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

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<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
E(4, 4) &= \frac34(E(2, 4)+1) + \frac14(E(4, 4)+1) \\
+
E(4, 4) &= \frac14(E(4, 4)+1) + \frac34(E(2, 4)+1)\\
−
E(2, 4) &= \frac14 + \frac12(E(2, 4)+1) + \frac18(E(2, 2)+1) + \frac18(E(4, 4)+1) \\
+
E(2, 4) &= \frac14 + \frac12(E(2, 4)+1) + \frac18(E(4, 4)+1) + \frac18(E(2, 2)+1) \\
 
E(2, 2) &= \frac12 + \frac14(E(2, 2)+1) + \frac14(E(2, 4)+1)  
 
E(2, 2) &= \frac12 + \frac14(E(2, 2)+1) + \frac14(E(2, 4)+1)  
 
\end{align*}</cmath>
 
\end{align*}</cmath>

Revision as of 11:55, 26 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 1

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)

Solution 1 Supplement (Markov Chain)

The above solution can be represented by the following Markov Chain:

Markov Chain Frog AIME.png
From state $(4, 4, 4)$ to state $(4, 4, 4)$, the $3$ frogs must jump in the same direction, $2 \cdot \frac18 = \frac14$.

From state $(4, 4, 4)$ to state $(2, 4, 6)$, $2$ frogs must jump in the same direction, and the other must jump in the opposite direction, $\binom32 \cdot 2 \cdot \frac18 = \frac34$.

From state $(2, 4, 6)$ to state $(4, 4, 4)$, the $2$ frogs with a distance of $4$ in between must jump in the same direction so that they will be further away from the other frog, and the other frog must jump in the opposite direction as those $2$ frogs, $\frac18$.

From state $(2, 4, 6)$ to state $(2, 4, 6)$, the $3$ frogs can all jump in the same direction, or the frogs with a distance of $2$ between jumps away from each other and the other frog jumps closer to the closest frog, or the $2$ frogs with a distance of $2$ between jumps in the same direction to make the distance $6$ to $5$ and the other frog jumps in the opposite direction, $(2 + 1 + 2)\frac18 = \frac12$.

From state $(2, 4, 6)$ to state $(2, 2, 8)$, the $2$ frogs with a distance of $2$ in between must jump closer to the other frog, and the other frog must jump close to those $2$ frogs, $\frac18$.

From state $(2, 2, 8)$ to state $(2, 4, 6)$, $2$ frogs that have no frogs in between must both jump in the same direction away from the other frog, the other frog must also jump away from those $2$ frogs, $2 \cdot \frac18 = \frac14$.

From state $(2, 2, 8)$ to state $(2, 2, 8)$, the $3$ frogs must all jump in the same direction, $2 \cdot \frac18 = \frac14$.

From state $(2, 2, 8)$ to state $(0, x, y)$, frogs with a distance of $2$ must jump closer to each other, the other frog can jump in any direction, $\frac12 \cdot \frac12 \cdot 2 = \frac12$.

From state $(2, 4, 6)$ to state $(0, x, y)$, the frogs with a distance of $2$ must jump closer to each other, the other frog can jump in any direction, $\frac12 \cdot \frac12 = \frac14$.


Because $a + b + c = 12$, we can use $(a, b)$ to represent the states. The following system of equations is the same as above.

\begin{align*} E(4, 4) &= \frac14(E(4, 4)+1) + \frac34(E(2, 4)+1)\\ E(2, 4) &= \frac14 + \frac12(E(2, 4)+1) + \frac18(E(4, 4)+1) + \frac18(E(2, 2)+1) \\ E(2, 2) &= \frac12 + \frac14(E(2, 2)+1) + \frac14(E(2, 4)+1) \end{align*}

~isabelchen

See Also

2021 AIME I (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

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