ACTIVITY: MAGIC TRIANGLES & MAGIC SQUARE 1

ACTIVITY: MAGIC TRIANGLES & MAGIC SQUARE 1 Place the numbers 1, 2, 3, 4, 5, 6 in the circles so that the sums of the three numbers on either side of t...
Author: Roland Jordan
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ACTIVITY: MAGIC TRIANGLES & MAGIC SQUARE 1 Place the numbers 1, 2, 3, 4, 5, 6 in the circles so that the sums of the three numbers on either side of the triangle are the same.

Like most of our activities, this one aims to introduce teachers to the underlying ideas of mathematics through a problem. The problem here requires only a simple knowledge of arithmetic but the process we go through demands a considerable use of ingenuity and creativity. In this activity we see how a mathematical theory might develop through experimentation, conjecturing, proving, generalising and extending. Encourage participants to try anything that comes into their minds. So encourage them to guess, to play around, and don’t worry about what method they use even if it seems totally unmathematical! When they start to get answers, you will need to get them to think about which answers differ from each other. The point is that once they get one answer they can get another five simply by using the rotations of the equilateral triangle. Because we can get any one of six from the others here, we might as well say that they are all the same. So we’ll say that the two answers below are the same: 1

6

5

3

2

4

4

6

2

1

5

3

So what answers exist and how many of them are there? Trial and error will produce four at least. We show them below. But are there any more that we haven’t shown yet? 1

6

5

3

1

4

2

5

2

2

6

3

B: Sum = 10

A: Sum = 9

4

5

3

6

4

1 C: Sum = 11

3

2

4

6

1

5

D: Sum = 12

Now what can we say about these answers? Is there any relation between them? Are there any patterns that we ought to see? Here are some solutions participants may find:

1. The corner circles contain 1, 2, 3; 4, 5, 6; 1, 3, 5; and 2, 4, 6. These are the smallest and largest consecutive numbers, the odd numbers and the even numbers. 2. If you move the numbers on one answer all round one circle, you get another answer. It doesn’t matter which way you move them: clockwise or anti-clockwise. Using this move, A becomes D becomes A and B becomes C becomes B. 3. Changing corner numbers to the middle and vice-versa changes one answer into another. 4. The difference between opposite corner and middle numbers is the same. Look at A: 4 is opposite 1 and 4 – 1 = 3. 5 is opposite 2 and 5 – 2 = 3. 6 is opposite 3 and 6 – 3 = 3. 5. If you replace every number m by 7 – m, you get another answer. So in B if you replace 1 by 6, 2 by 5, and so on, B turns into C.

Prove that there are only four answers to the Magic Triangles Problem So how and why are there only 4 answers to the original question? And how can we go about establishing this beyond all reasonable doubt? There are many ways to prove or justify things. In some way each of them is equally valid. However, there are some proofs that are ‘nicer’ or ‘more efficient’ or ‘more sophisticated’ than others are. We will provide some proofs below and comment on their quality. 1. “I have worked at this for 15 minutes now and can’t find any more. So there must be only four” This isn’t a proof. We can’t be sure that we couldn’t find another answer given some more time on the problem. Or maybe someone smarter than us could find another answer. Or maybe someone in another country could. There’s too much room for doubt here. This is not the basis for a mathematical proof. It may, however, be the best that we can do. For instance, at one stage we thought that there were just 7 planets. We had looked at the heavens for thousands of years and had seen only 7. But then along came John Adams and he said that if Newton’s equations of motion were correct, there was something funny going on up there. There was some ‘wobbling’ of the planets that suggested to him that there was another planet out there that we hadn’t come across by 1845. He calculated where the planet should be and then someone went looking and eventually found it in September 1846. If we haven’t got the tools to settle something we may have to rely on the ‘I can’t find any more’ approach. While this may constitute the best state of our knowledge at some point, in mathematics it is at best a conjecture – a guess at what the answer actually is. Mathematics requires a proof, a justification that cannot be faulted. The problem is that if you work for another 10 minutes you might just find another answer. 2. “I have worked out all of the possibilities and there are only four answers” This is a good enough proof provided it can be shown that all possible cases have been looked at. What would have to be done here is to list all possible ways of putting the numbers 1 to 6 into the 6 circles and then taking from this list the ones that give equal sums on all three sides. One way to do this would be to write a computer program. On the face of it there are 6 ways to put in the first number, 5 to put in the next and so on. Hence there are 6 × 5 × 4 × 3 × 2 × 1 = 720 ways of putting in the 6 numbers. This wouldn’t take a computer very long. You could even get the class to do it in a reasonable time by dividing up the cases between them. We give a way of doing this below that means we don’t have to consider all of the 720 possibilities. Think about where to put the number 1. Because of the symmetry of the triangular shape, there are only two places to put the 1: in a corner circle or in a circle in middle of a side, as shown below: 1

1

At the next stage, we want to put a 2 in all possible places. This can be done in three ways for each of the two ways of putting in a 1. Because it is tedious we will only do it for one case:

