Difference between revisions of "2020 AMC 10B Problems/Problem 13"
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<math>\textbf{(A)}\ (-1030, -994)\qquad\textbf{(B)}\ (-1030, -990)\qquad\textbf{(C)}\ (-1026, -994)\qquad\textbf{(D)}\ (-1026, -990)\qquad\textbf{(E)}\ (-1022, -994)</math> | <math>\textbf{(A)}\ (-1030, -994)\qquad\textbf{(B)}\ (-1030, -990)\qquad\textbf{(C)}\ (-1026, -994)\qquad\textbf{(D)}\ (-1026, -990)\qquad\textbf{(E)}\ (-1022, -994)</math> | ||
| − | + | == 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 <math>\boxed{\textbf{(B)}\ (-1030, -990)}.</math> ~happykeeper | |
| − | 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 <math>\boxed{\textbf{(B) } \ | ||
| − | === Video Solution | + | == Solution 2 (Detailed) == |
| + | Andy makes a total of <math>2020</math> moves: <math>1010</math> horizontal (left or right) and <math>1010</math> vertical (up or down). | ||
| + | |||
| + | The <math>x</math>-coordinate of Andy's final position is <cmath>-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.</cmath> | ||
| + | The <math>y</math>-coordinate of Andy's final position is <cmath>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.</cmath> | ||
| + | Together, we have <math>(x,y)=\boxed{\textbf{(B)}\ (-1030, -990)}.</math> | ||
| + | |||
| + | == Video Solution == | ||
https://youtu.be/t6yjfKXpwDs | https://youtu.be/t6yjfKXpwDs | ||
Revision as of 10:54, 15 May 2021
Contents
Problem
Andy the Ant lives on a coordinate plane and is currently at 
facing east (that is, in the positive 
-direction). Andy moves 
unit and then turns 
degrees left. From there, Andy moves 
units (north) and then turns degrees left. He then moves 
units (west) and again turns degrees left. Andy continues his progress, increasing his distance each time by unit and always turning left. What is the location of the point at which Andy makes the 
th left turn?
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 
~happykeeper
Solution 2 (Detailed)
Andy makes a total of moves: 
horizontal (left or right) and vertical (up or down).
The -coordinate of Andy's final position is ![\[-20+\overbrace{\underbrace{1-3}_{-2}+\underbrace{5-7}_{-2}+\underbrace{9-11}_{-2}+\cdots+\underbrace{2017-2019}_{-2}}^{\text{1010 numbers, 505 pairs}}=-20-2\cdot505=-1030.\]](http://latex.artofproblemsolving.com/7/2/8/7286600b24966168f7e21487ddcee568ec5f52e0.png)
The 
-coordinate of Andy's final position is ![\[20+\overbrace{\underbrace{2-4}_{-2}+\underbrace{6-8}_{-2}+\underbrace{10-12}_{-2}+\cdots+\underbrace{2018-2020}_{-2}}^{\text{1010 numbers, 505 pairs}}=20-2\cdot505=-990.\]](http://latex.artofproblemsolving.com/1/d/1/1d1480b85ee8c9a70b807255a7d834007668751a.png)
Together, we have 
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
