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Center POrthogonal Hyperbola on Circumcircle PCevian Triangle
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 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�ois
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
>
>
>
[Nontext 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
> François  On Sun, Nov 6, 2011 at 8:14 PM, Chris Van Tienhoven <van10hoven@...>wrote:
> **
[Nontext 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�ois
>
>
>
 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
[Nontext 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.
In "Applicationss d'analyse et de géométrie qui ont servi en 1822 de principal fondament au traite des proprietes projectives des figures" (MalletBachelier 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.
> 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.
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 (18131814)
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. JeanPierre  Dear friends
I wrote> In "Applicationss d'analyse et de géométrie qui ont servi en 1822 de principal fondament au traite des proprietes projectives des figures" (MalletBachelier 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
JeanPierre  Dear JeanPierre,
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 Pcevian 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 géométrie qui ont servi en 1822 de principal fondament au traite des proprietes projectives des figures" (MalletBachelier 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 (18131814)
>
> 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. JeanPierre
>  Dear JeanPierre
Is this just the theorem of Faure in case of the rectangular hyperbola?
Friendly
Fran�ois
On Mon, Nov 7, 2011 at 9:37 PM, Chris Van Tienhoven <van10hoven@...>wrote:
> **
>
>
> Dear JeanPierre,
>
> 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 Pcevian 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�om�trie qui ont servi en 1822 de
> principal fondament au traite des proprietes projectives des figures"
> (MalletBachelier 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 (18131814)
> >
> > 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. JeanPierre
> >
>
>
>
[Nontext portions of this message have been removed]  DearJeanPierre
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�ois
On Mon, Nov 7, 2011 at 10:01 PM, Francois Rideau
<francois.rideau@...>wrote:
> Dear JeanPierre
[Nontext portions of this message have been removed]
> Is this just the theorem of Faure in case of the rectangular hyperbola?
> Friendly
> Fran�ois
>
>
> On Mon, Nov 7, 2011 at 9:37 PM, Chris Van Tienhoven <van10hoven@...>wrote:
>
>> **
>>
>>
>> Dear JeanPierre,
>>
>> 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 Pcevian 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�om�trie qui ont servi en 1822 de
>> principal fondament au traite des proprietes projectives des figures"
>> (MalletBachelier 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
>> (18131814)
>> >
>> > 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. JeanPierre
>> >
>>
>>
>>
>
>
 Now if Gamma is a given rectangular hyperbola thru A, B, C and P a moving
point on it and let UVW be the Pcevian triangle wrt ABC.
The UVWcircumcircle is on the center O of Gamma but what is the envelope
of this circle when P moves on Gamma?
Friendly
Francois
[Nontext 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. JeanPierre  Dear Chris, JeanPierre and Francois!
> Let OH be the circumscribed orthogonal hyperbola of reference triangle ABC through random point P.
Synthetic proof of this theorem is in our book "Geometry of conics", AMS, 2007.
> Let Ce be the Center of OH.
> Proof that Ce is on the circumcircle of the cevian triangle of P wrt ABC.
>
Sincerely Alexey
[Nontext 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. JeanPierre  Dear Chris
> However the statement of Poncelet has the presumption that triangle UVW is selfpolar wrt the rectangular hyperbola.
It is exactly the same thing. I mean that
> In the case we are discussing this implies that the Pcevian triangle should be selfpolar wrt the rectangular hyperbola.
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. JeanPierre  Dear JeanPierre 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" <jeanpierre.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. JeanPierre
>  Dear Chris
> I am not so familiar with poles and polars.
See my previous message
> 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.
Theorem of Faure : If a triangle is selfpolar wrt a conic, its circumcircle is orthogonal to the Monge (or orthoptic) circle of the conic.
> I can't find it on internet. Somebody can tell me?
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^21, the Monge circle is the circle (O,root(a^2+b^2)
 for the hyperbola x^2/a^2y^2/b^21, the Monge circle is the circle (O,root(a^2b^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. JeanPierre  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.
