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Re: [EMHL] Re: Perspector

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  • Alexey.A.Zaslavsky
    Dear Antreas! ... These triangles are homothetic. It is proved in the paper of L.and T. Emelyanov in Math. Prosveschenie and the book of Sharygin. Sincerely
    Message 1 of 6 , Aug 31, 2004
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      Dear Antreas!
      >
      >I am wondering what the extraversion of this might be.
      >
      >That the cevian triangle of I and the Feuerbach triangle
      >FaFbFc are perspective?
      >
      These triangles are homothetic. It is proved in the paper of L.and T.
      Emelyanov in "Math. Prosveschenie" and the book of Sharygin.

      Sincerely Alexey
    • Darij Grinberg
      Dear Antreas, ... They are not perspective. This is not the way extraversion goes! Under extraversion, the triangle formed by the polars of the points A, B, C
      Message 2 of 6 , Sep 1, 2004
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        Dear Antreas,

        In Hyacinthos message #10348, you wrote:

        >> I am wondering what the extraversion of this
        >> might be.
        >>
        >> That the pedal triangle of I and the Feuerbach
        >> triangle FaFbFc are perspective?

        They are not perspective. This is not the way
        extraversion goes! Under extraversion, the
        triangle formed by the polars of the points A,
        B, C with respect to the A-excircle, B-excircle,
        C-excircle, respectively, is "mapped" to the
        triangle formed by the polars of the points A,
        B, C with respect to the incircle, C-excircle,
        B-excircle, respectively. And not all with
        respect to the incircle!

        Actually, in a message you have deleted, you
        asked whether the cevian triangle of I and the
        Feuerbach triangle FaFbFc are perspective. And
        in Hyacinthos message #10349, Alexey answered
        that "These triangles are homothetic".
        Actually, they are perspective but NOT
        homothetic. This matter is a real pitfall: In
        the paper

        Lev Emelyanov and Tatiana Emelyanova, "A
        Note on the Feuerbach Point", Forum
        Geometricorum, 1 (2001) pp. 121-124,

        http://forumgeom.fau.edu/FG2001volume1/FG200117index.html

        it has been proven that the cevian triangle
        of I and the Feuerbach triangle FaFbFc are
        1) perspective and 2) similar (even directly
        similar). It took me a long time until I
        realized that two triangles can be both
        perspective and directly similar without being
        homothetic! Actually, the quickest way to see
        that the cevian triangle of I is (in general)
        NOT homothetic to the Feuerbach triangle
        FaFbFc is to show that their corresponding
        sides are not parallel. In fact, the
        corresponding sides of both triangles pass
        through the feet of the exterior angle
        bisectors of triangle ABC, and hence they
        can't be parallel for a generic triangle ABC.

        Sincerely,
        Darij Grinberg
      • Antreas P. Hatzipolakis
        Dear Darij ... Well... If A1A2A3 is a triangle, and (I0),(I1),(I2),(I3) are the in/excircles and F0,F1,F2,F3,F4 the Feuerbach Points, then we have the nine
        Message 3 of 6 , Sep 1, 2004
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          Dear Darij

          [APH]:
          >> I am wondering what the extraversion of this
          >> might be.
          >>
          >> That the pedal triangle of I and the Feuerbach
          >> triangle FaFbFc are perspective?

          [DG]:
          >They are not perspective. This is not the way
          >extraversion goes! Under extraversion, the
          >triangle formed by the polars of the points A,
          >B, C with respect to the A-excircle, B-excircle,
          >C-excircle, respectively, is "mapped" to the
          >triangle formed by the polars of the points A,
          >B, C with respect to the incircle, C-excircle,
          >B-excircle, respectively. And not all with
          >respect to the incircle!

          Well... If A1A2A3 is a triangle, and (I0),(I1),(I2),(I3)
          are the in/excircles and F0,F1,F2,F3,F4 the Feuerbach Points,
          then we have the nine polars:
          Pij := the polar of Aj with respect the circle (Ii),
          forming C(9,3) (non-ordered) triangles
          and also we have C(4,3) triangles with vertices Fi

          Which ones are perspective? And which are the perspectors?

          [DG]:
          >Actually, in a message you have deleted, you
          >asked whether the cevian triangle of I and the
          >Feuerbach triangle FaFbFc are perspective.

          I deleted it because I meant to say pedal not cevian
          (since the sidelines of the pedal triangle of I
          are the polars of A,B,C wrt the incircle)


          Antreas
          --
        • Darij Grinberg
          Dear Antreas, ... ^^^^^^^^^^^^^^ Well, you probably wanted to say F0, F1, F2, F3. There are 4 Feuerbach points, not 5. ... Actually, you have twelve polars!
          Message 4 of 6 , Sep 1, 2004
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            Dear Antreas,

            In Hyacinthos message #10353, you wrote:

            >> Well... If A1A2A3 is a triangle, and
            >> (I0),(I1),(I2),(I3) are the in/excircles
            >> and F0,F1,F2,F3,F4 the Feuerbach Points,
            ^^^^^^^^^^^^^^

            Well, you probably wanted to say F0, F1, F2,
            F3. There are 4 Feuerbach points, not 5.

