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 2D-lattice 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 2D-envelope 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 2010-01-28 03:01:22