TITLE: Pigeonhole principle with cartesian product
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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.