2021 AIME I Problems/Problem 12

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

$E(3,3,3)=1+\frac{1}{4} E(3,3,3)+\frac{3}{4} E(1,3,5)$ , giving $E(3,3,3)=\frac{4}{3}+E(1,3,5)$ ; (1)

$E(1,3,5)=1+\frac{1}{8}E(1,1,7)+\frac{1}{2}E(1,3,5)+\frac{1}{8}E(3,3,3)$ , giving $E(1,3,5)=2+\frac{1}{4}E(1,1,7)+\frac{1}{4}E(3,3,3)$ ; (2)

$E(1,1,7)=1+\frac{1}{4}E(1,1,7)+\frac{1}{4}E(1,3,5)$ , giving $E(1,1,7)=\frac{4}{3}+\frac{1}{3}E(1,3,5)$ ; (3)

Plug in (1) and (3) into (2), we see that $E(1,3,5)=4$ . $E(3,3,3)=\frac{4}{3}+4=\frac{16}{3}$ . Each step is one minute. The answer is $16+3=\boxed{19}$ .

~Ross Gao (Solution)

~MRENTHUSIASM (Reformatting and Minor Revisions)

2021 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2021 AIME I Problems/Problem 12

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