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Define a convex polytope in $\mathbb{R}^d$ as totally rational (my terminology) if its vertex coordinates are rational, its edge lengths are rational, its two-dimensional face areas are rational, etc., and finally its (positive) volume is rational. So: rational coordinates, and the measure of every $k$-dimensional face, $1 \le k \le d{-}1$, is rational, and the $d$-dimensional volume is positive and rational. (Scaling could then convert all these rationals to integers.) For example, the hypercube with vertex coordinates $\{0,1\}^d$ is totally rational. Similarly an axis-aligned box with integral vertex coordinates is totally rational.

Q1. Are there other classes of totally rational polytopes, classes defined for all $d$?

In particular,

Q2. Do there exist totally rational simplices in $\mathbb{R}^d$ for arbitrarily large $d$?

Pythagorean triples yield totally rational triangles. I am not even certain that the Heronian tetrahedra described in this MathWorld article are totally rational, because it is unclear (to me) if they can be realized with rational vertex coordinates.

All this is likely known, in which case key search phrases or other pointers would be welcomed. Thanks!


Addendum. Gerry Myerson's useful summary of Problem D22 in Unsolved Problems In Number Theory answers Q2: The problem is open! Q1 remains (apparently) interesting; see the comments by Steve Huntsman and Gerhard Paseman.

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When I first saw this, I thought you were going to mention Euler bricks. You might mention that is a distinct problem you are not considering. (Or are you?) Gerhard "Ask Me About System Design" Paseman, 2011.08.02 –  Gerhard Paseman Aug 3 '11 at 3:26
    
Recalling fans and toric varieties and all that, this strikes me as likely to be (equivalent to) a nontrivial problem in algebraic geometry. –  Steve Huntsman Aug 3 '11 at 3:48
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Also, given a class of polytopes in dimension $d$ that are totally rational, rational prisms give totally rational polytopes in every higher dimension. So via the Pythagorean triples, the answer to Q1 in the strictest but trivial sense is yes. –  Steve Huntsman Aug 3 '11 at 3:50
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My immediate impression is that it makes more sense to allow a $k$-simplex in some $\mathbb R^d$ with $d$ larger, possibly much larger, than $k.$ The analogy is basis matrices for integral lattices. Or, equivalently, the Cholesky decomposition $C C^t = G$ for symmetric positive integral $G$ generally involves square roots in the entries of $C,$ but may involve all integers if $C$ is rectangular. –  Will Jagy Aug 3 '11 at 4:42
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Euler bricks can be converted into the special case of a "totally rational" tetrahedron with right angles at one vertex, i.e. with vertices at the origin and $(x,0,0)$, $(0,y,0)$, $(0,0,z)$. –  Noam D. Elkies Aug 14 '11 at 15:36
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1 Answer

up vote 16 down vote accepted

Guy, Unsolved Problems In Number Theory, problem D22: Simplexes with rational content. "Are there simplexes in any number of dimensions, all of whose contents (lengths, areas, volumes, hypewrvolumes) are rational?"

Guy notes the answer is "yes" in 2 dimensions, by Heron triangles. Also "yes" in three dimensions: "John Leech notes that four copies of an acute-angled Heron triangle will fit together to form such a tetrahedron, provided that the volume is made rational, and this is not difficult." The smallest example has three pairs of opposite edges of lengths 148, 195, and 203.

There is much more discussion, more examples, and several references. So far as I can see, there is no discussion of dimensions exceeding 3.

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As a side note it also doesn't seem hard to place isosceles tetrahedra with Heronian faces so that the coordinates of the vertices are rational. –  Gjergji Zaimi Aug 3 '11 at 6:08
    
@Gerry: Ah, so even without demanding rational coordinates, the question is open. Thanks for the apposite reference! –  Joseph O'Rourke Aug 3 '11 at 10:42
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This is an inappropriate comment, but I cannot resist: I just had the pleasure of verifing that the problem remains open in direct conversation with Richard Guy himself! :-) –  Joseph O'Rourke Aug 4 '11 at 2:14
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The nice coordinate parametrization of isosceles tetrahedra puts the vertices at $(\pm x, \pm y, \pm z)$ with an even number of minus signs, i.e. at alternating vertices of a brick. –  Noam D. Elkies Aug 14 '11 at 15:32
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