Dear friends Hyacinthos:
in

http://personal.us.es/rbarroso/trianguloscabri/

�

you can see 628 and 629 issues of the journal Research

TRIANGULOSCABRI

that the course has in yeartwelve.

This time it's a line that Professor

Izquierdo Asensi
Formulas and Geometric Properties
�

http://www.casadellibro.com/libro-formulas-y-propiedades-geometricas/97884933668\
41/1098595

�

and generalization I have raised.

Best regards from Sevilla

�

Ricardo Barroso

http://personal.us.es/rbarroso/

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

Hello:

The 629 trianguloscabri is:

Let ABC be a triangle. Let� sa, sb and sc be lines tangent to a circle
concentric with the circumcricle at points of intersection with rays OA, OB and
OC.Call A*, B*, C* the
intersection points of sa, sb and scwith lines BC, CA, AB respectively. Then A*,
B* and C* are collinear.

Greetings.

Ricardo



________________________________
De: Ricardo Barroso <ricardobca@...>
Para: "Hyacinthos@yahoogroups.com" <Hyacinthos@yahoogroups.com>
Enviado: martes 1 de noviembre de 2011 9:42
Asunto: Hello from Sevilla


Dear friends Hyacinthos:
in

http://personal.us.es/rbarroso/trianguloscabri/

�

you can see 628 and 629 issues of the journal Research

TRIANGULOSCABRI

that the course has in yeartwelve.

This time it's a line that Professor

Izquierdo Asensi
Formulas and Geometric Properties
�

http://www.casadellibro.com/libro-formulas-y-propiedades-geometricas/97884933668\
41/1098595

�

and generalization I have raised.

Best regards from Sevilla

�

Ricardo Barroso

http://personal.us.es/rbarroso/

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

Dear Randy, this point was finally included in the ETC under X(3613). How can I
contact Clark Kimberkling to let him know more properties about X(3613)?

Thanks in advance
Luis

--- In Hyacinthos@yahoogroups.com, Randy Hutson <rhutson2@...> wrote:
>
> Luis, I had also found this center interesting and submitted it to Clark
Kimberling in August with the following description:
>
> Perspector of ABC and intersection of tangents to nine-point circle at
> ﾂ�ﾂ�ﾂ� intersections with sidelines (side-triangle of tangential triangles
> ﾂ�ﾂ�ﾂ� of medial and orthic triangles).ﾂ� Isotomic conjugate of X(1078).
> ﾂ�ﾂ�ﾂ� Trilinear function: bc/[(b^2c^2 + c^2a^2 + a^2b^2 - a^4)]
>
> Peter Moses subsequently found the following:
> Can be writtenﾂ�as X(311) + (1 - 2 Cos[2 w]) X(325)
> on lines {{4, 160}, {5, 141}, {53, 232}, {66, 2548}, {157, 3425}, {184, 2980},
{311, 325}, {1316, 3447}, {1485, 3148}}
> A' = {(b^2-c^2)^2,-a^2 b^2-a^2 c^2-b^2 c^2+c^4,-a^2 b^2+b^4-a^2 c^2-b^2 c^2}
>
> Kimberling is hoping to update ETC this month, so hopefully it will find its
way in there.ﾂ� You should send him your note about the pole of the Lemoine axis
wrt NPC, and line (232,427).
>
> Regards,
>
> Randy

Hi Luis,

Clark Kimberling's email is ck6@... (from the bottom of the 'Links'
section in ETC).ﾂ� I gave him your additional info on X(3613) being the pole of
the Lemoine axis wrt NPC, and also lying on line 232,427 (referencing your
Hyacinthos message # and name as Luis G -- I did not know your last name).ﾂ� So
far he has not included it, but he might if you contact him directly.ﾂ� He is
working on more updates to ETC which he hopes to get out this month.

