A Unique Magic Hexagon
Charles W. Trigg (San Diego, California)
Recreational Mathematics Magazine (January-February 1964) 40-43
Six equal discs may be arranged around a seventh disc, and touching it,
to form a hexagon with two discs on a side and three along each of the
three long diagonals. Twelve more discs will build this second order
hexagon up into a third order hexagon of five parallel rows containing
19 discs. The addition process may be continued to obtain an nth order
hexagon of 2n-1 parallel rows containing
1 + 6 Sum_{k=2..n} (k-1) or 3n^2 - 3n + 1 discs.
Suppose that the discs are numbered with the first 3n^2 - 3n + 1 positive
integers. The sum of these integers is 1/2 (3n^2 - 3n + 1)(3n^2 - 3n + 1 + 1)
or 1/2 (9n^4 - 18n^3 + 18n^2 - 9n + 2), an integer. If the discs in the
hexagonal array can be rearranged so that the integers in the various rows
have the the same sum, then the sum of all the integers in the array must
be divisible by 2n-1, the number of rows parallel in one orientation. That is,
Q = 1/2 (9n^4 - 18n^3 + 18n^2 - 9n + 2)/(2n - 1)
= 1/2 (4n^3 - 7n^2 + 5n - 2 + n^2(n^2 + 1)/(2n - 1))
= 1/2 (4n^3 - 7n^2 + 5n - 2 + (n^2/4)(2n + 1 + 5/(2n - 1)))
must be an integer.
Clearly, n and 2n-1 have no common factor for any n but the trivial unity.
So, if Q is to be an integer, then 2n-1 must divide n^2 + 1 and 5/(2n - 1)
must be an integer. This is true only for n=1 or 3. Thus, only for n=3 can
the hexagon's row sums be equal. These sums each must be 190/5 or 38.
Necessary Conditions to be Magic
We now seek a distribution of the first 19 positive integers in the third
order hexagon such that the integers in every one of the 3*5 or 15 rows
have the magic constant sum of 38. Let the vertex elements be a_i,
the mid-side elements be b_i, the central element be d, and the other
elements be c_i, i=1, 2, 3, 4, 5, 6. Also, let Sum a_i = A, Sum b_i = B,
and Sum c_i = C.
a1 b1 a2
b6 c1 c2 b2
a6 c6 d c3 a3
b5 c5 c4 b3
a5 b4 a4
Summing the rows of 3 elements, 2A + B = 6(38).
Summing the rows of 4 elements, 2B + 2C = 6(38).
Summing the rows of 5 elements, A + C + 3d = 3(38).
Summing all of the elements, A + B + C + d = 190.
The indeterminate solution of this set of equations may be expressed in
several ways, namely:
B = 228 - 2A, or A = 76 - d, or A = 114 - 1/2 B, or A = 57 + 1/2 C,
C = 2A - 114, B = 76 + 2d, C = 114 - B, B = 114 - C,
d = 76 - A, C = 38 - 2d, d = 1/2 B - 38, d = 19 - 1/2 C.
Evidently B and C are even. The smallest possible value of C is
1+2+3+4+5+7 or 22. It follows that 1 <= d <= 8 and 67 < A < 76.
The Method of Search
We first proceed to identify all of the arrangements of 12 of the first 19
positive integers around the perimeter of the hexagon with the three elements
on each side totaling 38 and 67 < A < 76. There are 30 triads of distinct
integers with the proper sum, namely:
19 18 1 19 13 6 18 16 4 18 11 9 17 12 9 16 12 10
19 17 2 19 12 7 18 15 5 17 16 5 17 11 10 15 14 9
19 16 3 19 11 8 18 14 6 17 15 6 16 15 7 15 13 10
19 15 4 19 10 9 18 13 7 17 14 7 16 14 8 15 12 11
19 14 5 18 17 3 18 12 8 17 13 8 16 13 9 14 13 11
No integer may be a vertex integer, a_i, unless it appears in two of these
triads. So, the smallest possible vertex integer is 3. The four pentads
with central element 3 are 18 17 3 16 19, 18 17 3 19 16, 17 18 3 16 19,
and 17 18 3 19 16. With these as foundations and with the aid of the
table above, all of the complete associated perimeters may be
identified quickly. There are four starting pentads with central element 4,
twelve with 5, twelve with 6, twenty-four with 7, twenty-four with 8,
and forty with 9.
There is only one set of two-digit integers with a sum less than 76,
i. e., 10, 11, 12, 13, 14, 15. There are only six of the triads each of
which contains at least two of the integers of the set. They are the
following: 17 11 10; 16 12 10; 15 14 9; 15 13 10; 15 12 11; and 14 13 11.
However, collectively, these triads contain only nine distinct integers
so cannot be assembled into a complete perimeter with twelve distinct
elements. It follows that no perimeter with A<76 can have all vertices
occupied by two-digit integers.
