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Difference between revisions of "2021 AIME I Problems/Problem 12"

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~MRENTHUSIASM (Reconstruction)
 
~MRENTHUSIASM (Reconstruction)
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===Alternate solution===
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Without loss of generality, let's have each frog jump one space to their right after their usual jump. Then observe that half of the vertices will be blank due to parity reasons. Then, we can instead consider the frogs on a hexagon instead. Let them start at vertices <math>A_{0}</math>, <math>A_{2}</math>, and <math>A_{4}</math>. Each time, they either stay at the same lilypad, or jump right by one spot. Observe that configurations are still equivalent to their reflections. If some frogs jump in one reflection, we have all frogs jump in the other reflection except for the ones which jumped in the original reflection.
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 +
There are only four states on this hexagon. The first state, <math>x</math>, is where the frogs are arranged in an equilateral triangle. The second state, <math>y</math>, is where two frogs are adjacent. The third state, <math>z</math>, is where all three frogs are adjacent. And the fourth state, <math>H</math>, is the halting state. We can then consider how these states change.
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<math>x</math> moves to <math>x</math> with <math>\frac{1}{4}</math> probability, and to <math>y</math> with <math>\frac{3}{4}</math> probability.
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<math>y</math> moves to <math>y</math> with <math>\frac{1}{2}</math> probability, to <math>x</math> and <math>z</math> with <math>\frac{1}{8}</math> probability each, and to <math>H</math> with <math>\frac{1}{4}</math> probability.
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<math>z</math> moves to <math>z</math> and <math>y</math> with <math>\frac{1}{4}</math> probability each, and to <math>H</math> with <math>\frac{1}{2}</math> probability.
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<math>H</math> always moves to itself.
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Now, let <math>x</math>, <math>y</math>, <math>z</math>, and <math>H</math> be functions of time. We can create a state vector <math>\overrightarrow{S}=\begin{bmatrix}x\\y\\z\\H\end{bmatrix}</math>. Observe that we can then multiply <math>\overrightarrow{S}\left(t+1\right)=A\overrightarrow{S}\left(t\right)</math> for some matrix <math>A</math>. This matrix can be found from our previous examination of how the states move:
 +
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<math>A=\begin{bmatrix}\frac{1}{4} & \frac{1}{8} & 0 & 0\\\frac{3}{4} & \frac{1}{2} & \frac{1}{4} & 0\\0 & \frac{1}{8} & \frac{1}{4} & 0\\0 & \frac{1}{4} & \frac{1}{2} & 1\end{bmatrix}</math>
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From the problem, we have <math>\overrightarrow{S}\left(0\right)=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}</math>, and by induction, we have <math>\overrightarrow{S}\left(t\right)=A^{t}\overrightarrow{S}\left(0\right)</math>. The expected stopping time is <math>\sum_{t=1}^{\infty}t\left(H\left(t\right)-H\left(t-1\right)\right)</math>. To find <math>H\left(t\right)</math>, we need to be able to easily raise <math>A</math> to high powers. <math>PDP^{-1}</math> factorization is the tool we need for this job. We need the eigenvalues and eigenvectors for <math>A</math>. By definition, if <math>\overrightarrow{v}</math> is an eigenvector of <math>A</math>, then there exists an eigenvalue <math>\lambda</math> where <math>A\overrightarrow{v}=\lambda\overrightarrow{v}</math>. To subtract matrices from matrices, we insert a copy of the identity matrix:
 +
 +
<math>A\overrightarrow{v}=\left(\lambda I\right)\overrightarrow{v}</math>
 +
 +
Then we want values of <math>\lambda</math> where <math>A-\lambda I</math> has a null space with nonzero dimension. This happens when <math>\det\left(A-\lambda I\right)=0</math>. We know:
 +
 +
<math>A-\lambda I=\begin{bmatrix}\frac{1}{4}-\lambda & \frac{1}{8} & 0 & 0\\\frac{3}{4} & \frac{1}{2}-\lambda & \frac{1}{4} & 0\\0 & \frac{1}{8} & \frac{1}{4}-\lambda & 0\\0 & \frac{1}{4} & \frac{1}{2} & 1-\lambda\end{bmatrix}</math>
 +
 +
We need the determinant of that. One obvious way to proceed would be to do a cofactor expansion across the last column. However, it's easier to expand across the first column, because one cofactor is lower triangular, meaning that we can take the determinant of that cofactor by multiplying its diagonal entries, and the other cofactor has two zeroes in one of its columns, meaning we only have to evaluate one more determinant. Two of the entries in the first column are still 0, so we can ignore this.
 +
 +
<math>\det\begin{bmatrix}\frac{1}{4}-\lambda & \frac{1}{8} & 0 & 0\\\frac{3}{4} & \frac{1}{2}-\lambda & \frac{1}{4} & 0\\0 & \frac{1}{8} & \frac{1}{4}-\lambda & 0\\0 & \frac{1}{4} & \frac{1}{2} & 1-\lambda\end{bmatrix}=\left(\frac{1}{4}-\lambda\right)\det\begin{bmatrix}\frac{1}{2}-\lambda & \frac{1}{4} & 0\\\frac{1}{8} & \frac{1}{4}-\lambda & 0\\\frac{1}{4} & \frac{1}{2} & 1-\lambda\end{bmatrix}-\left(\frac{3}{4}-\lambda\right)\det\begin{bmatrix}\frac{1}{8} & 0 & 0\\\frac{1}{8} & \frac{1}{4}-\lambda & 0\\\frac{1}{4} & \frac{1}{2} & 1-\lambda\end{bmatrix}</math>
  
