Square board with $N\times N$ squares, at least $N(\sqrt N + {1\over 2})$ colored

Question

In a square board made of $N\times N$ equal squares, at least $N(\sqrt N + {1\over 2})$ squares are colored (all other squares are, say, white). Prove that there are four colored squares that form a rectangle (the four colored squares are at the corners of some rectangle found on the board; for example, in a chess board, squares b2, b7, e2 and e7 form a rectangle while squares a1, a5, b1 and b6 do not).

I have tried to count all rectangles in the board, which is pretty easy, and sum the number of corners found in it, then to count to number a corners defined by coloured squares and make same comparison between the two but I fell short.

Any ideas?

Answer 1

This problem is well-known. I first met it when I was a schoolboy, learning the following problem from All-Soviet-Union mathematical competition from the year when I was born. :-)

208 a) Given a big square consisting of $7\times 7$ squares. You should mark the centres of $k$ points in such a way, that no quadruple of the marked points will be the vertices of a rectangle with the sides parallel to the sides of the given squares. What is the greatest k such that the problem has solution? b) The same problem for $13\times 13$ square.

I scanned its solution from [VE] (in Russian). About the beginning of the millennium I met a similar problem related to graphs at Erich Friedman's Math Magic. Similarly to the scanned solution of the competition problem or to the proof of Proposition 1 from my paper, we can find an upper bound on the maximal number $s=s(m,n)$ of squares which can be colored in a rectangular $m\times n$ board without forming rectangles. Namely, $s\le \frac 12(m+\sqrt{D})$, where $D=4mn^2-4mn+m^2$. In particular, if $D$ is a perfect square, that is $m=\frac{n(n-1)}{k(k-1)}$ for some natural $k$ then the upper bound proposed above becomes $s(m,n)\le km$. See also hardmath’s answer to this question about Füredi’s paper. But the board case is a bit different from the graph case and the known result do not need such subtle analysis as that done by Füredi. Namely, if there exists a Steiner system $S(2,k,n) $ then it yields $s(m,n)\ge km$, so in this case the inequality becomes an equality. This yield us an exact solution for a rectangular board $q^2\times (q^2+q)$ and a square board with a side $q^2+q+1$, where $q$ is a power of a prime number we still cannot construct respective Steiner systems for other $q$ and a rectangular board $n\times n(n-1)/6$, where $n$ has a form $6q+1$ or $6q+3$ for some $q$.

References

[VE] Vasil'ev N.B, Egorov A.A. The problems of the All-Soviet-Union mathematical competitions, Moscow.:Nauka, 1988 (in Russian).

Answer 2

If you remove any row and column, and there are at least $(N-1)(\sqrt{N-1}+1/2)$ coloured squares left, then there is a rectangle by induction.
So you can assume that fewer coloured squares remain when any row and column are removed;
so each of the row and column you removed has at least a certain number of coloured squares, to make the grand total grow from below $(N-1)(\sqrt{N-1}+1/2)$ to at least $N(\sqrt{N}+1/2)$

Answer 3

I'm not sure if you are interested in a solution after almost 3 years but you can do it like this.

Imagine every row as a subset of $\{1,2,...n\}$, so we have $n$ subsets $R_1,...,R_n$ and let $r:= |R_1|+...+|R_n|$ be the number of colored squares, so $\boxed{r\geq n(\sqrt{n}+{1\over 2})}$. Now suppose the statment does not hold. Then each pair $\{i,j\}$ is contained in at most one row $R_k$ and every row $R_k$ contains ${r_k\choose 2}$ such a pairs (where $r_k=|R_k|)$. So we have $${r_1\choose 2}+{r_2\choose 2}+...+{r_n\choose 2}\leq {n\choose 2}$$

By Jensen inequality we have $${r_1\choose 2}+{r_2\choose 2}+...+{r_n\choose 2}\geq {{1\over n}r^2-r\over 2}$$ and thus we have $${{1\over n}r^2-r\over 2}\leq {n\choose 2}$$ Useing boxed relation we get a contradiction.