CORRECTED Geometry puzzle: --------------------------

(Posted to math-fun on Oct. 10, 2008)

Subject: Geometry puzzle

Let B = B_0 be a unit ball in 3-space that's tangent to the z-axis.

Let K > 0 be an unknown but fixed constant.

Now inductively, for all n > 0, place B_(n+1) so that it's tangent to both B_n and the z-axis, so that its center's z-coordinate exceeds that of B_n by K.

Assume this is done so there exists a rigid motion of 3-space that takes each B_n into B_(n+1).

QUESTION: --------- Find K_min := the least K such that all nonempty intersections between two B_n's are tangencies.

the smallest such value of K is sqrt(3) - 1 . let P_n be the center of the n-th unit sphere, and suppose that P_0 = (1, 0, 0) . it is easy to see that the rigid motion must preserve the orientation of the z-axis, and thus is a rotation about the z-axis, followed by a translation along the z-axis. this transformation can be expressed as (x, y, z) |---> (cos(2 alpha) x - sin(2 alpha) y , sin(2 alpha) x + cos(2 alpha) y , z + K) , where we may assume that 0 <= 2 alpha <= pi . then P_1 = (cos(2 alpha), sin(2 alpha), K) , and since its distance to P_0 is 2 , we find that K^2 = 4 cos^2(alpha) , so K = 2 cos(alpha) . we then get the formula P_n = (cos(2 n alpha), sin(2 n alpha), 2 n cos(alpha)) . the square of the distance between P_0 and P_2 is 32 cos^2(alpha) - 16 cos^4(alpha) = 8 K^2 - K^4 . for K in [0, sqrt(3) - 1) , this distance squared is < 4 , so the 0-th sphere and the 2nd sphere intersect transversally. for K = sqrt(3) - 1 , the 0-th sphere and the 2nd are tangent, while for n >= 3 , the z-coordinate of P_n is

= n ((sqrt(3) - 1)) > 2 , so the 0-th sphere does not intersect the n-th sphere for n >= 3 .

by the way, this is reminiscent of a problem posed a while back by dan, concerning a sequence of unit spheres such that each was tangent to the previous three. dan, did you ever send solution(s) to the list? i remember working on it and sending my solution off-list as requested. mike