Difference between revisions of "2020 AMC 10B Problems/Problem 23"

(Solution)
(Solution)
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Let (+) denote counterclockwise/starting orientation and (-) denote clockwise orientation.
 
Let (+) denote counterclockwise/starting orientation and (-) denote clockwise orientation.
 
Let 1,2,3, and 4 denote which quadrant A is in.
 
Let 1,2,3, and 4 denote which quadrant A is in.
 +
 
Realize that from any odd quadrant and any orientation, the 4 transformations result in some permutation of <math>(2+, 2-, 4+, 4-)</math>.
 
Realize that from any odd quadrant and any orientation, the 4 transformations result in some permutation of <math>(2+, 2-, 4+, 4-)</math>.
 +
 
The same goes that from any even quadrant and any orientation, the 4 transformations result in some permutation of <math>(1+, 1-, 3+, 3-)</math>.
 
The same goes that from any even quadrant and any orientation, the 4 transformations result in some permutation of <math>(1+, 1-, 3+, 3-)</math>.
We start our first 19 moves by doing whatever we want, 4 choices each time. Since 19 is odd, we must end up on an even quadrant. As said above, we know that exactly one of the four transformations will give us <math>(1+)</math>, and we must use that transformation.
+
 
 +
We start our first 19 moves by doing whatever we want, 4 choices each time. Since 19 is odd, we must end up on an even quadrant.  
 +
 
 +
As said above, we know that exactly one of the four transformations will give us <math>(1+)</math>, and we must use that transformation.
 +
 
 
Thus <math>4^{19}=\boxed{(C) 2^{38}}</math>
 
Thus <math>4^{19}=\boxed{(C) 2^{38}}</math>
  

Revision as of 01:17, 8 February 2020

Problem

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

Let (+) denote counterclockwise/starting orientation and (-) denote clockwise orientation. Let 1,2,3, and 4 denote which quadrant A is in.

Realize that from any odd quadrant and any orientation, the 4 transformations result in some permutation of $(2+, 2-, 4+, 4-)$.

The same goes that from any even quadrant and any orientation, the 4 transformations result in some permutation of $(1+, 1-, 3+, 3-)$.

We start our first 19 moves by doing whatever we want, 4 choices each time. Since 19 is odd, we must end up on an even quadrant.

As said above, we know that exactly one of the four transformations will give us $(1+)$, and we must use that transformation.

Thus $4^{19}=\boxed{(C) 2^{38}}$

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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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