IMC 2016 Q21

Let \(a\) be the side length of the octagon and \(b\) be as shown on the diagram. The square in the centre is \(a\) by \(a\), each rectangle is \(a\) by \(b\) and the triangles are each half of a \(b\) by \(b\) square. Applying Pythagoras' Theorem to a triangle shows that \(a^2 = 2b^2\). So the shaded area is \(b^2 + ab = b^2 + \sqrt{2}b^2 = b^2(1 + \sqrt{2})\).
Similarly the total area of the figure is \(a^2 + 4ab + 2b^2 = 4b^2 + 4\sqrt{2}b^2\)
\(= b^2(4 + 4\sqrt{2})\). Therefore the ratio required is \((1 + \sqrt{2}) : (4 + 4\sqrt{2}) = 1 : 4\).