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

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Combining both cases we get <math>6+6=\boxed{\textbf{(A)}  12}</math>.
 
Combining both cases we get <math>6+6=\boxed{\textbf{(A)}  12}</math>.
  
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==Solution 2 (Rewording Solution 1)==
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==Solution 2 (Rewording of Solution 1)==
  
 
As in the previous solution, note that we must have either <math>0</math> or <math>2</math> reflections because of orientation since reflection changes orientation that is impossible to fix by rotation. We also know we can't have the same reflection twice, since that would give a net of no change and would require an identity rotation.
 
As in the previous solution, note that we must have either <math>0</math> or <math>2</math> reflections because of orientation since reflection changes orientation that is impossible to fix by rotation. We also know we can't have the same reflection twice, since that would give a net of no change and would require an identity rotation.
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Suppose there are two reflections. As noted already, they must be different, and as a result will take the triangle to the opposite side of the origin if we don't do any rotation. We have <math>1</math> rotation left that we can do though, and the only one that will return to the original position is <math>2</math>, which is <math>180^{\circ}</math> AKA reflection across origin. Therefore, since all <math>3</math> transformations are distinct. The three transformations can be applied anywhere since they are commutative (think quadrants). This gives <math>6</math> ways.
 
Suppose there are two reflections. As noted already, they must be different, and as a result will take the triangle to the opposite side of the origin if we don't do any rotation. We have <math>1</math> rotation left that we can do though, and the only one that will return to the original position is <math>2</math>, which is <math>180^{\circ}</math> AKA reflection across origin. Therefore, since all <math>3</math> transformations are distinct. The three transformations can be applied anywhere since they are commutative (think quadrants). This gives <math>6</math> ways.
  
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<math>6+6=\boxed{(A) 12}</math>.
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<math>6+6=\boxed{\textbf{(A)}  12}</math>.
  
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== Video Solution 1 ==
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==Solution 3 (Group Theory)==
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Define <math>s</math> as a reflection, and <math>r</math> as a <math>90^{\circ}</math> counterclockwise rotation. Thus, <math>r^4=s^2=e</math>, and the five transformations can be represented as <math>{r, r^2, r^3, r^2s, s}</math>, and <math>rs=sr^{-1}</math>.
  
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Education, The Study of Everything
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Now either <math>s</math> doesn't appear at all or appears twice. For the former case, it's easy to see that only <math>r, r, r^2</math> and <math>r^2, r^3, r^3</math> will work. Both can be permuted in <math>3</math> ways, giving <math>6</math> ways in total.
  
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For the latter case, note that <math>s</math> can't appear twice, neither does <math>r^2s</math>, else we need to get <math>e</math> from <math>{r, r^2, r^3}</math>, which is not possible. So <math>r^2s</math> and <math>s</math> must appear once each. The last transformation must be <math>r^2</math>. A quick check shows that <math>{r^2, r^2s, s}</math> is permutable, since <math>r^2s=sr^{-2}=sr^2</math> (since <math>r^4=e</math>). This gives <math>6</math> ways.
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Thus the answer is <math>\boxed{\textbf{(A)} 12}</math>.
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~Xrider100
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==Video Solution by Brain Math Club==
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https://youtu.be/yAkj_5YMhhQ
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==Video Solution by Education, The Study of Everything==
 
https://youtu.be/SBhkM2frTUA
 
https://youtu.be/SBhkM2frTUA
  
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==Video Solution 2==
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==Video Solution by Richard Rusczyk==
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https://youtu.be/5TjrDCxTm7Q - Richard Rusczyk
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https://artofproblemsolving.com/videos/amc/2020amc10a/513
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== Video Solution 3 ==
 
  
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https://www.youtube.com/watch?v=iXwvTmFvo0c ~ MathEx
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==Video Solution by MathEx==
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https://www.youtube.com/watch?v=iXwvTmFvo0c
  
 
==See Also==
 
==See Also==

Latest revision as of 21:02, 30 October 2024

The following problem is from both the 2020 AMC 12A #20 and 2020 AMC 10A #23, so both problems redirect to this page.

Problem

Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$-axis, and reflection across the $y$-axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by a reflection across the $y$-axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by another reflection across the $x$-axis will not return $T$ to its original position.)

$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25$

Solution 1

[asy] size(10cm); Label f; f.p=fontsize(6); xaxis(-6,6,Ticks(f, 2.0)); yaxis(-6,6,Ticks(f, 2.0)); filldraw(origin--(4,0)--(0,3)--cycle, gray, black+linewidth(1)); [/asy]

First, any combination of motions we can make must reflect $T$ an even number of times. This is because every time we reflect $T$, it changes orientation. Once $T$ has been flipped once, no combination of rotations will put it back in place because it is the mirror image; however, flipping it again changes it back to the original orientation. Since we are only allowed $3$ transformations and an even number of them must be reflections, we either reflect $T$ $0$ times or $2$ times.


