This solution was sent in by Liwei Deng, aged 17, of The Latymer School in London.

First of all, join up the centres of the three circles. This
will form a triangle with side lengths $x+9$, $x+4$ and $13$.

Then draw vertical and horizontal lines as shown to create three
right-angled triangles (shaded on diagram).

Taking the unknown radius to be $x$, we can mark on the lengths
shown in green.

We can now use Pythagoras' Theorem on the two yellow right-angled triangles to find the missing sides. The upper one is $6 \sqrt{x}$,

and the lower one is $ 4 \sqrt{x}$ .

Now we move to the pale blue shaded triangle.

The horizontal side can be found by subtracting:

$$ 6\sqrt{x} - 4\sqrt{x} = 2\sqrt{x}.$$

The vertical side can also be found by subtracting: as we know
that the height of the rectangle is $2x$, this side must be $2x-9-4
= 2x-13$.

Using Pythagoras' Theorem on this triangle, we find that

$$ (2\sqrt{x})^2 + (2x-13)^2 = 13^2 $$ $$ 4x + 4x^2 - 52x + 169 =
169 $$ $$ 4x^2 - 48x = 0 $$ $$ 4x(x - 12) = 0 $$ $$ x = 12 $$.

Sine rule & cosine rule. Cartesian equations of circles. Pythagoras' theorem. Circle properties and circle theorems. Similarity and congruence. Regular polygons and circles. Sine, cosine, tangent. 2D shapes and their properties. Creating and manipulating expressions and formulae. Angles - points, lines and parallel lines.