By J.N. Coldstream

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**Sample text**

Note that the determinant of L(v) is ||v||. Lemma 4. For every d > 2 and every e > 0 there exists c(e) > 0 such that for large enough T, the number o f v ^ F * 1 with \\v\\ < Td~~l and n(v] > c(e}T is at most We will use {x} to denote the rational part of x G R: {x} = x — [x\. Lemma 5. For every £ > 0 there exists a 6 > 0 such that for any a > 0 and for large S, among the m G Z with ^S < ^/m < S all but s percent satisfy > 6. Proof of Theorem 2 for d > 3. The number of v G P^ satisfying \ Td~l < IHI < Td~l is ;> J1^-1) according to (1).

However, they will also point out that it has a refinement in an unexpected direction. All our examples consist of 6 lines t\ , t^ , • • • , t§ in P3 (though described in R3 C P3), and 6 "transversals" which we may denote ^,^2", • • • ,^j- according to the rule: Example 1. Let ^1,^2,^3 be three lines in a plane P meeting at different points pij = til\lj. Consider three non colinear points in P, called P4,ps,p6) and any point p not in the plane P. Define ii — p V Pi for i = 4,5,6. The 5 to 5 transversals are then uniquely determined by t^ = p V pjk for (z, j, k} = {1,2,3}, and if- — PJ V pk for (z, j, k} = {4,5,6}.

If p > 3^(v), and x G aff L(v), then the polytope K = conv{(a; + pB) fl L(v}} contains a lattice point in its relative interior. Proof. By the definition of//, y-\- /J-(v)B contains a lattice point for every y € aff L(v). In particular, there is a z 6 L(v} fl (x + n(v}B}. So it suffices to show that x -f n(v}B C K since then z is in the relative inte rior of K . Assume a point from the boundary of x + n(v}B is not contained in A'. Then a (d — 2)-dimesional hyperplane separates it from A'. This hyperplane splits x + pB into two parts.