We disprove Heilbronn's conjecture—that N points lying in the unit disc necessarily contain a triangle of area less than c/N 2.

We prove that every n-vertex graph G with pathwidth pw(G) has a three-dimensional straight-line grid drawing with O(pw(G) n) volume. Thus for

In a total order of the vertices of a graph, two edges with no endpoint in common can be crossing, nested, or disjoint. A k-stack (resp...

We consider the problem of simultaneous embedding of planar graphs. There are two variants

Let PuP2,...,Pn be a distribution of n points (where n ^ 3) in a closed convex region K of unit area such that the minimum of the areas of the triangles PjPjPk (taken over all selections of three out of n points) assumes its maximum possible value A(/C; n). We write A(n) = A(D; n) if D is a disc.

. In this paper we show a lower bound for the generalization of Heilbronn's triangle problem to d dimensions; namely, we show that there exists a set S of n points in the d-dimensional unit cube so that every d + 1 points of S define a simplex of volume## 1 n d ). We also show a constructive incremental positioning of n points in a unit 3-cube for which every tetrahedron defined by four of these points has volume## 1 n 4 ). Key words. Heilbronn's triangle problem, probabilistic method AMS subject classifications. 05D40, 51M16 PII. S0895480100365859 1.

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

Consider the following problem: how many collinear triples of points must a transversal of Zn×Zn have? This question is connected with venerable issues in discrete geometry. We show that the answer, for n prime, is between (n − 1)/4 and (n − 1)/2, and consider an analogous question for collinear quadruples. We conjecture that the upper bound is the truth and suggest several other interesting problems in this area. In [4], Erdős offered a construction concerning the “Heilbronn Problem”. What is the smallest A so that, for any choice of n points in the unit square, some triangle formed by three of the points has area at most A? His elegant construction of a point-set with large minimum-area triangle ( ∼ n −2) is as follows: take the smallest prime p ≥ n, and let the set of points be {p −1 (x, x 2 (mod p)) : x ∈ Zp}. (If necessary, throw out a few points so that there are n left.) It is easy to see that this set has no three collinear points, and therefore any three points form a nondegenerate lattice triangle – which must have area at least p −2 /2 ≫ n −2. Another area in which collinear triples of points on a lattice arise is in connection with the so-called “no-three-in-line ” problem, dating back at least to 1917 ([1]). Is it possible to choose 2n points on the n-by-n grid so that no three are collinear? Clearly, if this is the case, then 2n is best possible. Guy and Kelly ([2]) conjecture that, for sufficiently large n, not only is it true that every set of 2n points has a collinear triple, but that it is possible to avoid collinear triples in a set of size (α −ǫ)n and impossible to avoid them in a set of size (α + ǫ)n, where α = (2π 2 /3) 1/3 ≈ 1.874 and ǫ> 0. In this note, we address the question of when it is possible to avoid collinear triples modulo n, particularly in the case of transversals (i.e., graphs of permutations) and when n is prime. 1 1

www.elsevier.com/locate/comgeo It is proved that every k-colourable graph on n vertices has a grid drawing with O(kn) area, and that this bound is best possible. This result can be viewed as a generalisation of the no-three-in-line problem. A further area bound is established that includes the aspect ratio as a parameter. © 2004 Elsevier B.V. All rights reserved.