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Re: [EMHL] Re: Bilogic locus

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  • Bernard Gibert
    Dear François, ... please take a look at http://perso.orange.fr/bernard.gibert/Exemples/k003.html property 6 and also
    Message 1 of 10 , Oct 8, 2006
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      Dear François,

      > [FR] In my previous post, I ask:
      > What is the locus (if any) of the point P such that if lines AP,
      > BP, CP cut
      > again the ABC-circumcircle respectively in A', B', C', then
      > triangles ABC
      > and A'B'C' are orthologic ?
      > Of course as they are already perspective wrt perspector P, they
      > will be
      > also bilogic.
      >
      > I don't know yet what is this locus but I feel like this is also
      > the locus
      > of points P such that the pedal and circumcevian triangles of P wrt
      > ABC are
      > homothetic. Besides line at infinity must be part of the locus if
      > we look at
      > ABC and A'B'C' symmetric wrt some diameter of the circumcircle

      please take a look at

      http://perso.orange.fr/bernard.gibert/Exemples/k003.html
      property 6

      and also

      http://perso.orange.fr/bernard.gibert/Tables/table7.html

      Best regards

      Bernard



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    • Francois Rideau
      Dear friends In fact, I think this locus is well known as the isocubic with pivot the circumcenter. That s why the drawing is not so easy even with Cabri!
      Message 2 of 10 , Oct 8, 2006
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        Dear friends
        In fact, I think this locus is well known as the isocubic with pivot the
        circumcenter.
        That's why the drawing is not so easy even with Cabri!
        Friendly
        François


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      • Francois Rideau
        Thanks At the last moment, I remembered it was this cubic but I forgot the name. So it is the Mac Kay or Griffits cubic, good to know that! As for the
        Message 3 of 10 , Oct 8, 2006
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          Thanks
          At the last moment, I remembered it was this cubic but I forgot the name. So
          it is the Mac Kay or Griffits cubic, good to know that!
          As for the orthology centers, are you sure they are both on the Mac kay
          butterfly, (another species in danger at least in France) ? It seems to me
          that one of them must be the isogonal of the perspector and so also on the
          cubic?
          Friendly
          François


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        • Francois Rideau
          Dear Bernard I just check the MacKay drawing with Cabri. Let S be any point on this cubic and let A B C be the circumcevian triangle of S wrt ABC. Then we
          Message 4 of 10 , Oct 8, 2006
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            Dear Bernard
            I just check the MacKay drawing with Cabri.
            Let S be any point on this cubic and let A'B'C' be the circumcevian triangle
            of S wrt ABC. Then we know that ABC and A'B'C' are orthologic with orthology
            centers U and U'.
            Then U is conjugate isogonal with S as I said it and so also on the MacKay
            cubic and U' is on the butterfly. I use the tool Trace to get the locus of
            U' and I saw a beautiful butterfly flying high on the big blue screen.

            Notice that the perspectrix of ABC and A'B'C' is the polar of S wrt the
            circumcircle, hence orthogonal to OS , (where O is the circumcenter) as we
            already know it , courtesy of Sondat for S, O, U, U' are on a same line.
            Friendly
            François

            PS
            For Cabri users, here the way to get the MacKay cubic.
            1° Choose any point M on the circumcircle.
            2° Draw line OM
            3° Draw the rectangular hyperbola isogonal of line OM. To get it, it
            suffices to construct its intersection N with the circumcircle as the
            isogonal conjugate of the point at infinity of line OM. Then use the conic
            tool to get the hyperbola through A, B, C, N, H where H is the orthocenter.
            4° Draw the 2 intersections of the line OM with the hyperbola.
            5° Draw their locus when M moves on the circumcircle with the locus tool.

            After that, it is easy to check what I have said after choosing any point S
            on the MacKay cubic and so one...
            I wonder if GSP can do the same task? Please, overseas, let me know about
            it!


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          • Bernard Gibert
            Dear friends, let Q be a point and A B C its anticevian triangle. the perpendicular at Q to QA meets BC at U and V, W likewise. A1 = BV / CW and B1, C1
            Message 5 of 10 , Oct 10, 2006
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              Dear friends,

              let Q be a point and A'B'C' its anticevian triangle.

              the perpendicular at Q to QA meets BC at U and V, W likewise.

              A1 = BV /\ CW and B1, C1 likewise.

              the conic inscribed in both triangles A'B'C' and A1B1C1 has focus Q.

              I don't see any good synthetic reason for that.


              Best regards

              Bernard

              PS : naturally, there's a strong flavour of orthocorrespondence here...

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            • Jim Parish
              Bernard Gibert described a certain configuration, based on a point Q. As usual, I will attack it by taking polar reciprocals with respect to a ... The polar
              Message 6 of 10 , Oct 10, 2006
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                Bernard Gibert described a certain configuration, based on a point Q.
                As usual, I will attack it by taking polar reciprocals with respect to a
                circle c centered at Q; let DEF be the polar reciprocal of ABC. Then:

                > let Q be a point and A'B'C' its anticevian triangle.

                The polar reciprocal of A'B'C' is the medial triangle of DEF.

                > the perpendicular at Q to QA meets BC at U and V, W likewise.

                The pole of the perpendicular at Q to QA is the point at infinity
                perpendicular to EF; the polars of U, V, W are the altitudes of DEF.

                > A1 = BV /\ CW and B1, C1 likewise.

                The polars of A1, B1, C1 are the edges of the orthic triangle of DEF.

                > the conic inscribed in both triangles A'B'C' and A1B1C1 has focus Q.

                The polar reciprocal of this conic is thus the nine-point circle of DEF.
                Since it is a circle, its polar reciprocal with respect to c is a conic with
                focus at the center of c, i.e., Q.

                Jim Parish
              • Jim Parish
                ... In the second sentence, it refers to the nine-point circle of DEF. Jim Parish
                Message 7 of 10 , Oct 10, 2006
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                  In my last post, I wrote:
                  > The polar reciprocal of this conic is thus the nine-point circle of DEF.
                  > Since it is a circle, its polar reciprocal with respect to c is a conic with
                  > focus at the center of c, i.e., Q.

                  In the second sentence, "it" refers to the nine-point circle of DEF.

                  Jim Parish
                • Bernard Gibert
                  Dear Jim, ... the post refers to a very clever man. thank you Best regards Bernard [Non-text portions of this message have been removed]
                  Message 8 of 10 , Oct 10, 2006
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                    Dear Jim,

                    > In my last post, I wrote:
                    > > The polar reciprocal of this conic is thus the nine-point circle
                    > of DEF.
                    > > Since it is a circle, its polar reciprocal with respect to c is a
                    > conic with
                    > > focus at the center of c, i.e., Q.
                    >
                    > In the second sentence, "it" refers to the nine-point circle of DEF.

                    the post refers to a very clever man.

                    thank you

                    Best regards

                    Bernard



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