Let $A,B,C,D$ be the corners of a tetrahedron with positive volume and distinct sidelengths. Is there a positive $x$ and a planar straight-line embedding of a $K_4$ graph with distinct vertices $A’,B’,C’,D’$ such that each edge of the tetrahedron is greater than the corresponding edge of the graph by exactly $x$?
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1$\begingroup$ I tested this using the Cayley-Menger determinant (mathworld.wolfram.com/Cayley-MengerDeterminant.html) and it turned out true in all the examples I tried. $\endgroup$– user44143Sep 20, 2021 at 2:54
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$\begingroup$ A purely algebraic reformulation is: If $a,b,c,d,e,f$ are distinct positive reals, where $a-b-c$, $c-d-e$, $b-d-f$ and $a-e-f$ satisfy triangle inequalities, and the Cayley-Menger determinant $CM(a,b,c,d,e,f)$ is positive, then there is some $t$ with $0<t<\min(a,b,c,d,e,f)$ and $CM(a-t,b-t,c -t,d -t,e -t,f -t)=0$. $\endgroup$– user44143Sep 20, 2021 at 12:57
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$\begingroup$ @MattF. thanks for improving my question, but as $K_4$ graphs are always planar in the usual sense of graph theory, wouldn't it be less ambiguous to call it a straight-line embedding of a $K_4$ with the constraints on the line-lengths? $\endgroup$– Manfred WeisSep 20, 2021 at 14:28
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$\begingroup$ Your edit is good. Calling it a complete quadrangle would work too. $\endgroup$– user44143Sep 20, 2021 at 17:22
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$\begingroup$ Are three points in the $K_4$ allowed to be collinear? I assumed so in my answer, since you mentioned "distinct vertices" but not "no three collinear". $\endgroup$– Reid BartonSep 20, 2021 at 20:13
1 Answer
Yes, this is true. The main point is that the "first thing that goes wrong" cannot be two vertices coming together.
Let $a_0$, $b_0$, $c_0$, $d_0$, $e_0$, $f_0$ denote the edge lengths of the original tetrahedron, let $x$ be a variable, and let $a = a_0 - x$, $b = b_0 - x$, $c = c_0 - x$, $d = d_0 - x$, $e = e_0 - x$, $f = f_0 - x$. The set $U$ of possible 6-tuples of edge lengths of (nondegenerate) tetrahedra is cut out by some strict inequalities (described in the comment of Matt F.), and in particular it is an open subset of $\mathbb{R}^6$. So, there exists a smallest $x > 0$ such that $(a, b, c, d, e, f) \notin U$; fix this choice of $x$ and the corresponding quantities $a$, ..., $f$.
I claim that $a > 0$. Among the inequalities defining $U$ are the triangle inequalities for the faces. Since $(a, b, c, d, e, f)$ belongs to the closure of $U$, we must have $a + b - c \ge 0$, as well as $a + c - b \ge 0$. If $a = 0$, then these together imply $b = c$, but then $b_0 = c_0$, contradicting the assumption that the original edge lengths were distinct. (Without this assumption there are counterexamples, for example a tetrahedron with $AB = AC = BC < AD = BD = CD$.)
Now, we just have to show that $(a, b, c, d, e, f)$ are the pairwise distances between some 4 points $A$, $B$, $C$, $D$ of $\mathbb{R}^3$. Indeed, $A$, $B$, $C$, $D$ then lie in a plane because otherwise they would form a nondegenerate tetrahedron, contradicting $(a, b, c, d, e, f) \notin U$. On the other hand the points $A$, $B$, $C$, $D$ are distinct because $a > 0$ and likewise $b > 0$, ..., $f > 0$.
Presumably this follows easily from known facts about metric embeddings in $\mathbb{R}^n$, but here is a rather general proof using semialgebraic geometry/o-minimal geometry. The background theory can be found for example in van den Dries, Tame topology and o-minimal structures, chapter 6.
Fix the point $A$ at the origin of $\mathbb{R}^3$, and consider the function $\psi : \mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^6$ sending $(B, C, D)$ to the 6-tuple of distances formed by $A$, $B$, $C$, $D$. This function is semialgebraic, continuous, and proper: the preimage of a closed and bounded subset of $\mathbb{R}^6$ is bounded (because we fixed $A$, and using the triangle inequality). The set $U$ is contained in the image of $\psi$, by definition. Using the definable choice principle, there is a semialgebraic curve $\gamma : [0, x) \to \mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}^3$ which sends $t$ to a choice of tetrahedron realizing the distances $(a_0 - t, \ldots, f_0 - t)$. The curve $\gamma$ will be continuous after restriction to $[t_0, x)$ for some $0 \le t_0 < x$. By properness of $\psi$, since $\psi \circ \gamma$ converges (to $(a, \ldots, f)$) as $t$ approaches $x$, the curve $\gamma$ can be completed continuously to $[t_0, x]$. Then the value of this extension at $x$ is the required configuration $(A = 0, B, C, D)$.