Difference between revisions of "2004 AMC 8 Problems/Problem 23"
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==Solution 1== | ==Solution 1== | ||
| − | 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, | + | 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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:07, 10 November 2017
Problem
Tess runs counterclockwise around rectangular block 
. She lives at corner 
. 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]](http://latex.artofproblemsolving.com/c/f/a/cfa3151064b41051d0fd988b63693968a5f5531e.png)
Solution 1
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 
and 
with straight lines. Because 
is the diagonal of the rectangle, and 
is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in 
is at the end, and 
has two maximums, ruling both out. Thus the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
