Given two intersecting lines L1 and L2, and a initial point P0, I apply one of the three following algorithms:

(1) Let P1 be the closest point to P0 in L1; then let P2 be the closest point to P1 in L2; then, P3 the closest point to P2 in L1, P4 the closest point to P3 in L2, and so to infinity.

(2) Let P0' be the closest point to P0 in L1 and P0" the closest point to P0 in L2, and let P1 be the midpoint of P0' and P0"; now take P1' as the closest point to P1 in L1, and P1" as the closest point to P1 in L2, and let P2 be the midpoint of P1' and P1", and so to infinity.

(3) Let P0' be symmetrical to P0 with regard to L1, and P0" symmetrical to P0 with regard to L2, and let P1 be the midpoint of P0' and P0"; now let P1' be symmetrical to P1 with regard to L1, and P1" symmetrical to P1 with regard to L2, and let P2 be the midpoint of P1' and P1", and so to infinity.

Can you prove that no matter which algorithm I pick, the sequence P0, P1, P2, ... will converge to the same limit? Could you generalize that for three intersecting planes? For four three-dimensional spaces?

In each case the P's converge to the point of intersection of the lines.  The ratio between succesive distances from this point is determined by the angle between the two lines and is always <1.

Case 1:   The point can be anywhere and the ratio is .5+.5cos(2a) where a is the acute angle between the lines.

Case 2:  If P0 is on an angle bisector of the (acute or obtuse) angle formed by the lines the ratio is .5+.5cos(a) where a is the (acute or obtuse) angle between the lines.
If P0 is not on the angle bisector the ratio between succesive Pn's varies but still tends to the same value.

Case 3:  The point can be anywhere and the ratio is the absolute value of cos(a) where a is the angle formed by the lines that has P0 within it.

It may be that I've defined the angle in slightly strange ways in each case but it still works.  The point is this ratio is always <1 so the points converge.

Posted by Jer on 2007-06-11 12:46:43