1961 AMC 12 Problem 39

Question

Any five points are taken inside or on a square with side length

1.

Let

a

be the smallest possible number with the property that it is always possible to select one pair of points from these five such that the distance between them is equal to or less than

a.

Then

a

is:

\dfrac{\sqrt3}{3}

\dfrac{\sqrt2}{2}

\dfrac{2\sqrt2}{3}

1

\sqrt2

Answer: B

Concepts:pigeonhole principle extremal argument square (geometry)

Difficulty rating: 1670

Small hint:

Partition the unit square into four congruent smaller squares

Big hint:

To prove sharpness, look for five points whose closest-pair distance reaches the bound

Solution:

Divide the square into four squares of side

\frac{1}{2}.

Two of the five points lie in the same small square, so their distance is at most its diagonal,

\frac{\sqrt2}{2}.

This bound is attainable by placing points at the four corners and the center: the shortest distance is then

\frac{\sqrt2}{2}.

Hence the least guaranteed value is

\frac{\sqrt2}{2}.

Therefore, the correct answer is B.