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2D case

Let a convex quadrilateral $Q= \operatorname{conv}\{A_1,A_1',A_2,A_2'\}$ have vertex pairs $$(A_1,A_1'),\quad (A_2,A_2'),$$ and define the “diagonal vectors” $$d_i = \overrightarrow{A_iA_i'} = A_i' - A_i \in \mathbb{R}^2.$$ Form the diagonal triangle $$T = \operatorname{conv}\{O,d_1,d_2\}, \qquad \operatorname{Area}(T) = \tfrac{1}{2}|\det(d_1,d_2)|.$$ The “shoelace formula” gives $$ \operatorname{Area}(Q) =\tfrac12\bigl|\det(A_1'-A_1,\,A_2'-A_2)\bigr| =\tfrac12|\det(d_1,d_2)| =\operatorname{Area}(T) $$ No symmetry assumptions are needed.

$Q$ $T$
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3D case

Now take three vertex pairs $$ (A_1,A_1'),\ (A_2,A_2'),\ (A_3,A_3'), $$ and define $d_i=A_i'-A_i$.

Let $\mathcal{O} = \operatorname{conv}\{A_1,A_1',A_2,A_2',A_3,A_3'\}$.

If these six points form a polyhedron with 8 triangular faces, one vertex from each pair on every face, then the volume satisfies $$ \operatorname{Vol}(\mathcal O) =\frac{1}{6}\bigl|\det(d_1,d_2,d_3)\bigr| =\operatorname{Vol}(\operatorname{conv}\{O,d_1,d_2,d_3\}). $$

Sketch of proof:
Fix one “axis” edge $A_1A_1'$. The four faces incident to that edge are $$ \triangle(A_1, A_2, A_3),\ \triangle(A_1, A_2', A_3),\ \triangle(A_1, A_2, A_3'),\ \triangle(A_1, A_2', A_3'). $$ They bound four tetrahedra $T_{\varepsilon,\delta}=\operatorname{conv}\{A_1,A_1',A_2^{(\varepsilon)},A_3^{(\delta)}\}$
with $\varepsilon,\delta\in\{0,1\}$. These tetrahedra fill $\mathcal O$ disjointly.

Hence $$ \operatorname{Vol}(\mathcal O) =\frac{1}{6}\sum_{\varepsilon,\delta}\bigl|\det(d_1,\,A_2^{(\varepsilon)}-A_1,\,A_3^{(\delta)}-A_1)\bigr|. $$ Dropping absolute values (Swapping $A_2$ and $A_2'$ in a term gives the reverse orientation) Using multilinearity in the last two slots, $$ \sum_{\varepsilon,\delta}\det(d_1,\,A_2^{(\varepsilon)}-A_1,\,A_3^{(\delta)}-A_1) =\det(d_1,\,A_2'-A_2,\,A_3'-A_3) =\det(d_1,d_2,d_3). $$ Thus $$ \operatorname{Vol}(\mathcal O)=\frac{1}{6}|\det(d_1,d_2,d_3)|. $$

Question

For higher dimensions:

Let $A_i,A_i'\in\mathbb{R}^n$ for $i=1,\dots,n$, and set $d_i=A_i'-A_i$. Define the $n$-dimensional “cross-polytope” $$ \mathcal{C}=\operatorname{conv}\{A_1,A_1',\dots,A_n,A_n'\}. $$ If each face of $\mathcal C$ uses exactly one vertex from each pair, is it true that $$ \operatorname{Vol}_n(\mathcal C) =\frac{1}{n!}\,|\det(d_1,\dots,d_n)|, $$ the same as the simplex spanned by the diagonal vectors?

user1693987
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  • Please include your attempts. – Pranay Oct 05 '25 at 03:54
  • What's the source of this question? That is, are you looking to verify a known result or are you attempting to show something new? (If the latter, then restoring at least some of your numerical explorations from previous versions of the question would help readers gauge the plausibility of the result.) ... Cheers! – Blue Oct 05 '25 at 05:01
  • @Blue I'm looking for whether the higher dimensional analog of a known result is true. – user1693987 Oct 05 '25 at 06:40
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    @Pranay I included the proof for 2d and 3d cases in the new revision. – user1693987 Oct 05 '25 at 06:41

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