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 "2010 AMC 10A Problems/Problem 7"

m
(→‎Solution)
Line 17: Line 17:
  
  
−
Crystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from traveling North for one mile, and her current destination is <math>\sqrt{2}</math> miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to <math>\sqrt{((\sqrt{2})^2+1^2)}</math>, which is equal to <math>\sqrt{3}</math>. The answer is <math>\boxed{C}</math>
+
Crystal first runs north for one mile. Changing directions, she runs northeast for another mile. The angle difference between north and northeast is 45 degrees. She then switches directions to southeast, meaning a 90 degree angle change. The distance now from traveling north for one mile, and her current destination is <math>\sqrt{2}</math> miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to <math>\sqrt{((\sqrt{2})^2+1^2)}</math>, which is equal to <math>\sqrt{3}</math>. The answer is <math>\boxed{C}</math>
  
  

Revision as of 08:45, 4 July 2021

Problem 7

Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?

$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ \sqrt{2} \qquad \mathrm{(C)}\ \sqrt{3} \qquad \mathrm{(D)}\ 2 \qquad \mathrm{(E)}\ 2\sqrt{2}$

Solution

Crystal first runs north for one mile. Changing directions, she runs northeast for another mile. The angle difference between north and northeast is 45 degrees. She then switches directions to southeast, meaning a 90 degree angle change. The distance now from traveling north for one mile, and her current destination is $\sqrt{2}$ miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to $\sqrt{((\sqrt{2})^2+1^2)}$, which is equal to $\sqrt{3}$. The answer is $\boxed{C}$


[asy] import olympiad; draw((0,0)--(0,1)); draw((0,1)--(0,1.7071067811865476), dotted); draw((0,1)--(0.7071067811865476, 1.7071067811865476)); draw((0.7071067811865476, 1.7071067811865476)--(1.4142135623730951,1)); draw(anglemark((0.7071067811865476, 1.7071067811865476),(0,1),(0,1.7071067811865476))); label("$45^{\circ}$", (0,1.25), NE); draw((0, 1)--(1.4142135623730951,1), dotted); draw((1.4142135623730951,1)--(0,0), green); [/asy]

Video Solution

https://youtu.be/P7rGLXp_6es?t=132

~IceMatrix

See Also

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 AMC 10 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=2010_AMC_10A_Problems/Problem_7&oldid=157348"