Source: IMO-2017-P3
Category: Operations and Strategies

Problem:
A hunter and an invisible rabbit play a game in the plane.  
The rabbit and the hunter start at the same point $A_0 = B_0$.  
In the $n$th round of the game ($n \ge 1$), three events occur in order:

(i) The rabbit moves invisibly from $A_{n-1}$ to a point $A_n$ such that $A_{n-1}A_n = 1$.

(ii) The hunter has a tracking device (e.g., a dog) which reports an approximate location $P_n$ of the rabbit, satisfying $P_nA_n \le 1$.

(iii) The hunter moves visibly from $B_{n-1}$ to a point $B_n$ such that $B_{n-1}B_n = 1$.

Let $N = 10^9$. Can the hunter guarantee that $A_NB_N < 100$?

Solution:
No.

We will show how to increase the distance in the following way:

\begin{claim*}
  Suppose the rabbit is at a distance $d \ge 1$ from the hunter
  at some point in time.
  Then it can increase its distance to at least
  $\sqrt{d^2+\frac12}$ in $4d$ steps
  regardless of what the hunter already knows about the rabbit.
\end{claim*}
\begin{proof}
  Consider a positive integer $n > d$, to be chosen later.
  Let the hunter start at $B$ and the rabbit at $A$.
  Let $\ell$ denote line $AB$.

  Now, we may assume the rabbit reveals its location $A$,
  so that all previous information becomes irrelevant.

  The rabbit chooses two points $X$ and $Y$ symmetric about $\ell$
  such that $XY = 2$ and $AX = AY = n$. Let $M$ be the midpoint of segment $XY$; by symmetry, $M$ lies on $\ell$, and $MX = 1$.
  The rabbit can then hop to either $X$ or $Y$,
  pinging a point on $\ell$ each time.
  This takes $n$ hops.

  Now among all points $H$ the hunter can go to,
  $\min \max \{HX, HY\}$ is clearly minimized with $H \in \ell$ by symmetry.
  So the hunter moves to a point $H$ on the ray $BA$ such that $BH = n$ as well.
  In that case the new distance is $HX = HY$.

  Since $H$ and $M$ lie on $\ell$, we have $AH = BH - AB = n - d$. Furthermore, in the right triangle $AMX$, $AM = \sqrt{AX^2 - MX^2} = \sqrt{n^2 - 1}$. Thus, $HM = AM - AH = \sqrt{n^2-1} - (n-d)$.

  We now compute
  \begin{align*}
    HX^2 &= 1 + HM^2 = 1 + \left( \sqrt{AX^2-1}-AH \right)^2 \\
    &= 1 + \left( \sqrt{n^2-1}-(n-d) \right)^2 \\
    &\ge 1 + \left( \left( n-\frac1n \right) - (n-d) \right)^2 \\
    &= 1 + \left(d-\frac{1}{n}\right)^2
  \end{align*}
  which exceeds $d^2 + \frac12$ whenever $n \ge 4d$.
\end{proof}

In particular we can always take $n = 400$ even very crudely;
applying the lemma $2 \cdot 100^2$ times,
this gives a bound of $400 \cdot 2 \cdot 100^2 < 10^9$, as desired.

Answer: No

Grading guidelines:
(Partial) 1. Gave a correct rabbit strategy idea to increase distance: choose two symmetric candidate positions X,Y about the line AB, hop among them while reporting points on AB, with the aim that the hunter’s best response is forced onto AB by symmetry. 2. Established a quantitative distance-growth inequality of the form: starting from distance d, after some number of steps the rabbit can force the distance to be at least sqrt(d^2+c) for a fixed constant c>0 (even with a weaker constant or with an unspecified-but-correct dependence on d for the number of steps). 3. Correctly argued that if the rabbit can repeatedly increase distance by a fixed positive amount in squared distance (or by a multiplicative factor >1) within O(d) steps, then after sufficiently many iterations (and hence within N=10^9 steps) it can exceed 100, concluding the hunter cannot guarantee AB<100. (Almost) 1. Presented the full symmetric two-point construction (X,Y with XY=2, AX=AY=n, midpoint M on AB, rabbit pings on AB each hop) and the hunter optimality-by-symmetry reduction to a position H on AB, but made a minor computational slip in the final estimate for HX^2 (e.g., the bound sqrt(n^2-1) >= n-1/n or the concluding choice n>=4d). 2. Proved the key claim ‘from distance d the rabbit can force distance at least sqrt(d^2+1/2) in at most 4d steps’ but did not correctly finish the iteration to reach >100 within 10^9 steps (e.g., chose the number of iterations or total-step bound incorrectly, or omitted checking d>=1/base case). 3. Solution is almost complete, but made minor mistakes which are not negligible.