- Dear Paul,

You wrote:

> Talking about the center of the Kiepert hyperbola, I just recalled

Do we know any notable points lying on this circle with diameter

>that it is also the midpoint of the Fermat points.

the Fermats?

APH - Let's continue this interesting discussion !

On Wed, 29 Dec 1999, Bernard Gibert wrote:

[APH]:

>>J. R. Musselman published in the AMM 45 (1938) 482, two problems

>>on Kiepert (and Jerabek) hyperbola:

>>

>>The pedal circle of the centroid G of a triangle A_1A_2A_3 passes through

>>the centers of the hyperbolas of Kiepert and Jerabek.

>

> Those two points are called 1st and 2nd Schroeter points : they both lie on

> the 9pt circle. (Schroeter 1865)

> They have quite a lot of interesting properties.

> Among them, the 2nd point is the fourth intersection point of the 9pt circle

> and the ellipse K.

>

> May I recommend the excellent book (in French) by Collet & Griso called " le

> cercle d'Euler " where a whole chapter is dedicated to them.

I have not this remarkable book and I don't know where could one obtain a

copy from (great French on-line bookstores, like chapitre, have no copies).

Scrhoeter construction of Jerabek/Kiepert H. centers is this - if I remember

correctly the construction I read once somewhere:

Let A'B'C', A"B"C" be the orthic, medial triangles of a triangle ABC. Denote:

B'C' /\ B"C" := A*

C'A' /\ C"A" := B*

A'B' /\ A"B" := C*

Then:

A'A*, B'B* , C'C* concur at Jerabek H. center

A"A*, B"B* , C"C* concur at Kiepert H. center

I believe that Scrhoeter studied the general case of two arbitrary (?) points

P,Q, and their cevian triangles A'B'C', A"B"C".

Are in general the triads of lines (A'A*, B'B* , C'C*), (A"A*, B"B* , C"C*)

concurrent? (A*,B*,C* as defined above)

Also, where the points of concurrences are lying on ? (in which pedal/cevian

circles?)

[If P = H, Q = I, then do we obtain Feuerbach H. center?]

Antreas - I wrote:

>Let's continue this interesting discussion !

Yes, they are. I proved it by Menelaus/Ceva Theorems.

>

>On Wed, 29 Dec 1999, Bernard Gibert wrote:

>

>[APH]:

>>>J. R. Musselman published in the AMM 45 (1938) 482, two problems

>>>on Kiepert (and Jerabek) hyperbola:

>>>

>>>The pedal circle of the centroid G of a triangle A_1A_2A_3 passes through

>>>the centers of the hyperbolas of Kiepert and Jerabek.

>>

>> Those two points are called 1st and 2nd Schroeter points : they both lie on

>> the 9pt circle. (Schroeter 1865)

>> They have quite a lot of interesting properties.

>> Among them, the 2nd point is the fourth intersection point of the 9pt circle

>> and the ellipse K.

>>

>> May I recommend the excellent book (in French) by Collet & Griso called " le

>> cercle d'Euler " where a whole chapter is dedicated to them.

>

>I have not this remarkable book and I don't know where could one obtain a

>copy from (great French on-line bookstores, like chapitre, have no copies).

>

>Scrhoeter construction of Jerabek/Kiepert H. centers is this - if I remember

>correctly the construction I read once somewhere:

>

>Let A'B'C', A"B"C" be the orthic, medial triangles of a triangle ABC. Denote:

>B'C' /\ B"C" := A*

>C'A' /\ C"A" := B*

>A'B' /\ A"B" := C*

>

>Then:

>

>A'A*, B'B* , C'C* concur at Jerabek H. center

>A"A*, B"B* , C"C* concur at Kiepert H. center

>

>I believe that Scrhoeter studied the general case of two arbitrary (?) points

>P,Q, and their cevian triangles A'B'C', A"B"C".

>Are in general the triads of lines (A'A*, B'B* , C'C*), (A"A*, B"B* , C"C*)

>concurrent? (A*,B*,C* as defined above)

The theorem is:

Let P, Q be two points on the plane of ABC, and A'B'C', A"B"C" their cevian

triangles. Denote:

B'C' /\ B"C" := A*

C'A' /\ C"A" := B*

A'B' /\ A"B" := C*

Then:

A'A*, B'B* , C'C* concur.

A"A*, B"B* , C"C* concur.

(or in other words: The triangle A*B*C* is in perspective with both

triangles A'B'C', A"B"C")

[The proof in my next posting, with subject line: Another Fruitful Theorem]

Does anyone know a reference?

Had Schroeter discovered the theorem, or just applied it for P = H, Q = G,

and found that the concurrence points are the centers of Jerabek/Kiepert H.?

Has anyone access to his (1865) paper ?

Antreas

>Also, where the points of concurrences are lying on ? (in which pedal/cevian

>circles?)

>[If P = H, Q = I, then do we obtain Feuerbach H. center?]

>

>Antreas - ----- Message d'origine -----

De : <xpolakis@...>

À : <Hyacinthos@egroups.com>

Envoyé : vendredi 1 septembre 2000 02:10

Objet : Re: [EMHL] Asymptotes of Kiepert hyperbola

Dear Antreas and other Hyacinthists,

Antreas wrote :> The theorem is:

cevian

> Let P, Q be two points on the plane of ABC, and A'B'C', A"B"C" their

> triangles. Denote:

We can add that A', B', C', A", B", C" and the two perspectors lie on a same

> B'C' /\ B"C" := A*

> C'A' /\ C"A" := B*

> A'B' /\ A"B" := C*

>

> Then:

> A'A*, B'B* , C'C* concur.

> A"A*, B"B* , C"C* concur.

> (or in other words: The triangle A*B*C* is in perspective with both

> triangles A'B'C', A"B"C")

conic;

if Dp and Dq are the trilinear polars of P and Q, this conic is the

locus of the poles of Dp wrt the circumconics going through Q and the locus

of the poles of Dq wrt the circumconics going through P.

This is just a projective generalization of the particular case P = G, Q = H

where the conic is the NPC and (the poles of Dp wrt the circumconics going

through Q = the centers of the rectangular circumhyperbolas)

If P =(x1,y1,z1) and Q = (x2,y2,z2), the conic is

x x/x1/x2 - (1/y1/z2 + 1/y2/z1) y z + circular = 0.

Friendly from France. Jean-Pierre