The sum of the squares of the edges of a tetrahedron is S. Prove that the tetrahedron can be fitted between two parallel planes a distance √(S/12) apart.
Solution
Solution by Bazavan Eduard
Let the tetrahedron be PQRS. Put PQ = a, PR = b, PS = c, QR = d, RS = e, SQ = f. We find the distance x between the midpoints of PS and QR. Let M be the midpoint of QR and let PM = m. Using the median result (see below), we have m2 = a2/2 + b2/2 - d2/4. Similarly, if SM = n, then n2 = e2/2 + f2/2 - d2/4. Let N be the midpoint of PS, then MN is the median of MPS, so x2 = m2/2 + n2/2 - c2/4 = S/4 - (c2 + d2)/2.
Similarly, if y is the distance between the midpoints of PR and QS we have y2 = S/4 - (b2 + f2)/2, and if z is the distance between the midpoints of PQ and RS, then z2 = S/4 - (a2 + e2)/2. Hence x2 + y2 + z2 = S/4. Assume that x is the smallest of x, y, z. Then x2 ≤ S/12. But if we take a plane through PS parallel to QR and a parallel plane through QR, then the tetrahedron fits between the planes, and the distance between them is at most x.
Notes
The median result is as follows. If ABC is a triangle with M the midpoint of BC, then 2 AM2 + 2 BM2 = AB2 + AC2. The proof is immediate by applying the cosine formula to triangles AMB and AMC and using the fact that cos AMB + cos AMC = 0.
Note that we cannot always fit a tetrahedron between two planes a distance √(S/12) apart if we put a face flat on one of the planes. For consider a regular tetrahedron side 1. It has height √(2/3), whereas √(S/12) = √(1/2), which is smaller.
© John Scholes
jscholes@kalva.demon.co.uk
4 Sep 2002