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Source. Published pp. 358–359, Theorem 23 (published scan).

Statement. Let $d_{ij}=d_{ji}>0$, $1\le i<j\le4$, and suppose

$$d_{ik}^2+d_{jk}^2\ge d_{ij}^2$$

for every three distinct indices. There are four distinct vertices of a brick of dimension at most six realizing these distances. No prior Euclidean realization of the distance array is assumed.

Complete proof. Put $e_{ij}=d_{ij}^2$. Use the seven nontrivial cuts of $\{1,2,3,4\}$ up to taking complements: four singleton cuts and three partitions into two pairs. Write their real weights as $u_1,\ldots,u_4$ and $v_P$, respectively. The distance equations are

$$e_{ij}=u_i+u_j+\sum_{P:\,i,j\text{ on opposite sides of }P}v_P. \tag{1}$$

This is a linear map from seven weights to six edge values. For a partition $P=\{k,l\}\mid\{i,j\}$, its equations imply

$$\frac{e_{ki}+e_{kj}-e_{ij}}2=u_k+v_P. \tag{2}$$

Indeed, the singleton cut $k$ and the pair cut $P$ each contribute twice before division by two; every other cut contributes zero.

If all six edge values in (1) are zero, (2) gives $u_k+v_P=0$ for all four $k$ and all three $P$. Thus all $u_k$ have a common value $t$ and all $v_P=-t$. Conversely these weights do give zero edge values, since each edge is separated by exactly two pair cuts. The kernel therefore has dimension one, and (1) has rank six. Hence any prescribed six real edge values have a real solution to (1).

Take such a solution and let $u_{k_0}=\min_k u_k$. Replace all singleton weights by $u_k-u_{k_0}$ and all pair weights by $v_P+u_{k_0}$. Equations (1) do not change. All new singleton weights are nonnegative, and (2) with $k=k_0$ shows that every new pair weight is nonnegative too, by the assumed squared triangle inequalities. The new weight at $k_0$ is zero.

For each of the at most six positive cut weights $w$, introduce one orthogonal coordinate with possible values zero and $\sqrt w$, assigning them to the two sides of that cut. The squared distance between vertices $i$ and $j$ is the sum of weights of cuts separating them, exactly $e_{ij}$ by (1). All four vertices belong to the resulting brick and are distinct since the prescribed $d_{ij}$ are positive. This proves the statement, including equality in any of the squared inequalities. $\square$

This is the source's seven-coordinate, one-free-parameter construction written in cut notation. The rank and nonnegativity steps supply the consistency details compressed on p. 359. If more weights vanish, delete the corresponding coordinates. The result says at most six dimensions; it does not require every edge of the final brick to be positive in six separate coordinates.

Every angle determined by three vertices of a brick is nonobtuse, because in each coordinate the product of the two differences from a given vertex is nonnegative. Thus the squared inequalities are also necessary for a four-point brick subset. The corresponding condition fails to suffice for five points, as proved in nonobtuse_five_point_obstruction.

Bears on. #174.