Author 
Topic 

TchrWill
Advanced Member
USA
80 Posts 
Posted  04/19/2013 : 14:13:51

Determine the location of the integers 1 through 12 in the circles so that the sum of the numbers in each of the six rows is the same? Is there more than one solution?
.............O .......O..O..O..O .........O.....O .......O..O..O..O .............O

Edited by  TchrWill on 04/19/2013 14:19:19 


the_hill1962
Advanced Member
USA
1470 Posts 
Posted  04/20/2013 : 23:14:37

I only see 5 rows. There must be something that I am not seeing correctly because, as I see it, there is no solution. The 1st and 5th row only have one circle in each. The sum could not be the same because different integers would be put in them. Maybe the rows are verticle? However, I see seven 'columns' of circles.

Edited by  the_hill1962 on 04/20/2013 23:20:42 


someguy
Advanced Member
Canada
143 Posts 
Posted  04/21/2013 : 04:37:42

the_hill, I see it as a 6 pointed start made up of two triangles. The edges of the triangles are the six 'rows'. Each row has 4 numbers and each number is in two rows. The 4 numbers in each row must all sum to the same value.
Although I had not seen this one before, I have seen a similar but easier problem (figuring out how to create a 3 by 3 magic square without guessing or using the 'pattern' to fill in the numbers). The same technique works here. The first step is to figure out what each row sums to.
Good luck (there is more than one solution).



TchrWill
Advanced Member
USA
80 Posts 
Posted  04/21/2013 : 09:04:53

quote: Originally posted by someguy
the_hill, I see it as a 6 pointed start made up of two triangles. The edges of the triangles are the six 'rows'. Each row has 4 numbers and each number is in two rows. The 4 numbers in each row must all sum to the same value.
Although I had not seen this one before, I have seen a similar but easier problem (figuring out how to create a 3 by 3 magic square without guessing or using the 'pattern' to fill in the numbers). The same technique works here. The first step is to figure out what each row sums to.
Good luck (there is more than one solution).
Magic triangles, squares, stars, and polygons have been around for centuries and come in many varieties. This article provides some insight into the creation of 6 pointed Magic Stars. The subject of 5 pointed magic stars is covered in a seperate article. As with other typical magic sum figures, the object of the 6 pointed magic star problem is to place either a given set of 12 numbers in the figure so that all six rows of four numbers each sum to a constant value or to find a set of numbers to place in the figure so that all six rows of four numbers each sum to a given constant. One requires that you use the given set of numbers to create a constant sum while the other requires that you find a set of 12 numbers that will sum to the given constant. It is obvious that the 6 pointed magic star is considerably more difficult to derive than the 5 pointed magic star. With 12 different numbers, there are literally 479,001,600 different ways Of arranging the 12 numbers in the star pattern. Clearly, not all of them will meet the requirement of every row summing to the same constant but the task does seem overwhelming at first sight. As it turns out, for every set of 12 valid numbers, there are 80 basic solutions and 880 additional ones derived through rotations and reflections of the basic 80. This classic math puzzle is typically attacked by trial and error. It does not appear to lend itself to a solution by traditional analytical techniques. The recreational mathematics literature is filled with material that addresses magic squares but very little material exists addressing the "how too" for creating magic stars. As a natural extension of the 5 pointed Magic Star method, I set out to determine whether there was an analytical approach that could be used to simplify the seemingly overwhelming task. Lets start out by attempting to derive what might be called the most basic 6 pointed magic star using the first 12 integers from 1 to 12 with a sum of S = 12(13)/2 = 78. Based on the previously derived methods for magic triangles and magic stars, with six rows and each number appearing in two rows, the row sum s may be derived from s = 2S/6 = S/3, which in our starting case becomes s =78/3 = 26. We are faced with the selection of combinations of the numbers 1 through 12, in sets of four, in any way possible, such that they can be arranged in six rows of four numbers each with each row adding up to 26. You might already have recognized that this takes the form of two overlapping equilateral triangles, or the six pointed star figure (traditionally referred to as the Star of David), that enables us to locate the six rows of four numbers each, with each row adding up to our constant of 26.
Good luck.

