Let S be a set of n concentric circles in the plane. Prove that if a function f:S→S satisfies the property d(f(A),f(B))≥d(A,B) for all A,B∈S, then d(f(A),f(B))=d(A,B), where d is the euclidean distance function.
Problem
Source: Romanian IMO Team Selection Test TST 1996, problem 15
Tags: function, geometry proposed, geometry
27.09.2005 06:03
Consider the two circles D and E such that d(D,E) is maximal. Then d(f(D),f(E))=d(D,E), and either f(D)=D,f(E)=E or f(D)=E,f(E)=D. Now assume for two circles A,B, d(f(A),f(B))>d(A,B). Then d(f(D),f(E))=d(D,E)≤d(D,A)+d(A,B)+d(B,E)< min d(f(D),f(B))+d(f(B),f(A))+d(f(A),f(E))) = d(f(D),f(E)) . Absurd.
21.04.2010 01:19
julien_santini wrote: Consider the two circles D and E such that d(D,E) is maximal. Then d(f(D),f(E))=d(D,E) , and either f(D)=D, f(E)=E or f(D)=E, f(E)=D . Now assume for two circles A,B , d(f(A),f(B))>d(A,B) . Then d(f(D),f(E)) = d(D,E) \leq d(D,A)+d(A,B)+d(B,E) < \min(d(f(D),f(A))+d(f(A),f(B))+d(f(B),f(E)), d(f(D),f(B))+d(f(B),f(A))+d(f(A),f(E))) = d(f(D),f(E)) . Absurd. I don't think this argument works unless you can show that f is continuous. Anyway, I think we just first show that the function f, restricted on the largest circle, is an isometry. Then, get rid of this circle, and proceed inward to the smallest circle.
21.04.2010 07:20
(Due to Chung Ping Ngai, HK IMO team member) First off, f must be bijective, for otherwise there exists two distinct elements (say, X, Y) in S which are mapped to the same element, so that d(f(X), f(Y)) = 0 < d(X, Y), a contradiction. Note that since S is finite, for all A, B in S we have f^p(A)=A and f^q(B)=B for some positive integers p, q. Then f^{pq}(A)=A and f^{pq}(B)=B. But d(f^{pq}(A), f^{pq}(B)) \geq d(f^{pq-1}(A), f^{pq-1}(B)) \geq ... \geq d(A, B), so all \geq signs must become equal signs. QED.
21.04.2010 09:06
Ohhh. OK. I thought S was the set of points of n circles. Apologies to the first reply.