Here is a specific example of a tetrahedron which is continuously deformed with volume changing but surface area remaining constant. As 'a' varies from 0 to sqrt(2), the tetrahedron with vertices (-a,-a,1), (a,a,1), (b,-b,-1), (-b,b,-1) where b=sqrt(a^2+2)-sqrt(2)*a maintains a constant surface area 8, but the volume (8/3)*a*(sqrt(a^2+2) - sqrt(2)*a) varies from 0 to (8/3)*(sqrt(2)-1) = 1.10457and back to 0 again as 'a' varies from 0 to sqrt(2). The maximum volume occurs when a = b = sqrt(sqrt(2)-1) = 0.64359 Jim On Mon, Oct 20, 2008 at 2:18 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
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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