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Re: [EMHL] a Morley configuration
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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
I know only of one Stammler triangle. What are the other two?
> 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.
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.  Floor van Lamoen wrote:
>On 9 Nov 1998 Den Roussel wrote (as result of discussion) in
Thread:
>geometryresearch:
>
>"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."
http://forum.swarthmore.edu/epigone/geom.research/jahclumyax/
APH  Dear all,
> [SS]:
[FvL]:
> > 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.
>
> About the triangle I came up with. Those are the points of contact of
I almost forgot Peter Yff's circumtangential and circumnormal triangles.
> 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).
See TCCT 6.28 and 6.29.
Kind regards,
Sincerely,
Floor.  [FvL]:
>I almost forgot Peter Yff's circumtangential and circumnormal triangles.
.... and of course one can define infinitely many equilateral triangles,
>See TCCT 6.28 and 6.29.
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  on 1/4/01 1:10 AM, Bernard Gibert wrote:
> [SS]
John seems to think that if equilateral triangles are formed by anything
>> Are these triangles parallel to the original Morley one?
>
> Yes, they are.
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  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  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. JeanPierre  Dear Antreas and other Hyacinthists,
Antreas wrote :> And here is a related problem:
PBC,PCA,
>
> For which point(s) P the triangle formed with the circumcenters of
> PAB is equilateral?
The two Fermat points : for each one the three circumcenters are the
>
vertices of an imperial triangle.
Friendly. JeanPierre  Dear Steve and friends,
[SS]> Here is the construction. Create the Stammler triangle by drawing tangents
The triangle called gAgBgC in my first "Morley configuration" message is the
> 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.
>
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 nonpivotal 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  Dear JeanPierre and other Hyacinthists,
[JP]> I've computed the locus of the centers of the equilateral triangles
there's at least the symmetric of X76 (= isotomic
> 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.
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  Dear JeanPierre,
[JPE]:>I've computed the locus of the centers of the equilateral triangles
Another triangle ellipse is "Lemoine ellipse."
>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.
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/inellipse 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, 266268 (1950).
Toscano, L.: Sur l'ellipse de Lemoine.
Bull. Soc. Roy. Sci. Liege 23, 221229 (1954).
Greetings
Antreas  Dear Antreas,
>
The Lemoine ellipse is the inellipse with foci G & K.
> 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/inellipse centered
> at K, the Lemoine point.
>
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 inellipse centered at K is called "ellipse K" in Brocard & Lemoyne, the
contacts being the feet of the altitudes therefore the perspector is H.
The circumellipse centered at K is not mentionned in the book. Its
perspector is O.
Best regards
Bernard  On Thu, 4 Jan 2001, Steve Sigur wrote:
> John seems to think that if equilateral triangles are formed by anything
I think that's a garbled version of something I told Steve some
> 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.
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  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, AltschillerCourt,
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 quarterangles.
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(1uu)/(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 Mcircles 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 ccircles. 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 Mcircles 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(1u)
= (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 30 
    
2=0 +0 000 +0 =20 +
\ / \ / \ / \ /
2+ 0++ +0+ 2+ 
   
2 +0 0+ 2 +
/ \ / \ / \ / \
=22 +2 002 +2 2=2 
  
20= 11= 02= +
\ / \ /
03 03 
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 03,
30 with 03, 30 with 30 and 2=0 with =20,
So tan A/4 has the value
u u 1/u 1/u (1+u)/(1u) (1u)/(1+u) (u1)/(u+1) (u+1)/(u1)
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:
1rational: only the edges (and cosines of the angles) rational.
2rational: also the sines, area, inradii, circumradius,
3rational: the cosines of the onethird angles are rational,
the sines are rational multiples of root 3,
the sides of the Morley triangles are rational.
4rational: the tangents, but not necessarily the sines and
cosines, of the quarter angles are rational.
Automatically 2rational. 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+7pq12qq : 25pq
and with
p = q = 1 24 : 7 : 25 Pythagorean  I claim that a Pythagorean
triangle can't have any rational Malfatti radii. A Ptriangle
is 2rational, but never 4rational.
p = 2, q = 1 6 : 5 : 5  see above
p = 3, q = 1 40 : 39 : 25 w = 3root10, 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 4rational? 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 2rational (or 4rational).
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 4rational 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
`4rational' 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.  Richard Guy wrote:
>These ramblings are mainly talking to myself  there's always the
Two or three words before read the rest:
>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, AltschillerCourt,
>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.
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 CCied 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  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, AltschillerCourt,
[:]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 quarterangles.
[:]
[:]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(1uu)/(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 Mcircles 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 ccircles. 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 Mcircles 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(1u)
[:] = (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 30 
[:]     
[:] 2=0 +0 000 +0 =20 +
[:] \ / \ / \ / \ /
[:] 2+ 0++ +0+ 2+ 
[:]    
[:] 2 +0 0+ 2 +
[:] / \ / \ / \ / \
[:] =22 +2 002 +2 2=2 
[:]   
[:] 20= 11= 02= +
[:] \ / \ /
[:] 03 03 
[:]
[:]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 03,
[:]30 with 03, 30 with 30 and 2=0 with =20,
[:]
[:]So tan A/4 has the value
[:]
[:]u u 1/u 1/u (1+u)/(1u) (1u)/(1+u) (u1)/(u+1) (u+1)/(u1)
[:]
[:]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:
[:]
[:]1rational: only the edges (and cosines of the angles) rational.
[:]
[:]2rational: also the sines, area, inradii, circumradius,
[:]
[:]3rational: the cosines of the onethird angles are rational,
[:] the sines are rational multiples of root 3,
[:] the sides of the Morley triangles are rational.
[:]
[:]4rational: the tangents, but not necessarily the sines and
[:] cosines, of the quarter angles are rational.
[:] Automatically 2rational. 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+7pq12qq : 25pq
[:]and with
[:]p = q = 1 24 : 7 : 25 Pythagorean  I claim that a Pythagorean
[:]triangle can't have any rational Malfatti radii. A Ptriangle
[:]is 2rational, but never 4rational.
[:]p = 2, q = 1 6 : 5 : 5  see above
[:]p = 3, q = 1 40 : 39 : 25 w = 3root10, 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 4rational? 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 2rational (or 4rational).
[:]
[:]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 4rational 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
[:]`4rational' 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.
[:]
[:]
[:]
[:]  Dick Klingens wrote:
>There is a Malfatti reference in
Another reference:
>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).
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  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 ESPASACALPE, 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  In Hyacinthos@egroups.com, Richard Guy <rkg@c...> wrote:
> These ramblings are mainly talking to myself  there's always the
AltschillerCourt,
> 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,
> 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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