JeanPierre  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
>
>
> [Nontext portions of this message have been removed]
>  Dear Angel,
Firstly a remark: F* is perspector of circumconic passing through P and Q,
and also is the intersection of tripolar of P and Q.
Secondly, I'll look into the calculations later (sorry but, would you mind
giving the formula for orthojoin transformation? what does it map (x:y:z)
into?) but I generalized my initial conjecture in the following form by
taking G and H as general points:
There are circumconic with perspector R (generalized rectangular hyperbola)
and a point G (generalized centroid). Let one of the [two intersections]
of [tripolar of G] and [circumconic with perspector P] be I (generalized
infinite point). Consider line connecting I and G/R (generalized
asymptote). Let this line cut the line BC at Pa. Consider intersection of
APa and [circumconic with perspector R]. (generalized P) Define Pb and Pc
by BP.CA and CP.AB. Then PbPc passes through I (generalized statement for
'parallel with asymptote')
I give a proof of the above statement:
Firstly note that G/P is pole of [tripolar of G] with respect to
[circumconic with perspetor R]. (directly verifiable by barycentric
calculation)
Hence the line IPa is the tangent at I. Therefore polar of Pa passes
through I. Also note that PaPbPc is selfpolar with respect to [circumconic
with perspector R] (because they are formed by intersections of diagonals
of quadrilateral ABCP). Therefore polar of Pa is PbPc. Hence PbPcI all lie
on polar of Pa. Hence PbPcI are collinear.
Regards,
Sung Hyun
[Nontext portions of this message have been removed]  Sorry, I made a typo. I send a revised version:
Dear Angel,
Firstly a remark: F* is perspector of circumconic passing through P and Q,
and also is the intersection of tripolar of P and Q.
Secondly, I'll look into the calculations later (sorry but, would you mind
giving the formula for orthojoin transformation? what does it map (x:y:z)
into?) but I generalized my initial conjecture in the following form by
taking G and H as general points:
There are circumconic with perspector R (generalized rectangular hyperbola)
and a point G (generalized centroid). Let one of the [two intersections]
of [tripolar of G] and [circumconic with perspector R] be I (generalized
infinite point). Consider line connecting I and G/R (generalized
asymptote). Let this line cut the line BC at Pa. Let intersection of APa
and [circumconic with perspector R] be P. Define Pb and Pc by BP.CA and
CP.AB. Then PbPc passes through I (generalized statement for 'parallel with
asymptote')
I give a proof of the above statement:
Firstly note that G/P is pole of [tripolar of G] with respect to
[circumconic with perspetor R]. (directly verifiable by barycentric
calculation, though I think there might be a simple geometric proof) Hence
the line IPa is the tangent at I (to the circumconic with perspector R).
Therefore polar of Pa passes through I. Also note that PaPbPc is selfpolar
with respect to [circumconic with perspector R] (because they are formed by
intersections of diagonals of quadrilateral ABCP). Therefore polar of Pa is
PbPc. Hence PbPcI all lie on polar of Pa. Hence PbPcI are collinear.
Regards,
Sung Hyun
[Nontext portions of this message have been removed]  Dear Friends,
> Let OH be the circumscribed orthogonal hyperbola of reference triangle ABC through random point P.
I just found an addition to this property when I was strolling through older Hyacinthos messages.
> Let Ce be the Center of OH.
> Proof that Ce is on the circumcircle of the cevian triangle of P wrt ABC.
Ce = also on the PPedal Triangle Circle
(see Hyacinthos message 19260 dated Sep 13, 2010 from Linh Nguyen Van)
What is the (synthetic) proof for this feature?
Best regards,
Chris van Tienhoven  Dear Sung Hyun,
> would you mind giving the formula for orthojoin transformation? what > does it map (x:y:z) into?
orthojoin(X) is the orthopole of the trilinear polar of the isogonal conjugate of X. (see note above X(1512) in ETC), then (homogenous barycentric coordinates):
(x:y:z) > ( b^2*c^2xc^2 SC*y  b^2SB*z) (SB*SC*xa^2*(SB*y + SC*z)): .... : ....)
