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Re: Bilogic locus

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  • Francois Rideau
    Dear friends 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
    Message 1 of 10 , Oct 8, 2006
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      Dear friends
      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.
      Friendly
      François


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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 2 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 3 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 4 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 5 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 6 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 7 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 8 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 9 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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