XVII OBM

If $S=O$ then $AC=BD=2R$, where $R$ is the circumradius of $ABCD$ and $S$ is the midpoint of both $AC$ and $BD$. This means that $ABCD$ is an inscribed parallelogram. The sum of its opposite angles, which are congruent, is $180^{\circ }$, so all angles of $ABCD$ are right, that is, $ABCD$ is a rectangle. But there is an inscribed circle of radius $r$ in $ABCD$, so all sides are equal to $2r$ and so $ABCD$ is a square.

If $S=I$ then the diagonals $AC$ and $BD$ are bisectors of the angles of $ABCD$, which means that every side subtends the same arc of the circumcircle as its neighbouring sides. This means that the four vertices $A$, $B$, $C$, $D$ divide the circle in four equal parts, so $ABCD$ is a square.

If $I=O$ then the quadrilateral can be partitioned in eight congruent right triangles with hypotenuse equal to $R$ and one of the legs equal to $r$; the angle between $R$ and $r$ in those right triangles has always vertex on $O$, so all central angles $\angle AOB$, $\angle BOC$, $\angle COD$ and $\angle DOA$ are equal and again $A$, $B$, $C$ and $D$ divide the circumcircle in four equal arcs, leading again to $ABCD$ being a square.