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

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 $\boxed{\textbf{(D)}}$ .

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

Latest revision as of 16:26, 25 July 2021

Problem

Solution

See Also

@@ Line 2: / Line 2: @@
 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]]
+<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>
+<math>\textbf{(A)}</math>
+<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>
+<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
+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>
+<math>\textbf{(D)}</math>
+<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>
+<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==
-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>.
+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==
 {{AMC8 box|year=2004|num-b=22|num-a=24}}
+{{MAA Notice}}