Dear Bernard and Steve

> >> [JP] consider the rectangular hyperbola through G, O, K
> >> (symedian), X[110).
> >> His center is the midpoint of GX[110] and the hyperbola goes
through
> >> the infinite points of the Jerabek hyperbola, X[154], X[354], X
> >> [392], X
> >> [1201],...
> >> Consider a common point P (not X[110]) of the hyperbola with the
> >> circumcircle;
>
> These 3 points P lie on the Thomson cubic.

I don't think so

> >> let M be the homothetic of P in (O, -3) and A'B'C' the
> >> pedal triangle of M.
> >> Then AA', BB', CC' are parallel.
> >> Any explanation?

> there's a close analogy with the bottom of my web page on the
Darboux
> cubic.
>
> http://perso.wanadoo.fr/bernard.gibert/Exemples/k004.html

Yes, indeed. These three points are UVW in your Darboux page. many
thanks.
It is quite interesting to see that these points are the only points
P in the plane such as, if A'B'C' is the pedal triangle of P, the
lines AA', BB', CC' are parallel (of course they are parallel to the
asymptots of Lucas - or Thomson -)
Friendly. Jean-Pierre

Dear Jean-Pierre,

>>>> [JP] consider the rectangular hyperbola through G, O, K
>>>> (symedian), X[110).
>>>> His center is the midpoint of GX[110] and the hyperbola goes
> through
>>>> the infinite points of the Jerabek hyperbola, X[154], X[354], X
>>>> [392], X
>>>> [1201],...
>>>> Consider a common point P (not X[110]) of the hyperbola with the
>>>> circumcircle;
>>
>> These 3 points P lie on the Thomson cubic.
>
> I don't think so

THEY DO, dear Jean-Pierre !



Best regards

Bernard



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

Dear Bernard
> THEY DO, dear Jean-Pierre !
Of course, you're right - in fact I was thinking to your points UVW -
Looking at your nice page about Darboux, I have an obvious explanation
of the nice property of the pedal triangle of U,V,W.
Suppose that A'B'C' is the pedal triangle of M; then BB' and CC' are
parallel iff M lies on your hyperbola Ha.
As U,V,W are the common points of Ha,Hb,Hc, for these points, AA',
BB', CC' are parallel.
Thus, there is nothing new in what I said.
Friendly.Jean-Pierre

Sorry,I mixed up Steiner circumellipse with Steiner inellipse, oops!
I read your posts too quickly! That's why I went wrong!

Francois


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

Steve,

A minor correction below - perhaps even more interesting!



>
>
> On Mar 31, 2006, at 9:26 AM, Wilson Stothers wrote:
>
> >
> > Fact 1
> > The vertices of the cevian triangles of tF1 and tF2
> > lie in a circle C whose center R is on line tF1tF2
> > We know that L = X(20) is on this line.
> >
> > Conjecture 1 R = L.
>
[SS}> Wow -- very nice. My pictures support this conjecture.

Now proved by Bernard - a very nice proof.

> >
> > Fact 2
> > The inconic IR with centre R is bitangent with C above
> > with contact points on tF1tF2, so this is the axis of IR
> > containing the real foci G1,G2 (isogonal conjugates of
> > course, and on the Darboux Cubic if Conjecture 1 is true)
> > The perspector of IR is the cevapoint of tF1 and tF2.
> > The inconic has axes parallel to those of the Steiner
> > ellipses.
>
[SS}> Even more interesting. My pictures seem to indicate the G1 and
G2 are
> already defined intersections. The lines   f1--tf2 and f2--O meet
> at  these foci. If true, this is almost too nice.

I think there is a typo here.

What I see is that f1--H and f2--O meet at G1
and f2--H, f1--O at G2

Note that
G is the mid-point of f1--f2
L is the mid-point of G1--G2, parallel to f1--f2
so the lines fi--Gj , fj--Gi meet on the Euler Line as it is G--L,
but why at O, H?
Is it relevant that f1,f2, G1,G2, O,H are isogonal conjugates?
>


>
> >
> > Conjecture 2
> > The lines F1G1 and F2G2 meet on the Euler line.
> > The lines F1G2 and F2G1 meet on the Euler line.