Best regards,
Randy




>________________________________
>From: Luis <luisgeometria@...>
>To: Hyacinthos@yahoogroups.com
>Sent: Tuesday, November 1, 2011 2:40 PM
>Subject: Re: [EMHL] Isotomic conjugate of X(1078)
>
>
>ﾂ�
>
>
>Dear Randy, this point was finally included in the ETC under X(3613). How can I
contact Clark Kimberkling to let him know more properties about X(3613)?
>
>Thanks in advance
>Luis
>
>--- In Hyacinthos@yahoogroups.com, Randy Hutson <rhutson2@...> wrote:
>>
>> Luis, I had also found this center interesting and submitted it to Clark
Kimberling in August with the following description:
>>
>> Perspector of ABC and intersection of tangents to nine-point circle at
>> ﾃつ�ﾃつ�ﾃつ� intersections with sidelines (side-triangle of tangential
triangles
>> ﾃつ�ﾃつ�ﾃつ� of medial and orthic triangles).ﾃつ� Isotomic conjugate of
X(1078).
>> ﾃつ�ﾃつ�ﾃつ� Trilinear function: bc/[(b^2c^2 + c^2a^2 + a^2b^2 - a^4)]
>>
>> Peter Moses subsequently found the following:
>> Can be writtenﾃつ�as X(311) + (1 - 2 Cos[2 w]) X(325)
>> on lines {{4, 160}, {5, 141}, {53, 232}, {66, 2548}, {157, 3425}, {184,
2980}, {311, 325}, {1316, 3447}, {1485, 3148}}
>> A' = {(b^2-c^2)^2,-a^2 b^2-a^2 c^2-b^2 c^2+c^4,-a^2 b^2+b^4-a^2 c^2-b^2 c^2}
>>
>> Kimberling is hoping to update ETC this month, so hopefully it will find its
way in there.ﾃつ� You should send him your note about the pole of the Lemoine
axis wrt NPC, and line (232,427).
>>
>> Regards,
>>
>> Randy
>
>
>
>
>

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

Dear Hyacinthists,
an article concerning the 窶廢qual incircles theorem窶� has been put on my
website.
http://perso.orange.fr/jl.ayme ﾂ�ﾂ�ﾂ�vol. 20

Sincerely
Jean-Louis

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

Dear Jean-Louis,

incidentally, recently I've been doing problems similar (often, same)
to those in your paper and also came up with yours "4.  Un
quadrilatﾃｨre  circonscriptible" as a generalization of #5 (which
appeared in one of Russian national olympiads, I guess).
I'd like to draw your attention to the fact that #4 is, in turn, a
degenerate case of the following:

Let ABCD be a convex quad and let two points be taken on each of it's
sides. Joining the corresponding points on opposite sides, we divide
ABCD into 9 quads. Suppose that the corner quads are circumscriptible.
Then the central one is circumscriptible iff ABCD is circumscriptible.

This also is a slightly modified problem of one of old Russian
olympiads (from '80s, I think). I can't find an exact reference now,
but I suspect it was Igor Sharygin's (not surely). Your #4 emerges
from it by shrinking corner circles at, say, A and B into points and
pasting together the two dividing points on CD. In doing so, the proof
of the general problem turns into the proof of your #4 that you give.

Best regards,
Vladimir

2011/11/2 Jean-Louis Ayme <jeanlouisayme@...>
>
>
>
> Dear Hyacinthists,
> an article concerning the 窶廢qual incircles theorem窶� has been put on my
website.
> http://perso.orange.fr/jl.ayme ﾂ�ﾂ�ﾂ�vol. 20
>
> Sincerely
> Jean-Louis

Dear All My Friends,

The problem is 3.8 in page 59 of the book:
http://students.imsa.edu/~tliu/Math/planegeo.pdf

Best regards
Bui Quang Tuan


________________________________
From: Vladimir Dubrovsky <vndubrovsky@...>
To: Hyacinthos@yahoogroups.com
Sent: Wednesday, November 2, 2011 6:57 PM
Subject: Re: [EMHL] Equal incircles theorem

Let ABCD be a convex quad and let two points be taken on each of it's
sides. Joining the corresponding points on opposite sides, we divide
ABCD into 9 quads. Suppose that the corner quads are circumscriptible.
Then the central one is circumscriptible iff ABCD is circumscriptible.


