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    I'm looking for a proof of this theorem--namely that if R is a convex region symmetrical about the origin with area greater than 4, then R must contain a point with integral coordinates different from the origin. I'd be content to see the proof for the 2-dimensonal case.
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    (Original post by J.F.N)
    I'm looking for a proof of this theorem--namely that if R is a convex region symmetrical about the origin with area greater than 4, then R must contain a point with integral coordinates different from the origin. I'd be content to see the proof for the 2-dimensonal case.


    Suppose that R° is an open region including 0, that Rp is the congruent and similarly situated at region about any point P of A and that no two regions Rp overlap. Then area R° « 1
    If R° were the square bounded by the lines x= ±½ and y= ±½, then the theorem becomes obvious. The area of R° would be 1, and the region Rp with their boundaries would cover the plane.

    Now suppose A is the area of A and A the maximum distance of a point C° from 0 and that we consider the (2n+1)² regions Rp corresponding to points of λ whose coordinates are not greater numerically than n. All the regions lie in the square whose sides are parallel to the axes and at a distance n+A from 0. Hence (since the regions do not overlap)

    Ø((2n+1)²) « (2n+2A)² where Ø « (1+ ((A-½)/(n+½)))²

    the result follows when we make n tend to infinity.

    This result allows for Minkowski's theorem to be proved:

    Take the first definition and suppose R° is the result of contracting R about 0 to half its linear dimensions. Then the area of R° is greater than 1, so the two regions Rp of the above definition, overlaps and there is a lattice-point P such that R° and Rp overlap.
    Let Q be a point common to R° and Rp. If OQ' is equal and parallel to PQ and Q" is the image of Q' in 0, then Q' and therefore Q" lies in R° and by definition of convexity the mid point of QQ" lies in R. This point is the mid point of OP and therefore P lies in R.

    Note: this isn't the full theorem (its the first part of the theorem)

    http://www.cut-the-knot.com/Curricul...Addition.shtml
 
 
 
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