If [tripolar of G]/\[circumconic with perspector R] is not empty!!!:
>There are circumconic with perspector R (generalized rectangular >hyperbola) >and a point G (generalized centroid). Let one of the [two >intersections] of [tripolar of G] and [circumconic with perspector R] >be I (generalized infinite point). Consider line connecting I and G/R >(generalized asymptote). Let this line cut the line BC at Pa. Let >intersection of APa and [circumconic with perspector R] be P. Define >Pb and Pc by BP.CA and CP.AB. Then PbPc passes through I (generalized >statement for 'parallel with asymptote')
J is the point (other than I) of the [two intersections] of [tripolar of G] and [circumconic with perspector R] be J (generalized
infinite point). Consider line connecting J and G/R (generalized
asymptote). Let this line cut the line BC at Qa. Let intersection of AQa and [circumconic with perspector R] be Q. Define Qb and Qc by BQ.CA and CQ.AB. Then QbQc passes through J (generalized statement for 'parallel with asymptote')
A'=PbPc /\ QbQc is the point of intersection of the tangent lines to circumconic in B and C. (i.e. if the points B' and C' are define in the same way, then the triangles ABC and A'B'C' are perspective, with perspector R).
Regards,
Angel Montesdeoca
 In Hyacinthos@yahoogroups.com, Sung Hyun Lim <progressiveforest@...> wrote:
>
> Dear Angel,
>
> Firstly a remark: F* is perspector of circumconic passing through P and Q,
> and also is the intersection of tripolar of P and Q.
>
> Secondly, I'll look into the calculations later (sorry but, would you mind
> giving the formula for orthojoin transformation? what does it map (x:y:z)
> into?) but I generalized my initial conjecture in the following form by
> taking G and H as general points:
>
> There are circumconic with perspector R (generalized rectangular hyperbola)
> and a point G (generalized centroid). Let one of the [two intersections]
> of [tripolar of G] and [circumconic with perspector P] be I (generalized
> infinite point). Consider line connecting I and G/R (generalized
> asymptote). Let this line cut the line BC at Pa. Consider intersection of
> APa and [circumconic with perspector R]. (generalized P) Define Pb and Pc
> by BP.CA and CP.AB. Then PbPc passes through I (generalized statement for
> 'parallel with asymptote')
>
> I give a proof of the above statement:
> Firstly note that G/P is pole of [tripolar of G] with respect to
> [circumconic with perspetor R]. (directly verifiable by barycentric
> calculation)
> Hence the line IPa is the tangent at I. Therefore polar of Pa passes
> through I. Also note that PaPbPc is selfpolar with respect to [circumconic
> with perspector R] (because they are formed by intersections of diagonals
> of quadrilateral ABCP). Therefore polar of Pa is PbPc. Hence PbPcI all lie
> on polar of Pa. Hence PbPcI are collinear.
>
> Regards,
> Sung Hyun
>
>
> [Nontext portions of this message have been removed]
>  Dear Chris
> I just found an addition to this property when I was strolling through older Hyacinthos messages.
The first proof that I've seen of this result is due to Bobillier in "Les Annales de Gergonne"(18281829). See page 356 at
> Ce = also on the PPedal Triangle Circle
> (see Hyacinthos message 19260 dated Sep 13, 2010 from Linh Nguyen Van)
> What is the (synthetic) proof for this feature?
http://archive.numdam.org/ARCHIVE/AMPA/AMPA_18281829__19_/AMPA_18281829__19__349_0/AMPA_18281829__19__349_0.pdf
Bobillier uses two facts :
 the reciprocal polar of a parabola wrt a circle centered on the directrix is a rectangular hyperbola
 if a parabola is inscribed in a triangle, the circumcircle of the triangle goes through the focus
This result gives the one sentence proof due to Fontené of the Feuerbach theorem :
As P and P*(isogonal of P) have the same pedal circle,the common points of the 9Pcircle with the pedal circle of P are the centers of the rectangular circumhyperbolae through P and P*; thus,if P is the incenter or one excenter, these common points are the same ones.
