IMC 2011 Q21

Let 4\(x\) be the length of each side of the regular octagon. The diagram shows part of the figure. The four triangles shown in the diagram are all isosceles right-angled triangles. In such triangles the ratio of the length of the hypotenuse to the length of the shorter sides is \(\sqrt{2} : 1\).
So in the larger triangles which have hypotenuse of length 4\(x\), the length of the shorter sides is \(2\sqrt{2}x\), whilst the smaller triangles with hypotenuse 2\(x\) have shorter sides of length \(\sqrt{2}x\).
Therefore the shaded square in the question has side of length \((4 + 2\sqrt{2})x\).
The length of the side of the outer square is \((4\sqrt{2} + 4)x = \sqrt{2}(4 + 2\sqrt{2})x\).
Therefore the two squares have sides in the ratio \(1 : \sqrt{2}\), which means that their areas have ratio 1 : 2.