Let D be the set of all points in the real plane such that |x| + |y| <= 1,
where |x| (respectively |y|) denotes the absolute value of x (respectively y).
Prove that amongst every 5 points in D, there exist two points whose distance
from one another is at most 1.
Submitted by: AdministratorYes,D is a square with ends at (0,1),(1,0),(-1,0),(0,-1), and also we could draw a round which will go through these four points. Its equation is x^2+y^2=1. we could see the square is totally contained within the round. Also we will notice that any points in the round will have distance less than 1 from each other. So in conclusion, every points in D have distance at most 1 from each other .
Submitted by: Administrator
Submitted by: Administrator
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