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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| + | ==Solution 2== | ||
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| + | For any point in the square, the sum of its distance to the left edge and right edge is equal to <math>10</math>, and the sum of its distance to the up edge and down edge is also equal to <math>10</math>. Thus, the answer is <math>\boxed{\textbf{(C)}\ 5}</math>, and the moving progress is misguide at all. | ||
==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=0X3-nEEXHGo | https://www.youtube.com/watch?v=0X3-nEEXHGo | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/0J8-ufx4xf0 | ||
==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}} | ||
Latest revision as of 13:16, 29 October 2024
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 1
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.
Solution 2
For any point in the square, the sum of its distance to the left edge and right edge is equal to , and the sum of its distance to the up edge and down edge is also equal to . Thus, the answer is , and the moving progress is misguide at all.
Video Solution
https://www.youtube.com/watch?v=0X3-nEEXHGo
Video Solution by WhyMath
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. 
