Suppose we select a set X of 5 points at random from an equilateral traingle with side length 1
show that no matter how these points are selected there will always be at least 2 whose distance apart is at most 1/2
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Draw triangle ABC, equilateral with side length 1. Now at each vertex construct a circle of radius 1/2. Each circle intercepts two sides at their midpoints.
Without loss of generality we can say that the first point selected is on a radius of circle A. (We can rename the vertices if necessary). If another point is selected on a radius of A, then the two points will be at most 1/2 unit apart. So the next point must lie on the radius of another circle, say B.
Now if the third point selected lies on a radius of A or B it will be at most 1/2 unit from a previous point. So the third selected point must lie on a radius of C.
Now any further points selected that lie on a radius of A,B,or C are within 1/2 a unit from a previously selected point. So the next point must lie outside the three circles.
Now we have points in each of four regions; within circle A, within circle B,within circle C, and between the circles. There is no other region in which to put a fifth point without it being within 1/2 unit of another point. QED
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