IMO Shortlist 1975 problem 8
Dodao/la:
arhiva2. travnja 2012. In the plane of a triangle
![ABC,](/media/m/8/a/f/8afcbd6e815ca10256c79a5b310e3d67.png)
in its exterior
![,](/media/m/c/c/6/cc6a28b756c2449c071b10d734c2e4c7.png)
we draw the triangles
![ABR, BCP, CAQ](/media/m/3/8/0/380efd4d2833c29c77d19679f33b14b8.png)
so that
Prove that
a.)
![\angle QRP = 90\,^{\circ},](/media/m/8/2/1/821b4c730bac1de91446ec1f13b79156.png)
and
b.)
%V0
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