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

Dear Hyacinthists,
the line  X(2)X(32), containing the points X(83), X(251), X(315), X(626),
X(754), also passes through the point Q of homogenous barycentric coordinates:


            Q=   ((a^2+b^2)(a^2+c^2)(b^4+c^4-a^2(b^2+c^2)):...:...).


   Q is the barycentric product of X(83) and X(325).




GEOMETRIC PROPERTY of the point Q:
---------------------------------


Given a triangle ABC and a point P in its plane, let DEF be the cevian triangle
of point P, A_b is the point on BC such that the triangles ADD_b and A_bDA are
similar (common vertex D),  and the  point A_c on BC such that the triangles
ADD_c and A_cDA are similar (common vertex D).

Define B_c and B_a (on CA),  C_a and C_b (on AB) cyclically.


Consider the triangle A'B'C' of vertices  A'=A_bB_a/\A_cC_a,
B'=B_cC_b/\B_aA_b,and C'=C_aA_c/\C_bB_c.

Then, the triangles ABC and A'B'C' are perspective (or equivalently, the points
A_b, A_c, B_c, B_a, C_a and C_b are on the same conic) if P is on the
circumcircle of ABC or P is on the quartic (C) of barycentric equation


(C):   Cyclic sum
y*z[(b^2-c^2)(a^2+b^2+c^2)x^2-2SC(a^2+c^2)y^2-2SA(b^2-c^2)y*z+2SB(a^2+b^2)z^2] =
0.


SA=(b^2+c^2-a^2)/2, ....


The quartic (C) is a  circumquartic through G (point of inflection) and H.

The another intersection Q of the quartic (C)  with the tangent at G, have
homogenous barycentric coordinates

       Q=  ((a^2+b^2)(a^2+c^2)(b^4+c^4-a^2(b^2+c^2)):...:...)


=======

Remark:  The complement of the quartic (C) has a simpler expression (Bernard
Gibert, personal communication):

   Cyclic sum [yz((c^2-b^2)(b^2+c^2-a^2)x^2+a^2(y-z)(c^2y+b^2z))] = 0.



Sincerely

Angel Montesdeoca

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

http://forumgeom.fau.edu/FG2011volume11/FG201123index.html

The editors
Forum Geometricorum
Martin Josefsson, The area of the diagonal point trinangle,
Forum Geometricorum, 11 (2011) 213--216.

Abstract.   In this note we derive a formula for the area of the diagonal point
triangle belonging to a cyclic quadrilateral in terms of the sides of the
quadrilateral, and prove a characterization of trapezoids.


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

Dear Friends,

Let OH be the circumscribed orthogonal hyperbola of reference triangle ABC
through random point P.
Let Ce be the Center of OH.
Proof that Ce is on the circumcircle of the cevian triangle of P wrt ABC.
This is actually a generalization of the theorem where is stated that the center
of an orthogonal hyperbola is on the Ninepoint Circle. In that case P is the
Centroid.
I can't find a synthetic solution.
Best regards,

Chris van Tienhoven

Dear Chris
Maybe, you shoud say that P is the orthocenter?
Friendly
Fran輟is


On Sun, Nov 6, 2011 at 9:00 AM, Chris Van Tienhoven <van10hoven@...>wrote:

> **
>
>
> Dear Friends,
>
> Let OH be the circumscribed orthogonal hyperbola of reference triangle ABC
> through random point P.
> Let Ce be the Center of OH.
> Proof that Ce is on the circumcircle of the cevian triangle of P wrt ABC.
> This is actually a generalization of the theorem where is stated that the
> center of an orthogonal hyperbola is on the Ninepoint Circle. In that case
> P is the Centroid.
> I can't find a synthetic solution.
> Best regards,
>
> Chris van Tienhoven
>
>
>


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

Dear Francois,

The orthogonal hyperbola I mean is defined by A,B,C,H,P.
H = Orthocenter, P = random point unequal A,B,C,H.
Now the Center of this Orthogonal Hyperbola lies on the circumcircle of the
cevian triangle of P wrt triangle ABC.
Best regards,

Chris


>
> Dear Chris
> Maybe, you shoud say that P is the orthocenter?
> Friendly
> Fran輟is

Dear Alexey and triangle geometers,

Alexey has discovered that:
Let the incircle of triangle ABC touche its sides in A', B', C'; I and G
are the incenter and the Gergonne point. Then the hyperbolas ABCIG and
A'B'C'IG are homothetic.

