Six Circles: Proof
Two Justifications of Four Keys
Here we show in two ways that there are precisely four keys. In other words there are essentially only four ways that the numbers 1 to 6 can be put into the six circles so that the sums on each side of the triangular array are the same. The first way that we show is a nice combination of number knowledge and logic. This method can be seen by relatively young students – bright 10 and 11 year olds have come to see this. The other way involves algebra and so it is only open to older students.
Before we start, note that once have produced an answer to the problem we can immediately find 5 more answers by rotating and flipping the equilateral triangle with the first numbers. The rotations are about an axis through the centre of the equilateral triangle (and perpendicular to it) and the flips are about axes through a vertex and the centre of the equilateral triangle (and in its plane). We show these below. Starting from position I, position II is a clockwise rotation of 120; position III is a clockwise rotation of 240; position IV is a flip about the black vertex; position V is a flip about the red vertex; and position VI is a flip about the white vertex.
All of these six answers represent the same key. So we are now trying to show that there are only four keys or four possible different sets of answers.
Method 1, Logic and number sense: The first part of this process is to limit the possible side sums. To do this first note that 1 has to be used somewhere. The largest side sum that 1 can be included in is 12 because the largest two numbers that can be added to 1 are 6 and 5. This means that no side sum can be bigger than 12.
Now look at this from the other perspective. Certainly 6 has to be used somewhere. What is the smallest sum that 6 can be involved in? Surely it is 6 + 1 + 2 = 9.
From the above discussion we can see that the only possible side sums are 9, 10, 11 and 12. But trial and error shows that there is one of each of these keys. And most student groups will be able to find them. The problem now is to show that there are no more keys. In other words that there is only one key with a side sum of 9 and that the same is true for 10, 11 and 12. We will do this for the side sum of 9 and you can see how it is done for the other side sums.
Now how can 9 be made up using three different numbers from 1 to 6? We’ll be systematic and start by using 6. This leaves two numbers to produce a sum of 3. Clearly this can only be done using 2 and 1.
9 = 6 + 2 + 1
So we can’t get any more sums using 6. Let’s use 5. So now we have to make up 4 using two numbers. This can be done with 3 and 1 and then 2 and 2. But we can’t use the same number twice. So this gives us only
9 = 5 + 3 + 1.
Now we have used up 6 and 5 so try 4. How can we get 5 using only two numbers? Surely 4 + 1 and 3 + 2 are the only possibilities but 4 + 1 would mean a repetition of 4 so this is out. Therefore we get
9 = 4 + 3 + 2.
Having used 6, 5 and 4 we now start with 3. This requires us to get 6 using only two numbers. So this forces 5 + 1, 4 + 2, or 3 + 3. But we can’t use the first two because we have already considered all cases where 5 and 4 appear. What’s more we can’t use the last one because we can only use 3 once so there is no sum making up 9 that uses 3 and smaller numbers. The same can obviously be said for 2 and 1. Hence the only sums we can have are
9 = 6 + 2 + 1 = 5 + 3 + 1 = 4 + 3 + 2
But let’s look at this carefully. The numbers 1, 2 and 3 appear in two of these sums. If these sums are the side sum of a key, then 1, 2 and 3 have to be in the corners and 4, 5 and 6 have to be in the middle circle of a side. Clearly they give us one of the four keys we are looking for. What’s more they show us that there is only one way to get a key with a side sum of 9.
Method 2, Algebra, logic and number sense: Suppose that the numbers a to f are in the circles and that they make up a key with side sum s.
Now the only knowledge we have at this point is that the side sums are equal. So this means that
a + b + c = c + d + e = e + f + a = s
Now that looks like far too many variables and few too many equations. But add each one of the equations together.
So 3s = (a + b + c) + (c + d + e) + (e + f + a).
We can simplify this to get
3s = (a + b + c + d + e + f) + (a + c + e).
This is not as bad as it looks because the first bracket contains each of the numbers 1 to 6. So this bracket adds up to 21 whatever a, b, c, d, e and f are individually. And the second bracket is just the three corner numbers.
So 3s = 21 + (a + c + e).
The lowest that the corner numbers can be is 6 (1 + 2 + 3) and the highest is 15 (4 + 5 + 6). That means that
27 ≤ 3s ≤ 36 or 9 ≤ s ≤ 12.
Where have we seen that before?
From here we can continue as in the second half of Method 1 or we can go this way. Consider the case where s = 9. Then the corners must add to 6. This can only be done by using 1, 2 and 3. If we put these numbers in the corners the middle numbers are forced by the fact that the side sum is 9. And, as before, we get a unique key with side sum 9.
Repeating with s = 10, 11 and 12 the desired proof is completed.
Justification of a generalisation
In this section we show that a set {a, b, c, d, e, f} is good if and only if a – d = c – f = e – b.
(Note that with ‘if and only if’ proofs, we have to prove two things. In this case we have to show that (i) if a set is good, then the three equations follow and further that (ii) if the three equations hold, then the numbers are good.)
(i) Assume the set {a, b, c, d, e, f} is good. So it can form a key. Hence, from above, the numbers satisfy equation (1). Hence we have
a + b = d + e and c + d = f + a. By a little algebra, this clearly forces a – d = e – b and a – d = c – f.
This gives us precisely what we are trying to prove.
(ii) Now we assume that a – d = c – f = e – b and try to show that the numbers a, b, c, d, e and f with these restrictions can make a key. But this is straightforward as we can recover equation (i) from a – d = c – f = e – b. so we can put a, b, c, etc. into the circles in the order that we did above.
Note: Actually we can put the numbers back in another way. Look at this diagram
From the equations a – d = c – f = e – b we have a + b = d + e, so a + b + f = d = e + f and we have b + c = e + f, so b + c + d = d + e + f. So in the diagram above, d + e + f = f + a + b = b + c + d. So the diagram above is a key.
This shows that every good set makes two keys. Why are they different keys?
But {1, 2, 3, 4, 5, 6} makes four keys. Does every good set make four keys?
Just to show that it may be hard to determine if a set is good or not, look at {1, 2, 3, 4, 5, 7}. At first sight this may appear to be not at all good. But 7 – 5 = 4 – 2 = 3 – 1. So it is good. (And you will find that it can only produce two keys. So that should give you a proof of something else.)
