- 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 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:

> **

[Non-text 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 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

>

>

>