Multiplication arithmagons
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
Multiplication Arithmagons printable sheet
An arithmagon is a polygon with numbers at its vertices which determine the numbers written on its edges. An introduction to arithmagons can be found here.
Usually, we add the numbers at the vertices to find the numbers on the edges, but these arithmagons follow a different rule.
Can you work out how the values at the vertices determine the values on the edges in the arithmagons generated by the interactivity below?
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![Multiplication arithmagons Multiplication arithmagons](/sites/default/files/styles/large/public/thumbnails/content-id-7447-arith.jpg?itok=HsSX_yRb)
If you are given the values on the edges, can you find a way to work out what values belong at the vertices? Use the interactivity below to test out your strategies. There are three different challenge levels to try.
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![Multiplication arithmagons Multiplication arithmagons](/sites/default/files/styles/large/public/thumbnails/content-id-7447-arith.jpg?itok=HsSX_yRb)
Once you are confident that you can work out the values at the vertices efficiently, here are some questions you might like to consider:
- Can you describe a strategy to work out the values at the vertices irrespective of the values given for the edges?
- Is there a relationship between the product of the values at the vertices and the product of the values on the edges?
- What happens to the numbers at the vertices if you double (or treble, or quadruple...) one or more of the numbers on the edges?
- Can you create a multiplication arithmagon with fractions at some or all of the vertices and whole numbers on the edges?
Can you create a multiplication arithmagon where the numbers at the vertices are all irrational but the numbers on the edges are all rational?
What about where just one or two numbers at the vertices are irrational but the numbers on the edges are rational?
How does the product of the numbers on the edges relate to the product of the numbers at the vertices?
It may help to label the numbers at the vertices $A$, $B$ and $C$ and then express the edge numbers in terms of $A$, $B$ and $C$.
Alexander from Wilson's School described his method for finding the missing numbers:
There is a method to solve multiplication arithmagons that works every time. You first multiply all of the numbers in the red bordered boxes together, and then find the square root. This will find the product of all the numbers in the purple bordered boxes. This is because all of the numbers in the purple bordered boxes are multiplied together at some point, but they are multiplied two times. So you find the square root.
To find the numbers in the purple bordered boxes you divide one of the red-bordered boxes from the number that you got in the previous operation. This will find the number in the purple bordered box that is not connected to the red one. You can then find the rest of the numbers by dividing.
Brandon and Fran from Coombe Dean School, Alex and Rhys from Llandovery College, and William from Barnton Community Primary School found the same method.
Fionn from Thomas Hardye School explained how to find the numbers by solving simultaneous equations:
Let $x$, $y$ and $z$ be the vertex numbers and $A$, $B$ and $C$ be the edge numbers.
$A$, $B$, and $C$ are the product of $xy$, $xz$ and $yz$ respectively.
$A=xy, B=xz, C=yz$.
Then $y=\frac{A}{x}$ and $z=\frac{B}{x}$
We can rearrange to get everything in terms of $X$:
$yz=\frac{AB}{x^2}$
$\Rightarrow x^2yz=AB$
$\Rightarrow Cx^2=AB$
$\Rightarrow x^2=\frac{AB}{C}$
$\Rightarrow x=\sqrt{\frac{AB}{C}}$
From this, we can easily work out:
$y=\sqrt{\frac{AC}{B}}$
and
$z=\sqrt{\frac{BC}{A}}$
Francesco from Lecce in Italy worked in a similar way to Fionn but also pointed out that the values could each be multiplied by -1 to give two possible solutions to each arithmagon.
Oliver from Wilson's School pointed out:
It's quite obvious that the product of all of the edges is equal to all of the vertices multiplied together then squared as the edges are equivalent to $xy, yz, xz$: $xy\times yz\times xz=x^2y^2z^2$ and $\sqrt{x^2y^2z^2}=xyz$
Morgan from Llandovery College sent us this solution which considered what happens when you scale the numbers on the edges.
Finally, Niharika from Leicester High School for Girls sent us this solution, and Rajeev from Haberdashers' Aske's Boys' School sent us this solution.
Why do this problem?
This problem offers students the opportunity to explore numerical relationships algebraically, and use their insights to make generalisations that can then be proved.
Relating to this month's theme, think of the process of putting numbers in the vertices and then calculating the edge numbers as an action. Is it possible to undo that action uniquely, that is, to 'solve' the arithmagon?
Possible approach
This problem could follow on from work on Arithmagons.
If a computer room is available, students could use the interactivity to explore multiplication arithmagons and come up with a strategy for deducing the vertex numbers from the edge numbers.
Alternatively, students could create their own multiplication arithmagons and then give their partner the edge numbers to see if they can deduce the vertex numbers.
Start with vertex numbers in the range 1-12, then move on to 20-100, and finally simple fractions or decimals.
Once students have had time to explore a range of different arithmagons, bring the class together to discuss the strategies they have found to work out the vertex numbers.
"Can you see a relationship between the product of the three edge numbers and the product of the three vertex numbers?"
If students haven't given this any thought, give them time to try a few examples, and then encourage them to use algebra to explain any generalisations they make.
When students have devised an efficient method for solving any multiplication arithmagon, return to the more challenging arithmagons that may have taken them some time to solve before, to show the power of general thinking in solving problems.
Finally, the insights offered by algebraic thinking and general methods can be used to tackle these questions:
- What must be true about the edge numbers for the vertex numbers to be whole numbers?
- How does the strategy for finding a vertex number given the edges on an addition arithmagon relate to the strategy for a multiplication arithmagon?
- What happens to the numbers at the vertices if you double (or treble, or quadruple...) one or more of the numbers on the edges?
- Can you create a multiplication arithmagon with fractions at some or all of the vertices and whole numbers on the edges?
Key questions
Is it always possible to find numbers to go at the vertices given any three numbers on the edges?
What is the relationship between the product of the edge numbers and the product of the vertex numbers?
Possible support
Begin by spending some time looking closely at the structure of addition Arithmagons.
Possible extension
For solving the simpler multiplication arithmagons, finding the factors of each number is a useful method. Why is there no analagous method for addition Arithmagons?
Can students create a multiplication arithmagon where the numbers at the vertices are all irrational but the numbers on the edges are all rational?
What about where just one or two numbers at the vertices are irrational but the numbers on the edges are rational?
The stage 5 problem Irrational Arithmagons takes some of these ideas further.