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Jul 27, 2020 at 19:28 comment added Pietro Majer Since there can't be three co-linear points, in fact two points determine at most 1 square (either an edge or a diagonal). Since a square has 6 pairs of vertices, this gives a trivial upper bound $n(n-1)/12$.
Jul 27, 2020 at 18:42 answer added Will Brian timeline score: 6
Jul 27, 2020 at 17:30 history asked Mark Lewko CC BY-SA 4.0