Rectangle $ABCD$ has center $O$ and $AB/AD=k$. A point is randomly chosen from the interior of rectangle $ABCD$. What is the probability that it is closer to $O$ than to any of the four vertices?  [asy]
size(200);
draw((-250,100)--(250,100)--(250,-100)--(-250,-100)--cycle);
dot((0,0));
label("$O$",(0,0),N);
label("$A$",(-250,100),NW); label("$B$",(250,100),NE); label("$C$",(250,-100),SE); label("$D$",(-250,-100),SW);[/asy]
The original rectangle may be subdivided into four smaller congruent rectangles, all sharing $O$ as a vertex. Each of these rectangles is analogous, so we can consider our random point $P$ to be without loss of generality in the smaller rectangle with $A$ as a vertex. All points in this smaller rectangle are closer to $A$ than they are to $B$, $C$, or $D$, so we just need to determine the probability that $OP<AP$. [asy]
size(100);
draw((0,0)--(0,100)--(-250,100)--(-250,0)--cycle);
label("$A$",(-250,100),NW); label("$O$",(0,0),SE);
draw((-105,100)--(-145,0));
fill((-105,100)--(-145,0)--(0,0)--(0,100)--cycle, gray(.7));
[/asy] Since a $180^\circ$ rotation about the center of the smaller rectangle takes $O$ to $A$, it takes the shaded region to the unshaded region. Therefore, exactly half the area is shaded, and the overall probability is $\frac{1}{2}$, independent of $k$.