Dissecting trapezoids into triangles of equal area

[Lightly edited for copy and proper formatting of mathematics. -- Pete L. Clark]

The Background: Let $T$ be a trapezoid. Sherman Stein, using valuation theory, showed that if $T$ is dissectible into n triangles of equal area, then $\frac{n}{(r+1)}$, where $r$ is the ratio of the parallel sides, is an algebraic integer. See

Monsky and Jepsen, Constructing equidissections for certain classes of trapezoids, Discrete Mathematics, 308 23 (2008) 5672-5681.

In particular $r$ is algebraic.

If $r$ is in $\mathbb{Q}$, a diagonal of $T$ divides $T$ into two triangles of commensurable area, so $T$ has an "equidissection". Stein showed that if $r$ is a root of a degree $2$ element of $\mathbb{Q}[z]$ [i.e., the univariate polynomial ring over $\mathbb{Q}$ -- PLC] with both roots positive, then $T$ is equidissectible. Using ideas of Stein and Jepsen I showed that if $r$ is a root of a degree $3$ or $4$ element of $\mathbb{Q}[z]$ with all roots in the open right half-plane, then once more $T$ is equidissectible. (It's not clear that my proof can be adapted to degree greater than $4$). In contrast, whenever the algebraic $r$ has a conjugate over $\mathbb{Q}$ that's not in the open right half-plane, nothing is known about the equidissectibilty of $T$. Stein conjectured that when $r$ has a $\mathbb{Q}$-conjugate not in this half plane, then $T$ is not equidissectible. I ask:

The Question: Is it true that if $r$ has a $\mathbb{Q}$-conjugate not in the open right half-plane then $T$ cannot be dissected into triangles of equal area? (For no such $r$ has an equidissection of $T$ or a proof of non-equidissectibility been found).

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Is this an open problem? If so, it should have an "open problem" tag. – Victor Protsak May 29 '10 at 4:25
Prof. Monsky is the founder of this subject area: see math.uga.edu/~pete/Monsky.pdf. So, yes, if he's asking the question, I think we may assume it is an open problem. – Pete L. Clark May 29 '10 at 4:52
What is the meaning of the z in Q[z]? It doesn't seem to be defined anywhere. – Gerry Myerson May 29 '10 at 6:55
@Gerry: Apparently, the fact that Q isn't defined anywhere is OK by you? :) – Victor Protsak May 29 '10 at 7:12
Micro-comment-question: do we need axiom of choice to be able to extend p-adic valuation on rationals to reals? – Dror Speiser May 29 '10 at 8:46