Well, Gene beat me to it, and managed to say essentially the same thing without any algebra at all --- but just in case anybody is interested, here goes: Suppose we have $n+1$ notional "points" $P_i$, for which we know only the squared distances $U_{ij} = |P_i - P_j|^2$. If they are embeddable in Euclidean $n$-space, then the squared content of the simplex times 2^n (n!)^2 is given by the Cayley-Menger determinant D_{n+2} = | 0 1 1 ... 1 | | 1 -U_{00} -U_{01} ... -U_{0n} | | . | - | . | | . | | 1 -U_{n0} -U_{n1} ... -U_{nn} | So denote 288 x tetrahedron volume^2 V0 = D_5(U21, U31, U32, U41, U42, U43), 16 x face areas^2 A1 = D_4(U32, U42, U43), A2 = D_4(U31, U41, U43), A3 = D_4(U21, U41, U42), A4 = D_4(U21, U31, U32), where edges^2 are U21, U31, U32, U41, U42, U43. Substitute the special values U31 = U32 = U41 = U42 = 1; then A1 = A2 = (4 - U43)U43, A3 = A4 = (4 - U21)U21, V0 = 2 (4 - U43 - U21) U21 U43. Now setting U21 = U43 = 0, 4 in turn, V0 = 0, -128, but A1 = A2 = A3 = A4 = 0 in both cases: so V0 cannot be a function of the A_i. Of course, these do not represent tetrahedra; but it shouldn't be hard to concoct similar "proper" examples where all values involved are positive; also perhaps rational. Nice little puzzle, begging quite a few similar questions --- which I'm tring hard to ignore! WFL On 10/20/08, Eugene Salamin <gene_salamin@yahoo.com> wrote:
True or False Quickie: The volume of a tetrahedron is determined by the areas of its faces. _______________________________________________
False. There exists a face area preserving continuous deformation that alters the volume. Start with a regular tetrahedron. Perform a dilatation along the mutual perpendicular of two opposite edges, while simultaneously performing a dilatation in the transverse plane so as to preserve the face areas. This is possible because all four faces are inclined with respect to that mutual perpendicular by the same angle. The volume is not preserved, in particular the volume goes to zero as the tetrahedron is squeezed flat.
Gene