Pursuit Problem II: Surrounded in Marauders' Circular Cove

Question

(This is the sequel to this puzzle. It has a similar setup, but believe me, the solution is very different. Be careful! The answer is counterintuitive. It shocked me at first.)

You are a pirate. Previously, you captured a rogue mutineer. She begged for a second chance on the crew, in exchange for the location of her criminal headquarters: Marauders' Circular Cove.

In promise of great treasure, you raided the cove at night. But it was a trap! Now smugglers are coming out of the forest, more than you can count.

(Pictured: there are WAY MORE than this)

Your ship moves at the same speed as the smugglers. The smugglers cannot swim. If you manage to take even one step onto the shore, then you can instantly beach the ship and escape into the cover of darkness.

Right now (and only right now, i.e. not after the chase starts), you can bribe some of the smugglers to leave, so that you only have to evade $n$ of them. Your goal is to reach the shore without getting caught. The smugglers' goal is to catch you when you reach the shore, or starve you out on the lake forever.

How many smugglers can you evade, if the cove is shaped like a perfect circle?

The ship and smugglers are all points of zero radius. Everyone can always see everyone else's position. This is not a loophole-finding contest: there are no issues with reaction time, scurvy, et cetera.

Answer 1

It is possible to escape from

any finite number of smugglers.

Say the lake has unit radius. Label the boundary of the lake with angles $\theta$, measured in radians. We will begin by heading to a point near the boundary of the lake, by angle $\theta=0$. For the moment, pretend we actually reach the boundary but cannot escape due to a smuggler (moving inward just a bit will not change the answer).

Suppose we choose a point $0<\theta<\pi$ on the boundary and head straight there. The distance to this point is $2\sin(\theta/2)$, so to keep us from escaping, there must be a smuggler somewhere in the arc $$A_\theta:=\big[\theta-2\sin(\theta/2),\,\theta+2\sin(\theta/2)\big].$$ We will find an infinite sequence $\theta_1,\theta_2,\ldots$ of angles such that the arcs $A_{\theta_1},A_{\theta_2},\ldots$ are disjoint. This means that for any positioning of a finite number of smugglers, there will be some $\theta_i$ that we can reach before any of them.

We define the $\theta_i$ recursively. Start with $\theta_1=\pi$. For $i\geq 2$, define $$\theta_i=\frac{\theta_{i-1}-2\sin(\theta_{i-1}/2)}{2}.$$ Observe that $0<2\sin(\theta/2)<\theta$ for every $0<\theta<\pi$, so that $\theta_1>\theta_2>\ldots>0$. The arcs $A_{\theta_i}$ are disjoint because $\theta_{i-1}-2\sin(\theta_{i-1}/2)>\theta_i+2\sin(\theta_i/2)$, which comes from the definition of $\theta_i$ and the inequality $\theta_i+2\sin(\theta_i/2)<2\theta_i$.

If there are $N$ smugglers, the arcs $A_{\theta_1},\ldots,A_{\theta_{N+1}}$ are disjoint, so there must be one of these arc $A_{\theta_i}$ containing no smugglers. We then head directly toward the point $\theta=\theta_i$, and no smuggler can keep us from escaping.

Finally, suppose that instead of starting on the boundary of the lake, we instead start $\epsilon>0$ units in from the boundary. For each $0<\theta<\pi$, we define $A^\epsilon_\theta$ to be the arc of points on the boundary of the lake from which a smuggler could reach the point $\theta$ as quickly as we can. The arcs $A_{\theta_1},\ldots,A_{\theta_{N+1}}$ are disjoint, so for $\epsilon$ sufficiently small, $A_{\theta_1}^\epsilon,\ldots,A_{\theta_{N+1}}^\epsilon$ will also be disjoint. This means we can still find $i\in\{1,\ldots,N+1\}$ such that we can reach $\theta=\theta_i$ before any of the smugglers.

Answer 2

Surprisingly,

the marauder can always escape!

Suppose you have a very very high number of smugglers. The strategy to escape is approaching the coast as much as possible (the more the smugglers, the closest you have to be), then follow a straight line (a chord) which is always shorter than the smugglers' path to reach you.
Check the picture: when you come very very close to the coast, the other smugglers won't be able to stay near enough, so you can easily pass between them!

Answer 3

I think you can evade:

an infinite number of smugglers

Say you are 10 feet from the edge of the cove. A smuggler would have to be within 10 feet of the closest point on shore or you'll escape. If smugglers are spaced every 20 feet, with one on the shore closest to you, you can circle the ship until you are between 2 smugglers, then make a break for it.

Now take it further. Say you're 1 foot away. Now smugglers would have to be spaced every 2 feet. 1 inch leads to 2 inch spacing. Eventually you can get close enough to handle any gap between smugglers.

Answer 4

Since the answer has already been stated, I thought would provide another view as to why it takes

infinite smugglers.

Here the red dot is our boat, and the green dot is a smuggler. Since the shortest path our boat can take to shore (safely) is X length, and the green dot must move at least (sqrt)(x²+y²), X will always be shorter. If you were to add another smuggler onto the curve, the boat simply needs to move close enough to the shore such that only one dot is remaining (like the picture). [Think of it as you are trying to race your friend to a finishing point and only have to travel in a straight line (shortest possible distance) and your friend is trying to beat you to the finish line, but has to travel along a curve]