I made a generalization:
For point X, consider Y=barycentric product of X and [anticomplement of X].
Then [circumconic through X and Y] and [diagonal conic through X and Y]
share same points of infinity, and hence are homothetic.

And just yesterday I found a further projective generalization to this
property:
If a circumconic and diagonal conic intersect at X1 X2 X3 X4, then tripolar
of cevian product of X1 and X2 is line X3X4.

This seems to make a remarkable connection with the following known
properties:
If X and Y lies on a diagonal conic, then their cevian product is the pole
of line XY.
If X and Y are conjugates with respect to a circumconic, then their cevian
product lies on the circumconic.

Meanwhile, I'm not exactly sure how Alexey's generalization works:

This result can be generalised as next projective theorem.
> Let triangles ABC and A'B'C' be polar wrt conic k and P be their
> perspective center. Consider two conics passing through A, B, C, P and A',
> B', C', P respectively. Then the remaining common points of these conics
> form the triangle autopolar wrt k.
> Partially two conics passing through the center of k are homothetic. If
> also k is the circle then these conics are equilateral hyperbolas.
> V.Thebault indicate this fact in the paper concerned to Sondat theorem, but
> I suppose that it was known earlier. May be it was found by Sollertinsky?


Do you mind explaining, Alexey?

Sung Hyun


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

On Sun, Nov 6, 2011 at 8:14 PM, Chris Van Tienhoven <van10hoven@...>wrote:

> **
>
>
> Dear Francois,
>
> The orthogonal hyperbola I mean is defined by A,B,C,H,P.
> H = Orthocenter, P = random point unequal A,B,C,H.
> Now the Center of this Orthogonal Hyperbola lies on the circumcircle of
> the cevian triangle of P wrt triangle ABC.
> Best regards,
>
> Chris
>
>
> >
> > Dear Chris
> > Maybe, you shoud say that P is the orthocenter?
> > Friendly
> > Fran輟is
>
>
>


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

Dear all,

I don't think this helps a lot, but I discovered a slightly related
conjecture to Chris' problem:

consider rectangular hyperbola F that passes through triangle ABC. Take
point P from F, and let PaPbPc be its cevian triangle. if Pa lies on an
asymptote of F, then PbPc is parallel to that asymptote.

Sung Hyun


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

Dear Chris and Francois
[Chris]
> Let OH be the circumscribed orthogonal hyperbola of reference triangle ABC
through random point P.
> Let Ce be the Center of OH.
> Proof that Ce is on the circumcircle of the cevian triangle of P wrt ABC.
> This is actually a generalization of the theorem where is stated that the
center of an orthogonal hyperbola is on the Ninepoint Circle. In that case P is
the Centroid.
> I can't find a synthetic solution.

In "Applicationss d'analyse et de g駮m騁rie qui ont servi en 1822 de principal
fondament au traite des proprietes projectives des figures" (Mallet-Bachelier
Paris 1862), Poncelet states and proves the following result : if a triangle UVW
is selfpolar wrt a rectangular hyperbola, its circumcircle goes through the
center O of the hyperbola.

