A magic square is an arrangement like the one below where the vertical, horizontal and diagonal lines of numbers all add up to the same value. This ‘same value’ is called the sum of the magic square.
It’s a critical part of this problem that three times the centre square is equal to the sum of the magic square.
Magic squares are interesting objects in both mathematics proper and in recreational mathematics. It is likely that students have already encountered magic squares. The problems in this sequence give students the opportunity to use known numerical or algebraic concepts.
This problem is part of a series exploring magic squares. Little Magic Squares and A Square of Circles , Level 2, Big Magic Squares Level 3. Also at Level 4, Negative Magic Squares, . The Magic Square, Level 5 shows why three times the centre number is equal to the sum of the magic square. Difference Magic Squares at Level 6, looks at an interesting variation of the magic square concept.
Tui is getting the idea of magic squares. She decides to make all of the magic squares that she can using the fractions 7/6, 4/3, and 3/2.
How many can she make?
It takes her quite a while because she doesn’t know that the sum of a magic square is always three times the number in the centre.
- Show the students a magic square such as the one below, and ask why it is called a magic square.
- Have them to check that the rows all have the same sum (of 12); that the columns all have the same sum; and that the diagonals have the same sum.
- Pose Tui’s problem.
- Have them work in pairs to see how many magic squares they can find.
- Get some of the pairs to report back. Can they prove that the arrays they have produced are magic squares?
- How many different magic squares can they find? How many do they think there are?
- Ask the students to write up what they have discovered.
- Pose the Extension problem when appropriate.
Extension to the problem
Can you make up a magic square using any of the fractions 1/12, 1/6, 1/4, 1/3, 5/12, 1/2, and 7/12, with 1/6, 1/3 and 1/2, in that order, down the main diagonal?
Can you make up a magic square that has fractions in it?
There are four answers if you allow the same fraction to be used in each entry of the squares. The neat way to see this is to notice that 7/6, 4/3, 3/2 is the same as 7/6, 8/6, 9/6.
Being systematic in this problem means choosing different numbers for the centre square.
centre square = 7/6: This means that the sum of the magic square has to be 21/6. This sum can only be made by using three 7/6s. So this magic square consists of all 7/6s. Call this square A.
centre square = 4/3 (or 8/6): This means that the sum of the magic square has to be 24/6. Now 24/6 can only be made with 7/6 + 8/6 + 9/6 or 8/6 +8/6 + 8/6. One way to get a magic square here is for all of the entries to be 8/6. This gives another magic square that we will call square B.
Now suppose that the centre square (8/6) is used with 7/6 and 9/6 somewhere to get the sum of 24/6. Because of the symmetry of the square, we can assume without loss of generality that this is either done on the main diagonal or on the vertical column through the centre.
In the first case, the middle square in the top row is either a 8/6 or a 9/6. (It can’t be 7/6 because then the row sum would not be 24/6.) We follow through these two situations.
In the ‘8/6‘ case, we have to have a 9/6 in the top right-hand square. But then the last column can’t sum to 24/6.
In the ‘9/6’ case, the 8/6 in the top row and the 7/6 in the middle column are forced. This then means that there has to be a 7/6 in the middle square of the last column. This forces the 9/6 and 8/6 in the first column. A quick check shows that we have another magic square. Call this magic square C.
Now we think about 7/6, 8/6, 9/6 being in the centre row. Because of the symmetry of the square, we can assume that there is a 8/6 in the top left-hand square and a 9/6 in the top right-hand square. This forces the two 7/6s as shown and then the final 9/6 falls into place. A final check shows that this is a magic square.
If we rotate this last magic square through 90°, then it looks exactly the same as C. So we don’t get a new magic square this way.
centre square = 9/6: This means that the sum of the magic square has to be 27/6. This can only be done if the three numbers that make up a row or a column are all 9/6s. So we get another uninteresting magic square that we will call square D.
Tui should have found four magic squares. These are the ones we have called A, B, C and D.
Solution to the Extension
Using the techniques that should have become standard if you have done the earlier magic square problems, you should find two magic squares. First note that the sum here is 1. The magic squares are:
You can get these by putting in the entries along the main diagonal and then trying each of the fractions in the centre entry of the top row. By finding the fraction that makes the row (column/diagonal) sum to 1 you will either complete the magic square or show that some other number is need to do so.
One way to 'cheat' is to make up a magic square using whole numbers and then divide them all by the same number. The resulting fractions will make a magic square. You can see we did that with 7/6, 4/3 and 3/2.