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 20 transformations chosen from will send all of the labeled vertices back to their original positions? (For example, is one sequence of 4 transformations that will send the vertices back to their original positions.)
Answer Choices:
A.
B.
C.
D.
E.
Solution:
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:
The th
Transformation
Each sequence of transformations generates one valid sequence of transformations. Therefore, the answer is .
OR
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 (C) .
OR
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 C) .
OR
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
when or when which is again, .
Therefore, the answer is .
The problems on this page are the property of the MAA's American Mathematics Competitions