Subfield: Combinatorics


Question: A hunter and an invisible rabbit play a game in the Euclidean plane. The hunter's starting point $H_{0}$ coincides with the rabbit's starting point $R_{0}$. In the $n^{\text {th }}$ round of the game $(n \geqslant 1)$, the following happens.

(1) First the invisible rabbit moves secretly and unobserved from its current point $R_{n-1}$ to some new point $R_{n}$ with $R_{n-1} R_{n}=1$.

(2) The hunter has a tracking device (e.g. dog) that returns an approximate position $R_{n}^{\prime}$ of the rabbit, so that $R_{n} R_{n}^{\prime} \leqslant 1$.

(3) The hunter then visibly moves from point $H_{n-1}$ to a new point $H_{n}$ with $H_{n-1} H_{n}=1$.

Is there a strategy for the hunter that guarantees that after $10^{9}$ such rounds the distance between the hunter and the rabbit is below 100 ?

Solution:
If the answer were "yes", the hunter would have a strategy that would "work", no matter how the rabbit moved or where the radar pings $R_{n}^{\prime}$ appeared. We will show the opposite: with bad luck from the radar pings, there is no strategy for the hunter that guarantees that the distance stays below 100 in $10^{9}$ rounds.

So, let $d_{n}$ be the distance between the hunter and the rabbit after $n$ rounds. Of course, if $d_{n} \geqslant 100$ for any $n<10^{9}$, the rabbit has won — it just needs to move straight away from the hunter, and the distance will be kept at or above 100 thereon.

We will now show that, while $d_{n}<100$, whatever given strategy the hunter follows, the rabbit has a way of increasing $d_{n}^{2}$ by at least $\frac{1}{2}$ every 200 rounds (as long as the radar pings are lucky enough for the rabbit). This way, $d_{n}^{2}$ will reach $10^{4}$ in less than $2 \cdot 10^{4} \cdot 200=4 \cdot 10^{6}<10^{9}$ rounds, and the rabbit wins.

Suppose the hunter is at $H_{n}$ and the rabbit is at $R_{n}$. Suppose even that the rabbit reveals its position at this moment to the hunter (this allows us to ignore all information from previous radar pings). Let $r$ be the line $H_{n} R_{n}$, and $Y_{1}$ and $Y_{2}$ be points which are 1 unit away from $r$ and 200 units away from $R_{n}$, as in the figure below.

![](https://cdn.mathpix.com/cropped/2023_12_21_52ff42def30603016bd3g-1.jpg?height=440&width=1140&top_left_y=1642&top_left_x=458)

The rabbit's plan is simply to choose one of the points $Y_{1}$ or $Y_{2}$ and hop 200 rounds straight towards it. Since all hops stay within 1 distance unit from $r$, it is possible that all radar pings stay on $r$. In particular, in this case, the hunter has no way of knowing whether the rabbit chose $Y_{1}$ or $Y_{2}$.

Looking at such pings, what is the hunter going to do? If the hunter's strategy tells him to go 200 rounds straight to the right, he ends up at point $H^{\prime}$ in the figure. Note that the hunter does not have a better alternative! Indeed, after these 200 rounds he will always end up at a point to the left of $H^{\prime}$. If his strategy took him to a point above $r$, he would end up even further from $Y_{2}$; and if his strategy took him below $r$, he would end up even further from $Y_{1}$. In other words, no matter what strategy the hunter follows, he can never be sure his distance to the rabbit will be less than $y \stackrel{\text { def }}{=} H^{\prime} Y_{1}=H^{\prime} Y_{2}$ after these 200 rounds.

To estimate $y^{2}$, we take $Z$ as the midpoint of segment $Y_{1} Y_{2}$, we take $R^{\prime}$ as a point 200 units to the right of $R_{n}$ and we define $\varepsilon=Z R^{\prime}$ (note that $H^{\prime} R^{\prime}=d_{n}$ ). Then



$$
y^{2}=1+\left(H^{\prime} Z\right)^{2}=1+\left(d_{n}-\varepsilon\right)^{2}
$$

where

$$
\varepsilon=200-R_{n} Z=200-\sqrt{200^{2}-1}=\frac{1}{200+\sqrt{200^{2}-1}}>\frac{1}{400}
$$

In particular, $\varepsilon^{2}+1=400 \varepsilon$, so

$$
y^{2}=d_{n}^{2}-2 \varepsilon d_{n}+\varepsilon^{2}+1=d_{n}^{2}+\varepsilon\left(400-2 d_{n}\right)
$$

Since $\varepsilon>\frac{1}{400}$ and we assumed $d_{n}<100$, this shows that $y^{2}>d_{n}^{2}+\frac{1}{2}$. So, as we claimed, with this list of radar pings, no matter what the hunter does, the rabbit might achieve $d_{n+200}^{2}>d_{n}^{2}+\frac{1}{2}$. The wabbit wins.

Final answer: There is no such strategy for the hunter. The rabbit "wins".