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Christiane Eaves, The Mount School, York (Year 10) sent the
following solution.
Join the two centres together and drop a perpendicular from the
centre of the top circle. The line joining the centres will pass
through the point of contact of the two circles because the common
tangent at this point will meet both radii at right-angles.
The 3-4-5 triangle is shaded.
Now
$\begin{eqnarray} \\ a &=& \frac{1}{2}\\ b &=&
1-r\\ c &=& \frac{1}{2} + r. \end{eqnarray}$ So, since $a^2
+ b^2 = c^2$ then $b^2 = c^2 - a^2$ and
$\begin{eqnarray} \\ b^2 &=& (\frac{1}{2} + r)^2 -
(\frac{1}{2})^2 \\ &=& \frac{1}{4} + r + r^2 -
\frac{1}{4}\\ &=& r + r^2. \end{eqnarray}$
$ \begin{eqnarray} \\ (1 - r)^2 &=& r + r^2 \\ 1 - 2r + r^2
&=& r + r^2 \\ 1 &=& 3r \\ r &=&
\frac{1}{3} \end{eqnarray}$
So $a = {1\over 2}$, $b={2\over 3}$ and $c = {5\over 6}$, or $a =
{3\over 6}$, $b={4\over 6}$ and $c = {5\over 6}$. From this it can
be seen that the triangle's sides are in the ratio 3:4:5.