# Irrational arithmagons

Can you work out the irrational numbers that belong in the circles to make the multiplication arithmagon correct?

## Problem

In a multiplication arithmagon, the number on each edge of the arithmagon is the product of the numbers at the adjacent vertices.

*You may wish to look at Multiplication Arithmagons and develop a general strategy for working out the numbers at the vertices, given the edge numbers, before tackling the question below.*

Can you work out the numbers that belong in the circles to make this multiplication arithmagon correct?

Image

If you're not sure where to start, click for a series of hints:

Each circle will be of the form $a+b\sqrt2$ for some values of $a$ and $b$.

In general, a multiplication arithmagon can be solved by multiplying, dividing, and square-rooting.

Dividing an expression by $a+b\sqrt2$ is the same as saying, "What must I multiply $a+b\sqrt2$ by to get that expression?"

To find the square root of an expression such as $12-8\sqrt2$, consider the equation $(x+y\sqrt2)^2=12-8\sqrt2$, expand the brackets, and deduce $x$ and $y$.

## Getting Started

Start by exploring Multiplication Arithmagons.

Once you have made sense of the structure of multiplication arithmagons, can you find a general method to solve them?

Does the presence of surds change your method in any way?

## Student Solutions

Well done to everyone who sent a solution - we had a wonderful variety of creative solutions to this problem!

Ed from Wyoming Valley West in the USA mostly used a trial method:

I first created a system of equations rigorously:

$$\begin{align}\left( a + b \sqrt{2} \right) \times \left(c + d \sqrt{2} \right) &= -6 + 2\sqrt{2}\\

\left(a + b\sqrt2\right) \times \left(e + f\sqrt2\right) &= 2 - 4\sqrt2\\

\left(c + d\sqrt2\right) \times \left(e + f\sqrt2\right) &=7 + 7\sqrt2\end{align}$$

To get a set of 6 equations for 6 unknown variables:

$$\begin{align}ac + 2bd &= -6\\

bc + ad & = 2\\

ae + 2bf &= 2\\

be + af &= -4\\

ce + 2df& = 7\\

de + cf &= 7\end{align}$$ My matrix background is shaky at best, so while I know creating a matrix would net a unique solution, I was certain I'd make a math error along the way, so I tried instead to deduce some things logically.

First I noted that from equation 5, both $c$ and $e$ must be odd.

Then, from equation 6, either $d$ is odd and $f$ is even, or vice versa.

Then I took a leap of faith that for all these sums to be bounded by $-6$ and $7$, all 6 values should be fairly small. And since equations 5 and 6 are both positive sums, all coefficients are likely positive as well.

From that, I tried a couple guesses, but then I hit paydirt with $c=1$, $d=2$, $e=3$, $f=1.$

Once I managed success from that combination, I tested it back into equations 1 & 2.

This yielded 2 equations with only 2 unknown variables, a much more manageable task.

Equation 1 became $a + 4b = -6$

and Equation 2 became $2a + b = 2$

resulting in $a = 2$ and $b = -2$.

I checked that result into equations 3 & 4, and everything checked on the math, but to make sure I didn't violate any rules of algebra, I checked the arithmagon and the math worked out nicely.

Jamie from BSB in the UK began by considering a simpler version of the problem to find a sensible method:

Next, set up a system of equations: $$\begin{align}xz&=132\hspace{20mm}&(1)\\xy&=11&(2)\\yx&=12&(3)\end{align}$$

from $(1)$ $z=\frac{132}x$ from $(2)$ $y=\frac{11}{x}$ from $(3)$ $y=\frac{12}z$

$\therefore \frac{11}x=\frac{12}z$

but $z=\frac{132}x$, so $\frac{11}x=\frac{12x}{132}$

and so $\frac{11\times132}{12}=x^2$

By repeating this process you can also find that $y^2=\frac{11\times12}{132}$ and $z^2=\frac{132\times12}{11}$

Jamie went on to apply this method to the problem:

From our previous research of multiplication arithmagons, we can deduce that $$\begin{align}b^2&=\frac{\left(7+7\sqrt2\right)\left(2-4\sqrt2\right)}{-6+2\sqrt2}\\

&=\frac{14-28\sqrt2+14\sqrt2-28\times2}{-6+2\sqrt2}\\

&=\frac{-14\left(3+\sqrt2\right)}{-2\left(3-\sqrt2\right)}\\

&=\frac{7\left(3+\sqrt2\right)\left(3+\sqrt2\right)}{\left(3-\sqrt2\right)\left(3+\sqrt2\right)}\\

