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

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</asy>
 
</asy>
  
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[[Image:AMC8200423.gif]]
+
<math>\textbf{(A)}</math>
 +
<asy>
 +
size(80);defaultpen(linewidth(0.8));
 +
//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);
 +
</asy>
 +
 
 +
<math>\textbf{(B)}</math>
 +
<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>
 +
 
 +
<math>\textbf{(C)}</math>
 +
<asy>
 +
size(80);defaultpen(linewidth(0.8));
 +
//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));
 +
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>
 +
 
 +
<math>\textbf{(D)}</math>
 +
<asy>
 +
size(80);defaultpen(linewidth(0.8));
 +
//D
 +
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>
 +
<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==
 
==Solution==

Latest revision as of 16:26, 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 (Problems • Answer Key • Resources)
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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