Let $A{}$ be a point in the Cartesian plane. At each step, Ann tells Bob a number $0\leqslant a\leqslant 1$ and he then moves $A{}$ in one of the four cardinal directions, at his choice, by a distance of $a{}.$ This process cotinues as long as Ann wishes. Amongst every 100 consecutive moves, each of the four possible moves should have been made at least once. Ann's goal is to force Bob to eventually choose a point at a distance greater than 100 from the initial position of $A.{}$ Can Ann achieve her goal? Selected from an Argentine Olympiad
Problem
Source: Romania TST 2024 Day 1 P4
Tags: combinatorics, combinatorial geometry
SomeonesPenguin
01.08.2024 22:41
We will show that Ann can win.
Ann's strategy is to guarantee that Bob will move upwards a distance of at least $\frac{1}{2^{99}}$ in at most $100$ moves. Here is the strategy: Ann tells Bob $\frac{1}{2^{99}}$. If Bob moves upwards, she repeats the strategy from the start. If Bob moves in any other direction, Ann will say the number $\frac{1}{2^{98}}$ and so on, doubling the value that she previously said, whenever Bob doesnt move upwards. Generally, Ann will say $\frac{1}{2^{i}}$ and if Bob moves upwards, then she starts over from saying $\frac{1}{2^{99}}$, but if Bob moves in any other direction, on the next turn she will say $\frac{1}{2^{i-1}}$. If Bob moved upwards when Ann said $\frac{1}{2^{i}}$, he could've went downwards a maximum of $\frac{1}{2^{99}}+\frac{1}{2^{98}}+\dots+\frac{1}{2^{i-1}}$ so he goes upwards a minimum of $\frac{1}{2^{99}}$. Also, we know that Bob has to go upwards sometime in $100$ moves, so Ann isnt going to say a number greater than $1$ and hence can play this strategy. With this, we have shown that Bob is forced to go upwards a distance of $\frac{1}{2^{99}}$ in maximum $100$ moves so after enough time, he will go upwards a distance of $100$ which clearly means that he will be at a distance greater than $100$ away from the starting point.