I summarize the proof :
As the conjugate diameter of the direction of a line goes through the pole of
the line,
- the parallel at U to VW intersects the asymptots at E,E' with midpoint U
- the parallel at V to UW intersects the asymptots at F,F' with midpoint V
clearly, UO=UE and VO=VF,thus
<UWV = <VFO + <OEU = <FOV + <UOE = <UOV and the result

Poncelet discovered this result when he was in jail in Russia (1813-1814)

Personal remark : As the 6 vertices of two autopolar triangles lie on a conic,
UVWOIJ lie on a conic (I,J=circular points at infinity). This gives the same
result

Friendly. Jean-Pierre

Dear friends
I wrote
> In "Applicationss d'analyse et de g駮m騁rie qui ont servi en 1822 de principal
fondament au traite des proprietes projectives des figures" (Mallet-Bachelier
Paris 1862), Poncelet states and proves the following result : if a triangle UVW
is selfpolar wrt a rectangular hyperbola, its circumcircle goes through the
center O of the hyperbola.

See Page 508 Theorem IV in
http://gallica.bnf.fr/ark:/12148/bpt6k902142.r=.langFR
Jean-Pierre

Dear Jean-Pierre,

Thanks for the very nice proof and reference.
I noticed earlier that Poncelet and Brianchon did some remarkable findings.
Probably their time in prison gave them lots of time for doing their pioneering
work.
However the statement of Poncelet has the presumption that triangle UVW is
selfpolar wrt the rectangular hyperbola.
In the case we are discussing this implies that the P-cevian triangle should be
selfpolar wrt the rectangular hyperbola. I checked in Cabri that this is true
indeed.
Can you tell me what the theory is behind this? Probably it is obvious for you.
Best regards,

Chris van Tienhoven



> [JPE]
> In "Applicationss d'analyse et de g駮m騁rie qui ont servi en 1822 de principal
fondament au traite des proprietes projectives des figures" (Mallet-Bachelier
Paris 1862), Poncelet states and proves the following result : if a triangle UVW
is selfpolar wrt a rectangular hyperbola, its circumcircle goes through the
center O of the hyperbola.
>
> I summarize the proof :
> As the conjugate diameter of the direction of a line goes through the pole of
the line,
> - the parallel at U to VW intersects the asymptots at E,E' with midpoint U
> - the parallel at V to UW intersects the asymptots at F,F' with midpoint V
> clearly, UO=UE and VO=VF,thus
> <UWV = <VFO + <OEU = <FOV + <UOE = <UOV and the result
>
> Poncelet discovered this result when he was in jail in Russia (1813-1814)
>
> Personal remark : As the 6 vertices of two autopolar triangles lie on a conic,
UVWOIJ lie on a conic (I,J=circular points at infinity). This gives the same
result
>
> Friendly. Jean-Pierre
>

Dear Jean-Pierre
  Is this just the theorem of Faure in case of the rectangular hyperbola?
Friendly
Fran輟is

On Mon, Nov 7, 2011 at 9:37 PM, Chris Van Tienhoven <van10hoven@...>wrote:

> **
>
>
> Dear Jean-Pierre,
>
> Thanks for the very nice proof and reference.
> I noticed earlier that Poncelet and Brianchon did some remarkable
> findings. Probably their time in prison gave them lots of time for doing
> their pioneering work.
> However the statement of Poncelet has the presumption that triangle UVW is
> selfpolar wrt the rectangular hyperbola.
> In the case we are discussing this implies that the P-cevian triangle
> should be selfpolar wrt the rectangular hyperbola. I checked in Cabri that
> this is true indeed.
> Can you tell me what the theory is behind this? Probably it is obvious for
> you.
> Best regards,
>
> Chris van Tienhoven
>
> > [JPE]
>
> > In "Applicationss d'analyse et de g駮m騁rie qui ont servi en 1822 de
> principal fondament au traite des proprietes projectives des figures"
> (Mallet-Bachelier Paris 1862), Poncelet states and proves the following
> result : if a triangle UVW is selfpolar wrt a rectangular hyperbola, its
> circumcircle goes through the center O of the hyperbola.
> >
> > I summarize the proof :
> > As the conjugate diameter of the direction of a line goes through the
> pole of the line,
> > - the parallel at U to VW intersects the asymptots at E,E' with midpoint
> U
> > - the parallel at V to UW intersects the asymptots at F,F' with midpoint
> V
> > clearly, UO=UE and VO=VF,thus
> > <UWV = <VFO + <OEU = <FOV + <UOE = <UOV and the result
> >
> > Poncelet discovered this result when he was in jail in Russia (1813-1814)
> >
> > Personal remark : As the 6 vertices of two autopolar triangles lie on a
> conic, UVWOIJ lie on a conic (I,J=circular points at infinity). This gives
> the same result
> >
> > Friendly. Jean-Pierre
> >
>
>
>


