Difference between revisions of "2020 AMC 10B Problems/Problem 23"
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== Solution 3 == | == Solution 3 == | ||
− | Notice that any '''pair''' of transformations either swaps the <math>x</math> and <math>y</math>-coordinates, negates the <math>x</math> and <math>y</math>-coordinates, swaps and negates the <math>x</math> and <math>y</math>-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 <math> | + | Notice that any '''pair''' of transformations either swaps the <math>x</math> and <math>y</math>-coordinates, negates the <math>x</math> and <math>y</math>-coordinates, swaps and negates the <math>x</math> and <math>y</math>-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 <math>9</math> pairs of transformations, there is a <math>\frac{1}{4}</math> chance the last pair of transformations will return the figure to its original position. Therefore, the answer is <math>\frac{4^{20}}{4} = 4^{19} = \boxed{\textbf{(C)}\ 2^{38}}.</math> |
== Solution 4 == | == Solution 4 == | ||
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− | == Solution 6 ( | + | == Solution 6 (Fakesolve? Quick) == |
− | The number of possible ways to orient a square using this method 20 times would be <math>4^{20}</math>, as for each of the transformations we can only use the four transformations <math>L</math>, <math>V</math>, <math>H</math>, or <math>R</math>. | + | The number of possible ways to orient a square using this method <math>20</math> times would be <math>4^{20}</math>, as for each of the transformations we can only use the four transformations <math>L</math>, <math>V</math>, <math>H</math>, or <math>R</math>. |
There are only 4 possible outcomes, as each reflection can also be achieved by rotation. By symmetry, all <math>4</math> of these possibilities are equally likely. Thus, our answer is <math>\frac{4^{20}}{4} = 4^{19}=\boxed{\textbf{(C)}\ 2^{38}}.</math> | There are only 4 possible outcomes, as each reflection can also be achieved by rotation. By symmetry, all <math>4</math> of these possibilities are equally likely. Thus, our answer is <math>\frac{4^{20}}{4} = 4^{19}=\boxed{\textbf{(C)}\ 2^{38}}.</math> | ||
− | ~ cdm | + | ~cdm |
+ | |||
+ | == Solution 7 (Generating Functions) == | ||
+ | |||
+ | Let <math>x</math> represent rotating <math>\overline{AC}</math> <math>90^\circ</math> counterclockwise about the origin and let <math>y</math> represent rotating <math>\overline{BD}</math> <math>90^\circ</math> counterclockwise about the origin. Then, note that the transformations <math>L, V, H,</math> and <math>R</math> can be encoded by <math>xy, \tfrac{1}{xy}, \tfrac{y}{x},</math> and <math>\tfrac{x}{y},</math> respectively. Thus, we wish to compute the sum of the coefficients of the terms of the form <math>cx^{4a}y^{4b}</math> for <math>a, b, c \ge 0</math> in the polynomial <cmath>(xy+\tfrac{1}{xy}+\tfrac{y}{x}+\tfrac{x}{y})^{20}.</cmath> We can factor the polynomial to get <cmath>((x+\tfrac{1}{x})(y+\tfrac{1}{y}))^{20} = (x+\tfrac{1}{x})^{20}(y+\tfrac{1}{y})^{20}.</cmath> We can multiply the polynomial by <math>x^{20}y^{20}</math> and it wouldn't affect the sum of the coefficients of the terms of the form <math>cx^{4a}y^{4b},</math> so we can now work with <cmath>(x^2+1)^{20}(y^2+1)^{20}.</cmath> Now, it is very evident that we will get a term of the form <math>cx^{4a}y^{4b}</math> exactly when the first part of the polynomial contributes an even number of <math>x^2</math>'s and the second part of the polynomial contributes an even number of <math>y^2</math>'s, so the answer is | ||
+ | \begin{align*} | ||
+ | \left(\binom{20}{0} + \binom{20}{2} + \ldots + \binom{20}{20}\right)\left(\binom{20}{0} + \binom{20}{2} + \ldots + \binom{20}{20}\right) &= 2^{19} \cdot 2^{19} \ | ||
+ | &= \boxed{\text{(C)} \ 2^{38}}. | ||
+ | \end{align*} | ||
+ | <math>\square</math> | ||
+ | |||
+ | ~lpieleanu | ||
==Video Solutions== | ==Video Solutions== |
Latest revision as of 12:57, 10 November 2024
- The following problem is from both the 2020 AMC 10B #23 and 2020 AMC 12B #19, so both problems redirect to this page.
Contents
[hide]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
.
Solution 6 (Fakesolve? Quick)
The number of possible ways to orient a square using this method times would be
, as for each of the transformations we can only use the four transformations
,
,
, or
.
There are only 4 possible outcomes, as each reflection can also be achieved by rotation. By symmetry, all of these possibilities are equally likely. Thus, our answer is
~cdm
Solution 7 (Generating Functions)
Let represent rotating
counterclockwise about the origin and let
represent rotating
counterclockwise about the origin. Then, note that the transformations
and
can be encoded by
and
respectively. Thus, we wish to compute the sum of the coefficients of the terms of the form
for
in the polynomial
We can factor the polynomial to get
We can multiply the polynomial by
and it wouldn't affect the sum of the coefficients of the terms of the form
so we can now work with
Now, it is very evident that we will get a term of the form
exactly when the first part of the polynomial contributes an even number of
's and the second part of the polynomial contributes an even number of
's, so the answer is
~lpieleanu
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.