6 circles
A question to get you started.
This is not the entire task, just a starting point. See where it takes you!
Fill each of the six circles below with one of the integers 1, 2, 3, 4, 5 or 6, so that the triangle's sidesums are equal. Then, think beyond what you have been asked to do.
Here are some ideas students have had while working on this problem.
You may find inspiration in their ideas!
From student session, MATRIX Creswick, Nov 2018
This is a list of questions you can ask yourself or your students while working on this problem.
Naturally this isn't a complete list, but it will take you down one path.
 Can you find a solution?
 Can you find another solution?
 Are the solutions related in some way?
 Are any solutions `the same'?
 How do you define `the same'?
 How many `different' solutions are there?
 How do you know there aren't more?
 Are there any sidesums you can't get?
 How do you know?
 If I give you a sidesum, can you tell me which numbers have to be in the corners?
 Does it always have to be those ones in the corners?
This section will give you hints on some parts of the problem.
Expanding a section will reveal potential spoilers. If you only want a small hint, tread carefully.
What are you stuck on?
Deciding which solutions count as "the same"
Some solutions seem pretty similar. Some are definitely different. What's "the same" about the ones that are similar?
If you're like me, you've found that the ones with the same numbers in the corners look pretty similar. They might be rotated or flipped versions of each other. For example, there's some solutions that have $1$, $3$ and $5$ in the corners, that all can be rotated or flipped or both to get the same thing.
The answer to the question "Are these the same or are these different?" is "It's up to you." But if you decide they are the same, then you need to say what defines "a solution"  what makes two the same and another one different.
I want to say that all solutions related by flips and rotations are actually the same solution, and ones that aren't related by flips or rotations are different.
This is actually enough to define "same" and "different", but it may help to have another property to look for.
There's many properties that would work, but whatever it is it needs to follow these two rules:
 If the property is the same for two solutions, then one can be rotated or flipped to get the other
 If the property is different for two solutions, then there's no way to rotate or flip one to get the other
Let's take as an example the numbers in the corners, ignoring the order. The second rule is pretty easy: no amount of flipping or rotating will change what numbers are in the corners. The first rule is not so easy. For that we have to also be sure that if the corners are fixed then the sides can only be one thing. So until we've proved that, we can't use this property can use this property as long as we note that we might run into trouble later.
If we want to play it safe, we can instead use a list of the numbers on each of the three sides.
Working out how the corners are related to the sidesum
The corners are each on two sides, whereas the middles are only on one.
It seems like they have more 'weight' than the middles, that they contribute twice  but to what?
A corner doesn't contribute twice to the sidesum.
It contributes once to two sidesums.
So it contributes twice to the sum of the two sidesums.
So it also contributes twice to the sum of all three sidesums.
There's two ways to get the sum of all three sidesums. The first is clear: just sum all three side sums.
The sidesums are all the same, so it's really 3 times the sidesum.
The other way is to look at what it's made of. We already know that each corner contributes twice... what about the middles?
The middles each contribute once, since each is only on one side.
So '3 times the side sum' is the same as 'add the middles to two times the corners'
But there's another way to think about this. (at least one)
Each circle must be filled with the numbers $1$  $6$, and each number must appear once.
The sum of all the numbers is 21.
If we sum all the numbers once, what else do we need to add to make up '3 times the sidesum'?
The sum of all the numbers once is the middles once and the corners once, so we would need to add the corners once more.
So "3 times the sidesum" is "21 plus the corners".
[There are other ways to think about this.]
Determining the smallest and largest sidesums BAD
First, check to see which sidesums you've got.
Let's go small first. What sidesum do you think you can't get?
There are at least two options for how to be sure BAD:
 Brute force/Exhaustion: Try all combinations of numbers in the circles and, for each one check whether the sidesums are all equal to the number you're after. If none of them pass, then you've proved that sidesum isn't possible.
 Bestcase scenario: Work towards getting the smallest possible sidesum.
If you're trying for option 2, remember what the rules of the game are: you have to use all 6 numbers from $1$  $6$ in the circles.
In particular, you have to use $6$ somewhere.
