Five points in a unit square

Question

Five points are placed anywhere in a square of side 1, including on its edges.

Prove that some two of the points are at distance at most the square root of 2 divided by 2, about 0.707. Show also that this bound cannot be improved: exhibit five points with no two closer than that.

Accepted Answer

Two of the five points lie within a distance of root 2 over 2 of each other, and this is the best possible bound. The pigeonhole step. Divide the unit square into four squares of side 1/2 by cutting it in half horizontally and vertically. Five points in four small squares means some small square contains at least two of the points (a point on a dividing line can be assigned to either side; the argument is unaffected). The distance step. Two points inside the same square of side 1/2 can be no farther apart than that square's diagonal, which is the square root of (1/4 + 1/4) = root 2 over 2, ab…

Five points in a unit square

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