Difference between revisions of "2007 AMC 8 Problems/Problem 22"
Line 15: | Line 15: | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=21|num-a=23}} | {{AMC8 box|year=2007|num-b=21|num-a=23}} | ||
+ | {{MAA Notice}} |
Revision as of 00:29, 5 July 2013
Problem
A lemming sits at a corner of a square with side length meters. The lemming runs meters along a diagonal toward the opposite corner. It stops, makes a right turn and runs more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
Solution
Algebraic: The shortest segments would be perpendicular to the square. The lemming went meters horizontally and meters vertically. No matter how much it went, the lemming would have been and meters from the sides and and meters from the remaining two. To find the average, add the lengths of the four segments and divide by four: .
See Also
2007 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.