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

m (Solution 2)
m (Video Solution)
(3 intermediate revisions by 2 users not shown)
Line 4: Line 4:
 
<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 ==
+
== Solution 1 (Associative Property) ==
 
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 57: Line 57:
  
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 +
==Video Solution (HOW TO CRITICALLY THINK!!!)==
 +
https://youtu.be/gbnWUskoOuU
 +
 +
~Education, the Study of Everything
 +
 +
  
 
== Video Solution ==
 
== Video Solution ==
https://youtu.be/t6yjfKXpwDs
+
https://youtu.be/t6yjfKXpwDs?t=413
 
 
~IceMatrix
 
  
 
== Similar Problem ==
 
== Similar Problem ==

Revision as of 22:48, 1 September 2024

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 $90^{\circ}$ left. From there, Andy moves $2$ units (north) and then turns $90^{\circ}$ left. He then moves $3$ units (west) and again turns $90^{\circ}$ 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 (Associative Property)

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 (Pattern)

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 3 (Pattern)

Let's first mark the first few points Andy will arrive at:

Starting Point ($0^{\text{th}}$ move): $(-20,20)$

$1^{\text{st}}$ move: $(-19,20)$

$2^{\text{nd}}$ move: $(-19,22)$

$3^{\text{rd}}$ move: $(-22,22)$

$4^{\text{th}}$ move: $(-22,18)$

$5^{\text{th}}$ move: $(-17,18)$

$6^{\text{th}}$ move: $(-17,24)$

$7^{\text{th}}$ move: $(-24, 24)$

In the $3^{\text{rd}}$ move Andy lands on $(-22,22)$, in the $7^{\text{th}}$ move, Andy lands on $(-24, 24)$.

There is a pattern, for every $4$ moves (starting from the $3^{\text{rd}}$ move), Andy will arrive on a coordinate in the form of $(-2n, 2n)$. From this we can deduce:

$(1) \ 3^{\text{rd}}$ move: $(-22,22)$

$(2) \ 7^{\text{th}}$ move: $(-24, 24)$

$(3) \ 11^{\text{th}}$ move: $(-26, 26)$

$(4) \ 15^{\text{th}}$ move: $(-28, 28)$

$(n) \ (4n-1)^{\text{th}}$ move: $(-20-2n, 20+2n)$

We have $2019 = 4 \cdot 505 -1$, for which $n = 505$ and $20 + 2 \cdot 505 = 1030$. So, on the $2019^{\text{th}}$ move Andy is at $(-1030, 1030)$.

Because the problem asks for the $2020^{\text{th}}$ move, $1030-2020=-990$, on the $2020^{\text{th}}$ move, Andy will be on $\boxed{\textbf{(B)}\ (-1030, -990)}$.

~isabelchen

Video Solution (HOW TO CRITICALLY THINK!!!)

https://youtu.be/gbnWUskoOuU

~Education, the Study of Everything


Video Solution

https://youtu.be/t6yjfKXpwDs?t=413

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