&=\frac{7\left(9+6\sqrt2+2\right)}{9-2}\\

&=11+6\sqrt2\end{align}$$ $\left(11+6\sqrt2\right)$ can be expressed $\left(x+y\sqrt2\right)^2$ to allow for the square rooting of $b^2$. $$\begin{split} 11+6\sqrt2&=\left(x+y\sqrt2\right)^2\\&=x^2+2y^2+2xy\sqrt2\end{split}$$ Form simultaneous equations $$\begin{align}x^2+2y^2&=11\hspace{20mm}&(1)\\

2xy\sqrt2&=6\sqrt2&(2)\end{align}$$

From $(2)$, $2xy=6$ so $x=\frac3y$.

We can substitute this into $(1)$: $$\begin{align}x^2+2y^2&=11\\

\left(\frac3y\right)^2+2y^2&=11\\

\Rightarrow\frac{9+2y^4}{y^2}&=11\\

\Rightarrow2y^4-11y^2+9&=0\\

\Rightarrow(2y^2-9)(y^2-1)&=0\end{align}$$ Ignore $2y^2-1=0$ since $y$ should be an integer. If $y^2-1=0$ then $y=1$ or $y=-1$.

But $x=\frac3y$, so $x=3$ or $x=-3$ So $b=\pm\left(3+\sqrt2\right)$ $$\begin{align} ab&=7+7\sqrt2\\

\Rightarrow a&=\frac{7+7\sqrt2}{b}\\

\Rightarrow a&=\frac{7+7\sqrt2}{3+\sqrt2}\end{align}$$ Now, rationalise the denonimator $$\begin{align}a&=\frac{\left(7+7\sqrt2\right)\left(3-\sqrt2\right)}{\left(3+\sqrt2\right)\left(3-\sqrt2\right)}\\

\Rightarrow a&=\frac{21-7\sqrt2+21\sqrt2-14}7\\

\Rightarrow a&=\frac{7+14\sqrt2}7\\

\Rightarrow a&=1+2\sqrt2\end{align}$$ So $a=\pm\left(1+2\sqrt2\right)$

Now repeat for $c$.

KryÅ¡tof from The English College in Prague used the same method as Jamie, but without doing the 'research' first. This is KryÅ¡tof's work:

First we begin by assigning variables to the vertices of the triangle, $x$, $y$ and $z$ starting at the top and going clockwise.

Then we construct a system of three equations: $$\begin{align} xy&=7+7\sqrt{2}\hspace{20mm}&(1)\\

yz&=2-4\sqrt2&(2)\\

zx&=-6+2\sqrt2&(3)\end{align}$$ The method we will use is to make $y$ the subject of equation $(1)$ and $z$ the subject of equation $(3).$ Then we will substitute into equation $(2)$ to solve for $x$. $$\begin{align}y&=\frac{7+7\sqrt2}{x}\hspace{20mm}&(4)\\

z&=\frac{-6+2\sqrt2}{x}&(5)\end{align}\\$$

$$\begin{align}\frac{7+7\sqrt2}x.\frac{-6+2\sqrt2}x&=2-4\sqrt2\\

\Rightarrow\frac{\left(7+7 \sqrt {2}\right).(-6 +2 \sqrt {2})} {2-4 \sqrt {2}} &= x^2\end{align}$$ KryÅ¡tof then found $x$ using a calculator, but could have used Jamie's method shown above, or the method that Theodore uses below.

Theodore from Institut Saint-LÃ´ in France and Mitali from Delhi Public School Sushant Lok in India both used a very neat method. This is Theodore's working.

We have:$$\begin{align} xy&=7+7\sqrt{2}\hspace{20mm}&(1)\\

yz&=2-4\sqrt2&(2)\\

zx&=-6+2\sqrt2&(3)\end{align}$$

$$\begin{align} \Rightarrow xy\times zx&=\left(7+7\sqrt2\right)\left(-6+2\sqrt2\right)\\

\Rightarrow x^2(yx)&=-42+14\sqrt2-42\sqrt2+28\\

&=-14-28\sqrt2\\

\Rightarrow x^2\left(2-4\sqrt2\right)&=-\left(14+28\sqrt2\right)\\

\Rightarrow x^2&=-\frac{14+28\sqrt2}{2-4\sqrt2}=-\frac{14\left(1+2\sqrt2\right)}{2\left(1-2\sqrt2\right)}\\