Now obvious as above as G is mid-point of F1--F2

> > One of the points of intersection is
> > the anticomplement of L (so reflection of L in H)
> > (which is also the cross-point of F1 and F2)
> > The other cuts LH in the ratio 2:3(a strange ratio)
>
>

Wilson

A correction and some new stuff


On Mar 31, 2006, at 9:16 PM, Steve Sigur wrote:

> My pictures seem to indicate the G1 and G2 are
> already defined intersections. The lines   f1--tf2 and f2--O meet at
> these foci. If true, this is almost too nice.
>
> Oh yes, f1 and f2 are the foci of the Steiner inellipse.

A correction: f1--tf2 does not go though G1 or G2, but f1--O and f2--
O seem to.

If G1 and G2 are on the Darboux cubic, its two operations are the
isogonal conjugate and reflection in the circumcircle, which we call o.

Both the group table and observation says that oG1--oG2 goes through
H. There should be lines going though gL but I haven't figured out
what they are.

Not directly related but perhaps interesting. The tripolars of points
on the non-focal axis is an inparabola with perspector the
intersection of GK with Steiner ellipse and Focus the OK intersection
with the circumcircle.

The dual of this inparabola is a circumhyperbola with perspector an
end of a Steiner axis. There are corresponing ones for the other axis.

Steve who is liking this.




Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm





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

On Mar 31, 2006, at 12:40 PM, Bernard Gibert wrote:

>  ctP

what is "c" here?

Steve

Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm





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

Dear Steve ,

>
>>  ctP
>
> what is "c" here?

cP is the complement of P


Best regards

Bernard



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

Wilson, François, Jean-Pierre, Bernard, Peter, and all,

In checking Wilson's statements last night I noticed that there is a
point on each axis of the Steiner ellipses where the its tripolar and
its dual are perpendicular. The dual of course is always
perpendicular to an axis, so at these points these lines are parallel
to the Steiner axes. They seem to go through the Kiepert center mS
(the medial Steiner point). Hence  there seems to be a point on each
axis whose dual and tripolar line form the axes of the Kiepert
hyperbola. By intermediate values the points have to exist. That
these lines are the Kiepert axes is conjecture.

It will be easy to find these points but I am occupied by my conics
writeups so it will wait. I suspect the the existence of the points
is more interesting than the points themselves.

Steve




Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm

On Apr 1, 2006, at 8:24 AM, Wilson Stothers wrote:

> Is it relevant that f1,f2, G1,G2, O,H are isogonal conjugates?

Yes there is more to this story than we have found (and we have found
so much more than I ever thought was there!). G1, G2, O, H are on the
Darboux cubic, which should give more relations. f1, f2, G, O, H are
on Thomson cubic (at least I think they all are), so this will give
more.

Steve

Steve Sigur

Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm





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

If f1 is mF1 (i.e., on opposite side of non-focal axis) then   F1,
tf1, tF1, tG1 are colinear. This means that tG1 and tf1 can be
written as linear combinations of F1 and tF1, yet another project for
a future time.

I would like to comment that these Steiner discoveries required the
collaboration of many people with many ways of thinking about
geometry. I has been a pleasure working on this with all of you.

steve


Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm

On 4/1/06, Francois Rideau <francois.rideau@...> wrote:
>
>  Dear Steve
>
> I think I understand better the geometric situation.
> I call a1b1c1 the cevian triangle of tF1 and a2b2c2 the cevian triangle of
> tF2.
> Then after a possible change of notations, a1b1c1 is the pedal triangle of
> G1 and a2b2c2 the pedal triangle of G2.
> G1 (resp G2) is the pivot center of inscribed triangles directly similar
> to a1b1c1 (resp a2b2c2).
> The circumcircles of these inscribed triangles are bitangent to the inner
> conic of center L.
> That's what gives me my chuffchuffs theory.
> Can you check that?
> Friendly
> Francois
> PS Here in france these circles are called focus circles of 2th type.
>


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

Dear Bernard

It looks a nice proof, but I have one problem ...


-- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@...> wrote:
>

> >
> > Let F1, F2 be the real foci of the Steiner (circum)ellipse.
> > Write tX for the isotomic conjugate of a point X.
> >
> > Fact 1
> > The vertices of the cevian triangles of tF1 and tF2
> > lie in a circle C whose center R is on line tF1tF2
> > We know that L = X(20) is on this line.
> >
> > Conjecture 1 R = L.
>
> *** This is true.
>
> proof 1 :
>
> the cevian circle of a point P has center X(20)
> iff P lies on 3 sextics,
> iff tP lies on 3 cubics
> iff ctP lies on 3 circum-cubics in a same pencil containing the
> Thomson cubic K002 and the second Brocard cubic K018.
>
> These meet at 9 points A, B, C, G, K and the 4 foci of the Steiner
in-
> ellipse, etc
>
>

This appears to suggest that P with ctP = G, K have the property
that the cevian circle is centred at X(20). But here P = G, H..

