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

m (→‎Solution)
m (→‎Solution)
Line 14: Line 14:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
E(3,3,3)&=1+\frac{2}{8}E(3,3,3)+\frac{6}{8}E(1,3,5), \\
 
E(3,3,3)&=1+\frac{2}{8}E(3,3,3)+\frac{6}{8}E(1,3,5), \\
−
E(1,3,5)&=1+\frac{1}{8}E(1,1,7)+\frac{4}{8}E(1,3,5)+\frac{1}{8}E(3,3,3), \\
+
E(1,3,5)&=1+\frac{4}{8}E(1,3,5)+\frac{1}{8}E(3,3,3)+\frac{1}{8}E(1,1,7), \\
 
E(1,1,7)&=1+\frac{2}{8}E(1,1,7)+\frac{2}{8}E(1,3,5).
 
E(1,1,7)&=1+\frac{2}{8}E(1,1,7)+\frac{2}{8}E(1,3,5).
 
\end{align*}</cmath>
 
\end{align*}</cmath>
Line 20: Line 20:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
E(3,3,3)&=\frac{4}{3}+E(1,3,5), &(1) \\
 
E(3,3,3)&=\frac{4}{3}+E(1,3,5), &(1) \\
−
E(1,3,5)&=2+\frac{1}{4}E(1,1,7)+\frac{1}{4}E(3,3,3), &\hspace{12.75mm}(2) \\
+
E(1,3,5)&=2+\frac{1}{4}E(3,3,3)+\frac{1}{4}E(1,1,7), &\hspace{12.75mm}(2) \\
 
E(1,1,7)&=\frac{4}{3}+\frac{1}{3}E(1,3,5). &(3)
 
E(1,1,7)&=\frac{4}{3}+\frac{1}{3}E(1,3,5). &(3)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Substituting <math>(1)</math> and <math>(3)</math> into <math>(2),</math> we obtain <cmath>E(1,3,5)=2+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(1,3,5)\right]+\frac{1}{4}\left[\frac{4}{3}+E(1,3,5)\right],</cmath> from which <math>E(1,3,5)=4.</math> Substituting this into <math>(1)</math> gives <math>E(3,3,3)=\frac{16}{3}.</math>
+
Substituting <math>(1)</math> and <math>(3)</math> into <math>(2),</math> we obtain <cmath>E(1,3,5)=2+\frac{1}{4}\left[\frac{4}{3}+E(1,3,5)\right]+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(1,3,5)\right],</cmath> from which <math>E(1,3,5)=4.</math> Substituting this into <math>(1)</math> gives <math>E(3,3,3)=\frac{16}{3}.</math>
  
 
Therefore, the answer is <math>16+3=\boxed{019}.</math>
 
Therefore, the answer is <math>16+3=\boxed{019}.</math>

Revision as of 18:01, 7 July 2021

Problem

Let $A_1A_2A_3...A_{12}$ 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

The expected number of minutes depends on the distances between the frogs, in either clockwise or counterclockwise order.

Let $E(a,b,c)$ denote the expected number of minutes for two frogs to meet, such that the frogs are $a,b,$ and $c$ vertices apart, in either clockwise or counterclockwise order. Note that

  1. $a+b+c=9$ always holds.

  2. If at least one of $a,b,$ or $c$ is $0,$ then $E(a,b,c)=0.$

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

We have the following system of equations: \begin{align*} E(3,3,3)&=1+\frac{2}{8}E(3,3,3)+\frac{6}{8}E(1,3,5), \\ E(1,3,5)&=1+\frac{4}{8}E(1,3,5)+\frac{1}{8}E(3,3,3)+\frac{1}{8}E(1,1,7), \\ E(1,1,7)&=1+\frac{2}{8}E(1,1,7)+\frac{2}{8}E(1,3,5). \end{align*} Rearranging and simplifying each equation, we get \begin{align*} E(3,3,3)&=\frac{4}{3}+E(1,3,5), &(1) \\ E(1,3,5)&=2+\frac{1}{4}E(3,3,3)+\frac{1}{4}E(1,1,7), &\hspace{12.75mm}(2) \\ E(1,1,7)&=\frac{4}{3}+\frac{1}{3}E(1,3,5). &(3) \end{align*} Substituting $(1)$ and $(3)$ into $(2),$ we obtain \[E(1,3,5)=2+\frac{1}{4}\left[\frac{4}{3}+E(1,3,5)\right]+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(1,3,5)\right],\] from which $E(1,3,5)=4.$ Substituting this into $(1)$ gives $E(3,3,3)=\frac{16}{3}.$

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

~Ross Gao (Solution)

~MRENTHUSIASM (Reformatting and Minor Revisions)

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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