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52 Mathematical Olympiad

52nd International Mathematical Olympiad12 24 July 2011 AmsterdamThe NetherlandsInternationalMathematicalOlym piad Amsterdam 2011 IMO2011 AmsterdamProblem Shortlistwith Solutions52nd InternationalMathematical Olympiad12-24 July 2011 AmsterdamThe NetherlandsProblem shortlistwith solutionsIMPORTANTIMO regulation:these shortlist problems have tobe kept strictly confidentialuntil IMO problem selection committeeBart de Smit (chairman), Ilya Bogdanov, Johan Bosman,Andries Brouwer, Gabriele Dalla Torre, G eza K os,Hendrik Lenstra, Charles Leytem, Ronald van Luijk,Christian Reiher, Eckard Specht, Hans Sterk, Lenny TaelmanThe committee gratefully acknowledges the receipt of 142 problem proposalsby the following 46 countries:Armenia, Australia, Austria, Belarus, Belgium,Bosnia and Herzegovina, Brazil, Bulgaria, Canada, Colombia,Cyprus, Denmark, Estonia, Finland, France, Germany, Greece,Hong Kong, Hungary, India, Islamic Republic of Iran, Ireland, Israel,Japan, Kazakhstan, Republic of Korea, Luxembourg, Malaysia,Mexico, Mongolia, Montenegro, Pakistan, Poland, Romania,Russian Federation, Saudi Arabia, Serbia, Slovakia, Slovenia,Sweden, Taiwan, Thailand, Turkey, Ukraine, United Kingdom,United States of AmericaAlgebraProblem shortlist52nd IMO 2011 AlgebraA1A1 For any setA={a1, a2, a3, a4}of four distinct positive integers with sumsA=a1+a2+a3+a4,letpAdenote the number of pairs ()

Suppose also that the circumcenter O of the triangle ABC lies on the shorter arc B′C′ of ω. Prove that the circumcircle of ABC and ω meet at two points. G2 G2 Let A 1A 2A 3A 4 be a non-cyclic quadrilateral. Let O 1 and r 1 be the circumcenter and the circumradius of the triangle A 2A 3A 4. Define O 2, O 3, O 4 and r 2, r 3, r 4 in a ...

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