I later discovered a couple of solutions, (g,h,q) = (15,17,4)/12 and (12,17,12)/15 --- apologies for earlier false claim, and thanks to those who contributed! The upside to this fresh clanger was that, thinking there were no solutions, I generalised the construction to incorporate a new parameter. Instead of an outermost cuboid, there is a trapezohedron with 2 congruent e x f (empty) rectangles above and below, a band of 4 congruent isosceles trapezium faces of vertical depth g around the outside, and 16 interior triangular faces as before connecting to a quadrangle of height h and radius d (was q). This (homogeneous) 5-parameter polytore is isometric / origami / planar just when vanishes the quartic (2*d^2+3*e^2+3*f^2+2*e*f-4*e*d-4*f*d)*h^2 + (2*d^2-e^2-f^2-2*e*f)*g^2 + 2*(e-f)*(-e+2*d-f)*g*h + (d^2-e^2-f^2)*(-e-f+d)^2 ; conditions for embedding are unfortunately not as obvious as before, involving 3 new quadratic inequalities.
From a pedagogic point of view, we can now construct examples which are both rational, and easier to visualise than the cuboid special cases --- (d,e,f,g,h) = (9,16,15,8,16) is particularly nicely balanced.
Fred Lunnon On 9/1/09, Warut Roonguthai <warut822@gmail.com> wrote:
On Wed, Sep 2, 2009 at 4:24 AM, Fred W. Helenius<fredh@ix.netcom.com> wrote:
While sitting through a seminar this afternoon, I found that g = 20/9, h = 7/3, q = 8/9 works.
This proves that my parametric solution does not cover all the rational solutions! :P
Warut
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