1 Feb
2013
1 Feb
'13
7:11 a.m.
Actually RWG's samosa, and the locus of (*) below in [-1,1]^3, should be identical, both corresponding to all triples of cosines of 3 angles that add up to 2pi. But it seems that peculiarities of the way acos() is interpreted might mess things up. This is the case in the Mac Grapher utility, where only one face of the pillow is displayed for the equation (**) acos(x) + acos(y) + acos(z) = 2pi. PUZZLE: Determine whether any planar geometric circle lies on the locus of (*). --Dan On 2013-02-01, at 2:25 AM, Dan Asimov wrote:
(since (x,y,z) must lie in [-1,1]^3) so that leaves only the fact that
(*) 2xyz + 1 - (x^2 + y^2 + z^2) >= 0
whose locus fits RWG's samosa description to a T.