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

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Tess runs counterclockwise around rectangular block <math>JKLM</math>. She lives at corner <math>J</math>. Which graph could represent her straight-line distance from home?
 
Tess runs counterclockwise around rectangular block <math>JKLM</math>. She lives at corner <math>J</math>. Which graph could represent her straight-line distance from home?
  
[[Image:AMC8200423.gif]]
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<asy>
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unitsize(5mm);
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pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2);
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draw(J--K--L--M--cycle);
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label("$J$",J,NW);
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label("$K$",K,SW);
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label("$L$",L,SE);
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label("$M$",M,NE);
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</asy>
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<math>\textbf{(A)}</math>
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<asy>
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size(80);defaultpen(linewidth(0.8));
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//A
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draw((16,0)--origin--(0,16));
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draw(origin--(15,15));
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label("time", (8,0), S);
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label(rotate(90)*"distance", (0,8), W);
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</asy>
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<math>\textbf{(B)}</math>
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<asy>
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size(80);defaultpen(linewidth(0.8));
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//B
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draw((16,0)--origin--(0,16));
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draw((0,6)--(1,6)--(1,12)--(2,12)--(2,11)--(3,11)--(3,1)--(12,1)--(12,0));
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label("time", (8,0), S);
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label(rotate(90)*"distance", (0,8), W);
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</asy>
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<math>\textbf{(C)}</math>
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<asy>
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size(80);defaultpen(linewidth(0.8));
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//C
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draw((16,0)--origin--(0,16));
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draw(origin--(2.7,8)--(3,9)^^(11,9)--(14,0));
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draw(Arc((4,9), 1, 0, 180));
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draw(Arc((10,9), 1, 0, 180));
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draw(Arc((7,9), 2, 180,360));
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label("time", (8,0), S);
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label(rotate(90)*"distance", (0,8), W);
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</asy>
 +
 
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<math>\textbf{(D)}</math>
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<asy>
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size(80);defaultpen(linewidth(0.8));
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//D
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draw((16,0)--origin--(0,16));
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draw(origin--(2,6)--(7,14)--(10,12)--(14,0));
 +
label("time", (8,0), S);
 +
label(rotate(90)*"distance", (0,8), W);
 +
</asy>
 +
 
 +
<math>\textbf{(E)}</math>
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<asy>
 +
size(80);defaultpen(linewidth(0.8));
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//E
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draw((16,0)--origin--(0,16));
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draw(origin--(3,6)--(7,6)--(10,12)--(14,12));
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label("time", (8,0), S);
 +
label(rotate(90)*"distance", (0,8), W);
 +
</asy>
  
 
==Solution==
 
==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, rulling 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>.
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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>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2004|num-b=22|num-a=24}}
 
{{AMC8 box|year=2004|num-b=22|num-a=24}}
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{{MAA Notice}}

Latest revision as of 17:26, 25 July 2021

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]

$\textbf{(A)}$ [asy] size(80);defaultpen(linewidth(0.8)); //A draw((16,0)--origin--(0,16)); draw(origin--(15,15)); label("time", (8,0), S); label(rotate(90)*"distance", (0,8), W); [/asy]

$\textbf{(B)}$ [asy] size(80);defaultpen(linewidth(0.8)); //B draw((16,0)--origin--(0,16)); draw((0,6)--(1,6)--(1,12)--(2,12)--(2,11)--(3,11)--(3,1)--(12,1)--(12,0)); label("time", (8,0), S); label(rotate(90)*"distance", (0,8), W); [/asy]

$\textbf{(C)}$ [asy] size(80);defaultpen(linewidth(0.8)); //C draw((16,0)--origin--(0,16)); draw(origin--(2.7,8)--(3,9)^^(11,9)--(14,0)); draw(Arc((4,9), 1, 0, 180)); draw(Arc((10,9), 1, 0, 180)); draw(Arc((7,9), 2, 180,360)); label("time", (8,0), S); label(rotate(90)*"distance", (0,8), W); [/asy]

$\textbf{(D)}$ [asy] size(80);defaultpen(linewidth(0.8)); //D draw((16,0)--origin--(0,16)); draw(origin--(2,6)--(7,14)--(10,12)--(14,0)); label("time", (8,0), S); label(rotate(90)*"distance", (0,8), W); [/asy]

$\textbf{(E)}$ [asy] size(80);defaultpen(linewidth(0.8)); //E draw((16,0)--origin--(0,16)); draw(origin--(3,6)--(7,6)--(10,12)--(14,12)); label("time", (8,0), S); label(rotate(90)*"distance", (0,8), W); [/asy]

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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