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.

$\angle \mathrm{QPS}$ = ${72}^o$ (given), $\angle \mathrm{QRS}=\angle \mathrm{QPS}={72}^o$ as they are opposite angles of a rhombus. The diagonal of the rhombus bisects the angles of a rhombus and therefore $\angle \mathrm{QPO}=\angle \mathrm{SPO}=\angle \mathrm{QRO}=\angle \mathrm{SRO}={36}^o$.

Triangles OQR and OSR are isosceles triangles, therefore $\angle \mathrm{OSR}=\angle \mathrm{OQR}={36}^o$.

Using the fact that sum of angles of a triangle is 180 degrees, we can see that $\angle \mathrm{SOR}$ and $\angle \mathrm{QOR}$ are equal to 108 degrees. Since we have proved that POR is a straight line in Part 1, we can determine $\angle \mathrm{QOP}$ and $\angle \mathrm{SOP}$ to be 72 degrees.

Again using the fact that sum of angles of a triangle is 180 degrees, we can see that $\angle \mathrm{OQP}$ and $\angle \mathrm{OSP}$ are equal to 72 degrees.

Part 3 Let the side of the rhombus be x.

Consider triangle OPS in which PO=PS=x (since $\angle \mathrm{POS}=\angle \mathrm{PSO}$).

Applying the cosine rule

${\displaystyle \cos \mathrm{OPS}=[{\mathrm{PS}}^2+{\mathrm{PO}}^2-{\mathrm{OS}}^2]/[2\times \mathrm{PS}\times \mathrm{PO}]}$

therefore

${\displaystyle \cos 36=[2x^2-1]/[2x^2]\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1).}$

Splitting the isosceles triangle ORS into two right angled triangles gives

${\displaystyle \cos 36=x/2\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(2).}$

From (1) and (2),

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{[2x^2-1]/[2x^2]}& =x/2\\ \multicolumn{1}{c}{x^3-2x^2+1}& =0\\ \multicolumn{1}{c}{(x-1)(x^2-x-1)}& =0.\end{array}}$

Now $x\ne 1$ because triangle POS is not equilateral, therefore $x^2-x-1=0$ and hence

${\displaystyle x=[1+\sqrt{5}]/2\mathrm{or}x=[1-\sqrt{5}]/2.}$

Clearly $x\ne [1-\sqrt{5}]/2$ because the side length cannot be negative, therefore $x=[1+\sqrt{5}]/2$, the Golden Ratio.