Difference between revisions of "1985 AIME Problems/Problem 12"
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~Nafer | ~Nafer | ||
− | == Solution 5 ( | + | ==Solution 5 (Dynamic Programming)== |
− | |||
+ | Let <math>A(n)</math> be the probability the bug lands on vertex <math>A</math> after crawling <math>n</math> meters, <math>B(n)</math> be the probability the bug lands on vertex <math>B</math> after crawling <math>n</math> meters, and etc. | ||
+ | |||
+ | Note that <math>A(1)=0</math> and <math>B(1)=C(1)=D(1)=\frac13.</math> For <math>n\geq2,</math> the probability that the bug land on each vertex after <math>n</math> meters is <math>\frac13</math> the sum of the probability the bug land on other vertices after crawling <math>n-1</math> meters. So, we have | ||
+ | <cmath>\begin{align*} | ||
+ | A(n) &= \frac13 \cdot [B(n-1) + C(n-1) + D(n-1)], \\ | ||
+ | B(n) &= \frac13 \cdot [A(n-1) + C(n-1) + D(n-1)], \\ | ||
+ | C(n) &= \frac13 \cdot [A(n-1) + B(n-1) + D(n-1)], \\ | ||
+ | D(n) &= \frac13 \cdot [A(n-1) + B(n-1) + C(n-1)]. | ||
+ | \end{align*}</cmath> | ||
+ | It follows that <math>A(n) = B(n-1) = C(n-1) = D(n-1).</math> | ||
+ | |||
+ | We construct the following table: | ||
+ | <cmath>\begin{array}{c|cccc} | ||
+ | & & & & \\ [-2ex] | ||
+ | n & A(n) & B(n) & C(n) & D(n) \\ [1ex] | ||
+ | \hline | ||
+ | & & & & \\ [-1ex] | ||
+ | 1 & 0 & \frac13 & \frac13 & \frac13 \\ | ||
+ | & & & & \\ | ||
+ | 2 & \frac13 & \frac29 & \frac29 & \frac29 \\ | ||
+ | & & & & \\ | ||
+ | 3 & \frac29 & \frac{7}{27} & \frac{7}{27} & \frac{7}{27} \\ | ||
+ | & & & & \\ | ||
+ | 4 & \frac{7}{27} & \frac{20}{81} & \frac{20}{81} & \frac{20}{81}\\ | ||
+ | & & & & \\ | ||
+ | 5 & \frac{20}{81} & \frac{61}{243} & \frac{61}{243} & \frac{61}{243} \\ | ||
+ | & & & & \\ | ||
+ | 6 & \frac{61}{243} & \frac{182}{729} & \frac{182}{729} & \frac{182}{729} \\ | ||
+ | & & & & \\ | ||
+ | 7 & \frac{182}{729} & \frac{547}{2187} & \frac{547}{2187} & \frac{547}{2187} \\ [1ex] | ||
+ | \end{array}</cmath> | ||
+ | Therefore, the answer is <math>n = \boxed{182}.</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == Solution 6 (Generating Functions and Roots of Unity Filter / Casework) == | ||
+ | The generating function for a problem with this general form (<math>4</math> states, <math>n</math> steps) is <math>(x+x^2+x^3)^n</math>, so the generating function of interest for this problem is <math>(x+x^2+x^3)^7</math>. Our goal is to find the coefficients of every <math>x^{4n}</math> and add them up before dividing by <math>3^7</math>. Here we have two ways to proceed: | ||
+ | |||
+ | ===1. Roots of Unity Filter=== | ||
Let <math>\omega = e^{i\pi / 2}</math>. We have that if <math>G(x) = (x+x^2+x^3)^7</math>, then <cmath>\frac{G(1) + G(\omega) + G(\omega^2) + G(\omega^3)}{4} = \frac{2187-1-1-1}{4} = 546.</cmath> From here, the desired probability is <math>\frac{546}{2187} = \frac{182}{729}</math>. Therefore, the answer is <math>n=\boxed{182}</math>. | Let <math>\omega = e^{i\pi / 2}</math>. We have that if <math>G(x) = (x+x^2+x^3)^7</math>, then <cmath>\frac{G(1) + G(\omega) + G(\omega^2) + G(\omega^3)}{4} = \frac{2187-1-1-1}{4} = 546.</cmath> From here, the desired probability is <math>\frac{546}{2187} = \frac{182}{729}</math>. Therefore, the answer is <math>n=\boxed{182}</math>. | ||
~RedFlame2112 | ~RedFlame2112 | ||
− | == Solution | + | ===2. Casework=== |
+ | We can factor <math>(x+x^2+x^3)^7</math> as <math>x^7(1+x+x^2)^7.</math> The <math>x^{4n}</math> coefficients of <math>(x+x^2+x^3)^7</math> will be the same as the <math>x^{4n+1}</math> coefficients of <math>(1+x+x^2)^7.</math> The possible values for <math>4n+1</math> would then be <math>1,</math> <math>5,</math> <math>9,</math> and <math>13.</math> Because <math>1+13=5+9=14,</math> the coefficients of <math>x^1</math> and <math>x^{13}</math> are equal and so are the coefficients of <math>x^5</math> and <math>x^9.</math> To make an <math>x</math> term, we need <math>6</math> "<math>1</math>" terms and one "<math>x</math>" term multiplied together. There are <math>7</math> ways to arrange these numbers. The coefficient of the <math>x^5</math> term will be the sum of the number of the possible arrangements for these numbers: | ||