Am I missing something simple?

The gemeral theory is that there should be six points with
cevian circles centred on a given point, so we'd expect two
other than taF, F = focus of Steiner inellipse.

Wilson

Dear Steve,

You may choose not to mention this one again, but I'm hanging onto it
while thinking there might be something here.

Sincerely, Jeff


> This system of 3 points and six lines is affine invariant and
> invariant under the interchange of a point with its conjugate. The
> ancients called this the anallagmatic symmetry.
>
> Every structure I give will both affine and anallagmatic symmetric
so
> I will not mention this fact again.
>
> We now have 3 points G, P, P' and 6 lines %, ~P, ~P', G--P, G--P',
P--
> P'.
>
> Note that the P--P' is, by itself, anallagmatically symmetric.
Mineur
> made a big deal of this, so I call it the Mineur line.
>
> There are 5 intersections at infinity: (G--P).%, (G--P').%, (P'--
P).
> %, (~P).%,  (~P').%, where the dot represents intersection.
>
> The isotomic conjugates of the 5 points at infinity are very
> interesting. They are all on the Steiner elllipse.
>
> t((G--P).%)  and   t((~P).%) and G are collinear.
> t((G--P').%)  and   t((~P').%) and G are collinear.
> t((G--P').%)  and   t((~P).%) and P are collinear.
> t((G--P).%)  and   t((~P').%) and P' are collinear.
>
> There is a lot more that is interesting but I am trying to make
this
> short. Go to http://paideiaschool.org/TeacherPages/Steve_Sigur/
> geometryIndex.htm   for more (posted tomorrow).
>
> The point  t((~P).%) has a truly amazing property that is the
reason
> for this discussion:
>
> Suppose I "project" a point from the steiner ellipse to another
> circumconic, say the one with P as perspector. Let Q be a point on
> the Steiner ellipse, then this point is the barycentric product
PQ.
> (This is a foundation of conjugation, I think).
>
> The amazing thing is that Q--PQ goes through t((~P).%).
>
> So, for example, we can project a point Q from the Steiner ellipse
to
> the circumcircle by drawing a line from the Steiner point through
P
> the intersection with the circumcircle being the point.
>
> I figure this result must by known by the people that study
> conjugations, but it is my highly symmetric way of introducing the
> subject.
>
> Steve
>

Dear Steve

--- In Hyacinthos@yahoogroups.com, Steve Sigur <s.sigur@...> wrote:
>
>
> On Apr 1, 2006, at 8:24 AM, Wilson Stothers wrote:
>
> > Is it relevant that f1,f2, G1,G2, O,H are isogonal conjugates?
>
> Yes there is more to this story than we have found (and we have
found
> so much more than I ever thought was there!). G1, G2, O, H are on
the
> Darboux cubic, which should give more relations.

We can identify G1, G2 as follows

they lie on the Darboux Cubic (as isogonals with line through X(20))
they lie on Bernards K187 as the mid-point is X(20), on Euler Line.

The common intersections are O,H and the four foci of the inconic
centred on L = X(20)

Wilson

>

Dear Wilson,

>>> [WS] Let F1, F2 be the real foci of the Steiner (circum)ellipse.
>>> Write tX for the isotomic conjugate of a point X.
>>>
>>> Fact 1
>>> The vertices of the cevian triangles of tF1 and tF2
>>> lie in a circle C whose center R is on line tF1tF2
>>> We know that L = X(20) is on this line.
>>>
>>> Conjecture 1 R = L.
>>
>> [BG] *** This is true.
>>
>> proof 1 :
>>
>> the cevian circle of a point P has center X(20)
>> iff P lies on 3 sextics,
>> iff tP lies on 3 cubics
>> iff ctP lies on 3 circum-cubics in a same pencil containing the
>> Thomson cubic K002 and the second Brocard cubic K018.
>>
>> These meet at 9 points A, B, C, G, K and the 4 foci of the Steiner
>> in- ellipse, etc
>
> [WS] This appears to suggest that P with ctP = G, K have the property
> that the cevian circle is centred at X(20). But here P = G, H..
>
> Am I missing something simple?
>
> The general theory is that there should be six points with
> cevian circles centred on a given point, so we'd expect two
> other than taF, F = focus of Steiner inellipse.

