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$\newcommand{\man}{\text{man}}$ $\newcommand{\tiger}{\text{tiger}}$

Consider the following situation:

A man is standing in the center of a circle of radius $r$. On the circle there is a tiger. The man move in arbitrary piecewise smooth paths with a fixed speed, and the tiger can move only along the circle, where $V_{\tiger}=cV_{\man} $, $c>1$. I model the man and the tiger as "points" - they have no actual volume.

Question: For which values of $c$ can the man escape from the enclosing circle without meeting the tiger? (assuming the tiger does its best to catch him).

If $c < \pi$ then its trivial. The man just goes in a straight line in the opposite direction from the tiger: The man reaches the circle after a time of $\frac{r}{V_{\man}}$ and the tiger reaches that same point after $\frac{\pi r}{V_{\tiger}}$.

In fact, there is a more sophisticated idea, which works for any $c < \pi +1$, as I explain below.

Is $\pi+1$ the highest speeds ration when the game is solvable?

The man starts by going to an intermediate circle with sufficiently small radius $r_1$, in the sense that $$ \frac{r_1}{V_{\man}} < \frac{r}{V_{\tiger}}= \frac{r}{cV_{\man}} \Rightarrow r_1 < \frac{r}{c}. \tag{1}$$

The man than starts circling the small circle until it lies on the "opposite" point of the tiger, so the distance between them is $r+r_1$.

If $r_1$ satisfies $$ \frac{r-r_1}{V_{\man}} < \frac{\pi r}{V_{\tiger}}=\frac{\pi r}{cV_{\man}}, \tag{2}$$

then the man can win.

Combining $(1),(2)$ we see that $r_1$ must satisfy

$$ r(1-\frac{\pi}{c}) < r_1 < \frac{r}{c},$$ so this technique can work if and only if $ 1-\frac{\pi}{c} < \frac{1}{c} $, i.e. $c < \pi +1$.

I am also not entirely sure how to formalize this question mathematically. We are looking for a winning "strategy"-but what is a strategy? In my solution with the intermediate circle, what the man does (e.g. direction of movement) is affected by the tiger's moves. In that case one could say that there is a periodic path of the man such that no matter what the tiger does, there exist a "breaking point" where the man can run free.

But is this really the shape of the best solution?

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  • $\begingroup$ The duplicate shows the constant is about $4.6033$ $\endgroup$ – Ross Millikan Feb 11 '18 at 16:12