1961 AMC 12 Problem 39

Attempt Problem 39 of the 1961 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1961 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

39.

Any five points are taken inside or on a square with side length 1.1. Let aa 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.a. Then aa is:

33\dfrac{\sqrt3}{3}

22\dfrac{\sqrt2}{2}

223\dfrac{2\sqrt2}{3}

11

2\sqrt2

Answer: B
Concepts:pigeonhole principleextremal argumentsquare (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 12.\frac{1}{2}. Two of the five points lie in the same small square, so their distance is at most its diagonal, 22.\frac{\sqrt2}{2}. This bound is attainable by placing points at the four corners and the center: the shortest distance is then 22.\frac{\sqrt2}{2}. Hence the least guaranteed value is 22.\frac{\sqrt2}{2}.

Therefore, the correct answer is B.

← Problem 38#38
Full Exam

Problem 39 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12