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 "2020 AMC 10B Problems/Problem 13"

m (→‎Solution 2 (Detailed))
m (Prioritized the detailed solution.)
Line 5: Line 5:
  
 
== Solution 1 ==
 
== Solution 1 ==
−
You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you the answer of <math>\boxed{\textbf{(B)}\ (-1030, -990)}.</math> ~happykeeper
 
−
 
−
== Solution 2 (Detailed) ==
 
 
Andy makes a total of <math>2020</math> moves: <math>1010</math> horizontal (<math>505</math> left and <math>505</math> right) and <math>1010</math> vertical (<math>505</math> up and <math>505</math> down).
 
Andy makes a total of <math>2020</math> moves: <math>1010</math> horizontal (<math>505</math> left and <math>505</math> right) and <math>1010</math> vertical (<math>505</math> up and <math>505</math> down).
  
Line 15: Line 12:
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
== Solution 2 ==
 +
You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you the answer of <math>\boxed{\textbf{(B)}\ (-1030, -990)}.</math>
 +
 +
~happykeeper
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 15:38, 8 September 2021

Problem

Andy the Ant lives on a coordinate plane and is currently at $(-20, 20)$ facing east (that is, in the positive $x$-direction). Andy moves $1$ unit and then turns $90^{\circ}$ degrees left. From there, Andy moves $2$ units (north) and then turns $90^{\circ}$ degrees left. He then moves $3$ units (west) and again turns $90^{\circ}$ degrees left. Andy continues his progress, increasing his distance each time by $1$ unit and always turning left. What is the location of the point at which Andy makes the $2020$th left turn?

$\textbf{(A)}\ (-1030, -994)\qquad\textbf{(B)}\ (-1030, -990)\qquad\textbf{(C)}\ (-1026, -994)\qquad\textbf{(D)}\ (-1026, -990)\qquad\textbf{(E)}\ (-1022, -994)$

Solution 1

Andy makes a total of $2020$ moves: $1010$ horizontal ($505$ left and $505$ right) and $1010$ vertical ($505$ up and $505$ down).

The $x$-coordinate of Andy's final position is \[-20+\overbrace{\underbrace{1-3}_{-2}+\underbrace{5-7}_{-2}+\underbrace{9-11}_{-2}+\cdots+\underbrace{2017-2019}_{-2}}^{\text{1010 terms, 505 pairs}}=-20-2\cdot505=-1030.\] The $y$-coordinate of Andy's final position is \[20+\overbrace{\underbrace{2-4}_{-2}+\underbrace{6-8}_{-2}+\underbrace{10-12}_{-2}+\cdots+\underbrace{2018-2020}_{-2}}^{\text{1010 terms, 505 pairs}}=20-2\cdot505=-990.\] Together, we have $(x,y)=\boxed{\textbf{(B)}\ (-1030, -990)}.$

~MRENTHUSIASM

Solution 2

You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you the answer of $\boxed{\textbf{(B)}\ (-1030, -990)}.$

~happykeeper

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

Similar Problem

2015 AMC 10B Problem 24 https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_24

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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=2020_AMC_10B_Problems/Problem_13&oldid=161893"