Nonconvex manhole covers One common reason given for the circularity of manhole covers is that they can't fall through the manhole. For convex manhole covers, this property is equivalent to having constant width — if you have different widths, just orient the cover so that the shorter width slides through the larger one. Since convex polygons can't have constant width, this rules them out for manhole covers. However, for nonconvex shapes, a longer width does not necessarily give you a longer hole.
Is it possible to have a nonconvex polygon that cannot be moved through a hole of the same shape?
Some Clarifications:


*

*The hole has zero thickness.

*I was thinking of the polygon being simply connected. But if this matters I would like to know the answer both ways.

 A: Here is a construction of a polygon that cannot fall through the hole.
Begin with a regular $MN$-gon circumscribed around a unit circle, where $M\gg N\gg 1$ and $N$ is even. For every $M$th side, draw a segment of length 1 extending this side in the clockwise direction. Take a rectangular neighborhood of width $\varepsilon\ll 1/MN$ of each of these segments. The manhole is the union of the $MN$-gon and these $N$ narrow rectangles.
Suppose that the polygon can be moved through the hole. Then there is a moment when its center is on the ground level. Consider the intersection line of the two planes: the ground and the polygon. Its intersection $S$ with the polygon consists of the central segment of length $>2$ and several pairs of short segments arising from the pairs of symmetric narrow rectangles ($N$ should be chosen large enough so that there are at least 3 such pairs for any line through the center).
The same line on the ground must intersect the hole by a set that contains $S$ in the interior. The central segment must go very close to the center of the hole because there is no other place for a segment of length 2 (we can control how close is should be by choosing $M$ large enough). Now consider a symmetric pair of short segments in $S$ somewhere in the middle (to avoid borderline effects). If the line on the ground is sufficiently close to the center (and we can ensure that), this pair of segments must fit into a pair of symmetric rectangles in the hole. But it is easy to see that it cannot fit - locally these rectangles are just parallel strips of width $\varepsilon$ separated by a fixed distance $2-\varepsilon$, and however you rotate the intersecting line, you will get either shorter segments or shorter interval between them.
A: What about a square with four thin slices taken out of it?
 (source)
A: Here are some potentially relevant references,
which unfortunately I cannot pursue fully at the moment (sorry!).
There are a series of short publications based on Leo Moser's 1966 problem,
"Moving Furniture through a Hallway," Problem 66-11 in SIAM Review.
(I did not follow up the answers in later issues, one by Michael Goldberg.)
There is also a 1982 paper by Gilbert Strang, "The width of a chair," in Amer. Math. Monthly.
A: Having personally fallen halfway through a circular manhole in Somerville, MA, 
I can say that this excuse for circular  design is LOUSY.  I stepped on
the metol disc and it flipped up,
a bit like a see-saw, with a diameter being the hinge.
My wife grabbed me before the family jewels --so to say -- were destroyed on the rim of
the flipped up steel manhole -- and pulled me out in a rush of adrenaline. 
I believe a circle with a bit --eg a small square
or rectangle attached on the circumference  so as to make
the shape a bitnon convex would actually be a much safer design.
But, I am unclear of how to make a precise
restatement of your problem which would lead to such
a solution. 
A: In the non-simply connected case, an annulus gives an easy example. To get a simply connected example, just cut a thin strip out of the annulus along a radius.
