Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 3 / 5E (Algebra and Geometry)

Move the origin to the centroid and write the four vertex vectors as $r_1,r_2,r_3,r_4$. Their sum is zero. The midpoints of the first opposite-edge pair have vectors $(r_1+r_2)/2$ and $(r_3+r_4)/2=-(r_1+r_2)/2$, so their distances from the centroid are equal. The same argument applies to the other two opposite-edge pairs.

Choose the three midpoint distances as $u=|r_1+r_2|/2$, $v=|r_1+r_3|/2$, $w=|r_1+r_4|/2$. Expanding the dot products,Identifying these four norms with the vertex distances proves the tetrahedron centroid-to-midpoint sum of squares:The proof uses no orthogonality or regularity assumption on the tetrahedron.

Put $v=x-a$ and decompose it into axial and perpendicular parts, $v=(v\cdot n)n+v_{\perp }$. For the nondegenerate semi-angle $0<\alpha <\pi /2$, the equation is equivalent toThus each plane perpendicular to $n$ at axial distance $s$ from $a$ meets the surface in a circle of radius $|s|\tan \alpha$. Both signs of $s$ are allowed, and $s=0$ gives the vertex. This proves that it is a double circular cone with the specified vertex and axis; $\alpha$ is the angle each generator makes with either direction of the axis.

For the intersection calculation, define the symmetric matrix $Q={\cos }^2\alpha I-n{n}^T$. The two equations are $(x-a_i{)}^{T}Q(x-a_i)=0$. Their difference givesWith $b=a_1-a_2$ and $m=(a_1+a_2)/2$, symmetry of $Q$ makes the last two terms $2m\cdot Qb$. Hence the intersection plane of parallel congruent double cones isThe unnormalized normal vector is $v=Qb=b{\cos }^2\alpha -n(n\cdot b)$. Expanding its squared norm givesEquivalently, if $b=b_{\perp }+(n\cdot b)n$, this is ${\cos }^4\alpha |b_{\perp }{|}^2+{\sin }^4\alpha (n\cdot b{)}^2>0$. Thus normalization is legitimate and yieldsThe assumption of a genuine cone matters: at the degenerate limits $\alpha =0$ or $\pi /2$, the surface collapses respectively to an axis or a plane and this normal need not exist.