you're right. my proof is not precise enough and Thomson is just a
(misplaced) coincidence.

let us consider a point R and let us seek P such that the cevian
circle of P has center R.

the 3 circum-cubics as seen above are isogonal nKs with root at
infinity and they belong to a same pencil of isogonal nKs.

the point is that these cubics are ALL circular if and only if P = X
(20) in which case they meet at A, B, C, the 4 foci of the Steiner in-
ellipse and the circular points at infinity. This gives your six points.

if P is not X(20), the pencil contains only one circular nK and we
can find situations where the six points P (two by two cyclocevian
conjugates) are all real, hence six points ctP two by two isogonal
conjugates on each cubic of the pencil.

I will probably write a web page since the problem is interesting.


Best regards

Bernard



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

Dear All foci fans ...

Wonderful results Wilson ..

Various lines concerning the Steiner foci ...

f +/-, foci of the Steiner Inellipse
F +/-, foci of the Steiner Circumellipse
g = isogonal
t = isotomic
I+ = Steiner major axis at infinity = g X(1379)
I- = Steiner minor axis at infinity = g X(1380)
P = t g X(1379)

{tF-, tgtF-, tgF+}.
{tF+, tgtF+, tgF-}.

{tH, F-, tf+, tF-}.
{tH, F+, tf-, tF+}.

{f-, gtf-, X(1379)}.
{f+, gtf+, X(1379)}.

{gtF+, f+, gF+, O}.
{gtF-, f-, gF-, O}.

{gF-, gtF+, gtgF-, X(1379)}.
{gF+, gtF-, gtgF+, X(1379)}.

{F+, P, tF-, gF-}
{F-, P, tF+, gF+}

{F+, tH, tf-, tF+}
{F-, tH, tf+, tF-}

{tK, tf-, tf+, tgF-, tgF+, I+}.
{gtF-, gtf-, K, gtF+, gtf-, I+}, trilinear polar of X(1380).
{L, I+ ,tF-, tF+}
{G, X(1341), X(1348), X(2542), F+, F-, f+, f-, I+} .. major axis ..

Best regards,
Peter.

On Apr 1, 2006, at 11:03 PM, Jeff Brooks wrote:

> You may choose not to mention this one again, but I'm hanging onto it
> while thinking there might be something here.

Jeff, I definitely think there is something here. It is the basis for
my whole geometric theory!

Steve


Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm





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

Dear Steve,

Of course this is the basis for your geometric theory -- and a very
beautiful theory it is.

I am interested in learning more about the ancients views on
anallagmatic symmetry that you mentioned and the uses of your theory in
studying conjugates.

Sincerely, Jeff

[SS]
> Jeff, I definitely think there is something here. It is the basis
for
> my whole geometric theory!
>

Dear friends,
it is known that if ABC is a triangle and P is a
point in the plane of ABC not on the sides of ABC
then the isogonal conjugate of the circle PBC
is again a circle (the circle tPBC).
Hence the isogonal conjugate, of the circle IBC
that passes through Ia and has barycentric equation
aayz+bbzx+ccxy - (x+y+z)bcx = 0, is the same circle.
Similarly the isogonal conjugate of the circle IcBC
that passes through the point Ib and has equation
aayz+bbzx+ccxy + (x+y+z)bcx = 0 is the same circle.
So we have 6 invariant circles under isogonal conjugation
IBIaC, ICIbA, IAIcB, IcBCIb, IaCAIc, IbABIc.
Are there other circles with this property?

For points we know that there are only 4.
For lines we know that there are only 6.
For cubics there are infinite many.

Similarly is invariant under isogonal conjugation
a conic that passes through the points I, B, Ia, C
a conic that passes through the points I, C, Ib, A
a conic that passes through the points I, A, Ic, B
a conic that passes through the points Ic, B, C, Ib
a conic that passes through the points Ia, C, A, Ic
a conic that passes through the points Ib, A, B, Ia.
Of course these are infinite many but
Are there other such conics?

