= Solution
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>,
$$
4(u^2+v^2+w^2)=3|r_1|^2+\sum_{j=2}^4|r_j|^2+2r_1\cdot\sum_{j=2}^4r_j=\sum_{i=1}^4|r_i|^2.
$$
Identifying these four norms with the vertex distances proves the <tetrahedron centroid-to-midpoint sum of squares>:
$$
\boxed{u^2+v^2+w^2=\frac14(a^2+b^2+c^2+d^2)}.
$$
The proof uses no orthogonality or regularity assumption on the <tetrahedron>.
Back to article page
