A postscript --- it took some time to dawn on me that the constant coefficient of the polynomial in SABC is actually the square of 4*SA*SB*SC + SA^2 + SB^2 + SC^2 - 2*(SA*SB + SA*SC + SB*SC), and the vanishing of this polynomial is equivalent to A+B+C = k \pi. This considerably reduces the computational cost of proving the h versus q relations for a (cuboid) polytore; the bad news is that it fatally entices one (me) into attempting to find a similar relation for polytores constructed from prisms based on regular even n-gons, rather than just square cuboids (n = 4). It seems we got lucky with the cuboidal case. For n = 6 (and any n > 4?), the (numerator of square root of) constant coefficient fails to factorise, has degree 16 in q and 12 in h, and numerical coefficients of order 10^8. WFL [21/08/09] On 8/16/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
Er, make that a quartic in SABC with 71 terms in SA,SB,SC. To wit,
With SA -> sin^2(A), SB -> sin^2(B), SC -> sin^2(C), SABC -> sin^2(A+B+C),
0 = SABC^4
+ SABC^3 * ( -16*SA*SB*SC + 8*(SA*SB + SA*SC + SB*SC) - 4*(SA + SB + SC) )
+ SABC^2 * ( + 24*SA*SB*SC + 16*(SA^2*SB^2 + SA^2*SC^2 + SB^2*SC^2) - 16*(SA^2*SB + SA^2*SC + SA*SB^2 + SA*SC^2 + SB^2*SC + SB*SC^2) + 6*(SA^2 + SB^2 + SC^2) + 4*(SA*SB + SA*SC + SB*SC) )
+ SABC * ( - 40*SA*SB*SC - 16*(SA^3*SB*SC + SA*SB^3*SC + SA*SB*SC^3) + 8*(SA^3*SB + SA^3*SC + SA*SB^3 + SA*SC^3 + SB^3*SC + SB*SC^3) + 24*(SA^2*SB*SC + SA*SB^2*SC + SA*SB*SC^2) - 16*(SA^2*SB^2 + SA^2*SC^2 + SB^2*SC^2) - 4*(SA^3 + SB^3 + SC^3) + 4*(SA^2*SB + SA^2*SC + SA*SB^2 + SA*SC^2 + SB^2*SC + SB*SC^2) )
+ ( 16*SA^2*SB^2*SC^2 + 8*(SA^3*SB*SC + SA*SB^3*SC + SA*SB*SC^3) - 16*(SA^2*SB^2*SC + SA^2*SB*SC^2 + SA*SB^2*SC^2) + (SA^4 + SB^4 + SC^4) - 4*(SA^3*SB + SA^3*SC + SA*SB^3 + SA*SC^3 + SB^3*SC + SB*SC^3) + 6*(SA^2*SB^2 + SA^2*SC^2 + SB^2*SC^2) + 4*(SA^2*SB*SC + SA*SB^2*SC + SA*SB*SC^2) ) ;
Intriguingly, tidying the above up for publication produces a reduction in numerical rounding error by an order of magnitude! WFL