# # Solve the 6-Pointed Magic Star Problem # # Use the numbers 1-12 to give equal sums in each of 6 lines making # up the star shape below. Eg: B+C+D+E = 26, B+F+I+L = 26, etc. # # A # # B C D E # # F G # # H I J K # # L #

This is a very old puzzle, first proposed by Henry E. Dudeney and catalogued by Donald E. Knuth with the entry:

+50(15)210 X261* A star puzzle: Magic 6-star of {1,2,…,12} with sums 26*

meaning: Volume 50, (1915), page 210, puzzle number X261.

This can be considered a problem in 12 unknowns with only 6 equations (the 6 straight lines making up the figure). There are therefore apparently 6 degrees of freedom. A straight-forward approach would be to choose A, B, C, D, F and G, (E must be equal to 26 – B – C – D, of course) and then calculate the other letters from the obvious relations. This would appear to give 12 x 11 x 10 x 9 x 8 x 7 solutions (12 choices for A, 11 for B etc) or 665,280 in all.

Unfortunately, many of these choices give values which are less than 1 or greater than 12 for some letters E or H – L. When these are eliminated, there are still many illegal combinations left where some numbers are duplicated. Finally the number of solutions is whittled down to 960.

But the figure has 12-fold symmetry (6 rotations x 2 reflections), so we find the number of unique solutions to be (960 / 12) or just 80.

Hardly any deep mathematical knowledge was required, just brute-force and symmetry.

We have written a simple *Perl* program to produce the 80 solutions in 5 seconds of PC elapsed time and invite reader solutions, too.

Here are the first few we have found:

N a b c d e f g h i j k l P V 1: 1 2 4 12 8 10 6 11 5 3 7 9 Peak: 38 Valley: 40 2: 1 2 6 10 8 12 4 7 3 5 11 9 Peak: 38 Valley: 40 3: 1 2 7 11 6 8 5 10 4 3 9 12 Peak: 40 Valley: 38 4: 1 2 7 12 5 10 4 8 3 6 9 11 Peak: 36 Valley: 42 5: 1 2 8 9 7 11 4 6 3 5 12 10 Peak: 38 Valley: 40 6: 1 2 8 10 6 12 4 5 3 7 11 9 Peak: 34 Valley: 44 7: 1 2 9 12 3 6 8 10 7 4 5 11 Peak: 32 Valley: 46 8: 1 2 10 11 3 8 5 7 4 6 9 12 Peak: 34 Valley: 44 9: 1 3 4 8 11 12 7 9 5 2 10 6 Peak: 40 Valley: 38 10: 1 3 5 11 7 12 4 8 2 6 10 9 Peak: 38 Valley: 40 11: 1 3 6 12 5 11 4 8 2 7 9 10 Peak: 36 Valley: 42 12: 1 3 7 11 5 6 10 12 8 2 4 9 Peak: 34 Valley: 44 13: 1 3 7 11 5 12 4 6 2 8 10 9 Peak: 34 Valley: 44 14: 1 3 7 12 4 8 11 10 9 5 2 6 Peak: 26 Valley: 52 * ....

A little curiosity: solution number 14 shows the sum of the Peaks also to be 26. Six solutions have this property; another six have the sum of the Valleys to be 26.

[ Peak Sum = A + E + K + L + H + B, Valley Sum = D + G + J + I + F + C ]

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with the same 6 point magic star , is it possible to have sum of each side as 20?

The numbers 1-12 sum to 78, so the sum of all six lines must be twice that (156) since each number occurs in two lines.

Hence the line sum must be 1/6th of that total, i.e. 26.

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I see how u got the number of solutions and how you got the solutions by pc, but is there a way to solve it without a pc’s help and without guessing, what is the logic to solving it and getting those 80 solutions.

Good question.

It is possible that you could generate all solutions from an initial solution by some sort of “powering” technique. Think of nth complex roots of 1, for example. Given a “fundamental” root, all the others are powers of this one.

It sounds like a good research project and may open up a new field of study (maybe).

Thanks for the help on the math problem

Makasih atas infonya , Aerith