), then there
                                    [:]are 4 choices for the direct common tans, leading to 12,
                                    [:]possibly 24 solutions of equations of shape
                                    [:]
                                    [:] tc = a + tb + 2 root(rb.rc)
                                    [:] tc = b + ta
                                    [:] tb + 2 root(ra.rb) = c + ta
                                    [:]
                                    [:]where ta, tb, tc are the lengths of the tans from A, B, C to
                                    [:]the repective Malfattis. E.g., ta = ra cot A/4, for some
                                    [:]suitable value of A (k pi + or - A).
                                    [:]
                                    [:]If the edges are chosen (in one of 8 ways) from the 3 pairs
                                    [:]of direct common tans, then we have the classical case.
                                    [:]32 solutions.
                                    [:]
                                    [:]Comments welcome, if anyone's still awake! R.
                                    [:]
                                    [:]
                                    [:]
                                    [:]
                                  • xpolakis@otenet.gr
                                    ... Another reference: Jacques Hadamard: Lecons de geometrie elementaire. 13e ed. Paris: Armand Colin, 1947. In Appendice: Probleme de Malfatti, pp. 312 - 316
                                    Message 17 of 24 , Jan 8, 2001
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                                      Dick Klingens wrote:

                                      >There is a Malfatti reference in
                                      >Julian Coolidge, A Treatise on the Circle and the Sphere (Chelsea Publishing
                                      >Company, 1971)
                                      >illustrating Lemoine's simplicity test of geometrical constructions
                                      > and in
                                      >George E. Martin, Geometric Constructions (Springer, 1998)
                                      >with (a part of) Malfatti's equations.
                                      >Steiner's construction is given as a problem (p. 96).

                                      Another reference:
                                      Jacques Hadamard: Lecons de geometrie elementaire. 13e ed.
                                      Paris: Armand Colin, 1947. In Appendice: Probleme de Malfatti, pp. 312 - 316
                                      (Reprint: Sceaux : Gabay, 1988)

                                      APH
                                    • Ignacio Larrosa Cañestro
                                      ... From: To: Sent: Monday, January 08, 2001 7:42 PM Subject: RE: [EMHL] Malfatti musings. ... 316 ... Another
                                      Message 18 of 24 , Jan 8, 2001
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                                        ----- Original Message -----
                                        From: <xpolakis@...>
                                        To: <Hyacinthos@egroups.com>
                                        Sent: Monday, January 08, 2001 7:42 PM
                                        Subject: RE: [EMHL] Malfatti musings.

                                        > Another reference:
                                        > Jacques Hadamard: Lecons de geometrie elementaire. 13e ed.
                                        > Paris: Armand Colin, 1947. In Appendice: Probleme de Malfatti, pp. 312 -
                                        316
                                        > (Reprint: Sceaux : Gabay, 1988)
                                        >
                                        > APH

                                        Another one:

                                        The 'venerable' Enciclopedia ESPASA, Edit ESPASA-CALPE, Madrid (over one
                                        hundred volumes),
                                        curiously under the epigrah 'Circunferencia'. It include the euclidean
                                        construction (justified).

                                        Greatings from a flooded Galicia,

                                        Ignacio Larrosa Cañestro
                                        A Coruña (España)
                                        ilarrosa@...
                                        ICQ #94732648
                                      • Frans Gremmen
                                        ... Altschiller-Court, ... Dear Richard and Hyacinthers, In november, december 1998 a question about the Malfatti problem were answered by me in
                                        Message 19 of 24 , Jan 9, 2001
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                                          --- In Hyacinthos@egroups.com, Richard Guy <rkg@c...> wrote:
                                          > These ramblings are mainly talking to myself -- there's always the
                                          > delete key. Andrew Bremner and I are interested in those rational
                                          > triangles which have Malfatti circles of rational radius.
                                          >
                                          > First of all, perhaps a Hyacinthian can put me right about *the* (?)
                                          > construction. Coxeter, Coxeter & Greitzer, Johnson,
                                          Altschiller-Court,
                                          > none of them have Malfattri in their Index.