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

DearJean-Pierre
I think we have the following projective theorem:
The cevian triangle UVW of P wrt ABC is self polar wrt every conic through
the four points A, B, C, P.
So by the Faure theorem, the UVW circumcircle is on the center of the
rectangular hyperbola thru A, B, C, PN
Is that correct?
Friendly
Fran輟is

On Mon, Nov 7, 2011 at 10:01 PM, Francois Rideau
<francois.rideau@...>wrote:

> Dear Jean-Pierre
>  Is this just the theorem of Faure in case of the rectangular hyperbola?
> Friendly
> Fran輟is
>
>
> On Mon, Nov 7, 2011 at 9:37 PM, Chris Van Tienhoven
<van10hoven@...>wrote:
>
>> **
>>
>>
>> Dear Jean-Pierre,
>>
>> Thanks for the very nice proof and reference.
>> I noticed earlier that Poncelet and Brianchon did some remarkable
>> findings. Probably their time in prison gave them lots of time for doing
>> their pioneering work.
>> However the statement of Poncelet has the presumption that triangle UVW
>> is selfpolar wrt the rectangular hyperbola.
>> In the case we are discussing this implies that the P-cevian triangle
>> should be selfpolar wrt the rectangular hyperbola. I checked in Cabri that
>> this is true indeed.
>> Can you tell me what the theory is behind this? Probably it is obvious
>> for you.
>> Best regards,
>>
>> Chris van Tienhoven
>>
>> > [JPE]
>>
>> > In "Applicationss d'analyse et de g駮m騁rie qui ont servi en 1822 de
>> principal fondament au traite des proprietes projectives des figures"
>> (Mallet-Bachelier Paris 1862), Poncelet states and proves the following
>> result : if a triangle UVW is selfpolar wrt a rectangular hyperbola, its
>> circumcircle goes through the center O of the hyperbola.
>> >
>> > I summarize the proof :
>> > As the conjugate diameter of the direction of a line goes through the
>> pole of the line,
>> > - the parallel at U to VW intersects the asymptots at E,E' with
>> midpoint U
>> > - the parallel at V to UW intersects the asymptots at F,F' with
>> midpoint V
>> > clearly, UO=UE and VO=VF,thus
>> > <UWV = <VFO + <OEU = <FOV + <UOE = <UOV and the result
>> >
>> > Poncelet discovered this result when he was in jail in Russia
>> (1813-1814)
>> >
>> > Personal remark : As the 6 vertices of two autopolar triangles lie on a
>> conic, UVWOIJ lie on a conic (I,J=circular points at infinity). This gives
>> the same result
>> >
>> > Friendly. Jean-Pierre
>> >
>>
>>
>>
>
>


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

Now if Gamma is a given rectangular hyperbola thru A, B, C and P a moving
point on it and let UVW be the P-cevian triangle wrt ABC.
The UVW-circumcircle is on the center O of Gamma but what is the envelope
of this circle when P moves on Gamma?
Friendly
Francois


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

Dear Francois

>  Is this just the theorem of Faure in case of the rectangular hyperbola?

Yes (the orthoptic circle of a rectangular hyperbola is reduced to the center)
But, the Faure's theorem is about 1864 and Poncelet discovered his result about
1814 and his proof is very clever.
More over, I'm not sure that Faure has published a proof of his result : it
appears as a question in Nouvelles Annales in 1861.
Of course, this leads to the fact that the cevian circle of P goes through the
center of the rectangular circumhyperbola going through P (the cevian triangle
of P is clearly selfpolar wrt the hyperbola)
Friendly. Jean-Pierre

Dear Chris, Jean-Pierre and Francois!

