smc

Let the side length of the regular octagon be $s$, and let the length of the legs of the isosceles right triangles be $x$. The triangles are $45-45-90$ triangles, so $s=x\sqrt{2}$. Because each side of the square is length $1$ and is composed of two legs of the triangles and one side of the octagon, $2x+s=1$. Substituting $s=x\sqrt{2}$ into this equation, we can now solve for $x$ to get our desired answer: $$\begin{array}{rl}2x+s& =1\\ 2x+x\sqrt{2}& =1\\ x(2+\sqrt{2})& =1\\ x& =\frac{1}{2+\sqrt{2}}\cdot \frac{2-\sqrt{2}}{2-\sqrt{2}}\\ x& =\frac{2-\sqrt{2}}{4-2}=\frac{2-\sqrt{2}}{2}\end{array}$$ Thus, our answer is $\boxed{\frac{2-\sqrt{2}}{2}}$.