Edited by  TchrWill on 04/21/2013 13:17:37 


Ultraglide
Advanced Member
Canada
299 Posts 
Posted  04/21/2013 : 12:52:39

Here's a hint: Each circle appears in exactly two lines so each number is used twice. 


Ultraglide
Advanced Member
Canada
299 Posts 
Posted  04/22/2013 : 17:20:28

You beat me to it, that was my next hint: the lines have to add up to 26. There are more than 6 groupings that add up to 26 so this is more difficult than the 3x3 magic square where there is only one solution (sum 15)  all other solutions are rotations or reflections of the same square. 


TchrWill
Advanced Member
USA
80 Posts 
Posted  05/17/2013 : 09:11:29

quote: Originally posted by TchrWill
Determine the location of the integers 1 through 12 in the circles so that the sum of the numbers in each of the six rows is the same? Is there more than one solution?
.............O .......O..O..O..O .........O.....O .......O..O..O..O .............O
Draw yourself a cup of coffee and enjoy.
How can you locate 12 numbers in six rows of four numbers each (two equilateral triangles intersecting one another as the Star of David) such that each row adds up to a constant?
Magic triangles, squares, stars, and polygons have been around for centuries and come in many varieties. Articles covering magic triangles, magic squares, and magic stars may be found elsewhere in the Knowledge Database. This article provides some insight into the creation of 6 pointed Magic Stars. The subject of 5 pointed magic stars is covered in a seperate article. As with other typical magic sum figures, the object of the 6 pointed magic star problem is to place either a given set of 12 numbers in the figure so that all six rows of four numbers each sum to a constant value or to find a set of numbers to place in the figure so that all six rows of four numbers each sum to a given constant. One requires that you use the given set of numbers to create a constant sum while the other requires that you find a set of 12 numbers that will sum to the given constant. It is obvious that the 6 pointed magic star is considerably more difficult to derive than the 5 pointed magic star. With 12 different numbers, there are literally 479,001,600 different ways af arranging the 12 numbers in the star pattern. Clearly, not all of them will meet the requirement of every row summing to the same constant but the task does seem overwhelming at first sight. As it turns out, for every set of 12 valid numbers, there are 80 basic solutions and 880 additional ones derived through rotations and reflections of the basic 80. This classic math puzzle is typically attacked by trial and error. It does not appear to lend itself to a solution by traditional analytical techniques. The recreational mathematics literature is filled with material that addresses magic squares but very little material exists addressing the "how too" for creating magic stars. As a natural extension of the 5 pointed Magic Star method, I set out to determine whether there was an analytical approach that could be used to simplify the seemingly overwhelming task. The following defines a process through which one can logically derive the many solutions that do exist for each set of numbers. There might be other ways, other twists and turns that could be taken, and perhaps, a simpler approach all together. Lets start out by attempting to derive what might be called the most basic 6 pointed magic star using the first 12 integers from 1 to 12 with a sum of S = 12(13)/2 = 78. Based on the previously derived methods for magic triangles and magic stars, with six rows and each number appearing in two rows, the row sum s may be derived from s = 2S/6 = S/3, which in our starting case becomes s =78/3 = 26.
We are faced with the selection of combinations of the numbers 1 through 12, in sets of four, in any way possible, such that they can be arranged in six rows of four numbers each with each row adding up to 26. You might already have recognized that this takes the form of two overlapping equilateral triangles, or the six pointed star figure (traditionally referred to as the Star of David), that enables us to locate the six rows of four numbers each, with each row adding up to our constant. The following picture illustrates the problem.
.................................................X
...................................X.......X........X.......X
.......................................X.................X
...................................X.......X........X.......X
.................................................X
As you can see, there are six rows, each containing 4 numbers each. Now, for the purpose of the analysis, we will use the following nomenclature throughout:
.................................................a
.....................................f......F........A......b
.......................................E................B
....................................e......D........C......c
.................................................d
The numbers 1 through 12 offer us the following 33 combinations for adding up to 26:
121112.....23912.....34712.....45611.....5678 131012.....231011....34811.....45710 14912......24 812.....34910.....4589 141011.....24 911.....35612.....4679 15812......25712......35711.. 15911......25811......35810 16712.......25910......36710 16811.......26711......3689 16910.......26810 17810.......2789
Of course, these are merely the 33 combinations of the 12 digits that can add up to 26. There are 24 permutations of each combination set making a total of 792 different permutations of the digits 1 through 12 that add up to 26. Sounds like an overwhelming task, but lets see what we can uncover.
Using the designations of the numbers from the above figure, we can write A + B + a + c = 26 from which we can obtain the sum of A + B as A + B = 26  a  c. Doing this for the other 5 rows we obtain the following: A + B = 26  a  c B + C = 26  b  d C + D = 26  c  e D + E = 26  d  f E + F = 26  e  a F + A = 26  f  b
From A + B = 26  a  c we get B = 26  A  a  c. Substituting into the expression for B + C, we derive 26  A  a  c + C = 26  b  d from which we get C = A + a + c  b  d. Substituting into the expression for C + D, and following the same process all the way through all six expressions, we end up with the startling indentity a + c + e = b + d + f. How simple. The sums of the two sets of three corner numbers of our overlapping equilateral triangles must equal each other. But, we have no idea as to their order or placement, numerically speaking.
Using the same basic expressions from above, we can also write:
a + c = 26  A  B b + d = 26  B  C c + e = 26  C  D d + f = 26  D  E e + a = 26  E  F f + b = 26  F  A
Solving for c in a + c = 26  A  B and substituting into c + e we get e = A + B + a  C  D. Substituting into the expression for e + a and collecting terms, we get A + B + E + F + 2a = 26 + C + D. Doing the same thing with the expressions for b + d, d + f, and f + b, we get A + B + C + F + 2b = 26 + D + E. Subtracting the two expressions, we get E  C + 2a  2b = C  E which simplifies to E + a = C + b. Carrying this to its ultimate, we find that
E + a = C + b F + b = D + c A + c = E + d B + d = F + e C + e = A + f D + f = B + a
With these six expressions and a + c + e = b + d + f, how can we utilize them to zero in on the solutions? For the sake of starting someplace, lets start out at ground zero by assuming that the outer six numbers are 1 through 6, the simplest of starting points. Unfortunately, we discover that the sum of the numbers 1 through 6 is 21 and therefore cannot work. Why? One side of our boundry condition, a + c + e = b + d + f, adds to 10 and the other side 11. We can now deduce that the least sum of the outer six numbers must be 12, the sum of the six outer numbers, must be even, and 12 sequential numbers will not work. So lets try 1, 2, 3, 4, 5, and 7 which add up to 22. How can we split them up so that both sets of three add up to 11 each. Only one way I'm afraid; 137 and 245. Lets assume that a = 1, c = 3, and e = 7 for starters. We must find two sets of 4 digits from our 33 candidates that both start with 1, where one row contains a 3 and the other row contains a 7. What do you know, we have 131012 and 17810 made to order. Of course in our trianglular pattern, they would be ordered as 110123 and 18107. Are we home free? Sorry. We have a 10 in both rows meaning that the 137 combination cannot be made to work and thus the 245 combination is meaningless also. We must therefore move up to the next even number. Lets try 1, 2, 3, 4, 5, and 9 adding up to 24. We can split them up into 129 and 345. Place the numbers 129 in a, c, and e. We have six ways of locating the 3, 4, and 5 in b, d, and f. We have only one set of numbers that will include the 1 and 2, that being 121112 making row aABc 111122 or 112112. Lets try the 111122 first.
Row aFEe can only be 16109. Row fFAb can be made from 45611 or 34811 or 35711. The only candidate of these three that has commonality with the required 6 or 10 in aFEe is 45611 with a common 6 in the F location. Therefore, fFAb is either 46115 or 56114. From A + c = E + d and A + c = 13, E + d = 13. Since we have a 10 at E, d must be 3. What do we have so far?
.................................................1
.....................................f......6.......11......b
.......................................10...............12
....................................9.......D.......C......2
..................................................3
Having B + a = 13 and D + f = B + a, D + f = 13 = 1+12, 2+11, 3+10, 4+9, 5+8, or 6+7. Since 1, 3, 6, 9, 11, and 12 are already used, D + f = 5 + 8 or 8 + 5. Since fFAb can have only a 4 or 5 at f, D + f = 8 + 5 giving us
.................................................1
....................................5......6......11......b
......................................10...............12
...................................9.......8.......C......2
................................................3
Clearly b = 4 and C = 7 for our final answer of
.................................................1
.....................................5......6.......11......4
......................................10...............12............... all rows summing to 26.
.....................................9......8........7.......2