> Let OH be the circumscribed orthogonal hyperbola of reference triangle ABC
through random point P.
> Let Ce be the Center of OH.
> Proof that Ce is on the circumcircle of the cevian triangle of P wrt ABC.
>
Synthetic proof of this theorem is in our book "Geometry of conics", AMS, 2007.

Sincerely                                     Alexey

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

Dear Sung Hyun!

Me result is the generalization of Thebault theorem and can be reduced to it by
projective map. But I don't know  can it be used for the proof of your result
because I don't understand the projective interpretation of cevian product.

Sincerely                                Alexey

And just yesterday I found a further projective generalization to this
property:
If a circumconic and diagonal conic intersect at X1 X2 X3 X4, then tripolar
of cevian product of X1 and X2 is line X3X4.

This seems to make a remarkable connection with the following known
properties:
If X and Y lies on a diagonal conic, then their cevian product is the pole
of line XY.
If X and Y are conjugates with respect to a circumconic, then their cevian
product lies on the circumconic.

Meanwhile, I'm not exactly sure how Alexey's generalization works:

This result can be generalised as next projective theorem.
> Let triangles ABC and A'B'C' be polar wrt conic k and P be their
> perspective center. Consider two conics passing through A, B, C, P and A',
> B', C', P respectively. Then the remaining common points of these conics
> form the triangle autopolar wrt k.
> Partially two conics passing through the center of k are homothetic. If
> also k is the circle then these conics are equilateral hyperbolas.
> V.Thebault indicate this fact in the paper concerned to Sondat theorem, but
> I suppose that it was known earlier. May be it was found by Sollertinsky?



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

Dear Chris and Francois
Just a remark :
if ABC is selfpolar wrt a rectangular hyperbola, the hyperbola is member of the
pencil of conics going through the incenter and the 3 excenters. So, the center
of the hyperbola is on the circumcircle of ABC.
Friendly. Jean-Pierre

Dear Chris
> However the statement of Poncelet has the presumption that triangle UVW is
selfpolar wrt the rectangular hyperbola.
> In the case we are discussing this implies that the P-cevian triangle should
be selfpolar wrt the rectangular hyperbola.

It is exactly the same thing. I mean that
A triangle UVW is the cevian triangle of a point of a conic wrt three other ones
if and only if UVW is selfpolar wrt the conic.
=> If you consider 4 points ABCD upon the conic, the classical construction of
the polar of a point wrt the conic shows that the cevian triangle of D wrt ABC
is selfpolar
<= Take any point D upon the conic; if ABC is the precevian triangle of D wrt
UVW, then A,B,C lie on the conic (the equation of the conic with UVW as triangle
of reference is pxx+qyy+rzz = 0)

Friendly. Jean-Pierre

Dear Jean-Pierre and Francois,

I am not so familiar with poles and polars.
I just know the definitions and some theorems.
Interesting enough for me to study.
Is it possible to state in a simple way why the cevian triangle of P is
selfpolar wrt the rectangular hyperbola?
And of course I would like to know what the theorem of Faure says.
I can't find it on internet. Somebody can tell me?

Best regards,

Chris van Tienhoven


--- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-pierre.ehrmann@...> wrote:
>
> Dear Francois
>
> >  Is this just the theorem of Faure in case of the rectangular hyperbola?
>
> Yes (the orthoptic circle of a rectangular hyperbola is reduced to the center)
> But, the Faure's theorem is about 1864 and Poncelet discovered his result
about 1814 and his proof is very clever.
> More over, I'm not sure that Faure has published a proof of his result : it
appears as a question in Nouvelles Annales in 1861.
> Of course, this leads to the fact that the cevian circle of P goes through the
center of the rectangular circumhyperbola going through P (the cevian triangle
of P is clearly selfpolar wrt the hyperbola)
> Friendly. Jean-Pierre
>

Dear Chris

> I am not so familiar with poles and polars.
> I just know the definitions and some theorems.
> Interesting enough for me to study.
> Is it possible to state in a simple way why the cevian triangle of P is
selfpolar wrt the rectangular hyperbola?