Whichever side you put it on, how can you make sure that side sums to the smallest possible total?
The smallest total with $6$ will be if you put it on the same side as the smallest two numbers, $1$ and $2$.
$6+1+2=9$, meaning that the smallest possible sidesum is $9$.
Now let's think about the biggest. You can use similar logic:
You have to use $1$ somewhere.
Whichever side you put it on, how can you make sure that side sums to the largest possible total?
The largest total with $1$ will be if you put it on the same side as the largest two numbers, $5$ and $6$.
$1+5+6=12$, meaning that the largest possible sidesum is $12$.
[There are other ways to be sure of this.]
Knowing I've got all possible solutions
There's a few ways to be sure you have all the solutions. The one that requires the least thinking, and the most elbowgrease, is to simply try all the combinations of 6 numbers in 6 circles and throw out the ones that don't give three equal sides. There's only $6! = 720$ combinations to try, get cracking!
If you don't want to do that, you can find ways to reduce your search. There's many many ways.
It may help to first determine the smallest and largest sidesums.
It may also help to then decide which solutions count as the same. Though you probably will also want to know how many candidates were secretly the same, and that depends on how you've defined "the same".
It may also help to then work out how the corners are related to the sidesums.
If you know all of that, then you know that:
 There are only four sidesums
 Solutions are related by flips and rotations
 Each sidesum corresponds to a cornersum
Since there are only four sidesums, there are only four numbers the corners could sum to. If you've gotten this far it's not giving anything away to say that those numbers are $6$, $9$, $12$, and $15$ (corresponding to sidesums of $9$, $10$, $11$ and $12$).
How many ways are there for the corners to sum to $6$ (giving a sidesum of $9$)?
There is only one possible choice for the corners, $1$, $2$ and $3$. My definition counts two solutions as the same if one can be rotated or flipped to get another, so I don't care which corner is which.
Since the corners sum to $6$, the sidesum must be $9$.
So we can fill the middles with the numbers that make up $9$. For example, between the corners with $1$ and $3$ there must be a $5$.
The solution is
which is valid, since it uses $1$  $6$ once each. Since there is only one choice for the corners, and only one choice for the middles, there is only one solution with corners that sum to $6$ (and therefore a sidesum of $9$).
How many ways are there for the corners to sum to $15$ (giving a sidesum of $12$)?
Follow the same logic as before.
How many ways are there for the corners to sum to $9$ (giving a sidesum of $10$)?
Three: $\{1,3,5\}$, $\{1,2,6\}$, and $\{2,3,4\}$.
Since the corners sum to $9$, the sidesum must be $10$.
So for each set of corners, we can fill the middles with numbers that make up $10$.
We then need to make sure the solutions this gives us are valid.
For the 2nd and 3rd corner sets, we get
and 
both of which are invalid, because they don't use the numbers $1$  $6$ exactly once each.
For the other one, $\{1,3,5\}$, we get
and again since there's only one choice for the corners and the middles, there's only one solution with a cornersum of $9$ (and therefore a sidesum of $10$).
How many ways are there for the corners to sum to $12$ (giving a sidesum of $11$)?
Follow the same logic as before.
This section contains a "complete" path through this problem.
It covers only one path of many that could be taken, and there are still plenty of unanswered questions.
If you haven't tried the problem for yourself yet, and don't want it spoiled, don't read this section!
Download as a PDF
Trying out numbers in the circles, we find that there are actually quite a few solutions. Here are three, along with their sidesum $S$:
$S=10$  $S=10$  $S=9$ 
Notice that there are some that have the same $S$, despite being "different"... are they secretly the same?
Some solutions definitely are related, e.g.
and 
are simply rotations of each other, and $S=10$ for both. This makes sense, because a rotation won't change the sidesum! Neither will a flip, e.g.:
$\longrightarrow$ 
So if we've found one solution, we can rotate it or flip it to get another without changing $S$. How many can we relate in this way?
Well, there are 3 rotations of a solution, and each one can be flipped to get another solution, so there's 6 in total that are related by rotating and flipping. For example,



These all have a sidesum of $9$.