In constructing complete perimeters from each of the key one-digit vertices,
duplication was avoided by not permitting any smaller element than the key
element at any vertex. Out of the 1896 perimeters identified, only 121 are
such that 67 < A < 76.
The method of testing inside each perimeter to see if it belonged to a
magic hexagon is exemplified by the case shown below.
8 19 11
16 c1 c2 17
14 c6 2 c3 10
6 c5 c4 15
18 7 13
Since A=74, d=2, then c2 + c5 = 7 = 1 + 6 = 2 + 5 = 3 + 4. Neither c2 nor c5
can be 2 or 6 since they duplicate present elements. If c2=3 then c1=2 which
duplicates d. If c5=3 then c4=14 which duplicates a6.
Hence, this perimeter does not belong to a magic hexagon. In this way 120
perimeters were discarded. So the only possible magic hexagon is the one
shown below.
3 19 16
17 7 2 12
18 1 5 4 10
11 6 8 13
7 14 15
Rotations or reflections of this hexagon are not considered to be different
hexagons.
Background of the Problem
About 1910, Clifford W. Adams, a 19-year old Floridian saw an intriguing
proposition in "The Pathfinder", a small weekly newspaper published in
Washington, D. C. This was a challenge to place the first 19 positive
integers in the nineteen blank cells of a hexagon so that any straight
line would add up to 35. Adams soon understood why no solution was
published in the paper, for it became evident to him that the correct
sum was 38.
Throughout his career as a freight handler and clerk with the Reading
Railroad in Philadelphia he made sporadic attempts to find a solution.
He obtained a numbered set of hexagonal ceramic tiles to facilitate his
search. Finally, while convalescing from an operation in 1957, Adams
found a solution - the one given above. The record of the solution was
misplaced and not recovered until December 1962. Its uniqueness was in
doubt.
Adams sent a copy of his magic hexagon to Martin Gardner for his column,
"Mathematical Games", in Scientific American. Gardner asked me if I knew
of any reference to magic hexagons. This precipitated the investigation
reported above.
On April 25, 1963, these results were transmitted to Gardner, and they
were acknowledged in his column on page 116 of the August 1963 Scientific
American.
On September 7, 1963, Gardner informed me that the uniqueness of Adams'
hexagon hab been confirmed by William M. Daly on a Honeywell-800 computer
which analyzed 196729 configurations in 3 minutes, 20 seconds. The
uniqueness was also verified by G. W. Anderson on an IBM 1620 in 42
minutes. Furthermore, Eduardo Esperón, starting with 15 equations in
19 unknowns, demonstrated without the aid of a computer that the hexagon
is unique.
On page 291 of the Mathematical Gazette (London, England) for December
1958 there appears without comment a magic hexagon by T. Vickers which
is essentially the mirror image of the one given above.
Properties of the Magic Hexagon
Adams has identified 48 "constellations" consisting of from 3 to 7
non-collinear elements in his hexagon that add up to the magic constant
38. These are, of course, particular partitions of 38 into distinct
integers less than or equal to 19. There are many more. Indeed, as shown
above, there are 30 partitions of order 3. There are also 148 partitions
of order 4, and so on up to 2 partitions of order 8.
Gardner, in his column, shows that the hexagon has a curious bilateral
symmetry that provides a neat mnemonic for recalling how to number the
cells.
Daly called attention to the hexagons of order 2 which are contained in
the larger one. Three of the alternate ones around the periphery sum to
54 and the other three sum to 65. The elements of the central one add up
to 33. (So Sum c_i + 2d = 38.)
It may also be noted that all of the elements of the central hexagon are
digits. The central element is the same as the central element of the
unique third order magic square.
The sums of the elements in the six rhombi cornered on the vertices of
the hexagon form three number pairs: 39 (= 10 + 13 + 4 + 12);
40 (= 9 + 11 + 6 + 14); 46, 47; and 49, 50.
The sum of the vertices of each equilateral triangle centered on the
sides less the central element equals the magic sum. That is,
19 + 13 + 11 - 5 = 38 = 17 + 12 + 14 - 5. It follows that the sum of
all the elements in each of these equilateral triangles is the same,
namely 76, which is twice the magic constant. One-half of the vertex
elements are odd, and the others even. The elements of only two rows
are all even, and these two rows are connected in a "T".
The elements 2, 7, 12, 17, of the second horizontal row are in arithmetic
progression. The first differences, 6, 4, 8, of the elements 1, 7, 11, 19,
of the second left-slanted row are in A. P. In the second and fourth
horizontal rows, the difference of the means equals the difference of
the extremes.
Other isolated curiosities exist. In general, the distribution of the
integers has a random character. The path joining the consecutive
integers is wildly knotted.
15
14 13
9 8 10
6 4
11 5 12
1 2
18 7 16
17 19
3