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

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 2 (Detailed)

Andy makes a total of $2020$ moves: $1010$ horizontal (left or right) and $1010$ vertical (up or down).

The $x$-coordinate of Andy's final position is $$-20+\overbrace{\underbrace{1-3}_{-2}+\underbrace{5-7}_{-2}+\underbrace{9-11}_{-2}+\cdots+\underbrace{2017-2019}_{-2}}^{\text{1010 numbers, 505 pairs}}=-20-2\cdot505=-1030.$$ The $y$-coordinate of Andy's final position is $$20+\overbrace{\underbrace{2-4}_{-2}+\underbrace{6-8}_{-2}+\underbrace{10-12}_{-2}+\cdots+\underbrace{2018-2020}_{-2}}^{\text{1010 numbers, 505 pairs}}=20-2\cdot505=-990.$$ Together, we have $(x,y)=\boxed{\textbf{(B)}\ (-1030, -990)}.$

~IceMatrix

## Similar Problem

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

## See Also

 2020 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 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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