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>. | ||
− | ==Solution 2 (Rewording Solution 1)== | + | ==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. | ||
Line 49: | Line 49: | ||
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. | ||
− | <math>6+6=\boxed{(A) 12}</math>. | + | <math>6+6=\boxed{\textbf{(A)} 12}</math>. |
==Solution 3 (Group Theory)== | ==Solution 3 (Group Theory)== | ||
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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. | 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. | ||
− | Thus the answer is <math>\boxed{(A) 12}</math>. | + | Thus the answer is <math>\boxed{\textbf{(A)} 12}</math>. |
~Xrider100 | ~Xrider100 | ||
− | + | Note- this may be a tad bit overkill | |
− | |||
− | |||
− | + | ==Video Solution by Brain Math Club== | |
− | + | https://youtu.be/yAkj_5YMhhQ | |
+ | ==Video Solution by Education, The Study of Everything== | ||
https://youtu.be/SBhkM2frTUA | https://youtu.be/SBhkM2frTUA | ||
− | + | ==Video Solution by Richard Rusczyk== | |
− | https:// | + | https://artofproblemsolving.com/videos/amc/2020amc10a/513 |
− | + | ==Video Solution by MathEx== | |
− | https://www.youtube.com/watch?v=iXwvTmFvo0c | + | https://www.youtube.com/watch?v=iXwvTmFvo0c |
==See Also== | ==See Also== |
Latest revision as of 04:07, 2 July 2024
- The following problem is from both the 2020 AMC 12A #20 and 2020 AMC 10A #23, so both problems redirect to this page.
Contents
Problem
Let be the triangle in the coordinate plane with vertices
and
Consider the following five isometries (rigid transformations) of the plane: rotations of
and
counterclockwise around the origin, reflection across the
-axis, and reflection across the
-axis. How many of the
sequences of three of these transformations (not necessarily distinct) will return
to its original position? (For example, a
rotation, followed by a reflection across the
-axis, followed by a reflection across the
-axis will return
to its original position, but a
rotation, followed by a reflection across the
-axis, followed by another reflection across the
-axis will not return
to its original position.)
Solution 1
First, any combination of motions we can make must reflect an even number of times. This is because every time we reflect
, it changes orientation. Once
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
transformations and an even number of them must be reflections, we either reflect
times or
times.
Case 1: reflections on
.
In this case, we must use rotations to return
to its original position. Notice that our set of rotations,
, contains every multiple of
except for
. We can start with any two rotations
in
and there must be exactly one
such that we can use the three rotations
which ensures that
. That way, the composition of rotations
yields a full rotation. For example, if
, then
, so
and the rotations
yields a full rotation.
The only case in which this fails is when would have to equal
. This happens when
is already a full rotation, namely,
or
. However, we can simply subtract these three cases from the total. Selecting
from
yields
choices, and with
that fail, we are left with
combinations for case
.
Case 2: reflections on
.
In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps back to itself, inserting a rotation before, between, or after these two reflections would change
's final location, meaning that any combination involving two reflections across the x-axis would not map
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 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
combinations for case 2.
Combining both cases we get .
Solution 2 (Rewording of Solution 1)
As in the previous solution, note that we must have either or
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 as
,
as
, and
as
, just for simplification purposes. We want a combination of
of these that will sum to either
or
(
and
are impossible since the minimum is
and the max is
).
can be achieved with any permutation of
and
can be achieved with any permutation of
. This case can be done in
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 rotation left that we can do though, and the only one that will return to the original position is
, which is
AKA reflection across origin. Therefore, since all
transformations are distinct. The three transformations can be applied anywhere since they are commutative (think quadrants). This gives
ways.
.
Solution 3 (Group Theory)
Define as a reflection, and
as a
counterclockwise rotation. Thus,
, and the five transformations can be represented as
, and
.
Now either doesn't appear at all or appears twice. For the former case, it's easy to see that only
and
will work. Both can be permuted in
ways, giving
ways in total.
For the latter case, note that can't appear twice, neither does
, else we need to get
from
, which is not possible. So
and
must appear once each. The last transformation must be
. A quick check shows that
is permutable, since
(since
). This gives
ways.
Thus the answer is .
~Xrider100
Note- this may be a tad bit overkill
Video Solution by Brain Math Club
Video Solution by Education, The Study of Everything
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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