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{\textbf{(C)}\ 5}</math>.
 
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>.
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*Note that from any point in the square, the average distance from one vertex to the other corner is half of the side length of the square.
  
 
==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 14:25, 29 October 2017

Problem

A lemming sits at a corner of a square with side length $10$ meters. The lemming runs $6.2$ meters along a diagonal toward the opposite corner. It stops, makes a $90^{\circ}$ right turn and runs $2$ 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?

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6.2 \qquad \textbf{(E)}\ 7$

Solution

Algebraic: The shortest segments would be perpendicular to the square. The lemming went $x$ meters horizontally and $y$ meters vertically. No matter how much it went, the lemming would have been $x$ and $y$ meters from the sides and $10-x$ and $10-y$ meters from the remaining two. To find the average, add the lengths of the four segments and divide by four: $\frac {\cancel{x}+10-\cancel{x}+\cancel{y}+10-\cancel{y}}{4} = 5$ $\boxed{\textbf{(C)}\ 5}$.

  • Note that from any point in the square, the average distance from one vertex to the other corner is half of the side length of the square.

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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