# Multiplication arithmagons

Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?

## Problem

*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?

**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?

**A challenging extension to think about:**

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?

## Getting Started

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 $C$.

## Student Solutions

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.

## Teachers' Resources

### 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.

### Possible approach

- 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?

### 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?