colonel@sunybcs.UUCP (Col. G. L. Sicherman) (11/05/86)
> If we must think of it in terms of pure geometry, why this silly > preference for a space with zero curvature? If we assume positive > curvature, as for Riemannian spherical geometry, we lend an entirely > new twist to the problem. Why consider 3-space at all? I would guess that the original problem was meant for 2-space. So how about 3 random vertices on a sphere? That's meaningful. Shouldn't be too hard to figure out the integral. -- Col. G. L. Sicherman UU: ...{rocksvax|decvax}!sunybcs!colonel CS: colonel@buffalo-cs BI: colonel@sunybcs, csdsiche@sunyabvc
colonel@sunybcs.UUCP (Col. G. L. Sicherman) (11/06/86)
> So how about 3 random vertices on a sphere? That's meaningful. Shouldn't > be too hard to figure out the integral. And I'll bet it comes out less than 1/4. -- Col. G. L. Sicherman UU: ...{rocksvax|decvax}!sunybcs!colonel CS: colonel@buffalo-cs BI: colonel@sunybcs, csdsiche@sunyabvc