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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 {x+10-x+y+10-y}{4} = 5</math>
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==Problem==
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Solution: 5 '''(C)'''
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A lemming sits at a corner of a square with side length <math>10</math> meters. The lemming runs <math>6.2</math> meters along a diagonal toward the opposite corner. It stops, makes a <math>90^{\circ}</math> right turn and runs <math>2</math> 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?
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<math>\textbf{(A)}\ 2 \qquad
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\textbf{(B)}\ 4.5 \qquad
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\textbf{(C)}\ 5 \qquad
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\textbf{(D)}\ 6.2 \qquad
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\textbf{(E)}\ 7</math>
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==Solution==
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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{\text{(C)}}</math>.

Revision as of 18:10, 22 November 2011

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: $\frac {\cancel{x}+10-\cancel{x}+\cancel{y}+10-\cancel{y}}{4} = 5$ $\boxed{\text{(C)}}$.

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