The other day I was constructing new mathematical problems for my pupils and thought of something like this:
Given the infinite sequence of "stairs" $n\times n$, constructed from $1\times1$ squares:
$\hspace{130pt}$
$\cdots$
tesselate whole plane with it, using $\underline{\text{every}}$ "stair" and $\underline{\text{only once}}$.
Unfortunately, neither I could prove it, nor my pupils. If you know a solution, please post it here, or any reference link will be much applied. But actually, if we weaken either of two initial conditions it becomes very easy. Let me demonstrate:
1. If we are allowed to use every stair, but not only once, then the solution comes from the idea of merging every $(2k-1)\times(2k-1)$ stair with $2k\times2k$ stair into square $2k\times2k$ like this:
$\hspace{110pt}$
$\cdots$
Our plane tesselation is constructed like this: we stack $2\times2$ squares in an infinte vertical strip (roughly speaking $2\times\infty$), then to the right of it we put $4\times\infty$ infinte strip of stacked $4\times4$ squares, to the left we put $6\times\infty$, then to the right we put $8\times\infty$, then again to the left we put $10\times\infty$ and so on. Plane tesselation complete!
2. If instead we are allowed to use only once, but not every stair, then the solution comes from idea of squaring the square: just like in previous solution, merging every $(2k-1)\times(2k-1)$ stair with $2k\times2k$ stair give us a sequence of squares $2k\times2k$.
Plane tesselation constructed like this: we join some squares to make a bigger square $S_1$ with the help of squaring the square method, then we make new huge square $S_2$, consisting of obtained square $S_1$ as a smallest one, and others to complete construction and so on.
To point it out, we can solve the weaken problem, but is there a solution for a stronger one with both conditions kept? If you have an answer, post it here please, or at least give me a link.
