Difference between revisions of "2020 AMC 12B Problems/Problem 19"
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Joshuamh111 (talk | contribs) (→Solution 2(Similar to solution 1)) |
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Therefore, <math>\frac{2^{40}}{2\cdot2}=\boxed{\mathbf(C) 2^{38}}</math> | Therefore, <math>\frac{2^{40}}{2\cdot2}=\boxed{\mathbf(C) 2^{38}}</math> | ||
+ | ~joshuamh111 | ||
{{AMC12 box|year=2020|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2020|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:00, 8 February 2020
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
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 A can only be in first quadrant
or third quadrant
. Only the one in the first quadrant works, so divide by
. Now, C must be in the opposite quadrant as A. B 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 A and B satisfy the conditions, C and D will also be at their original spot.
. The answer is
~Kinglogic
Solution 2(Similar to solution 1)
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 orginal position is the probability that
![$|n(R)-n(L)|\equiv2\pmod{4}$](http://latex.artofproblemsolving.com/5/e/f/5efc205ae74ac19a12cb2b5f8643e09dfd6e371e.png)
which is again, .
Therefore,
~joshuamh111
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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