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Six Circles: Students' Notes


These notes are designed to provide you with hints to help you while you work towards solving this problem and also to help you see a much larger picture of both the problem and of mathematics itself. The activity is available online at:
The notes below are split into four phases:
  1. Understanding and solving specific problems
  2. Generalising
  3. Proving
  4. Extending
The four phases are the same as those used by mathematicians when they are working on research problems.  Of course, if you only want to go as far as finding the number of possible doors you can get through, you won’t go through all of these phases. However, we hope that you will want to go further than this.
Below we have provided a number of hints for you to use if you need help. Try to do as much of the problem as you can yourself, and only use the hints if you get stuck.

Phase 1: Understanding and solving specific problems

This activity asks you to arrange the numbers from 1 to 6 into a triangle so that the three numbers on each side of the triangle add to make the same number. Once you have found one answer you are asked to find more until you have found all the possible answers. Being able to determine these will allow you to proceed through a series of doors. The problem is how many doors can you go through? Try to solve the problem without using the hints, but if you get stuck, hopefully a hint or two will help you.

Hint 1

If you are not sure how to solve the problem, the first thing you should try is just playing (experimenting) with the numbers. 
  • Put three numbers on one side and see what they add up to.
  • Now see if you can place two more numbers to make another side add up to the same amount.
  • That should leave one number left – does it work to put it in the last space?
  • Try this a few times with different combinations of three starting numbers and see if you can find a solution that works.
  • Or is it better to put numbers in the corners?

Hint 2

You may have noticed that the computer is not accepting a solution that you think is a new one. The reason for this is probably that it is actually just the same solution flipped or rotated into a different position. There are actually six different ways you can place any one solution (see below) We will call any one of these a key because it will get you through a door, but you can only use one key from each set of six.
Circles diagram.

Hint 3

You may have found one, or several, keys by playing with the numbers, but it is unlikely that you found them all, and if you did, how will you be able to be sure that you haven’t missed any? To settle this question you need to take a systematic approach. Think about what numbers each side could add up to.

Hint 4

Hopefully you worked out that the lowest number that you can make by adding three of the numbers from 1 to 6 together is 6 (1 + 2 + 3). But since you only have one of each number if you make 6 on one side, the other sides will have larger sums. How can you make the lowest possible sum on all three sides? 

Hint 5

You should have noticed that the numbers on the corners are each used twice (once for each of the sums of the sides that meets at that corner). So if you want to make the lowest possible sums then you should probably put the smallest numbers in the corners. Try that and see if you can find a key.

Hint 6

If you put 1, 2 and 3 in the corners then the only numbers left to place are 4, 5 and 6. One side has a 1 and a 2 on it. This side has a lower sum so far than the other two so you should put the largest remaining number (the 6) in the middle of it. Since that side now adds to 9, you need to make the other two sides add to 9 too. Once you have done this, see if you can use the same method to find the key with the largest sum on each side.

Hint 7

Putting the three largest numbers in the corners (4, 5 and 6), and then putting the smallest remaining number (the 1) on a side, should have allowed you to find a key with all the sides adding to 12. So, if the key with the greatest sum is 12 and the key with the smallest sum is 9, what other keys might exist, and can you find them?

Hint 8

If you are having trouble finding keys with sums of 10 and 11, try making a list of all the ways that you can get to those numbers using the numbers from 1 to 6. For example:
6 + 3 + 1 = 10 and 5 + 3 + 2 = 10 etc.

Hint 9

If you have listed all the ways to produce 10 and all the ways to produce 11 you should be able to work out how to make triangles so that the sides add to those numbers. If you are still having trouble, look at your lists and think about the role of the numbers that occur more than once in a sum. Those are the numbers that should go in the corners.

Hint 10

So you have found four keys, one adding to each of 9, 10, 11 and 12. Are you sure that there are no more?
After you have tried for a long while you may not be able to produce any more ways of getting the numbers into the circles so that they produce the same sums on each side. So at this point we hope that you would try to show that there is only that number of possibilities.
Let’s take this as our conjecture.
Conjecture: There are only 4 keys.
How can we disprove this? We can do that by finding at least one more key. But surely there must be a small finite number of keys. After all, there is a limit to the number of different ways that we can put the numbers into the circles. So even if 4 is wrong, some other number, 5, 23, 720, must be right.
Then how can we prove this finite number? It turns out that there are at least two approaches. One method is by algebra and one is not.

Hint 11

Is there a counter-example? That is, can you find another, fifth key?

Hint 12

Can you show that there is only one key with a side sum of 9?

Hint 13

Or if you are an algebraist, start with a, b, c, d, e, f in the circles starting with a in the top circle and moving round one circle at a time. If s is the side sum of that key, what can be said about 3s? How does 21 come in here?

Phase 2: Generalising

The hints above should give you enough information to complete the online Bright Sparks activity. However, the notes below provide some support if you would like to explore this problem further. One possible generalisation here is to think about what sets of 6 different numbers can be used to form a key. For example, 2, 4, 6, 8, 10, 12 can be put into the circles so that each side has a sum of 18, 20, 22, or 24. Let’s call a set of 6 different numbers good, if it can make a key. So {1, 2, 3, 4, 5, 6} is a good set.
What sets are good? Can you find a simple way of deciding?

Hint 14

Experiment! This is always best done with a friend. Choose sets at random and see what you get. Then choose sets more carefully.
Call a set good, if it produces an answer with all three sides having the same sum.

Hint 15

Can you tell in advance which sets are going to work and which won’t? What properties do all good sets have?
Is {1, 2, 3, 4, 5, 100} good? How about {1, 2, 3, 4, 5, 10}? How about {1, 2, 3, 4, 5, 7}? How can you tell by just looking at the numbers?
Can you build up a good set starting with arbitrary numbers? Put 20, 30 and 42 on the left side of the equilateral triangle. What other numbers do you need to complete the good set?

Hint 16

What can you say about the numbers in each of the corner circles and the numbers in the centre circle in the opposite side?

Hint 17

Can you formulate and solve a conjecture for good sets?

Phase 3: Proving

Conjecture: If {a, b, c, d, e, f} is a good set then a – d = c – f = e – b.
How can you prove that?

Hint 18

What properties does a good set have? Write down some algebraic equations and experiment.
Sometimes conjectures ‘go the other way’. Let’s try the converse of the previous conjecture.
Conjecture: If a – d = c – f = e – b, then {a, b, c, d, e, f} is a good set.

Hint 19

How could you prove this? What properties do good sets have?
That all should lead you to a Theorem.
Theorem: Let a, b, c, d, e, f be different numbers. The set {a, b, c, d, e, f} is good if and only if a – d = c – f = e – b.
Mathematicians try to get these ‘if and only if’ theorems because they tell them all there is to know about the situation. In this case we no longer have to struggle to put the numbers into the circles to see if they are good. It is enough to see if they satisfy the equations. And hopefully that is easier.

Phase 4: Extending the problem

Now we get back to some fun stuff again after the hard work of finding and proving conjectures. One way that mathematics develops is by mathematicians trying to take a problem and extending it in some way. What other similar problems can we build around this one? Come up with your own ideas and see where they lead you. We have given a few examples below to start you off.
  • What if the 6 numbers to have repetitions?
  • What if there are only 3 circles in the array and not 6?
  • What if there are 9 circles in the array?
  • What if we use a square and not an equilateral triangle?
  • What if we use a regular polygon with n sides?
  • What if we use any old shape at all – a W say? Or the Olympic rings?