There is a generalization of this question, where "area" and "circumference" are replaced by arbitrary "nice" measures (for the purpose of this answer, say absolutely continuous measures) $\mu$ and $\nu$ on $\mathbb{R}^2$. Bárány and Matoušek have a nice paper on the subject.

Even more generally, fix nice probability measures $\mu_1,\ldots,\mu_i$ on $\mathbb{R}^2$.
A *$k$-fan* in $\mathbb{R}^2$ consists of $k$ rays (semi-infinite lines) $r_1,\ldots,r_k$ emanating from a point, listed in some clockwise order. (In fact $k$-fans are also allowed to emanate from the point at infinity, i.e., a set of $k$ parallel lines is considered to be a $k$-fan.) Write $C_k$ for the region proceeding $r_k$ in the clockwise order.

Given a vector $\alpha=(\alpha_1,\ldots,\alpha_k)$ with non-negative entries summing to one,
say that $\mu_1,\ldots,\mu_r$ can be *simultaneously $\alpha$-partitioned* if
there exists a $k$-fan such that $\mu_i(C_j)=\alpha_j$ for each $i=1,\ldots,r$ and $j=1,\ldots,k$. (If $\alpha_1=\ldots=\alpha_k=1/k$ say that the measures can be *simultaneously equipartitioned*. This case, with $k=2$, is closest to the original )

Bárány and Matoušek have a whole host of results about when such partitions exist and do not exist. Here are just a couple:

- For any $k \geq 5$ and any $\alpha$, there are two measures that can not be simultaneously $\alpha$-partitioned.
- For any $\alpha=(\alpha_1,\alpha_2)$, any two measures can be simultaneously $\alpha$-partitioned, even if the center of the fan is specified in advance.

~~No one knows, for example, if any two measures can be simultaneously equipartitioned into four parts.~~ Karasev seems to have a paper where he proves that any two measures can be simultaneously equipartitioned into $q$ *convex* parts, whenever the number of parts is a prime power. (This was first achieved for three parts -- this is the result by Bárány et al that Joseph O'Rourke mentioned.) I am unclear on the relation between this and the result of Hubard and Aronov, mentioned by Joseph O'Rourke in his answer.

Higher-dimensional versions have also been considered but much is open. For example, for any three measures in $\mathbb{R}^3$ can one always find a convex $3$-partition of space so that each measure has measure $1/3$ on each part? (I heard Bárány say in a seminar that the version with $3$ replaced by a power of $2$, is known to be true; but I didn't note down a reference.)

Using the Borsuk-Ulam theorem, because it mentions similar problems. I don't think it talks about convex sets, but instead cutting arbitrary shapes to equal pieces with straight lines. Expect to find lots of hard and unsolved questions. – Zsbán Ambrus May 12 '10 at 9:45