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Re: [EMHL] Please help me with a reference... (Third)

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  • Gilles Boutte
    ... Precisely, the three pairwise radical axis of the 3 circles , i.e. the sidelines of ABC, are concurrant, this fact is impossible for ABC non-degenerate. Y,
    Message 1 of 6 , May 16, 2004
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      >>Dear Hyacinthians,
      >>
      >>The following theorem is an example of a
      >>non-standard triangle geometry result because
      >>the proof (at least, the proof I know) uses a
      >>reductio ad absurdum with the help of the
      >>fact that the three pairwise radical axes of
      >>three circles always concur:
      >>
      >> If ABC is a non-degenerate triangle, X and X'
      >> are two points on the line BC, Y and Y' are
      >> two points on the line CA, and Z and Z' are
      >> two points on the line AB, such that the
      >> points Y, Y', Z and Z' lie on one circle, the
      >> points Z, Z', X and X' lie on one circle, and
      >> the points X, X', Y and Y' lie on one circle,
      >> then all six points X, X', Y, Y', Z and Z'
      >> lie on one circle.
      >>
      >>I remember having seen this result in a paper
      >>by Third in the Edinburgh Math. Proceedings.
      >>Can anybody give an *exact* reference with
      >>year, page nos, etc.?
      >>
      >> Thanks a lot!
      >> Sincerely,
      >> Darij Grinberg
      >>
      >>
      >
      >That result looks false to me. Start with distinct overlapping
      >circles c1, c2 and c3, and define {X,X'} = c2 /\ c3, {Y, Y'} =
      >c3 /\ c1, {Z, Z'} = c1 /\ c2. Then take BC = line XX' etc.
      >
      >
      Precisely, the three pairwise radical axis of the 3 circles , i.e. the
      sidelines of ABC, are concurrant, this fact is impossible for ABC
      non-degenerate.

      Y, Y', Z, Z' lie on c1
      Z, Z', X, X' lie on c2
      X, X', Y, Y' lie on c3

      The common power of A wrt c1 and c3 is AY.AY'
      The common power of A wrt c1 and c2 is AZ.AZ'

      Hence A as the same power wrt c2 and c3. But c2/\c3 = {X,X'} and A don't
      lie on the line XX' = BC (since ABC is non-degenerate).

      So the locus of the points with the same power wrt c2 and c3 is not a
      line and c2 = c3 and X, X', Y, Y', Z, Z' lie on the circle c1=c2=c3.

      Friendly

      Gilles Boutte
    • Darij Grinberg
      Dear Gilles, ... Yes, this is the proof of Third s theorem. By the way, if I remember correctly, Third formulated his theorem with the help of antiparallels.
      Message 2 of 6 , May 17, 2004
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        Dear Gilles,

        In Hyacinthos message #9794, you wrote:

        >> Precisely, the three pairwise radical axis of the 3
        >> circles , i.e. the sidelines of ABC, are concurrant,
        >> this fact is impossible for ABC non-degenerate.

        Yes, this is the proof of Third's theorem.

        By the way, if I remember correctly, Third formulated his
        theorem with the help of antiparallels.

        Sincerely,
        Darij Grinberg
      • Ken Pledger
        ... Was Third s reference earlier than 1949? I remember striking this problem as part of a longer proof of an old text-book exercise: E. A. Maxwell, Geometry
        Message 3 of 6 , Jun 1, 2004
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          >Dear Hyacinthians,
          >
          >The following theorem is an example of a
          >non-standard triangle geometry result because
          >the proof (at least, the proof I know) uses a
          >reductio ad absurdum with the help of the
          >fact that the three pairwise radical axes of
          >three circles always concur:
          >
          > If ABC is a non-degenerate triangle, X and X'
          > are two points on the line BC, Y and Y' are
          > two points on the line CA, and Z and Z' are
          > two points on the line AB, such that the
          > points Y, Y', Z and Z' lie on one circle, the
          > points Z, Z', X and X' lie on one circle, and
          > the points X, X', Y and Y' lie on one circle,
          > then all six points X, X', Y, Y', Z and Z'
          > lie on one circle.
          >
          >I remember having seen this result in a paper
          >by Third in the Edinburgh Math. Proceedings.
          >Can anybody give an *exact* reference with
          >year, page nos, etc.? ....


          Was Third's reference earlier than 1949? I remember striking
          this problem as part of a longer proof of an old text-book exercise:
          E. A. Maxwell, "Geometry for Advanced Pupils," (1949), p.170, ex.14.
          The first stage of the proof I know is to show that certain specially
          defined points like your Y, Y', Z, Z' are concyclic; which prepares
          the way for the radical axes argument which you mention. Maxwell
          ascribes his problem to the Oxford and Cambridge Schools Examination
          Board, presumably some time before 1949.

          Ken Pledger.
        • Darij Grinberg
          Dear Ken Pledger, Thanks for the information in Hyacinthos message #9836. Lately, I have succeeded to localise the Third reference: J. A. Third, Systems of
          Message 4 of 6 , Jun 2, 2004
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            Dear Ken Pledger,

            Thanks for the information in Hyacinthos message
            #9836. Lately, I have succeeded to localise the
            Third reference:

            J. A. Third, "Systems of circles analogous to
            Tucker circles", Proc. Edinburgh Math. Soc.,
            17 (1898) p. 70-99.

            In fact, the fact I called "Third theorem" is
            mentioned and proved on the very first page of
            the above article. Third states it using the
            notion of antiparallels (instead of saying that
            the points Y, Y', Z and Z' lie on one circle,
            he says that the lines YZ' and Y'Z are
            antiparallel to each other with respect to the
            angle CAB).

            Thanks to all for the replies to my inquiry.

            Sincerely,
            Darij Grinberg
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