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

m (→‎Solution 3)
(→‎Solution 3)
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  <math>7^{\text{th}}</math> move: <math>(-24, 24)</math>
 
  <math>7^{\text{th}}</math> move: <math>(-24, 24)</math>
  
−
In the <math>3^{rd}</math> move Andy lands on <math>(-22,22)</math>, in the <math>7^{th}</math> move, Andy lands on <math>(-24, 24)</math>.
+
In the <math>3^{\text{rd}}</math> move Andy lands on <math>(-22,22)</math>, in the <math>7^{\text{th}}</math> move, Andy lands on <math>(-24, 24)</math>.
  
−
There is a pattern, for every <math>4</math> moves (starting from the <math>3^{rd}</math> move), Andy will arrive on a coordinate in the form of <math>(-2n, 2n)</math>. From this we can deduce:
+
There is a pattern, for every <math>4</math> moves (starting from the <math>3^{\text{rd}}</math> move), Andy will arrive on a coordinate in the form of <math>(-2n, 2n)</math>. From this we can deduce:
  
 
  <math>(1) \ 3^{\text{rd}}</math> move: <math>(-22,22)</math>
 
  <math>(1) \ 3^{\text{rd}}</math> move: <math>(-22,22)</math>
Line 41: Line 41:
 
  <math>(n) \ (4n-1)^{\text{th}}</math> move: <math>(-20-2n, 20+2n)</math>
 
  <math>(n) \ (4n-1)^{\text{th}}</math> move: <math>(-20-2n, 20+2n)</math>
  
−
We have <math>2019 = 4 \cdot 505 -1</math>, for which <math>n = 505</math> and <math>20 + 2 \cdot 505 = 1030</math>. So, on the <math>2019^{th}</math> move Andy is at <math>(-1030, 1030)</math>.
+
We have <math>2019 = 4 \cdot 505 -1</math>, for which <math>n = 505</math> and <math>20 + 2 \cdot 505 = 1030</math>. So, on the <math>2019^{\text{th}}</math> move Andy is at <math>(-1030, 1030)</math>.
  
−
Because the problem asks for the <math>2020^{th}</math> move, <math>1030-2020=-990</math>, on the <math>2020^{th}</math> move, Andy will be on <math>\boxed{\textbf{(B)}\ (-1030, -990)}</math>.
+
Because the problem asks for the <math>2020^{\text{th}}</math> move, <math>1030-2020=-990</math>, on the <math>2020^{\text{th}}</math> move, Andy will be on <math>\boxed{\textbf{(B)}\ (-1030, -990)}</math>.
  
 
~isabelchen
 
~isabelchen

Revision as of 18:16, 24 December 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 $90^{\circ}$ 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

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

Solution 3

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

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

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