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This solution is from Arun Iyer of SIA High School and Junior College.
Part 1
First I will show that POR is a straight line. For this I would
like to state the perpendicular bisector theorem.
PERPENDICULAR BISECTOR THEOREM: Every point equidistant from the
two ends of a line segment lies on the perpendicular bisector of
the line segment.
Now consider the line segment QS.
OQ=OS=1 therefore by the perpendicular bisector theorem, O must
lie on the perpendicular bisector of QS.
PQ=PS (as the sides of the rhombus are equal), therefore by the
perpendicular bisector theorem, P must lie on the perpendicular
bisector of QS.
RQ=RS (as the sides of the rhombus are equal), therefore by the
perpendicular bisector theorem, R must lie on the perpendicular
bisector of QS.
Now the perpendicular bisector of a line segment is unique and
hence P, O, R must lie on the same perpendicular bisector and
hence POR is a straight line.
Part 2
Now I will get all the angles of the rhombus.
\angle QPS = 72^{\circ} (given), \angle QRS = \angle QPS =
72^{\circ} as they are opposite angles of a rhombus. The diagonal
of the rhombus bisects the angles of a rhombus and therefore
\angle QPO = \angle SPO = \angle QRO = \angle SRO =
36^{\circ}.
Triangles OQR and OSR are isosceles triangles, therefore
\angle OSR = \angle OQR = 36^{\circ}.
Using the fact that sum of angles of a triangle is 180 degrees, we can see that \angle SOR and \angle QOR are equal to 108 degrees. Since we have proved that POR is a straight line in Part 1, we can determine \angle QOP and \angle SOP to be 72 degrees.
Again using the fact that sum of angles of a triangle is 180 degrees, we can see that \angle OQP and \angle OSP are equal to 72 degrees.
Part 3
Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.
The diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.