<s>[INST] Covering curved 1/8 of a circle A circle of radius 1 is given, and 8 semicircles of radius 1/2, like in this picture:

What is the radius of the smallest circle that can cover shaded area?
There was another problem involving the same picture, and this problem occurred to me while working on it. I find the formulation of this problem simple, yet it seems difficult. I used software simulation to find the desired radius, however, this does not count, a pure math solution is needed.
 [/INST] Such a region is enclosed by the circular arcs defined by the parametric equations $$A:  (x,y) = \left( \tfrac{1}{2} (1 + \cos \theta), \tfrac{1}{2} \sin \theta \right), \quad 0 \le \theta \le \pi,$$ $$B : (x,y) = (\cos \theta, \sin \theta), \quad -\frac{\pi}{4} \le \theta \le 0,$$ $$C : (x,y) = \left( \tfrac{1}{2}\left(\cos \tfrac{\pi}{4} + \cos \theta \right), \tfrac{1}{2} \left(-\sin \tfrac{\pi}{4} + \sin \theta \right) \right), \quad -\frac{\pi}{4} \le \theta \le \frac{3\pi}{4}.$$  It should be visually obvious that the maximum diameter of this shape must be measured from the point $(x,y) = \left(\cos \tfrac{-\pi}{4}, \sin \tfrac{-\pi}{4} \right)$, and the other endpoint must lie on the arc $A$.  So it is simply a calculus problem:  For a given $x$, locate $y = f(x) = \sqrt{x-x^2}$ such that $$P^2 = \left(x - \tfrac{1}{\sqrt{2}}\right)^2 + \left(y + \tfrac{1}{\sqrt{2}}\right)^2$$ is maximized.  It is a simple but computationally tedious matter to compute the derivative with respect to $x$ and obtain the point $$(x,y)_{\text max} = \left( \frac{1}{2}\left(1 - \sqrt{\frac{7 - 4\sqrt{2}}{17}}\right), \sqrt{\frac{5 + 2\sqrt{2}}{34}} \right).$$  This corresponds to a distance of $$P = \frac{1}{4}(1 + \sqrt{5 - 2\sqrt{2}}).$$  Thus a lower bound for the disk of minimal radius that will enclose this shape is $P/2 \approx 0.618406$.  Indeed, this lower bound is attainable, since the resulting disk has curvature strictly less than the curvature of the arc $A$, and it is not hard (but again, a bit tedious) to show that the disk does in fact cover.

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