MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.
share|cite|improve this question
Do you care if the polygon is simply connected or can it have holes? – Gjergji Zaimi Jul 14 '10 at 9:12
up vote 7 down vote accepted

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.

share|cite|improve this answer
I do not find sergei's answer to be entirely adequate. The transformations you are allowed to perform on the intersection line are not exclusively rotations, no matter how large you make $M$, an arbitrarily tiny translation, along with some arbitrarily tiny rotations of the polygon about two axes, could easily cause it to fit inside the hole. For me, this question is still very much open. – user35132 Jun 20 '13 at 5:29

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.

share|cite|improve this answer

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.

share|cite|improve this answer
Now that I've had a chance to look at Strang's paper, I see it is not the same problem. Sorry about the misdirection! He considers the question of determining necessary and sufficient conditions for a compact set $C$ to pass through a closed interval (in 2D) and an analogous question in 3D. He considers nonconvex sets, and ends with a nice conjecture. But, again, this seems not to be directly addressing the (interesting!) question you raise. – Joseph O'Rourke Jul 14 '10 at 14:31

What about a square with four thin slices taken out of it?

alt text

share|cite|improve this answer
Zack Kenyon ( says: @MichaelBiro do you have a proof that this fails to fit? – Anton Geraschenko Jun 26 '13 at 4:43
Not a rigorous one, but there is no unblocked segment of length $1$ in the hole, and in order for the cover to fit, it must pass through a region of diameter at least $1$. – Michael Biro Jun 26 '13 at 22:03

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.

share|cite|improve this answer
an annulus is not a polygon. – Richard Dore Jul 22 '10 at 19:26
This is not convincing. Removing a "thin" strip is similar to chopping a "small" piece from a circle: it's enough to make the manhole fall into the hole. – André Henriques Nov 12 '14 at 22:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.