Best regards
Nikos Dergiades

On Apr 2, 2006, at 12:15 PM, peter_mows wrote:

> Dear All foci fans ...
>
> Wonderful results Wilson ..
>
> Various lines concerning the Steiner foci ...
>
> f +/-, foci of the Steiner Inellipse
> F +/-, foci of the Steiner Circumellipse
> g = isogonal
> t = isotomic
> I+ = Steiner major axis at infinity = g X(1379)
> I- = Steiner minor axis at infinity = g X(1380)
> P = t g X(1379)

I am adding things to this for my benefit. I am adding the new
results as well as why we know the statements are true.   G+ is taken
to be in the same direction as F+.   the "o" operation represents
reflection in the circumcircle. It is one of the two generating
operations of the Darboux cubic.


>
> {tF-, tgtF-, tgF+}.
> {tF+, tgtF+, tgF-}.

Don't know why this is true.
>
> {tH, F-, tf+, tF-}.      and tG+
> {tH, F+, tf-, tF+}.     and tG-

Some of this (D, F-, tF-)  comes from cubic properties.

>
> {f-, gtf-, X(1379)}.
> {f+, gtf+, X(1379)}.

gt is the projective operation that takes G to K and the Steiner
ellipse to the Circumcircle. This operation is ubiquitous in this
schema.

>
> {gtF+, f+, gF+, O}.      and oG-
> {gtF-, f-, gF-, O}.       and oG+

Dont know why.
>
> {gF-, gtF+, gtgF-, X(1379)}.
> {gF+, gtF-, gtgF+, X(1379)}.

dont know why

>
> {F+, P, tF-, gF-}
> {F-, P, tF+, gF+}

gt X1379 = P this line is gt of above line.

>
> {tK, tf-, tf+, tgF-, tgF+, I+}.

this very interesting results follows from two things. Jean-Pierre's
(I think) demonstration that the tF+ , tF- line is parallel to focal
axis. Also that gt of this line is the focal axis.

> {gtF-, gtf-, K, gtF+, gtf-, I+}, trilinear polar of X(1380).

gtG = K, this line is the gt of the focal axis and parallel to it
(perhaps)

> {L, I+ ,tF-, tF+}   add   G+, G-

  From group properties for two cubics

> {G, X(1341), X(1348), X(2542), F+, F-, f+, f-, I+} .. major
> axis ..    and infinite point
>

The x pts here are centers of similitude with the Brocard circle and
other circles.

Some others with the new points

G-   tG-    f+    H
G+  tG+   f-    H

oG+  oG-   H

f-  D   G-  gF-
f+  D   G+  gF+

G+ F+ dL
G- F-  dL





Steve Sigur

Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm

On Apr 2, 2006, at 2:59 PM, Jeff Brooks wrote:

> I am interested in learning more about the ancients views on
> anallagmatic symmetry that you mentioned and the uses of your
> theory in
> studying conjugates.

Jeff,

The ancients I refer to were 100 year old ancients. I leaned it from
the article by Mineur called Anallagmatic Cubics which is translated
from French on my web page. I think Mineur was greatly influenced by
Neuberg and deLongchamps.

Steve in pretty Atlanta



Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm





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

On Apr 2, 2006, at 4:07 PM, Nikolaos Dergiades wrote:

> Similarly the isogonal conjugate of the circle IcBC
> that passes through the point Ib and has equation
> aayz+bbzx+ccxy + (x+y+z)bcx = 0 is the same circle.
> So we have 6 invariant circles under isogonal conjugation
> IBIaC, ICIbA, IAIcB, IcBCIb, IaCAIc, IbABIc.
> Are there other circles with this property?

Hello Nakolaos,

I wrote an FG article that answered some of these questions a year or
so ago. The title was something like "Where are the conjugates?"

Your approach is very interesting.

Friendly,

Steve

Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm





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

Peter has used the gt and tg operations in his post so I want to say
a few things about those since they are not so well known.

John turned me onto them. He calls gt the "pro" operation (short for
"projective.") and the tg operation "retro."

They are projective and John likes them so much that he was thinking
of replacing the conjugate t and g operations with them, but decided
that, although desirable, it was too radical a change.

The pro operation is the projective operation  ABCG -> ABCK. It takes
the line GK to OK and the Steiner ellipse to the Circumcircle. Once
you start looking for it you see it everywhere. It preserves
prjective type so that points go to points, lines to lines, conics to
conics. It preserves strength.

