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