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Stage: 4 Challenge Level: Challenge Level:1

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

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