Excellent solutions were sent in by Michael from Wilmslow High School, Cheshire, UK and Andrei from Tudor Vianu National College, Bucharest, Romania.

(1) As triangles AOD and AEC are similar ${\displaystyle \frac{\mathrm{OA}}{\mathrm{AC}}=\frac{\mathrm{OD}}{\mathrm{EC}}=\frac{2}{3}\mathrm{so}\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{2}{5}.}$ As triangles BOF and BCG are similar ${\displaystyle \frac{\mathrm{OB}}{\mathrm{BC}}=\frac{\mathrm{OF}}{\mathrm{GC}}=\frac{3}{1}\mathrm{so}\frac{\mathrm{OB}}{\mathrm{OC}}=\frac{3}{4}.}$ Hence: ${\displaystyle \frac{\mathrm{AB}}{\mathrm{OC}}=\frac{\mathrm{OB}-\mathrm{OA}}{\mathrm{OC}}=\frac{3}{4}-\frac{2}{5}=\frac{7}{20}.}$

2) From triangles OAD and AXE ${\displaystyle \frac{\mathrm{OA}}{\mathrm{AX}}=\frac{\mathrm{OD}}{\mathrm{EX}}=\frac{2}{1.5}=\frac{4}{3}}$ and hence ${\displaystyle \frac{\mathrm{OA}}{\mathrm{OX}}=\frac{4}{7}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{and}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{2}{7}.}$ Again ${\displaystyle \frac{\mathrm{OB}}{\mathrm{OC}}=\frac{3}{4}.}$ Hence: ${\displaystyle \frac{\mathrm{AB}}{\mathrm{OC}}=\frac{\mathrm{OB}-\mathrm{OA}}{\mathrm{OC}}=\frac{3}{4}-\frac{2}{7}=\frac{13}{28}.}$

(3) The reasoning in this part is identical to part (1).