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

m (Solution 1)
m (minor edit)
Line 1: Line 1:
==Problem==
+
== Problem ==
 
Andy the Ant lives on a coordinate plane and is currently at <math>(-20, 20)</math> facing east (that is, in the positive <math>x</math>-direction). Andy moves <math>1</math> unit and then turns <math>90^{\circ}</math> degrees left. From there, Andy moves <math>2</math> units (north) and then turns <math>90^{\circ}</math> degrees left. He then moves <math>3</math> units (west) and again turns <math>90^{\circ}</math> degrees left. Andy continues his progress, increasing his distance each time by <math>1</math> unit and always turning left. What is the location of the point at which Andy makes the <math>2020</math>th left turn?
 
Andy the Ant lives on a coordinate plane and is currently at <math>(-20, 20)</math> facing east (that is, in the positive <math>x</math>-direction). Andy moves <math>1</math> unit and then turns <math>90^{\circ}</math> degrees left. From there, Andy moves <math>2</math> units (north) and then turns <math>90^{\circ}</math> degrees left. He then moves <math>3</math> units (west) and again turns <math>90^{\circ}</math> degrees left. Andy continues his progress, increasing his distance each time by <math>1</math> unit and always turning left. What is the location of the point at which Andy makes the <math>2020</math>th left turn?
 +
 
<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>
  
==Solution 1==
+
== Solutions ==
 +
=== 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) } \text{(-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
  
==See Also==
+
=== Video Solution ===
2015 AMC 10B Problem 24
 
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_24
 
 
 
==Video Solution==
 
 
https://youtu.be/t6yjfKXpwDs
 
https://youtu.be/t6yjfKXpwDs
  
 
~IceMatrix
 
~IceMatrix
  
==See Also==
+
== Similar Problem ==
 +
2015 AMC 10B Problem 24
 +
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_24
  
 +
== See Also ==
 
{{AMC10 box|year=2020|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2020|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:37, 19 October 2020

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?

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

Solutions

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 $\boxed{\textbf{(B) } \text{(-1030,-990)}}$ ~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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png