1961 AHSME Problems/Problem 39

Problem

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:

$\textbf{(A)}\ \sqrt{3}/3\qquad \textbf{(B)}\ \sqrt{2}/2\qquad \textbf{(C)}\ 2\sqrt{2}/3\qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ \sqrt{2}$

Solution

Partition the unit square into four smaller squares of sidelength $\frac{1}{2}$ . Each of the five points lies in one of these squares, and so by the Pigeonhole Principle, there exists two points in the same $\frac{1}{2}\times \frac{1}{2}$ square - the maximum possible distance between them being $\frac{\sqrt{2}}{2}$ by Pythagoras.

Hence our answer is $\fbox{B}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 38	Followed by Problem 40
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

Page

Toolbox

Search

1961 AHSME Problems/Problem 39

Problem

Solution

See Also