Difference between revisions of "2003 AMC 8 Problems/Problem 24"
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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 semicircle (prove this by dividing <math>\triangle{XCB}</math> into two congruent triangles using the perpendicular bisector from vertex <math>\text{X}</math>). Since the point on line <math>\text{BC}</math> and the perpendicular bisector from vertex <math>\text{X}</math> is the shortest distance between <math>\text{X}</math> and <math>\text{BC}</math> as well as the midpoint of line <math>\text{BC}</math> it will represent the shortest point on the semicircle in the graph as well as the midpoint of the semicircle. Using the information found, the answer choice that fits them all is <math>\boxed{\text{(B)}}</math>. | 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 semicircle (prove this by dividing <math>\triangle{XCB}</math> into two congruent triangles using the perpendicular bisector from vertex <math>\text{X}</math>). Since the point on line <math>\text{BC}</math> and the perpendicular bisector from vertex <math>\text{X}</math> is the shortest distance between <math>\text{X}</math> and <math>\text{BC}</math> as well as the midpoint of line <math>\text{BC}</math> it will represent the shortest point on the semicircle in the graph as well as the midpoint of the semicircle. Using the information found, the answer choice that fits them all is <math>\boxed{\text{(B)}}</math>. | ||
| + | ==See Also== | ||
{{AMC8 box|year=2003|num-b=23|num-a=25}} | {{AMC8 box|year=2003|num-b=23|num-a=25}} | ||
Revision as of 03:20, 24 December 2012
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?
![[asy]size(150); pair X=origin, A=(-5,0), B=(5,0), C=(0,5); draw(Arc(X, 5, 180, 360)^^B--C); dot(X); label("$X$", X, NE); label("$C$", C, N); label("$B$", B, E); label("$A$", A, W); [/asy]](http://latex.artofproblemsolving.com/9/1/9/91933f8be27c24174af480d964c31782da9f3bb4.png)
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 semicircle (prove this by dividing 
into two congruent triangles using the perpendicular bisector from vertex ). Since the point on line and the perpendicular bisector from vertex is the shortest distance between and as well as the midpoint of line it will represent the shortest point on the semicircle in the graph as well as the midpoint of the semicircle. Using the information found, 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 | ||
