# two shapes in a $2n\times 2n$ grid sheet, can we pick third one?

Can anyone help me with this problem? It just popped to my mind!!!

we have a $2n\times 2n$ grid sheet and a connected shape $L$ consisting of $2n-1$ grid squares. we've cut two copies of $L$ out of the sheet. Is it always possible to cut a third copy of $L$?

I think the answer is yes, but I couldn't solve it. any Ideas?

• Why insist that the grid size be even? Seems more natural to ask this for $(n-1)$-ominos in an $n \times n$ grid, as long as there's no immediate counterexample with $n$ odd. – Noam D. Elkies Feb 20 '12 at 5:47
• If the shape is at most 2n/5 squares wide, the answer is yes. For n =11, consider an S shape that fits in a 5 by 9 rectangle. It is possible to cut out 3 copies of s, but just barely. Perhaps someone can tweak this into a counterexample. Gerhard "Ask Me About System Design" Paseman, 2012.02.19 – Gerhard Paseman Feb 20 '12 at 6:43
• I assume you mean the shape can be rotated, but can it be reflected? – Zack Wolske Feb 20 '12 at 7:28
• why not four copies? – alberto.bosia Feb 20 '12 at 8:02
• @Noam D.Elkies: I made a figure for all $2n\times 2n$ grid sheets that could be placed such that no other figure could be cut out, that's why I'm intrested for the grid size to be even. @Zack Wolske: yes, it can be also rotated and reflected. @alberto.bosia: I could make a counter example for that case. – Goodarz Feb 20 '12 at 8:53

EDIT: I was hoping that my answer for the size 8 grid would generalize readily to larger squares, but so far the only ones I can handle are size $6k+2$, $k=1,2,\dots$. The piece is a cross with mutually bisecting lines of $4k+1$ and $2k+1$ squares. The two crosses are placed in the same orientation and in such a way that the short lines of squares meet at a single point, that point being the center of the big square. An argument similar to the one given at m.se shows that there are very few places to put a third line of $4k+1$ squares:
1. You can put it too close to the edge of the big square to allow for the line of $2k+1$, or
• yes, I asked the question there, but that answer is for the case when $n=4$, I want a solution (or counter-example) for all $n$. – Goodarz Feb 20 '12 at 11:42