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 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 we are in 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 sequences is
Note that there can only be an even number of reflections since they result in the same anti-clockwise orientation of the vertices Therefore, the probability of having the same anti-clockwise orientation with the original arrangement after the transformation is
Next, the even number of reflections means that there must be an even number of rotations since their sum is even. Even rotations result in only the original position or a rotation of it.
Since rotation and rotation cancel 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, the answer is
~joshuamh111
~Edits by Eric X
Solution 5 (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
Equations , , are equivalent. Here I will prove that is equivalent to .
From , . We have From , . We have So, It follows that , 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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