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9854Re: [EMHL] The Brocard points of a quadrilateral (Geoff Millin)

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  • Alexey.A.Zaslavsky
    Jun 4, 2004
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      Dear Darij!
      >This is very interesting! At first, consider
      >the Brocard points R and R' of the harmonic
      >quadrilateral ABCD, and the symmedian point
      >K (this is the point whose distances to the
      >sides of the quadrilateral are proportional
      >to the lengths of the sides). Also consider
      >the circumcenter O of our quadrilateral.
      >Then, the points O, K, R and R' lie on one
      >circle, which is analogous to the "Brocard
      >circle" of a triangle. The points O and K
      >are the endpoints of a diameter of this
      >circle. The points R and R' lie
      >symmetrically with respect to this diameter.
      >The circle through the points O, K, R and R'
      >also passes through the midpoints of the
      >diagonals AC and BD. The line joining these
      >midpoints meets the diameter OK at the
      >midpoint of the segment RR'. The angles
      >< ROK and < R'OK are equal to the Brocard
      >angle of the quadrilateral ABCD.
      >Would be very nice if anybody had a proof to
      >all of this...
      I can prove some of this facts. Let P is the first Brocard point and the
      lines AP, BP, CP, DP intersect the circumcircle in B', C', D', A'. Then the
      quadrilateral A'B'C'D' is the rotation of ABCD with center O and angle 2x,
      where x is the Brocard angle. So P is the second Brocard point Q' of
      A'B'C'D', OP=OQ and angle POQ=2x. Now let K is the common point of AC and
      BD, K' - the common point of A'C' and B'D'. Then P is in KK' (see my paper
      in "Kvant" N.1, 1996). Similarly Q is in KK'', where K'' is the reflection
      of K' in OK. As K''Q=KP and POQ=K'OK=KOK''=2x, P and Q are the midpoints of
      KK' and KK" respectively. So the angles OPK and OQK are direct and P, Q are
      in the circle with diameter OK. Also it is easy to find x from equation
      sin^4(x)=sin(A-x)sin(B-x)sin(C-x)sin(D-x), but I don't understand why ABCD
      must be harmonic.

      Sincerely Alexey
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