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I Partitions of Euclidean space, cubic lattice, convex sets

  1. Mar 2, 2016 #1
    If the Euclidean plane is partitioned into convex sets each of area A in such a way that each contains exactly one vertex of a unit square lattice and this vertex is in its interior, is it true that A must be at least 1/2?
    If not what is the greatest lower bound for A?
    The analogous greatest lower bounds for En obviously form a non increasing sequence (ordered by n). What is the value for En?
  2. jcsd
  3. Mar 2, 2016 #2


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    I would have thought that A must be at least 1, since there's a natural bijection between lattice vertices and the unit squares delimited by the lattice.
    Proving it may be somewhat tricky though.
    What's the context of the question?
    It's not homework is it?
  4. Mar 3, 2016 #3
    No it's not homework (bit old for that).

    I was doing some number theory revision a couple of years ago and the question occurred to me after reading Minkowski's theorem on convex regions symmetric about a lattice point (Hardy and Wright 3.9/3.10), having noticed that Theorem 38, used in the proof, referred to congruent and similarly situated regions about the lattice points, but needed congruence only to show uniform boundedness of the regions (he only considers bounded regions) and similar situation apparently not at all (though the proof of Minkowski's theorem which follows on does deal with convex regions which are congruent and similarly situated).

    In fact I first considered the question with the partition members being pathwise connected sets with a defined area, not necessarily convex. In this case [itex]A[/itex] can be any value in [itex](0,\infty][/itex], I believe, but I couldn't find convex examples.

    I went on to look for examples with convex partition members. I should have spent more time remembering what I did before typing the question, because for this case I dropped the requirement that a lattice vertex is in the interior of its corresponding partition member (because I suspected this requirement might force [itex]A=1[/itex]) but the requirement has still found its way into what I wrote. Unfortunately I can't edit my original post any more, so I'll write it bold here:

    Please ignore the phrase "... and this vertex is in its interior" in the original question.

    With that proviso I think [itex]A[/itex] can be anywhere in [itex][\frac{1}{2},\infty)[/itex]. The question is can this range be extended?

    I didn't follow the reasoning in your reply. If the diameters of the partition members have a common finite upper bound, then I think [itex]A=1[/itex] (with essentially the same proof as Theorems 38/41 in Hardy and Wright), but this is not a premiss of the question. Neither is it stated that the partition members are congruent. I think I have a construction for a suitable partition with [itex]A=\frac{1}{2}[/itex].
    Last edited: Mar 3, 2016
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