6
$\begingroup$

Given sets $A=\{1,2,\dots ,10\}, B=\{1,2,\dots,12\}$. Let $S\subset A\times B$ s.t. $|S|=61$.

Prove that there exist three pairs $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ in $S$ which fulfill:$$ x_1=x_2,\quad |y_1-y_2|=1,\quad |x_2-x_3|=1,\quad y_2=y_3 $$

Progress:
I figured that there are at most $4$ legitimate patterns which a randomly picked pair (e.g $(a,b)$) can fit into:

$1.$ $(a,b-1),(a,b),(a-1,b)$
$2.$ $(a,b-1),(a,b),(a+1,b)$

$3.$ $(a,b+1),(a,b),(a-1,b)$
$4.$ $(a,b+1),(a,b),(a+1,b)$

If those patterns contribute, I'm struggling to define proper sets which relate to them in order to apply the pigeonhole principle.

$\endgroup$
6
$\begingroup$

Let's partition $A\times B$ into $30$ little $2\times 2$ squares : explicitely, we write the disjoint union as follows :

$$ \begin{align} A\times B &= \bigcup_{a=1}^5\bigcup_{b=1}^6 \{(2a-1,2b-1),(2a-1,2b),(2a,2b-1),(2a,2b)\} \\ &=: \bigcup_{a=1}^5\bigcup_{b=1}^6 X_{a,b}. \end{align} $$

By the pigeonhole principle, there is a square $X_{a,b}$ that contains at least $3$ elements of $S$ (if it was not the case, $\lvert S \rvert$ would be less than $2\times 30=60$).

Now among those $3$ elements, apply again the pigeonhole principle : two of them must have the same $x$-coordinate. These two will be $(x_1,y_1)$ and $(x_2,y_2)$, and the third element of $S$ inside the square is $(x_3,y_3)$.

$\endgroup$
2
  • $\begingroup$ How did you get $30$ little $2\times 2$ squares? I keep getting $99$ $\endgroup$ Dec 23 '17 at 14:05
  • $\begingroup$ @SlavikEgorov : You have $120$ elements total, divide that by $4$ and you get $30$ (disjoint) little squares. Your $99$ squares are not disjoint. $\endgroup$
    – tristan
    Dec 23 '17 at 14:10
4
$\begingroup$

enter image description here

We are trying to avoid one of the four pieces above. Consider the $30$ red $2$ by $2$ squares, each of these can have at most two squares filled in.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.