Hamilton 2020 Q3
3. The diagram shows two semicircles with a common centre \(O\) and a rectangle \(OABC\). The line through \(O\) and \(C\) meets the small semicircle at \(D\) and the large semicircle at \(E\). The lengths \(CD\) and \(CE\) are equal.
What fraction of the large semicircle is shaded?
Hamilton Olympiad 2020, Question 3. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: \(\tfrac{9}{25}\)
Say \(OA = r\) and \(OB = R\). Then \(OC = \tfrac{R + r}{2}\).
By Pythagoras’ Theorem,
\(\displaystyle \begin{aligned} r^2 + \left(\frac{R + r}{2}\right)^2 &= R^2\\ 3R^2 - 2Rr - 5r^2 &= 0\\ (3R - 5r)(R + r) &= 0 \end{aligned}\)
So (discounting \(r = -R\), which is not practicable) \(r = \tfrac{3}{5}R\).
Hence the fraction of the larger semicircle that is shaded is \(\left(\tfrac{3}{5}\right)^2 = \tfrac{9}{25}\).
Solution © UK Mathematics Trust, from the Hamilton Olympiad 2020 solutions.