17

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.
Joseph O'Rourke
  • 149,182
  • 34
  • 342
  • 933
  • 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 03 '11 at 03:26
  • 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 03 '11 at 03:48
  • 2
    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 03 '11 at 03:50
  • 2
    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 03 '11 at 04:42
  • @Gerhard: I am pleased to be introduced to Euler bricks, with whom I was not previously acquainted. – Joseph O'Rourke Aug 03 '11 at 13:45
  • A regular octahedron (cube dual) is not totally rational, but scaled along one axis (square prism dual) strikes me as having potential. You might also ask for which n dimensional rational polytopes there is a rational n+1 dimensional cone as the scaled octahedron can be a rational bicone for the square. I sense that knowing about sums of squares being a square may be useful; consult Will Jagy for further suggestions. Gerhard "Ask Me About System Design" Paseman, 2011.08.03 – Gerhard Paseman Aug 03 '11 at 17:30
  • Unless I miss my guess, there is a result that says there are at most four (maybe only three?) rational squares in an arithmetic progression, which in turn suggests to me that regular hypercubes of large dimension will not have mostly regular rational cones of the next higher dimension. So one must turn to less regular prisms or less regular cones to get objects of higher dimension. Gerhard "Let Us Talk About... Irregularity" Paseman, 2011.08.03 – Gerhard Paseman Aug 03 '11 at 17:41
  • 3
    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
  • Noam, does this mean there is an Euler brick iff there is a certain rational tetrahedron? I can believe there is an Euler brick iff [(there is a certain rational tetrahedron) and (some condition implying a brick space diagonal is rational)]. What I am not yet seeing is that a (totally) rational tetrahedron has implications for the length of the space diagonal of a brick. Gerhard "Head Seems Full of Brick" Paseman, 2011.08.14 – Gerhard Paseman Aug 14 '11 at 19:48
  • @G.Paseman: as I understand the terminology, "Euler brick" = rational box with rational face diagonals; Euler brick + rational space diagonal = perfect cuboid. "Euler brick" requires 6 rational lengths, which is exactly what a rational tetrahedron provides if all three angles at one vertex are right angles. To get a similar bijection with perfect cuboids you'd need a polyhedron with at least 7 edges. – Noam D. Elkies Aug 15 '11 at 03:11
  • Ah. I am confusing Euler brick with perfect cuboid. Maybe rightly so, but now that I know what you mean by Euler brick, I defer to your presentation. Noam, thanks for the clarity! Gerhard "Head Full Of Imperfect Cuboids" Paseman, 2011.08.15 – Gerhard Paseman Aug 15 '11 at 16:28
  • @Noam: The triangle $(x,0,0),(0,y,0),(0,0,z)$ ought to have rational area, for a "totally rational" tetrahedron. I checked the first two examples on the Wikipedia page for Euler bricks, and they failed this test. For example, the triangle with side lengths 125,244,267 does not have rational area. – Günter Rote Jan 30 '13 at 01:12

1 Answers1

17

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.

Gerry Myerson
  • 39,024
  • 1
    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 03 '11 at 06:08
  • @Gerry: Ah, so even without demanding rational coordinates, the question is open. Thanks for the apposite reference! – Joseph O'Rourke Aug 03 '11 at 10:42
  • 2
    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 04 '11 at 02:14
  • 2
    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
  • @JosephO'Rourke It seems that Heronian tetrahedra can be placed with integer coordinates, see this article of Marshall and Perlis http://www.maa.org/sites/default/files/pdf/upload_library/2/Marshall2-Monthly-2014.pdf . Getting rational coordinates is much easier, see Fig. 7 of that paper. – j.c. Apr 23 '16 at 01:11