Same way,we have the generalization (Mac Cay or Griffith, I don't know) : the pedal circla of P touches the 9Pcircle if and only if O,P,P* are colinear :
the pedal circle of P touches the 9Pcircle if and only if the rectangular circumhyperbola through P goes through P*,which means that ABCHPP* lie on a conic,ie (with isogonality) that O,P,P* are colinear
Friendly. JeanPierre  Dear Chris
I've removed my previous message because a typo.
> I just found an addition to this property when I was strolling through older
Hyacinthos messages.
> Ce = also on the PPedal Triangle Circle
A,B,C,P are 4 points of a rectangular hyperbola with center O;A'B'C' is the
> (see Hyacinthos message 19260 dated Sep 13, 2010 from Linh Nguyen Van)
> What is the (synthetic) proof for this feature?
pedal triangle of P wrt ABC.
Proof of Bobillier (1828) (in short) :
Ci is a circle centered at P;f=inversion wrt Ci. The reciprocal polar of the
hyperbola wrt Ci is a parabola with focus f(P) (and directrix the tangent at P
to the hyperbola)
As the poles wrt Ci of the lines BC,CA,AB are f(A'),f(B'),f(C') and as the
circumcircle of a triangle circumscribing a parabola goes through the focus of the
parabola, it follows that f(O),f(A'),f(B'),f(C') are concyclic; so do
O,A',B',C'.
A proof with angle chasing :
I,J,K are the midpoints of BC,BA,BP
<C'B'A'=<C'B'P+<PB'A'=<C'AP+<PCA' (APB'C' and CPA'B' are concyclic)
= <C'JK+<KIA' = <C'OK+<KOA' (the circles C'JK and A'IK are the 9Pcircles of PAB
and PBC,so they pass by O)
=<C'OA' and the result
Friendly.
JeanPierre  Dear JeanPierre
l like your last remark about the condition for a self polar conic to be a
rectangular hyperbola.
I just find an analytic proof of this fact but do you know a synthetic one?
Friendly
Fran�ois
On Fri, Nov 11, 2011 at 11:55 AM, jpehrmfr
<jeanpierre.ehrmann@...>wrote:
> **
[Nontext portions of this message have been removed]
>
>
> Dear Chris
> I've removed my previous message because a typo.
>
>
> > I just found an addition to this property when I was strolling through
> older
> Hyacinthos messages.
> > Ce = also on the PPedal Triangle Circle
> > (see Hyacinthos message 19260 dated Sep 13, 2010 from Linh Nguyen Van)
> > What is the (synthetic) proof for this feature?
>
> A,B,C,P are 4 points of a rectangular hyperbola with center O;A'B'C' is the
> pedal triangle of P wrt ABC.
>
> Proof of Bobillier (1828) (in short) :
> Ci is a circle centered at P;f=inversion wrt Ci. The reciprocal polar of
> the
> hyperbola wrt Ci is a parabola with focus f(P) (and directrix the tangent
> at P
> to the hyperbola)
> As the poles wrt Ci of the lines BC,CA,AB are f(A'),f(B'),f(C') and as the
> circumcircle of a triangle circumscribing a parabola goes through the
> focus of the
>
> parabola, it follows that f(O),f(A'),f(B'),f(C') are concyclic; so do
> O,A',B',C'.
>
> A proof with angle chasing :
> I,J,K are the midpoints of BC,BA,BP
> <C'B'A'=<C'B'P+<PB'A'=<C'AP+<PCA' (APB'C' and CPA'B' are concyclic)
>
> = <C'JK+<KIA' = <C'OK+<KOA' (the circles C'JK and A'IK are the 9Pcircles
> of PAB
> and PBC,so they pass by O)
> =<C'OA' and the result
>
> Friendly.
> JeanPierre
>
>
>
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