+ | <cmath>\begin{align*} | ||
+ | 0000122&=105\text{ arrangements}, \\ | ||
+ | 0001112&=140\text{ arrangements}, \\ | ||
+ | 0011111&=21\text{ arrangements}. | ||
+ | \end{align*}</cmath> | ||
+ | Thus, the coefficient of the <math>x^5</math> term is <math>105+140+21=266.</math> From here, the desired probability is <math>\frac{2(7+266)}{2187}=\frac{546}{2187}=\frac{182}{729}.</math> Thus, our answer is <math>\boxed{182}.</math> | ||
+ | |||
+ | ~[[BS2012]] | ||
+ | |||
+ | == Solution 7 (Partitions) == | ||
We can find the number of different times the bug reaches vertex <math>A</math> before the <math>7</math>th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at <math>A.</math> | We can find the number of different times the bug reaches vertex <math>A</math> before the <math>7</math>th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at <math>A.</math> | ||
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Finally, this is a probability question, so we divide by <math>3^7,</math> or <math>\frac{546}{3^7} = \frac{182}{3^6}.</math> The answer is <math>n=\boxed{182}.</math> | Finally, this is a probability question, so we divide by <math>3^7,</math> or <math>\frac{546}{3^7} = \frac{182}{3^6}.</math> The answer is <math>n=\boxed{182}.</math> | ||
+ | == Solution 8 (Sequences) == | ||
+ | |||
+ | Note that this problem is basically equivalent to the following: | ||
+ | How many distinct sequences of <math>8</math> integers <math>a_1, a_2, a_3, \ldots, a_8</math> are there such that <math>a_1 = a_8 = 1,</math> <math>a_i \in \{1, 2, 3, 4\}</math> for all <math>2 \leq i\leq 8,</math> and <math>a_i \neq a_{i+1}</math> for all <math>1 \leq i \leq 7</math>? | ||
+ | |||
+ | Now consider the <math>8</math> integers modulo <math>4.</math> | ||
+ | Let <math>b_1, b_2, b_3, \ldots, b_7</math> be a new sequence of integers such that <math>b_i = a_{i+1} - a_i \mod 4</math> for all <math>1 \leq i \leq 7.</math> | ||
+ | |||
+ | Note that the condition is equivalent to that <math>b_i \in \{1, 2, 3\}</math> for all <math>1 \leq i \leq 7,</math> and since <math>a_1 \mod 4 = a_8 \mod 4,</math> <math>b_1 + b_2 + \cdots + b_7</math> must be a multiple of <math>4.</math> | ||
+ | |||
+ | Thus, our desired number of paths is equivalent to the number of ordered septuples of positive integers <math>(b_1, b_2, \ldots, b_7)</math> such that <math>b_i \in \{1, 2, 3\}</math> for all <math>1 \leq i \leq 7</math> and <math>b_1 + b_2 + \cdots + b_7</math> is congruent to <math>0 \mod 4.</math> | ||
+ | |||
+ | We will now proceed with casework on the number of <math>2</math>s in the septuple. | ||
+ | |||
+ | One <math>2</math>: There are <math>\dbinom{7}{1} = 7</math> ways to arrange the <math>2</math>, and <math>\dbinom{6}{0} + \dbinom{6}{2} + \dbinom{6}{4} + \dbinom{6}{6} = 2^5</math> valid ways (the proof of this combinatorial identity is left as an exercise to the reader) to arrange the <math>1</math>s and <math>3</math>s. | ||
+ | |||
+ | Three <math>2</math>s: There are <math>\dbinom{7}{3} = 35</math> ways to arrange the <math>2</math>s, and <math>\dbinom{4}{1} + \dbinom{4}{3} = 2^3</math> valid ways to arrange the <math>1</math>s and <math>3</math>s. | ||
+ | |||
+ | Five <math>2</math>s: There are <math>\dbinom{7}{5} = 21</math> ways to arrange the <math>2</math>s, and <math>\dbinom{2}{0} + \dbinom{2}{2} = 2</math> valid ways to arrange the <math>1</math>s and <math>3</math>s. | ||
+ | |||
+ | Adding up our cases, we obtain <math>7 \cdot 32 + 35 \cdot 8 + 21 \cdot 2 = 546</math> valid sequences. | ||
