National Math Olympiad – First Round, 2015

Denote by $x$ the side length of the inscribed octagon. The four triangles that are formed at the vertices of the square are isosceles right-angled triangles. Since their hypotenuse is of length $x$, their legs are of length $\frac{x}{\sqrt{2}}$. Thus
$$a = x + 2\frac{x}{\sqrt{2}} = x + x\sqrt{2} = x(\sqrt{2} + 1).$$
From this we deduce

$$x = \frac{a}{\sqrt{2} + 1} = \frac{a(\sqrt{2} - 1)}{2 - 1} = \sqrt{2}a - a.$$