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Difference between revisions of "2020 AMC 10A Problems/Problem 23"
m (→‎Solution 1)
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Case 1: 0 reflections on T
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Case 1: <math>0</math> reflections on T.
  
  
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In this case, we must use <math>3</math> rotations to return <math>T</math> to its original position. Notice that our set of rotations, <math>\{90^\circ,180^\circ,270^\circ\}</math>, contains every multiple of <math>90^\circ</math> except for <math>0^\circ</math>. We can start with any two rotations <math>a,b</math> in <math>\{90^\circ,180^\circ,270^\circ\}</math> and there must be exactly one <math>c \equiv -a - b \pmod{360^\circ}</math> such that we can use the three rotations <math>(a,b,c)</math> which ensures that <math>a + b + c \equiv 0^\circ \pmod{360^\circ}</math>. That way, the composition of rotations <math>a,b,c</math> yields a full rotation. For example, if <math>a = b = 90^\circ</math>, then <math>c \equiv -90^\circ - 90^\circ = -180^\circ \pmod{360^\circ}</math>, so <math>c = 180^\circ</math> and the rotations <math>(90^\circ,90^\circ,180^\circ)</math> yields a full rotation.
 
In this case, we must use <math>3</math> rotations to return <math>T</math> to its original position. Notice that our set of rotations, <math>\{90^\circ,180^\circ,270^\circ\}</math>, contains every multiple of <math>90^\circ</math> except for <math>0^\circ</math>. We can start with any two rotations <math>a,b</math> in <math>\{90^\circ,180^\circ,270^\circ\}</math> and there must be exactly one <math>c \equiv -a - b \pmod{360^\circ}</math> such that we can use the three rotations <math>(a,b,c)</math> which ensures that <math>a + b + c \equiv 0^\circ \pmod{360^\circ}</math>. That way, the composition of rotations <math>a,b,c</math> yields a full rotation. For example, if <math>a = b = 90^\circ</math>, then <math>c \equiv -90^\circ - 90^\circ = -180^\circ \pmod{360^\circ}</math>, so <math>c = 180^\circ</math> and the rotations <math>(90^\circ,90^\circ,180^\circ)</math> yields a full rotation.
  
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The only case in which this fails is when <math>c</math> would have to equal <math>0^\circ</math>. This happens when <math>(a,b)</math> is already a full rotation, namely, <math>(a,b) = (90^\circ,270^\circ),(180^\circ,180^\circ),</math> or <math>(270^\circ,90^\circ)</math>. However, we can simply subtract these three cases from the total. Selecting <math>(a,b)</math> from <math>\{90^\circ,180^\circ,270^\circ\}</math> yields <math>3 \cdot 3 = 9</math> choices, and with <math>3</math> that fail, we are left with <math>6</math> combinations for case 1.
+
The only case in which this fails is when <math>c</math> would have to equal <math>0^\circ</math>. This happens when <math>(a,b)</math> is already a full rotation, namely, <math>(a,b) = (90^\circ,270^\circ),(180^\circ,180^\circ),</math> or <math>(270^\circ,90^\circ)</math>. However, we can simply subtract these three cases from the total. Selecting <math>(a,b)</math> from <math>\{90^\circ,180^\circ,270^\circ\}</math> yields <math>3 \cdot 3 = 9</math> choices, and with <math>3</math> that fail, we are left with <math>6</math> combinations for case <math>1</math>.
  
  
  
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Case 2: 2 reflections on T
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Case 2: <math>2</math> reflections on T.
  
  
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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 <math>180^\circ</math> 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 <math>3! = 6</math> combinations for case 2.
 
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 <math>180^\circ</math> 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 <math>3! = 6</math> combinations for case 2.
  
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Combining both cases we get <math>6 + 6 = \boxed{\textbf{(A) } 12}</math>
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Combining both cases we get <math>6+6=\boxed{\textbf{(A)} 12}</math>.
  
 
==Solution 2 (Rewording solution 1)==
 
==Solution 2 (Rewording solution 1)==

Revision as of 14:40, 29 December 2020

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 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 is 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.

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

Video Solution 1

Education, The Study of Everything

https://youtu.be/SBhkM2frTUA

Video Solution 2

https://youtu.be/5TjrDCxTm7Q - Richard Rusczyk

Video Solution 3

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

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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