Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 8 Problems/Problem 24"

(→‎Solution)
(→‎Video Solution 2)
 
(8 intermediate revisions by 5 users not shown)
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 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 Island <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 Island <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 <math>B</math> to <math>C</math>, the distance between the ship and Island <math>X</math> will first decrease until it reaches the point <math>Y</math> so that <math>\overline{XY}</math> is perpendicular to <math>\overline{BC}</math>, and then increase afterwards. Hence the answer choice that fits them all is <math>\boxed{\text{(B)}}</math>.
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=ibhy_qcQXiw  ~David
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2003|num-b=23|num-a=25}}
 
{{AMC8 box|year=2003|num-b=23|num-a=25}}
 +
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 09:47, 14 June 2024

Problem

A ship travels from point $A$ to point $B$ along a semicircular path, centered at Island $X$. Then it travels along a straight path from $B$ to $C$. Which of these graphs best shows the ship's distance from Island $X$ 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]

2003amc8prob24ans.png

Solution

The distance from Island $\text{X}$ to any point on the semicircle will always be constant. On the graph, this will represent a straight line. The distance between Island $\text{X}$ and line $\text{BC}$ will not be constant though. We can easily prove that the distance between $\text{X}$ and line $\text{BC}$ will represent a curve. As the ship travels from $B$ to $C$, the distance between the ship and Island $X$ will first decrease until it reaches the point $Y$ so that $\overline{XY}$ is perpendicular to $\overline{BC}$, and then increase afterwards. Hence the answer choice that fits them all is $\boxed{\text{(B)}}$.

Video Solution

https://www.youtube.com/watch?v=ibhy_qcQXiw ~David

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems/Problem_24&oldid=220531"