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We draw 4 equilateral triangles on the plane (not necessarily congruent) with no repeating vertices. Let $S$ denote the set of the 12 vertices of the triangles. Suppose $S$ has the property of not having a subset of 3 aligned points. What's the maximum number of squares that can exist with vertices from $S$?

A construction with 3 squares in it

This photo shows a construction that yields 3 squares. Can anyone do better?

Edit: 4 squares are achievable as proved by heropup. What about 5?

Luca T. Castrillón
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    I can't tell what the restrictions are. Please [edit] the question to clarify, Show us a picture of how you arrange for three squares. – Ethan Bolker Oct 29 '23 at 19:50

1 Answers1

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Here is an example with four squares:

enter image description here

heropup
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  • I should also point out that the non-collinearity rule implies that no two squares can share an edge, which might suggest (but not prove) that four squares is maximal. The case where the diagonal of one square is the side of another square also doesn't work because then the smaller square has a vertex that is located at the center of the larger square, making three collinear points. Thus any two squares cannot share more than one vertex in common. – heropup Oct 29 '23 at 21:14
  • In fact 5 triangles let us build 6 squares. Remove the top triangle, add two copies of the big square around to give it symmetry, and observe that you can then form 2 triangles with the 6 external vertices of the 3 biggest squares, in a manner similar to that used for the inner triangles. Recursively, one can build $3n$ squares out of $2n+1$ triangles. – Luca T. Castrillón Oct 30 '23 at 09:19