3
$\begingroup$

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 $x$?

$\endgroup$
7
  • 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. $\endgroup$
    – user44143
    Sep 20, 2021 at 2:54
  • $\begingroup$ 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 $CM(a-t,b-t,c -t,d -t,e -t,f -t)=0$. $\endgroup$
    – user44143
    Sep 20, 2021 at 12:57
  • $\begingroup$ @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? $\endgroup$ Sep 20, 2021 at 14:28
  • $\begingroup$ Your edit is good. Calling it a complete quadrangle would work too. $\endgroup$
    – user44143
    Sep 20, 2021 at 17:22
  • $\begingroup$ 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". $\endgroup$ Sep 20, 2021 at 20:13

1 Answer 1

1
$\begingroup$

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 $(A = 0, B, C, D)$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.