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9824The Brocard points of a quadrilateral (Geoff Millin)

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  • Darij Grinberg
    Jun 1, 2004
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      An interesting thread was started in the
      geometry-college group:

      http://mathforum.org/epigone/geometry-college/pringkheldeh

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      Subject: Brocard points of quadrilaterals
      Author: geoff millin <cgmillin*tiscali.co.uk>
      replace the sign * by @

      Correction:

      ABCD is a cyclic quadrilateral. I have proved
      that a necessary and sufficient condition for
      it to have Brocard points is that the products
      of the opposite sides are equal. Ie
      AB.CD = AD.BC.

      I expect this is a well-known fact, but I can
      find no reference to it on the web.

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      Subject: Re: Brocard points of quadrilaterals
      Author: Darij Grinberg

      Dear Geoff Millin,

      You wrote:

      >> ABCD is a cyclic quadrilateral. I have proved
      >> that a necessary and sufficient condition for
      >> it to have Brocard points is that the products
      >> of the opposite sides are equal. Ie
      >> AB.CD = AD.BC.

      Congratulations for having discovered a nice new
      fact in Euclidean Geometry! A cyclic
      quadrilateral ABCD satisfying AB * CD = AD * BC
      is called a "harmonic quadrilateral"; such
      quadrilaterals were studied in the 19th century,
      and many properties of them were uncovered. Just
      two examples: The line AC is a symmedian in the
      triangles DAB and BCD simultaneously, and the
      line BD is a symmedian in the triangles ABC and
      CDA simultaneously. And: There exists an
      inversion mapping the points A, B, C, D to the
      vertices of a square. But I have never seen your
      result about Brocard points mentioned anywhere,
      it is most likely new.

      I have tried to explore your Brocard points by
      dynamic geometry software. One of the nice
      properties (I have no proof yet) of these points
      is that the two "Brocard angles" are equal, much
      like in a triangle. In other words, if you take
      a harmonic quadrilateral ABCD and its two
      Brocard points R and R', with

      < RAB = < RBC = < RCD = < RDA and
      < R'BA = < R'CB = < R'DC = < R'AD,

      then

      < RAB = < RBC = < RCD = < RDA
      = < R'BA = < R'CB = < R'DC = < R'AD.

      Equivalently, the Brocard points R and R' are
      "isogonal conjugates" with respect to the
      quadrilateral ABCD, i. e. the lines AR and AR'
      are symmetric with respect to the angle
      bisector of the angle DAB, and similarly for
      the other pairs of lines. Generally, not every
      point in the plane of our quadrilateral ABCD
      has an isogonal conjugate, but these Brocard
      points have isogonal conjugates, namely the
      isogonal conjugate of one Brocard point is the
      other one.

      I don't see, however, any formula of the kind
      cot w = cot A + cot B + cot C for the Brocard
      angle of a harmonic quadrilateral.

      May I ask you how you proved your original
      result? Did you perform a calculation using
      coordinates or do you have a synthetic proof?
      Do you also have an idea about my above
      conjectures?

      Sincerely,
      Darij Grinberg

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