IMO Shortlist 1991 problem 6
Dodao/la:
arhiva2. travnja 2012. %V0
$ABCD$ is a terahedron: $AD+BD=AC+BC,$ $BD+CD=BA+CA,$ $CD+AD=CB+AB,$ $M,N,P$ are the mid points of $BC,CA,AB.$ $OA=OB=OC=OD.$ Prove that $\angle MOP = \angle NOP =\angle NOM.$
Izvor: Međunarodna matematička olimpijada, shortlist 1991