The smallest disk containing all sides of an $n$-gon Start with a regular $n$-gon of side 1 and consider its sides as open segments that can be moved around in the plane, allowing only translations. Two segments may not intersect. 

What is the radius of the smallest disk that can contain (cover) all segments? 

For $n=3$ it is clear that the equilateral triangle with $r=\dfrac1{\sqrt{3}}$ is best possible.
But already for $n=4$, we can do better than a square, by shifting two perpendicular segments as shown on the left below and placing the two other ones at small distances parallel to them. So we can realize $r={\sqrt{\dfrac38}}+\epsilon< \dfrac1{\sqrt{2}}$.
Likewise for $n=5$, where the three black edges of the pentagon can be moved inside such that the whole fits in the blue circle with diameter $\dfrac{\sqrt{5}+1}2$, thus $r=\dfrac{\sqrt{5}+1}4<\sqrt{\dfrac{5+\sqrt{5}}{10}}$ (the circumradius of the pentagon, orange circle). And there is still room for improvement, as the blue circle can be made smaller after pushing one of the grey segments diagonally upwards.

For even $n$, the segments come in pairs of same slope, so it makes sense to ignore one of each pair. Then e.g. for $n=4$, there is no need for epsilons or $r_{inf}$, and we can state $r_{min}={\sqrt{\dfrac38}}$. Further, $n=6$ reduces to the $n=3$ case.
For $n>6$, the unit circle does the job for a star-like constellation, shifting each segment in a way that $O$ is one of its endpoints. (Well, it also does it for smaller $n$, but worse than the above constructions.)

Is $r=1$ best possible for $n\ge7$? And what is $r_{min}$ for $n=5$? 

 A: (This should be rather a comment to Yoav's answer, but a picture says more than 1000 comments.)  
What about talking the two wedges and splitting the blue one into smaller wedges as shown, starting at the bottom and pushing each part rightmost against the circle? The result will be best if the blue parts have infinitesimally small angles. The blue area would then fill an area of $\pi/4$ between the circle on the right and a smooth curve obeying a certain differential equation on the left, which depends only on $r$ and the angle of the wedge (for which I have chosen 90° here). The angle and $r$ would then have to be optimized under the condition that there remains just enough space for the red wedge, probably as a whole.
I guess this is close to Gerhard's most recent idea. The missing link: finding the equation of that curve or at least some bounds...
EDIT: Thanks to Will Brian for alluding to the tractrix. After calculations by Robert Bryant in the tractrix thread, for an angle of 90° this construction can be done with $r\approx 0.8250033$ - which is within Gerhard Paseman's estimated range.
Of course the question remains whether 90° is best.

A: Here is a variant on Will Brian's idea, but inspired by the fan arrangement in the other answer.
As before, we treat only odd n, arrange them in a fan, but now we start moving the edges to the bottom.  The horizontal edge is the bottommost edge, the next edge with least positive slope goes above this segment and slightly to the right (so its midpoint is above and to the right of the midpoint of the horizontal segment), its mirror image above and slightly to the left.  As above, I won't do the algebra, but I imagine at least n/4 of the edges can be stacked this way before reaching the top half of the semicircle, leaving a depression into which the rest of the edges can sink.  I suspect for n=7 one can beat the cosine pi/7 value reached by the pinwheel construction.
Gerhard "Don't Close The Suitcase Yet!" Paseman, 2015.11.03
A: Start by arranging all the segments into a single unit-radius half-disk. Then cut a wedge of about $39$ degrees and slide it down along with its contents as in the figure. You did not cut or rotate any of the segments and the circumscribing circle has radius of about $0.94$.

A: Indeed you can do better than r=1.  The small and even cases are handled in the post, so I focus on the case of an odd number of sides.
Instead of arranging sides in a star, arrange them in a fan or in open pages of a book, taking up less than a semicircle (semidisk?).  Move the horizontal edge down near the bottom out of the way. Split the remaining edges in half, each occupying a quarter disk.  Ease one of the quarter disks down slightly, and the other quarter disk toward the center.
As n gets large, not much movement will be available, but more than 0.  ( For large n, less than .01 this way.  Sorry, Anton.)
Gerhard "Exact Radius Will Cost More" Paseman, 2015.11.03
A: (This should be a comment on Wolfgang's answer/comment, but there's too much to say.) Consider unit length chords approaching the center of a circle of radius $r$. The chord part would actually be longer than unit length, but we will have endpoints lie on other chords.
Suppose we start with one chord below the circle at a distance of $-y_0$, so we are dealing with chords at the bottom.  The new chord to be stacked has an endpoint on the current chord and makes an angle $\alpha$ with the current chord, and is pushed so that its other endpoint lies on the circle, so the chord intercepts the circle at abscissa $y_1 = y_0 + \sin \alpha$ .  However, we are interested in how far this chord is from the center, so we rotate the whole diagram by $\alpha$ to make the new chord horizontal.  This gives $y_2 = y_1\cos \alpha - \sqrt{r^2 - (y_1)^2}\sin \alpha$, so this second chord is at distance $-y_2 \lt -y_0$ from the circle center.
Starting with $y_0= -\sqrt{r^2 - 1/4}$, we iterate this for $\alpha=\pi/n$, giving a construction similar to the camera shutter but here the chords go only halfway around and are at different distances from the center of the circle.  For those wishing to investigate further, iterate for $n-1$ times the map
$$f_{r,n}(x) = x\cos \pi/n + \frac{\sin 2\pi/n}{2} - 
 r \sin \pi/n \sqrt{1 - \big(\frac{x + \sin \pi/n}{r}\big)^2},$$
starting with values of $r$ near $0.94$ (thanks to Yoav Kallus).  I suspect for $n=5$ we can get $r_{min}$ below $0.8$.
Gerhard "Round And Round We Go" Paseman, 2015.11.05
