A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
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.