## Re: [EMHL] Center P-Orthogonal Hyperbola on Circumcircle P-Cevian Triangle

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• Dear Sung Hyun, I noticed something about the property you mention in the rectangular circumhyperbolas: Consider rectangular hyperbola (F) that passes through
Message 1 of 26 , Nov 8, 2011
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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]
>
• 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
Message 2 of 26 , Nov 9, 2011
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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 self-polar 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

[Non-text 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
Message 3 of 26 , Nov 9, 2011
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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 self-polar
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

[Non-text portions of this message have been removed]
• Dear Friends, ... I just found an addition to this property when I was strolling through older Hyacinthos messages. Ce = also on the P-Pedal Triangle Circle
Message 4 of 26 , Nov 9, 2011
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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.

I just found an addition to this property when I was strolling through older Hyacinthos messages.
Ce = also on the P-Pedal 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, ... orthojoin(X) is the orthopole of the trilinear polar of the isogonal conjugate of X. (see note above X(1512) in ETC), then (homogenous
Message 5 of 26 , Nov 10, 2011
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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^2x-c^2 SC*y - b^2SB*z) (SB*SC*x-a^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 self-polar 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
>
>
> [Non-text portions of this message have been removed]
>
• Dear Chris ... The first proof that I ve seen of this result is due to Bobillier in Les Annales de Gergonne (1828-1829). See page 356 at
Message 6 of 26 , Nov 10, 2011
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Dear Chris

> I just found an addition to this property when I was strolling through older Hyacinthos messages.
> Ce = also on the P-Pedal Triangle Circle
> (see Hyacinthos message 19260 dated Sep 13, 2010 from Linh Nguyen Van)
> What is the (synthetic) proof for this feature?

The first proof that I've seen of this result is due to Bobillier in "Les Annales de Gergonne"(1828-1829). See page 356 at
http://archive.numdam.org/ARCHIVE/AMPA/AMPA_1828-1829__19_/AMPA_1828-1829__19__349_0/AMPA_1828-1829__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 9P-circle 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 9P-circle if and only if O,P,P* are colinear :
the pedal circle of P touches the 9P-circle 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. Jean-Pierre
• Dear Chris I ve removed my previous message because a typo. ... Hyacinthos messages. ... A,B,C,P are 4 points of a rectangular hyperbola with center O;A B C
Message 7 of 26 , Nov 11, 2011
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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 P-Pedal 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 9P-circles of PAB
and PBC,so they pass by O)
=<C'OA' and the result

Friendly.
Jean-Pierre
• Dear Jean-Pierre 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
Message 8 of 26 , Nov 18, 2011
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Dear Jean-Pierre
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
<jean-pierre.ehrmann@...>wrote:

> **
>
>
> 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 P-Pedal 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 9P-circles
> of PAB
> and PBC,so they pass by O)
> =<C'OA' and the result
>
> Friendly.
> Jean-Pierre
>
>
>

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