            >> then we have the nine polars:
            >> Pij := the polar of Aj with respect the
            >> circle (Ii),
            >> forming C(9,3) (non-ordered) triangles

            Actually, you have twelve polars! There
            are four in/excircles, and three vertices
            of triangle ABC. And you thus have C(12,3)
            non-ordered triangles.

            But extraversion is not just replacing
            *some* incircle by *some* excircle. If you
            extravert with respect to, say, A, then
            you replace

            incircle --> A-excircle;
            A-excircle --> incircle;
            B-excircle --> C-excircle;
            C-excircle --> B-excircle.

            This is a strict rule, and if you have a
            theorem, then its extraversion with
            respect to A is uniquely defined. You can
            also extravert with respect to B or to C
            (just analogously to the A-extraversion
            explained above). If you have a triangle
            formed by the polars of the points A, B, C
            with respect to the A-excircle, B-excircle,
            C-excircle, respectively, then you can
            never extravert it into a triangle formed
            by the polars of A, B, C with respect to
            the incircle, since an extraversion
            "transforms" (I put this word in quotes
            since extraversion is not a real
            geometrical transformation) different
            in/excircles into different in/excircles,
            and thus it cannot "transform" all three
            excircles into one and the same incircle.

            >> Which ones are perspective?

            Hmm, well, this is a difficult question.
            Of course, you have the perspector of the
            triangles P11P22P33 (meaning: the triangle
            formed by the lines P11, P22 and P33) and
            F1F2F3. This is the perspector found by
            Juan Carlos. You also have the three
            extraversions of this perspector, namely

            - its A-extraversion, i. e. the perspector
            of triangles P01P32P23 and F0F3F2.
            - its B-extraversion, i. e. the perspector
            of triangles P31P02P13 and F3F0F1.
            - its C-extraversion, i. e. the perspector
            of triangles P21P12P03 and F2F1F0.

            [I hope I have done no typos or mistakes
            in extraverting.]

            But concerning the other combinations of
            polars and Feuerbach points - I don't know.
            Maybe somebody could write a computer
            program to check all possible cases...

            >> I deleted it because I meant to say pedal
            >> not cevian

            Yes, that's clear to me. But with "cevian"
            it's correct (it's not a correct
            extraversion, but the triangles are
            perspective), while with "pedal" it's wrong.

            Please forgive me that, as I'm just noting,
            the above comments sounded rude. They were
            actually by no means meant in this way. I
            wanted to clear up some fundamental things
            about extraversion. Finally, extraversion,
            when applied properly, can be of great help
            in understanding relationships between
            theorems from triangle geometry.

            Sincerely,
            Darij Grinberg
          • Antreas P. Hatzipolakis
            Dear Darij ... Sorry for the stupid errors! (I should have delete it and resend a correct one) ... [...] Thanks for your analysis ! Now, how about a
            Message 5 of 6 , Sep 1, 2004
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              Dear Darij

              [APH]:
              >>> Well... If A1A2A3 is a triangle, and
              >>> (I0),(I1),(I2),(I3) are the in/excircles
              >>> and F0,F1,F2,F3,F4 the Feuerbach Points,

              [DG]:
              >Well, you probably wanted to say F0, F1, F2,
              >F3. There are 4 Feuerbach points, not 5.

              [APH]:
              >>> then we have the nine polars:
              >>> Pij := the polar of Aj with respect the
              >>> circle (Ii),
              >>> forming C(9,3) (non-ordered) triangles

              [DG]:
              >Actually, you have twelve polars! There
              >are four in/excircles, and three vertices
              >of triangle ABC. And you thus have C(12,3)
              >non-ordered triangles.

              Sorry for the stupid errors!
              (I should have delete it and resend a correct one)

              [DG]:
              >But extraversion is not just replacing
              [...]

              Thanks for your analysis !

              Now, how about a generalization of JCS's problem?

              I think that a possible one is this:

              Let ABC be a triangle, P a point, A'B'C'
              the antipedal triangle of P, and A'aA'bA'c,
              B'aB'bB'c, C'aC'bC'c the pedal triangles
              of A',B',C' resp.

              Let Pa, Pb, Pc be the points of contact of the
              circle which touches externally the pedal circles
              of A', B', C' (ie the circles (A'aA'bA'c),
              (B'aB'bB'c), (C'aC'bC'c)) .

              Which is the locus of P such the triangle PaPbPc and
              the triangle bounded by the lines A'bA'c, B'cB'a, C'aC'b
              are perspective?

              PS: Probably are interesting the cases
              (1) ABC, PaPbPc (2) ABC, bounded by (A'bA'c, B'cB'a, C'aC'b)
              be perspective.


              Greetings

              Antreas

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