Difference between revisions of "2020 AMC 10B Problems/Problem 13"

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<math>\textbf{(A)}\ (-1030, -994)\qquad\textbf{(B)}\ (-1030, -990)\qquad\textbf{(C)}\ (-1026, -994)\qquad\textbf{(D)}\ (-1026, -990)\qquad\textbf{(E)}\ (-1022, -994)</math>
 
<math>\textbf{(A)}\ (-1030, -994)\qquad\textbf{(B)}\ (-1030, -990)\qquad\textbf{(C)}\ (-1026, -994)\qquad\textbf{(D)}\ (-1026, -990)\qquad\textbf{(E)}\ (-1022, -994)</math>
  
== Solutions ==
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== Solution 1 ==
=== Solution 1 ===
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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
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) } \text{(-1030,-990)}}</math> ~happykeeper
 
  
=== Video Solution ===
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== Solution 2 (Detailed) ==
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Andy makes a total of <math>2020</math> moves: <math>1010</math> horizontal (left or right) and <math>1010</math> vertical (up or down).
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The <math>x</math>-coordinate of Andy's final position is <cmath>-20+\overbrace{\underbrace{1-3}_{-2}+\underbrace{5-7}_{-2}+\underbrace{9-11}_{-2}+\cdots+\underbrace{2017-2019}_{-2}}^{\text{1010 numbers, 505 pairs}}=-20-2\cdot505=-1030.</cmath>
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The <math>y</math>-coordinate of Andy's final position is <cmath>20+\overbrace{\underbrace{2-4}_{-2}+\underbrace{6-8}_{-2}+\underbrace{10-12}_{-2}+\cdots+\underbrace{2018-2020}_{-2}}^{\text{1010 numbers, 505 pairs}}=20-2\cdot505=-990.</cmath>
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Together, we have <math>(x,y)=\boxed{\textbf{(B)}\ (-1030, -990)}.</math>
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== Video Solution ==
 
https://youtu.be/t6yjfKXpwDs
 
https://youtu.be/t6yjfKXpwDs
  

Revision as of 11:54, 15 May 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

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

Solution 2 (Detailed)

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

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

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 (ProblemsAnswer KeyResources)
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

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