Difference between revisions of "2004 AMC 8 Problems/Problem 23"

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[[Image:AMC8200423.gif]]
 
[[Image:AMC8200423.gif]]
  
==Solution 1==
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==Solution==
 
For her distance to be represented as a constant horizontal line, Tess would have to be running in a circular shape with her home as the center. Since she is running around a rectangle, this is not possible, ruling out <math>B</math> and <math>E</math> with straight lines. Because <math>JL</math> is the diagonal of the rectangle, and <math>L</math> is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in <math>A</math> is at the end, and <math>C</math> has two maximums, ruling both out. Thus the answer is <math>\boxed{\textbf{(D)}}</math>.
 
For her distance to be represented as a constant horizontal line, Tess would have to be running in a circular shape with her home as the center. Since she is running around a rectangle, this is not possible, ruling out <math>B</math> and <math>E</math> with straight lines. Because <math>JL</math> is the diagonal of the rectangle, and <math>L</math> is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in <math>A</math> is at the end, and <math>C</math> has two maximums, ruling both out. Thus the answer is <math>\boxed{\textbf{(D)}}</math>.
  

Revision as of 21:01, 12 October 2020

Problem

Tess runs counterclockwise around rectangular block $JKLM$. She lives at corner $J$. Which graph could represent her straight-line distance from home?

[asy] unitsize(5mm); pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2);  draw(J--K--L--M--cycle); label("$J$",J,NW); label("$K$",K,SW); label("$L$",L,SE); label("$M$",M,NE); [/asy]

AMC8200423.gif

Solution

For her distance to be represented as a constant horizontal line, Tess would have to be running in a circular shape with her home as the center. Since she is running around a rectangle, this is not possible, ruling out $B$ and $E$ with straight lines. Because $JL$ is the diagonal of the rectangle, and $L$ is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in $A$ is at the end, and $C$ has two maximums, ruling both out. Thus the answer is $\boxed{\textbf{(D)}}$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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

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