&=\frac{-7\left(1+2\sqrt2\right)^2}{1-8}=\left(1+2\sqrt2\right)^2\end{align}$$ Hence $x=1+2\sqrt2$

$$\begin{align}zx&=-6+2\sqrt2\\

\Rightarrow z&=\frac{-6+2\sqrt2}{1+2\sqrt2}=\frac{\left(-6+2\sqrt2\right)\left(1-2\sqrt2\right)}{1-8}\\

&=\tfrac{1}{-7}\left(-6+12\sqrt2+2\sqrt2-8\right)=\tfrac{1}{-7}\left(-14+14\sqrt2\right)=2-\sqrt2\end{align}$$

$$\begin{align}xy&=7+7\sqrt2\\

\Rightarrow y=&\frac{7+7\sqrt2}{1+2\sqrt2}=\frac{\left(7+7\sqrt2\right)\left(1-2\sqrt2\right)}{1-8}\\

&=-\tfrac17\left(7-14\sqrt2+7\sqrt2-28\right)=-\tfrac17\left(-21-7\sqrt2\right)=3+\sqrt2\end{align}$$

Se Sun from City College Norwich in the UK, Junsu (David) from Bangkok Patana School in Thailand, Julian from British School Manila in the Philippines and Pablo from King's College Alicante in Spain used a slightly different very neat method, and Julian generalised further. This is Pablo's work:

Let $x$ be the top vertex, $y$ be the bottom left vertex and $z$ the remaining one

Therefore:

$$\begin{align}xy &= -6 + 2\sqrt2\\

xz &= 7 + 7\sqrt2\\

yz &= 2 - 4\sqrt2\end{align}$$

If we multiply them all together:

$$\begin{align}&(xy)(xz)(yz) = \left(-6 + 2\sqrt2\right)\left(7 + 7\sqrt2\right)\left(2 - 4\sqrt2\right)\\

\Rightarrow&x^2y^2z^2 = 2\left(-3 + \sqrt2\right)\times7\left(1 + \sqrt2\right)\times2\left(1 - 2\sqrt2\right)\\

\Rightarrow&(xyz)^2 = 28\times\left(-3 - 3\sqrt2 + \sqrt2 + 2\right)\left(1 - 2\sqrt2\right)\\

\Rightarrow&(xyz)^2 = 28\times\left(-1 - 2\sqrt2\right)\left(1 - 2\sqrt2\right)\\

\Rightarrow&(xyz)^2 = 28\times\left(-1 + 2\sqrt2 - 2\sqrt2 + 4\sqrt2\sqrt2\right)\\

\Rightarrow&(xyz)^2 = 28\times(-1+8)\\

\Rightarrow&(xyz)^2=196\\

\Rightarrow &xyz = \pm14\end{align}$$

Now we can replace the values of $xy$, $xz$ and $yz$ in that expression to find the values of the three variables, ie the vertices

Let $x = a + b\sqrt2$, $yz = 2 - 4\sqrt2$

$$\begin{align}\left(a + b\sqrt2\right)\left(2 - 4\sqrt2\right) &= \pm14\\

\Rightarrow2a - 4a\sqrt2 + 2b\sqrt2 - 8b &= \pm14\\

\Rightarrow (2a - 8b) + (-4a + 2b)\sqrt2 &= \pm14 + 0\sqrt2\end{align}$$

By comparison of coefficients:

Eq 1: $2a - 8b = \pm14$

Eq 2: $-4a + 2b = 0 \Rightarrow -2a + b = 0$

$$\begin{align}2a - 8b &= \pm14\\

+(-2a + b &= 0)\\

-7b& = \pm14

\Rightarrow b = ±2\end{align}$$

$$\begin{align}-2a + b &= 0\\

\Rightarrow2a &= b\\

\Rightarrow2a &= \pm2\\

\Rightarrow a &= \pm1\end{align}$$

So $x = 1 + 2\sqrt2$ or $-1 - 2\sqrt2$

Let $y = c + d\sqrt2$, $xz = 7 + 7\sqrt2$

$$\begin{align}\left(c + d\sqrt2\right)\left(7 + 7\sqrt2\right) &= ±14\\

\Rightarrow (c + 2d) + (c+d)\sqrt2 &= ±2 + 0\sqrt2\end{align}$$

By comparison of coefficients:

Eq 1: $c + 2d = ±2$

Eq 2: $c + d = 0$

Solving these equations gives:

$d = ±2$, $c = âˆ“2$ (in other words, whenever $d$ is +ve $c$ is -ve, and vice versa)

So $y = 2 - 2\sqrt2$ or $-2 + 2\sqrt2$

Let $z = e + f\sqrt2$, $xy = -6 + 2\sqrt2$

$$\begin{align}\left(e + f\sqrt2\right)\left(-6 + 2\sqrt2\right) &= ±14\\

\Rightarrow (-3e + 2f) + (e - 3f)\sqrt2 &= ±7 + 0\sqrt2\end{align}$$

By comparison of coefficients:

Eq 1: $-3e + 2f = ±7$

Eq 2: $e - 3f = 0$

Solving these equations gives:

$f = ±1$, $e = ±3$

So $z = 3 + \sqrt2$ or $-3 - \sqrt2$

Tommy from Costello Technology College in England used a very similar method, but that removed all division by surds:

Because the surd expressions are products, it is clear that multiplication/division is needed to solve the problem. There was nothing to divide by, so multiplying them together is the only option. The only relevant two are:

$$x^2yz = -28\sqrt2 - 14\\

x^2y^2z^2 = 196$$ Because the last one can be rephrased into $(xyz)^2$ and it is a square number, the square rooting gives the integer $14$, which is also $xyz$.

$\dfrac{x^2yz}{xyz}=x$, and as both of these variables are available, we can calculate the value of $x$. $$\frac{-28\sqrt2-14}{14}=-2\sqrt2-1=x$$

Tommy then divided $xy$ and $xz$ by $x$ by expressing them as surd fractions and rationalising the denominators, but this original method would also to find $y$ and $z$ without any division.

Amrit from Hymers College in the UK used a trick to remove the surds:

$$\begin{align} ab&=-6+2\sqrt2\\bc&=7+7\sqrt2\\ac&=2-4\sqrt2\end{align}$$ These equations can be written as $$\begin{align}\frac{ab+6}2&=\sqrt2\\\frac{bc-7}7&=\sqrt2\\\frac{2-ac}4&=\sqrt2\\\Rightarrow&\frac{ab+6}2=\frac{bc-7}7=\frac{2-ac}4\end{align}$$ Taking the left hand side of the three equations, $$7ab+42=2bc-14\Rightarrow b=\frac{56}{2c-7a}$$ This value of $b$ can be substitued into right hand side of the three equations above: $$\begin{align}&\frac{\tfrac{56c}{2c-7a}-7}7=\frac{2-ac}4\\

\Rightarrow&\frac{8c}{2c-7a}-1=\frac{2-ac}{4}\\

\Rightarrow&=\frac{6c+7a}{2c-7a}=\frac{2-ac}4\\

\Rightarrow&24c+28a=4c-2ac^2-14a+7a^2c\\

\Rightarrow&2ac^2-7a^2c+42a+20c=0\end{align}$$ Solving this quadratic for $c$ then multiplying both sides by $a$ $$ac=\frac{7a^2-20\pm\sqrt{49a^4-616a^2+400}}4$$ But we know that $ac=2-4\sqrt2$, therefore $$\begin{align}\frac{7a^2-20\pm\sqrt{49a^4-616a^2+400}}4&=2-4\sqrt2\\

\Rightarrow7a^2-20\pm\sqrt{49a^4-616a^2+400}&=8-16\sqrt2\\

\Rightarrow7a^2+16\sqrt2-28&=\pm\sqrt{49a^4-616a^2+400}\\

\Rightarrow\left(7a^2+\left(16\sqrt2-28\right)\right)^2&=49a^4-616a^2+400\\

\Rightarrow49a^2+14a^2\left(16\sqrt2-28\right)+\left(16\sqrt2-28\right)^2&=49a^4-616a^2+400\\

\Rightarrow\left(14\left(16\sqrt2-28\right)+616\right)a^2&=400-\left(16\sqrt2-28\right)^2\\

\Rightarrow 224\left(\sqrt2+1\right)a^2&=896\left(\sqrt2-1\right)\\

\Rightarrow a^2&=4\left(\sqrt2-1\right)^2\\

\Rightarrow a&=\pm\left(\sqrt2-1\right)\end{align}$$

And Amrit then found the values of $b$ and $c$ by substituting this back into the original equations.

