Try one more time; please ignore my earlier posts

We can find M>0, such that each of the three sets have elements in (0,M], to simplify our notation, let X,Y,Z be the intersection of A,B,C with (0,M] respectively, and let x,y,z be the corresponding supremums. Clearly, (0,M] = union(X,Y,Z)

Without loss of generality, we may assump x
case 1. x so, for any x1 in X, we can find y1 in Y1 and z1 in Z1, such that |y1-z1| = x1, z1 >= x1, so x1,y1,z1 can form a triangle.


case 2. x=y, then inf(Z1) = x, so we can find x1,y1,z1 from X,Y,Z respecively, so that they are close to x, with maximum difference less than x/9. Then it can be easily shown that x1,y1,z1 can be the lengths of the edges of a triangle.

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• 回複：Try one more time; please ignore my earlier posts -HF:- ♀ (101 bytes) () 01/03/2009 postreply 20:58:27

• 應該沒有這麽複雜。本來題中還有一問，貼題時少貼了： -康MM- ♀ (67 bytes) () 01/04/2009 postreply 10:10:55

• 回複：應該沒有這麽複雜。本來題中還有一問，貼題時少貼了： -HF:- ♀ (107 bytes) () 01/04/2009 postreply 10:55:22