 
==See Also==
 
==See Also==

Revision as of 10:14, 18 February 2022

Problem

Let $A_1A_2A_3\ldots A_{12}$ be a dodecagon ($12$-gon). Three frogs initially sit at $A_4,A_8,$ and $A_{12}$. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Define the distance between two frogs as the number of sides between them that do not contain the third frog.

Let $E(a,b,c)$ denote the expected number of minutes until the frogs stop jumping, such that the distances between the frogs are $a,b,$ and $c$ (in either clockwise or counterclockwise order). Without the loss of generality, assume that $a\leq b\leq c.$

We wish to find $E(4,4,4).$ Note that:

  1. At any moment before the frogs stop jumping, the only possibilities for $(a,b,c)$ are $(4,4,4),(2,4,6),$ and $(2,2,8).$

  2. $E(a,b,c)$ does not depend on the actual positions of the frogs, but depends on the distances between the frogs.

  3. At the end of each minute, each frog has $2$ outcomes. So, there are $2^3=8$ outcomes in total.

We have the following system of equations: \begin{align*} E(4,4,4)&=1+\frac{2}{8}E(4,4,4)+\frac{6}{8}E(2,4,6), \\ E(2,4,6)&=1+\frac{4}{8}E(2,4,6)+\frac{1}{8}E(4,4,4)+\frac{1}{8}E(2,2,8), \\ E(2,2,8)&=1+\frac{2}{8}E(2,2,8)+\frac{2}{8}E(2,4,6). \end{align*} Rearranging and simplifying each equation, we get \begin{align*} E(4,4,4)&=\frac{4}{3}+E(2,4,6), &(1) \\ E(2,4,6)&=2+\frac{1}{4}E(4,4,4)+\frac{1}{4}E(2,2,8), &\hspace{12.75mm}(2) \\ E(2,2,8)&=\frac{4}{3}+\frac{1}{3}E(2,4,6). &(3) \end{align*} Substituting $(1)$ and $(3)$ into $(2),$ we obtain \[E(2,4,6)=2+\frac{1}{4}\left[\frac{4}{3}+E(2,4,6)\right]+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(2,4,6)\right],\] from which $E(2,4,6)=4.$ Substituting this into $(1)$ gives $E(4,4,4)=\frac{16}{3}.$

Therefore, the answer is $16+3=\boxed{019}.$

~Ross Gao (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Alternate solution

Without loss of generality, let's have each frog jump one space to their right after their usual jump. Then observe that half of the vertices will be blank due to parity reasons. Then, we can instead consider the frogs on a hexagon instead. Let them start at vertices $A_{0}$, $A_{2}$, and $A_{4}$. Each time, they either stay at the same lilypad, or jump right by one spot. Observe that configurations are still equivalent to their reflections. If some frogs jump in one reflection, we have all frogs jump in the other reflection except for the ones which jumped in the original reflection.

There are only four states on this hexagon. The first state, $x$, is where the frogs are arranged in an equilateral triangle. The second state, $y$, is where two frogs are adjacent. The third state, $z$, is where all three frogs are adjacent. And the fourth state, $H$, is the halting state. We can then consider how these states change.

