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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 \angleQAP=\angleOBR holds if and only if \anglePDQ=\angleRCO.

(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\angle+BRQ\angle=(PAB\angle+QAP\angle)+(BOP\angle-OBR\angle)=

=(PAB\angle+BOP\angle)+(QAP\angle-OBR\angle)=180o+(QAP\angle-OBR\angle).

Therefore, QAP\angle=OBR\angle holds if and only if the quadrilateral ABRQ is cyclic, which is equivalent to HQ.HR=HA.HB.

Similarly, QAP\angle=OBR\angle holds if and only if HQ.HR=HC.HD.

Combining with (1),


QAP\angle=OBR\angle 
\Longleftrightarrow
HQ\cdot HR = HA\cdot HB
\Longleftrightarrow
HQ\cdot HR = HC\cdot HD
\Longleftrightarrow
PDQ\angle=RCO\angle.


Statistics:

7 students sent a solution.
5 points:Ágoston Tamás, Gyarmati Máté, Janzer Olivér, Machó Bónis, Mester Márton, Omer Cerrahoglu, Strenner Péter.

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