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IMO 2014/2

Let $$n\geq2$$ be an integer. Consider an $$n \times n$$ chessboard consisting of $$n^2$$ unit squares. A configuration of $$n$$ rooks on this board is peaceful if every row and every column contains exactly one rook. Find the greatest positive integer $$k$$ such that, for each peaceful configuration of $$n$$ roots, there is a $$k \times k$$ square which does not contain a rook on any of its $$k^2$$ unit squares.

Note by Calvin Lin
2 years, 10 months ago

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The answer is $$\left \lfloor \sqrt{n-1} \right \rfloor.$$ Rather simple for an IMO P2; I suppose it's a C2 in the shortlist. The main idea is to tile the grid into several disjoint $$j^2 \times j^2$$ tiles where $$j= \left \lfloor \sqrt{n-1} \right \rfloor.$$ I have posted my solution in AoPS, I can post it here if necessary. :) · 2 years, 10 months ago

Could you please post the full solution for this problem? · 2 years, 10 months ago

I didn't like the phrasing of this question because it was slightly ambiguous.
- It wasn't clear to me that $$k$$ should depend on $$n$$. Would have preferred if they called it $$k_n$$ instead.
- It wasn't clear if the square have to be consecutive rows and columns, or could be a sub-matrix (i.e. you get to pick $$k$$ rows and $$k$$ columns). I'm guessing the answer is "have to be consecutive". Staff · 2 years, 10 months ago

Since $$k\times k$$ is a square, it has to be consecutive. · 2 years, 10 months ago

Conjecture: $k=\left\lfloor\frac n2\right\rfloor$ · 2 years, 10 months ago

$$\begin{array} {| l | l | l | l |} \hline x & & & \\ \hline & & x & \\ \hline & x & & \\ \hline & & & x \\ \hline \end{array}$$

seems to be a counterexample to $$n=4$$. There is no $$2 \times 2$$ (consecutive) square. Staff · 2 years, 10 months ago

My mistake, I thought we're finding the maximum value of $$k$$. · 2 years, 10 months ago

It is "for each peaceful configuration of $$n$$ roots". I've noticed that you made a similar interpretation error in another problem, which asked for the "smallest needed", which could be interpreted as "the most ever".

Otherwise, your solution works for the problem that you're solving. Staff · 2 years, 10 months ago

If $$k>\left\lfloor\frac n2\right\rfloor$$, there will be $$(n-k)$$ rows and $$(n-k)$$ columns left, which is not enough to put all rooks, because $$2(n-k)=2n-2k<n$$. · 2 years, 10 months ago