Can the boy escape the teacher for a regular $n$-gon?

This is related to Prove that the boy cannot escape the teacher

Suppose there is a boy in the center of a regular $n$-gon. The teacher is on the edge of the $n$-gon (but cannot leave the edge) and wants to capture the boy. At the beginning he is on a vertex. The teacher is $v(n)$ times faster than the boy. Which is the maximum $v(n)$ such that the boy can escape? (By escaping means he reaches the edge of the $n$-gon and the teacher is not there)

From the linked question we know $3 \le v(4) < 6$, and for $n= \infty$ (a circle) then I know a strategy such that $v(\infty) = \pi + 1$ suffice; I don't know if this is optimal. I also put convergence in the tags because my wild hypothesis is that the maximum $v(n)$ will converge to some value and it would be interesting to know which one!

Any cool way to solve this?

Solution 1:

This is an interesting question! I thought I would get the ball rolling by looking first at the case for the square pool (4-gon).

A diagram:

The teacher is located at point $A$, the boy is in the center of the pool at point $E$. The square has the size $2x2$, the speed of the boy is arbitrarily set at $1$ and the speed of the teacher is $K > 1$.

It is easy to check that if the boy swims directly to the corner furthest from the teacher (point $C$), $K=2 \sqrt2 \approx 2.83$ is sufficient for the teacher to catch the boy. A slightly better strategy for the boy is to swim perpendicular to the nearest pool side (furthest from $A$) which requires $K=3$.

A much better strategy (although not quite the optimal, as I will show later) is the one illustrated in green in the diagram. The boy starts off by heading for the point furthest from the teacher (point $C$). The teacher must now also head for $C$ and can either go left towards $D$ or down towards $B$. Both paths are equally long. Let's assume he goes down (red line in diagram). The boy continues heading for $C$ until the teacher reaches point $G$. At this time, the boy and the teacher are equidistant from the horizontal centerline of the pool. More importantly, the teacher is now at the maximum distance from the point $H$ (mirror point of $G$). The boy now heads for $H$. It doesn't matter at this time whether the teacher turns around or continues in the direction he was going. The point $H$ is equally far away.

Let us set up the equations for this scenario. Let $d_1$ be the distance to the horizontal centerline at the moment the boy heads for $H$. Let $t_1$ be the time since start. Then, for the teacher $$t_1 = \frac{1-d_1}{K}$$

and for the boy $$t_1= \frac{d_1}{\frac{1}{\sqrt2}}= \sqrt2d_1$$

Let the time for the boy to swim from $F$ to $H$ be $t_2$. Then $$t_2 = 1-d_1$$

For the teacher, the time to reach $H$ from $G$ would be $$t_2 = \frac{4}{K}$$

Solving these equations for $d_1$ and $K$, we find that $$K=2+\sqrt{2+2\sqrt2}\approx4.61$$

A substantial improvement of $K=3$. However, as I mentioned earlier, a further improvement exists.

Suppose the boy doesn't wait for the teacher to reach $G$ before changing direction, but changes his direction as soon as he sees whether the teacher moves left towards $D$ or down towards $B$. Suppose he changes his direction (continuously) such that he is always heading towards the point on the square furthest from the teacher. The boy thus starts heading for the point $C$ but then heads more and more towards the pool side $CD$. What can the teacher do? He can turn around and head for the pool side $CD$ as well, but if he does, the boy will again change direction and head for the pool side $CB$, always aiming for the point on the square furthest from the teacher. The teacher gains nothing and is best served continuing his progress towards $B$.

It was not intuitively clear to me what form the boy's curve would take in this scenario, so I simulated it. A close-up of the new path of the boy is shown below:

The green curve is the same as the green curve in the first diagram, while the purple curve is the new scenario.

The result of the simulation is that in this scenario, $K\approx4.782$.

The above is not a proof that no better scenario is possible and I invite suggestions on how such a proof could be made. And if anyone can work out what the analytical form of the curve is (it doesn't seem to be a circle or a parabola) that would also be much appreciated.

