Given two labeled tetrahedra $$ V_0V_1V_2V_3 \quad\text{and}\quad V_0'V_1'V_2'V_3', $$ define their opposite-edge pairs $$ (01,23),\quad (02,13),\quad (03,12). $$ Let $$ m_{ij}=\frac{|V_i'V_j'|}{|V_iV_j|}\quad(0\le i<j\le3) $$ be the six edge–length ratios.
If the second tetrahedron is the image of the first under a sphere inversion with center $O$ and radius $R$, then for each pair of vertices $$ |V_i'V_j'|=\frac{R^2}{r_i r_j}|V_iV_j|,\qquad r_i=|OV_i|. $$ Hence $$ m_{ij}=\frac{|V_i'V_j'|}{|V_iV_j|}=\frac{R^2}{r_i r_j}. $$ Taking any two opposite edges, say $(01,23)$ and $(02,13)$, we have $$ m_{01}m_{23}=\frac{R^4}{r_0r_1r_2r_3}=m_{02}m_{13}=m_{03}m_{12}. $$ Thus the three opposite–edge products are all equal — this is the “product-triangle similarity” condition from the ending note of this article.
The question is find sufficient condition for an inversion to exist.
1. Algebraic characterization of an inversion
A single inversion centered at $O$ with radius $R$ maps $V_i\mapsto V_i'$ iff there exist positive constants $r_i=|OV_i|$ such that $$ |V_i'V_j'|=\frac{R^2}{r_i r_j}\,|V_iV_j|. $$ Equivalently, $$ m_{ij}=K\,y_i\,y_j,\qquad K>0,\ y_i>0, $$ where $K=R^2/r_0^2$ and $y_i=r_0/r_i$.
Taking logarithms gives $$ \log m_{ij}=t+u_i+u_j,\qquad (t,u_0,u_1,u_2,u_3\in\mathbb R). $$ This is a linear model of dimension 4 inside $\mathbb R^6$, so the space of possible $\{m_{ij}\}$ coming from an inversion is cut out by two independent equations.
2. The two independent constraints
(A) Opposite-edge product equality $$ m_{01}m_{23}=m_{02}m_{13}=m_{03}m_{12}. $$ This expresses the product-triangle similarity.
(B) Vertex balance condition
For example at vertex 0: $$ \frac{m_{01}m_{02}}{m_{12}} =\frac{m_{01}m_{03}}{m_{13}} =\frac{m_{02}m_{03}}{m_{23}}. $$ Equivalent relations hold at vertices 1, 2, 3.
From $\log m_{ij}=t+u_i+u_j$, subtract the equations corresponding to edges from a fixed vertex, say vertex 0: $$ \log m_{01}-\log m_{02}=u_1-u_2,\quad \log m_{01}-\log m_{03}=u_1-u_3. $$ These imply the equalities in (B). Conversely, if (A) and (B) hold, one can solve for $u_i$ and $t$, recovering positive $y_i=e^{u_i}$ and $K=e^{t}$.
3. The additional geometric condition
Even if (A) and (B) hold, the derived distances $r_i$ must come from a single geometric point $O$.
That requires the four spheres
$$
S_i:\ |X-V_i|=r_i
$$
to intersect in one real point.
Let $$ A=2[(V_1-V_0),(V_2-V_0),(V_3-V_0)]^\top$$$$ b(s)=\big(|V_j|^2-|V_0|^2+s(\tfrac1{y_0^2}-\tfrac1{y_j^2})\big)_{j=1,2,3} $$ where $s=r_0^2$. The linearized intersection point is $$ O_*(s)=A^{-1}b(s). $$ The four spheres intersect iff the remaining equation $$ F(s)=\|O_*(s)-V_0\|^2-\frac{s}{y_0^2}=0 $$ has a positive real root $s>0$. If $F(s)$ has no positive roots, there is no Euclidean inversion center.
4. Counterexample
Take the orthogonal tetrahedron $$ V_0=(0,0,0),\quad V_1=(1,0,0),\quad V_2=(0,1,0),\quad V_3=(0,0,1), $$ and choose vertex weights $$ y_0=1,\qquad y_1=y_2=y_3=t. $$ Then $m_{ij}=K\,y_i y_j$ satisfies (A) and (B) exactly for all $K>0$.
For this configuration, the function $F(s)$ is quadratic: $$ F(s)=\alpha s^2+\beta s+\gamma, $$ where $$ \alpha=\frac{3}{4}\left(1-\frac{2}{t^2}+\frac{1}{t^4}\right),\qquad \beta=\frac{1}{2}\left(1-\frac{3}{t^2}\right),\qquad \gamma=\frac{3}{4}. $$ The discriminant is $$ \Delta=\beta^2-4\alpha\gamma=-2+\frac{3}{t^2}. $$ Hence:
- For $t=1$, $\Delta=1>0$: inversion exists (identity case).
- For $t>\sqrt{3/2}\approx1.225$, $\Delta<0$: no real intersection.
Taking $t=\tfrac{7}{5}=1.4$ gives
$$
\Delta=-\frac{23}{49}<0.
$$
Thus (A) and (B) hold perfectly, yet the four spheres have no common point.
No real inversion center exists — this is an exact symbolic counterexample.
5. Corrected characterization
For two labeled tetrahedra $V_i,V_i'$:
- (A) Opposite-edge products equal.
- (B) Vertex balance equalities hold.
- (C) The four spheres $|X-V_i|=r_i$ (with $r_i$ from $m_{ij}=K y_i y_j$) intersect in a real point.
Then and only then does a single real Euclidean inversion exist mapping $V_i\mapsto V_i'$.
If (A)+(B) hold but (C) fails, the inversion exists only in a complex-projective or formal algebraic sense — not as a real geometric transformation in $\mathbb R^3$.
6. Summary table
| Type of condition | Description | Required for… |
|---|---|---|
| (A) | $m_{01}m_{23}=m_{02}m_{13}=m_{03}m_{12}$ | algebraic necessity (product-triangle similarity) |
| (B) | $(m_{01}m_{02})/m_{12}=(m_{01}m_{03})/m_{13}=(m_{02}m_{03})/m_{23}$ | algebraic sufficiency for consistent vertex factors |
| (C) | $F(s)=0$ has a positive real root | geometric sufficiency for a real inversion center |
Question.
A single real inversion between two tetrahedra exists iff (A), (B), and (C) all hold.
The product-triangle similarity (A) is necessary but not sufficient.
Even (A)+(B) can fail geometrically, as shown by the $t=\tfrac{7}{5}$ example.
