Difference between revisions of "2007 AMC 8 Problems/Problem 22"
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− | 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 {\cancel{x}+10-\cancel{x}+\cancel{y}+10-\cancel{y}}{4} = 5 </math> <math>\boxed{\ | + | 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 {\cancel{x}+10-\cancel{x}+\cancel{y}+10-\cancel{y}}{4} = 5 </math> <math>\boxed{\textbf{(C)}\ 5}</math>. |
Revision as of 23:45, 9 December 2012
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:
.