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Difference between revisions of "2020 AMC 12B Problems/Problem 19"

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==Solution==
 
==Solution==
  
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Hopefully someone will think of a better one, but here is an indirect (make-sense) answer, use only if you are really desperate.
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Hopefully, someone will think of a better one, but here is an indirect answer, use only if you are really desperate.
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<math>20</math> moves can be made, and each move have <math>4</math> choices, so a total of <math>4^{20}=2^{40}</math> moves. First, after the <math>20</math> moves, Point A can only be in first quadrant <math>(1,1)</math> or third quadrant <math>(-1,-1)</math>. Only the one in the first quadrant works, so divide by <math>2</math>. Now, C must be in the opposite quadrant as A. B can be either in the second (<math>(-1, 1)</math>) or fourth quadrant (<math>(1, -1)</math>) , but we want it to be in the second quadrant, so divide by <math>2</math> again. Now as A and B satisfy the conditions, C and D will also be at their original spot. <math>\frac{2^{40}}{2\cdot2}=2^{38}</math>. The answer is \boxed{(C)}
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<math>20</math> moves can be made, and each move have <math>4</math> choices, so a total of <math>4^{20}=2^{40}</math> moves. First, after the <math>20</math> moves, Point A can only be in first quadrant <math>(1,1)</math> or third quadrant <math>(-1,-1)</math>. Only the one in the first quadrant works, so divide by <math>2</math>. Now, C must be in the opposite quadrant as A. B can be either in the second (<math>(-1, 1)</math>) or fourth quadrant (<math>(1, -1)</math>) , but we want it to be in the second quadrant, so divide by <math>2</math> again. Now as A and B satisfy the conditions, C and D will also be at their original spot. <math>\frac{2^{40}}{2\cdot2}=2^{38}</math>. The answer is <math>\boxed{C}</math>
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~Kinglogic
 
{{AMC12 box|year=2020|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2020|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:16, 8 February 2020

Square $ABCD$ in the coordinate plane has vertices at the points $A(1,1), B(-1,1), C(-1,-1),$ and $D(1,-1).$ Consider the following four transformations: $L,$ a rotation of $90^{\circ}$ counterclockwise around the origin; $R,$ a rotation of $90^{\circ}$ clockwise around the origin; $H,$ a reflection across the $x$-axis; and $V,$ a reflection across the $y$-axis.

Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying $R$ and then $V$ would send the vertex $A$ at $(1,1)$ to $(-1,-1)$ and would send the vertex $B$ at $(-1,1)$ to itself. How many sequences of $20$ transformations chosen from $\{L, R, H, V\}$ will send all of the labeled vertices back to their original positions? (For example, $R, R, V, H$ is one sequence of $4$ transformations that will send the vertices back to their original positions.)

$\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}$

Solution

Hopefully, someone will think of a better one, but here is an indirect answer, use only if you are really desperate. $20$ moves can be made, and each move have $4$ choices, so a total of $4^{20}=2^{40}$ moves. First, after the $20$ moves, Point A can only be in first quadrant $(1,1)$ or third quadrant $(-1,-1)$. Only the one in the first quadrant works, so divide by $2$. Now, C must be in the opposite quadrant as A. B can be either in the second ($(-1, 1)$) or fourth quadrant ($(1, -1)$) , but we want it to be in the second quadrant, so divide by $2$ again. Now as A and B satisfy the conditions, C and D will also be at their original spot. $\frac{2^{40}}{2\cdot2}=2^{38}$. The answer is $\boxed{C}$ ~Kinglogic

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AMC 12 Problems and Solutions

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