Combinatorial geometry

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Helly's theorem

If is a family of more than bounded closed convex sets in Euclidean -space , and if every (where is the Helly number) members of have at least one point in common, then all the members of have at least one point in common.

Pick's theorem

Let be the area of a simply closed lattice polygon. Let denote the number of lattice points on the polygon edges and the number of points in the interior of the polygon. ThenThe formula has been generalized to three- and higherdimensions using Ehrhart polynomials.

Five disks problem

Given five equal disks placed symmetrically about a given center, what is the smallest radius for which the radius of the circular area covered by the five disks is 1? The answer is , where is the golden ratio, and the centers of the disks , ..., 5 are located atThe golden ratio enters here through its connection with the regular pentagon. If the requirement that the disks be symmetrically placed is dropped (the general disk covering problem), then the radius for disks can be reduced slightly to 0.609383... (Neville 1915).

Disk covering problem

Given a unit disk, find the smallest radius required for equal disks to completely cover the unit disk. The first few such values are(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)Here, values for , 8, 9, 10 are approximate values obtained using computer experimentation by Zahn (1962).For a symmetrical arrangement with (known as the five disks problem), , where is the golden ratio. However, rather surprisingly, the radius can be slightly reduced in the general disk covering problem where symmetry is not required; this configuration is illustrated above (Friedman). Neville (1915) showed that the value is equal to , where and are solutions to(11)(12)(13)(14)These solutions can be found exactly as(15)(16)where (17)(18)are the smallest positive roots of the given polynomials, with denoting the th root of the polynomial in the ordering of the Wolfram Language. This gives (OEIS A133077) exactly as(19)where the root is the smallest positive one of the..

Kepler conjecture

In 1611, Kepler proposed that close packing (either cubic or hexagonal close packing, both of which have maximum densities of ) is the densest possible sphere packing, and this assertion is known as the Kepler conjecture. Finding the densest (not necessarily periodic) packing of spheres is known as the Kepler problem.Buckminster Fuller (1975) claimed to have a proof, but it was really a description of face-centered cubic packing, not a proof of its optimality (Sloane 1998). A second putative proof of the Kepler conjecture was put forward by W.-Y. Hsiang (Cipra 1991, Hsiang 1992, 1993, Cipra 1993), but was subsequently determined to be flawed (Conway et al. 1994, Hales 1994, Sloane 1998). According to J. H. Conway, nobody who has read Hsiang's proof has any doubts about its validity: it is nonsense.Soon thereafter, Hales (1997a) published a detailed plan describing how the Kepler conjecture might be proved using a significantly..

Illumination problem

In the early 1950s, Ernst Straus asked 1. Is every region illuminable from every point in the region? 2. Is every region illuminable from at least one point in the region? Here, illuminable means that there is a path from every point to every other by repeated reflections.In 1958, a young Roger Penrose used the properties of the ellipse to describe a room with curved walls that would always have dark (unilluminated) regions, regardless of the position of the candle. Penrose's room, illustrated above, consists of two half-ellipses at the top and bottom and two mushroom-shaped protuberances (which are in turn built up from straight line segments and smaller half-ellipses) on the left and right sides. The ellipses and mushrooms are strategically placed as shown, with the red points being the foci of the half-ellipses. There are essentially three possible configurations of illumination. In this figure, lit regions are indicated in white, unilluminated..


A family of nonempty subsets of whose union contains the given set (and which contains no duplicated subsets) is called a cover (or covering) of . For example, there is only a single cover of , namely . However, there are five covers of , namely , , , , and .A minimal cover is a cover for which removal of one member destroys the covering property. For example, of the five covers of , only and are minimal covers. There are various other types of specialized covers, including proper covers, antichain covers, -covers, and -covers (Macula 1994).The number of possible covers for a set of elements arethe first few of which are 1, 5, 109, 32297, 2147321017, 9223372023970362989, ...(OEIS A003465).

Combinatorial geometry

Combinatorial geometry is a blending of principles from the areas of combinatorics and geometry. It deals with combinations and arrangements of geometric objects and with discrete properties of these objects. It is concerned with such topics as packing, covering, coloring, folding, symmetry, tiling, partitioning, decomposition, and illumination problems. Combinatorial geometry includes aspects of topology, graph theory, number theory, and other disciplines.Although combinatorial geometry was studied by classical mathematicians such as Euler and Kepler, many advances have been made since the middle of the 20th century. This topic was one which drew the interest of the late prolific mathematician Paul Erdős. The term "Combinatorial Geometry" was apparently first used in 1955 by H. Hadwiger (Hadwiger and Debrunner 1964).David Eppstein's "Geometry Junkyard," a collection of geometry related..

Borsuk's conjecture

Borsuk conjectured that it is possible to cut an -dimensional shape of generalized diameter 1 into pieces each with diameter smaller than the original. It is true for , 3 and when the boundary is "smooth." However, the minimum number of pieces required has been shown to increase as . Since at , the conjecture becomes false at high dimensions.Kahn and Kalai (1993) found a counterexample in dimension 1326, Nilli (1994) a counterexample in dimension 946. Hinrichs and Richter (2003) showed that the conjecture is false for all .

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