Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 3 / 6E (Algebra and Geometry)

Set $w=(x-y)/2$. The sum condition then gives $x=a+w$ and $y=a-w$, and their dot product isTherefore $|w{|}^2=|a{|}^2-c$. Since $c<|a{|}^2$, this is a strictly positive squared radius. The complete solution isEvery choice of unit vector $n$ satisfies both conditions, and every solution has this form. The endpoints lie on the sphere with center $a$ and radius $R$, and their midpoint is $a$. They are therefore opposite ends of one diameter of that sphere.

Translate one vertex of the tetrahedron to the origin and denote the other vertex vectors by $a,b,c$. Translation preserves edge vectors and their dot products. We may label the two known perpendicular pairs of opposite edges of a tetrahedron as the pairs represented by $a,c-b$ and $b,a-c$. Their orthogonality givesHence all three pairwise dot products have the same value $s$. The third pair satisfiesso the third pair of opposite edges is also perpendicular.

Expanding squared lengths using the dot product givesThus the sums of squared opposite-edge lengths are equal. These two conclusions are the basic identities for orthogonal opposite edges of a tetrahedron.