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Given a tetrahedron with all edges having integer length, is it always possible to increase all of the edge lengths by one?

More precisely: Let $P_1, P_2, P_3, P_4$ be four distinct non-coplanar points in $\mathbb{R}^3$, such that $d(P_i, P_j) \in \mathbb{Z}$ for all $i,j$, Must there exist non-coplanar points $Q_1, Q_2, Q_3, Q_4$ in $\mathbb{R}^3$ such that $d(Q_i, Q_j) = d(P_i, P_j) + 1$ for all $i < j$?

I don't know if the integer length condition is necessary. My intuition is that if all of the edges of a tetrahedron are increasing at the same rate, then the tetrahedron will approach a regular tetrahedron in the limit, so it should not become degenerate. But I have no proof of this.

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If you try to increase just one edge at a time, you will run into problems with a unit regular tetrahedron. I believe however that there is a nicely classifiable set of tetrahedra (with integral edge lengths; etc.) which become degenerate after increasing a small edge by one unit, and that these are (I am guessing) precisely those which have a unit equilateral triangle as a face. If you take things one (or three maybe) edges at a time, I think you will be rewarded. Gerhard "Fondly Remembering Integral Edge Tetrahedra" Paseman, 2012.10.19 –  Gerhard Paseman Oct 20 '12 at 3:54
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This would be implied by the convexity of the cone of realizable edge lengths by taking the midpoint of the line connecting the regular tetrahedron with edge length $1$ with the tetrahedron given. However, this cone is not convex. If you use the squares of the edge lengths instead, then that cone is convex. I learned these form Igor Rivin: arxiv.org/abs/math/0308239 –  Douglas Zare Oct 20 '12 at 4:17
    
This question is very annoying, since in addition to @Douglas' comment, there is the amazing theorem of I. J. Schoenberg, which states that the set of logs of the edges of tetrahedra is starshaped with respect to $0,$ so, in other words, if you raise all edges to some power $\alpha \leq 1,$ you still get a legitimate tetrahedron (or, in general, simplex, there is no dimensional restriction). But this helps not at all for the OP's question. –  Igor Rivin Oct 20 '12 at 5:04
    
Four distinct points in $\mathbb{R}^3$ with no three coplanar? I must be missing something. –  Vidit Nanda Oct 20 '12 at 5:37
    
@Vel: Oops, that was dumb of me. Thanks for catching my mistake. –  Dave R Oct 20 '12 at 6:01

1 Answer 1

It's possible to completely characterize when a six-tuple $(a, b, c, d, e, f)$ forms the lengths of edges of a (non-degenerate) tetrahedron. Namely, by a quick google search I found for example a paper Edge lengths determining tetrahedrons by Wirth and Dreiding, Elem. Math. 64 (2009) 160 – 170.

By their Lemma 2.1, such a 6-tuple (taken as determining not only the lengths of the edges, but also the positions of the edges in the tetrahedron) actually comes from a tetrahedron if and only if each face triangle satisfies triangle inequality and there is a vertex such that the 3 angles on faces around it are all acute and satisify the triangle inequality.

Now we can try to check that if $(a, b, c, d, e, f)$ come from a tetrahedron, then so does $(a+1, b+1, c+1, d+1, e+1, f+1)$: for the faces, just note that adding 1 to the edge lengths makes the triangle more equilateral, and so the triangle inequality will again hold (if $a < b+c$, then $(a+1) < (b+1)+(c+1)$).

EDIT: Ok, for the angles it's more tricky, as Igor Rivin says in his comment.

First, seeing that acute angles remain acute is easy, just from the fact that the angle opposite side $x$ in a triangle with sides $x, y, z$ is acute iff $x^2 < y^2+z^2$ (by the cosine rule).

As Anton Petrunin pointed out, the counterexample was nonsense. Sorry. For the rest we'll need some lower bound on the lengths of edges, the statement in fact does not hold with short edges.

Here is a counterexample: Label the edge lengths so that the faces are $abc$, $dec$, $dbf$ and $aef$. We'll be looking at the vertex $V$ where $a, b, f$ meet. Choose some small $\varepsilon$ and set $a=b=\varepsilon$, $c=\varepsilon^2$, and choose $d, e, f$ very large ($> \varepsilon^{-1}$) so that the angles at $V$ in triangles $dbf$ and $aef$ are small (and the edges come from a tetrahedron). However, after I increase all edge lenths by 1, the triangles $(d+1),(b+1),(f+1)$ and $(a+1),(e+1),(f+1)$ have almost the same shape as $dbf$ and $aef$, hence their angles at $V$ have almost not changed. But the triangle $(a+1),(b+1),(c+1)$ is now almost equilateral, and so its angle at $V$ is almost $\pi/3$. Thus the triangle inequality between angles at $V$ has no chance of holding.

It seems intuitively clear that if we assume that all the sides are $\geq C$ for a constant $C$, than the bigger $C$ is, the less will the angles change when we enlarge the sides by 1. But to make this into a proof of the statement, we'd need some quantitative estimate on how much the angles can change depending on the size of the angle (and $C$). It just seems to me like very technical and not very enjoyable calculations, so I'll be happy to leave them to someone else :) At least, then the solution of the integral problem would follow just by checking the finitely many remaining cases.

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These computations are usually harder than you think, so unless you can actually do it, this is not really an answer. –  Igor Rivin Oct 20 '12 at 5:02
    
You're right, thanks! I've edited the answer; the statement even doesn't hold when some of the edges are very short. –  Vita Kala Oct 20 '12 at 16:13
    
The counterexample does not work, the angles in dbf and aef change badly. –  Anton Petrunin Oct 20 '12 at 17:31
    
@Vita, check 100, 99.99 and 0.02 –  Anton Petrunin Oct 20 '12 at 18:07
    
@Anton, sorry, I think I finally see where I'm making a mistake. Thanks! –  Vita Kala Oct 20 '12 at 18:16

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