If one vertex is moved with constant tet volume, it HAS to move parallel to the opposite face (vol = 1/3 height x altitude). Next, one would have to prove or disprove that any tet deformation can be performed by a series of single-vertex moves. Conjecture: There is a closed plane curve parallel to the base such that if the top vertex moves on that curve, the 3 face areas will sum to a constant. Arm-wave "proof": let the base be equilateral and let the top vertex NOT be over its center. By symmetry, there must be 3 positions of the top vertex that give the same area of the 3 moving triangles. Then there also must be a continuous path from one of those points to one of the others. Now deform the equilateral - more or less the same thing must hold, etc. End of armwave. I think this works for the top vertex over and not over the base. There is some surface U above the fixed base such that if the top vertex moves U, the sum S of the 3 triangle areas is constant. U can't be a plane parallel to the base, as I showed before. Then U has to vary in its height above the base. If we move the top vertex in U but not parallel to the base, the volume will change but S will be constant. Pseudo-QED. Maybe I'll get a chance to plot this in Mathematica. Steve -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Monday, October 20, 2008 1:54 PM To: math-fun Subject: Re: [math-fun] tetrahedron volume ----- Original Message ---- From: Stephen Gray <stevebg@roadrunner.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Monday, October 20, 2008 1:32:32 PM Subject: Re: [math-fun] tetrahedron volume More simply, if a vertex moves parallel to the opposite face, volume is preserved. Move a vertex in that manner so that it goes way off to one side. When the moving vertex is no longer above a point on the opposite side, further motion will increase the altitudes of the 3 changing triangles, so their areas increase also. QED Steve Gray ________________________________ This shows that the areas can change while the volume remains fixed. But RWG was asking if it is possible to change the volume while holding the four face areas fixed. One possible way to complete your proof is to scale the tetrahedron. But it is not clear that such a scaling is possible that preserves all four face areas. Gene -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Monday, October 20, 2008 12:18 PM To: math-fun Subject: Re: [math-fun] tetrahedron volume ----- Original Message ---- From: "rwg@sdf.lonestar.org" <rwg@sdf.lonestar.org> To: math-fun <math-fun@mailman.xmission.com> Cc: math-fun <math-fun@mailman.xmission.com> Sent: Monday, October 20, 2008 3:37:08 AM Subject: [math-fun] tetraroller volume 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun