Alice and Bob are play a game in a $(2n)*(2n)$ chess boared.Alice starts from the top right point moving 1 unit in every turn.Bob starts from the down left square and does the same.The goal of Alice is to make a closed shape ending in its start position and cannot reach any point that was reached before by any of players .if a players cannot move the other player keeps moving.what is the maximum are of the shape that the first player can build with every strategy of second player.
Problem
Source: Iran third round combinatorics problem 1
Tags: combinatorics
04.05.2019 16:38
the answer is n * n Does anybody want my solution??
12.07.2019 21:23
alirezabgi wrote: the answer is n * n Does anybody want my solution?? ? Yeah post it. The answer is $n(n+1)$ btw.
17.05.2021 18:18
Does anyone have a solution to this one?
02.07.2021 22:42
Bumping it again. Solution?
03.07.2021 11:44
The minimum amount of turn in which B can block A is $4n-3$, since it suffices for him to arrive to the cell next to the top right that A didn't initially go through, and so A has at most $4n-3+1=4n-2$ moves he can use. Assume A does $a$ moves to the left (and necessarily $a$ to the right) and $b$ moves down (and necessarily $b$ up); if so, we must have $2a+2b=4n-2\implies a+b=2n-1$. Horizontally, he can at most cover $a+1$ columns and vertically at most $b+1$ columns, so the maximum theoretical area is $(a+1)(b+1)$. Now by amgm $(a+1)(b+1)\leq (n-\frac 12)^2$, and the closest case to equality is $a=n-1$ and $b=n$ (or viceversa), which gives an area of $n(n+1)$, so we now must only show that it can be done. Let us use coordinates $(x,y)$ ($1\leq x,y\leq 2n$) to describe the cells. With no problem A can get in a straight line to $(n+1,2n)$ and then to $(n+1,n+1)$, since in the meanwhile B can only cover a Manhattan distance of $2n-3$. Now, A can go down (that cell couldn't have been reached with $2n-2$ moves by B). But now A has always the advantage over B (since he moves first) and so he can just return in a straight line going to $(2n,n)$ and finally back to $(2n,2n)$. Therefore the maximum area A can cover is $n(n+1)$.
30.07.2022 20:08
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