$x$ moves to $x$ with $\frac{1}{4}$ probability, and to $y$ with $\frac{3}{4}$ probability.

$y$ moves to $y$ with $\frac{1}{2}$ probability, to $x$ and $z$ with $\frac{1}{8}$ probability each, and to $H$ with $\frac{1}{4}$ probability.

$z$ moves to $z$ and $y$ with $\frac{1}{4}$ probability each, and to $H$ with $\frac{1}{2}$ probability.

$H$ always moves to itself.

Now, let $x$, $y$, $z$, and $H$ be functions of time. We can create a state vector $\overrightarrow{S}=\begin{bmatrix}x\\y\\z\\H\end{bmatrix}$. Observe that we can then multiply $\overrightarrow{S}\left(t+1\right)=A\overrightarrow{S}\left(t\right)$ for some matrix $A$. This matrix can be found from our previous examination of how the states move:

$A=\begin{bmatrix}\frac{1}{4} & \frac{1}{8} & 0 & 0\\\frac{3}{4} & \frac{1}{2} & \frac{1}{4} & 0\\0 & \frac{1}{8} & \frac{1}{4} & 0\\0 & \frac{1}{4} & \frac{1}{2} & 1\end{bmatrix}$

From the problem, we have $\overrightarrow{S}\left(0\right)=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$, and by induction, we have $\overrightarrow{S}\left(t\right)=A^{t}\overrightarrow{S}\left(0\right)$. The expected stopping time is $\sum_{t=1}^{\infty}t\left(H\left(t\right)-H\left(t-1\right)\right)$. To find $H\left(t\right)$, we need to be able to easily raise $A$ to high powers. $PDP^{-1}$ factorization is the tool we need for this job. We need the eigenvalues and eigenvectors for $A$. By definition, if $\overrightarrow{v}$ is an eigenvector of $A$, then there exists an eigenvalue $\lambda$ where $A\overrightarrow{v}=\lambda\overrightarrow{v}$. To subtract matrices from matrices, we insert a copy of the identity matrix:

$A\overrightarrow{v}=\left(\lambda I\right)\overrightarrow{v}$

Then we want values of $\lambda$ where $A-\lambda I$ has a null space with nonzero dimension. This happens when $\det\left(A-\lambda I\right)=0$. We know:

$A-\lambda I=\begin{bmatrix}\frac{1}{4}-\lambda & \frac{1}{8} & 0 & 0\\\frac{3}{4} & \frac{1}{2}-\lambda & \frac{1}{4} & 0\\0 & \frac{1}{8} & \frac{1}{4}-\lambda & 0\\0 & \frac{1}{4} & \frac{1}{2} & 1-\lambda\end{bmatrix}$

We need the determinant of that. One obvious way to proceed would be to do a cofactor expansion across the last column. However, it's easier to expand across the first column, because one cofactor is lower triangular, meaning that we can take the determinant of that cofactor by multiplying its diagonal entries, and the other cofactor has two zeroes in one of its columns, meaning we only have to evaluate one more determinant. Two of the entries in the first column are still 0, so we can ignore this.

$\det\begin{bmatrix}\frac{1}{4}-\lambda & \frac{1}{8} & 0 & 0\\\frac{3}{4} & \frac{1}{2}-\lambda & \frac{1}{4} & 0\\0 & \frac{1}{8} & \frac{1}{4}-\lambda & 0\\0 & \frac{1}{4} & \frac{1}{2} & 1-\lambda\end{bmatrix}=\left(\frac{1}{4}-\lambda\right)\det\begin{bmatrix}\frac{1}{2}-\lambda & \frac{1}{4} & 0\\\frac{1}{8} & \frac{1}{4}-\lambda & 0\\\frac{1}{4} & \frac{1}{2} & 1-\lambda\end{bmatrix}-\left(\frac{3}{4}-\lambda\right)\det\begin{bmatrix}\frac{1}{8} & 0 & 0\\\frac{1}{8} & \frac{1}{4}-\lambda & 0\\\frac{1}{4} & \frac{1}{2} & 1-\lambda\end{bmatrix}$

See Also

2021 AIME I (Problems • Answer Key • Resources)
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Problem 11 		Followed by
Problem 13
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