+ | Dividing by the total number of paths without restrictions, <math>3^7 = 2187,</math> our desired probability is <math>\frac{546}{2187} = \frac{182}{729},</math> giving an answer of <math>\boxed{182}.</math> | ||
+ | |||
+ | ~fidgetboss_4000 | ||
+ | |||
+ | ==Solution 9== | ||
+ | We instead find the probability that the bug is NOT at vertex <math>A</math> after crawling <math>n</math> meters (equivalent to moving <math>n</math> times). Call <math>A_n</math> the probability that the bug IS at vertex <math>A</math> after <math>n</math> moves; call <math>O_n</math> the probability that the bug is on some other vertex. We have the following recurrence relations. <cmath>A_n = \frac{1}{3}O_{n-1}</cmath> <cmath>O_n = A_{n-1} + \frac{2}{3}O_{n-1}</cmath> However, we can calculate <math>A_{n-1}</math> in terms of <math>O_n</math>; take <math>n = k-1</math>, and we have <cmath>A_{k-1} = \frac{1}{3}O_{k-2}</cmath>. Substituting this into our equation for <math>O</math>, we have that <cmath>O_n = \frac{1}{3}O_{n-2} + \frac{2}{3}O_{n-1}</cmath>. We also have the conditions that <math>O_0 = 0</math> (the bug will not be at vertex other than <math>A</math> on the "0th" move) and <math>O_1 = 1</math> (the bug will be at a vertex other than <math>A</math> after the <math>1st</math> move). Iteratively calculating <math>O_7</math>, we find that it is equal to <math>\frac{547}{729}</math> (I will not do this calculation here; you can do it manually if you wish to check). However, this is the probability that the ant is NOT at vertex <math>A</math> after <math>7</math> moves; then its complement, <math>\frac{182}{729}</math> is the probability that the ant IS at vertex <math>A</math> after <math>7</math> moves, so our answer is <math>\boxed{182}</math>. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
− | ==Solution | + | ==Solution 10 (Linear Algebra)== |
− | < | + | Think of the problem as a state machine with a transition matrix. State 1 is if the bug is at A, State 2 is if the bug is not at A. In vector form we can represent state 1 as <math>\begin{bmatrix} 1 \\ 0 \end{bmatrix}</math> and state 2 as <math>\begin{bmatrix} 0 \\ 1 \end{bmatrix}</math>. Our transition matrix <math>T =\begin{pmatrix} 0 & \frac{1}{3} \\ 1 & \frac{2}{3} \end{pmatrix}</math>. Thus the probability of being in each state after k moves is <math>T^{k}\begin{bmatrix} 1 \\ 0 \end{bmatrix}=\begin{pmatrix} 0 & \frac{1}{3} \\ 1 & \frac{2}{3} \end{pmatrix}^{k}\begin{bmatrix} 1 \\ 0 \end{bmatrix}</math>. Those familiar with linear algebra will recognize the need to diagonalize the transition matrix through eigendecomposition. In short, the eigenvalues of the transition matrix are <math>-\frac{1}{3}</math> and 1, with corresponding eigenvector basis of <math>\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \begin{pmatrix} \frac{1}{3} \\ 1 \end{pmatrix}</math>. Thus <math>T = Q \Lambda Q^{-1} = \begin{pmatrix} 1 & \frac{1}{3} \\ -1 & 1 \end{pmatrix}\begin{pmatrix} -\frac{1}{3} & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & \frac{1}{3}\\ -1 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & \frac{1}{3} \\ -1 & 1 \end{pmatrix}\begin{pmatrix} -\frac{1}{3} & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} \frac{3}{4} & -\frac{1}{4}\\ \frac{3}{4} & \frac{3}{4} \end{pmatrix}</math> |
− | |||
− | & | ||
− | 1 & & & | ||
− | |||
− | |||
− | |||
− | 3 & | ||
− | |||
− | 4 | ||
− | \end{ | ||
− | + | Thus <math>T^{k}\begin{bmatrix} 1 \\ 0 \end{bmatrix}= \begin{pmatrix} 1 & \frac{1}{3} \\ -1 & 1 \end{pmatrix}\begin{pmatrix} (-\frac{1}{3})^{k} & 0\\ 0 & 1^{k} \end{pmatrix}\begin{pmatrix} \frac{3}{4} & -\frac{1}{4}\\ \frac{3}{4} & \frac{3}{4} \end{pmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{pmatrix} \frac{3}{4}(-\frac{1}{3})^{k}+\frac{1}{4} & -\frac{1}{4}(-\frac{1}{3})^{k}+\frac{1}{4}\\ -\frac{3}{4}(-\frac{1}{3})^{k}+\frac{3}{4} & \frac{1}{4}(-\frac{1}{3})^{k}+\frac{3}{4} \end{pmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{3}{4}(-\frac{1}{3})^{k}+\frac{1}{4} \\ -\frac{3}{4}(-\frac{1}{3})^{k}+\frac{3}{4} \end{bmatrix}</math>. | |
