How prove that for any curve $\alpha(s)$ of length $L=1$ in the real plane, there is a semicircle of diameter $2R=1$ that contains it.
Any hints would be appreciated.
I gave up, googled and found an extremely elegant agrument (due to A. Meir) here. I guess the hard part was to locate the center of the semicircle.
This (real analysis) is not really a subject I'm good at, but I think I have a solution.
Suppose $\alpha$ has length two, so the radius of the circle is $1$. Take the midpoint of $\alpha$ (the point for which the two curves you get when cutting $\alpha$ have the same length) and call it $M$. Now, at both sides of $M$, the curve has length $1$. Now look at the (closed) disk with midpoint $M$ and radius $1$. Because this disk is per definition the set of points with distance at most $1$ from the center, all points not in this disk have a distance to $M$ strictly larger than $1$. Such a point (say $P$) cannot be part of $\alpha$, because then, the distance $|PM|$ would be larger than $1$ and a straight line segment is shorter than any other curve connecting $P$ and $M$ (this follows intuitively from the triangle inequality) the part $PM$ of the curve must be longer than $1$, but it has length $1$, so this is a contradiction. Therefore, we can always draw a disk with radius $1$ that fully covers a curve with length $2$.
The only possible flaws I see are that a straight line segment is the shortest possible connection and that the boundary of the disk (set of points $P$ with $d(P,M)\leq 1$) is not trivially equal to the circle with radius $1$ around $M$.