IMC 2011 Q23

Points \(A\), \(B\), \(C\), \(D\) and \(E\) are, respectively: the point where the large semicircle and the circle touch, the centre of the circle, the centre of the left-hand semicircle, the centre of the large semicircle and the centre of the right-hand semicircle. The radius of the circle is \(r\) m. In triangle \(BCD\), \(CD\) has length 1m, \(BD\) has length \((2 - r)\) m, since \(AD\) is a radius of the semicircle of diameter 4m, and BC has length \((1 + r)\) m, since it is the sum of the radii of the left-hand semicircle and the circle. Therefore, by Pythagoras' Theorem: \((1 + r)^2 = 1^2 + (2 - r)^2\), that is \(1 + 2r + r^2 = 1 + 4 - 4r + r^2\). So \(6r = 4\) and the radius of the circle is \(\tfrac{2}{3}\)m.
(Note that triangle BCD is a 3,4,5 triangle with sides \(\tfrac{3}{3}\)m, \(\tfrac{4}{3}\)m and \(\tfrac{5}{3}\)m.)