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 == | + | == Solution 1 (Associative Property) == |
Andy makes a total of <math>2020</math> moves: <math>1010</math> horizontal (<math>505</math> left and <math>505</math> right) and <math>1010</math> vertical (<math>505</math> up and <math>505</math> down). | Andy makes a total of <math>2020</math> moves: <math>1010</math> horizontal (<math>505</math> left and <math>505</math> right) and <math>1010</math> vertical (<math>505</math> up and <math>505</math> down). | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Solution 2 == | + | == Solution 2 (Pattern) == |
You can find that every four moves both coordinates decrease by <math>2.</math> Therefore, both coordinates need to decrease by two <math>505</math> times. You subtract, giving you the answer of <math>\boxed{\textbf{(B)}\ (-1030, -990)}.</math> | You can find that every four moves both coordinates decrease by <math>2.</math> Therefore, both coordinates need to decrease by two <math>505</math> times. You subtract, giving you the answer of <math>\boxed{\textbf{(B)}\ (-1030, -990)}.</math> | ||
~happykeeper | ~happykeeper | ||
− | == Solution 3 == | + | == Solution 3 (Pattern) == |
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 (<math>0^{\text{th}}</math> move): <math>(-20,20)</math> | |
− | |||
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− | + | <math>1^{\text{st}}</math> move: <math>(-19,20)</math> | |
− | + | <math>2^{\text{nd}}</math> move: <math>(-19,22)</math> | |
− | + | <math>3^{\text{rd}}</math> move: <math>(-22,22)</math> | |
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− | |||
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− | <math>2019 = 4 \cdot 505 -1</math>, <math>n = 505</math> | + | <math>4^{\text{th}}</math> move: <math>(-22,18)</math> |
+ | |||
+ | <math>5^{\text{th}}</math> move: <math>(-17,18)</math> | ||
+ | |||
+ | <math>6^{\text{th}}</math> move: <math>(-17,24)</math> | ||
+ | |||
+ | <math>7^{\text{th}}</math> move: <math>(-24, 24)</math> | ||
+ | |||
+ | In the <math>3^{\text{rd}}</math> move Andy lands on <math>(-22,22)</math>, in the <math>7^{\text{th}}</math> move, Andy lands on <math>(-24, 24)</math>. | ||
+ | |||
+ | There is a pattern, for every <math>4</math> moves (starting from the <math>3^{\text{rd}}</math> move), Andy will arrive on a coordinate in the form of <math>(-2n, 2n)</math>. From this we can deduce: | ||
+ | |||
+ | <math>(1) \ 3^{\text{rd}}</math> move: <math>(-22,22)</math> | ||
+ | |||
+ | <math>(2) \ 7^{\text{th}}</math> move: <math>(-24, 24)</math> | ||
+ | |||
+ | <math>(3) \ 11^{\text{th}}</math> move: <math>(-26, 26)</math> | ||
+ | |||
+ | <math>(4) \ 15^{\text{th}}</math> move: <math>(-28, 28)</math> | ||
+ | |||
+ | <math>(n) \ (4n-1)^{\text{th}}</math> move: <math>(-20-2n, 20+2n)</math> | ||
+ | |||
+ | We have <math>2019 = 4 \cdot 505 -1</math>, for which <math>n = 505</math> and <math>20 + 2 \cdot 505 = 1030</math>. So, on the <math>2019^{\text{th}}</math> move Andy is at <math>(-1030, 1030)</math>. | ||
+ | |||
+ | Because the problem asks for the <math>2020^{\text{th}}</math> move, <math>1030-2020=-990</math>, on the <math>2020^{\text{th}}</math> move, Andy will be on <math>\boxed{\textbf{(B)}\ (-1030, -990)}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Video Solution (HOW TO CRITICALLY THINK!!!)== | ||
+ | https://youtu.be/gbnWUskoOuU | ||
+ | |||
+ | ~Education, the Study of Everything | ||
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== Video Solution == | == Video Solution == | ||
https://youtu.be/t6yjfKXpwDs | https://youtu.be/t6yjfKXpwDs | ||
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== Similar Problem == | == Similar Problem == |
Latest revision as of 14:51, 6 June 2023
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
left. From there, Andy moves
units (north) and then turns
left. He then moves
units (west) and again turns
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 (Associative Property)
Andy makes a total of moves:
horizontal (
left and
right) and
vertical (
up and
down).
The -coordinate of Andy's final position is
The
-coordinate of Andy's final position is
Together, we have
~MRENTHUSIASM
Solution 2 (Pattern)
You can find that every four moves both coordinates decrease by Therefore, both coordinates need to decrease by two
times. You subtract, giving you the answer of
~happykeeper
Solution 3 (Pattern)
Let's first mark the first few points Andy will arrive at:
Starting Point ( move):
move:
move:
move:
move:
move:
move:
move:
In the move Andy lands on
, in the
move, Andy lands on
.
There is a pattern, for every moves (starting from the
move), Andy will arrive on a coordinate in the form of
. From this we can deduce:
move:
move:
move:
move:
move:
We have , for which
and
. So, on the
move Andy is at
.
Because the problem asks for the move,
, on the
move, Andy will be on
.
Video Solution (HOW TO CRITICALLY THINK!!!)
~Education, the Study of Everything
Video Solution
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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