Math Goodies is a free math help portal for students, teachers, and parents.
|
Interactive Math Goodies Software

testing left nav
Math Forums @ Math Goodies
Home | Profile | Active Topics | Members | Search | FAQ
 All Forums  Homework Help Forums  Miscellaneous Math Topics  SIx Pointed Magic Stars New Topic  Reply to Topic  Printer Friendly
Author  Topic

TchrWill

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

USA
1469 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

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 buteasier problem (figuring out how to create a 3 by 3 magic squarewithout 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

USA
80 Posts

 Posted - 04/21/2013 :  09:04:53 quote:Originally posted by someguythe_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 buteasier problem (figuring out how to create a 3 by 3 magic squarewithout 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

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

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

USA
80 Posts