Question

If the angles between the three pairs of opposite sides of a tetrahedron are equal, then these angles are

• A. equal to ${90}^0$
• B. All are $>{90}^0$
• C. All are $<{90}^0$
• D. some are $>{90}^0$ and some are $<{90}^0$

Answer 1

The correct option is A equal to ${90}^0$

Let ABCD be our tetrahedron; we will denote by $\overrightarrow{AB}\cdot \overrightarrow{CD}$ the dot product of the two vectors and by $AB\cdot CD$ the product of the lengths of the two segments. Let $a=\overrightarrow{OA},b=\overrightarrow{OB},c=\overrightarrow{OC},c=\overrightarrow{OD}$, where O is an arbitrary point and letting $\alpha$ be the common angle, we will have
$(a-d)(b-c)=\pm AD\cdot BCcos\alpha$
$(a-b)(c-d)=\pm AD\cdot DCcos\alpha$
$(a-c)(d-b)=\pm AC\cdot BDcos\alpha$
Summing these up yields
$0=cos\alpha (\pm AD.BC\pm AB.CD\pm AC.BD)$ ............. (i)
But we will show $\pm CD.BC+\pm AB.DC+\pm AC.BD\ne 0$ whatever the signs; if the signs were all equal, then this is obvious. Otherwise, we would need to have something of the form
$AD\cdot BC=AB\cdot DC+AC\cdot BD$
Take an inversion around A, and the images of the points B, C, D will not be collinear (otherwise, our four points would be on a circle). Then, we would have
$B^{\prime }{C}^{\prime }<B^{\prime }{D}^{\prime }+C^{\prime }{D}^{\prime }\Rightarrow \frac{BC}{AB.AC}<\frac{BD}{AB.AD}+\frac{CD}{AC.AD}\Rightarrow AD.BC<AB.CD+AC.BC$
So the bracket in the RHS of (i) will always be non-zero, so we must have
$\alpha =\frac{\pi }{2}$