See my previous message

> And of course I would like to know what the theorem of Faure says.
> I can't find it on internet. Somebody can tell me?


Theorem of Faure : If a triangle is selfpolar wrt a conic, its circumcircle is
orthogonal to the Monge (or orthoptic) circle of the conic.
The Monge circle of a conic is the locus of the points M for which the tangents
from M to the conic are perpendicular.
In rectangular coordinates
- for the ellipse x^2/a^2+y^2/b^2-1, the Monge circle is the circle
(O,root(a^2+b^2)
- for the hyperbola x^2/a^2-y^2/b^2-1, the Monge circle is the circle
(O,root(a^2-b^2) not necessarily real. For a rectangular hyperbola, the Monge
circle is reduced to the center (This gives the Poncelet theorem)
- for a parabola, the Monge circle is the directrix (So, if a triangle is
selfpolar wrt a parabola, its circumcenter lies on the directrix)

There exists a  reciprocal of the theorem of Faure : if a circle is orthogonal
to the Monge circle of a conic, there exists infinitely many triangles inscribed
in the circle and selfpolar wrt the conic

More generally, consider two conics C1 and C2; if there exists a triangle
inscribed in C1 and selfpolar wrt C2, there exists infinitely many such
triangles. If A and B are the matrices of C1 and C2 in any system of projective
coordinates, this happens if and only if trace(A^(-1)B)=0

Friendly. Jean-Pierre

Dear Friends
I wrote
> More generally, consider two conics C1 and C2; if there exists a triangle
inscribed in C1 and selfpolar wrt C2, there exists infinitely many such
triangles. If A and B are the matrices of C1 and C2 in any system of projective
coordinates, this happens if and only if trace(A^(-1)B)=0

The condition is trace(A.B^(-1)) = 0
I apologize for the confusion.
Jean-Pierre

Dear Sung Hyun,

I noticed something about the property you mention in the rectangular
circumhyperbolas:



Consider rectangular hyperbola (F) that passes through triangle ABC.  If p and q
are the asymptotes of F,  we consider the points Pa = p /\ BC and Qa = q /\ BC.

  Let P be the another intersection of the rectangular hyperbola (F) with the
line APa, and  let PaPbPc be the cevian triangle of P, then PbPc is parallel to
that asymptote p.

Let Q be the another intersection of the rectangular hyperbola (F) with the line
AQa, and  let QaQbQc be the cevian triangle of Q, then QbQc is parallel to that
asymptote q.

Let A' be the point A'= PbPc /\ QbQc. Similarly we define the points B', C'.
Then the triangles ABC and A'B'C' are perspective. The perspector F* of ABC and
A'B'C' lies on line X(230)X(231), and the orthojoin(F*) is the center of the
rectangular hyperbola (F).

( orthojoin(X) is the orthopole of the trilinear polar of the isogonal conjugate
of X,  as indicated just before X(1512) in ETC ).



In particular:

If (F) is the Feuerbach hyperbola (center X(11)), then F*=X(650).

If (F) is the Jerabek hyperbola (center X(125)), then F*=X(647).

If (F) is the Kiepert hyperbola (center X(115)), then F*=X(523).

If (F) is the X(5)- rectangular circumhyperbola (center X(137)), then F* (search
value 0.8276910284).

If (F) is the X(32)- rectangular circumhyperbola (center X(2679)), then
F*=X(2491).



Angel Montesdeoca

--- In Hyacinthos@yahoogroups.com, Sung Hyun Lim <progressiveforest@...> wrote:
>
> Dear all,
>
> I don't think this helps a lot, but I discovered a slightly related
> conjecture to Chris' problem:
>
> consider rectangular hyperbola F that passes through triangle ABC. Take
> point P from F, and let PaPbPc be its cevian triangle. if Pa lies on an
> asymptote of F, then PbPc is parallel to that asymptote.
>
> Sung Hyun
>
>
> [Non-text portions of this message have been removed]
>