----- 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 __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
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
On 10/20/08, Fred lunnon <fred.lunnon@gmail.com> wrote:
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.
I didn't look very hard ... Taking U43 = 1, and U21 = 1, 3 in turn: the first tetrahedron is equilateral, the second flat, and both have all 16 x face areas^2 = 3. WFL Sufficient conditions for embeddability are that every possible D_k determinant is non-negative; or else that there is some increasing sequence of elements ("flag manifold") for which the D_k are all positive. WFL
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 -----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
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:
----- 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
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
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
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
----- 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
Yes. Or drawn to a long needle. There are formulas germane to other bits of this thread in the Mathworld tetrahedron article. It was surprisingly easy to write vol_polyhedron(faces) in terms of vol_pyramidpts(apex,face), in turn in terms of vol_tetrahedron(pts). It even flips your faces for you. For the Szilassi holeyheptahedron [[[-90,50,40],[75,75,-60],[-40,100,-160],[-240,0,240],[240,0,240],[140,50,40]], [[-140,0,40],[-140,-50,40],[-240,0,240],[-40,100,-160],[0,252,-240],[0,-252, -240]],[[-75,-75,-60],[75,75,-60],[-90,50,40],[-140,0,40],[0,-252,-240],[40, -100,-160]],[[75,75,-60],[-75,-75,-60],[90,-50,40],[140,0,40],[0,252,-240], [-40,100,-160]],[[140,0,40],[140,50,40],[240,0,240],[40,-100,-160],[0,-252, -240],[0,252,-240]],[[90,-50,40],[-75,-75,-60],[40,-100,-160],[240,0,240], [-240,0,240],[-140,-50,40]],[[140,50,40],[140,0,40],[90,-50,40],[-140,-50,40], [-140,0,40],[-90,50,40]]] it claims vol_polyhedron(%) = 22124800/3, if anyone cares to check. --rwg INCONSISTENT NONSCIENTIST PS, Mma 6.0 displays and tumbles the Szilassi perfectly! But I'm surprised it seems to have no AreaPolygon[pts_List], let alone VolumePolyhedron[faces_List].
Oops. I think the original question was: if each face of the tet has constant area, can the tet have different volumes. (Without rearranging the faces.) I somehow changed the question to: if the sum of the face areas is constant, can the volume vary. The answer to the second question is yes, but I don't know the answer to the original problem except as Gene gave it. Steve Gray -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of rwg@sdf.lonestar.org Sent: Wednesday, October 22, 2008 3:28 AM 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
Yes. Or drawn to a long needle. There are formulas germane to other bits of this thread in the Mathworld tetrahedron article. It was surprisingly easy to write vol_polyhedron(faces) in terms of vol_pyramidpts(apex,face), in turn in terms of vol_tetrahedron(pts). It even flips your faces for you. For the Szilassi holeyheptahedron [[[-90,50,40],[75,75,-60],[-40,100,-160],[-240,0,240],[240,0,240],[140,50,40 ]], [[-140,0,40],[-140,-50,40],[-240,0,240],[-40,100,-160],[0,252,-240],[0,-252, -240]],[[-75,-75,-60],[75,75,-60],[-90,50,40],[-140,0,40],[0,-252,-240],[40, -100,-160]],[[75,75,-60],[-75,-75,-60],[90,-50,40],[140,0,40],[0,252,-240], [-40,100,-160]],[[140,0,40],[140,50,40],[240,0,240],[40,-100,-160],[0,-252, -240],[0,252,-240]],[[90,-50,40],[-75,-75,-60],[40,-100,-160],[240,0,240], [-240,0,240],[-140,-50,40]],[[140,50,40],[140,0,40],[90,-50,40],[-140,-50,40 ], [-140,0,40],[-90,50,40]]] it claims vol_polyhedron(%) = 22124800/3, if anyone cares to check. --rwg INCONSISTENT NONSCIENTIST PS, Mma 6.0 displays and tumbles the Szilassi perfectly! But I'm surprised it seems to have no AreaPolygon[pts_List], let alone VolumePolyhedron[faces_List]. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (6)
-
Eugene Salamin -
Fred lunnon -
James Buddenhagen -
rwg＠sdf.lonestar.org -
Schroeppel, Richard -
Stephen Gray