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You are going to draw $6$ congruent squares to make as many squares as you can!

What is the maximum amount of squares (except the original squares) you can create by drawing 6 congruent squares?

If this question was asked for $2$ squares the answer will be $1$ as shown below:

enter image description here

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Here is my attempt:
enter image description here
The six squares are arranged evenly along a diagonal line.
Any two of the 6 squares overlap to make a large square. There are 6*5/2=15 of these.
There are also the tiny squares arranged in triangles at the top right and bottom left. There are 10 on each side. There are also 3 squares on each side made from 2x2 of the tiny squares.
This makes for a total of 15+2*(10+3)=41 extra squares.

I have a feeling there is a better solution, so I'll keep trying.

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    $\begingroup$ nope, you got it already :) $\endgroup$
    – Oray
    Mar 2, 2018 at 19:54
  • $\begingroup$ Can we prove maximality or is that too complicated? $\endgroup$
    – greenturtle3141
    Mar 2, 2018 at 20:05
  • $\begingroup$ @greenturtle3141: I think it is locally optimal (i.e. moving any of the 6 squares in my pattern reduces the number of squares). Proving it is a global maximum may be trickier. Maybe one could prove that the size and coordinates of the 6 squares can be integers, and then it becomes a finite problem that can be exhaustively checked by computer. $\endgroup$
    – Jaap Scherphuis
    Mar 2, 2018 at 20:51

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