Ed from Wyoming Valley West in the USA mostly used a trial method:

I first created a system of equations rigorously:

$$\begin{align}\left( a + b \sqrt{2} \right) \times \left(c + d \sqrt{2} \right) &= -6 + 2\sqrt{2}\\

\left(a + b\sqrt2\right) \times \left(e + f\sqrt2\right) &= 2 - 4\sqrt2\\

\left(c + d\sqrt2\right) \times \left(e + f\sqrt2\right) &=7 + 7\sqrt2\end{align}$$

To get a set of 6 equations for 6 unknown variables:

$$\begin{align}ac + 2bd &= -6\\

bc + ad & = 2\\

ae + 2bf &= 2\\

be + af &= -4\\

ce + 2df& = 7\\

de + cf &= 7\end{align}$$ My matrix background is shaky at best, so while I know creating a matrix would net a unique solution, I was certain I'd make a math error along the way, so I tried instead to deduce some things logically.

First I noted that from equation 5, both $c$ and $e$ must be odd.

Then, from equation 6, either $d$ is odd and $f$ is even, or vice versa.

Then I took a leap of faith that for all these sums to be bounded by $-6$ and $7$, all 6 values should be fairly small. And since equations 5 and 6 are both positive sums, all coefficients are likely positive as well.

From that, I tried a couple guesses, but then I hit paydirt with $c=1$, $d=2$, $e=3$, $f=1.$

Once I managed success from that combination, I tested it back into equations 1 & 2.

This yielded 2 equations with only 2 unknown variables, a much more manageable task.

Equation 1 became $a + 4b = -6$

and Equation 2 became $2a + b = 2$

resulting in $a = 2$ and $b = -2$.

I checked that result into equations 3 & 4, and everything checked on the math, but to make sure I didn't violate any rules of algebra, I checked the arithmagon and the math worked out nicely.

Jamie from BSB in the UK began by considering a simpler version of the problem to find a sensible method:

Image

Next, set up a system of equations: $$\begin{align}xz&=132\hspace{20mm}&(1)\\xy&=11&(2)\\yx&=12&(3)\end{align}$$

from $(1)$ $z=\frac{132}x$ from $(2)$ $y=\frac{11}{x}$ from $(3)$ $y=\frac{12}z$

$\therefore \frac{11}x=\frac{12}z$

but $z=\frac{132}x$, so $\frac{11}x=\frac{12x}{132}$

and so $\frac{11\times132}{12}=x^2$

By repeating this process you can also find that $y^2=\frac{11\times12}{132}$ and $z^2=\frac{132\times12}{11}$

Jamie went on to apply this method to the problem:

From our previous research of multiplication arithmagons, we can deduce that $$\begin{align}b^2&=\frac{\left(7+7\sqrt2\right)\left(2-4\sqrt2\right)}{-6+2\sqrt2}\\

&=\frac{14-28\sqrt2+14\sqrt2-28\times2}{-6+2\sqrt2}\\

&=\frac{-14\left(3+\sqrt2\right)}{-2\left(3-\sqrt2\right)}\\

&=\frac{7\left(3+\sqrt2\right)\left(3+\sqrt2\right)}{\left(3-\sqrt2\right)\left(3+\sqrt2\right)}\\

&=\frac{7\left(9+6\sqrt2+2\right)}{9-2}\\

&=11+6\sqrt2\end{align}$$ $\left(11+6\sqrt2\right)$ can be expressed $\left(x+y\sqrt2\right)^2$ to allow for the square rooting of $b^2$. $$\begin{split} 11+6\sqrt2&=\left(x+y\sqrt2\right)^2\\&=x^2+2y^2+2xy\sqrt2\end{split}$$ Form simultaneous equations $$\begin{align}x^2+2y^2&=11\hspace{20mm}&(1)\\

2xy\sqrt2&=6\sqrt2&(2)\end{align}$$

From $(2)$, $2xy=6$ so $x=\frac3y$.

We can substitute this into $(1)$: $$\begin{align}x^2+2y^2&=11\\

\left(\frac3y\right)^2+2y^2&=11\\

\Rightarrow\frac{9+2y^4}{y^2}&=11\\

\Rightarrow2y^4-11y^2+9&=0\\

\Rightarrow(2y^2-9)(y^2-1)&=0\end{align}$$ Ignore $2y^2-1=0$ since $y$ should be an integer. If $y^2-1=0$ then $y=1$ or $y=-1$.

