A regular octagon of side length $1$ is inscribed in a circle. Each minor arc of the circle determined by a side of the octagon is reflected over that side. What is the area of the region bounded by these $8$ reflected arcs?

1. The octagon has $8$ sides, each subtending a central angle $\theta=45^\circ=\frac\pi4$ at the circle's center $O$. For a side of length $1$, the circumradius is $R=\dfrac1{2\sin(\pi/8)}$.
2. Reflecting the minor arc of a side over that side moves it from the outer side of the chord to the inner side, so it cuts a circular segment out of the octagon instead of adding one. The segment area (sector minus triangle) for one side is $\dfrac{R^2}2(\theta-\sin\theta)$.
3. The octagon's own area is $\dfrac82R^2\sin\theta=4R^2\sin\theta$ (eight congruent triangles from the center).
4. So the total bounded area is $4R^2\sin\theta-8\cdot\dfrac{R^2}2(\theta-\sin\theta)=4R^2(2\sin\theta-\theta)$.
5. With $R^2=\dfrac1{4\sin^2(\pi/8)}=\dfrac1{2-\sqrt2}=\dfrac{2+\sqrt2}2$, $\sin\theta=\sin45^\circ=\dfrac{\sqrt2}2$, and $\theta=\dfrac\pi4$: Area $=4\cdot\dfrac{2+\sqrt2}2\left(\sqrt2-\dfrac\pi4\right)=2(2+\sqrt2)\left(\sqrt2-\dfrac\pi4\right)$.
6. Expanding: Area $=4\sqrt2+4-\dfrac{(2+\sqrt2)\pi}2$.
7. The answer is $\boxed{4+4\sqrt2-\dfrac{(2+\sqrt2)\pi}2}\approx4.294$.