A Math Forum Web Unit
Mutsumi Suzuki's

Constructing a Magic Star

Begin with a number 1 at the top.


If we use the first set {1,2,11,12} as
the line ACFH, then we cannot use
the second set {1,3,10,12} as the other
line ADGK because the number 12
would be used twice.

Thus the sets that work for ADGK are {1,6,9,10} and {1,7,8,10}.

Listing such combinations gives:

{1,2,11,12} --- {1,6,9,10} or {1,7,8,10}
{1,3,10,12} --- {1,5,9,11} or {1,6,8,11}
{1,4,9,12} --- {1,6,8,11} or {1,7,8,10}
{1,4,10,11} --- and so on.
{1,5,8,12} ---
{1,5,9,11} --- {1,6,7,12} or {1,7,8,10}
{1,6,7,12} --- NIL
{1,6,8,11} --- NIL
{1,6,9,10} --- NIL
{1,7,8,10} --- NIL

Understanding the patterns: first trial

First let's try using {1,2,11,12} for A,C,F,H and {1,6,9,10} for A,D,G,K to see if these combinations might work.

The possible combinations for B,C,D,E will be

B,2,6,E
B,2,9,E
B,2,10,E
B,11,6,E
B,11,9,E
B,11,10,E
B,12,6,E
B,12,9,E
B,12,10,E

But the rest of the numbers {3,4,5,7,8} are used for {B,E,I,J,L}, so the possible combinations of B+E = 3+4, 3+5, 3+7, 3+8, ... , 7+8. Thus B+E = 7 or 8 or 9 or 10 or 11 or 12 or 13 or 15.

Let's check possibilities for the set.

1. Let's try the first possible combination: B,2,6,E
   2 + 6 = 8; 26 - 8 = 18; therefore B + E = 18.
   But B + E cannot equal 18, so you cannot use 2,6 for C,D.

2. The next combination to try is B,2,9,E
   2 + 9 = 11; 26 - 11 = 15; therefore B + E = 15, which is possible:
   {B,E} = {7,8}
   {I,J,L} = {3,4,5}.

   The possible combinations for H,I,J,K are:

   11,I,J,6
   11,I,J,10
   12,I,J,6
   12,I,J,10
   

   But the condition of the sum = 26 is satisfied only by

   11,I,J,6 and {I,J} = {4,5} or
   12,I,J,6 and {I,J} + {3,5}

   Thus we get
   

3. B,2,10,E
   2 + 10 = 12, 26 - 12 = 14; therefore, B + E = 14.
   This case cannot be used.

4. B,11,6,E
   11 + 6 = 17, 26 - 17 = 9, therefore, B + E = 9.
   This case is possible: {B,E} = {4,5].

   . . . . and so on.

Such procedures can be repeated one by one. They are very good for using up time on a long train trip.

Transforming Stars

The next problem is how to construct a star with the number 1 not at the top but at an inner corner.

The transformation (or exchanging rule) yields such stars.

On to Transforming stars
Back to What is a magic star?