IMO 1967 LL CZS12

Origin: CZS

Problem

Given a segment AB of the length 1, define the set M of points in the following way: it contains the two points A, B, and also all points obtained from A, B by iterating the following rule: (∗) for every pair of points X, Y in M, the set M also contains the point Z of the segment XY for which Y Z = 3XZ. (a) Prove that the set M consists of points X from the segment AB for which the distance from the point A is either AX = 3k 4n or AX = 3k −2 4n , where n, k are nonnegative integers. (b) Prove that the point X0 for which AX0 = 1/2 = X0B does not belong to the set M.

Solution

Let us denote by Mn the set of points of the segment AB obtained from A and B by not more than n iterations of (∗). It can be proved by induction that Mn = . X \inAB | AX = 3k 4n or 3k −2 4n for some k \inN ; . Thus (a) immediately follows from M = % Mn. It also follows that if a, b \inN and a/b \inM, then 3 | a(b −a). Therefore 1/2 ̸\inM.