And now for the inevitable corrections ...
Wlog, we can assume that all of d,e,f,g,h are in |N.
I posited this casually on physical grounds --- but even if it's true, it's not obvious. For example, one symmetry of the constraint is h -> -h, e -> f, f -> e.
A necessary condition on the embedding region is 0 < 2*d < e+f [more required for sufficiency, but not relevant --- yet].
For completeness, required for sufficiency are in addition d*g+h*(e-f) > 0, (e+f-2*d)*g + (e-f)*h > 0, (e+f-2*d)*g - (e-f)*h > 0, arising from various tetrahedral volumes needing to be positive.
To establish the sufficiency of planar_trapeze constraint, I need to bound the "right_angle" polynomial e^2 + f^2 - (e+f)*d + g^2 - h^2 + e*f away from zero in this region, subject to constraint.
There are (at least) two goofs here, which should have read --- To establish the sufficiency of planar_trapeze constraint, I need to bound the "right_angle" polynomial e^2 + f^2 - (e+f)*d + g^2 - h^2 away from -e*f in this region, subject to constraint. In fact, it appears that right_angle > -e*f/2 is true. [The constraint right_angle = 0 ensures that the 3-space solid has those irritating right-angles, which cause it to impersonate planarity even when it ain't. No, I don't understand why it shows up in this new context either!] Fred Lunnon On 9/4/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
I've thought off and on about the info from John Cremona reported by Victor; but I must confess that I don't really understand its implications. I'm used to thinking about extensions by irrationals, but not by rationals ...
What can be said in general about the structure of the rational solution sets of these equations (the cube case with e = f = g (= 1), the cuboid with e = f (= 1), or the (most general) trapezoidal with all 5 (homogeneous) parameters?
For example, what's their dimension?
Current presentation (with old q -> new d) of "planar_trapeze" constraint :
+ ((e-f)^2+2*(e+f-d)^2)*h^2 - ((e+f)^2-2*d^2)*g^2 - 2*(e-f)*(e+f-2*d)*g*h - (e^2+f^2-d^2)*(e+f-d)^2 = 0 .
Wlog, we can assume that all of d,e,f,g,h are in |N. A necessary condition on the embedding region is 0 < 2*d < e+f [more required for sufficiency, but not relevant --- yet].
To establish the sufficiency of planar_trapeze constraint, I need to bound the "right_angle" polynomial e^2 + f^2 - (e+f)*d + g^2 - h^2 + e*f away from zero in this region, subject to constraint.
I can do this for the cuboid case. For the general trapezoidal case, I can show that right_angle > 1/2 on the boundary in the same way, by expressing it as a sum of squares --- but I cannot see how to establish that there are no turning points in the interior. Any ideas, anybody?
[Lagrange lambda shows only that there is a maximum outside, on the boundary of the immersion region at e = f = 1 (say), d = 2+2sqrt2, g = h = 0.]
Fred Lunnon
On 9/4/09, Warut Roonguthai <warut822@gmail.com> wrote:
Here's another parametric solution:
q = 4r/(2*r^2+1),
h = (2*s^2+1)*(2*r^2-1)/((2*s^2-1)*(2*r^2+1)),
g = 4*s*(2*r^2-2*r+1)/((2*s^2-1)*(2*r^2+1)),
where r, s are rational numbers. This one includes Fred W. Helenius's solution, but not my previous solutions. Fred Lunnon's solution is not included in any of my parametric solutions.
Warut
On Fri, Sep 4, 2009 at 3:32 AM, victor miller<victorsmiller@gmail.com> wrote:
I gave this to John Cremona who has an algorithm for finding a point (if it exists) on a diagonal conic over a function field. The equation in question is an example of such -- if we clear denominators we get:
(2-q)^2*h^2 = (2-q^2)*g^2 + 1/2*(2-q)^2*(2-q^2) (*)
So this is a conic in (g,h) over the field Q(q).
It's well know that if you have one solution to a conic, all the others are generated by a rational parametrization from that one. However, John's program showed that (*) had no points in Q(q). Warut's parametrized solutions essentially amount to finding one point on (*) in a quadratic extension of Q(q) (his parameter, r, satisfies a quadratic over Q(q)), and then using the general recipe to generate all other solutions over that field. An alternative approach would be to look at (*) as an elliptic curve in (h,q) over Q(g), which must have positive rank since we know that there are an infinite number of solutions.
Victor
On Tue, Sep 1, 2009 at 11:13 AM, Fred lunnon <fred.lunnon@gmail.com> wrote:
It's always nice to have examples of geometrical configurations with rational parameters --- apart from anything else, they're useful for testing software.
A polytore with square section edge 2, outer cuboid depth g, inner quadrangle height h, inner radius q, is planar just when
h^2/(2-q^2) = g^2/(2-q)^2 + 1/2
but this equation appears to have no small rational solutions, at any rate when 0 < q < 1.
Is it soluble over the rationals? WFL
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