Yes, the equation has infinitely many rational solutions. Let q = (r^2+2*r-1)/(r^2+1), h/sqrt(2-q^2) + g/(2-q) = s, h/sqrt(2-q^2) - g/(2-q) = 1/(2*s), where r, s are rational numbers. Therefore, h = +/- (2*s^2+1)*(r^2-2*r-1)/(4*s*(r^2+1)), g = +/- (2*s^2-1)*(r^2-2*r+3)/(4*s*(r^2+1)). For example, when r = 1/2 and s = 1, we have q = 1/5, h = 21/20, g = 9/20. Warut On Tue, Sep 1, 2009 at 10:13 PM, 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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