Problem A. 554. (February 2012)
A. 554. The circumcenter of the cyclic quadrilateral ABCD is O. The second intersection point of the circles ABO and CDO, other than O, is P, which lies in the interior of the triangle DAO. Choose a point Q on the extension of OP beyond P, and a point R on the extension of OP beyond O. Prove that QAP=OBR holds if and only if PDQ=RCO.
(5 pont)
Deadline expired on 12 March 2012.
Solution. Let H be the radical center of the circles ABCD, ABOP and CDPO. Then the radical axes of any two of there cirlces, i.e. the lines AB, CD és OP pass through H. Since P lines on the shorter arcs AO and DO, it follows that H lies on the extension of OP beyond P. The radical center satisfies
HA^{.}HB=HC^{.}HD=HO^{.}HP.  (1) 
Since the quadrilateral ABOP is cyclic,
QAB+BRQ=(PAB+QAP)+(BOPOBR)=
=(PAB+BOP)+(QAPOBR)=180^{o}+(QAPOBR).
Therefore, QAP=OBR holds if and only if the quadrilateral ABRQ is cyclic, which is equivalent to HQ^{.}HR=HA^{.}HB.
Similarly, QAP=OBR holds if and only if HQ^{.}HR=HC^{.}HD.
Combining with (1),
Statistics:
