51 Points in Unit Square

In any set of 51 points inside a unit square, there are always three points that can be covered by a circle of radius 1/7.

Solution

In any set of 51 points inside a unit square, there are always three points that can be covered by a circle of radius 1/7.

Solution

Cut the square into 25 equal subsquares. By the pigeonhole principle, at least one of the squares contains at least 3 = ⌈51/25⌉ points. (To remind, ⌈x⌉ denotes the ceiling function: ⌈x⌉ is the least integer not less than x.) The diagonal of a subsquare equals √2/5 (≈ .283) < 2/7 (≈ 285). It follows that a subsquare and, the three points within it, can be covered by a circle of diameter 2/7, i.e., of radius 1/7.

(Instead of using approximations, we might have squared the two numbers: 2/25 (= 4/50) < 4/49,) with the same conclusion.

Reference

C. Alsina, R. B. Nelsen, Charming Proofs, MAA, 2010, p. 46

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51 Points in Unit Square

Solution

Reference

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