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Re: [EMHL] Gossard again

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  • Antreas P. Hatzipolakis
    ... No. It is ZEEMAN. Julio s message in HM (Fri, 03 Dec 1999) on Z. reads: _____________________________________ Dear John, Antreas, & all, 1. If a,
    Message 1 of 83 , Aug 14, 2003
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      Richard Guy wrote:

      >Perhaps `Zeman' ??? R.

      No. It is ZEEMAN.

      Julio's message in HM (Fri, 03 Dec 1999) on Z. reads:

      _____________________________________

      <quote>
      Dear John, Antreas, & all,

      1. If a, b, c, d denote four straight lines (of the same plane) such
      that $d$ be parallel to the Euler line of the triangle $abc$, then
      the three Euler lines

      a' of bcd, b' of cda, c' of dab,

      are parallel to a, b, c, respectively.

      This theorem of professor P. Zeeman Gz [from Delft, renowned all over
      the world as the city of Delft blue earthenware, in the Netherlands] was
      published early in 1903 in the Dutch journal _Wiskundige Opgaven_ [vol
      VIII (1899-1902), p 305]. I first heard of this theorem thanks to a brief
      note [_Mathesis_ (3) 3 (1903) p 60] written by the famous triangle guru
      Joseph Neuberg (1840-1926) of Luxembourg:

      "Si quatre droites a, b, c, d d'un plan sont telles
      que l'une d'elles est parall\ele \a la droite d'Euler
      du triangle des trois autres, cette propri/et/e a
      encore lieu pour chacune des autres droites."

      Neuberg noticed he could easily prove this proposition by using a formula
      he had discussed in an earlier volume of _Wiskundige Opgaven_ [vol III
      (1886-1889), p 372].


      2. An interesting complement to Zeeman's theorem is that of professor
      C.A. Cikot [from Bois-le-Duc (s'Hertogenbosch) capital of North Brabant
      province, south central Netherlands], a proposition first published in
      an article written also by Neuberg [_Mathesis_ (3) 8 (1908) p 233]:

      Let a, b, c, d denote four straight lines (of the same plane). If $d$
      is the Euler line of the triangle $abc$, then the three Euler lines a',
      b', c' respectively of the triangles $bcd$, $cda$, $dab$ form a triangle
      congruent to $abc$, and the Euler line d' of $a'b'c'$ coincides with $d$.

      What about "The Cikot Point"? ...

      John, this theorem implicitly includes the congruence-by-reflection
      statement you've independently discovered. Do you agree? ...

      By the way, many thanks for your kindest words in your previous letter
      to me on this thread!

      With best regards,
      Julio Gonzalez Cabillon

      </quote>

      I don't know what the initial P. stands for
      but the name appears as:

      ZEEMAN GZ., P.
      ZEEMAN GZN, P.

      Probably the P. stands for PIETER (must
      not be confused with the 1902 Physics nobelist
      PIETER ZEEMAN)

      P. ZEEMAN GZN wrote several geometrical articles
      in Dutch periodicals.

      In my next message I will give the German abstract
      of an article of his about triangle cubics.

      Antreas


      ------------------------------------------------------------------------



      --
    • Antreas P. Hatzipolakis
      Dear Peter ... To paraphrase one of the greatest stylists of the English language: I was working on the proof of one of my perspectors all the morning, and
      Message 83 of 83 , Oct 19, 2004
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        Dear Peter

