IMO Shortlist 1975 problem 8


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2. travnja 2012.
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In the plane of a triangle ABC, in its exterior, we draw the triangles ABR, BCP, CAQ so that \angle PBC = \angle CAQ = 45\,^{\circ}, \angle BCP = \angle QCA = 30\,^{\circ}, \angle ABR = \angle RAB = 15\,^{\circ}.

Prove that

a.) \angle QRP = 90\,^{\circ}, and

b.) QR = RP.
Izvor: Međunarodna matematička olimpijada, shortlist 1975