Difference between revisions of "1961 AHSME Problems/Problem 39"
(Added solution) |
Rockmanex3 (talk | contribs) m |
||
| Line 1: | Line 1: | ||
| − | == Problem | + | == Problem == |
Any five points are taken inside or on a square with side length <math>1</math>. Let a be the smallest possible number with the | Any five points are taken inside or on a square with side length <math>1</math>. Let a be the smallest possible number with the | ||
| Line 15: | Line 15: | ||
Hence our answer is <math>\fbox{B}</math>. | Hence our answer is <math>\fbox{B}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1961|num-b=38|after=num-a=40}} | ||
| + | |||
| + | [[Category:Intermediate Geometry Problems]] | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 13:55, 15 July 2018
Problem
Any five points are taken inside or on a square with side length 
. 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 
. Then is:
Solution
Partition the unit square into four smaller squares of sidelength 
. Each of the five points lies in one of these squares, and so by the Pigeonhole Principle, there exists two points in the same 
square - the maximum possible distance between them being 
by Pythagoras.
Hence our answer is 
.
See Also
| 1961 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by num-a=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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
