Can the thief escape (from a smooth, simple closed curve)?

Question

Let $C\subset \mathbb{R}^2$ be a smooth, simple closed curve. The thief is inside $C$. Before he starts to move, the police bureau of the $\mathbb{R}^2$ world can freely place countably infinite officers on $C$. We know that

• The thief and the officers move simultaneously and continuously. Maximum speed is $1$ for everyone.
• The officers are restricted to move on $C$. They can pass right through each other without collision.
• The thief is caught if his coordinates coincide with those of an officer.

There're 3 possibilities:

1. The thief always has a plan to get out of $C$.
2. The officers always have a plan to prevent the thief from getting out.
3. It depends on the shape of $C$.

Which one is true?

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Response to comments:

1. If $m(t)$ is a path of the thief, continuous movement means $\vert m(t)-m(s)\vert \leq \vert t-s\vert, \forall t,s$. In particular, we do not require the path to be differentiable. Similarly for an officer path $l(t)$.
2. Let $Q(t)\subset C$ be the set of officers at time $t$. The thief escapes if $\ \exists_t\,m(t)\,\in\, C\setminus Q(t)$.
3. The thief and the officers have perfect information about everybody's current position and move according to that information. For example, officer 1 may adopt a strategy like "if $\vert m_x-l_x\vert\gt 0$, move at maximum speed in the direction that decreases it, otherwise stay still".

Answer 1

Claim. The thief $T$ can escape if $C$ is a circle, with a simple strategy of dribbling left and right each policeman at a time in such a way that he is left out of reach of the thief no matter what the future dribbles will be.

Proof. The key insight is due to Pietro Majer: the thief can approach $C$ in such a way that its shadow $S$ (closest point) on $C$ always moves at speed faster than $1$.

Assume $C$ has radius $1$ and pick a number $r$ once and for all, with $1/2<r<1$.

At any time there are exactly two circles $C_1$, $C_2$ of radius $r$, tangent to $C$ and contaning $T$. The strategy of the the thief is to always run at speed $1$, alternating between these two ways:

$-$ either move right on $C_1$ towards $A$ (dragging $C_2$ and $B$ along)

$-$ or move left on $C_2$ towards $B$ (dragging $C_1$ and $A$ along).

No matter how $T$ zigzags left and right, the symmetry between $C_1$ and $C_2$ with respect to $T$ guarantees that the path followed by $T$ has the same total length $\overset{\frown}{TA}=\overset{\frown}{TB}$. Therefore $T$ will land on $C$ in finite time.

As for the shadow $S$, $r>1/2$ implies that it always moves at speed $>1$, i.e. the (infinitesimal) arc length inequality $\delta_1<\delta_2$ always holds (see figure). This is a tedious but elementary trigonometric inequality, better left to the reader.

Before detailing how the thief's zigzags are decided, we need to notice that any policeman to the right of $A$, or to the left of $B$, by strictly more than $\overset{\frown}{SA}$, is inactive, in other words he can never catch $T$, even if $T$ runs towards him all the way to $C$.

Finally, enumerate all the policemen $P_1, P_2, P_3 \dots$

Take the first active policeman on the list, say $P_{i_1}$. If $P_{i_1}$ is to the left of $S$ (or on $S$) $T$ chooses to run right towards $A$ at speed $1$. (Similarly if $P$ were to the right, $T$ would choose to run left towards $B$.) Because $S$ moves at speed $>1$ at some point $P_{i_1}$ permanently falls behind $S$ by some amount $\epsilon_1$ (it doesn't matter how small). However $P_{i_1}$ may still be within the active range, so $T$ keeps running towards $A$ until $\overset{\frown}{SA}<\epsilon_1/3$. Since $\overset{\frown}{SB}=\overset{\frown}{SA}$, now $P_{i_1}$ is behind $B$ by strictly more than $\epsilon_1/3$. This renders $P_{i_1}$ permanently inactive!

Similarly, take the next active policeman $P_{i_2}$ on the list. Again $T$ runs in the direction away from him, until $P_{i_2}$ is inactive too. There at most countably many active policemen and one by one they all become inactive, guaranteeing that $T$ lands on a police-free point of $C$.$\quad \blacksquare$

This proof should easily extend to any curve $C$, since $T$ can first move close to a point of positive curvature, where locally the curve can be approximated well by a circle.

Answer 2

The cops should always win. Here is a sketch of a proof.

1. One cop (for short) trying to catch the thief crossing a segment: the cop can run on the segment at top speed towards the thief (provided the thief starts far enough and the cop's reaction time is 0). Clearly they will meet.

2. Likewise a dense set of cops on a convex curve $C$: they can run towards the thief only when the latter is "close" to them, while keeping dense all the time. To make this precise: let $r$ be the minimum radius of curvature, which exists because $C$ is smooth and compact; whenever the thief is closer to $C$ than $r$ there is a unique point $P\in C$ to which the thief is closest, with distance $d_P<r$; if $d\ge d_P$ is the distance of a cop from the thief, that cop should run towards $P$ at speed $\text{max}(1-d/r,0)$. This strategy is continuous on $C$ and guarantees that the cops will stay a dense set, and also approaches the strategy in 1) as the thief gets closer to $C$, guaranteeing an eventual catch.

3. This works on a non convex curve too. On a curve with a singularity the thief can try to split the cops and trap them in a cul de sac, but on a smooth curve that is thwarted by the cops remaining dense always.

Update. See comments for where this argument goes wrong.