[seqfan] Re: How many squares can you make from n points in the plane?

[Benoît Jubin]

Another proof that a(7) = 3 (that is, a planar figure with 7 points
contains at most 3 squares, and there are figures realizing this
number):

First, note that two different squares cannot share three vertices.
If a 7-point figure contains (at least) 3 squares, then by simple
counting there is a pair of squares sharing two vertices.  There are
only two possibilties:
(1) they share an edge;
(2) an edge of one of them is a diagonal of the other.
None of these 6-point figures contain a third square.  Therefore, if
the 7-point figure contains other squares, they are formed from a 7th
point, and three points in the 6-point figure (and not all three
belonging to the same square of that figure, by the preliminary
remark).  The triples of points in these 6-point figures that are in
square-position are few: two (one up to symmetry) in case (1), and
three (two up to symmetry) in case (2).  They lead to the two 7-point
3-square configurations (since one of the cases in (2) yields the same
figure as the one from (1)).  None of these two figures contain a
fourth square.

Benoît

On Wed, Sep 29, 2021 at 6:54 AM <hv at crypt.org> wrote:
>
> I should have said: I worked below _without_ the assumption that the
> points had integer coordinates, so if accurate this proves something
> slightly stronger. _With_ that assumption, it would not have been
> safe to assume we can scale any arrangement to make the smallest
> square be a unit square.
>
> Hugo
>
> I wrote:
> :Neil Sloane <njasloane at gmail.com> wrote:
> ::It seems that surprisingly little is known - see A051602. Even a(7) is an
> ::open question, although it is easy to see that a(7) >= 3.
> :
> :Unless I'm missing something, the 7-point case doesn't seem too hard.
> :
> :Fix 4 points on the standard unit square, and wlog assume this is the
> :smallest square in the final combination. There are then four
> :possibilities.
> :
> :A) There's another square of the same size, sharing two points with the
> :first. Then the 7th point can only give us a third square in two
> :symmetrically placed ways, of side sqrt(2). No fourth is possible.
> :
> :B) There's another square of the same size, sharing one point with the
> :first. Then all 7 points are accounted for. Any third square must use
> :two points from each of the existing two squares, so again can only
> :have a side of sqrt(2); that is possible, but there can't be a fourth.
> :
> :C) There's another square of a different size, sharing two points with
> :the first; this can only have side sqrt(2). The only ways to select
> :three of those points such that the 7th point can be added to give a
> :square in each case reduce us to case A or B.
> :
> :D) There's another square of a different size, sharing one point with
> :the first. All points are accounted for, and any additional square
> :must share two points with each of the first two squares; but then
> :any such third square must reduce us to case A or C.
> :
> :So the possibilities are:
> :
> :.XX
> :XXX
> :XX.
> :
> :and
> :
> :.X.
> :XXX
> :XXX
> :
> :Hugo
>
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