Playing with a pencil on a checkered sheet I encountered this construction:

1) take a point $A$ on the grid and a point $B$ that is distant from $A$ $n=2,3,4...$ horizontal steps and $1$ vertical step, so that $\overline{AB}=\sqrt{n^2+1}$.

2) complete a square $ABCD$ on the grid starting from the segment $AB$. This is the square $Q_0$.

3) Starting from $A$ take a point $A_1$ at the first intersection of $AB$ with a line of the grid. And do the same starting from the other points $B,C,D$ finding the points $B_1,C_1,D_1$ .

4) connect the points $A_1, B_1,C_1,D_1$, so that we have a new square $Q_1$.

5) redo the same points $3$ and $4$ and find the square $Q_2$ and so again...

The result is a sequence $S_n$ of squares whose sides, orientation, and number of different elements depend from $n$.

The figures show the construction for $n=2$, and for $n=3$.

enter image description here

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Note that it seems that the sequence $S_2$ has infinite different elements that converge to a square of the grid, but $S_3$ has a finite number of different elements.

So a first question is:

How we can prove that the number of different elements of $S_n$ is finite or not and, if finite, what is this number?

A second question is to find the length of the last square in the sequence, or the limit of the lengths if the sequence has infinite different elements.

More general, the side lengths of all the squares in a sequence $S_n$ can be determined in some way that does not require a terribly boring calculation?

Added: construction of the first six elements of $S_4$.

enter image description here

  • $\begingroup$ I already love this question before understanding it. But to help the understanding: could you please mark the original $A$ and $B$ (and perhaps a few of the mentioned other points) in the drawing? $\endgroup$ – Vincent Nov 21 '18 at 22:17
  • $\begingroup$ Given the symmetry of the construction you can mark as $A$ and $B$ any two extremes of a side of the greatest square, and $C, D$ are the other two vertices. $A_1$ is than the vertex of the second square that stay on $AB$ , $B_1$ the vertex on $BC$ and so one... $\endgroup$ – Emilio Novati Nov 21 '18 at 22:23
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    $\begingroup$ I added the construction for $S_4$ with some marks. $\endgroup$ – Emilio Novati Nov 21 '18 at 22:59
  • $\begingroup$ Yes, thank you! Now it is completely clear. I somehow thought you were working form inside to outside and got confused, but now that I see you work from outside in, it all makes sense $\endgroup$ – Vincent Nov 22 '18 at 9:48

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