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Arithmagons
By Helen Bailey (hjb37) on November 4, 1998:

Dear All,
I'm a trainee teacher doing some work on arithmagons and was wondering if anyone could help. An arithmagon looks like a triangle. Each corner of the triangle there is a number, can be a positive, negative or a fraction. On the sides of each triangle is the sum of its two adjacent numbers. On example would be to have 5, 6, 7 as the sums on the sides of the triangles and 2,3,4 as the numbers which need to be summed. I have found some patterns but can't explain why they work. For example if the two smaller numbers on the sides of the triangle are added together and they are still smaller than the other number on the triangle side, at least one negative number will be used for the numbers on the corners of the triangle. I was wondering if there were any general rules which i may have not noticed and what good extension ideas they may be, i've thought about having squares and hexagons, and also instead of adding the corner numbers perhaps taking them away or multiplying them.

Thanks

Helen Bailey

By Richard Dwight (rpd25) on November 4, 1998:

Dear Helen,

Before looking for rules we must usually state the problem in a general form. First take the numbers at the vertices of the triangle to be variables, x, y and z say. Now without restricting the problem any further we can say that x >= y >= z. Think about why we can do this.

You can see that the arithmagon will retain its properties if it is rotated. It's perhaps harder to see that the same will be true when it is reflected, but if you think of the rules for the arithmagon, you can see that none of the depend upon the orientation of the triangle. If you draw an arithmagon and look at the paper from the other side it will still be the same arithmagon.

(Note that you can not do this for an arithmagon with any number of sides. Infact it is only possible for a triangle, see if you can do rigid operations on a square and get all the possible orderings of vertices).

So using rotations and reflections it is possible to transpose any two of x,y,z and so it is possible to write x,y,z in any order round the triangle you like. Hence the restriction x >= y >= z is not really a restriction at all.

Using this notation we can now explain the rule you have given above.

If x >= y >= z then the biggest number on a side is x+y, so the remaining two are the smaller sums. The condition you gave was that the sum of the two smaller sides is less than that of the bigger. We can now write that as:
x+y > (x+z) + (y+z)
x+y > (x+y) + 2z
0 > z.
And so z must be a negative number.

Some other rules:

The sum of all the numbers at the vertices is half the sum of all the numbers on the edges. Note that this is true for an arithmagon with any number of vertices, you don't need the condition that x>y>z as you did before.

If you specify all three vertices, there is a unique arithmagon with these vertices, i.e. you can work out the edges and there is no choice as to what numbers they take (this is obvious).

If you specify all three sides there is a unique arithmagon with these sides, (you can work out the numbers at the vertices.)

If you specify a combination of a total of three sides and vertices then, you may or may not have a unique arithmagon. e.g. If you specify x and y and x+y, then you can take z to be any value, you will be able to work out an arithmagon for any z you choose. There is no unique arithmagon.

These last two are quite tricky to prove, but demonstrating that they are true in specific cases could be a useful exercise.

I think you will find less rules for arithmagons with more than three sides, though the last three rules I gave will work for general n-sided arithmagons, provided you specify n vertices/edges.

As for multiplication and subtraction you can but experiment. If you find some interesting results for multiplication in a three sided arithmagon you can prove them using the same method as used above, label the vertices x,y,z etc.

Think about how you will define the sides if you use subtraction. You will need to specify which vertex is subtracted from which, clockwise or anti-clockwise. Note that this isn't invariant under reflection, so you will not be able generalize to x>y>z as before.

Hope this helps,

Richard.