Are there any well-known continuous area-preserving maps between fat convex polygons and fat rectangles? Specifically, given a fat convex polygon $C$, is there a natural way to choose a fat rectangle $R$ and a continuous mapping $f:C\rightarrow R$ such that area of sub-regions is preserved? I've avoided giving a precise definition of "fatness", but two common definitions are:


\*The aspect ratio of the minimum bounding box of $C$ is bounded by some constant (see page 5 of <http://portal.acm.org/citation.cfm?id=1137901>).


\*The ratio between the diameters of the smallest circle containing $C$ and the largest circle contained in $C$ is bounded by some constant (see <http://books.google.com/books?id=QS6vnl8WlnQC&pg=PA588>).



Area-preserving map between rectangles and fat polygons
There are lots of ways to do this; without more constraints (or indication of what is desired), it's hard to pick a best one. But here's one.
We'll exploit the fact that for any two triangles, there is a unique affine linear map that takes one to the other (with vertices in a specified order); if the two triangles have the same area, this will be area-preserving.


Start by chopping up $C$ into triangles $T\_1,\dots,T\_k$ by drawing lines connecting the vertices to the center.  

Now take $R$ to be a square with the same area as $C$. Divide up the boundary of $R$ into $k$ intervals $I\_1,\dots,I\_k$ whose lengths are proportional to the area of the $T\_i$. (Some of the intervals may make a turn at the corner.) Divide the square into regions $S\_i$ by connecting the ends of $I\_i$ to the center of the square. Most of the $S\_i$ will be triangles, except for the four at the corners, which are quadrilaterals.


Now, the area of $S\_i$ is equal to the area of $T\_i$, since the area of $S\_i$ is proportional to the length of its base (as all $S\_i$ have the same height), which is proportional to the area of $T\_i$. Now just map $T\_i$ to $S\_i$ in an area preserving way. For the $S\_i$ that are not at the corners, just take the affine linear map from above. For the $S\_i$ that do go around the corner, cut both $S\_i$ and $T\_i$ into two triangles in the same proportion of area, and map the corresponding triangles to each other.


