Take any four points in space. Draw all lines connecting pairs of them. Then draw all lines connecting pairs of points on those lines.

Can the resulting set of points cover all of space?

(In reply to re(3): Detailed Solution by David Shin)

Thanks.

I should have seen that since the planes of the parallelepiped were x=-y, x=-z, y=-z, x=1-y, x=1-z, y=1-z, that the last three could be solved simultaneously as (1/2,1/2,1/2), and that indeed did not lie on the equilateral face.

Posted by Charlie on 2005-02-09 13:35:18