But $x=\frac3y$, so $x=3$ or $x=-3$ So $b=\pm\left(3+\sqrt2\right)$ $$\begin{align} ab&=7+7\sqrt2\\

\Rightarrow a&=\frac{7+7\sqrt2}{b}\\

\Rightarrow a&=\frac{7+7\sqrt2}{3+\sqrt2}\end{align}$$ Now, rationalise the denonimator $$\begin{align}a&=\frac{\left(7+7\sqrt2\right)\left(3-\sqrt2\right)}{\left(3+\sqrt2\right)\left(3-\sqrt2\right)}\\

\Rightarrow a&=\frac{21-7\sqrt2+21\sqrt2-14}7\\

\Rightarrow a&=\frac{7+14\sqrt2}7\\

\Rightarrow a&=1+2\sqrt2\end{align}$$ So $a=\pm\left(1+2\sqrt2\right)$

Now repeat for $c$.

KryÅ¡tof from The English College in Prague used the same method as Jamie, but without doing the 'research' first. This is KryÅ¡tof's work:

First we begin by assigning variables to the vertices of the triangle, $x$, $y$ and $z$ starting at the top and going clockwise.

Then we construct a system of three equations: $$\begin{align} xy&=7+7\sqrt{2}\hspace{20mm}&(1)\\

yz&=2-4\sqrt2&(2)\\

zx&=-6+2\sqrt2&(3)\end{align}$$ The method we will use is to make $y$ the subject of equation $(1)$ and $z$ the subject of equation $(3).$ Then we will substitute into equation $(2)$ to solve for $x$. $$\begin{align}y&=\frac{7+7\sqrt2}{x}\hspace{20mm}&(4)\\

z&=\frac{-6+2\sqrt2}{x}&(5)\end{align}\\$$

$$\begin{align}\frac{7+7\sqrt2}x.\frac{-6+2\sqrt2}x&=2-4\sqrt2\\

\Rightarrow\frac{\left(7+7 \sqrt {2}\right).(-6 +2 \sqrt {2})} {2-4 \sqrt {2}} &= x^2\end{align}$$ KryÅ¡tof then found $x$ using a calculator, but could have used Jamie's method shown above, or the method that Theodore uses below.

Theodore from Institut Saint-LÃ´ in France and Mitali from Delhi Public School Sushant Lok in India both used a very neat method. This is Theodore's working.

We have:$$\begin{align} xy&=7+7\sqrt{2}\hspace{20mm}&(1)\\

yz&=2-4\sqrt2&(2)\\

zx&=-6+2\sqrt2&(3)\end{align}$$

$$\begin{align} \Rightarrow xy\times zx&=\left(7+7\sqrt2\right)\left(-6+2\sqrt2\right)\\

\Rightarrow x^2(yx)&=-42+14\sqrt2-42\sqrt2+28\\

&=-14-28\sqrt2\\

\Rightarrow x^2\left(2-4\sqrt2\right)&=-\left(14+28\sqrt2\right)\\

\Rightarrow x^2&=-\frac{14+28\sqrt2}{2-4\sqrt2}=-\frac{14\left(1+2\sqrt2\right)}{2\left(1-2\sqrt2\right)}\\

&=\frac{-7\left(1+2\sqrt2\right)^2}{1-8}=\left(1+2\sqrt2\right)^2\end{align}$$ Hence $x=1+2\sqrt2$

$$\begin{align}zx&=-6+2\sqrt2\\

\Rightarrow z&=\frac{-6+2\sqrt2}{1+2\sqrt2}=\frac{\left(-6+2\sqrt2\right)\left(1-2\sqrt2\right)}{1-8}\\

&=\tfrac{1}{-7}\left(-6+12\sqrt2+2\sqrt2-8\right)=\tfrac{1}{-7}\left(-14+14\sqrt2\right)=2-\sqrt2\end{align}$$

$$\begin{align}xy&=7+7\sqrt2\\

\Rightarrow y=&\frac{7+7\sqrt2}{1+2\sqrt2}=\frac{\left(7+7\sqrt2\right)\left(1-2\sqrt2\right)}{1-8}\\

&=-\tfrac17\left(7-14\sqrt2+7\sqrt2-28\right)=-\tfrac17\left(-21-7\sqrt2\right)=3+\sqrt2\end{align}$$

Se Sun from City College Norwich in the UK, Junsu (David) from Bangkok Patana School in Thailand, Julian from British School Manila in the Philippines and Pablo from King's College Alicante in Spain used a slightly different very neat method, and Julian generalised further. This is Pablo's work:

Let $x$ be the top vertex, $y$ be the bottom left vertex and $z$ the remaining one

Therefore:

$$\begin{align}xy &= -6 + 2\sqrt2\\

xz &= 7 + 7\sqrt2\\

yz &= 2 - 4\sqrt2\end{align}$$

If we multiply them all together:

$$\begin{align}&(xy)(xz)(yz) = \left(-6 + 2\sqrt2\right)\left(7 + 7\sqrt2\right)\left(2 - 4\sqrt2\right)\\

\Rightarrow&x^2y^2z^2 = 2\left(-3 + \sqrt2\right)\times7\left(1 + \sqrt2\right)\times2\left(1 - 2\sqrt2\right)\\

\Rightarrow&(xyz)^2 = 28\times\left(-3 - 3\sqrt2 + \sqrt2 + 2\right)\left(1 - 2\sqrt2\right)\\

\Rightarrow&(xyz)^2 = 28\times\left(-1 - 2\sqrt2\right)\left(1 - 2\sqrt2\right)\\

\Rightarrow&(xyz)^2 = 28\times\left(-1 + 2\sqrt2 - 2\sqrt2 + 4\sqrt2\sqrt2\right)\\

\Rightarrow&(xyz)^2 = 28\times(-1+8)\\

\Rightarrow&(xyz)^2=196\\

\Rightarrow &xyz = \pm14\end{align}$$

Now we can replace the values of $xy$, $xz$ and $yz$ in that expression to find the values of the three variables, ie the vertices

Let $x = a + b\sqrt2$, $yz = 2 - 4\sqrt2$

$$\begin{align}\left(a + b\sqrt2\right)\left(2 - 4\sqrt2\right) &= \pm14\\

\Rightarrow2a - 4a\sqrt2 + 2b\sqrt2 - 8b &= \pm14\\

\Rightarrow (2a - 8b) + (-4a + 2b)\sqrt2 &= \pm14 + 0\sqrt2\end{align}$$

By comparison of coefficients:

Eq 1: $2a - 8b = \pm14$

Eq 2: $-4a + 2b = 0 \Rightarrow -2a + b = 0$

$$\begin{align}2a - 8b &= \pm14\\

+(-2a + b &= 0)\\

-7b& = \pm14

\Rightarrow b = ±2\end{align}$$

$$\begin{align}-2a + b &= 0\\

\Rightarrow2a &= b\\

\Rightarrow2a &= \pm2\\

\Rightarrow a &= \pm1\end{align}$$

So $x = 1 + 2\sqrt2$ or $-1 - 2\sqrt2$

Let $y = c + d\sqrt2$, $xz = 7 + 7\sqrt2$

$$\begin{align}\left(c + d\sqrt2\right)\left(7 + 7\sqrt2\right) &= ±14\\

\Rightarrow (c + 2d) + (c+d)\sqrt2 &= ±2 + 0\sqrt2\end{align}$$

By comparison of coefficients:

Eq 1: $c + 2d = ±2$

Eq 2: $c + d = 0$

Solving these equations gives:

$d = ±2$, $c = âˆ“2$ (in other words, whenever $d$ is +ve $c$ is -ve, and vice versa)

So $y = 2 - 2\sqrt2$ or $-2 + 2\sqrt2$

Let $z = e + f\sqrt2$, $xy = -6 + 2\sqrt2$

$$\begin{align}\left(e + f\sqrt2\right)\left(-6 + 2\sqrt2\right) &= ±14\\

\Rightarrow (-3e + 2f) + (e - 3f)\sqrt2 &= ±7 + 0\sqrt2\end{align}$$

By comparison of coefficients:

Eq 1: $-3e + 2f = ±7$

Eq 2: $e - 3f = 0$

Solving these equations gives:

$f = ±1$, $e = ±3$

So $z = 3 + \sqrt2$ or $-3 - \sqrt2$

Tommy from Costello Technology College in England used a very similar method, but that removed all division by surds:

Because the surd expressions are products, it is clear that multiplication/division is needed to solve the problem. There was nothing to divide by, so multiplying them together is the only option. The only relevant two are:

$$x^2yz = -28\sqrt2 - 14\\

x^2y^2z^2 = 196$$ Because the last one can be rephrased into $(xyz)^2$ and it is a square number, the square rooting gives the integer $14$, which is also $xyz$.