1

1 d

2 a

b

c

2 1

2

Now in the first of these three situations, we can put 3 in four places. The symmetry of the triangle doesn’t help us now at all. If we put 3 in the ‘a’ circle then the side we have produced has a sum of 6. There is no other way to make up 6 so there is no answer possible here. If we put 3 in the ‘b’ circle, then we get a sum of a + 3 on the left side and a + 3 + c on the bottom side. They can never be equal. If we put 3 in the ‘c’ circle, then a similar argument shows that the left side has a sum of a + 3 and the bottom has a sum of a + b + 3. These too can never be equal. If we put a 3 in the ‘d’ circle we have to put 4, 5, 6 along the bottom in some order. This gives a sum that is bigger than can be made on the other two sides. We leave you to go through all the other cases. Perhaps the only consolation of doing it this way is that this is a lot easier than doing 720 different cases! We would hope that there was a better way. As much as anything, the above is an exercise in being systematic. It’s important to do things carefully and in some order so that every possibility is covered. 3. Odds and evens: “Where can I put the even numbers and where can I put the odd numbers?” The argument here is based on the fact that the sum of two even numbers is even; the sum of two odd number is even and the sum of an even and an odd number is odd. Now there are just three odd numbers and we can either have an odd number of odd numbers on one side of the triangle or we can have an even number. By careful consideration, we see that there are just four possible arrangements for the odds and evens. We show these below. The even numbers are in the blank circles and the odd ones in the circles marked with an ‘O’: O

O

O O

O

O

O

O O O

O

O

So we now know where the 1, 3, 5 go. Because of the symmetry of the equilateral triangle, there is only one way that the numbers 1, 3, 5 can be put into the first and fourth possibilities above. From there it is reasonably quick to see where the even numbers have to go. There are three possible arrangements of 1, 3, 5 in the other two cases. Working systematically will reveal the arrangements that work and those that don’t. When this has all been done we get the four answers we found earlier. 4. Reduce the sums and produce the sums In this method of proof we first of all show that the sums that we can get on each side of the triangle in an answer, are fairly restricted. After all, can you have a side sum of 24, 18, 13, 12? To do this think first about the 1. What is the biggest sum that 1 can be in? The biggest sum will be with 5 and 6 to give a sum of 1+ 6 + 5 = 12. So this is the largest sum we can get. Now turn this around. What is the smallest sum that a 6 can be in? Surely this is with 1 and 2 to give a smallest sum of 6 = 1 + 2 = 9. So the sums can only lie between 9 and 12 inclusive. OK so how can we get 9? Do this systematically. If we used a 6 we’d have to make up 3 using two numbers. This can only be done with 1 and 2. If we used a 5 we’d have to make up 4 using two numbers. This can only be done with 1 and 3. (2 and 2 isn’t allowed.) If we used a 4 we’d have to make up 5 using two numbers. This can only be done with 2 and 3. (1 and 4 isn’t allowed.) So there are only three possibilities: 6 + 2 + 1;

5 + 3 + 1;

4 + 3 + 2.

There are only three sides to the triangle so we have them now. The only question is, which numbers go in the corners? But that is easy. These are the numbers that occur twice in the three sums. So we quickly get the four answers that we got by trial and error in the first session. This method of proving that there are exactly four answers may be the nicest one of the lot. It certainly has a nice argument that shows that the sums are restricted to between 9 and 12 inclusive. 5. A little bit of algebra Here we try to use algebra to see what progress can be made. We start by putting the letters a, b, c, d, e, f into the circles. As shown in the diagram below.

a f

b c

d

e

Suppose the sum along each side is s. Then we have the following equation: 3s = (a + b + c) + (c + d + e) + (e + f + a) = (a + b + c + d + e + f) + (a + c + e) But the sum in the first bracket is 1 + 2 + 3 + 4 + 5 + 6 in some order, so that sum is 21. The sum in the second bracket is just the sum of the three corners. So we now have 3s = 21 + the corners. But the smallest value of the corners is 1 + 2 + 3 = 6 and the largest value of the corners is 4 + 5 + 6 = 15. So we have 27 = 21 + 6 < 3s < 21 + 15 = 36. So the sum has to lie between 9 and 12, just as we discovered in Method 4 above. But now we have a bonus, because the equation here tells us what the corners sum to. If s = 9, the corners sum to 6 and so are 1, 2, 3. If s = 10, the corners sum to 9 and so are 1, 2, 6; 1, 3, 5; or 2, 3, 4. If s = 11, the corners sum to 12 and so are 1, 5, 6; 2, 4, 6; 3, 4, 5. If s = 12, the corners sum to 15 and so are 4, 5, 6. For s = 9 and 12, the answers now just fall out. For s = 10 and 11 there is a bit of work to be done but it is clear that with, for instance, 1, 5 and 6 in the corners, there is no way to get a sum of 11.

Extension – MAGIC SQUARES 1 Place the numbers 1, 2, 3, 4, 5, 6, 7, 8 in the circles so that the sums of the three numbers on either side of the square are the same? Find all its answers and show there are no others.

The techniques that we have used with Magic Triangles Problem can be used again here. Any of the methods that we noted will work here. Clearly though, some are more efficient. For instance, if your teachers have a reasonable algebraic ability, we suggest that they try Method 5 to get the side sums. This method cuts down the number of options more effectively than looking at the biggest sum that 1 can be in. So, your class should find that the only possible sums are 12, 13, 14 and 15. This problem takes a little more work than the Magic Triangles Problem but in the end, because your teachers will have approached it systematically, they will know that they have found all possible answers. In the end they should find a total of six answers:

Copyright © 2010 New Zealand Ministry of Education Source: http://nzmaths.co.nz/resource/six-circles