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Re: [EMHL] Re: Perspector

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  • 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 1 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 2 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 3 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 4 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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