Problem 1934

We claim the answer is $\frac{1}{\sqrt{2}}$.

Firstly, we prove that there is a configuration of landmines such that the rabbit cannot be closer than $\frac{1}{\sqrt{2}}$ to the origin.
Consider a square packing of circles, with centers $\left(\frac{2m+1}{\sqrt{2}}, \frac{2n+1}{\sqrt{2}}\right)$ for all $m, n \in \mathbb{Z}$ and radii $\frac{1}{\sqrt{2}}$.

If the rabbit begins on one of the tangency points, it can only move to other tangency points and thus will always be at least $\frac{1}{\sqrt{2}}$ away from the origin.

Now we prove that $\frac{1}{\sqrt{2}}$ can always be achieved. Firstly, we reduce the distance to $\le 1$. Denote the origin by $O$ and the rabbit's position by $A$. Let $d = OA$ and suppose that $d > 1$. Consider the circles $(O, d)$ and $(A, 1)$ which intersect at two points, $X$ and $Y$.

Claim. No landmine can fully contain the red arc $XY$.
Proof. If a landmine were to exist, consider the center of the landmine. Since it is closer to $X$ and $Y$ than $A$, it must be in the blue region bounded by the perpendicular bisectors of $XA$ and $YA$. But since it is also closer to $X$ and $Y$ than $O$, it must be in the green region bounded by the perpendicular bisectors of $XO$ and $YO$.

Notice that the two regions are disjoint, meaning the center of the landmine cannot exist. $\Box$

Hence, the rabbit can hop onto the arc. Notice that the maximum distance from the arc to $O$ is at $XO = YO = \sqrt{d^2-1}$. Hence the squared distance between the rabbit and $O$ decreases by at least $1$ every move and thus the rabbit can eventually achieve a distance of at most $1$ from $O$.

Now, if the distance $OA$ is still greater than $\frac{1}{\sqrt{2}}$, then $\frac{1}{\sqrt{2}} < OA \le 1$. Thus we consider the circles $(O, \frac{1}{\sqrt{2}})$ and $(A, 1)$. Since $OP^2 + OA^2 > 1 = AP^2$, $\angle POA$ is acute, thus $O$ and $A$ lie on opposite sides of the line $PQ$. Therefore, similar to before, the perpendicular bisectors form two disjoint regions, no landmine fully contains the red arc $PQ$.

Therefore, the rabbit can hop onto the red arc and will be of distance at most $\frac{1}{\sqrt{2}}$ from $O$.