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 1 ==
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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} = </math> <math>\boxed{\textbf{(C)}\ 5}</math>.
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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.
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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.
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==Video Solution==
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https://www.youtube.com/watch?v=0X3-nEEXHGo
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==Video Solution by WhyMath==
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https://youtu.be/0J8-ufx4xf0
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==See Also==
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{{AMC8 box|year=2007|num-b=21|num-a=23}}
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{{MAA Notice}}

Latest revision as of 13:16, 29 October 2024

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 1

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

  • 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 $10$, and the sum of its distance to the up edge and down edge is also equal to $10$. Thus, the answer is $\boxed{\textbf{(C)}\ 5}$, and the moving progress is misguide at all.

Video Solution

https://www.youtube.com/watch?v=0X3-nEEXHGo

Video Solution by WhyMath

https://youtu.be/0J8-ufx4xf0

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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