..................................................3
Not too bad for the first version. What about the version where aABc is 112112? By the same process, which moved much faster the second time around, we get
.................................................1
.....................................5......6.......12......3
.......................................10...............11 all rows summing to 26.
...................................9........7........8.......2
.................................................4
Note that with the switch of the 11 and 12, only the 3 and 4 plus the 7 and 8 switched places.
That was easier than I expected and having gone through the process once, it will be easier to find other solutions rather quickly. All you need to do is select another set of values for a + c + e = b + d + f and check out the possible options and paths, one by one. After one or two more, you will see the definite patterns that emerge which will aid you in selecting compatible options for subsequent selections. Knowing that the minimum outer point sum is 12, we can create a list of possible outer set combinations as shown in the following table
.................1....2....3....4....5....6....7....8....9....10....11....12 Sum #1.............1....2.......................................9............................12 < a + c + e ............................3....4....5...................................................12 < b + d + f #2.............1....2.............................................10....................13 ............................3....4..........6.............................................13 (No 1211 as 11 and 12 must be with 1 and 2) #3.............1..........3...........................8.................................12 ......................2..........4..........6.............................................12 #4.............1..........3................................9............................13 ......................2..........4...............7........................................13 #5.............1..........3................................9............................13 ......................2...............5....6..............................................13 #6.............1..........3.......................................10....................14 ......................2..........4.....................8..................................14 #7.............1..........3.......................................10....................14 ......................2...............5.........7.........................................14 #8.............1..........3..............................................11.............15 ......................2..........4..........................9.............................15 #9.............1..........3..............................................11.............15 ......................2...............5................8..................................15 #10...........1..........3..............................................11.............15 ......................2....................6....7.........................................15 and so on.
The table only defines the possible number combinations for ace and bdf but does not identify the order in which the numbers appear in the locations. There are six different arrangements for each set as illustrated below. ............a..........a..........b..........b..........c..........c ..........c..b......b..c.......a..c......c..a......a..b......b..a (a, b, and c representing any set three numbers)
...........................a..................a Take note that b..c is merely c..b rotated about a vertical axis such that if a solution is found for the first, the second is simply the first rotated about the vertical axis. This applies to the other sets too.
Lets try and find another solution using the #2 corner sums. We start out with
.................................................1
.....................................f......F.......A......b
.......................................E................B
..................................10......D.......C......2
................................................d
Using our method, we have
AB must be 1112 or 1211 again, try 11 and 12 aFEe can only use 16910 or 17810 Trying 16910 for aFEe, d = 4 from A + c = E + d fFAbcan pnly use 45611 but 4 is already used so aFEe cannot be 16910 Try aFEe = 19610 There are no candidates for fFAb so aFEe cannot be 19610 Let aFEe be 17810 making d = 5 fFAb can only use 35711 but 5 is already used so aFEe cannot be 17810 Lastly, try aFEe = 18710 making d = 6 fFAb can only use 34811 Trying fFAb = 38114, C = 4 from C + b = E + a but b = 4 Trying fFAb = 48113, C = 5 making D = 9 We end up with
.................................................1
.....................................4......8.......11.....3
........................................7................12
...................................10.......9.......5......2
..................................................6
Switching A and B to 12 and 11 we ultimately derive
.................................................1
.....................................3......7.......12.....4
