Let there be a cube with edge length of 9 units. Inside this cube there are 1981 points.
1. Prove that among these points there are at least 2 with the distance between them of less than 1 unit.
2. What's the biggest number of points that do not have the property stated at (1)?
2.1. How many points for the distance of 2  8 units?
The height of an equilateral triangle of unit base length is
SQRT(3)/2 units. A 2Dlattice of 160 of these equilateral
triangles will fit within the face of the 9 by 9 unit cube, with a total of 99 distinct vertices. The 2Denvelope is slightly greater than 8.66 by slightly infinitismally less than 8.5. This extra space allows 11 parallel planes of these equilateral lattices to fit within the cube, with 1 additional that may fit on each plane plus at least 5 additional points per plane that may be offset from the 11 planes. Therefore, a total of 1155 points will fit within the cube at a distance of one unit from each other. This number of points is a little over half the number of points given (1981) within the cube, thus there must exist at least 827 points that are within a unit distance from another point.
The spherical packing volume approximation for a 9x9x9 cube is
539.81, and can hold around 1030 unit diameter spheres.
The spherical packing volume approximation for a 10x10x10 cube is 740.48, and can hold around 1414 unit diameters spheres. The value 1155 falls between these two.
Edited on January 28, 2010, 3:05 pm
From a further examination of the lattice structure, I believe the maxmimum number of points that can exist a unit distance from other points of 1981 may be 1167.
Edited on January 31, 2010, 2:35 am

Posted by Dej Mar
on 20100128 03:01:22 