Magic Hexagon of Hexagons
Can we arrange consecutive numbers (1, 2, 3..) in hexagonal cells forming a hexagon, so that all rows add up to the same number?
Indeed we can. There is a unique solution, which has been discovered independently several times. Here is the magic hexagon, with all rows etc adding up to 38:-
Here are some of the people who found the solution, starting with the latest:-
Apparently, Clifford W. Adams started work on this problem in 1910, and finally solved it in 1957. This is a rather nice story, as you can imagine the guy slaving away methodically year after year, and just as he solves it along comes the computer...
Martin Kühl, of Hannover, Germany is reported to have discovered the solution in 1940.
H. Lulli is also quoted as the discoverer.
Well before this, in 1895, William Radcliffe from the Isle of Man (which is in the Irish Sea). laid claim to having found it.
(picture courtesy of Jerry Slocum and The Slocum Puzzle Foundation - click to enlarge)
But the earliest known discovery has recently come to light in Germany. Ernst von Haselberg (photo below), the chief architect in a town called Stralsund, left his notes of the problem. They date from 1887. The story, and von Haselberg's technique, were written up in a German magazine article by Hans-Friedrich Bauch. I have recently translated the article which will be available here shortly.
One can't help feeling a little sympathy for Clifford W Adams. I hope he never found out about his forerunners!
Proof of uniqueness
The proof that there is only one magic hexagon is straightforward, if a little messy, and was given by C W Trigg in 1964. In essence, we take a hexagon with n hexagon cells on each side:
The total number of cells is given by the hexagon centred number, 3n(n-1)+1. Each cell has a number in it 1, 2, 3 ... 3n(n-1)+1
The sum of these numbers is x(x+1)/2 where x = 3n(n-1)+1.
As we said, the total is x(x+1)/2; or, in terms of n, it is (3n(n-1)+1).(3n(n-1)+1+1)/2 = (3n(n-1)+1).(3n(n-1)+2)/2
Note that there are 2n-1 (vertical) columns, which each add up to the same magic constant M. Adding all these columns together we get:
(2n-1)M = (3n(n-1)+1).(3n(n-1)+2)/2, so:
M has to be integer, so 5/(2n-1) must be integer, which only happens (n>0) for n=1 (the trivial case of a single hexagon) and n=3.
So any magic hexagons, if they exist, have 3 cells to a side. We can now use a computer program (for example this straightforward C program) to search all possibilities; we find there is just the one shown above.
Reference: Clifford W. Adams' discovery: - Aug. 1963 Sci. Am., Math. Games
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