........................................8................11
...................................10.......9.......5......2
..................................................6
Now that we have a reasonably simple and organized way to derive 6 pointed Magic Stars with row sums of 26, the next logical question is whether we can derive a Magic star with larger row sums? If we draw on the knowledge gained from the articles on Magic Triangles and 5 pointed Magic Stars, we soon see that we can derive any row sum we wish. For 12 numbers in sequence, n1 through n12, N being the highest, we can use the following relationships in any way necessary to establish the boundries of our problem. Let s = the row sum, S = the sum of the 12 numbers, and N = the highest of the 12 numbers. Then, s = S/3 = 4N  22, S = 3s = 12N 66, and N = (3s + 66)/12 = (S + 66)/12. Lets search for a given row sum of 50. For s = 50, S = 150, N = 18, making n1 through n12 7 through 18. For the most basic solution, we can make use of our minimal sum basic solution derived earlier. In the solution for s = 26 using the numbers 1 through 12, we had the 1st, 2nd, and 9th numbers located in the ace positions with the 3rd, 4th, and 5th numbers located in the dbf positions. Lets place the 7, 8, and 15 in the ace positions and the 9, 10, and 11 numbers in the dbf positions giving us
.................................................7
...................................11......F.......A.....10
........................................E...............B
...................................15......D.......C......8
..................................................9
Lets place the remaining numbers in the same relative positions as the sequence of numbers from our basic solution which gives us
.................................................7
.................................11......12.......17.....10
......................................16...............18 all rows summing to 50.
.................................15......14.......13......8
................................................9
How much simpler can we get? We could also have derived the solution by the step by step process we used earlier but would have had to create the list of 33 possible row sums using the numbers 7 through 18 first. Clearly, we also have 80 basic solutions for this row sum and 880 others derived from rotations and mirror imaging.
What if we had been asked to create a star using the numbers 7 through 18 where all the rows added up to some constant? We would then derive S = 12(18)  66 = 150 s = 150/3 = 50 and proceed as before.
As with the Magic Triangles, we can also use 12 numbers that are not in sequence but rather in arithmetic progression. The smallest base solution for 12 numbers in arithmetic progression is based on the 12 odd numbers from 1 to 23 with constant difference of 2. The sum of these 12 numbers is S = 12(1 + 23)/2 = 144. The row sum is s = S/3 = 48. Using our earlier procedure, or the basic solution pattern for the numbers 1 through 12 from above, we end up with the minimal odd number arithmetic progression star of
.................................................1
...................................9......11.......21.....7
......................................19...............23 all rows summing to 48.
.................................17......15.......13......3
................................................5
The next smallest solution is for the 12 numbers 2 to 24 with the same constant difference of 2. The sum of these 12 numbers is S = 12(2 + 24)/2 = 156. The row sum is s = 156/3 = 52. We now have the minimal even number arithmetic progression star of
.................................................2
.................................10......12.......22.....8
......................................20...............24 all rows summing to 52.
.................................18......16.......14......4
................................................6
In general, 6 point Magic Stars using 12 numbers in arithmetic progression can be derived from the following rules. Given the desired row sum s. The sum of the 12 numbers is then S = 3s. Selecting a common difference d, the highest number of the 12 numbers is given by N = (s + 22d)/4. The lowest number is given by n = N  11d. If the choice of d does not result in an integral n and N, try another. When integral values of n and N are achieved, the sum can be checked from S = 12(n + N)/2 = 6(n + N).
Given the 12 numbers in arithmetic progression. S = 6(n + N) and s = S/3.
The row sums follow a distinct pattern with the common difference. d............2......3.....4.......5 1 to N s...........48....70....92....114 2 to N s...........52....74....96....118 3 to N s...........56....78...100...122 4 to N s..........60....82....104...126 5 to N s..........64...86.....108...130
I hope the above material has been interesting and informative for you and that you will enjoy deriving magic stars for your friends.
quote: Originally posted by the_hill1962
It has been a long time since I have heard of the 6 Pointed Star problem and I forget what it is.




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