I have a script in Sketchpad that draws a line from P to gtP so I can
move the mouse over the picture and see the effects of the operation.

There is a picture of the pro operation on my web site.

The invariant curve is what I call "the orbit of the incenter" and it
defines the path around which point patterns that I call "sweeps"
occur. On my website there is an article on the invariant curves of
projective operations called "The non-Euclidean Geometry of Euclidean
points."

In short I really like this operation since it taught me so much.

Applications to the Steiner foci picture.

pro takes (among many others)

tH to O,  G to K,   tK to G
P  to  x1379

t of anything to g of anything so that
tG+ to G-    and   tf+ to f-
tF+ tF- line to focal axis.

Steve



Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm

Dear Francois,

I agree with the pedal triangles of Gi and Cevian triangles of tFi
being the same, but I am lost on the inscribed triangles with pivots
Gi.  Can you please help me understand this?

This configuration is absolutely wonderful in so many ways!

Sincerely, Jeff
PS I hope you have not left yet for your trip to Tokyo and if you
have then I wish you the very best and look forward to hearing from
you when you return.



>
> On 4/1/06, Francois Rideau <francois.rideau@...> wrote:
> >
> >  Dear Steve
> >
> > I think I understand better the geometric situation.
> > I call a1b1c1 the cevian triangle of tF1 and a2b2c2 the cevian
triangle of
> > tF2.
> > Then after a possible change of notations, a1b1c1 is the pedal
triangle of
> > G1 and a2b2c2 the pedal triangle of G2.
> > G1 (resp G2) is the pivot center of inscribed triangles directly
similar
> > to a1b1c1 (resp a2b2c2).
> > The circumcircles of these inscribed triangles are bitangent to
the inner
> > conic of center L.
> > That's what gives me my chuffchuffs theory.
> > Can you check that?
> > Friendly
> > Francois
> > PS Here in france these circles are called focus circles of 2th
type.
> >
>
>
> [Non-text portions of this message have been removed]
>

Hello Steve,

> I wrote an FG article that answered some of these questions a year or
> so ago. The title was something like "Where are the conjugates?"

many thanks.

And what I wrote   "(the circle tPBC)"
must be corrected as "(the circle gPBC)"
Best regards
Nikos Dergiades

Dear Jeff
Here, I explain the situation:

For a conic with center, a bitangent circle is called focal circle ( cercle
focal in french).
Those centered on the major axis are of the first type and those centered on
the minor axis are of the second type. Focal circles of the second type are
similar between themselves in similarities where foci of the conic are
pivots (i.e fixed points).

In your configuration of this inner conic with foci G1 and G2 and  their
pedal triangles a1b1c1 and a2b2c2, also the cevian triangles of tF1 and tF2,
the circle a1b1c1a2b2c2 of center L is the principal circle, (cercle
principal in french), of the inner conic. Of course this  bitangent circle
has  both type for it is centered at L.
Now if abc is any inscribed triangle directly similar to a1b1c1 in a
similarity of center G1, the circle abc cuts again the sides of ABC in
a',b',c'and this new inscribed triangle a'b'c' is directly similar to a2b2c2
in a similarity of center G2 and the circle abca'b'c' is a focal circle of
the second type of the inner conic.
I must add a precision I have not given to Steve: Triangles abc and a'b'c'
define a(n?) homography on their common circumcircle. Then the axis of this
homography is just the contact chord of circle abca'b'c' with the inner
conic centered at L.

This geometric situation is very well described in Lalesco book, La
Géométrie du triangle, Chapitre 11, page 83 and next. I don't know if this
book was translated in English but it would deserve it despite some minor
errors and its old fashioned style. In fact, you can draw this configuration
beginning with any pair of isogonal points G1 and G2!

My chuffchuff theory is just an affine generalisation (metric is not needed)
of this situation where equicenters (affine concept created by Neuberg)
replace foci (metric concept). In this theory, I am interested in affine
bijections between sides of triangle ABC. Equicenters, inner and circum
conics, signed areas play a key role in it  as well as all the machinery of
Linear Algebra with trace, determinant, characteritic polynomials and all
the stuff!
   In Steve configuration, foci G1 and G2 were also equicenters but the
converse is false of course since in affine geometry there are no foci!

