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 $x$?



Deflating a tetrahedron to a $K_4$ graph with equal changes to sidelengths
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 $(A = 0, B, C, D)$.


