Difference between revisions of "2007 AMC 8 Problems/Problem 22"
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*Note that from any point in the square, the average distance from one side to the other is half of the square's side length. | *Note that from any point in the square, the average distance from one side to the other is half of the square's side length. | ||
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| + | ==Video Solution== | ||
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| + | https://www.youtube.com/watch?v=0X3-nEEXHGo | ||
==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}} | {{MAA Notice}} | ||
Revision as of 23:24, 10 September 2022
Contents
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
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: 
.
- Note that from any point in the square, the average distance from one side to the other is half of the square's side length.
Video Solution
https://www.youtube.com/watch?v=0X3-nEEXHGo
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. 
