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

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• Jun 4, 2004
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
>
>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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