Difference between revisions of "2003 AMC 8 Problems/Problem 24"
Hanningyan (talk | contribs) (→Solution) |
|||
Line 18: | Line 18: | ||
==Solution== | ==Solution== | ||
− | The distance from <math>\text{X}</math> to any point on the semicircle will always be constant. On the graph, this will represent a straight line. The distance between <math>\text{X}</math> and line <math>\text{BC}</math> will not be constant though. We can easily prove that the distance between <math>\text{X}</math> and line <math>\text{BC}</math> will represent a | + | The distance from <math>\text{X}</math> to any point on the semicircle will always be constant. On the graph, this will represent a straight line. The distance between <math>\text{X}</math> and line <math>\text{BC}</math> will not be constant though. We can easily prove that the distance between <math>\text{X}</math> and line <math>\text{BC}</math> will represent a curve. As the ship travels from B to C, the distance between the ship and X will first decrease until it reaches the point Y so that XY is perpendicular to BC, and then increase afterwards. Hence the answer choice that fits them all is <math>\boxed{\text{(B)}}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2003|num-b=23|num-a=25}} | {{AMC8 box|year=2003|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:26, 9 October 2020
Problem
A ship travels from point to point along a semicircular path, centered at Island . Then it travels along a straight path from to . Which of these graphs best shows the ship's distance from Island as it moves along its course?
Solution
The distance from to any point on the semicircle will always be constant. On the graph, this will represent a straight line. The distance between and line will not be constant though. We can easily prove that the distance between and line will represent a curve. As the ship travels from B to C, the distance between the ship and X will first decrease until it reaches the point Y so that XY is perpendicular to BC, and then increase afterwards. Hence the answer choice that fits them all is .
See Also
2003 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.