On folding a polygonal sheet Consider a polygonal sheet $P$ of area $A$ with $N$ vertices (it material is not stretchable or tearable). Let $n$ be a positive integer >=2.
Question: Let $P$ lie on a flat plane. We need to fold up $P$ so that it now occupies only an area $A/n$ of the plane. It is also needed that the folding is as uniform as possible - ie the number of layers of the material above any given point should be as close to $n$ as possible. We need an algorithm that does it and an estimate of its complexity.
Example: If $P$ is a rectangle of area $A$ and $n$ is an integer, it is easy to see that we can fold it to an area $A/n$ such that it is exactly $n$ layers thick throughout - the 'creases' could simply be $n-1$ equally spaced parallel lines. It appears that no other shape of $P$ has  this property of 'perfectly uniform foldability'.  Which is the shape(s) of $P$ that causes greatest variation in the number of layers for a given $n$?
Further possibilities: One can further ask: Minimize the perimeter of the area $A/n$ region that is covered by the folded polygon. Alternatively, We could require $P$ to be folded as uniformly as possible so that it can be packed into a rectangular or square box of some specified dimensions - and area not necessarily equal to $A/n$ with n an integer.
 A: For every $k\ge 2$ there is a non-rectangular shape allowing a uniform folding for all $n$ that are multiplies of $k$:

The reason is that you can uniformly fold it into a rectangle with $k$ layers.

If you are looking for convex shapes, then $k=2$ above is convex.
Here is a non-rectangular convex shape admitting a uniform folding with three layers.
If you want to make it rectangular you will get six layers, but then you can proceed in all multiples of 6.

More generally, every regular $n$-gon admits a $2n$-layer folding. And it can be further made into a rectangular $4n$-layer folding (and then every multiple thereof).

The differently colored creases in the pentagon are to be understood as alternatingly folded upwards and downwards.
Or even better, for every $n$ there is a non-rectangular convex shape admitting an $n$-layered folding, or a $2n$-layerd rectangular folding (and then every multiple thereof).

So an interesting question might be whether for every $n$ there is a non-rectangular convex shape admitting a rectangular $n$-layer folding.
