(Written before the problem statement was corrected)
$AK>KB$ is simply not true in all cases:
In this case $AD>BC$ but $AK<KB$.
Some additional thoughts:
Extend $DK$ and $BC$ and denote the intersection point with $E$. Triangle $EDC$ is isosceles so:
$$EB=EC-BC=CD-BC$$
It's easy to see that triangles BKE and AKD are similar. So we have:
$$\frac{AK}{KB}=\frac{AD}{EB}$$
If $AK>KB$ then :
$$\frac{AK}{KB}=\frac{AD}{EB}=\frac{AD}{CD-BC}>1\tag{1}$$ which is true for
$$AD+BC>CD$$
So this problem needs some additional condtion. Otherwise, the premise is not true in a general case.
EDIT: It turns out that the problem statement was wrong. We have to suppose that $AD>CD$.
In that case, from (1) it is obvious that:
$$\frac{AK}{KB}=\frac{AD}{CD-BC}\gt \frac{AD}{CD}\gt1$$
...or:
$$AK>KB$$
Let $C$ and $D$ be the midpoints of $AO$ and $OB$ respectively. Draw $PC$ and $PD$. Drop perpendiculars from point $P$ on $OC$ and $OB$ meeting them at point $E$ and $F$ respectively.
Notice that, if one vertex of any such equilateral triangle is between $C$ and $E$, then the other one(vertex $P$ is fixed ) has to be between $O$ and $F$. Every such triangle will have a congruent triangle which will have one of its vertices between $E$ and $O$ and the other one between $F$ and $D$. Also notice that there can't be two congruent triangles with both of them having one of their vertices between $C$ and $E$ and the other one between $O$ and $F$. Hence only those triangles are to be counted.
Choose a random point $Q$ on segment $CE$. Now choose a point $R$ on segment $OF$ such that $OR=10-OQ$. Draw $PQ$ and $PR$.
Observe that, in $\triangle PCQ$ and $\triangle POR$,
$\left(i\right)$ $CQ=10-OQ=OR$
$\left(ii\right)$ $\angle PCQ=60^{\circ}=\angle POR$
$\left(iii\right)$ $PC=PO$ [ Since $\triangle PCO$ is equilateral ]
Hence, $\triangle PCQ\cong \triangle POR$ by $S-S-A$ criterion of congruence.
Thus, $\angle QPR=\angle QPO+\angle OPR=\angle QPO+\angle CPQ=\angle CPO=60^{\circ}$ and $PQ=PR$; Which indicates that $\triangle PQR$ is equilateral.
This is true for every point $Q$ on $CE$. Since there are an infinite number of such points on $CE$, $k\to\infty$.
Therefore, $\boxed {\lfloor\left(\frac{200}{k}\right)\rfloor \to 0}$
Best Answer
Here I could find a justification,
Assume the larger base is $AB$ and $G$ is its midpoint. The trapezoids wich have the legs equal to half of the larger base can be obtained by drawing a parallel line to $AB$ and its intersections with the circles are endpoints of the smaller base. Here we can see that the length of the smaller base can vary from $0$ to length of $AB$ (exclusively). So there is only one such trapezoid with smaller base equals to half of the larger one (which is the one that can be constructed by putting three equilateral triangles together).