$\dfrac{x^2yz}{xyz}=x$, and as both of these variables are available, we can calculate the value of $x$. $$\frac{-28\sqrt2-14}{14}=-2\sqrt2-1=x$$

Tommy then divided $xy$ and $xz$ by $x$ by expressing them as surd fractions and rationalising the denominators, but this original method would also to find $y$ and $z$ without any division.

Amrit from Hymers College in the UK used a trick to remove the surds:

$$\begin{align} ab&=-6+2\sqrt2\\bc&=7+7\sqrt2\\ac&=2-4\sqrt2\end{align}$$ These equations can be written as $$\begin{align}\frac{ab+6}2&=\sqrt2\\\frac{bc-7}7&=\sqrt2\\\frac{2-ac}4&=\sqrt2\\\Rightarrow&\frac{ab+6}2=\frac{bc-7}7=\frac{2-ac}4\end{align}$$ Taking the left hand side of the three equations, $$7ab+42=2bc-14\Rightarrow b=\frac{56}{2c-7a}$$ This value of $b$ can be substitued into right hand side of the three equations above: $$\begin{align}&\frac{\tfrac{56c}{2c-7a}-7}7=\frac{2-ac}4\\

\Rightarrow&\frac{8c}{2c-7a}-1=\frac{2-ac}{4}\\

\Rightarrow&=\frac{6c+7a}{2c-7a}=\frac{2-ac}4\\

\Rightarrow&24c+28a=4c-2ac^2-14a+7a^2c\\

\Rightarrow&2ac^2-7a^2c+42a+20c=0\end{align}$$ Solving this quadratic for $c$ then multiplying both sides by $a$ $$ac=\frac{7a^2-20\pm\sqrt{49a^4-616a^2+400}}4$$ But we know that $ac=2-4\sqrt2$, therefore $$\begin{align}\frac{7a^2-20\pm\sqrt{49a^4-616a^2+400}}4&=2-4\sqrt2\\

\Rightarrow7a^2-20\pm\sqrt{49a^4-616a^2+400}&=8-16\sqrt2\\

\Rightarrow7a^2+16\sqrt2-28&=\pm\sqrt{49a^4-616a^2+400}\\

\Rightarrow\left(7a^2+\left(16\sqrt2-28\right)\right)^2&=49a^4-616a^2+400\\

\Rightarrow49a^2+14a^2\left(16\sqrt2-28\right)+\left(16\sqrt2-28\right)^2&=49a^4-616a^2+400\\

\Rightarrow\left(14\left(16\sqrt2-28\right)+616\right)a^2&=400-\left(16\sqrt2-28\right)^2\\

\Rightarrow 224\left(\sqrt2+1\right)a^2&=896\left(\sqrt2-1\right)\\

\Rightarrow a^2&=4\left(\sqrt2-1\right)^2\\

\Rightarrow a&=\pm\left(\sqrt2-1\right)\end{align}$$

And Amrit then found the values of $b$ and $c$ by substituting this back into the original equations.

## Teachers' Resources

### Why do this problem?

This problem offers students the opportunity for lots of practice at manipulating surds in order to complete an intriguing challenge. Students will need to make sense of the structure of the problem to find a route to the solution.

### Possible approach

This problem follows on from Arithmagons and Multiplication Arithmagons.

The algebra required to analyse the simple addition arithmagons in the first problem is very straightforward and should not take long to deduce, but it is worth starting here in order to make the links between the additive structure and the multiplicative structure in the second problem.

In order to solve Irrational Arithmagons, students will need to spend time making sense of the structure of multiplication arithmagons - the interactivity in the second problem could be useful here.

Once students have an analytical method for solving multiplication arithmagons in general, they can be given the irrational arithmagon to work on. Lots of important points for discussion might be raised, for example:

"What form does the product of $a + b\sqrt{2}$ and $c + d\sqrt{2}$ take?"

"How can we divide $a + b\sqrt{2}$ by $c + d\sqrt{2}$?"

"How can we find the square root of an expression of the form $a + b\sqrt{2}$?"

### Key questions

What is the relationship between the product of the edge numbers and the product of the vertex numbers?

Given a multiplication arithmagon with edge numbers $A$, $B$ and $C$, how can we calculate the vertex numbers?

If I multiply two numbers of the form $a + b\sqrt{c}$ together, what can you say about the form of the product?### Possible extension

Possible questions to extend students' thinking could be:

Is the solution unique?

Can any three numbers be placed on the edges of a multiplication arithmagon and yield a solution?

Is it possible to create arithmagons where some/all of the vertex numbers are irrational but the edges are rational?