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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

*The ratio between the diameters of the smallest circle containing $C$ and the largest circle contained in $C$ is bounded by some constant (see

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Perhaps not omitting a definition of fatness would be useful to those who don't know what it means. – Daniel Litt Jun 1 '11 at 19:57
Good point, put in two examples. – John Gunnar Carlsson Jun 1 '11 at 20:03
Do you want your map $f$ to be continuous? Do you want it to be piecewise linear? – André Henriques Jun 1 '11 at 20:23
Thanks. Continuity is required and I edited the question to reflect that. Piecewise linearity would be nice but not necessary. – John Gunnar Carlsson Jun 1 '11 at 20:33
up vote 8 down vote accepted

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.

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Thanks! Just what I was looking for. – John Gunnar Carlsson Jun 1 '11 at 21:07
It seems you did not use the fatness constraint. Does the construction work when the polygon $C$ is convex but not fat? – Erel Segal-Halevi May 20 '14 at 8:36
Correct, the map I described just uses that $C$ is star-shaped, no fatness required. – Dylan Thurston May 21 '14 at 4:02

This problem is treated in section 18 ("Monge problem for polytopes") of Igor Pak's book: (without any reference to fatness).

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Thanks! I wasn't aware of that book. – John Gunnar Carlsson Jun 1 '11 at 21:07

It is easy to construct a volume preserving map between any two shapes of equal volume with triangular Jacobian matrix. (Gromov used such map from the given domain to ball to prove isoperemetric inequality.)

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