XXVI OBM

If $ABCD$ is a rhombus the incircle touch the respective triangle in the midpoint of the diagonals, which belongs to all four circles.

Now suppose the circles have a common point $P$. Since the incircle are contained in its triangle and the intersection of the triangles $ABC$, $BCD$, $CDA$, $DAB$ is the intersection of the diagonals, $P$ must coincide with $O$.

Let's prove that the diagonals are perpendicular. Suppose that it's not the case. So we may suppose without loss of generality that $\angle AOB=\angle COD>90^{\circ }$. Let $l_1,l_2,l_3$ and $l_4$ be the incenters of $ABC$, $BCD$, $CDA$ and $DAB$, respectively. Thus $l_1$ and $l_4$ are in the interior of $\angle ABC$ and $l_2$ and $l_3$ are in the interior of $\angle COD$.



Let $E$ and $F$ be the intersection of $A{I}_4$, $B{I}_1$ and $C{I}_2$, $D{I}_3$, respectively. So $\angle BEA>\angle AOB>90^{\circ }\Rightarrow 180^{\circ }-\frac{\angle DAB}{2}-\frac{\angle ABC}{2}>90^{\circ }\iff \angle DAB+\angle ABC<180^{\circ }$. Using a similar argument on $F$ we obtain $\angle BCD+\angle CDA<180^{\circ }$. Summing these two inequalities, we obtain that the sum of the internal angles of the quadrilateral $ABCD$ is less than $360^{\circ }$, contradiction. So $AC$ and $BD$ are perpendicular and, moreover, the four incenters $l_1,l_2,l_3$ and $l_4$ lie in the diagonals. This means that the bisectors and altitudes from vertices $A,B,C,D$ with respect to the triangles $DAB$, $ABC$, $BCD$, $CDA$ coincide and hence these triangles are isosceles. Thus $DA=AB=BC=CD$ and $ABCD$ is a rhombus.