So are these all the same or are they different?
Which ones are "the same"?
Here there is a choice to be made: A choice of definition of "the same". No choice is right or wrong, but some may be more helpful than others, depending on what we're trying to do.
For now, let's say they are the same if they have the same sidesum. There may be more that give the same sidesum than the $6$ you get from rotating and flipping, but we'll put that question on the shelf for now.
The next question is, how many different (what we're now calling different) solutions are there? In other words, how many different sidesums are there?
So far we've found solutions with sidesums of $9$ and $10$. Working a bit harder we can find one with $S=11$ and one with $S=12$:
$S=11$  $S=12$ 
Now we have 4 different solutions. Since we don't care about orientation let's put the smallest corner at the top and the largest at the bottomright:
$S=9$  $S=10$  $S=11$  $S=12$ 
We don't know we've got everything yet, but we've got enough to make some observations.
Observations
Looking at all the ones we've got, here are some things available to be noticed.
 We haven't found a sidesum smaller than $9$ or bigger than $12$
 $S=9$ has the smallest numbers in the corners, $S=12$ has the largest numbers in the corners
 $S=10$ has the odd numbers in the corners, $S=11$ has the even numbers in the corners
 $6$ is always next to $1$
 The 3 small triangles inside each big one also sum to the same thing:and that total seems to be the "opposite" of whatever $S$ is: if $S=12$ this total is $9$, if $S=9$ it's $12$, and similarly for $10$ and $11$
 Some seem to be a weird sort of notquiterotation of each other, e.g.:
as if they live on a bicycle chain going around a sprocket. "Sprocketing" a solution seems to turn the sidesum of $9$ into $12$, $10$ into $11$, and viceversa$\longrightarrow$  etc...
Observation 1: Is it possible to get 8 or 13?
Maybe we should try to find the smallest possible sidesum. We know that $6$ has to go somewhere, and if we want the smallest sidesum it makes sense to put $6$ on the same side as the two smallest possible numbers, namely $1$ and $2$. Doing this we get a sidesum of $9$, so we've shown that we can't get $S$ smaller than $9$. To show that we can get $9$ all we need is the example we already found. So $9$ actually is the smallest.
We can use a similar argument to show that $12$ is the highest: $1$ has to go somewhere, and to get the biggest we should put it with $5$ and $6$, which gives $12$. And we already have an example with a sidesum of $12$ that works, so $12$ is the biggest.
Observations 2 & 3: What goes in the corners?
It sort of feels like the numbers in the corners contribute twice. But they don't contribute twice to the side sum. What do they contribute twice to?
Each corner contributes to two sidesums. For example, here the top corner contributes to the blue and red sidesums:
So if we add the blue and red ones together (this gives $2S$, since they're both $S$), the top corner will contribute twice to this.
Aha! Instead of just two, we can add all three sidesums together:
Then each corner will contribute twice to this sum of sidesums (let's call it $T$ for total), which is
\[T=3S.\]
Let's take stock and think about what $T$ is made of. Now may be a nice time to bring in some notation for the values in the circles:
$T$ is made using the middle numbers once, and the corner numbers twice:
\[ T=(b+c+e)+(2a+2d+2f), \]
or, to put it another way, we have all the numbers once and then the corner numbers once again,
\[ T=(a+b+c+d+e+f)+(a+d+f). \]
This form might be more useful to us, because while we may not know which letter corresponds to which number, we know that they are each one of $1$  $6$. That means we know that in some order,
\begin{align*}
&a+b+c+d+e+f\\
&=1+2+3+4+5+6\\
&=21
\end{align*}
So that means
\[ T=21+(a+d+f), \]
in other words, the total (sum of the side sums) is equal to $21$ plus the sum of the corners!
Cool, but so what?
What have we actually found?
Remember that $T=3S$, since it's the sum of the sidesums and the sidesums are equal. That means that
\[ 21+(a+d+f)= 3S. \]
Let's look at this equation for a bit and see what we can see.
 This equation gives a direct relationship between the sidesum and the sum of the numbers in the corners.
