Difference between revisions of "1975 IMO Problems/Problem 3"
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Revision as of 13:28, 20 October 2013
On the sides of an arbitrary triangle ABC; triangles ABR;BCP;CAQ are constructed externally with CBP = CAQ = 45°; BCP = ACQ = 30°; ABR = BAR = 15°: Prove that QRP = 90° and QR = RP:
