The opposite edges of a tetrahedron ABCD are mutually…

The opposite edges of a tetrahedron $ABCD$ are mutually perpendicular. Prove that $\cos(AB)\cos(CD)=\cos(AC)\cos(BD)=\cos(AD)\cos(BC)$ where $(XY)$ is the dihedral angle at the edge $XY$.

Step 1: The perpendicularity conditions

With $\vec p=\overrightarrow{AB}$, $\vec q=\overrightarrow{AC}$, $\vec r=\overrightarrow{AD}$, the conditions

\[ \vec p\cdot(\vec r-\vec q)=0,\qquad \vec q\cdot(\vec r-\vec p)=0,\qquad \vec r\cdot(\vec q-\vec p)=0 \] give $\vec p\cdot\vec q=\vec q\cdot\vec r=\vec r\cdot\vec p=k$. Write $P=|\vec p|^2$, $Q=|\vec q|^2$, $R=|\vec r|^2$.

Step 2: Dihedral angles via face vectors

Let $\vec S_X$ be the outward area vector of the face opposite the vertex $X$, with length $s_X$ (the face area). The interior dihedral angle along an edge equals $180^\circ$ minus the angle between the outward normals of the two faces meeting there. Hence

\[ \cos(AB)=-\frac{\vec S_C\cdot\vec S_D}{s_Cs_D},\qquad \cos(CD)=-\frac{\vec S_A\cdot\vec S_B}{s_As_B}, \]

\[ \cos(AB)\cos(CD)=\frac{(\vec S_A\cdot\vec S_B)(\vec S_C\cdot\vec S_D)}{s_As_Bs_Cs_D}, \] and similarly for the other two pairs. The denominators are the same, so it is enough to compare the three numerators.

Step 3: The face vectors

Put $\vec X=\vec q\times\vec r$, $\vec Y=\vec r\times\vec p$, $\vec Z=\vec p\times\vec q$. Up to a common sign, which cancels in all products, the face vectors are

\[ \vec S_B=\tfrac12\vec X,\quad \vec S_C=\tfrac12\vec Y,\quad \vec S_D=\tfrac12\vec Z,\quad \vec S_A=-\tfrac12(\vec X+\vec Y+\vec Z)\quad(\text{Problem 139}). \]

By Problem 140, $(\vec u\times\vec v)\cdot(\vec w\times\vec z)=(\vec u\cdot\vec w)(\vec v\cdot\vec z)-(\vec u\cdot\vec z)(\vec v\cdot\vec w)$, so

\[ \vec Y\cdot\vec Z=k(k-P),\qquad \vec Z\cdot\vec X=k(k-Q),\qquad \vec X\cdot\vec Y=k(k-R), \]

\[ |\vec X|^2=QR-k^2,\qquad |\vec Y|^2=RP-k^2,\qquad |\vec Z|^2=PQ-k^2. \]

Step 4: The products

\[ \vec S_A\cdot\vec S_B=-\tfrac14\big(|\vec X|^2+\vec X\cdot\vec Y+\vec X\cdot\vec Z\big)=-\tfrac14\big(QR-k^2+k^2-kR+k^2-kQ\big)=-\tfrac14(Q-k)(R-k), \] and cyclically $\vec S_A\cdot\vec S_C=-\tfrac14(P-k)(R-k)$, $\vec S_A\cdot\vec S_D=-\tfrac14(P-k)(Q-k)$. Therefore

\[ (\vec S_A\cdot\vec S_B)(\vec S_C\cdot\vec S_D)=-\tfrac14(Q-k)(R-k)\cdot\tfrac14k(k-P)=\tfrac1{16}\,k(P-k)(Q-k)(R-k), \]

\[ (\vec S_A\cdot\vec S_C)(\vec S_B\cdot\vec S_D)=-\tfrac14(P-k)(R-k)\cdot\tfrac14k(k-Q)=\tfrac1{16}\,k(P-k)(Q-k)(R-k), \]

\[ (\vec S_A\cdot\vec S_D)(\vec S_B\cdot\vec S_C)=-\tfrac14(P-k)(Q-k)\cdot\tfrac14k(k-R)=\tfrac1{16}\,k(P-k)(Q-k)(R-k). \]

All three numerators are equal (the expression is symmetric in $P,Q,R$), hence

\[\cos(AB)\cos(CD)=\cos(AC)\cos(BD)=\cos(AD)\cos(BC)=\frac{k(P-k)(Q-k)(R-k)}{16\,s_As_Bs_Cs_D}\qquad\blacksquare\]

Step 5: Numerical check (added)

For $\vec p=\{2,0,0\}$, $\vec q=\{1,2,0\}$, $\vec r=\{1,\tfrac12,\tfrac32\}$ ($k=2$), all three products come out as $0.10541\ldots$. For a tetrahedron without perpendicular opposite edges they differ (e.g. $0.205$, $-0.002$, $0.262$), so the hypothesis is really needed.