− | + | The probability we seek is the top entry of this vector for k = 7, or <math>\frac{182}{729}</math>. | |
== Remark == | == Remark == | ||
Here is a similar problem from another AIME test: [[2003 AIME II Problems/Problem 13|2003 AIME II Problem 13]], in which we have an equilateral triangle instead. | Here is a similar problem from another AIME test: [[2003 AIME II Problems/Problem 13|2003 AIME II Problem 13]], in which we have an equilateral triangle instead. | ||
+ | |||
+ | There is another similar problem from the AMC8: [[2022 AMC 8 Problems/Problem 25]], where we have the same question, just with less steps | ||
== See also == | == See also == |
Revision as of 00:53, 22 June 2024
Contents
- 1 Problem
- 2 Solution 1 (Single Variable Recursion)
- 3 Solution 2 (Multivariable Recursion by Algebra)
- 4 Solution 3 (Multivariable Recursion by Table)
- 5 Solution 4 (Single Variable Version of Solution 2)
- 6 Solution 5 (Dynamic Programming)
- 7 Solution 6 (Generating Functions and Roots of Unity Filter / Casework)
- 8 Solution 7 (Partitions)
- 9 Solution 8 (Sequences)
- 10 Solution 9
- 11 Solution 10 (Linear Algebra)
- 12 Remark
- 13 See also
Problem
Let ,
,
and
be the vertices of a regular tetrahedron, each of whose edges measures
meter. A bug, starting from vertex
, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let
be the probability that the bug is at vertex
when it has crawled exactly
meters. Find the value of
.
Solution 1 (Single Variable Recursion)
For all nonnegative integers let
be the probability that the bug is at vertex
when it has crawled exactly
meters. We wish to find
Clearly, we have For all
note that after
crawls:
- The probability that the bug is at vertex
is
and the probability that it crawls to vertex
on the next move is
- The probability that the bug is not at vertex
is
and the probability that it crawls to vertex
on the next move is
Together, the recursive formula for is
Two solutions follow from here:
Solution 1.1 (Recursive Formula)
We evaluate recursively:
Therefore, the answer is
~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 1.2 (Explicit Formula)
Let for some function
and constant
For all
the recursive formula for
becomes
Solving for
we get
For simplicity purposes, we set
which gives
Clearly,
is a geometric sequence with the common ratio
Substituting
and
into
produces
the first term of the geometric sequence.
For all nonnegative integers the explicit formula for
is
and the explicit formula for
is
Finally, the requested probability is
from which
~MRENTHUSIASM
Solution 2 (Multivariable Recursion by Algebra)
Denominator
There are ways for the bug to make
independent crawls without restrictions.
Numerator
Let denote the number of ways for the bug to crawl exactly
meters starting from vertex
and ending at vertex
where
and
is a positive integer. We wish to find
Since the bug must crawl to vertex or
on the first move, we have
where
More generally, we get
For
note that
- Base Case:
- Recursive Case:
Clearly, is a geometric sequence with the first term
and the common ratio
Therefore, its explicit formula is
Recall that
By
and
we rewrite
recursively:
Answer
The requested probability is from which
~MRENTHUSIASM
Solution 3 (Multivariable Recursion by Table)
Define notation as Solution 2 does.
In fact, we can generalize the following relationships for all nonnegative integers
Using these equations, we recursively fill out the table below:
Note that the paths from
to
and the paths from
to
have one-to-one correspondence. So, we must get
for all values of
The requested probability is from which
~MRENTHUSIASM
Solution 4 (Single Variable Version of Solution 2)
Let denotes the number of ways that the bug arrives at
after crawling
meters, then we have
.
