2007 AMC 8 Problems/Problem 22: Difference between revisions
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5 | Algebraic: The shortest segments would be perpendicular to the square. The lemming went <math>x</math> meters horizontally and <math>y</math> meters vertically. No matter how much it went, the lemming would have been <math>x</math> and <math>y</math> meters from the sides and <math>10-x</math> and <math>10-y</math> meters from the remaining two. To find the average, add the lengths of the four segments and divide by four: <math>\frac {x+10-x+y+10-y}{4} = 5</math> | ||
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Solution: 5 '''(C)''' | |||
Revision as of 19:55, 21 November 2011
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: 
Solution: 5 (C)
