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Re: [EMHL] Asymptotes of Kiepert hyperbola

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  • xpolakis@otenet.gr
    Dear Paul, ... Do we know any notable points lying on this circle with diameter the Fermats? APH
    Message 1 of 4 , Aug 21, 2000
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      Dear Paul,

      You wrote:

      > Talking about the center of the Kiepert hyperbola, I just recalled
      >that it is also the midpoint of the Fermat points.

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

      APH
    • xpolakis@otenet.gr
      Let s continue this interesting discussion ! On Wed, 29 Dec 1999, Bernard Gibert wrote: ... I have not this remarkable book and I don t know where could one
      Message 2 of 4 , Aug 30, 2000
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        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
      • xpolakis@otenet.gr
        ... Yes, they are. I proved it by Menelaus/Ceva Theorems. The theorem is: Let P, Q be two points on the plane of ABC, and A B C , A B C their cevian
        Message 3 of 4 , Aug 31, 2000
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          I wrote:

          >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)


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

          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
        • Jean-Pierre.EHRMANN
          ... De : À : Envoyé : vendredi 1 septembre 2000 02:10 Objet : Re: [EMHL] Asymptotes of Kiepert hyperbola Dear
          Message 4 of 4 , Sep 2, 2000
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            ----- 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:
            > 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")

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