Find (for n=2,3,4,5,6,7,8,9,10) a configuration of n points on a regular tetrahedron ABCD such that the minimal
distance between two distinct points out of these n is as great as possible.
MY OWN SOLUTIONS:
We can formulate the problem in terms of disjoint geodesic circles with maximal radius. Such a circle has normally
area pi*r2, but half of that if the center is a vertex of the tetrahedron. So it seems best to place the first
space traveler in a vertex. The incenter of the opposite face is the best second point if n=2. From now on, I place at
least one pirate in a vertex.
If there are more than two settlers, we first consider vertices and incenters, but sometimes there is room in an other
face, so that we can improve the configuration by shifting points in the direction of free room.
We strive after symmetry and equal distances.
I also did some attempts of exploration with a computer program.
n=2
Choose D and the incenter Z of triangle ABC. The distance between these points over the tetrahedron is (2/3)*sqrt(3) =
about 1.15470054.
n=3
Choose for example the vertices of triangle ABC. The distance between two points is always 1.
(Proof of optimality: we must choose one vertex, say A. If we want to place the other two pirates at a distance greater
than 1 from A, they both come within triangle BCD, so they are at a distance smaller than 1 from each other.)
n=4
Choose the vertices of the tetrahedron. The distance between two points is always 1.
n=5
Choose D and the incenter Z of triangle ABC, and the points on the sides DA, DB and DC at distance 2/3 from D.
For each of the five points, the distance to a nearest other point is 2/3 = about 0.66666667.
(Lay the triangles ABD, BCD and ABD
in the plane of triangle ABC. See sketch1.)
n=6
First choose the four vertices, and the incenters of triangles ABC and ABD, but shift the space travelers in vertices
C and D somewhat closer to each other, and the space travelers in the above-mentioned incenters somewhat further away
from side AB.
The minimum distance is about 0.58986472.
(See sketch2. Here you find the exact solution.)
n=7
Choose the four vertices and three out of the four incenters. The minimum distance is sqrt(3)/3 = about 0.57735027.
(Proof of optimality: First choose A,B,C,D and the incenters of triangles ABD, ACD and BCD. If we shift the pirates in
these three incenters away from D, they always come closer to those in A,B,C or there about.)
n=8
Choose the four vertices and the four incenters. The minimum distance is again sqrt(3)/3 = about 0.57735027.
n=9
Choose the four vertices and the midpoints of five out of the six edges. The minimum distance is 0.500..
(Proof of optimality: exercise for the reader.)
n=10
Choose the four vertices and the midpoints of the six edges. The minimum distance is 0.500..
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