Difference between revisions of "2007 AMC 8 Problems/Problem 22"

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

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 9: / Line 9: @@
 \textbf{(E)}\ 7</math>
-==Solution==
+==Solution 1 ==
-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>.
+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>.
-*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.
+*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 <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.
+==Video Solution==
+https://www.youtube.com/watch?v=0X3-nEEXHGo
+==Video Solution by WhyMath==
+https://youtu.be/0J8-ufx4xf0
 ==See Also==
 {{AMC8 box|year=2007|num-b=21|num-a=23}}
 {{MAA Notice}}

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