Notice that there is respectively way to arrive at
for each of the different routes after the previous
crawls, excluding the possibility that the bug ends up at
after the
th crawl (as it will be forced to move somewhere else.). Thus, we get the recurrence relation
Quick calculations yield
Thus,
.
~Nafer
Solution 5 (Dynamic Programming)
Let be the probability the bug lands on vertex
after crawling
meters,
be the probability the bug lands on vertex
after crawling
meters, and etc.
Note that and
For
the probability that the bug land on each vertex after
meters is
the sum of the probability the bug land on other vertices after crawling
meters. So, we have
It follows that
We construct the following table:
Therefore, the answer is
Solution 6 (Generating Functions and Roots of Unity Filter / Casework)
The generating function for a problem with this general form ( states,
steps) is
, so the generating function of interest for this problem is
. Our goal is to find the coefficients of every
and add them up before dividing by
. Here we have two ways to proceed:
1. Roots of Unity Filter
Let . We have that if
, then
From here, the desired probability is
. Therefore, the answer is
.
~RedFlame2112
2. Casework
We can factor as
The
coefficients of
will be the same as the
coefficients of
The possible values for
would then be
and
Because
the coefficients of
and
are equal and so are the coefficients of
and
To make an
term, we need
"
" terms and one "
" term multiplied together. There are
ways to arrange these numbers. The coefficient of the
term will be the sum of the number of the possible arrangements for these numbers:
Thus, the coefficient of the
term is
From here, the desired probability is
Thus, our answer is
Solution 7 (Partitions)
We can find the number of different times the bug reaches vertex before the
th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at
Define to be the number of paths of length
which start and end at
but do not pass through
otherwise. Obviously
In general for
the bug has three initial edges to pick from. From there, since the bug cannot return to
by definition, the bug has exactly two choices. This continues from the
nd move up to the
th move. The last move must be a return to
so this move is determined. So
Now we need to find the number of cycles by which the bug can reach at the end. Since
we know that
cannot be used, as on the
th move the bug cannot move from
to
So we need to find the number of partitions of
using only
and
These are
and
We can calculate these and sum them up using our formula. Also, order matters, so we need to find the number of ways to arrange each partition:
Finally, this is a probability question, so we divide by
or
The answer is
Solution 8 (Sequences)
Note that this problem is basically equivalent to the following:
How many distinct sequences of integers
are there such that
for all
and
for all
?
Now consider the integers modulo
Let
be a new sequence of integers such that
for all
Note that the condition is equivalent to that for all
and since
must be a multiple of
Thus, our desired number of paths is equivalent to the number of ordered septuples of positive integers such that
for all
and
is congruent to
We will now proceed with casework on the number of s in the septuple.
One : There are
ways to arrange the
, and
valid ways (the proof of this combinatorial identity is left as an exercise to the reader) to arrange the
s and
s.
Three s: There are
ways to arrange the
s, and
valid ways to arrange the
s and
s.
Five s: There are
ways to arrange the
s, and
valid ways to arrange the
s and
s.
Adding up our cases, we obtain valid sequences.
Dividing by the total number of paths without restrictions,
our desired probability is
giving an answer of
~fidgetboss_4000
Solution 9
We instead find the probability that the bug is NOT at vertex after crawling
meters (equivalent to moving
times). Call
the probability that the bug IS at vertex
after
moves; call
the probability that the bug is on some other vertex. We have the following recurrence relations.
However, we can calculate
in terms of
; take
, and we have
. Substituting this into our equation for
, we have that
. We also have the conditions that
(the bug will not be at vertex other than
on the "0th" move) and
(the bug will be at a vertex other than
after the
move). Iteratively calculating
, we find that it is equal to
(I will not do this calculation here; you can do it manually if you wish to check). However, this is the probability that the ant is NOT at vertex
after
moves; then its complement,
is the probability that the ant IS at vertex
after
moves, so our answer is
.
~ cxsmi
Solution 10 (Linear Algebra)
Think of the problem as a state machine with a transition matrix. State 1 is if the bug is at A, State 2 is if the bug is not at A. In vector form we can represent state 1 as and state 2 as
. Our transition matrix
. Thus the probability of being in each state after k moves is
. Those familiar with linear algebra will recognize the need to diagonalize the transition matrix through eigendecomposition. In short, the eigenvalues of the transition matrix are
and 1, with corresponding eigenvector basis of
. Thus
Thus .
The probability we seek is the top entry of this vector for k = 7, or .
Remark
Here is a similar problem from another AIME test: 2003 AIME II Problem 13, in which we have an equilateral triangle instead.
There is another similar problem from the AMC8: 2022 AMC 8 Problems/Problem 25, where we have the same question, just with less steps
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |