2011 AMC 10B Problem 16

Let the side length be $1.$ Then, the area of the center is $1.$ Then, we must find the area of the octagon. It can be found as a square with $4$ isosceles right triangles taken out. The side length of this square is $$1 + 2\cdot \dfrac 1{\sqrt 2} = 1 + \sqrt 2 .$$ It has an area of $$(1+\sqrt 2)^2 = 3 + 2 \sqrt 2.$$ Then, the side length of the right triangles is $\dfrac {1}{\sqrt 2} ,$ making the area of one equal to $$\dfrac {\frac {1}{\sqrt 2}^2}{2} = \dfrac 14 .$$ This makes them have a total combined area of $1,$ so the area of the octagon is $2+ 2 \sqrt 2.$ Thus, the ratio is $$\begin{aligned}&\dfrac 1{2+ 2 \sqrt 2} \\&= \dfrac {-2+ 2 \sqrt 2}{(2+ 2 \sqrt 2)(-2+ 2 \sqrt 2)}\\ &=\dfrac {2(\sqrt 2-1)}{4} \\&=\dfrac {\sqrt 2-1}{2}. \end{aligned}$$ Thus, the correct answer is A .