What is the minimum number of pennies that can be placed upon a table so that each penny touches three, and only three, others?

(All the pennies must lie flat on the table.)

(In reply to Big table! by Jack Squat)

Jack Squat and SliverKnight are indeed correct. I just wanted to add a few comments:

A simple trial indicates that the problem is ever expanding for an open form. In that to reach a 3 penny [valence] shell, you must always add another penny. And every penny you add requires at least another penny to complete it.
Therefore a circular or polygonal solution seems required. So that the one penny you add is an existing one. Thus a closed form is required.

Given that we need a polygon, we need a basic building block for each side. The 080 cluster of four pennies is the smallest unit where the ends of the "cluster" need one and only one penny to complete them. And every other non-linking penny already has its [valence] shell complete.

Obviously silverknight has shown that a 4 sided diamond or square [quadrilateral] is a solution. A pentagon is of course also a solution, etc. However the proof of why a triangle is not a solution may be interesting to a few.

If you draw out a triangle solution with the 080 cluster, and then connect the centers of every penny to it's nearest neighbors' center. You will have a lattice of triangles. Now every penny that should make a connection to a neighboring penny has a line of length "one" between their centers. Where the unit of length is equal to one penny diameter. Now every line that cannot form a connection [i.e. is not allowed due to it's valence shell being already filled], must be assigned a variable. To show that a triangle is possible, you must show that all of the variables must be greater and not equal to one.

I have not done this problem, however if you try... I think you will show that the solution to this geometry problem is a lattice of equilateral triangles.

00
00 00
0 0
080

Triangle of 080 clusters.

Posted by Juan on 2004-01-02 22:43:00