Jian Cedric from Chatham Grammar School for Boys solved the problem in this way:

Call the numbers in the three circles $A$, $B$ and $C$. Then their products are:

$AB= 7+7\sqrt2$

$AC= -6+2\sqrt2$

$BC= 2-4\sqrt2$

$$AB+AC= 1+9\sqrt2$$ $$\Rightarrow A(B+C)=1+9\sqrt2$$ $$\Rightarrow A= \frac{1+9\sqrt2}{B+C}$$

$$AC+BC= -4-2\sqrt2$$ $$\Rightarrow C(A+B)= -4-2\sqrt2$$ $$\Rightarrow C= \frac{-4-2\sqrt2}{A+B}$$

$$AB+BC= 9+3\sqrt2$$ $$\Rightarrow B(A+C)=9+3\sqrt2$$ $$\Rightarrow B= \frac{9+3\sqrt2}{A+C}$$

$$AB= 7+7\sqrt2 = \frac{1+9\sqrt2}{B+C} \times \frac{9+3\sqrt2}{A+C}$$ $$\Rightarrow 7+7\sqrt2=\frac{63+84\sqrt2}{(B+C)(A+C)}$$ $$\Rightarrow (B+C)(A+C) = \frac{63+84\sqrt2}{7+7\sqrt2}$$

$$\Rightarrow (B+C)(A+C)=15-3\sqrt2$$ $$\Rightarrow AB+BC+AC+C^2=15-3\sqrt2$$ $$\Rightarrow (7+7\sqrt2)+(2-4\sqrt2)+(-6+2\sqrt2)+(C^2)=15-3\sqrt2$$ $$\Rightarrow C^2=15-3\sqrt2-(7+7\sqrt2)-(2-4\sqrt2)-(-6+2\sqrt2)$$ $$\Rightarrow C^2= 12-8\sqrt2=(2-2\sqrt2)(2-2\sqrt2)$$ $$\Rightarrow C=2-2\sqrt2$$

$$B(2-2\sqrt2)=2-4\sqrt2$$ $$\Rightarrow B=\frac{2-4\sqrt2}{2-2\sqrt2}= 3+\sqrt2$$

$$A(2-2\sqrt2)= -6+2\sqrt2$$ $$\Rightarrow A= \frac{-6+2\sqrt2}{2-2\sqrt2} = 1+2\sqrt2$$

So $A$ (top circle) $= 1+2\sqrt2$, $B$(bottom right circle) $= 3+\sqrt2$ and $C$ (bottom left circle) $= 2-2\sqrt2$

Aurimas who is also from Chatham Grammar School for Boys solved it in a different way:

Well, first I started by naming each circle with letters ABC - A being bottom left, B being top and C being bottom right.

Then, I formed 3 simultaneous equations :

$AB = -6 + 2\sqrt2$

$BC = 7 + 7\sqrt2$

$AC = 2 - 4\sqrt2$

By making $A$ the subject in the first equation and $C$ the subject of the second, I had a new equation :

$AC = \frac{(-6 + 2\sqrt2)(7 + 7\sqrt2)}{B^2} = 2 - 4\sqrt2$

Solving for $B$ gave me $B^2 = 9 + 4\sqrt2$ , and square rooting that led me to $B = 1 + 2\sqrt2$

Now, it was only a matter of substituting $B$ into two equations to get the value of $A$ and $C$ so : $A = 2 - 2\sqrt2$ $B = 1 + 2\sqrt2$ $C = 3 +\sqrt2$

Finally, Edward from The Leys school sent us this solution. Well done to all of you! But is this solution unique???