This suggests changing the question a little: (a) Given the areas of the four faces, what's the maximum volume? (b) What extra information is needed (beyond the face areas) to determine the volume? One or two edges? One of the lines connecting centers of opposite edges? Recall that altitude + opposite face determines the volume. Also that three face areas of a right tetrahedron determine the volume. So maybe we should be thinking about angles, or solid angles. Rich ________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] On Behalf Of James Buddenhagen [jbuddenh@gmail.com] Sent: Tuesday, October 21, 2008 11:21 AM To: math-fun Subject: Re: [math-fun] tetrahedron volume Yes, I realized shortly after posting that I was holding total surface area constant, but not individual faces. Thanks for pointing that out, WFL, and providing a yet simpler example where the individual face areas remain constant. Jim On Tue, Oct 21, 2008 at 11:25 AM, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 10/21/08, James Buddenhagen <jbuddenh@gmail.com> wrote:
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
While correct as stated, this example misses the point of the problem --- all the face areas can remain individually constant (which the above example fails to achieve) while the volume varies!
While correctly computed, this misses the point of the problem: all the face areas need to remain individually constant (which the above fails to achieve) while the volume varies!
Inscribe a tetrahedron in a cuboid (as suggested earlier), with vertices [a,0,0], [0,b,0], [0,0,c], [a,b,c]; all faces are congruent, with (via Cayley-Menger) 16*area^2 = 4 (a^2 b^2 + b^2 c^2 + c^2 a^2), 288*volume^2 = 32 a^2 b^2 c^2.
Now choose b = a, c^2 = (1 - a^4)/(2*a^2); then as a varies from 0 to 1, each face area = 1/2 is constant, but volume = sqrt((1-a^4) a^2/18),
WFL
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