case 1 again:

來源: 2009-01-03 20:47:08 [舊帖] [給我悄悄話] 本文已被閱讀:

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.