# 2020 AMC 10B Problems/Problem 13

## 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 $\boxed{\textbf{(B)}\ (-1030, -990)}$.

~IceMatrix

## Similar Problem

2015 AMC 10B Problem 24 https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_24

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