TITLE: Can the boy escape the teacher for a regular $n$-gon?
QUESTION [14 upvotes]: 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?

REPLY [5 votes]: 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.