 Since $S$ is a whole number, $21+(a+d+f)$ must be a multiple of $3$. Since $21$ is a multiple of $3$, we come to the conclusion that the sum of the corners must be a multiple of $3$.
 The smallest that $a$, $d$ and $f$ can be are $1$, $2$, and $3$, so that $a+d+f=6$. This gives $27$ on the LHS, so $S=9$. This tells us, in a different way from before, that $9$ is the smallest sidesum.
 The largest that $a$, $d$ and $f$ can be are $4$, $5$, and $6$, so that $a+d+f=15$. This gives $36$ on the LHS, so $S=12$. Again we have a new way of knowing that $12$ is the biggest sidesum.
 Taking the three observations above and putting them together, we have that $a+d+f$ must be a multiple of $3$, but it can't be smaller than $6$ or larger than $15$. So the only options for $a+d+f$ are $6$, $9$, $12$ and $15$, which give sidesums of $9$, $10$, $11$ and $12$, respectively.
 Let's rearrange the equation slightly, to make $S$ more obvious:
\[ S=7+\frac{a+d+f}{3} \]
(This may have been helpful earlier, oh well!) With the right knowledge at hand we can realise that this says the side sum is $7$ more than the average of the corners! Weird  why $7$ more?
Constructing solutions with sidesums of 9 and 12
From the observations above, we know that if we want the smallest $S$, we want the smallest numbers in the corners. And this makes sense when we remember that the corners contribute twice to $T$ (which is 3 lots of the sidesum).
Let's assume $a$, $d$, and $f$ are $1$, $2$, and $3$:
Now, if this can form a valid solution, it should have $S=9$. To test it we can fill in each blank to make up $9$ on each side:
and check that we've used each number once  voila, a valid solution!
Similarly for $S=12$ and having the biggest numbers in the corners:
To test it we fill in the blanks to make up $12$:
Again, voila!
Constructing solutions with sidesums of 10 and 11
It felt like we relied on a bit of luck for the last two, and it worked out. (Though really we knew that it had to work, since we'd already found the solutions.)
But what about $S=10$ and $11$? They're not the biggest or smallest so it's not obvious what to choose for the corners. So we'll need to be a bit more careful.
From the observations above we know that for $S=10$, the corners should add to $9$:
\[ a+d+f=9. \]
How can we make 9 using three of the numbers from $1$  $6$?
We know from examples that it's possible to use $1+3+5$. But let's see if we can find another way. One way is $2+3+4$. Let's try it:
If it works we should have a sidesum of $10$. So we fill in the gaps to make $10$ on each side:
Darn. This isn't a valid solution, because it uses $3$ and $4$ twice and doesn't use $1$ or $6$.
Ok, so we now know that there are some choices for corners that don't work, even if they satisfy the equation we found. We could also have chosen $1+2+6$ for the corners. Since the examples we've got each seem have a nice symmetry to them (smallest, odd, even, largest) it seems unlikely that this unbalanced choice will work, but let's try it anyway. Put $\{1, 2, 6\}$ in the corners, and fill the gaps to add to $10$:
This time $2$ is used twice, $4$ and $5$ are missing, and we have a new problem: $7$ isn't allowed! So this isn't valid either. The only valid solution for $S=10$ has $1$, $3$, and $5$ in the corners.
Now for $S=11$: in this case $a+d+f=12$. There are three ways to pick three numbers from $1$  $6$ that add to $12$:
\[ 1+5+6,\qquad 2+4+6,\qquad 3+4+5. \]
Putting these numbers in the corners, and filling in the blanks to make sure the sides add to 11:
The first and the last are not valid solutions, and the middle is the one we've already found, so this is the only way to get $S=11$.
Wrapping up (but not finished!)
We have managed to:
 Find four solutions to the 6circle problem
 Recognise that they can be rotated and reflected
 Show that they are the only solutions
 Understand why there can't be any smaller or larger sidesums
 Understand why there aren't any different solutions that give the same sidesum (but maybe not in the most satisfying way)
 Understand how the sidesums relate to the choice of corners
We have not:
 Figured out if there is a deeper reason why $6$ is always next to $1$
 Understood why the 3 small triangles in each solution sum to the same thing
 Understood the "Sprocketing" phenomenon
 Figured out if there is a deeper reason why the sidesum is $7$ plus the average of the corners
Some of these may be harder than others, some may not have solutions. But we would not be true to ourselves if we left you nothing to do!
