Does every convex polyhedron have a combinatorially isomorphic counterpart whose all faces have rational areas?

• Does every convex polyhedron have a combinatorially isomorphic counterpart whose faces all have rational areas?
• Does every convex polyhedron have a combinatorially isomorphic counterpart whose edges all have rational lengths?
• Does every convex polyhedron have a combinatorially isomorphic counterpart whose vertices all have rational $x,y,z$ coordinates?

Can multiple conditions above be combined?

Update: all polyhedra in question are in $\mathbb{R}^3$.

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Somewhat related to this question: the answer mathoverflow.net/questions/23478/… and comments below. –  Vladimir Reshetnikov May 4 '13 at 21:48
According to Ziegler's Lectures on Polytopes, question 2 is an open problem. See page 123, question 4.18. But I don't know if it is still open. –  Gregor Samsa Jan 10 at 16:14

The answer to the third question is no. This is a rather counter-intuitive discovery of Micha Perles from the sixties. See this paper of Ziegler, for a simpler construction and other pertinent information.

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In case the original question was for dimension $3$, however, the answer is yes, as I think is mentioned in that paper. –  Douglas Zare May 4 '13 at 23:10
Thanks Douglas. Yes, it is indeed mentioned in Ziegler's paper. –  Tony Huynh May 4 '13 at 23:12
However, note that there are nonrational, nonconvex polyhedral surfaces in $\mathbb{R}^3$, due to Ulrich Brehm. –  Joseph O'Rourke May 5 '13 at 1:09
@Joseph O'Rourke: Could you please give a reference? –  Liu Jin Tsai May 5 '13 at 1:25
I only know this via G. Ziegler's paper already cited, "Non-rational conﬁgurations, polytopes, and surfaces." I am not certain if Ulrich published it separately, as he was more concentrated on his universality theorem ("A universality theorem for realization spaces of maps"). –  Joseph O'Rourke May 5 '13 at 2:15