APMO

Let the four incenters be $I_1$, $I_2$, $I_3$, and $I_4$ with inradii $r_1$, $r_2$, $r_3$, and $r_4$ respectively (in the order given in the question). Without loss of generality, let $I_1$ be closer to $AB$ than $I_2$. Let the acute angle between $I_1{I}_2$ and $AB$ (and hence also the angle between $I_3{I}_4$ and $CD$) be $\theta$. Then $$r_2-r_1=I_1{I}_2\sin \theta =I_3{I}_4\sin \theta =r_4-r_3,$$ which implies $r_1+r_4=r_2+r_3$. Similar arguments show that $r_1+r_2=r_3+r_4$. Thus we obtain $r_1=r_3$ and $r_2=r_4$. Now let's consider the possible positions of $W$, $X$, $Y$, $Z$. Suppose $AZ\ne CX$. Without loss of generality assume $AZ>CX$. Since the incircles of $AWZ$ and $CYX$ are symmetric about the centre of the parallelogram $ABCD$, this implies $CY>AW$. Using similar arguments, we have $$CY>AW⟹BW>DY⟹DZ>BX⟹CX>AZ,$$ which is a contradiction. Therefore $AZ=CX⟹AW=CY$ and $WXYZ$ is a parallelogram.