Shortlist JBMO, 2008

Solution:
From given condition it is clear that $\text{A O B + C O D = B O C + A O D = 180}$.
Let $E$ be a point such that $A E = D O$ and $B E = C E$. Clearly, $\triangle A E B \equiv \triangle D O C$ and from that $A E \parallel D O$ and $B E \parallel C O$. Also, $\text{A E B = C O D}$ so $\text{A O B + A E B = A O B + C O D = 180}$. Thus, the quadrilateral $A O B E$ is cyclic.
So $\triangle A O B$ and $\triangle A E B$ have the same circumcircle, therefore the circumcircles of the triangles $\triangle A O B$ and $\triangle C O D$ have the same radius.
Also, $A E \parallel D O$ and $A E = D O$ gives $A E O D$ is a parallelogram and $\triangle A O D \equiv \triangle O A E$. So $\triangle A O B$, $\triangle C O D$ and $\triangle D O A$ have the same radius of their circumcircle (the radius of the cyclic quadrilateral $A E B O$). Analogously, triangles $\triangle A O B$, $\triangle B O C$, $\triangle C O D$ and $\triangle D O A$ have the same radius $R$.
Obviously, the circle with center $O$ and radius $2R$ is externally tangent to each of these circles, so this will be the circle $k$.