Solution by: Andrew Ross, age 17, Dr. Challoner's Grammar School

The octagon has side lengths $a$ and $b$.

By Pythagoras Theorem:

${\displaystyle r=\frac{s\sqrt{2}}{2}}$

Because $s=r+d$ so

${\displaystyle d=s(1-\frac{\sqrt{2}}{2}).}$

One side of the octagon is given by:

${\displaystyle a=s-2d=s-2s(1-\frac{\sqrt{2}}{2})=s(\sqrt{2}-1).}$

By Pythagoras Theorem $b^2=2d^2$ so $b=d\sqrt{2}$. Hence

${\displaystyle b=s\sqrt{2}(1-\frac{\sqrt{2}}{2})=s(\sqrt{2}-1)=a.}$

The octagon has rotational symmetry of order 4 as it was constructed from a square and all the sides are the same length, so it is a regular octagon.