                                          Dear Richard and Hyacinthers,

                                          In november, december 1998 a question about the Malfatti problem
                                          were answered by me in geometry.college. I gave a 'recipe' to compute
                                          the radii of the Malfatti circles in case they are all within the
                                          triangle. A remark about more than one solution was made and how they
                                          could be obtained, but it was not very clear to me. I said something
                                          "as far as I remember...".
                                          It was a little later mentioned in a posting, that they are 32
                                          solutions found by Steiner. I did not know that.
                                          John Conway presented his short formula. It was checked against my
                                          'recipe' and found to be in accordance. A little later I checked up my
                                          notes and saw a much easier formula found earlier. From that formula I
                                          tried to find all 32 solutions. First by hand, subsequently aided by
                                          computer use and at last completely with a computer program to run
                                          through all cases which could be possible with all combinations of +
                                          and - in some places.
                                          Cases were rejected, that did not 'restore' a sidelength.
                                          Three segmentlengths, 2 times from a vertex to a point of tangency and
                                          the length of a common tangent should make up a sidelength of triangle
                                          A1A2A3.
                                          32 possibilities remained, the solutions.


                                          Triangle A1A2A3 with sidelengths a1, a2 and a3, DELTA = area(A1A2A3)
                                          i1, i2 and i3 are +1 or -1.

                                          cosbeta1 = 2*u2*u3/u1 - 1
                                          cosbeta2 = 2*u1*u3/u2 - 1
                                          cosbeta3 = 2*u1*u2/u3 - 1

                                          cosbeta1, cosbeta2 and cosbeta3 are cosines between a Malfatti circle
                                          and a nontangent side of A1A2A3 if it intersects it really, otherwise
                                          these quantities are < -1 or > 1,

                                          where u1, u2, u3 are if

                                          k = 0 u1 = 1 + i1*cos(A1/2) k = 1 u1 = 1 + i1*cos(A1/2)
                                          u2 = 1 + i2*cos(A2/2) u2 = 1 + i2*sin(A2/2)
                                          u3 = 1 + i3*cos(A3/2) u3 = 1 + i3*sin(A3/2)

                                          k = 2 u1 = 1 + i1*sin(A1/2) k = 3 u1 = 1 + i1*sin(A1/2)
                                          u2 = 1 + i2*cos(A2/2) u2 = 1 + i2*sin(A2/2)
                                          u3 = 1 + i3*sin(A3/2) u3 = 1 + i3*cos(A3/2)

                                          j1, j2 and j3 are +1 or -1.

                                          2*DELTA
                                          r1 = -------------------------------- (radius of a Malfatti circle)
                                          (j2*a2 + j3*a3 + j1*a1*cosbeta1)

                                          and accompanying r2 and r3 by cyclic permutation.

                                          Each triple of Malfatti circles corresponds to a unique a combination
                                          of k, i1, i2 and i3.
                                          The values of j1, j2 and j3 depend on k, i1, i2 and i3.

                                          The 32 combinations:
                                          k i1 i2 i3 j1 j2 j3
                                          ------------------------
                                          0 -1 -1 -1 1 1 1
                                          0 -1 -1 1 1 1 1
                                          0 -1 1 -1 1 1 1
                                          0 -1 1 1 1 1 1
                                          0 1 -1 -1 1 1 1
                                          0 1 -1 1 1 1 1
                                          0 1 1 -1 1 1 1
                                          0 1 1 1 1 1 1
                                          1 -1 -1 -1 1 -1 -1
                                          1 -1 -1 1 1 -1 -1
                                          1 -1 1 -1 1 -1 -1
                                          1 -1 1 1 1 -1 -1
                                          1 1 -1 -1 -1 1 1
                                          1 1 -1 1 -1 1 1
                                          1 1 1 -1 -1 1 1
                                          1 1 1 1 -1 1 1
                                          2 -1 -1 -1 -1 1 -1
                                          2 -1 -1 1 -1 1 -1
                                          2 -1 1 -1 1 -1 1
                                          2 -1 1 1 1 -1 1
                                          2 1 -1 -1 -1 1 -1
                                          2 1 -1 1 -1 1 -1
                                          2 1 1 -1 1 -1 1
                                          2 1 1 1 1 -1 1
                                          3 -1 -1 -1 -1 -1 1
                                          3 -1 -1 1 1 1 -1
                                          3 -1 1 -1 -1 -1 1
                                          3 -1 1 1 1 1 -1
                                          3 1 -1 -1 -1 -1 1
                                          3 1 -1 1 1 1 -1
                                          3 1 1 -1 -1 -1 1
                                          3 1 1 1 1 1 -1

                                          These combinations were found stable over trials with some acute and
                                          obtuse angled triangles and yielding positive values for r1, r2 and r3
                                          .
                                          There remains a question open concerning the dependence of j1, j2 and
                                          j3.
                                          In the j1 (/j2/j3) column you find 12 times -1 and 20 times +1.

                                          Best Regards,

                                          Frans Gremmen, University of Nijmegen, The Netherlands.
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