Another thought
Throughout there has been an idea that keeps popping up, that we haven't focussed on. The solutions with sidesums of $9$ and $12$ feel similar in some way, and $10$ and $11$ feel related as well. In fact, in this theme, if we take the solution for $S=9$
and swap $1\leftrightarrow 6$, $2\leftrightarrow 5$, and $3\leftrightarrow 4$, we get
which is the solution for $S=12$!
So maybe there are really only two solutions to the 6circle problem...
An alternative approach
We know we only have 6 numbers to work with, and we're trying to get 3 sets of three that sum to the same thing. So let's think about what sets of three from $1$  $6$ there can be. Going through the options systematically, we get a large list:
\begin{array}{lr}
Set & Total \\ \hline
1,2,3 & 6 \\
1,2,4 & 7 \\
1,2,5 & 8 \\
1,2,6 & 9 \\
1,3,4 & 8 \\
1,3,5 & 9 \\
1,3,6 & 10 \\
\dots &
\end{array}
and so on. Since we're looking for ones that add to the same thing, it might be more worthwhile to arrange them by total.
\[\begin{array}{lr}
Set & Total \\ \hline
1,2,3 & 6 \\
& \\
1,2,4 & 7 \\
& \\
1,2,5 & 8 \\
1,3,4 & 8 \\
& \\
1,2,6 & 9 \\
1,3,5 & 9 \\
2,3,4 & 9 \\
& \\
1,3,6 & 10 \\
1,4,5 & 10 \\
2,3,5 & 10
\end{array}
\qquad\qquad\begin{array}{lr}
Set & Total \\ \hline
1,4,6 & 11\\
2,3,6 & 11\\
2,4,5 & 11\\
& \\
1,5,6 & 12\\
2,4,6 & 12\\
3,4,5 & 12\\
& \\
2,5,6 & 13\\
3,4,6 & 13\\
& \\
3,5,6 & 14\\
& \\
4,5,6 & 15
\end{array}
\]
Using one of these sets on a side of the triangle means putting two of the numbers in the corners and the third in the middle. Let's see how it works.
First take the sidesum of $8$ as an example, our options are the sets
\begin{align*}
1,2,5\\
1,3,4
\end{align*}
Beyond the fact that there are only two options to use on three sides, there is no $6$, so this will not give us a valid solution.
Now let's look at a sidesum of $9$. Our only options are:
\begin{align*}
1,2,6 \\
1,3,5 \\
2,3,4
\end{align*}
For each set, we want two numbers to be in corners, so they each have to appear in a second set (different for each). This is true:
\begin{align*}
\color{blue}{1},\color{red}{2},6 \\
\color{blue}{1},\color{green}{3},5 \\
\color{red}{2},\color{green}{3},4
\end{align*}
We want the other number in each set to be in the middle of a side, so that number shouldn't appear anywhere else. This is also true: the numbers that appear only once are $6$, $5$, and $4$.
This shows us not only that there is a solution for $S=9$, but that there is only one. We can also find solutions for $S=10$, $11$, and $12$ this way.
It's also a systematic approach that may be useful if we want to take the problem any further...
Think beyond
To take this further, we could:
 Use different numbers instead of $1$  $6$ (we already have a couple of solutions if we relax this rule, that might be an interesting place to start)
 Use different numbers of circles in a triangle shape
 Use a different shape instead of a triangle
Some of these changes may be too relaxed, some may be too strict. Only one way to find out.
The second one is usually a fruitful track. But there's more than one way to make the triangle bigger.
 First choice:
using $1$  $10$ But then we have to rethink the sidesum restriction  what about that middle circle? (Could we treat it as a garbage bin  put a number in which we then ignore?)

Second choice:
using $1$  $9$ Seems a bit cleaner, and easy to see how to generalise further.
Good luck!