We show that an old but not well-known lower bound for the crossing number of a graph yields short proofs for a number of bounds in discrete plane geometry which were considered hard before: the number of incidences among points and lines, the maximum number of unit distances among n points, the minimum number of distinct distances among n points.

From among � � n triangles with vertices chosen from n points in the unit square, 3 let T be the one with the smallest area, and let A be the area of T. Heilbronn’s triangle problem asks for the maximum value assumed by A over all choices of n points. We consider the average-case: If the n points are chosen independently and at random (with a uniform distribution), then there exist positive constants c and C such that c/n3 <µ n < C/n3 for all large enough values of n, where µ n is the expectation of A. Moreover, c/n3 <A<C/n3, with probability close to one. Our proof uses the incompressibility method based on Kolmogorov complexity; it actually determines the area of the smallest triangle for an arrangement in

Heilbronn’s triangle problem asks for the least ∆ such that n points lying in the unit disc necessarily contain a triangle of area at most ∆. Heilbronn initially conjectured ∆ = O(1/n 2). As a result of concerted mathematical effort it is currently known that there are positive constants c and C such that c log n/n 2 ≤ ∆ ≤ C/n 8/7−ǫ for every constant ǫ> 0. We resolve Heilbronn’s problem in the expected case: If we uniformly at random put n points in the unit disc then (i) the area of the smallest triangle has expectation Θ(1/n 3); and (ii) the smallest triangle has area Θ(1/n 3) with probability almost one. Our proof uses the incompressibility method based on Kolmogorov complexity. 1

We survey some results and problems arising from a classic problem of Steinhaus: Is there a subset S of R² such that each isometric copy of Z² (the lattice points in the plane) meets S in exactly one point.

Given k 3, denote by t 0 k (N) the largest integer for which there is a set of N points in the plane, no k + 1 of them on a line such that there are t 0 k (N) lines, each containing exactly k of the points. Erd}os (1962) raised the problem of estimating the order of magnitude of t 0 k (N ). We prove that: t 0 k (N) c 0 k N log(k+4)=log(k) improving a previous bound of Grunbaum for all 5 k 35. 1

In this paper several unconnected old and new problems in combinatorial geometry are discussed. The reference list is as complete as possible, including two papers dealing with similar subjects [6, 7]. Many interesting problems and results are in Grünbaum [11], which also has an extensive and useful bibliography and interesting historical remarks.

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. It is shown that there is a subset S of R 2 such that each isometric copy of Z 2 (the lattice points in the plane) meets S in exactly one point. This provides a positive answer to a problem of H. Steinhaus. 1. Introduction Sometime in the 1950's, Steinhaus posed the following problem. Do there exist two sets A and B in the plane such that every set congruent to A has exactly one point in common with B? The trivial case where one of the sets is the plane and the other consists of a single point is ruled out. The rst appearance of this problem in the literature seems to be in a 1958 paper of Sierpinski [11]. In this paper, he showed the answer is yes, a result later rediscovered by Erd}os [5]. Of course, there are many variants of this problem. For example, one could specify the set A. In this direction, Komjath showed that such a set exists if A = Z, the set of all integers [10]. Steinhaus also asked about the specic case where A = Z 2 . The rst reference to this problem a...

From among \Gamma n 3 \Delta triangles with vertices chosen from among n points in the unit square, U , let T be the one with the smallest area, and let A be the area of T . If the n points are chosen independently and at random (uniform distribution) then there exist positive c and C such that c=n 3 ! n ! C=n 3 for all large enough n, where n is the expectation of A. Moreover, with probability close to one c=n 3 ! A ! C=n 3 . Our proof uses the incompressibility method based on Kolmogorov complexity. The related Heilbronn problem asks for the maximum value assumed by A over all choices of n points. 1 Introduction From among \Gamma n 3 \Delta triangles with vertices chosen from among n points in the unit circle, let T be the one of least area, and let A be the area of T . Let \Delta n be the maximum assumed by A over all choices of n points. H.A. Heilbronn (1908--1975) 1 asked for the exact value or approximation of \Delta n . The list [1, 2, 3, 5, 8, 9, 10, 11,...

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