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

m (→‎Problem: Degrees is redundant. Corrected wording: https://ivyleaguecenter.files.wordpress.com/2020/01/2020-amc-10b-problems-and-answers-1.pdf)
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~happykeeper
 
~happykeeper
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== Solution 3 (Find the Pattern)==
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Let's first mark the first few points Andy will arrive at:
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Starting Point (Zero move): <math>(-20,20)</math>
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<math>1^{st}</math> move: <math>(-19,20)</math>
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<math>2^{nd}</math> move: <math>(-19,22)</math>
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<math>3^{rd}</math> move: <math>(-22,22)</math>
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<math>4^{th}</math> move: <math>(-22,18)</math>
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<math>5^{th}</math> move: <math>(-17,18)</math>
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<math>6^{th}</math> move: <math>(-17,24)</math>
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<math>7^{th}</math> move: <math>(-24, 24)</math>
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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:
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<math>(1)</math> <math>3^{rd}</math> move: <math>(-22,22)</math>
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<math>(2)</math> <math>7^{th}</math> move: <math>(-24, 24)</math>
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<math>(3)</math> <math>11^{th}</math> move: <math>(-26, 26)</math>
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<math>(4)</math> <math>15^{th}</math> move: <math>(-28, 28)</math>
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<math>(n)</math> <math>(4n-1)^{th}</math> move: <math>(-20-2n, 20+2n)</math>
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<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>
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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>
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~isabelchen
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 02:17, 19 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 (Find the Pattern)

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

Starting Point (Zero move): $(-20,20)$
$1^{st}$ move: $(-19,20)$
$2^{nd}$ move: $(-19,22)$
$3^{rd}$ move: $(-22,22)$
$4^{th}$ move: $(-22,18)$
$5^{th}$ move: $(-17,18)$
$6^{th}$ move: $(-17,24)$
$7^{th}$ 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: 

$(1)$ $3^{rd}$ move: $(-22,22)$
$(2)$ $7^{th}$ move: $(-24, 24)$
$(3)$ $11^{th}$ move: $(-26, 26)$
$(4)$ $15^{th}$ move: $(-28, 28)$
$(n)$ $(4n-1)^{th}$ 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 (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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