Case 1: $0$ reflections on $T$.


In this case, we must use $3$ rotations to return $T$ to its original position. Notice that our set of rotations, $\{90^\circ,180^\circ,270^\circ\}$, contains every multiple of $90^\circ$ except for $0^\circ$. We can start with any two rotations $a,b$ in $\{90^\circ,180^\circ,270^\circ\}$ and there must be exactly one $c \equiv -a - b \pmod{360^\circ}$ such that we can use the three rotations $(a,b,c)$ which ensures that $a + b + c \equiv 0^\circ \pmod{360^\circ}$. That way, the composition of rotations $a,b,c$ yields a full rotation. For example, if $a = b = 90^\circ$, then $c \equiv -90^\circ - 90^\circ = -180^\circ \pmod{360^\circ}$, so $c = 180^\circ$ and the rotations $(90^\circ,90^\circ,180^\circ)$ yields a full rotation.

The only case in which this fails is when $c$ would have to equal $0^\circ$. This happens when $(a,b)$ is already a full rotation, namely, $(a,b) = (90^\circ,270^\circ),(180^\circ,180^\circ),$ or $(270^\circ,90^\circ)$. However, we can simply subtract these three cases from the total. Selecting $(a,b)$ from $\{90^\circ,180^\circ,270^\circ\}$ yields $3 \cdot 3 = 9$ choices, and with $3$ that fail, we are left with $6$ combinations for case $1$.


Case 2: $2$ reflections on $T$.


In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps $T$ back to itself, inserting a rotation before, between, or after these two reflections would change $T$'s final location, meaning that any combination involving two reflections across the x-axis would not map $T$ back to itself. The same applies to two reflections across the y-axis.

Therefore, we must use one reflection about the x-axis, one reflection about the y-axis, and one rotation. Since a reflection about the x-axis changes the sign of the y component, a reflection about the y-axis changes the sign of the x component, and a $180^\circ$ rotation changes both signs, these three transformation composed (in any order) will suffice. It is therefore only a question of arranging the three, giving us $3! = 6$ combinations for case 2.

Combining both cases we get $6+6=\boxed{\textbf{(A)} 12}$.

Solution 2 (Rewording of Solution 1)

As in the previous solution, note that we must have either $0$ or $2$ reflections because of orientation since reflection changes orientation that is impossible to fix by rotation. We also know we can't have the same reflection twice, since that would give a net of no change and would require an identity rotation.

Suppose there are no reflections. Denote $90^{\circ}$ as $1$, $180^{\circ}$ as $2$, and $270^{\circ}$ as $3$, just for simplification purposes. We want a combination of $3$ of these that will sum to either $4$ or $8$ ($0$ and $12$ are impossible since the minimum is $3$ and the max is $9$). $4$ can be achieved with any permutation of $(1-1-2)$ and $8$ can be achieved with any permutation of $(2-3-3)$. This case can be done in $3+3=6$ ways.

Suppose there are two reflections. As noted already, they must be different, and as a result will take the triangle to the opposite side of the origin if we don't do any rotation. We have $1$ rotation left that we can do though, and the only one that will return to the original position is $2$, which is $180^{\circ}$ AKA reflection across origin. Therefore, since all $3$ transformations are distinct. The three transformations can be applied anywhere since they are commutative (think quadrants). This gives $6$ ways.

$6+6=\boxed{\textbf{(A)} 12}$.

Solution 3 (Group Theory)

Define $s$ as a reflection, and $r$ as a $90^{\circ}$ counterclockwise rotation. Thus, $r^4=s^2=e$, and the five transformations can be represented as ${r, r^2, r^3, r^2s, s}$, and $rs=sr^{-1}$.

Now either $s$ doesn't appear at all or appears twice. For the former case, it's easy to see that only $r, r, r^2$ and $r^2, r^3, r^3$ will work. Both can be permuted in $3$ ways, giving $6$ ways in total.

For the latter case, note that $s$ can't appear twice, neither does $r^2s$, else we need to get $e$ from ${r, r^2, r^3}$, which is not possible. So $r^2s$ and $s$ must appear once each. The last transformation must be $r^2$. A quick check shows that ${r^2, r^2s, s}$ is permutable, since $r^2s=sr^{-2}=sr^2$ (since $r^4=e$). This gives $6$ ways.

Thus the answer is $\boxed{\textbf{(A)} 12}$.

~Xrider100

Video Solution by Brain Math Club

https://youtu.be/yAkj_5YMhhQ

Video Solution by Education, The Study of Everything

https://youtu.be/SBhkM2frTUA

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2020amc10a/513

Video Solution by MathEx

https://www.youtube.com/watch?v=iXwvTmFvo0c

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
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
2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 12 Problems and Solutions

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