        [APH]:
        >>> Definition: (1,2,3) := the triangle bounded by the lines 1,2,3
        >>>
        >>> We know:
        >>>
        >>> Lemma:
        >>> Let ABC be a triangle, and L its Euler line.
        >>> La := the Euler Line of (AB,AC,L)
        >>> ==> BC // La
        >>>
        >>> Gossard Perspector:
        >>> The triangles ABC, (La,Lb,Lc) are perspective (homothetic).
        >>> The perspector is called Gossard perspector.
        >>>
        >>> Now, let P = (x:y:z) be a point.
        >>>
        >>> L1,L2,L3 := the Euler lines of PBC,PCA,PAB, resp.
        >>>
        >>> L11,L22,L33 := the Euler lines of (PB,PC,L1), (PC,PA,L2),
        >>> (PA,PB,L3), resp.
        >>>
        >>> The triangles ABC, (L11,L22,L33) and (La,Lb,Lc), (L11,L22,L33)
        >>> are obviously perspective (homothetic).
        >>>
        >>> QUESTION:
        >>> Which are the perspectors in terms of x,y,z?
        >>>
        >>> [PM]:
        >>> For ABC, (L11,L22,L33) I get a rather nasty equation.
        >>> However for P = I, I reckon on the perspector as
        >>> a (3 + (3 (a + b) (a + c))/(a + 2 b - c) (a - b + 2 c))) : :
        >>>
        >>> [APH]
        >>> If I read correctly the expression (too many parentheses :-),
        >>> then the 3 simplifies ie
        >>>
        >>> a (1 + (a + b) (a + c)) / (a + 2 b - c) (a - b + 2 c)) : :
        >>> in barycentrics I guess.
        >>>
        >>> (1 + (a + b) (a + c)) / (a + 2 b - c) (a - b + 2 c)) : :
        >>> In trilinears

        [PM]:
        >Still not right! It is (trilinear)
        >1 + ((a + b) (a + c))/((a + 2 b - c) (a - b + 2 c)) : :
        >ETC search = 1.6459643667906088


        To paraphrase one of the greatest stylists of the English language:

        "I was working on the proof of one of my perspectors all the
        morning, and took out a parenthesis. In the afternoon I put
        it back again." (or reversely :-)

        This is, right?

        (a + b) (a + c)
        1 + ----------------------------- :: (in trilinears)
        (a + 2 b - c) (a - b + 2 c)


        [APH]:
        >> How about to construct an other point, possibly lying on
        >> a simple line?
        >>
        >> Let A'B'C' be the antimedial triangle of ABC.
        >>
        >> L1,L2,L3 := the Euler lines of A'BC,B'CA,C'AB, resp.
        >>
        >> L11,L22,L33 := the Euler lines of (A'B,A'C,L1), (B'C,B'A,L2),
        >> (C'A,C'B,L3), resp.
        >>
        >> Then I think that the perspector of ABC, (L11,L22,L33)
        >> lies on a very simple line of ABC. Doesn't it?

        [PM]:
        >From before, ABC perspector to (L11,L22,L33) = X(402) = U
        >
        >Medial triangle perspector to (L11,L22,L33) = V =
        >(2 a^4 - a^2 b^2 - b^4 - a^2 c^2 + 2 b^2 c^2 - c^4) (2 a^8 - 2a^6
        >b^2 - 7 a^4 b^4 + 12 a^2 b^6 - 5 b^8 - 2 a^6 c^2 + 16 a^4 b^2 c^2 -
        >12a^2 b^4 c^2 - 2 b^6 c^2 - 7 a^4 c^4 - 12 a^2 b^2 c^4 + 14 b^4 c^4
        >+ 12 a^2 c^6 - 2 b^2 c^6 - 5 c^8)::
        >
        >AntiMedial triangle perspector to (L11,L22,L33) = W =
        >(2 a^4 - a^2 b^2 - b^4 - a^2 c^2 + 2 b^2 c^2 - c^4) (a^8 -a^6 b^2 -
        >5 a^4 b^4 + 9 a^2 b^6 - 4 b^8 - a^6 c^2 + 11 a^4 b^2 c^2 - 9 a^2 b^4
        >c^2 - b^6 c^2 - 5 a^4 c^4 - 9 a^2 b^2 c^4 + 10 b^4 c^4 + 9 a^2 c^6 -
        >b^2 c^6 - 4 c^8)::
        >
        >U,V and W are on the Euler line.
        >Also |WV|/|VU| = 2/5
        >|WG|/|GU| = 4/5
        >|VG|/GU| = 2/7
        >G is the {U,V}-harmonic conjugate of W.

        Very good!

        I think that, in general: if we have a triangle A'B'C' homothetic to ABC,
        and (ABC, A'B'C') share the same Euler line, then the perspector
        of ABC, (L11,L22,L33) lies on the [common] Euler line.

        Anyway, two other interesting triangles that are homothetic to ABC,
        and share the same Euler line, are the midway triangles of G and H.
        That is: the triangles formed by the midpoints of {AG,BG,CG} and
        {AH,BH,CH} = Euler triangle.


        Greetings

        Antreas


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