Yes, I am leaving for Tokyo if there is no general strike, next Wednesday,
for a 3 week break but I could have a glance at Hyacinthos occasionally and
I will try to visit sangakus in Tokyo area if any!
Friendly
François


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

Dear Nikolaos,

Isogonal conjugation (can be generalized) is a 2nd degree transform
with 3 singular points A, B, C.

It transforms a curve (C) of degree n into a curve (C*) of degree 2n
less the number of times (C) contains a singular point.

Hence, a general conic will give a circum-quartic. If you want a
conic, (C) must contain two distinct singular points (unless it is
degenerate).

Suppose (C) contains B and C. If you want (C) invariant under
isogonal transformation, it must contain either I and Ia or Ib and Ic.
This gives you two pencils of invariant conics say BC+ and BC- and
obviously 4 other similar pencils

BC+ : bcx^2+a^2yz + k bx(cy+bz) = 0
BC- : bcx^2-a^2yz + k bx(cy-bz) = 0 (k any real or infinity)

Each pencil contains :
- one circle hence all the conics have parallel axes,
- one rectangular hyperbola,
- two (real or not) parabolas,
- three decomposed conics (diagonals of the quadrangle)

Best regards

Bernard

> Dear friends,
> it is known that if ABC is a triangle and P is a
> point in the plane of ABC not on the sides of ABC
> then the isogonal conjugate of the circle PBC
> is again a circle (the circle tPBC).
> Hence the isogonal conjugate, of the circle IBC
> that passes through Ia and has barycentric equation
> aayz+bbzx+ccxy - (x+y+z)bcx = 0, is the same circle.
> Similarly the isogonal conjugate of the circle IcBC
> that passes through the point Ib and has equation
> aayz+bbzx+ccxy + (x+y+z)bcx = 0 is the same circle.
> So we have 6 invariant circles under isogonal conjugation
> IBIaC, ICIbA, IAIcB, IcBCIb, IaCAIc, IbABIc.
> Are there other circles with this property?
>
> For points we know that there are only 4.
> For lines we know that there are only 6.
> For cubics there are infinite many.
>
> Similarly is invariant under isogonal conjugation
> a conic that passes through the points I, B, Ia, C
> a conic that passes through the points I, C, Ib, A
> a conic that passes through the points I, A, Ic, B
> a conic that passes through the points Ic, B, C, Ib
> a conic that passes through the points Ia, C, A, Ic
> a conic that passes through the points Ib, A, B, Ia.
> Of course these are infinite many but
> Are there other such conics?
>
> Best regards
> Nikos Dergiades

The following paper has been published in Forum Geometricorum. It can be
viewed at

http://forumgeom.fau.edu/FG2006volume6/FG200613index.html

The editors
Forum Geometricorum
==================================
Wilson Stothers, Grassmann cubics and desmic structures,
Forum Geometricorum, 6 (2006) 117--138.

Abstract:  We show that each cubic of type nK which is not of type cK can
be described as a Grassmann cubic. The geometry associates with each such
cubic a cubic of type pK.  We call this the parent cubic. On the other
hand, each cubic of type pK has infinitely many child cubics. The key is
the existence of a desmic structure associated with parent and child. This
extends work of Wolk by showing that, not only do (some) points of a desmic
structure lie on a cubic, but also that they actually generate the cubic as
a locus. Along the way, we meet many familiar cubics.

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

Hello all in geometry,

I have always found the Tarry point mysterious. It is opposite the
Steiner point in the circumcircle and the intersection of the
circumcircle with the Kiepert hyperbola. Both of these are nice, if
disconnected properties.

I have a new construction which both explains the above and shows
that the Tarry point is one example of a universal phenomenon.

We begin as I always begin, with a single point in the plane of the
triangle. We will choose this point to be K, the symmedian point.

There are now two natural lines, the dual of K and G—K. Natural in
this context means that the structure is affine invariant. The
endpoints of these lines map to antipodal points on the Steiner ellipse.

These lines meet at P. The isotomic of the two lines are
circumconics, one, the Kiepert hyperbola, has its perspector tS at
infinity, the second has perspector K. The isotomic of P is the
intersection between these conics. This is the Tarry point.

The constuction shows that this is construction is general. Slightly
more analysis shows that S and T are opposite on the K circumconic
and that this too is general.

Friendly,

Steve




Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm