Let A=\(x,y) :|x+y| 3\ and B=\(x,y) :|x|+|y| 3\. If C=\(x,y) :x=0 or y=0\, then (x,y) |x+y| is :
(4) 12
|x+y| |x|+|y| 3 inside B, so A (|x+y| 3) meets B only where |x+y|=3: the edges in the first and third quadrants.
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On the axes this gives ( 3,0) and (0, 3) (origin excluded): C has 4 points, each with |x+y|=3.
|x+y|=4 3=12
Answer: (4)
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