IMO 2014/2

Let n2n\geq2 be an integer. Consider an n×n n \times n chessboard consisting of n2n^2 unit squares. A configuration of nn rooks on this board is peaceful if every row and every column contains exactly one rook. Find the greatest positive integer kk such that, for each peaceful configuration of nn roots, there is a k×k k \times k square which does not contain a rook on any of its k2k^2 unit squares.

Note by Calvin Lin
5 years, 3 months ago

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The answer is n1.\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 j2×j2j^2 \times j^2 tiles where j=n1.j= \left \lfloor \sqrt{n-1} \right \rfloor. I have posted my solution in AoPS, I can post it here if necessary. :)

Sreejato Bhattacharya - 5 years, 3 months ago

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Could you please post the full solution for this problem?

Saksham Bansal - 5 years, 3 months ago

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I didn't like the phrasing of this question because it was slightly ambiguous.
- It wasn't clear to me that kk should depend on nn. Would have preferred if they called it knk_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 kk rows and kk columns). I'm guessing the answer is "have to be consecutive".

Calvin Lin Staff - 5 years, 3 months ago

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Since k×kk\times k is a square, it has to be consecutive.

Kenny Lau - 5 years, 3 months ago

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Conjecture: k=n2k=\left\lfloor\frac n2\right\rfloor

Kenny Lau - 5 years, 3 months ago

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

seems to be a counterexample to n=4n=4. There is no 2×2 2 \times 2 (consecutive) square.

Calvin Lin Staff - 5 years, 3 months ago

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My mistake, I thought we're finding the maximum value of kk.

Kenny Lau - 5 years, 3 months ago

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@Kenny Lau It is "for each peaceful configuration of nn 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.

Calvin Lin Staff - 5 years, 3 months ago

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If k>n2k>\left\lfloor\frac n2\right\rfloor, there will be (nk)(n-k) rows and (nk)(n-k) columns left, which is not enough to put all rooks, because 2(nk)=2n2k<n2(n-k)=2n-2k<n.

Kenny Lau - 5 years, 3 months ago

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