2020 AMC 10B Problems/Problem 23
- The following problem is from both the 2020 AMC 10B #23 and 2020 AMC 12B #19, so both problems redirect to this page.
Contents
Problem
Square in the coordinate plane has vertices at the points and Consider the following four transformations:
a rotation of counterclockwise around the origin;
a rotation of clockwise around the origin;
a reflection across the -axis; and
a reflection across the -axis.
Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying and then would send the vertex at to and would send the vertex at to itself. How many sequences of transformations chosen from will send all of the labeled vertices back to their original positions? (For example, is one sequence of transformations that will send the vertices back to their original positions.)
Solution 1
For each transformation:
- Each labeled vertex will move to an adjacent position.
- The labeled vertices will maintain the consecutive order in either direction (clockwise or counterclockwise).
- and will retain the direction of the labeled vertices, but and will alter the direction of the labeled vertices.
After the th transformation, vertex will be at either or All possible configurations of the labeled vertices are shown below: Each sequence of transformations generates one valid sequence of transformations. Therefore, the answer is
~MRENTHUSIASM
Solution 2
Let denote counterclockwise/starting orientation and denote clockwise orientation. Let and denote which quadrant is in.
Realize that from any odd quadrant and any orientation, the transformations result in some permutation of
The same goes that from any even quadrant and any orientation, the transformations result in some permutation of
We start our first moves by doing whatever we want, choices each time. Since 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 and we must use that transformation.
Thus, the answer is
Solution 3
Notice that any pair of two of these transformations either swaps the and -coordinates, negates the and -coordinates, swaps and negates the and -coordinates, or leaves the original unchanged. Furthermore, notice that for each of these results, if we apply another pair of transformations, one of these results will happen again, and with equal probability. Therefore, no matter what state after we apply the first pairs of transformations, there is a chance the last pair of transformations will return the figure to its original position. Therefore, the answer is
Solution 4
The total number of sequence is
Note that there can only be even number of reflections since they result in the same anti-clockwise orientation of the verices Therefore, the probability of having the same anti-clockwise orientation with the original arrangement after the transformation is
Next, even number of reflections mean that there must be even number of rotations since their sum is even. Even rotations result only in the original position or rotation of it.
Since rotation and rotation cancels each other out, the difference between the numbers of them define the final position. The probability of the transformation returning the vertices to the original position given that there are even number of rotations is equivalent to the probability that
which is again,
Therefore,
~joshuamh111
Solution 5
Hopefully, someone will think of a better one, but here is an indirect answer, use only if you are really desperate. moves can be made, and each move have choices, so a total of moves. First, after the moves, Point can only be in first quadrant or third quadrant Only the one in the first quadrant works, so divide by Now, must be in the opposite quadrant as Note that can be either in the second () or fourth quadrant () , but we want it to be in the second quadrant, so divide by again. Now as and satisfy the conditions, and will also be at their original spot. The answer is
~Kinglogic
Solution 6 (Quick)
We can rotate and reflect the square freely until the th step. There are ways to do this, (because for each one you could do which is equal to . Once you reach the th step, we have 1 choice of reflection/rotation. Therefore, we have which is
Side Note
MRENTHUSIASM's Solution 1 has the explanation to why on the th step, we have choice of reflection and rotation. You might think "What if it is diagonally opposite from our original position or on the original position on the th step?" The answer is that because on the th step is the (mod th) step, we can never go to the (mod th) step because we cannot reflect across a diagonal because we are merely rotating and reflecting. The reflection only takes place over the or axis, and therefore will never reflect across it diagonally. If we could reflect diagonally, it would distort the whole mod idea and this would not be a valid answer.
~Arcticturn
Solution 7 (Group Theory)
This problem is a Dihedral Group problem, , in Group Theory.
The transformation has associativity, for ,
Let be the initial state of the square and .
It's not hard to see that after a series of transformations from initial state to initial state , the number of transformations must be even. Denote be the number of sequences of transformations from initial state to initial state . We are going to prove
For each transformation composite operator, there are replacements.
For example, when :
From , , so can be replaced with , , without changing the result. Suppose we choose , then From , , so can be replaced with , , without changing the result. Suppose we choose , then From , , so can be replaced with , , without changing the result. Suppose we choose , then
So, we have : With transformations, it will go from initial state to initial state. There are transformation composite operators between the transformations, and each pair of transformations surrounding the transformations composite operator have options. So, we have , from which .
Side Note
, , are equivalent. Here I will prove is equivalent to .
From ,
From ,
So,
, which is
Video Solutions
Video Solution 1
https://www.youtube.com/watch?v=XZs9DHg6cx0
~MathEx
Video Solution 2 by The Beauty of Math
https://youtu.be/Bl2kn9oVxQ8?t=348
~Icematrix
See Also
2020 AMC 10B (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 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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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