The above scenario, by the way, would be directly applicable to all regular polygons with an even number of sides.

EDIT

An old question revived!

After reading the link provided by @David K in the comments, I think I can improve on my answer above.

Suppose we make an inner square in the pool with side length $\frac{2}{K}-\epsilon$, where $\epsilon$ approaches $0$. The boy's angular speed along the perimeter of this inner square will slightly outpace the teacher's angular speed along the perimeter of the outer square. Suppose further that the boy swims to the midpoint of a side on this inner square. The teacher will then run to the matching midpoint on the outer square and thus be as close to the boy as possible. If the boy now starts swimming along the perimeter of the inner square, he will slightly outpace the teacher as the teacher attempts to match the boy's position on the outer square. In fact, the boy can make shortcuts within the inner square which are not available to the teacher on the outer square. It therefore seems clear that the boy can eventually position himself wherever he wants on the perimeter of the inner square, in relation to where the teacher is. Let us now consider two scenarios for the teacher's position:

1. The teacher is at the midpoint of side

The boy then positions himself at the midpoint of the inner square's side which is furthest from the teacher. The boy then swims straight for the nearest poolside. The time it takes him to reach this is: $$t_3 = 1-\frac{1}{K}$$

The time it takes for the teacher to reach the boy is: $$t_3= \frac{4}{K}$$

Setting these two equal gives: $$K=5$$

An improvement of my previous answer!

2. The teacher is at a corner

The boy then positions himself at the opposite corner on the inner square. He then uses the basic strategy I gave in my old answer, i.e. he swims straight for the corner until the teacher reaches a position which is equidistant from the boy's nearest poolside, and then the boy swims straight for this poolside. Looking at the equations from my old answer, only one needs to be changed, namely the one for $t_1$: $$t_1=\frac{d_1-\frac{1}{K}}{\frac{1}{\sqrt{2}}}$$

which results in $$K^2-5K-\frac{4}{\sqrt{2}}=0$$

which gives a value of $K\gt 5$. The teacher will therefore not position himself at a corner.

Summing up, the optimal answer for a square pool appears to be $$K = 5$$

Solution 2:

See all the details here; they are too long for a MSE answer. In summary, suppose the pool occupies a regular $n$-gon $P$. Let $\theta=\pi/n$, and let $K$ be the largest integer between $0$ and $n$ such that $$ \sin(K\theta)-(K+n)\tan\theta\cos(K\theta) $$ is negative. Equivalently, $K$ is the floor of the unique root of $\tan(x\theta)-(x+n)\tan\theta$ in $[1,n/2)$. Then $$ \cos((K+2)\theta)\leq\frac{2\sin(K\theta)}{(K+n)\tan\theta}-\cos(K\theta)<\cos(K\theta) $$ so we can define $$ \alpha=\frac12\left(K\theta+\cos^{-1}\left( \frac{2\sin(K\theta)}{(K+n)\tan\theta}-\cos(K\theta)\right)\right), $$ Then the cutoff speed is $\lambda=1/\cos\alpha$. Explicitly, for speeds $<\lambda$ the student wins regardless of starting positions. For speeds $>\lambda$ the teacher wins provided the student starts in a regular $n$-gon $Q$, where $Q$ is obtained from $P$ by rotating by $K\theta$ and scaling by $$ s=\frac{\cos\alpha}{\cos(\alpha-K\theta)} $$ about the center of $P$.

As for a circular pool, the student must consider two cases in which the teacher chases clockwise or anti-clockwise, and find the optimal path in each. In fact from each vertex of $Q$ there are two equally optimal paths in each direction, as shown below (follow the link for other values of $n$).

For a square we have $K=1$, $\alpha\approx1.397$ and $\lambda\approx5.789$.

For large $n$, $\theta$ is small and $n\tan\theta\approx\pi$. Thus $\alpha\approx K\theta\approx\mu$, where $\mu$ satisfies $$ \tan\mu=\mu+\pi. $$ This agrees with the cutoff for a circular pool.