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
Short solution by Federico Kereki)
Consider a regular tetrahedron (the result can be stretched for nonregular tetrahedra).
The tetrahedron (whose vertices are the four points given) fits in a cube with one edge of the tetrahedron per face of the cube, as one of that face's two diagonals. Four vertices are common to the tetrahedron and the cube; the other four are of the cube only.
Take your "any point" to be one of the vertices of the cube that's not a vertex of the tetrahedron. The plane determined by the point and any of three of the lines of the tetrahedron will be adjacent sides of the cube, extended. All the other lines meet that chosen plane at one of two points, determining a line (the diagonal of the face of the cubeone edge of the tetrahedron) that does not go through the point in question. The remaining edge is on the opposite side of the cube parallel to that face and so never meeting it.
In a nonregular tetrahedron the same is done in a parallelepiped, as opposing skew lines can be put into parallel planes.

Posted by Charlie
on 20050207 20:41:35 