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

(Solution 2)
m (Solution 3 (Find the Pattern))
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~happykeeper
 
~happykeeper
  
== Solution 3 (Find the Pattern)==
+
== Solution 3 ==
  
 
Let's first mark the first few points Andy will arrive at:
 
Let's first mark the first few points Andy will arrive at:
  
  Starting Point (Zero move): <math>(-20,20)</math>
+
  Starting Point (<math>0^{\text{th}}</math> move): <math>(-20,20)</math>
  <math>1^{st}</math> move: <math>(-19,20)</math>
+
  <math>1^{\text{st}}</math> move: <math>(-19,20)</math>
  <math>2^{nd}</math> move: <math>(-19,22)</math>
+
  <math>2^{\text{nd}}</math> move: <math>(-19,22)</math>
  <math>3^{rd}</math> move: <math>(-22,22)</math>
+
  <math>3^{\text{rd}}</math> move: <math>(-22,22)</math>
  <math>4^{th}</math> move: <math>(-22,18)</math>
+
  <math>4^{\text{th}}</math> move: <math>(-22,18)</math>
  <math>5^{th}</math> move: <math>(-17,18)</math>
+
  <math>5^{\text{th}}</math> move: <math>(-17,18)</math>
  <math>6^{th}</math> move: <math>(-17,24)</math>
+
  <math>6^{\text{th}}</math> move: <math>(-17,24)</math>
  <math>7^{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^{rd}</math> move Andy lands on <math>(-22,22)</math>, in the <math>7^{th}</math> move, Andy lands on <math>(-24, 24)</math>.
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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^{rd}</math> move), Andy will arrive on a coordinate in the form of <math>(-2n, 2n)</math>. From this we can deduce:
  
  <math>(1)</math> <math>3^{rd}</math> move: <math>(-22,22)</math>
+
  <math>(1)</math> <math>3^{\text{rd}}</math> move: <math>(-22,22)</math>
  <math>(2)</math> <math>7^{th}</math> move: <math>(-24, 24)</math>
+
  <math>(2)</math> <math>7^{\text{th}}</math> move: <math>(-24, 24)</math>
  <math>(3)</math> <math>11^{th}</math> move: <math>(-26, 26)</math>
+
  <math>(3)</math> <math>11^{\text{th}}</math> move: <math>(-26, 26)</math>
  <math>(4)</math> <math>15^{th}</math> move: <math>(-28, 28)</math>
+
  <math>(4)</math> <math>15^{\text{th}}</math> move: <math>(-28, 28)</math>
  <math>(n)</math> <math>(4n-1)^{th}</math> move: <math>(-20-2n, 20+2n)</math>
+
  <math>(n)</math> <math>(4n-1)^{\text{th}}</math> move: <math>(-20-2n, 20+2n)</math>
  
 
<math>2019 = 4 \cdot 505 -1</math>, <math>n = 505</math>, <math>20 + 2 \cdot 505 = 1030</math>, So on the <math>2019^{th}</math> move Andy is at <math>(-1030, 1030)</math>
 
<math>2019 = 4 \cdot 505 -1</math>, <math>n = 505</math>, <math>20 + 2 \cdot 505 = 1030</math>, So on the <math>2019^{th}</math> move Andy is at <math>(-1030, 1030)</math>

Revision as of 16:57, 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 $3^{rd}$ move Andy lands on $(-22,22)$, in the $7^{th}$ move, Andy lands on $(-24, 24)$.

There is a pattern, for every $4$ moves (starting from the $3^{rd}$ 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: $(-20-2n, 20+2n)$

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

Because the problem asks for the $2020^{th}$ move, $1030-2020=-990$, on the $2020^{th}$ 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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